System Dynamics for Engineering Students [2 ed.] 9780128045596

Table of contents :
Front-Matter_2018_System-Dynamics-for-Engineering-Students
System Dynamics for Engineering StudentsConcepts and ApplicationsSecond EditionNicolae LobontiuUniversity of Alaska Anchorage?
Copyright_2018_System-Dynamics-for-Engineering-Students
Copyright
Dedication_2018_System-Dynamics-for-Engineering-Students
Dedication
Foreword-to-the-First-Edition_2018_System-Dynamics-for-Engineering-Students
Foreword to the First Edition
Preface_2018_System-Dynamics-for-Engineering-Students
Preface
Resources-That-Accompany-This-Boo_2018_System-Dynamics-for-Engineering-Stude
Resources That Accompany This Book
Available to All
For Instructors Only
Chapter-1---Introduction_2018_System-Dynamics-for-Engineering-Students
1. Introduction
1.1 ENGINEERING SYSTEM DYNAMICS
1.2 MODELING ENGINEERING SYSTEM DYNAMICS
1.2.1 Modeling Variants
1.2.2 Dynamical Systems Lumped-Parameter Modeling and Solution
Modeling Methods
Solution Methods
System Response
1.3 ELEMENTS, SYSTEM, INPUT, AND OUTPUT
1.4 COMPLIANT MECHANISMS AND MICROELECTROMECHANICAL SYSTEMS
1.5 SYSTEM ORDER
1.5.1 Zero-Order Systems
1.5.2 First-Order Systems
1.5.3 Second- and Higher-Order Systems
1.6 COUPLED-FIELD (MULTIPLE-FIELD) SYSTEMS
1.7 LINEAR AND NONLINEAR DYNAMIC SYSTEMS
1.8 TIME- AND FREQUENCY-DOMAIN SYSTEM DYNAMICS
1.9 FEEDBACK CONTROL OF DYNAMIC SYSTEMS
Chapter-2---Mechanical-Elements_2018_System-Dynamics-for-Engineering-Student
2. Mechanical Elements
INTRODUCTION
2.1 SPRING ELEMENTS
2.1.1 Basic Spring Elements and Stiffness
2.1.2 Series and Parallel Spring Connections
2.2 INERTIA ELEMENTS
2.2.1 Basic Inertia Elements
2.2.2 Lumped-Parameter Inertia of Distributed-Parameter (Elastic) Members
2.3 VISCOUS DAMPING ELEMENTS
2.3.1 Basic Viscous Damping Elements
2.3.2 Series and Parallel Damper Connections
2.4 EQUIVALENT MECHANICAL ELEMENTS THROUGH GEAR AND LEVER TRANSFER
2.4.1 Toothed Gears
Toothed Gears and Springs
Toothed Gears and Inertia
Gears With Dampers
2.4.2 Levers of Small Rotation
Levers With Springs
Levers and Inertia
Levers With Dampers
SUMMARY
Suggested Reading
Chapter-3---Mechanical-Systems_2018_System-Dynamics-for-Engineering-Students
3. Mechanical Systems
INTRODUCTION
3.1 CONFIGURATION, DEGREES OF FREEDOM
3.2 SINGLE–DOF SYSTEMS
3.2.1 Free Response
Conservative Systems and the Free Undamped (Natural) Response
Mathematical Model (Differential Equation)
Mathematical Model (Differential Equation)
Newton's Second Law of Motion
Newton's Second Law of Motion
The Energy Method
The Energy Method
Solution of Mathematical Model (Differential Equation)
Solution of Mathematical Model (Differential Equation)
Systems With Losses and the Free Damped Response
Mathematical Model (Differential Equation)
Mathematical Model (Differential Equation)
Solution of Mathematical Model (Differential Equation)
Solution of Mathematical Model (Differential Equation)
3.2.2 Forced Response
3.3 MULTIPLE–DOF SYSTEMS
3.3.1 Free Response
Conservative Systems and the Natural Response
Mathematical Model (Differential Equation)
Mathematical Model (Differential Equation)
Newton's Second Law of Motion Method
Newton's Second Law of Motion Method
Lagrange's Equations
Lagrange's Equations
Solution of Mathematical Model (Differential Equation)
Solution of Mathematical Model (Differential Equation)
Analytical Approach
Analytical Approach
MATLAB Approach—the Dynamic Matrix and the Eigenvalue Problem
MATLAB Approach—the Dynamic Matrix and the Eigenvalue Problem
Free Damped Response
Mathematical Model Derivation
Mathematical Model Derivation
Solution of the Mathematical Model Differential Equations
Solution of the Mathematical Model Differential Equations
3.3.2 Forced Response
SUMMARY
Suggested Reading
Chapter-4---Electrical-Systems_2018_System-Dynamics-for-Engineering-Students
4. Electrical Systems
INTRODUCTION
4.1 ELECTRICAL ELEMENTS: VOLTAGE AND CURRENT SOURCES, RESISTOR, CAPACITOR, INDUCTOR, AND OPERATIONAL AMPLIFIER
4.1.1 Voltage and Current Sources
4.1.2 Resistor Elements
Mechanical Displacement Sensing
4.1.3 Capacitor Elements
Actuation and Sensing in Microelectromechanical Systems
4.1.4 Inductor Elements
4.1.5 Operational Amplifiers
4.2 ELECTRICAL SYSTEMS: CIRCUITS OR NETWORKS
4.2.1 Kirchhoff's Laws
4.2.2 Configuration, Degrees of Freedom
4.2.3 Free Response
Natural (Free Lossless) Response
Single-DOF Conservative Electrical Systems
Single-DOF Conservative Electrical Systems
Multiple-DOF Conservative Electrical Systems
Multiple-DOF Conservative Electrical Systems
Lagrange's Equations
Lagrange's Equations
Mesh Analysis
Mesh Analysis
Node Analysis
Node Analysis
Free Damped Response
4.2.4 Forced Response
Single-DOF Systems
Operational Amplifier Circuits
Operational Amplifier Circuits
Inverting Amplifier Circuits
Inverting Amplifier Circuits
Mathematical Operations With Operational Amplifier Circuits
Mathematical Operations With Operational Amplifier Circuits
Multiple-DOF Systems
Lagrange's Equations
Lagrange's Equations
Mesh Analysis
Mesh Analysis
Node Analysis
Node Analysis
4.3 MECHANICAL–ELECTRICAL ANALOGY
SUMMARY
Suggested Reading
Chapter-5---Fluid-and-Thermal-Syst_2018_System-Dynamics-for-Engineering-Stud
5. Fluid and Thermal Systems
INTRODUCTION
5.1 LIQUID SYSTEMS MODELING
5.1.1 Liquid Elements
Inertance
Capacitance
Resistance
Nonlinear Resistance
Nonlinear Resistance
Linear Resistance, Lost Head, and Hagen–Poiseuille Resistance
Linear Resistance, Lost Head, and Hagen–Poiseuille Resistance
Sources of Hydraulic Energy
5.1.2 Liquid Systems
Natural Response
Single-DOF Conservative Liquid Systems
Single-DOF Conservative Liquid Systems
Multiple-DOF Conservative Liquid Systems
Multiple-DOF Conservative Liquid Systems
Forced Response of Liquid-Level Systems
Single-DOF Liquid-Level Systems
Single-DOF Liquid-Level Systems
Multiple-DOF Liquid-Level Systems
Multiple-DOF Liquid-Level Systems
5.2 PNEUMATIC SYSTEMS MODELING
5.2.1 Gas Laws
5.2.2 Pneumatic Elements
Inertance
Capacitance
Resistance
Sources of Pneumatic Energy
5.2.3 Pneumatic Systems
Natural Response
Forced Response
5.3 THERMAL SYSTEMS MODELING
5.3.1 Thermal Elements
Capacitance
Resistance
Conduction
Conduction
Convection
Convection
Radiation
Radiation
Mixed Heat Flow Through Composite Walls
Mixed Heat Flow Through Composite Walls
Convection and Radiation
Convection and Radiation
Conduction, Convection, and Radiation
Conduction, Convection, and Radiation
Heat Sources
5.3.2 Thermal Systems
Single-DOF Systems
Multiple-DOF Systems
5.4 ELECTRICAL–FLUID–THERMAL SYSTEM ANALOGY
SUMMARY
Suggested Reading
Chapter-6---The-Laplace-Transfor_2018_System-Dynamics-for-Engineering-Studen
6. The Laplace Transform
INTRODUCTION
6.1 DIRECT AND INVERSE LAPLACE TRANSFORMATIONS
6.1.1 Laplace Transform Pairs
6.1.2 Properties of the Laplace Transform
Linearity
Frequency Shift Theorem
Time-Shift Theorem
Laplace Transform of Piecewise Continuous Functions
Laplace Transform of Piecewise Continuous Functions
Laplace Transform of Derivatives
Laplace Transform of Indefinite Integrals
Initial-Value and Final-Value Theorems
Periodic Functions
The Convolution Theorem
Partial Fraction Expansion
Analytical Partial Fraction Expansion
Analytical Partial Fraction Expansion
MATLAB Partial Fraction Expansion
6.2 LAPLACE TRANSFORM SOLUTION OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODE)
6.2.1 Linear ODE With Constant Coefficients
6.2.2 Systems of Linear ODE With Constant Coefficients
6.3 LAPLACE TRANSFORM SOLUTION OF INTEGRAL AND INTEGRAL–DIFFERENTIAL EQUATIONS
6.3.1 Laplace Transform Approach
6.3.2 Convolution Theorem Approach
6.4 TIME-DOMAIN SYSTEM IDENTIFICATION FROM LAPLACE-DOMAIN INFORMATION
SUMMARY
Suggested Reading
Chapter-7---Transfer-Function-Appro_2018_System-Dynamics-for-Engineering-Stu
7. Transfer Function Approach
INTRODUCTION
7.1 THE TRANSFER FUNCTION CONCEPT AND DEFINITION
7.1.1 SISO Systems
7.1.2 MIMO SYSTEMS
7.2 TRANSFER FUNCTION MODEL FORMULATION
7.2.1 Transfer Function From the Time-Domain Mathematical Model
SISO Systems
MIMO SYSTEMS
7.2.2 Transfer Function From the Zero-Pole-Gain Mathematical Model
7.2.3 Impedance Transfer Function
Electrical Systems
Nonloading and Loading Cascading (Series) Electrical Systems
Nonloading and Loading Cascading (Series) Electrical Systems
Thermal Systems
Fluid Systems
Mechanical Systems
7.3 TRANSFER FUNCTION AND SYSTEM STABILITY
7.4 TRANSFER FUNCTION AND THE TIME RESPONSE
7.4.1 SISO Systems
Analytical Approach
MATLAB and Simulink Approach
Transforming a System's Transfer Function to Account for Nonzero Initial Conditions
Transforming a System's Transfer Function to Account for Nonzero Initial Conditions
Response to Unit Impulse, Unit Step, Arbitrary Input—The MATLAB Impulse, Step, lsim Functions
Response to Unit Impulse, Unit Step, Arbitrary Input—The MATLAB Impulse, Step, lsim Functions
7.4.2 MIMO SYSTEMS
Analytical Approach
MATLAB and Simulink Approach
SUMMARY
Suggested Reading
Chapter-8---State-Space-Modeling_2018_System-Dynamics-for-Engineering-Studen
8. State-Space Modeling
INTRODUCTION
8.1 THE CONCEPT AND MODEL OF THE STATE-SPACE APPROACH
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Nonuniqueness of a State-Space Model
Nonuniqueness of a State-Space Model
Solution of the State-Space Equations
Solution of the State-Space Equations
8.2 STATE-SPACE MODEL FORMULATION
8.2.1 State-Space Model From the Time-Domain Mathematical Model
Dynamic Systems Without Input Time Derivative
SISO Systems
SISO Systems
MIMO Systems
MIMO Systems
Dynamic Systems With Input Time Derivative
Nonlinear Systems
8.2.2 State-Space Model From Other Models
Conversions Between Transfer Function and State-Space Models
Transformation of a Transfer Function Model Into a State-Space Model
Transformation of a Transfer Function Model Into a State-Space Model
Transformation of a State-Space Model Into a Transfer Function Model
Transformation of a State-Space Model Into a Transfer Function Model
Conversion Between Zero-Pole-Gain and State-Space Models
8.3 STATE-SPACE MODEL AND THE TIME-DOMAIN RESPONSE
8.3.1 Analytical Approach: The State-Transition Matrix Method
Homogeneous State-Space Model
Nonhomogeneous State-Space Model
8.3.2 MATLAB Approach
Free Response With Nonzero Initial Conditions
Forced Response
8.3.3 Simulink Approach
SUMMARY
Suggested Reading
Chapter-9---Frequency-Domain-Appro_2018_System-Dynamics-for-Engineering-Stud
9. Frequency-Domain Approach
INTRODUCTION
9.1 THE CONCEPT OF COMPLEX TRANSFER FUNCTION IN STEADY-STATE RESPONSE AND FREQUENCY-DOMAIN ANALYSIS
9.2 STEADY-STATE RESPONSE OF DYNAMIC SYSTEMS TO HARMONIC INPUT
9.2.1 Analytical Approach
SISO Systems
Steady-State Solution Under Harmonic (Sinusoidal) Input
Steady-State Solution Under Harmonic (Sinusoidal) Input
Stability and Steady-State Response
Stability and Steady-State Response
Frequency Response Parameters of First-Order Systems
Frequency Response Parameters of First-Order Systems
Frequency Response Parameters of Second-Order Systems
Frequency Response Parameters of Second-Order Systems
Asymptote Representation of Bode Plots
Asymptote Representation of Bode Plots
MIMO Systems
Complex Transfer Function Matrix Approach
Complex Transfer Function Matrix Approach
Linear Superposition for Steady-State Time Response
Linear Superposition for Steady-State Time Response
9.2.2 MATLAB Approach
Bode Plots
Frequency Response Data Handling
Frequency Response Model Conversion
9.3 FREQUENCY-DOMAIN APPLICATIONS
9.3.1 Mechanical Vibration Transmission
Transmissibility for Motion Input; Mass Detection by the Frequency Shift Method in MEMS
Transmissibility for Force Input
Vibration Absorption and Vibration Isolation
Measuring Vibration Displacement and Acceleration Amplitudes
9.3.2 Steady-State Response of Nonloading Cascading Systems
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Single-Input Systems
Single-Input Systems
Multiple-Input Systems
Multiple-Input Systems
9.3.3 Filters
Electrical Filter Systems
Mechanical Filters
SUMMARY
Suggested Reading
Chapter-10---Coupled-Field-System_2018_System-Dynamics-for-Engineering-Stude
10. Coupled-Field Systems
INTRODUCTION
10.1 CONCEPT OF SYSTEM COUPLING
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Sensing and Actuation
Sensing and Actuation
10.2 THERMOMECHANICAL AND ELECTROTHERMOMECHANICAL COUPLING
10.2.1 Thermomechanical Coupling: The Bimetallic Strip
10.2.2 Electrothermomechanical Coupling
10.3 ELECTROMECHANICAL COUPLING
10.3.1 Electrostatic-Mechanical Coupling
10.3.2 Mechanical Strain and Electrical Voltage Coupling
10.3.3 Electromagnetomechanical Coupling
 
Rotary Direct-Current Electric Motor With Mechanical Load
Rotary Direct-Current Electric Motor With Mechanical Load
Translatory Direct-Current Electric Motor With Mechanical Load
Translatory Direct-Current Electric Motor With Mechanical Load
10.3.4 Electromagnetomechanical Coupling With Optical Detection in MEMS
10.3.5 Piezoelectric Coupling
Brief Introduction to Piezoelectricity
Longitudinal Actuation and Sensing With Piezoelectric Block
Actuation
Actuation
Piezoelectric Block Actuator and Load Spring
Piezoelectric Block Actuator and Load Spring
Sensing
Sensing
Piezoelectric and Strain Gauge Sensory-Actuation
SUMMARY
Suggested Reading
Chapter-11---Block-Diagrams-and-Feedback-Con_2018_System-Dynamics-for-Engine
11. Block Diagrams and Feedback Control System Modeling
INTRODUCTION
11.1 CONCEPT OF FEEDBACK CONTROL OF DYNAMIC SYSTEMS
11.2 BLOCK DIAGRAMS AND SISO FEEDBACK SYSTEMS
11.2.1 Transfer Functions and Basic Block Diagrams
Components of Block Diagrams
11.2.2 Controllers
Proportional, Integral, Derivative, and Combined Controllers
Hydraulic Controllers
Dashpot
Dashpot
Hydraulic Servomotor
Hydraulic Servomotor
Pneumatic Controllers
Compensators: Lag, Lead, and Lag–Lead
Compensators: Lag, Lead, and Lag–Lead
Lag Compensators
Lag Compensators
Lead Compensators
Lead Compensators
Lag–Lead Compensators
Lag–Lead Compensators
11.2.3 Feedback-Control Physical Systems
11.2.4 Sensitivity Analysis
11.2.5 State-Space Modeling of SISO Feedback Systems
11.3 BLOCK DIAGRAMS AND MIMO FEEDBACK SYSTEMS
11.3.1 MISO Feedback Systems With Disturbances
11.3.2 MIMO Feedback Systems Transfer-Function Matrix Modeling
SUMMARY
Suggested Reading
Chapter-12---Stability-of-Feedback-Cont_2018_System-Dynamics-for-Engineering
12. Stability of Feedback Control Systems
INTRODUCTION
12.1 CONCEPT OF STABILITY APPLIED TO FEEDBACK CONTROL SYSTEMS
12.1.1 SISO Systems
Closed-Loop Poles at the Origin
12.1.2 MIMO Systems
State-Space Stability
12.2 THE ROUTH–HURWITZ STABILITY TEST
12.2.1 Construction of the Routh–Hurwitz Array
12.2.2 Excepted Cases
Zero Element in First Column
Method of the Polynomial With Reciprocal Roots
Method of the Polynomial With Reciprocal Roots
Method of Epsilon
Method of Epsilon
All Elements in a Row Are Zero
Method of the Auxiliary Polynomial
Method of the Auxiliary Polynomial
12.2.3 Design Problems
12.3 STABILITY OF FEEDBACK SYSTEMS BY THE ROOT LOCUS METHOD
12.3.1 Basic Rules for Sketching the Root Locus
12.3.2 Using MATLAB to Plot the Root Locus
12.4 NYQUIST PLOT AND BODE PLOTS FOR STABILITY OF FEEDBACK SYSTEMS
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Gain Margin and Phase Margin
SUMMARY
Suggested Reading
Chapter-13---Time--and-Frequency-Domain-Contr_2018_System-Dynamics-for-Engin
13. Time- and Frequency-Domain Controls of Feedback Systems
INTRODUCTION
13.1 TIME-DOMAIN RESPONSE OF SISO FEEDBACK CONTROL SYSTEMS
13.1.1 Transient Time-Domain Response and Specifications
First-Order Systems
Second-Order Systems
Underdamped Second-Order Feedback Systems
Underdamped Second-Order Feedback Systems
13.1.2 Steady-State Time-Domain Response and Errors
Unity-Feedback Control Systems
Nonunity-Feedback Systems
13.1.3 Transitory and Steady-State Time Response of Feedback Control Systems
First-Order Plants
Proportional (P) Control
Proportional (P) Control
Derivative (D) Control
Derivative (D) Control
Integral (I) Control
Integral (I) Control
Second-Order Plants
Proportional–Derivative (P+D) Control—Additional Zero
Proportional–Derivative (P+D) Control—Additional Zero
Additional Zero
Additional Zero
Proportional–Integral (P+I) Control—Additional Zero and Additional Pole
Proportional–Integral (P+I) Control—Additional Zero and Additional Pole
Dominant Poles
Dominant Poles
Zero-Pole Cancellation or Near–Zero-Pole Cancellation
Zero-Pole Cancellation or Near–Zero-Pole Cancellation
Proportional–Integral–Derivative (P+I+D) Control
Ziegler–Nichols P+I+D Controller Tuning Algorithm
Nonlinear Control Systems
13.2 TIME-DOMAIN RESPONSE OF MIMO FEEDBACK CONTROL SYSTEMS
13.2.1 MISO Feedback Systems With Disturbances
13.2.2 MIMO Feedback Systems by the Transfer Function Matrix and State-Space Methods
13.3 FEEDBACK CONTROL SYSTEMS IN THE FREQUENCY DOMAIN
13.3.1 Frequency-Domain and Time-Domain Connections
Frequency Response and Transient Time-Response Characteristics
Frequency Response and Steady-State Error Constants
Position Constant Kp
Position Constant Kp
Velocity Constant Kv
Velocity Constant Kv
Acceleration Constant Ka
Acceleration Constant Ka
Minimum-Phase and Nonminimum-Phase Angle Systems
Minimum-Phase and Nonminimum-Phase Angle Systems
13.3.2 Lead- and Lag-Phase Compensation Design
Phase-Lead Compensation
Phase-Lag Compensation
SUMMARY
Suggested Reading
Appendix-A---Complex-Numbers_2018_System-Dynamics-for-Engineering-Students
Complex Numbers
Appendix-B---Matrix-Algebra_2018_System-Dynamics-for-Engineering-Students
Matrix Algebra
SPECIAL-FORM MATRICES
BASIC MATRIX OPERATIONS
Appendix-C---Solutions-to-Linear-Homogeneous-Ordina_2018_System-Dynamics-for
Solutions to Linear Homogeneous Ordinary Differential Equations With Constant Coefficients
Appendix-D---Basics-of-Simulink_2018_System-Dynamics-for-Engineering-Student
Basics of Simulink
AN EXAMPLE
SOLVING LINEAR ORDINARY DIFFERENTIAL EQUATIONS AND SYSTEMS
First-Order Differential Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
Second- and Higher-Order Differential Equations
SECOND- AND HIGHER-ORDER DIFFERENTIAL EQUATIONS
Systems of Ordinary Differential Equations
SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS
Suggested Reading
Appendix-E---Essentials-of-MATLAB-and-System-_2018_System-Dynamics-for-Engin
Essentials of MATLAB and System Dynamics–Related Toolboxes
MATHEMATICAL CALCULATIONS
VISUALIZATION AND GRAPHICS
LINEAR SYSTEM MODELING
TIME-DOMAIN ANALYSIS
FREQUENCY-DOMAIN ANALYSIS
CONTROLS
LINEAR TIME INVARIANT (LTI) OBJECTS
Appendix-F---Deformations--Strains--and-Stresse_2018_System-Dynamics-for-Eng
Deformations, Strains, and Stresses of Basic Line Mechanical Members
BARS UNDER AXIAL FORCE OR TORQUE
BEAMS IN BENDING
Index_2018_System-Dynamics-for-Engineering-Students
Index

Citation preview

System Dynamics for Engineering Students Concepts and Applications Second Edition

Nicolae Lobontiu

University of Alaska Anchorage

Academic Press is an imprint of Elsevier 125 London Wall, London EC2Y 5AS, United Kingdom 525 B Street, Suite 1800, San Diego, CA 92101-4495, United States 50 Hampshire Street, 5th Floor, Cambridge, MA 02139, United States The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, United Kingdom Copyright © 2018 Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. ISBN: 978-0-12-804559-6 Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library For information on all Academic Press publications visit our website at https://www.elsevier.com/books-and-journals

Publisher: Katey Birtcher Acquisition Editor: Steve Merken Developmental Editor: Nate McFadden Production Project Manager: Mohanapriyan Rajendran Designer: Vicky Pearson Esser Typeset by TNQ Books and Journals

Dedication

To all readers coming across this book, with friendly consideration.

Foreword to the First Edition This text is a modern treatment of system dynamics and its relation to traditional mechanical engineering problems as well as modern microscale devices and machines. It provides an excellent course of study for students who want to grasp the fundamentals of dynamic systems, and it covers a significant amount of material also taught in engineering modeling, systems dynamics, and vibrations, all combined in a dense form. The book is designed as a text for juniors and seniors in aerospace, mechanical, electrical, biomedical, and civil engineering. It is useful for understanding the design and development of micro- and macroscale structures, electric and fluidic systems with an introduction to transduction, and numerous simulations using MATLAB and SIMULINK. The creation of machines is essentially what much of engineering is all about. Critical to almost all machines imaginable is a transient response, which is fundamental to their functionality and needs to be our primary concern in their design. This might be in the form of changing voltage levels in a sensor, the deflection of a spring supported mass, or the flow of fluid through a device. The phenomena that govern dynamics are not simply its mechanical components but often involve the dynamics of transducers as well, which are often electromechanical or fluidic based. This text not only discusses traditional electromagnetic type actuators but also ventures into electrostatics, which are the dominant form of actuators in microelectromechanical systems. This book presents an opportunity for introducing dynamic systems to scientists and engineers who are concerned with the engineering of machines both at the micro- and macroscopic scale. Mechanism and movement are considered from the types of springs and joints that are critical to micromachined, lithographic-based devices to traditional models of macroscale electrical, fluidic, and electromechanical systems. The examples discussed and the problems at the end of each chapter have applicability at both scales. In essence this is a more modern treatment of dynamical systems, presenting views of modeling and substructures more consistent with the variety of problems that many engineers will face in the future. Any university with a substantive interest in microscale engineering would do well to consider a course that covers the material herein. Finally, this text lays the foundation and framework for the development of controllers applied to these dynamical systems. Professor Ephrahim Garcia Sibley School of Mechanical and Aerospace Engineering Cornell University Ithaca, New York

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Preface Engineering system dynamics is a discipline that focuses on deriving mathematical models based on simplified physical representations of actual systems, such as mechanical, electrical, fluid, or thermal, and on solving the mathematical models (most often consisting of differential equations). The resulting solution (which reflects the system response or behavior) is utilized in design or analysis before producing and testing the actual system. Because dynamic systems are characterized by similar mathematical models, a unitary approach can be used to characterize individual systems pertaining to different fields as well as to consider the interaction of systems from multiple fields as in coupled-field problems. This book was designed to be utilized as a one-semester system dynamics text for upper-level undergraduate students with emphasis on mechanical, aerospace, or electrical engineering. Comprising important components from these areas, the material should also serve cross-listed courses (mechanicaleelectrical) at a similar study level. In addition to the printed chapters, the book contains an equal number of chapter extensions that have been assembled into a companion website section; and this makes it useful as an introductory text for more advanced courses, such as vibrations, controls, instrumentation, or mechatronics. The book can also be useful in graduate coursework or in individual study as reference material. The material contained in this book most probably exceeds the time allotted for a one-semester course lecture, and therefore topical selection becomes necessary, based on particular instruction emphasis and teaching preferences. While the book maintains its focus on the classical approach to system dynamics, a new feature of this text is the introduction of examples from compliant mechanisms and micro- and nanoelectromechanical systems (MEMS/NEMS). As demonstrated in the book, and for the relatively simple examples that have been selected here, this inclusion can really be treated within the regular system dynamics lumped-parameter (pointlike) modeling; therefore, the students not so familiar with these topics should face any major comprehension difficulties. Another central point of this book is proposing a chapter on coupled-field (or multiple-field) systems, whereby interactions between the mechanical, electrical, fluid, and thermal fields occur and generate means for actuation or sensing applications, such as in thermomechanical, electrothermomechanical, electromechanical, electromagnetomechanical, or piezoelectric applications. Another key objective was to assemble a text that is structured, balanced, cohesive, and providing a fluent and logical sequence of topics along the following lines: 1. It starts from simple components (the elements), proceeds to the objects’ assembly (the individual system), and arrives at the system interaction level (coupled-field systems).

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Preface

2. It uses modeling and solution techniques that are familiar from other disciplines, such as physics or ordinary differential equations, and subsequently introduces new modeling and solution procedures. 3. It provides a rather even coverage (space) to each book chapter. 4. While various chapter structures are possible in a system dynamics text, this book proposes a sequence that was intended to be systematic and consistent with the logical structure and progression of the presented material. As such, the book begins with an introductory Chapter 1, which offers an overview of the main aspects of a system dynamics course for engineering students. The next four chaptersdChapters 2e5dare dedicated to mechanical (Chapters 2 and 3), electrical (Chapter 4), and fluid and thermal (Chapter 5) system modeling. They contain basic information on components, systems, and the principal physical and mathematical tools enabling to model a dynamic system and determine its solution. Dynamic systems modeling is performed both by means of Lagrange’s equations and of methods designed for a particular system, such as Newton’s second law of motion for mechanical system or Kirchhoff’s laws for electrical systems. These chapters also define and illustrate analogies between the different systems. Once the main engineering dynamic systems have been studied, Chapter 6 presents the Laplace transform technique, a mathematical tool that allows simplifying the differential equation solution process for any of the individual systems. This chapter is directly connected to the next segment of the book, containing Chapters 7e9. Chapter 7 introduces the transfer function approach, which facilitates modeling a dynamic system directly in the Laplace domain, by expressing the output as the product between the transfer function and the input. The complex impedance, which is actually a transfer function connecting the Laplace-transformed input and output of a specific system element, is also introduced and thoroughly treated in this chapter. Chapter 8 studies the state space modeling and solution approach, which is also related to the Laplace transform of Chapter 6 and the transfer function of Chapter 7. Chapter 9 discusses modeling system dynamics in the frequency domain by means of the sinusoidal (harmonic) transfer function. Chapter 10 analyzes coupled-field (or multiple-field) dynamic systems, which are combinations of mechanical, electrical, magnetic, piezoelectric, fluid, or thermal systems. In this chapter, dynamic models are formulated and solved by means of the procedures studied in previous chapters. Because of the partial and natural overlap between system dynamics and controls, the majority of textbooks on either of these two areas contain coverage of material from the adjoining domain. Consistent with this approach, Chapters 11e13 cover basic concepts of feedback controls, as follows: Chapter 11 discusses components and block diagrams for feedback control system modeling; Chapter 12 covers the stability of control systems; Chapter 13 presents notions of feedback controls in the time and frequency domains. The book also includes six appendixes: Appendix A is a refresher on complex numbers, Appendix B is a review of matrix algebra, Appendix C gives a summary

Preface

of solutions to linear homogeneous ordinary differential equations with constant coefficients, Appendix D is an introduction to SimulinkdMATLAB’s graphical application, which allows diagram/graphical solutions of system dynamics problems, Appendix E contains basic MATLAB commands that have been used throughout this text, and Appendix F comprises a summary of equations for calculating deformations, strains, and stresses of deformable mechanical components such as bars and beams. The book introduces several topics that are new to engineering system dynamics, as highlighted here: Chapter 2, Mechanical Elements • Lumped-parameter inertia properties of basic compliant (flexible) members. Chapter 3, Mechanical Systems • Lumped-parameter dynamic modeling of simple compliant mechanical microsystems. • Mass detection in MEMS by the resonance shift method. Chapter 4, Electrical Systems • Capacitive sensing and actuation in MEMS. Chapter 5, Fluid and Thermal Systems • Natural response of fluid systems. • Analogies between electrical, fluid, and thermal systems. Chapters 3e5 • Notion of degrees of freedom (DOFs) for defining the system configuration of dynamic systems. • Application of the energy method to calculate the natural frequencies of single- and multiple-DOF conservative systems. • Utilization of the vectorematrix method to calculate the eigenvalues either analytically or using MATLAB. • Application of Lagrange’s equations to derive the mathematical models of the free and forced response. • Model derivation and solution of dynamic system nonlinear mathematical models. Chapter 6, Laplace Transform • Laplace transformation of vectorematrix differential equations. • Use of the convolution theorem to solve integral and integraledifferential equations. • Time-domain system identification. Chapter 7, Transfer Function Approach • Extension of the single-input, single-output transfer function approach to multiple-input, multiple-output (MIMO) systems by means of the transfer function matrix. • Application of the transfer function approach to solve the forced and the free responses with nonzero initial conditions.

xix

xx

Preface

•

Systematic introduction and comprehensive application of the complex impedance approach to electrical, mechanical, and fluid and thermal systems. • Stability analysis of dynamic systems by means of the transfer function. Chapter 8, State Space Approach • Application of the state space approach to solve the forced and free responses with nonzero initial conditions. Chapter 9, Frequency Response Approach • Simplified (asymptote) Bode plots. • Actuation, sensing, transmission, and reduction/isolation of mechanical vibrations. • Steady-state response of cascading unloading systems. • Mechanical and electrical filters. Chapter 10, Coupled-Field Systems • Formulation of the coupled-field (multiple-field) problem. • Principles and applications of sensing and actuation. • Strain gauge and Wheatstone bridge circuits for measuring mechanical deformation. • Thermomechanical and electrothermomechanical coupling, including nonlinear problems. • Applications of electromagnetomechanical system dynamics. • Principles and applications of piezoelectric coupling with mechanical deformable systems. Chapter 11, Block Diagrams and Feedback Control System Modeling • Multiple examples of physical systems that operate as components in feedback control systems. • Systematic formulation of lead-, lag-, and lag-lead compensators in the frequency domain. Chapter 12, Stability of Feedback Control Systems • Stability of MIMO control systems by the transfer function matrix method. Chapter 13, Time- and Frequency-Domain Controls of Feedback Systems • Time-domain modeling of MIMO feedback control systems. Within this printed book’s space limitations, attention has been directed at generating a balanced coverage of minimally necessary theory presentation, many solved examples, and end-of-chapter problems. Whenever possible, examples are solved analytically, using hand calculation, so that any mathematical software can be used in conjunction with any model developed here. The book is not constructed on MATLAB, but it uses this software to determine numerical solutions and to solve symbolically mathematical models too involved to be obtained by hand. It would be difficult to overlook the built-in capabilities of MATLAB’s tool boxes (really programs within the main program, such as the ones designed for symbolic calculation or controls), which many times use one-line commands to solve complex system dynamics problems and which have been used in this text. Equally appealing solutions

Preface

to system dynamics problems are the ones provided by Simulink, the graphical user interface program built atop MATLAB, and applications are included in almost all the chapters of solved and proposed exercises that can be approached by Simulink. Through a companion website, the book comprises more ancillary support material, including companion book chapters with extensions to the printed book (with more advanced topics, details of the printed book material, and additional solved examples, this section could be of interest and assistance to both the instructor and the motivated student). Whenever possible, alternative solution methods have been provided in the text to enable using the algorithm that best suits various individual approaches to the same problem. The ancillary material also comprises an instructor’s manual, an image bank of figures from the book, MATLAB code for the book’s solved examples, and PowerPoint lecture slides. After publication and as a result of specific requirements or suggestions expressed by instructors who adopted the text and feedback from students, additional problems resulting from this interaction will be provided on the website, as well as corrections of the unwanted but possible errors. To make distinction between variables, small-cap symbols are used for the time domain (such as f for force, m for moment, or v for voltage), whereas capital symbols denote Laplace transforms (such as F for force, M for moment, or V for voltage). With regard to matrix notation, the probably old-fashioned symbols { } for vectors and [ ] for matrices are used here, which can be replicated easily on the board. Several solved examples and end-of-chapter problems in this book resulted from exercises that I have used and tested in class over the years while teaching system dynamics, and I am grateful to all the students who contributed to enhancing the scope and quality of the original variants. I am indebted to the anonymous academic reviewers who critically analyzed this text. They have made valid suggestions for improvement, which were well taken and applied to this current second edition. I am very thankful to Steven Merken, Senior Acquisition Editor at Elsevier Engineering & Computing Textbooks, whose commitment to this project and quality support has been instrumental in realizing this second edition. I am also grateful to Nate McFadden, Senior Development Editor at Elsevier Engineering & Computing Textbooks, and Mohanapriyan Rajendran, Production Project Manager, for kind and efficient assistance, as well as for timely converting this project from its draft form to its refined print state. In closing, I would like to acknowledge and thank the unwavering support of my wife, Simona, who painstakingly reviewed, checked, and edited the manuscript for the second edition. She definitely made this project possible. As always, my thoughts and profound gratitude for everything they gave me go to my daughters Diana and Ioana and to my parents Ana and Nicolae.

xxi

Resources That Accompany This Book System dynamics instructors and students will find additional resources at textbooks. elsevier.com

AVAILABLE TO ALL Additional Online Content linked to specific sections of the book, extra content includes advanced topics, additional worked examples, and more. Downloadable MATLAB Code for the book’s solved examples.

FOR INSTRUCTORS ONLY Instructor’s Manual The book itself contains a comprehensive set of end-of-chapter problems. Worked-out solutions to the problems are available online to instructors who adopt this book. Image Bank that provides adopting instructors with various electronic versions of the figures from this book that may be used in lecture slides and class presentations. Power Point Lecture Slides Use the available set of lecture slides in your own course as provided, or edit and reorganize them to meet your individual course needs. Instructors should contact their Elsevier textbook sales representative at textbooks@ elsevier.com to obtain a password to access the instructor-only resources.

xxiii

CHAPTER

Introduction

1

This chapter discusses the notion of modeling or simulation of dynamic engineering systems as a process that involves physical modeling of an actual (real) system, mathematical modeling of the resulting physical representation (which generates differential equations), and solution of the mathematical model, followed by interpretation of the result (response). Modeling in this text uses lumped (or pointlike) parameters and is placed in the context of either analysis or design. The dynamic system mathematical model is studied in connection to its input and output signals, such that single-input, single-output (SISO) and multiple-input, multiple-output (MIMO) systems can be formed. Linear systems are categorized depending on the order of the governing differential equations as zero-, first-, second-, or higherorder systems. In addition to the applications usually encountered in system dynamics texts, examples of compliant (or flexible) mechanisms that are incorporated in micro- or nanoelectromechanical systems (MEMS or NEMS) are included here. The nature of presentation is mainly descriptive in this chapter, as it attempts to introduce a few of the concepts that are covered in more detail in subsequent chapters.

1.1 ENGINEERING SYSTEM DYNAMICS Engineering system dynamics is a discipline that studies the dynamic behavior of various systems, such as mechanical, electrical, fluid, and thermal, either as isolated entities or in their interaction, when they are coupled-field (or multiple-field) systems. One trait specific to this discipline is that systems belonging to different physical fields are described by similar mathematical models (expressed most often as differential equations); therefore, the same mathematical apparatus can be utilized to analyze or design different-field systems. This similitude also enables migration between systems in the form of analogies, as well as application of a unitary approach to coupled-field problems. System dynamics relies on previously studied subject matter, such as differential equations, matrix algebra, and physics, and the dynamics of systems (mechanical, electrical, and fluid or thermal), which it integrates in probably the first engineering-oriented material in the undergraduate course work. Engineering system dynamics is concerned with physically and mathematically modeling dynamic systems, which means deriving the mathematical models that govern the behavior System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00001-4 Copyright © 2018 Elsevier Inc. All rights reserved.

1

2

CHAPTER 1 Introduction

(response) of these systems, as well as solving these mathematical models and obtaining the system response. In addition to known modeling procedures, such as Newton’s second law of motion for mechanical systems or Kirchhoff’s laws for electrical systems, the student will learn or reinforce new techniques, such as Lagrange’s equations, direct and inverse Laplace transforms, the transfer function, the state-space approach, and frequency-domain analysis. This course teaches the use of simplified physical models for real-world engineering applications to design or analyze a dynamic system. Once an approximate, yet sufficiently accurate, mathematical model has been derived, one can employ MATLAB, a software program possessing numerous built-in functions, to solve system dynamics mathematical models. Simulink, a graphical user interface computing environment that is built atop MATLAB and that allows using blocks and signals to perform various mathematical operations, can also be used to model, solve, and plot the time response of engineering system dynamics problems. At the end of this course, the student should feel more confident in approaching an engineering design project from the model-based standpoint, rather than the empirical one; this approach should enable selecting the key physical parameters of an actual system, combining them into a relevant mathematical model and finding the solution (either time response or frequency response). Complementing the classical examples encountered in previous courses (such as the rigid body, the spring, and the damper in mechanical systems), new examples are offered in this course of compliant (flexible) mechanisms and MEMS or NEMS. These devices can be modeled using the approach used for regular systems, which is the lumped-parameter procedure (according to which system parameters are pointlike). In addition to being designed as an introduction to actual engineering course work, and as a subject matter that studies various systems through a common prism, engineering system dynamics is also valuable to subsequent courses in the engineering curricula, such as vibrations, controls, instrumentation, or mechatronics.

1.2 MODELING ENGINEERING SYSTEM DYNAMICS The modeling process of engineering system dynamics starts by identifying the fundamental properties of an actual system and its parameters. The minimum set of variables necessary to fully define the system configuration is formed of the degrees of freedom (DOF). Key to this selection is simplifying assumptions that enable retaining the essential features of the system. This process is usually aided by a schematic representation or diagram, which pictorially identifies the parameters and the variables, such as the free-body diagram that corresponds to the dynamics of a pointlike body in mechanical systems with forces and moments shown, and which plays the role of a physical model for the actual system. It is then necessary to utilize an appropriate modeling procedure or method that will result in the mathematical model of the system. Generally, a mathematical

1.2 Modeling Engineering System Dynamics

Actual system

Physical model

Simplifying assumptions

System response

Mathematical model

Mathematical procedure

Solving algorithm

FIGURE 1.1 Flow in the Dynamic System Modeling Process.

model describing the dynamic behavior of an engineering system consists of a differential equation (or a system of differential equations) combining parameters with known functions, unknown functions, and their derivatives. For relatively simple systems, the mathematical model consists of a set of linear ordinary differential equations (ODE), whereas more complex systems may be described by nonlinear or/and partial differential equations. The next step involves solving the mathematical model through adequate mathematical algorithms, analytical or numerical. When the solution is analytically available, it is formed of equations depending on system parameters and time (or frequency) and that reflect the system response or behavior. Figure 1.1 gives a graphical depiction of this process that connects an actual dynamic system under the action of external forcing to its response. There are also situations when interrogation of the system response results in information that is fed back to the actual system at the start of the chain, so it allows for corrections to be applied; this is a topic pertaining to feedback-control systems, which is briefly covered in Chapters 11e13 of this text.

1.2.1 Modeling Variants Various steps can be adopted in transitioning from the actual system to a simplified physical model, then from a physical model to a mathematical one, as sketched in Figure 1.1. Several physical models can be developed, starting from an actual system, depending on the severity (or laxity) of the simplifying assumptions applied. Once a physical model has been selected, several modalities are available to mathematically describe that physical model. The application of different algorithms to one mathematical model should produce the same result or solution, as the system response is unique. In the case of a car that runs on even terrain, the car vertical motion has a direct impact on its passengers. A basic physical model is shown schematically in Figure 1.2, which indicates the car mass is lumped at its center of gravity (CG) and the front and rear suspensions are modeled as springs. Because the interest here lies only in the car vertical motion, and the terrain is assumed even (perfectly flat), it is safe to consider, as a rough approximation resulting in a first-iteration physical model, that the impact points between the wheels and the road surface

3

4

CHAPTER 1 Introduction

Car motion l2

l1

x

CG

θ

Original position

m, J Front suspension

k1

Car body

k2 Rear suspension

Road surface

FIGURE 1.2 Simplified Physical Model of a Car That Moves Over Even Terrain.

are fixed points. Under these simplifying assumptions, the parameters that define the car’s properties are its mass, its mechanical moment of inertia about an axis passing through the CG and perpendicular to the drawing plane, and the spring stiffness of the two suspensions. What is the minimum number of variables fully describing the state (or configuration) of this simplified system at any moment in time? If we attach the system motion to the CG, it follows that the vertical motion of the CG (measured by the variable y) and the rotation (pitch) of the rigid rod (which symbolizes the car body) about a horizontal axis and measured by an angle q are sufficient to specify the position of the car body at any time moment. Of course, we have used another simplifying assumption that the rotations and vertical displacements are relatively small, and therefore the motions of the suspensions at their joining points with the car body (modeled here as a rigid rod) are purely vertical. As a consequence, the system parameters are the car mass m and its moment of inertia J, the suspension spring constants (stiffnesses) k1 and k2, as well as the distances l1 and l2, which position the CG of the car. Generally, all these parameters have known values. The variables (unknowns or DOF) are y, the vertical motion of the CG, and q, the rotation of the body car about its CG. The next step is deriving the mathematical model corresponding to the identified physical model, and this phase can be achieved using a specific modeling technique, such as Newton’s second law of motion, the energy method, Lagrange’s equations, or the state-space representation for this mechanical systemdall these modeling techniques are discussed in subsequent chapters. The result consists of two linear ODE containing the system parameters m, J, k1, k2, l1, l2, and the unknowns y, q, together with their time derivatives. Solving for y and q in terms of initial conditions (for this system, these are the initial displacements when t ¼ 0, namely y(0), q(0), and the initial _ _ velocities yð0Þ; qð0Þ) provides explicitly the functions y(t) and q(t), and this constitutes the system’s response. The system behavior can be studied by plotting y and q as functions of t.

1.2 Modeling Engineering System Dynamics

l1 y

θ

CG

ks1

y1

l2

cs1

m, J

u1

cs2 y 2 m2

m1 kt

ks2

kt ct

ct u2

FIGURE 1.3 Simplified Physical Model of a Car Moving Over Uneven Terrain, With the Degrees of Freedom of the Suspensions Shown.

More complexity can be added to the simple car physical model of Figure 1.2, for instance, by considering that the wheels are separate from the mechanical suspension through the tire elasticity and damping. The assumption of an uneven terrain surface can also be introduced. Figure 1.3 depicts the physical model of the car when all these system properties are taken into accountdplease note that the masses of wheels, tires, and suspensions are included and combined together (they are denoted by m1 and m2 in Figure 1.3), and when the two wheels are considered identical. It can now be seen that two more DOF are added to the existing ones, so that the system becomes a four-DOF system (they are y, q, y1, and y2), whereas the input is formed by the two displacements applied to the front and rear tires, u1 and u2. Dynamic modeling is involved in two apparently opposite directions: the analysis and the design (or synthesis) of a specific system. Analysis starts from a given system whose parameters are known. The dynamic analysis objective is to establish the response of a system through its mathematical model. Conversely, the design needs to find an actual dynamic system capable of producing a specified performance or response. In analysis we start from a real-world, well-defined system, which we attempt to characterize through a mathematical model, whereas in design (synthesis) we embark with a set of requirements and use a model to obtain the skeleton of an actual system. Figure 1.4 gives a graphical representation of the two processes.

Actual system

Mathematical model

System response

(a) Specified system response

Mathematical model

Designed system

(b)

FIGURE 1.4 Processes Utilizing Dynamic Models: (a) Analysis; (b) Design (Synthesis).

5

6

CHAPTER 1 Introduction

1.2.2 Dynamical Systems Lumped-Parameter Modeling and Solution It is convenient from the modeling viewpoint to consider that the parameters defining the dynamic behavior of a system are located at well-specified spatial stations, so they can be considered pointlike. Lumping parameters result in dynamic system models consisting of ODE. The mass of a rigid body, for instance, is considered to be concentrated at the center of mass (gravity) of that body, so that the center of mass becomes representative for the whole body, which simplifies the modeling task substantially, without diminishing the modeling accuracy. Similar lumping considerations can be applied to springs or dampers in the mechanical realm, and also in the electrical domain, where resistances, capacitances, and inductances are considered lumped-parameter system properties. Also, in some cases, the lumped-parameter modeling can be used for components that have inherently distributed properties. Take the example of a cantilever, such as the one sketched in Figure 1.5(a). Both its inertia and elastic properties are distributed, as they are functions of the position x along the length of the cantilever. Chapter 2 shows how to transform the actual distributed-parameter model into an equivalent lumped-parameter model, as in Figure 1.5(b). That approach provides the tip mass meq and stiffness kt (the subscript denotes translation) that are equivalent to the dynamic response of the original cantilever. y

y meq

x

(a)

kt

(b)

FIGURE 1.5 Cantilever Beam: (a) Actual, Distributed-Parameter Inertia and Stiffness; (b) Equivalent, Lumped-Parameter Inertia and Stiffness.

Caution should be exercised when studying complex flexible systems, where the lumping of parameters can yield results that are sensibly different from the expected and actual results, as measured experimentally or simulated by more advanced (numerical) techniques, such as the finite element method. However, for the relatively simple compliant device configurations analyzed in subsequent chapters, lumpedparameter modeling yields results with relatively small errors.

Modeling Methods Several procedures or methods are available for deriving the mathematical model of a specified lumped-parameter physical model. Some of them are specific to a certain

1.3 Elements, System, Input, and Output

system (such as the Newton’s second law of motion, which is applied to mechanical systems or the Kirchhoff’s laws, which are used in electrical systems). Others can be utilized more across the board for all dynamic systems, such as the energy method, the Lagrange’s equations, the transfer function method, and the state-space approach. These methods are detailed in subsequent chapters.

Solution Methods Once the mathematical model of a dynamic system has been derived, and consists of one or more differential equations (linear or nonlinear), the solution can be obtained mainly using two methods. One method is the direct integration of the differential equations, and the other method uses the direct and inverse Laplace transforms. The big advantage of the Laplace method, as will be shown in Chapter 6, consists in the fact that the original, time-defined differential equations are transformed into algebraic equations, whose solution can be found by simpler means. The Laplacedomain solutions are subsequently converted back into the time-domain solutions by means of the inverse Laplace transform. The transfer function and the statespace methods are also employed to determine the time response in Chapters 7 and 8, respectively.

System Response The majority of the systems studied in this text are linear and are modeled with constant-coefficient ODE as initial-value problems. The solution to a linear ODE that describes the system behavior is the sum of two parts: one is the complementary (or homogeneous) solution (which is the solution when no input or excitation is applied to the system), and the other is the particular solution (which is one solution of the equation when a specific forcing or input acts on the system). The complementary solution is representative of the free response, and usually vanishes after a period of time with dissipation present, so it is indicative of the transient response. The particular solution, on the other hand, persists in the overall solution, and therefore defines the forced or steady-state response of the system to a specific input.

1.3 ELEMENTS, SYSTEM, INPUT, AND OUTPUT A system in general (and an engineering one in particular in this text) is a combination of various elements (or components), which together form an entity that can be studied in its entirety. Take for instance a resistor, an inductor, a capacitor, and a voltage source, as shown in Figure 1.6(a); they are individual electrical elements that can be combined in the series connection of Figure 1.6(b) to form an electrical system. Similarly, mechanical elements such as inertia (mass), stiffness, damping, and forcing can be combined in various ways to generate mechanical systems. There are also fluid systems, thermal systems, and systems that combine elements from at least two different fields (or domains) to generate coupled-field (or multiple-field) systems, such as electromechanical or thermoelectromechanical, to mention just two possibilities.

7

8

CHAPTER 1 Introduction

R

+

+ R

L

C

i

v

v

L

–

–

C

(a)

(b)

FIGURE 1.6 (a) Individual Electrical Elements; (b) Electrical System Formed of These Components. y1

y2

k1

k2 m1

c

m2 f1

f2

FIGURE 1.7 MIMO Mechanical System With Linear Motion.

The response of a dynamical system is generated by external causes, such as forcing or initial conditions. It is customary to name the cause that generates the change in the system as input, whereas the resulting response is known as output. A system can have one input and one output, in which case it is a SISO system or it can have several inputs or several outputs, consequently known as MIMO system. A SISO example is the single-mesh series-connection electrical circuit of Figure 1.6(b). For this example the input is the voltage v, whereas the output is the current i. A MIMO mechanical system is sketched in Figure 1.7, where there are two inputs, the forces f1 and f2, and two outputs, the displacements y1 and y2. The car models just analyzed are also MIMO systems, as they all have more than one input or output. The input signals (or forcing functions) that are applied to dynamic systems can be deterministic or random in nature. This text is concerned with deterministic input signals only, which are known, analytic functions of time. Elementary input signals include the step, ramp, parabolic, sine (cosine), pulse, and impulse functions; Figure 1.8 plots these functions in terms of time.

1.4 COMPLIANT MECHANISMS AND MICROELECTROMECHANICAL SYSTEMS In addition to examples that are somewhat classical for dynamics of engineering systems, this text discusses several applications from the fields of compliant mechanisms and MEMS or NEMS. This book demonstrates that, under regular circumstances, simple applications from compliant mechanisms and MEMS can be

1.4 Compliant Mechanisms and Microelectromechanical Systems

u

u

u

A

u= u=

0, t < 0

u=

A, t ≥ 0

t

0, t < 0

A ⋅ t 2, t ≥ 0

A ⋅ t, t ≥ 0

t

(a)

t

(b)

(c)

u

u A τ

0, t < 0

u ∞

A , 0≤ t ≤ τ u= τ 0, t < 0, t > τ

0←τ

t

τ

(d)

0, t < 0

u=

A u = τ→0 τ 0, t < 0, t > τ lim

A ⋅ sin ( ω ⋅ t ) , t ≥ 0

A

t

t

(e)

(f)

FIGURE 1.8 A Few Input Functions: (a) Step; (b) Ramp; (c) Parabolic; (d) Pulse; (e) Impulse; (f) Sinusoidal.

reduced to lumped-parameter (most often) linear systems that are similar to other well-established system dynamics examples. Compliant (flexible) mechanisms are devices that use the elastic deformation of slender, springlike portions, instead of classical rotation or sliding pairs to create, transmit, or sense mechanical motion. The example of Figure 1.9 illustrates the relationship between a classical translation (sliding) joint with regular springs and the corresponding compliant joint formed of flexure hinges (slender portions that bend and enable motion transmission). The compliant device of Figure 1.9(a) is Fixed support Flexure hinges y k Motion direction

Body

k

m k

k

Flexure hinges Fixed support (a)

(b)

FIGURE 1.9 Realizing Translation: (a) Compliant Mechanism With Flexure Hinges; (b) Equivalent Lumped-Parameter Model.

9

10

CHAPTER 1 Introduction

Compliant frame

Flexure hinge Rotation joint

Amplified output

Input

Input

Amplified output Piezoelectric actuator (a)

(b)

FIGURE 1.10 Flexure-Based Planar Compliant Mechanism for Motion Amplification: (a) Photograph of Actual Device; (b) Schematic Representation With Pointlike Rotation Joints.

constrained to move horizontally because the four identical flexure hinges bend identically (in pairs of two), whenever a mechanical excitation is applied along the motion direction. The lumped-parameter counterpart is drawn in Figure 1.9(b), where the four identical flexure hinges have been substituted by four identical translation springs, each of stiffness k. Another compliant mechanism example is the one of Figure 1.10(a), which pictures a piezoelectrically actuated, displacement-amplification device. Figure 1.10(b) is the schematic representation of the actual mechanism, where the flexure hinges are replaced by classical pointlike rotation joints. The schematic shows that the input from the two piezoelectric (PZT) actuators is amplified twice by means of two lever stages. The mechanism is clamped to and offset above the base centrally, as indicated in Figure 1.10(a), and is free to deform and move in a plane parallel to the base plane. As monolithic (single-piece) devices, compliant mechanisms present several advantages over their classical counterparts, such as lack of assembly, no moving parts, and therefore no losses due to friction between adjacent parts, no need for maintenance, and simplicity of fabrication (although at costs that are higher generally compared to classical manufacturing procedures). Their main drawback is that the range of motion is reduced because of the constraints posed by limited deformations of their compliant joints. Compliant mechanisms are encountered in both macroscale applications (with dimensions larger than millimeters) and micro- or nanoscale ones (when the device dimensions are in the micrometer or nanometer range (1 mm ¼ 106 m, 1 nm ¼ 109 m)), particularly in microelectromechanical systems. In many situations, compliant mechanisms are built as single-piece (monolithic) devices with techniques such as wire electro-discharge machining (wire EDM), through water jet machining or by microfabrication techniques (for MEMS), such as surface or bulk micromachining.

1.4 Compliant Mechanisms and Microelectromechanical Systems

Flexure hinges

Output motion Electrostatic sensing

Thermal actuation

Anchor

Input motion Fixed armature Anchor

Output motion

FIGURE 1.11 Top View of Compliant MEMS With Thermal Actuation and Electrostatic Sensing of Motion.

MEMS applications, such as sensors, actuators, pumps, motors, accelerometers, gyroscopes, electrical or mechanical filters, electronic or optical switches, GPS devices (to mention just a few), are encountered in the automotive, defense, medical, biology, computing, and communications domains. Figure 1.11 is the microphotograph of a flexure-hinge thermal microactuator, whose motion is sensed electrostatically by several pairs of capacitors. The entire device floats over a substrate on which it is attached by four anchors (two are shown in the figure). Another MEMS application is sketched in Figure 1.12(a) and 1.12(b). It represents a torsional micromirror, which can be used in several applications such as dynamic redirectioning of incoming optical signals. Attraction/repulsion forces, which can be produced electrostatically or magnetically, act on the side of and underneath the central plate (the mirror) and generate partial rotation of the plate about the axis that passes through the two end elastic hinges. The hinges deform in torsion, hence Fixed support Torsion hinge A

Mobile plate

kr

Fixed plate Mobile plate

Jt

kr

Force

(a)

(b)

(c)

FIGURE 1.12 Torsional Micromirror: (a) Top View; (b) Side View (From A); (c) Equivalent LumpedParameter Model.

11

12

CHAPTER 1 Introduction

the name torsional mirror. Figure 1.12(c) shows the equivalent lumped-parameter model of the actual torsional mirror. Several other MEMS devices are analyzed in subsequent chapters as mechanical, electrical, or coupled-field systems.

1.5 SYSTEM ORDER As mentioned previously, the simplest linear dynamic system is described by a constant-coefficient linear ODE. The order of the differential equation(s) gives the order of the system, as is shown next. For a SISO system, the relationship between the input u(t) and the output y(t) is described by a differential equation of the type n X i¼0

ai $

d i yðtÞ ¼ b $ uðtÞ dti

(1.1)

where ai (i ¼ 0 to n) and b are constant factors, and the input function u(t) can also include derivatives. The maximum derivation order of the output function y(t) in a system of the nth order is n, such that a second-order system, for instance, is defined by a maximum-order input derivative of 2, and so on.

1.5.1 Zero-Order Systems A zero-order system is defined by the equation a0 $ yðtÞ ¼ b $ uðtÞ or yðtÞ ¼ K $ uðtÞ

(1.2)

where K ¼ b/a0 is the constant gain or static sensitivity. The static sensitivity constant reflects the storage nature of a zero-order system, and this is illustrated in the following example. A force f is applied slowly (quasistatically) to a massless body that is elastically supported by two identical springs, as illustrated in Figure 1.13. The static equilibrium requires f ¼ fe, the elastic force being produced by two springs as fe ¼ 2k$y. As a consequence, the following equation results: 1 $f 2k where the static sensitivity is K ¼ 1/(2k), and the input is f ¼ u. y¼

y k

k

FIGURE 1.13 Massless Body With Springs.

f Massless body

(1.3)

1.5 System Order

1.5.2 First-Order Systems First-order systems are described by a differential equation: dyðtÞ dyðtÞ þ a0 $ yðtÞ ¼ b $ uðtÞ or s $ þ yðtÞ ¼ K $ uðtÞ (1.4) dt dt where s is the time constant, and it is defined as s ¼ a1/a0. It can be seen that a firstorder system is described by two constants: the static sensitivity K and the time constant sdthis latter one displays the dissipative side of a first-order system. The thermal system sketched in Figure 1.14 is a first-order system, as shown next. a1 $

θ

Thermometer

Bath θb

FIGURE 1.14 Bath-Thermometer Thermal System.

As detailed in Chapter 5, the heat flow rate stored by the thermometer during contact is dqðtÞ (1.5) dt where c is the specific heat of the thermometer, m is its mass, and q is the thermometer temperature. At the same time, it is known that convective heat transfer between the bath and thermometer is governed by the equation q ¼ m$c$

q ¼ h $ A $ ½qb  qðtÞ

(1.6)

where h is the convection heat transfer (film) coefficient, and A is the thermometer area in contact with the fluid. Equating the heat flow rates of Eqs. (1.5) and (1.6) results in m $ c dqðtÞ $ þ qðtÞ ¼ qb (1.7) h $ A dt which indicates that s ¼ m$c/(h$A) and K ¼ 1. The input is the bath temperature qb, and the output is the thermometer temperature q(t). Figure 1.15 displays a typical first-order system response for the particular case where s ¼ 10 s and qb ¼ 80 C. For a first-order system, the response to a step input can be characterized by the steady-state response, y(N), the rise time (time after which the response gets to 90%

13

14

CHAPTER 1 Introduction

FIGURE 1.15 Thermometer Temperature as a Function of Time (Typical First-Order System Time Response).

of the steady-state response, but other definitions are also applicable), and the settling time (time necessary for the response to stay within 2% of the steadystate response values); more details on this topic are given in the website Chapter 13, which studies the dynamics of control systems.

1.5.3 Second- and Higher-Order Systems Second-order systems are defined by the following differential equation: d2 yðtÞ dyðtÞ þ a0 $ yðtÞ ¼ b $ uðtÞ þ a1 $ (1.8) dt2 dt Division of Eq. (1.8) by a2 and rearrangement of the resulting equation yields a2 $

d2 yðtÞ dyðtÞ þ u2n $ yðtÞ ¼ u2n $ K $ uðtÞ þ 2x $ un $ (1.9) dt2 dt where the new constants, the natural frequency un, and the damping ratio x, are defined as a0 a1 b b a0 ¼ u2n ; ¼ 2x $ un ; ¼ $ ¼ K $ u2n a2 a0 a2 a2 a2

(1.10)

Let us prove that the mechanical system of Figure 1.16 operates as a secondorder system.

1.5 System Order

y

k m

f c

FIGURE 1.16 Mechanical System With Mass, Spring, and Damper.

As detailed in Chapters 2 and 3, the equation of motion is derived by means of Newton’s second law of motion as m$

d2 yðtÞ dyðtÞ d 2 yðtÞ c dyðtÞ k 1  k $ yðtÞ or þ $ yðtÞ ¼ $ f ðtÞ ¼ f ðtÞ  c $ þ $ 2 2 dt dt dt m dt m m (1.11)

where f ¼ u; therefore, the three coefficients defining the second-order system are k c 1 u2n ¼ ; 2x $ un ¼ ; K ¼ (1.12) m m k Based on this mechanical system, Figure 1.17 shows the time response of a typical second-order system for f ¼ 1 N, x ¼ 0.5, un ¼ 100 rad/s, and K ¼ 1. The main characteristics of the time response of a second-order system to a step input are the steady-state response, the rise time, the peak time (time required for the

FIGURE 1.17 Mass Displacement as a Function of Time (Typical Second-Order System Time Response).

15

16

CHAPTER 1 Introduction

Ra

+

Rotor armature

ia vb

va

– La

Jl θ, ma Stator

cr Load

FIGURE 1.18 Schematic of a DC Motor as an Electromechanical System.

response to reach its maximum value), the peak response (the maximum response), and the settling time; all these parameters are studied in more detail in Chapter 13 in the context of controls. Systems of orders larger than two are also encountered in engineering applications, as in the following example, which results in a third-order system model. The electromechanical system of Figure 1.18 consists of a dc (direct-current) motor and load. The dynamic model of this rotary system consists of equations that describe the mechanical, electrical, and mechanicaleelectrical (coupled-field) behavior. Essentially, the electromechanical system sketched in Figure 1.18 is formed of a mobile part (the rotor armature), which rotates under the action of a magnetic field produced by the electrical circuit of an armature (the stator). The electrical circuit is formed of a resistor Ra, an inductor La, and a voltage source va. As discussed in Chapters 4 and 10, the electrical system’s behavior is governed by Kirchhoff’s second law, according to which dia ðtÞ ¼ va ðtÞ  vb ðtÞ (1.13) dt where the subscript a indicates the armature, and vb is the back electromotive force (voltage). The mechanical part of the system is formed of the rotary load inertia Jl and a rotary damper cr; according to Newton’s second law of motion and as detailed in Chapters 3 and 10, the system is governed by the equation Ra $ ia ðtÞ þ La $

d2 qðtÞ dqðtÞ (1.14) ¼ ma ðtÞ  cr $ 2 dt dt where ma is the torque developed due to the statorerotor interaction. It is also known that the following equations couple the mechanical and electrical fields: 8 > < ma ðtÞ ¼ Km $ ia ðtÞ (1.15) dqðtÞ > : vb ðtÞ ¼ Ke $ dt Jl $

1.6 Coupled-Field (Multiple-Field) Systems

with Km and Ke being constants. By combining Eqs. (1.13)e(1.15), the following third-order differential equation is produced:    2  La $ Jl d3 qðtÞ La $ cr Ra $ Jl d qðtÞ Ra $ c r dqðtÞ ¼ va ðtÞ $ þ þ þ þ Ke $ $ dt3 dt2 dt Km Km Km Km (1.16) where va(t) is the input, and q(t) is the output. The order of Eq. (1.16) can be reduced to two by using the substitution u(t) ¼ dq(t)/dt.

1.6 COUPLED-FIELD (MULTIPLE-FIELD) SYSTEMS The previous example illustrated the interaction between mechanical and electrical elements and systems that resulted in an electromechanical system. The corresponding mathematical model is formed of equations pertaining to a single domain or field (either mechanical or electrical) and equations combining elements from both fields. Such a mathematical model is representative of coupled-field (or multiple-field) systems. Another example is presented here in a descriptive manner, and more details on coupled-field systems are given in Chapter 10. Consider the system of Figure 1.19, which is formed of a PZT block with a strain gauge attached to it. Each of these two subsystems has its own electrical circuit. PZT materials essentially deform when an external voltage is applied to them due to the inverse piezoelectric effect. A voltage vi applied between the end points of the bloc sketched in Figure 1.19 generates an axial deformation, which is proportional to the applied voltage. Assuming now that the strain gauge (which is a resistor) is affixed longitudinally to the block, the resistor incurs the same axial deformation as the block;

+ –

Mechanical deformation Strain gauge

Actuation circuit vi

Mechanical deformation

+ vo

PZT linear actuator

– Sensing circuit

FIGURE 1.19 Coupled-Field System With Mechanical, Electrical, and Piezoelectric Elements. PZT, piezoelectric.

17

CHAPTER 1 Introduction

therefore, its resistance changes by a quantity proportional to the mechanical deformation. This resistance change can be sensed in the external circuit and measured as vo. To summarize, an equation is obtained that combines electrical, PZT, and mechanical elements; therefore, this system is a coupled-field one. As illustrated by this example, the PZT block behaves as an actuator when supplied with a voltage generating the mechanical motion. The dc motor is another actuator (or motor) example, where the armature voltage is the source of shaft angular rotation. More generically, an actuator transforms one form of energy (such as electrical, most often) into mechanical energy. A PZT block can also produce voltage when subjected to mechanical pressure or deformation through direct PZT effect. In this case it behaves as a sensor (or generator) by converting one form of energy (mechanical in this particular situation) into electrical energy to perform a measurement (quantitative assessment) operation. A sensor generally aims at measuring the variation of a physical parameter, such as displacement, velocity, acceleration, pressure, electrical resistance, also named measurand, by converting that variation into another parameter’s variation, which can subsequently be processed more easily. The two parameters are usually connected by a linear relationship, which is typical of zero-order systems (see Eq. 1.2), and where the input is the quantity to be measured, while the output is the parameter that measures (the converted quantity). The two parameters are related by the static sensitivity K, as illustrated in Figure 1.20. Being methods of converting one form of energy into another one, actuation and sensing are known collectively as transduction (although in many instances transduction substitutes for sensing). More on transduction is discussed in Chapter 10, and in specialized measurement and instrumentation texts. Converted quantity

18

K Quantity to be measured

FIGURE 1.20 Sensing Characteristics in a Linear Measurement Process.

1.7 LINEAR AND NONLINEAR DYNAMIC SYSTEMS Linearity or nonlinearity of a dynamic system is mathematically associated with the differential equation that defines the behavior of that specific system. A SISO system, for instance, is linear when the coefficients ai of Eq. (1.1) do not depend on

1.7 Linear and Nonlinear Dynamic Systems

the unknown function (response) y(t) and when the unknown function and its derivatives in the left-hand side of Eq. (1.1) are first-degree polynomial functions. Systems where the coefficients ai are not constant (are time variable, for instance) still preserve their linear character. Consider a linear system whose output y(t) is formally expressed in terms of the input u(t) as O(u) ¼ y where O is a linear operator. For an input c$u (c being a constant factor), the resulting output is O(c$u) ¼ c$O(u) ¼ c$y; this property of linear systems is known as scalability or homogeneity. Linear systems are also additive, which means that when the input is a sum of multiple inputs, such as u1 þ u2, the corresponding output becomes O(u1 þ u2) ¼ O(u1) þ O(u2) ¼ y1 þ y2. Combining scaling and additivity results in O(c1$u1 þ c2$u2) ¼ O(c1$u1) þ O(c2$u2) ¼ c1$O(u1) þ c2$O(u2) ¼ c1$y1 þ c2$y2, where c1 and c2 are constant factors. This linear superposition feature allows calculating the output of a dynamic system to a linear combination of multiple inputs, which is c1$u1 þ c2$u2 þ ., by superimposing the system outputs to individual inputs, namely c1$y1 þ c2$y2 þ .. An important consequence of linearity of systems and of their mathematical models is that of enabling direct and inverse Laplace transformations. For instance, Laplace transforming the linear Eq. (1.1) results in Y(s) ¼ G(s)$U(s), where Y(s) and U(s) are the Laplace transforms of the output y(t) and input u(t), while G(s)dthe transfer functiondplays the role of a gain function. The output y(t) can simply be found by inverse Laplace transforming Y(s). In a SISO mechanical system, nonlinearity can be produced by several factors connected to either mass, stiffness, damping, or motion features. Consider the body of mass m of Figure 1.21, which is attached by two identical springs of stiffness k and of undeformed length l. When the body moved a distance x to the right from the equilibrium position, the pffiffiffiffiffiffiffiffiffiffiffiffiffiffi elongation of each of the two identical springs is equal to l2 þ y2  l. By applying

y

k

l m

k

l

y

FIGURE 1.21 Mass With Two Springs in Deformed Position as a Nonlinear Mechanical System.

19

CHAPTER 1 Introduction

Newton’s second law of motion and projecting the two identical spring forces on the horizontal motion direction, the following equation results: d2 yðtÞ þ 2keq $ yðtÞ ¼ 0 dt2 with the equivalent stiffness being 1 0 m$

(1.17)

l C B keq ¼ k $ @1  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA 2 l2 þ yðtÞ

(1.18)

which indicates the stiffness is nonlinear, and therefore the whole mechanical system is nonlinear. The stiffness increase is produced by a “hardening” effect, whereby the stiffness and its slope increase with y. Another variant, where the slope decreases with y, is generated by a “softening” effect. Both behaviors are nonlineardthey are sketched in Figure 1.22, alongside the characteristic of a linear spring. Hardening spring Spring force

20

Linear spring Softening spring Deformation

FIGURE 1.22 Linear, Hardening, and Softening Spring Characteristics.

Another nonlinearity type, particularly encountered in electrical components is saturation. Figure 1.23 shows the voltageecurrent characteristic curve of an inductor; it can be seen that the linear relationship between the voltage and the current changes past a central zone, as the voltage decreases (saturates) when the Voltage

Current – i lin

FIGURE 1.23 Saturation-Type Nonlinearity.

i lin

1.9 Feedback Control of Dynamic Systems

current increases past the linearity limits. As known from electromagnetism, the voltage across an inductor is defined as vL ðtÞ ¼ L $ diðtÞ=dt. Obviously, when the current is confined within the ilin and þilin bounds, the slope di/dt is constant, and the voltageecurrent relationship is linear. This actually is the range utilized in the majority of electrical system calculations.

1.8 TIME- AND FREQUENCY-DOMAIN SYSTEM DYNAMICS Dynamic systems are preponderantly modeled by means of equations that provide the time-domain response, which is y(t) for a SISO system whose input is u(t). Using the Laplace transform (as discussed in Chapter 6), time-domain linear dynamic systems are converted into frequency-domain models, whose independent variable is sdconventionally called frequency. For instance, a constant-coefficient linear ordinary differential equation with the input u(t) and the output y(t) is Laplace transformed into a frequency-domain algebraic equation where the input is U(s), and the output is Y(s). This equation is solved for Y(s), which enables calculating its original, time-domain counterpart y(t). A related frequency-domain modeling approach utilizes the transfer function G(s), which relates the frequency input and output as Y(s) ¼ G(s)$U(s) for a SISO systemdmore details are given in Chapter 7. Frequency-domain modeling is also necessary for harmonic inputs, such as u(t) ¼ U$sin(u$t). The resulting steady-state response (when t / N) in those cases is also harmonic, namely:  Y ¼ U $ jGðu $ jÞj yðtÞ ¼ Y $ sinðu $ t þ 4Þ with (1.19) 4 ¼ ;Gðu $ jÞ As Eq. (1.19) indicates, both the output amplitude Y and phase angle 4 depend on the complex number G(u$j), which, for a SISO system, results from the transfer function G(s) ¼ Y(s)/U(s). Because Y and 4 depend on the input frequency u solely (which is identical to the output frequency), formulating Y(u) and 4(u) is known as frequency-domain response, and the resulting two plots are called Bode plots (see Chapters 9 and 13). This text studies both the time- and the frequency-domain responses of engineering dynamic systems.

1.9 FEEDBACK CONTROL OF DYNAMIC SYSTEMS Both time-domain and frequency-domain approaches fundamentally assume that the response (output) of a dynamic system can be predicted reliably and solely based on an accurate mathematical model of a system for any forcing (input), as illustrated in Figure 1.24(a). In actuality, the system response can and does vary from the modelpredicted one. The main sources of differences between model (ideal) and actual

21

22

CHAPTER 1 Introduction

Input

System model

Output

Input

System model

Control model

Output

Feedback

(a)

(b)

FIGURE 1.24 (a) Uncontrolled Dynamic System; (b) Feedback-Control Dynamic System.

(real-life) system responses are the errors due to physical/mathematical modeling simplifying assumptions, to numerical procedures, and to the presence of unwanted factors, such as noise or disturbances. It therefore becomes necessary to apply some corrective measure to minimize or eliminate the errors. The most common procedure of ensuring that the response of a dynamic system is the desired one uses an additional systemdthe controllerdto interact with the original dynamic system based on output information that is back fed to the controller. Figure 1.24(b) is a simplified, principle representation of this corrective technique, known as feedback control, which is introduced with more details in Chapters 11e13 of this text.

CHAPTER

Mechanical Elements

2

OBJECTIVES In this chapter you will learn about: • Springs, heavy bodies, and viscous dampers, which are the main mechanical elements of systems undergoing translation or/and rotation. • Lumped-parameter properties of and energy associated with translatory and rotary mechanical elements: stiffness, inertia, and damping. • Transforming distributed-parameter inertia properties of elastic members such as bars and beams into equivalent lumped-parameter properties. • Transforming lumped-parameter mechanical properties by series/parallel connection, gears, and levers.

INTRODUCTION This chapter introduces the basic mechanical elements that are used as building blocks for translatory and/or rotary mechanical systems. These elements are the springs, which describe elastic properties through stiffness (or spring constants), the heavy solids, which carry inertial or mass properties, and the viscous dampers, which qualify a widely used type of energy losses through damping coefficients. The stiffness, mass (or mass moment of inertia), and damping coefficient are formulated as lumped-parameter (or pointlike) element properties in terms of the related forces/moments, as well as in terms of the associated energy. Flexible, distributed-parameter stiffness, and inertia of members such as bars and beams are rendered into equivalent, lumped-parameter elements. Examples are also offered of transferring basic lumped-parameter mechanical element properties into equivalent ones through motion transformers such as levers and gears.

2.1 SPRING ELEMENTS Springs serving as elastic support for translatory and rotary motion are studied in this section in relation to their lumped-parameter stiffness (or spring constant) k. The stiffness and elastic energy of basic springs and springs connected in series or parallel are presented. System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00002-6 Copyright © 2018 Elsevier Inc. All rights reserved.

23

24

CHAPTER 2 Mechanical Elements

2.1.1 Basic Spring Elements and Stiffness Springs are mechanical elements that generate elastic forces in translatory motion and elastic torques in rotary motion; these elastic reactions oppose the spring deformation and are proportional to the spring linear or angular deformation. Figure 2.1 sketches one of the most common spring configurationsdthe helical springdwhich can be used for either axial (translatory) deformation or torsiongenerated (rotary) deformation. The parameters defining the helical spring are the mean diameter D, the wire diameter d, the number of active turns N, and the material shear and Young’s moduli, G and E. For a general spring whose end points undergo the displacements x1 and x2 (as shown in Figure 2.1), the elastic force developed in the spring, fe, is proportional to the spring deformation, which is the difference between the two end point displacements. The elastic force is expressed in Table 2.1, where kt is the translatory spring stiffness. Similarly, an elastic torque is generated by a spring in rotation whose end points undergo the rotations q1 and q2 (see Figure 2.1). The elastic torque me is also provided in Table 2.1 where kr is the rotary spring stiffness. These equations assume the springs are linear and therefore the stiffness is constant. A deformed spring stores

D

d

x1

kt

Translation

Rotation

x2

θ1

kr

θ2

FIGURE 2.1 Helical Spring With Symbols for Translatory and Rotary Motion.

Table 2.1 Elastic Force/Moment and Elastic Potential Energy for Translatory and Rotary Springs Translation

Rotation

Elastic force/ moment

fe ¼ kt $ Dx ¼ kt $ ðx1  x2 Þ

me ¼ kr $ Dq ¼ kr $ ðq1  q2 Þ

Elastic potential energy

Ue ¼ 12 $ kt $ ðDxÞ2 ¼ 12 $ kt $ ðx1  x2 Þ2

Ue ¼ 12 $ kr $ ðDqÞ2 ¼ 12 $ kr $ ðq1  q2 Þ2

2.1 Spring Elements

elastic potential energy (Ue) and equations corresponding to a translatory spring as well as to a rotary spring are included in the same Table 2.1. Because we are interested only on the relative displacement of the end pointsd that connect the spring to other elementsdthe stiffness is lumped and is known as lumped-parameter stiffness. Tables 2.2 and 2.3 give the description, symbols, and corresponding stiffnesses of a few of the most utilized springs in translation/rotation applications, whereas the companion website Chapter 2 provides mathematical demonstrations of these stiffnesses. In these tables, A represents the member’s cross-sectional area, Iz is the cross-sectional (or second) moment of area about the Table 2.2 Basic Translatory Springs and Their Lumped-Parameter Stiffnesses Configuration

Model

Stiffness

fy, uy

E, Iz l

Free end

Fixed end

Cantilever beam

kt ¼ 3El3$ Iz l

fy, uy kt

fy, uy

Gudied end

Fixed end

Fixed-guided beam

E, I z l

kt ¼ 12El3 $ Iz

fy, uy

E, I z l/2

Fixed end

Fixed end

Bridge (fixed-fixed beam)

l

fy, uy

l/2

$ Iz kt ¼ 192E l3

kt

l

Fixed-free bar Free end

Fixed end E, A

kt ¼ E $l A

fx, ux

l

kt

Fixde end

Free end

Fixed-free helical spring G, I t

l

fx, ux

fx, ux

l 4

G$d kt ¼ 8N $ D3

25

CHAPTER 2 Mechanical Elements

Table 2.3 Basic Rotary Springs and Their Lumped-Parameter Stiffnesses Configuration

Model

Stiffness mz, θz

mz, θz

E, Iz l

kr

Free end

Fixed end

Cantilever beam with end bending moment

kr ¼ E $l Iz

l

E, I z, l Fixed end

Free end

Fixed-free spiral spring mz, θz kr

Fixed end

Fixed-free helical spring in torsion E, I z l Free end

mx, θx

l

mx, θx G, I p I

E $d kr ¼ Elt$ I ¼ p $64l t lt, total wire length; d, cross-section diameter 4

4

E $d kr ¼ p E$ N$ I$z D ¼ 64N $D

kr

mx, θx

l

Free end

Fixed-free bar in torsion Fixed end

26

kr ¼

G $ Ip l

z-axis (axis perpendicular on the xy plane of drawings), and Ip is the polar moment of area of the circular cross section. Note that the rotary stiffness of a helical spring is identical to that of a spiral spring in Table 2.3 because the total wire length of a helical spring is lt ¼ N$p$D.

2.1.2 Series and Parallel Spring Connections To vary the stiffness properties of a mechanical system, springs can be combined in series or in parallel, as sketched in Figure 2.2(a) and 2.2(b) where it is assumed that the serial and parallel translatory spring chains are fixed at one end and acted on at the opposite end by a force f. However, the analysis that follows is also valid when both ends are mobile, as well as for rotary springs. For springs connected in series, the force is the same in each component and is equal to f, whereas the total deformation is the sum of individual spring deformations. Conversely, for parallel spring combinations, the spring displacements are

2.1 Spring Elements

x2

x1

k1

k2 (a)

k1 f

x

x k1

k2

f k2

f

(b)

(c)

FIGURE 2.2 Translatory Spring Combinations: (a) Series; (b) and (c) Parallel.

identical, whereas the sum of individual spring forces equals the force f. The spring combination of Figure 2.2(c) is a parallel one with respect to the reference x because the two springs undergo the same deformation at that point. The equivalent series stiffness ks and the parallel stiffness kp corresponding to n spring elements of stiffnesses k1, k2, ., kn are derived in the companion website Chapter 2 as: 1 1 1 1 ¼ þ þ / þ ; kp ¼ k1 þ k2 þ / þ kn ks k1 k2 kn

(2.1)

nn

Example 2.1 Four identical translatory helical springs are combined in two arrangements, and in each of the two combinations there are both series and parallel connections. When the same force is applied separately to each spring arrangement at the free end (the other one being fixed), the free-end displacement of one arrangement is maximum, whereas the free-end displacement of the other arrangement is minimum. Identify the two spring configurations, determine the ratio of their free-end displacements, and calculate the equivalent stiffness for each when d ¼ 1 mm, D ¼ 12 mm, N ¼ 10, and G ¼ 160 GPa.

Solution The largest displacement is obtained when all four springs are coupled in series because the stiffness is minimum, namely ks ¼ k/4dsee Eq. (2.1). Conversely, the smallest displacement corresponds to a full parallel spring connection when the stiffness is maximum, kp ¼ 4k, as also indicated in Eq. (2.1). However, these connections are not allowed in this example. Two spring configurations (1 and 2) are sketched in the top row of Figure 2.3, which are candidates satisfying this example’s requirement to mix series and parallel pairs. Configuration 1 results in the maximum stiffness among all possible designs of four springs, with at least two connected in series. Similarly, Configuration 2 yields the minimum stiffness of all four-spring variants with at least two springs coupled in parallel. The two springs of the top branch in Figure 2.3(a) are combined in series and the two springs in the portion at the free end at the top in Figure 2.3(b) are connected in parallel; as a consequence, the original designs on the top row of Figure 2.3 become the equivalent configurations shown in the bottom row of the same Figure. The equivalent stiffnesses of the spring connections shown in Figure 2.3 are:

8 k 5 > > > < k1 ¼ 2 þ k þ k ¼ 2 $ k > 1 1 1 1 5 2 > > or k2 ¼ $ k : ¼ þ þ ¼ k2 k k 2k 2k 5

(2.2)

27

28

CHAPTER 2 Mechanical Elements

Configuration 1 x1 k k

Configuration 2

k k

f

k

x2

k

f k

k

k/2 k

x1 x2 f

k

k

2k

f

k

(a)

(b)

FIGURE 2.3 Translatory Spring Combinations With: (a) Three Parallel Branches; (b) Three Series Portions. where the subscripts 1 and 2 identify configurations 1 and 2, respectively. Because f ¼ k1$x1 ¼ k2$x2, Eq. (2.2) yield:

x2 k1 ð5kÞ=2 25 ¼ ¼ ¼ 4 x1 k2 ð2kÞ=5

(2.3)

By using the equation of a translatory helical spring given in Table 2.2 and the specified numerical values, the following results are obtained: k ¼ 1157.4 N/m, k1 ¼ 5k/2 ¼ 2893.5 N/m, and k2 ¼ 2k/5 ¼ 462.96 N/m. nn

nn

Example 2.2 The symmetric shaft-disk system of Figure 2.4(a) consists of identical and mirrored portions, each formed of two shaft segments that are anchored at their distant ends and connected to the middle rigid gear at their opposite ends. The fixed-end shafts have a diameter d1 ¼ 0.004 m and a length l1 ¼ 0.1 m, whereas the other shafts have a diameter d2 ¼ 0.005 m and a length l2 ¼ 0.14 m. The shear modulus is G ¼ 80 GPa. (i) Calculate the torsion moment mx that has to be applied at O to produce a maximum rotation angle qx,max ¼ 5 degrees at the same point. (ii) Calculate the number of active turns N of a rotary helical spring with fixed-free ends to replace all the actual shafts of Figure 2.4(a). The single spring produces the angle qx,max of (i) under the action of the same moment mx. The spring is formed of a wire with diameter d ¼ 0.005 m, Young’s modulus E ¼ 210 GPa, and mean diameter cannot exceed Dmax ¼ 0.03 m.

Solution

(i) As illustrated in Figure 2.4(b), the two shaft segments to the left of point O are serially connected, and the same applies for the two mirrored and identical shafts to the right of O; the stiffness of

2.1 Spring Elements

Flexible shaft d2 d1

Flexible shaft d2 d 1

O

l2

l2

k1 x

Gear l1

Anchor

Anchor

θx, mx θx, mx k2

O

keq

k1 k2

θx, mx

x

x O

l1

(a)

(b)

(c)

FIGURE 2.4 Symmetric Shaft-Disk System: (a) Schematic Configuration; (b) Lumped-Parameter Spring Model; (c) Equivalent Rotary Spring Model. each of these two segments is calculated with respect to O based on Eq. (2.1) as: ks ¼ k1 $ k2 =ðk1 þ k2 Þ. Because these two serial portions are jointed at O, they undergo the same rotation and therefore are coupled in parallel with respect to O. A single-spring model is shown in Figure 2.4(c) and the equivalent stiffness with respect to O is:

keq ¼ 2ks ¼ 2 $

k1 $ k2 k1 þ k2

(2.4)

Based on Table 2.3 (the particular design of a fixed-free bar in torsion), the rotary stiffnesses of the   two flexible shaft segments are expressed as k1 ¼ G $ Ip1 l1 ; k2 ¼ G $ Ip2 l2 where the polar   area moments of the two bars are (see Appendix F): Ip1 ¼ pd14 32; Ip2 ¼ pd24 32. As a consequence, the equivalent spring stiffness of Eq. (2.4) becomes:

keq ¼

2G $ Ip1 $ Ip2 pG $ d14 $ d24  ¼  Ip1 $ l2 þ Ip2 $ l1 16 l $ d 4 þ l $ d4 1 2 2 1

(2.5)

As per Table 2.1, the maximum torque to be applied at the midpoint O is maximum because the rotation angle at that point is maximum, namely:

mx ¼ mx;max ¼ keq $ qx;max ¼

pG $ d14 $ d24 $ qx;max   16 l1 $ d24 þ l2 $ d14

(2.6)

With the numerical values of this example, the equivalent stiffness is keq ¼ 25.557 N m and the maximum moment is mx,max ¼ 2.23 N m. (ii) The single rotary spring to replace all the elastic shaft portions of the original system connects to the shaft at one end and is fixed at the other end, as shown in Figure 2.4(c). The equivalent stiffness value (determined at (i)) should be matched by the stiffness of a torsional helical spring as provided in Table 2.3, namely: keq ¼ kr ¼ Ed4/(64ND). The number of active turns of the helical spring is therefore evaluated as:

N¼

E $ d4 E $ d4 or Nmin ¼ 64D $ keq 64Dmax $ keq

(2.7)

With D ¼ Dmax and the numerical value of keq obtained at (i), the minimum number of active turns is Nmin ¼ 2.675. The first integer larger than Nmin is 3; as a consequence, N ¼ 3 is a value that satisfies the requirements. nn

29

30

CHAPTER 2 Mechanical Elements

nn

Example 2.3 The microscale mechanism of Figure 2.5(a) is formed of a serpentine springdwhose details are provided in Figure 2.5(b)dand a bridge (a beam that is fixed at both ends). The two components are initially separated by a gap Dy. (i) Find the minimum force fmin acting at A along the y axis, which needs to be applied to the serpentine spring to close the gap Dy to the midpoint B of the bridge. (ii) Calculate the deflection at the bridge midpoint when a force f1 ¼ 3$fmin is applied at A on the serpentine spring along the y direction with the device in its original (detached) position. All flexible segments are of identical circular cross section with diameter d and of identical material properties. Known are: d ¼ 2 mm, l ¼ 100 mm, Young’s modulus E ¼ 160 GPa, and Dy ¼ 5 mm. 2l y

y

l

l B

Δy

fmin Rigid connector

ks

Long beam

f A

y

A kl

Bridge

ks Short beam

Serpentine spring O

l

Anchor

(a)

(b)

(c)

FIGURE 2.5 Translatory Microspring System: (a) Serpentine Spring and Bridge; (b) Serpentine Spring; (c) Equivalent Lumped-Parameter Model of the Serpentine Spring.

Solution (i) The three beams of the serpentine unit of Figure 2.5(b) act as springs in series and the lumpedparameter model of the serpentine spring alone is that of Figure 2.5(c). The stiffness of the serpentine spring is:

1 1 1 1 2 1 ¼ þ þ ¼ þ kse ks kl ks ks kl

(2.8)

where ks and kl are the stiffnesses of the short and long beams, respectively. It should be noted that in terms of their end-point deformations, the three beams are fixed-guided, and therefore, their stiffnesses are calculated based on Table 2.2 and Appendix F as:

ks ¼ 12 $ ¼ 12 $

E $ Iz E p $ d 4 3p $ E $ d 4 E $ Iz ¼ ¼ 12 $ $ ; kl ¼ 12 $ 3 3 3 l l 64 16l ð2lÞ3 E ð2lÞ

$ 3

p $ d 4 ks ¼ 64 8

(2.9)

By combining Eqs. (2.8) and (2.9), the stiffness of the serpentine spring becomes:

1 2 8 10 ks 6E $ Iz 3p $ E $ d4 ¼ ¼ þ ¼ or kse ¼ ¼ kse ks ks ks 10 5l3 160l3

(2.10)

The force fmin to be applied at the free end of the serpentine spring (as shown in Figure 2.5(a)) to close the gap Dy is fmin ¼ kse $ Dy . The numerical values of the serpentine spring stiffness and minimum force are kse¼ 0.151 N/m and fmin ¼ 0.754 mN.

2.2 Inertia Elements

f f1

kb A kse

Applied at A

y B

fmin O

(a)

N

f1 M

P

Δy Δy1

Δy

(b)

FIGURE 2.6 (a) Lumped-Parameter Model of Coupled Serpentine Spring and Bridge; (b) Spring Characteristics Before and After Contact. (ii) Once point A on the serpentine has been moved to coincide with the midpoint B of the bridge, the resulting spring system is shown in Figure 2.6(a). From this stage on, the force f1 works against both the serpentine spring and the bridge (because the two springs are coupled in parallel with respect to point A h B) and therefore the resulting equivalent stiffness is:

keq ¼ kse þ kb ¼

6E $ Iz 192E $ Iz 126E $ Iz þ ¼ ¼ 21 $ kse 5l3 5l3 ð2lÞ3

(2.11)

where kb is the bending stiffness of the bridge with respect to its midpoint and is provided in Table 2.2. The two spring characteristics, corresponding to the serpentine alone (deformation between 0 and Dy) and then to the coupled serpentine-bridge springs (deformation between Dy and Dy1), are illustrated in Figure 2.6(b). The total deformation Dy1 is calculated as:

  fmin f1  fmin fmin 2fmin 1 2 Dy1 ¼ Dy þ kMPk ¼ þ ¼ þ ¼ fmin $ þ kse kse þ kb kse keq kse keq   2kse 23 ¼ 1þ $ Dy ¼ $ Dy 21 kse þ kb (2.12)

which is Dy1 ¼ 5.476 mm. Eq. (2.12) took into account that the extra deformation equal to the length of segment MP is produced by a force equal to f1  fmin ¼ 3fmin  fmin ¼ 2fmin. nn

2.2 INERTIA ELEMENTS For rigid bodies, inertia properties can be considered pointlike; therefore, inertia features corresponding to either translatory or rotary motion are naturally lumped. Inertia is represented by mass (usually denoted by m) in translatory motion and mechanical (or mass) moment of inertia (generally symbolized by J to avoid confusion with the cross-section moment of area identified by I in this book) in rotary motion, as sketched in Figure 2.7. Lumped-parameter inertia can also be obtained from elastically deformable members such as bars or beams, whose inertia is distributed, by means of energy-equivalence methods.

31

32

CHAPTER 2 Mechanical Elements

Guide Body

θ

Rotor

Shaft

x

m

J Bearings Central axis

l

θ

x

d

θ

Δ

dm m

(a)

r

Offset parallel axis

J

m

(b)

(c)

(d)

FIGURE 2.7 (a) Lumped-Parameter Mechanical Inertia and Related Symbol for Translatory Motion; (b) and (c) Lumped-Parameter Mechanical Inertia and Related Symbol for Rotary Motion; (d) Schematic for the Parallel-Axis Theorem.

2.2.1 Basic Inertia Elements The mechanical moment of inertia of a body of mass m rotating about an axis, Figure 2.7(c), is calculated as: Z J ¼ r 2 dm (2.13) m

where r is the distance from the rotation axis to an element of mass dm. Table 2.4 provides the mass moments of inertia for a mass particle, a heavy rod, a solid cylinder/disk, and a plate (out-of-plane thickness is h)dfor the latter two bodies, the x-axis is the central axis. The parallel-axis theorem, which is illustrated in Figure 2.7(d), calculates the mechanical moment of inertia of a body rotating around an axis D that is offset from the body’s central axis in terms of the distance d between axes and the body mass m as: JD ¼ J þ m $ d 2

(2.14)

where J is the mechanical moment of inertia of the body about its central axis. Table 2.4 Mass Moments of Inertia for a Few Mechanical Parts Mechanical part

Mass particle O

Mass moment of inertia

Heavy rod O

l m

JO ¼ m $ l 2

l/2

Cylinder

l/2

1 $ m $ l2 JO ¼ 12

Plate

m

m x

Thick: h 2R

Jx ¼ 12 $ m $ R2

x

m w

  1 $ m $ w 2 þ h2 Jx ¼ 12

2.2 Inertia Elements

Table 2.5 Inertia Force/Moment and Kinetic Energy for Translatory and Rotary Motion Translation

Rotation

Inertia force/moment

fi ¼ m $ a ¼ m $ x€

€ mi ¼ J $ a ¼ J $ q

Kinetic energy

T ¼ 12 $ m $ v2 ¼ 12 $ m $ x_2

T ¼ 12 $ J $ u2 ¼ 12 $ J $ q_

2

For translation, the mass is included in the inertia force, which is proportional to the acceleration (the second time derivative of displacement) and is calculated as shown in Table 2.5 where fi has the same direction as x in Figure 2.7(a). Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)dsee Table 2.5 where mi has the same direction as q in Figure 2.7(b) and 2.7(c). Bodies in motion possess kinetic energy, denoted by T, which is expressed for translation and rotation as a function of the corresponding velocity in the same Table 2.5. The mass moment of inertia is an additive/subtractive amount, and this property is used in calculating the mass moment of inertia of compound bodies formed of several elementary bodies whose moments of inertia are readily available. nn

Example 2.4 (i) Calculate the mass moment of inertia about the centroidal (symmetry) axis of the solid right circular cone frustum shown in Figure 2.8(a) in side view and in a cross-section. The frustum is defined by d1, d2, and l. The material is homogeneous with a mass density r. Use the obtained result to also calculate the mass moment of inertia of a cylinder, both about its centroidal axis and about a parallel axis that is offset at a distance d ¼ d2 from the centroidal axis. l Section A– A

A

d1

dx

d2

rx

d22 d21

r

d3

dr x

dx

l2

l1

A

(a)

(b)

FIGURE 2.8 (a) Frustum of Right Circular Conical Solid in Side View and Cross-Section; (b) Simplified Longitudinal Section of a Bevel Gear.

d1

33

34

CHAPTER 2 Mechanical Elements

(ii) Calculate the mass moment of inertia of the simplified bevel gear of Figure 2.8(b) with respect to its longitudinal axis in terms of the given dimensions. Numerical application: d1 ¼ 0.015 m, d21 ¼ 0.016 m, d22 ¼ 0.024 m, d3 ¼ 0.008 m, l1 ¼ 0.01 m, l2 ¼ 0.015 m, and r ¼ 7800 kg/m3.

Solution

(i) For a homogeneous cone frustum of mass density r, such as the one of Figure 2.8(a), the mechanical moment of inertia is expressed as:

Z J ¼ r$

Z V

r 2 dV ¼ r $

h

0 @

1

Z

0

A

r 2 dAAdx

(2.15)

As shown in Figure 2.8(a), the area of an elementary circular strip of width dr and inner radius r is dA ¼ 2prdr. By substituting dA into Eq. (2.15), the mass moment of inertia of the cone frustum becomes:

Z

l

J ¼ 2pr $

0 @

Z

0

0

rx

1

r 3 dr Adx ¼

pr $ 2

Z 0

l

rx4 dx

(2.16)

The variable external radius rx of Figure 2.8(a) is calculated as:

rx ¼

d1 d2  d1 þ $x 2 2l

(2.17)

Symbolic mathematical calculations can be performed by using MATLAB Symbolic Math Toolbox. Integrals are evaluated by means of the int(expr, x, a, b) MATLAB command, where expr is the expression to be integrated, x is the integration variable, and a and b are the integration limits. The following MATLAB sequence is used to calculate the mass moment of inertia of Eq. (2.16): >> syms d1 d2 r l x rho %defines r1, r2, r, x, l, rho as symbolic >> rx = d1/2+(d2-d1)/(2*l)*x; >> J = simplify(pi*rho/2*int(rx^4, x, 0, l)) which returns:

J¼

 pr $ l  4 $ d1 þ d13 $ d2 þ d12 $ d22 þ d1 $ d23 þ d24 160

(2.18)

When d1 ¼ d2 ¼ d, the cone frustum becomes a cylinder and Eq. (2.18) simplifies to:

  2 pr $ l  4  1 d4 1 pd2 d 1 1 $ 5d ¼ $ pr $ l $ ¼ $ l $ r $ ¼ $ m $ d2 ¼ $ m $ r2 J¼ 160 2 8 2 16 2 4 4 (2.19) with m being the mass of the cylinder. The cylinder’s mass moment of inertia about an axis situated at d ¼ d2 from its centroidal axis is found from Eq. (2.19) by means of the parallel-axis theorem, Eq. (2.14), as:

1 9 JD ¼ J þ m $ d22 ¼ $ m $ d2 þ m $ d2 ¼ $ m $ d2 8 8

(2.20)

2.2 Inertia Elements

(ii) The bevel gear sketched in Figure 2.8(b) is geometrically formed by adding a cylinder to a conical frustum and subtracting another cylinder. Similarly, the mass moment of inertia of the gear shown in Figure 2.8(b) is calculated by algebraic addition of the moments of inertia defining the three abovementioned elements, namely:

J ¼ J1 þ J2  J3    d12 1 pr $ l2  4 3 2 2 3 4 d21 þ d21 ¼ $ r p $ $ l1 $ d12 þ $ d22 þ d21 $ d22 þ d21 $ d22 þ d22 8 4 160

2 d 1  $ r p $ 3 $ ðl1 þ l2 Þ $ d32 8 4 (2.21) and its numerical value is J ¼ 2.29$106 kg m2.

nn

2.2.2 Lumped-Parameter Inertia of Distributed-Parameter (Elastic) Members Elastic members possess distributed inertia properties; for them, the distributed inertia can be transformed into approximate equivalent lumped-parameter inertia parameters (either a point mass, meq, if the member’s deformation is translatory or a mechanical moment of inertia, Jeq, if the elastic member undergoes rotary deformations) by equating the kinetic energies of the actual and equivalent systems when the two systems undergo identical motions. The distributed inertia of beams in bending and bars under axial load or in torsion can be substituted by equivalent pointlike (lumped-parameter) inertia properties, which are located at convenient points on the original flexible members. Tables 2.6 and 2.7 illustrate this process. The companion website Chapter 2 gives the derivations of these lumped-parameter inertia properties. In Table 2.6 m is the member’s actual mass, l is the length, and r is the mass density. In Table 2.7, Jp is the polar mass moment of inertia of the bar undergoing torsion with respect to its longitudinal axis. nn

Example 2.5 The mechanical system of Figure 2.9(a) uses a piezoelectric actuator in the form of a fixed-free, circularecross-section flexible bar (diameter is d1 ¼ 0.008 m) of length l1 ¼ 0.03 m and a pointlike ball of mass m3 ¼ 0.005 kg at its tip. By linear expansion, the bar actuator pushes the midpoint of a bridge with rectangular cross section (dimensions are h ¼ 0.001 m and w ¼ 0.014 m) and of length l2 ¼ 0.04 m. Find the lumped-parameter mass meq that is inertially equivalent to the actual system and that is placed at A when considering the x-axis translatory motiondsee Figure 2.9(b). Determine the error in this equivalent mass when ignoring the inertia contribution from the bridge. Known are also the mass densities of the bar r1 ¼ 7500 kg/m3 and bridge r2 ¼ 7800 kg/m3.

35

CHAPTER 2 Mechanical Elements

Table 2.6 Equivalent Mass Fractions of Elastic Members Under Axial Load or Bending Configuration

Model

Free end

Fixed end

ux

ρ, A

Equivalent Inertia ux

Fixed-free bar

l

meq

meq ¼ 13 $ m ¼ 13 $ r $ A $ l

l

Free end

Fixed end

Cantilever beam under end force uy

ρ, A l

33 $ m ¼ 33 $ r $ A $ l meq ¼ 140 140

meq l

uy

ρ, A l

13 meq ¼ 13 35 $ m ¼ 35 $ r $ A $ l

Fixed end

Fixed end

Bridge (fixed-fixed beam) uy

ρ, I l/2

uy

Guided end

Fixed end

Fixed-guided beam

uy

meq

l/2 l

l

Table 2.7 Equivalent Mechanical Moments of Inertia of Bars in Torsion Configuration

Model

l

Free end

Fixed end

θx

Jp

l

l/2 l

Fixed end

θx

θx Jeq, x

Fixed-fixed bar in torsion Jp

Equivalent Inertia Jeq,x

Fixed-free bar in torsion

Fixed end

36

l/2 l

θx

1 Jeq;x ¼ $ Jp 3

 Jp ¼ r $ l $ p $ d4 32

2.2 Inertia Elements

Anchor Bridge

Anchor

Flexible bar

l2/2

Ball d1

meq

x

x

A

A l1

l2/2

l1

h Anchor

(a)

(b)

FIGURE 2.9 (a) Flexible Bar With Axial Inertia and Bridge With Bending Inertia; (b) Inertially Equivalent Lumped-Parameter (Point) Mass.

Solution

The equivalent mass to be placed at A when all inertia contributions are considered is the sum of the equivalent mass of the bar, the equivalent mass of the bridge, and the actual mass of the ball, namely:

1 13 meq ¼ meq;1 þ meq;2 þ m3 ¼ $ m1 þ $ m2 þ m3 3 35

(2.22)

Eq. (2.22) considers the equivalent masses of a fixed-free bar and of a bridge as given in Table 2.6. Because the masses of the bar and of the bridge are actual m1 ¼ r1 $ p $ d12 $ l1 4 and m2 ¼ r2 $ w $ h $ l2 , the equivalent mass of Eq. (2.22) becomes:

meq ¼

pd12 $ l1 $ r1 13w $ h $ l2 $ r2 þ þ m3 12 35

(2.23)

with a numerical value of meq ¼ 0.0104 kg. When the bridge inertia is not included, the resulting equivalent mass is obtained from Eq. (2.23) as:

meq ¼

pd12 $ l1 $ r1 þ m3 12

(2.24)

 ¼ 0:0088 kg. The relative error between the equivalent masses of Eqs. (2.23) and and its value is meq .   meq ¼ 15:61%. (2.24) is meq  meq nn

nn

Example 2.6 The microsensor of Figure 2.10, which is formed of two flexible end segments and a central rigid plate, is used to sense rotation around the x axis and translation along the y axis. Calculate the equivalent mass moment of inertia around the x axis and the equivalent mass in translation along the y axis, taking into account inertia contributions from all system’s parts; compare these values to the ones that do not consider the flexible members inertia. The circular cross-section bars have a diameter d ¼ 20 mm; the other parameters are l ¼ 310 mm, a ¼ 150 mm, b ¼ 320 mm, h ¼ 50 mm (plate thickness), and r ¼ 5600 kg/m3 (mass density) for plate and bars.

37

38

CHAPTER 2 Mechanical Elements

y Rigid plate a/2

Flexible bar

O

x

a/2 d b/2 l

b

l

FIGURE 2.10 Mechanical Microsensor With Flexible Segments and Central Rigid Plate.

Solution

The plate rotation about the x-axis involves torsion of the two end bars; therefore, the equivalent mass moment of inertia is composed of the plate’s own mass moment of inertia and inertia contributions from the 2 bars undergoing torsion, as given in Table 2.7:

Jeq;x

  Jb;x r $ a $ b $ h $ a2 þ h2 1 r $ l $ p $ d4 þ 2$ $ ¼ ¼ Jp;x þ 2 $ 3 12 3 32 4

 2  r p $ l $ d ¼ $ a $ b $ h $ a þ h2 þ 12 4

(2.25)

where the subscripts p and b denote plate and bar, respectively; the parallelipedic plate’s mass moment of inertia is given in Table 2.4. When the bar’s inertia is not accounted for, the x-axis rotation inertia comes only from the plate and is Jp,x of Eq. (2.25). With the numerical values of the example, the mass moments of inertia given in Eq. (2.25) are Jeq,x ¼ 2.802$1017 kg m2 and Jp,x ¼ 2.8$1017 kg m2. The bars inertia contribution is very smalldapproximately 0.074% as it results from calculating (Jeq,x  Jp,x)/Jeq,x$100. The total inertia in the system’s translation along the y-axis results from the two end beams bending around the z axis (which is perpendicular to the xy plane of the figure), as well as from the plate’s own mass, namely:

13 26 p $ d2 meq ¼ mp þ 2 $ $ mb ¼ r $ a $ b $ h þ $ r $ l $ 35 35 4

13 ¼ r $ a $ b $ h þ $ p $ l $ d2 70

(2.26)

where the equivalent mass corresponding to the bending of a fixed-guided beam is given in Table 2.6. The y-axis translation inertia without the beams considered is simply the mass of the plate mp, which is expressed in Eq. (2.26). With the given numerical values, the equivalent masses of Eq. (2.26) are meq ¼ 1.3845$108 kg and mp ¼ 1.344$108 kg. The relative error between these values is (meq  mp)/meq$100 ¼ 2.92%, which indicates a contribution of the beams in the total translatory inertia that is slightly larger than the one in rotation. nn

2.3 VISCOUS DAMPING ELEMENTS Damping is associated with energy losses, and the main mechanical damping types are viscous, frictional, drag, and internal/material (hysteretic). The present chapter studies the viscous-type losses in relation to basic translatory and/or rotary dampers, as well as with series- and/or parallel-connection dampers.

2.3 Viscous Damping Elements

2.3.1 Basic Viscous Damping Elements In viscous damping, damping forces (or torques) that are proportional to the relative velocity are set whenever there is relative motion between a structural member and the surrounding fluid (liquid or gas). Fixed hub

Rotor

Fluid Fixed surfaces

v

fd

ω

Fluid

md

End bearings

(a) x1

(b) x2

θ1

ct

θ2

cr

(a)

(b)

FIGURE 2.11 Damping Through StructureeFluid Relative Motion and Related Symbols: (a) Translatory; (b) Rotary.

Figure 2.11 illustrates the principle and symbols of translatory and rotary viscous damping. The symbol is a piston-cylinder system with relative motion between the two parts in the presence of viscous fluid inside the cylinder. A force fd results in translatory damping, and a moment (torque) md is generated in rotary damping. They both oppose the relative motion and are proportional to the corresponding relative velocity by means of (usually) constant damping coefficients (ct for translation and cr for rotation), as shown in Table 2.8. Table 2.8 Viscous Damping Force/Moment and Energy for Translatory and Rotary Dampers Damping force/ moment Damping (lost) energy

Translation

Rotation

fd ¼ ct $ vðtÞ ¼ ct $ x_

md ¼ cr $ uðtÞ ¼ cr $ q_

R R Ud;t ¼ ct $ x_ $ dx ¼ ct $ x_2 $ dt

R R 2 Ud;r ¼ cr $ q_ $ dq ¼ cr $ q_ $ dt

The energy dissipated through viscous damping is equal to the work done by the damping force in translation and the damping torque in rotationdtheir equations are also included in Table 2.8.

39

40

CHAPTER 2 Mechanical Elements

The viscous damping coefficient c can be determined as a function of geometrical and material parameters for either translatory or rotary dampers. In both cases, Newton’s law of viscous flow is applied, which establishes that shear stresses s are set up between adjacent layers of fluid. Consider a plate that moves parallel to another fixed plate with fluid in between, as sketched in Figure 2.12(a). The stresses are expressed as: s ¼ m$

dvðzÞ dz

(2.27)

where m is the coefficient of dynamic viscosity and dv(z)/dz is the gradient of the relative velocity between the moving surface and the fixed one. nn

Example 2.7

Derive the coefficient of viscous damping ct corresponding to: (i) The plates of Figure 2.12(a). Known are the two-plate superposition area A ¼ 1000 mm2, the plates gap g ¼ 0.1 mm, and the fluid coefficient of dynamic viscosity m ¼ 0.001 N s/m2. Assume the fluid velocity varies linearly from zero at the interface with the fixed plate to the velocity of the mobile plate at the interface with it (Couette’s flow model). (ii) A solid cylinder of diameter d ¼ 0.008 m and length l ¼ 0.05 m that translates inside a hollow cylinder of inner diameter D ¼ 0.01 m and of infinite lengthdsee Figure 2.12(b). Use the same coefficient of dynamic viscosity and assume Couette’s flow model. Ignore fluid resistance on the front circular face of the cylinder (which can be realized, for instance, with a number of through holes parallel to the cylinder longitudinal axis).

Distance Mobile surface v

v

v

l

d

g

v(z)

Fluid

v Mobile cylinder

z Fixed surface

D

Velocity

(a)

Fluid

Fixed hollow cylinder

(b)

FIGURE 2.12 Couette’s Flow Between: (a) Parallel Plates With Linear Profile of Velocity; (b) Parallel Cylinders.

Solution (i) Figure 2.12(a) illustrates the velocity gradient under the assumptions of Couette’s flow model. For this type of flow, the velocity v(z) at a distance z from the fixed surface is determined as a function of the maximum velocity v and the gap g as:

z vðzÞ ¼ $ v g

(2.28)

2.3 Viscous Damping Elements

Because Eq. (2.28) yields dv(z)/dz ¼ v/g, Newton’s law for viscous flow, Eq. (2.27), becomes s ¼ m $ v =g . By considering now that the contact area between the moving plate and the fluid is A, the damping (dragging) force that acts on the moving plate is shear stress multiplied by area:

fd ¼ s $ A ¼

m$A $v g

(2.29)

Comparison of Eq. (2.29) with fd of Table 2.8 indicates that the damping coefficient is:

ct ¼

m$A g

(2.30)

The numerical value of the damping coefficient is ct ¼ 0.01 N s/m is obtained. (ii) In the case of the solid cylinder translating in a stationary fluid inside an infinite-length hollow cylinder, as pictured in Figure 2.12(b), because the gap is g ¼ (D  d)/2, the velocity distribution of Eq. (2.28) becomes:

vðzÞ ¼ 2 $

z $v Dd

(2.31)

Because dv(z)/dz ¼ 2v/(D  d), see Eq. (2.31), the shear stress of Eq. (2.27) is constant and equal to: s ¼ 2m $ v =ðD  dÞ. Taking into account that the contact area between the moving cylinder and the surrounding fluid is A ¼ p $ d $ l, the damping force opposing the cylinder motion is obtained as:

fd ¼ s $ A ¼

2p $ m $ d $ l $v Dd

(2.32)

and because fd ¼ ct$v, the damping coefficient becomes:

ct ¼

2p $ m $ d $ l Dd

with a numerical value of ct ¼ 0.00126 N s/m.

(2.33) nn

nn

Example 2.8

Assuming Couette’s flow, determine the rotational damping coefficient cr for: (i) the bearing-shaft system of Figure 2.13(a) (also known as journal bearing or hydrostatic radial bearing) when the shaft rotates with an angular velocity u inside the fluid-filled bearing. The shaft diameter is d ¼ 0.004 m, the bearing inner diameter is D ¼ 0.0046 m, the superposition length is l ¼ 0.01 m, and the liquid dynamic viscosity is m ¼ 0.0015 N s/m2. (ii) the bearing-cylinder system of Figure 2.13(b) when only considering the interaction between the top (frontal) face of the rotating cylinder and the parallel face of the fixed wall. The cylinder diameter is d ¼ 0.03 m, the cylinder wall gap is g ¼ 0.001 m, and the dynamic viscosity of the fluid between the cylinder and the wall is m ¼ 0.0012 N s/m2.

Solution (i) Because the tangential velocity of any point on the piston’s lateral surface is angular velocity times the radius, the damping force that is produced at the rotary pistonefluid interface is:

d m$A d $u$ fd ¼ ct $ v ¼ ct $ u $ ¼ 2 g 2

(2.34)

41

CHAPTER 2 Mechanical Elements

Cylinder

Shaft d

D d Fluid

Fluid

Bearing

ω

Fixed Wall

42

ω Radial bearing g

l

(a)

(b)

FIGURE 2.13 Rotary-Motion Damping for: (a) Journal Bearing; (b) Cylinder With Front Face Against Fixed Wall. where ct of Eq. (2.30) was used. The interface area A and the radial clearance g are: A ¼ p $ d $ l and g ¼ ðD  dÞ=2. They are substituted in Eq. (2.34), which results in:

fd ¼

2p $ m $ d $ l d p $ m $ d2 $ l $u$ ¼ $u Dd 2 Dd

(2.35)

The resulting damping torque is calculated as:

d d p $ m $ d2 $ l p $ m $ d3 $ l $u ¼ $u md ¼ $ fd ¼ $ 2 2 Dd 2ðD  dÞ

(2.36)

Comparing the definition of the rotary damping coefficient of Table 2.8 and Eq. (2.36), it follows that:

cr ¼

p $ m $ d3 $ l 2ðD  dÞ

(2.37)

with a numerical value cr ¼ 2.513$109 N m s. (ii) For any given radius r (0 < r < d/2), the shear stress is constant and based on Newton’s law of viscous dampingdEq. (2.27), is expressed as s ¼ m $ v ðr Þ=g . The gap g is shown in Figure 2.13(b). Figure 2.14 shows an elementary annulus of thickness dr at a radius r from the cylinder’s center. Its elementary area is dA ¼ 2pr $ dr. An elementary damping force dfd corresponds to this element, which generates an elementary damping torque dmd:

dmd ¼ r $ dfd ¼ r $ ðs $ dAÞ ¼

2p $ m $ u 3 $ r $ dr g

(2.38)

with s and dA previously expressed. The total damping torque acting on the cylinder frontal face contacting the fluid is found by integrating the elementary damping moment of Eq. (2.38) in terms of r between 0 and d/2:

Z

md ¼

0

d=2

dmd ¼

p $ m $ d4 $u 32g

(2.39)

2.3 Viscous Damping Elements

ω dr dA

r d

dmd

FIGURE 2.14 Cylinder Front Face With Elementary Area and Damping Torque.

Based on the damping torque definition of Table 2.8, Eq. (2.39) indicates that the damping coefficient is:

cr ¼

p $ m $ d4 32g

(2.40)

For the numerical values provided here, cr ¼ 9.543$108 N m s.

nn

The website companion Chapter 2 comprises the equations defining ct or cr for a few other pistonecylinder dampers and gives the derivations for these coefficients.

2.3.2 Series and Parallel Damper Connections Translatory viscous dampers can be combined in series, in parallel or in seriesparallel to alter the damping properties of single dampers. Figure 2.15 shows two translatory dampers defined by the coefficients c1 and c2, which are connected in series, Figure 2.15(a), and in parallel, Figure 2.15(b). The design of Figure 2.15(c) sketches a rotating rigid shaft that is supported on two fixed bearings, which are considered rotary dampers. Since the relative velocity between the piston (shaft) and the two dampers is the same, it follows that the two cylinders (bearings) are connected in parallel. To simplify notation, the subscripts t and r are dropped whenever translationerotation distinction is clear.

x1

c1

c2

(a)

x2

x

c1

Rigid shaft c1

θ c 2

f

f c2

(b)

Bearings

(c)

FIGURE 2.15 Damper Combinations: (a) Translatory Series; (b) Translatory Parallel; (c) Rotary Parallel.

43

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CHAPTER 2 Mechanical Elements

The website companion Chapter 2 demonstrates that, similarly to springs, n different translatory dampers of coefficients c1, c2, ., cn result in an equivalent damping coefficient cs when combined in series and in a coefficient cp when coupled in parallel, which are: 1 1 1 1 ¼ þ þ / þ ; cp ¼ c1 þ c2 þ / þ cn cs c1 c2 cn

(2.41)

nn

Example 2.9 The massless body of Figure 2.16(a), which is connected to two identical dampers of damping coefficient c ¼ 150 N s/m, has a velocity v1 ¼ 0.1 m/s at a given time moment. By conveniently adding three other identical dampers c, find the design which will generate a 25% reduction in the body velocity while preserving the total damping force of the original two-damper design. Also calculate the resulting equivalent damping coefficient ceq and the damping force fd.

Massless body

x

x c

c

c c

v1 (a)

v1

(b)

FIGURE 2.16 Massless Body With Two Dampers in Translatory Motion: (a) System With Actual Damper Connection; (b) System With Equivalent Parallel Damper Connection.

Solution

The two dampers act as a parallel connection since their end points have the same velocity vdthis ð1Þ feature is illustrated in Figure 2.16(b). The equivalent damping coefficient is simply ceq ¼ 2c by Eq. (2.41). Adding three dampers in parallel with the existing ones, as illustrated in Figure 2.17(a), will ð2Þ generate an equivalent stiffness of: ceq ¼ 2c þ 3c ¼ 5c. Because the same total damping force is produced in the two designs, it follows that: ð2Þ cð1Þ eq $ v1 ¼ ceq $ v2 or v2 ¼

ð1Þ

ceq

ð2Þ ceq

$ v1 ¼

2c $ v1 ¼ 0:4v1 5c

(2.42)

but the reduction in velocity (60%) exceeds the 25% figure needed. On the opposite end of the spectrum, when the three identical dampers are added as a series chain connected in series with the original damper, as shown in Figure 2.17(b), the equivalent damping coefficient would be: ð3Þ

ceq ¼ 2c þ c=3 ¼ 7c=3. As a consequence, the same-damping-force condition results in:

cðeq1Þ $ v1 ¼ cðeq3Þ $ v3 or v3 ¼

ð 1Þ

ceq

ð 3Þ ceq

$ v1 ¼

2c 6 $ v1 ¼ $ v1 ¼ 0:86 $ v1 7 7 $c 3

and generates a 14% velocity reduction, which is smaller than the one needed.

(2.43)

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

c

x

x c

c

c

c

c

c

c c

v2

v3

c (a)

(b)

FIGURE 2.17 Possible Damper Combinations With Additional Dampers Connected in: (a) Parallel; (b) Series.

x c

c c

c

v4

c

FIGURE 2.18 Damper Combination Realizing the Required Velocity Reduction. The design solution seems to consist of the three dampers being coupled in parallel þ series; as a consequence another candidate design is proposed in Figure 2.18. In this case, the equivalent  ð4Þ damping coefficient is: ceq ¼ 2c þ ð2c $ cÞ=ð2c þ cÞ ¼ 8c 3, which results in: ð4Þ cð1Þ eq $ v1 ¼ ceq $ v4 or v4 ¼

ð1Þ

ceq

ð4Þ ceq

$ v1 ¼

3 $ 2c 3 $ v1 ¼ $ v1 ¼ 0:75v1 8c 4

(2.44)

Eq. (2.44) indicates that, indeed, a velocity reduction of 25% is achieved by using the design of Figure 2.18. The value of the equivalent damping coefficient results from Eq. (2.44) as ð4Þ

ð4Þ

ceq ¼ 400 N s=m and the corresponding damping force is fd ¼ ceq $ v4 ¼ 30 N.

nn

2.4 EQUIVALENT MECHANICAL ELEMENTS THROUGH GEAR AND LEVER TRANSFER This section studies the transfer (relocation) of mechanical elements through gears and levers. By equating the energy of an actual element to the energy of the moved element, equivalent element properties are obtained.

2.4.1 Toothed Gears Rotary motion can be transmitted through contact, by means of toothed gears or by using pulleys from one shaft to another in many mechanical engineering

45

46

CHAPTER 2 Mechanical Elements

mt 1, ω1 mt1, ω1 R1

O1

v

N1

Pitch circles

R1

O1

Driving disk

P

Pitch circles

Belt

v

P R2

R1

p O2

N2

R2 R2

mt 2, ω2

O2

Driven disk mt 2, ω2

(a)

(b)

(c)

FIGURE 2.19 (a) Schematic of Meshing Gears (Also of Contact Disks); (b) Detail of a Toothed Gear Pair at Contact Point P; (c) Pulley Disks With Belt Connection.

applications, as depicted in Figure 2.19(a)e2.19(c). Tooth-gear, contact, and pulley transmissions are very similar, and the focus of this section is formally on toothed gear transmission. Part of a geared wheel connection is shown in Figure 2.19(b). The median circle is called the pitch circle (of radius R) and the distance measured on this circle between two consecutive teeth is known as circular pitch, denoted by p. Two gears that mesh (or engage) are schematically shown in Figure 2.19(a) with their pitch circles being tangent. Instead of the two meshing gears, two contact disks can be utilized to achieve the similar function of transmitting rotary motion from one shaft to another, and Figure 2.19(a) is illustrating this principle. The length of the pitch circle is l ¼ 2p $ R ¼ p $ N where N represents the number of teeth. It is necessary that two meshing toothed gears have the same circular pitch, and therefore, for the meshing gears 1 and 2, p ¼ 2pR1 =N1 ¼ 2pR2 =N2 or R1 =R2 ¼ N1 =N2 . Since there is no slippage between the two pitch circles (the actual gear engaging is equivalent to a pure rolling of the two pitch circles), the following is true R1 $ q1 ¼ R2 $ q2 . Because of nonslippage, the two engaging gears have the same velocity at the contact point, namely: v ¼ R1 $ q_ 1 ¼ R2 $ q_ 2 . Under ideal conditions, the power is fully transmitted from one gear to the other meshing gear. Assuming the power torque on gear 1 is mt1 and that a torque mt2 is transmitted to gear 2 through meshing, the power equality is: mt1 $ q_ 1 ¼ mt2 $ q_ 2 . All these relationships are combined as follows: R1 N1 q2 q_ 2 mt1 ¼ ¼ ¼ ¼ R2 N2 q1 q_ 1 mt2

(2.45)

Eq. (2.45) is instrumental in transferring mechanical element properties between various shafts that interact through meshing gears.

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

Toothed Gears and Springs Consider the two elastic shafts (represented as rotary springs) of Figure 2.20(a). They mesh through a pair of gears having N1 and N2 teeth. The elastic potential energy stored by the two deforming springs is: 1 1 Ue ¼ k1 $ q21 þ k2 $ q22 (2.46) 2 2 where q1 and q2 are relatively small angular deformations of the two rotary springs. From Eq. (2.45) it follows that q2 ¼ N1 $ q1 =N2 which, substituted in Eq. (2.46), results in: " #  2 1 N1 $ k2 $ q21 (2.47) Ue ¼ k1 þ 2 N2 Eq. (2.47) suggests that, from the viewpoint of stiffness, the actual system is equivalent to another system that is reduced (or transferred) to the first shaft and whose equivalent stiffness is:  2 N1 keq;1 ¼ k1 þ $ k2 (2.48) N2 and which is represented at the top of Figure 2.20(b). When q1 is expressed in terms of q2 from Eq. (2.45) and substituted into Eq. (2.46), a similar result is obtained and the equivalent stiffness transferred to the second shaft is:  2 N2 keq;2 ¼ k2 þ $ k1 (2.49) N1 That equivalent system is sketched at the bottom of Figure 2.20(b). It is concluded that for two gear-meshing shafts the transferred stiffness is equal to the original stiffness multiplied by the square of the ratio of the destination to the origin gear teeth number.

N1 Elastic shaft

keq,1

Gear

θ1

θ1 Anchor

N1

k1 θ2 N2

(a)

k2

Anchor

θ2 N2

keq,2

(b)

FIGURE 2.20 Meshing Gears With Associated Rotary Springs: (a) Actual System; (b) EquivalentStiffness (Transferred) Systems.

47

48

CHAPTER 2 Mechanical Elements

nn

Example 2.10 Identify the equivalent torsional stiffness transferred to the intermediate rigid shaft axis of the gear train sketched in Figure 2.21. The other three shafts are flexible, are constructed from the same material with G ¼ 75 GPa, and are defined by d1 ¼ 0.009 m, l1 ¼ 0.04 m, d2 ¼ 0.008 m, l2 ¼ 0.035 m, d3 ¼ 0.012 m, and l3 ¼ 0.05 m. Known are also the number of teeth of the four gears N1 ¼ 20, N2 ¼ 30, N3 ¼ 18, and N4 ¼ 32.

l1

l3 N1

N4 Anchor

Shaft

d1

d3

Gear Rigid shaft N2

N3

d2 l2

FIGURE 2.21 Gear Train With Flexible Shafts.

Solution The gear train of Figure 2.21 is shown with its flexible shafts represented as torsional springs in Figure 2.22. N1

N4

k1

k3 keq,1

k2

k2 + keq,3

keq,1 + k2 + keq,3 θx, mx

N2

N3

FIGURE 2.22 Sequence of Lumped-Parameter Stiffness Models of the Gear Train With Flexible Shafts. The three shafts’ torsional stiffnesses are calculated based on Table 2.3 as:

k1 ¼

G $ Ip1 p $ G $ d14 G $ Ip2 p $ G $ d24 G $ Ip3 p $ G $ d34 ¼ ; k2 ¼ ¼ ; k3 ¼ ¼ l1 32l1 l2 32l2 l3 32l3 (2.50)

The stiffnesses k1 and k3 are transferred from their original locations to the intermediate rigid shaft (according to Eq. 2.49), and they become:

keq;1

 2  2 N2 N3 ¼ $ k1 ; keq;3 ¼ $ k3 N1 N4

(2.51)

The transferred stiffness keq,3 adds to the stiffness k2 residing on the rigid shaft, as indicated in the intermediate equivalent spring stiffness of Figure 2.22. Eventually, keq is obtained as a parallel combination of the two springs in the middle model of Figure 2.22 by adding up these

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

stiffnesses, as indicated in the final equivalent configuration of Figure 2.22: keq ¼ keq;1 þ k2 þ keq;3 eventually becoming:

keq

 2  2 p $ G $ d34 p $ G $ d14 p $ G $ d24 N2 N3 ¼ $ þ þ $ N1 32l1 32l2 N4 32l3

(2.52)

after utilizing the intermediate stiffnesses of Eqs. (2.50) and (2.51). Its numerical value is keq ¼ 4545.28 N m. nn

Toothed Gears and Inertia For shafts interacting through toothed gears, lumped-parameter inertia properties can be transferred between shafts similarly to moving lumped-parameter stiffness. Consider the two rigid and massless shafts of Figure 2.23(a) that mesh through a pair of gears, having the mass moments of inertia J1 and J2 and N1 and N2 teeth. Jeq,1 Gear

J1, N1

Bearing

θ1

θ1

Shaft

N1

Jeq,2 θ2

θ2 J2,N2

(a)

N2

(b)

FIGURE 2.23 Meshing Gears With Associated Mass Moments of Inertia: (a) Actual System; (b) Equivalent-Inertia (Transferred) Systems.

The kinetic energy of the shaft-gear system of Figure 2.23(a) is: 1 1 1 1 2 2 T ¼ $ J1 $ u21 þ $ J2 $ u22 ¼ $ J1 $ q_ 1 þ $ J2 $ q_ 2 (2.53) 2 2 2 2 The angular velocity u2 is expressed as a function of u1 and the number of teeth N1 and N2 from Eq. (2.45) as q_ 2 ¼ q_ 1 $ ðN1 =N2 Þ and substituted in Eq. (2.53), which results in: " #  2 1 N1 2 $ J2 $ q_ 1 (2.54) T¼ J1 þ 2 N2 Eq. (2.54) indicates that the actual system’s inertia is equivalent to the following inertia when reduction (transfer) to the first (top) shaft is utilized:  2 N1 Jeq;1 ¼ J1 þ $ J2 (2.55) N2

49

50

CHAPTER 2 Mechanical Elements

A similar procedure can be applied to transfer the mass moment of inertia corresponding to the gear J1 to the other (bottom) shaft, which results in the following mass moment of inertia:  2 N2 Jeq;2 ¼ J2 þ $ J1 (2.56) N1 Jeq,1 and Jeq,2 are illustrated in Figure 2.23(b). Eqs. (2.55) and (2.56) show that the equivalent (transferred) mass moment of inertia is equal to the original moment of inertia multiplied by the square of the ratio of the destination to the origin gear teeth number. This is similar to the transfer of stiffness between shafts connected through gear meshing. nn

Example 2.11 The electric motor of Figure 2.24 connects through a belt to an input shaft, which transmits the motion to a middle shaft through a gear pair. This shaft further connects through another gear pair to the output shaft, where a cam contacts a follower and transforms the rotary motion into a translatory one. Known are the following parameters: Jmo ¼ 2000$106 kg m2, Jd ¼ 72$106 kg m2, J1 ¼ 75$106 kg m2, J2 ¼ 610$106 kg m2, J3 ¼ 15$106 kg m2, J4 ¼ 1300$106 kg m2, Jc ¼ 12$106 kg m2, N1 ¼ 32, N2 ¼ 48, N3 ¼ 24, N4 ¼ 36, Rp ¼ 0.02 m (pulley radius), and Rd ¼ 0.03 m (belt disk radius). Using the rotary inertia properties of this system, calculate the equivalent mass moment of inertia reduced to the input shaft. Input shaft Belt disk

N1 J1

Belt

Jd Jmo

Gears J2

Bearing

Middle shaft N 3 J3

Motor Pulley

Cam disk

J4

N2 Follower Guide

Jc N4

Output shaft

FIGURE 2.24 Gear-Shaft-Cam Rotary System.

Solution The relevant inertia properties are first transferred from the output (lower) shaft to the middle shaft, which results in the following total mass moment of inertia on the middle shaft:

Jeq;m ¼ Jm þ

N32 N32 $ J ¼ ðJ þ J Þ þ $ ðJc þ J4 Þ o 2 3 N42 N42

(2.57)

where the subscript o indicates the output shaft and the subscript m denotes the middle shaft. The mass moment of inertia of the input shaft combines the actual mass moments of inertia of the

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

rotors located on it, the belt disk and the gear N1, and the mass moments of inertia transferred from the motor shaft and from the middle shaft, which means:

Jeq;i ¼ Ji þ

R2d N2 $ Jmo þ 12 $ Jeq;m 2 Rp N2

(2.58)

with the subscript i denoting the input shaft and the subscript mo identifying the motor. Taking into account that Ji ¼ Jd þ J1 (Jd being the disk mass moment of inertia), as well as Jeq,m of Eq. (2.57), the total equivalent mass moment of inertia corresponding to the input shaft of Eq. (2.58) becomes:

Jeq;i

" # R2d N32 N12 ¼ Jd þ J1 þ 2 $ Jmo þ 2 $ J2 þ J3 þ 2 $ ðJc þ J4 Þ Rp N2 N4

(2.59)

The numerical values of the equivalent mass moment of inertia expressed in Eq. (2.59) is Jeq,i ¼ 5.184$103 kg m2. nn

Gears With Dampers Damping properties can be transferred between shafts that connect through meshing gears similarly to rotary stiffnesses and inertia. Consider the two meshing gears of Figure 2.25(a) that have a number of teeth N1 and N2. The rigid shafts carrying the two gears are supported each on two bearings, whose damping properties can be modeled by the lumped-parameter rotary dampers of coefficients c1 and c2 (the subscript r has been dropped for simplicity) that are shown in Figure 2.25(b).

Gear

J1,N1

N1

N1

ceq,1

c1

Bearing θ1

Shaft θ2

θ1

θ1 θ2 c2

J2, N2

N2

(a)

(b)

θ2 ceq,2 N2

(c)

FIGURE 2.25 Two-Shaft Gear System With Bearing Damping: (a) Actual Model; (b) Simplified LumpedParameter Model; (c) Equivalent-Damping (Transformed) Systems.

Eq. (2.45) provides the relationship between the angular velocities and the numbers of teeth for two meshing gears, whereas Table 2.8 gives the energy lost through viscous damping. The total viscous-damping energy lost by the mechanical system of Figure 2.25(b) is therefore:  2 Z t Z t Z t Z t N1 2 2 2 2 _ _ _ Ud ¼ c1 $ $ q1 dt þ c2 $ q2 dt ¼ c1 $ q1 dt þ c2 $ q_ 1 dt N2 0 0 0 0 "  2 # Z t N1 2 ¼ c1 þ c2 $ (2.60) q_ 1 dt $ N2 0

51

52

CHAPTER 2 Mechanical Elements

The total equivalent viscous-damping energy reduced to the first shaft is: Z t 2 Ud;eq ¼ ceq;1 $ q_ 1 dt

(2.61)

0

By equating the energies of Eqs. (2.60) and (2.61), the following equivalent damping coefficient is obtained:  2 N1 (2.62) ceq;1 ¼ c1 þ c2 $ N2 and is shown at the top of Figure 2.25(c). By following a similar reasoning, it can simply be shown that the equivalent damping coefficient transferred to the second (bottom) shaft in Figure 2.25(c) is:  2 N2 ceq;2 ¼ c2 þ c1 $ (2.63) N1 and is shown at the bottom of Figure 2.25(c). Eqs. (2.62) and (2.63) indicate that damping can be transferred from one shaft to another similarly to the rule that applied for inertia or stiffness relocation. As such, the transferred damping coefficient is equal to the original damping coefficient multiplied by the square of the ratio of the destination to the source gear teeth number. nn

Example 2.12 The gear train of Figure 2.26 is formed of an input shaft whose rotary motion is transmitted to an output shaft through an intermediate shaft and two pairs of meshing gears. The input shaft is supported by two identical bearings of damping coefficient c1 ¼ 0.02 N m s; similarly, the output shaft is supported by two other identical bearings of coefficient c3 ¼ 0.03 N m s, whereas the intermediate shaft sits on a single long bearing of damping coefficient c2 ¼ 0.01 N m s. Knowing the number of gear teeth N1 ¼ 24, N2 ¼ 32, N3 ¼ 20, N4 ¼ 36, as well as the constant angular velocity of the input shaft, n1 ¼ 120 rpm, determine the equivalent damping coefficient transferred to the output shaft, as well as the total energy dissipated through the bearings in 1 hour.

Input shaft

Bearing

θ1

N1

Gear

N4

θ3

Output shaft θ2

N3 N2 Intermediate shaft

FIGURE 2.26 Gear Train With Bearings.

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

N1

2c1

N4

N4 2c3

2c3 θ3 ceq,3 N4

c2

ceq,2 N2

N3

N3

FIGURE 2.27 Sequence of Lumped-Parameter Damping Models for the Gear Train.

Solution The gear train of Figure 2.27 is shown in a succession of reduction steps in terms of its lumpedparameter damping properties from an initial model to the final variant, with one equivalent damper transferred to the output shaft. The original model to the left in Figure 2.27 uses a damper for each of the three shafts and takes into considerations that two dampers add in parallel for both the input and the output shafts (which results in the damping coefficients 2c1 and 2c3). The damper from the input shaft is transferred to the intermediate (connecting) shaft, as indicated in the next model in Figure 2.27, so that the equivalent damping coefficient on this shaft becomes:

ceq;2 ¼ c2 þ 2c1 $

 2 N2 N1

(2.64)

based on Eq. (2.62). The final equivalent damping coefficient reduced to the output shaft can now be calculated based on the last transition of Figure 2.27 as:

ceq;3

"  2  2 #  2 N4 N2 N4 ¼ ceq;2 $ þ 2c3 ¼ c2 þ 2c1 $ þ 2c3 $ N3 N1 N3

(2.65)

With the given numerical values, this yields ceq,3 ¼ 0.323 N m s. The energy dissipated through damping in the bearings can be calculated using the appropriate equation of Table 2.8 as:

Z

Ud ¼ ceq;3 $

0

t

2 2 q_ 3 dt ¼ ceq;3 $ q_ 3 $ t

(2.66)

Eq. (2.66) took into account that the output-shaft angular velocity is constant given the input-shaft angular velocity is also constant. Using the connection Eq. (2.45), the input and output angles and velocities are related as:

q3 q3 q2 ¼ $ ¼ q1 q2 q1

        N3 N1 N3 N1 $ or q_ 3 ¼ $ $ q_ 1 N4 N2 N4 N2

(2.67)

which actually proves that the output angular velocity is constant. Also taking into consideration that n (in revolutions per minute, rpm) and u ¼ q_ are connected as u ¼ q_ ¼ p $ n=30, Eqs. (2.66) and (2.67) yield:

Z

Ud ¼ ceq;3 $

0

t

 2  2 p2 $ n21 N3 N1 2 $t $ $ q_ 3 dt ¼ ceq;3 $ N4 N2 900

(2.68)

With the value of the equivalent damping coefficient and the other provided parameters, the dissipated energy corresponding to t ¼ 3600 s is Ud ¼ 31,859.083 J. nn

53

54

CHAPTER 2 Mechanical Elements

2.4.2 Levers of Small Rotation Mechanical levers are employed to change motion characteristics, such as displacements and forces or moments. Springs, point masses, or dampers that are attached to mechanical levers that undergo small-angle rotations around a fixed-pin axis can be relocated to different positions along the lever by preserving their original-element energy. This process allows expressing new properties of the moved elements in terms of the original-element properties and the relocation distance. For a rotation angle q of the rigid rod, the free-end point C displaces to the position C0 in Figure 2.28(a) by following an arc of the circle centered at O. For small angles (normally smaller than 5 degrees), the length of the arc CC0 can be approximated to the vertical distance yc (or the length CC00 ), which is yC ¼ CC 00 zCC 0 ¼ lC $ q. C' C" O

yC

θ

A'

O lA

yB

fB B

A

C

lC

B'

yA

fA lB

(a)

(b)

FIGURE 2.28 Rigid Lever: (a) Small-Rotation Approximation; (b) With Two Opposing Forces.

This approximation, which is known as the small-angle (or small-rotation) approximation, enables transforming the actual rotation motion of the rod into a translation of the end point. Consider the rigid rod of Figure 2.28(b), which is pinned at O and acted on by the forces fA and fB (where B is not necessarily at the end of the rod). Under the assumption of small angular motions, the lever position (after the loads fA and fB have been applied quasi-statically and reached equilibrium) is determined by points A0 and B0 , whose displacements from the original rod position are related as: yB lB ¼ ¼a yA lA

(2.69)

where a is the displacement amplification. Eq. (2.69) indicates that the displacement at B is a times larger than the one at A. Since the forces fA and fB act antagonistically, the work done by the two forces is equal: fA $ yA ¼ fB $ yB , which means that the force fB is a times smaller than the force fA: fB ¼

yA 1 $ fA ¼ $ fA a yB

(2.70)

such that the lever mechanism of Figure 2.28(b) operates as a force-reduction device. It is necessary that lB > lA for the pinned lever to act as a displacement amplifier. When lB < lA, the lever becomes a displacement-reduction (or force-amplification) device.

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

Levers With Springs When springs connect to a pinned lever, the actual stiffness properties can be moved (transferred) from their original positions to any other location on the lever. Consider, for instance, that the spring of stiffness kA of Figure 2.29(a) is relocated

kA O

kB

yA

A

yB

yB

O

B

lB

B

lA lB

(a)

(b)

FIGURE 2.29 Equivalent Lever-Spring Systems With: (a) Original Spring; (b) Transferred Spring.

from point A on the lever that pivots around O to another position, such as point Bd see Figure 2.29(b)dby preserving the motion/displacement and energy characteristics of the original lever system. It therefore follows that the new spring is defined by an unknown stiffness kB. The original and equivalent lever-spring systems of Figure 2.29 possess the same elastic potential energy and their displacement fields are identical. Taking into account Eq. (2.69), the elastic potential energy of the original system shown in Figure 2.29(a) is:  2 1 1 lA Ue ¼ $ kA $ ðyA Þ2 ¼ $ kA $ $ ðyB Þ2 (2.71) 2 2 lB whereas the elastic potential energy of the equivalent system is: 1 Ue ¼ $ kB $ ðyB Þ2 2

(2.72)

Comparison of Eqs. (2.71) and (2.72) results in:  2 lA kB ¼ $ kA lB

(2.73)

nn

Example 2.13 For small rotations of the rigid lever sketched in Figure 2.30(a), determine the stiffness of a single spring to be placed at point C, as in Figure 2.30(b), and deforming along the y direction that is equivalent to the actual springs connected to the lever. What is the rotation angle of the lever when a force fy acts at C? Known are k1 ¼ 80 N/m, k2 ¼ 130 N/m, k3 ¼ 60 N/m, l ¼ 0.08 m, and fy ¼ 0.6 N.

55

56

CHAPTER 2 Mechanical Elements

3l

k2 k3

k1

O

fy

A

C

fy y

C

B

l

l

O

l

y

kC

(a)

(b)

FIGURE 2.30 Springs Attached to a Small-Rotation Pinned Lever: (a) Actual Lumped-Parameter Spring Model; (b) Equivalent Single-Spring Model. The springs of stiffnesses k1 and k2 are connected in series and their equivalent stiffness is k12 ¼ k1 $ k2 =ðk1 þ k2 Þ with the equivalent spring still placed at A. When transferred to point C, this stiffness becomes: 0 k12 ¼

 2 l k1 $ k2 $ k12 ¼ 3l 9ðk1 þ k2 Þ

(2.74)

as per Eq. (2.73). Similarly, the original stiffness k3 is transferred from B to C and transforms into:

k30 ¼

 2 2l 4 $ k3 ¼ $ k3 3l 9

(2.75)

The two equivalent springs that are now located at point C are connected in parallel (they undergo identical motions), and therefore the total stiffness at point C is:

  1 k1 $ k2 0 kC ¼ k12 þ k30 ¼ $ þ 4k3 9 k1 þ k2

(2.76)

Its numerical value is kC ¼ 32.17 N/m. The displacement at C is yC ¼ fy/kC z 0.019 m and the rotation of the lever is q ¼ yC/(3l) ¼ 4.45 . nn

Levers and Inertia Inertia (masses) can be equivalently transferred on lever systems between various points. The point mass mA for instance is placed at point A on the heavy lever of mass m and shown in Figure 2.31(a); it is of interest to determine the point mass

O

⋅ θ

mA

yA A

yB

m B

lA

O

⋅ θ

mB Massless rod lB

yB

B

lB

(a)

(b)

FIGURE 2.31 Equivalent Lever-Mass Systems With: (a) Original Point Mass; (b) Transferred Point Mass.

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

mB located at point B in Figure 2.31(b), which is dynamically equivalent to the original masses mA and m. The equivalence between the levers of Figure 2.31 requires that the two systems have identical displacement fields and the same kinetic energy at all times. The kinetic energy of the original lever system is:  2 1 1 2 T ¼ $ mA $ y_A þ $ JO $ q_ (2.77) 2 2 Taking into account that the velocity at A is y_A ¼ lA $ q_ and  the lever mass moment of inertia with respect to the pivot point O is JO ¼ m $ l2B 3, the kinetic energy of Eq. (2.77) becomes:      2 1 1 1 1 1 2 2 $ m $ l2B $ q_ ¼ $ mA $ l2A þ $ m $ l2B $ q_ T ¼ $ mA $ lA $ q_ þ $ 2 2 3 2 3 (2.78) On the other hand, the kinetic energy of the equivalent lever system with a point mass mB of Figure 2.31(b) is:  2 1  2 1 1 2 T ¼ $ mB $ y_B ¼ $ mB $ lB $ q_ ¼ $ mB $ l2B $ q_ 2 2 2

(2.79)

The two systems have the same kinetic energy; therefore, the transferred mass is obtained from Eqs. (2.78) and (2.79):  2 lA 1 (2.80) mB ¼ $ mA þ $ m 3 lB

nn

Example 2.14

The two identical heavy levers (each of mass m) of Figure 2.32(a) undergo small rotations being connected at their ends by a short, rigid, and massless rod. A point mass m is placed on the left lever, as shown in the same figure. Determine the position of another point mass m to be placed on the right lever that results in an equivalent point mass meq ¼ 3m/2 located on the connecting rod, as shown in Figure 2.32(b).

θ

m

y

y

θ m

θ

l/2 l

θ

meq

x l

(a)

(b)

FIGURE 2.32 (a) Actual Lever System; (b) Equivalent Lumped-Parameter Mass Lever System.

57

58

CHAPTER 2 Mechanical Elements

Solution The equivalent point mass that is located on the connecting rigid rod of Figure 2.32(b) is determined by moving the lumped masses m from both levers and the two levers rotary inertia, as per Eq. (2.80), which results in:



meq

 x2 l=2 2 m ¼ meq;1 þ meq;2 þ 2meq;3 ¼ $m þ $m þ 2$ l l 3     2 2 1 x 2 11 x þ þ þ ¼ $m ¼ $m 4 l2 3 12 l2

(2.81)

with meq,1 and meq,2 being the transferred point masses from the left and right levers, respectively, to the connecting vertical rod, and meq,3 ¼ m/3 being the lumped mass equivalent to one lever’s rotary inertia. Substituting meq ¼ 3m=2 in Eq. (2.81) yields:

pffiffiffi 7 x ¼ pffiffiffi $ l 2 3

(2.82) nn

Levers With Dampers Similarly to springs and inertia, lumped translatory dampers can be equivalently transferred on lever systems between various points. For instance, the damper cA, which has its mobile end attached at A to the lever of Figure 2.33(a), needs to be lB

lA O

yA A

yB

yB B

B

O

cB

cA

lB

(a)

(b)

FIGURE 2.33 Equivalent Lever-Damper Systems With: (a) Original Damper; (b) Displaced Damper.

transformed into an equivalent damper cB located at B (which does not have to necessarily be at the free end of the lever), as shown in Figure 2.33(b). For the two levers of Figure 2.33 to be equivalent they need to have identical velocity fields and dissipated energy at all times. Based on Table 2.8, the damping energies of the two damped lever systems are: Z t Z t  2  2 Ud ¼ cA $ (2.83) y_A dt; Ud ¼ cB $ y_B dt 0

0

2.4 Equivalent Mechanical Elements Through Gear and Lever Transfer

Using Eq. (2.69), which results in y_A ¼ y_B $ ðlA =lB Þ, as well as Eq. (2.83), the following damping coefficient is obtained:  2 lA cB ¼ $ cA (2.84) lB which is similar to Eqs. (2.73) and (2.80) relating transfer stiffnesses and masses, respectively. nn

Example 2.15

A linear-motion damper of damping coefficient cA ¼ 120 N s/m is attached at A to the lever of Figure 2.34(a). What is the damping coefficient cB of the additional damper connected at B to the original lever systemdsee Figure 2.34(b)dthat reduces the velocity at point C by 25%? The lever undergoes small-angle rotations with constant angular velocities under the action of the same external force fy applied at C.

fy

l

fy

l

y

y O

A

O

C

cA

A

B

cA

C

cB

3l (a)

(b) fy y O

C ceq

(c)

FIGURE 2.34 Lever With Dampers: (a) Original System; (b) Modified System; (c) Equivalent System.

Solution

Since the velocity at point C is of interest, it is advantageous to transfer the damping properties of the original and modified systems from their actual positions to the free end of the lever, as illustrated in Figure 2.34(c), namely:

ceq ¼

 2  2 l 1 2l 1 $ cA ¼ $ cA ; ceq ¼ ceq þ $ cB ¼ $ ðcA þ 4cB Þ 3l 9 3l 9

(2.85)

59

60

CHAPTER 2 Mechanical Elements

In Eq. (2.85), the star indicates the equivalent modified system of Figure 2.34(b); its damping coefficient took into account that the original dampers were transferred from their positions to the end C where they act in parallel. For small deformations, the damping force resulting from the equivalent dampers whose coefficients are given in Eq. (2.85) is expressed as:

fy ¼ fd ¼ ceq $ vC ; fd ¼ ceq $ vC

(2.86)

Because the velocity of the modified system is 75% of the original system’s velocity, the velocities at C are related as: vC ¼ 0:75 $ vC . Combining now this relationship with Eq. (2.86) allows solving for the damping coefficient cB as cB ¼ cA/12 ¼ 10 N s/m. nn

SUMMARY This Chapter has introduced the basic mechanical elements: springs, heavy bodies, and viscous dampers, as the main components of mechanical systems. The mechanical elements properties of stiffness, inertia (mass and mass moment of inertia), and viscous damping coefficients were also presented together with their associated energy. Distributed-parameter inertia of elastic bars and beams has been converted into equivalent lumped-parameter inertia. You have also learned to transform or transfer actual mechanical element properties into equivalent ones, when using motion transformers such as gears and levers by means of energy approaches.

PROBLEMS 2.1 Several springs are connected as illustrated in Figure 2.35(a). Knowing the individual springs stiffnesses k1 ¼ 20 N/m, k2 ¼ 30 N/m, k3 ¼ 15 N/m, k4 ¼ 25 N/m, and k5 ¼ 35 N/m, calculate the equivalent stiffness keq of this spring ensemble, see Figure 2.35(b), which corresponds to point A of the original chain. k3

k2

x A

k1

x k5

A

keq

k4

(a)

(b)

FIGURE 2.35 (a) Translatory Spring Combination; (b) Equivalent Spring.

2.2 The five springs of unknown stiffnesses k1 and k2 are combined as illustrated in Figure 2.36(a). A force f is applied at A and produces a displacement x1 at the same point. When the end B of the spring arrangement is set free and the same force f is applied at that point, as indicated in Figure 2.36(b), the

Problems

k1

k1

k1

x1

k1

f

x2 A

A k 2

k2

k2

k2

B f

k1

k1

(a)

(b)

FIGURE 2.36 Translatory Spring Combinations With: (a) Fixed Ends; (b) Fixed-Free Ends.

corresponding displacement at B is x2. Calculate k1 and k2 knowing f ¼ 100 N, x1 ¼ 0.005 m, x2 ¼ 0.05 m, as well as k1 > k2. 2.3 The shaft of Figure 2.37 is supported by two end bearings and carries a central disk, which is keyed on it. When the shaft accelerates or decelerates, an inertia torque (moment) is applied to it by the disk. Assume the inertia torque mmax ¼ 125 N m is applied as shown in the figure at the same time with suddenly locking (fixing) the shaft ends, which produces the maximum rotation angle qmax ¼ 3 at the inertia moment application point. Knowing the following dimensions: d1 ¼ 0.01 m, d2 ¼ 0.02 m, d3 ¼ 0.025 m, l1 ¼ 0.03 m, t ¼ 0.01 m, and the shaft’s shear modulus G ¼ 1.6 GPa, calculate the necessary length l2. Bearing Anchor

d1

Inertia torque Bearing d1 Shaft

Disk d2

Anchor

d3

l1

l2

l2 – t

l1

t

FIGURE 2.37 Shaft-Disk System Rotating in Two End Bearings.

2.4 Two identical elastic shaft segments are connected to a rigid disk acted on by a moment m, as illustrated in Figure 2.38(a). To reduce the rotation angle at O by 40%, the design of Figure 2.38(b) is used instead of the original one. Assuming all shaft segments are built from the same material and have constant circular cross sections, express the diameter d2 in terms of the diameter d1, and the lengths l1 and l2. Also calculate the value of d2 for d1 ¼ 0.012 m, l1 ¼ 0.24 m, and l2 ¼ 0.16 m. 2.5 The microsensor of Figure 2.39(a) consisting of two identical beams and a central rigid link is redesigned into the one of Figure 2.39(b). Consider that the two rigid pushrods in Figure 2.39(b) are attached to the corresponding identical bridges and the central link. The stiffness of the

61

CHAPTER 2 Mechanical Elements

m

d1

Flexible shaft

Flexible shaft

O Rigid disk

d1

Anchor

m

Anchor

62

d2

d1

O l2

l1

l1

d2 d 1

l2 l1

l1

(a)

(b)

FIGURE 2.38 Fixed-Fixed Shaft With: (a) Two Identical Segments; (b) Two Pairs of Identical Segments. 2l2 l2 Motion direction

Anchor

Anchor

Anchor

Anchor

Bridge

Pushrod

Beam

l1

Moving link

l1

Anchor

Anchor l2

(a)

2l2

(b)

FIGURE 2.39 Micromechanism With Rigid Link and (a) Two Beams and (b) Six Beams.

modified design is five (5) times larger than the original design stiffness in the indicated motion direction. Express the length l2 in terms of l1 using lumped-parameter stiffness modeling and considering that all beam segments have identical cross sections and material properties. 2.6 The microdevice of Figure 2.40 is formed of two identical sets of springs connected symmetrically to the mobile rigid link. Each set comprises two pairs of beams. Determine the corresponding lumped-parameter physical model and calculate the equivalent spring stiffness in terms of the motion direction of the central link. Known are l1 ¼ 120 mm, l2 ¼ 150 mm and all the beams are made of the same material with a Young’s modulus E ¼ 160 GPa. Considering that all beams have identical circular cross sections, calculate the beam diameter d that will result in a static displacement of the central link of 2 mm when a force f ¼ 100 mN is applied at the link’s center along the motion direction.

Problems

Beams

Motion direction

Anchor

l2

Moving link l1

FIGURE 2.40 Symmetric Microdevice for Translatory Motion.

2.7 Determine the equivalent lumped-parameter spring model of the six-beam microdevice sketched in Figure 2.41, and evaluate the stiffness of the single equivalent spring to be attached to the moving rigid link along the y direction. All beams have identical square cross section with side length h ¼ w ¼ 1 mm. Known also are l1 ¼ 100 mm, l2 ¼ 140 mm, l3 ¼ 200 mm, and the beams Young’s modulus E ¼ 155 GPa. The allowable displacement of the moving link is 10 mm. Calculate the corresponding force fy applied at the moving link center along the y direction. Anchor

l2

Beam

Rigid link

l3

Anchor

Anchor

Moving link y

l1 l3

l1 l2

Anchor

FIGURE 2.41 Microdevice With Six Beams for Translatory Motion.

2.8 Answer the same question as in Problem 2.7 for the microdevice of Figure 2.42 with respect to the x-motion direction. Calculate the force fx to be applied at O to displace the moving link by x ¼ 1 mm. All the beams have identical circular cross sections of diameter d ¼ 1.6 mm. The other known parameters are l1 ¼ 110 mm, l2 ¼ 45 mm, and Young’s modulus E ¼ 165 GPa. 2.9 The micromechanism of Figure 2.43 is a test setup for investigating the Young’s modulus of the upper bridge. Using lumped-parameter stiffness modeling, propose a spring model that is equivalent to the actual beam system to describe the motion of the moving rigid link along the y axis. Determine the force fy,min to be applied at O to close the gap g ¼ 1 mm

63

CHAPTER 2 Mechanical Elements

Serpentine spring l1

Anchors

Anchor O x

l2

2l2

l1 Moving link Rigid link

FIGURE 2.42 Microdevice With Four Beams and a Serpentine Spring for Translatory Motion.

Bridge

B

g

Anchor

Anchor

l2

Beam

l1

y O

l1

Anchor

A Pushrod

Anchor

64

Moving link

FIGURE 2.43 Micromechanism for Testing Material Properties in Translatory Motion.

between the pushrod point A and the bridge midpoint B. Also, based on a stiffness lumped-parameter model, evaluate the bridge material Young’s modulus Eb when a force fy ¼ 2$fy,min applied at O produces a displacement y ¼ 1.5 mm at the same point. All beams have identical circular cross section of diameter d ¼ 2 mm. The two length parameters are l1 ¼ 60 mm and l2 ¼ 150 mm. The beams Young’s modulus is E ¼ 100 GPa. 2.10 The plate of Figure 2.44(a) has a width w ¼ 35 mm and a thickness h ¼ 3 mm. The mass moment of inertia of the plate with respect to the central axis x is 20 times smaller than it needs to be. It is determined that the necessary increase in the mass moment of inertia can be realized in three ways: (1) by enlarging the width (no more than 50% width increase is allowed); (2) by offsetting the plate parallel to its central axis at a distance d (d can be no larger than w); or (3) by a combination of the methods (1) and (2)dsee Figure 2.44(b). Demonstrate that only method (3) is valid, and determine the corresponding parameter changes to realize this condition.

Problems

l Central axis x

w w + Δw d Offset axis Δ (b)

(a)

FIGURE 2.44 (a) Original Plate Rotating Around Central Axis; (b) Modified Plate Rotating Around an Offset Axis.

2.11 Calculate the mass moment of inertia of the pulley shown in Figure 2.45 about its longitudinal axis. A hole of diameter d extends over the whole length of the pulley. Known are d1 ¼ 0.05 m, d2 ¼ 0.03 m, d ¼ 0.02 m, l1 ¼ 0.006 m, l2 ¼ 0.01 m, l3 ¼ 0.009 m, and r ¼ 7800 kg/m3. l1

d2

l2

l3

l2

l1

d

d1

FIGURE 2.45 Longitudinal View of Pulley.

2.12 Calculate the mass moment of inertia about the longitudinal axis of the rotary mechanical system shown in Figure 2.46. The system comprises a stepped shaft, a bevel (conical) gear, and a spur (cylindrical) gear. Known are: d1 ¼ 0.006 m, d2 ¼ 0.01 m, d3 ¼ 0.02 m, d4 ¼ 0.014 m, l1 ¼ 0.01 m, l2 ¼ 0.02 m, l3 ¼ 0.1 m, l4 ¼ 0.02 m, db1 ¼ 0.03 m, db2 ¼ 0.05 m, lb ¼ 0.01 m, ds ¼ 0.04 m, ls ¼ 0.01 m, as well as the shaft and gears mass density r ¼ 7100 kg/m3. 2.13 The cylindrical hub of length l2 can slide along the rod of length l1. A disk of radius R and out-of-plane thickness t is attached at one end of the rod, as illustrated in Figure 2.47. The hub is relocated from the left end of the rod to the right end, where it becomes tangent to the disk. Calculate the change in the mass moment of inertia of this compound pendulum with respect to an

65

66

CHAPTER 2 Mechanical Elements

Bevel gear lb

Spur gear ls

db1 d1

db2

d2

d3

d1

d4

ds

l1

Shaft

l1 l2

l3

l4

FIGURE 2.46 Shaft With Bevel and Spur Gears.

d1

Relocatable hub

Rod

R

d2

O

Disk x

l2 l1

FIGURE 2.47 Compound Pendulum With Relocatable Hub.

end axis passing through O and perpendicular to the plane of the pendulum. Known are d1 ¼ 0.005 m, d2 ¼ 0.01 m, l1 ¼ 0.3 m, l2 ¼ 0.1 m, R ¼ 0.04 m, t ¼ 0.005 m, and the mass density of all parts forming the pendulum, r ¼ 6800 kg/m3. 2.14 A cantilever has an equivalent mass at its free end that is equal to the equivalent mass of a bridge (at its midpoint) in terms of out-of-plane bending. Is it possible that the two members also have identical equivalent mechanical torsion-related moments of inertia about their respective points if they are made of the same material and have circular cross sections? 2.15 The microaccelerometer of Figure 2.48 is formed of a rigid link (plate) defined by l1 ¼ 30 mm and a constant thickness (out-of-plane dimension) h¼ 4 mm and four beams that support the central link. The beams have identical circular cross section of diameter d ¼ 4 mm. Calculate the maximum value of the length l2 such that the beams inertia contribution to the total (equivalent) inertia with respect to the motion about the y direction does not exceed 10%. The plate and beams are constructed from the same material.

Problems

2l2

l1

l2

Anchors

2l2

l1

l2

Beam

y Moving link

FIGURE 2.48 Four-Beam Microaccelerometer.

2.16 The shaft of Figure 2.38(a) is replaced by the shaft of Figure 2.38(b), as described in Problem 2.4. Known are d1 ¼ 0.01 m, l1 ¼ 0.12 m, l2 ¼ 0.06 m, as well as the disk diameter D ¼ 0.08 m and its thickness h ¼ 0.008 m. Find the diameter d2 for the design of Figure 2.38(b) that will increase the mass moment of inertia of the original system of Figure 2.38(a) with respect to the longitudinal axis by 10%. Consider that all portions are constructed of the same homogeneous material. 2.17 Several thin and identical beams are affixed in a radially symmetric fashion to the free end of a circular cross-section piezoelectric bar, as shown in Figure 2.49. The beams form a spring opposing the axial deformations of the bar. Known are the following geometric parameters: d1 ¼ 0.008 m, l1 ¼ 0.07 m, w ¼ 0.003 m, h ¼ 0.0004 m, l2 ¼ 0.05 m, as well as the mass densities of the bar r1 ¼ 6500 kg/m3 and the thin beams r1 ¼ 7500 kg/m3. Evaluate the number of thin beams that generate an increase of 5% in the lumped-parameter mass of the piezoelectric flexible bar at A. Anchor w l2

Anchor

d1

l2

Beam

Flexible bar x

A h l1

Beam

Radial symmetry

Anchor (a)

(b)

FIGURE 2.49 Flexible Bar Under Axial Deformation With End Beams: (a) Side View; (b) End View.

2.18 The parallelipipedic plate of Figure 2.50 slides in a fluid within a similar channel of very large out-of-plane width. Study graphically the variation of the viscous damping coefficient with the transverse position of the plate in the

67

68

CHAPTER 2 Mechanical Elements

Channel v h

Fluid

g0

l

FIGURE 2.50 Body Sliding in a Channel With Viscous Damping.

channel and determine the specific positions that yield extreme values of the damping coefficient. Known are the dynamic viscosity m ¼ 0.002 N s/m2, the plate length l ¼ 0.06 m, the width w ¼ 0.01 m, the thickness h ¼ 0.002 m, and the channel thickness g0 ¼ 0.015 m. It is also known that the minimum gap between the body and the internal wall of the channel is (g0  h)/20. Ignore damping due to fluidebody interaction on the body frontal face. 2.19 Figure 2.51 shows a translatory-motion bearing. It is formed of a rectangular cross-section rod placed symmetrically in and translating inside a similar box filled with viscous fluid. Calculate the total damping coefficient based on the geometric parameters shown in Figure 2.51 and the fluid dynamic viscosity m. Fluid

Case Cap

w Case

Cap Rod

v

hi

Rod

l

wi

(a)

(b)

h

FIGURE 2.51 Translatory Bearing System as a Viscous Damper: (a) Side View; (b) Cross-Section.

2.20 A piston of length l ¼ 30 mm and diameter Di, which is not known precisely, can move in a very long cylinder of inner diameter Do ¼ 10 mm. When the piston cylinder translates into the cylinder with a velocity of 0.5 m/s, the damping force is 0.05 N, and when the piston rotates with n ¼ 2000 rot/ min, the damping torque is 9.2$104 N m. Evaluate the dynamic viscosity of the liquid in the cylinder, as well as the piston diameter. Ignore fluid resistance on the cylinder frontal face. 2.21 Calculate the total rotary damping coefficient of the Houdaille damper shown in Figure 2.52, which consists of a rotating shaft-disk in a case filled with fluid. Known are ds ¼ 0.006 m, Di ¼ 0.02 m, Do ¼ 0.026 m, l ¼ 0.01 m, g ¼ 0.002 m, and the liquid dynamic viscosity m ¼ 0.0015 N s/m2. Consider fluid resistance on both the front and lateral faces.

Problems

Case Disk Di ds Do Shaft

Bearing

g

g l

FIGURE 2.52 Cross Section of Houdaille Damper System.

2.22 Two translatory dampers of unknown damping coefficients ct1 and ct2 are connected in the two arrangements of Figure 2.53. Using a velocity sensor and a force sensor in each of the two setups, the damping forces and velocities are measured at points A and B, which are fdA, vA and fdB, vB, respectively. It is noticed that fdB / fdA ¼ 4vB / vA. Determine ct1 and ct2 based on these experimental measurements. xB

xA

Anchor

ct1

ct2

A

ct1

(a)

B

ct2

(b)

FIGURE 2.53 Translatory-Motion Dampers With: (a) Fixed-Free Ends; (b) Fixed-Fixed Ends.

2.23 The translatory dampers of damping coefficients ct1 ¼ 0.01 N s/m, ct2 ¼ 0.012 N s/m, ct3 ¼ 0.013 N s/m, ct4 ¼ 0.009 N s/m, ct5 ¼ 0.008 N s/m, and ct6 ¼ 0.015 N s/m are connected as pictured in Figure 2.54. Calculate the force that needs to be applied at point A to produce a velocity v ¼ 1 m/s at that point.

c t3

x

ct4 c t1

ct2

A ct5

FIGURE 2.54 Translatory Damper Combination.

ct 6

69

70

CHAPTER 2 Mechanical Elements

2.24 Determine the equivalent torsional stiffness transferred to the intermediate rigid shaft axis of the gear transmission sketched in Figure 2.55. All other shafts are flexible; they are constructed from the same material with G ¼ 90 GPa N/m2 and are defined by d1 ¼ 0.005 m, l1 ¼ 0.15 m, d2 ¼ 0.004 m, l2 ¼ 0.2 m, d3 ¼ 0.0045 m, l3 ¼ 0.1 m, d4 ¼ 0.006 m, and l4 ¼ 0.12 m. Known are also the number of teeth of the gears N1 ¼ 36, N2 ¼ 44, N3 ¼ 32, and N4 ¼ 48. l1

l4 N4

N1

Shaft

Anchor d1 Gear d2

d4

d3

N3

N2 l2

l3 Rigid shaft

FIGURE 2.55 Rotary-Motion Gear Transmission With Flexible Shafts.

2.25 Three rigid shafts mesh through their respective gears as illustrated in Figure 2.56. Calculate the total mass moment of inertia of this system transferred to the middle shaft. The shafts have a mass density r ¼ 7800 kg/m3 and are defined by d1 ¼ 0.012 m, l1 ¼ 0.15 m, d2 ¼ 0.01 m, l2 ¼ 0.1 m, d3 ¼ 0.009 m, and l3 ¼ 0.13 m. The following parameters define the four gears: the number of teeth N1 ¼ 18, N2 ¼ 22, N3 ¼ 28, N4 ¼ 14, their mass moments of inertia J1 ¼ 6$104 kg m2, J2 ¼ 8$104 kg m2, J3 ¼ 9$104 kg m2, J4 ¼ 4$104 kg m2, as well as the mass moments of inertia of the two end flywheels Jw1 ¼ 1$103 kg m2 and Jw2 ¼ 9$104 kg m2. l1 Wheel

Bearings

Jw1

N3, J3

Shaft

N1, J1

Gear d1

l2

N2, J2

d2 Jw 2 N4, J4 d3 l3

FIGURE 2.56 Rotary-Motion Gear Transmission With Heavy, Rigid Shafts.

Problems

2.26 Calculate the mass moment of inertia on the input shaft of Figure 2.57, and which is inertially equivalent to the gear train system. The shafts have a mass density r ¼ 7400 kg/m3. The following geometrical parameters are known: d1 ¼ 0.009 m, l1 ¼ 0.05 m, d2 ¼ 0.008 m, l2 ¼ 0.035 m, d3 ¼ 0.012 m, and l3 ¼ 0.04 m. Known are also the number of gear teeth N1 ¼ 20, N2 ¼ 30, N3 ¼ 18, and N4 ¼ 32 and the moments of inertia J1 ¼ 5$105 kg m2, J2 ¼ 7$105 kg m2, J3 ¼ 6$105 kg m2, and J4 ¼ 8$105 kg m2. l1

l3 N4, J4 N1, J1

Bearing

d1

Shaft

Gear

d3 d2 N3, J3

N2, J2 l2

FIGURE 2.57 Gear Train With Inertia.

2.27 A long shaft is supported by three identical hydraulic bearings and meshes through gears with a shorter shaft supported by two identical bearings, as shown in Figure 2.58. Calculate the total damping moment acting on the short shaft, knowing N1 ¼ 64, N2 ¼ 48, the bearing dynamic viscosity m ¼ 0.003 N s/m2, shaft diameters Di ¼ 1 cm, interior bearing diameters Do ¼ 1.1 cm, the length of long shaft bearings l1 ¼ 4 cm, the length of short shaft bearings l2 ¼ 2 cm, and angular velocity of long shaft n1 ¼ 250 rot/min. Compare this moment with the one acting on the long shaft. N1 Gear

Shaft

Bearings

Shaft Gear N2

Bearings

FIGURE 2.58 Meshing Gears With Shafts and Hydraulic Bearings.

2.28 Three shafts mesh through their respective gears as illustrated in Figure 2.59. Calculate the damping coefficient of a rotary damper attached to the lower shaft, which is equivalent to all rotary dampers. Known are: the number

71

72

CHAPTER 2 Mechanical Elements

Bearings

N1 Shaft

Gear

cr1

cr1 N3

cr2

cr 2

N2

cr 3 N4

cr3

FIGURE 2.59 Rotary-Motion Shaft-Gear Transmission With Dampers.

of teeth N1 ¼ 18, N2 ¼ 22, N3 ¼ 28, and N4 ¼ 14, and the damping coefficients of the three sets of rotary dampers: cr1 ¼ 0.0012 N m s, cr2 ¼ 0.0015 N m s, and cr3 ¼ 0.001 N m s. Assuming an external torque ma ¼ 0.05 N m balances the damping torque on the lower shaft, what is the constant angular velocity of this shaft? 2.29 The rigid lever OC can pivot around the axis passing through O, as illustrated in Figure 2.60. The lever is supported by three translation springs, which are parallel, and whose deformations are along the y axis that is perpendicular to the lever’s longitudinal axis. Knowing k1 ¼ 120 N/m, k2 ¼ 100 N/m, and k3 ¼ 50 N/m, determine the stiffness of a single spring to be placed at point B and parallel to y that is equivalent to the actual springs acting on the lever. Assume the lever undergoes small rotations. Calculate the vertical displacement at C when a force fy ¼ 5 N is applied at B.

k1

k3

O

A l

y C

B k2

l

l

FIGURE 2.60 Small-Rotation Pinned Lever With Attached Springs.

2.30 Calculate the force fy applied perpendicularly to the rigid lever OC at Bdsee Figure 2.61dthat generates a displacement yc ¼ 0.01 m for k1 ¼ 80 N/m, k2 ¼ 150 N/m, k3 ¼ 40 N/m, and k4 ¼ 30 N/m. The lever undergoes small rotations around the axis passing through the pivot O.

Problems

l fy

k3

O

C

A

B

k4

y

k1 k2

l

l

FIGURE 2.61 Small-Rotation Pinned Lever With Attached Springs and Force.

2.31 The rigid lever of Figure 2.62 undergoes small rotations around an axis passing through O. The lever is supported by several parallel springs that are perpendicular to it and have the following stiffnesses: k1 ¼ 180 N/m, k2 ¼ 160 N/m, k3 ¼ 70 N/m, and k4 ¼ 80 N/m. Knowing also l ¼ 0.4 m and assuming small rotations of the lever, determine the stiffness of an equivalent rotary spring to be placed at O and replacing all original springs. Hint: Equate the elastic potential energy of the actual springs to the one of the single rotary spring. l

2l O A

B

k1 k4

k3 k2

FIGURE 2.62 Lever With Several Attached Springs.

2.32 The rigid lever of length l ¼ 0.8 m shown in Figure 2.63 undergoes small rotations about the pivot axis O. A known point mass m1 ¼ 0.6 kg is placed on it at a distance l1 ¼ 0.6 m from O. Find the position of another Lever O

m2

mr l2 l1

l

FIGURE 2.63 Inertia System With Lever and Two Point Masses.

m1

A

y

73

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CHAPTER 2 Mechanical Elements

point mass m2 ¼ 0.4 kg attached to the lever to the left of m1 such that the equivalent mass at A increases by 10% when the lever is considered: (i) Massless (ii) Heavy, with a mass mr ¼ 0.5 kg. 2.33 The two identical rigid and massless levers of Figure 2.64(a) undergo small rotations around their pivot axes and are connected at their ends by a short, rigid rod of mass m. A point mass m is placed on the left lever, as shown in the same figure. Determine the value of another point mass m1 to be placed on the right lever that results in an equivalent point mass meq ¼ 4m/3 located on the connecting rod, as shown in Figure 2.64(b). θ

m

y m

l/2

y

θ m1

θ

l

meq

θ

x l

(a)

(b)

FIGURE 2.64 (a) Actual Lever System; (b) Equivalent Lumped-Parameter Mass.

2.34 Several identical translatory dampers are connected to the small-rotation lever of Figure 2.65. Find the damping coefficient value ct for which a force fy ¼ 20 N applied at the lever’s free-end results in a constant angular velocity u ¼ 20 rad/s for l ¼ 0.5 m.

ct

ct

ct fy

O y ct l

ct l

l

ct

FIGURE 2.65 Small-Rotation Pinned Lever With Attached Dampers and Force.

Suggested Reading

2.35 The torsional-vibration absorber of Figure 2.66 needs to increase its power dissipation capability by 30%. If two dampers identical to the ones already attached to the small-rotation lever are available, determine their position, symmetric with respect to the pivot point, that would realize this requirement. Assume the actual and modified lever systems undergo small rotations with identical and constant angular velocities.

l

l O

ct

ct

FIGURE 2.66 Torsional-Vibration Absorber With Lever and Attached Translatory Dampers.

Suggested Reading R.C. Hibbeler, Engineering Mechanics: Dynamics, 14th Ed. Pearson, Upper Saddle River, 2015. W.C. Young and R.G. Budynas, Roark’s Formulas for Stress and Strain, 8th Ed. McGraw-Hill, New York, 2011. S.S. Rao, Mechanical Vibrations, 6th Ed. Pearson, Upper Saddle River, 2016. R.G. Budynas and K. Nisbett, Shigley’s Mechanical Engineering Design, 10th Ed. McGraw-Hill, New York, 2014. W. Trylinski, Fine Mechanisms and Precision Instruments, Pergamon Press, Oxford, 1971. I. Cochin, Analysis and Design of Dynamic Systems, Harper&Row, New York, 1980.

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Mechanical Systems

3

OBJECTIVES This chapter studies mechanical systems as combinations of mechanical elements, which are introduced in Chapter 2. The following subjects are discussed: • Degrees of freedom (DOF) or configuration of a mechanical system. • Conservative, free damped, and forced single-DOF mechanical systems: mathematical model formulation by means of the energy method and Newton’s second law of motion; time-domain solutions by MATLAB and Simulink. • Application of Newton’s second law of motion and the Lagrange’s equations to formulate the mathematical models of conservative, free damped, and forced multiple-DOF mechanical systems. • Free undamped (natural) vibrations of multiple-DOF mechanical systems; calculation of natural frequencies (eigenvalues) and natural (modal) motions of multiple-DOF mechanical systems, analytically and by means of MATLAB. • Free damped and forced vibrations of multiple-DOF, linear and nonlinear mechanical systems, MATLAB and Simulink solutions.

INTRODUCTION One key objective of this chapter is to derive time-domain mathematical models of mechanical systems that are formed of inertia, spring, and damping elements. The lumped-parameter modeling approach is utilized to study single-DOF systems that depend on one independent variable, and whose mathematical model consists of one differential equation. Multiple-DOF systems with mathematical models formed of several differential equations are also analyzed. Translatory and rotary mechanical systems are modeled to study the free-undamped (natural), free damped, and forced responses by using Newton’s second law of motion, the conservation of energy method or Lagrange’s equations. Analytical methods as well as MATALB and Simulink are used to solve and analyze several examples.

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00003-8 Copyright © 2018 Elsevier Inc. All rights reserved.

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3.1 CONFIGURATION, DEGREES OF FREEDOM The notion of degrees of freedom (DOF) is very important in correctly modeling dynamic systems. The DOF represent the minimum number of independent parameters that fully define the configuration (or state) of a system. The number of DOF is equal to the number of equations defining the mathematical model of a dynamic system, as detailed in this chapter and in Chapters 4e13. Consider for instance the mechanical system of Figure 3.1(a) consisting of a mass, spring, and damper. The configuration of this system is fully defined by the parameter x, which locates the mass m at all times. As a consequence, this system has one DOF (it is called a single-DOF system).

y

Car path

y x

Point mass

Car

P

k m

C

θ

yP yC

c

x xP

x xC

(a)

(b)

(c)

FIGURE 3.1 (a) SingleeDOF Mechanical System; (b) Two-DOF Point Mass System in Planar Motion; (c) Top View of Car in Planar Motion as a Three-DOF Body System.

The point mass of Figure 3.1(b) has two DOF because its location in the xy plane is determined by two coordinates, xP and yP. The planar motion of a rigid body, such as the car of Figure 3.1(c), can be defined by the linear coordinates of the car center C, xC and yC, and the angular coordinate q; as a consequence, this system possesses three DOF. While a particle or a rigid body undergoing translation is positioned by a single point, springs, and dampers, whose various points are in relative motion and that are connected at their ends to other elements, require their two end-point parameters to be specified. Figure 3.2 shows schematically springs and dampers that are utilized in either translatory or in rotary motion. Each of these elements attach at their ends to either mobile elements or to fixed supports (anchors). When both ends are moving, each of the components shown in Figure 3.2 has two DOF. When only one end is movable (the other being fixed), the respective spring or damper has one DOF.

3.1 Configuration, Degrees of Freedom

ROTATION

TRANSLATION x1

x2

k

θ1

k

θ2

Springs x2

x1

θ1

θ2

Dampers c

c

FIGURE 3.2 Springs and Dampers With Mobile Ends in Translatory and Rotary Motion. x1 x1

x3

x2

k

k

c 1

1

x2

2

2

3 c

(a)

(b)

FIGURE 3.3 Movable Elements Connected in: (a) Series; (b) Parallel.

Often times, springs and/or dampers are connected in series or in parallel to achieve motion features that are unavailable when produced by individual springs or dampers. The total displacement at the end of a serial chain is the sum of individual displacements, whereas the displacement of components coupled in parallel is identical, as illustrated in Figure 3.3. For the series connection of Figure 3.3(a), the absolute displacements at points 1, 2, and 3 are denoted by x1, x2, and x3, respectively. The spring and damper share the displacement x2 at point 2, and therefore, this system has three DOFdthe coordinates x1, x2, and x3. The end points of the spring and the damper of the parallel connection sketched in Figure 3.3(b) have the same displacements at points 1 and 2, which are x1 and x2. As a consequence, the system has two DOFdthe coordinates x1 and x2. nn

Example 3.1 Evaluate the DOF of the translatory mechanical system shown in Figure 3.4.

Solution Figure 3.5 redraws the mechanical system of Figure 3.4 by identifying the relevant points with differing displacements. The damper c1, spring k1, point mass m1, and damper c2 are connected in series. The damper c1 and spring k1 share the displacement x1 at 1, whereas the spring k1 and point mass m1 share the displacement x2 at 2. The mass m1 and damper c2 have the same displacement x2 at 2. The

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CHAPTER 3 Mechanical Systems

k2 Fixed support

c4 m2

k1

Fixed support

m1 c1

c2

k4 k3

c3

FIGURE 3.4 Translatory Mechanical System. x4

x3 x2

k2

x1 m1 0

c3 m2

k1 c1

x6

4

3

2

6

5

c2

1

7 k4 k3

c4 x5

FIGURE 3.5 Translatory Mechanical System With Points of Different Motions. displacement x3 is shared by the right end of the damper c2, the left ends of the spring k2, and the damper c4. The spring k2, point mass m2, and damper c3 are connected in series, which identifies the displacements x3, x4, and x6. The damper c4 and spring k3 are serially connected as well, which identifies the common displacement x5. The serial group formed of the spring k2, point mass m2, and damper c3 is in parallel with the series chain comprising the damper c4 and spring k3. This parallel formation is in series with the damper c2 (with which it shares the displacement x3 at point 3) and with the spring k4 (they have the same displacement x6 at point 6). Note that the end points 0 and 7 are fixed, which results in x0 ¼ 0 and x7 ¼ 0. Combining all this information, it is concluded that this mechanical system has six DOFda possible choice being the coordinates x1, x2, x3, x4, x5, and x6. nn

The previous example illustrates how movable connections, which are connections realized through spring or damping elements, actually do not alter the DOF in a mechanical system. There are also mechanical systems with adjacent elements that are rigidly connected, such as the translating mechanical system of Figure 3.6(a)dwhere a rigid rod attaches to the two point masses, or as the rotary system of Figure 3.6(b), which depicts two disks that are coupled by a rigid shaft. x2 = x1

x1 Point mass

Rigid rod

θ2 = θ1

θ1 Rigid shaft

Point mass Disk

Disk

(a)

(b)

FIGURE 3.6 Rigid Connections for: (a) Two Translatory Point Masses; (b) Two Rotary Disks.

3.1 Configuration, Degrees of Freedom

Because of the rigid rod, the two point masses of Figure 3.6(a) undergo identical displacements x1 ¼ x2 at all moments in time. Similarly, the two disks of Figure 3.6(b) rotate identically due to the rigid shaft connecting them, which means q1 ¼ q2. As a consequence, instead of possessing two DOF (x1 and x2 or q1 and q2), the two systems of Figure 3.6 are single-DOF systems. There is a similar situation when the apparent DOF can be reduced due to adjacent mechanical elements that realize motion transmission. Consider the fixed-axis rotating pinion of Figure 3.7(a) which contacts tangentially the rack and, by imparting its tangential velocity to the rack, sets it into translatory motion. For no slippage between θ

Pinion C

N1 C

R

R P

Rack

xP

(a)

θ1

θ Pinion x C

P Guide

Fixed rack

(b)

C1

θ2

yP

R1

C2 P

R2

N2

Pinion 2 Pinion 1

(c)

FIGURE 3.7 Related Coordinates in Motion Transmission: (a) Fixed-Axis Pinion and Translating Rack; (b) Rolling Pinion and Fixed Rack; (c) Two Fixed-Axis Meshing Pinions.

the two adjacent mechanical elements, xP and q are connected by xP ¼ R$q. As a result of this linear relationship, the pinion-rack mechanical transmission needs only one coordinate to define its state (which can be either xP or q) instead of two coordinates (xP and q); it follows that this system has one DOF. A similar case is the one depicted in Figure 3.7(b) where the pinion roles without slippage on the fixed rack (surface). The planar motion of the pinion can be defined by two coordinates: the rotation angle q and the center displacement xC. However, the planar motion of the pinion is equivalent to a rotation around an axis passing through the instantaneous center of zero velocities P (the contact point between the pinion and the fixed rack), which means xC ¼ R$q. As a consequence, the mechanical system shown in Figure 3.7(b) has only one independent parameter instead of the two coordinates xC and q. This system is also a single-DOF one. The mechanical system of Figure 3.7(c) consisting of two fixedaxis pinions transfers the rotation q1 from one shaft to the rotation q2 of the other shaft. As discussed in Chapter 2, the two coordinates are related as: q2/q1 ¼ R1/R2 ¼ N1/N2 (where N1 and N2 are the number of teeth). Because of this relationship between the two coordinates, the system possesses one DOF. The examples of Figures 3.6 and 3.7 illustrate the rule of calculating the DOF of a mechanical system: the number of actual DOF is the difference between the potential (candidate or apparent) DOFdwhich are the free coordinates of the systemdand the constraint equations reflecting coordinate connections (as a direct result of either rigid connections or motion transmission between adjacent mechanical elements).

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CHAPTER 3 Mechanical Systems

nn

Example 3.2 Establish the DOF of the mechanical system sketched in Figure 3.8. The system is formed of a twocylinder pinion rolling without slippage on a fixed rack and a fixed-axis pinion, which interacts through a translating rack with the inner pinion cylinder. Another rack meshes with the outer pinion cylinder and attaches to a block at its end. A horizontal spring connects to the rolling pinion’s center.

2R Pinion θ1

D

O Rack

Guide A

Rack

x2 θ2

R1

B C

Block

x1 Spring

R2

x3 x2

D

x3

x3

B C θ2

x1

D’ B’

C’

Rolling pinion P Fixed rack

(a)

P

(b)

FIGURE 3.8 (a) Mechanical System With Pinions, Racks, Spring and Block; (b) Cylinder Rolling Without Slippage on a Fixed Surface.

Solution The free coordinates that describe the motion of the mechanical system of Figure 3.8 are: the rotation angles q1 of the top pinion, which rotates around a fixed axis passing through O and q2 of the rolling pinion, as well as the translatory displacements x1 of the rolling pinion center, x2 of the rack connecting the two pinions, and x3 of the block. There are five free coordinates, which are candidate DOF. The following constraints can be written between these coordinates under the assumption of small motions and based on the fact that various points on the rolling pinion undergo pure rotations with respect to the instantaneous center of velocities P (the contact point between the pinion and the fixed rack):

x2 ¼ R $ q1 ; x2 ¼ ðR1 þ R2 Þ $ q2 ; x1 ¼ R1 $ q2 ; x3 ¼ 2R1 $ q2

(3.1)

The last three in Eq. (3.1) are related to the rolling without slippage of the bottom pinion along the fixed horizontal rack, as illustrated in Figure 3.8(b). The pinion rotates with respect to the instantaneous center of velocities P. As a consequence, the small displacements of the roller center can be considered to be horizontal as points C, B, and D move to C0 , B0 , and D0 , respectively. The horizontal distances traveled by these points, CC0 , BB0 , and DD0 are expressed as arc lengths (corresponding radius times rotation angle q1), which explains the last three of Eq. (3.1). Because the difference between the 5 (five) apparent DOF and the 4 (four) constraints is 1 (one), this system is a single-DOF system. nn

nn

Example 3.3 The rotary gear system of Figure 3.9 comprises an input shaft, which receives its motion from an electric motor through a disk pulley and transmits it further to a cam and to a flywheel located on two output shafts through the gear pairs N1eN2 and N3eN4. Determine the DOF of this system when the shafts are assumed: (i) Rigid (ii) Torsionally flexible.

3.2 SingleeDOF Systems

From motor Input shaft N1 Pulley

Bearing N3 Output shaft #2

Output shaft #1 Gear

Cam

N4

Flywheel

N2 To follower

FIGURE 3.9 Rotary Motion Transmission With Gears, Pulley, Cam, and Shafts.

Solution (i) Figure 3.10(a) shows the original rotary system with all the shafts considered rigid. In this situation all rotary components attached to any of the three shafts have the same angular velocity with their shaft. As a result, three rotation angles: qi (for the input shaft), qo1, and qo2 (for the output shafts) define the system’s motion and these are candidate DOF. However, the following connections do exist between these three would-be DOF due to gear meshing:

qo1 ¼

N1 N3 $ qi ; qo2 ¼ $ qi N2 N4

(3.2)

The DOF equals the number of candidate DOF, which is three, minus the two constraint relationships of Eq. (3.2), which is 3  2 ¼ 1. As a consequence, the rigid-shaft system of Figure 3.10(a) has one DOF. (ii) When the shafts are considered flexible in torsion, each of them can be modeled as a rotary spring connecting two adjacent rigid disksdsee Figure 3.10(b). As a consequence, the input rotation angle qi is no longer retrieved at any point along the input shaft; an angle qi1, different from qi, defines the angular position of gear N1 and, similarly, the angle qi2 measures the angular position of gear N3 because of the rotary springs connecting these gears on the input shaft. Similarly, the distinct angles qo1 and q1 define the rotary positions at the ends of the left output shaft, whereas the right output shaft is defined by the end angles qo2 and q2 as indicated in Figure 3.10(b). Counting all these angular variables, it follows that the candidate DOF are seven. However, the following two constraint equations apply:

qo1 ¼

N1 N3 $ qi1 ; qo2 ¼ $ qi2 N2 N4

(3.3)

which results in 7  2 ¼ 5. The rotary system with torsionally flexible shafts has therefore five DOF. nn

3.2 SINGLEeDOF SYSTEMS In this section, single-DOF mechanical systems are modeled and analyzed for both translatory and rotary motion. Newton’s second law of motion and the energy method are used here to derive mathematical models for free undamped systems. Newton’s second law of motion is applied to obtain mathematical models of free damped systems, as well as of forced systems.

83

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CHAPTER 3 Mechanical Systems

Rigid shaft

θ

(a)

(b)

FIGURE 3.10 Rotary Motion Transmission With Gears, Pulley, Cam, and: (a) Rigid Shafts; (b) Flexible Shafts.

3.2.1 Free Response In the free response, no forcing is applied to a mechanical system, which consequently undergoes free vibrations when excited externally by an initial displacement or initial velocity. The natural (undamped) response and the damped response are studied separately.

Conservative Systems and the Free Undamped (Natural) Response Free undamped vibrations are characteristic to the simplest mechanical systems that are formed of a rigid body with inertia and a spring, as shown in Figure 3.11(a) for translation, and in Figure 3.11(b) for rotation. PHYSICAL MODEL

FREE-BODY DIAGRAM

PHYSICAL MODEL

FREE-BODY DIAGRAM

x

x

θ

k

k

fe

m

θ

me

m J

J

(a)

(b)

FIGURE 3.11 Physical Model of Inertia-Spring Systems With Free-Body Diagrams for: (a) Translation; (b) Rotation.

Mathematical Model (Differential Equation) The mathematical model of a conservative mechanical system is derived in this section by means of Newton’s second law of motion and the conservation of energy method. Newton’s Second Law of Motion According to this method, the acceleration
of a
€ ¼ d2 qðtÞ dt2 particle or rigid body (a ¼ x€ ¼ d2 xðtÞ dt2 in translation and a ¼ q in rotation) is proportional to the force/moment applied to it. The proportionality constant is the mass m for a particle or a rigid body undergoing translation and is the mass moment of inertia J of the body (or particle) undergoing rotation. When

3.2 SingleeDOF Systems

several forces/moments act on the particle/body, the acceleration is proportional to the algebraic sum of forces fj (for translation) or the algebraic sum of moments mj (for rotation), namely: ! ! n n n n X X X X € fj or fj þ fi ¼ 0; J $ q ¼ mj or mj þ mi ¼ 0 m $ x€ ¼ j¼1

j¼1

j¼1

j¼1

(3.4) The last form of the two in Eq. (3.4) uses the inertia force fi ¼ m $ x€ and the inertia € These equations are known as d’Alembert equations. They exmoment mi ¼ J $ q. press the dynamic equilibrium between all the external and the inertia forces/moments acting on a body. A necessary component in the process of correctly using these equations is the free-body diagram, which isolates the mechanical element of interest from its system, and applies external forcing, as well as reactions from the elements that have been separated from the studied element. According to Newton’s second law of motion of first Eq. (3.4) and based on the free-body diagram of Figure 3.11(a), the equation of motion for the translating mass is: k $x ¼ 0 (3.5) m where fe is the elastic force. The mathematical model of the free undamped response for a single-DOF translatory system is represented by either of the last two of Eq. (3.5). For the rotary system of Figure 3.11(b), the second Eq. (3.4) results in the following equation and mathematical model: m $ x€ ¼ fe or m $ x€ ¼ k $ x or x€ þ

€ ¼ me or J $ q € ¼ k $ q or q € þ k$q ¼ 0 (3.6) J $q J where me is the elastic moment. The Energy Method The energy method can be used to derive the mathematical model of conservative systems, as an alternative to Newton’s second law of motion. For a mechanical system with no external forcing acting on it and no losses (such as those owing to damping), the total energy E (which comprises kinetic T, elastic potential Ue and gravitational potential Ug energy fractions) is constant and such a system is a conservative system. As a consequence, the time derivative of the total energy is zero: dE d ¼ ðT þ Ue þ Ug Þ ¼ 0 (3.7) dt dt In the case of the particular system of Figure 3.11(a), the total energy of the system consists of kinetic and elastic potential: 1 1 E ¼ T þ Ue ¼ $ m $ x_2 þ $ k $ x2 2 2

(3.8)

85

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CHAPTER 3 Mechanical Systems

After applying the chain rule of differentiation to E of Eq. (3.8), the generic Eq. (3.7) becomes: dE d ¼ ðT þ Ue Þ ¼ m $ x_ $ x€ þ k $ x $ x_ ¼ x_ $ ðm $ x€ þ k $ xÞ ¼ 0 (3.9) dt dt The last product has to be zero at all moments in time, but the velocity of this system, which is v ¼ x,_ cannot always be zero unless the system is at rest. The only variant is when the other factor of the product is zero, namely: m $ x€ þ k $ x ¼ 0, which is the mathematical model of the mass-spring system also derived by means of Newton’s second law in Eq. (3.5). The reader is encouraged to follow a similar procedure, and derive the mathematical model for the rotary system of Figure 3.11(b) by the energy method, taking into
account
that the total rotary-motion constant energy is 2 E ¼ T þ Ue ¼ J $ q_ 2 þ k $ q2 2.

Solution of Mathematical Model (Differential Equation) The solution to a differential equation with no forcing (a homogeneous equation) consists only of the natural part, and the theory of ordinary differential equations shows that this solution is of the type: x ¼ X $ sinðu $ t þ 4Þ

(3.10)

where X is the amplitude, u is the rotation frequency, and 4 is the (initial) phase angledall quantities being unknown. Eq. (3.10) produces the following relationship between acceleration and displacement: x€ ¼ u2 $ x. Comparing this relationship to Eq. (3.5) yields: pffiffiffiffiffiffiffiffiffi u ¼ k=m ¼ un and x€ þ u2n $ x ¼ 0 (3.11) The parameter u ¼ un, which is based on the physical characteristics of the singleDOF mass-spring system, is named natural frequency, and plays an important role in defining the free undamped response of mechanical systems. Eqs. (3.10) and (3.11) indicate that the body motion, also known as natural or modal motion, is harmonic and has a frequency of oscillation equal to the natural frequency un. The second Eq. (3.11) indicates that, once the mathematical model was derived, the natural frequency is simply identified from the second-order differential equation. A similar approach is applied to the rotary system of Figure 3.11(b) by using q ¼ Q $ sinðut þ 4Þ, which enables finding a natural frequency: pffiffiffiffiffiffiffiffi € þ u2 $ q ¼ 0 u ¼ k=J ¼ un and q (3.12) n The amplitude X and phase angle 4 of Eq. (3.10) can be determined from two known initial conditions (the displacement x0 and velocity v0 at t ¼ 0), which are expressed as:  x0 ¼ X $ sinðun $ t þ 4Þjt¼0 ¼ X $ sinð4Þ (3.13) v0 ¼ un $ X $ cosðun $ t þ 4Þjt¼0 ¼ un $ X $ cosð4Þ

3.2 SingleeDOF Systems

x(t) X x0 0

X

Tn

Time

FIGURE 3.12 Free Undamped Time Response Plot of SingleeDOF Mechanical System.

The system of two equations with two unknowns (X and 4)dEq. (3.13)dhas the following solution: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   u2n $ x20 þ v20 1 un $ x0 ; 4 ¼ tan X¼ (3.14) un v0 The time response of Eq. (3.10) is plotted in Figure 3.12, which highlights the sinusoidal shape with its amplitude X, initial displacement x0 (incidentally at t ¼ 0 here), and the natural period Tn, which is connected to the natural frequency un as Tn ¼ 2p/un. nn

Example 3.4 Calculate the natural frequency of the lever system of Figure 3.13(a) by transferring the lumpedparameter stiffness and inertia from their locations to the free end point of the rigid lever, as shown in Figure 3.13(b) considering that (i) The lever is massless. (ii) The lever has a mass ml ¼ mB/4. Known are lA ¼ 0.5 m, lB ¼ 1 m, lC ¼ 1.5 m, kA ¼ 1000 N/m, and mB ¼ 1 kg. Assume small lever rotations in a horizontal plane.

Solution

(i) According to Eqs. (2.73) and (2.80) of Chapter 2, the equivalent stiffness and mass at point C are kC ¼ ðlA =lC Þ2 $ kA ; mC ¼p ðlBffiffiffiffiffiffiffiffiffiffiffiffiffiffi =lC Þ2 ffi$ mB . The natural frequency of the equivalent mass-spring ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi system is therefore un ¼ kC =mC ¼ ðlA =lB Þ $ kA =mB ¼ 15:81 rad=s. (ii) The mathematical model of the system needs to be derived first when the lever mass is accounted for. Let us use both Newton’s second law of motion and the conservation of energy method. Newton’s Second Law of Motion The free-body diagram for the rod of Figure 3.13(b) is shown in Figure 3.13(c), where fe is the elastic force produced by the equivalent spring kC. The following equation describes the rotation of the lever around the pivot point O:

€ ¼ fe $ lC JO $ q

(3.15)

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88

CHAPTER 3 Mechanical Systems

yA O

kA

mB

A

B

kC yB

yC

yC

O

C

lC

lA

C

O

JO

θ

mC

fe

lC

lB lC

(a)

(b)

(c)

FIGURE 3.13 Lever-Mass-Spring System: (a) Actual; (b) Transformed; (c) Free-Body Diagram of a Rotating Rod.

The total mechanical moment of inertia with respect to point O (JO) results from the heavy rod and the lumped-mass mC, while the elastic force uses the small-angle approximation; they are expressed as:

8   > > ml $ l2C mB $ l2C 1 2 2 2 2 > > J $ l þ m þ m ¼ $ l ¼ $ l ¼ þ l C C > C C B $ mB < O 12 C 3 12 > > l2A l2A > > f ¼ k $ y ¼ k $ l $ q ¼ k $ $ l $ q ¼ k $ $q e C C C C A C A > : lC l2

(3.16)

C

Substituting Eq. (3.16) in Eq. (3.15), the latter equation becomes:

l2 1 þ 12 2B lC

!

€ þ 12 $ lA $ kA $ q ¼ 0 $ mB $ q l2C 2

(3.17)

which is the mathematical model for the heavy-lever system motion. Energy Method The total energy of this system is:

1 1 2 E ¼ T þ Ue ¼ $ JO $ q_ þ $ kC $ y2C 2 2   1 1 2 1 2 $ l þ l2B $ mB $ q_ þ $ kA $ l2A $ q2 ¼ $ 2 12 C 2

(3.18)

where Eq. (3.16), as well as the given relationship ml ¼ mB/4, have been used. Applying the time derivative to E of Eq. (3.18) and equating the result to zero (because E is constant) yields:

  1 2 2 2 € $ l þ lB $ mB $ q þ kA $ lA $ q $ q_ ¼ 0 12 C " ! # 2 l l2B A € þ 12 $ ,kA $ q $ q_ ¼ 0 or 1 þ 12 2 $ mB $ q lC l2C 

(3.19)

Eq. (3.19) is always complied with when the bracket is equal to zero. This condition generates Eq. (3.17), which is the mathematical model of the heavy-lever.

3.2 SingleeDOF Systems

Eq. (3.17) indicates that the natural frequency of the analyzed system is:

un

rffiffiffiffiffiffi pffiffiffi 2 3ðlA =lC Þ kA ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ mB 1 þ 12ðlB =lC Þ2

(3.20)

Its numerical value is 14.51 rad/s, which is obviously smaller than un obtained at (i), where less mass was accounted for. nn

nn

Example 3.5 Two elastic shafts of negligible inertia are connected through two meshing gearsdsee Figure 3.14(a) and 3.14(b). The top gear also meshes with a translating rack, which connects at its ends with two identical springs. Known are N1 ¼ 32, N2 ¼ 26, R1 ¼ 0.032 m, J1 ¼ 0.001 kg m2, J2 ¼ 0.0008 kg m2, k1 ¼ 80 N m, k2 ¼ 200 N m, m ¼ 0.1 kg, and k ¼ 100 N/m. (i) Demonstrate that the system has one DOF. (ii) Derive the mathematical model of this system and find its natural frequency.

Anchor k1

N 1, J1 θ1

Spring

xA

k

Rack

m θ1

Gears

θ2

Rack

θ2

k

2R1 P C2

k

m

A C1

xA

k

θ1

keq,1

Jeq,1

2R2

k2

N 2 , J2

(a)

(b)

(c)

FIGURE 3.14 Two Shaft-Gear-Rack Lumped-Parameter Model: (a) Side View; (b) Front View; (c) Front View of Equivalent-Gear-Rack System.

Solution (i) There are three coordinates defining the state of this mechanical system, namely: the rotation angles of the two meshing gears q1 and q2 and the translatory displacement xA of the rack. The two angles are related as: q2 ¼ (N1/N2)$q1 ¼ (R1/R2)$q1 due to gear meshing. The contact between the top gear and the rack being without slippage, it follows that: xA ¼ R1$q1. Subtracting thus the number of constraint equations (which is 2) from the number of free coordinates (which is 3), the DOF are 3  2 ¼ 1, so this is a single-DOF system. (ii) The inertia and stiffness corresponding to the lower shaft are transferred to the upper shaft, which changes the original system to the one of Figure 3.14(c). As detailed in Chapter 2 and according to Eq. (2.48), the total stiffness corresponding to the upper shaft (which contains the resident stiffness k1 and the transferred stiffness k2) is keq,1 ¼ k1 þ (N1/N2)2$k2. Similarly, the total inertia of the same shaft is Jeq,1 ¼ J1 þ (N1/N2)2$J2 by Eq. (2.55). The total energy of the reduced equivalent system of Figure 3.14(c) is:

    1 1 1 1 2 2 2 2 _ _ $ Jeq;1 $ q1 þ $ m $ xA þ $ keq;1 $ q1 þ 2 $ $ k $ xA E ¼ T þ Ue ¼ 2 2 2 2



1 1 2 ¼ Jeq;1 þ m $ R21 $ q_ 1 þ keq;1 þ 2k $ R21 $ q21 2 2 (3.21)

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CHAPTER 3 Mechanical Systems

Eq. (3.21) took into account that the two translatory springs k are in parallel and so their constants add up. The same equation also used xA ¼ R1$q1 and x_A ¼ R1 $ q_ 1 . Because the system is conservative and E ¼ constant, it follows that dE/dt ¼ 0. Applying the time derivative to E of Eq. (3.21) yields:



€1 þ keq $ q1 $ q_ 1 ¼ 0 or Jeq $ q €1 þ keq $ q1 ¼ 0 Jeq $ q

(3.22)

since the time derivative of q1 (which is the angular velocity u1) cannot be zero perpetually. The last form of Eq. (3.22) is the mathematical model of the original mechanical system with:

(

Jeq ¼ Jeq;1 þ m $ R21 ¼ J1 þ ðN1 =N2 Þ2 $ J2 þ m $ R21

(3.23)

keq ¼ keq;1 þ 2k $ R21 ¼ k1 þ ðN1 =N2 Þ2 $ k2 þ 2k $ R21

Comparing Eqs. (3.12) and (3.22) indicates that the natural frequency of the mechanical system of pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi Figure 3.14 is un ¼ keq =Jeq . The following numerical values are obtained: Jeq ¼ 0.0023 kg m2, keq ¼ 383.1634 N m, and un ¼ 406.9 rad/s. nn

nn

Example 3.6 A particle attaches to the paddle portion of a nanobridge as shown in Figure 3.15(a).Through the particle mass addition, the original natural frequencies of this system change with respect to both translation along the y axis and rotation around the x axis. Consider the paddle is rigid and parallelipipedic, with an out-of-plane thickness h. The two identical flexible wires have circular cross section of diameter d. Known are l1 ¼ 100 mm, h ¼ d ¼ 20 mm, l ¼ 200 mm, w ¼ 100 mm, E ¼ 180 GPa, G ¼ 140 GPa, and r ¼ 5000 kg/m3. The altered y-axis translatory natural frequency is 3,998,671 rad/s, and the altered x-axis rotary natural frequency is 5,014,600 rad/s. Determine the mass of the attached particle mp, as well as its position parameter b on the paddle, when considering inertia contributions from the two end flexible wires.

y

meq + mp y Rigid paddle

Flexible wire

kb

Particle Anchor

90

w/2

kb

d O

Anchor

x

b

w/2

(b)

l/2 l1

l1 l

(a)

Anchor

θx

kt

kt Jeq + Jp

(c)

FIGURE 3.15 (a) Top View of Paddle Nanobridge With Attached Particle; (b) Equivalent LumpedParameter Nanobridge Model in Bending; (c) Equivalent Lumped-Parameter Nanobridge Model in Torsion.

3.2 SingleeDOF Systems

Solution

Translation Along the y Axis Figure 3.15(b) shows the equivalent lumped-parameter model of the bridge-particle system undergoing bending vibrations with the paddle translating along the y axis. The natural frequency of the original system (without the particle attached to the paddle) un,b and the one of the system with the attached particle un;b are:

un;b

sffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2kb 2kb ; un;b ¼ ¼ meq meq þ mp

(3.24)

where kb is the stiffness of one of the two nanowires. They are coupled in parallel and, due to the paddle’s y-axis translation, deformed as fixed-guided beams; as a consequence, kb ¼ 12$E$Iz/(l1)3 from Table 2.2 with Iz ¼ pd4/64. The equivalent mass is meq ¼ mpa þ 2$(13/35)$ mwdTable 2.6dwith mpa ¼ r$l$w$h being the paddle mass and mw ¼ r$(pd2/4)$l1 being the mass of one wire. The original system’s bending-related natural frequency is therefore un,b ¼ 4,003,671 rad/s. The two in Eq. (3.24) yield the mass of the particle as:

2

2  un;b  un;b mp ¼ $ meq  2 un;b

(3.25)

with a numerical value of mp ¼ 5.297$1012 kg. Rotation Around the x Axis Rotation of the paddle around the x axis subjects the two nanowires to torsion. The torsion-related natural frequency of the original system (without the particle attached to the paddle), un,t, and the natural frequency of the system with the attached particle, un;t , are:

un;t

sffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2kt 2kt  ; un;t ¼ ¼ Jeq Jeq þ Jp

(3.26)

The two nanowires act as two rotary springs in paralleldsee Figure 3.15(c)dand the stiffness of one spring is kt ¼ G$Ip/l1 as per Table 2.3 with Ip ¼ pd4/32. The equivalent mass moment of inertia with respect to the x axis is Jeq ¼ Jpa þ 2$Jw. Jpa ¼ mpa$(w2 þ h2)/12 (as provided in Table 2.4) is the paddle mass moment of inertia, and Jw ¼ (1/3)$(1/2)$mw$(d/2)2 is the nanowire mass moment of inertia in terms of its longitudinal axis (which is aligned with the x axis), as given in Table 2.7. The original system’s torsion-related natural frequency has a value of un,t ¼ 5,029,710 rad/s. Eq. (3.26) allows solving for Jp as:

Jp ¼

un;t

 2  un;t $ Jeq  2 un;t

2

(3.27)

Because Jp ¼ mp$b2 (see Figure 3.16(a)), it follows that:

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 2 

2  u  $ Jeq u un;t  un;t 1 t b¼  $ un;t mp

(3.28)

The numerical value of the offset is therefore b ¼ 44.5 mm, which indicates the particle deposited almost at the edge of the middle plate (whose half width is of 50 mm). nn

91

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CHAPTER 3 Mechanical Systems

Systems With Losses and the Free Damped Response Mechanical systems as the ones sketched in Figure 3.16 result when viscous damping is added to an inertia-spring pair. The mathematical models of these systems are derived here and the solution to the resulting differential equation is provided for one particular set of values of the damping coefficient. PHYSICAL MODEL

FREE-BODY DIAGRAM

x

PHYSICAL MODEL

x

k

θ

k

fe

FREE-BODY DIAGRAM

c

me θ md

m

m fd

c

J

(a)

J

(b)

FIGURE 3.16 Lumped-Parameter Models of Inertia-Spring-Damper Systems With Free-Body Diagrams for: (a) Translation; (b) Rotation.

Mathematical Model (Differential Equation) The mathematical model is derived by means of Newton’s second law of motion. Application of the principle of energy and work to derive the same mathematical model is exemplified in the companion website Chapter 3. Using Newton’s second law of motion in conjunction with the free-body diagram of the translatory mechanical system of Figure 3.16(a), we have the following equation: m $ x€ ¼ fd  fe or m $ x€ ¼ c $ x_  k $ x or m $ x€ þ c $ x_ þ k $ x ¼ 0

(3.29)

With fd being the damping force and fe the elastic force, the last Eq. (3.29) is the mathematical model of the translatory mechanical system. For the rotary system of Figure 3.16(b) and the corresponding free-body diagram, Newton’s second law of motion results in: € ¼ md  me or J $ € € þ c $ q_ þ k $ q ¼ 0 (3.30) J $q q ¼ c $ q_  k $ q or J $ q where md is the damping moment and me is the elastic moment. The last Eq. (3.30) is the mathematical model of the rotary system.

Solution of Mathematical Model (Differential Equation)

As mentioned in Appendix C, solutions of the form x ¼ el$t are sought for Eq. (3.29); they generate the characteristic equation m$l2 þ c$l þ k ¼ 0. When the discriminant of this equation D ¼ c2  4m$k ¼ 0, the critical damping coeffipffiffiffiffiffiffiffiffiffiffi cient ccr ¼ 2 m $ k ¼ 2m $ un is obtained and the associated damping is known as critical damping. For c > ccr, we have overdamping while c < ccr leads to underdamping. After dividing the left- and right-hand sides of Eq. (3.29) by m and utilizing the definitions of the natural frequency un and damping ratio x (c/m ¼ 2x$un, as introduced in Chapter 1), Eq. (3.29) changes to: x€ þ 2x $ un $ x_ þ u2n $ x ¼ 0

(3.31)

3.2 SingleeDOF Systems

Note that x ¼ c/(2m$un) ¼ c/ccr, which justifies the name of damping ratio. Therefore, x ¼ 1 for critical damping, x > 1 for overdamping, and 0 < x < 1 for underdamping. The cases of overdamping and critical damping are analyzed in the companion website Chapter 3. Underdamping, which is frequently encountered in engineering applications, refers to situations where 0 < x < 1. As known from ordinary differential equations theory, the solution to Eq. (3.31) is: qffiffiffiffiffiffiffiffiffiffiffiffiffi  xðtÞ ¼ X $ ex $ un $ t $ sinðud $ t þ 4Þ ¼ X $ ex $ un $ t $ sin 1  x2 $ un $ t þ 4 (3.32) pffiffiffiffiffiffiffiffiffiffiffiffiffi The quantity ud ¼ 1  x2 $ un is the damped frequency of the mass-dashpot system, and it can be seen that ud < un since x < 1. The solution of Eq. (3.32) has an exponentially decaying amplitude X $ ex $ un $ t .Using known initial conditions for this solution, such as initial displacement x(0) ¼ x0 and velocity vð0Þ ¼ dxðtÞ=dtjt¼0 (after expressing dx(t)/dt from Eq. (3.32)), the amplitude X and phase angle 4 are determined from Eq. (3.32) as: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi! v20 þ 2x $ v0 $ un $ x0 þ ðun $ x0 Þ2 1 1  x2 1 un $ x0 $ X¼ ; 4 ¼ tan $ un v0 þ x $ un $ x0 1  x2 (3.33) Figure 3.17 is the plot of the free underdamped response given in Eq. (3.32) for some arbitrary values of x, un, x0, and v0. The plot also displays the two exponentially decaying envelope curves (X $ ex $ un $ t and X $ ex $ un $ t ) of the sine curve and its damped period Td equal to: Td ¼

2p 2p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ud 1  x2 $ un

(3.34)

x

Xe−ξω t x(t)

x0 0

Time Xk–1

−Xe−ξω t tk–1

Td tk

FIGURE 3.17 Plot of Free Underdamped Response.

Xk

93

94

CHAPTER 3 Mechanical Systems

The same approach can be applied to the rotary system of Figure 3.16(b) based on its mathematical modeldEq. (3.30). The result is an equation similar to Eq. (3.31), namely: € þ 2x $ un $ q_ þ u2 $ q ¼ 0 q n

(3.35)

with c/J ¼ 2x$un and k/J ¼ (un) . In the case of underdamping (0 < x < 1), the solution to Eq. (3.35) is similar to Eq. (3.32)done has to use Q instead of X for amplitude. An amount that characterizes the viscous underdamping is the logarithmic decrement d, which is defined as the natural logarithm of the ratio of any two successive decaying amplitudes or displacements situated at Td intervals apart. For the translatory system of Figure 3.16(a), d is 2

d ¼ ln

Xk1 Xk

(3.36)

The two successive amplitudes of Eq. (3.36), as seen in Figure 3.17, are offset by a time Td and can be written by means of Eq. (3.32) as: Xk1 ¼ X $ ex $ un $ tk1 ; Xk ¼ X $ ex $ un $ ðtk1 þTd Þ

(3.37)

The logarithmic decrement of Eq. (3.36) becomes: 2p $ x d d ¼ x $ un $ Td ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi or x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1x 4p2 þ d2

(3.38)

Eq. (3.38) indicates that the logarithmic decrement depends only on the damping ratio. This equation is instrumental in assessing the damping ratio from available (usually experimentally) logarithmic decrement values. nn

Example 3.7 The gear-shaft system of Figure 3.18(a) comprises two identical, massless, and flexible shafts with l ¼ 0.15 m, d ¼ 0.01 m, G ¼ 80 GPa, four identical journal bearings with length lb ¼ 0.02 m, inner diameter db ¼ 0.0101 m, oil dynamic viscosity m ¼ 0.7 N s/m2, and two gears with N1 ¼ 36, J1 ¼ 25$106 kg m2 and N2 ¼ 48, J2 ¼ 37$106 kg m2. Two brakes are applied suddenly at the shafts’ ends, which results in free damped rotary vibrations. Knowing that the first (maximum) angular amplitude of the bottom shaft is Q2,0 ¼ 6 , determine the amplitude of the same shaft at t ¼ 6$Td after the brakes have been applied, where Td is the damped period.

Solution The brakes lock the two shaft ends, and the shafts become fixed-free bars under torsion from inertia action. The corresponding lumped-parameter physical model of this system is sketched in Figure 3.18(b) with the elastic shafts modeled as rotary springs, and the effect of the journal bearings being played by the rotary dampers. Due to the rigid connection through the meshing gears, the lumped-parameter stiffness, inertia, and damping pertaining to the top shaft can be transferred to the bottom shaft. The resulting system behaves as the single-DOF spring-disk-damper system illustrated in Figure 3.18(c).

3.2 SingleeDOF Systems

l

Brake

N1, J1

k

N1, J1

c

θ1

d Gear

θ2

Shaft

N2, J2

(a)

θ1

θ2

Bearing

c Brake

θ2

k

keq

ceq

N2, J2

Jeq

(b)

(c)

FIGURE 3.18 Two-Shaft Gear System: (a) Actual Representation Before Braking; (b) Transformed Lumped-Parameter System After Braking; (c) Equivalent Lumped-Parameter System After Braking. The equivalent stiffness, mass moment of inertia, and damping coefficient are:

keq ceq

"  2  2 #  2 N2 N2 N2 ¼ k$ þk ¼ 1þ þ J2 ; $ k; Jeq ¼ J1 $ N1 N1 N1 "  2  2 # N2 N2 ¼ 2c $ þ 2c ¼ 2 1 þ $c N1 N1

(3.39)

The last Eq. (3.39) took into account that there are two journal bearings associated with each of the two shafts (and disks), which result in two rotary dampers connected in parallel for each disk. The individual lumped stiffness and damping coefficient of Eq. (3.39) are .
calculated as: k ¼ G $ Ip =l ¼ p $ G $ d 4 ð32 $ lÞ ¼ 523:599 Nm and c ¼ p $ m $ db3 $ lb ½2ðdb  d Þ ¼ 0:000226575 N m sdsee Eq. (2.37). The parameters of Eq. (3.39) are determined as: keq ¼ 1454.441 N m, Jeq ¼ 0.0000814444 kg m2, and ceq ¼ 0.00125875 N m s. The mathematical model of the mechanical system of Figure 3.18(c) is obtained by means of Newton’s second law of motion as:

€2 ¼ ceq $ q_ 2  keq $ q2 ¼ 0 or Jeq $ q €2 þ ceq $ q_ 2 þ keq $ q2 ¼ 0 Jeq $ q

(3.40)

Division in Eq. (3.40) by Jeq transforms this equation into:

€2 þ 2x $ un $ q_ 2 þ q

u2n $ q2

sffiffiffiffiffiffi keq ceq ceq ; 2x $ un ¼ ¼ 0 with un ¼ or x ¼ Jeq Jeq 2un $ Jeq (3.41)

Eq. (3.41) and the numerical values of the three equivalent parameters result in un ¼ 4225.882 rad/s and x ¼ 0.0018. The amplitude ratio of the first amplitude to the sixth one can be expressed as:

Q2;0 ¼ Q2;6



           Q2;5 Q2;0 Q2;1 Q2;2 Q2;3 Q2;4 $ $ $ $ $ Q2;1 Q2;2 Q2;3 Q2;4 Q2;5 Q2;6

(3.42)

95

96

CHAPTER 3 Mechanical Systems

Application of the natural logarithm to Eq. (3.42) results in:

          Q2;0 Q2;0 Q2;1 Q2;2 Q2;3 ln ¼ ln þ ln þ ln þ ln Q2;6 Q2;1 Q2;2 Q2;3 Q2;4     Q2;5 Q2;4 þ ln þ ln ¼ 6$d Q2;5 Q2;6

(3.43)

where d ¼ 0.01149 as per Eq. (3.38). As a consequence, the unknown amplitude after six periods, Q2,6, is determined from Eq. (3.43) as:

Q2;6 ¼ Q2;0 $ e6d ¼ 5:6

(3.44) nn

nn

Example 3.8

A body of mass m ¼ 0.25 kg can move along the x direction being supported by two identical springs of stiffness k ¼ 1000 N/m and undeformed length l ¼ 0.4 m. The spring axes are initially perpendicular to the x motion direction. Two identical dampers of damping coefficient c ¼ 0.2 N s/m, placed along the x direction, are connected to the body, as shown in Figure 3.19(a). For an initial displacement of the body x(0) ¼ 0.2 m and zero initial velocity v (0) ¼ 0 m/s: (i) Derive the mathematical model of the body motion. (ii) Calculate the time history x(t) and plot it by using MATLAB’s ode45 operator. (iii) Design the Simulink block diagram and plot x(t).

x x k

β

l

Anchor

x c

l fe m

m k

c

fd

l

Anchor pin (a)

fd fe

(b)

FIGURE 3.19 (a) Translatory Body With Dampers and Springs; (b) Free-Body Diagram.

Solution

(i) For an arbitrary displacement x to the right, Figure 3.19(b) sketches the free-body diagram of the body with the damping forces fd and elastic (spring) forces fe identified. Newton’s second law of motion is used based on this free-body diagram, which results in the equation:

m $ x€ ¼ 2fd  2fe $ cosb

(3.45)

3.2 SingleeDOF Systems

where cosb ¼ x

.pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi l2 þ x2 ; fd ¼ c $ x;_ fe ¼ k $ Dl ¼ k $ l2 þ x2  l . Using these relationships

into Eq. (3.45) yields:

    l 2k l 2c m $ x€ þ 2c $ x_ þ 2k $ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ x ¼ 0 or x€ ¼  $ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ x  $ x_ 2 2 2 2 m m l þx l þx (3.46) Eq. (3.46) is the mathematical model of the body motion. The equation is nonlinear in x due to the character of the spring rate (the factor multiplying x). (ii) To use MATLAB’s ode45 operator, the second-order nonlinear homogeneous Eq. (3.46) has to be transformed into an equivalent set of two first-order differential equations. Using x1 ¼ x and x2 ¼ x_ in conjunction with the second form of Eq. (3.46), results in the equivalent equations:

8 > < x_1 ¼ x2   2k l 2c > ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi p _ $ 1  ¼  x $ x1  $ x2 : 2 2 2 m m l þx

(3.47)

The following MATALB function script named nonlineq is created: function dx = nonlineq(t,x) % a function file used to define the output vector dx in terms of. input vectors t and x dx=zeros(2,1); %defines the 2-row, 1-column vector dx m=0.25;c=0.2;k=1000;l=0.4; x0=0.2; v0=0;%numerical values of. mechanical parameters dx(1)=x(2);%first equation: x(1)=x and x(2)=dx/dt dx(2)=-2*k/m*(1-l/sqrt(l^2+x(1)^2))*x(1)-2*c/m*x(2);%second. differential equation end The following commands are in MATLAB’s workspace: >> [t,x]=ode45(@nonlineq,[0:0.001:8],[x0,v0]); >> %solves the two nonlinear,first-order equations for x(1) and x(2) >> plot(t,x(:,1))%plots the first unknown x(1)=x which solve the nonlinear equation over a time interval of 8 s with a time step of 0.001 s and produces the plot of Figure 3.20. While the profile of the curve is the one typical to the underdamped free response displaying an exponential decay in time, an interesting feature is the variable damped period, which increases over timedas highlighted by two consecutive periods, Td,kþ1 > Td,k. This shows that the equivalent stiffness of the system decreases with time because Td varies inversely proportionally with un and therefore with k, which suggests that the spring behaves as a softening spring. (iii) The last form of the differential equation in Eq. (3.46) is utilized in Simulink to design the block diagram of Figure 3.21(a).

97

CHAPTER 3 Mechanical Systems

0.2 0.15

Td,k

Td,k+1

0.1 x (m)

98

0.05 0 -0.05 -0.1 -0.15 -0.2 0

1

2

3

4 5 Time (s)

6

7

8

FIGURE 3.20 Time Response x(t) With Increasing Damped Period.

FIGURE 3.21 (a) Simulink Block Diagram to Solve and Export Solution to MATLAB; (b) Function Block Parameters.

The central piece of the diagram that directly transposes Eq. (3.46) is the Add block where the inputs are the two terms in the side of Eq. (3.46) and the output is the left-hand side of
right-hand

the same equation d 2 xðt Þ dt 2 . The two integrators produce sequentially dxðt Þ=dt and x(t). Note that the second integratordIntegrator 1 in Figure 3.22(a)drequires the initial condition x(0) ¼ 0.2 m. The Function block of the Simulink diagram of Figure 3.21(a) is taken from the UserDefined Functions library. The input to the block is x(t) and the output is the nonlinear factor that is specified in the block diagram. Figure 3.21(b) shows the format of this block. The block denotes its first input as u(1), which is actually the value of x corresponding to the first time step. The numerical values corresponding to k, m, and l have been used in this block’s parameter definition per the algebra of its output, as shown in Figure 3.21(a). Once x(t) has been obtained, it is exported to MATALB’s workspace to be plotted by using a fixedstep integration with a time step of 0.001 s. Full details on exporting Simulink data to MATLAB’s

3.2 SingleeDOF Systems

workspace for plotting purposes are provided in Appendix D. The MATLAB plot is identical to the one of Figure 3.20 that was obtained by means of MATLAB’s ode45 command structure. That should be expected since Simulink uses the same ode45 numerical integration algorithm. nn

3.2.2 Forced Response This section studies the forced response of single-DOF mechanical systems under the action of external forcing. The mathematical model of such a system is derived here by using Newton’s second law of motion. Time-domain solutions and solution plots of the resulting mathematical models can be obtained analytically or in MATLAB with the dsolve command for linear systems. Nonlinear systems can be solved numerically in MATLAB by means of ode45 based scripts, for instance, or by using Simulink. PHYSICAL MODEL

PHYSICAL MODEL

x k

k

f

m

θ

c J

c

m

FREE-BODY DIAGRAM x

FREE-BODY DIAGRAM

me

fe

f

m

θ

md

J

fd m

(a)

(b)

FIGURE 3.22 Lumped-Parameter Models and Free-Body Diagrams of Forced Damped Mechanical Systems: (a) Translatory; (b) Rotary.

Two basic mechanical systems, one translatory and the other rotary, are illustrated in Figure 3.22. Based on their free-body diagrams, Newton’s law of motion is expressed as follows: (

m $ x€ ¼ f  fd  fe € ¼ m  md  me J $q

or

m $ x€ ¼ f  c $ x_  k $ x or

€ q ¼ m  cq_  kq

or m $ x€ þ c $ x_ þ k $ x ¼ f

€ þ c $ q_ þ k $ q ¼ m or J $ q (3.48)

99

100

CHAPTER 3 Mechanical Systems

As discussed in Chapter 1, Eq. (3.48) can also be written in the form: ( x€ þ 2x $ un $ x_ þ u2n $ x ¼ u2n $ K $ f € þ 2x $ un $ q_ þ u2 $ q ¼ u2 $ K $ m q n n

(3.49)

where x is the damping ratio, un is the natural frequency, and K is the static sensitivity. The input (or forcing) is the force f or the moment m and the output (or response) is the translatory displacement x or the rotary one q in Eq. (3.49); they are all functions of time. As per the analysis of the free damped motiondsee Eq. (3.32)dthe solution to Eq. (3.49) is of the form: qffiffiffiffiffiffiffiffiffiffiffiffiffi  8 x $ u 2 n $t > sin 1  x $ un $ t þ 4 þ xp ðtÞ < xðtÞ ¼ xh ðtÞ þ xp ðtÞ ¼ X $ e qffiffiffiffiffiffiffiffiffiffiffiffiffi  > : qðtÞ ¼ q ðtÞ þ q ðtÞ ¼ Q $ ex $ un $ t sin 2 1  x $ un $ t þ 4 þ qp ðtÞ p h (3.50) where the term xh(t) is the solution to the homogeneous equation whereas the term xp(t) is a particular solution of the nonhomogeneous Eq. (3.49). The unknown amplitudes X or Q and phase angles 4 are determined using the initial conditions:

dxðtÞ

dqðtÞ

¼ v0 or qðt0 Þ ¼ q0 ; ¼ u0 (3.51) xðt0 Þ ¼ x0 ; dt dt t¼t0

t¼t0

in conjunction with the displacements of Eq. (3.50) and their time derivatives (the velocities). nn

Example 3.9 Derive the mathematical model of the mechanical system shown in Figure 3.23(a), which contains a two-disc rigid pinion, detailed in Figure 3.23(b), a translating rigid heavy rack with a spring and a damper attached to it, and another rigid massless rack that is connected to identical springs at its ends. An external force acts on the heavy rack along its motion direction.

Heavy rack

k1

Guide

f

m3 c

m1 R 1 C

k2

2R2

A

m2

R2

m2 Pinion

m1 k2

B

2R1

Massless rack

(a)

(b)

FIGURE 3.23 Mechanical System With Compound Pinion, Two Racks, Translatory-Motion Springs and Damper: (a) Schematic Diagram; (b) Rotated Side View of Compound Pinion.

3.2 SingleeDOF Systems

Solution Figure 3.24 presents the free-body diagrams of the pinion and of the two racks. Newton’s second law of motion is applied to these three bodies by taking into account the coordinates and forces shown in Figure 3.24, which results in

8 € ¼ f A $ R 1  f B $ R2 > J $q > < m3 $ x€A ¼ f  fA  fd  fe1 > > : 0 ¼ fB  2fe2

(3.52)

where the forces acting on the pulley, fA and fB, are reactions from the two racks. Similarly, fd and fe1 are the damping and elastic forces acting on the heavy rack, whereas fe2 is the elastic force applied by one of the two identical springs on the massless rack.

Heavy rack xA

Pinion xA

fe1

A C

R1

fA

fd

fA

A

R2

θ

Massless rack fB B

B xB

f m3

fB

fe2

xB

fe2

FIGURE 3.24 Free-Body Diagrams of the Pinion and the Two Racks. The interaction forces fA and fB are expressed from the last two in Eq. (3.52) and then substituted in the first Eq. (3.52), which becomes:

€ ¼ ð f  fd  fe1  m3 $ x€A Þ $ R1  2fe2 $ R2 J $q

(3.53)

The tangential displacements xA and xB are xA ¼ R1 $ q; xB ¼ R2 $ q. As a consequence, the elastic and damping forces of Eqs. (3.52) and (3.53) become

8 > < fe1 ¼ k1 $ xA ¼ k1 $ R1 $ q fe2 ¼ k2 $ xB ¼ k2 $ R2 $ q > : fd ¼ c $ x_A ¼ c $ R1 $ q_

(3.54)

 . The mass moment of inertia of the two-cylinder pinion is J ¼ m1 R12 þ m2 R22 2, and the € By using these relationships in conjunction with the forces of Eq. acceleration at A is x€A ¼ R1 $ q. (3.54) into Eq. (3.53) changes the latter equation to:



 

1 1 2 2 € þ c $ R2 $ q_ þ k1 $ R2 þ 2k2 $ R2 $ q ¼ f $ R1 m1 þ m3 $ R1 þ $ m2 $ R2 $ q 1 1 2 2 2 (3.55)

101

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Eq. (3.55) can be written in short form as

a1 a0 1 a2 $ € q þ a1 $ q_ þ a0 $ q ¼ ma or €q þ $ q_ þ $ q ¼ $ ma a2 a2 a2 where ma ¼ f$R1 is the actuation moment.

(3.56) nn

nn

Example 3.10 Determine the static sensitivity, the natural frequency, and the damping ratio values corresponding to the mechanical system of Example 3.9 and sketched in Figure 3.23 for m1 ¼ 0.5 kg, m2 ¼ m3 ¼ 1 kg, k1 ¼ 120 N/m, k2 ¼ 100 N/m, R1 ¼ 0.02 m, R2 ¼ 0.04 m, c ¼ 25 N s/m, and the equivalent moment ma* ¼ ma/a2 ¼ 3.846$sin(2p$t) rad/s2dsee Eq. (3.56). Use Simulink to plot the solution q(t) for both the regular input and an input with a dead zone extending from 1 to 1 N of the maximum input, which ranges from 3.846 to 3.846 rad/s2. The dead zone is due to the electric motor driving the actuator, which cannot respond to small voltages. The result is an input moment that is zero between the dead zone limits.

Solution The natural frequency, damping ratio, and static sensitivity are obtained by comparing the second Eq. (3.49) with Eq. (3.56):

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi v u 8 a0 u k1 $R21 þ 2k2 $R22   u u ; ¼ ¼ > n > > 1 a2 t 1 > 2 2 > $m $m þ m þ $R $R > 1 3 2 1 2 > 2 2 > > > > > rffiffiffiffiffi < 1 a1 a2 a1 c$R21 x¼ $ $ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi;  > 2 a2 a0 2 a0 $a2

1 > 1 > 2 2 2 2 > $m1 þ m3 $R1 þ $m2 $R2 2 k1 $R1 þ 2k2 $R2 > > 2 2 > > > > > > : 1 1 . K¼ ¼ 2 a0 k1 $R1 þ 2k2 $R22 (3.57) Eq. (3.57) yields the following values for the parameters of interest: un ¼ 16.825 rad/s, x ¼ 0.229, and K ¼ 2.717. The coefficients of Eqs. (3.55) and (3.56) assume the values a0 ¼ 0.368, a1 ¼ 0.01, and a2 ¼ 0.0013. Since 0 < x < 1, Eq. (3.56) becomes:

€ ¼ 3:846 $ sinð2p $ tÞ  7:692 $ q_  283:077 $ q q

(3.58)

and is of the general form given in Eq. (3.49). Figure 3.25(a) is the Simulink diagram that integrates the differential Eq. (3.58) for both a regular input and an input with dead zone. Except for the two different inputs, the diagram comprises two identical structures based on the Add block, whose inputs are the terms in the right hand-side of Eq. (3.58). The output of this block is integrated twice and then conditioned by a Gain function to yield the angle q in degrees and to export it to MATLAB’s workspace to plot it against time. The Dead Zone block is in the Discontinuities library of Simulink, and its effect is that it transmits the regular input unchanged only outside the dead zone. Inside the dead zone (denoted by DZ in

DZ Input 45° m (t )

45°

m (t ) – dead zone .. θ (t )

. θ (t )

θ (t )

Input

DZ θ(t ) [deg]

. (a1 / a2) θ(t )

(a0 / a2) θ(t )

.. θ (t )

S E

θ(t ) [deg] . θ (t )

θ (t )

. (a1 / a2) θ(t )

(a)

FIGURE 3.25 (a) Simulink Block Diagram Solving for q(t); (b) Dead Zone Discontinuity.

(b)

3.2 SingleeDOF Systems

(a0 / a2) θ(t )

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CHAPTER 3 Mechanical Systems

(a) 4 Regular 3

1 Deadzone

ma (N·m)

2

0 -1 -2 -3 -4 0

0.5

1

1.5

2

2.5 Time (s)

3

3.5

4

4.5

5

2.5

3

3.5

4

4.5

5

(b) 1 0.8

Regular

0.6 0.4 0.2 Deadzone

θ (degrees)

104

0

-0.2 -0.4 -0.6 -0.8 -1 0

0.5

1

1.5

2

Time (s)

FIGURE 3.26 Regular and Dead Zone Signal Time-Domain Plots: (a) Input Moment; (b) Output Pinion Rotation Angle. Figure 3.25(b) and where S and E stand for start and end respectively), the output from this block is zero. The To Workspace and To Workspace1 blocks export the regular and dead zone inputs to MATLAB, and they are plotted in Figure 3.26(a). Note the change effect in both input magnitude and time range over one period for the particular regular sine input. The angle q is plotted against time in

3.3 MultipleeDOF Systems

Figure 3.26(b), after being exported to MATLAB through the To Workspace2 and To Workspace3 blocks. The output plots highlight the differences between the effects of the two inputs. nn

3.3 MULTIPLEeDOF SYSTEMS This section derives mathematical models for conservative and nonconservative multiple-DOF mechanical systems. The free undamped (or natural), free damped, and forced responses of multiple-DOF mechanical systems are further investigated by analytical formulation, MATLAB, and Simulink.

3.3.1 Free Response Multiple-DOF systems are studied in this section either without dampingdresulting in the free, undamped (or natural) responsedor with dampingdproducing the free damped response.

Conservative Systems and the Natural Response The mathematical models of multiple-DOF conservative (free undamped) mechanical systems can be obtained by several methods. This chapter employs Newton’s second law of motion, as well as Lagrange’s equations method, which is an energy-based method derived using principles of analytical mechanics. The resulting mathematical model is further utilized to calculate the natural frequencies and qualify the corresponding natural (or modal) motions of the system at the respective natural frequencies either analytically or with MATLAB.

Mathematical Model (Differential Equation) Newton’s Second Law of Motion Method To apply Newton’s second law of motion, the mechanical system has to be partitioned into a number of subsystems equal to the DOF. Using subsystem connecting reaction forces/moments and free-body diagrams for each subsystem, a differential (or algebraic) equation results for each subsystem. All individual subsystem equations are collected into a set comprising a number of equations (differential and/or algebraic) that is equal to the number of DOF. Lagrange’s Equations Mathematical models for conservative mechanical systems can also be derived by means of energy methods based on kinetic and potential energies. Energy methods are particularly advantageous to apply to mechanical systems formed of several bodies where all kinetic energy contributions are collected in one total kinetic energy and all potential energy fractions are summed in one potential energy. On the other hand, application of Newton’s second law of motion in such cases appears to be somewhat more difficult, as it implies fractioning the

105

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system into subsystems (one for each DOF), using the subsystem free-body diagrams, incorporating the reactions resulting from adjacent DOF, and formulating a number of equations equal to the number of DOF. While the companion website Chapter 3 presents the conservation of energy method, this chapter discusses another energy-based method that enables formulation of mathematical models for dynamic systems, namely the Lagrange’s equations, a demonstration of which is not included here. In the case of conservative systems, Lagrange’s equations are: ! ! d vL vL d vT vU ¼ 0 with L ¼ T  U ¼ T  ðUe þ Ug Þ or ¼0  þ dt vq_j vqj dt vq_j vqj (3.59) where L is the system’s Lagrangian, T is the kinetic energy, and U is the potential energy, which is the sum of elastic energy Ue and gravitational energy Ug. The variables qj ( j ¼ 1, 2, ., n) are the generalized coordinates. For mechanical systems the generalized coordinates are the minimum-number geometric coordinates (linear and/or angular) that fully define the state of the systems and can be identical to the DOF. The second set of equations in Eq. (3.61) takes into account that T depends only on the time derivatives of the generalized coordinates, while U depends only on the generalized coordinates. nn

Example 3.11 The rotor shaft mechanical system shown in Figure 3.27 consists of two rigid rotors (disks) and three flexible shafts, which behave as rotary springs. The end shafts are supported on two end bearings that are assumed ideal and therefore lossless. The system is equipped with two end brakes, which can stop the rotation. Consider known the rotor mass moments of inertia J1 and J2, as well as the rotary spring stiffnesses of the three shafts: k1 for the left shaft, k for the middle shaft, and k2 for the right shaft. Using Lagrange’s equations and Newton’s second law of motion, derive the mathematical model of this system when: (i) The brakes are not applied and the system rotates freely. (ii) The brakes are applied and the extremities of the left and right shafts become fixed.

Bearing Brake

Middle shaft ω

J1

J2

Bearing

Brake Right shaft

Left shaft Rotors

FIGURE 3.27 Rotary System With Two Disk Rotors, Three Flexible Shafts, Ideal Bearings and Brake System.

3.3 MultipleeDOF Systems

Solution

The system is a two-DOF system because the two angular parameters q1 and q2, which indicate the absolute angular positions of the rotors, fully define the configuration of the system. (i) When the brakes are not applied, the end shafts meet no external resistance and are free to rotate. As a result, they do not generate any elastic moment on the two rotors they connect to. Consequently, the lumped-parameter model is the one of Figure 3.28(a) and consists of the two disks J1 and J2, and the connecting rotary spring k.

θ1

θ2 θ1

k J1

θ2

J2

J2

J1

me*

me

(a)

(b)

FIGURE 3.28 (a) Lumped-Parameter Model Without Brakes Being Applied; (b) Free-Body Diagrams of the Two Rotors. Lagrange’s Equations Method The kinetic and potential energy expressions for this system are:

1 1 1 2 2 T ¼ $ J1 $ q_ 1 þ $ J2 $ q_ 2 ; U ¼ Ue ¼ $ k $ ðq1  q2 Þ2 2 2 2

(3.60)

The two explicit Lagrange’s equations are:

8   8   > > d vL vL d vT vU > > > > > > < dt vq_  vq1 ¼ 0 < dt vq_ þ vq1 ¼ 0 1 1 or     > > d vL vL d vT vU > > > > > > ¼0 ¼0  þ : dt : dt _ _ vq vq vq 2 vq2 2 2

(3.61)

The following derivatives are required in Eq. (3.61):

8  

> d vT d > € ; > J1 $ q_ 1 ¼ J1 $ q ¼ > 1 < dt vq_ dt 1   >

d vT d > > € ; > J2 $ q_ 2 ¼ J2 $ q ¼ : 2 dt vq_ 2 dt

vU ¼ k $ ðq1  q2 Þ vq1 vU ¼ k $ ðq1  q2 Þ vq2

(3.62)

Substituting Eq. (3.62) into Eq. (3.61) results in

(

€ þ k $ q1  k $ q2 ¼ 0 J1 $ q 1 €  k $ q1 þ k $ q2 ¼ 0 J2 $ q 2

(3.63)

which is the mathematical model of the two-DOF mechanical system of Figure 3.28(a). Newton’s Second Law of Motion Method Newton’s second law is now applied to the two rotors, based on the free-body diagrams of Figure 3.28(b). The differential equations describing the dynamics of the rotors are

107

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CHAPTER 3 Mechanical Systems

(

€ ¼ me J1 $ q 1

(3.64)

€ ¼ m J2 $ q e 2

The magnitudes of the elastic moments, whose directions oppose the positive directions indicated by the two angular coordinates q1 and q2, are expressed as me ¼ k $ ðq1  q2 Þ and me ¼ me ¼ k $ ðq2  q1 Þ. Note that they are actually action and reaction being generated by the same spring connected to the two rotors. Eq. (3.63)dthe mathematical model derived by Lagrange’s equations methoddis retrieved after substituting these moments in Eq. (3.64). θ1

k1

θ2

k

me2 θ1

k2

J1

J2 J1

Anchor

Anchor

θ2

me1

J2

me*

me

(a)

(b)

FIGURE 3.29 (a) Lumped-Parameter Model With Applied Brakes; (b) Free-Body Diagrams of the Two Rotors. (ii) When the brakes are applied, the end shafts become fixed at their extremities. As a result, the shafts behave as springs and apply elastic torques on the two disks they are connected to. The lumped-parameter model of this system is represented in Figure 3.29(a). Lagrange’s Equations Method While the kinetic energy is provided in Eq. (3.60), the elastic potential energy results from all three deforming springs, namely:

1 1 1 U ¼ Ue ¼ $ k1 $ q21 þ $ k $ ðq1  q2 Þ2 þ $ k2 $ q22 2 2 2

(3.65)

The partial derivatives of U are obtained from Eq. (3.65) as:

vU vU ¼ k1 $ q1 þ k $ ðq1  q2 Þ; ¼ k $ ðq1  q2 Þ þ k2 $ q2 vq1 vq2

(3.66)

The kinetic energy derivatives of Eq. (3.62) and the potential energy derivatives of Eq. (3.66) are substituted into the Lagrange’s equations, Eq. (3.61). The following mathematical model results for the system shown in Figure 3.30(a):

(

€1 þ ðk1 þ kÞ $ q1  k $ q2 ¼ 0 J1 $ q €2  k $ q1 þ ðk þ k2 Þ $ q2 ¼ 0 J2 $ q

(3.67)

Newton’s Second Law of Motion Method The free-body diagrams of the two rotors that are part of the system sketched in Figure 3.29(a) are drawn in Figure 3.29(b). Applying Newton’s second law of motion results in:

(

€ ¼ me  me1 J1 $ q 1

€ ¼ m  me2 J2 $ q e 2

(3.68)

The magnitudes of the new elastic moments are me1 ¼ k1 $ q1 and me2 ¼ k2 $ q2 .They are substituted into Eq. (3.68) together with the moments me and me that were expressed previously; the result is the differential equations system of Eq. (3.67), which is the mathematical model, also obtained using Lagrange’s equations. nn

3.3 MultipleeDOF Systems

nn

Example 3.12 Derive the mathematical model for the cart-pendulum shown in Figure 3.30 using Lagrange’s equations under the assumption of small motions. The system is in a vertical plane and the cart can move horizontally. Known are the masses of the cart, m1, and of the bob, m2, the spring stiffness k, the pendulum length l, and the gravitational acceleration g.

x Fixed guide

k m1 Cart

l

l θ Massless Rigid Rod

h Datum line

v

vr

Bob

m2

vt

FIGURE 3.30 Cart-Pendulum Mechanical System.

Solution

The system has two DOF: the linear motion coordinate x, which locates the cart, and the angular coordinate q, which defines the pendulum angular position. The total (absolute) velocity v of the bob has two components: the tangential (horizontal) velocity of the cart vt and the one resulting from the pendulum rotation about the cart, denoted vr, which is perpendicular to the pendulum rod, as illustrated in Figure 3.30. The kinetic energy of the mechanical system is:

1 1 T ¼ $ m1 $ x_2 þ $ m2 $ v2 2 2

(3.69)

According to the cosine theorem applied to one of the two velocity triangles, the total bob velocity is:

v2 ¼ v2t þ v2r  2vt $ vr $ cosðp  qÞ ¼ v2t þ v2r þ 2vt $ vr $ cosq

(3.70)

where p  q is the angle between the directions of vt and vr in the obtuse-angle velocity triangle. It is _ These relationships are combined with Eq. (3.70), and the result is _ vr ¼ l $ q. also known that vt ¼ x; substituted into the kinetic energy of Eq. (3.69). For small angular displacements (which lead to the approximation cos q y 1  q2/2), the following expression of the kinetic energy (where the term in q2, which is very small, is neglected) results:

1  2 T ¼ $ ðm1 þ m2 Þ $ x_2 þ 2m2 $ l $ x_ $ q_ þ m2 $ l2 $ q_ 2

(3.71)

The total potential energy comprises elastic and gravitational parts, namely:

1 1 1 U ¼ Ue þ Ug ¼ $ k $ x2 þ m2 $ g $ l $ ð1  cosqÞx $ k $ x2 þ $ m2 $ g $ l $ q2 2 2 2 (3.72)

109

110

CHAPTER 3 Mechanical Systems . where h ¼ l$(1  cosq) in Figure 3.30 and cosqx1  q2 2. The system is a two-DOF system, and the generalized coordinates are the two DOF x and q, which means that the explicit Lagrange’s equations are:

8   8   > > d vL vL d vT vU > > > > > > < dt vx_ þ vx ¼ 0 < dt vx_  vx ¼ 0 or     > > d vL vL d vT vU > > > > > > ¼ 0 ¼0  þ : dt : dt vq vq vq_ vq_

(3.73)

The following derivative terms are needed in the two of Eq. (3.73):

8    d vT d > > € vU ¼ k $ x > ¼ ðm1 þ m2 Þ $ x_ þ m2 $ l $ q_ ¼ ðm1 þ m2 Þ $ x€ þ m2 $ l $ q; > < dt vx_ dt vx   >  > d vT d > € vU ¼ m2 $ g $ l $ q > ¼ m2 $ l $ x_ þ m2 $ l2 $ q_ ¼ m2 $ l $ x€ þ m2 $ l2 $ q; : _ dt vq dt vq (3.74) Substituting Eq. (3.74) into Eq. (3.73) yields:

(

€ þ k$x ¼ 0 ðm1 þ m2 Þ $ x€ þ m2 $ l $ q € þ g$q ¼ 0 x€ þ l $ q

(3.75)

The two in Eq. (3.75) form the mathematical model of the cart-pendulum mechanical system.

nn

Solution of Mathematical Model (Differential Equation) For multiple-DOF mechanical systems, the number of natural frequencies is equal to the number of DOF, except for the repeated natural frequencies. With particular initial conditions associated with any of the natural frequencies, all the bodies in the system vibrate at that natural frequency. Moreover, the amplitudes of all the individual vibrations preserve well-defined proportions. The overall system motion corresponding to a natural frequency is called natural (or modal) motion, or mode shape. However, for arbitrary initial conditions, the motion of each subsystem (DOF) is a superposition of harmonic terms, one for each natural frequency. Next are discussed the analytical and the MATLAB approaches to defining the actual response of multiple-DOF conservative systems. Analytical Approach Consider a generic two-DOF free and lossless mechanical system described by the differential equations system: ( a11 $ q€1 ðtÞ þ a12 $ q€2 ðtÞ þ a13 $ q1 ðtÞ þ a14 $ q2 ðtÞ ¼ 0 (3.76) a21 $ q€1 ðtÞ þ a22 $ q€2 ðtÞ þ a23 $ q1 ðtÞ þ a24 $ q2 ðtÞ ¼ 0 The unknown functions q1(t), q2(t) in Eq. (3.76) are translatory or rotary coordinates, whereas the coefficients in the same equations are constants depending on the inertia and stiffness properties of the conservative mechanical system. Because the terms with the second derivative of the two unknown functions appear in both equations, the system has inertia coupling. Similarly, due to the terms with the unknown functions being present in both equations, the system is displaying stiffness coupling. As

3.3 MultipleeDOF Systems

111

shown in ordinary differential equations theory, the solution to the set of homogeneous Eq. (3.76) is: 8 > q1 ðtÞ ¼ Q11 $ sinðun1 $ t þ 41 Þ þ Q12 $ sinðun2 $ t þ 42 Þ > > < q2 ðtÞ ¼ Q21 $ sinðun1 $ t þ 41 Þ þ Q22 $ sinðun2 $ t þ 42 Þ ¼ r1 $ (3.77) > > > : Q11 $ sinðun1 $ t þ 4 Þ þ r2 $ Q12 $ sinðun2 $ t þ 4 Þ 1 2 where r1 and r2 are constant ratios. The four initial conditions are q1 ð0Þ ¼ q1;0 ; q2 ð0Þ ¼ q2;0 ; dq1 ðtÞ=dtjt¼0 ¼ v1;0 ; dq2 ðtÞ=dtjt¼0 ¼ v2;0 : Solving these equations, as shown in the companion website Chapter 3, yields the amplitudes Q11, Q12 and the initial (phase) angles 41, 42: 8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > u2n1 $ ðr2 $ q10  q20 Þ2 þ ðr2 $ v10  v20 Þ2 > un1 $ ðr2 $ q10  q20 Þ > > Q ; 41 ¼ tan1 ¼ > < 11 r2 $ v10  v20 un1 $ ðr2  r1 Þ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > > u2n2 $ ðr1 $ q10  q20 Þ2 þ ðr1 $ v10  v20 Þ2 > un2 $ ðr1 $ q10  q20 Þ > > ; 42 ¼ tan1 : Q12 ¼ r1 $ v10  v20 un2 $ ðr1  r2 Þ (3.78) The solution of Eq. (3.77) is expressed by linearly superimposing two natural motions, each being a simple, harmonic motion, corresponding to one of the two natural frequencies un1 and un2. The first natural motion is:  q11 ðtÞ ¼ Q11 $ sinðun1 $ t þ 41 Þ (3.79) q21 ðtÞ ¼ Q21 $ sinðun1 $ t þ 41 Þ ¼ r1 $ Q11 $ sinðun1 $ t þ 41 Þ and the second natural motion is:  q12 ðtÞ ¼ Q12 $ sinðun2 $ t þ 42 Þ (3.80) q22 ðtÞ ¼ Q22 $ sinðun2 $ t þ 42 Þ ¼ r2 $ Q12 $ sinðun2 $ t þ 42 Þ Each natural motion is uniquely defined by a natural frequency, un1 or un2, and by the two amplitude ratios r1 or r2, which are ri ¼ Q2i/Q1i (i ¼ 1 or i ¼ 2). Formally, each of the two amplitude ratios can be used to form an eigenvector. The unit-norm eigenvectors corresponding to the two natural frequencies are: 8 8 9 9 > > > > 1 1 > > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > > > > > 2> 2>    >  > < < = = 1 þ r 1 þ r Q11 Q12 1 2 ; (3.81) ¼ ¼ ¼ ¼ fQ1 g fQ1 g > > > > r1 r2 Q21 Q 22 > > > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > > > > > > : 1 þ r 21 > : 1 þ r 22 > ; ; These vectors result from solving the set of equations Q2i ¼ ri$Q1i and Q21i þ Q22i ¼ 1dwhich is the unit-norm conditiondfor i ¼ 1, 2. MATLAB Approachdthe Dynamic Matrix and the Eigenvalue Problem The natural response of multiple-DOF systems can also be formulated as an eigenvalue

112

CHAPTER 3 Mechanical Systems

problem (see Appendix B). The individual Eq. (3.76) are reformulated in matrix form as:     a11 a12 a13 a14 € ½M $ fqðtÞg þ ½K $ fqðtÞg ¼ f0g with ½M ¼ ; ½K ¼ a21 a22 a23 a24 (3.82) where [M] and [K] are the mass (inertia) and stiffness matrices. The vector {q(t)} collects the unknown coordinatesdfor instance, the vector of a two-DOF system is {q(t)} ¼ {q1(t) q2(t)}T. The individual solutions are sinusoidal and can be written in the general form: q1(t) ¼ Q1$sin(u$t), q2(t) ¼ Q2$sin(u$t), or {q(t)} ¼ {Q}$ sin(u$t)dwith {Q} ¼ {Q1 Q2}T being the amplitude vector. The second time deriv€ € ative of {q(t)} is fqðtÞg ¼ u2 $ fqðtÞg. Substituting {q(t)} and fqðtÞg into Eq. (3.82) and left multiplying by [M]1 results in: ½M1 $ ½K $ fQg ¼ u2 $ fQg after factoring and eliminating sin (u$t). Eq. (3.83) can be written as: ( ½D ¼ ½M1 $ ½K ½D $ fQg ¼ l $ fQg with l ¼ u2

(3.83)

(3.84)

Eq. (3.84) expresses an eigenvalue problem applied to the matrix [D], which is known as the dynamic matrix and whose eigenvalues are li (i ¼ 1, 2 for the system of Eq. 3.76 and i ¼ 1, 2, ., n for a system with n DOF). The theory of ordinary differential equations shows that li ¼ u2ni , where uni is a natural frequency. The vector {Qi} ¼ {Q1i$Q2i}T is the eigenvector corresponding to li and uni. MATLAB’s command eig(D) returns the eigenvalues of a given square matrix. A MATLAB command [Q, Dn] = eig(D) returns the modal matrix [Q], whose columns are the unit-norm eigenvectors (each containing the relative vibration amplitudes), and the diagonal matrix Dn, which has the eigenvalues on the main diagonal. The examples discussed next are a continuation of Examples 3.11 and 3.12 that derived the mathematical models of two conservative mechanical systems. They take you through the main steps involved in formulating and solving the natural problem. nn

Example 3.13 Consider the mechanical system of Example 3.11(ii) and shown in Figure 3.29(a). Find its natural frequencies, determine the system’s unit-norm eigenvectors and express/plot the natural motions corresponding to each natural frequency for k1 ¼ 800 N m, k2 ¼ 600 N m, k ¼ 400 N m, J1 ¼ 0.002 kg m2, and J2 ¼ 0.0015 kg m2.

Solution

Analytical Approach The solution q1 and q2 of Eq. (3.67) needs to be harmonic:

q1 ¼ Q1 $ sinðu $ t þ 4Þ; q2 ¼ Q2 $ sinðu $ t þ 4Þ

(3.85)

3.3 MultipleeDOF Systems

€1 ¼ u2 $ Q1 sinðu $ t þ 4Þ and €q2 ¼ u2 $ The two angles q1 and q2 and their derivatives, q Q2 $ sinðu $ t þ 4Þ, are substituted in Eq. (3.67), which yields the equations:

(

 u2 $ J1 þ k1 þ k $ Q1  k $ Q2 ¼ 0

k $ Q1 þ  u2 $ J2 þ k þ k2 $ Q2 ¼ 0

(3.86)

after canceling the sine factors. The following amplitude ratio results from Eq. (3.86):

r¼

Q2 u2 $ J1 þ k1 þ k k ¼ ¼ k u2 $ J2 þ k þ k2 Q1

(3.87)

Cross-multiplying in the last proportion of Eq. (3.87) produces the algebraic equation:

J1 $ J2 $ u4  ½J1 $ ðk þ k2 Þ þ J2 $ ðk1 þ kÞ $ u2 þ k1 $ k2 þ k $ ðk1 þ k2 Þ ¼ 0 (3.88) Eq. (3.88) is the characteristic equation that can also be obtained from Eq. (3.86), which are two homogeneous algebraic equations in Q1 and Q2. These two amplitudes are nonzero when the determinant of their system is zero, namely:

u2 $ J þ k þ k

1 1

k

k

¼0 2 u $ J2 þ k þ k2

(3.89)

Expanding the determinant of Eq. (3.89) results in Eq. (3.88), whose roots are the natural frequencies. For the particular parameters of the example, Eq. (3.88) yields the following positive roots, which are the system’s natural frequencies: un1 ¼ 632.455 rad/s and un2 ¼ 930.949 rad/s. The amplitude ratio of Eq. (3.87) assumes a value for each of the two natural frequencies, namely:

8

Q2

Q21 u2n1 $ J1 þ k1 þ k k > > > r ¼ ¼ ¼ ¼ ¼1 1 >

2 < k Q1 u¼un1 Q11 un1 $ J2 þ k þ k2

> > Q2

Q22 u2n2 $ J1 þ k1 þ k k > > ¼ ¼ ¼ ¼ ¼ 1:333 r : 2

k Q1 u¼un2 Q12 u2n2 $ J2 þ k þ k2

(3.90) As a consequence, the two unit-norm eigenvectors of Eq. (3.81) become:

 fQ1 g ¼

¼

Q11 Q21

 ¼

9 8 > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > = < 1 þ r 21 > > > r1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > 1 þ r2 > ; :

8 9 > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > < 1 þ r 22 > = > > r2 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > 1 þ r2 > > : ; 2

 ¼

0:707 0:707



 ; fQ2 g ¼

Q12



Q22

1

 ¼

0:6 0:8

 (3.91)

113

114

CHAPTER 3 Mechanical Systems

The two natural motions result from the two generic Eqs. (3.79) and (3.80). The first natural motion is in this case:



q11 ðtÞ ¼ Q11 $ sinðun1 $ t þ 41 Þ ¼ 0:707 $ sinðun1 $ t þ 41 Þ q21 ðtÞ ¼ r1 $ Q11 $ sinðun1 $ t þ 41 Þ ¼ 0:707 $ sinðun1 $ t þ 41 Þ

(3.92)

whereas the second natural motion is:



q12 ðtÞ ¼ Q12 $ sinðun2 $ t þ 42 Þ ¼ 0:6 $ sinðun2 $ t þ 42 Þ q22 ðtÞ ¼ r2 $ Q12 $ sinðun2 $ t þ 42 Þ ¼ 0:8 $ sinðun2 $ t þ 42 Þ

(3.93)

They are plotted in Figure 3.31, with the nonrestrictive condition 41 ¼ 0 and 42 ¼ 0.

Time(s)

Time(s)

(b)

(a)

FIGURE 3.31 Natural Vibrations of Two Disks at the: (a) First Natural Frequency; (b) Second Natural Frequency. Because r1 ¼ 1, the two disks rotate identically during the first natural motion, as illustrated by Eq. (3.92) and the plot of Figure 3.31(a). This is a particular motion, called synchronous motion, where the two disks move in unison. The effect of the connecting spring k is completely removed as a rigid connector placed between the disks instead of the spring, which would result in the same motion. Unlike the first modal motion, the two disks rotate in opposition during the second modal motion (which is the result of the minus sign of the amplitude ratio r2). It is also seen from both Eq. (3.93) and Figure 3.31(b) that the amplitude of the second disk (which is 0.8) is larger than the amplitude of the first disk (which is 0.6). MATLAB Approach The mathematical modeldEq. (3.67)dis written in matrix form as:

"

J1 0

½M  ¼

# (

€ q 1 $ € q J2 2 " # J1 0 0

0

J2

)

"

þ

k1 þ k

; ½K  ¼

k

# (

q1

$ k k þ k2 q2 " # k k1 þ k k

)

¼

( ) 0 0

with (3.94)

k þ k2

which identifies the inertia matrix [M] and the stiffness matrix [K]. With the numerical parameters of this example and MATLAB’s eig command applied to the dynamic matrix [D] ¼ [M]1$[K], the eigenvectors are the ones of Eq. (3.91), whereas the eigenvalues are l1 ¼ 400,000 pffiffiffiffiffi (which results in un1 ¼ l1 ¼ 632:455 rad=s) and l2 ¼ 866,666 (which results in un2 ¼ pffiffiffiffiffi l2 ¼ 930:949 rad=s)dthese are also the values obtained analytically. nn

3.3 MultipleeDOF Systems

nn

Example 3.14 Consider the mechanical system of Example 3.11(i) and sketched in Figure 3.28(a). Find its natural frequencies, determine the system’s unit-norm eigenvectors, and express the natural motions corresponding to each natural frequency for J1 ¼ 0.002 kg m2, J2 ¼ 0.0015 kg m2, and k ¼ 400 N m.

Solution Analytic Approach Sinusoidal solutions of the form given in Eq. (3.85) of the previous Example 3.13 together with their second time derivatives are substituted in Eq. (3.63)dthe mathematical model of the studied systemdwhich results in:

(

 u 2 $ J1 þ k $ Q 1  k $ Q 2 ¼ 0

k $ Q1 þ  u2 $ J2 þ k $ Q2 ¼ 0

(3.95)

Q2 u2 $ J1 þ k k ¼ ¼ 2 k u $ J2 þ k Q1

(3.96)

The amplitude ratio:

r¼

results from Eq. (3.95). Cross-multiplying in the last proportion of Eq. (3.96) yields the characteristic equation:

  u2 $ J1 $ J2 $ u2  ðJ1 þ J2 Þ $ k ¼ 0

(3.97)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðJ1 þ J2 Þ $ k=ðJ1 $ J2 Þ ¼

whose solutions are the following natural frequencies: un1 ¼ 0, un2 683:13 rad=s. The amplitude ratios corresponding to the two natural frequencies result from Eq. (3.96):

8 Q2

Q21 > > r ¼ ¼ ¼1 1 >

< Q1 u¼un1 Q11

> Q2

Q22 > > : r2 ¼ ¼ ¼ 1:333 Q1 u¼un2 Q12

(3.98)

As per Eq. (3.81), the eigenvectors are exactly the ones found in the previous Example 3.13 and provided in Eq. (3.91). The first natural motion corresponds to a zero natural frequency and is expressed as:



q11 ðtÞ ¼ Q11 $ sinð41 Þ ¼ 0:707 $ sinð41 Þ q21 ðtÞ ¼ r1 $ Q11 $ sinð41 Þ ¼ 0:707 $ sinð41 Þ

(3.99)

This type of nonvibratory motion that corresponds to a zero natural frequency is known as rigid-body motion. Should the phase angle be zero as well, Eq. (3.99) simplify to q11(t) ¼ q22(t) ¼ 0, which indicates a stationary state. The second natural motion is identical to the one of the previous Example 3.13, which is expressed in Eq. (3.93) and plotted in Figure 3.31(b). MATLAB Approach The matrix-form mathematical model corresponding to the original Eq. (3.63) is:

"

J1 0

½M  ¼

# (

€ q 1 $ € q2 J2 " # J1 0 0

0

J2

)

"

þ

k

k

k "

k

; ½K  ¼

# ( $

k

k

k

k

#

q1 q2

)

¼

( ) 0 0

; with (3.100)

With [M] and [K] of Eq. (3.100), the dynamic matrix [D] ¼ [M]1$[K] is formed and used in MATLAB in conjunction with the eig command. The analytical eigenvectors of Eq. (3.91) are retrieved while

115

116

CHAPTER 3 Mechanical Systems

pffiffiffiffiffi the eigenvalues are l1 ¼ 0 (leading to un1 ¼ l1 ¼ 0 rad=s) and l2 ¼ 466,666.667 (which results in pffiffiffiffiffi un2 ¼ l2 ¼ 683:13 rad=s). The same values resulted through the analytic approach. nn

nn

Example 3.15 Use analytic calculation to determine the natural frequencies, describe the corresponding modes, and calculate the unit-norm eigenvectors for the cart-pendulum mechanical system of Example 3.12, and shown in Figure 3.30. Verify the results using the matrix approach and MATLAB. In addition, find the analytical solutions x(t), q(t) and plot them for the following initial conditions: x(0) ¼ 0 m, v(0) ¼ 0 m/s, q(0) ¼ 6 , u(0) ¼ 0 rad/s. Known are m1 ¼ 1 kg, m2 ¼ 0.2 kg, k ¼ 210 N/m, l ¼ 0.8 m, and g ¼ 9.8 m/s2.

Solution

The two variables of Eq. (3.75) are of different nature, as x is a length and q is an angle; this makes their comparison difficult. Let us introduce the variable y ¼ l$q (where y is actually the bob displacement measured with respect to the cart) instead of q. Eq. (3.75) becomes:

8 < ðm1 þ m2 Þ $ x€ þ m2 $ y€ þ k $ x ¼ 0 g : x€ þ y€ þ $ y ¼ 0 l The sinusoidal solution:



x ¼ X $ sinðu $ t þ 4Þ y ¼ Y $ sinðu $ t þ 4Þ

(3.101)

(3.102)

is substituted in Eq. (3.101), which becomes:

(

  ðm1 þ m2 Þ $ u2 þ k $ X  m2 $ u2 $ Y ¼ 0

u2 $ X þ u2 þ g=l $ Y ¼ 0

(3.103)

after canceling the nonzero sine factors. The characteristic equation corresponding to Eq. (3.103) is:

ðm þ m Þ $ u2 þ k

1 2

u2

m2 $ u2

¼0 u2 þ g=l

(3.104)

which is written as:

m1 $ l $ u4  ½ðm1 þ m2 Þ $ g þ k $ l $ u2 þ k $ g ¼ 0

(3.105)

The eigenvalues (and corresponding natural frequencies) resulting from Eq. (3.105) are:

l1;2 ¼ u2n1;2 ¼

ðm 1 þ m 2 Þ $ g þ k $ l 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½ðm1 þ m2 Þ $ g þ k $ l2  4m1 $ l $ k $ g 2m1 $ l (3.106)

The numerical values of the natural frequencies are un1 ¼ 3.4785 rad/s and un2 ¼ 14.5808 rad/s. From the first Eq. (3.103), the following amplitude is found:

r¼

Y k  ðm1 þ m2 Þ $ u2 ¼ X m2 $ u2

(3.107)

3.3 MultipleeDOF Systems

which results in:

Y Y1 Y Y2 r1 ¼

¼ ¼ 80:777; r2 ¼

¼ ¼ 1:061 X u¼un1 X1 X u¼un2 X2

The two eigenvectors are therefore:

 fV1 g ¼

 ¼

X1 Y1

 ¼

8 9 > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > < 1 þ r 21 > = > > r1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > 1 þ r2 > > : ; 1



0:6858

 ¼

(3.108) 8 9 > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > < 1 þ r 22 > =

   0:0124 X2 ; fV2 g ¼ ¼ > > r2 0:9999 Y2 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > 1 þ r2 > > : ; 2

0:7278 (3.109)

The first ratio of Eq. (3.108) indicates that during the first natural motion, the rotation of the pendulum is in the direction shown in the problem’s figure and that Y ¼ l$Q > X. During the second natural motion, the pendulum rotates in a direction opposite to the positive direction, and the magnitude of Y ¼ l$Q is slightly larger than X, as shown by the second amplitude ratio of Eq. (3.109). The mathematical modeldEq. (3.101)dis written in vector-matrix form as:



m1 þ m2 1

    k x€ $ þ € y 1 0

m2 

with ½M ¼

m1 þ m2

m2

1

1

     x 0 $ ¼ g=l y 0 0



 ; ½K ¼

k

0

0 g=l

 (3.110)

The dynamic matrix [D] is calculated as [D] ¼ [M]1$[K]. The corresponding MATLAB code returns the two eigenvectors of Eq. (3.109), as well as the eigenvalues l2 ¼ 212.5998 ¼ u2n2 and l1 ¼ 12.1002 ¼ u2n1. The solution of the mathematical model given in Eq. (3.101) together with the initial conditions (denoted as I.C. below) resulting from Eq. (3.78) are:

8 > x ¼ X1 $ sinðun1 $ t þ 41 Þ þ X2 $ sinðun2 $ t þ 42 Þ > > > > < y ¼ r1 $ X1 $ sinðun1 $ t þ 41 Þ þ r2 $ X2 $ sinðun2 $ t þ 42 Þ with I.C. > > > 1 > > :q ¼ $y l 8 xð0Þ ¼ q ð0Þ ¼ 0 1 > > > > > > yð0Þ ¼ q2 ð0Þ ¼ l $ qð0Þ > > > >
> dt t¼0 > > > >

> >

> dqðtÞ

> : dyðtÞ ¼ v2;0 ¼ l $ ¼0 dt t¼0 dt t¼0 (3.111)

117

θ (degrees)

CHAPTER 3 Mechanical Systems

x (m)

118

Time (s)

Time (s)

(b)

(a)

FIGURE 3.32 Displacements During Free Undamped Vibrations of the: (a) Cart; (b) Bob. Using Q11 ¼ X1, Q12 ¼ X2 and the I.C. of Eq. (3.111) generates the following amplitudes and phase angles:

yð0Þ l $ qð0Þ X1 ¼  ¼ ¼ 0:001; X2 ¼ X1 ¼ 0:001; 41 ¼ 42 ¼ p=2 r2  r1 r2  r1 (3.112) For these values, the plots of x and of q given in Eq. (3.111) are shown in Figure 3.32.

nn

Free Damped Response Mathematical Model Derivation Newton’s second law of motion and Lagrange’s equations are applied to derive the mathematical models of multiple-DOF mechanical free systems with viscous damping. To use Newton’s second law of motion, each subsystem needs free-body diagrams that help deriving the differential or algebraic equations of the mathematical model. For free damped systems, Lagrange’s equations are: ! ! d vL vL vF d vT vU vF þ ¼ 0 or þ ¼ 0; L ¼ T  U (3.113)  þ dt vq_j vqj vq_j dt vq $ j vqj vq_j where T is the system’s total kinetic energy, U is the total potential energy (both gathered into the Lagrangian L) and F is the Rayleigh dissipation function. F incorporates all the viscous-damping losses and is defined as: n1  2 1 X cj $ q_j  q_jþ1 F¼ $ 2 j¼1

(3.114)

3.3 MultipleeDOF Systems

Eq. (3.114) assumes there is a damper cj connecting any two consecutive generalized coordinates qj and qjþ1 (translation or rotation displacements), of either translatory or rotary nature. nn

Example 3.16

A body of mass m1 is connected to a heavy cylinder of mass m2 and radius R by a shock absorber consisting of a spring of stiffness k and a damper of viscous damping coefficient c, as illustrated in Figure 3.33(a). Ignore the friction between the two moving bodies and the fixed contact surfaces, while assuming the cylinder rolls without slippage on the fixed surface. Establish the DOF of this system, and derive its mathematical model using Newton’s second law of motion, as well as Lagrange’s equations.

θ

m2

yC C P

k

fe m1

R

θ

y

y

fd

fe* m1

fd*

c

m2 yC CR P

(a)

(b)

FIGURE 3.33 (a) MultipleeDOF Mechanical System With Rotary and Translatory Components; (b) FreeBody Diagrams of The Translating Body and Rolling Cylinder.

Solution

The relevant coordinates of this system are yC, q (related to the cylinder motion), and the translating body displacement y. The coordinates yC and q are connected as yC ¼ R $ q, which results from the cylinder rolling about its instantaneous center of velocities P. As a consequence, the mechanical system possesses 3  1 ¼ 2 DOF. Newton’s Second Law of Motion The free-body diagrams of the two separated bodies are shown in Figure 3.33(b). Applying Newton’s second law of motion yields:

(

m1 $ y€ ¼ fe  fd € ¼ f  $ R  f  $ R JP $ q e d

(3.115)

The second Eq. (3.115) expresses the cylinder rotation around an axis passing through Pdthe instant center of zero velocities. The elastic (spring) forces, the damping forces, and the cylinder mass moment of inertia with respect to the instant center P axis are:

8 > fe ¼ k $ ðy  yC Þ ¼ k $ ðy  R $ qÞ; fe ¼ fe > > > >



< fd ¼ c $ y_  y_C ¼ c $ y_  R $ q_ ; fd ¼ fd > > > m2 $ R2 3m2 $ R2 > > þ m 2 $ R2 ¼ : J P ¼ J C þ m 2 $ R2 ¼ 2 2

(3.116)

119

120

CHAPTER 3 Mechanical Systems

Substituting the parameters of Eq. (3.116) into Eq. (3.115) results in:

8 > < m1 $ y€ þ c $ y_  R $ c $ q_ þ k $ y  R $ k $ q ¼ 0 3m $ R € > : 2 $ q þ R $ c $ q_  c $ y_ þ R $ k $ q  k $ y ¼ 0 2

(3.117)

Eq. (3.117) forms the mathematical model of the studied mechanical system. Lagrange’s Equations The kinetic energy, potential energy, and the dissipation function of this system are:

1 1 1 1 2 T ¼ $ JP $ q_ þ $ m1 $ y_2 ; U ¼ $ k $ ðyC  yÞ2 ¼ $ k $ ðR $ q  yÞ2 ; 2 2 2 2

2

2 1 1 F ¼ $ c $ y_C  y_ ¼ $ c $ R $ q_  y_ 2 2 (3.118) and the two Lagrange’s equations corresponding to the generalized coordinates y and q are:

8   > d vT vU vF > > > < dt vy_ þ vy þ vy_ ¼ 0   > d vT vU vF > > > þ ¼0 þ : dt _ vq vq_ vq

(3.119)

The following terms are required in Eq. (3.119):

8   > d vT d > > _ ¼ m1 $ y; € > < dt vy_ ¼ dt ðm1 $ yÞ   >

d vT d > > € > JP $ q_ ¼ JP $ q; ¼ : dt dt vq_



vU vF ¼ k $ ðR $ q  yÞ; ¼ c $ R $ q_  y_ vy vy_

vU vF ¼ k $ R $ ðR $ q  yÞ; ¼ c $ R $ R $ q_  y_ vq vq_ (3.120)

Substituting Eq. (3.120) into Eq. (3.119) with JP of the third Eq. (3.116) results in Eq. (3.117)dthe mathematical model derived by Newton’s second law. nn

Solution of the Mathematical Model Differential Equations The differential equations system describing the free undamped response of a multiple-DOF vibratory system can be solved by direct integration (either using solutions available from Calculus or mathematical software dedicated code, such as MATLAB) or by using Simulink. nn

Example 3.17 Determine and plot the time response of the mechanical system shown in Figure 3.33(a) of the previous Example 3.16 for an initial displacement of the translating body of y(0) ¼ 0.1 m. Known are also

3.3 MultipleeDOF Systems

m1 ¼ 5 kg, m2 ¼ 2 kg, R ¼ 0.1 m, k ¼ 200 N/m, and c ¼ 10 N s/m. Use both MATLAB and Simulink solution procedures.

Solution MATLAB Solution The MATLAB dsolve command can be used to symbolically solve several differential equations in terms of time as the independent variable with specified initial conditions. As detailed in Appendix E, the basic command is: >> dsolve('eq1, eq2, ., ' , 'initial condition1, initial condition2,., ') where eq1, eq2, . are the differential equations and initial condition1, initial condition2, . are the initial conditions. For our example, we use the following MATLAB script with the mathematical model of Eq. (3.117): .>> [th,y]= dsolve('3*m2*r/2*D2th+r*c*Dth-c*Dy+r*k*th-k*y =0,m1*D2y+c*. Dy-r*c*Dth+k*y-r*k*th=0','Dth(0)=0,Dy(0)=0,th(0)=0,y(0)=0.1'); %finds. symbolic q(t) and y(t) >> m1=5;m2=2;k=200;r=0.1;c=10; >> y=subs(y);th=subs(th);%attributes numeric values to the functions. of y and q >> t=0:0.00001:2; %defines a time vector over the [0; 2]s range and a. 0.00001s time step >> y1=eval(y);th1=eval(th);%evaluates y and q resulting from the t. vector >> subplot(2,1,1) >> plot(t,y1) >> subplot(2,1,2) >> plot(t,th1*180/pi) % q is converted from radians into degrees and the resulting plots are shown in Figure 3.34. Simulink Solution Eq. (3.117) is rewritten in the following form, which keeps the larger derivative from each of the two equations in the left-hand side:

(

y€ ¼ a1 $ y_ þ a2 $ q_ þ a3 $ y þ a4 $ q € ¼ b1 $ y_ þ b2 $ q_ þ b3 $ y þ b4 $ q q

(3.121)

with:

c k 2c ; a2 ¼ R $ a1 ; a3 ¼  ; a4 ¼ R $ a3 ; b1 ¼ ; m1 m1 3R $ m2 2k ; b4 ¼ R $ b3 b2 ¼ R $ b1 ; b3 ¼ 3R $ m2

a1 ¼ 

(3.122)

whose numerical values are: a1 ¼ 2, a2 ¼ 0.2, a3 ¼ 40, a4 ¼ 4, b1 ¼ 33.333, b2 ¼ 3.333, b3 ¼ 666.667, and b4 ¼ 66.667. Each of the two in Eq. (3.121) requires two integrations to find the variables y and q from their corresponding second derivatives. The right-hand sides of Eq. (3.121) contain four terms each, which need to be algebraically added after adequate multiplication by the factors just determined and which are input in the corresponding gain blocks. All these graphical operations are shown in the Simulink block diagram of Figure 3.35. Note that in

121

CHAPTER 3 Mechanical Systems

y (m)

(a)

Time (s)

(b) θ (degrees)

122

Time (s)

FIGURE 3.34 Free Damped Time Response of TwoeDOF Mechanical System With Rotary and Translatory Elements. a4θ( t ) .

b4θ( t )

a4

b2θ( t )

b4 b2

..

θ( t )

θ( t ) - in degrees

.

θ( t )

θ( t )

b3 y (t )

.

a2

b1y (t ) ..

.

b3 b1

.

y (t )

y (t )

a2θ( t )

y (t )

.

a1y (t ) a3 y (t )

a1 a3

FIGURE 3.35 Simulink Block Diagram for Plotting the Free Damped Response of a TwoeDOF Mechanical System With Rotary and Translatory Elements.

3.3 MultipleeDOF Systems

Integrator 4 the initial condition y(0) needs to be specified. After selecting an integration time of 2 s with a constant time step of 0.0001 s and running the Simulink code, we get the plots of Figure 3.34. nn

3.3.2 Forced Response The mathematical models of multiple-DOF mechanical systems with forcing are derived by using Newton’s second law of motion and Lagrange’s equations. While Newton’s second law of motion is based on free-body diagrams of the subsystems whose DOF were identified and on the procedure already described in this chapter, Lagrange’s equations are: ! ! nj nj d vL vL vF X d vT vU vF X þ ¼ Qj;i or þ ¼ Qj;i  þ dt vq_j vqj vq_j dt vq_j vqj vq_j i¼1 i¼1 (3.123) vWnc with Qj;i ¼ vqj with T being the system’s total kinetic energy, U the total potential energy, F the dissipation function, and Qj,i one of the nj generalized forces or moments that are applied at the point with the generalized coordinate qj. Qj,i is calculated as the partial derivative of the total work Wnc produced by nonconservative forces/ moments (which are all forces/moments except the ones of elastic and gravitational nature). nn

Example 3.18

The mechanical filter microsystem of Figure 3.36(a) is driven by a force f ¼ 2e10t mN. Propose a lumped-parameter physical model of the original system, derive its mathematical model, and use Simulink to plot the system response consisting of the time-domain displacements of the two shuttle masses for m1 ¼ 3.6$1010 kg, m2 ¼ 2.5$1010 kg, l1 ¼ 160 mm, l2 ¼ 100 mm, l ¼ 80 mm, d1 ¼ d2 ¼ d ¼ 2 mm (circular cross-section diameter of all flexible members), and E ¼ 165 GPa. Consider zero initial conditions.

Anchor Beam spring

l1 l

f

m1

l2

x1 k1

m2 Shuttle mass l

l2

l1

x2

m1 k1

k2

k m2 f

k2

Serpentine spring

(a)

(b)

FIGURE 3.36 Linear-Motion Mechanical Filter Microsystem: (a) Schematic of System; (b) LumpedParameter Model.

123

124

CHAPTER 3 Mechanical Systems

Solution The beam springs act as two parallel springs on each of the shuttle masses of Figure 3.36(a), and the serpentine spring realizes the elastic coupling between m1 and m2. The lumped-parameter physical model equivalent to the actual microdevice is shown in Figure 3.36(b). This system has two DOFs, the displacements x1 and x2.

x1 fe

fe1 m1

* e

f

x2 fe2 m2

f

fe1

fe2

FIGURE 3.37 Shuttle Masses Free-Body Diagrams. Newton’s Second Law of Motion Method Based on the free-body diagrams of Figure 3.37 and applying Newton’s second law of motion, the mathematical model corresponding to the lumped-parameter model of Figure 3.37(b) is:

(

m1 $ x€1 ¼ f  2fe1  fe m2 $ x€2 ¼ fe  2fe2

(3.124)

By taking into account that fe1 ¼ k1 $ x1 ; fe ¼ fe ¼ k $ ðx1  x2 Þ; fe2 ¼ k2 $ x2 , Eq. (3.124) result in the following mathematical model:

(

m1 $ x€1 ¼ f  ð2k1 þ kÞ $ x1 þ k $ x2 m2 $ x€2 ¼ k $ x1  ðk þ 2k2 Þ $ x2

(3.125)

Lagrange’s Equations Method To use Lagrange’s equations, the kinetic energy T and the elastic potential energy Ue ¼ U of the mechanical system of Figure 3.36(b) are:

1 1 T ¼ $ m1 $ x_21 þ $ m2 $ x_22 ; 2 2 1 1 1 Ue ¼ U ¼ 2 $ $ k1 $ x21 þ $ k $ ðx1  x2 Þ2 þ 2 $ $ k2 $ x22 2 2 2

(3.126)

The derivatives and the generalized forces needed in Lagrange’s equations are:

8   d vT d > > > < dt vx_1 ¼ dt ðm1 $ x_1 Þ ¼ m1 $ x€1 ;   > d vT d > > : ¼ ðm2 $ x_2 Þ ¼ m2 $ x€2 ; dt vx_2 dt

vU vF ¼ 2k1 $ x1 þ k $ ðx1  x2 Þ; ¼ 0; Q1 ¼ f vx1 vx_1 vU vF ¼ k $ ðx1  x2 Þ þ 2k2 $ x2 ; ¼ 0; Q2 ¼ 0 vx2 vx_2 (3.127)

3.3 MultipleeDOF Systems

Note that there is no damping, which means that F ¼ 0 and the corresponding derivatives of F in Eq. (3.121) are also zero. Eq. (3.127) is combined with the definition Eq. (3.123) and they result in Eq. (3.125). Eq. (3.125) is rewritten in the form:

8 8 2k1 þ k k 1 > > > $ x1 þ $ x2 þ $f > < x€1 ¼  m < x€ ¼ a11 $ x1 þ a12 $ x2 þ 1 $ f m1 m1 1 1 m1 or > > k 2k þ k 2 > : > x€2 ¼ a21 $ x1  a22 $ x2 $ x1  $ x2 : x€2 ¼ m2 m2 (3.128) The stiffnesses k1 and k2 are given in Table 2.2 for fixed-guided beams, whereas the stiffness of the . . serpentine spring k is expressed in Eq. (2.10); they are k1 ¼ 12E $ Iz l13 ; k2 ¼ 12E $ Iz l23 ;


k ¼ 6E $ Iz 5l 3 with Iz being the circular cross-sectional area moment of inertia, Iz ¼ p$d4/64.

t

a11x1(t )

f

f / m1

.. x1(t )

a11

. x1(t )

x1(t )

1/ m1

a21x1(t ) a12 x2 (t )

a12

a21

a22 x2 (t )

.. x 2 (t )

a22

. x 2 (t )

x 2 (t )

FIGURE 3.38 Simulink Diagram for Solving the Differential Eq. (3.128). With the numerical values of this example, the parameters of Eq. (3.128) are k1 ¼ 0.380 N/m, k2 ¼ 1.555 N/m, k ¼ 0.304 N/m, a11 ¼ 2.952913017$109 m2/s2, a12 ¼ 8.436894333$108 m2/s2, a21 ¼ 1.214912784$109 m2/s2, a22 ¼ 1.365561969$1010 m2/s2, 1/m1 ¼ 2.777777777$109 kg1. The block diagram of Figure 3.38 integrates Eq. (3.128).The input force on the mass m1 is produced by combining the Ramp block from the Source library and the Math Function from the Math Operations library. While the unit ramp (Slope parameter is 1) provides the time input (which is interpreted as a variable u), the following expression needs to be specified for the Math Function: 0.000002*exp(e10*u(1)), which constitutes f of the example. The remaining blocks are typical ones and can be configured easily. As can be seen in Figure 3.38, the stiffness coupling (which is realized through k) is reflected in the cross-connections between x1 and x2. The Variable names in the two To Workspace blocks are x1 and x2 and their Save Format is Array. Under Simulation/Configuration Parameters and Data Import/Export the (checked) Time name is tout, while the (checked) Output name is x1dthis will ensure that both x1 and x2 are exported to MATLAB’s Workspace. As a consequence, a typical command

125

CHAPTER 3 Mechanical Systems

x1 Shuttle displacements (m)

126

x2

Time (s)

FIGURE 3.39 Time Variation of the Shuttle Mass Displacements (m) for the Mechanical System of Figure 3.36(b).

plot(tout,x1,tout,x2) in Workspace produces the raw plots that are illustrated in a more refined form in Figure 3.39. nn

nn

Example 3.19 (i) Establish the DOF of the gear-train lumped-parameter mechanical system of Figure 3.40(a). An actuation torque ma drives the input shaft and a load torque ml acts on the output shaft. Using Lagrange’s equations, derive the mathematical model of this system. Assume the gear rotary inertia is negligible. (ii) Use Simulink to generate the corresponding simulation diagram and plot the angular velocities u2(t), u(t), u3(t) of the three disks placed on the middle shaft in Figure 3.40(a) for 0  t  10 s, N1 ¼ 44, N2 ¼ N3 ¼ 38, N4 ¼ 24, J ¼ 0.002 kg m2, c ¼ 240 N m s, k ¼ 3500 N m, ml ¼ 50$sin(t) N m and ma defined as in Figure 3.40(b).

Solution

(i) The five rotation angles q1, q2, q, q3, and q4 are the parameters defining the state of the mechanical system shown in Figure 3.40(a). The following two constraint relationships result from the two meshing gear pairs: q1/q2 ¼ N2/N1 and q3/q4 ¼ N4/N3. As a consequence, the system has

N1 θ1

c

c

c

ma Gear

k

θ2

Rotor k

θ

N3 θ3

J N2

ma (N m)

Spring

ma,eq ml

Damper

θ4 N4

c

θ

J

θ2

k

ceq,2 θ3

ml,eq

4000 3000

2

45

(b)

Time (s)

(c)

FIGURE 3.40 (a) Gear-Shaft Rotary Mechanical System; (b) Actuation Moment Definition; (c) System Reduced to the Middle Shaft.

3.3 MultipleeDOF Systems

0

(a)

k

ceq,1

127

128

CHAPTER 3 Mechanical Systems

5  2 ¼ 3 DOF and therefore, three equations are needed in its mathematical model. The equivalent system of Figure 3.40(c) results after transferring the damping properties of the input and output shafts of Figure 3.40(a) to the middle shaft. This system has three DOF and is defined by the variables q2, q, and q3, which are also the generalized coordinates in the sense of Lagrange’s equations. These equations are:

8   8   d vL vL vF d vT vU vF > > > > > > þ ¼ ma;eq þ ¼ ma;eq  þ > > > > _2 _2 _2 dt vq dt vq > > v q v q v q vq_ 2 2 2 > > > >   > > < d vT  vU vF < d vL vL vF þ þ or  ¼0 þ ¼0 > dt vq_ > dt vq_ vq vq_ vq vq_ > > > > > >     > > > > > > d vL vL vF d vT vU vF > > > > þ ¼ ml;eq þ ¼ ml;eq  þ : dt : dt _ _ _ vq vq vq 3 vq 3 vq 3 vq_ 3 3 3

(3.129) where the moments on the right-hand sides of Eq. (3.129) are associated with the three identified DOF. With the transfer relationships of Eq. (2.45), these moments become:

ma;eq ¼

N2 N3 $ ma ; ml;eq ¼ $ ml N1 N4

(3.130)

The kinetic energy, potential energy, and dissipation function of the system are:

1 1 1 T ¼ $ J $ q2 ; Ue ¼ U ¼ $ k $ ðq  q2 Þ2 þ $ k $ ðq  q3 Þ2 ; 2 2 2 1 1 1 2 2 2 F ¼ $ ceq;1 $ q_ 2 þ $ c $ q_ þ $ ceq;2 $ q_ 3 2 2 2

(3.131)

The derivatives that are required by Eq. (3.129) are calculated from Eq. (3.131) and are substituted in Eq. (3.129) together with the moments of Eq. (3.130), which yields the following mathematical model:

8  2 > > N2 N2 > >0 ¼ $ ma  $ c $ q_ 2  k $ ðq2  qÞ > > N N > 1 1 > < € ¼ c $ q_  k $ ðq  q2 Þ  k $ ðq  q3 Þ J $q > > >  2 > > N3 N3 > > > $ c $ q_ 3  k $ ðq3  qÞ : 0 ¼ N $ ml  N 4 4

(3.132)

Eq. (3.132) takes into account that ceq,1 ¼ c$(N2/N1)2 and ceq,2 ¼ c$(N3/N4)2. (ii) Eq. (3.132) can also be reformulated in a way that would render use of Simulink applicable:

8 > < q_ 2 ¼ a11 $ q  a11 $ q2 þ a12 $ ma € ¼ b11 $ q_  b12 $ q þ b13 $ q2 þ b13 $ q3 q > : q_ ¼ c $ q  c $ q  c $ m 3 11 11 3 12 l

(3.133)

3.3 MultipleeDOF Systems

with

8  2 > k N1 1 N1 > > ; a12 ¼ $ > a11 ¼ $ > > c c N2 N 2 > > > < c 2k k b11 ¼ ; b12 ¼ ; b13 ¼ > J J J > > >  2 > > > k N4 1 N4 > > ; c12 ¼ $ : c11 ¼ $ c c N3 N3

(3.134)

The numerical values of the coefficients given in Eq. (3.134) are a11 ¼ 19.552, a12 ¼ 0.00482, b11 ¼ 120,000, b12 ¼ 3,500,000, b13 ¼ 1,750,000, c11 ¼ 5.817, c12 ¼ 0.00263. The Simulink diagram corresponding to the differential Eq. (3.133) is shown in Figure 3.41(a). Each of the three differential equations is represented by a horizontal chain plus the connections between chains because of the coupling. The input to the first chain (differential equation) is a Signal Builder block, which can be dragged from the Source library. This block is configured as per Figure 3.41(b) by clicking on each individual line segment of the predefined signal and by specifying start/end values of the time and of the function (ma) as required in Figure 3.40(b). The first segment, for instance, has a Left Point time of 0 and a Right Point value of 2. Correspondingly, the output (ma) values are: 0 at the Left Point and

θ

θ

θ θ

θ θ

θ θ θ θ θ

θ

(a)

(b)

FIGURE 3.41 (a) Simulink Diagram for Solving the Mathematical Model of a Gear-Shaft Rotary System; (b) Signal Builder Block.

129

ω (rad/s)

CHAPTER 3 Mechanical Systems

ω2 (rad/s)

Time (s)

Time (s)

ω3 (rad/s)

130

Time (s)

FIGURE 3.42 Time Variation of the Angular Velocities for the Mechanical System of Figure 3.40. 3800 at the Right Pointdsee Figure 3.41(b) for the fully formatted ma(t) signal. The three outputs are exported and plotted in MATLAB as shown in the plots of Figure 3.42. nn

SUMMARY By using the lumped-parameter approach, you have learned to model the dynamics of single- and multiple-DOF mechanical systems. Newton’s second law of motion, the conservation of energy method, and Lagrange’s equations have been applied to derive the mathematical models for the natural, free damped, and forced responses of translatory and rotary mechanical systems. The MATLAB code has been used to solve linear and nonlinear differential equations. You have also learned how to use Simulink to model the dynamics of mechanical systems, to solve the corresponding differential equations, and to plot the resulting time response for various examples.

PROBLEMS 3.1 Establish the DOF of the lumped-parameter mechanical system of Figure 3.43. 3.2 Evaluate the DOF of the planar mechanical system of Figure 3.44. There is no slippage between the roller and the fixed horizontal surface.

Problems

k2

c1

k1 Anchor

c2

k5

m1

k4

k6

m2

Anchor

m3 k3

c3

FIGURE 3.43 Translatory Mechanical System.

Roller Spring

k1

C m1 Rod

m2 Body k2

R

FIGURE 3.44 Mechanical System With Roller (Without Slippage), Translating Body, Rod, and Springs.

3.3 The mechanical system of Figure 3.45 consists of two bodies and a continuous inextensible wire passing around two pulleys. Determine the DOF of this system assuming that: (i) The pulleys do not rotate and the wire slides around the pulleys. (ii) The pulleys do rotate and there is no slippage between pulleys and wire. Fixed support

C2 R2

Pulley

Wire

R1 C1

Body

FIGURE 3.45 Mechanical System With Pulleys, Inextensible Wire, and Bodies.

3.4 The planar slider-crank mechanism of Figure 3.46 is used to convert the rotary motion of the crank around the fixed pin O into the intermittent

131

132

CHAPTER 3 Mechanical Systems

Crank

O

Connecting rod

r

Slider

l

Fixed support

FIGURE 3.46 Slider-Crank Mechanism.

translatory motion of the slider. Determine the DOF of this mechanism, knowing the crank length r and the connecting-rod length l. 3.5 The rotary gear system of Figure 3.47 transmits the rotary motion from an input shaft to an intermediate shaft and then to the output shaft, by means of the gear pairs N1eN2, N3eN4, and a high-stiffness elastic coupling. Determine the DOF of this system when the shafts are assumed: (i) Rigid (ii) Torsionally flexible From motor θi

Input shaft

Bearing Pulley

N4

N1

Disk

θo

Gears Elastic coupling

Output shaft

N3 N2 Intermediate shaft

FIGURE 3.47 Rotary Motion Transmission With Gears, Pulley, Coupling, and Shafts.

3.6 The planar mechanical system of Figure 3.48 is formed of two rigid levers that undergo small rotations around their pivot points O1 and O2. Attached to these levers are several springs and dampers. Establish the DOF of this system. 3.7 A mechanical vibration absorber consists of four rigid levers, of which two are identical, and several springs attached to the levers, as illustrated in Figure 3.49. Determine the DOF of this system assuming small rotations of the levers. 3.8 Determine the DOF of the mechanical system shown in Figure 3.50, which consists of a pulley, a body, two rigid rods, a rigid lever, and three springs. Consider small rotations of the rigid lever and the pulley.

Problems

l1

l2

l3

l5

O1 O2

Rigid lever Damper

l4

Spring

Anchor

FIGURE 3.48 Small-Rotation, Two-Lever Mechanical System With Attached Springs and Dampers.

l1 k1

m2 k2

l3

l1

m1

k1

l2

Rigid lever

m2 l3

l2

Spring

k2

m3

k3

Anchor

FIGURE 3.49 Small-Rotation, Four-Lever Mechanical System With Attached Springs.

Pulley

m R Rod

k

Rod Rigid lever

l 2l

k Body

m k Spring

Anchor

FIGURE 3.50 Mechanical System With Pulley, Lever, Rods, and Body.

133

134

CHAPTER 3 Mechanical Systems

3.9 A fixed-center cylindrical pinion of mass m1 and radius R1 transmits its rotation to a translating rack of mass m3 as illustrated in Figure 3.51. The mobile rack further interacts with another cylindrical pinion of mass m2 and radius R2, which has its center connected to a spring of stiffness k, and rolls on a fixed rack. For no slippage between pinions and racks, demonstrate that this system has one DOF, formulate its mathematical model, and determine its natural frequency.

Rotating pinion

m1

Translating rack

Guide

R1

m3

m2 k

R2

Anchor

Rolling pinion

Fixed rack

FIGURE 3.51 Mechanical System With Pinions and Racks.

3.10 The pendulum of Figure 3.52(a) can oscillate in a vertical plane; it is formed of a rigid heavy rod and a bob of mass M ¼ 0.5 kg. The pendulum is placed in a horizontal plane, where its original vertical-plane natural frequency is preserved by two identical springs of stiffness k ¼ 230 N/m, which compensate for the absence of gravitydsee Figure 3.52(b). What is the distance x where the two springs need to be placed? Assume small displacements and consider known the rod’s length l ¼ 0.03 m, its cross-sectional

Anchor

x

Spring Rigid rod

M

l

k

k

Bob

(a)

(b)

FIGURE 3.52 Pendulum With: (a) Rigid Rod and Point Mass in Vertical Plane; (b) Rigid Rod, Point Mass, and Two Identical Springs in Horizontal Plane.

Problems

diameter d ¼ 0.005 m, the rod’s material mass density r ¼ 7800 kg/m3, and the gravitational acceleration g ¼ 9.8 m/s2. 3.11 Two parallel and identical rigid massless levers undergo small rotations in a horizontal plane around the fixed pins O1 and O2. They are connected by means of a short, rigid, and massless pushrod, as sketched in Figure 3.53. The system also comprises a spring of stiffness k and a particle of mass m. (i) Demonstrate that this system has one DOF and calculate its natural frequency. (ii) The particle m is relocated to a different position on the top lever to increase the natural frequency of this single-DOF system by 15%. Determine the new position of the mass m. Known is l ¼ 1 m. Anchor

l l /4 Rigid lever

k

m

O2

Particle

O1

l /2

Push rod

l

FIGURE 3.53 Two-Lever Mechanical System.

3.12 Consider the mechanical system sketched in Figure 3.54, consisting of a rigid lever of mass ml that connects at one end (through a massless rod) to a disk of radius R and mass m, as well as to a spring of stiffness k1. The

Spring

k1 Rigid lever

m

k2 ml

Rod

l m R

Wheel

Fixed wall

FIGURE 3.54 Mechanical System With Planar Motion.

l

l

135

136

CHAPTER 3 Mechanical Systems

wheel can roll without slippage on a fixed wall under small angular motions of the rod. A heavy particle of mass m is also placed on the lever. At the rod’s pivot point is a rotary (spiral) spring of stiffness k2. Demonstrate the system has one DOF, and derive its mathematical model when ignoring gravitational effects. 3.13 Calculate the natural frequency of the mechanical system of Problem 3.12 and shown in Figure 3.54 for m ¼ 0.5 kg, ml ¼ 0.3 kg, l ¼ 0.4 m, k1 ¼ 120 N/m, and k2 ¼ 60 N/m. What change in the stiffness k1 increases the natural frequency by 20%, when all other parameters remain the same? 3.14 The massless lever of Figure 3.55(a) undergoes small rotations around the pin O in a horizontal plane with a point mass m and four different springs attached to it. A design miscalculation results in a natural frequency of this system that is 5% less than intended. It is decided to add a rotary (spiral) spring at the lever’s pinned enddsee Figure 3.55(b)dto remedy the error. Calculate the stiffness k5 knowing l ¼ 0.5 m, M ¼ 0.02 kg, k1 ¼ 100 N/m, k2 ¼ 120 N/m, k3 ¼ 80 N/m, and k4 ¼ 60 N/m. Anchor

k2 k1 O

k3

Spring

k5

Rigid lever

M k4

l l

(a)

(b)

FIGURE 3.55 (a) Rigid Lever With Translation Springs and Point Mass; (b) System With Additional Rotary Spring.

3.15 The mechanical system of Figure 3.56 is formed of a point mass m ¼ 0.1 kg and two springs of stiffnesses k1 ¼ 100 N/m and k2 ¼ 120 N/m. Its natural frequency is evaluated by means of a cantilever sensor, which is attached to the point mass. Knowing the cantilever has a constant circular cross section, a length l ¼ 0.08 m, the cantilever’s Young’s modulus is E ¼ 200 GPa, and mass density is r ¼ 7600 kg/m3, determine its diameter d, such that the altered natural frequency does not increase by more than 10% when considering the cantilever’s inertia and stiffness. 3.16 Obtain a lumped-parameter model for the gear train mechanical system of Figure 3.57, by transferring the relevant parameters to the connecting rigid shaft and show this is a single-DOF system. Find its natural frequency

Problems

Anchor

Spring Cantilever

k1 m

y

Point mass

k2 l

FIGURE 3.56 Spring-Mass Mechanical System With Cantilever Sensor.

l1 Bearing

l2 N4, J4

d1

d2

N1, J1 Gears

Anchor Flexible shaft

Flexible shaft

N3, J3

N2, J2 Rigid shaft

FIGURE 3.57 Three-Shaft Gear Train.

knowing that N1 ¼ 24, N2 ¼ 48, N3 ¼ 32, N4 ¼ 40, J1 ¼ 0.001 kg m2, J2 ¼ 0.008 kg m2, J3 ¼ 0.002 kg m2, J4 ¼ 0.006 kg m2. The two flexible shafts lengths are l1 ¼ 0.1 m and l2 ¼ 0.08 m, the shafts constant circular cross-section diameters are d1 ¼ 0.02 m and d2 ¼ 0.015 m, the flexible shafts shear modulus is G ¼ 150 GPa, and the mass density is r ¼ 7800 kg/ m3. Include the inertia contributions of the flexible shafts. 3.17 A particle can attach either at the free end of a microcantilever, or at the midpoint of and axially on a microbridge of identical dimensions and material properties with the microcantileverdsee Figure 3.58. Which of these two detectors produces the larger natural frequency change with respect to their out-of-plane bending vibrations? Consider the design with r ¼ 5600 kg/m3, w ¼ 10 mm, h ¼ 1 mm (w and h are dimensions of the rectangular cross section, with h being the out-of-plane thickness), and l ¼ 80 mm. 3.18 The microbridge system of Figure 3.59 can rotate around the x axis and translate along the y axis. It is composed of a rigid central plate and two identical flexible hinges of circular cross section. The plate and hinges are

137

138

CHAPTER 3 Mechanical Systems

Anchor

Particle

w l

l /2 l

(a)

(b)

FIGURE 3.58 Top View of: (a) Microcantilever With Attached Particle at Its Free End; (b) Microbridge With Attached Particle at Its Midpoint. y Anchor

Rigid plate

l 2 /2

d l2 x

Hinge

l1

l1

l1

FIGURE 3.59 Microbridge With Rigid Plate and Two Side Flexible Hinges.

constructed from the same material, and the two elastic moduli are related as E ¼ 2.7 G. Known also are d ¼ 3 mm, l2 ¼ 220 mm, h ¼ 20 mm (the plate thickness), as well as the ratio between the natural frequency corresponding to the y-axis translation and the natural frequency of the x-axis rotation, which is 1.8. Consider the effect of added inertia of the two hinges, and find the length l1. 3.19 The rigid plate of Figure 3.60 is elastically supported by four identical hinges of constant circular cross section. A particle deposits at the plate’s center, which changes the natural frequency of the system with respect to the plate translation (out of the plane) along the z axis by 30 Hz. Find the mass of the attached particle knowing E ¼ 150 GPa and r ¼ 5800 kg/m3. Also known are d ¼ 2 mm, l1 ¼ 50 mm, l2 ¼ 40 mm, and h ¼ 10 mm (the plate thickness). Ignore the inertia contribution of the four hinges. 3.20 Two particles of unknown masses attach to the midsection of a rectangular cross-section microbridge, which is supported by two flexible hinges of identical material and circular cross-section diameter d but of different lengths l1 and l2dsee Figure 3.61. Determine the two particles masses m1 and m2, knowing that the in-plane bending natural frequency shift with respect to the y-axis translation is 4000 rad/s, and the torsion natural frequency shift in terms of the x-axis rotation is 10,000 rad/s. The following

Problems

d

l1 l1

l 1 /2

Rigid plate

Anchor

l 2/2 l 2 z

Hinge

l1

l1

l1

FIGURE 3.60 Microbridge With Central Plate and Four Flexible Hinges. y Rigid plate

Anchor

w/2

Flexible hinge

m1

w/2

d

b1

O

x

b2

m2 l/2

l1

l2 l

FIGURE 3.61 Top View of Paddle Microbridge With Two Attached Particles.

parameters are known: l ¼ 30 mm, w ¼ 20 mm, h ¼ 8 mm (out-of-plane plate thickness), l1 ¼ 50 mm, l2 ¼ 40 mm, d ¼ 5 mm, b1 ¼ 6 mm, b2 ¼ 7 mm, E ¼ 190 GPa, G ¼ 130 GPa, and r ¼ 6000 kg/m3. The inertia of the two hinges is not considered. 3.21 The shuttle mass of Figure 3.62 can translate along the y axis and is symmetrically supported by three pairs of beams, all of constant circular cross-section diameters. Knowing the shuttle mass m ¼ 35$1013 kg, the beams’ Young modulus E ¼ 150 GPa, and mass density r ¼ 5500 kg/m3, as well as l1 ¼ 10 mm, l2 ¼ 3 mm, l3 ¼ 5 mm, d1 ¼ 0.8 mm (diameter of longer beams), d2 ¼ d3 ¼ 0.4 mm (diameters of shorter beams), calculate the natural frequency of this system: (i) Without considering the beams inertia. (ii) By taking into account the long beam inertia; in this case, determine the difference between the two natural frequency evaluations.

139

140

CHAPTER 3 Mechanical Systems

y Anchors l3

l3

l2

O

Beam

l2

Anchor

l1

l1 Shuttle mass

FIGURE 3.62 Microdevice With Three Sets of Beams and a Shuttle Mass for Translatory Motion.

3.22 Derive the mathematical model for the lever mechanical system of Figure 3.63, which undergoes small rotations around the pivot O in a horizontal plane. Determine the equivalent natural frequency and the damping ratio knowing the mass of the rigid lever ml ¼ 0.5 kg, the mass of the end body m ¼ 0.4 kg, the stiffnesses k1 ¼ 150 N/m, k2 ¼ 90 N/m, the damping coefficient c ¼ 20 N s/m, and the length l ¼ 0.2 m. Use Simulink to model and plot the time response of the system when an initial rotation angle q(0) ¼ 5 is applied to the lever. Anchor

O

k2

k1

θ

ml

Rigid lever

l

m

l c

FIGURE 3.63 Lever With Attached Springs, Damper, and Point Mass.

3.23 A massless lever undergoes small-angle rotations around the fixed pivot O in a horizontal plane, as in Figure 3.64. Known are the stiffnesses and damping coefficients k1 ¼ 1000 N/m, k2 ¼ 300 N/m, c1 ¼ 1 N s/m, c2 ¼ 2 N s/m for the springs and dampers connected to the lever. A particle of unknown mass m is attached at the end point A on the lever. The damped frequency of this mechanical system is ud ¼ 38 rad/s. Calculate the mass m by transferring all mechanical properties at A. 3.24 The rotary mechanical system shown in Figure 3.65 is composed of a cylinder, a torsional damper, and a torsional spring. The damped

Problems

l

l Massless lever

l y m

O

k1 c1

A

c2 k2

Fixed surface Fixed surface

FIGURE 3.64 Lever Mechanical System With Attached Springs, Dampers, and Point Mass. θ

c J

k

FIGURE 3.65 Mechanical System Undergoing Rotary Vibrations.

frequency of this system is 14 Hz, and the number of vibration cycles of the cylinder necessary to reduce the amplitude of the free vibrations by a factor of 10 is 6. Knowing J ¼ 0.01 kg m2, determine the damping coefficient c and the spring stiffness k. 3.25 Consider a translatory mechanical system that is composed of rigid body of mass m and a helical spring with n ¼ 8 turns, G ¼ 1.5 GPa, d ¼ 1 mm, and R ¼ 6 mm. The spring is anchored at one end and connects to the body at the other end. When placed in vacuum, the natural period of the system is 0.72 s, and when placed in a fluid of unknown viscous damping coefficient, the period of the free vibrations becomes 0.78 s. Calculate the viscous damping coefficient of the fluid. 3.26 After applying the brakes to the rotating shaft-disk system of Figure 3.66, its angular velocity decreases by a factor of n ¼ 20, after s ¼ 0.2 s. Knowing the mass moment of inertia of the disk J ¼ 0.00004 kg m2, the shaft length

l

l d Brake

J

d Brake

Bearing θ Disk

FIGURE 3.66 Shaft-Disk System With Rotary Damping.

Flexible shaft

141

142

CHAPTER 3 Mechanical Systems

l ¼ 0.8 m, its constant cross-section diameter d ¼ 0.01 m, the shaft mass density r ¼ 7600 kg/m3, and the shaft shear modulus G ¼ 72.5 GPa, evaluate the rotary viscous-damping coefficient c of the two identical bearings supporting the shaft-disk system. Include the inertia of the two shafts, and demonstrate that the stiffness does not affect the calculation of c. 3.27 Consider the shaft-gear system of Problem 3.16 and shown in Figure 3.57. In addition to the parameters provided in Problem 3.16, we also know the damping coefficients of the identical dampers supporting the top left shaft, c1 ¼ 1 N m s, of the identical dampers supporting the top right shaft, c2 ¼ 24 N m s, as well as the logarithmic decrement of the entire system d ¼ 0.9. Calculate the damping coefficient c of the identical dampers supporting the bottom rigid shaft. Ignore the inertia contributions of the shafts. 3.28 A translatory mechanical system, as the one of Figure 3.16(a), is formed of a body of mass m ¼ 1 kg, a damper of viscous damping coefficient c ¼ 20N s/m, and a spring whose elastic force is expressed in terms of its deformation as fe ¼ 250$x3/(x5 þ 1) N. For an initial body displacement x(0) ¼ 0.6 m and zero initial velocity, v(0) ¼ 0 m/s, (i) Obtain the system’s mathematical model, calculate the body displacement x(t) using MATLAB ode45, and plot it. (ii) Use Simulink to confirm the plot of x(t). 3.29 Derive a mathematical model for the translatory mechanical system of Figure 3.67(a). Use MATLAB and Simulink to plot the time variation of coordinate x for m ¼ 0.5 kg, k1 ¼ 100 N/m, k2 ¼ 60 N/m, k3 ¼ 90 N/m, c1 ¼ 10 N s/m, and c2 ¼ 6 N s/m, when the system is subjected to: (i) The initial conditions x(0) ¼ 0.03 m and v(0) ¼ 0.5 m/s. (ii) The force f defined as in Figure 3.67(b) and acting on the mass m along the positive x direction (the force is not shown in Figure 3.67(a)). Consider zero initial conditions. f (N) k1

x

k2

c1

120 100 80

c2

0

m k3

(a)

2

45 (b)

Time (s)

FIGURE 3.67 (a) Translatory Mechanical System With Mass, Springs, and Dampers; (b) Time Variation of Force.

Problems

3.30 A pinion with fixed rotation axis passing through C2 in Figure 3.68 meshes with a translatory rack of mass m ¼ 0.8 kg. The rack, which is supported on two identical guides (bearings), each of damping coefficient c ¼ 12 N s/m, meshes with another solid pinion that, as a result, rolls without slippage on another fixed rack. (i) Demonstrate the system has one DOF and derive the equation of motion of the mobile rack. Use Simulink to plot the rack’s time displacement when we know the mass moments of inertia of the two pinions, J1 ¼ 0.001 kg m2 and J2 ¼ 0.0003 kg m2, the radii R1 ¼ 0.05 m, R2 ¼ 0.07 m, R ¼ 0.06 m, the stiffness k ¼ 160 N/m, and the force fA ¼ 80$(1  e0.5$t) N. (ii) Find the system’s damped natural frequency when the force fA is removed. Pinion

J2

C2 R Rack

fA

k

A

m

R1 C1 Pinion

R2

Guide

J1

Fixed rack

FIGURE 3.68 Mechanical System With Disc-Pulleys, Rack, and Spring.

3.31 Derive the mathematical model for the cart-pendulum of Example 3.12 and shown in Figure 3.30 by using Newton’s second law of motion. 3.32 The two-DOF mechanical system shown in Figure 3.69 consists of two point masses m1 and m2 connected by a rigid rod of length l and mass mr and two linear springs of stiffnesses k1 and k2. Using Lagrange’s equations, find the system’s mathematical model corresponding to the following coordinate selection: (i) The coordinates are the mass displacements, x1 and x2. (ii) The coordinates are the displacement x of the center of gravity C and the tilt angle q of the rod. 3.33 Calculate the natural frequencies and determine the associated modes for the mechanical system shown in Figure 3.69 of Problem 3.32, both analytically

143

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CHAPTER 3 Mechanical Systems

l l2

l1

x2

θ

x

x1

C

m1

m2

mr

k2 k1

FIGURE 3.69 TwoeDOF Lumped-Parameter Model of a Car’s Suspension.

and using MATLAB. Consider the following numerical values: m1 ¼ 0.6 kg, m2 ¼ 0.8 kg, mr ¼ 1 kg, k1 ¼ 110 N/m, k2 ¼ 120 N/m, l ¼ 0.15 m; ignore gravity effects. 3.34 Calculate the natural frequencies of the mechanical system of Problem 3.2 and sketched in Figure 3.44 if m1 ¼ 0.5 kg, m2 ¼ 0.3 kg, k1 ¼ k2 ¼ 120 N/m, R ¼ 0.06 m. Also plot the solution, q(t)drotation angle of rollerdand x2(t)dtranslation of body of mass m2 for an initial displacement x2(0) ¼ 0.01 m. 3.35 The mechanical microdevice of Figure 3.70 comprises two identical shuttle masses elastically supported by several identical beams. Propose a lumpedparameter mass-spring system for the x-direction motion of this system and derive the mathematical model of the natural response using the Lagrange’s equations method. Calculate the natural frequencies and the corresponding unit-norm eigenvectors, both analytically and by means of MATLAB. The identical beams have a length l ¼ 30 mm, rectangular cross section with a Rigid link Anchor

Anchor

Beam Shuttle mass

m

m

Symmetry axis x

FIGURE 3.70 Spring-Mass Mechanical Microsystem for Translatory Motion.

Problems

width (in-plane dimension) w ¼ 5 mm and thickness (out-of-plane dimension) h ¼ 1 mm. In addition, we know m ¼ 1.2 $ 1010 kg, Young’s modulus E ¼ 150 GPa, and mass density r ¼ 5600 kg/m3. 3.36 Derive the mathematical model of the mechanical system shown in Figure 3.71, which is formed of a pinion, two racks, a body, and several springs. The pinion comprises two concentric disks forming a single solid part that rotates around a fixed axis passing through its center C. The pinion meshes (without slippage) with two racks that consequently undergo translatory and parallel motions. Calculate the eigenvalues, unit-norm eigenvectors, and describe the modes of the mechanical system. Use the MATLAB specialized command to check the analytical eigenvalues and eigenvectors. Known are m1 ¼ 1 kg, m2 ¼ 2.1 kg, JC ¼ 0.05 kg m2, m3 ¼ 0.6 kg, k1 ¼ 120 N/m, k2 ¼ 100 N/m, k3 ¼ 110 N/m, R2 ¼ 2R1 ¼ 0.03 m. Rack

k1

k1

Spring

Body

Pinion

C

R1

JC R2

k2

k3 m3

m1

m2

Anchor

k2

Rack

FIGURE 3.71 Conservative Mechanical System With Pinion, Racks, Translating Mass, and LinearMotion Springs.

3.37 Consider the three-DOF system of Figure 3.72 with m1 ¼ m2 ¼ m3 ¼ m and k1 ¼ k12 ¼ k23 ¼ k3 ¼ k. (i) Derive the mathematical model of this system. (ii) Demonstrate that if bodies 1 and 2 move with equal-amplitude motions of opposite directions, the middle body 2 has to be at rest during one of the three natural motions of this system. Use the matrix-vector x1 k12

k1 m1

x3

x2 k23 m2

FIGURE 3.72 ThreeeDOF Spring-Mass Mechanical Translatory System.

k3 m3

145

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CHAPTER 3 Mechanical Systems

formulation and MATLAB to calculate the natural frequencies, their corresponding eigenvectors, and to confirm the assertion of this point for m ¼ 1 kg and k ¼ 100 N/m. (iii) Find and plot the motions x1(t), x2(t), and x3(t) when the only nonzero initial condition is x1(0) ¼ 0.01 m, and for m and k defined at (ii). 3.38 The simplified model of an airplane’s vertical vibrations is pictured in Figure 3.73(a) with the wings represented as identical beams of constant rectangular cross section (thickness h and width w) and of length l. Lumped masses model the wings inertia, resulting from both the actual engines placed at the wings free ends and from wings distributed inertia. The fuselage is modeled as a heavy body. The corresponding lumped-parameter model is shown in Figure 3.73(b). Knowing the wings parameters l ¼ 10 m, h ¼ 0.08 m, w ¼ 0.4 m, E ¼ 200 GPa, r ¼ 7500 kg/m3, one engine’s mass me ¼ 220 kg, as well as the fuselage mass M¼ 2300 kg, determine the natural frequencies and the corresponding eigenvectors of the z-direction vibration. Include the wings’ distributed inertia and ignore gravity.

m

z2

Engine

h

Fuselage

Wing

h z1

l

Wing

Engine

z3

M

z3

z1

l m

(a)

z2

(b)

FIGURE 3.73 (a) Simplified Airplane Wings and Fuselage in Vertical Vibrations; (b) Equivalent ThreeeDOF Lumped-Parameter Model.

3.39 The microcantilever represented in top view in Figure 3.74(a) vibrates out of the plane along the z direction. The equivalent, lumped-parameter model consists of the springs and heavy particles shown in Figure 3.74(b) that reproduce the z-axis motion of the three heavy, flexible segments. Knowing l1 ¼ 130 mm, l2 ¼ 80 mm, w ¼ 20 mm, h ¼ 20 mm (out-of-plane constant thickness), E ¼ 170 GPa, and r ¼ 6000 kg/m3, calculate the natural frequencies and determine the corresponding eigenvectors. 3.40 (i) Find the mathematical model of the double pendulum of Figure 3.75, which comprises two massless rods, two bobs, and two springs. The system performs small rotations in a vertical plane under the action of

Problems

l2

l2

z2

w z2

k2

k2 l1

z3

m2 m2

z3

z1

m1

w Beam

z1

k1

Anchor

(a)

(b)

FIGURE 3.74 (a) Top View of Three-Segment Microcantilever Vibrating Out-of-the-Plane; (b) Equivalent, Lumped-Parameter Conservative Model.

O2

O1 l 2l

k

Anchor

Rod Spring

k

Bob

m1

m2

f

FIGURE 3.75 Double Pendulum With Springs, Viscous Damping, and External Force.

a force f. Consider that the bobs motions encounter viscous-damping resistance; the coefficient of damping is c. (ii) Use Simulink to plot the time response of the system for m1 ¼ 0.5 kg, m2 ¼ 1.2 kg, k ¼ 190 N/m, c ¼ 110 N s/m, l ¼ 0.8 m, g ¼ 9.8 m/s2, f ¼ 150$sin(10t) N. 3.41 In the mechanical system of Problem 3.36 and shown in Figure 3.71, the spring k3 is replaced by a damper of viscous damping coefficient c ¼ 32 N s/m, as indicated in Figure 3.76. A force f ¼ 180 þ 20$e4$t$sin(10p$t) N acts now on the body of mass m3. All other parameters provided in Problem 3.36 being the same, plot the system’s time response by means of Simulink. 3.42 For the geared system shown in Figure 3.77, where ma is the actuation torque and ml is the load torque,

147

148

CHAPTER 3 Mechanical Systems

Rack

k1

c

k1 m3

m1 Spring Pinion

C

R1

JC Anchor

R2

k2

Body Damper

f

k2

m2 Rack

FIGURE 3.76 Mechanical System With Pinion, Racks, Translating Mass, Linear-Motion Springs, Damper, and External Force.

c

N 2 , J2

N3, J3

k Spring

c

ma Damper

c

Gear

ml c

N 1 , J1 N4, J4

FIGURE 3.77 Gear System With Inertia, Spring, Dampers, Actuation, and Load Moments.

(i) Establish the DOF and formulate the mathematical model. (ii) Use Simulink to plot the relevant rotation angles, as well as the angular velocities in terms of time when: N1 ¼ 26, N2 ¼ 20, N3 ¼ 30, N4 ¼ 36, J1 ¼ 4$105 kg m2, J2 ¼ 3$105 kg m2, J3 ¼ 5$105 kg m2, J4 ¼ 7$105 kg m2, c ¼ 2 N m s, k ¼ 20 N m, ma ¼ 250 N m, ml ¼ 30$sin(6p$t) N m. Consider the load moment ml has a dead zone that affects the moment over the 5 to þ5 N m range.

Suggested Reading L. Meirovitch, Fundamentals of Vibrations, McGraw Hill, New York, 2000. J.P. DenHartog, Mechanical Vibrations, Dover Publications, Inc., New York, 1985. D.J. Inman, Engineering Vibration, 4th Ed. Prentice Hall, Upper Saddle River, 2014. S.S. Rao, Mechanical Vibrations, 6th Ed. Prentice Hall, Upper Saddle River, 2016. D.W. Jordan and P. Smith, Mathematical Techniques, 4th Ed. Oxford University Press, Oxford, UK, 2008.

CHAPTER

Electrical Systems

4

OBJECTIVES With an approach similar to that of Chapters 2 and 3, this chapter covers the electrical systems by studying the following topics: • Electrical elementsdvoltage and current source, resistor, capacitor, inductor, and operational amplifier (op amp). • Electrical principles and methods of resistive sensing, as well as of electrostatic actuation/sensing used in microelectromechanical systems (MEMS). • Application of Ohm’s law, Kirchhoff’s laws, the energy method, the Lagrange’s equations, the mesh analysis method, and the node analysis method to derive mathematical models for dynamic electrical systems. • Analytical and MATLAB calculation of the electrical systems’ natural frequencies and eigenvectors, of the free response with losses, and of the forced response. • Application of Simulink to modeling, solving differential equations, and plotting the time response of electrical systems, including systems with nonlinearities. • Element and system mechanicaleelectrical analogy.

INTRODUCTION Electrical systems, also named circuits or networks, are designed as combinations of mainly three fundamental components: resistor, capacitor, and inductor. They are correspondingly defined by resistance, capacitance, and inductancedgenerally considered to be lumped-parameter properties. In addition to these primary electrical components, we also discuss the op amp in this chapter. The voltage or current sources, which produce the electron motion in an electrical circuit, are the counterparts of forces or moments in mechanical systems. The focus in this chapter is on formulating mathematical models with methods of electrical circuit analysis and on finding and characterizing the natural, free damped, and forced responses of electrical systems by using MATLAB and Simulink. Analogies between electrical and mechanical systems are also studied.

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00004-X Copyright © 2018 Elsevier Inc. All rights reserved.

149

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CHAPTER 4 Electrical Systems

4.1 ELECTRICAL ELEMENTS: VOLTAGE AND CURRENT SOURCES, RESISTOR, CAPACITOR, INDUCTOR, AND OPERATIONAL AMPLIFIER This section introduces the voltage and current sources, resistor, capacitor, inductor, and op amp as the basic electrical elements relating voltage to current. In addition to these electrical components, the companion website Chapter 4 presents the electrical transformer.

4.1.1 Voltage and Current Sources Similar to mechanical systems, where forces for translation and moments for rotation generate the system dynamics (mechanical motion), voltage and current sources are the inputs producing the dynamic response of electrical systems. We discuss here ideal (also named independent) sources in the following sense: An ideal voltage source delivers a prescribed voltage at its terminals that is independent of the current it provides to the other electrical elements. Conventionally, the positive direction of the current outside the voltage source is from the positive terminal. Similarly, an ideal current source produces a specified current that does not affect the voltage across the source and is not altered by the voltages across other electrical components in the circuit. Figure 4.1 shows schematic representations of voltage and current sources. Both sources can provide constant signals (in which case, they are termed direct current or dc sources) or time-varying ones, particularly sinusoidal (known as alternating current or ac sources). +

Current

v

V

–

i

(b)

(a)

+

Current

+ –

– (c)

FIGURE 4.1 Electrical Source Representations: (a) Variable Voltage; (b) Constant Voltage (Battery); (c) Current.

4.1.2 Resistor Elements The resistor, symbolized as in Figure 4.2, is an electrical element for which, ideally, the voltage across it, v, is proportional to the current passing through it, i, according to Ohm’s law: vðtÞ ¼ R $ iðtÞ (4.1) i

R

+ v

(a)

−

R

R

(b)

(c)

FIGURE 4.2 Resistor Representation: (a) In a Circuit Segment; (b) Symbol for Constant Resistance; (c) Symbol for Variable Resistance.

4.1 Electrical Elements: Voltage and Current Sources, Resistor

The proportionality constant is the resistance, whose unit is the ohm (U) in the International System (SI) with 1 U ¼ 1 V/A. For a wire of length l and crosssectional area A, the electrical resistance is calculated as R¼

r$l A

(4.2)

where r denotes the electric resistivity.

Mechanical Displacement Sensing Variable (or active) resistances are encountered in potentiometers, where a contact wiper can move over a resistor and change the resistor length connected to a (voltage) source, resulting in a change in resistance, as seen in Eq. (4.2). When connected in an electrical circuit, potentiometers can be used as sensors for translatory or rotary mechanical displacement, by producing a voltage that is proportional to the input displacement. nn

Example 4.1

Express the voltage detected by a voltmeter as a function of the translation displacement x, with the potentiometer of Figure 4.3(a). Similarly, find a relationship between the voltage measured and the input angular displacement q with the rotary potentiometer of Figure 4.3(b). Conductive guide Voltmeter Wiper

vx

i

Conductive guide Wiper

A x

P

B

Resistor

Voltmeter

P

i

θl r

θ

l v

(a)

A Voltage source B

–

+

Voltage source

vθ

+

v

–

Resistor

(b)

FIGURE 4.3 (a) Translation Potentiometer; (b) Rotation Potentiometer.

Solution

Figure 4.3(a) illustrates a variable-resistance translation potentiometer. The source voltage v is supplied to the resistor of length l, and the translatory-motion wiper, through its mobile contact, separates a voltage vx that is read by a voltmeter. Two voltageecurrent relationships can be written according to Ohm’s law:

v ¼ R$i ¼

r$l r$x $ i; vx ¼ Rx $ i ¼ $i A A

(4.3)

151

152

CHAPTER 4 Electrical Systems

Eliminating the current i between the two Eq. (4.3) yields

vx ¼

Rx v $v ¼ $x l R

(4.4)

which indicates that the output voltage vx varies proportionally to the input mechanical translation x. Similarly, a voltage v is applied to the circular conductor of total length l ¼ r$ql, which is sketched in Figure 4.3(b). The circular-motion wiper divides a voltage vq that is applied to the conductor segment of length s ¼ r$q. The source voltage v and measured voltage vq are

v ¼ R$i ¼

r$l r $ r $ ql r$s r$r$q $i ¼ $i ¼ $i $ i; vq ¼ Rq $ i ¼ A A A A

(4.5)

which result in

vq ¼

Rq v $v ¼ $q ql R

(4.6)

As indicated in Eq. (4.6), the rotary sensor generates a voltage vq that is proportional to the input mechanical rotation q. The conductive guide has small electrical resistance for both designs of Figure 4.3. nn

The resistor dissipates energy in a circuit, similar to damping elements in mechanical systems. The electric power dissipated by a resistor is vðtÞ2 (4.7) R where Ohm’s law, Eq. (4.1), has been used. The energy ER dissipated in time t by a resistor transforms into heat, according to the Joule effect: PR ðtÞ ¼ R $ iðtÞ2 ¼ vðtÞ $ iðtÞ ¼

ER ðtÞ ¼ PR ðtÞ $ t ¼ vðtÞ $ iðtÞ $ t ¼ R $ iðtÞ2 $ t ¼

vðtÞ2 $t R

(4.8)

Eqs. (4.7) and (4.8) considered the voltage v is constant. Resistors can be combined in series (as sketched in Figure 4.4(a)), in parallel (as illustrated in Figure 4.4(b)), or in a mixed (serieseparallel) manner. The companion website Chapter 4 contains the derivation of the equivalent series (Rs) and parallel (Rp) resistances, indicated collectively as equivalent resistance (Req) in Figure 4.4(c). For n different resistors, Rs and Rp are as follows: 8 > < R s ¼ R1 þ R2 þ / þ R n (4.9) 1 1 1 1 > :R ¼ R þ R þ / þ R p n 1 2

+

i

R1

i

v1

R2

i1

R1

i2

R2

i

–

+

–

i

+

Req

v

v2 v

(a)

(b)

FIGURE 4.4 Resistor Connections: (a) Series; (b) Parallel; (c) Equivalent Resistance.

(c)

–

4.1 Electrical Elements: Voltage and Current Sources, Resistor

nn

Example 4.2

The resistive Wheatstone bridge is a circuit used in sensing instrumentation. It contains four resistors R1, R2, R3, and R4das illustrated in Figure 4.5dof which one or more can be variable. For a source voltage v applied between points A and B of the bridge, calculate the voltage vCD between nodes C and D of the circuit in terms of v and the four resistances. In addition, determine R4 in terms of R1, R2, R3, when vCD ¼ 0 and the bridge is balanced.

+

v

–

C R1 A

R2 i1

i1

i2

i2

B

R4

R3 D

FIGURE 4.5 Resistive Wheatstone Bridge Circuit.

Solution

The currents i1 and i2 are expressed taking into account that the source voltage v is applied across both the R1eR2 series couple and the R3eR4 series pair, namely:

i1 ¼

v v ; i2 ¼ R1 þ R2 R3 þ R4

(4.10)

The voltage between nodes C and D is calculated as follows:



vCD ¼ vCA þ vAD ¼ R1 $ i1 þ R4 $ i2 ¼ ¼

 R4 R1  $v R3 þ R4 R 1 þ R 2

R2 $ R4  R 1 $ R3 $v ðR1 þ R2 Þ $ ðR3 þ R4 Þ

(4.11)

The same voltage can also be calculated as follows:

 vCD ¼ vCB þ vBD ¼ R2 $ i1  R3 $ i2 ¼ ¼

 R2 R3  $v R1 þ R2 R3 þ R 4

R2 $ R 4  R1 $ R3 $v ðR1 þ R2 Þ $ ðR3 þ R4 Þ

(4.12)

The currents of Eq. (4.10) have been used to derive vCD of Eqs. (4.11) or (4.12). When vCD ¼ 0, the Wheatstone bridge is called balanced and Eq. (4.11) or Eq. (4.12) gives

R4 ¼

R1 $ R 3 R2

(4.13) nn

153

154

CHAPTER 4 Electrical Systems

4.1.3 Capacitor Elements Capacitor elements in electrical systems store electrostatic energy and are functionally similar to springs in mechanical systems, which store elastic potential energy. The defining parameter of a capacitor (typical representations of which are given in Figure 4.6) is the capacitance, C, which can be constant or variable. In the SI units, capacitance is measured in farads (F).

+

i

C

C

C

C

C

(b)

(c)

−

v

(a)

FIGURE 4.6 Capacitor Representation: (a) In a Circuit Segment; (b) Symbols for Constant Capacitance; (c) Symbols for Variable Capacitance.

The capacitance relates the charge q that is separated on the two capacitor terminals to the voltage v as follows: qðtÞ ¼ C $ vðtÞ (4.14) Applying time derivation in Eq. (4.14) under the assumption that C is constant, and considering that the current is the time derivative of charge, iðtÞ ¼ dqðtÞ=dt results in dvðtÞ (4.15) dt Eq. (4.15) indicates that, when a capacitor is connected to a constant-voltage source, the current passing through it is zero. Integration of Eq. (4.15) yields iðtÞ ¼ C $

1 vðtÞ ¼ $ C

ðt iðtÞdt

(4.16)

0

Eq. (4.16) shows that the voltage reaches a value v over a period of time t, and that the voltage across a capacitor cannot change abruptly (instantaneously). A common capacitor configuration has two parallel-plate terminals and a dielectric between them. For this configuration, the capacitance is calculated as ε$A (4.17) g where ε is the dielectric permittivity, A is the plates’ superposition area, and g (the gap) is the distance between the plates. The capacitor electric power is C¼

PC ðtÞ ¼ vðtÞ $ iðtÞ ¼ vðtÞ $ C $

dvðtÞ dt

(4.18)

4.1 Electrical Elements: Voltage and Current Sources, Resistor

The energy stored by a capacitor in an elementary time dt is dEC ¼ PC$dt, which, by integration and consideration of Eq. (4.18), yields ðt

ðt

0

0

vðtÞ ð

dvðtÞ dt ¼ EC ðtÞ ¼ PC ðtÞdt ¼ C $ vðtÞ $ dt

1 C $ vðtÞdvðtÞ ¼ $ C $ vðtÞ2 2

0

2

1 1 qðtÞ (4.19) ¼ $ qðtÞ $ vðtÞ ¼ $ 2 2 C Eq. (4.19) assumed that C is constant. Introducing the electric flux l, which is related to voltage as v(t) ¼ dl(t)/dt, the electrostatic energy of Eq. (4.19) is also written as follows: 1 _ 2 EC ðtÞ ¼ $ C $ lðtÞ 2

+

+q

C1

–q + q

v1

C2

v2

–q

+ q1 –

+q

+ q2

C1 C2

(4.20)

– q1 – q2

–q

+

+q

Ceq

–q

–

v v

(a)

(b)

(c)

FIGURE 4.7 Capacitor Connections: (a) Series; (b) Parallel; (c) Equivalent Capacitance.

Similar to resistors, capacitors can be connected in series or in parallel, as illustrated in Figure 4.7. The series (Cs) and parallel (Cp) equivalent capacitances resulting from connecting n capacitors are derived in the companion website Chapter 4 as 1 1 1 1 ¼ þ þ / ; Cp ¼ C1 þ C2 þ / þ Cn Cs C1 C2 Cn

(4.21)

Actuation and Sensing in Microelectromechanical Systems Variable capacitors are utilized for either actuation or sensing purposes in MEMS. Variable plate capacitors can be designed using two principles, both having one plate that moves with respect to the other, which is fixed. According to one design principle (named parallel-plate or transverse-motion), the mobile plate moves along a direction perpendicular to the two plates, such that the common area A remains constant, but the gap varies, as shown in Figure 4.8(a). In contrast, the other principle (known as comb drive or longitudinal motion) utilizes designs where the gap is preserved, but the common area variesdsee Figure 4.8(b). Each of these design principles is illustrated through a few examples where either actuation or sensing (also referred to as transduction) is studied.

155

156

CHAPTER 4 Electrical Systems

Mobile plate

Mobile plate

Fixed plate

Fixed plate

(a)

(b)

FIGURE 4.8 Capacitive Transduction in MEMS: (a) Parallel-Plate or Transverse; (b) Comb-Drive or Longitudinal. nn

Example 4.3

Determine the force f that acts on the mobile plate of the capacitor shown in Figure 4.9. A constant voltage v is applied externally between the two plates, and the mobile plate moves by remaining parallel to the fixed one. The initial gap is g0, the common plate area is A, and the electric permittivity of the dielectric is ε. Plot the f/f0 ratio in terms of the x/g0 ratio (f0 is the initial force, and x is the mobile plate displacement). Fixed plate

– v g0

g

+ f

x Initial position Mobile plate

FIGURE 4.9 Variable-Gap, Transverse-Motion (Parallel-Plate) Variable Capacitor.

Solution The attraction electrostatic force between the two plates generates the motion of the mobile plate toward the fixed one. In MEMS, this principle is used to generate actuation by utilizing several parallel-connected pairs of the type shown in Figure 4.9 (a microphotograph of several such pairs is shown in Figure 4.8(a)) to multiply the force generated by a single pair. The capacitance after the mobile plate has displaced a distance x (and the gap is g) is

C¼

ε$A ε$A ¼ g g0  x

(4.22)

The electrostatic force can be expressed as the partial derivative of the electrostatic energy of Eq. (4.19) in terms of mechanical displacement (see details in the companion website Chapter 4) as

    vEC v C $ v2 v2 vC v2 v ε$A ε $ A $ v2 ¼ $ ¼ f ¼ ¼ $ ¼ vx vx 2 2 vx 2 vx g0  x 2ðg0  xÞ2

(4.23)

4.1 Electrical Elements: Voltage and Current Sources, Resistor

f / f0

x/g0

FIGURE 4.10 Plot of Force Ratio in Terms of Nondimensional Gap. As Eq. (4.23) indicates, the actuation force is proportional to the applied voltage squared and increases nonlinearly when the gap decreases. Initially, when x ¼ 0, the electrostatic force is

f0 ¼

ε $ A $ v2 2g20

(4.24)

Combination of Eqs. (4.23) and (4.24) results in

f =f0 ¼

g20 ðg0  xÞ

2

¼

1 ð1  x=g0 Þ2

(4.25)

and the corresponding plot is shown in Figure 4.10. For values of x close to g0, the force f reaches very large values (it is infinite when x / g0, just before the mobile plate collides with the fixed one). To avoid capacitor plate collision, the mobile plate in MEMS is connected to a spring that limits its motion range. nn

nn

Example 4.4

Consider that the variable capacitor of Figure 4.8(a) was designed to sense the displacement x of the mobile plate. Find the relationship between the voltage variation and the mechanical displacement. Assume that a bias dc voltage vb is applied to the variable capacitor circuit, as indicated in Figure 4.11, when external mechanical motion is the input to the mobile plate.

Solution

With no displacement of the capacitor mobile plate and the switch S in closed position, the bias voltage loads the capacitor with the charge

q ¼ C 0 $ vb

(4.26)

The original capacitance is determined by taking x ¼ 0 in Eq. (4.22):

C0 ¼

ε$A g0

(4.27)

157

158

CHAPTER 4 Electrical Systems

vb

–

+ S

C

Voltmeter

FIGURE 4.11 Electrical Circuit With Variable-Gap Capacitor and Bias Voltage Used as a Displacement Sensor. Combining Eqs. (4.26) and (4.27) results in

q¼

ε $ A $ vb g0

(4.28)

After the mobile plate of the capacitor moves a distance Dx (Dx is the same as x in the previous example; it is used to highlight the relationship with Dv, the voltage variation) toward the fixed plate, the voltage changes by a quantity Dv, and the capacitance changes from C0 to C. As a consequence, the total voltage sensed by the low-resistance meter of Figure 4.11 is

vb þ Dv ¼

q C

(4.29)

By substituting the capacitance C from Eq. (4.22) and the charge from Eq. (4.28) into Eq. (4.29), the latter equation becomes

vb þ Dv ¼ Eq. (4.30) results in

ðg0  DxÞ $ vb g0

vb Dv ¼  $ Dx g0

(4.30)

(4.31)

which shows that the voltage variation Dv is proportional to the mobile plate displacement Dx. The minus sign in Eq. (4.31) indicates that the voltage variation opposes the bias voltage; that is, the total voltage vb þ Dv decreases. nn

nn

Example 4.5 Find the relationship between the electrostatic attraction force and the displacement of the mobile plate in a comb-drive (constant-gap, longitudinal-motion) capacitive actuator, such as the one sketched in Figure 4.12, assuming a constant dc voltage is applied between the two capacitor plates.

Solution

When the two plates are superimposed over a length x, the capacitance is

C¼

ε$w$x g

(4.32)

4.1 Electrical Elements: Voltage and Current Sources, Resistor

where g is the constant gap and w is the width (dimension perpendicular to the drawing plane, assuming the plates are of rectangular shape) of the two capacitor plates. The electrostatic force f is produced by the electrical field lines, as shown in Figure 4.12, and is expressed based on EC of Eq. (4.19) as:

f ¼

  vEC v C $ v2 v2 vC ε $ w 2 ¼ $v ¼ ¼ $ vx 2g vx 2 2 vx

(4.33)

As Eq. (4.33) indicates, the electrostatic force is constant for a comb-drive capacitive actuator and is proportional to the square of the applied voltage. nn

nn

Example 4.6

The MEMS of Figure 4.12 is considered to function as a detector and sense the displacement x of the mobile capacitor plate, which is produced externally. Determine the connection between the mechanical displacement and the voltage variation produced by it. Assume that a bias dc voltage vb is applied to the capacitor circuit, as shown in Figure 4.11, and that an initial superposition of length x0 between the two plates occurs before the mechanical motion is to be measured.

Fixed plate

– v

+

x

Field line

g f

Mobile plate

FIGURE 4.12 Fixed-Gap, Longitudinal-Motion (Comb-Drive) Variable Capacitor.

Solution

The initial capacitance C0 and the capacitance C after the mobile plate moved a distance Dx are as follows:

C0 ¼

ε $ w $ x0 ε $ w $ ðx0 þ DxÞ ; C¼ g g

(4.34)

The constant charge can be expressed as

q ¼ C0 $ vb ¼

ε $ w $ x0 $ vb g

(4.35)

The total voltage that results when the mobile plate moved a distance Dx is the sum of vb and Dv, which is equal to the ratio of the charge q to the modified capacitance C:

vb þ Dv ¼

q ε $ w $ x0 $ vb g x0 ¼ $ ¼ $ vb C ε $ w $ ðx0 þ DxÞ x0 þ Dx g

(4.36)

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CHAPTER 4 Electrical Systems

Eq. (4.36) can be written as

  Dx x0 $ vb ¼  1  Dv ¼  $ vb x0 þ Dx x0 þ Dx

(4.37)

which shows that the voltage variation Dv increases with the mobile plate displacement Dx. The minus sign (as was the case with the transverse capacitive sensing) indicates that Dv opposes the bias voltage vb. nn

4.1.4 Inductor Elements An inductor, schematic representations of which are shown in Figure 4.13, consists of a conductive coil with N turns. In the case of a linear inductor, the voltage across the inductor is proportional to the current rate: vðtÞ ¼ L $

+

i

L

diðtÞ dt

(4.38)

L

L

(b)

(c)

– v

(a)

FIGURE 4.13 Inductor Representation: (a) In a Circuit Segment; (b) Symbols for Constant Inductance; (c) Symbols for Variable Inductance.

The proportionality constant, the inductance L, is a measure of this electrical component’s capacity to store magnetic energy. For an inductor (or solenoid) with N turns, coil diameter D, total wire length l, and magnetic permeability m, the inductance is expressed, as demonstrated in the companion website Chapter 4, as p $ m $ N 2 $ D2 (4.39) 4l In SI, the inductance unit is the henry (H). Eq. (4.38) indicates that the voltage across the inductor is zero for a constant current source connected to it. Eq. (4.38) also enables expressing the current as a function of voltage: ðt 1 iðtÞ ¼ $ vðtÞdt (4.40) L 0 L¼

Eq. (4.40) shows that the current through an inductor cannot change instantaneously, as it needs a time interval to modify its value. The power related to an inductor is expressed as PL ðtÞ ¼ vðtÞ $ iðtÞ ¼ L $

diðtÞ $ iðtÞ dt

(4.41)

4.1 Electrical Elements: Voltage and Current Sources, Resistor

The magnetic energy stored by the inductor over a period of time t  ðt 2 ðt ðt 1 1 2 $ EL ðtÞ ¼ PL ðtÞdt ¼ L $ iðtÞdiðtÞ ¼ $ L $ iðtÞ ¼ vðtÞdt 2 2L 0 0 0  ðt 2 1 dlðtÞ 1 $ $ dt ¼ $ lðtÞ2 ¼ 2L dt 2L 0

(4.42)

where the flux l, introduced in Section 4.1.3, has been used. Inductors, too, can be connected in series or in parallel. An equivalent inductance can be calculated for n different inductors, as derived in the companion website Chapter 4: 8 > < Ls ¼ L1 þ L2 þ / þ Ln (4.43) 1 1 1 1 > :L ¼ L þ L þ / þ L p n 1 2 where the subscripts s and p stand for series and parallel, respectively.

4.1.5 Operational Amplifiers Operational amplifiers (or op amps in technical jargon) are multielements (usually produced by integrated circuit technology) that can be connected with other electrical components in circuits to amplify voltage, isolate circuits, filter signals, or perform arithmetical and mathematical operations (addition, integration, differentiation, etc.). The simplified symbol of an op amp, which is shown in Figure 4.14, indicates its main feature of having two input ports (a negative one and a positive one) and therefore a differential input voltage. Voltages are usually measured with respect to the ground (which has zero voltage). It should be mentioned that an op amp has also a power circuit of its own, which in the simplified representation of Figure 4.14, as well as in all subsequent op amp system analysis is not employed. Inverting input v1 v2 Noninverting input

– +

Output

vo

Ground,v = 0

FIGURE 4.14 Simplified Symbol Representation of an Operational Amplifier.

As its name suggests, the op amp amplifies the differential input voltage v2  v1 to an output voltage by means of a factor K, known as gain or amplification: vo ¼ K $ vi ¼ K $ ðv2  v1 Þ ¼ K $ ðv1  v2 Þ

(4.44)

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CHAPTER 4 Electrical Systems

Eq. (4.44) is only valid for a limited range of the input voltage, beyond which the relationship between the output and input voltages is not linear as the op amp saturatesdmore details will follow shortly in this chapter. The op amp has some basic properties: •

• •

High input impedance (the notion of impedance is studied more thoroughly in Chapter 7, but it mainly refers to the electrical resistance posed to the passing of current), which is ideally equal to infinity. Low output impedance (which is ideally zero). High gain (normally in excess 104, but infinity ideally; actually the gain depends on frequency, when harmonic or sinusoidal signals are involved, and it can decrease significantly with the frequency increase).

4.2 ELECTRICAL SYSTEMS: CIRCUITS OR NETWORKS Electrical elements are connected by means of electrical conductors in electrical systems, known as circuits or networks, whose dynamic behavior is described by mathematical models expressed as differential and/or algebraic equations, similarly to mechanical systems. To obtain mathematical models of electrical systems, the Kirchhoff’s laws are mainly utilized in this chapter together with the mesh analysis method (the most popular probably) and the node analysis method. Alternatively, the energy method is applied to conservative electrical systems, as well as the Lagrange’s equations method for both conservative and nonconservative systems. Electrical networks are formed of circuit branches connecting at nodes. A branch is a continuous conductor line linking two nodes. A node is the intersection locus of at least three branch lines. Branches and nodes form closed contours, known as meshes or loops. While a loop in an electrical network is any closed contour of circuit branches, a mesh is a loop that contains no other loop within it. Figure 4.15 illustrates these definitions by a simple sketch showing just the circuit branches (lines) and not the electrical elements it usually contains. According to these definitions, the network has two nodes (B and E) and three branches (BAFE, BE, and BCDE), which are assembled in two meshes (ABEF and BCDE) and three loops (the two meshes and the external contour ACDF).

Node #1

F

Branch #2 Branch #1 A

D

E

Mesh #1

Mesh #2 B Node #2

FIGURE 4.15 Two Meshes and Three Loops in an Electrical Network.

Branch #3 C

4.2 Electrical Systems: Circuits or Networks

4.2.1 Kirchhoff’s Laws The two Kirchhoff’s laws are basic theorems/procedures for deriving mathematical models of electrical systems. The first Kirchhoff’s law, also known as the node or current law (with the acronym KCL from Kirchhoff’s current law), states that the algebraic sum of currents corresponding to all branches that converge into a node is zero. This principle is illustrated in Figure 4.16, and this is basically the mathematical expression of charge conservation. For the particular case of Figure 4.16, the first Kirchhoff’s law is ði1 þ i2 Þ  ði3 þ i4 Þ ¼ 0. i2

i1

i4

i3

FIGURE 4.16 Illustration of Kirchhoff’s (Node or Current) Law. nn

Example 4.7

Using KCL, confirm that the input current ii is identical to the output current io for the electrical network of Figure 4.17. F

io

D i8 i7

i5

i6

E

C i3

i4 B

i1

i2

ii

A

FIGURE 4.17 Electrical Network With Nodes and Currents.

Solution

Based on KCL, the following current balance equations can be written at nodes A, B, C, D, E, and F of Figure 4.17:

ii ¼ i1 þ i2 ; i2 ¼ i3 þ i4 ; i3 ¼ i5 þ i6 ; i8 ¼ i1 þ i5 ; i7 ¼ i4 þ i6 ; io ¼ i7 þ i8 (4.45) Successive substitutions, going from the last Eq. (4.45) to the second one, result in

io ¼ ði4 þ i6 Þ þ ði1 þ i5 Þ ¼ i4 þ i1 þ i3 ¼ i1 þ i2 ¼ ii

(4.46)

Eq. (4.46) shows that, as expected, the current entering the electrical network ii is equal to the current that exits the network io. nn

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CHAPTER 4 Electrical Systems

The second Kirchhoff’s law, also known as the mesh or voltage law (its acronym being KVL from Kirchhoff’s voltage law), states that the algebraic sum of voltages across individual electrical components is equal to the algebraic sum of source voltages in a mesh. nn

Example 4.8 Use KVL to derive the mathematical model of the single-mesh electrical circuit of Figure 4.18, which is formed of two voltage sources, a resistor, an inductor, and a capacitor. vR

v1

+ –

R

i

vL L

C vC

–

v2

+

FIGURE 4.18 Single-Mesh Electrical Circuit to Illustrate Kirchhoff’s Voltage Law.

Solution The arrow curve inside the circuit defines the (arbitrary) positive reference direction and is similar to the displacement coordinate in a mechanical system indicating the positive motion direction. The source v1 produces a positive voltage conforming to the direction of the arbitrarily chosen positive direction, which is also the direction of the current i. The other source, v2, opposes the positive direction; it therefore has a minus sign in the corresponding KVL equation:

ð diðtÞ 1 þ $ iðtÞdt ¼ v1  v2 vR þ vL þ vC ¼ v1  v2 or R $ iðtÞ þ L $ dt C

(4.47)

where Eqs. (4.1), (4.38), and (4.16) have been used for vR, vL, and vC, respectively. Eq. (4.47) is the mathematical model for the electrical system of Figure 4.18. A model that is expressed in terms of charge instead of current is obtained from Eq. (4.47) when using the chargeecurrent relationship iðt Þ ¼ dqðt Þ=dt:

L$

d2 qðtÞ dqðtÞ 1 1 þ $ qðtÞ ¼ v1  v2 or L $ q€ þ R $ q_ þ $ q ¼ v1  v2 þR$ 2 dt dt C C (4.48) nn

It should be noted that KVL is the electrical-domain counterpart of Newton’s second law of motion for mechanical systems. Indeed, forcing and voltage are similar, and voltages across inductors, resistors, and capacitors are similar to inertia, damping, and spring forces/moments, respectivelydmore on these similarities (analogies) at the end of this chapter.

4.2 Electrical Systems: Circuits or Networks

4.2.2 Configuration, Degrees of Freedom The configuration of an electrical system is determined by the minimum number of independent physical parameters (they can be currents, charges, voltages, voltage rates, etc.) that are sufficient to define the state of the electrical system, as well as the values of all other variables of interest. These independent parameters are named system coordinates, and they actually are degrees of freedom (DOF), similar to the DOF introduced for mechanical systems in Chapter 3. Figure 4.19 illustrates an electrical system formed of a voltage source, a resistor, and an inductor. vR

+

i

v

R L

vL

– FIGURE 4.19 Electrical Circuit as a Single-DOF System.

Assume the values of the voltage v, resistance R, and inductance L are known. The variables characterizing this system are the current i, as well as the voltages across the resistor and the inductor, vR and vL. It is known, however, that diðtÞ (4.49) dt These two relationships indicate that vR and vL are dependent parameters. As a consequence, the number of DOF (or independent parameters) is 3 (number of free variables)  2 (number of parameter connections) ¼ 1, so this system is a single-DOF one. The current i can be selected as the system’s DOF. A common approach to establishing the DOF of an electrical network is to represent the state of the system by a set of independent currents (or associated charges), such that KCL is automatically complied with. After that, a number of equations that is equal to the number of independent variables (DOF) are obtained by using KVL. These equations, which constitute the mathematical model of the electrical system, can further be solved for the unknown currents (or charges). This procedure is based on the mesh analysis approach. An alternate method is to find a set of independent voltages as DOF, such that KVL requirements are inherently satisfied. These voltages are determined by using KCL, and this approach follows the node analysis method. Both the mesh analysis and node analysis procedures are discussed in this Chapter. vR ¼ R $ i; vL ¼ L $

nn

Example 4.9 Determine the configuration (number of DOF) of the electrical network shown in Figure 4.20 in terms of mesh currents and then in terms of node voltages.

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CHAPTER 4 Electrical Systems

i1

+

R1

vA A

vR1

i2

R2

L i3

v

–

vL

vR2

B vB

FIGURE 4.20 Two-Mesh Electrical Network.

Solution The first approach attempts to identify a set of independent currents as DOF. For the circuit of Figure 4.20, the branch currents i1, i2, and i3 are DOF candidates. However, the three currents are connected at an adjacent node (either A or B) by means of KCL as follows:

i1 ¼ i2 þ i3

(4.50)

Eq. (4.50) plays the role of a constraint on i1, i2, and i3. As a consequence, the number of DOF for the electrical system of Figure 4.20 is 3  1 ¼ 2; the currents i1 and i2, for instance, can be the DOF of this electrical system. The second approach aims to find a set of independent voltages as DOF. The voltage across the inductor is expressed in terms of the voltages at nodes A and B as follows:

vL ¼ vA  vB

(4.51)

The KVL is formulated for the two meshes of Figure 4.20:



v R1 þ v L ¼ v v R2  v L ¼ 0



or

vR1 ¼ v  ðvA  vB Þ v R2 ¼ v A  v B

(4.52)

Eqs. (4.51) and (4.52) are three constraint relationships between the five would-be DOF: vR1 ; vR2 ; vL ; vA ; vB . As a result, the number of DOF is 5  3 ¼ 2; the voltages vA and vB can be selected as DOF. As seen in Eqs. (4.51) and (4.52), the voltage difference vA  vB is actually sufficient to calculate all other variable of interest. nn

A rule of thumb provides the number of independent coordinates (DOF) for an electrical network. The number of DOF, which is the number of independent currents or voltages (fluxes), is equal to the number of meshes in the electrical network and a quick proof follows here. It is demonstrated through network architecture that given a system with B branches and N nodes, there are B  N þ 1 linearly independent KVL equations in any network. Therefore, there are B  N þ 1 independent parameters (which are the DOF) these equations yield. At the same time, it is shown through similar electrical network topology reasoning that the number of meshes is also B  N þ 1. As a result, the number of meshes is identical to the number of DOF, and both are equal to B  N þ 1. According to these principles, the network of Figure 4.21(a) has B ¼ 3 branchesddenoted as (1), (2), and (3)dand N ¼ 2 nodes (A and D). It follows that the number of meshes, which is the same as the number of independent KVL equations and the number of DOF, is 3  2 þ 1 ¼ 2. The particular system of Figure 4.20 is of the general type (depicted in Figure 4.21(a)) and was shown to have two DOF by different approaches. Similarly, the network of Figure 4.21(b)

4.2 Electrical Systems: Circuits or Networks

A A

Mesh #1

Mesh #1

Mesh #2

Mesh #2

(2)

(1) (2)

(1)

C

(5)

Mesh #3

(3)

(4)

(3) D

(6)

D

E

(7)

F

(8)

G

Mesh #4 (9)

(a)

(b)

FIGURE 4.21 Electrical Network Configurations With: (a) Two Meshes; (b) Four Meshes.

is defined by B ¼ 9 branches and N ¼ 6 nodes. This results in 9  6 þ 1 ¼ 4 meshes and an equal number of independent KVL equations and of DOF. Simple inspection shows that the network of Figure 4.21(a) has 2 meshes, and the one of Figure 4.21(b) has 4 meshes.

4.2.3 Free Response This section studies the free, lossless, or natural response and the free response of electrical systems with losses, both describing the behavior of electrical networks in the absence of voltage or current sources.

Natural (Free Lossless) Response Electrical systems without voltage or current sources and with no resistors (with no energy dissipation) have a natural response. These systems contain only capacitors and inductors and are conservative systems; they display a natural response consisting of one or more natural frequencies, depending on the number of DOF. The differential equation(s) defining the natural response of an electrical system can be derived using Kirchhoff’s laws, applying the conservation of energy method or Lagrange’s equations, similarly to mechanical systems. MATLAB can also be utilized to determine the natural frequencies and eigenvectors associated with the natural response. These methods are discussed next for single- and multiple-DOF electrical systems.

Single-DOF Conservative Electrical Systems KVL and the conservation of energy method are used to derive the mathematical models of single-DOF conservative electrical systems. The natural frequency is subsequently calculated by searching for sinusoidal (harmonical) solutions of the mathematical model differential equation, as also discussed in Chapter 3. nn

Example 4.10

Consider the LC (resonant) circuit sketched in Figure 4.22(a), which is formed of an inductor and a capacitor. Derive its mathematical model using KVL and also by the conservation of energy method.

167

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CHAPTER 4 Electrical Systems

S1 i

L

C

v

(a)

+

S2

C

–

L

(b)

FIGURE 4.22 LC (InductoreCapacitor) System: (a) Actual Circuit; (b) Schematic for Charging the Circuit.

Determine the natural frequency of this electrical system for L ¼ 1 H and C ¼ 4 mF. In addition, express the charge variation with time, q(t), for the initial conditions q(0) ¼ q0 ¼ 0.0001 C and i(0) ¼ i0 ¼ 0 A.

Solution The single-DOF electrical system shown in Figure 4.22(a) is a conservative one, as no voltage or current sources injects energy into the system and no resistors draw energy from the system. We can assume that the capacitor is charged separately from a voltage source v (when switch S1 is closed and switch S2 is open) then disconnected from the source and connected to the inductor (by opening S1 and closing S2), as illustrated in Figure 4.22(b). After that, the capacitor discharges on the inductor, and a current i is produced through the circuit. Application of KVL to the electrical circuit of Figure 4.22(a) yields

L$

ð diðtÞ 1 þ $ iðtÞdt ¼ 0 dt C

(4.53)

which can be written in terms of the charge q as

L $ q€ þ

1 1 $ q ¼ 0 or q€ þ $q ¼ 0 C L$C

(4.54)

The total electrical energy stored by the inductor and the capacitor is

1 1 q2 1 1 q2 E ¼ EL þ EC ¼ $ L $ i2 þ $ ¼ $ L $ q_2 þ $ 2 2 C 2 2 C

(4.55)

Because the total electrical energy is conserved, the time derivative of the energy is zero; therefore, the following equation results from Eq. (4.55):

  dE 1 ¼ q_ $ L $ q€ þ $ q ¼ 0 dt C

(4.56)

The zero condition of Eq. (4.56) should be valid at all times, but the charge rate (the current) is not zero at all times; therefore, Eq. (4.56) is satisfied only when the parenthesis is zero, but this is identical to Eq. (4.54), which is the mathematical model of the electrical system obtained by KVL. By comparing Eq. (4.54) with the generic equation that modeled the free undamped response of a single-DOF mechanical system and was of the form: x€ þ u2n $ x ¼ 0, it follows that the natural frequency is ffi pffiffiffiffiffiffiffiffiffi un ¼ 1 L $ C ¼ 500 rad=s. The free response of the electrical system consists of a harmonic vibration at the natural frequency, quite similar to the case of a spring-mass mechanical system:

q ¼ Q $ sinðun $ t þ 4Þ

(4.57)

4.2 Electrical Systems: Circuits or Networks

The amplitude Q and phase angle 4 are calculated from initial conditions as

8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > u2n $ q20 þ i20 > > > 

> > 1 un $ q0 > > 4 ¼ tan : i

(4.58)

0

For the given initial conditions, Eqs. (4.57) and (4.58) result in q ¼ 0.0001$sin (500$t þ p/2) ¼ 0.0001$cos (500$t)C. nn

Multiple-DOF Conservative Electrical Systems Kirchhoff’s laws or Lagrange’s equations are applied here to derive the mathematical models of multiple-DOF conservative electrical systems and to determine their natural frequencies. The companion website Chapter 4 presents the conservation of energy method. According to KVL, all the voltages across inductors and capacitors add up to zero because there are no source voltages. Lagrange’s Equations Depending whether mesh analysis (and mesh charge DOF) or node analysis (and node voltage or node flux DOF) is used, the Lagrange’s equations are formulated differently, as discussed below. Mesh Analysis In mesh analysis, Lagrange’s equations for a conservative electrical system are: ! ! d vL vL d vEL vEC ¼ 0 or ¼0 (4.59)  þ dt vq_j vqj dt vq_j vqj where qj (j ¼ 1, 2, ., n) are the electrical system’s generalized coordinates or DOF, and L is the system Lagrangian (introduced in Chapter 3). The generalized coordinates are the independent charges, and therefore the same symbol qj is employed for both the generalized coordinates and the charges. The Lagrangian is expressed as L ¼ EL  EC, where EL is the energy term adding up all inductor contributions, and EC is the term collecting the energy of all capacitors.  The energy of an inductor is calculated as L $ q_2 2, which indicates dependence on charge derivative and justifies inductive energy selection as the positive term in L; this term is used with the time derivative operator and will result to the second time derivative of the generalized coordinate. The energy stored by a capacitor is  q2 ð2CÞ, this indicates dependence on the generalized coordinate and justifies capacitive energy selection as the other (negative) terms in L. Note that EL and EC are counterparts of the mechanical kinetic energy T and potential energy U, respectively, that are used in Chapter 3 to enable application of Lagrange’s equations to mechanical systems. Node Analysis In node analysis, the generalized coordinates qj are the independent fluxes lj, and the Lagrangian L is expressed as L ¼ EC  EL. The reason of this energy reversal in comparison to the Lagrangian that used mesh charges as DOF and where L ¼ EL  EC is that the energy of a capacitor is expressed as

169

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CHAPTER 4 Electrical Systems  2 C $v2 2 ¼ C $ l_ 2dEq. (4.20), whereas the energy of an inductor is formulated as l2 ð2LÞdEq. (4.42). It can be seen that the capacitive energy depends on the time derivative of the generalized coordinate l and therefore plays the role of the positive term in L, whereas the inductive energy depends on the generalized coordinate and is selected as the negative energy term in L. For j ¼ 1, 2, ., n, Lagrange’s equations are: ! ! d vL vL d vEC vEL ¼ 0 or ¼0 (4.60)  þ dt vl_ j vlj dt vl_ j vlj nn

Example 4.11 Derive the mathematical model of the two-mesh electrical system sketched in Figure 4.23 using the mesh analysis in conjunction with Kirchhoff’s laws and the Lagrange’s equations.

A

vA

C1

C2

vB B L2

L1

i2

i 3 = i1 – i 2 C vC = 0 i1

FIGURE 4.23 Two-Mesh, Conservative Electrical Circuit.

Solution This is a two-DOF system due to its two meshes. KVL is written for the two meshes of the electrical circuit of Figure 4.23 as

  8 ð di1 ðtÞ 1 di1 ðtÞ di2 ðtÞ > > L þ  $ $ i ðtÞdt þ L $ ¼0 > 1 1 2 < dt C1 dt dt   ð > 1 di1 ðtÞ di2 ðtÞ > > : $ i2 ðtÞdt  L2 $  ¼0 C2 dt dt

(4.61)

Substituting the currentecharge relationships, i1 ¼ q_1 ; i2 ¼ q_2 , in Eq. (4.61) results in the following mathematical model of the given circuit:

8 1 > > > < ðL1 þ L2 Þ $ q€1  L2 $ q€2 þ C $ q1 ¼ 0 1 > 1 > > : L2 $ q€1 þ L2 $ q€2 þ $ q2 ¼ 0 C2

(4.62)

The electric energy corresponding to the four elements of the circuit is

2 1 1 1 q2 1 q2 EL ¼ L1 $ q_21 þ L2 $ q_1  q_2 ; EC ¼ $ 1 þ $ 2 2 2 2 C1 2 C2

(4.63)

4.2 Electrical Systems: Circuits or Networks

The partial derivatives needed in Lagrange’s equations are determined using Eq. (4.63) as follows:

vEL vEL vEC 1 ¼ L1 $ q_1 þ L2 $ ðq_1  q_2 Þ; ¼ L2 $ ðq_1  q_2 Þ; ¼ $ q1 ; C1 vq_1 vq_2 vq1 vEC 1 ¼ $ q2 C2 vq2 (4.64) Substituting Eq. (4.64) into the generic Lagrange’s equations Eq. (4.59) corresponding to mesh analysis (for j ¼ 1, 2) results in Eq. (4.62)dthe mathematical model of the electrical system of Figure 4.23. Eq. (4.62) is written in vector-matrix form as follows:

"

€ þ ½B1  $ fqg ¼ f0g with ½A1  ¼ ½A1  $ fqg 2

1 6 C1 6 ½ B1  ¼ 6 4 0

3

L1 þ L 2

L2

L2

L2

#

( ) 0 7 q1 7 7 ; fqg ¼ 1 5 q2 C2

; (4.65)

where [A1] is an inductance-based matrix, and [B1] is a capacitance-defined matrix. Note that the dynamic matrix [D1] corresponding to the differential Eq. (4.65) is calculated in a manner similar to the one used in Chapter 3 to calculate the dynamic matrix corresponding to a mechanical system, which was shown to be

€ þ ½K $ fxg ¼ f0g; ½D ¼ ½M1 $ ½K ½M $ fxg

(4.66)

With [A1] and [B1] being the counterparts of [M] and [K], respectively, the dynamic matrix corresponding to Eq. (4.65) is calculated:

2

1 6L $C 1 1 6 ½D1  ¼ ½A1 1 $ ½B1  ¼ 6 4 1 L1 $ C1

1 L1 $ C2 L1 þ L2 L1 $ L2 $ C1

3 7 7 7 5

(4.67)

nn nn

Example 4.12 Derive the mathematical model of the two-mesh electrical system sketched in Figure 4.23 of Example 4.11 by the node analysis method in conjunction with Kirchhoff’s laws and the Lagrange’s equations.

Solution As shown in Example 4.11, the electrical system has two DOF. In node analysis, the DOF are fluxes or voltages, and we can use the voltages at nodes A and B in Figure 4.23, which are vA and vB, to be the two DOF. After selecting C as the reference node, with vC ¼ 0, the two voltages vA and vB are sufficient to express the other variables of the problem, such as currents or charges. The currents through the four elements of Figure 4.23 are:

ð 1 i1 ¼ $ ð0  vA Þdt ¼ C1 $ ðv_A  v_B Þ; i2 ¼ C2 $ ðv_B  0Þ; L1 ð 1 i3 ¼ $ ðvB  0Þ $ dt L2

(4.68)

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CHAPTER 4 Electrical Systems

Noting that, according to KCL, i1 ¼ i2 þ i3, Eq. (4.68) can be written as

8 ð 1 > > > < C1 $ ðv_A  v_B Þ ¼ L $ vA dt 1 ð > 1 > > : C1 $ ðv_A  v_B Þ ¼ C2 $ v_B þ $ vB dt L2

(4.69)

Using the voltageeflux relationship v(t) ¼ dl(t)/dt, Eq. (4.69) becomes

8 1 € € > < C1 $ lA  C1 $ lB þ L $ lA ¼ 0 1

> :

(4.70)

€ þ ðC1 þ C2 Þ $ l € þ 1 $ lB ¼ 0 C1 $ l A B L2

Eq. (4.70) form the mathematical model of the electrical system shown in Figure 4.23, with fluxes as the unknowns. Eq. (4.70) can be written in matrix form as:

 € þ ½B2  $ flg ¼ f0g with ½A2  ¼ ½A2  $ l

C1

C1

C1

C1 þ C2

#

;

3

2

1 6L 6 1 ½B2  ¼ 6 4 0

"

0 7  lA 7 ; l ¼ 7 lB 1 5 L2

(4.71)

The corresponding dynamic matrix is

2

C1 þ C2 6L $C $C 1 2 6 1 ½D2  ¼ ½A2 1 $ ½B2  ¼ 6 4 1 L1 $ C2

3 1 L2 $ C2 7 7 7 1 5 L2 $ C2

(4.72)

To derive the mathematical model by means of Lagrange’s equations, the capacitive energy and the inductive energy are expressed based on Eqs. (4.20) and (4.42), namely:

2 1 1 1 1 2 EC ¼ $ C1 $ l_ A  l_ B þ $ C2 $ l_ B ; EL ¼ $ l2 þ $ l2 2 2 2L1 A 2L2 B

(4.73)

The following partial derivatives are needed in Lagrange’s Eq. (4.60) with the assumption that the two generalized coordinates are q1 ¼ lA and q2 ¼ lB:



vEC

vEC ¼ C1 $ l_ A  l_ B ; ¼ C1 $ l_ A  l_ B þ C2 $ l_ B ; vl_ A vl_ B vEL 1 vEL 1 ¼ $ lA ; ¼ $ lB vlA L1 vlB L2

(4.74)

Eq. (4.74) is now used in the generic flux-defined Lagrange’s equations of Eq. (4.60), resulting in Eq. (4.70)dthe mathematical model obtained by Kirchhoff’s laws. nn

4.2 Electrical Systems: Circuits or Networks

nn

Example 4.13 Determine the natural frequencies (eigenvalues) and the eigenvectors for the electrical circuit of Figure 4.23 and Example 4.11, both analytically and using vector-matrix formulation method with MATLAB’s eig command. Consider L1 ¼ L2 ¼ L ¼ 0.05 H, C1 ¼ C2 ¼ C ¼ 0.0008 F.

Solution The free response of a conservative system requires solution of the harmonic type:

q1 ¼ Q1 $ sinðu $ t þ 4Þ; q2 ¼ Q2 $ sinðu $ t þ 4Þ

(4.75)

Substitution of Eq. (4.75) into Eq. (4.62) yields the following algebraic equations system:

8  > 1 > 2 >  2u $ L þ $ Q1 þ u2 $ L $ Q2 ¼ 0 > < C   > 1 > 2 2 > > : u $ L $ Q1 þ  u $ L þ C $ Q2 ¼ 0

(4.76)

after canceling the sine factors out. The determinant of the system above needs to be zero to get nontrivial solutions Q1 and Q2. With the particular expression of the electrical components of this example, the resulting characteristic equation is:

L2 $ C 2 $ u4  3L $ C $ u2 þ 1 ¼ 0

(4.77)

whose roots, the natural frequencies, are:

un1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 3 5 3þ 5 ¼ ; un2 ¼ 2L $ C 2L $ C

(4.78)

with the numerical values un1 ¼ 97.720 rad/s and un2 ¼ 255.834 rad/s. The following amplitude ratio is obtained from the second Eq. (4.76):

r¼

Q2 u2 $ L u2 $ L $ C ¼ 2 n ¼ 2 n Q1 un $ L  1=C un $ L $ C  1

(4.79)

For the two natural frequencies found above, this ratio becomes

r1 ¼

  Q2  Q21 Q2  Q22 ¼ ¼ 0:618; r ¼ ¼ ¼ 1:618 2 Q1 un ¼un1 Q11 Q1 un ¼un2 Q12

(4.80)

Using the ratios of Eq. (4.80) in conjunction with the unit-norm conditions:

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Q211 þ Q221 ¼ 1; Q212 þ Q222 ¼ 1

(4.81)

173

174

CHAPTER 4 Electrical Systems

results in the following unit-norm eigenvectors:

 fQgjun ¼un1 ¼

¼

Q21 Q11

¼

8 9 > > 1 > > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > 2 < 1 þ r1 > = > > r1 > > > qffiffiffiffiffiffiffiffiffiffiffiffi ffi> > > > > > : 1 þ r2 > ;

8 9 > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > < 1 þ r 22 > = > > r2 > > > qffiffiffiffiffiffiffiffiffiffiffiffi ffi> > > > > 1 þ r2 > > : ;

 ¼

0:851 0:526

 ; fQgjun ¼un2 ¼

Q22

Q12

1

 ¼

0:526

0:851

2

(4.82) The dynamic matrix [D1] of Eq. (4.67) is used with the particular numerical values of the four electrical elements and with MATLAB’s eig command written as follows: >>[V,Dn]=eig(D1) The resulting matrix [V] contains the two eigenvectors of Eq. (4.82) as columns, whereas the diagonal matrix [Dn] has the eigenvalues l1 ¼ u2n1 ¼ 9549:151 and l2 ¼ u2n2 ¼ 65; 450:850 on the main diagonal. These eigenvalues result in the natural frequencies that have been obtained analytically. Eq. (4.82) indicate that during the first modal motion, the charges q1 and q2 flow in opposite directionsdas illustrated by the opposite signs of the first eigenvector components. The amplitude of q1 is larger than that of q2. During the second modal motion, the charges vibrate along identical directions. The amplitude of q1 is smaller than the amplitude of q2. nn

nn

Example 4.14 Calculate the natural frequencies (eigenvalues) and the eigenvectors for the electrical circuit of Figure 4.23 using the flux-based mathematical model of Example 4.12, both analytically and with the dynamic matrix. Utilize the numerical parameters of Example 4.13: L1 ¼ L2 ¼ L ¼ 0.05 H, C1 ¼ C2 ¼ C ¼ 0.0008 F.

Solution

The harmonic definition of the two fluxes lA and lB is as follows:

lA ¼ LA $ sinðu $ t þ 4Þ; lB ¼ LB $ sinðu $ t þ 4Þ

(4.83)

Substituting them into the mathematical model of Eq. (4.70) results in the following:

 8 2

1 > 2 > > <  u $ C þ L $ LA þ u $ C $ LB ¼ 0   >

2 1 > 2 > : u $ C $ LA þ  2u $ C þ $ LB ¼ 0 L

(4.84)

4.2 Electrical Systems: Circuits or Networks

after canceling the sine factors out. The system of Eq. (4.84) has nontrivial solutions LA and LB when its determinant is zero. With L1 ¼ L2 ¼ L, C1 ¼ C2 ¼ C, the zero-determinant condition applied to Eq. (4.84) results in the characteristic equation given in Eq. (4.77), which was obtained with the chargedbased mathematical model of the electrical system of Figure 4.23. As a consequence, the natural frequencies of Example 4.13 are also calculated here. The expectation is that the flux model results in the same eigenvectors determined with the charge model. The following flux amplitude ratio results from the first Eq. (4.84):

r¼

LA u2 $ C u2 $ L $ C ¼ 2 n ¼ 2 n LB un $ C  1=L un $ L $ C  1

(4.85)

The amplitude ratio of Eq. (4.85) is identical to the charge amplitude ratio derived in Eq. (4.79). Because of that, and of the same natural frequencies, the two values of r resulting from Eq. (4.85) are the ones obtained in Example 4.13, namely: r1 ¼ 0.618 and r2 ¼ 1.618. The unit-normal eigenvectors resulting from Eqs. (4.85) and (4.81) are as follows:

 fLgjun ¼un1 ¼

¼

LA1 LB1

¼

9 8 r1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > 2> > = < 1 þ r1 > > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > ; : 1 þ r2 >

8 9 r2 > > > > > > qffiffiffiffiffiffiffiffiffiffiffiffi2ffi > > > > > < 1 þ r2 > = > > 1 > > > qffiffiffiffiffiffiffiffiffiffiffiffiffi > > > > > > : 1 þ r2 > ;

 ¼

0:526 0:851

 ; fLgjun ¼un2 ¼

LA2

LB2

1

 ¼

0:851 0:526

2

(4.86) Due to particular manner of defining r in Eq. (4.85), the ratio of the first component amplitude to the second component amplitude, there is a reversal of components in Eq. (4.85) in comparison to Eq. (4.82), which were formulated with the ratio of the second component amplitude to the first component amplitude, see Example 4.13. Using MATLAB’s eig command with the dynamic matrix [D2] of Eq. (4.72) and the given numerical values of the electrical element parameters, the natural frequencies of Example 4.13 and the eigenvectors of the analytical Eq. (4.86) are retrieved. nn

Free Damped Response Let us consider an RLC series circuit without a source, such as the one of Figure 4.24. The mathematical model is obtained by means of KVL as ð diðtÞ 1 R 1 þ R $ iðtÞ þ $ iðtÞdt ¼ 0 or q€ þ $ q_ þ $q ¼ 0 (4.87) L$ dt C L L$C

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CHAPTER 4 Electrical Systems

L

C

i

R

FIGURE 4.24 Single-Mesh Electrical Circuit With Resistor, Inductor, and Capacitor.

_ has been used. Eq. (4.87) can be written in standard form as where iðtÞ ¼ qðtÞ q€ þ 2x $ un $ q_ þ u2n $ q ¼ 0

(4.88)

where un is the natural frequency, and x is the electric damping ratio. These two parameters result from Eqs. (4.87) and (4.88) as follows: 1 R ; 2x $ un ¼ (4.89) L$C L Eq. (4.88) and Eq. (3.31), which defines the free damped vibrations of a single-DOF mechanical system, are similar. More about similarity (and analogy) between systems of different kinds is discussed at the end of this Chapter, but it should be mentioned that the electrical damping ratio was introduced in a way that is similar to the mechanical damping ratio. The dissipative nature of x is highlighted by the presence of R (which accounts for energy dissipation in an electrical system), and the value of x can be determined directly in terms of the actual electrical parameters, by combining the two Eq. (4.89) in rffiffiffiffi rffiffiffiffi R C L x¼ (4.90) ; for x ¼ 1/Rcr ¼ 2 $ 2 L C u2n ¼

Eq. (4.90) also shows that for critical damping (x ¼ 1), there is a critical resistance value Rcr that depends on L and C. It has been shown in mechanical systems that for 0 < x < 1 there is underdamping, for x > 1 there is overdamping, and for x ¼ 1 the damping is critical. The same definitions apply for electrical damping. For 0 < x < 1, the solution to Eq. (4.88) is q ¼ Q1 $ es $ t $ cosðud $ tÞ þ Q2 $ es $ t $ sinðud $ tÞ ¼ Q $ es $ t $ sinðud $ t þ 4Þ (4.91) with s ¼ x $ un ; ud ¼ un $

qffiffiffiffiffiffiffiffiffiffiffiffiffi 1  x2

(4.92)

where ud is the damped frequency, and Q1, Q2 (or Q and 4) are found from the initial conditions. The cases of critical damping (x ¼ 1) and overdamping are presented in the companion website Chapter 4.

4.2 Electrical Systems: Circuits or Networks

nn

Example 4.15

In the electrical circuit of Figure 4.24, the resistor has a resistance R ¼ 250 U. It is determined experimentally that the system’s damping ratio is 0.45, and its natural frequency is 1200 Hz. How should the system’s capacitance C and inductance L change for the damping ratio to decrease by 25% and the natural frequency to increase by 30%, when the same resistance is used? Plot the charge variation with time in its latter configuration when the capacitor is initially charged by a voltage source of v ¼ 100 V, which is subsequently removed.

Solution Eq. (4.89) enables expressing the inductance and capacitance as

L¼

R R R 1 2x x ¼ ¼ ; C¼ ¼ ¼ 2 2x $ un 2x $ ð2p $ fn Þ 4p $ x $ fn L $ un R $ un p $ R $ fn (4.93)

where fn is the natural frequency measured in Hz. With the original-design numerical values, these parameters are L1 ¼ 0.0368 H and C1 ¼ 4.77  107 F. The changed values of the damping ratio and natural frequency are x2 ¼ 0:75x1 ¼ 0:3375; fn2 ¼ 1:3fn1 ¼ 1560 Hz. Eq. (4.93) is used again with these new parameter values, which yield L2 ¼ 0.0378 H and C2 ¼ 2.75  107 F. Eqs. (4.91) and (4.92) give the charge equation as a function of time, and the constants Q1 and Q2 are determined by using the initial conditions:

 dqðtÞ qð0Þ ¼ v $ C; ¼0 dt t¼0

(4.94)

Eq. (4.94) states that the initial charge is furnished by the voltage v applied to the capacitor, and the initial current through the circuit is zero. The charge derivative results from Eq. (4.91) as follows:

q_ ¼ Q $ est $ ½s $ sinðud $ t þ 4Þ þ ud $ cosðud $ t þ 4Þ

(4.95)

With q of Eq. (4.91), its time derivative of Eq. (4.95) and the initial conditions of Eq. (4.94), the following system is obtained:



Q $ sin4 ¼ v $ C s $ sin4 þ ud $ cos4 ¼ 0

whose solution is:

4 ¼ tan

1

 u  d  ¼ tan1 s

pffiffiffiffiffiffiffiffiffiffiffiffiffi! v$C v$C 1  x2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; Q¼ sin4 x 1  x2

(4.96)

(4.97)

or 4 ¼ 1.2265 rad and Q ¼ 3.08456$105 C for the values of x2 and C2. The charge is plotted as a function of time in Figure 4.25; it can be seen that the charge settles to the steady-state value of zero after approximately 2 ms. The MATLAB dsolve command can also be employed here to solve a second-order differential equation. For the present example, the command line that generates q(t) is >> q = dsolve(’D2q+2*xi*omn*Dq+omn^2*q=0’,’q(0)=v*c’,’Dq(0)=0’) The charge is returned symbolically, so the pertinent numerical values have to be subsequently substituted in (with the subs and eval commands, for instance) and the result plotted as shown in Figure 4.25.

177

CHAPTER 4 Electrical Systems

Charge (C)

178

Time (s)

FIGURE 4.25 Charge as a Function of Time. nn

4.2.4 Forced Response Electrical systems with voltage or current sources result in forced dynamic models. This section studies such systems by first deriving mathematical models using Kirchhoff’s laws and Lagrange’s equations. Solutions to these models are subsequently determined by MATLAB or Simulink. Single-DOF and multiple-DOF electrical networks are studied.

Single-DOF Systems Several examples are analyzed in this section, including a few op amp applications.

Operational Amplifier Circuits As mentioned previously, circuits with op amps can realize various functions, such as inversion, amplification, addition, integration, differentiation, and filtering, to mention a few capabilities. We discuss next some of these functions. Chapters 7e13 include other op amp applications and examples. Inverting Amplifier Circuits In a circuit with two resistors, such as the one of Figure 4.26, the op amp can change the sign and value of the input voltage at the output port. Based on KCL, the currents identified in Figure 4.26 are connected as i1 ¼ i2 þ i3 . However, it has been shown that the input impedance of an ideal op amp is infinity; as a result, no current passes through the op amp, that is, i3 ¼ 0, which means that i1 ¼ i2. The two currents can be expressed by means of Ohm’s law as vi  vA vA  vo i1 ¼ ; i2 ¼ (4.98) R1 R2

4.2 Electrical Systems: Circuits or Networks

R2 R1 vi i1

i2

i3 A B

– +

vo

Ground

FIGURE 4.26 Basic Operational Amplifier in a Negative Feedback Circuit.

Because terminal B is grounded, it follows that vB ¼ 0; but vB ¼ v2 (see Figure 4.14), which means v2 ¼ 0 and vi ¼ v1. The inputeoutput voltage relationship introduced in Eq. (4.44) becomes 1 vo ¼ K $ vA or vA ¼  $ vo K Combining Eqs. (4.98) and (4.99) yields vi vo vo vo þ ¼  R 1 K $ R1 R2 K $ R 2

(4.99)

(4.100)

Regularly, the gain K of an op amp is high (larger than 104); therefore, the two terms in Eq. (4.100) with K to the denominator are very small and can be neglected; as a consequence, Eq. (4.100) changes to vo R2 ¼ vi R1

(4.101)

Eq. (4.101) indicates that the input and output voltages have different signs, which proves the inverting effect of this particular op amp circuit. In addition, if R2 > R1, there is an amplification effect as well, which justifies the name of inverting amplifier. The effect of amplifying and changing the sign of the input voltage at the output can be used as a logical operation in analog computing, illustrative examples are included in the companion website Chapter 4. If we now analyze Eq. (4.98) in conjunction with Eq. (4.101) by taking into consideration that i1 ¼ i2, we conclude that vA ¼ 0. At the same time, as Figure 4.26 shows, the positive input terminal is connected to the ground; therefore, its voltage is also zero. It follows that the voltages at the two input terminals, vA and vB, are zero. Actually, this property can be extended for all the op amp circuits, where the output is fed back to the negative input terminal. These systems have negative feedback, with the important consequence that the voltages at the two input terminals are always equal.

179

180

CHAPTER 4 Electrical Systems

nn

Example 4.16

A sinusoidal input voltage vi ¼ 20$sin (5t) V is applied to the two-stage op amp circuit of Figure 4.27, where R2 ¼ 2$R1 and R4 ¼ 5$R3. Knowing that the output voltage saturates for voltages smaller than 100 V and larger than 150 V, use Simulink to plot the output voltage and compare it to the output voltage for the ideal amplifier (without saturation).

R2

R4

R1 vi

R3

– +

– +

v

vo

FIGURE 4.27 Two-Stage Operational Amplifier System.

Solution

The output voltage vo, input voltage vi, and interstage voltage v are related as

vo v vo ¼ $ ¼ vi vi v





   R2 R4 R2 $ R4 ¼K $  ¼ R1 R3 R1 $ R3

(4.102)

which results in a gain (amplification) of K ¼ 10. For an ideal amplifier, the relationship between the output voltage and the input voltage is a linear one, as illustrated in Figure 4.28(a). For input voltages exceeding some threshold values, vi < vi,min and vi > vi,max, the output voltage no longer increases due to saturationdthis situation is also illustrated in Figure 4.28(a) and is introduced in Chapter 1. Saturation itself does not occur in reality at such precise limits, and there is a transition

vo

Ideal amplifier

Output,y Saturation

vo,max ymax = umax

Saturation vi,min

u min

K O

vi,max

vi

O

Saturation

umax

ymin = umin vo,min

(a)

Saturation

(b)

FIGURE 4.28 Saturation-Type Nonlinearity in (a) Amplifier; (b) Simulink Saturation Block.

Input, u

4.2 Electrical Systems: Circuits or Networks

vi

vo

Saturated vo

Nonsaturated vo

FIGURE 4.29 Simulink Diagram for Plotting the Output Voltage From an Operational Amplifier With and Without Saturation. between the unsaturated and saturated regions. However, for the ideal saturated amplifier, the output voltage is defined in terms of the input voltage as

8 > < vo;min ¼ K $ vi;min ; vi < vi;min vo ¼ K $ vi ; vi;min  vi  vi;max > : vo;max ¼ K $ vi;max ; vi > vi;max

(4.103)

where K is a constant gain factor. The nonlinear output voltage of Eq. (4.103) can be modeled in Simulink. Simulink possesses a generic saturation block, whose inputeoutput behavior is shown in Figure 4.28(b). For input values comprised between two limit values, such as umin and umax, the output is equal to the input, whereas for inputs outside this range, the output becomes saturated and constant. The Simulink block diagram that graphically solves this example is shown in Figure 4.29. The Saturation block is selected from the Discontinuities library. Under Main, an Upper limit of 150 V and a Lower limit of 100 V need to be specified. These limits are attained by multiplying the limits of the input voltage by the gain of 10. Figure 4.30 shows the plot of the simulation solution; the result of saturation is a truncation of the ideal sinusoidal output. The peak regions of the ideal sine wave are eliminated in the saturated response between 100 V and 150 V.

Output voltage (V)

Saturation

Saturation

Time (s)

FIGURE 4.30 Simulink Plot of Unsaturated and Saturated Output Voltages. nn

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182

CHAPTER 4 Electrical Systems

Mathematical Operations With Operational Amplifier Circuits We study here addition and integration; other op amp examples are presented in the companion website Chapter 4. nn

Example 4.17

Determine the relationship between the output voltage vo and the two input voltages v1 and v2 for the op amp circuit of Figure 4.31; establish the function produced by this circuit.

R i1

i

R1

v1

– +

v2 i2

R2

vo

FIGURE 4.31 Operational Amplifier in a Negative Feedback Circuit With Two Input Resistors.

Solution The currents shown in Figure 4.31 are expressed by means of Ohm’s law as

i1 ¼

v1  0 v2  0 0  vo ; i2 ¼ ; i¼ R1 R2 R

(4.104)

The op amp circuit has negative feedback; and because the positive input voltage is zero, it follows that the negative input terminal voltage is zero as well. No current is drawn into the op amp due to infinite input impedance; therefore i1 þ i2 ¼ i. Substituting Eq. (4.104) into this current connection, the following relationship is established between the output and input voltages:



R R vo ¼  $ v1 þ $ v2 R1 R2



(4.105)

In the case where the three resistors are identical, that is, R1 ¼ R2 ¼ R, Eq. (4.105) reduces to vo ¼ ðv1 þ v2 Þ, which indicates the function of the op amp circuit of Figure 4.31 is that of an inverting adder. The particular case with two input resistors can be generalized. When n resistors of corresponding resistances R1, R2, ., Rn are connected to the negative input terminal of Figure 4.31, having v1, v2, ., vn the corresponding voltages with respect to the ground, the output voltage becomes



R R R vo ¼  $ v1 þ $ v2 þ / þ $ vn R1 R2 Rn



(4.106)

For identical resistances, R1 ¼ R2 ¼ . ¼ Rn ¼ R, Eq. (4.106) reduces to vo ¼ ðv1 þ v2 þ / þ vn Þ.

nn

4.2 Electrical Systems: Circuits or Networks

nn

Example 4.18 Formulate the relationship between the output and input voltages for the op amp circuit of Figure 4.32.

i R vi

C

i

– +

vo

FIGURE 4.32 Operational Amplifier in a Negative Feedback Circuit With Input Resistor and Feedback Capacitor.

Solution This circuit, too, has negative feedback and the positive input connected to the ground; as a consequence, the two input ports have zero voltages applied to them. The current i can be expressed in two ways, for the input branch and for the feedback branch:

i¼

vi  0 d ; i ¼ C $ ð0  vo Þ R dt

(4.107)

Eq. (4.107) is rewritten as

Integration of Eq. (4.108) yields

dvo vi ¼ dt R$C

(4.108)

ð 1 $ vi dt vo ¼  R$C

(4.109)

which indicates that the role of the op amp circuit of Figure 4.32 is that of an integrator.

nn

nn

Example 4.19

The electrical circuit of Figure 4.33 comprises a resistor R ¼ 50 U, an inductor L ¼ 0.2 H, a capacitor C ¼ 1 mF, a voltage source vi ¼ 20$e0.3$t V, and two switches S1 and S2. The S1 is open, whereas S2 is closed for the first s ¼ 5 s. After that, S1 closes and S2 opens, actually removing the voltage source from the circuit. Derive the mathematical model for this system, which expresses the voltage across the resistor (which is considered the output voltage, vo) in terms of the input voltage vi for the two switch states. Use Simulink to solve this mathematical model, and plot vo as well as vi in terms of time for zero initial conditions.

183

184

CHAPTER 4 Electrical Systems

L

i

C

R

vo

S1 S2

vi

FIGURE 4.33 Electrical Circuit With Resistor, Inductor, Capacitor, Switches, and Variable Voltage Source.

Solution

When the switch S2 is closed and the switch S1 is open, the following equation is obtained by using KVL:

L$

ð diðtÞ 1 d2 iðtÞ dvo ðtÞ 1 dvi ðtÞ þ vo ðtÞ þ $ iðtÞdt ¼ vi ðtÞ or L $ þ $ iðtÞ ¼ þ 2 dt C dt dt C dt (4.110)

The voltage across resistor is:

vo ðtÞ ¼ R $ iðtÞ or

d2 vo ðtÞ d 2 iðtÞ ¼ R $ dt2 dt2

(4.111)

Substituting the second derivative of the current from Eq. (4.111) into Eq. (4.110) results in:

d2 vo ðtÞ R dvo ðtÞ 1 dvi ðtÞ  $ vo ðtÞ þ ¼ $ dt2 L dt L$C dt

(4.112)

which is valid for 0  t  s. For t > s when the switch S2 is open and the switch S1 is closed, there is no input voltage applied to the circuit, and Eq. (4.112) becomes

d2 vo ðtÞ R dvo ðtÞ 1  $ vo ðtÞ ¼ $ 2 dt L dt L$C

(4.113)

Eqs. (4.112) and (4.113) form the mathematical model of the studied electrical system. Simulink has the block called Wrap To Zero in the Discontinuities library to model the physical action of the switch, which, in this example, forces the input (source) voltage to drop to zero suddenly. The input to the block is the regular signal that needs to be chopped. The block output is identical to the input for a defined period of time, after which becomes zero, similarly to the operation of an actual switch. By specifying the Threshold value as the time s, the action of the block allows using the derived mathematical model for both 0  t  s (when Eq. 4.112 is inherently valid) and for t > s (when Eq. 4.113 governs the system behavior). Figure 4.34 is the Simulink block diagram that evaluates vo and plots both vi and vo in terms of time. The Function (Fcn) block is formatted as 20*exp(0.3*u(1)) in the block’s Expression, where u is the time, and is modeled by means of a Ramp source block with a Slope of 1. The Threshold value in the Wrap To Zero block is found as vi (s) ¼ vi (5) ¼ 89.63 V. The gains shown in Figure 4.34 are: Gain ¼ R/L ¼ 250, Gain 1 ¼ 1/(L$C) ¼ 5,000,000, and

4.2 Electrical Systems: Circuits or Networks

Zero-switched vi vo vi

t

dvi (t ) dt

R . dvi (t ) L dt

–

d2vo (t ) dt 2

dv o(t ) dt

R . dvo (t ) L dt 1 . vo (t ) LC

FIGURE 4.34 Simulink Block Diagram to Calculate and Plot the Resistor and Input Voltages.

Voltage (V)

Unaltered vi

Zero-switched vi

vo

Time (s)

FIGURE 4.35 Input and Output Voltages as Functions of Time.

Gain 2 ¼ Gain ¼ 250. Also note that before adding the three functions in the right-hand side of Eq. (4.112), the time derivative of the input voltage is calculated. The input voltage (both unaltered and with the switch applied after s s) and the output voltages are exported in MATLAB through the three To Workspace blocks, where they are plotteddFigure 4.35 shows their time history. The output voltage is zero, except for a negative spike at t ¼ s ¼ 5 s, this response is typical to a very narrow band pass filter, topic which is analyzed in more detail in Chapter 9. nn

Multiple-DOF Systems Mathematical models of multiple-DOF electrical systems are derived by means of Kirchhoff’s laws and Lagrange’s equations. MATLAB and Simulink can be utilized to solve the resulting differential equations and to plot the time-domain response.

185

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CHAPTER 4 Electrical Systems

The KCL and KVL methods have been discussed previously in this chapter, and they can be applied to the examples of this section. Lagrange’s equations are analyzed in the following.

Lagrange’s Equations Mesh Analysis In mesh analysis the Lagrange’s equations are: ! ! d vL vL vF d vEL vEC vF þ ¼ vj or þ ¼ vj  þ dt vq_j vqj vq_j dt vq_j vqj vq_j

(4.114)

for j ¼ 1, 2, ., n. As detailed in the section dedicated to the natural response of multiple-DOF electrical systems, the Lagrangian is L ¼ EL  EC, where EL is the total inductive energy, and EC is the total capacitive energy. The function F is the Rayleigh dissipation function (it is introduced in Chapter 3 in conjunction with dissipative mechanical systems), which, for electrical systems results from resistor contributions and is: n  2 1 X F¼ $ Rj $ q_j 2 j¼1

(4.115)

Eq. (4.115) is general, as it assumes there is one resistor for each branch current that is associated with a generalized coordinate (charge) qj. Oftentimes, the current of Eq. (4.115) is a difference of two currents related to generalized coordinates. In the right-hand side of Eq. (4.114), vj is the generalized voltage, actually a source voltage that is associated with the branch j, which corresponds to the generalized charge qj. Node Analysis In node analysis the Lagrange’s equations are as follows: ! ! d vL vL vF d vEC vEL vF þ ¼ ij or þ ¼ ij (4.116)  þ dt vl_ j vlj vl_ j dt vl_ j vlj vl_ j for j ¼ 1, 2, ., n. The Lagrangian in this case is L ¼ EC  EL, and explanations were provided in the section covering the natural response of multiple-DOF electrical systems. The dissipation function F is expressed from Eq. (4.115) as follows: n n ðvj Þ2 1 X 1 X 1 _ 2 $ F¼ $ Dl j Rj $ ¼ 2 2 j¼1 2 j¼1 Rj ðRj Þ

(4.117)

Eq. (4.117) assumes there is a resistor connected between any two nodes defined by a flux difference Dlj. In the right-hand side of Eq. (4.116), ij is the generalized current, actually a source current that is associated with the node j that is characterized by the generalized coordinate lj. A source current is considered positive if it enters a node and negative when it exits the node.

4.2 Electrical Systems: Circuits or Networks

187

nn

Example 4.20 Derive the mathematical model of the electrical system shown in Figure 4.36 using Kirchhoff’s laws and then Lagrange’s equations, in conjunction with mesh currents and the mesh analysis.

i1

v

+ –

vR1

vR3

R1 i 1 – i 2 R2 C

i2 vR2

R3

L vL

vC

FIGURE 4.36 Two-Mesh Electrical Network.

Solution Using KVL, the following equations are obtained for the two-mesh electrical system of Figure 4.36:

8 ð 1 > > >  < R1 $ i1 þ R2 $ ði1  i2 Þ þ C $ ði1  i2 Þdt ¼ v v R1 þ v R2 þ v C ¼ v or ð > v R3 þ v L  v C  v R2 ¼ 0 di2 1 > >  $ ði1  i2 Þdt  R2 $ ði1  i2 Þ ¼ 0 : R3 $ i 2 þ L $ dt C (4.118) In terms of the corresponding charges, Eq. (4.118) becomes

8 1 1 > > < ðR1 þ R2 Þ $ q_1  R2 $ q_2 þ $ q1  $ q2 ¼ v C C > 1 > : L $ q€  R2 $ q_1 þ ðR2 þ R3 Þ $ q_2  $ q1 þ 1 $ q2 ¼ 0 2 C C

(4.119)

Eq. (4.119) represents the mathematical model of the studied electrical system. In the mesh analysis approach, Lagrange’s equations are expressed in terms of the two charges q1 and q2 as generalized coordinates. The inductive and capacitive energies, as well as the Rayleigh dissipation function are:

1 1 $ ðq1  q2 Þ2 ; EL ¼ $ L $ q_22 ; Ec ¼ 2 2C

2 1 1 1 F ¼ $ R1 $ q_21 þ $ R2 $ q_1  q_2 þ $ R3 $ q_22 2 2 2

(4.120)

The derivatives needed in the general Lagrange’s Eq. (4.114) are:

vEL vEL vEC 1 vEC 1 ¼ 0; ¼ L $ q_2 ; ¼ $ ðq1  q2 Þ; ¼  $ ðq1  q2 Þ; C C vq_1 vq_2 vq1 vq2

vF

vF ¼ R1 $ q_1 þ R2 $ q_1  q_2 ; ¼ R2 $ q_1  q_2 þ R3 $ q_2 vq_1 vq_2 (4.121)

188

CHAPTER 4 Electrical Systems

Also note that v1 ¼ v (the source voltage associated with loop 1 and charge q1) and v2 ¼ 0 (because there is no source voltage in loop 2). These values, together with the derivatives of Eq. (4.121), are substituted in Eq. (4.114), and they result in Eq. (4.119), the mathematical model of the electrical system as obtained by Kirchhoff’s laws. nn

nn

Example 4.21 Use Simulink and the mathematical model derived in Example 4.20 for the electrical system of Figure 4.36 to plot the currents i1 and i2, when R1 ¼ 1000 U, R2 ¼ 910 U, R3 ¼ 740 U, L ¼ 0.9 H, C ¼ 0.8 mF, and a square-wave source voltage as the one of Figure 4.37(a). The source voltage is defined by an amplitude V ¼ 10 V, a period T ¼ 0.002 s, of which the pulse width is Tp ¼ 0.001 s, and the space width is Ts ¼ 0.001 s (Tp and Ts are not to scale in Figure 4.37(a)).

Voltage V

O

Time

Ts

Tp T

(a)

(b)

FIGURE 4.37 (a) Generic Square-Wave Source Voltage; (b) Simulink Pulse Generator Block Parameters.

Solution Eq. (4.119) is written as follows:



q_1 ¼ a1 $ q_2 þ a2 $ q1 þ a3 $ q2 þ a4 $ v ¼ a1 $ i2 þ a2 $ q1 þ a3 $ q2 þ a4 $ v q€2 ¼ b1 $ q_1 þ b2 $ q_2 þ b3 $ q1 þ b4 $ q2 ¼ b1 $ i1 þ b2 $ i2 þ b3 $ q1 þ b4 $ q2 (4.122)

with

a1 ¼

R2 1 ¼ 654:45; ¼ 0:476; a2 ¼ a3 ¼  R1 þ R2 ð R1 þ R 2 Þ $ C a4 ¼

1 ¼ 5:236 $ 104 ; R1 þ R2

R2 R2 þ R3 b1 ¼ ¼ 1011:111; b2 ¼  ¼ 1833:333; L L b3 ¼ b4 ¼

1 ¼ 1; 388; 888:889 L$C

(4.123)

4.2 Electrical Systems: Circuits or Networks

a2·q1(t) i (t )

a4·v(t)

v(t)

q1(t )

i1(t )

b1 · i1(t )

b3·q1(t)

i (t )

i 2 (t )

q2 (t )

b2 · i 2 (t )

a1 · i 2 (t )

a3·q2(t) b4·q2(t)

FIGURE 4.38 Simulink Block Diagram to Calculate and Plot Currents i1 and i2. The block diagram that integrates Eq. (4.122) with the gains of Eq. (4.123) is given in Figure 4.38. The Pulse Generator block is taken from the Source library and is formatted with the parameters shown in Figure 4.37(b). Figure 4.39 plots the two currents i1 and i2 for approximately four periods. The current i1 that passes through the resistor R1 follows the square-wave profile of the input voltage, whereas the current i2 through the resistor R3 and the inductor L displays an almost triangular profile, gradually increasing and decreasing in a repetitive manner.

Currents (A)

i1

i2

Time (s)

FIGURE 4.39 Output Currents i1 and i2 as Functions of Time. nn

189

190

CHAPTER 4 Electrical Systems

nn

Example 4.22 Obtain the mathematical model for the electrical system shown in Figure 4.40 using Kirchhoff’s laws and then Lagrange’s equations with node fluxes as DOF and nodal analysis.

vD

–

D

v

C

+ vA

R1

A

vB B

L1

R2

i

vC C L2

vO = 0 O

FIGURE 4.40 Three-Mesh Electrical Network With Voltage and Current Sources.

Solution

The bottom node O is selected as the reference node (with vO ¼ 0). The node voltages vA and vD are related through the source voltage v as vD ¼ vA  v. As a consequence, the node voltages vA, vB, and vC are independent parameters and can be the DOF of this electrical system. KCL is applied at the nodes defined by these three voltages:

8 ð 1 1 > > $ vA dt þ $ ðvA  vB Þ þ C $ ðv_A  v_  v_B Þ ¼ 0; node A > > > L R 1 1 > > > < 1 1 $ ðvB  vA Þ þ $ vB þ C $ ðv_B þ v_  v_A Þ  i ¼ 0; node B > R1 R2 > > > ð > > 1 > > : $ vC dt þ i ¼ 0; node C L2

(4.124)

Eq. (4.124) are an alternate mathematical formulation of the node analysis, which is introduced in Example 4.12. This new formulation states that the algebraic sum of currents corresponding to a node is zero. All currents converging to the node through branches carrying electrical components are considered positive (with positive voltage at that node). Similarly, source currents that exit the _ Eq. (4.124) transforms into node are taken positive. By using the voltageeflux relationship, v ¼ l, the following:

8 > > €  C$l € þ 1 $ l_ A  1 $ l_ B þ 1 $ lA ¼ C $ v_ > C$l > A B > > R1 R1 L1 > > >   < 1 _ 1 1 € € C $ lA þ C $ lB  $ lA þ þ $ l_ B ¼ i  C $ v_ > R1 R1 R2 > > > > > 1 > > > : L $ lC ¼ i 2 Eq. (4.125) is the mathematical model of the electrical system.

(4.125)

4.3 MechanicaleElectrical Analogy

With lA, lB, lC being the DOF, the three Lagrange’s equations are as follows:

8   > d vL vL vF > > > þ ¼ 0;  > > _ dt vl vl A vl_ A > A > > > < d  vL  vL vF þ ¼ i;  _ > dt vl v l v l_ B B > B > >   > > > d vL vL vF > > þ ¼ i  > : dt vl_ vlC vl_ C C

(4.126)

Note that source currents entering a node are positive according to Lagrange’s equations, which is unlike the node analysis convention. The Lagrangian of this system is:

ð 2 ð 2 1 1 1 2 vA dt  vC dt L ¼ EC  EL ¼ $ C $ ðvA  v  vB Þ  $ $ 2 2L1 2L2

2 1 1 1 ¼ $ C $ l_ A  v  l_ B  $ ðlA Þ2  $ ðlC Þ2 2 2L1 2L2 (4.127)

and the resistance-related dissipation function is:

F¼

2 1 1 1 _ 1 _ 2 $ ðvA  vB Þ2 þ $ ðvB Þ2 ¼ $ lA  l_ B þ $ lB 2R1 2R2 2R1 2R2

(4.128)

The following derivatives are substituted in Eq. (4.126):



vL

vL 1 vF 1 ¼ C $ l_ A  v  l_ B ; ¼  $ lA ; ¼ $ l_ A  l_ B ; vlA L1 vl_ A vl_ A R1

vL

1 vL vF 1 ¼ C $ l_ A  v  l_ B ; ¼ 0; ¼  $ l_ A  l_ B þ $ l_ B ; _ _ vl R R2 vlB vlB B 1 vL vL 1 vF ¼ 0; ¼  $ lC ; ¼0 vlC L1 vl_ C vl_ C (4.129) and the result is Eq. (4.125)dthe mathematical model derived by KCL.

nn

4.3 MECHANICALeELECTRICAL ANALOGY Two dynamic systems, one mechanical and the other electrical, are analogous if they have similar mathematical models expressed not only by differential equations, mainly, but also by other models, as discussed in Chapters 7 and 8. Mechanicale electrical system analogy enables testing a system that is simpler to design and test/monitor (usually an electrical system) then extrapolating the results to the analogous system (mechanical, for instance). Two analogous systems have the same order and structure of their differential equations; they also have their corresponding energies expressed similarly.

191

192

CHAPTER 4 Electrical Systems

nn

Example 4.23 Demonstrate that the mechanical system of Figure 4.41(a) and the electrical systems of Figure 4.41(b) and 4.41(c) are analogous.

(a)

(b)

(c)

FIGURE 4.41 Analogous First-Order Systems: (a) Mechanical; (b) Electrical; (c) Electrical With Operational Amplifiers.

Solution

The mechanical system of Figure 4.41(a) consists of the damper of coefficient c and the spring of stiffness k. The equation of motion is formulated for coordinate y as

c $ y_ þ y ¼ u k

0 ¼ cy_  k $ ðy  uÞ or

(4.130)

For the electrical system of Figure 4.41(b), the input and output voltages are:



vi ¼ R $ i þ vo or R $ C $ v_o þ vo ¼ vi i ¼ C $ v_o

(4.131)

with i being the current through the electrical resistance R and capacitance C, and vi and vo being the input and output voltages. The same current i passes through all elements of the two-stage op amp system of Figure 4.41(c). For the two resistors of the first stage, this current is as follows:

i¼

vi  v1 0  v2 di 1 ¼  $ v_2 ¼ or dt R R R

(4.132)

The currentevoltage relationship for the inductor is:

v1  0 ¼ L $

di dt

(4.133)

The second Eqs. (4.132) and (4.133) result in:

L v1 ¼  $ v_2 R

(4.134)

Substituting v1 from Eq. (4.134) into the first Eq. (4.132) yields

L $ v_2 þ v2 ¼ vi R

(4.135)

4.3 MechanicaleElectrical Analogy

Because of the two identical resistors connected to it, the second stage produces the following voltage ratio: vo/v2 ¼ 1. Combining this result with Eq. (4.135) produces the following differential equation:

L $ v_o þ vo ¼ vi R

(4.136)

The first-order differential Eqs. (4.130), (4.131), and (4.136) are very similar, which indicates the mechanical system and the two electrical systems of Figure 4.41 are analogous. nn

Forceevoltage or forceecurrent analogies can be established between mechanical and electrical systems on a more systematic basis, as discussed next. y

R

k m

f

v

+

R

i

–

c

C

L i L C

(a)

(b)

(c)

FIGURE 4.42 Analogous Second-Order Systems: (a) Mechanical; (b) Series Electrical; (c) Parallel Electrical.

The differential equations of the mechanical system shown in Figure 4.42(a) and of the electrical system illustrated in Figure 4.42(b) are as follows: m $ y€ þ c $ y_ þ k $ y ¼ f ; L $ q€ þ R $ q_ þ

1 $q ¼ v C

(4.137)

which illustrates the analogy between the two systems. Table 4.1 shows the pairing of physical quantities, according to this particular analogy known as forceevoltage analogy. It should be noted that rotary mechanical systems are also analogous to the electrical system of Figure 4.42(b), where angular displacement, velocity, and acceleration replace the translational counterparts, and moments (torques) are needed instead of forces (so the momentevoltage analogy is the correct denomination in this case). Consider now the electrical system of Figure 4.42(c). The source current is equal to the sum of currents through the resistor, inductor, and capacitor, which means that ð dvðtÞ 1 1 C$ þ $ vðtÞ þ $ vðtÞdt ¼ iðtÞ (4.138) dt R L Table 4.1 ForceeVoltage Analogy Pairs Mechanical Electrical

f v

y q

m L

c R

k 1/C

193

194

CHAPTER 4 Electrical Systems

Table 4.2 ForceeCurrent Analogy Pairs Mechanical Electrical

f i

y l

m C

c 1/R

k 1/L

where v is the voltage across each of the three electrical components. When using the relationship between voltage v and electrical flux l, v ¼ dl(t)/dt, Eq. (4.138) changes to € þ 1 $ l_ þ 1 $ l ¼ i (4.139) C$l R L which is similar to the first Eq. (4.137) and indicates the analogy between the mechanical system of Figure 4.42(a) and the electrical one of Figure 4.42(c). This forceecurrent analogy is summarized by the pairs illustrated in Table 4.2. Again, rotary mechanical systems can be used instead of translatory ones with the corresponding changes in displacement, velocity, accelerations (angular instead of linear), and forcing (moment instead of force), which results in a momentecurrent analogy. nn

Example 4.24 Using the forceevoltage analogy, find an electrical system that is analogous to the mechanical system of Figure 4.43. The three bodies of masses m1, m2, and m3 move along parallel directions. They are connected by a massless rod, which is pinned to the middle body, and has end slots that slide on the pins attached to the top and bottom bodies, which can move independently.

f

m1

x1

k

x3 m3

A

A l

C

Massless rod

x1

C x3

A’

C’

l B m2

c x2

B’ x2

B

Fixed surface

FIGURE 4.43 Three-Mass Mechanical System.

Solution

For independent and parallel displacements x1 and x2 of the top and bottom bodies along the positive directions indicated in Figure 4.43, the fork massless rod forces the body of mass m3 to displace a distance x3 parallel to x1 and x2 and whose value is x3 ¼ (x1  x2)/2dthis can easily be determined from the displacement sketch at the right of Figure 4.43. As a result of this constraint, the mechanical system has two DOF, which can be x1 and x2. To find the mathematical model of this system,

4.3 MechanicaleElectrical Analogy

Lagrange’s equations are used, as per the general Eq. (3.113), with the generalized coordinates being q1 ¼ x1 and q2 ¼ x2. The kinetic energy T, potential energy U, and dissipation function F are:

195

  1 1 1 1 1 1 x_1  x_2 2 2 2 2 2 2 T ¼ $ m1 $ x_1 þ $ m2 $ x_2 þ $ m3 $ x_3 ¼ $ m1 $ x_1 þ $ m2 $ x_2 þ $ m3 $ ; 2 2 2 2 2 2 2 1 1 U ¼ $ k $ x21 ; F ¼ $ c $ x_22 2 2 (4.140)

The partial derivatives needed in Lagrange’s equations are:

vT 1 vT 1 ¼ m1 $ x_1 þ $ m3 $ ðx_1  x_2 Þ; ¼ m2 $ x_2  $ m3 $ ðx_1  x_2 Þ; vx_1 4 vx_2 4 vU vU vF vF ¼ k $ x1 ; ¼ 0; ¼ 0; ¼ c $ x_2 ; vx1 vx2 vx_1 vx_2

(4.141)

Substituting these derivatives into Lagrange’s Eq. (3.113) yields in the following mathematical model of the mechanical system of Figure 4.43:

8 1 > > < m1 $ x€1 þ $ m3 $ ðx€1  x€2 Þ þ k $ x1 ¼ f 4 > > : m2 $ x€  1 $ m3 $ ðx€  x€ Þ þ c $ x_2 ¼ 0 2 1 2 4

(4.142)

Using the forceevoltage analogy and Table 4.1, Eq. (4.142) transforms into the following electrical equations:

8

1 1 > > < L1 $ q€1 þ $ L3 $ q€1  q€2 þ $ q1 ¼ v 4 C

> > : L2 $ q€  1 $ L3 $ q€  q€ þ R $ q_2 ¼ 0 2 1 2 4

(4.143)

An electrical system that might be the physical model identified from Eq. (4.143) is the one of Figure 4.44.

L2

L1

C

L3 /4

+ FIGURE 4.44 Two-Mesh Electrical Network.

i2

i1 – i2

i1

v

–

R

196

CHAPTER 4 Electrical Systems

Using KVL and mesh analysis for the two-DOF electrical system of Figure 4.44 results in the following equations:

8   ð > di1 1 di1 di2 1 > > > < L1 $ dt þ 4 $ L3 $ dt  dt þ C $ i1 dt ¼ v   > di2 1 di1 di2 > > > $ L   $ $ L þ R $ i2 ¼ 0 : 2 dt 4 3 dt dt

(4.144)

When substituting i1 ¼ dq1/dt and i2 ¼ dq2/dt in Eq. (4.144), we find Eq. (4.143). As a consequence, the electrical system proposed in Figure 4.44 is, indeed, analogous to the original mechanical system of Figure 4.43. nn

Note: General Procedure for MechanicaleElectrical Analogy Identification As seen from Example 4.24, the following steps need to be followed in finding an analog system: (1) Derive the mathematical model of the given system (mechanical or electrical, in this case); (2) Utilize the forceevoltage or forceecurrent analogy pairs of Tables 4.1 or 4.2 to write the mathematical model of the analog system to be identified using the equations derived at (1); (3) Based on the equations obtained at (2), propose the schematic diagram of the analog system to be identified; and (4) Independently derive the mathematical model of the system proposed at (3) and verify whether this model is identical to that resulted at (2). nn

Example 4.25 Determine a rotary mechanical system that is analogous to the electrical system of Figure 4.45, by using the momentecurrent analogy and the node analysis.

vA i1 i

C

R2

i2

i3

i4

L1

R1

vB

L2

v=0

FIGURE 4.45 Two-Mesh Electrical System With Current Source.

Solution

The source current splits into the four currents shown in Figure 4.45 as follows: i ¼ i1 þ i2 þ i3 þ i4. Expressing the currents on the right-hand side of this equation based on the components they pass through results in the following:

ð dvA 1 vA vA  vB þ $ vA dt þ þ i ¼ C$ L1 dt R1 R2

(4.145)

4.3 MechanicaleElectrical Analogy

The following equation expresses that the current through R2 and L2 is the same:

ð vA  vB 1 ¼ $ vB dt L2 R2

(4.146)

Using the fluxevoltage relationships vA ¼ dlA/dt and vB ¼ dlB/dt into Eqs. (4.145) and (4.146), changes these equations into

8

1 1 1 _ > > € _ _ > < C $ lA þ L $ lA þ R $ lA  lB þ R $ lA ¼ i 1 2 1

> 1 1 > > : $ l_ A  l_ B ¼ $ lB R2 L2

(4.147)

A one-to-one transformation of Eq. (4.147) by using the momentecurrent analogy synthesized in Table 4.2 and rotary mechanical elements results in:

(



€ þ k1 $ qA þ c2 $ q_ A  q_ B þ c1 $ q_ A ¼ m J $q A

c2 $ q_ A  q_ B ¼ k2 $ qB

(4.148)

Figure 4.46(a) identifies a mechanical system candidate to comply with Eq. (4.148).

c1 m

θA A

c2

(a)

.

.

) B

J

k2 .

c1

(

c2 θ B − θ A

m A

B

J

k1

k1θA θA

θB

c1θ A

(

.

.

c2 θ A − θ B

)

k2θ B θB

(b)

FIGURE 4.46 (a) Two-DOF Mechanical System as an Analogous to the Electrical System of Figure 4.45; (b) Free-Body Diagrams. The free-body diagrams associated with the two DOF qA and qB are shown in Figure 4.46(b) with the moments acting at those points. Using Newton’s second law of motion in conjunction with the freebody diagrams yields:

(



€ ¼ c1 $ q_ A  k1 $ qA  c2 $ q_ A  q_ B þ m J $q A

0 ¼ c2 $ q_ B  q_ A  k2 $ qB

(4.149)

which are identical to Eq. (4.148) that was obtained indirectly, from the mathematical model of the original electrical system. As a result, the proposed mechanical system of Figure 4.46(a) is a valid analogue to the electrical system depicted in Figure 4.45. nn

197

198

CHAPTER 4 Electrical Systems

SUMMARY This chapter discusses the electrical elements of voltage and current source, resistor, capacitor, inductor, and op amp. It also introduces the methods of electrostatic actuation and sensing in MEMS. Ohm’s law, Kirchhoff’s laws, the energy method, Lagrange’s equations, the mesh analysis method, and the node analysis method are applied to derive mathematical models for electrical systems. Examples are studied of how to apply MATLAB for symbolic and numerical calculation of the natural, free, and forced responses of electrical systems. Simulink is used to model and solve differential equations for electrical systems that might include nonlinearities, such as saturation. The chapter also studies analogies between mechanical and electrical elements/systems. The next chapter analyzes, in a similar manner, the modeling of fluid systems and of thermal systems.

PROBLEMS 4.1 Find the unknown resistance R1 of the electrical system shown in Figure 4.47, knowing that a voltage v ¼ 200 V applied between nodes A and B results in a lost resistive power PR ¼ 400 W for R ¼ 80 U.

R

R

R

R A

R1

R

R

R

B

FIGURE 4.47 Electrical Circuit Segment With Unknown Resistor.

4.2 Five identical resistors need to be combined using both series and parallel connections, with the constraint that any configuration should use at least two resistors coupled in series and at least two resistors in parallel. Determine the combinations that produce the maximum resistance Rmax and minimum resistance Rmin, respectively, between two end points; demonstrate that the Rmax/Rmin ratio is equal to 49/4. 4.3 Three resistors, of which two are identical, can be placed in any of the three positions of the circuit shown in Figure 4.48, which uses a battery of constant voltage v ¼ 100 V, and a wattmeter to measure the power P between nodes A and B. Determine the unknown resistances R1 and R2, knowing that the power consumed by one configuration is P1 ¼ 80 W, and the power for another configuration is P2 ¼ 110 W. In addition, identify the two circuit configurations.

Problems

A

B

v

FIGURE 4.48 Electrical Circuit Segment With Unknown Resistors.

(a)

(b)

FIGURE 4.49 (a) D-to-Y Resistance Transformation; (b) Resistance Circuit With D-to-Y Transformation.

4.4 Using the D-to-Y resistance transformation of Figure 4.49(a), together with the equations expressing the resistances of the Y connection in terms of those of the original D connection, determine the equivalent resistance between nodes A and B for the electric circuit of Figure 4.49(b) knowing R1 ¼ 200 U, R2 ¼ 170 U, R3 ¼ 150 U. Hint: Transform the D connections of the two triangles comprising nodes A and B into Y connections. 4.5 Determine the equivalent resistance between nodes A and B of the electrical system shown in Figure 4.50 for Rx and Ry defined by their lengths lx and ly, respectively, and having identical cross sections A and resistivities r. Assuming lx ¼ c$ly (with 0.5  c  2), find the value of c that minimizes the power of this circuit for a constant voltage vAB ¼ 50 V and calculate this minimum power value.

Rx ly

Ry

Rx Ry

Rx

Ry

Rx

Ry

Ry lx

A

FIGURE 4.50 Electrical Resistances in a Mixed Connection.

B

199

200

CHAPTER 4 Electrical Systems

4.6 A source voltage vi is applied to the resistor electrical system of Figure 4.51. Express the voltage ratio vo/vi in terms of the resistances R1 and R2. Assuming you have six identical resistors, three to use instead of R1 and three to replace R2, how would you connect them to maximize vo/vi, and what is this maximum voltage ratio? Generalize when 2n identical resistors are utilized, n instead of R1 and n instead of R2.

R1 vi

R2

vo

FIGURE 4.51 Two-Resistor Electrical Circuit.

4.7 A constant voltage vAB ¼ 20 V is applied between nodes A and B of the capacitive circuit shown in Figure 4.52, where C1 ¼ 0.8 mF and C2 ¼ 0.6 mF. To reduce the total electrostatic energy by 20% with the same applied voltage, it is decided to change the capacitor C2. What is the new value of C2? C2

A

C1

C2

C1 B

C1

FIGURE 4.52 Electric Circuit With Two Different Capacitors.

4.8 The capacitive Wheatstone bridge of Figure 4.53 is formed of four capacitors C1, C2, C3, and C4. For a source voltage v applied between points A and B of the bridge: (i) Calculate the voltage vCD between nodes C and D of the circuit in terms of v and the four capacitances. (ii) Determine C1 in terms of C2, C3, and C4 for vCD ¼ 0, when the bridge is balanced. 4.9 Utilizing the D-to-Y capacitor transformation illustrated in Figure 4.54(a), calculate the equivalent capacitance between nodes A and B of the circuit shown in Figure 4.54(b) for C1 ¼ C2 ¼ 20 mF, C3 ¼ 15 mF, C4 ¼ 30 mF, and C5 ¼ 40 mF.

Problems

+

v

C

C1

–

C2 B

A C4

D

C3

FIGURE 4.53 Capacitive Wheatstone Bridge Circuit.

(a)

(b)

FIGURE 4.54 (a) D-to-Y Capacitance Transformation; (b) Capacitance Circuit With D-to-Y Transformation.

4.10 A MEMS capacitive actuator with one fixed plate and two mobile plates is sketched in Figure 4.55. The distance between the two mobile plates is d ¼ 10 mm, the plate overlapping area is 9000 mm2, and ε0 ¼ 8.8$1012 F/ m. An external voltage v ¼ 30 mV is applied between the fixed and the mobile plates. A net force results from the two capacitive actuators, which displaces the mobile pair whenever the fixed plate is not placed symmetrically between the mobile plates. Express this force as a function

–

Motion

v

+

d x

Conductive guide Fixed plate Mobile pair

FIGURE 4.55 MEMS Capacitive Actuator With Transverse Displacement.

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CHAPTER 4 Electrical Systems

of the variable distance x and plot it in terms of x. Also calculate the resulting force corresponding to a minimum capacitive gap gmin ¼ d/200. 4.11 Compare the sensing performance of a transverse-motion capacitive unit with that of a longitudinal motion one for MEMS displacement detection, in terms of the initial gap g0 and overlap x0. Specifically, study the ratio of transverse voltage variation Dvt to the longitudinal voltage variation Dvl in terms of displacement Dx. Assume the two physical units are identical (same initial gap, plate areas, and dielectric properties). 4.12 The motion of the block of Figure 4.56 is sensed both longitudinally and transversely, simultaneously but separately by two capacitive units. Knowing g0 ¼ 12 mm, x0 ¼ 37 mm, vb ¼ 2 mV (the bias voltage of both units), Dvt ¼ 0.85 mV (reading by the transverse unit), and Dvl ¼ 0.21 mV (reading by the longitudinal unit), calculate the block displacement corresponding to this state. If the two capacitive readings result in different block displacements, consider the transverse unit is the precise one, indicate what design parameter of the longitudinal unit might have been erroneously specified and determine its exact valuedstudy all parameters. Fixed plate

g0 Transverse-motion Motion voltmeter Mobile block

g0 x0 Fixed plate

Longitudinal-motion voltmeter

FIGURE 4.56 MEMS Capacitive Sensor With Transverse and Longitudinal Pickup.

4.13 A longitudinal MEMS capacitive actuator is replaced with a transverse one to obtain a 12-fold increase in the actuation force. Consider that the two actuators’ dimensions and operation conditions are identical, the initial gap of the transverse actuator is equal to the longitudinal actuator gap, and the maximum distance traveled is one fourth of that gap. For a capacitor plate

Problems

that is rectangular with an in-plane side length of 120 mm and an out-ofplane side length of 50 mm, calculate the initial gap of the transverse actuator. 4.14 A constant current source powers a group of five identical inductors. Determine the manner of connecting these inductors such that the actual total electromagnetic energy EL be within the 3$EL,min and EL,max/3 range, where EL,min is the energy of the configuration with the five inductors connected in parallel, and EL,max is the energy of the network with the five inductors coupled in series. 4.15 Use the equations allowing conversion of an inductor D connection into a Y connection (as sketched in Figure 4.57(a)) to calculate the equivalent inductance between nodes A and B of the network shown in Figure 4.57(b). Known are L1 ¼ 130 mH, L2 ¼ 100 mH, L3 ¼ 70 mH, L4 ¼ 80 mH, and L5 ¼ 60 mH. Δ connection

Y connection

L1 La

L3

Lc

La =

L1L2 L1 + L2 + L3

Lb =

L2 L3 L1 + L2 + L3

Lc =

L3 L1 L1 + L2 + L3

Lb L2

(a)

L1

A

L3

L2 L4 B

L5

(b)

FIGURE 4.57 (a) D-to-Y Inductance Transformation; (b) Inductance Circuit With D-to-Y Transformation.

4.16 An electrical network is formed of eight branches and four nodes. Design/ draw the schematics of the network identifying branches, nodes, and meshes. Determine the number of DOF of this system. 4.17 Using voltages as parameters, demonstrate that the electrical system of Figure 4.58 has two DOF.

+ v

L2

–

R1

Multiple-Element Electrical Circuit.

R3

R2 L1

FIGURE 4.58

L3

C1

C2

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CHAPTER 4 Electrical Systems

4.18 Use the conservation of energy method to determine the mathematical model of the electrical system shown in Figure 4.59 and calculate its natural frequency for L1 ¼ 0.3 H, L2 ¼ 0.6 H, C1 ¼ 45 mF, and C2 ¼ 10 mF. C1 C2

C2

L1

L2

L1

C1

FIGURE 4.59 Single-Mesh Conservative Electrical Circuit.

4.19 A single-mesh (single-DOF) circuit comprises series and parallel combinations of three identical capacitors and two identical inductors. Identify the two electrical systems for which the natural frequencies ratio is maximum and calculate that ratio. 4.20 Establish the DOF of the electrical system sketched in Figure 4.60. Derive its mathematical model by using Kirchhoff’s laws with mesh charges as DOF. Calculate the corresponding natural frequencies and unit-norm eigenvectors. Use L1 ¼ 0.1 H, L2 ¼ 0.5 H, and C ¼ 35 mF. i2 i1 – i2

i1 L1

C

L2

FIGURE 4.60 Two-Mesh Conservative Electrical Circuit.

4.21 Derive the mathematical model of the electrical system of Figure 4.60 by using Kirchhoff’s laws with node fluxes as DOF. Find the corresponding natural frequencies and unit-norm eigenvectors, by using the numerical parameters given in Problem 4.20. 4.22 Demonstrate that the electrical circuit of Figure 4.61 has two DOF. Derive its mathematical model using Lagrange’s equations and mesh charges as DOF. Find the ratio of the two natural frequencies.

L C

L C

C

FIGURE 4.61 Two-Mesh Electrical Circuit With Coupling Inductor and Capacitor.

Problems

4.23 Obtain the mathematical model of the electrical system shown in Figure 4.61 by means of node fluxes and Lagrange’s equations. 4.24 Use the Lagrange’s equations and mesh charges to determine the mathematical model for the electrical circuit of Figure 4.62 in matrix form. Find the system’s natural frequencies and the unit-norm eigenvectors by means of MATLAB’s eig command for L1 ¼ 100 mH, L2 ¼ 95 mH, C1 ¼ 200 mF, and C2 ¼ 260 mF. 4.25 Use Kirchhoff’s laws and node fluxes to derive the mathematical model for the electrical system of Figure 4.62 in matrix form. Find the system’s natural frequencies and the unit-norm eigenvectors by means of MATLAB’s eig command using the parameter values given in Problem 4.24. L1

C1

C1

C2

L2

FIGURE 4.62 Two-Mesh Electrical Circuit With Coupling Capacitor.

4.26 An electrical circuit comprising a resistor, an inductor, and a capacitor connected in series is tested experimentally. It is determined that the natural frequency is 200 Hz, and the damped period is 0.006 s. It is also known that Rcr ¼ 230 U. Calculate the unknown parameters R, L, and C. 4.27 A single-mesh circuit is formed of a voltage source vs ¼ 80$sin (15$t) V connected in series with a capacitor C ¼ 32 mF, a nonlinear resistor R1 whose current is defined in terms of the voltage across as i ¼ 0.0001$(1.1  e0.3$v) A, and a solenoid that acts as an inductor L in series with a resistor R2. For the solenoid known are the wire diameter d ¼ 0.2 mm, the coil median diameter D ¼ 50 mm, the number of turns N ¼ 2000, the electric resistivity r ¼ 17  109 U m, and magnetic permittivity m ¼ 0.01 H/m. Derive the mathematical model of the electrical system. Use MATLAB’s ode45 command and also Simulink to solve for the mesh current and plot it as a function of time. Hint: The nonlinear resistance is defined as 1/R1 ¼ di/dv. 4.28 Derive the mathematical model of the circuit shown in Figure 4.63. Consider that, in a first phase, the switch S1 is closed, whereas switch S2 is open. In a second phase, the two switches change their states. Plot the current through R for the second phase. Known are R ¼ 180 U, L ¼ 0.8 H, C ¼ 30 mF, and v ¼ 120 V.

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CHAPTER 4 Electrical Systems

S1

S2

+ v

C

R L

– FIGURE 4.63

Two-Mesh Electrical Circuit With Two Switches Swapping States.

4.29 Determine the relationship between the output and input voltages for the op amp circuit of Figure 4.64 and establish the function produced by this circuit. Plot the output voltage vo for vi ¼ 100$t V, R ¼ 200 U, and C ¼ 28 mF. i

R

i

C vi

– +

vo

FIGURE 4.64 Operational Amplifier in a Negative Feedback Circuit With Input Capacitor and Feedback Resistor.

4.30 Derive the mathematical model for the op amp circuit shown in Figure 4.65 by expressing output voltage vo for R1 ¼ 60 U, R2 ¼ 45 U, R3 ¼ 30 U, L ¼ 0.5 H, C ¼ 70 mF, and an input voltage defined as vi ¼ 90/ (1 þ 0.015$t) V, where t is time. Plot both vi and vo in terms of time using Simulink.

R3

C

R2 vi

R1 L

FIGURE 4.65 Electrical Circuit With Operational Amplifier.

– +

vo

Problems

4.31 Apply the mesh analysis method to derive the mathematical model for the electrical circuit of Figure 4.66. Use MATLAB to find the currents in the circuit for R1 ¼ 35 U, R2 ¼ 50 U, R3 ¼ 60 U, R4 ¼ 20 U, R5 ¼ 45 U, R6 ¼ 10 U, R7 ¼ 28 U, R8 ¼ 40 U, and R9 ¼ 45 U. The source voltage is v ¼ 100 V. R1

R8

R9

R4

R2

+

R7 R5

R3

v

–

R6

FIGURE 4.66 Electrical Circuit With Resistors and Voltage Source.

4.32 Calculate the currents produced by the current sources in the circuit sketched in Figure 4.67, by employing the nodal analysis method and MATLAB for i1 ¼ 1 A, i2 ¼ 0.6 A, R1 ¼ 30 U, R2 ¼ 50 U, R3 ¼ 20 U, R4 ¼ 40 U, and R5 ¼ 70 U. R1

+ i1

R2

R3

R5

R4

–

+ i2

–

FIGURE 4.67 Electrical Network Comprising Resistors and Current Sources.

4.33 Utilizing Lagrange’s equations with node fluxes and node analysis, formulate the mathematical model for the electrical system of Example 4.20 and shown in Figure 4.36. Apply KCL to obtain the same mathematical model. 4.34 With the numerical parameters of Example 4.21, use Simulink to solve the mathematical model derived in Problem 4.33. Plot the time variations of the voltages across R1, C, and L of Figure 4.36. 4.35 For the electrical network of Example 4.22 and illustrated in Figure 4.40, known are R1 ¼ 120 U, R2 ¼ 140 U, L1 ¼ 0.1 H, L2 ¼ 0.15 H, C ¼ 3 mF, i ¼ 15 mA, and v ¼ 50$sin (20$t) V. Using MATLAB’s dsolve command, solve the mathematical model derived in Example 4.22 and plot the voltages across R1, L1, and C.

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4.36 Use mesh charges to derive the mathematical model for the electrical system of Example 4.22 and of Figure 4.40 by means of KVL and then by applying Lagrange’s equations with branch charges. 4.37 Solve the mathematical model of Problem 4.36 with the numerical parameters of Problem 4.35 by using Simulink. Plot the time-varying currents through R1, L1, and C. 4.38 Establish the DOF for the electrical circuit of Figure 4.68. Use the mesh analysis and mesh charges in conjunction with KVL to determine the mathematical model. Also apply Lagrange’s equations to obtain the same mathematical model. Find the different branch currents and plot them using MATLAB for L1 ¼ 85 mH, L2 ¼ 70 mH, L3 ¼ 55 mH, R1 ¼ 210 U, R2 ¼ 160 U, v ¼ 90$sin (6$t) V. L2 R1 v

L1

R2

L3

FIGURE 4.68 Three-Mesh Electrical Circuit.

4.39 Derive the mathematical model of the circuit shown in Figure 4.69 when the switch is closed (position 1), then when the switch is opened. Plot the current through R2 and L when the switch is in the open position. Consider R1 ¼ 170 U, R2 ¼ 200 U, C ¼ 20 mF, L ¼ 78 mH, and v ¼ 120 V.

L

2 S

+

1

–

R1

v

C

R2

FIGURE 4.69 Two-Mesh Electrical Circuit With Switch.

4.40 Establish the DOF of the electrical circuit depicted in Figure 4.70. Derive the mathematical model using mesh charges and the Lagrange’s equations. Confirm the validity of the mathematical model by applying KVL.

Problems

4.41 Use Simulink to model the circuit shown in Figure 4.70 of Problem 4.40. Plot the time variation of the branch currents directly connected to the charges that were determined as DOF. The initial charge on capacitor C1 is q1(0) ¼ 0.0015 C; known also are R1 ¼ 320 U, R2 ¼ 260 U, R3 ¼ 300 U, R4 ¼ 210 U, L1 ¼ 1.2 H, L2 ¼ 2 H, L3 ¼ 1.8 H, C1 ¼ 350 mF, C2 ¼ 280 mF, and v ¼ [180/(t þ 1)]$cos (10t) V. R1

C2

R4

R2

L1 R3

L2

C1

v

+

L3

–

FIGURE 4.70 Electrical Network Comprising Resistors, Capacitors, Inductors, and a Voltage Source.

4.42 By means of the momentevoltage analogy, use the charges q1 and q2 to find a rotary mechanical system that is analogous to the electrical system of Figure 4.71. L1

q2 C 2 q1 L2

C1 v

–

+

R

FIGURE 4.71 Two-Mesh Electrical Network With Voltage Source.

4.43 Utilize the forceevoltage analogy to identify a mechanical system that is analogous to the electrical system depicted in Figure 4.72. Hint: Apply a reasoning similar to that of Example 4.24 and use the charges q1 and q2 as DOF for the electrical system. L1

L2

q1

+

q2

L3

C1

+

C2

v1

–

– R1

R2

FIGURE 4.72 Two-Mesh Electrical Network With Voltage Sources.

v2

209

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CHAPTER 4 Electrical Systems

4.44 Using forceevoltage analogy, identify a mechanical system that is analogous to the electrical system of Problem 4.38 and illustrated in Figure 4.68. 4.45 Identify a translatory mechanical system that is analogous to the electrical system of Figure 4.73, by using the forceecurrent analogy and node analysis. R2

i

L1

C1

R1

L2

C2

FIGURE 4.73 Electrical System With Current Source.

4.46 Using forceecurrent analogy, propose an electrical system that is analogous to the lumped-parameter model of the MEMS sketched in Figure 4.74. The MEMS is formed of two shuttle masses, m1 and m2, six identical beams connecting the masses to the fixed base, and a serpentine spring coupling the two shuttle masses. The system is acted on capacitively by the forces f1 and f2; viscous damping of coefficient c exists between the two masses and the underneath supporting base.

Fixed base

k

Beams

Serpentine spring Shuttle mass

f1

f2 m1

m2 k12

Fixed base

FIGURE 4.74 Microelectromechanical System with Capacitive Actuation.

Suggested Reading

4.47 Determine a mechanical system and an electrical system that are analogous by using forceevoltage analogy under the assumption that the two systems are defined by the same mathematical model: ( x€1 þ 15x_1  15x_2 þ 1000x1  400x2 ¼ 70 $ sinð100 $ tÞ x€2 þ 20x_2  15x_1 þ 800x2  400x1 ¼ 180

Suggested Reading R. C. Dorf, and J. A. Svoboda, Introduction to Electric Circuits, 9th Ed. John Wiley & Sons, New York, 2013. K. C. A. Smith, and R. E. Alley, Electrical CircuitsdAn Introduction, Cambridge University Press, Cambridge, UK, 1992. A. E. Fitzgerald, D. E. Higginbotham, and A. Grabel, Basic Electrical Engineering, 5th Ed. McGraw-Hill, New York, 1981. J. W. Nilsson, and S. A. Riedel, Electric Circuits, 10th Ed. Prentice Hall, Upper Saddle River, New Jersey, 2014. A. B. Carlson, Circuits, Brooks/Cole, Pacific Grove, California, 2000. D. Bell, Fundamentals of Electric Circuits, 7th Ed. Oxford University Press, Oxford, 2009. F. L. Ryder, The Lagrange Equations in Electrical Networks, Journal of the Franklin Institute, Vol. 266, Issue 1, 1958, pp. 27e38.

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CHAPTER

Fluid and Thermal Systems

5

OBJECTIVES The chapter models, analyzes, and designs fluid and thermal systems similarly to Chapters 2e4 that focus on mechanical and electrical systems. The main topics addressed here are: • Fluid (liquid and pneumatic) elementsdinertance, capacitance, resistance, and fluid energy sources. • Thermal capacitance and resistance elements. • Physical and mathematical modeling of the natural and forced responses of fluid and thermal systems by energy methods and Lagrange’s equations. • Analogies between electrical, fluid, and thermal systems. • Use of MATLAB and Simulink in evaluating and plotting the time response of fluid and thermal systems.

INTRODUCTION This chapter is dedicated to modeling the dynamics of fluid (liquid and pneumatic) systems, as well as thermal systems. These systems are formed of elements similar to those of mechanical and electrical systems: inertance, capacitance, and resistance. In case only inertance and capacitance properties are present, the natural response of fluid systems is studied. In many applications, the inertia properties of fluid and thermal systems can be neglected. The corresponding mathematical models are based on only capacitive and resistive properties resulting in first-order systems. The chapter also studies the analogies between electrical, fluid, and thermal systems. The use of MATLAB and Simulink in solving for the natural and forced responses of fluid and thermal systems is illustrated by several solved examples.

5.1 LIQUID SYSTEMS MODELING In liquid (or hydraulic) systems, the medium of energy transmission is a liquid. We introduce the basic liquid elements of inertance, capacitance, and resistance, together with the sources generating liquid system motion. We formulate the natural System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00005-1 Copyright © 2018 Elsevier Inc. All rights reserved.

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CHAPTER 5 Fluid and Thermal Systems

(free) response of inertanceecapacitance liquid systems, as well as the forced response of capacitanceeresistance liquid systems. Related hydraulic applications are presented in Chapter 2din the context of translatory and rotary dampersdand in Chapter 11das examples of controllers. The notion of flow rate is utilized in liquid and pneumatic systems with different meanings. In liquid systems, the volume flow rate is employed, which is denoted here by qv. Pneumatic systems use the mass flow rate, denoted by qm. The two amounts are connected by means of the mass density r, which is constant for incompressible fluids: Dvol Dx $ A Dm Dvol ¼ ¼ v $ A; qm ¼ ¼ r$ ¼ r $ qv qv ¼ (5.1) Dt Dt Dt Dt where vol stands for volume, v is the fluid velocity, Dx is the distance traveled by the fluid, and A is area perpendicular on flow direction.

5.1.1 Liquid Elements Similar to mechanical or electrical systems, which are formed of elements with inertia, storage capacity, losses, and energy input, liquid systems combine similar elements. Inertance elements portray liquid inertia, whereas capacitances and resistances characterize liquid storage and loss features, respectively. The pressure or the difference of level (head) among various components of a liquid system represents source elements that set the liquid into motion. As a consequence of the dual manner of generating liquid motion, the elements’ definitions can be provided by either using the pressure or the head, in addition to other amounts of interest. The subscript p (for pressure) or h (for head) accompanies the letter l (standing for liquid) in the following. The elements are defined for the laminar (linear and relatively slow) flow regime here, but nonlinearities, such as those pertaining to turbulent flow (relatively fast), are also discussed.

Inertance The inertance quantifies the inertia effects in liquid systems and is particularly important in long conduits such as pipes. For laminar flow, the inertance (which is denoted by Il) is defined as the pressure (or head) difference necessary to produce a unit change in the rate of change of the volume flow rate: p p p $ dt h h h $ dt ¼ ¼ ¼ ; Il;h ¼ ¼ (5.2) dq dq dqv dqv q_y q_y v v dt dt The International System (SI) units are kg/m4 for Il,p and s2/m2 for Il,h. By taking into account that a static pressure difference is connected to the head difference as p ¼ r $ g $ h, it follows that the two inertance definitions of Eq. (5.2) are related as Il;p ¼ r $ g $ Il;h (5.3) Il;p ¼

5.1 Liquid Systems Modeling

The kinetic energy that corresponds to inertance is very similar to the kinetic energy of mechanical systems and the inductive energy of electrical systems, being defined as
: 2
: 2 1 1 1 1 EI;p ¼ $ Il;p $ q2v ¼ $ Il;p $ vol ; EI;h ¼ $ Il;h $ q2v ¼ $ Il;h $ vol (5.4) 2 2 2 2 The pressure-defined energy of Eq. (5.4) has units of actual energy (N m or J in SI), whereas the head energy of the same Eq. (5.4) is measured in m4 in SI units. nn

Example 5.1 (i) Derive the inertance equation of the liquid flowing through the variable cross-section channel segment of Figure 5.1(a). Apply the general equation to calculate the inertance of a constant cross-section segment. (ii) Use the result at (i) to express the fluid inertance of the conical frustum sketched in Figure 5.1(b) knowing the end diameters d1 and d2, the length l, and the liquid mass density r.

Solution

(i) The inertance of a liquid element of length dx is first determined based on Figure 5.1(a). The total inertance is calculated by summing all similar prismatic elements (which means integration over the segment length l). Newton’s second law of motion is used for the prismatic element of length dx:

dm $

dv ¼ AðxÞ $ ½pðxÞ  pðx þ dxÞ ¼ AðxÞ $ dpðxÞ dt

(5.5)

where it has been considered that the pressures acting on the left face, p(x), and on the right face, p(x þ dx), are related as p(x) ¼ p(x þ dx) þ dp(x). Eq. (5.1) indicates that q_v ¼ A $ ½dv ðt Þ=dt and Eq. (5.5) enables expressing the inertance of the element as

dIl;p

dv dm $ dpðxÞ dt ¼ rAðxÞdx ¼ r $ dx ¼ ¼ q_v AðxÞ 2 dv AðxÞ2 AðxÞ $ dt

p(x)

(5.6)

p(x + dx)

v d(x) = d

A(x)

A1 x

dx

d1

d2

x

dx

l

l

(a)

(b)

FIGURE 5.1 (a) Variable Cross-Section Pipe Segment With Laminar Flow; (b) Conical Frustum Pipe Segment.

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CHAPTER 5 Fluid and Thermal Systems

Eq. (5.6) took into account that dm ¼ r$d(vol ) ¼ r$A(x)$dx. The inertance of the entire segment is found by integrating the element inertance of Eq. (5.6):

Z

Il;p ¼

Z

l

dIl;p ¼ r $ 0

dx AðxÞ

(5.7)

For a constant cross-section segment, Eq. (5.7) becomes:

Il;p ¼

Il;p r$l l ; Il;h ¼ ¼ A r$g A$g

(5.8)

Eq. (5.8), where A is the inner cross-sectional area, is valid for any constant cross-section pipe. (ii) The variable diameter d(x) of Figure 5.1(b) is:

dðxÞ ¼ d ¼ d1 þ

d2  d1 $x l

(5.9)

MATLAB’s symbolic capability and the following script are used to perform the integration of Eq. (5.7), considering that the area is A(x) ¼ p$d(x)2/4: >> syms d1 d2 l x rho >> dx = d1+(d2-d1)/l*x; Ax = pi*dx^2/4; >> in = int(1/Ax,x); >> inu = limit(in,x,l,’left’); inl = limit(in,x,0,’right’); >> inertance = rho*simplify(inu-inl) The last MATLAB command in the previous sequence returns the following:

Il;p ¼

4r $ l p $ d1 $ d2

(5.10)

It can be seen that the integration between the limits of 0 and l is performed in two steps: first the indefinite integral of 1/A(x) is calculated then the definite integral is evaluated as the difference between the upper and lower limits of the indefinite integral; on attempting to directly calculate the definite integral by means of the command int(1/Ax,x,0,l), no meaningful result is returned. nn

As demonstrated in the companion website of this chapter, two inertances Il1 and Il2 can be connected in series or in parallel, resulting in the equivalent inertances Ils and Ilp: Ils ¼ Il1 þ Il2 ;

1 1 1 ¼ þ Ilp Il1 Il2

(5.11)

Capacitance The capacitance in the liquid domain reflects the storage capacity by a tank-type device. The capacitance is defined as the ratio of the volume flow rate to the rate of pressure (or head): Cl;p

qv qv $ dt ¼ ¼ : ¼ dp p

dðvolÞ dðvolÞ $ dt dðvolÞ $ dt d ðvolÞ qv qv $ dt dt ; Cl;h ¼ : ¼ ¼ dt ¼ ¼ dp dh dh dp dh h (5.12)

5.1 Liquid Systems Modeling

where vol is the volume. The SI units of Cl,p are m5/N ¼ m4 s2/kg, whereas Cl,h is measured in m2. Due to the connection between head and static pressure, dp ¼ r$g$dh, the two capacitances of Eq. (5.12) are related as Cl;h ¼ r $ g $ Cl;p . Similar to springs in mechanical systems and capacitors in electrical systems, the liquid capacitance is incorporated into potential energy as
: 2 1 ðvolÞ2 1 1 EC;p ¼ $ ¼ $ Cl;p $ p2 ¼ $ Cl;p $ lp ; 2 Cl;p 2 2
: 2 1 ðvolÞ2 1 1 EC;h ¼ $ ¼ $ Cl;h $ h2 ¼ $ Cl;h $ lh 2 Cl;h 2 2

(5.13)

In Eq. (5.13), lp and lh are the pressure flux and head flux, respectively, which are defined as: dlp/dt ¼ p, dlh/dt ¼ h. These fluxes are similar to the electric flux utilized in Chapter 4. The pressure-defined capacitive energy EC,p is measured in N m (or J), whereas the SI unit for EC,h is m4. nn

Example 5.2 (i) Determine the capacitance of the variable cross-section cylindrical tank of Figure 5.2(a). Assume the tank is placed with its longitudinal axis vertically and is filled with incompressible liquid over the height h. (ii) Use the result to calculate the capacitance of the conical frustum of Figure 5.2(b). Known are the end diameters d1 and d2, the height h, as well as the liquid mass density r, and the gravitational acceleration g.

d2

dx

dx

h

A(x)

h

x

x

d1 d(x) = d (a)

(b)

FIGURE 5.2 (a) Variable Cross-Section Vertical Tank; (b) Conical-Segment Vertical Tank.

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CHAPTER 5 Fluid and Thermal Systems

Solution (i) The pressure-defined capacitance can be expressed as

Cl;p

dðvolÞ ¼ ¼ dp

dðvolÞ dh dp dh

(5.14)

The volume occupied by the liquid in a variable cross-section tank with a head (height) of h and the volume derivative in terms of h are

Z vol ¼

h

0

dðvolÞ d ¼ Að xÞdx; dh dh

Z

h

 Að xÞdx

(5.15)

0

where A(x) is the tank variable cross-sectional area at a distance x, measured from the bottom of the tank. The pressure at the bottom of the tank and its derivative in terms of h are

p ¼ r $ g $ h;

dp ¼ r$g dh

(5.16)

Substituting the derivatives of Eqs. (5.15) and (5.16) into Eq. (5.14), the pressure-defined and head-defined capacitances become

Cl;p

1 dðvolÞ $ ¼ ¼ r$g dh

d dh

Z

h

AðxÞdx ; Cl;h ¼ r $ g $ Cl;p

0

r$g

2 3 Z d 4 h ¼ AðxÞdx5 dh 0 (5.17)

Note: Eq. (5.17) is general, and they are also valid for tanks (containers) oriented with their longitudinal axes in directions other than the gravity-bearing vertical direction. As long as the difference p between the two tank ends’ pressures is equal to a gravitational pressure generated by an equivalent head h as p ¼ r$g$h, one can simply rederive Eq. (5.17). When the tank cross section is constant, the general Eq. (5.17) simplifies to:

Cl;p

d ðA $ hÞ A dh ; Cl;h ¼ r $ g $ Cp ¼ A ¼ ¼ r$g r$g

(5.18)

(ii) The diameter defining the variable area A(x) is the one given in Eq. (5.9) with h instead of l. The tank volume is therefore expressed as

p vol ¼ $ 4

h

Z 0

d2  d1 $x d1 þ h

2 dx ¼

p$h  2 $ d1 þ d1 $ d2 þ d22 12

(5.19)

Calculating the derivative of vol with respect to h in Eq. (5.19) and substituting it into Eq. (5.17) yields

Cl;p ¼


 p $ d12 þ d1 $ d2 þ d22 12r $ g

; Cl;h ¼


 p $ d12 þ d1 $ d2 þ d22 12

(5.20)

5.1 Liquid Systems Modeling

For a cylindrical tank with d1 ¼ d2 ¼ d, Eq. (5.20) become

Cl;p ¼

p $ d2 p $ d2 ; Cl;h ¼ 4r $ g 4

(5.21)

which are also directly obtained from Eq. (5.18).

qv

qv qv1

qv2

Cl1

qv

Cl1

Cl2

Cl2 qv qv

(a)

(b)

FIGURE 5.3 Liquid Storage Tanks Connected in (a) Series; (b) Parallel. nn

Liquid capacitances can be connected in series and in parallel in the same way as electrical capacitances. The companion website Chapter 5 gives the derivation of the equivalent series and parallel capacitances Cls and Clp, based on the capacitances Cl1 and Cl2 (which can be either pressure- or head-defined) connected as in Figure 5.3: 1 1 1 ¼ þ ; Clp ¼ Cl1 þ Cl2 Cls Cl1 Cl2

(5.22)

Resistance The liquid motion through pipes or conduits encounters resistance from inner wall friction, changes in the conduit direction, the existence of valves, or other constrictions, which act as energy dissipators. The dissipative action is quantified by means of resistances, very similar to electrical systems. Figure 5.4 shows the portion of a pipeline with a valve on it. The valve changes the area of flow, and the net result of it is a pressure drop from p1 to p2. Mathematically, the liquid resistance is defined as the ratio of the pressure or head drop to the volume flow rate: Rl;p ¼

p p1  p2 h ¼ ; Rl;h ¼ qv qv qv

(5.23)

219

220

CHAPTER 5 Fluid and Thermal Systems

Valve R

p1

p2

qv

FIGURE 5.4 Pipeline With Valve and Pressure Drop.

The SI unit for Rl,p is N s/m5 ¼ kg/(m4s), whereas Rl,h is measured in s/m2 in SI. Again, when the static pressureehead relationship is considered, the two resistances of Eq. (5.23) are connected as Rl;p ¼

p r$g$h ¼ ¼ r $ g $ Rl;h qv qv

(5.24)

Both definitions assume a linear relationship between pressure (or head) variation and volume flow rate.

Nonlinear Resistance While linearity covers the laminar flow domain, the relationship between these variables becomes nonlinear for turbulent flow; it is accepted that pressure varies with the square of the volume flow rate in turbulent flow: p ¼ k $ q2v

(5.25)

where k is a constant depending on the design configuration through which the flow occurs and on other physical properties of the fluid. Figure 5.5 shows the variation of the pressure in terms of volume flow rate for the laminar and turbulent regimes. This relationship can be linearized using the Taylor series expansion about a given nominal (or operational) point, such as the one indicated with the letter N p

Laminar (linear)

Turbulent (nonlinear)

P

δ ( pn )

(qv , p)

N pn

qv qv,n

δqv,n

FIGURE 5.5 Laminar- and TurbulenteRegime Relationships Between Pressure Drop and Volume Flow Rate.

5.1 Liquid Systems Modeling

in Figure 5.5 and defined by qv,n and pn. For a point P in the vicinity of N, and defined by the coordinates qv and p, the Taylor series expansion of p that keeps the first two (linear) terms is as follows:

 vp

p ¼ pn þ (5.26) $ qv  qy;n

vqv qv ¼qy;n As seen in Figure 5.5, p  pn ¼ dðpn Þ; qv  qy;n ¼ dqy;n

(5.27)

represents small variations in pressure and volume flow rate. As a consequence, Eq. (5.26) can be written as

vp

$ dqy;n (5.28) dðpn Þ ¼ vqv qv ¼qy;n Eq. (5.28) suggests the linearized resistance about the nominal point is

dð pn Þ vp

Rl;p ¼ ¼ dqy;n vqv qv ¼qy;n

(5.29)

Eq. (5.29) indicates that the linearized liquid resistance is the ratio of a small change of the pressure variation to the small change of the volume flow rate and can be calculated as the partial derivative of the pressure variation, in terms of the volume flow rate at the nominal point. By taking into account Eq. (5.25), the linearized resistance defined in Eq. (5.29) becomes

  vp

p Rl;p ¼ ¼ 2k $ qy;n ¼ 2 (5.30)

vqv qv ¼qy;n qv qv ¼qy;n In other words, the linearized liquid resistance for turbulent flow is twice the liquid resistance of laminar (linear) flow. nn

Example 5.3 The pressure variation in a pipe with turbulent liquid flow is expressed in terms of volume flow rate as p ¼ qv þ 3qv2 . Compare the linearized hydraulic resistance corresponding to this relationship to the .  ¼ p q by plotting the two-resistance ratio in terms linear resistance, which can be defined as Rl;p v of qv. Calculate this ratio for qv ¼ 0.01 m3/s and qv ¼ 2 m3/s.

Solution According to Eq. (5.30), the linearized hydraulic resistance is obtained as

Rl;p ¼

vp ¼ 1 þ 6qv vqv

(5.31)

 is calculated as While the linear resistance Rl;p

Rl;p ¼

p qv þ 3q2v ¼ ¼ 1 þ 3qv qv qv

(5.32)

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CHAPTER 5 Fluid and Thermal Systems

cR

qv (m3/s)

FIGURE 5.6 Linear-to-Nonlinear Hydraulic Resistance Ratio as a Function of Volume Flow Rate.

The following ratio can be used to compare the resistances of Eqs. (5.31) and (5.32)

cR ¼

Rl;p 1 þ 6qv 1 ¼ ¼2 1 þ 3qv Rl;p 1 þ 3qv

(5.33)

which indicates the linearized resistance is almost twice the linear resistance of Eq. (5.32). Eq. (5.33) also shows that the ratio increases with the increasing volume flow rate, and the limit is

cR;max ¼ lim cR ¼ 2 qv /N

(5.34)

The numerical values of the resistance ratio of Eq. (5.35) are cR ¼ 1.029 for qv ¼ 0.01 m3/s and cR ¼ 1.857 for qv ¼ 2 m3/s. Figure 5.6 is the plot of cR as a function of qv. nn

Linear Resistance, Lost Head, and HagenePoiseuille Resistance For a pipe segment of length l with liquid flowing through it at velocity v in laminar regime, the head that is lost due to fluid friction with the conduit inner wall, hf, is calculated as hf ¼ f $

l v2 $ dh 2g

(5.35)

In Eq. (5.35) f is the Moody friction factor, and dh is the hydraulic diameter, which is defined as dh ¼

4A Pw

(5.36)

5.1 Liquid Systems Modeling

with Pw being the wetted perimeter of the pipe internal cross section of area A. For laminar flow, the Moody friction factor is determined as f ¼ 64=Re

(5.37)

The Reynolds number Re of Eq. (5.37) has the following definition: Re ¼

r $ v $ dh v $ d h ¼ m y

(5.38)

where m is the dynamic viscosity, and n is the kinematic viscosity of the fluiddthey are connected as m ¼ n$g. The friction head is found using Eqs. (5.35), (5.37), and (5.38) as hf ¼

32m $ l $ v 32m $ l $ v 32m $ l 2m $ l $ P2w ; p ¼ r $ g $ h ¼ ¼ ¼ $ qv $ q   v f A3 r $ g $ dh2 dh2 4A 2 $A Pw (5.39)

Eq. (5.39) shows that the friction head is equivalent to a pressure drop p ¼ r$g$hf, and it takes into account that the volume flow rate is cross-sectional area times velocity: qv ¼ A$v; the same Eq. (5.39) also used Eq. (5.36). Comparing now Eq. (5.39) with the definition Eq. (5.23), it follows that the liquid resistance of a pipe segment with constant cross section of area A, wetted perimeter Pw, and length l through which liquid of dynamic viscosity m passes is as follows: 2m $ l $ P2w (5.40) A3 When the constant cross section is circular of diameter, Eq. (5.40) becomes as follows: Rl;p ¼

128m $ l (5.41) p $ d4 which is known as the HagenePoiseuille equation. Eq. (5.41) was obtained with Pw ¼ p$d and A ¼ p$d2/4. Rl;p ¼

nn

Example 5.4

Determine the hydraulic resistance of a tapered (conic frustum) pipe having a length l, and end diameters d1 and d2, as illustrated in Figure 5.7. Consider laminar flow and calculate the resistance numerical value for d1 ¼ 0.5 m, d2 ¼ 0.3 m, l ¼ 10 m, and m ¼ 0.001 N s/m2.

Solution

If an elementary portion of length dx of the tapered pipe is studied, the respective portion is approximately a cylinder of diameter d, as shown in Figure 5.7; therefore, its pressure difference is given by the HagenePoiseuille Eq. (5.41) as:

dp ¼

128m $ ðdxÞ $ qv pd4

(5.42)

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CHAPTER 5 Fluid and Thermal Systems

qv

p1

p2 d1

d

d2

x

dx l

FIGURE 5.7 Tapered Pipe With Laminar Flow. The diameter d is expressed geometrically in terms of x:

d ¼ d2 þ ðd1  d2 Þ $

lx l

(5.43)

Substitution of Eq. (5.43) into Eq. (5.42) and integration between the limits of 0 and l with respect to x yields the pressure difference between the input and the output of the pipe:

Z

l

p1  p2 ¼ p ¼ 0

 128m $ l $ d12 þ d1 $ d2 þ d22 dp ¼ $ qv 3pd13 $ d23

(5.44)

which indicates that the fluid resistance of the tapered pipe of Figure 5.7 is

Rl;p ¼


 128m $ l $ d12 þ d1 $ d2 þ d22 3pd13 $ d23

(5.45)

Obviously, when d1 ¼ d2, and therefore the taper is zero (the tapered pipe becomes a cylindrical one), Eq. (5.45) reduces to the classical HagenePoiseuille equation resistance, Eq. (5.41), which corresponds to a constant cross-section cylindrical pipe. For the numerical parameters of this example, the hydraulic resistance is Rl,p ¼ 19.72 N s/m5. nn

Similar to dampers in mechanical systems and resistors in electrical systems, energy is lost through liquid resistances, and this energy loss can be expressed in pressure or head form as:
: 2 2 1 1 1 p 1 lp 1 1 ER;p ¼ $ p $ qv ¼ Rl;p $ q2v ¼ $ ¼ $ ; ER;h ¼ $ h $ qv ¼ $ Rl;h $ q2v 2 2 2 Rl;p 2 Rl;p 2 2
: 2 1 h2 1 lh ¼ $ ¼ $ 2 Rl;h 2 Rl;h (5.46) The SI unit of ER,p is N m/s ¼ W (which is the unit for power), whereas the SI unit of ER,h is m4/s.

5.1 Liquid Systems Modeling

Rl2

Rl1

Rl1 Rl2

(a)

(b)

FIGURE 5.8 Liquid Resistances Connected in (a) Series; (b) Parallel.

Liquid resistances can be connected in series (as shown in Figure 5.8(a)) or in parallel (as illustrated in Figure 5.8(b)), and the equivalent resistances Rs and Rp, which are derived in the companion website Chapter 5, are expressed in terms of pressure/head-defined resistances as Rls ¼ Rl1 þ Rl2 ;

1 1 1 ¼ þ Rlp Rl1 Rl2

(5.47)

nn

Example 5.5 A microfluidic channel network is shown in Figure 5.9. Calculate the equivalent hydraulic resistance between points 1 and 2 by considering that the liquid losses are produced according to the Hagene Poiseuille equation. In addition, calculate the power lost in the microsystem. All the pipe segments are identical. Known are l ¼ 100 mm, m ¼ 0.0005 N s/m2, d ¼ 20 mm (constant cross-section pipe diameter), and p ¼ p1  p2 ¼ 103 N/m2.

l

l

p1 qv

1

p2 l

l l

2

l

FIGURE 5.9 Six-Component Microfluidic Network.

Solution Figure 5.10(a) shows the microchannel system of Figure 5.9 with the corresponding hydraulic resistances, which are identical. The aim is to obtain the equivalent, one-resistance system of Figure 5.10(b). To achieve that, the resistance to replace the four actual resistances in the middle of the system of Figure 5.10(a) can be calculated by combining in parallel two groups of series connected resistances, which yields the following:

R1 ¼

ð2RÞ $ ð2RÞ ¼R ð2RÞ þ ð2RÞ

(5.48)

The equivalent resistance is formed by connecting in series the end resistances of the original system of Figure 5.10(a) to the middle resistance of Eq. (5.48):

Req ¼ R þ R1 þ R ¼ 3R ¼ 3 $

128m $ l m$l ¼ 384 $ 4 pd 4 pd

(5.49)

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CHAPTER 5 Fluid and Thermal Systems

R p1

R

R

R

1 qv

R

p2

p1

2

1

R (a)

Req

p2 2

qv (b)

FIGURE 5.10 (a) Actual Microfluidic Network With Liquid Resistances; (b) Equivalent, SingleResistance Hydraulic System. With the numerical values of this example, the equivalent resistance is found to be Req ¼ 3.82$1013 N s/m5. Eq. (5.46) is used to calculate the power dissipated through the pipeline system: ER,p ¼ (p)2/(2Req) ¼ 1.31$108 W. nn

Sources of Hydraulic Energy For liquid-level systems, as is discussed shortly in this chapter, the liquid motion is generated through the head difference among various components, such as tanks and piping. Hydraulic actuators are components that convert high input fluid pressure into kinetic energy at the output. In many liquid applications, the energy necessary to generate flow in a liquid network is provided by pumps, which transform the input electric energy into output liquid work, generally manifested as flow rate or equivalent head. Pumps can be of several configurations, such as centrifugal, axial, rotary, or reciprocating in regularscale applications, as well as diaphragm (or membrane) in micro- and nanoapplications. The characteristic curve of a pump shows the head variation or the pressure variation in terms of the flow rate. For a centrifugal pump, which is widely used in hydraulic systems, the equations expressing the head or pressure variation with the flow rate are generally nonlinear: h ¼ hg  Kh $ q2v ; p ¼ p0  Kp $ q2v

(5.50)

where hg, the geometric head, is the maximum head-type energy a pump produces when no energy is lost through friction, and Kh is a coefficient related to the energetic losses. Similarly, p0 is the maximum attainable pressure at zero flow rate, and Kp is a constant depending on the pump construction. Eq. (5.50) indicates that as the flow rate increases (together with the corresponding losses), the head/ pressure of the pump decreases.

5.1.2 Liquid Systems Assembling several of the liquid elements presented thus far in this section leads to the formation of liquid systems. When only inertances and capacitances are present in a liquid system, and no forcing source is considered, the natural response can be

5.1 Liquid Systems Modeling

formulated. When external action or energy is applied in any form in a liquid system, the response is forced. Both liquid system responses are studied next. The notions of system configuration and degrees of freedom (DOF) are analyzed in detail for mechanical systems (in Chapter 3) and electrical systems (in Chapter 4). Due to similarities of liquid elements/systems with mechanical and electrical counterparts, the DOF are not studied separately in this chapter; however, the examples discussed herein identify and are based on the DOF of liquid systems.

Natural Response We study the natural response of free lossless liquid systems that are described by one single variable (single-DOF systems) as well as for systems whose response needs to be formulated in terms of more than one liquid-system variable (multiple-DOF systems). The natural frequencies and corresponding modes (eigenvectors) of multiple-DOF systems can be calculated analytically or by MATLAB, as also shown in Chapters 3 and 4.

Single-DOF Conservative Liquid Systems Consider a lumped-parameter liquid system that is defined by inertance Il and capacitance Cl. This system possesses a natural frequency, which can easily be found using the energy method, similar to the modality used to determine the natural frequencies of single-DOF mechanical or p electrical ffiffiffiffiffiffiffiffiffiffiffiffi systems. The demonstration of the following natural frequency: un ¼ 1= Il $ Cl is given in the companion website of this chapter. This natural frequency is very similar to the natural frequency of an electrical system formed of an impedance L and a capacitance C, as discussed in Chapter 4. It can be checked that un is measured in s1, which is the unit of natural frequency.

Multiple-DOF Conservative Liquid Systems The natural response of multiple-DOF conservative liquid systems is studied employing the analytical approach and MATLAB. Schematic liquid circuits can be drawn and analyzed similarly to electrical systems. It should be mentioned that, while relative agreement exists in terms of the symbols used for hydraulic elements such as pumps, actuators, or resistances, there is no consensus on the graphical representation of liquid inertances and capacitances. Due to the similitude between electrical and hydraulic systems, the symbol used for electrical inductances is used for hydraulic inertances, whereas hydraulic capacitances are symbolized similar to electrical capacitances. Lagrange’s equations are utilized in this section to formulate the mathematical model of multiple-DOF conservative systems in a variant that compares to the mesh analysis applied to electrical systems in Chapter 4. Lagrange’s equations are: ! d vL vL ¼ 0; L ¼ EI  EC (5.51)  dt vq_j vqj

227

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CHAPTER 5 Fluid and Thermal Systems

where L is the Lagrangian, EI is the energy pertaining to all lumped-parameter inertances of the liquid system, and EC is the energy comprising all capacitive contributions. The variables qj ( j ¼ 1, 2, ., n) are the liquid system’s generalized coordinates. The energy of an inertance element Il can be calculated
: 2 as EI ¼ Il $ vol 2, and the energy of a capacitive element Cl is 2 EC ¼ ðvolÞ ð2Cl Þdboth Il and Cl are pressure defined. It follows that the generalized coordinates are the independent volumesdthey can also be the DOF of the conservative liquid system. nn

Example 5.6 Figure 5.11(a) is the top view of a liquid system formed of two identical containers (tanks) of pressuredefined capacitances Cl connected by two short pipe segments. A third short pipe segment and two other identical long pipe segments, each of the latter ones defined by a lumped-parameter pressure inertance Il, complete the network. Assuming all pipes are in a horizontal plane at the tanks base and are lossless, also assuming the two short pipe segments are of negligible inertia: (i) Determine the DOF of this conservative liquid system. (ii) Utilize Lagrange’s equations to derive the system’s mathematical model.

Short pipes

q2

q1

Cl Long pipe

Tanks

Long pipe

Il

Cl

Il

q3 = q1 – q2

Short pipe (a)

(b)

FIGURE 5.11 (a) Top-View Schematics of a Conservative Liquid System With Two Tanks and Connecting Lossless Piping; (b) Equivalent Lumped-Parameter Two-Mesh System Representation.

Solution (i) The equivalent lumped-parameter model of the actual liquid system is illustrated in Figure 5.11(b) where, in addition to the two tanks capacitances Cl, the inertances Il corresponding to the two long pipe segments have been added. If the volume flow rates through the long pipes are denoted by q1 and q2, and considering there are no other sources or drains affecting the two tanks, two conclusions are drawn: (1) the input and output flow rates corresponding to the two tanks are identical, and (2) the flow rate through the tank on the middle line is q3 ¼ q1  q2 because of flow conservation at any node (such as the one above or the one underneath this tank). Similarly to an electric circuit, two parameters, the flow rates q1 and q2, for instance, are sufficient to determine the state of the liquid system of Figure 5.11(b)das a result, this is a two-DOF system. (ii) If instead of volume flow rates volumes are used as generalized coordinates (because the volume : flow rate is related to the volume as qv ¼ vol ), it follows that vol1 for q1 and vol2 for q2 are the generalized coordinates.

5.1 Liquid Systems Modeling

Using the volume flows indicated in Figure 5.11(b), the Lagrangian of the system is

L ¼ EI  EC     1 _ 1 2 þ 1 $ Il $ vol _ 2 2  1 $ ðvol1  vol2 Þ2 þ 1 $ ðvol2 Þ2 ¼ $ Il $ vol 2 2 2Cl 2Cl (5.52) The following partial derivatives are needed in Lagrange’s equations:

vL _ 1 ; vL ¼  1 $ ðvol1  vol2 Þ; ¼ Il $ vol _ vvol1 Cl vvol1 vL _ 2 ; vL ¼ 1 $ ðvol1  vol2 Þ  1 $ vol2 ¼ Il $ vol _2 vvol2 Cl Cl vvol

(5.53)

Eq. (5.53) is used with the generic Lagrange’s Eq. (5.51), while considering that the two generalized coordinates are vol1 and vol2; the result is the following mathematical model of our system

8 1 1 > > € > < Il $ vol1 þ C $ vol1  C $ vol2 ¼ 0 l l > > €  1 $ vol1 þ 2 $ vol2 ¼ 0 > : Il $ vol 2 Cl Cl

(5.54)

nn nn

Example 5.7 Calculate the natural frequencies and determine the corresponding eigenvectors of the liquid system modeled in Example 5.6, both analytically and by means of MATLAB’s eig command. Consider Il ¼ 2$106 kg/m4 and Cl ¼ 3$108 m4 s2/kg.

Solution The solution to Eq. (5.54) is harmonic of the form:

vol1 ¼ V1 $ sinðu $ t þ 4 Þ; vol2 ¼ V2 $ sinðu $ t þ 4 Þ

(5.55)

where V1 and V2 are volume amplitudes. Eq. (5.55) is substituted into Eq. (5.54), and the result is the following algebraic equations system, as sin(u$t) cannot be zero at all times:

8  > > > 1  u 2 $ I l $ V 1  1 $ V2 ¼ 0 > < Cl Cl   > 1 2 > 2 > > :  C $ V1 þ C  u $ I l $ V2 ¼ 0 l l

(5.56)

The homogeneous equations system (Eq. 5.56) has nontrivial solutions V1 and V2 when the determinant of the system is zero:



1

 u 2 $ Il

C

l



1



Cl







¼0

2 2  u $ Il

Cl 1  Cl

(5.57)

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230

CHAPTER 5 Fluid and Thermal Systems

which results in the characteristic equation

Il2 $ Cl2 $ u4  3 $ Il $ Cl $ u2 þ 1 ¼ 0

(5.58)

The solution of this equation in u provides the two natural frequencies of the hydraulic system:

un1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 3 5 3þ 5 ; un2 ¼ ¼ 2Il $ Cl 2Il $ Cl

(5.59)

With the numerical data of the problem, the natural frequencies are un1 ¼ 2.5 rad/s and un2 ¼ 6.6 rad/s. The following amplitude ratio is obtained from the first Eq. (5.56):

V2 ¼ 1  u2 $ Il $ Cl V1

(5.60)

Substituting un1 in Eq. (5.60), and requiring that the corresponding eigenvector be unit-norm, we have the following equations:

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V2 V21 ¼ ¼ 0:618; V 211 þ V 221 ¼ 1 V1 u¼un1 V11

(5.61)

The eigenvector corresponding to Eq. (5.61) is

fV gu¼un1 ¼

V11



V21

¼

0:851

 (5.62)

0:526

As seen in Eq. (5.62), the two flows v1 and v2 have identical directions, and the amplitude V1 is larger than the amplitude V2 during the modal motion at the natural frequency un1. The second natural frequency of Eq. (5.59) is now substituted into the amplitude ratio of Eq. (5.60), and a unit-norm eigenvector is again sought, which results in the following equations:

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V2 V22 ¼ ¼ 1:618; V 212 þ V 222 ¼ 1 V1 u¼un2 V12

(5.63)

whose corresponding eigenvector is

fV gu¼un2 ¼

V12 V22



¼

0:526 0:851

 (5.64)

Inspection of Eq. (5.64) shows that, during the second modal motion at un2, the two flows have opposite directions (this is indicated by the minus sign), and the amplitude V1 is smaller than the amplitude V2. Similar to mechanical and electrical systems, the natural response of a multiple-DOF hydraulic system can be formulated in vector-matrix form as an eigenvalue problem, and MATLAB can be employed to solve for eigenvalues and eigenvectors. The equation describing the free vibrations of a multiple-DOF hydraulic system can be written as

  € þ ½Cl  $ fvolg ¼ f0g ½Il  $ vol

(5.65)

where [Il] is the inertance matrix, [Cl] is the hydraulic capacitance matrix, and {vol} is the volume vector. It can be shown by following a procedure similar to the one applied to mechanical systems

5.1 Liquid Systems Modeling

in Chapter 3 that when sinusoidal solution of the type {v } ¼ {V } sin(u$t) is sought for Eq. (5.65), the following equation results:


 det ½Il 1 $ ½Cl   l $ ½I ¼ 0

(5.66)

where l ¼ u2 are the eigenvalues, [I ] is the identity matrix, and [Dl] ¼ [Il]1[Cl] is the hydraulic dynamic matrix. Once [Dl] has been determined, the MATLAB command [V, d] = eig(Dl), which was utilized in Chapters 3 and 4, returns the modal matrix V, whose columns are the eigenvectors, and the diagonal matrix d, whose diagonal elements are the eigenvalues. Eq. (5.54) can be arranged in the matrix-vector form of Eq. (5.65) with

 ½Il  ¼

Il 0

2 1 6 Cl 0 ; ½Cl  ¼ 6 4 1 Il  Cl 

13

 Cl 7 vol1 7; fvolg ¼ vol 2 5



Cl

The dynamic matrix is calculated using MATLAB symbolic calculation:

½Dl  ¼ ½Il 1 $ ½Cl  ¼

(5.67)

2

 1 1 $ Il $ Cl 1

1 2

 (5.68)

As expected, the eigenvalues returned by MATLAB are the squares of the natural frequencies of Eq. (5.59), and the eigenvectors are

fV gu¼un1 ¼



 0:851 0:526 ; fV gu¼un2 ¼ 0:526 0:851

(5.69)

which are essentially the eigenvectors obtained analytically (the minus signs in the first eigenvector of Eq. (5.69) shows that the two components move in the same direction). nn

Forced Response of Liquid-Level Systems As mentioned previously in this chapter, one modality of generating the forced response of liquid systems is by means of tanks, where the flow is gravitygenerated, and connects in and out with pipelines, for instance. Often, in such liquid systems known as liquid-level systems, hydraulic calculations can be performed without considering the inertia effects because the liquid flow is relatively slow; as a consequence, such systems consist of only capacitances and resistances.

Single-DOF Liquid-Level Systems Consider the liquid-level system of Figure 5.12, which is formed of a tank of constant capacitance Cl communicating with a pipe segment equipped with a valve of constant resistance Rl. We want to derive a mathematical model for this system by connecting the output (either the flow rate qo or the head h) to the input flow rate qi when the pressure at the input and output ports is zero. For this system the total volume flow rate conserves and, as a result, the flow rate passing through the tank q is expressed in terms of the input and output flow rates (qi, qo) as q ¼ qi  qo.

231

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CHAPTER 5 Fluid and Thermal Systems

qi

Tank

h Cl

Rl

p

qo

Valve

FIGURE 5.12 Liquid-Level System With Tank and Valve.

If we use the pressure definitions of resistances and capacitances, the following equations can be written: ( : : p ¼ Rl $ qo and p ¼ Rl $ qo : (5.70) qi  qo ¼ C l $ p where the pressure p is at the bottom of the tank just before the valve. Eliminating the pressure time derivative between the two Eq. (5.70), produces the mathematical model: :

Rl $ Cl $ qo þ qo ¼ qi

(5.71)

where qi is the input, and qo is the output. Since the order of the differential equation is one, the physical system is a first-order one. An alternate model can be obtained by considering that the pressure and its time derivative are: :

:

p ¼ r $ g $ h and p ¼ r $ g $ h

(5.72)

Eqs. (5.70) and (5.72) result in the following: qo ¼

: r$g Rl $ qi $ h and Rl $ Cl $ h þ h ¼ Rl r$g

(5.73)

which represents another mathematical model, where the output is the head h, and the input is the flow rate qi.

Multiple-DOF Liquid-Level Systems Two methods are suggested in this section to derive mathematical models of liquidlevel systems for multiple-DOF systems: one method simply uses the equations

5.1 Liquid Systems Modeling

233

defining each liquid component and is, essentially, an energy balance method; the other method employs Lagrange’s equations. Lagrange’s equations are applied in a manner similar to the one that utilizes node voltages/fluxes and the node analysis method for electrical systems in Chapter 4. Using the pressure flux lp (p ¼ dlp/dt, as introduced in Section 5.1.1) the generalized coordinates qj are the independent fluxes lpj. The Lagrangian is: Lp ¼ EC,p  EL,p ¼ EC,p because the capacitive energy EC,p contains terms of the  2 form Cp $ p2 2 ¼ Cp $ dlp =dt 2 (similarly to the electrical capacitance energy), whereas EL,p ¼ 0 since there are no liquid inertia contributions. The Rayleigh dissipation function Fp adds all resistance energies, as provided in Eq. (5.46), and is: n n 1 X 1  _ 2 1 X 1  2 Fp ¼ $ $ Dlpj ¼ $ $ pj 2 j¼1 Rpj 2 j¼1 Rpj

(5.74)

and Lagrange’s equations are: ! !   vLp vFp vFp vFp d vLp d vEC;p d vEC;p þ ¼ qj or ¼ qj or ¼ qj þ  þ _ dt vl_ pj dt vl_ pj dt vlpj vl_ pj vp vpj vlpj j (5.75) for j ¼ 1, 2, ., n. The term in the right-hand side of Eq. (5.75) is the liquid forcing function and is actually a volume flow rate. An alternative model can be obtained by Lagrange’s equations method if one uses the head instead of pressures; the final Lagrange’s equations in this case are: ! d vEC;h vFh ¼ qj (5.76) þ dt vhj vhj for j ¼ 1, 2, ., n. The generalized coordinates are the n heads this time. The capacitive energy EC,h sums up contributions of the type Ch$h2/2, whereas the dissipation function Fh (the head-related resistance energy) is: n 1 X 1  2 Fh ¼ $ $ hj 2 j¼1 Rhj

(5.77)

nn

Example 5.8

The liquid system sketched in Figure 5.13 consists of two tanks of capacitances Cp1 and Cp2 and three pipe segments whose resistances are Rp1, Rp2, and Rp3. All five element properties are defined in terms of pressure. The inputs to the system are the pump-delivered pressure p, the atmospheric pressure pa, and the volume flow rate q. Derive the pressure-defined mathematical model of this system by expressing the pressure p1 and p2 at the bottoms of the two tanks. Use both the element definition (energy balance) approach and Lagrange’s equations.

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CHAPTER 5 Fluid and Thermal Systems

q

h1 p

Rp1

p1

h2 Cp1

p1

Rp2

p2

Cp2

p2

Rp3

pa

Tank

Pump

q1

q2

q3

FIGURE 5.13 Liquid-Level System With Pump, Two Tanks, and Three Valves.

Solution Based on the hydraulic capacitance and resistance definitions, as well as on Figure 5.13, the following relationships can be formulated as follows:

Rp1 ¼

p  p1 p1  p2 p2  pa q1  q2 q þ q2  q3 ; Rp2 ¼ ; Rp3 ¼ ; Cp1 ¼ ; Cp2 ¼ : : q1 q2 q3 p1 p2 (5.78)

The flow rates q1, q2, and q3 are expressed from the first three Eq. (5.78) and substituted into the last two Eq. (5.78). The following first-order differential equations result after a few algebraic manipulations:

8   > 1 1 1 1 : > > > < Cp1 $ p1 þ Rp1 þ Rp2 $ p1  Rp2 $ p2 ¼ Rp1 $ p   > 1 1 1 1 : > > > C $ p  $ p þ þ $ pa $ p2 ¼ q þ 1 : p2 2 R Rp2 Rp3 Rp3 p2

(5.79)

which form the mathematical model in pressure form, obtained by using the definition equations of the five elements. Considering that the first generalized coordinate is p1 and the second one is p2, the two Lagrange’s equations resulting from Eq. (5.75) are:

    vFp vFp d vEC;p d vEC;p ¼ 0; ¼q þ þ dt vp1 dt vp2 vp1 vp2

(5.80)

The capacitive energy and the dissipation function are formulated:

1 1 EC;p ¼ $ Cp1 $ p21 þ $ Cp2 $ p22 ; 2 2 1 1 1 $ ðp  p1 Þ2 þ $ ðp1  p2 Þ2 þ $ ðp2  pa Þ2 Fp ¼ 2Rp1 2Rp2 2Rp3

(5.81)

5.1 Liquid Systems Modeling

The following derivatives are needed in Lagrange’s Eq. (5.80):

vEC;p vEC;p ¼ Cp1 $ p1 ; ¼ Cp2 $ p2 ; vp1 vp2 vFp 1 1 ¼ $ ðp  p1 Þ þ $ ðp1  p2 Þ; Rp1 Rp2 vp1

(5.82)

vFp 1 1 ¼ $ ðp1  p2 Þ þ $ ðp2  pa Þ Rp2 Rp3 vp2 Substituting the derivatives of Eq. (5.82) into Lagrange’s Eq. (5.80), the two Eq. (5.79) result, which represent the mathematical model of the studied liquid system. nn

nn

Example 5.9 Derive the head-related mathematical model for the liquid system of Example 5.8 and illustrated in Figure 5.13 by expressing the heads h1 and h2 in terms of the pump pressure p, the atmospheric pressure pa, and the volume flow rate q. Apply both the element definition (energy balance) equations and Lagrange’s equations. Known are Ch1, Ch2, Rh1, and Rh2.

Solution

Assume that the pump pressure p is related to a head h as p ¼ r$g$h and, similarly, the atmospheric pressure pa relates to a head ha as pa ¼ r$g$ha. The head-defined resistances and capacitances shown in Figure 5.13 are

Rh1 ¼ Ch2 ¼

h  h1 h1  h2 h2  ha q1  q2 ; Rh2 ¼ ; Rh3 ¼ ; Ch1 ¼ ; : q1 q2 q3 h1 q þ q 2  q3

(5.83)

:

h2

Substitution of the q1, q2, and q3 from the first three Eq. (5.83) into the last two Eq. (5.83) results in the head-defined mathematical model of the liquid system:

  8 : 1 1 1 1 > > > < Ch1 $ h1 þ Rh1 þ Rh2 $ h1  Rh2 $ h2 ¼ Rh1 $ h   > : 1 1 1 1 > > : Ch2 $ h2  $ h1 þ þ $ ha $ h2 ¼ q þ Rh2 Rh2 Rh3 Rh3

(5.84)

The pressure-defined mathematical model of Eq. (5.79) is retrieved when using the following substitutions in Eq. (5.84): p ¼ r$g$h, pa ¼ r$g$ha, p1 ¼ r$g$h1, dp1/dt ¼ r$g$dh1/dt, p2 ¼ r$g$h2, dp2/dt ¼ r$g$dh2/dt, Ch1 ¼ r$g$Cp1, Ch2 ¼ r$g$Cp2, Rh1 ¼ Rp1/(r$g), Rh2 ¼ Rp2/(r$g), and Rh3 ¼ Rp3/(r$g). Considering that the generalized coordinates are h1 and h2, the two Lagrange’s equations resulting from the general Eq. (5.76) are:

    d vEC;h vFh d vEC;h vFh ¼ 0; ¼q þ þ dt vh1 dt vh2 vh1 vh2

(5.85)

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The capacitive energy and the dissipation function are:

1 1 EC;h ¼ $ Ch1 $ h21 þ $ Ch2 $ h22 ; 2 2 1 1 1 $ ðh  h1 Þ2 þ $ ðh1  h2 Þ2 þ $ ðh2  ha Þ2 Fh ¼ 2Rh1 2Rh2 2Rh3

(5.86)

The following derivatives are needed in Lagrange’s Eq. (5.85):

vEC;h vEC;h ¼ Ch1 $ h1 ; ¼ Ch2 $ h2 ; vh1 vh2 vFh 1 1 ¼ $ ðh  h1 Þ þ $ ðh1  h2 Þ; Rh1 Rh2 vh1

(5.87)

vFh 1 1 ¼ $ ðh1  h2 Þ þ $ ðh2  ha Þ Rh2 Rh3 vh2 Substituting the derivatives of Eq. (5.87) into Lagrange’s Eq. (5.85), we obtain the two Eq. (5.84), which represent the mathematical model of the studied liquid system. nn

5.2 PNEUMATIC SYSTEMS MODELING In pneumatic systems, the motion agent is a gas, most often air. Gases are compressible, particularly at large velocities; therefore, pneumatic systems produce responses that are slower than liquid systems, but for velocities that are smaller than the sound velocity, they are nearly incompressible. We discuss some basic gas laws and then introduce the pneumatic elements and modeling of pneumatic systems.

5.2.1 Gas Laws Gas laws describe either the state or the transformation (process) between different states of a gaseous substance. The perfect (or ideal) gas law postulates that, for a given gas state that is defined by pressure p, volume V, and absolute (Kelvinscale) temperature q, the following relationship applies for a gas mass of m: m p $ V ¼ $ R $ q or p $ V ¼ m $ Rg $ q (5.88) M where R is the universal gas constant, M is the gas molecular mass, and Rg is the gas constant, which is defined as Rg ¼ R/M. Eq. (5.88) allows expressing the gas mass density: r¼

m p$M p ¼ ¼ V R $ q Rg $ q

(5.89)

5.2 Pneumatic Systems Modeling

Transformation or process gas laws connect two states using specific conditions. The polytropic transformation is defined by the following equation:
mn (5.90) p ¼ a $ rn ¼ a $ V where a is a constant, and n is the polytropic exponent. Several particular transformations relating to actual physical conditions can be derived from the general polytropic transformation, each defined by a specific exponent n. All assume the gas mass m is constant. The adiabatic transformation, which considers no heat exchange between the gas and its surroundings, has an exponent defined as n ¼ cp/cv, where cp is the constant-pressure specific heat and cv is the constant-volume specific heat. The specific heat is defined as the heat (energy) Q necessary to raise the temperature of the mass unit by 1 degree: c ¼ Q/(m$Dq). Transformations that keep the temperature constant are called isothermal, and for such processes, the exponent is n ¼ 1. That can easily be checked as follows: Eq. (5.88) shows that pV ¼ constant, a condition that also results from Eq. (5.90) when n ¼ 1. Constant-pressure transformations (also called isobaric) satisfy the condition that the ratio V/q is a constant, as it results from Eq. (5.88). This means the polytropic exponent needs to be n ¼ 0, as can be checked in Eq. (5.90). Eventually, constant-volume transformations result in equations of the type p/q ¼ constant, as seen in Eq. (5.88). It can also be verified that such processes imply n / N because V ¼ constant when n / N.

5.2.2 Pneumatic Elements The pneumatic elements are defined similarly to liquid systems, particularly the inertance and the resistance, but as mentioned in the introduction to this chapter, the mass flow rate is used instead of the volume flow rate that operates for liquids. The pneumatic capacitance is discussed in terms of the specific gas transformation.

Inertance A column of gas moving in a duct possesses kinetic energy; therefore, its inertance is defined as: Ig ¼

p p $ dt ¼ q_m dqm

(5.91)

The SI unit of Ig is m1. It can be shown that Ig ¼ Il,p/r by comparing Eqs. (5.2) and (5.91). The kinetic energy of a gas is 1 EI;g ¼ $ Ig $ q2m 2 2 2 The energy’s SI unit is kg /(m s ).

(5.92)

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Capacitance The pneumatic capacitance of a container is defined as follows: Cg ¼

qm dm : ¼ dp p

(5.93)

The SI unit for gas capacitance is m s2. Comparing Eqs. (5.12) and (5.93) indicates that Cg ¼ r$Cl,p when the gas is incompressible. For a polytropic process, where the pressure is defined as in Eq. (5.90), p (5.94) dp ¼ n $ a $ rn1 $ dr ¼ n $ $ dr r which indicates the capacitance of Eq. (5.93) can be written for a constant-pressure gas as: Cg ¼

dm V $ dr V $ r V ¼ ¼ ¼ dp dp n $ p n $ Rg $ q

(5.95)

where Eqs. (5.89) and (5.93) have been used. It can be seen that, for a constantpressure transformation where the polytropic coefficient is n ¼ 0, the pneumatic capacitance is infinity, whereas for a constant-volume process with n / N, the pneumatic capacitance is equal to zero. In general, the pneumatic capacitance is variable, as it depends on temperature. For an isothermal transformation (where the temperature is constant), the pneumatic capacitance is constant and results from Eq. (5.95) with n ¼ 1: V Cg ¼ (5.96) Rg $ q The energy stored by a pneumatic capacitive element is 1 EC;g ¼ $ Cg $ p2 2 2 2 3 2 2 and its SI unit is N s /m ¼ kg /(m s ).

(5.97)

Resistance The pneumatic resistance is defined in terms of pressure variation and mass flow rate as p Rg ¼ (5.98) qm Gas resistance is measured in me1 se1 in the SI units. For incompressible gases, Rg ¼ Rl,p/rdcompare Eqs. (5.23) and (5.98). The energy dissipated through pneumatic resistance is 1 1 1 p2 ER;g ¼ $ p $ qm ¼ $ Rg $ q2m ¼ $ 2 2 2 Rg and the SI energy unit is N2 s/m3 ¼ kg2/(m s3).

(5.99)

5.2 Pneumatic Systems Modeling

Sources of Pneumatic Energy Similar to liquid systems, energy needs to be supplied to a pneumatic system to allow operation. The main pneumatic energy sources are the fans or blowers, for which the delivered pressure is a parabolic function of the air volume. This is expressed by means of an equation that also includes the energy losses, such as those due to impeller friction or shock:
c 2 2 p ¼ c1 $ 1 (5.100) V In Eq. (5.100), c1 and c2 are constants depending on the type of pneumatic source and performance.

5.2.3 Pneumatic Systems The natural response of conservative pneumatic systems is briefly discussed next, followed by an example illustrating the forced response of a pneumatic system with losses.

Natural Response Pneumatic systems with no losses, therefore defined by only inertance and capacitance properties, can be analyzed in terms of their natural response similarly to liquid, mechanical, or electrical systems. For a single-DOF pneumatic system, the natural frequency is calculated with an equation similar to the one of liquid systems, where the subscript g (gas) should be used instead of l, for liquid: 1 un;g ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi Ig $ Cg

(5.101)

The companion website Chapter 5 includes examples of calculating the natural frequencies of single- and multiple-DOF pneumatic systems, using the energy method and MATLAB, but the procedure is identical to that used for the natural response of liquid systems and is not pursued here. However, a few problems dedicated to this topic are proposed at end of this chapter.

Forced Response When pneumatic sources are included in a pneumatic system, the dynamic response is forced. We study the forced response of a two-DOF pneumatic system with resistance losses and negligible inertia, by using the energy balance methoddsimilar to the method used for liquid system modeling. nn

Example 5.10

A fan is used to pressurize the container of capacitance Cg 2 as in Figure 5.14, where another vessel of capacitance Cg1 is connected to the target vessel and the fan. Derive the mathematical model of this pneumatic system that connects the output pressure po to the input pressure created by the fan, pi, and considering the duct losses Rg1 and Rg 2.

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po Cg2

p po

Rg2

p

p

Cg1

qmo

Rg1

pi

qmi

fan

FIGURE 5.14 Pneumatic System With Fan, Two Containers, and Two Valves.

Solution The following equations are written for the four pneumatic components:

Rg1 ¼

pi  p p  po qmi  qmo qmo ; Rg2 ¼ ; Cg1 ¼ ; Cg2 ¼ : : qmi qmo p qo

(5.102)

The mass flow rate of the second Eq. (5.102) is substituted into the fourth Eq. (5.102), which enables expressing the pressure p and its time derivative as :

:

:

::

p ¼ po þ Rg2 $ Cg2 $ po ; p ¼ po þ Rg2 $ Cg2 $ po

(5.103)

The first and the third Eq. (5.102) are used in conjunction with Eq. (5.103), which yields the following differential equation:

 :: : Rg1 $ Rg2 $ Cg1 $ Cg2 $ po þ ½Rg1 $ Cg1 þ Rg1 þ Rg2 $ Cg2  $ po þ po ¼ pi (5.104) nn

5.3 THERMAL SYSTEMS MODELING In thermal systems, the focus is on heat and mass exchange among various states of a medium or different media. The analysis in this section is restricted to heat exchange. We introduce the thermal elements of capacitance and resistance, followed by the mathematical modeling of thermal systems. Since thermal inertia can safely be neglected, thermal systems behave as first-order systems. To keep notation unitary with that used for electrical, fluid, and pneumatic systems, the symbol qth or simply q is used here to indicate the heat flow rate, which is defined as the time derivative of the heat flow, or thermal energy Q: dQðtÞ dt As mentioned in Section 5.2, the symbol used here for temperature is q. qth ðtÞ ¼ qðtÞ ¼

(5.105)

5.3 Thermal Systems Modeling

5.3.1 Thermal Elements When neglecting thermal inertia, the elements of interest are the thermal capacitance (involved with energy storing) and thermal resistance (responsible for energy losses). These amounts are assumed to be of a lumped-parameter nature. The role of the electrical charge rate (current) in electrical systems or flow rate in fluid systems is played by the heat flow rate q in defining these thermal quantities.

Capacitance Thermal capacitance is connected to the energy storage capacity. It is defined as the heat flow necessary to change the temperature rate of a medium by one unit in one second: Cth ¼

q

¼

dQ dq

(5.106) q The SI unit for thermal capacitance is N m/deg ¼ J/deg, where deg stands for degree. As known from physics, the heat quantity Q is related to a change of temperature q in a medium (solid or fluid) of mass m as follows: :

dQ ¼ m$c (5.107) dq where c is the specific heat. The specific heat can be defined under process conditions of constant pressure, when denoted by cp, or constant volume, and then the symbol cv is used. Comparison of Eqs. (5.106) and (5.107) shows that dQ ¼ m $ c $ dq

or

Cth ¼ m $ c

(5.108)

Similarly to fluid capacitive elements, thermal capacitive elements carry potential energy, which is expressed as:
: 2 1 1 (5.109) EC;th ¼ $ Cth $ q2 ¼ $ Cth $ lth 2 2 In Eq. (5.109), lth is the thermal flux, which is defined as q ¼ d lth /dt. The SI unit for EC,th is N m deg.

Resistance Thermal resistance is formulated with regard to energy losses. Similar to electrical or fluid systems, thermal resistance is defined as the temperature variation produced by a unit heat flow rate: Rth ¼

q or q ¼ Rth $ q q

(5.110)

and its SI unit is deg s/J ¼ deg/W. To find a specific expression for Rth in terms of physical parameters, the types of heat transfer processes (conduction, convection, and radiation) need to be considered separately, as each process is governed by a specific law that connects the heat flow rate to temperature variation.

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Conduction In conduction, the heat is transmitted through one single medium (solid or fluid) as energy released by particles that possess more energy to adjacent particles having less energy. Our study of conduction thermal resistance is limited here to onedirectional heat flow through solids, under steady-state conditions. Fourier’s law of heat conduction governs the conduction process. q

Δθ x dx

dr

k

A

r

θ1

θ(x) θ2

ri

d ( θ) dx

θ1 ro

q x

θ2 q

q

l

(a)

(b)

FIGURE 5.15 Conduction Heat Transfer Through: (a) Planar Wall; (b) Hollow Cylinder.

For a planar wall, such as the one of Figure 5.15(a), Fourier’s law of heat conduction is: dq or q $ dx ¼ k $ A $ dq (5.111) dx where k is the thermal conductivity of the wall material, A is the area of the surface normal to the heat flow direction, dx is the thickness of an elementary layer (of total thickness l), and q is the temperature at the surface determined by the abscissa x. The temperature to the left of the wall, q1, is assumed higher than the temperature to the right of the wall, q2; therefore, the heat flow direction is from left to right. The minus sign in Eq. (5.111) indicates that the temperature decreases in the wall from left to right; therefore, the slope of the temperature as a function of distance is negative. As a consequence, a positive heat flow rate going from the higher temperature to the lower one is possible only with a minus sign, as in Eq. (5.111). Integrating the differential Eq. (5.111), whose variables are separable, with the limits 0 and l for x and q1 and q2 for the temperature, results in an equation similar to Eq. (5.110) with the thermal resistance: q ¼ k $ A $

l $ q / Rcond ¼ l=ðk $ AÞ (5.112) k$A Heat flow is also one-dimensional for the hollow cylinder of Figure 5.15(b), which is defined by the inner radius ri, outer radius ro, total out-of-the-plane cylinder q $ l ¼ k $ A $ ðq1  q2 Þ or q1  q2 ¼

5.3 Thermal Systems Modeling

length l, and thermal conductivity k. When the temperature in the cylindrical cavity, q1, is higher than the temperature outside the cylinder, q2, the heat flows radially from the inside toward the outside. Consider the heat flow rate through an annular strip of thickness dr. The area of this elementary strip, which is normal to the radial flow direction, is A ¼ 2p$r$l. In the case of the cylinder the length variable is r, and Fourier’s law of heat conduction is as follows: q ¼ kð2p $ r $ lÞ $

dq dr 2p $ l $ k or ¼ $ dq dr r q

(5.113)

which is a differential equation with the variables r and q separated. Integration of it, with r ranging between ri and ro, and q varying within the q1  q2 interval, results in an equation similar to Eq. (5.110), with the thermal resistance being as follows: ro 2p $ l $ k $ ðq1  q2 Þ or ¼ q ri   1 ro 1 ro $ ln $ ln q1  q2 ¼ $ q/Rcond ¼ 2p $ l $ k 2p $ l $ k ri ri ln

(5.114)

Convection In convection, the heat is transmitted between two media, one of them a fluid in motion, and the second, in most of the cases, (the surface of) a solid. In general, the solid has a higher temperature; therefore, heat transfers from the solid to the moving fluid. The governing law defining the heat flow, known as Newton’s law of cooling, is expressed as qconv ¼ hconv $ A $ q / Rconv ¼ 1=ð hconv $ AÞ

(5.115)

where hconv is the convection heat transfer coefficient, A is the area of the surface of the solid in contact with the moving fluid, and Dq ¼ qs  qN. The temperature at the heat exchange surface is qs, and qN is the fluid temperature at a distance far enough from the surface. Values of hconv are generally obtained experimentally, as h depends on many factors, such as surface geometry, fluid and solid properties, and fluid velocity. Comparison of Eqs. (5.110) and (5.115) shows that the thermal resistance corresponding to convection is the one provided in Eq. (5.115).

Radiation Radiation involves emission of heat by matter that is at a finite temperature, transmission of the heat by electromagnetic waves, and heat absorption by other bodies. The radiative heat absorption by a solid body of exposed surface A can be expressed similarly to the convection Eq. (5.115) and results in the following radiation resistance: qrad ¼ hrad $ A $ q/Rrad ¼ 1=ð hrad $ AÞ

(5.116)

In Eq. (5.116), q is the difference between the radiation source temperature and the recipient body surface temperature. The coefficient hrad is the radiation heat transfer coefficient, which depends on the emission and absorption properties of the source and target, as well as on their temperaturesdin a nonlinear fashion.

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The thermal resistive energy is expressed similarly to the lost energy in electric or fluid systems as:
$ 2 2 lth 1 1 1 q 1 (5.117) ER;th ¼ $ qth $ q ¼ $ Rth $ q2th ¼ $ ¼ $ 2 2 2 Rth 2 Rth where lth is the thermal flux. The resistive energy is measured in J deg/s ¼ W deg.

Mixed Heat Flow Through Composite Walls Composite walls, planar or cylindrical, result from assembling layers of different geometric and material parameters to achieve thermal properties, such as better insulation, which cannot be obtained readily through single-material layers. 1 2

1 q1 θi

θo

q

q

θi

q12 q

θo

θi

θo q3

q2 1

2

2

3

q1 R 1 θi

R1

R2

θo

q (a)

q

R1 θo

θi

q

θi

R2

q12 q3

q2 R2

R3

(b)

(c)

θo

FIGURE 5.16 Connection of Planar Walls in (a) Series; (b) Parallel; (c) Mixed Series/Parallel.

Figure 5.16 illustrates three composite wall configurations that are formed of different planar sheets. Two sheets are combined in series in Figure 5.16(a), as the same heat flow rate q passes through them, whereas Figure 5.16(b) illustrates two sheets that are mixed in parallel with the incoming heat flow rate q splitting into q1 that passes through layer 1 and q2 passing through layer 2. Figure 5.16(c) depicts a wall design with three panels: panels 1 and 2 are connected in series, and both are placed in parallel with panel 3. Correspondingly, the conduction thermal resistances pertaining to these layers are connected in series, in parallel, or in series/parallel. The equivalent series and parallel resistances of Figure 5.16(a) and 5.16(b)dfull derivation is provided in the companion website Chapter 5dare calculated as: R s ¼ R1 þ R 2 ;

1 1 1 R1 $ R2 ¼ þ or Rp ¼ Rp R1 R2 R1 þ R2

(5.118)

5.3 Thermal Systems Modeling

Noting the similarity between electrical resistance and thermal resistance series/parallel connection, the equivalent thermal resistance for the wall of Figure 5.16(c) is: Req ¼

R12 $ R3 ðR1 þ R2 Þ $ R3 ¼ R12 þ R3 R 1 þ R2 þ R3

(5.119)

The principles illustrated in Figure 5.16 for planar composite walls also apply for cylindrical composite walls. Imagine that instead of being planar, the layers of Figure 5.16 are cylindrical. With the same direction of the heat flow rate, which is now radial as it passes through concentric annular strata, it can easily be seen that the different cylindrical layers are combined, as indicated in the same Figure 5.16. Oftentimes, one-direction conduction, convection, and radiation heat transfer occur simultaneously through composite walls, and thermal resistances pertaining to each of these mechanisms have to be considered, as discussed in the following. Convection and Radiation With the assumption that the same temperature difference q governs both heat transfer mechanisms, the convective and radiative heat absorbed by a solid body through an area A is as follows:   1 1 q ¼ qconv þ qrad ¼ ðhconv þ hrad Þ $ A $ q ¼ þ $q (5.120) Rconv Rrad which indicates that when radiation is taken into account alongside convection, an overall heat transfer coefficient hcr ¼ hconv þ hrad needs to be utilized. Moreover, when taking into account into account Eqs. (5.115) and (5.116), the result of Eq. (5.120) indicates that the two resistances are connected in parallel, namely 1 1 $ Rconv $ Rrad 1 1 h $ A hrad $ A ¼ (5.121) ¼ conv Rcr ¼ ¼ 1 1 ðhconv þ hrad Þ $ A hcr $ A Rconv þ Rrad þ hconv $ A hrad $ A Conduction, Convection, and Radiation Consider the wall of Figure 5.17(a), which separates an indoor conditioned environment (such as a room) characterized by a temperature qi from the outdoors environment, which is defined by a temperature qo < qi. l θi

θo A

hconv,i + hrad

hconv,o q

Rconv,i θi

Rcond Rconv,o

q

θo

q

Rrad k

(a)

(b)

FIGURE 5.17 Planar Wall With Convection, Radiation, and Conduction Heat Transfer (a) Schematic Diagram; (b) Thermal Resistance Connection.

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The wall has a surface area A and a thickness l, and its thermal conductivity is k. Assume that the wall surface in contact with the room absorbs heat through both convection (defined by a coefficient hconv,i) and radiation from a source (defined by a coefficient hrad). Similarly, there is convective heat exchange between the wall external surface and the outdoors characterized by a coefficient hconv,o. Figure 5.17(b) shows the thermal resistances corresponding to the mixed heat transfer and their connections. The inner convection and radiation resistances, as just demonstrated, are coupled in parallel. The incoming flow rate passes unaltered through the planar wall; and, as a consequence, the conduction resistance and the outer convection resistance are connected in series, as depicted in Figure 5.17(b). As a result, the equivalent resistance substituting the four actual resistances is calculated as:   Rconv;i $ Rrad 1 1 l 1 R¼ þ Rcond þ Rconv;o ¼ $ þ þ (5.122) A hconv;i þ hrad k hconv;o Rconv;i þ Rrad

nn

Example 5.11 Consider the heat conduction through the composite wall sketched in Figure 5.18(a). The external layer is a plywood panel, the middle insulation layer is made of fiberglass, and the inner layer is a gypsum board. Known are the inside and outside temperatures qi, qo (qi > qo), the common surface area A, the wall component thicknesses l1, l2, and l3, and the layer’s thermal conductivities k1, k2, and k3. There is convection on the outside surface (film coefficient is hconv,o), while the inside one there is mixed convection (film coefficient is hconv,i) and radiation (film coefficient is hrad). (i) Calculate the equivalent resistance of the composite wall accounting for conduction, convection, and radiation; evaluate the heat flow rate through the composite wall. (ii) Calculate the temperatures at the inner surface and at the outer surface.

θo

θi hi

ho

Rconv,i

q q A

k1

k2

θi

θsi Rrad

k3

R2 R1

q

θso R3 Rconv,o

θo

Plywood

Gypsum Fiberglass l3

l1 l2

(a)

(b)

FIGURE 5.18 Three-Panel Composite Wall: (a) Schematic Representation; (b) Thermal Resistance Connection.

5.3 Thermal Systems Modeling

Numerical application: l1 ¼ 0.012 m, l2 ¼ 0.15 m, l3 ¼ 0.02 m, A ¼ 30 m2, k1 ¼ 0.17 W/ (m deg), k2 ¼ 0.035 W/(m deg), k3 ¼ 0.16 W/(m deg), hconv,o ¼ 52 W/(m2 deg), hconv,i ¼ 18 W/ (m2 deg), hrad ¼ 10 W/(m2 deg), qi ¼ 23 C, and qo ¼ 15 C.

Solution (i) The thermal resistance connectivity is illustrated in Figure 5.18(b). The interior convection and radiation result in the parallel-connection resistances Rconv,i and Rrad. They are further coupled in series, with the conduction resistors corresponding to the three structural layers, which are denoted by R1, R2, R3, and eventually to the convective resistor Rconv,o. The equivalent thermal resistance is therefore:

Req ¼

Rconv;i $ Rrad þ R1 þ R2 þ R3 þ Rconv;o Rconv;i þ Rrad

(5.123)

Substituting Eqs. (5.112), (5.115), (5.116), and (5.121) into Eq. (5.123) results in:

Req

  1 1 l1 l2 l3 1 ¼ $ þ þ þ þ A hconv;i þ hrad k1 k2 k3 hconv;o

(5.124)

The heat flow rate is

q¼

qi  qo Req

(5.125)

With the given numerical parameters, the thermal resistance of Eq. (5.124) is Req ¼ 0.1512 deg/ W, and the heat flow rate of Eq. (5.125) is q ¼ 251.309 W. (ii) The temperature on the inner surface of the wall is evaluated based on the heat flow of Eq. (5.125) and the parallel resistance resulting from convection and radiation:

 qi  qsi ¼ hcond;i þ hrad $ A $ ðqi  qsi Þ or Rcond;i $ Rrad Rcond;i þ Rrad q qsi ¼ qi   hcond;i þ hrad $ A q¼

(5.126)

A similar reasoning allows expressing the temperature on the outer surface of the wall by taking into account the resistances corresponding to interior convection, radiation, as well as the three conduction resistances as:

q¼

qi  qso A $ ðqi  qso Þ ¼ Rcond;i $ Rrad 1 l1 l2 l3 þ R1 þ R2 þ R2 þ þ þ Rcond;i þ Rrad hcond;i þ hrad k1 k2 k3

or qso ¼ qi 



q 1 l1 l2 l3 $ þ þ þ A hcond;i þ hrad k1 k2 k3

(5.127)



Numerically, the temperatures of Eqs. (5.126) and (5.127) are qsi ¼ 22.70 C and qso ¼ 14.84 C. nn

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CHAPTER 5 Fluid and Thermal Systems

nn

Example 5.12 Figure 5.19(a) illustrates a joint between two identical steel pipe segments, denoted by 1, which are welded frontally and are used for undersea liquid transportation. A polypropylene coating 2 layers the pipe segments. Solid polyurethane 3 is used between the two pipes respective coatings, and a polyethylene sleeve 4 encases the resulting joint. Consider that the radial heat flow is from the warmer liquid flowing through the pipe at temperature qi to the cooler outer sea water of temperature qo. Calculate the following amounts for the joint segment defined by l1 and: (i) The equivalent resistance. (ii) The heat flow rate, as well as the temperatures qsi on the pipe inner surface and qso on the sleeve outer surface. Known are d1 ¼ 0.25 m, d2 ¼ 0.27 m, d3 ¼ 0.37 m, d4 ¼ 0.39 m, l1 ¼ 0.16 m, and l2 ¼ 0.09 m; the thermal conductivities k1 ¼ 30 W/(m deg), k2 ¼ 0.12 W/(m deg), k3 ¼ 0.22 W/(m deg), and k4 ¼ 0.33 W/(m deg); and the inner and outer convection coefficients hi ¼ 40 W/(m2 deg) and ho ¼ 100 W/(m2 deg), also qi ¼ 60 C, qo ¼ 5 C. θo

l1 4

l3

l2

3

2

Welding 1

d4

d1 A

B

(a)

Ro θo R4 R3

q θi

d2

θo

q θso q2

q3

q

R2

R3

qA

RA1

Ri

q θso

RA4

d3 R1 θsi

Ro

RB4 qB

RB1

θsi Ri

R2

q

θi

θi

(b)

(c)

FIGURE 5.19 Multilayer Pipe Joint: (a) Schematic Representation; (b) Model 1 Thermal Resistance Connection; (c) Model 2 Thermal Resistance Connection.

Solution (i) The resistance connection model illustrated in Figure 5.19(b) and called Model 1 assumes that with respect to the radial flow direction, the conduction portion has the pipe resistance R1 in series with the parallel connection of R2 (the resistance of the coating) and R3 (the polyurethane resistance) and then in series with the external sleeve resistance R4. Another resistance connectivity is suggested in Model 2 of Figure 5.19(c), which considers two regions, A and B in Figure 5.19(a), defined by the longitudinal separation between layers 2 and 3. Each region is formed of three resistances coupled in series, the two resulting branches being connected in paralleldsee Figure 5.5(c). In the heat transfer literaturedsee Incropera and DeWitt [1] or Cengel [2]dthe basic assumption of Model 1 is that surfaces perpendicular to the heat flow direction (which is the radial direction in this problem) are isothermal surfacesdwith the same temperature, whereas in Model 2 all surfaces parallel to the heat flow direction are adiabatic surfacesdwith no heat exchange along them. In both Model 1 and Model 2, the resistances owing to convection, Ri and Ro, add in series to the conductive portion of the resistance circuit. As per Figure 5.19(b) and 5.19(c), the equivalent resistances provided by the two models are as follows:

R2 $ R 3 þ R4 þ R o ; R2 þ R3 ðRA1 þ R3 þ RA4 Þ $ ðRB1 þ R2 þ RB4 Þ ¼ Ri þ þ Ro RA1 þ R3 þ RA4 þ RB1 þ R2 þ RB4

Req;1 ¼ Ri þ R1 þ Req;2

(5.128)

5.3 Thermal Systems Modeling

The conduction resistances of Eq. (5.128) are calculated as per the definition Eq. (5.114) and the convection ones with Eq. (5.115) as follows:

R1 ¼

lnðd2 =d1 Þ lnðd3 =d2 Þ lnðd3 =d2 Þ lnðd4 =d3 Þ ; R2 ¼ ; R3 ¼ ; R4 ¼ ; 2p $ l1 $ k1 2p $ l2 $ k2 2p $ l3 $ k3 2p $ l1 $ k4

RA1 ¼ Ri ¼

lnðd2 =d1 Þ lnðd4 =d3 Þ lnðd2 =d1 Þ lnðd4 =d3 Þ ; RA4 ¼ ; RB1 ¼ ; RB4 ¼ ; 2p $ l3 $ k1 2p $ l3 $ k4 2p $ l2 $ k1 2p $ l2 $ k4

4 4 ; Ro ¼  2 hi $ p $ d1 ho $ p $ d42 

(5.129) The numerical values of the equivalent resistances are as follows: Req,1 ¼ 2.6682 deg/W, Req,2 ¼ 2.6819 deg/W, which are not very different. (ii) The flow rates for the two models are therefore:

q1 ¼

qi  qo qi  qo ; q2 ¼ Req;1 Req;2

(5.130)

and numerically: q1 ¼ 20.6128 W and q2 ¼ 20.5078 W. The temperatures qsi on the pipe inner surface and qso on the sleeve outer surface are calculated based on the two-resistance models of Figure 5.19(b) and 5.19(c):



qsi;1 ¼ qi  Ri $ q1 ; qso;1 qsi;2 ¼ qi  Ri $ q2 ; qso;2

 R2 $ R3 ¼ q i  Ri þ R1 þ þ R4 $ q 1 ; R2 þ R 3   ðRA1 þ R3 þ RA4 Þ $ ðRB1 þ R2 þ RB4 Þ ¼ q i  Ri þ $ q2 RA1 þ R3 þ RA4 þ RB1 þ R2 þ RB4 (5.131)

with the following numerical values: qsi,1 ¼ 49.502 C, qsi,2 ¼ 49.556 C, qso,1 ¼ 6.726 C, and qso,2 ¼ 6.717 C. nn

Heat Sources Thermal or heat sources represent the forcing agents in dynamic thermal systems. For a thermally conditioned space, such as a room, heating can be achieved by means of an electric heater, or through hot water or hot air; these sources provide an input flow rate qi. Conversely, the room can be cooled by means of heat sinks in the form of cold water or cold air being circulated in it; this is the equivalent of removing (negatively adding) the input flow rate qi. Other sources of positive qi are exothermal chemical reactions or nuclear reactions, whereas negative qi are obtained by endothermal chemical reactions. Boundary conditions may also be heat sources, as detailed in more advanced heat transfer texts.

5.3.2 Thermal Systems Thermal systems, as we saw in the few examples discussed thus far in this section, are used for heating or cooling (either as simple processes or as parts of the more

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complex indoor air conditioning equipmentdHVAC, heating, ventilating, and air conditioning), in sensing devices (such as thermometers) or in thermal homogenizing equipment. Systems formed of thermal capacitors and resistors are described by differential equations that are the mathematical models of the analyzed thermal systems. These equations are usually first-order differential equations because thermal mass transfer (which would be responsible for second time derivatives) is not included. Similar to mechanical, electrical, or fluid systems, thermal systems may have one or more DOF, thus resulting in single or multiple differential equations. While a formal analysis of DOF for thermal systems can be carried out, it is simply stated here that temperatures are thermal DOF. Consider the elementary example of a wall of resistance R that separates a room of temperature qi from the outdoors, whose temperature is qo. While qi and qo are candidate DOF, they are related by means of the flow rate q as qi  qo ¼ q$R. This relationship indicates that qi and qo are dependent; as a result, the system has one DOF.

Single-DOF Systems Mathematical models of single-DOF thermal systems consist of one first-order differential equation, mainly derived by using the thermal energy balance. According to the first law of thermodynamics, for a system with pure heat transfer (such as the ones studied here), the heat quantity defining the system at any moment in time is the difference between the heat entering the system and the heat exiting the system, which is Q ¼ Qin  Qout. A similar energy balance equation can be written with heat flow rates used instead of heat quantities: q ¼ qin  qout. nn

Example 5.13

A long cylindrical body is immersed into a bath with hot liquid of temperature qb ¼ 150 C. The cylinder has a circular cross section of radius r ¼ 0.16 m, mass density r ¼ 7800 kg/m3, and specific heat c ¼ 435 J/(kg deg). Considering that heat transfer between the bath and the body occurs by convection through the lateral cylinder surface (convection coefficient is h ¼ 290 W/(m2 deg)), and assuming the temperature of the rod is uniform, obtain a mathematical model of the body temperature time variation, calculate this temperature, and plot it versus time. Assume that the body’s initial temperature is q0 ¼ q(0) ¼ 23 C.

Solution The energy equation for the cylinder takes into account that the body’s heat flow rate is gained through convection from the surrounding hot fluid: :

q ¼ qin or C $ q ¼

qb  q R

(5.132)

where C is the cylinder thermal capacitance, R is the convection resistance at the fluidebody interface, and calculated as follows: C ¼ m$c ¼ pr2$l$r$c and R ¼ 1/(h$A) ¼ 1/(2pr$l$h). With these expressions, Eq. (5.132) becomes: :

R $ C $ q þ q ¼ qb or

: r$r$c : $ q þ q ¼ qb or s $ q þ q ¼ qb 2h

(5.133)

θ [oC]

5.3 Thermal Systems Modeling

Time [s]

FIGURE 5.20 Time Variation of Cylinder Temperature.

where, as discussed in Chapter 1, the time constant is s ¼ r$r$c/(2h) ¼ 936 s. The solution to the differential Eq. (5.133) is typical for a first-order system, namely: t

qðtÞ ¼ qb  ðqb  q0 Þ $ es

(5.134)

which is plotted in Figure 5.20. It can be seen that q reaches the bath temperature in roughly 4 times the s value. nn

nn

Example 5.14

The room sketched in Figure 5.21(a) has a thermal capacity C ¼ 185,000 J/deg, a total wall thermal resistance R ¼ 1.135 deg/W, and an initial temperature q0 ¼ q(0) ¼ 22 C. Assume it is wintertime and a constant outside temperature qo ¼ 12 C. The room heating system provides the input heat flow rate shown in Figure 5.21(b), where s is the time constant of the room thermal system, and qmin is the minimum heat flow rate that balances the effect of the outdoors temperature. Derive the mathematical model of the room temperature q(t) and determine the time constant of this system s. Solve the mathematical model to find q as an algebraic expression in terms of t; plot q(t) for the [0; 6s] time interval; and calculate the steady-state room temperature value qðNÞ ¼ lim qðt Þ. t/N

Solution The energy equation for the room equates the stored capacitive room energy with the gained energy (qi) and the energy lost through the walls (qo): :

q ¼ qi  qo or C $ q ¼ qi 

q  qo R

(5.135)

When the right-hand side of Eq. (5.135) is zero, the input flow rate balances out the outgoing flow rate. In that situation, the indoors temperature would be constant (because dq(t)/dt ¼ 0), which means that

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CHAPTER 5 Fluid and Thermal Systems

qo

θo

qi (W)

θ qo

qo

qi

q1 = 1.5·qmin

Heater q2 = 1.1·qmin

R

C

0

τ

6τ Time (s)

qo

(a)

(b)

FIGURE 5.21 (a) One-Room Thermal System With Internal Heating; (b) Input Heat Flow Rate Profile.

q(t) ¼ q(0). As a consequence, the minimum input heat flow rate is qmin ¼ (q(0)  qo)/R ¼ 29.9559 W. The two values defining the actual flow ratedas per Figure 5.21(b)dare q1 ¼ 1.5$qmin ¼ 44.9339 W and q2 ¼ 1.1$qmin ¼ 32.9515 W. Eq. (5.135) is written as the first-order differential equation: :

:

R $ C $ q þ q ¼ R $ qi þ qo or s $ q þ q ¼ R $ qi þ qo

(5.136)

which is the mathematical model of the room system. The constant s ¼ R$C is the time constant, and its value is s ¼ 209,975 s. The general solution to Eq. (5.136) is: t

q ¼ R $ qi þ qo þ K $ es or 8 > t > ( > < R $ q1 þ qo þ K1 $ es ; q1 ; 0  t  s ¼ q¼ t > q2 ; s  t  6s > > : R $ q þ q þ K $ es ; 2

o

2

0ts

(5.137)

s  t  6s

with qi defined as in Figure 5.21(b). The constants K1 and K2 are found by using the initial conditions: q1(0) ¼ q0 ¼ 22 C and q1(s) ¼ q2(s) in conjunction with Eq. (5.137). After determining K1 and K2 and some algebraic manipulation, Eq. (5.137) becomes as follows:

q¼

8 > t > < R $ q1 þ qo  ½R $ q1 þ qo  qð0Þ $ es ; > > : R $ q2 þ q o þ

t ðq1  q2 Þ $ R $ e1 s

0ts t

 ½R $ q1 þ qo  qð0Þ $ es ;

s  t  6s (5.138)

The plot of q(t), as it results from Eq. (5.138), is shown in Figure 5.22. The steady-state room temperature is qðNÞ ¼ lim qðt Þ ¼ 25.4 C, whereas the maximum temperature is q(s) ¼ 32.75 C. t/N

Temperature [ oC]

5.3 Thermal Systems Modeling

θ1 θ2

τ Time [s·105]

FIGURE 5.22 Time Variation of Room Temperature. The plot of Figure 5.22 can also be obtained using Simulink to integrate the first-order differential Eq. (5.136). For qi a Step source block is needed with a Step Time of s, an Initial Value of q1 and a Final Value of q2 as Parameters. nn

Multiple-DOF Systems Mathematical models can be derived using of the thermal energy balance method or the Lagrange’s equations. The method of energy was introduced in the previous section dedicated to single-DOF thermal systems. Lagrange’s equations are applied in a manner similar to the one that used node voltages/fluxes and the node analysis method for electrical systems. The generalized coordinates qj are the independent fluxes lj (qj ¼ dlth,j/dt ¼ dlj/dt) and the Lagrangian L is: L ¼ EC  EL ¼ EC. We take into account that the capacitive energy EC contains terms of the form Cth $ q2/2 ¼ Cth $ (dlth/dt)2/2 (as in Eq. (5.109)), whereas EL ¼ 0 since there are no thermal inertia contributions. The Rayleigh dissipation function F adds all resistive energy contributionsdsee Eq. (5.117)dand is expressed as: n 1 X 1  _ 2 F¼ $ $ lj 2 j¼1 Rj

(5.139)

Lagrange’s equations are: ! ! ! d vL vL vF d vEC vF d vEC vF  þ þ þ ¼ qj or ¼ qj or ¼ qj dt vl_ j vlj vl_ j dt vl_ j dt vqj vqj vl_ j (5.140)

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for j ¼ 1, 2, ., n. The term in the right-hand side of Eq. (5.140) is the thermal forcing function and is actually a heat flow rate. nn

Example 5.15 Figure 5.23 shows two identical rooms, with six identical external walls. A cooling system operates in only one of the rooms. Consider it is summertime, and heat flow infiltrates from the outside. Heat also flows through the wall separating the two rooms. Known are the following amounts: heat flow absorbed by the cooler qi, the outside temperature qo, the walls thermal resistances Ro (total resistance of one room’s external walls), R, and the two rooms’ thermal capacitances C. Find the mathematical model of this thermal system that expresses the two indoor temperatures q1 and q2. Use both the energy balance method and Lagrange’s equations.

qo1

θo Ro

qo2

θ1

θo

θ2

qo1

R

C

qo2

qi

q

Cooler Ro

C

qo1

qo2

FIGURE 5.23 Two-Room Thermal System With Internal Cooling.

Solution Conservation of Energy Method The following equations can be written for the left and right rooms, in terms of thermal capacitances: :

:

C $ q1 ¼ qo1  q; C $ q2 ¼ qo2 þ q  qi

(5.141)

The room walls behave as thermal resistances; therefore, the following relationships can be formulated:

Ro ¼

qo  q1 qo  q2 q1  q2 ; Ro ¼ ; R¼ qo1 qo2 q

(5.142)

which correspond to the three external walls on the left room, to the three external walls of the right room, and to the room separating wall, respectively. The three heat flow rates are expressed from Eq. (5.142) and substituted into Eq. (5.141) adequately, such that the latter ones become

  8 : 1 1 1 1 > > > < C $ q1 þ Ro þ R $ q1  R $ q2 ¼ R0 $ qo   > : 1 1 1 1 > > : C $ q2 þ þ $ q 2  $ q 1 ¼ $ q o  qi Ro R R R0

(5.143)

5.3 Thermal Systems Modeling

The two differential Eq. (5.143) form the mathematical model of the thermal system of Figure 5.23. Lagrange’s Equations Method The capacitive energy and the Rayleigh dissipation function are:

1 1 EC ¼ $ C $ q21 þ $ C $ q22 ; 2 2 1 1 1 $ ðq1  q2 Þ2 $ ðqo  q1 Þ2 þ $ ðqo  q2 Þ2 þ F¼ 2Ro 2Ro 2R

(5.144)

With q1 and q2 being the DOF, the Lagrange’s equations resulting from the general Eq. (5.140) are:

    d vEC vF d vEC vF ¼ 0; ¼ qi þ þ dt vq1 vq1 dt vq2 vq2

(5.145)

The partial derivatives needed in Eq. (5.145) are obtained from Eq. (5.144) as:

8 vEC > > > < vq ¼ C $ q1 ; 1 > vE > C > ¼ C $ q2 ; : vq2

vF 1 1 ¼ $ ðq1  qo Þ þ $ ðq1  q2 Þ vq1 Ro R vF 1 1 ¼ $ ðq2  qo Þ þ $ ðq2  q1 Þ vq2 Ro R

(5.146)

which, on substitution into Eq. (5.145), result in Eq. (5.143)dthe mathematical model of the thermal system of Figure 5.23. nn

nn

Example 5.16 Consider the following numerical values for the two-room thermal system of Example 5.15: R ¼ 0.8889 deg / W, Ro ¼ 1.2 deg/W, C ¼ 250,000 J/deg, and qi ¼ 20 W. Assume the outside temperature varies as indicated in Figure 5.24(a). Use Simulink and a time interval of 9$105 s to

θo (oC) 46 38 30 0

1

2

3

4 4.5

(a)

Time (s·103)

(b)

FIGURE 5.24 (a) Outside Temperature Variation With Time; (b) Parameters of Simulink’s Repeating Signal Source Block.

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CHAPTER 5 Fluid and Thermal Systems

plot the two room temperatures considering the initial temperatures are q1(0) ¼ 29.9 C in the left room and q2(0) ¼ 27 C in the right room.

Solution The input (outside) temperature can be modeled in Simulink by various sources, such as Signal Builder (which is used in Example 3) or Repeating Sequence, which is used in this example. This latter block is in the Sources library, and its Parameters window is shown in Figure 5.24(b). Note that each relevant point on the plot of qo(t) is represented by a pair of coordinates: one in the Time Values table and the other (the temperature) in the Output Values table, as indicated in Figure 5.24(b). The mathematical model Eq. (5.143) can be written as

(

where

a11

q_ 1 ¼ a11 $ q1 þ a12 $ q2 þ a13 $ qo q_ 2 ¼ a12 $ q1 þ a11 $ q2 þ a13 $ qo þ a21

  1 1 1 1 1 qi ; a13 ¼ ; a21 ¼  ¼ $ þ ; a12 ¼ C Ro R R$C Ro $ C C

(5.147)

(5.148)

a12·θ2 θo

a13·θo

dθ1/dt

θ1

a11·θ1 a12·θ1 a11·θ2

dθ2/dt

θ2

a21

FIGURE 5.25 Simulink Diagram for Integrating the Differential Eq. (5.147).

The parameters above have the following numerical values: a11 ¼ 7.8$106, a12 ¼ 4.5$106, a13 ¼ 3.33$106, and a21 ¼ 0.00008. The Simulink diagram, which integrates Eq. (5.147), is shown in Figure 5.25, whereas the time-domain room temperatures are plotted in Figure 5.26. As Figure 5.26 indicates, the temperature in the room without a cooling unit slightly increases initially and reaches a stable value of approximately 29.5 , whereas the temperature in the cooled room slightly decreases toward a steady-state value a little above 23 .

5.4 ElectricaleFluideThermal System Analogy

Room Temperatures (oC)

θ1

θ2

Time (s·103)

FIGURE 5.26 Room Temperatures Versus Time. nn

5.4 ELECTRICALeFLUIDeTHERMAL SYSTEM ANALOGY As seen in this chapter, elements of fluid and thermal systems are defined similarly with electrical-system elements, and, therefore, there are analogies between these systems. Analogous systems from different fields have to be defined by similar mathematical models, as was the case with the mechanicaleelectrical analogy examples studied in Chapter 4. The discussion of this section centers on systems where inertia properties is neglected. nn

Example 5.17 Demonstrate that the electrical, operational-amplifier, and pneumatic systems of Figure 5.27 are analogous.

Solution

The electrical circuit sketched in Figure 5.27(a) is a voltage divider; for a current i passing through the two components, the following equations are written:

: vi  vo ¼ R $ i : or R $ C $ yo þ vo ¼ vi C $ yo ¼ i

(5.149)

with R and C being the electrical resistance and capacitance, whereas vi and vo are the input and output voltages.

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CHAPTER 5 Fluid and Thermal Systems

R R

vi

L

– +

v

R

C

vo

vi

po Cg

vo

qm pi Rg

(a)

(c)

(b)

FIGURE 5.27 Analogous Systems: (a) Electrical; (b) Electrical With Operational Amplifier; (c) Pneumatic. The three elements of the operational-amplifier circuit of Figure 5.27(b) are in series; and therefore, the current that passes through each of them is as follows:

vi  y 1 ¼ $ R L

Z

ðy  0Þdt ¼

0  vo R

(5.150)

Using the last equality of Eq. (5.150) and applying the time derivative to it, the intermediate voltage is found and substituted in the equality [vi(t)  v(t)] ¼ vo(t):

L : L : $ yo þ vo ¼ vi y ¼  $ yo ; R R

(5.151)

Eventually, the following equations apply to the two elements of the pneumatic system of Figure 5.27(c):

(

: p i  p o ¼ Rg $ q m : or Rg $ Cg $ po þ po ¼ pi Cg $ po ¼ qm

(5.152)

where Rg and Cg are the pneumatic (gas) resistance and capacitance, whereas pi and po are the input and output pressures. Eqs. (5.149), (5.151), and (5.152) are similar first-order differential equations. As a consequence, the three systems of Figure 5.27 are analogous. The quantities R$C in Eq. (5.149), L/R in Eq. (5.151), and Rg$Cg in Eq. (5.152) are all time constants (as introduced in Chapter 1), as they relate to first-order systems. nn

Table 5.1 shows the physical amounts that are analogous for electrical, liquid, gas, and thermal elements. A more systematic approach in identifying analog systems starting from a given system and based on the groups of Table 5.1 requires applying the four steps mentioned in Chapter 4: (1) Derive the mathematical model of the given system; (2) Using the pairs of Table 5.1, write equations of the analog system to be identified; (3) Based on the equations obtained at (2) propose the structure and schematic of the

Table 5.1 ElectricaleLiquidePneumaticeThermal Analogy Electrical Liquid Gas Thermal

v p p q

i qv qm qth

R Rl Rg Rth

C Cl Cg Cth

5.4 ElectricaleFluideThermal System Analogy

analog system; and (4) Derive independently the mathematical model of the system proposed at (3) and verify if it is identical to the model resulted at (2). nn

Example 5.18 Find a liquid system and a thermal system that are analogous to the electrical system of Figure 5.28(a).

vi

i1

i2

v

R1

C1

i1 − i2

R2

θi vo

θ Cl1 Rl1 pi

Ground,vg = 0

Rl2 p

qv1

(a)

Cl2

Rth1

θo qth2

qth1 C2

Adiabatic walls

qth1

Cth1

Cth2

Rth2

po qv2

(b)

qth1

(c)

FIGURE 5.28 Analogous Systems: (a) Electrical; (b) Liquid; (c) Thermal.

Solution The following equations model the four electrical elements of the system sketched in Figure 5.28(a):

vi  y ¼ R1 $ i1 ; y  vo ¼ R2 $ i2 ; : : C1 $ y ¼ i1  i2 ; C2 $ yo ¼ i2

(5.153)

To identify a liquid system analogous to the electrical system, Eq. (5.153) is converted into liquidelement formulation, with the aid of Table 5.1, as:

pi  p ¼ Rl1 $ qy1 ; p  po ¼ Rl2 $ qy2 ; : : Cl1 $ p ¼ qy1  qy2 ; Cl2 $ po ¼ qy2 ðtÞ

(5.154)

Based on Eq. (5.154), we suggest the liquid system of Figure 5.28(b), which is formed of two capacitive tanks and two resistance conduits. The mathematical model for this system is actually represented by Eq. (5.154), which are not written again, but can be derived easily starting from Figure 5.28(b)das a consequence this liquid system is analogous to the original electrical system. In a similar manner, Table 5.1 is used to obtain the following thermal equations, based on Eq. (5.153):

(

qi  q :¼ Rth1 $ qth1 ; q  qo ¼: Rth2 $ qth2 ; Cth1 $ q ¼ qth1  qth2 ; Cth2 $ qo ¼ qth2

(5.155)

Eq. (5.155) hints us to the thermal system of Figure 5.28(c), which is formed of two rooms of different thermal capacitances and separated by a wall of thermal resistance Rth2. The total resistance of the three walls separating the left room from the outside is Rth1, whereas the three outside walls of the right room are perfectly insulated and do not exchange heat with the exterior. Assuming that the outside temperature qi is higher than the indoor temperature of the left room q, which, in turn, is higher than the right room’s temperature qo, the equations that define the four thermal elements are those of Eq. (5.155). As a result, the thermal system of Figure 5.28(c) is, indeed, analogous to the electrical system of Figure 5.28(a). This indicates that all three systems of Figure 5.28 are nn analogous.

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SUMMARY With an approach similar to the one utilized in Chapters 2e4 that focuses on modeling mechanical and electrical systems, this chapter introduces the fluid elements of inertance, capacitance, resistance, as well as the pumps and fans. The thermal elements of capacitance and resistance are also studied. Based on these elements, mathematical models are formulated for the natural and forced responses of fluid systems, as well as to derive mathematical models for the forced response of thermal systems by means of energy methods and the Lagrange’s equations. MATLAB is applied to run symbolic and numerical calculations and evaluate the eigenvalues and eigenvectors of fluid systems. There is also an example on how to apply Simulink to model and solve for the time response of a thermal system. Analogies between electrical, fluid, and thermal first-order systems are included in this chapter. Subsequent chapters employ fluid and thermal system models, along with mechanical and electrical system models, to identify various responses in the time or frequency domains.

PROBLEMS 5.1 Determine the liquid inertance of the parabolic-shaped conduit segment defined by the end radii r1 and r2 and length l, as sketched in Figure 5.29. The tangent to the parabola is perpendicular to the radius r1 direction at their intersection. Find the inertance numerical value for r ¼ 900 kg/m3, l ¼ 2 m, r1 ¼ 0.06 m, and r2 ¼ 0.09 m. l

r2 r1

FIGURE 5.29 Parabolic-Profile Pipe Segment.

5.2 Calculate algebraically the liquid inertance of the pipeline segment shown in Figure 5.30 and its numerical value for l ¼ 1.6 m, inner diameters d1 ¼ 40 mm, d2 ¼ 70 mm, and fluid density r ¼ 1800 kg/m3.

Problems

d1

d2

l

l

l

FIGURE 5.30 Variable Cross-Section Pipeline Segment. l1

l2 d2

d1 qv

p2

p1 d3

l3

FIGURE 5.31 Inertance Pipe Connection.

5.3 Using inertances, calculate the pressure difference p1  p2 for the pipe connection of Figure 5.31 for dqv/dt ¼ 0.1 m/s2 and r ¼ 1000 kg/m3. Known are also the lengths and interior diameters of the three pipe segments: l1 ¼ 10 m, d1 ¼ 0.05 m, l2 ¼ 8 m, d2 ¼ 0.08 m, l3 ¼ 18 m, and d3 ¼ 0.06 m. 5.4 Determine the hydraulic capacitance of a semispherical tank of inner radius R given the liquid mass density r and the gravitational acceleration g. 5.5 Calculate the capacitance of a cylindrical vertical tank whose cross section varies according to a decreasing exponential: rðxÞ ¼ c1 $ ec2 $ x (x being the distance measured from the upper end of the tank). The tank has its maximum radius r1 at the top and minimum radius r2 at the bottom; its total height is h. Numerical application: r1 ¼ 1.6 m, r2 ¼ 1 m, h ¼ 7 m, r ¼ 1200 kg/m3, and g ¼ 9.8 m/s2. 5.6 Calculate the capacitance of the tank shaped as in Figure 5.32. Known are also the liquid mass density r and gravitational acceleration g. 5.7 Using the HagenePoiseuille approach determine the hydraulic resistance of the parabolic-shaped pipe segment shown in Figure 5.29 of Problem 5.1. The liquid dynamic viscosity is m. 5.8 Answer the same question of Problem 5.7 for a pipe segment having the exponential decreasing profile defined in Problem 5.5 and being placed horizontally.

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CHAPTER 5 Fluid and Thermal Systems

r1

h

r2

FIGURE 5.32 Vertical Tank With Tapered and Semispherical Segments.

5.9 In turbulent pffiffiffi flow, the volume flow rate is expressed in terms of head as q ¼ c $ h, where c is a constant. Determine the pressure-defined hydraulic resistance knowing the liquid mass density r and the gravitational acceleration g. 5.10 The energy loss is measured for a valve corresponding to three values of the flow rate, and the following values are obtained: qV1 ¼ 0.2 m3/s, ER1 ¼ 2300 W; qV2 ¼ 0.3 m3/s, ER2 ¼ 2900 W; qV3 ¼ 0.5 m3/s, and ER3 ¼ 3400 W. Determine the relationship between the pressure drop at the valve p and the volume flow rate q. 5.11 Liquid with a volume flow rate qv ¼ 1.8 m3/s flows through the pipe system sketched in Figure 5.33. Knowing that l ¼ 2.4 m, di ¼ 35 mm (the inner diameter of the circular piping), and m ¼ 0.05 N s/m2, determine the loss of pressure between points 1 and 2. Hint: Liquid resistances connect similarly to electrical resistances. l

l

1 qv

2

2⋅l l

l l

qv

l

FIGURE 5.33 Pipeline System With Seven Lines Containing Flowing Liquid.

5.12 Calculate the hydraulic resistance of the pipeline shown in Figure 5.30 of Problem 5.2 and the energy that is lost. Known also are qV ¼ 3.1 m3/s and m ¼ 0.0012 N s/m2. 5.13

(i) Derive a mathematical model for the conservative liquid system sketched in Figure 5.34 by using the energy balance method and then by the Lagrange’s equations.

Problems

Il

Il

Cl

Cl

Cl

FIGURE 5.34 Two-Mesh Conservative Hydraulic Circuit With Identical Inertances and Capacitances.

(ii) Calculate the natural frequencies and the eigenvectors using the analytical method and the matrix method with MATLAB. 5.14 Formulate a mathematical model of the conservative liquid system of Figure 5.35 by using the energy balance method and then by the Lagrange’s equations. Calculate the corresponding natural frequencies. Is it possible that the amplitudes of the flow rates in the two meshes are equal during either of the two modal motions, and the directions of these flows are identical (i.e., either clockwise or counterclockwise)? Known are Cl1 ¼ 1.6$106 m5/N, Cl2 ¼ 2$106 m5/N, and Il ¼ 3.6$104 N s2/m5.

Cl1

Il

Cl2

FIGURE 5.35 Two-Mesh Conservative Hydraulic Circuit.

5.15 Water flows in a tank of diameter d ¼ 2.3 m with an input volume flow rate qi ¼ 0.12 m3/s. At the base of the tank is a discharge pipe of length l ¼ 3.2 m and inner diameter di ¼ 50 mm. Find the output flow rate exiting the pipe after t ¼ 15 s and the difference between the input and output flow rates as time goes to infinity (the steady-state error). Known are also m ¼ 0.001 N s/m2, r ¼ 1000 kg/m3, and g ¼ 9.8 m/s2. 5.16 Use the energy method and also Lagrange’s equations to derive a mathematical model for the liquid-level system shown in Figure 5.36 when the input to the system is formed of the volume flow rates qi1 and qi2 and the output is the volume flow rate qo. 5.17 Plot the output flow rate qo for the system shown in Figure 5.36 of Problem 5.16 when known are Cl1 ¼ 2.1$106 m5/N, Cl2 ¼ 1.5$106 m5/N, Rl1 ¼ 5.3$1011 N s/m5, Rl2 ¼ 7.2$1011 N s/m5, qi1 ¼ 0.1 m3/s, 3 5 2 qi2 ¼ 0.07 m /s, and pa ¼ 10 N/m .

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CHAPTER 5 Fluid and Thermal Systems

qi1 pa

qi2 pa

Rl1

Cl1

Rl2

Cl2

pa

qo

FIGURE 5.36 Liquid-Level System Consisting of Two Tanks and Two Valves.

5.18 Use Simulink to plot the output flow rate q3 of the liquid-level system of Example 5.8 and shown in Figure 5.13. Known are p ¼ 2.3$105 N/m2, pa ¼ 105 N/m2, Cp1 ¼ 3$106 m5/N, Cp2 ¼ 2.2$106 m5/N, Rp1 ¼ 3.8$1011 N s/m5, Rp2 ¼ 2.5$1011 N s/m5, Rp3 ¼ 4$1011 N s/m5, and q ¼ 0.15$ (1  e0.6$t) m3/s. 5.19 Find the capacitance of a conical vessel of height h and base circle radius r containing a polytropic gas (of coefficient n and gas constant Rg) in it for a specified temperature q. 5.20 Calculate the gas capacitance related to an incompressible gas that flows through the conduit segment of Figure 5.37. Known are the gas mass density r and the gravitational acceleration g. Hint: Calculate Cl,p first. Parabola d2

d1

l

l

FIGURE 5.37 Parabolic-Profile Pipe Segment.

5.21 Using the HagenePoiseuille approach determine the resistance of an incompressible gas that flows through the conduit segment shown in Figure 5.37 of Problem 5.20 when the gas dynamic viscosity m is known. Hint: Calculate Rl,p first. 5.22 Disregarding the pneumatic resistances determine the natural frequency of the system sketched in Figure 5.38. The conduits are tubular of inner diameter di ¼ 30 mm and l ¼ 2 m. The vertical-plane tapered tank is defined by h ¼ 1.8 m, d1 ¼ 0.7 m, and d2 ¼ 1.2 m. Hint: Calculate Il,p and Cl,p first.

Problems

Conduit

l

d2

l

l h

Tank

d1

FIGURE 5.38 Pneumatic System With Tapered Vessel and Conduits.

5.23 Neglect pneumatic resistances and determine the natural frequencies of the pneumatic system sketched in Figure 5.39. Use lumped-parameter inertances and capacitances. The middle conduit has an inner diameter d2 that is three times larger than the inner diameter of all other conduits, d1. Known are g ¼ 9.8 m/s2 and l ¼ 5 m. Conduit

d1

d2

l

l

l

FIGURE 5.39 Conduit Pneumatic System.

5.24 Derive a mathematical model for the pneumatic system of Figure 5.40, which is formed of two tanks and a connecting conduit. The input is the d

d

po h

h pi l

FIGURE 5.40 Pneumatic System With Two Containers and Conduit.

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CHAPTER 5 Fluid and Thermal Systems

pressure pi and the output is the pressure in the conical tank, po. Known are d ¼ 0.6 m, h ¼ 2 m, l ¼ 5 m, and the conduit inner diameter is di ¼ 0.01 m. The pneumatic agent is air with r ¼ 1.2 kg/m3 and m ¼ 1.3$105 N s/m2. Ignore the capacitance of the conduit. 5.25 An insulation layer is added to an original structural wall to reduce the heat flow rate through the original wall loss by m%. The thermal conductivity of the insulation layer is n times smaller than the thermal conductivity of the structural layer. Find the thickness of the insulation layer as a fraction of the structural layer thickness. Consider that the two layers have the same area, and that the temperature difference across the wall does not change through the addition of the insulation layer. 5.26 A double-pane window is formed of two identical glass panels that enclose still air space. The thickness of the glass panels is lg ¼ 0.005 m and that of the air space is la ¼ 0.012 m. The wall is 1 m high and 0.8 m wide and separates a conditioned space from the outdoors. The thermal conductivity of the glass is kg ¼ 0.80 W/(m deg) and that of the air space is ka ¼ 0.025 W/ (m deg). During wintertime, an indoor temperature of 25 C is to be maintained, whereas the outside temperature is 14 C. There is indoor convection of coefficient hi ¼ 11 W/(m2 deg), as well as radiation defined by hr ¼ 8 W/(m2 deg). In the presence of wind, there is an outside convection coefficient of ho ¼ 50 W/(m2 deg). For steady-state heat transfer, find the temperature of window inner surface. 5.27 A sandwich wall is formed of two identical wood panels and an insulation layer in-between. The wall separates a conditioned space defined by a temperature 23 C from an outdoors space of temperature 20 C. The thickness of the panels is 0.025 m and that of the insulation is 0.1 m. The thermal conductivity of the wood is 0.15 W/(m deg) and that of the insulation is 0.035 W/(m deg). The convection coefficient and radiation coefficient of the conditioned space are equal to 12 W/(m2 deg). Evaluate the outdoors convection coefficient if a thermal sensor placed on the outer wall surface indicates a temperature of 18.5 C under steady-state conditions. 5.28 A plane wall of known conduction thermal resistance R and area A separates two environments defined by the constant temperatures q1 > q2. Two temperature sensors that are placed on the wall indicate qs1 on the q1 wall side and qs2 on the q2 opposite wall side. Evaluate the convection þ radiation coefficients h1 and h2 at the two wall faces assuming steady-state heat transfer between the two environments through the wall. 5.29 The composite wall sketched in Figure 5.41 is formed of six panels. Sketch the thermal resistance connection for the outside-to-inside heat flow. Knowing that l3 ¼ 0.015 m, l4 ¼ 0.04 m, l5 ¼ 0.018 m, A1 ¼ 0.9 m2,

Problems

l1 = l2 A1

k1

A2

k2

θi

k3

k4

k5

θo

A3 = A4 = A5 l4

l3 A6

l5

k6

FIGURE 5.41 Six-Panel Composite Planar Wall.

A2 ¼ 0.8 m2, A3 ¼ A4 ¼ A5 ¼ 30 m2, A6 ¼ 1.1 m2, k1 ¼ 0.16 W/(m deg), k2 ¼ 0.4 W/(m deg), k3 ¼ 0.18 W/(m deg), k4 ¼ 0.02 W/(m deg), k5 ¼ 0.1 W/ (m deg), k6 ¼ 0.2 W/(m deg), qo ¼ 35 C (outside temperature), and qi ¼ 10 C (inside temperature), find the equivalent thermal resistance and the heat flow rate penetrating from outside under steady-state conditions. 5.30 The following geometrical parameters are known for the composite planar wall of Figure 5.42: l3 ¼ 0.01 m, l3 ¼ 0.08 m, l4 ¼ 0.06 m, A1 ¼ 27.2 m2, A2 ¼ 1.2 m2, A3 ¼ A4 ¼ 25 m2, A5 ¼ 1 m2. Known are also the thermal conductivities k1 ¼ 0.3 W/(m deg), k2 ¼ 0.12 W/(m deg), k4 ¼ 0.35 W/ (m deg), k5 ¼ 0.15 W/(m deg), the inner-surface convection coefficient

k2

A2

θi

θo

A1 k1

k3

k4

l3

l4 k5

l1

FIGURE 5.42 Five-Panel Composite Planar Wall.

A3 = A4

A5

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CHAPTER 5 Fluid and Thermal Systems

hi ¼ 14 W/(m2 deg), the outer-surface convection coefficient ho ¼ 62 W/ (m2 deg), as well as the inside temperature qi ¼ 22 C, the outside temperature qo ¼ 5 C, and the inner surface temperature qsi ¼ 21 C. Evaluate the thermal conductivity k3 using a resistance model with planes perpendicular to the heat flow direction considered isothermal. Hint: Consider that the resistance of layer 1 is connected in series with the resistance of the other layers. 5.31 Solve Problem 5.30 using the assumption that planes parallel to the heat flow direction are adiabatic. Hint: Split layer 1 into three different layers of identical length l1 and areas A2, A3, and A5, respectively. 5.32 Figure 5.43 illustrates a planar wall section, which is formed of a hollow concrete block sandwiched between two identical plaster layers; an extra foam layer is added to the interior wall face. Evaluate the heat flow rate through the wall with and without the foam layer. The inside temperature is qi ¼ 22 C, and the outside temperature is qo ¼ 13 C. The mixed convection and radiation coefficients are hi ¼ 12 W/(m2 deg) for the indoors wall face and ho ¼ 46 W/(m2 deg) for the outdoors wall face. The out-of-plane wall segment depth is w ¼ 0.16 m. The other geometric and material properties are as follows: l1 ¼ 0.04 m, l2 ¼ 0.02 m, l3 ¼ 0.03 m, l4 ¼ 0.1 m, l ¼ 0.16 m, b1 ¼ 0.04 m, b2 ¼ 0.2 m, b ¼ 0.28 m, k1 ¼ 0.025 W/(m deg), k2 ¼ 0.23 W/(m deg), k3 ¼ 0.7 W/(m deg), and k4 ¼ 0.02 W/(m deg). q l2

l2

Plaster

l3 Concrete θi

θo

Air

hi k1 k2 k3 Foam

k4

ho k3 k2

b2 b

l4 b1

Plaster l1

l

FIGURE 5.43 Composite Planar Wall.

5.33 A stainless steel pipe (k ¼ 17 W/(m deg)) of length l ¼ 20 m, inner diameter di ¼ 0.6 m, and outer diameter do ¼ 0.65 m is coated with an insulation material (ki ¼ 0.14 W/(m deg)) to allow a total heat loss of q ¼ 15 W

Problems

when the outdoor temperature is 8 C, and the temperature of the fluid inside the pipe is 35 C. The inner-surface convection coefficient is hi ¼ 5 W/(m2 deg), and the outer-surface convection coefficient ho ¼ 95 W/ (m2 deg). What is the thickness of the insulating layer? 5.34 A 30 m long stainless steel cylindrical pipe with k ¼ 18 W/(m deg), inner diameter di ¼ 0.1 m, and outer diameter do ¼ 0.12 m carries steam. The pipe is coated with an insulation material of kins,1 ¼ 0.14 W/(m deg) and of radial thickness 0.04 m. What is the thickness of an extra insulation layer with kins,2 ¼ 0.06 W/(m deg) that, added concentrically and atop the original coating layer reduces the radial heat transfer by 20%? The outdoors temperature is qo ¼ 20 C and the outer-surface convection coefficient is ho ¼ 58 W/(m2 deg). Consider that the inner pipe surface is identical to the steam temperature qi ¼ 300 C. 5.35 Figure 5.44 is the longitudinal section through a composite cylinder portion with fluid flowing through. The cylinder is formed of four different structural components defined by the following parameters: k1 ¼ 30 W/ (m deg), k2 ¼ 0.18 W/(m deg), k3 ¼ 0.12 W/(m deg), k4 ¼ 25 W/(m deg), d1 ¼ 0.44 m, d2 ¼ 0.42 m, d3 ¼ 0.32 m, d4 ¼ 0.30 m, l2 ¼ 3 m, and l3 ¼ 1 m. Known are also the outer temperature qo ¼ 16 C, the internal convection coefficient hi ¼ 18 W/(m2 deg), and the external convection coefficient ho ¼ 47 W/(m2 deg). A temperature sensor attached to the cylinder outer surface provides a steady-state value qso ¼ 15 C. Evaluate the temperature qi of the fluid inside the cylinder. l1 l2

l3 4 3

2 d1

1 d4

θi

d3 d2

θo

FIGURE 5.44 Longitudinal Section of a Composite Cylinder Section.

5.36 Several cylinders are combined as in the longitudinal section of Figure 5.45. A liquid of constant temperature qi ¼ 92 C flows inside the cylinders; the temperature outside the cylinders is qo ¼ 2 C. The inner convection coefficient is hi ¼ 30 W/(m2 deg), whereas the outside convection coefficient is ho ¼ 76 W/(m2 deg). Known are also k1 ¼ 0.230 W/(m deg), k2 ¼ 0.018 W/(m deg), k3 ¼ 21 W/(m deg), d1 ¼ 0.062 m, d2 ¼ 0.048 m, d3 ¼ 0.032 m, d4 ¼ 0.020 m, l1 ¼ 5 m, l2 ¼ 1 m, and l3 ¼ 6 m. Find the

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CHAPTER 5 Fluid and Thermal Systems

k1

k2 k3

d2

d4

d1 d3

l1

l2

l3

FIGURE 5.45 Longitudinal Section Through Composite Cylinder Tubing.

total heat flow rate in the radial direction for the entire cylinder sector. Hint: Divide the length into three portions of lengths l1, l2, and l3. 5.37 A hot rigid sphere is placed in air of temperature qa ¼ 20 C. The sphere has a radius r ¼ 0.04 m, mass density r ¼ 7600 kg/m3, and specific heat c ¼ 432 J/(kg deg). The heat transfer between the body and the air occurs by convection through the sphere outer surface with a convection coefficient h ¼ 25 W/(m2 deg). Derive the mathematical model of the body temperature variation with time, calculate this temperature, and evaluate the time after which the sphere’s temperature reaches the steadystate temperature. Assume that the temperature of the sphere is uniform through the volume and consider the body’s initial temperature is q0 ¼ q(0) ¼ 225 C. 5.38 A refrigerating room has a thermal capacity C ¼ 100,000 J/deg, a total wall thermal resistance R ¼ 1.2 deg/W, and an initial temperature q0 ¼ q(0) ¼ 5 C. A cooling unit with a power qi ¼ 250 W is placed in the room with the aim of reducing the indoors temperature to a prescribed value qp. Assuming a constant outside temperature qo ¼ 25 C, derive the mathematical model of the refrigerating room temperature q(t) and find q as an algebraic expression in terms of t. Also determine the time constant of this system, as well at the time tp after which q reaches the value qp ¼ 3 C. 5.39 Solve Example 5.14 for the following numerical values: C ¼ 160,000 J/deg, R ¼ 0.96 deg/W, q0 ¼ q(0) ¼ 25 C, and qo ¼ 17 C. The input flow rate of Figure 5.7(b) is defined by q1 ¼ 0.9$qmin (defined from t ¼ 0 to t ¼ 3s) and q2 ¼ 1.1$qmin (defined from t ¼ 3s to t ¼ 6s). 5.40 Two identical rooms (as illustrated in Figure 5.23) are defined by C ¼ 220,000 J/deg, Ro ¼ 1.115 deg/W, R ¼ 0.75 deg/W, and the initial temperatures q1 ¼ 23 C and q2 ¼ 18 C. Knowing the constant outdoors

Problems

temperature is qo ¼ 35 C and that there is no heat source/sink inside the rooms: (a) Derive the mathematical model of the two-room thermal system by means of the energy method and Lagrange’s equations. (b) Obtain the analytical solution of this model and plot q1(t) and q2(t). 5.41 The two rooms sketched in Figure 5.46 are heated differently during wintertime by means of two heat sources of known heat flow rates q1 and q2 (q1 > q2). Each room has three identical wall panels separating the conditioned space from the outsidedthe thermal resistances are R1 and R2 for one panel of the left and right rooms, respectively. Heat also flows through the wall separating the two rooms, whose thermal resistance is R. Knowing the outside temperature is qo, use the energy method and Lagrange’s equations to derive the mathematical model of this thermal system where the output amounts are the two room temperatures q1 and q2. θo

R1

R2

θ1

θ2 q1

R1

θo

q2

R2

R

C1

C2 R1

R2

FIGURE 5.46 Two-Room Thermal System With Internal Heating.

5.42 Known are the following amounts for the thermal system of Figure 5.46 in Problem 5.41: qo ¼ 13 C, R1 ¼ 1.25 deg/W, R2 ¼ 1.03 deg/W, R ¼ 0.67 deg/W, C1 ¼ 170,000 J/deg, C2 ¼ 184,000 J/deg, q1 ¼ 1900 þ 15$sin(105$t) W, and q2 ¼ 1600$e0.00002$t W. The initial temperature in both rooms is q1(0) ¼ q2(0) ¼ 21 C. Plot the two rooms’ temperatures as a function of time using Simulink. 5.43 Identify an electrical system that is analogue to the liquid-level system of Example 5.8 and illustrated in Figure 5.13. 5.44 Identify a thermal system that is analogue to the pneumatic system of Problem 5.24 and sketched in Figure 5.40. 5.45 Identify a liquid system that is analogue to the thermal system of Problem 5.41 and shown in Figure 5.46.

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Suggested Reading T.L. Bergman, A.S. Lavine, F.P. Incropera, and D.P. DeWitt, Introduction to Heat Transfer, 6th Ed. John Wiley & Sons, New York, 2011. Y.A. Cengel and A. Ghajar, Heat and Mass Transfer: Fundamentals and Applications, McGraw-Hill, Boston, 2014. I. Cochin, Analysis and Design of Dynamic Systems, Harper & Row, New York, 1980. K. Ogata, System Dynamics, 4th Ed. Pearson Prentice Hall, Upper Saddle River, NJ, 2004. F.C. McQuiston, J.D. Parker, and J.D. Spitler, Heating, Ventilating and Air Conditioningd Analysis and Design, 6th Ed. John Wiley & Sons, New York, 2005. N. Cheremisinoff, Fluid Flow Pocket Handbook, Gulf Publishing Company, Houston, 1984. D. McCloy and H.R. Martin, Control of Fluid Power: Analysis and Design, 2nd Ed. Ellis Horwood Limited, Chichester, UK, 1980. C.A. Belsterling, Fluidic System Design, Wiley Interscience, New York, 1971. B. Eck, Fans, Pergamon Press, Oxford, UK, 1973. H. Klee and R. Allen Simulation of Dynamic Systems with MATLAB and Simulink, 2nd Ed. CRC Press, Boca Raton, FL, 2011.

CHAPTER

The Laplace Transform

6

OBJECTIVES The chapter introduces the Laplace transform, which can be used to solve linear ordinary differential equations (ODE) modeling mechanical, electrical, fluid, thermal, or coupled dynamic systems, as derived in Chapters 2e5 and 10. The Laplace transform is also instrumental in developing other system dynamics formulations, such as the transfer function, the state space, or the frequency-domain models, as discussed in Chapters 7e9. The following topics are studied: • Direct and inverse Laplace transforms. • Laplace pairs and main Laplace transform properties. • Techniques of partial fraction expansion. • Application of direct and inverse Laplace transforms to solving linear ODE, integral equations, and integraledifferential equations with constant coefficients. • Use of MATLAB specialized commands for calculations related to direct or inverse Laplace transformations.

INTRODUCTION This chapter presents the main properties and applications of the Laplace transform. One important applicative strength of the Laplace method is the conversion of linear ODE into algebraic equations by means of the direct Laplace transform, which enables calculation of the transformed solution with relative ease. Subsequent use of the inverse Laplace transform allows retrieving the original, time-domain solution. The Laplace transform is utilized for linear ODE with constant coefficients in this chapter. The chapter includes the application of Laplace transform to dynamic models defined by systems of linear ODE, which are expressed in vectorematrix form. Integral and integraledifferential equations and systems can also be solved by the same two-phase Laplace approach. Another topic is time-domain system identification from Laplace-domain information.

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00006-3 Copyright © 2018 Elsevier Inc. All rights reserved.

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6.1 DIRECT AND INVERSE LAPLACE TRANSFORMATIONS Transformations, also named operators, convert functions into other functions; this is schematically illustrated in Figure 6.1. Transformations change the original function, but they can either keep the independent variable or modify it. An operator O that is applied to a function f(t) and multiplies it by a number n, for instance, results in another function g(t) that preserves the independent variable of the original function. Such an operator can formally be expressed as O½ f ðtÞ ¼ n $ f ðtÞ ¼ gðtÞ (6.1) If, for example, f(t) is cos (u$t) in Eq. (6.1), the result, n$cos (u$t), depends on time t, as well as on the input (original) function. Another set of transformations (or simply, transforms) modify both the original function and its independent variable through their application. Such transforms are the integral ones, which change a function and its variable using integration, and the Laplace transform is one of them. Other integral transforms are the Fourier transform, the Melin transform, the Hankel transform, the Kantorovich transform, and the MehlereFock transform, all of which operate over an infinite domain. Still other transforms apply over a finite domain and are known as finite transforms, including specific techniques such as those of SturmeLiouville, Legendre, and Tchebycheff. The direct Laplace transform takes a function f(t) that depends on time t and transforms into another function F(s) that depends on the complex variable s, as sketched in Figure 6.2. The inverse Laplace transform, also indicated in Figure 6.2, operates on F(s) to obtain f(t). Mathematically, the Laplace transform is defined as Z N f ðtÞ $ es $ t dt (6.2) FðsÞ ¼ L½ f ðtÞ ¼ 0

The integral defining the Laplace transform needs to be convergent for the transformation to exist, which requires the original function f(t) to satisfy the following conditions: 1. f(t) has to be piecewise continuous over the interval 0 < t < N, which means it has to be continuous on any partition of that interval (provided the interval can be divided into a finite number of nonintersecting intervals), and it needs to have finite limits at the ends of each subinterval.

FUNCTION

TRANSFORMATION (OPERATOR)

FUNCTION

FIGURE 6.1 Relationships Between Functions and Transformations (Operators).

6.1 Direct and Inverse Laplace Transformations

DIRECT LAPLACE TRANSFORMATION

TIME DOMAIN f (t)

s (LAPLACE) DOMAIN F(s)

INVERSE LAPLACE TRANSFORMATION

FIGURE 6.2 Direct and Inverse Laplace Transformations.

2. f(t) needs to be of exponential order, which means it has to satisfy the following equality: lim j f ðtÞj $ es $ t ¼ 0

t/N

(6.3)

for values of the real constant s that are larger than a threshold value sc, which is known as the abscissa of convergence.

6.1.1 Laplace Transform Pairs Formally, the inverse Laplace transform, which is often needed to return the original (generally unknown) time-domain function f(t), is calculated as f ðtÞ ¼ L1 ½FðsÞ

(6.4)

but the integration technique allowing direct calculation of f(t) from F(s) is a bit more involved for this introductory text. Suffice to say that application of the direct Laplace transform through integration as per the definition Eq. (6.2) enables formulating f(t) 4 F(s) pairs for elementary functions that satisfy the requirements for obtaining the Laplace transform. Table 6.1 contains several Laplace transform pairs for quick translation between the two domains; this table can be used to either find F(s) from f(t)deliminating thus the direct integration of Eq. (6.2)dor to identify the original, time-domain function f(t) from a given counterpart F(s). One of the simplest Laplace transforms is the one applied to the unit-step function sketched in Figure 6.3. Instead of simply using a value of 1 (or the more generic symbol u(t), as in other texts) to define it, the symbol h(t) is utilized here (in honor of the English scientist Oliver Heaviside and to conform with MATLAB notation) with the function definition indicated in Figure 6.3, which highlights that the unit-step function is zero for t < 0.

275

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CHAPTER 6 The Laplace Transform

Table 6.1 Laplace Transform Pairs f(t)

F(s)

d(t)

1

h(t) ¼ 1

1/s

t

1/s2

tn

n!/snþ1

eat

1/(s þ a)

sin (u$t)

u/(s2 þ u2)

cos (u$t) pffiffi 1 t

s/(s2 þ u2) pffiffiffiffipffiffiffi p s

tn$ea$t

n!/(s þ a)nþ1

e

a$t

a$e

e a$t

(a  b)/[(s  a) $ (s  b)]

b$t

 b$e

b$t

(a  b)$s/[(s  a) $ (s  b)]

(ea$t  eb$t)/t

ln[(s  a)/(s  b)]

t  (1  ea$t)/a

a/[s2 $ (s þ a)]

sinh (u$t)

u/(s2  u2)

cosh (u$t)

s/(s2  u2)

sin (u$t þ 4)

[u$cos4 þ (sin4)$s]/(s2 þ u2)

t$sin (u$t)

2u$s/(s2 þ u2)2

sin (u$t)/t

tan1(u/s)

t$cos (u$t)

(s2  u2)/(s2 þ u2)2

t$sinh (u$t)

2u$s/(s2  u2)2

t$cosh (u$t)

cosðu1 $ tÞHcosðu2 $ tÞ

(s2 þ u2)/(s2  u2)2     u1 s2 þ u21 Hu2 s2 þ u22      s $ 1 s2 þ u21 H1 s2 þ u22

1  cos (u$t)

u2/[s$(s2 þ u2)]

ut  sin (u$t)

u3/[s2$(s2 þ u2)]

sin (u$t)  u$t$cos (u$t)

2u3/(s2 þ u2)2

sin (ut) þ u$t$cos (ut)

2u$s2/(s2 þ u2)2

sin (u$t)$sinh (u$t)

2u2$s/[s4 þ (2u2)2]

cos (u$t)$cosh (u$t)

s3/[s4 þ (2u2)2]

sin (u$t)$cosh (u$t)

u$(s þ 2u2)/[s4 þ (2u2)2]

sinh (u$t)  sin (u$t)

2u3/(s4  u4)

cosh (u$t)  cos (u$t)

2u2$s/(s4  u4)

sinh (u$t) þ sin (u$t)

2u$s2/(s4  u4)

cosh (u$t) þ cos (u$t)

2s3/(s4  u4)

sinðu1 $ tÞHsinðu2 $ tÞ

6.1 Direct and Inverse Laplace Transformations

f 1

⎧0, t < 0 h( t) = ⎨ ⎩ 1, t ≥ 0

0

t

FIGURE 6.3 Unit-Step (Heaviside) Function h(t).

This special symbol allows truncating various other functions by adequate combination with h(t), as will be shown shortly. The Laplace transform of h(t) is N Z N Z N es $ t  1 s $ t s $ t (6.5) hðtÞ $ e dt ¼ 1$e dt ¼  ¼ L½hðtÞ ¼  s s 0 0 0 and the pair is included in Table 6.1. MATLAB’s Symbolic Math Toolbox uses the heaviside(t) dedicated command to represent the unit-step function h(t)dapplication of this command will be demonstrated in a subsequent example of this chapter. nn

Example 6.1 Calculate the Laplace transform of the unit-ramp function, which is sketched in Figure 6.4. Using the principle of mathematical induction also derive the Laplace transform of the function f (t) ¼ t n, where n is an integer and n > 1.

f

⎧ 0, t < 0 f (t ) = ⎨ ⎩ t, t ≥ 0 0

t

FIGURE 6.4 Unit-Ramp Function.

Solution The Laplace transform of the unit-ramp function is calculated through integration by parts as

Z

N

L½t ¼ 0

t$e

s $ t

N N Z N es $ t  1 es $ t  1 s $ t dt ¼ t $ þ $ e dt ¼  2  ¼ 2  s s s 0 s 0 0 (6.6)

277

278

CHAPTER 6 The Laplace Transform

which is also shown in Table 6.1. Application of L’Hospital rule

lim

t/N

f ðtÞ df ðtÞ=dt ¼ lim t/N gðtÞ dgðtÞ=dt

to Eq. (6.6) results in



lim

t/N

es $ t t$ s



for lim f ðtÞ ¼ N and lim gðtÞ ¼ N t/N



t ¼ lim t/N s $ es $ t

t/N



¼ lim

t/N

dð tÞ=dt dðs $ es $ t Þ=dt



(6.7)



1 ¼ lim 2 s $ t t/N s $ e



¼0 (6.8) The MATLAB Symbolic Math Toolbox enables calculation of the direct Laplace transform by using the laplace (f(t)) command, as shown in the code lines below: >> syms t % this command declares t as a symbolic variable. >> laplace(t) which is used to find the analytical result of Eq. (6.6) To calculate the Laplace transform of t 2, the procedure is similar to the one applied previously, which yields

  L t2 ¼

Z

N

t2 $ es $ t dt ¼ t2 $

0

N Z N es $ t  2 2 $ þ t $ es $ t dt ¼ 3 s s s 0 0

(6.9)

Eq. (6.9) took into account both the result of Eq. (6.6) and the fact that



lim

t/N

t2 $

es $ t s





¼ lim

t/N

t2 s $ es $ t



¼ lim

2t

t/N s2 $ es $ t

¼ lim

2

t/N s3 $ es $ t

¼0 (6.10)

based on applying twice L’Hospital rule. Eqs. (6.6) and (6.9) suggest that

L½tn  ¼

n! snþ1

(6.11)

as in Table 6.1, but this has to be demonstrated. If the relationship of Eq. (6.11) is true, the following relationship needs to be valid as well:

 ðn þ 1Þ!  L tnþ1 ¼ nþ2 s

(6.12)

and, according to the principle of mathematicalinduction, Eq. (6.11) is proven. This can be shown by  using integration by parts again to calculate L t nþ1 :

  L tnþ1 ¼

Z

N

t

nþ1

$e

s $ t

dt ¼ t

0

¼

nþ1

N Z N es $ t  nþ1 $ $ þ tn $ es $ t dt s s 0 0

ðn þ 1Þ $ n! ðn þ 1Þ! ¼ nþ2 s $ snþ1 s (6.13)

6.1 Direct and Inverse Laplace Transformations

This demonstrates that Eq. (6.12) is valid; therefore, Eq. (6.11) is valid too, as also indicated in Table 6.1. In Eq. (6.13) the following limit is zero (as can be demonstrated through L’Hospital rule and mathematical induction again):



lim

t/N

t

nþ1

es $ t $ s



¼ lim

t/N

ðn þ 1Þ $ tn ðn þ 1Þ $ n $ tn1 ¼ lim t/N s2 $ es $ t s3 $ es $ t

¼ / lim

t/N

ðn þ 1Þ $ n $ / $ 2 $ 1 ¼0 snþ1 $ es $ t

(6.14) nn

6.1.2 Properties of the Laplace Transform Table 6.2 synthesizes the main Laplace transform properties, some of which are demonstrated next. Several other properties and examples are included in the companion website Chapter 6.

Linearity An important property with applications in solving differential equations and determining transfer functions is the linearity character of both the direct and inverse Laplace operators. Consider a function f(t) that is a linear combination of two time-dependent functions, f1(t) and f2(t) by means of two constants, a1 and

Table 6.2 Main Laplace Transforms Properties (Theorems) Theorem

f(t)

F(s)

Linearity

a1$f1(t) þ a2$f2(t)

a1$F1(s) þ a2$F2(s)

at

Frequency shift

e

Time-shift

f(t  a)$h(t  a)

Time derivatives

dfðtÞ dt

$f(t)

d2 fðtÞ dt2

Time integral

Final value

Rt

0 fðtÞdt R fðtÞdt

lim fðtÞ

t/N

Initial value

lim fðtÞ

F(s þ a) eas$F(s) s$F(s)  f(0) 

s2

  $ FðsÞ  s $ fð0Þ  dfðtÞ dt 

FðsÞ s FðsÞ s

lim

þ t/0

R

fðtÞdt s

lim s $ FðsÞ

s/0

lim s $ FðsÞ

t/0 t>0

s/N

Time multiplication

t $ fðtÞ

Time division

fðtÞ t

dFðsÞ ds RN s FðsÞds

t¼0

279

280

CHAPTER 6 The Laplace Transform

a2: f(t) ¼ a1$f1(t) þ a2$f2(t). Application of the Laplace transform to this linear combination results in L½ f ðtÞ ¼ L½a1 $ f1 ðtÞ þ a2 $ f2 ðtÞ ¼ L½a1 $ f1 ðtÞ þ L½a2 $ f2 ðtÞ ¼ a1 $ L½ f1 ðtÞ þ a2 $ L½ f2 ðtÞ ¼ a1 $ F1 ðsÞ þ a2 $ F2 ðsÞ

(6.15)

The inverse Laplace transform operates on Eq. (6.15) as L1 ½L½ f ðtÞ ¼ f ðtÞ

(6.16)

Eqs. (6.15) and (6.16) result in L1 ½a1 $ F1 ðsÞ þ a2 $ F2 ðsÞ ¼ L1 ½a1 $ F1 ðsÞ þ L1 ½a2 $ F2 ðsÞ ¼ a1 $ L1 ½F1 ðsÞ þ a2 $ L1 ½F2 ðsÞ ¼ a1 $ f1 ðtÞ þ a2 $ f2 ðtÞ

(6.17)

The linearity property illustrated by Eqs. (6.15) and (6.17) can be extended to n functions combined with n constants.

Frequency Shift Theorem The frequency shift theorem (as shown in Table 6.2) yields the Laplace transform of the product between a function f(t), which is Laplace transformable, and eat, where a is a real constant. It can be shown that Z N Z N  a $ t  a $ t s $ t L e $ f ðtÞ ¼ e $ f ðtÞe dt ¼ f ðtÞ $ eðsþaÞ $ t dt ¼ Fðs þ aÞ 0

0

(6.18) The effect of multiplying the original function by an exponential on applying the Laplace transform to this product is a translation (shift) of the variable s into the Laplace domain. Because the Laplace domain is closely connected to the frequency response, as we see in Chapters 9 and 11e13, the theorem is known as the frequency shift theorem. nn

Example 6.2

The Laplace transform of the time-domain response y(t) of a single-degree of freedom (DOF) dynamic system is Y(s) ¼ 3s/(s2 þ 2s þ 5). Find y(t) by applying definition-based analytical calculation with the aid of Tables 6.1 and 6.2, and the frequency shift theorem.

Solution In checking the functions in the right column of Table 6.1, the closest format to the one of the problem appears to be the pair corresponding to cos (u$t), but some modifications are needed before finding y(t). By completing the square in the denominator, the function Y(s) can be written as

YðsÞ ¼

3s ðs þ 1Þ2 þ 22

¼

3ðs þ 1Þ  3 ðs þ 1Þ2 þ 22

¼ 3$

sþ1 ðs þ 1Þ2 þ 22

3 2  $ 2 ðs þ 1Þ2 þ 22 (6.19)

6.1 Direct and Inverse Laplace Transformations

Considering now the linearity propriety, the frequency shift theorem, together with the rows in Table 6.1 giving the Laplace transforms of cos (u$t) and sin (u$t), it follows that

"

1

y ðt Þ ¼ L

½ Y ðsÞ ¼ 3 $ L

1

#

sþ1

ð s þ 1Þ 2 þ 22  sinð2tÞ ¼ 3et $ cosð2tÞ  2

" # 3 2 1  $L 2 ðs þ 1Þ2 þ 22

(6.20)

Similarly to the symbolic evaluation of the direct Laplace transform, the MATLAB Symbolic Math Toolbox allows calculation of the inverse Laplace transform by means of the ilaplace command as shown in the code below: >> syms s >> Y = 3*s/(s^2+2*s+5); >> pretty (ilaplace(Y)) which returns the result obtained analytically in Eq. (6.20).

nn

nn

Example 6.3 Use the known Laplace transforms of sin (u$t) and cos (u$t) together with the frequency shift theorem to determine the Laplace transforms of the functions: f1(t) ¼ t$ea$t$sin (u$t) and f2(t) ¼ t$ea$t$cos pffiffiffiffiffiffiffi (u$t) by starting from the complex function f(t) ¼ t$e(aþu$j )$t where j ¼ 1. Check the results by using Table 6.1 and MATLAB symbolic calculation.

Solution The following relationship results from the frequency shift theorem:

h i L t $ eðaþu $ jÞ $ t ¼

1 ðs þ a þ u $ jÞ2

(6.21)

The complex number of Eq. (6.21) is expressed in standard form as

1 ðs þ a þ juÞ

2

¼

1 2

ðs þ aÞ  2

u2

ðs þ aÞ 

þ 2u $ ðs þ aÞ $ j

u2

¼h

 2u $ ðs þ aÞ $ j i2 ðs þ aÞ2  u2 þ 4u2 $ ðs þ aÞ2

¼h

ðs þ aÞ2  u2 2u $ ðs þ aÞ i2  h i2 $ j ðs þ aÞ2 þ u2 ðs þ aÞ2 þ u2

(6.22)

Combining now Eqs. (6.21) and (6.22) yields

h i ðs þ aÞ2  u2 2u $ ðs þ aÞ L t $ eðaþu $ jÞ $ t ¼ h i2  h i2 $ j ðs þ aÞ2 þ u2 ðs þ aÞ2 þ u2

(6.23)

281

282

CHAPTER 6 The Laplace Transform

At the same time, the original function of Eq. (6.21) can be formulated based on Euler’s identity

eu $ t $ j ¼ cosðu $ tÞ  sinðu $ tÞ $ j

(6.24)

as

t $ eðaþu $ jÞ $ t ¼ t $ ea $ t eu $ t $ j ¼ t $ ea $ t $ ½cosðu $ tÞ  sinðu $ tÞ $ j ¼ t $ ea $ t $ cosðu $ tÞ  t $ ea $ t $ sinðu $ tÞ $ j

(6.25)

Applying the Laplace operator to Eq. (6.25) results in

h i     L t $ eðaþu $ jÞ $ t ¼ L t $ ea $ t $ cosðu $ tÞ  L t $ ea $ t $ sinðu $ tÞ $ j (6.26)

Comparison of the real and imaginary parts of Eqs. (6.23) and (6.26) shows that

  2u $ ðs þ aÞ L½ f1 ðtÞ ¼ L t $ ea $ t $ sinðu $ tÞ ¼ h i2 ; ðs þ aÞ2 þ u2   ðs þ aÞ2  u2 L½ f2 ðtÞ ¼ L t $ ea $ t $ cosðu $ tÞ ¼ h i2 ðs þ aÞ2 þ u2

(6.27)

Note that Eq. (6.27) are also obtained using Table 6.1  in particular the Laplace transforms of the functions t$sin (u$t) and t$cos (u$t) in conjunction with the frequency shift theorem and the exponential ea$t operator. MATLAB commands similar to the ones of Example 6.1 allow symbolic calculation of the Laplace transforms of f1(t) and f2(t), resulting in Eq. (6.27). nn

Time-Shift Theorem The previous theorem has a counterpart, which, as shown in Table 6.2, indicates that multiplication of the function F(s) by an exponential ea$s, followed by application of the inverse Laplace transform, results in a function that is shifted in the time domain: L½ f ðt  aÞ $ hðt  aÞ ¼ ea $ s $ FðsÞ (6.28) This property is known as the time-shift theorem. Before demonstrating this property, a brief preamble is needed to better understand the functions f(t  a) and h(t  a) on the left-hand side of Eq. (6.28). An original arbitrary function f(t) is plotted in Figure 6.5(a), together with the f

a

f

f(t – a)

f(t)

h(t)

h(t – a)

1

f(t) = f(t1)

0

t1

a (a)

t

t

0

a (b)

FIGURE 6.5 Horizontally Translated: (a) Arbitrary Function; (b) Unit-Step Function.

t

6.1 Direct and Inverse Laplace Transformations

function f(t  a). As it can easily be verified, the function f(t  a) is shifted (translated) horizontally to the right by a quantity a and therefore, f(t  a) ¼ f(t1) with t ¼ t1 þ a. A similar situation applies to the functions h(t) and h (t  a), as h (t  a) translates to the right by the quantity a, as illustrated in Figure 6.5(b). The product between an arbitrary function f(t) and h(t) is ( f ðtÞ $ 0 ¼ 0; t < 0 (6.29) f ðtÞ $ hðtÞ ¼ f ðtÞ $ 1 ¼ f ðtÞ; t  0 In other words, the original function exists only for positive values of t and is zero to the left of the origin. Multiplication of a function by h(t) is omitted altogether in many applications, as this particular operation does not change the original function over the zero-to-infinity interval, where Laplace integration is being applied. It can simply be shown that multiplication of f(t) by h(t  a) results in a function that is zero to the left of a and is f(t) to the right of a. Let us analyze what is the result of multiplying f(t  a) by h(t  a). Without doing any math, the functions f(t) and h(t) are first translated to the right by a quantity a to yield the functions f(t  a) and h(t  a), respectively. The product of the two functions is zero, again, to the left of a, and f(t  a) to the right of a. The combined result is translating the product function f(t)$h(t) to the right by the quantity a: ( f ðt  aÞ $ 0 ¼ 0; t < a (6.30) f ðt  aÞ $ hðt  aÞ ¼ f ðt  aÞ $ 1 ¼ f ðt  aÞ; t  a To demonstrate the time-shift theorem, which is expressed in Eq. (6.28), we start from the Laplace transform definition: Z N L ½ f ð t  aÞ $ hð t  aÞ  ¼ f ðt  aÞ $ hðt  aÞ $ es $ t dt (6.31) 0

By using the change of variables t  a ¼ t1, Eq. (6.31) can be written as Z

0

f ðt1 Þ $ hðt1 Þ $ es $ ðt1 þaÞ dt1 Z N þ f ðt1 Þ $ hðt1 Þ $ es $ ðt1 þaÞ dt1 0 Z N a $ s ¼e $ f ðt1 Þ $ hðt1 Þ $ es $ t1 dt1 ¼ ea $ s $ F ðsÞ

L½ f ð t  a Þ $ h ð t  a Þ  ¼

a

0

(6.32) Eq. (6.32) takes into account that the integral between a and 0 is zero because f(t1)$ h(t1) is zero for t1 less than zero, as indicated in Eq. (6.29). nn

Example 6.4 Determine the Laplace transforms of the pulse and impulse functions plotted in Figure 6.6 using the time-shift theorem.

283

284

CHAPTER 6 The Laplace Transform

f

f ∞

A τ

Rectangle area is A

τ

0←τ

t

(a)

t (b)

FIGURE 6.6 (a) Pulse Function; (b) Impulse Function.

Solution

The pulse function, as can be seen in Figure 6.6(a), is nonzero between 0 and s, and zero elsewhere. For the zero-to-infinity time interval, it can be written as

8 > < A; 0  t  s pðtÞ ¼ s > : 0; t > s

(6.33)

The branch form of this function is not directly amenable to using the time-shift theorem; therefore, reformulation in an adequate manner is needed. We saw that one convenient way of defining a function starting from a point on the time axis that is different from zero is by using the translated unit-step function as a multiplier of the original function. As such, the pulse function consists of two step functions: one is equal to A/s, it starts from zero and reaches infinity; the other one is equal to A/s and is defined from t ¼ s to infinity. As a consequence, the pulse function is rewritten as

pðtÞ ¼

A A  $ hðt  sÞ s s

(6.34)

Applying the linearity principle together with the time-shift theorem, the following is the Laplace transform of the original pulse function:

 A A A  $ 1  es $ s PðsÞ ¼ L½pðtÞ ¼ L  $ hðt  sÞ ¼ s s s$s 

(6.35)

The impulse function (plotted in Figure 6.6(b)) is a particular case of the pulse function, occurring when the time interval over which the pulse is nonzero reduces to zero; in that case, the height of the needle-like rectangle must go to infinity to combine with a width that shrinks to zero, to keep a constant area A, as shown in Figure 6.6. Mathematically, the impulse is defined as

8 > < lim A; t ¼ s iðtÞ ¼ s/0 s > : 0; t < s and t > s

(6.36)

Eqs. (6.34) and (6.36) indicate that

iðtÞ ¼ lim pðtÞ s/0

(6.37)

6.1 Direct and Inverse Laplace Transformations

Therefore,

  A $ 1  es $ s IðsÞ ¼ L½iðtÞ ¼ L lim pðtÞ ¼ lim L½pðtÞ ¼ lim s/0 s/0 s/0 s$s    s $ s s $ s d A$ 1  e ds s$A$e ¼ lim ¼A ¼ lim s/0 s/0 d½s $ s=ds s 



(6.38)

Interchangeability between the limit and the Laplace operators, as well as L’Hospital’s rule (with derivations in terms of s), is applied in Eq. (6.38) because the original limit was of the form zero over zero. When A ¼ 1, the impulse function becomes the unit-impulse function, also known as Dirac delta function, denoted by d(t). In MATLAB symbolic calculation, dirac(t) is used for d(t), whereas dirac(t,n) is the n-th order derivative of d(t). Table 6.1 includes L½dðt Þ ¼ 1. nn

Laplace Transform of Piecewise Continuous Functions Rather than being defined as a single continuous time-domain function, the excitation or input to a dynamic system may be a piecewise continuous function, such as the generic one of Figure 6.7, which is formed of n different functions, each defined over one subinterval over the zero-to-infinity time interval. The generic function segment indicated with an asterisk in Figure 6.7 is defined over the [ti1, ti] subinterval and is part of the function fi(t), which spans the N to þN time interval. f

fi * (t )

f 2* (t )

f n*–1(t )

* 1

f (t )

t0 = 0

t1

t2

ti–1

ti

tn–2

f n* (t )

tn–1

t

FIGURE 6.7 Piecewise Continuous Function.

While the time-shift theorem can be applied for Laplace transformations of piecewise continuous functions, a direct approach is presented here. The Laplace transform of the piecewise continuous function f(t) of Figure 6.7 is calculated by splitting the zero-to-infinity definition integral into a sum of n integrals, each corresponding to one of the n different segment functions making up f(t), namely Z

N

L½ f ðtÞ ¼

e

s $ t

0

þ/þ

Z

Z $ f ðtÞdt ¼

t1

e 0

N

tn1

s $ t

Z $ f1 ðtÞdt þ / þ

ti

ti1

es $ t $ fi ðtÞdt

es $ t $ fn ðtÞdt (6.39)

285

286

CHAPTER 6 The Laplace Transform

Eq. (6.39) calculates the Laplace transform of piecewise continuous functions directly, as shown in the following example. nn

Example 6.5 Determine the Laplace transform of the function drawn with thick solid line in Figure 6.8.

f f1(t)

f2(t)

f3(t)

a

0

a

2a

t

FIGURE 6.8 Partial Train of Triangular Pulses.

Solution

Figure 6.8 indicates that the function f (t) is

8 8 > > < f1 ðtÞ; 0  t  a < t; 0  t  a f ðtÞ ¼ f2 ðtÞ ¼ f1 ðt  aÞ; a  t  2a ¼ t  a; a  t  2a > > : : a; t  2a f3 ðtÞ; t  2a

(6.40)

The Laplace transform applied to f (t) of Eq. (6.40) is calculated as

Z L½ f ðtÞ ¼

a

f1 ðtÞ $ es $ t dt þ

0

Z ¼

0

a

t $ es $ t dt þ

Z

Z

2a a

2a a

f2 ðtÞ $ es $ t dt þ

ðt  aÞ $ es $ t dt þ

Z

Z

N

2a N

f3 ðtÞ $ es $ t dt

a $ es $ t dt

(6.41)

2a

whose result (obtained with MATLAB symbolic calculation) is

 1 a a $ s 1 2a $ s 1 1 1 2a $ s a $ s L½f ðtÞ ¼ 2  $ e  2$e ¼ $  a$e  $e (6.42) s s s s s s the first two integrals of Eq. (6.41) can be calculated either analytically (integration by parts) or by symbolic integration with MATLAB. The third integral of Eq. (6.41) is calculated directly as a$e-2a$s/s. nn

6.1 Direct and Inverse Laplace Transformations

Laplace Transform of Derivatives The Laplace transform of the first derivative of a function f(t) can be approached using the definition and integration by parts:  Z N Z N Z N df ðtÞ df ðtÞ s $ t $e L dt ¼ u $ dv ¼ u $ vjN  v $ du (6.43) ¼ 0 dt dt 0 0 0 where u ¼ es$t, du ¼ s$e-s$tdt and the rest of the function under the integral of Eq. (6.43) being dv. The result is  Z N     df ðtÞ s $ t s $ t L f ðtÞ $ es $ t dt ¼ lim f ðtÞ $ e  lim f ðtÞ $ e þ s$ t/N t/0 dt 0 ¼ s $ FðsÞ  f ð0Þ 

Eq. (6.44) took into account the fact that lim f ðtÞ $ e t/N

 s $ t

(6.44) ¼ 0 because the func-

tion f(t) is Laplace transformable and therefore is of exponential order as per Eq. (6.3). Laplace transforms of higher-order derivatives of f(t) can be found in a similar manner. Problem 6.18 addresses this topic. nn

Example 6.6

Find the Laplace transform of the function f(t) ¼ sin (u$t) by using the relationship

 d2 f ðtÞ df ðtÞ 2 L . ¼ s $ L½f ðtÞ  s $ f ð0Þ  dt2 dt t¼0 

Solution The given relationship can be reformulated as

 

 2 1 d f ðtÞ df ðtÞ L½ f ðtÞ ¼ 2 $ L þ s $ f ð0Þ þ s dt2 dt t¼0

(6.45)

d 2 sinðu $ tÞ ¼ u2 $ sinðu $ tÞ dt2

(6.46)

 2 d f ðtÞ L ¼ u2 $ L½ f ðtÞ dt2

(6.47)

It is known that

therefore,

for f (t) ¼ sin (u$t). Combining Eqs. (6.45) and (6.47) results in

 

u2 1 df ðtÞ L½ f ðtÞ $ 1 þ 2 ¼ 2 $ s $ f ð0Þ þ s dt t¼0 s

(6.48)

Because

 df ðtÞ d½sinðu $ tÞ df ðtÞ ¼ ¼ u $ cosðu $ tÞ; ¼ u $ cosð0Þ ¼ u dt dt dt t¼0

(6.49)

287

288

CHAPTER 6 The Laplace Transform

and f (0) ¼ sin (0) ¼ 0, Eq. (6.48) transforms to the expected result:

L½ f ðtÞ ¼

u þ u2

s2

(6.50) nn

Laplace Transform of Indefinite Integrals Integral equations, as well as integraledifferential equations are oftentimes mathematical models of dynamic systems. Examples of applying the direct Laplace transform to such equations will be studied later on in this chapter. There are two modalities of formulating indefinite integrals: one that specifies the limits of integration as zero and t (a generic time station), and another form that does not use the limits of integration at all. As a consequence, there are two theorems related to the Laplace transformation of indefinite integrals. It will be shown next that 2 3 Z t FðsÞ L4 (6.51) f ðtÞdt5 ¼ s 0 By applying the definition of the Laplace transform, it follows that 2 0 3 1 Z t Z N Z t @ L4 f ðtÞdt5 ¼ f ðtÞdtA $ es $ t dt 0

0

(6.52)

0

Rt We integrate Eq. (6.52) by parts by taking u ¼ 0 f ðtÞdt and dv ¼ es$t$dt. Note that  Z Z t Z Z Z  f ðtÞdt ¼ f ðtÞdt  lim (6.53) f ðtÞdt ¼ f ðtÞdt  f ðtÞdt t/0

0

This means 2 d du ¼ 4 dt

Z 0

t

t¼0

3

Z  Z d 5 df ðtÞ dt ¼ f ðtÞdt f ðtÞdt dt ¼ f ðtÞdt dt ¼ dt

(6.54)

   R R Eq. (6.54) took into account that d f ðtÞdtt¼0 dt ¼ 0 since f ðtÞdtt¼0 is constant. Also taking dv ¼ es$tdt, yields 1 v ¼  $ es $ t (6.55) s Eq. (6.52) becomes 2 0 3 1 Z N Z t Z t 1 L4 f ðtÞdt5 ¼ u $ vjN v $ du ¼ lim @  $ es $ t $ f ðtÞdtA 0  t/N s 0 0 0 0 1 Z N Z t 1 1 FðsÞ  lim @  $ es $ t $ f ðtÞdtA þ $ f ðtÞ $ es $ t dt ¼ t/0 s s s 0 0 (6.56)

6.1 Direct and Inverse Laplace Transformations

which demonstrates Eq. (6.51). Note that the first limit in Eq. (6.56) is zero because the exponential becomes zero. The second limit in Eq. (6.56) is also zero because the integral in it is zero. We also demonstrate that  R Z f ðtÞdtt¼0 FðsÞ f ðtÞdt ¼ þ (6.57) L s s Applying again integration by parts to the Laplace definition equation and selecting the variables u and v as we did previously, it can be shown that Z   Z Z 1 s $ t 1 s $ t f ðtÞdt ¼ lim  $ e L $ f ðtÞdt  lim  $ e $ f ðtÞdt t/N t/0 s s Z N 1 þ f ðtÞ $ es $ t dt s 0  Z FðsÞ 1 s $ t þ lim $ e ¼ $ f ðtÞdt t/0 s s  R f ðtÞdtt¼0 FðsÞ þ ¼ s s

(6.58) R which took into account that lim  1s $ es $ t $ f ðtÞdt ¼ 0. t/N

Initial-Value and Final-Value Theorems The initial- and final-value theorems can be used to interrogate the behavior of a dynamic system at the initial moment and at the final one, without necessarily having to calculate the system’s response over the full time interval. Both the initial- and final-value theorems are demonstrated based on the theorem providing the Laplace transform of the first derivative of a function f(t). Let us first calculate the limit:  Z N Z N  df ðtÞ df ðtÞ s $ t df ðtÞ  lim L $e $ lim es $ t dt ¼ 0 dt ¼ ¼ lim s/N s/N 0 s/N dt dt dt 0 (6.59) In this equation, advantage has been taken again of the interchangeability of the limit and integration operators with respect to the s variable. However, according to Eq. (6.44), it also follows that  df ðtÞ lim L ¼ lim ½s $ FðsÞ  f ð0Þ ¼ lim ½s $ FðsÞ  lim ½ f ðtÞ s/N s/N s/N t/0 dt Comparison of Eqs. (6.59) and (6.60) yields lim ½s $ FðsÞ  lim ½ f ðtÞ ¼ 0 or lim ½ f ðtÞ ¼ lim ½s $ FðsÞ

s/N

t/0

t/0

s/N

(6.60)

(6.61)

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CHAPTER 6 The Laplace Transform

which is known as the initial-value theorem because lim ½ f ðtÞ ¼ f ð0Þ is the initial t/0 value of the function f(t). The final-value theorem is demonstrated similarly by considering the following limit first:   Z N Z N   df ðtÞ df ðtÞ s $ t df ðtÞ $e $ lim es $ t dt lim L dt ¼ ¼ lim s/0 s/0 0 s/0 dt dt dt 0 Z N df ðtÞ dt ¼ lim f ðtÞ  lim f ðtÞ (6.62) ¼ t/N t/0 dt 0 Based on Eq. (6.44), it is also true that  df ðtÞ ¼ lim ½s $ FðsÞ  f ð0Þ ¼ lim ½s $ FðsÞ  lim ½ f ðtÞ lim L s/0 s/0 s/0 t/0 dt

(6.63)

Because the algebraic sums in the right-hand sides of Eqs. (6.62) and (6.63) need to be equal, it is necessary that lim ½ f ðtÞ ¼ lim ½s $ FðsÞ

t/N

(6.64)

s/0

which is the final-value theorem because lim ½ f ðtÞ ¼ f ðNÞ is the final (also known t/N

as steady-state) value of the function f(t). nn

Example 6.7

A dynamic system is represented by the first-order differential equation a1 $ y_ðt Þ þ a0 $ y ðt Þ ¼ b, where a0, a1 are constant coefficients, b is the constant (step) input, and y(t) is the unknown (response) function. Calculate the initial value and the final value of y(t).

Solution The Laplace transform is applied to the given differential equation based on Eq. (6.44) and the linearity property, which results in

a1 $ ½s $ YðsÞ  yð0Þ þ a0 $ YðsÞ ¼

b s

(6.65)

Solving for Y(s), the Laplace transform of the function f (t), in Eq. (6.65) yields

b b a1 $ yð0Þ þ ða1 $ s þ a0 Þ $ YðsÞ ¼ þ a1 $ yð0Þ or YðsÞ ¼ s s $ ða1 $ s þ a0 Þ a1 $ s þ a0 (6.66) The initial value of y (t) is determined by means of the initial-value theorem as



yð0Þ ¼ lim yðtÞ ¼ lim s $ Y ðsÞ ¼ lim t/0

s/N

s/N

b a1 $ y ð 0Þ $ s þ ¼ yð0Þ (6.67) a1 $ s þ a0 a 1 $ s þ a0

as expected. The final (steady-state) value of y(t) is found using the final-value theorem as



yðNÞ ¼ lim yðtÞ ¼ lim s $ YðsÞ ¼ lim t/N

s/0

s/0

b a1 $ yð0Þ $ s b þ (6.68) ¼ a1 $ s þ a 0 a1 $ s þ a0 a0

6.1 Direct and Inverse Laplace Transformations

The following MATLAB code: >> >> >> >>

syms s y0 a1 a0 b Y = b/(s*(a1*s+a0))+a1*y0/(a1*s+a0); limit(s*Y,s,0,’right’) limit(s*Y,s,inf,’left’)

returns the results of Eqs. (6.67) and (6.68).

nn

Note on final-value theorem: For harmonic functions involving sin (u$t) or cos (u$t) whose Laplace transforms are u/(s2 þ u2) and s/(s2 þ u2), respectively, the final-value theorem predicts constant steady-state values, as it can easily be verified. However, this is not true because neither sin (u$t) nor cos (u$t) have a unique final value and, as a consequence, caution should be exercised when using this theorem in conjunction with harmonic functions.

Periodic Functions When the input to a dynamic system is periodic, it is necessary to determine the Laplace transform of such a function whenever the Laplace transform method is applied to investigate the behavior (response) of the system. Periodic functions have the property f ðtÞ ¼ f ðt þ TÞ ¼ f ðt þ 2TÞ ¼ / ¼ f ðt þ n $ TÞ

(6.69)

where T is the period and n is any positive integer. Let us calculate the Laplace transform of such a periodic function. The zero-to-infinity time interval can be split in an infinite number of subintervals, namely 0 to T, T to 2T, ., n$T to (n þ 1)$T, ., which means the Laplace transform of f(t) can be written as Z L½ f ðtÞ ¼

N

e

Z

s $ t

$ f ðtÞdt ¼

0

Z

T

e

s $ t

0 ðnþ1Þ $ T

es $ t $ f ðtÞdt þ ::: ¼

n$T

Z $ f ðtÞdt þ

T ðnþ1Þ $ T

N Z X n¼0

2T

n$T

es $ t $ f ðtÞdt þ / es $ t $ f ðtÞdt (6.70)

The following change of variable t ¼ t1 þ n $ T transforms the limits of integration of the last integral in Eq. (6.70) into 0 (the lower limit) and T (the upper limit) so that Eq. (6.70) becomes L½ f ðtÞ ¼

N Z X

e

T

e

$ 0

s $ ðt1 þn $ TÞ

0

n¼0

Z

T

$ f ðt1 Þdt1 ¼

N X n¼0

s $ t1

$ f ðt1 Þdt1

! e

n $ T $ s

(6.71)

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CHAPTER 6 The Laplace Transform

The infinite sum in Eq. (6.71) represents an infinite geometric series sum whose ratio is r ¼ eT$s, and whose first term is a ¼ 1. It is known that the sum of terms in a geometric series is calculated as     N X a $ 1  r nþ1 1 $ 1  eðnþ1Þ $ T $ s 1 n $ T $ s ¼ lim e ¼ lim ¼ T $ s T $ s n/N n/N 1  e 1  r 1  e n¼0 (6.72) Combining Eqs. (6.71) and (6.72) and taking into account that the time variable in the integral of Eq. (6.71) is arbitrary, the Laplace transform of the periodic function f(t) becomes R T s $ t R T s $ t 1 $ f ðt1 Þdt1 e $ f ðtÞdt 0 e ¼ 0 (6.73) L½ f ðtÞ ¼ 1  es $ T 1  es $ T nn

Example 6.8 Find the Laplace transform of the periodic function drawn with thick solid line in Figure 6.9. Consider that time is measured in seconds.

f Sinusoidal function 1

0

1

2

3

4

6

8

10

t

–1

FIGURE 6.9 Train of Sinusoidal Pulses.

Solution

Figure 6.9 indicates the function’s period is T ¼ 4 s. The sinusoidal portions have an amplitude of 1; and because this maximum value occurs at t ¼ 1 s for the first time, it follows that the sinusoidal factor is sin (u$t) with u ¼ p/2. As a consequence, f(t) is written in branch form as

8

p  > 1 > > sin $ t ; n $ T  t  n þ $T > < 2 2 f ðtÞ ¼

> 1 > > > $ T  t  ðn þ 1Þ $ T 0; n þ : 2 8

p  > 1 > > sin $ t ; 4n  t  4 n þ > < 2 2 ¼

> 1 > > > : 0; 4 n þ 2  t  4ðn þ 1Þ

(6.74)

6.1 Direct and Inverse Laplace Transformations

Because the function f(t) is zero for the second interval (for t between 2 and 4 s) of the first period, the following results when applying Eq. (6.73)

R2 L½ f ðtÞ ¼

0

es $ t $ sin

p

2 1  e4s

 $ t dt

(6.75)

The integral in the numerator of Eq. (6.75) can be solved by parts, and the final result is

  2p 1 þ e2s L½ f ðtÞ ¼ 2 ðp þ 4s2 Þ $ ð1  e4s Þ

(6.76)

The result of Eq. (6.76) is also found if MATLAB’s dedicated symbolic-integration command int(g, t, a, b) is used with g ¼ es $ t $ sinðp $ t =2Þdthe function to be integrated, tdthe integration variable, a ¼ 0 (the lower integration limit), and b ¼ 2 (the upper integration limit). nn

The Convolution Theorem The convolution theorem offers an elegant alternative to finding the inverse Laplace transform of an s-domain function that can be written as the product of two functions. The convolution theorem is based on the convolution of two functions f(t) and g(t). According to the definition, the convolution of f(t) and g(t)ddenoted by the symbol “*”dis Z

t

f ðtÞ  gðtÞ ¼

f ðsÞ $ gðt  sÞds

(6.77)

0

It is straightforward to demonstrate that the convolution of two functions is a commutative operation. If the change of variable s1 ¼ t  s is used in Eq. (6.77), this equation changes to Z 0

t

Z f ðsÞ $ gðt  sÞds ¼ 

t

0

Z

t

f ðt  s1 Þ $ gðs1 Þds1 ¼

gðs1 Þ $ f ðt  s1 Þds1

0

¼ gðtÞ  f ðtÞ (6.78) The convolution theorem states that L½ f ðtÞ  gðtÞ ¼ FðsÞ $ GðsÞ

(6.79)

To demonstrate Eq. (6.79), let us apply the Laplace transform to Eq. (6.77) according to the definition: 2 3 Z N Z t 4es $ t $ L½ f ðtÞ  gðtÞ ¼ f ðt  sÞ $ gðsÞds5dt (6.80) 0

0

While the variable t ranges between 0 and N, the variable s is bounded by 0 and t. When the order of integration is switched in Eq. (6.80)dwhich means ds is outside

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the bracket and dt is inside the bracketdthe limits of the inside integral (whose variable becomes t) are s and N, while the limits of the outside integral (whose variable is s now) remain 0 and N; as a consequence Eq. (6.80) becomes 2 3 Z N Z N 4 L½ f ðtÞ  gðtÞ ¼ es $ t $ f ðt  sÞ $ gðsÞdt5ds (6.81) 0

s

The change of variable t ¼ s þ t1 is used in the internal integral of Eq. (6.81), where t1 is the new variable, and s behaves as a constant with respect to the integral. Eq. (6.81) changes to Z N Z N es $ s $ gðsÞ $ es $ t1 $ f ðt1 Þdt1 ds L½ f ðtÞ  gðtÞ ¼ 0 0 Z N Z N s $ s s $ t1 e $ gðsÞds $ e $ f ðt1 Þdt1 ¼ GðsÞ $ F ðsÞ ¼ 0

0

¼ F ðsÞ $ GðsÞ (6.82) which proves the convolution theorem. Applying now the inverse Laplace transform to Eq. (6.79) yields Z t f ðsÞ $ gðt  sÞds L1 ½FðsÞ $ GðsÞ ¼ L1 ½L½ f ðtÞ  gðtÞ ¼ f ðtÞ  gðtÞ ¼ 0

(6.83) Eq. (6.83) indicates that the inverse Laplace transform of the product of two functions in the s (Laplace) domain is equal to the convolution of the original (time-domain) functions. nn

Example 6.9

Calculate the inverse Laplace transform of the function Y(s) ¼ 1/[(s2 þ a2)$(s2 þ b2)] using analytical calculation and the convolution theorem.

Solution The given function can be expressed in product form as

YðsÞ ¼

1 1 1 1 $ ¼ FðsÞ $ GðsÞ with FðsÞ ¼ 2 ; GðsÞ ¼ 2 s 2 þ a2 s 2 þ b2 s þ a2 s þ b2 (6.84)

6.1 Direct and Inverse Laplace Transformations

therefore, the time-domain counterparts of the functions in Eq. (6.84) are

1 1 f ðtÞ ¼ L1 ½FðsÞ ¼ $ sinða $ tÞ; gðtÞ ¼ L1 ½GðsÞ ¼ $ sinðb $ tÞ a b

(6.85)

By applying the convolution theorem and integral, the function y (t ) is calculated as

yðtÞ ¼ f ðtÞ  gðtÞ ¼

1 ab

Z

t

sinða $ sÞ $ sin½b $ ðt  sÞds

(6.86)

0

The sine product under the integral can be written as

1 sinða $ sÞ $ sin½b $ ðt  sÞ ¼ fcos½ða þ bÞ $ s  b $ t  cos½ða  bÞ $ s þ b $ tg 2 (6.87) Substituting the sine product of Eq. (6.87) into Eq. (6.86) and solving the two integrals yields

yðtÞ ¼

1 ½a $ sinðb $ tÞ  b $ sinða $ tÞ a $ b $ ða2  b2 Þ

(6.88)

The result of Eq. (6.88) is confirmed through symbolic MATLAB calculation with the ilaplace command applied to the original function Y(s). nn

Partial Fraction Expansion Most often, rather complicated functions are produced in the s domain by application of the direct Laplace transform, functions that cannot be directly converted into the time domain using relatively simple transformation pairs and adequate properties. As a consequence, simplifications need to be performed before the inverse Laplace transformation can be utilized conveniently. Laplace-domain functions are obtained in fraction form: YðsÞ ¼

NðsÞ RðsÞ ¼ QðsÞ þ DðsÞ DðsÞ

(6.89)

where N(s) (N standing for numerator) and D(s) (D meaning denominator) are polynomials in s. When the degree of N(s) is larger than or equal to the degree of D(s), it is necessary to divide N(s) by D(s) until a quotient Q(s) is produced together with another polynomial fraction where the degree of the new numeratordthe remainder (or residue) R(s)dis smaller than the degree of the denominator. nn

Example 6.10

Determine the original function y(t) whose Laplace transform is Y(s) ¼ (s4 þ 3s3 þ 6s2 þ 6)/ (s2 þ 2s þ 5).

Solution Division of the numerator to the denominator yields the following sum:

YðsÞ ¼ s2 þ s  1 

s2

3s  11 þ 2s þ 5

(6.90)

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CHAPTER 6 The Laplace Transform

Eq. (6.90) can be obtained by direct analytical calculation or by using symbolic MATLAB and its deconv command, which allows division of two polynomials. For the current example, the following MATLAB code: >> N = [1 3 6 0 6]; >> D = [1 2 5]; >> [Q,R] = deconv(N,D) %Q is the quotient and R is the remainder returns Q = 1 1 -1 R = 0 0 0 -3 11 Note that a polynomial, such as N or D above, is defined in MATLAB as a row vector comprising the polynomial coefficients in a complete sequence, starting from the largest-power term to the free termda zero is used in N for the coefficient of s1 because that term is missing. MATLAB returns two lists with the coefficients of Q and R, following the same rule of coefficient sequence. Y(s) can further be conditioned to the form

YðsÞ ¼ s2 þ s  1  3 $

sþ1 2

ðs þ 1Þ þ

22

þ 7$

2 ðs þ 1Þ2 þ 22

(6.91)

which enables application of the inverse Laplace transform resulting in

yðtÞ ¼

d2 dðtÞ ddðtÞ  dðtÞ  ½3 cosð2tÞ  7 sinð2tÞ $ et þ dt2 dt

Eq. (6.92) took into account that

L



 2 ddðtÞ d dðtÞ ¼ s; L ¼ s2 dt dt2

(6.92)

(6.93)

which are demonstrated in the companion website Chapter 6. Alternatively, the result of Eq. (6.92) can be found by using MATLAB symbolic calculation that incorporates the ilaplace command. nn

Still, division of the numerator to the denominator of Eq. (6.89) may not be sufficient because the resulting polynomial fraction R(s)/D(s) is too complex (particularly due to the nature of its denominator) and the procedure known as partial fraction expansion needs to be applied, as we discuss next. Expansion of complicated s-domain fractions is also needed to assess the behavior of a system’s response by analyzing the simpler expansion fractions and the roots of their denominators. Partial fraction expansion can be performed analytically or using specialized MATLAB commands.

Analytical Partial Fraction Expansion There are four cases, as determined by the nature of the denominator roots, as follows: • •

Real and simple (distinct) roots. Complex (or imaginary) and simple (distinct) roots.

6.1 Direct and Inverse Laplace Transformations

• •

Real and multiple (repeated) roots. Complex and multiple (repeated) roots.

In each of these cases, different partial fraction expansion rules apply, and several unknown coefficients in the numerators of these fractions need to be determined by employing three methods. For simple roots, the method of fraction combination (which is presented in the companion website Chapter 6) or the cover-up method can be used to determine the unknown fraction coefficients. For multiple roots, the method of the s derivative can be utilized in conjunction with either of the other two methods. A few examples are studied next; the full treatment of the four cases mentioned previously can be found in the companion website Chapter 6.

MATLAB Partial Fraction Expansion Partial fraction expansion of the function Y(s) given in Eq. (6.89) results in NðsÞ r1 r2 ri riþ1 ¼ kðsÞ þ þ þ/þ þ þ/ DðsÞ s  p1 s  p2 s  pi ðs  pi Þ2 (6.94) riþm1 rn þ / þ þ ðs  pi Þm s  pn where k(s)dalso denoted Q(s) in Eq. (6.89)dis the direct-terms polynomial, r1, r2, ., rn are residues, and p1, p2, ., pn are poles. The poles (the roots of the denominator) can be either real or complex. Please note that the pole pi is a pole of multiplicity m. MATLAB’s residue command enables partial fraction expansion in a simple way. FðsÞ ¼

nn

Example 6.11

Calculate the inverse Laplace transform of the function Y(s) ¼ (s þ 3)/D(s) using analytical partial fraction expansion and the cover-up method. The denominator D(s) has the following roots: 2, 1 as simple roots, as well as the imaginary simple pair þj and j. Also utilize MATLAB to identify the partial fractions.

Solution

Given the fact that D(s) ¼ (s þ 2)$(s þ 1)$(s  j )$(s þ j ) ¼ (s þ 1)$(s þ 2)$(s2 þ 1), the function Y(s) can be written as

YðsÞ ¼

sþ3 ðs þ 1Þ $ ðs þ 2Þ $ ðs2 þ 1Þ

(6.95)

The denominator of Y(s) in Eq. (6.95) involves two simple real roots, 1 and 2, and two simple imaginary roots, j and þj. The particular fraction expansion of Y(s) is of the form

sþ3 a b c$s þ d þ þ 2 ¼ 2 ðs þ 1Þ $ ðs þ 2Þ $ ðs þ 1Þ s þ 1 s þ 2 s þ1

(6.96)

It can be seen that, in any of the simple fractions in the right-hand side of Eq. (6.96), the difference between the degree of the polynomial in the denominator and the degree of the polynomial in the numerator is 1. Note also that the two complex conjugated (imaginary) roots have been kept in their original polynomial s2 þ 1.

297

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CHAPTER 6 The Laplace Transform

The cover-up method determines the coefficients individually and sequentially; the method is advantageous to use in applications involving a large number of coefficients (fractions). To find the coefficient a, Eq. (6.96) is multiplied by (s þ 1) then s is made equal to 1, which results in

  sþ3  a¼ ¼1 2 ðs þ 2Þ $ ðs þ 1Þs¼1

(6.97)

This process is equivalent to “covering” (s þ 1) on the left-hand side of Eq. (6.96) and evaluating the remainder of the fraction for the value of s that zeroes the denominator of the fraction that contains a on the right-hand side of Eq. (6.96), i.e., s ¼ 1. By using this procedure, b is found by covering (s þ 2) on the left-hand side of Eq. (6.96), and taking s ¼ 2 for the remainder:

  sþ3 1  b¼ ¼  2 ðs þ 1Þ $ ðs þ 1Þ s¼2 5

(6.98)

The coefficients c and d are determined similarly, by first multiplying through Eq. (6.96) by (s2 þ 1), then considering that s2 þ 1 ¼ 0. The last operation can be done for either of the two complex roots of s2 þ 1 ¼ 0; that is, s ¼ j or s ¼ j. For s ¼ j, for instance, the result is

  sþ3  cs þ d ¼ ðs þ 1Þ $ ðs þ 2Þs¼j

(6.99)

which is equivalent to the following identity:

d  3c þ ð3d þ cÞ $ j ¼ 3 þ j

(6.100)

By equating the real parts and then the imaginary parts in the two sides of Eq. (6.100), b and c are obtained as c ¼ 4/5 and d ¼ 3/5. As a consequence, Y(s) can be written as

YðsÞ ¼

1 1 1 4 s 3 1  $  $ 2 þ $ 2 sþ1 5 sþ2 5 s þ1 5 s þ1

(6.101)

In MATLAB, the coefficients of N(s), the numerator of Y(s), and D(s), the denominator of Y(s) in Eq. (6.95) need to be first defined; while the coefficients of N(s) are known, the coefficients of D(s) are found by performing the multiplications of the original factored form, which yields D(s) ¼ s4 þ 3$s3 þ 3$s2 þ 3$s þ 2. The following MATLAB script is utilized: >> N = [1,3]; >> D = [1,3,3,3,2]; >> [r,p,k] = residue(N,D) and MATLAB returns r = -0.2000 + 0.0000i -0.4000 - 0.3000i -0.4000 + 0.3000i 1.0000 + 0.0000i p = -2.0000 + 0.0000i -0.0000 + 1.0000i -0.0000 - 1.0000i -1.0000 + 0.0000i k = [] Note: MATLAB also operates with the symbol i having the same significance as j, which is used consistently in this text for the square root of 1.

6.1 Direct and Inverse Laplace Transformations

The interpretation of these results is that the partial fraction expansion of Y(s) is

YðsÞ ¼

0:2 0:4  0:3 $ j 0:4 þ 0:3 $ j 1 þ þ þ sþ2 sj sþj sþ1

(6.102)

Note the correspondence between the positions of a residue in the r list and of a pole in the p list. The two middle fractions of Eq. (6.102) are combined using their common denominator s2 þ 1 ¼ (s  j)$(s þ j), and then Y(s) of this equation becomes Y(s) of Eq. (6.101). The inverse Laplace transform of Y(s) from Eq. (6.101) is

1 4 3 yðtÞ ¼ et  $ e2t  $ cosðtÞ þ $ sinðtÞ 5 5 5

(6.103)

The result of Eq. (6.103) is also found using MATLAB’s ilaplace command and symbolic calculation applied directly to the original Y(s) of Eq. (6.95). nn

nn

Example 6.12

Determine the inverse Laplace transform of the function Y(s) ¼ 1/[s$(s þ 1)2] using partial fraction expansion, together with the cover-up and s-derivative methods.

Solution

The roots of the denominator are s ¼ 0 and s ¼ 1, the latter one is real and of the order 2 of multiplicity. The partial fraction expansion of Y(s) is

1 s $ ðs þ 1Þ

2

¼

a b c þ þ 2 s ðs þ 1Þ sþ1

(6.104)

The coefficients a, b, and c can be determined with a method that combines the cover-up procedure with a derivation technique, as shown next. The coefficient a is calculated using the cover-up method as

   a¼ 2  ðs þ 1Þ 1

¼1

(6.105)

s¼0

Similarly, the coefficient b, the one corresponding to the largest degree denominator of the (two) fractions with s þ 1 in the denominator is directly found by means of the cover-up method as

 1 b¼  ¼ 1 s s¼1

(6.106)

To find coefficient c, Eq. (6.104) is multiplied by (s þ 1)2, which results in

1 a ¼ $ ðs þ 1Þ2 þ b þ c $ ðs þ 1Þ s s

(6.107)

By applying the derivation with respect to s in Eq. (6.107), we obtain the following equation:



1 2a $ ðs þ 1Þ a $ ðs þ 1Þ2  ¼ þc s2 s s2

(6.108)

Taking s ¼ 1 in Eq. (6.108) yields the value of c:

 1  c ¼  2 ¼ 1 s s¼1

(6.109)

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300

CHAPTER 6 The Laplace Transform

As a consequence, the function Y(s) is

YðsÞ ¼

1 1 1   s ðs þ 1Þ2 s þ 1

(6.110)

The expression of Y(s) of Eq. (6.110) is also determined by means of MATLAB’s residue command sequence. The inverse Laplace transform is applied to Eq. (6.110) producing the original function:

yðtÞ ¼ 1  ðt þ 1Þ $ et

(6.111)

The result of Eq. (6.111) is confirmed through MATLAB symbolic calculation with the ilaplace command. nn

nn

Example 6.13

Determine the inverse Laplace transform of the function Y(s) ¼ (3s3 þ 1)/[s$(s2 þ 1)2] using partial fraction expansion, together with the cover-up and coefficient identification methods.

Solution

The denominator has one real root, s ¼ 0, and two complex conjugated imaginary roots, s ¼ j and s ¼ j, which are of the order 2 of multiplicity. The partial fraction expansion of Y(s) is

YðsÞ ¼

3s3 þ 1 s $ ðs2

a b$s þ c d$s þ e ¼ þ þ 2 s ðs2 þ 1Þ2 s þ1 þ 1Þ 2

(6.112)

The coefficient a is determined by the cover-up method directly:

 3s3 þ 1  a¼ 2 ðs2 þ 1Þ 

¼1

(6.113)

s¼0

Multiplying Eq. (6.112) by s$(s2 þ 1)2 yields

 2   3s3 þ 1 ¼ a $ s2 þ 1 þ ðb $ s þ cÞ $ s þ ðd $ s þ eÞ $ s $ s2 þ 1

(6.114)

Eq. (6.114) is an identity, which holds true for any value of s. Selecting s ¼ j (which is a root of the equation s2 þ 1 ¼ 0), results in

1  3j ¼ b þ c $ j

(6.115)

Identifying the real and imaginary parts in the two sides of Eq. (6.115) gives b ¼ 1 and c ¼ 3. The values of a, b, and c are now substituted in Eq. (6.114), which becomes

s4 þ 3s3  s2 þ 3s ¼ d $ s4 þ e $ s3 þ d $ s2 þ e $ s

(6.116)

Comparing the coefficients of like powers in the two sides of Eq. (6.116) shows that d ¼ 1 and e ¼ 3. Eq. (6.112) can now be written as

1 s 3 s 3 þ YðsÞ ¼    s ðs2 þ 1Þ2 ðs2 þ 1Þ2 s2 þ 1 s2 þ 1

(6.117)

The inverse Laplace transforms of the second and third terms in the right-hand side of Eq. (6.117) are determined based on Table 6.1 as

"

1

L

#

s ðs2 þ 1Þ

2

" # t $ sin t 1 sin t  t $ cos t 1 ; L ¼ ¼ 2 2 2 2 ðs þ 1Þ

(6.118)

6.2 Laplace Transform Solution of Linear Ordinary Differential Equations

The Laplace transforms of the first, fourth, and last terms in the right-hand side of Eq. (6.117) are direct, and therefore the original time-domain function y(t) results from Y(s) of Eq. (6.117) in combination with Eq. (6.118):

yðtÞ ¼ 1 

t $ sinðtÞ 3 $ t $ cosðtÞ 3 $ sinðtÞ þ þ  cosðtÞ 2 2 2

(6.119)

The result of Eq. (6.119) is also obtained by directly applying MATLAB’s ilaplace symbolic calculation command. nn

6.2 LAPLACE TRANSFORM SOLUTION OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODE) The theorem that applies the direct Laplace transform to derivatives of functions is fundamental in solving linear ODE and linear ODE systems. While the Laplace transform excels in solving linear ODE with constant coefficients, there are cases where the Laplace transform can be applied to solve linear ODE with timedependent coefficients. For constant-coefficient linear ODE, application of the Laplace transform changes a time-domain differential equation into an algebraic one in the s domain, where the unknown (the Laplace transform of the time-dependent unknown function of the differential equation) can be determined easily. After finding the unknown function in the s domain, the process of partial (or simple-) fraction expansion needs to be applied to simplify the usually complex functions. The simplification enables migration back to the time domain using regular Laplace pairs and properties or symbolic MATLAB. To summarize, the steps that are needed to solve linear ODE by means of the Laplace transforms are •

• • •

Formulate the dynamic system’s mathematical model consisting of linear ODE in the time domaindthis step is detailed in Chapters 3e5 and 10; the present chapter assumes the mathematical model is provided. Apply the direct Laplace transform(s) to the time-dependent differential equation(s). Solve the resulting s-domain algebraic equation(s) for the Laplace transform(s) of the time-domain unknown function(s). Apply the inverse Laplace transform(s) to the simplified s-domain unknown function(s), and thus determine the original time-domain unknown functions.

6.2.1 Linear ODE With Constant Coefficients A linear, constant-coefficient ODE used to model a dynamic system is of the general form an $

d n yðtÞ dn1 yðtÞ dyðtÞ þ a0 $ yðtÞ ¼ uðtÞ þ an1 $ þ / þ a1 $ n n1 dt dt dt

(6.120)

301

302

CHAPTER 6 The Laplace Transform

where t (time) is the independent variable, u(t) is the known input (or excitation) function, y(t) is the unknown output function (or dependent variable), and the coefficients a0 to an are constant. This n-th order differential equation is an initial-value model requiring knowledge of n initial conditions, which are the values of y and of its first n  1 derivatives at t0. Application of the Laplace transform to the ODE of Eq. (6.120) under the assumption that the n initial conditions are nonzero, with the aid of the derivative transforms _ L½yðtÞ ¼ YðsÞ; L½yðtÞ ¼ s $ YðsÞ  yð0Þ; /  ðnÞ ðn 2Þ ðn1Þ L y ¼ sn $ YðsÞ  sn1 $ yð0Þ  /  s $ y ð0Þ  y ð0Þ

(6.121)

NY ðsÞ results in YðsÞ ¼ D where Y ðsÞ

NY ðsÞ ¼ U ðsÞ þ an $ yð0Þ $ sn1 þ ½an $ y_ð0Þ þ an1 $ yð0Þ $ sn2 þ /þ h i an $ yðn1Þ ð0Þ þ an1 $ yðn2Þ ð0Þ þ / þ a1 $ yð0Þ $ s0 DY ðsÞ ¼ an $ sn þ an1 $ sn1 þ / þ a1 $ s þ a0 (6.122) The denominator DY(s) is named characteristic polynomial and DY(s) ¼ 0 is the corresponding characteristic equation. Note that the degree of the characteristic polynomial (and of the characteristic equation) is equal to the order of the corresponding ODE, which is the order of the dynamic system as well. Once Y(s) was identified, the time-domain response y(t) is calculated as  1 1 NY ðsÞ yðtÞ ¼ L ½YðsÞ ¼ L (6.123) DY ðsÞ The steady-state (final) value of the time response is calculated based on the finalvalue theorem: s $ NY ðsÞ s $ UðsÞ ¼ lim (6.124) s/0 s/0 DY ðsÞ s/0 DY ðsÞ For systems with no input, i.e., U(s) ¼ 0, which are known as free systems, the steady-state response is zero provided a0 s 0. For a0 ¼ 0, as it can be checked from Eqs. (6.122)e(6.124), y(N) is of the form 0/0; by applying the L’Hospital rule it becomes yðNÞ ¼ lim yðtÞ ¼ lim s $ YðsÞ ¼ lim t/N

yðNÞ ¼ lim yðtÞ ¼ lim s $ Y ðsÞ ¼ an $ yðn1Þ ð0Þ þ an1 $ yðn2Þ ð0Þ þ / þ a1 $ yð0Þ t/N

s/0

a1 (6.125) Definitely, y(N) s 0 when a1 s 0, as Eq. (6.125) indicates. However, y(N) / N for a1 ¼ 0, as also seen in Eq. (6.125), which means that the system’s free response

6.2 Laplace Transform Solution of Linear Ordinary Differential Equations

grows without bound (to infinity) under finite initial conditions and therefore, the dynamic system is unstableda condition studied more thoroughly in Chapters 7 and 11. nn

Example 6.14 Solve the differential equation

d 2 yðtÞ dyðtÞ þ yðtÞ ¼ uðtÞ þ 2$ 2 dt dt which is the mathematical model of a dynamic system. Use the Laplace transform analytical method based on partial fraction expansion, and check the results by means of MATLAB’s ilaplace command. Consider the initial conditions are zero and (i) u(t) ¼ t$e2t (ii) u(t) ¼ e2t$cos (2t)

Solution (i) The Laplace transform is applied to the differential equation of this problem, which results in



 s2 þ 2s þ 1 $ YðsÞ ¼ UðsÞ or YðsÞ ¼

UðsÞ UðsÞ ¼ s2 þ 2s þ 1 ðs þ 1Þ2

(6.126)

The Laplace transform of u(t) ¼ t$e2t is U(s) ¼ 1/(s þ 2)2 and Eq. (6.126) becomes

YðsÞ ¼

1 ðs þ 1Þ2 $ ðs þ 2Þ2

¼

a ðs þ 1Þ2

þ

b c d þ þ s þ 1 ðs þ 2Þ2 s þ 1

(6.127)

The coefficients a and c are found by the cover-up method as

   a¼ 2  ðs þ 2Þ 1

   ¼ 1; c ¼ 2  ðs þ 1Þ 1

s¼1

¼1

(6.128)

s¼2

Eq. (6.127) is multiplied by (s þ 1)2, which results in

"

d ¼ a þ b $ ðs þ 1Þ þ ðs þ 1Þ $ þ 2 2 s þ 2 ðs þ 2Þ ðs þ 2Þ 1

2

c

# (6.129)

Derivation in terms of s is applied to Eq. (6.129), which yields

"

# " # d d c d  ¼ b þ 2 $ ðs þ 1Þ $ þ þ þ ðs þ 1Þ2 $ ds ðs þ 2Þ2 s þ 2 ðs þ 2Þ3 ðs þ 2Þ2 s þ 2 2

c

(6.130) Using s ¼ 1 in the identity of Eq. (6.130) determines b:

   b¼  ðs þ 2Þ3  2

¼ 2 s¼1

(6.131)

303

304

CHAPTER 6 The Laplace Transform

The coefficient d is found with a similar procedure: Eq. (6.127) is multiplied by (s þ 2)2, the s derivative of the resulting equation is calculated and s ¼ 2 is used in the resulting equation, which yields

   d¼  ðs þ 1Þ3  2

¼2

(6.132)

s¼2

As a consequence, Y(s) of Eq. (6.127) becomes

YðsÞ ¼

1 ðs þ 1Þ2



2 1 2 þ þ s þ 1 ðs þ 2Þ2 s þ 1

(6.133)

and therefore

yðtÞ ¼ t $ et  2et þ t $ e2t þ 2e2t

(6.134)

Applying MATLAB’s ilaplace command directly to Y(s) of Eq. (6.127) yields y(t) of Eq. (6.134), as it can easily be checked. (ii) The Laplace transform of u(t) ¼ e2t$cos (2t) is U(s) ¼ (s þ 2)/[(s þ 2)2 þ 4], and Y(s) is

YðsÞ ¼

sþ2 sþ2 h i¼ 4 3 þ 17s2 þ 20s þ 8 2 s þ 6s ðs þ 1Þ $ ðs þ 2Þ þ 4 2

(6.135)

Using MATLAB’s residue command, the following partial fraction expansion is obtained:

YðsÞ ¼

0:06 þ 0:08 $ j 0:06  0:08 $ j 0:12 0:2 þ þ þ 2 s  ð2 þ 2jÞ s  ð2  2jÞ s þ1 ðs þ 1Þ

1 3s þ 14 0:12 0:2 ¼ $ þ þ 2 2 25 ðs þ 2Þ þ 4 ðs þ 1Þ sþ1 " # 1 sþ2 2 0:12 0:2 ¼  $ 3$ þ þ 4$ þ 25 ðs þ 2Þ2 þ 4 ðs þ 2Þ2 þ 4 ðs þ 1Þ2 s þ 1

(6.136)

When applying the Laplace transform to Eq. (6.136), the solution is

yðtÞ ¼ 0:04 $ ½3 $ cosð2tÞ þ 4 $ sinð2tÞ $ e2t þ 0:2 $ ð0:6 $ t þ 1Þ $ et (6.137) result that is also obtained by applying MATLAB’s ilaplace command to Y(s) of Eq. (6.135).

nn nn

Example 6.15

A single-DOF mechanical system is described by the differential equation y€ þ 10y_ þ 100y ¼ u. Calculate the system response y(t), plot it in terms of time, and determine the steady-state value of the time-domain solution for zero initial conditions and u(t) of Figure 6.10.

Solution The Laplace transform is applied to the given differential equation; this leads to

YðsÞ ¼

UðsÞ s2 þ 10s þ 100

(6.138)

6.2 Laplace Transform Solution of Linear Ordinary Differential Equations

u (N) 2 u2(t)

u1(t) 1

0

2

u3(t)

3

Time (s)

FIGURE 6.10 Time-Domain Definition of the Input u.

The time-domain input function shown in Figure 6.10 is expressed as

8 8 > < u1 ðtÞ; 0  t  2 > < t; 0  t  2 uðtÞ ¼ u2 ðtÞ; 2  t  3 ¼ t þ 4; 2  t  3 > > : : u1 ðtÞ; t  3 1; t  3

The Laplace transform of u(t) of Eq. (6.139) results in

Z

UðsÞ ¼ ¼

2

t $ es $ t dt þ

0 3s e

Z

3

ðt þ 4Þ $ es $ t dt þ

Z

2

N

(6.139)

1 $ es $ t dt

3

 2e2s þ 1 s2

(6.140)

The integrations of Eq. (6.140) were performed symbolically with MATLAB’s int command. Combining Eqs. (6.138) and (6.140) yields

YðsÞ ¼

e3s  2e2s þ 1 s2 $ ðs2 þ 10s þ 100Þ

(6.141)

The inverse Laplace transform of Y(s) from Eq. (6.141) is

t þ y1 ðtÞ $ hðtÞ  2 $ y2 ðtÞ $ hðt  2Þ þ y3 ðtÞ $ hðt  3Þ with yðtÞ ¼ 100 pffiffiffi 8    pffiffiffi  pffiffiffi  3 e5t 1 > > y ; $ cos 5 3 $ t  ðtÞ ¼ $ sin 5 3 $t  > 1 > > 1000 1000 3 > > > > >   <  pffiffiffi   pffiffi3ffi pffiffiffi t e5ðt2Þ 21 y2 ðtÞ ¼ $ sin 5 3 $ ðt  2Þ  þ ; $ cos 5 3 $ ðt  2Þ  > 3 100 1000 1000 > > > > >   >  pffiffi3ffi  pffiffiffi  > pffiffiffi > t e5ðt3Þ 31 > : y3 ðtÞ ¼ þ ; $ cos 5 3 $ ðt  3Þ  $ sin 5 3 $ ðt  3Þ  100 1000 1000 3 (6.142)

305

CHAPTER 6 The Laplace Transform

y (m)

306

Time (s)

FIGURE 6.11 Time-Domain Response y(t).

Eqs. (6.142) were obtained using MATLAB’s ilaplace command and Figure 6.11 graphs y as a function of t. As per Eq. (6.124), the steady-state response of this system is

  s $ e3s  2e2s þ 1 yðNÞ ¼ lim yðtÞ ¼ lim s $ YðsÞ ¼ lim 2 t/N s/0 s/0 s $ ðs2 þ 10s þ 100Þ ¼ lim

s/0

e3s  2e2s þ 1 ¼ 0:01 s $ ðs2 þ 10s þ 100Þ

(6.143)

The value of 0.01 was obtained using the following MATLAB code: >> syms s >> limit((exp(-3*s)-2*exp(-2*s)+1)/s/(s^2+10*s+100),s,0)

nn

6.2.2 Systems of Linear ODE With Constant Coefficients A system of ODE with constant coefficients (either with or without forcing) contains several unknown functions y1(t), y2(t), ., yn(t) together with their derivatives in n different equations. The method of Laplace transform can be applied to solve such ODE systems. They can be transformed into algebraic equations that form a system whose unknowns are the Laplace transforms of the unknown timedependent functions. These Laplace-transformed unknowns are solved for in the s domain algebraically, and then inverse Laplace transforms are applied to those to determine the original time-dependent unknown functions. An alternative method is to use a vectorematrix formulation of the ODE system and to apply the approach described above to this format. It is actually quite

6.2 Laplace Transform Solution of Linear Ordinary Differential Equations

straightforward to extend the Laplace transform from functions to collections of functions (which are vectors). Let us look at a vector {y(t)} consisting of two components, y1(t) and y2(t). The Laplace transforms of the two components (assuming they exist) are determined as Z N Z N Y1 ðsÞ ¼ y1 ðtÞ $ es $ t dt; Y2 ðsÞ ¼ y2 ðtÞ $ es $ t dt (6.144) 0

0

and they can be assembled into the vector Y(s) as fYðsÞg ¼ f Y1 ðsÞ Y2 ðsÞ gt

(6.145)

We can therefore define the Laplace transform of the vector {y(t)} into the vector {Y(s)} as L½fyðtÞg ¼ fYðsÞg

(6.146)

In conclusion, application of the Laplace transform to a vector results in another vector whose components are the Laplace transforms of the original (time-dependent) vector components. Extensions can be made, and it can simply be shown that, given the vectors {y1(t)} and {y2(t)}, as well as the matrices [A1] and [A2] (made up of constants only), the following linear relationship holds L½½A1  $ fy1 ðtÞg þ ½A2  $ fy2 ðtÞg ¼ ½A1  $ fY1 ðsÞg þ ½A2  $ fY2 ðsÞg

(6.147)

We thus can conclude that the Laplace pairs and properties pertaining to scalar functions are also valid for vector functions, which could be particularly helpful in solving differential equations written in vectorematrix form. For instance, taking the first and second time derivatives of a vector {y(t)} results in  d L fyðtÞg ¼ s $ fYðsÞg  fyð0Þg; dt (6.148)  2 d _ L 2 fyðtÞg ¼ s2 $ fYðsÞg  s $ fyð0Þg  fyð0Þg dt where the vectors having the argument zero indicate initial conditions. Consider a dynamic system defined by the following vector-form differential equation € _ ½A2  $ fyðtÞg þ ½A1  $ fyðtÞg þ ½A0  $ fyðtÞg ¼ fuðtÞg Applying the Laplace transform to Eq. (6.149) results in   _ ½A2  $ s2 $ fYðsÞg  s $ fyð0Þg  fyð0Þg þ ½A1  $ ðs $ fYðsÞg  fyð0ÞgÞ þ ½A0  $ fYðsÞg ¼ fUðsÞg

(6.149)

(6.150)

where L½fyðtÞg ¼ fYðsÞg; L½fuðtÞg ¼ fUðsÞg

(6.151)

307

308

CHAPTER 6 The Laplace Transform

Eq. (6.150) can also be written as 

 _ þ s $ fyð0ÞgÞ þ ½A1  $ fyð0Þg s2 $ ½A2  þ s $ ½A1  þ ½A0  $ fYðsÞg ¼ ½A2  $ ðfyð0Þg þ fUðsÞg (6.152)

The solution to Eq. (6.152) is therefore  1 _ $ ð½A2  $ ðfyð0Þg þ s $ fyð0ÞgÞ fYðsÞg ¼ s2 $ ½A2  þ s $ ½A1  þ ½A0  þ ½A2  $ fyð0Þg þ fUðsÞgÞ

(6.153)

and the time-domain solution vector {y(t)} is simply found by inverse Laplace transforming {Y(s)} of Eq. (6.153). nn

Example 6.16 The mathematical model of a two-DOF dynamic system is

y€1 ðtÞ þ 2y_1 ðtÞ þ y€2 ðtÞ ¼ sinðtÞ y_1 ðtÞ þ 2y1 ðtÞ þ y€2 ðtÞ þ 3y_2 ðtÞ þ 2y2 ðtÞ ¼ e2t

Solve this ODE system by using the Laplace transform approach and verify the result by MATLAB. Consider all initial conditions are zero.

Solution Directly applying the Laplace transforms to the two given ODE results in

8   > > > s2 þ 2s $ Y1 ðsÞ þ s2 $ Y2 ðsÞ ¼
  > > : ðs þ 2Þ $ Y1 ðsÞ þ s2 þ 3s þ 2 $ Y2 ðsÞ ¼

1 sþ2

(6.154)

These equations are collected in the vectorematrix form

"

s2 þ 2s sþ2

9 8 > 1 > # > >  = < 2 Y1 ðsÞ s2 s þ1 $ ¼ > > Y2 ðsÞ s2 þ 3s þ 2 > ; : 1 > sþ2

(6.155)

Alternatively, the original ODE system of this example can be written directly in the vectorematrix form of Eq. (6.149) with

 ½A2  ¼

1

1

0

1



 ; ½A1  ¼

2 1



 sinðtÞ ; ½A0  ¼ ; fuðtÞg ¼ (6.156) 3 2 2 e2t 0





0 0

6.3 Laplace Transform Solution of Integral

and therefore, the general Eq. (6.152) becomes

 s2 $

1 0





1 2 þ s$ 1 1





0 0 0 þ 3 2 2

!

9 1 > > = s2 þ 1 $ fYðsÞg ¼ > > > ; : 1 > sþ2

309

8 > >
> [Y1s, Y2s] = collect(inv([s^2+2*s,s^2;s+2,s^2+3*s+2])*[1/(s^2+1);. 1/(s+2)]) The inverse Laplace transform is subsequently applied to the two components of Eq. (6.158), by means of MATLAB’s ilaplace command, which yields

8 1 1 t 2 t 1 2t 1 3 > 2t > < y1 ðtÞ ¼  $ e $ sinðtÞ  $ e $ cosðtÞ þ $ e þ t $ e  $ sinðtÞ  $ cosðtÞ 2 5 5 2 5 5 > > : y2 ðtÞ ¼ 1 þ 2 $ et $ sinðtÞ  1 $ et $ cosðtÞ þ 1 $ e2t þ 2 $ sinðtÞ þ 1 $ cosðtÞ 2 5 5 2 5 5 (6.159) nn

6.3 LAPLACE TRANSFORM SOLUTION OF INTEGRAL AND INTEGRALeDIFFERENTIAL EQUATIONS Integral equations contain terms where the time-domain unknown function is under the integral operator, whereas integraledifferential equations have the unknown function comprised in both differential terms and integral terms. These two types of differential equations can be solved by the Laplace transform approach or by means of the convolution theorem, as discussed in this section.

6.3.1 Laplace Transform Approach Consider the following integraledifferential equation: Z t dyðtÞ þ a0 $ y ð t Þ ¼ uð t Þ a2 $ yðtÞdt þ a1 $ dt 0

(6.160)

310

CHAPTER 6 The Laplace Transform

For zero initial conditions, application of the Laplace transform to this equation yields s $ UðsÞ YðsÞ ¼ (6.161) a1 $ s 2 þ a0 $ s þ a2 and therefore y(t) can be determined by inverse Laplace transforming Y(s) of Eq. (6.161). nn

Example 6.17 A two-DOF dynamic system is modeled by means of the following integraledifferential equation system:

( a $ y_ ðtÞ þ b $ y ðtÞ þ b $ y ðtÞ ¼ u 1Z 2 1 a $ y_1 ðtÞ þ c $

t

y2 ðtÞdt ¼ 0

0

Solve this system considering zero initial conditions for a ¼ 20, b ¼ 200, c ¼ 10,000 and u ¼ 100.

Solution The following algebraic equations system results after applying the Laplace transform to the two differential equations

8 u > > < ða $ s þ bÞ $ Y1 ðsÞ þ b $ Y2 ðsÞ ¼ s c > > : a $ s $ Y1 ðsÞ þ $ Y2 ðsÞ ¼ 0 s

(6.162)

which can be written in vectorematrix form as

2 4

a$s þ b a $ s

8 9 3

 >  > = < = < s $ ða $ b $ s2 þ a $ c $ s þ b $ c > b 5 c $ s ¼ > : ; > a$u$s > > 0 ; : s a $ b $ s2 þ a $ c $ s þ b $ c (6.164)

The time-domain solution is obtained by inverse Laplace transforming Y1(s) and Y2(s) of Eq. (6.164) with the given numerical parameters; the solution is

8  pffiffiffi  pffiffiffi  pffiffiffi i 1 1 h > 25t > < y1 ðtÞ ¼  $ cosh 5 5 t þ 5 sinh 5 5 t $ e 2 2 h  pffiffiffi  pffiffiffi  pffiffiffi i > > : y2 ðtÞ ¼ 1 $ cosh 5 5 t  5 sinh 5 5 t $ e25t 2

(6.165)

nn

6.3 Laplace Transform Solution of Integral

6.3.2 Convolution Theorem Approach The convolution theorem can be used to solve integral and integraledifferential equations. Let us assume the mathematical model of a system consists of the following integral equation: Z t yðtÞ þ yðsÞ $ f ðt  sÞds ¼ uðtÞ (6.166) 0

where the functions f and u are known time-dependent functions and y(t) is the unknown function. Application of the Laplace transform to Eq. (6.166) results in YðsÞ þ YðsÞ $ FðsÞ ¼ UðsÞ (6.167) which is YðsÞ ¼

UðsÞ 1 þ FðsÞ

The inverse Laplace transform of Y(s) is yðtÞ ¼ L1 ½YðsÞ ¼ L1



UðsÞ 1 þ FðsÞ

(6.168)

(6.169)

nn

Example 6.18 The mathematical model of a single-DOF dynamic system is

Z

yðtÞ þ

t

yðt  sÞ $ et ds ¼ sinðtÞ

0

Solve this integral equation knowing that y (0) ¼ 0:

Solution

In this case the functions f(t) and u(t) are

f ðtÞ ¼ et ; uðtÞ ¼ sinðtÞ

(6.170)

therefore,

FðsÞ ¼

1 1 ; UðsÞ ¼ 2 sþ1 s þ1

(6.171)

so that, in accordance with Eq. (6.168), Y(s) is

YðsÞ ¼

sþ1 ðs þ 2Þ $ ðs2 þ 1Þ

(6.172)

whose inverse Laplace transform is

yðtÞ ¼ L1 ½Y ðsÞ ¼

 1 3 sin t þ cos t  e2t 5

(6.173) nn

311

312

CHAPTER 6 The Laplace Transform

Integraledifferential equations can be solved under certain conditions using the convolution theorem. nn

Example 6.19 Solve the following integral-differential equation, which models a single-DOF dynamic system:

Z

_ þ yðtÞ

t

yðsÞ $ ðt  sÞds ¼ 4

0

with y (0) ¼ 0.

Solution In this example

2 L4

Z

t

3 yðsÞ $ ðt  sÞds5 ¼ YðsÞ $ L½t ¼

0

YðsÞ s2

(6.174)

As a consequence, application of the Laplace transform to the given integraledifferential equation yields

s $ YðsÞ þ

YðsÞ 4 4s ¼ or YðsÞ ¼ 3 s2 s s þ1

(6.175)

Inverse Laplace transforming Y(s) of Eq. (6.175) results in

4 y ðt Þ ¼ $ 3



pffiffiffi

pffiffiffi  pffiffiffi 3 3 0:5t t cos t þ 3 sin t $e  e 2 2

(6.176) nn

6.4 TIME-DOMAIN SYSTEM IDENTIFICATION FROM LAPLACE-DOMAIN INFORMATION A time-domain dynamic system can be identified by analyzing it in the Laplace domain. The topic is particularly important when the frequency response of a system is knowndthis can be achieved through experimental means, for instance, and Chapter 9 studies the frequency response approach in more detail. In this section, we aim to identify a time-defined dynamic system from a given Laplace-domain function. Let us assume a dynamic system is defined by the generic differential equation: € þ a1 $ yðtÞ _ þ a0 $ yðtÞ ¼ uðtÞ a2 $ yðtÞ

(6.177)

with a2, a1, and a0 being constants and the initial conditions being nonzero. Application of the Laplace transform to Eq. (6.177) enables expressing the Laplace transform of the unknown y(t) as YðsÞ ¼

a2 $ yð0Þ $ s þ a2 $ y_ $ ð0Þ þ a1 $ yð0Þ þ UðsÞ a2 $ s 2 þ a1 $ s þ a 0

(6.178)

6.4 Time-Domain System Identification From Laplace-Domain Information

where U(s) is the Laplace transform of the forcing function u(t). Eq. (6.178) can be written in the generic form b $ s þ c þ UðsÞ YðsÞ ¼ (6.179) a2 $ s 2 þ a 1 $ s þ a 0 Eqs. (6.178) and (6.179) are general, as they encompass all possible parameters, forcing, and initial conditions for a second-order system. Particular cases can be derived from them by annulling various term(s) in either the numerator or denominator, terms that correspond to missing parameters, forcing functions, or initial conditions. nn

Example 6.20 Identify a single-DOF dynamic system knowing the Laplace-domain transform of the system’s response is

  s $ 5s2  s þ 5 YðsÞ ¼ 2 ðs þ 1Þ $ ð2s2 þ s þ 1Þ

Solution

The expression of Y(s) can be transformed as follows:

1   ð5s  1Þ $ s2 þ 1 þ 1 ð5s  1Þ þ s2 þ 1 5s3  s2 þ 5s  1 þ 1 ¼ ¼ YðsÞ ¼ 2 ðs þ 1Þ $ ð2s2 þ s þ 1Þ ðs2 þ 1Þ $ ð2s2 þ s þ 1Þ 2s2 þ s þ 1 (6.180) Comparison of Eqs. (6.178) and (6.180) shows that U(s) ¼ 1/(s2 þ 1), which yields u(t) ¼ sin (t). The same comparison also indicates that a2 ¼ 2, a1 ¼ 1, a0 ¼ 1, and

a2 $ yð0Þ ¼ 5 _ þ a1 $ yð0Þ ¼ 1 a2 $ yð0Þ

(6.181)

The equation system (Eq. 6.181) is solved for the two initial conditions, which are

5 dyðtÞ 7 _ ¼  yð0Þ ¼ ; ¼ yð0Þ j 2 dt t¼0 4

(6.182)

As a consequence, the time-domain differential equation that corresponds to Y(s) of this example is

€ þ yðtÞ _ þ yðtÞ ¼ sinðtÞ 2yðtÞ and the initial conditions are given in Eq. (6.182).

(6.183) nn

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SUMMARY This chapter introduces the direct and inverse Laplace transforms as tools for solving differential equations that are mathematical models of dynamic systems. By using Laplace transform pairs as well as properties of the Laplace transforms, differential equations are transformed into algebraic equations in the Laplace (s) domain. The solutions to these algebraic equations are calculated and used to find the original time-domain unknown functions through inverse Laplace transformation. Using the technique of partial fraction expansion, the direct and inverse Laplace transforms can be used to solve linear differential equations with constant coefficients. MATLAB symbolic calculation of direct and inverse Laplace transforms is presented here. The Laplace transformation is used extensively in Chapters 7e13, which are dedicated to transfer functions, state space modeling, frequency-domain analysis, coupled systems, and controls.

PROBLEMS 6.1 Use the definition to calculate the Laplace transforms of the hyperbolic sine and cosine functions, sinh (u$t) and cosh (u$t), knowing that sinh (u$t) ¼ (eu$t  eu$t)/2 and cosh(u$t) ¼ (eu$t þ eu$t)/2. Check the results using Table 6.1. 6.2 Use the definition to calculate the Laplace transforms of t$sin (u$t) and t$cos (u$t). Check the results using Table 6.1. 6.3 Use the definition of the Laplace transform and the principle of mathematical   induction to calculate L tn $ ea $ t where n is a positive integer and a is a real number. Check the result against Table 6.1. 6.4 Find the Laplace transforms of the following functions based on the known Laplace transforms of sin (u$t) and cos (u$t); check the results with MATLAB (i) f(t) ¼ sin (u1$t)$cos (u2$t). (ii) f(t) ¼ cos (u1$t)$cos (u2$t). (iii) f(t) ¼ sin (u1$t)$sin (u2$t). Hint: Use trigonometric identities to transform the products into sums/ differences of trigonometric functions.

Problems

6.5 Same question as in Problem 6.4 for the functions (i) f(t) ¼ cosh (u1$t)$sinh (u2$t). (ii) f(t) ¼ cosh (u1$t)$cosh (u2$t). (iii) f(t) ¼ sinh (u1$t)$sinh (u2$t). Use the known Laplace transforms of sinh (u$t) and cosh (u$t). 6.6 Find the Laplace transforms of sin (u$t) and cos (u$t) based on the known Laplace transforms of sinh (u$t) and cosh (u$t). Hint: Use the exponential formulation to connect trigonometric and hyperbolic functions. 6.7 Use the theorem that expresses the Laplace transform of the first time derivative of a function to calculate the Laplace transforms of the functions f1(t) ¼ t$sinh (u$t) and f2(t) ¼ t$cosh (u$t) based on the known Laplace transforms of sinh (u$t) and cosh (u$t). 6.8 Calculate the Laplace transform of the function f(t) ¼ cosh (u$t)$ sin (u$t)  sinh (u$t)$cos (u$t) analytically and with MATLAB. 6.9 Use the frequency shift theorem and Table 6.1 to analytically determine the Laplace transforms of (i) f(t) ¼ et$cos (u$t þ 4). (ii) f(t) ¼ e5t$t1/2. (iii) f(t) ¼ e4t$sin (2t)$cos (3t). Check the results by MATLAB. 6.10 Use the frequency shift theorem to analytically determine the inverse Laplace transforms of (i) F(s) ¼ (2s3 þ 14s2 þ 48s þ 88)/[(s2 þ 4s þ 20)$(s2 þ 4s þ 4)]. (ii) F(s) ¼ 6(s þ 2)/(s2 þ 2s  5)2. (iii) F(s) ¼ (48s þ 240)/[(s2  2s þ 17)$(s2 þ 22s þ 137)]. Check the results by MATLAB. 6.11 Use analytical calculation to find the Laplace transform of the function shown in Figure 6.12. Check the result by MATLAB. 6.12 Same question as in Problem 6.11 for the function shown in Figure 6.13. 6.13 Same question as in Problem 6.11 for the function shown in Figure 6.14.

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CHAPTER 6 The Laplace Transform

f 2b b 0

a

3a

t

FIGURE 6.12 Multistep Function. f

b 2a 3a 0

4a t

a

–b

FIGURE 6.13 Truncated Multisegment Function. f

Sinusoidal function

3

0

4

8

t

FIGURE 6.14 Truncated Sinusoidal Function.

6.14 Use the time-shift theorem and Table 6.1 (if needed) to analytically determine the inverse Laplace transforms of (i) F(s) ¼ e3s$(s2 þ 1)/(s2  5)2. (ii) F(s) ¼ e2s$(s þ 1)/(s2 þ 3s þ 2). (iii) F(s) ¼ e5s$(3s2 þ 12)/(s4 þ 16). Check the results by MATLAB.

Problems

6.15 Use the theorem that expresses the Laplace transforms of the first and second time derivatives of functions to calculate the Laplace transforms of the functions f1(t) ¼ t$sin (u$t), f2(t) ¼ t$cos (u$t), f3(t) ¼ t2$sin (u$t), and f4(t) ¼ t2$cos (u$t), when known are the Laplace transforms of sin (u$t) and cos (u$t). Check the results with MATLAB. 6.16 Same question as in Problem 6.15 for the functions f1(t) ¼ t$e a$t, f2(t) ¼ t2$ea$t, when known is the Laplace transform of ea$t (a is a positive real constant). 6.17 Same question as in Problem 6.15 for the functions f1(t) ¼ t$sinh (u$t), f2(t) ¼ t$cosh (u$t), f3(t) ¼ t2$sinh (u$t), and f4(t) ¼ t2$cosh (u$t), when we know the Laplace transforms of sinh (u$t) and cosh (u$t) 6.18 Use Table 6.2 and mathematical induction to demonstrate that    n  d f ðtÞ df n1 ðtÞ n n1 n2 df ðtÞ L $ f ð0Þ  s $ / ¼ s $ FðsÞ  s dtn dt t¼0 dtn1 t¼0 when the function f(t) and its first (n  1) derivatives are all of exponential order.    6.19 Use Table 6.2 to demonstrate that L t2 $ f ðtÞ ¼ d2 FðsÞ ds2 ; use the mathematical induction to calculate L½tn $ f ðtÞ. 6.20 Use the property



Z N f ðtÞ L FðsÞds ¼ t s

(which is given in Table 6.2 and derived in the companion website Chapter 6) to analytically calculate the Laplace transforms of sin (u$t)/t and cos (u$t)/t based on the known Laplace transforms of sin (u$t) and cos (u$t). 6.21 The mathematical model of a rotary mechanical system is represented by the € þ c $ q_ þ k $ q ¼ uðtÞ where J, c, and k are differential equation J $ q constant coefficients. Calculate the final value of the rotation angle q when the input is u(t) ¼ 10d(t) þ 25h(t). Use symbolic calculation and MATLAB to confirm the result. 6.22 A constant voltage v ¼ 80 V is applied to a series electrical system formed of a resistor R, an inductor L, and a capacitor C. It is determined that the steadystate current in the circuit is i(N) ¼ 0.02 A. Calculate the capacitance C using the final-value theorem and check the result by MATLAB symbolic calculation.

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CHAPTER 6 The Laplace Transform

6.23 The mathematical model of a dynamic system is represented by the differential equations system ( 2y€1 þ 40y_1 þ 10y1  20y_2  5y2 ¼ 5e4t y€2 þ 40y_2 þ 10y2  20y_2  5y2 ¼ 0 Use the initial-value theorem to calculate y1(0) and y2(0) knowing that the initial values of the first time derivatives of y1 and y2 are zero. 6.24 Calculate the Laplace transform of the periodic function sketched in Figure 6.15. f 2b b 0

a

3a

4a

5a

7a

8a

t

FIGURE 6.15 Periodic Multistep Function.

6.25 Same question as in Problem 6.24 for the periodic function sketched in Figure 6.16. f

b 2a 3a 0

a

4a

7a 5a 6a

8a t

–b

FIGURE 6.16 Periodic Multisegment Function.

6.26 Same question as in Problem 6.24 for the periodic function shown in Figure 6.17. 6.27 The input force f applied to a single-input and single-output mechanical system is represented as a function of time t in Figure 6.18 by the solid thicker line. Calculate the Laplace transform of f (t).

Problems

f Parabolic function 2

0

2

4

6

8

10

12

14 16

6

7

9

10

t

FIGURE 6.17 Periodic Multiline Function. f 3

0

3

4

12

t

FIGURE 6.18 Delayed Periodic Function.

6.28 Calculate the inverse Laplace transforms of the following functions analytically by means of the convolution theorem and verify the results by MATLAB: (i) Y(s) ¼ 2u$s/(s2 þ u2)2. (ii) Y(s) ¼ 1/(s3 þ a$s2). 6.29 Same question as in Problem 6.28 for the following functions: (i) Y(s) ¼ s2/(s2  u2)2. (ii) Y(s) ¼ s/[(s2 þ a2)$(s2 þ b2)]. 6.30 By using partial fraction expansion and the inverse Laplace transform, calculate the original functions y(t) for the following s-domain functions. Check the results with MATLAB for both partial fraction expansion and inverse Laplace calculations: (i) Y(s) ¼ 5/(s2 þ 3s þ 2). (ii) Y(s) ¼ (s þ 1)/(s3 þ 5s2 þ 4s). (iii) Y(s) ¼ s/(s3 þ 6s2 þ 11s þ 6).

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CHAPTER 6 The Laplace Transform

6.31 Same question as in Problem 6.30 for the following functions: (i) Y(s) ¼ (s þ 2)/(s3 þ 7s2 þ 15s þ 9). (ii) Y(s) ¼ s/(s4 þ 6s3 þ 13s2 þ 12s þ 4). (iii) Y(s) ¼ 5/(s3 þ 4s2). 6.32 Same question as in Problem 6.30 for the following functions: (i) Y(s) ¼ s3/(s4  1). (ii) Y(s) ¼ 1/(s3 þ 2s2 þ s þ 2). (iii) Y(s) ¼ (s þ 1)/[s$(s2 þ 2)2]. 6.33 Find y(t) for the function Y(s) ¼ (s þ 1)$e2s/[(s þ 2)$(s þ 3)2] by using the time-shift theorem and partial fraction expansion. Verify the result by MATLAB. 6.34 Same question as in Problem 6.33 for the function Y(s) ¼ 4$e5s/ [s2$(s2 þ 16)]. 6.35 Solve the following differential equations subject to the initial conditions that are specified next to the equations. Use direct and inverse Laplace transforms: _ ¼ 0. (i) y€  4y_ þ 2y ¼ 0; y(0) ¼ 1; yð0Þ _ ¼ 0. (ii) 2y€ þ 5y_ þ 3y ¼ sinð10tÞ; y(0) ¼ 0; yð0Þ _ ¼ 5. (iii) 3y€ þ y_ þ y ¼ t $ e4t ; y(0) ¼ 0; yð0Þ 6.36 Use Laplace and inverse Laplace transforms to solve the differential equation  d4 yðtÞ d3 yðtÞ d2 yðtÞ dyðtÞ dyðtÞ þ yðtÞ ¼ 0 for þ2$ þ2$ þ2$ ¼1 dt4 dt3 dt2 dt dt t¼0 all other initial conditions being zero. 6.37 Same question as in Problem 6.36 for

 d5 yðtÞ d4 yðtÞ d 3 yðtÞ d 2 yðtÞ dyðtÞ dyðtÞ3  þ 2 $ yðtÞ ¼ 0 and þ 2$ þ 2$ þ 4$ þ  dt4 dt3 dt2 dt dt3  dt5

¼2 t¼0

all other initial conditions being zero. 6.38 Solve the differential equation of Problem 6.36 for an input u(t) ¼ e0.1$t$sin (60$t) all initial conditions being zero. Use the Laplace transform approach.

Problems

321

6.39 Solve the differential equation of Problem 6.37 for an input u(t) ¼ e3t$t2$cos (100$t) all initial conditions being zero. Use the Laplace transform approach. 6.40 Solve the following systems of differential equations and initial conditions:

3y_1  y1  2y_2 þ y2 ¼ 0 ðiÞ ; y1 ð0Þ ¼ 1; y2 ð0Þ ¼ 0: y_1 þ 2y_2 þ y1 ¼ cosðtÞ ( y€1 þ 2y_1 þ y_2  3y2 ¼ 1 ; y1 ð0Þ ¼ 1; y2 ð0Þ ¼ 0; y_1 ð0Þ ¼ 0; y_2 ð0Þ ¼ 0: ðiiÞ 2y€1  y€2 þ 2y1 ¼ sinð20 $ tÞ 6.41 Same question as in Problem 6.40 for

y€1 þ 200y1  80y2 ¼ 0 ; y1 ð0Þ ¼ 2; y2 ð0Þ ¼ 0; y_1 ð0Þ ¼ 0; y_2 ð0Þ ¼ 0: y€2  120y1 þ 80y2 ¼ dðtÞ 6.42 Use the convolution theorem to solve the integral equation Z t yðsÞ $ eðtsÞ sin½10 $ ðt  sÞds ¼ e3t $ sinð100 $ tÞ 3 $ yðtÞ þ 2 $ 0

knowing that y(0) ¼ 0 and check the result by using MATLAB. 6.43 Same question as in Problem 6.42 for the equation Z t 2yðtÞ þ yðsÞ $ eðtsÞ cos½3ðt  sÞds ¼ et 0

and zero initial conditions. 6.44 Solve the following system of integraledifferential equations: 8 Z t Z t > > > y1 ðtÞdt  5 y2 ðtÞdt ¼ 120 < 2y_1 ðtÞ þ y1 ðtÞ þ 5 0 0 Z t Z t > > > y2 ðtÞdt  5 y1 ðtÞdt ¼ 0 : 4y_2 ðtÞ þ 5 0

0

using the Laplace transform approach and check the result by MATLAB. All initial conditions are zero. 6.45 Identify a dynamic system whose Laplace-domain transform coordinate is defined as Y(s) ¼ 10/[s$(s2 þ 2s þ 100)]. 6.46 A dynamic system has its time-domain coordinate transformed into the Laplace domain as Y(s) ¼ (s2 þ 3s þ 25)/[s2$(s2 þ 2s þ 0.001)]. Determine the system components, forcing function, and initial conditions.

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CHAPTER 6 The Laplace Transform

Suggested Reading T.F. Bogart, Jr., Laplace Transforms and Control Systems Theory for Technology, John Wiley & Sons, New York, 1982. I.H. Sneddon, The Use of Integral Transforms, McGraw-Hill, New York, 1972. W.R. LePage, Complex Variables and the Laplace Transform for Engineers, Dover Publications, 2010. C.J. Savant, Jr., Fundamentals of the Laplace Transformation, McGraw-Hill, New York, 1962. W.T. Thomson, Laplace Transformation, 2nd Ed. Prentice-Hall, New York, 1960. R.V. Churchill, Operational Mathematics, McGraw-Hill, New York, 1958. E.J. Watson, Laplace Transforms and Applications, Van Nostrand, New York, 1981. L. Debnath, Integral Transforms and Their Applications, 3rd Ed. CRC Press, Boca Raton, FL, 2014. D.W. Jordan and P. Smith, Mathematical Techniques, 4th Ed. Oxford University Press, Oxford, UK, 2008. S.L. Campbell, An Introduction to Differential Equations and their Applications, Wordsworth Publishing, Belmont, CA, 1990. W.E. Boyce and R.C. DiPrima, Elementary Differential Equations, 10th Ed. John Wiley & Sons, New York, 2012.

CHAPTER

Transfer Function Approach

7

OBJECTIVES Based on the mathematical models developed in Chapters 2e5 for mechanical, electrical, fluid, and thermal systems, and using the Laplace transform of Chapter 6, this chapter introduces you to the transfer function approach. The following topics are studied here: • The transfer function as a connector between input and output into the Laplace domain for single-input, single-output (SISO) and multiple-input, multiple-output (MIMO) dynamic systems. • Deriving the transfer function from other mathematical models of dynamic systems. • Introducing the concept of stability of dynamic systems through transfer functions. • Using the complex impedance as a tool for modeling mechanical, electrical, fluid, and thermal elements to derive the transfer function of a dynamic system directly in the Laplace domain. • The transfer function as a method of obtaining the time-domain forced response and the free response with nonzero initial conditions. • Solving transfer function applications with MATLAB and Simulink.

INTRODUCTION This chapter introduces the transfer function as a Laplace-domain operator, which characterizes the properties of a given dynamic system and connects the input to the output. The transfer function for single-input, single-output (SISO) systems and transfer function matrix for multiple-input, multiple-output (MIMO) systems can be used to determine the forced time response and the free time response with nonzero initial conditions. The complex impedance is introduced as a transfer function at the element level, which enables deriving the transfer function of dynamic systems directly in the Laplace domain. The transfer function can be utilized to study the stability of dynamic systems. MATLAB and Simulink examples illustrating the use of transfer function by means of built-in capabilities are incorporated alongside many solved examples that use the standard transfer function formulation

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00007-5 Copyright © 2018 Elsevier Inc. All rights reserved.

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and solution. The transfer function concept is also important in studying system dynamics by means of the state-space modeling approach (the topic of Chapter 8) and is the basis of frequency-domain analysis (the focus of Chapter 9). Block diagrams, as those employed in Chapters 11e13 to model control systems, are designed based on transfer functions.

7.1 THE TRANSFER FUNCTION CONCEPT AND DEFINITION The concept of transfer function is briefly presented here followed by the transfer function definition as it applies to SISO and MIMO systems. A major objective in system dynamics is finding the system time-domain response (or output) that corresponds to a specific forcing (or input). This normally means deriving and then solving the mathematical model of a dynamic system. The mathematical model consists of differential (and algebraic) equations, which are most often linear and with time-invariant coefficients. These equations are expressions that mix the input (known) function, the output (unknown) function, and its time derivatives with system parameters representing element properties. Consider a mechanical system formed of a mass m connected to a parallel combination between a damper of coefficient c and a spring of stiffness k. Under the action of an input force u(t), the mass displacement y(t)dthe outputdis expressed by the following differential equation: € þ c $ yðtÞ _ þ k $ yðtÞ ¼ uðtÞ m $ yðtÞ

(7.1)

As discussed in Chapter 6, linear ODE with constant coefficients is converted into algebraic equations through Laplace transformation. As such, Laplace transforming Eq. (7.1) with zero initial conditions results in:
 m $ s2 þ c $ s þ k $ YðsÞ ¼ UðsÞ or (7.2) 1 YðsÞ ¼ GðsÞ $ UðsÞ with GðsÞ ¼ 2 m$s þ c$s þ k Unlike the time-domain mathematical model of Eq. (7.1), the s-domain model of Eq. (7.2) clearly separates the input U(s), the output Y(s), and the system parameters that are gathered in the factor function G(s), which is called the system’s transfer function. Eq. (7.2) states that the output is calculated as the product of the transfer function and the inputdwhich is a fundamental trait of the transfer function approach and is valid for both SISO and MIMO systems. Some key characteristics of the transfer function are as follows: 1. The transfer function approach is applicable to dynamic systems described by linear differential equations with constant coefficients. 2. Although it is expressed as the ratio of the output to the input (in the s-domain), the transfer function solely describes the system. In other words, for a specified system, with unique characteristics, the transfer function is also unique.

7.1 The Transfer Function Concept and Definition

Transfer functions, known as impedances, can be defined at the element level for the different dynamic systems. Consider, for instance, the voltageecurrent relationship for an inductor of constant inductance L: v(t) ¼ L$di(t)/dt where v is the voltage, and i is the current Applying the Laplace transform to this equation with zero initial conditions yields:  VðsÞ ¼ L½vðtÞ (7.3) VðsÞ ¼ ðL $ sÞ $ IðsÞ with IðsÞ ¼ L½iðtÞ If I(s) is the input and V(s) is the output, it follows that the factor L$s plays the role of a transfer function (called the inductor’s impedance, as we shall see later in this chapter). Impedances can be defined for all basic elements in the mechanical, electrical, thermal, and fluid fields. These impedances, in conjunction with fundamental physical laws connecting them, can be used to derive mathematical models and system transfer functions directly into the s-domain by circumventing the time-domain modeling altogether. System transfer functions can also be obtained by converting other existing models, such as the state-space model (which is the focus of Chapter 8) or zero-pole-gain models, which is studied in this chapter. In the remainder of this section, the transfer function is defined more systematically for SISO and MIMO systems.

7.1.1 SISO Systems For a SISO system, the transfer function G(s) is the ratio of the Laplace transform of the output Y(s) to the Laplace transform of the input U(s) for zero initial conditions, which is mathematically expressed as  YðsÞ  or YðsÞ ¼ GðsÞ $ UðsÞ (7.4) GðsÞ ¼ UðsÞIC¼0 with IC standing for initial conditions. Eq. (7.4), which is graphically represented in Figure 7.1(a), separates the three factors characterizing a SISO system in the s-domain, namely: the input U(s), the output Y(s), and the system’s features, which are incorporated into the transfer function G(s). As also indicated in both Eq. (7.4) and Figure 7.1(a), the output is functionally proportional to the inputdthe proportionality function factor being the transfer function. U(s)

Y(s)

{U(s)}

{Y(s)}

G(s)

[G(s)]

(a)

(b)

FIGURE 7.1 InputeOutput Connections in the Laplace Domain for the Following: (a) SISO Systems by Scalar Transfer Function; (b) MIMO Systems by Transfer Function Matrix.

325

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CHAPTER 7 Transfer Function Approach

7.1.2 MIMO SYSTEMS In the case of MIMO systems, the Laplace-domain input and output are the vectors {U(s)} and {Y(s)}, respectively. For a SISO system, the transfer function is an s-function that relates the input U(s) to the output Y(s) in the proportional form of the definition Eq. (7.4). A similar connection between {U(s)} and {Y(s)} requires a transfer function matrix to accomplish the task. Consequently, for a MIMO system, the transfer function matrix [G(s)] connects the Laplace transform of the output vector {Y(s)} to the Laplace transform of the input vector {U(s)} with zero initial conditions: fYðsÞg ¼ ½GðsÞ $ fUðsÞg

(7.5)

and this relationship is illustrated in Figure 7.1(b).

7.2 TRANSFER FUNCTION MODEL FORMULATION The principal methods of deriving the transfer function of a dynamic system are detailed in this section. They are: Laplace transformation of the time-domain differential equations, conversion from other models, such as the zero-pole-gain model, and direct formulation in the Laplace domain by means of element impedances.

7.2.1 Transfer Function From the Time-Domain Mathematical Model This section analyzes the modalities of calculating the transfer function for SISO systems and the transfer function matrix for MIMO systems using time-domain mathematical models of dynamic systems.

SISO Systems The mathematical model of a SISO dynamic system is frequently the linear ODE with constant coefficients dn yðtÞ dn1 yðtÞ dyðtÞ þ a0 $ yðtÞ ¼ uðtÞ þ an1 $ þ / þ a1 $ (7.6) n dt dtn1 dt Applying the Laplace transform to this equation, considering zero initial conditions, and using the transfer function definition of Eq. (7.4) results in: an $

ðan $ sn þ an1 $ sn þ / þ a1 $ s þ a0 Þ $ YðsÞ ¼ UðsÞ or GðsÞ ¼

YðsÞ 1 ¼ n n1 UðsÞ an $ s þ an1 $ s þ / þ a1 $ s þ a0

(7.7)

Note that the order of the system (which is also the order of the differential equation describing the inputeoutput relationship) is identical to the degree of the s

7.2 Transfer Function Model Formulation

polynomial in the denominator of the transfer function G(s) of Eq. (7.7), which is named the characteristic polynomial. While obtaining the transfer function from a differential equation where the input and output are clearly defined is straightforward, as seen in Eqs. (7.6) and (7.7), there are also time-domain mathematical models consisting of several equations (some may be algebraic) with more than two dependent variables. If only the ratio of two specific s-dependent variables is sought, then successive substitutions should be performed to eliminate all variables other than the two variables of interest in one Laplace-domain algebraic equation. This will ensure that the target transfer function can be obtained. The following example illustrates this procedure. nn

Example 7.1 Derive the time-domain mathematical model of the electrical system of Figure 7.2 by relating the output voltage vo(t) to the input voltage vi(t). Use the model to calculate the transfer function Vo(s)/Vi (s).

i

vi

i1

C1 i

i – i 1 R1

i2 C2

i – i2 vo R2

FIGURE 7.2 Electrical System With Resistors and Capacitors.

Solution The following equations can be written for the electrical system of Figure 7.2 by applying Kirchhoff’s voltage law (KVL) (Chapter 4):

8 Z > 1 > > $ i1 ðtÞdt ¼ R1 $ ½iðtÞ  i1 ðtÞ > > > C1 > > > Z > < 1 $ i2 ðtÞdt ¼ R2 $ ½iðtÞ  i2 ðtÞ C2 > > > > > > vi ðtÞ ¼ R1 $ ½iðtÞ  i1 ðtÞ þ R2 $ ½iðtÞ  i2 ðtÞ > > > > : vo ðtÞ ¼ R2 $ ½iðtÞ  i2 ðtÞ

(7.8)

Laplace transforming the first two Eq. (7.8) with zero initial conditions yields

8 > > > < I1 ðsÞ ¼ > > > : I2 ðsÞ ¼

R1 $ C1 $ s $ IðsÞ R1 $ C1 $ s þ 1 R2 $ C2 $ s $ IðsÞ R2 $ C2 $ s þ 1

(7.9)

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CHAPTER 7 Transfer Function Approach

   R i1 ðt Þdt =s ¼ I1 ðsÞ þ lim dq1 ðt Þ =s ¼ I1 ðsÞþ t/0 t/0 R q1 ð0Þ=s ¼ I1 ðsÞ because q1(0) = 0. The same reasoning applies to L i2 ðt Þdt . Substitution of I1(s) and I2(s) of Eq. (7.9) into the equations produced by Laplace transforming the last two Eq. (7.8) results, after some algebra, in the following s-domain voltages: Eq. (7.9) used:

R L½ i1 ðt Þdt  ¼ I1 ðsÞ þ



lim

R

8 R1 $ R2 $ ðC1 þ C2 Þ $ s þ R1 þ R2 > > > $ IðsÞ < Vi ðsÞ ¼ ðR1 C1 s þ 1Þ $ ðR2 C2 s þ 1Þ > R2 > > : Vo ðsÞ ¼ $ IðsÞ R2 $ C2 $ s þ 1

(7.10)

The required transfer function is obtained from Eq. (7.10) as

GðsÞ ¼

Vo ðsÞ R1 $ R2 $ C1 $ s þ R2 ¼ Vi ðsÞ R1 $ R2 $ ðC1 þ C2 Þ $ s þ R1 þ R2

(7.11) nn

MIMO SYSTEMS For a large majority of physical systems with MIMO systems the time-domain mathematical model is the following general system of linear ODE with constant coefficients  n  n1 



d yðtÞ d yðtÞ dyðtÞ  $  $ ½an  $ þ ½a0  $ fyðtÞg þ ½a þ / þ ½a n1 1 dtn dtn1 dt ¼ ½bfu ðtÞg ¼ fuðtÞg (7.12) where {u*(t)} is the actual input vector, {y(t)} is the output vector, and [a0], [a1], ., [an1], [an], [b] are matrices with constant coefficients. Consider the general case with p input components: u1(t), u2(t), ., up(t) that are collected in the {u*(t)}vector, m output components: y1(t), y2(t), ., ym(t) making up the vector {y(t)}, and an equal number m of differential equations. The vector {u(t)} ¼ [b]${u*(t)} is the modified input vector that has the same dimension as the output vector, which is m  1. Applying the Laplace transform to Eq. (7.12) with zero initial conditions yields the following:
 ½an  $ sn þ ½an1  $ sn1 þ / þ ½a1  $ s þ ½a0  $ fYðsÞg ¼ ½b $ fU  ðsÞg ¼ fUðsÞg (7.13) If the matrix sum is nonsingular, left multiplication by its inverse in Eq. (7.13) results in:
1 fYðsÞg ¼ ½an  $ sn þ ½an1  $ sn1 þ / þ ½a1  $ s þ ½a0  $ fUðsÞg or (7.14) 1
n n1 þ / þ ½a1  $ s þ ½a0  ½GðsÞ ¼ ½an  $ s þ ½an1  $ s The transfer function matrix [G(s)] of Eq. (7.14), which is a square matrix of dimension m, is identified from Eq. (7.5) that defined [G(s)].

7.2 Transfer Function Model Formulation

nn

Example 7.2 The microelectromechanical systems (MEMS) actuator system of Figure 7.3(a) consists of two shuttle masses m1 and m2 that can move along the same axis and are coupled by means of a serpentine spring of stiffness k2. The mass m1 is also supported by two identical beam springs of total stiffness k1 and is subjected to frontal damping (damping coefficient c) at its left by the air squeezed between mass and a fixed wall. Electrostatic transverse actuation is applied to both masses, which results in the forces u1 and u2. Express the transfer function matrix using the lumped-parameter model of Figure 7.3(b) where the output vector is formed of the two masses’ displacements y1 and y2, and the input vector is defined by the two electrostatic forces u1 and u2. v1

Actuation

–v2

Spring

Motion axis

Actuation m1

m2

Anchor

v1

y2 k2

m1 c

Shuttle mass

Fixed wall

y1 k1

m2 u2

u1

Spring

(a)

(b)

FIGURE 7.3 Translatory-Motion MEMS Device: (a) Physical Model; (b) Lumped-Parameter Model With Two-Force Input and Two-Displacement Output.

Solution

As indicated in Figure 7.3(b), the system has two inputs (the forces u1 and u2) and two outputs (the mass displacements y1 and y2); therefore, it is a MIMO system with m ¼ p ¼ 2. Newton’s second law of motion yields the mathematical model of this mechanical system:

(

" # 

m1 0 m1 $ y€1 ¼ u1  c $ y_1  k1 $ y1  k2 $ ðy1  y2 Þ y€1 or $ m2 $ y€2 ¼ u2  k2 $ ðy2  y1 Þ y€2 0 m2 " #  " # ( ) ( ) c 0 k1 þ k2 k2 y1 u1 y_1 þ $ þ $ ¼ y_2 0 0 k2 k2 y2 u2

(7.15) Application of the Laplace transform to the two Eq. (7.15) with zero initial conditions yields

(


 m1 $ s2 þ c $ s þ k1 þ k2 $ Y1 ðsÞ  k2 $ Y2 ðsÞ ¼ U1 ðsÞ 
k2 $ Y1 ðsÞ þ m2 $ s2 þ k2 $ Y2 ðsÞ ¼ U2 ðsÞ

(7.16)

Eq. (7.16) is expressed in vector-matrix form:

"

m1 $ s2 þ c $ s þ k1 þ k2 k2

# 



Y1 ðsÞ U1 ðsÞ $ ¼ Y2 ðsÞ U2 ðsÞ m2 $ s2 þ k2 k2

(7.17)

329

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CHAPTER 7 Transfer Function Approach

which can be reformulated as



Y1 ðsÞ

"



Y2 ðsÞ

¼

#1 

U1 ðsÞ $ U2 ðsÞ m2 $ s2 þ k2

m1 $ s2 þ c $ s þ k1 þ k2

k2

k2

(7.18)

If {Y(s)} ¼ {Y1(s), Y2(s)}T is the output vector and {U(s)} ¼ {U1(s), U2(s)}T is the input vector, Eq. (7.18) shows, by comparison to the definition Eq. (7.14), that the transfer function matrix of this example is the matrix connecting {Y(s)} to {U(s)} in Eq. (7.18). Applying the Laplace transform (with zero initial conditions) to the matrix mathematical model of Eq. (7.15) yields:

"

m1

0

0

m2

(

#

þ

$s $ 2

"

Y1 ðsÞ

)

Y2 ðsÞ

k1 þ k2

k2

k2

k2

"

þ # ( $

c

0

0

0

#

Y1 ðsÞ

(

$s$ )

Y2 ðsÞ

( ¼

Y1 ðsÞ Y2 ðsÞ U1 ðsÞ

) )

(7.19)

U2 ðsÞ

Eq. (7.19) reduces to Eq. (7.17) after right-factoring the vector {Y1(s) Y2(s)}T and adding up the corresponding matrix components in the left-hand side of Eq. (7.19). The transfer matrix [G(s)] is obtained by inverting the matrix of Eq. (7.18):

2

m2 $ s2 þ k2 6 DðsÞ 6 ½GðsÞ ¼ 6 6 4 k2 DðsÞ

k2 DðsÞ

3

7  7 G11 ðsÞ 7¼ 7 2 G12 ðsÞ m1 $ s þ c $ s þ k1 þ k2 5 DðsÞ

G12 ðsÞ G22 ðsÞ

 (7.20)

We can use MATLAB symbolic calculation and the inv command to get [G(s)] of Eq. (7.20). The characteristic polynomial D(s) is

DðsÞ ¼ m1 $ m2 $ s4 þ m2 $ c $ s3 þ ½m1 $ k2 þ m2 $ ðk1 þ k2 Þ $ s2 þ c $ k2 $ s þ k1 $ k2

(7.21)

G11(s), G12(s), and G22(s) in Eq. (7.20) are regular (scalar) transfer functions that connect specific input and output components. Thus, combining Eqs. (7.18) and (7.20) yields the following:



Y1 ðsÞ ¼ G11 ðsÞ $ U1 ðsÞ þ G12 ðsÞ $ U2 ðsÞ Y2 ðsÞ ¼ G12 ðsÞ $ U1 ðsÞ þ G22 ðsÞ $ U2 ðsÞ

(7.22)

Eq. (7.22) reflects the principle of linear superposition as each output component is expressed as a linear (functional) combination of the two input components U1(s) and U2(s) by means of individual transfer function components of [G(s)]. nn

nn

Example 7.3

The liquid system of Figure 7.4 is formed of two tanks of capacitances C1 and C2 and two valves of resistances R1 and R2. Using a time-domain mathematical model, where the flow rate qi is the input, and the two tank pressures p1 and p2 are the system’s outputs, determine the transfer function matrix relating the input to the output. Assume the four liquid element properties are defined in terms of pressure.

7.2 Transfer Function Model Formulation

qi

R1

p1

C1

p2

p2

C2

q

R2

p3 = 0

qo

FIGURE 7.4 Liquid System With Two Tanks and Two Valves.

Solution The liquid resistances and capacitances are expressed as

R1 ¼

p1  p2 p2  p3 p2 qi  q q  qo ; R2 ¼ ¼ ; C1 ¼ ; C2 ¼ p_1 q qo qo p_2

(7.23)

The flow rates q and qo are substituted from the first and second Eq. (7.23) into the third and fourth Eq. (7.23), which results in the following time-domain mathematical model:

8 1 1 > > C $ p_ þ $ p1  $ p2 ¼ qi > < 1 1 R R 1

"

1

 > 1 1 1 > > : C2 $ p_2  $ p1 þ þ $ p2 ¼ 0 R1 R1 R2  þ

1=R1 1=R1

C1

0

0

C2

or

# ( $



  

qi 1 p1 ¼ $ qi ¼ $ 0 1=ðR1 þ R2 Þ p2 0 1=R1

p_1

)

p_2

(7.24)

The Laplace transform with zero initial conditions is applied to the first-order differential Eq. (7.24), which yields the algebraic equations system:

8 1 1 > > > C1 $ s þ $ P1 ðsÞ  $ P2 ðsÞ ¼ Qi ðsÞ > < R1 R1  > > 1 1 1 > >  $ P1 ðsÞ þ C2 $ s þ þ $ P2 ðsÞ ¼ 0 : R1 R1 R2 2

1 6 C1 $ s þ R 1 6 or 6 4 1  R1



1 R1

3

7  P ðsÞ  Q ðsÞ

i 1 7 ¼ 7$ P2 ðsÞ 0 1 1 5 C2 $ s þ þ R1 R2

(7.25)

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CHAPTER 7 Transfer Function Approach

Considering that the input vector is {U(s)} ¼ {Qi (s) 0}T and the output vector is {Y(s)} ¼ {P1(s) P2(s)}T in Eq. (7.25), the corresponding transfer function matrix is found as

31 1 1 C $ s þ  1 7 6 R1 R1 7 6 7 6 ½GðsÞ ¼ 6 7 4 1 1 1 5  C2 $ s þ þ R1 R 1 R2 3 2 R1 $ R2 $ C2 $ s þ R1 þ R2 R2 7 2 6 DðsÞ DðsÞ G11 ðsÞ 7 6 7 4 6 ¼6 7¼ 7 6 G12 ðsÞ R2 R1 $ R2 $ C1 $ s þ R2 5 4 DðsÞ DðsÞ 2

G12 ðsÞ G22 ðsÞ

DðsÞ ¼ R1 $ R2 $ C1 $ C2 $ s2 þ ½R1 $ C1 þ R2 $ ðC1 þ C2 Þ $ s þ 1 (7.26) The two outputs can therefore be expressed in terms of the individual transfer functions of Eq. (7.26) and the input as:

P1 ðsÞ ¼ G11 ðsÞ $ Qi ðsÞ; P2 ðsÞ ¼ G12 ðsÞ $ Qi ðsÞ

(7.27)

Note that [G(s)] of Eq. (7.26) can also be obtained by applying the Laplace transform to the matrix Eq. (7.24) directly. After right factoring the output vector {P1(s) P2(s)}T on the left-hand side of the resulting equation, the matrix multiplying the output vector is the one of Eq. (7.25). nn

7.2.2 Transfer Function From the Zero-Pole-Gain Mathematical Model The transfer function can be obtained by conversion from other models of linear systems with constant element properties. Chapter 8, for instance, which studies statespace models, discusses conversion between this model and the transfer function model. The transfer function G(s) of a SISO system (or any component of the transfer function matrix [G(s)] defining a MIMO system) is expressed as the ratio of two polynomials, N(s) and D(s). If these two polynomials are expressed in factor form, G(s) can be written in the factor form: GðsÞ ¼

NðsÞ k $ ðs  z1 Þ $ ðs  z2 Þ $ / $ ðs  zm Þ ¼ DðsÞ ðs  p1 Þ $ ðs  p2 Þ $ / $ ðs  pn Þ

(7.28)

This factor form of G(s) is known as a zero-pole-gain (or zpk in MATLAB) model. In Eq. (7.28) the roots of the numerator N(s), which are z1, z2, ., zm, are called zeros, the roots of the denominator D(s), which are p1, p2, ., pn, are the poles, whereas the factor k is the gain. It is evident that conversion is always possible between the two forms of the transfer function given in Eq. (7.28): if the zeros, poles, and gain are

3 5

7.2 Transfer Function Model Formulation

known, then multiplications in the factored numerator and denominator will generate the polynomials N(s) and D(s)dthis is equivalent to converting a zeropole-gain model into a (typical) transfer function model. The other route of factoring N(s) and D(s) leads to transforming a typical transfer function model into the corresponding zero-pole-gain model. MATLAB utilizes four different linear time invariant (LTI) models (or objects) to define linear systems with constant (time invariant) element properties. One such model is the zero-pole-gain model. Whenever the zeroes, poles, and gain are known, the MATLAB command zpk(z, p, k) generates a zero-pole-gain LTI object that can be stored for subsequent operations. In this command, z and p are lists with the zeros and the poles, whereas k is a factor. Another LTI model is the transfer function model. MATLAB has two different ways of defining a transfer function: the command tf (num, den), where the coefficients of the numerator and denominator polynomials have to be defined in descending order of powers, can be used; the alternative procedure is by means of the command s = tf(‘s’), which allows to input the transfer function with regular MATLAB syntax. Both methods are illustrated in the following example. The other two LTI models are the state-space (MATLAB-dedicated command is ss) model, which is studied in Chapter 8, and the frequency response data (MATLAB-dedicated command is frd) model analyzed in Chapter 9. MATLAB enables conversion between different LTI models. nn

Example 7.4

An s-domain function is defined by the zeroes 1 and 2; the poles 1, 2, and 3; and a gain of 5. Build the MATLAB zero-pole-gain object corresponding to these data and then convert this object into a transfer function model. Redefine the obtained transfer function model by using the s = tf(‘s’) method and reconvert the resulting transfer function into a zero-pole-gain model to confirm that the original zero-pole-gain model is retrieved.

Solution The following MATLAB command sequence is used to generate the zero-pole-gain model from the given data, convert that model into a transfer function model, redefine the transfer function by means of the s = tf(‘s’) option, and reconvert this latter object into the original zero-pole-gain model: >> f = zpk([1,2],[-1,-2,-3],5) Zero/pole/gain: 5 (s-1) (s-2) ----------------(s+1) (s+2) (s+3) >> g = tf(f) Transfer function: 5 s^2 - 15 s + 10 ---------------------s^3 + 6 s^2 + 11 s + 6 >> s=tf(‘s’);

333

334

CHAPTER 7 Transfer Function Approach

>> g1=(5*s^2-15*s+6)/(s^3+6*s^2+11*s+6); >> zpk(g1) Zero/pole/gain: 5 (s-2) (s-1) ----------------(s+3) (s+2) (s+1)

nn

7.2.3 Impedance Transfer Function The complex impedance (or simply impedance) is a tool derived by means of the transfer function approach, which models the inputeoutput relationship at the element level and enables deriving system transfer functions directly in the Laplace domain. In electrical systems, for instance, the transfer function between current and voltage defines the impedance pertaining to a resistor, an inductor, or a capacitor. The qualifier complex indicates that the impedance is a function of the variable s, which is a complex number, as learned in Chapter 6. In(s)

Out(s) Z(s)

FIGURE 7.5 Complex Impedance Z(s) as a Transfer Function for a Generic Dynamic System Element.

The complex impedance, denoted by Z(s) in Figure 7.5, relates an element input In(s) to the corresponding output Out(s) through a structural relationship, depending on the type of system (mechanical, electrical, fluid, or thermal) and the type of element. Its defining equation is  OutðsÞ ZðsÞ ¼ (7.29) InðsÞ IC¼0 As Eq. (7.29) and Figure 7.5 do suggest, the impedance is a transfer function formulated at the individual element level. Table 7.1 synthesizes the definition and formulation of the element impedances for mechanical, electrical, liquid, gas, and thermal systems with some subsequent derivation presented separately for each system in this chapter shortly; full details of the derivation are provided in the companion website Chapter 7. Several examples are also studied illustrating the use of complex impedances in deriving transfer functions for SISO and MIMO dynamic systems.

Electrical Systems Element impedances can easily be formulated considering that the output is the voltage and the input is the current, both Laplace-transformed, which means  VðsÞ ZðsÞ ¼ or VðsÞ ¼ ZðsÞ $ IðsÞ (7.30) IðsÞ IC¼0

Table 7.1 Impedances of Mechanical, Electrical, Fluid, and Thermal Systems (pdPressure, hdHead) System

Element Property Mass Translation

Mechanical Rotation

h

Force F(s)

v(t) ¼ (1/k)$df(t)/dt

k/s

Mass moment of inertia

m(t) ¼ J$du(t)/dt

J$s

Damping

m(t) ¼ c$u(t) v(t) ¼ L$di(t)/dt

Resistance

v(t) ¼ R$i(t) p(t) ¼ I$dqv(t)/dt

Resistance

p(t) ¼ R$qv(t)

Capacitance

qv(t) ¼ C$dp(t)/dt

Inertance

h(t) ¼ I$dqv(t)/dt

Resistance

h(t) ¼ R$qv(t)

Capacitance

qv(t) ¼ C$dh/dt

Resistance Capacitance

Moment M(s)

Voltage V(s)

R 1/(C$s) I$s

Pressure P(s)

R 1/(C$s)

Volume flow rate Qv(s)

I$s Head H(s)

R 1/(C$s) I$s

Mass flow rate Qm(s)

Pressure P(s)

qm(t) ¼ C$dp(t)/dt

R 1/(C$s)

q(t) ¼ R$q(t) q(t) ¼ C$dq(t)

c L$s

Current I(s)

p(t) ¼ I$dqm(t)/dt p(t) ¼ R$qm(t)

c

k/s

i(t) ¼ C$dv(t)/dt

Inertance

Resistance

Velocity U(s)

u(t) ¼ (1/k)$dm(t)/dt

Inductance

Capacitance Thermal

m$s Velocity V(s)

f(t) ¼ c$v(t)

Inertance Gas

f(t) ¼ m$dv(t)/dt

Z(s)

Thermal flow rate Q(s)

Temperature Q(s)

R 1/(C$s)

7.2 Transfer Function Model Formulation

Liquid

Out(s)

Stiffness

Capacitance p

In(s)

Damping

Stiffness Electrical

TimeeDomain Relationship

335

336

CHAPTER 7 Transfer Function Approach

I 1(s)

I(s) Z1

I(s)

Z1

I(s) Z

Z2

Z2 I 2(s)

V(s)

V(s)

V(s) (a)

(b)

(c)

FIGURE 7.6 Connection of Electrical Impedances: (a) Series; (b) Parallel; (c) Equivalent Impedance.

The Laplace-domain equation V(s) ¼ Z(s)$I(s) is similar to the time-domain Ohm equation: v(t) ¼ R$i(t). This means that impedance elements defined by Z(s) in the Laplace domain behave similarly to resistances in the time domain. Because of that, electrical impedances are connected in series, parallel, and in mixed formations exactly like resistances. According to Figure 7.6, the series and parallel equivalent impedances of two electrical impedances Z1 and Z2 are Zs ðsÞ ¼ Z1 ðsÞ þ Z2 ðsÞ;

1 1 1 ¼ þ Zp ðsÞ Z1 ðsÞ Z2 ðsÞ

(7.31)

where the subscripts s and p stand for series and parallel, respectively. The companion website Chapter 7 demonstrates Eq. (7.31). Another important consequence of using the complex impedance is that relationships based on Ohm’s law can be used to apply other laws, such as Kirchhoff’s first and second laws, by considering that all electrical components behave as resistors through their corresponding complex impedances. Applying Laplace transforms with zero initial conditions to the voltageecurrent relationships for a resistor, inductor, and capacitor, the following impedances are obtained: vðtÞ ¼ R $ iðtÞ;

V ðsÞ ¼ R $ I ðsÞ;

V ðsÞ ¼ ZR ¼ R I ðsÞ

vðtÞ ¼ L $

diðtÞ ; dt

V ðsÞ ¼ ðL $ sÞ $ I ðsÞ;

V ðsÞ ¼ ZL ¼ L $ s I ðsÞ

iðtÞ ¼ C $

dvðtÞ ; dt

I ðsÞ ¼ ðC $ sÞ $ V ðsÞ;

V ðsÞ 1 ¼ ZC ¼ I ðsÞ C$s

(7.32)

nn

Example 7.5

Using complex electrical impedances, determine the transfer function Vo(s)/Vi(s) of the following: (i) voltage divider system (also known as ladder network) represented in Figure 7.7(a) and, based on it, evaluate the transfer function of the electrical system shown in Figure 7.2 of Example 7.1. (ii) operational amplifier (op-amp) system sketched in Figure 7.7(b) and, based on it, calculate the transfer function of the electrical system of Example 4.18 and depicted in Figure 4.32 of Chapter 4.

7.2 Transfer Function Model Formulation

I(s) Z2(s)

I(s)

I(s) Z1(s)

– +

Z1(s)

Vi (s)

Z2(s)

Vo(s)

I(s)

Vi (s)

(a)

Vo(s)

(b)

FIGURE 7.7 Basic Electrical Systems With Two Generic Complex Impedances in: (a) Voltage Divider (Ladder Network) Configuration; (b) Operational Amplifier Configuration.

Solution

(i) As mentioned in this section, impedances in the s-domain behave as resistances in the time domain; therefore the s-domain KVL can be expressed similarly to the known time-domain KVL; one has to consider a regular (“time-domain”) circuit with resistances R1 and R2 instead of the actual impedances Z1 and Z2, with i(t) instead of I(s), vi(t) instead of Vi(s), and vo(t) for Vo(s). For such a circuit, the actual KVL yields:



vi ðtÞ ¼ R1 $ iðtÞ þ R2 $ iðtÞ vo ðtÞ ¼ R2 $ iðtÞ

(7.33)

The s-domain KVL second law applied to the two loops of the voltage divider shown in Figure 7.7(a) results in:



Vi ðsÞ ¼ Z1 ðsÞ $ IðsÞ þ Z2 ðsÞ $ IðsÞ Vo ðsÞ ¼ Z2 ðsÞ $ IðsÞ

(7.34)

The transfer function is therefore

GðsÞ ¼

Vo ðsÞ Z2 ðsÞ ¼ Vi ðsÞ Z1 ðsÞ þ Z2 ðsÞ

(7.35)

The electrical system of Figure 7.2 is of the form sketched in the generic Figure 7.7(a), where the two impedances are

1 R1 C1 $ s ; ¼ 1 R1 $ C1 $ s þ 1 R1 þ C1 $ s 1 R2 $ R2 C2 $ s ¼ ¼ 1 R2 $ C2 $ s þ 1 R2 þ C2 $ s

ZR1 $ ZC1 Z1 ðsÞ ¼ ¼ ZR1 þ ZC1

Z2 ðsÞ ¼

ZR2 $ ZC2 ZR2 þ ZC2

R1 $

(7.36)

each being calculated as a parallel-connection impedance. By substituting Eq. (7.36) into the transfer function of Eq. (7.35) yields the transfer function of Eq. (7.11). (ii) The negative feedback in the system of Figure 7.7(b) ensures that the positive and negative input ports have the same zero voltage because the positive input port is connected to the ground.

337

338

CHAPTER 7 Transfer Function Approach

At the same time, due to the high input impedance of the op-amp, the current I(s) that passes through Z1(s) is identical to the one passing through Z2(s):

IðsÞ ¼

Vi ðsÞ  0 0  Vo ðsÞ ¼ Z1 ðsÞ Z2 ðsÞ

(7.37)

The circuit transfer function results from Eq. (7.37) as:

GðsÞ ¼

Vo ðsÞ Z2 ðsÞ ¼ Vi ðsÞ Z1 ðsÞ

(7.38)

The minus sign in Eq. (7.38) indicates the inverting effect in the s-domain, which is similar to the time-domain inverting effect demonstrated in Chapter 4 for the system with two resistors R1 and R2. The particular op-amp system of Figure 4.32 that is studied in the solved Example 4.18 is formed of a resistor R and a capacitor C in the locations of Z1(s) and Z2(s) of Figure 7.7(b), respectively. As a consequence, Z1 ¼ R and Z2(s) ¼ 1/(C$s), substituting these particular impedance expressions in Eq. (7.38), results in the following:

GðsÞ ¼

Vo ðsÞ Z2 ðsÞ 1 ¼ ¼ Vi ðsÞ Z1 ðsÞ R$C$s

(7.39)

The time-domain mathematical model of the electrical system pictured in Figure 4.32 is given in Eq. (4.109) and is formulated below together with its Laplace transform with zero initial conditions (initial values of voltage and flux: vi (0) ¼ 0, li (0) ¼ 0):

Z 1 1 $ vi ðtÞdt / Vo ðsÞ ¼  $ Vi ðsÞ or R$C R$C$s Vo ðsÞ 1 GðsÞ ¼ ¼ Vi ðsÞ R$C$s

vo ðtÞ ¼ 

Note

that

R L½ vi ðt Þdt  ¼ Vi ðsÞ þ



lim

t /0

R

(7.40)

   R vi ðt Þdt =s ¼ Vi ðsÞ þ lim dli ðt Þ =s ¼ Vi ðsÞþ t /0

li ð0Þ=s ¼ Vi ðsÞ taking into account that li(0) ¼ 0.

nn

Nonloading and Loading Cascading (Series) Electrical Systems Two electrical systems are connected in cascade (or in series) when the output terminal(s) of one system are connected directly to the input terminal(s) of the second system. Particularly, the two electric systems of Figure 7.7, which are examples of a more general category of two-port systems, have voltages as their input and output dependent variables. Connecting the output voltage terminals from any of the two circuits shown in Figure 7.7 to the input voltage terminals of a similar circuit will result in a cascading system. When the output voltage of the first stage is identical to the input voltage of the second stage, the system is a nonloading cascading system, whereas when the output voltage from the first stage does not convert 100% into the input voltage to the second stage due to interaction between the two stages (also known as interstage loading), the system is named a loading cascading system. Figure 7.8 gives two schematic representations of loading and of nonloading cascading systems. Note that the schematic representation of a loading cascading system shown in Figure 7.8(a), where V1(s) s V2(s), could also symbolize a nonloading system when V1(s) ¼ V2(s), as indicated in the transfer function block diagram of

7.2 Transfer Function Model Formulation

System 1

System 2 Vo(s)

V1(s) V2(s)

Vi (s)

Vi (s)

G1(s)

V1(s) = V2(s)

(a)

G2(s)

Vo(s)

(b)

FIGURE 7.8 Cascading Systems: (a) Two-Port Schematic Representation of a Loading System; (b) Transfer Function Representation of Nonloading System.

Figure 7.8(b). For the nonloading system of Figure 7.8(b), the overall transfer function is: Vo ðsÞ V1 ðsÞ Vo ðsÞ V1 ðsÞ Vo ðsÞ ¼ $ ¼ $ ¼ G1 ðsÞ $ G2 ðsÞ Vi ðsÞ Vi ðsÞ V1 ðsÞ Vi ðsÞ V2 ðsÞ

GðsÞ ¼

(7.41)

which shows that the transfer function of a nonloading cascading system is equal to the transfer function product of component subsystems. On the other hand, for a loading cascading system, the condition of Eq. (7.41) is not satisfied. nn

Example 7.6 Analyze the loading/nonloading condition of the two-stage, impedance-based electrical systems illustrated in Figure 7.9(a) and 7.9(b).

Solution Due to infinite input impedance of the two op-amps of the electrical system shown in Figure 7.9(a), there are no currents passing through the op-amps, and therefore the same current I(s) passes through all four components. Taking into account the negative feedback and the ground connection of the positive input terminals for both op-amps, the current I(s) is expressed as follows:

IðsÞ ¼

Vi ðsÞ  0 0  VðsÞ VðsÞ  0 0  Vo ðsÞ ¼ ¼ ¼ Z1 ðsÞ Z2 ðsÞ Z3 ðsÞ Z4 ðsÞ

Stage 1

Stage 2

I(s) I(s)

Z2(s)

Z1(s)

– +

Vi (s)

(7.42)

I(s) Z3(s)

(a)

Z4(s)

– +

Stage 2

Stage 1

I(s)

I 1(s)

I 3(s)

Z1(s)

Z3(s)

Vi (s) Vo(s)

Z2(s) I 2(s)

Z4(s)

Vo(s) I 3(s)

(b)

FIGURE 7.9 Two-Stage Electrical Systems With Generic Complex Impedances in the Following: (a) Operational Amplifier Configuration; (b) Voltage Divider Configuration.

339

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CHAPTER 7 Transfer Function Approach

Substituting the voltage V(s) from the first proportion of Eq. (7.42) into the last proportion generates the ratio of the output voltage to the input voltage:

GðsÞ ¼

Vo ðsÞ Z2 ðsÞ $ Z4 ðsÞ ¼ Vi ðsÞ Z1 ðsÞ $ Z3 ðsÞ

(7.43)

To check whether the system of Figure 7.9(a) is nonloading, the stage transfer functions are multiplied to get the system’s would-be transfer function:



G1 ðsÞ $ G2 ðsÞ ¼

   Z2 ðsÞ Z4 ðsÞ  $  ¼ GðsÞ Z1 ðsÞ Z3 ðsÞ

(7.44)

Since G(s) is indeed equal to the product G1(s)$G2(s)das Eqs. (7.43) and (7.44) yielded the same resultdit follows that the cascading system of Figure 7.9(a) is a nonloading system. Consider now the system of Figure 7.9(b). If we assume the two stages as being separated, the individual transfer functions of the two subsystems are formulated based on Eq. (7.35) as:

G1 ðsÞ ¼

Z2 Z4 Z2 $ Z4 ; G2 ðsÞ ¼ ; G1 ðsÞ $ G2 ðsÞ ¼ Z1 þ Z2 Z3 þ Z4 ðZ1 þ Z2 Þ $ ðZ3 þ Z4 Þ (7.45)

An analysis of the original electrical system can be performed based on KVL applied to the three meshes depicted in Figure 7.9(b). The three equations are as follows:

8 > < Vi ðsÞ ¼ Z1 $ I1 ðsÞ þ Z2 $ ½I1 ðsÞ  I3 ðsÞ 0 ¼ Z3 $ I3 ðsÞ þ Z4 $ I3 ðsÞ  Z2 $ ½I1 ðsÞ  I3 ðsÞ > : Vo ðsÞ ¼ Z4 $ I3 ðsÞ

(7.46)

which also used Kirchhoff’s current law with I2(s) ¼ I1(s)  I3(s). Substituting I1(s) and I3(s) from two of Eq. (7.46) into the third equation, the following transfer function is obtained:

GðsÞ ¼

Vo ðsÞ Z2 $ Z4 ¼ Vi ðsÞ ðZ1 þ Z2 Þ $ ðZ3 þ Z4 Þ þ Z1 $ Z2

(7.47)

Comparing Eqs. (7.45) and (7.47) it can be seen that the actual transfer function of the cascading system, Eq. (7.47), is not obtained in Eq. (7.45), which calculated the product of the two component transfer functions. Because G(s) s G1(s)$G2(s), the cascading network of Figure 7.9(b) is a loading system. nn

Thermal Systems The generic thermal impedance is defined as:  QðsÞ ZðsÞ ¼ or QðsÞ ¼ ZðsÞ $ QðsÞ QðsÞ IC¼0

(7.48)

where, as before, IC ¼ 0 means zero initial conditions. Q(s) and Q(s) are the Laplace transforms of the temperature q(t) and heat flow rate q(t), respectively. Applying Laplace transforms with zero initial conditions to the temperatureeflow rate

7.2 Transfer Function Model Formulation

relationships for a resistor and capacitor (as also indicated in Table 7.1), the following impedances are obtained: qðtÞ ¼ R $ qðtÞ;

QðsÞ ¼ R $ QðsÞ;

dqðtÞ ; qð t Þ ¼ C $ dt

Q ðsÞ ¼ ZR ¼ R Q ðsÞ

Q ðsÞ 1 QðsÞ ¼ ðC $ sÞ $ QðsÞ; ¼ ZC ¼ Q ðsÞ C$s

(7.49)

nn

Example 7.7 Find the transfer function matrix of the one-room thermal system of Figure 7.10 using the complex impedance approach. Consider that the input components are the thermal flow rate qi of the heater and the outside temperature q2, whereas the output is the inside temperature q1. The enclosed space has a thermal capacity C, and the four identical walls are defined by a total thermal resistance R.

qo

θ2

θ1 qo

qo

qi Heater R

C

qo

FIGURE 7.10 Four-Wall, One-Room Thermal System With Internal Heating.

Solution

The thermal impedances related to the total wall resistance R and the room capacity C are:

ZR ¼ R ¼

Q1 ðsÞ  Q2 ðsÞ 1 Q1 ðsÞ ¼ ; ZC ¼ Qo ðsÞ C $ s Qi ðsÞ  Qo ðsÞ

(7.50)

By expressing Qo(s) of the first Eq. (7.50) and substituting it in the second Eq. (7.50), the following equation is obtained:

ðR $ Cs þ 1Þ $ Q1 ðsÞ ¼ Q2 ðsÞ þ R $ Qi ðsÞ which can be reformulated in vector form as



Q1 ðsÞ ¼

1 R$C$s þ 1

R R$C$s þ 1



Q2 ðsÞ $ Qi ðsÞ

(7.51)

(7.52)

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CHAPTER 7 Transfer Function Approach

The transfer function connecting the output Q1(s) to the input vector {Q2(s) Qi (s)}T is the row vector of Eq. (7.52) as:

 ½GðsÞ ¼

1 R R$C$s þ 1 R$C$s þ 1

(7.53) nn

Fluid Systems The fluid impedance is expressed for zero initial conditions (IC ¼ 0) as follows:  PðsÞ  ZðsÞ ¼ or PðsÞ ¼ ZðsÞ $ QðsÞ (7.54) QðsÞIC¼0 where P(s) is Laplace transform of pressure p(t), and Q(s) is the Laplace transform of the fluid flow rate q(t). As discussed in Chapter 5, for liquids one has to use the volume flow rate qv(t) instead of q(t), whereas for gases the mass flow rate qm(t) is employed. The alternative head formulation leads to the following liquid impedance definition:  HðsÞ ZðsÞ ¼ or HðsÞ ¼ ZðsÞ $ QðsÞ (7.55) QðsÞIC¼0 with H(s) being the Laplace transform of the head h(t). Table 7.1 comprises all fluid element impedances, but the head defined element impedances (which are needed in the next example) are derived below through application of Laplace transforms with zero initial conditions to the headeflow rate relationships: hðtÞ ¼ R $ qðtÞ;

H ðsÞ ¼ R $ QðsÞ;

H ðsÞ ¼ ZR ¼ R QðsÞ

H ðsÞ ¼ ðI $ sÞ $ QðsÞ;

H ðsÞ ¼ ZI ¼ I $ s QðsÞ

hð t Þ ¼ I $

dqðtÞ ; dt

qð t Þ ¼ C $

dhðtÞ H ðsÞ 1 ; QðsÞ ¼ ðC $ sÞ $ H ðsÞ; ¼ ZC ¼ dt QðsÞ C$s

(7.56)

nn

Example 7.8 Use an impedance-based mathematical model of the liquid system shown in Figure 7.11 and determine the transfer function matrix considering that the input to the system consists of the volume flow rates qi1 and qi2, whereas the liquid heads h1 and h2 are the system’s output. The system is formed of two tanks of capacitances C1 and C2 and two valves of resistances R1 and R2. These properties are defined in terms of head. The pressure and its related head are zero past R2.

7.2 Transfer Function Model Formulation

qi1 qi2

h1

h2 C1

R1

C2

R2

q

qo

FIGURE 7.11 Two-Tank Liquid-Level System.

Solution Using the fluid impedance definitions of Eq. (7.56), the following equations can be written for the fluid resistances and capacitances indicated in Figure 7.11 as:

H1 ðsÞ H1 ðsÞ  H2 ðsÞ ; ; ZR1 ðsÞ ¼ Qi1 ðsÞ  QðsÞ QðsÞ H2 ðsÞ H2 ðsÞ ZC2 ðsÞ ¼ ; ZR2 ðsÞ ¼ QðsÞ þ Qi2 ðsÞ  Qo ðsÞ Qo ðsÞ

ZC1 ðsÞ ¼

(7.57)

Referring to Eq. (7.57), the flow rate Q(s) is substituted from the second equation in the first one, and the flow rate Qo(s) is substituted from the fourth equation in the third one; the following equations are obtained:

8 > Z ðsÞ ZC ðsÞ > > 1 þ C1 $ H1 ðsÞ  1 $ H2 ðsÞ ¼ ZC1 ðsÞ $ Qi1 ðsÞ > < ZR1 ðsÞ ZR1 ðsÞ  > ZC2 ðsÞ ZC2 ðsÞ ZC2 ðsÞ > > > H1 ðsÞ þ 1 þ þ $ H2 ðsÞ ¼ ZC2 ðsÞ $ Qi2 ðsÞ : ZR1 ðsÞ ZR1 ðsÞ ZR2 ðsÞ

(7.58)

The impedances of Eq. (7.58) are

ZC1 ðsÞ ¼

1 1 ; ZC2 ðsÞ ¼ ; ZR1 ðsÞ ¼ R1 ; ZR2 ðsÞ ¼ R2 C1 $ s C2 $ s

(7.59)

By substituting them in Eq. (7.58), the latter ones are written in matrix form as:

2 6 6 4

1þ

1 R1 $ C1 $ s

1  R1 $ C2 $ s 2 1 6 C1 $ s ¼6 4 0

 1þ 3 0

1 C2 $ s

1 R1 $ C 1 $ s

1 1 þ R1 $ C 2 $ s R2 $ C 2 $ s

7 7$ 5



Qi1 ðsÞ Qi2 ðsÞ

3 7 7$ 5



H1 ðsÞ H2 ðsÞ



(7.60)

343

344

CHAPTER 7 Transfer Function Approach

Because the head vector of the left-hand side of Eq. (7.60) is the output and the volume flow rate vector of the right-hand side of Eq. (7.60) is the input, the transfer function matrix is

2 6 ½GðsÞ ¼ 6 4

1þ

1 R1 $ C1 $ s



1 R1 $ C 1 $ s

31 2 7 7 5

1 6 C1 $ s $6 4 0 3

1 1 1  1þ þ R1 $ C2 $ s R1 $ C2 $ s R2 $ C2 $ s 2 R1 $ R2 $ C2 $ s þ R1 þ R2 R2 6 7 D ð s Þ D ðsÞ 6 7 ¼6 7 4 R2 R1 $ R2 $ C1 $ s þ R2 5 DðsÞ D ðsÞ

3 0 1 C2 $ s

7 7 5

(7.61) the denominator polynomial being D(s) ¼ R1$R2$C1$C2$s þ [R2$(C1 þ C2) þ R1$C1]$s þ 1. 2

nn

Mechanical Systems Mechanical element impedances Z(s) can be defined in the s-domain as the ratio of the force or moment, F(s) or M(s), to the translation/rotary velocity variation, DV(s) or DU(s), with zero initial conditions:  FðsÞ  Translation: ZðsÞ ¼ or FðsÞ ¼ ZðsÞ $ DVðsÞ DVðsÞIC¼0 (7.62)  MðsÞ  or MðsÞ ¼ ZðsÞ $ DUðsÞ Rotation: ZðsÞ ¼ DUðsÞIC¼0 Application of Laplace transforms with zero initial conditions to the time-domain equations defining the forceevelocity relationships for damping, inertia, and elastic force produces the mechanical element impedances corresponding to translation: f ðtÞ ¼ c $ DvðtÞ; f ðt Þ ¼ m $

dvðtÞ ; dt

1 df ðtÞ ; DvðtÞ ¼ $ k dt

F ðsÞ ¼ c $ DV ðsÞ;

F ðsÞ ¼ Zd ¼ c DV ðsÞ

F ðsÞ ¼ ðm $ sÞ $ V ðsÞ;

F ðsÞ ¼ Zi ¼ m $ s V ðsÞ

 DV ðsÞ ¼

1 $ s $ F ðsÞ; k

(7.63)

F ðsÞ k ¼ Ze ¼ DV ðsÞ s

Note that for a spring or a damper, the end points experience different velocities, which leads to using the relative velocity DV(s) in the corresponding impedance definitions. For inertia pointlike elements, the inertia impedance is defined based on the velocity V(s) of the point mass. Rotation-related impedances similar to those of Eq. (7.62) can be obtained when using time-domain momentevelocity relationshipsd see Table 7.1, which lists velocities without the D symbol. Mechanical impedances can be combined in series and in parallel; therefore, equivalent impedances can be calculated for either case. Figure 7.12 illustrates

7.2 Transfer Function Model Formulation

V2(s)

V1(s) Z1

V1(s) F1(s)

V3(s)

F(s)

V2(s)

V1(s)

Z1

Z2

V2(s) Z

F(s)

Z2

F(s)

F2(s)

ΔV(s)

ΔV(s)

ΔV(s) (a)

(b)

(c)

FIGURE 7.12 Connections of Mechanical Impedances: (a) Series for Elastic and Damping Elements; (b) Parallel; (c) Equivalent.

the serial and the parallel connections of translatory-motion spring and damper impedances. For a serial combination of deformable elements, the force transmitted through the components is the same, whereas the relative velocity between the chain end points is the sum of relative velocity across components. Conversely, in parallel connections of any elements (deformable and inertia), the relative velocities across components are identical while the total force splits between components. These principles are used in the companion website Chapter 7 to derive the two equivalent mechanical impedances, which are 1 1 1 ¼ þ ; Zp ¼ Z1 þ Z2 Zs Z1 Z2

(7.64)

with s standing for series and p for parallel. Note that Eq. (7.64) is valid as long as there are no other external forces applied at internal points within the considered chain. This is also valid for an inertia element combined serially with other deformable elements, particularly when an inertia element is located between deformable elementsdsee Figure 7.13(a). In that situation, an inertia force has to be placed at the node where the inertia element lies, force

V1(s)

V(s) Z1

V(s) Zi

Z2

Inertia element

V(s)

V2(s) F1(s)

Zi

F2(s)

Fi (s)

Deformable element (a)

(b)

FIGURE 7.13 (a) Series Connection of One Inertia and Two Deformable Mechanical Impedances; (b) Laplace-Domain Free-Body Diagram for Inertia Element.

345

346

CHAPTER 7 Transfer Function Approach

that can be considered external. The s-domain d’Alembert principle (derived from Newton’s second law of motion, as discussed in Chapter 3) can be written based on the free-body s-domain diagram of Figure 7.13(b) as follows:  Fi ðsÞ  F1 ðsÞ  F2 ðsÞ ¼ 0 or  Zi ðsÞ $ VðsÞ  Z1 ðsÞ $ ½VðsÞ  V1 ðsÞ  Z2 ðsÞ $ ½VðsÞ  V2 ðsÞ ¼ 0 or

(7.65)

½Zi ðsÞ þ Z1 ðsÞ þ Z2 ðsÞ $ VðsÞ ¼ Z1 ðsÞ $ V1 ðsÞ þ Z2 ðsÞ $ V2 ðsÞ nn

Example 7.9

Determine the transfer function G(s) ¼ Y(s)/F(s) for the mechanical system of Figure 7.14(a) using element impedances. Y(s) and F(s) are the Laplace transforms of the mass displacement y (t) and of the external force f(t) applied to the mass, respectively. Assume c1 ¼ c2 ¼ c and k1 ¼ k2 ¼ k3 ¼ k.

Y(s)

y

c1 k2

f

k3

F(s)

Zc1

c2

m

Zk2

Zi

Zk3

Zc2

Zk1 k1 (a)

(b)

FIGURE 7.14 Translatory Mechanical System: (a) Lumped-Parameter Model; (b) Impedance Model.

Solution The impedance diagram is shown in Figure 7.14(b), and it can be reduced to the diagram of Figure 7.15(a) where Zc1, Zk1, and Zk2 are combined into Z1, whereas Zk3 and Zc2 form Z2. The three impedances of Figure 7.15(a) are expressed as:

Z1 ¼

ðZC1 þ Zk1 Þ $ Zk2 ðc1 þ k1 =sÞ $ ðk2 =sÞ k2 $ ðc1 $ s þ k1 Þ ¼ ¼ c1 þ k1 =s þ k2 =s s $ ðc1 $ s þ k1 þ k2 Þ Z C 1 þ Zk 1 þ Zk 2

Z2 ¼

ðk3 =sÞ $ c2 Zk3 $ ZC2 k3 $ c2 ¼ ¼ Zk3 þ ZC2 k3 =s þ c2 c2 $ s þ k3

(7.66)

Zi ¼ m $ s Y(s)

Y(s)

F(s) Z1

F(s) Zi

Z2

F1(s)

Zi

F2(s)

Fi (s) (a)

(b)

FIGURE 7.15 (a) Series Connection of One Inertia and Two Deformable Mechanical Impedances; (b) Laplace-Domain Free-Body Diagram for Inertia Element.

7.3 Transfer Function and System Stability

The s-domain free-body diagram of Figure 7.15(b) results in the dynamic equilibrium equation:

FðsÞ  F1 ðsÞ  F2 ðsÞ  Fi ðsÞ ¼ 0 or FðsÞ ¼ ðZ1 þ Z2 þ Zi Þ $ VðsÞ

(7.67)

Taking into account that the velocity is related to displacement as v(t) ¼ dy(t)/dt in the domain or as V(s) ¼ s$Y(s) in the Laplace domain, the transfer function is found by combining Eqs. (7.66) and (7.67):

GðsÞ ¼ ¼

YðsÞ VðsÞ 1 ¼ ¼ FðsÞ s $ FðsÞ s $ ðZ1 þ Z2 þ Zi Þ c2 $ s2 þ 3c $ k $ s þ 2k2 m $ c2 $ s4 þ 3m $ c $ k $ s3 þ 2k $ ðc2 þ m $ kÞ $ s2 þ 4c $ k2 $ s þ k3 (7.68)

Eq. (7.68) can be obtained using MATLAB’s collect command applied to G(s) ¼ 1/[(Z1 þ Z2 þ Zi)$s].

nn

7.3 TRANSFER FUNCTION AND SYSTEM STABILITY The characteristic polynomial, which is the denominator D(s) of a system’s transfer function, can be utilized to study the stability of that system by establishing the nature (real or complex) and multiplicity of the characteristic equation roots (the transfer function poles). Stability essentially requires that a system’s response does not grow unbounded (does not go to infinity) for any specific bounded input. Conversely, a system is unstable when its response grows unboundedly for a bounded input. The total time-domain response y(t) of a system is formed of the natural and forced responses: yðtÞ ¼ yn ðtÞ þ yf ðtÞ

(7.69)

The natural response yn(t) is determined by the system’s inherent characteristics, whereas the forced response yf(t) results from the forcing (input) action on the system. Considering a SISO system whose s-domain output Y(s) is obtained as the product of the transfer function G(s) and the input U(s), the following partial fraction expansion can be utilized:     NG ðsÞ NU ðsÞ YðsÞ ¼ GðsÞ $ UðsÞ ¼ (7.70) $ ¼ Yn ðsÞ þ Yf ðsÞ DG ðsÞ DU ðsÞ where N(s) and D(s) are the numerators and denominators corresponding to G(s) and U(s), respectively. Each of the two N(s)/D(s) ratios of Eq. (7.70) is expanded as a sum of partial fractionsdthe one denoted by Yn(s) corresponding to the natural response generated by the transfer function G(s) and the other one, Yf (s), owing to the particular input U(s). Assume that we are only interested in Yn(s)

347

348

CHAPTER 7 Transfer Function Approach

reflecting the system (natural) contribution to the overall response, which is as follows: Yn ðsÞ ¼

n NG ðsÞ X ai ¼ DG ðsÞ i¼1 ðs  pi Þmi

(7.71)

Eq. (7.71) shows that DG(s) has n roots pideach of the order of multiplicity mi. These roots pi are also the poles of the transfer function G(s). Before generalizing, consider the particular case where one term in the sum of Eq. (7.71) is 1/(s  2). Its inverse Laplace transform is e2$t, and this is one term of the natural response yn(t). As time goes to infinity, this particular response grows unboundedly, and therefore, the system is unstable. The peculiarity of this system was that its pole p ¼ 2 is a positive number, so any positive pole would result in an unstable system. On the contrary, a negative pole, such as in the related simple fraction 1/(s þ 2), whose inverse Laplace transform is e2$t, would lead to a solution term that becomes 0 (a bounded value) when time goes to infinity. This is the result of the negative pole p ¼ 2; and if all the poles of G(s) are negative, the partial natural responses would all be bounded leading to a stable system. With respect to stability, there are three different types of systems as described by their natural responses in connection with the transfer function pole location, namely: stable systems, unstable systems, and marginally stable (or neutrally stable) systems. Systems whose natural response is bounded in time (does not grow indefinitely) are stable systems. Such responses are generated by real negative poles (simple or multiple), as well as by simple or multiple complex conjugated poles with negative real parts as shown below and illustrated in Figure 7.16(a): c1 sþs 1 Yn;2 ðsÞ ¼ ðs þ sÞm c3 Yn;3 ðsÞ ¼ s þ s  u$j Yn;1 ðsÞ ¼

Yn;4 ðsÞ ¼

1 ðs þ s  u $ jÞm

/

yn;1 ðtÞ ¼ c1 $ es $ t

/

yn;2 ðtÞ ¼ c2 $ tm1 $ es $ t

/

yn;3 ðtÞ ¼ c3 $ es $ t $ sinðu $ t þ 4Þ

/

yn;4 ðtÞ ¼ c4 $ tm1 $ es $ t $ sinðu $ t þ 4Þ (7.72)

In Eq. (7.72) s > 0 (this is also valid subsequently in this section), and m is the order of multiplicity (m > 1) of the pole. Systems whose natural response is unbounded (it grows indefinitely with time increasing) are, on the other hand, unstable systems. The transfer functions of unstable systems have real positive poles (simple or multiple), simple or multiple complex conjugated poles with positive real parts, as well as multiple

7.3 Transfer Function and System Stability

yn,2 yn,1

yn,3

yn,4

Time (a)

yn,6

yn,5

yn,9 yn,7 yn,8 Time (b)

yn,10

Time (c)

FIGURE 7.16 Time Response of the Following: (a) Stable Systems; (b) Unstable Systems; (c) Marginally Stable Systems.

349

350

CHAPTER 7 Transfer Function Approach

imaginary polesdthis is indicated in the following equation and illustrated in Figure 7.16(b): c5 Yn;5 ðsÞ ¼ / yn;5 ðtÞ ¼ c5 $ es $ t ss 1 Yn;6 ðsÞ ¼ / yn;6 ðtÞ ¼ c6 $ tm1 $ es $ t ðs  sÞm c7 / yn;7 ðtÞ ¼ c7 $ es $ t $ sinðu $ t þ 4Þ Yn;7 ðsÞ ¼ s  s  u$j Yn;8 ðsÞ ¼

1 ðs  s  u $ jÞm

/

yn;8 ðtÞ ¼ c8 $ tm1 $ es $ t $ sinðu $ t þ 4Þ

Yn;9 ðsÞ ¼

1 ðs  u $ jÞm

/

yn;9 ðtÞ ¼ c9 $ tm1 $ sinðu $ t þ 4Þ (7.73)

Poles

Stability Behavior

Negative Simple/Multiple Real Stable Simple/Multiple Complex Conjugated with Negative Real Part

Positive Simple/Multiple Real

Simple/Multiple Complex Conjugated with Positive Real Part

Unstable

Multiple Imaginary

Simple Imaginary

Marginally Stable

FIGURE 7.17 System Stability as a Function of Transfer Function Poles Nature.

7.3 Transfer Function and System Stability

Eventually, a third category in stability is the one comprising marginally stable systems. Marginally stable systems display a natural response that is oscillatory and that neither becomes constant (as the case is with stable system responses) nor grows unbounded (as with unstable systems). The transfer function poles are simple imaginary ones in this case as shown in the following equation and as pictured in Figure 7.16(c). c10 Yn;10 ðsÞ ¼ / yn;10 ðtÞ ¼ c10 $ sinðu $ t þ 4Þ (7.74) s  u$j Figure 7.17 synthesizes the transfer function pole nature in connection to the system stability behavior. nn

Example 7.10 Discuss the stability of the inverted pendulum sketched in Figure 7.18 by analyzing its transfer function G(s) ¼ Q(s)/M(s) where M(s) is the Laplace transform of a fictitious moment m(t), and Q(s) is the Laplace transform of the actual rotation angle q(t). The pendulum is in a vertical plane and is formed of a rigid, massless rod of length l, a bob of mass m, and a rotary (torsional) spring of stiffness k that elastically restrains the rotation of the pendulum around its pivot point O. Assume small rotation angles q and consider known the gravitational acceleration g.

h

Datum line

m Bob m·g

θ

Massless Rigid Rod me(t) Rotary spring k

l m(t) (=0) O

FIGURE 7.18 Inverted Pendulum.

Solution Newton’s second law of motion for rotation is applied to the inverted pendulum shown in a displaced position:

€ ¼ mðtÞ  me ðtÞ þ m $ g $ l $ q or m $ l2 $ q € þ ðk  m $ g $ lÞ $ q ¼ mðtÞ Jo $ q (7.75) which took into account that sinq z q (for the small-angle approximation), Jo ¼ m$l 2, and me ¼ k$q. Laplace transforming the final-form Eq. (7.75) with zero initial conditions results in the transfer function:

GðsÞ ¼

QðsÞ NG ðsÞ 1 ¼ ¼ MðsÞ DG ðsÞ m $ l2 $ s2 þ k  m $ g $ l

(7.76)

351

352

CHAPTER 7 Transfer Function Approach

The solution to the characteristic equation DG(s) ¼ 0 resulting from Eq. (7.76) is:

1 s¼  $ l

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m$g$l  k m

(7.77)

When m$g$l < k, the two poles of Eq. (7.77) are simple imaginary, and the inverted pendulum is marginally stable. For m$g$l ¼ k, the two poles become s ¼ 0, which means they are multiple imaginary poles (since the origin belongs to the imaginary axis); therefore the mechanical system is € ¼ 0; the unstable. This condition also results by integrating the resulting differential equation q solution is q(t) ¼ c1$t þ c2, (c1 and c2 are integration constants), which is unbounded. The last possibility is m$g$l > k, case where the two poles of Eq. (7.77) are real and distinct. Because one solution is positive, the inverted pendulum is unstable. nn

nn

Example 7.11

(i) A dynamic system’s transfer function has the following known poles: 1 (simple), 1 þ j (double), and 2$j (double). Determine the minimum-degree characteristic polynomial DG(s) and establish the stability nature of the system. (ii) The transfer function of a dynamic system is as follows:

GðsÞ ¼

1 s7 þ 8s6 þ 27s5 þ 52s4 þ 66s3 þ 60s2 þ 40s þ 16

Determine the stability of this system.

Solution

(i) Having 1 þ j as a root, the characteristic equation also has its complex conjugate 1  j as a root, both roots being double roots. Similarly, 2$j is a double root of the characteristic equation due to the double root 2$j. As a consequence, the characteristic polynomial is as follows:

DG ðsÞ ¼ ðs þ 1Þ $ ðs þ 1  jÞ2 $ ðs þ 1 þ jÞ2 $ ðs þ 2jÞ2 $ ðs  2jÞ2 ¼ s9 þ 5s8 þ 20s7 þ 56s6 þ 124s5 þ 212s4 þ 288s3 þ 288s2 þ 192s þ 64 (7.78) Because the characteristic equation possesses multiple imaginary roots, the system is unstable. (ii) MATLAB has the pole command that can be used in conjunction with the tf command to determine the poles of the given transfer function; the following MATLAB script >> g=tf(1,[1,8,27,52,66,60,40,16]) >> pole(g) returns ans = -0.0000 + 1.0000i -0.0000 - 1.0000i -2.0000 + 0.0000i

7.4 Transfer Function and the Time Response

-2.0000 - 0.0000i -2.0000 + 0.0000i -1.0000 + 1.0000i -1.0000 - 1.0000i Alternatively, the roots command generates the roots of the characteristic polynomial in the denominator of the transfer function with a MATLAB command line: >> roots([1,8,27,52,66,60,40,16]) Except the simple imaginary poles, all the poles have negative real parts; as a consequence, the system is marginally stable. nn

7.4 TRANSFER FUNCTION AND THE TIME RESPONSE With the transfer functions derived by one of the methods presented in Section 7.2, we determine the time response of SISO and MIMO systems analytically, by means of MATLAB or using Simulink. The generic formulation considers that both forcing and nonzero initial conditions can be applied to a dynamic system.

7.4.1 SISO Systems The transfer function approach to determine the time response of SISO systems is discussed in this section.

Analytical Approach The general case of a dynamic system subjected to both forcing and nonzero initial conditions is considered here. Nonzero initial conditions are compounded with the actual input (forcing) to produce an equivalent input, which is denoted by Ueq(s) in the s-domain, as it will be exemplified shortly for typical first- and second-order systems. As a consequence, the s-domain output Y(s) is calculated with the aid of the transfer function G(s) as follows: Y(s) ¼ G(s)$Ueq(s). The time-domain response is found by applying the inverse Laplace transform to Y(s) as: yðtÞ ¼ L1 ½YðsÞ ¼ L1 ½GðsÞ $ Ueq ðsÞ

(7.79)

According to the initial value theorem, the initial value of the output is calculated as yð0Þ ¼ lim yðtÞ ¼ lim s $ YðsÞ ¼ lim s $ GðsÞ $ Ueq ðsÞ t/0

s/N

s/N

(7.80)

Similarly, the final value of the time-domain response can be determined by using the final value theorem: yðNÞ ¼ lim yðtÞ ¼ lim s $ YðsÞ ¼ lim s $ GðsÞ $ Ueq ðsÞ t/N

s/0

s/0

(7.81)

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CHAPTER 7 Transfer Function Approach

To evaluate the equivalent input Ueq(s) for a first-order system, consider the differential equation _ þ a0 $ yðtÞ ¼ uðtÞ a1 $ yðtÞ

(7.82)

with the initial condition: y(0) s 0. Laplace transforming Eq. (7.82) yields 1 $ ½UðsÞ þ a1 $ yð0Þ or YðsÞ ¼ GðsÞ $ Ueq ðsÞ with a1 $ s þ a0 1 GðsÞ ¼ ; Ueq ðsÞ ¼ UðsÞ þ a1 $ yð0Þ a1 $ s þ a0 YðsÞ ¼

(7.83)

where Ueq(s) combines the effects of the actual input function U(s) and the nonzero initial condition. Note that G(s) is obtained for zero initial conditions, as per the transfer function definition, but this is not the actual situation of the problem studied here since y(0) s 0. In addition, observe that when U(s) ¼ 0, which means no external forcing, the equivalent s-domain input resulting from Eq. (7.83) is Ueq(s) ¼ a1$y(0), which converts the nonzero initial condition into equivalent forcing. The mathematical model of a second-order system is: € þ a1 $ yðtÞ _ þ a0 $ yðtÞ ¼ uðtÞ a2 $ yðtÞ

(7.84)

_ with the initial conditions: yð0Þs0; y(0) s 0. The s-domain inputeoutput relationship results from applying the Laplace transform to Eq. (7.84), namely 1 _ þ a1 $ yð0Þ or $ ½UðsÞ þ a2 $ yð0Þ $ s þ a2 $ yð0Þ a2 þ a1 $ s þ a0 1 YðsÞ ¼ GðsÞ $ Ueq ðsÞ with GðsÞ ¼ ; a2 $ s 2 þ a1 $ s þ a0 _ þ a1 $ yð0Þ Ueq ðsÞ ¼ UðsÞ þ a2 $ yð0Þ $ s þ a2 $ yð0Þ YðsÞ ¼

$ s2

(7.85) where, again, Ueq(s) incorporates the actual input and the nonzero initial conditions. When U(s) ¼ 0, which means no external forcing, Eq. (7.85) yields _ þ a1 $ yð0Þ, which only incorporates the nonzero Ueq ðsÞ ¼ a2 $ yð0Þ $ sþ a2 $ yð0Þ initial conditions. nn

Example 7.12

The pneumatic system of Figure 7.19 is formed of a container with a capacity C ¼ 1.2$106 m s2, which is supplied with gas whose pressure depends on time as pi ¼ (1 þ 4 e0.2t)$105 N/m2. The gas passes through a constriction whose resistance is R ¼ 30,000 m1 s1. Determine the transfer function Po(s)/Pi (s) of this system; based on it, calculate the output (container) pressure po(t) and plot it against time for zero initial conditions. Also determine the final (steady-state) value of po(t).

7.4 Transfer Function and the Time Response

po po

C

R

pi

qm

FIGURE 7.19 Pneumatic System With Valve and Container.

Solution The gas resistance and capacitance are expressed based on their definitions given in Chapter 5 and on Figure 7.19 as

pi ðtÞ  po ðtÞ ¼ R $ qm ðtÞ; C $ p_o ¼ qm ðtÞ

(7.86)

Eliminating qm(t) between the two Eq. (7.86) results in the time-domain mathematical model

R $ C $ p_o ðtÞ þ po ðtÞ ¼ pi ðtÞ

(7.87)

Eq. (7.87) indicates that this is a first-order system. Because the initial conditions are zero, the equivalent input is identical to the actual input: Ueq(s) ¼ U(s). The system transfer function is obtained by applying the Laplace transform with zero initial conditions to Eq. (7.87):

GðsÞ ¼

Po ðsÞ 1 ¼ Pi ðsÞ 1 þ R $ C $ s

(7.88)

The Laplace transform of the output is

po (N/m2)

 105 1 4 þ $ Po ðsÞ ¼ GðsÞ $ Pi ðsÞ ¼ s s þ 0:2 1 þ R$C$s

Time (sec)

FIGURE 7.20 Container Pressure as a Function of Time.

(7.89)

355

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CHAPTER 7 Transfer Function Approach

The steady-state value of the container pressure is calculated as

 po ðNÞ ¼ lim po ðtÞ ¼ lim s $ Po ðsÞ ¼ 105 N m2 t/N

s/0

(7.90)

The container pressure is calculated by inverse Laplace transforming Po(s) of Eq. (7.89) and is approximately equal to


po ðtÞx 1 þ 4:029e0:2 $ t  5:029e27:778 $ t $ 105 N=m2

(7.91)

The output pressure is plotted in Figure 7.20.

nn

nn

Example 7.13

A single-mesh electrical system is formed of a resistor R, a capacitor C, an inductor L, and a voltage source providing a ramp voltage v(t) ¼ b$t. Find the current i(t) in the circuit when 4L ¼ R2$C, knowing the initial value of the current, i(0) ¼ 0.1 A. Plot i(t) for b ¼ 100 V/s, R ¼ 200 U, and L ¼ 3 H; also find the final value of the current.

Solution The differential equation of the series electrical system is

€ þ R $ qðtÞ _ þ L $ qðtÞ

1 $ qðtÞ ¼ vðtÞ C

(7.92)

with the charge q(t) being the output and v(t) being the input. According to Eq. (7.85), the transfer function and the equivalent Laplace-transformed input resulting from Laplace transforming Eq. (7.92) are

GðsÞ ¼

QðsÞ C b ¼ ; Ueq ðsÞ ¼ V ðsÞ þ L $ ið0Þ ¼ 2 þ L $ ið0Þ Ueq ðsÞ L $ C $ s2 þ R $ C $ s þ 1 s (7.93)

Eq. (7.93) used V(s) ¼ V/s2 as the Laplace transform of v (t) ¼ V$t. As a consequence and for the particular condition of this example, the Laplace transform of the output is

QðsÞ ¼ GðsÞ $ Ueq ðsÞ ¼

C$b L $ C $ ið0Þ þ s2 $ ðL $ C $ s2 þ R $ C $ s þ 1Þ L $ C $ s2 þ R $ C $ s þ 1 (7.94)

The Laplace transform of the current is expressed in terms of the Laplace transform of the charge as

IðsÞ ¼ s $ QðsÞ ¼

C$b L $ C $ ið0Þ $ s þ (7.95) s $ ðL $ C $ s2 þ R $ C $ s þ 1Þ L $ C $ s2 þ R $ C $ s þ 1

The steady-state value of the current can be determined by means of the final value theorem as

iðNÞ ¼ lim iðtÞ ¼ lim s $ IðsÞ ¼ C $ b ¼ t/N

s/0

4L $ b ¼ 0:03 A R2

(7.96)

i (A)

7.4 Transfer Function and the Time Response

Time (sec)

FIGURE 7.21 Current as a Function of Time.

By inverse Laplace transforming I(s) of Eq. (7.95), the following time-dependent current is obtained:

iðtÞ ¼

R 4L $ b 4L $ b þ R2 $ ið0Þ 4L $ b  R2 $ ið0Þ  R $ t 2L $t $ t $ e   $ e 2L R2 2R $ L R2

(7.97)

Eq. (7.97) used the relationship 4L ¼ R2$C provided in the example statement. The current i(t) is plotted in Figure 7.21, which shows that, indeed, the final value (steady-state) value is 0.03 A. nn

MATLAB and Simulink Approach MATLAB enables calculation and plotting of the forced response of dynamic systems by means of the transfer function. Once the mathematical model and the corresponding transfer function are obtained through modeling, MATLAB has the capability of obtaining the system response (and plotting it) for various inputs, such as unit step, unit impulse, or of an arbitrary form. In addition, Simulink has a transfer function block that allows modeling SISO systems in the time domain and plotting the output as a function of time. A particular feature of using this block is that while the transfer function is defined in the s-domain, the input signal to it and the output signal from it are time-domain signals.

Transforming a System’s Transfer Function to Account for Nonzero Initial Conditions As we shall see shortly, the MATLAB built-in commands dedicated to finding the time response by means of the transfer function are designed for problems with zero initial conditions, which is a limiting factor. However, it is possible to convert a system’s original transfer function into another equivalent transfer function that incorporates the actual nonzero initial conditions while preserving the original input function. Consider, for instance, a second-order system whose time-domain differential equation and transfer function are provided in Eqs. (7.84) and (7.85).

357

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CHAPTER 7 Transfer Function Approach

The Laplace-transformed output of Eq. (7.85) can be reformulated through multiplication and division by U(s) as: _ þ a1 $ yð0Þ UðsÞ þ a2 $ yð0Þ $ s þ a2 $ yð0Þ $ UðsÞ ¼ Geq ðsÞ $ UðsÞ or ða2 $ s2 þ a1 $ s þ a0 Þ $ UðsÞ GðsÞ _ þ a1 $ yð0Þ with $ ½UðsÞ þ a2 $ yð0Þ $ s þ a2 $ yð0Þ Geq ðsÞ ¼ UðsÞ 1 GðsÞ ¼ 2 ða2 $ s þ a1 $ s þ a0 Þ $ UðsÞ (7.98) YðsÞ ¼

Note that the equivalent transfer function Geq(s), which allows including nonzero initial conditions while utilizing the original input U(s), is a combination of the original transfer function G(s), the nonzero initial conditions, and the input. Two built-in MATLAB functions enable finding the time response for unit impulse d(t) and unit step h(t) inputs. Taking into account that U(s) ¼ 1 is the Laplace transform of u(t) ¼ d(t) and U(s) ¼ 1/s for u(t) ¼ h(t), the equivalent transfer function of Eq. (7.98) becomes: _ þ a1 $ yð0Þ þ 1 UðsÞ ¼ 1 / Geq ðsÞ ¼ GðsÞ $ ½a2 $ yð0Þ $ s þ a2 $ yð0Þ ¼

_ þ a1 $ yð0Þ þ 1 a2 $ yð0Þ $ s þ a2 $ yð0Þ 2 a2 $ s þ a1 $ s þ a0

1 _ þ a1 $ yð0Þ UðsÞ ¼ / Geq ðsÞ ¼ s $ GðsÞ $ ½1=s þ a2 $ yð0Þ $ s þ a2 $ yð0Þ s ¼

_ þ a1 $ yð0Þ $ s þ 1 a2 $ yð0Þ $ s2 þ ½a2 $ yð0Þ 2 a2 $ s þ a1 $ s þ a0 (7.99)

Response to Unit Impulse, Unit Step, Arbitrary InputdThe MATLAB Impulse, Step, lsim Functions The MATLAB function impulse enables finding and/or plotting the time response of a dynamic system defined through a transfer function when the input is a unit impulse. The MATLAB command sequence >> G = tf (N, D)%N and D are lists of numerator and denominator coefficients; >> impulse(G)

produces the plot of y(t) corresponding to a transfer function G for a unit-impulse input d(t). For a nonunity impulse input, such as c$d(t), the last command changes to impulse(c*G). If a command impulse(G,T) is used, MATLAB will generate the response plot for a time interval ranging from 0 to T; alternatively, T can be defined as a vector with initial value, time step, and final value.

7.4 Transfer Function and the Time Response

To plot the time response of a system defined by a transfer function to a unit step input h(t), the MATLAB step function needs to be used, which operates similarly to the previously described impulse function. Arbitrary input (forcing) can be generated in MATLAB by the lsim function under zero initial conditions. The arbitrary input function has to be defined before launching the calculating and plotting lsim command. A typical MATLAB script used to plot the time response of a system defined by the transfer function G under the action of an arbitrary time-domain input u ¼ f(t) for a time ranging between t_l and t_u with a time step t_s is as follows: >> G = tf (N, D); >> t = [t_l:t_s:t_u]; >> u = f(t)%time-domain input definition; >> lsim(G,u,t)

nn

Example 7.14

A second-order dynamic system is defined by the transfer function G(s) ¼ 1/(5s 2 þ s þ 4). Using the MATLAB tf, impulse and step commands, plot the time response corresponding to G(s) and (i) a unit impulse u(t) ¼ d(t) (ii) a unit step input u(t) ¼ h(t) Consider both the case with zero initial conditions y_ð0Þ ¼ 0; y ð0Þ ¼ 0 and the case with y_ð0Þ ¼ 0:2; y ð0Þ ¼ 1.

Solution

Inspecting the given second-order transfer function indicates that a2 ¼ 5, a1 ¼ 1, a0 ¼ 4. (i) When the initial conditions are different from zero, an equivalent transfer function is obtained corresponding to the first Eq. (7.99) and to a unit-impulse input, which is

Geq;1 ðsÞ ¼

_ þ a1 $ yð0Þ þ 1 a2 $ yð0Þ $ s þ a2 $ yð0Þ 5s þ 3 ¼ 2 2 a2 $ s þ a1 $ s þ a0 5s þ s þ 4

(7.100)

The following MATLAB script is used to generate the unit-impulse response corresponding to the original transfer function G(s) for zero initial conditions and the equivalent transfer function Geq,1(s) of Eq. (7.100), which accounts for the nonzero conditions: >> G = tf(1, [5, 1, 4]); >> Geq1 = tf([5, 3], [5, 1, 4]); >> impulse(G, Geq1)%both responses are on the same graph The two plots are shown in Figure 7.22(a). (ii) The equivalent transfer function that incorporates nonzero initial conditions corresponding to a unit step input is given in the second Eq. (7.99); for the current example, this function becomes as:

Geq;2 ðsÞ ¼

_ þ a1 $ yð0Þ $ s þ 1 5s2 þ 2s þ 1 a2 $ yð0Þ $ s2 þ ½a2 $ yð0Þ (7.101) ¼ 2 2 a2 $ s þ a1 $ s þ a0 5s þ s þ 4

359

360

CHAPTER 7 Transfer Function Approach

y(t) for Geq,1(s)

y(t) for G(s)

(a)

y(t) for Geq,2(s)

y(t) for G(s)

(b)

FIGURE 7.22 MATLAB Time Response With Zero and Nonzero Initial Conditions for the Following: (a) Unit-Impulse Input; (b) Unit Step Input. The following MATLAB script is used to generate the unit step response corresponding to the original transfer function G(s) for zero initial conditions and the equivalent transfer function Geq,2(s) of Eq. (7.101), which includes the nonzero conditions: >> G = tf(1, [5, 1, 4]); >> Geq2 = tf([5, 2, 1], [5, 1, 4]); >> step(G, Geq2)%both responses are on the same graph The resulting plots are illustrated in Figure 7.22(b).

nn

7.4 Transfer Function and the Time Response

nn

Example 7.15

Consider the system defined by the transfer function G(s) ¼ 1/(5s 2 þ s þ 4) and the nonzero initial conditions y_ð0Þ ¼ 0:2; y ð0Þ ¼ 1 of Example 7.14. Plot the system’s time response for an input u(t) ¼ e0.2$t$cos(5$t) by using (i) MATLAB’s lsim command. (ii) Simulink

Solution (i) Based on Eq. (7.98), the equivalent transfer function is

Geq ðsÞ ¼

_ þ a1 $ yð0Þ UðsÞ þ a2 $ yð0Þ $ s þ a2 $ yð0Þ UðsÞ þ 5s þ 2 ¼ ða2 $ s2 þ a1 $ s þ a0 Þ $ UðsÞ ð5s2 þ s þ 4Þ $ UðsÞ (7.102)

with

UðsÞ ¼

s þ 0:2 ðs þ 0:2Þ2 þ 52

(7.103)

One can use the following MATLAB script to obtain the time response plotted in Figure 7.23: >> s = tf(’s’)%allows defining transfer functions by regular algebra; >> us = (s + 0.2)/((s + 0.2)^2 + 25); >> Geq = (us + 5*s + 2)/us/(5*s^2 + s + 4); >> t = 0:0.001:25; >> u = exp(0.2*t). * cos(5*t); >> lsim (Geq, u, t) The lsim command also plots the input (shown with thin line in Figure 7.23) alongside the thick line output.

FIGURE 7.23 MATLAB Time Response With Nonzero Initial Conditions for Arbitrary Input.

361

362

CHAPTER 7 Transfer Function Approach

t

u(t)

y(t)

FIGURE 7.24 Simulink Block Diagram to Plot Dynamic System Response With Transfer Function Block. (ii) Combining Eqs. (7.102) and (7.103) results in the following standard transfer function:

Geq ðsÞ ¼

5s5 þ 6s4 þ 253:8s3 þ 201:2s2 þ 3200s þ 1259 5s5 þ 4s4 þ 130:2s3 þ 52:56s2 þ 105:5s þ 20:3

(7.104)

Figure 7.24 shows the block diagram allowing to plot the response of the system that is represented by the transfer function of Eq. (7.104) under the input u(t) ¼ e0.2$t$cos(5$t). This input is formulated by means of the Ramp block (which is in the Source library), which has a Slope of 1 and models the time independent variable. The time is fed into a Function block, which is taken from the User-Defined Functions library. Under Expression the following input is used: cos(5*u(1))*exp (0.2*u(1)) where u(1) is the first value of the time vector, which is interpreted here as the vector u. This function is directed as the input to the Transfer Function block that is taken from the Continuous library. The numerator and denominator coefficients of the transfer function have to be specified in the Function Block Parameters as follows: Numerator coefficient: [5,6,253.8,201.2,3200,1259] Denominator coefficient: [5,4,130.2,52.56,105.5,20.3] A simulation time of 25 s is selected in the Stop time of the Configuration Parameters, and then the model is rundthe simulation results in the plots shown in Figure 7.23. nn

7.4.2 MIMO SYSTEMS The forced response of MIMO systems is studied in this section using the analytical approach, MATLAB and Simulink.

Analytical Approach The forced response of MIMO systems is analyzed by considering the general situation where both forcing and nonzero initial conditions are applied to the system. Similar to SISO systems, an equivalent s-domain equivalent input vector {Ueq(s)} can be formed that combines the actual s-domain input vector {U(s)} with a vector comprising the nonzero initial conditions. Given the transfer function matrix [G(s)] has been determined, the output vector {Y(s)} is calculated as: {Y(s)} ¼ [G(s)]$ {Ueq(s)}. The time-domain output vector {y(t)} results by inverse Laplace transforming {Y(s)} as: fyðtÞg ¼ L1 ½fYðsÞg ¼ L1 ½½GðsÞ $ fUeq ðsÞg

(7.105)

7.4 Transfer Function and the Time Response

The initial value of the time-domain output can be calculated based on the initial value theorem as fyð0Þg ¼ lim fyðtÞg ¼ lim s $ fYðsÞg ¼ lim s $ ½GðsÞ $ fUeq ðsÞg t/0

s/N

s/N

(7.106)

The final value of the time-domain response is determined by using the final value theorem, namely fyðNÞg ¼ lim fyðtÞg ¼ lim s $ fYðsÞg ¼ lim s $ ½GðsÞ $ fUeq ðsÞg t/N

s/0

s/0

(7.107)

Let us evaluate the equivalent input vector for arbitrary first- and second-order systems. The differential equation of a first-order system is as follows: _ ½a1  $ fyðtÞg þ ½a0  $ fyðtÞg ¼ ½b $ fu ðtÞg ¼ fuðtÞg

(7.108)

with the initial condition: {y(0)} s {0}. The matrices [a1] and [a0] are square matrices of dimension m (m being the number of outputs in {y}), whereas the matrix [b] is of dimension m  p, with p being the number of components in the input vector {u}. As discussed in Section 7.2.1, the modified input vector {u(t)} ¼ [b]$ {u*(t)} has now the same dimension m  1 with the output vector {y(t)}. Laplace transforming Eq. (7.108) results in fYðsÞg ¼ ð½a1  $ s þ ½a0 Þ1 $ ðfUðsÞg þ ½a1  $ fyð0ÞgÞ or fYðsÞg ¼ ½GðsÞ $ fUeq ðsÞg with ½GðsÞ ¼ ð½a1  $ s þ ½a0 Þ1 ; fUeq ðsÞg ¼ fUðsÞg þ ½a1  $ fyð0Þg

(7.109)

The mathematical model of a second-order system is as follows: € _ þ ½a1  $ fyðtÞg þ ½a0  $ fyðtÞg ¼ ½b $ fu ðtÞg ¼ fuðtÞg ½a2  $ fyðtÞg

(7.110)

_ with the initial conditions: fyð0Þgs0; {y(0)} s 0. The s-domain inputeoutput relationship results from applying the Laplace transform to Eq. (7.110), namely
1 _ $ ðfUðsÞg þ ½a2  $ ðfyð0Þg $ s þ fyð0ÞgÞ fYðsÞg ¼ ½a2  $ s2 þ ½a1  $ s þ ½a0  þ ½a1  $ fyð0ÞgÞ or fYðsÞg ¼ ½GðsÞ $ fUeq ðsÞg with 1
½GðsÞ ¼ ½a2  $ s2 þ ½a1  $ s þ ½a0  ; _ þ ½a1  $ fyð0Þg fUeq ðsÞg ¼ fUðsÞg þ ½a2  $ ðfyð0Þg $ s þ fyð0ÞgÞ (7.111) nn

Example 7.16

The two-tank liquid system of Figure 7.4 analyzed in Example 7.3 is defined by C1 ¼ 20 m5/N, C2 ¼ 16 m5/N, R1 ¼ 2 N s/m5, and R2 ¼ 1.2 N s/m5. Knowing the initial pressures are p1(0) ¼ 3$105 N/m2, p2(0) ¼ 1$105 N/m2 and that the input volume flow rate is qi ¼ 0.01 m3/s, use the transfer function matrix determined in Example 7.3 to find the tank pressures p1(t) and p2(t). Also determine the steady-state pressure values.

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CHAPTER 7 Transfer Function Approach

Solution This is a first-order system, and the transfer function matrix [G(s)] was derived in Eq. (7.26). The Laplace-domain system response is found from Eqs. (7.105) and (7.109):

  fPðsÞg ¼ ½GðsÞ $ Qeq ðsÞ ¼ ½GðsÞ $ ðfQðsÞg þ ½a1  $ fpð0ÞgÞ

with

 fQðsÞg ¼

Qi ðsÞ 0



; ½a1  ¼

C1

0

0

C2



 ; fpð0Þg ¼

p1 ð0Þ

(7.112)



p2 ð0Þ

(7.113)

where {Q(s)} and [a1] are given in Eqs. (7.24) and (7.25), and [G(s)] is provided in Eq. (7.26). Combining Eqs. (7.26), (7.112), and (7.113), the s-domain pressures are calculated as:



P1 ðsÞ ¼ G11 ðsÞ $ ðQi ðsÞ þ C1 $ p1 ð0ÞÞ þ G12 ðsÞ $ C2 $ p2 ð0Þ P2 ðsÞ ¼ G12 ðsÞ $ ðQi ðsÞ þ C1 $ p1 ð0ÞÞ þ G22 ðsÞ $ C2 $ p2 ð0Þ

(7.114)

where Qi(s) ¼ qi/s. The final (steady-state) values of the pressures are calculated using the actual scalar transfer functions of Eq. (7.26):

 p1 ðNÞ ¼ lim p1 ðtÞ ¼ lim s $ P1 ðsÞ ¼ qi $ ðR1 þ R2 Þ ¼ 0:032 N m2 ; t/N s/0  p2 ðNÞ ¼ lim p2 ðtÞ ¼ lim s $ P2 ðsÞ ¼ qi $ R2 ¼ 0:012 N m2 t/N

(7.115)

s/0

The time-dependent pressures are found by inverse Laplace transforming Eq. (7.114) as

(

p1 ðtÞ ¼ 3 $ 105 $ e0:054 $ t $ ½coshð0:04 $ tÞ þ 0:928 $ sinhð0:04 $ tÞ þ 0:032 p2 ðtÞ ¼ 1 $ 105 $ e0:054 $ t $ ½coshð0:04 $ tÞ þ 1:6 $ sinhð0:04 $ tÞ þ 0:012 (7.116)

and they are plotted against time in Figure 7.25.

Pressure (N/m2)

364

p1

p2

Time (sec)

FIGURE 7.25 Time Variation of Tank Pressures. nn

7.4 Transfer Function and the Time Response

MATLAB and Simulink Approach By using the tf function, it is possible to model and determine the time response of MIMO systems with MATLAB. MIMO systems can also be modeled with Simulink by using transfer function matrices. nn

Example 7.17

Determine the system response produced by the input u(t) ¼ {8 sin(5t) 10/(tþ 1)}T and the transfer 2 3 2 4s þ 20   6 2 7 2 G11 ðsÞ G12 ðsÞ 6 s þ 2s þ 100 s þ 2s þ 100 7 function matrix ½GðsÞ ¼ ¼6 7 by using 4 5 G21 ðsÞ G22 ðsÞ 1 2s  2 2 s þ 2s þ 100 s þ 2s þ 100 MATLAB and its tf function.

Solution

The transfer function matrix is of a 2  2 dimension, so both the input and output vectors have two components. The s-domain output is calculated multiplying the transfer function matrix and the s-domain input as follows:

(

(

Y1 ðsÞ Y2 ðsÞ

)

"

¼

G11 ðsÞ

G12 ðsÞ

G21 ðsÞ

G22 ðsÞ

# ( $

U1 ðsÞ U2 ðsÞ

)

or

Y1 ðsÞ ¼ Y11 ðsÞ þ Y12 ðsÞ ¼ G11 ðsÞ $ U1 ðsÞ þ G12 ðsÞ $ U2 ðsÞ Y2 ðsÞ ¼ Y21 ðsÞ þ Y22 ðsÞ ¼ G21 ðsÞ $ U1 ðsÞ þ G22 ðsÞ $ U2 ðsÞ

The following MATLAB sequence >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >>

g11 = tf([2],[1,2,100]); g12 = tf([4,20],[1,2,100]); g21 = tf([-1],[1,2,100]); g22 = tf([-2,0],[1,2,100]); t = 0:0.01:10; u1 = 8*sin(5*t); u2 = 10./(t+1); [y11,t] = lsim(g11,u1,t); [y12,t] = lsim(g12,u2,t); [y21,t] = lsim(g21,u1,t); [y22,t] = lsim(g22,u2,t); y1 = y11+y12; y2 = y21+y22; subplot(2,1,1); plot(t,y1) subplot(2,1,2); plot(t,y2)

produces the plots of Figure 7.26.

(7.117)

365

y1

CHAPTER 7 Transfer Function Approach

y2

366

Time (sec)

FIGURE 7.26 Response Curves for a Two-Input, Two-Output Dynamic System. nn

nn

Example 7.18 A MIMO system is defined by the transfer function matrix

3 1 2 1 2 2 6 2s þ s þ 1 2s þ s þ 1 2s þ s þ 1 7 7. ½GðsÞ ¼ 6 5 4 1 3 0 2s2 þ s þ 1 2s2 þ s þ 1 2

2

Use Simulink to determine and plot the time response of this system under the action of the input fuðt Þg ¼ f 2 1 sinðt Þ gT .

Solution The number of output components is two, as indicated by the number of rows of the transfer function, which connects the input to the output Laplace-transformed vectors according to {Y(s)} ¼ [G(s)]$ {U(s)}. Simulink has the capability of modeling MIMO systems and transfer function matrices by means of the LTI System block, which is in the Control System Toolbox library. The given matrix has to be introduced in the Block Parameters window as [tf(1,[2,1,1]), tf(2,[2,1,1]), tf(1,[2,1,1]); tf(0,[2,1,1]), tf(1,[2,1,1]), tf(3,[2,1,1])] Because the LTI System block accepts formally one input and generates one output, the Mux block (taken from the Signal Routing library and shown in the Simulink diagram of Figure 7.27) has to be used to model the three input components. This block, as the name suggests, combines several input signals into a single output signal; in our case, three inputs need to be specified under Parameters in the Function Block: Mux. The two resulting time-domain outputs, y1 and y2, can be plotted separately through scopes using a Demux block (taken from the Signal Routing library). The Demux block is the opposite of the Mux block as it decomposes an input signal into several output signals. For this example, we need to specify two outputs under Parameters in the Function Block: Demux. Alternately, the two output signals can be exported into MATLAB through the two To Workspace blocks of Figure 7.27. The resulting plots shown in Figure 7.28 are drawn with a plot command in MATLAB.

Summary

u1(t)

y1(t)

u2(t)

y2(t) u3(t)

FIGURE 7.27 Simulink Transfer Function Matrix Model of a MIMO System With Three Inputs and Two Outputs.

y

y

Time (sec)

FIGURE 7.28 MATLAB Time Plot of the Simulink Outputs. nn

SUMMARY This chapter introduces the transfer function, which enables connecting the input to the output of a dynamic system into the Laplace domain for SISO and MIMO systems. It is shown how to derive the transfer function for SISO systems and the transfer function matrix for MIMO systems by starting from the system time-domain mathematical model, by conversion from zero-pole-gain models, or using complex impedances. The transfer function is further utilized to study the system stability and to determine the free and forced time responses with nonzero initial conditions of dynamic systems. Examples that apply built-in MATLAB commands and Simulink to model and solve dynamic systems problems by means of the transfer function are discussed. The concept of transfer function is used to study the state-space approach (Chapter 8), the frequency-domain analysis (Chapter 9), and to introduce the control systems (Chapters 11e13).

367

368

CHAPTER 7 Transfer Function Approach

PROBLEMS 7.1 Calculate the transfer function of a dynamic system whose output y(t) and input u(t) are connected as follows: € þ 0:2yðtÞ _ þ 100yðtÞ ¼ 6uðtÞ € þ uðtÞ _ þ 3uðtÞ 2yðtÞ 7.2 The output is y(t) in the following differential equations. Express the righthand side of these equations in terms of an input u(t) and its time derivatives. Also determine the corresponding transfer functions: € þ yðtÞ _ þ 30yðtÞ ¼ 4 $ cosð30tÞ  sinð30tÞ ðiÞ yðtÞ € þ a1 $ yðtÞ _ þ a0 $ yðtÞ ¼ t2 $ e2t þ e2t ðiiÞ a3 $ 0yðtÞ þ a2 $ yðtÞ 7.3 A two-input, two-output dynamic system is defined by the following differential equations system:  20 y1 ðtÞ þ y€1 ðtÞ þ 30y_1 ðtÞ þ 3y_2 ðtÞ þ 80y1 ðtÞ  60y2 ðtÞ ¼ u1 ðtÞ . y€2 ðtÞ þ 7y_2 ðtÞ  5y_1 ðtÞ þ 60y2 ðtÞ  60y1 ðtÞ ¼ u2 ðtÞ Determine its transfer function matrix considering that the input is {u1(t) u2(t)}T and the output is {y1(t) y2(t)}T. 

7.4 A dynamic system is modeled as:  



   5 0 120 1 2 y€1 ðtÞ y_1 ðtÞ $ þ þ $ y€2 ðtÞ y_2 ðtÞ 0 5 80 6 3

 



150 y1 ðtÞ u1 ðtÞ $ ¼ . 200 y2 ðtÞ u2 ðtÞ

Calculate the transfer function matrix connecting the input {u1(t) u2(t)}T to the output {y1(t) y2(t)}T. 7.5 Determine the transfer function Q1(s)/M(s) for the shaft-gear mechanical system of Figure 7.29, where Q1(s) and M(s) are the Laplace transforms of the angle q1(t) and of the moment m(t). Use the time-domain mathematical model of this system. Known are J1, k1, J2, c, k2, N1, and N2.

k2

N2 c

J2

θ2, m θ1

k1

J1 N1

FIGURE 7.29 Rotary Mechanical Shaft-Gear System.

Problems

7.6 Apply the node analysis method in the time domain to determine the transfer function matrix between the input source current and the relevant output voltages for the electrical circuit of Figure 7.30. Consider known R1, R2, L, and C. R2

i

R1

C

L

FIGURE 7.30 Electrical Circuit With Current Source.

7.7 A dynamic system has 0 as its zero; 1 (order three of multiplicity), 2 (order two of multiplicity) as poles, and a gain of 14. Use MATLAB to calculate the corresponding transfer function of this system then reconvert the obtained transfer function model into a zero-pole-gain model. 7.8 The following factorized function models a dynamic system: 20$(s  1)3$ (s  10)/[s2$(s  2)$(s  5)2]. Find the corresponding transfer function expressed as a ratio of 2 s polynomials. 7.9 It was determined that the poles of a MIMO system with two inputs and two outputs are 9, 7 (double), and 4. The zeroes and gains obtained when applying individual inputs u1 and u2 in correspondence with the individual outputs y1 and y2 are given as follows: u1 u2

y1

y2

gain: 6, zeroes: 3 (double) gain: 10, zeroes: 5, 1

gain: 1, zeroes: none gain: 2, zeroes: 2

Use MATLAB to determine the corresponding transfer function matrix of this system. 7.10 Using the complex impedance approach, find the transfer function Vo(s)/Vi(s) of the op-amp system shown in Figure 7.31(a) and demonstrate that the system is a noninverting amplifier. Apply the result to calculate the transfer function of the system of Figure 7.31(b) in terms of R1, R2, L, and C. 7.11 Using complex impedances, solve Example 7.1. 7.12 Use the complex impedance approach to formulate the transfer function matrix for the electrical circuit of Figure 7.32, considering that the input components are the voltages v1 and v2, and the output is formed of the meaningful currents. Known are R1, R2, L, and C.

369

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CHAPTER 7 Transfer Function Approach

+ –

+ – Vi

Z2

L

vi

Vo

Z1

R1

C

(a)

R2

vo

(b)

FIGURE 7.31 Noninverting Operational Amplifier: (a) Generic Impedance Model; (b) Actual Circuit. R2

i2

R1

+

i3 L

C

v2

–

i1

–

v1

+

FIGURE 7.32 Two-Mesh Electrical System With Voltage Sources.

7.13 A thermometer defined by a thermal capacitance C and a thermal resistance R is placed in a bath whose temperature is qb(t). Use complex impedances to determine the transfer function Q(s)/Qb(s), where Q(s) and Qb(s) are Laplace transforms of the thermometer and bath temperatures. 7.14 Derive the transfer function matrix of the thermal system shown in Figure 5.23 of Example 5.15 using the element properties C, R, Ro. Consider the input consists of the cooler thermal flow rate and the outdoors temperature while the output is formed of the two rooms’ temperatures. 7.15 The tank-pipe liquid system of Figure 7.33 is defined by a resistance Rl and a capacitance Cl. Use complex impedances to determine the transfer functions

Cl h po

Rl

pi q

FIGURE 7.33 Liquid System With Tank and Pipe.

Problems

Po(s)/Pi(s) and H(s)/Pi(s) with Po(s) and Pi(s) being the Laplace transforms of the output (tank) and input pressures, and H(s), being the Laplace transform of the head h(t). 7.16 Consider that the input to the tank-pipe liquid system of Figure 7.34 is formed of a source pressure p1, the atmospheric pressure pa, and the input flow rate qi, whereas the output is the output flow rate qo. Derive the corresponding transfer function matrix by means of complex impedances considering that known are the element properties Cl and Rl.

qi pa Tank h p1

Pipe Rl

Cl

Pressure source

pa

qo

FIGURE 7.34 Liquid System With Pressure Source, Tank, and Pipe.

7.17 Derive the transfer function matrix for the liquid-level system of Example 5.8 and depicted in Figure 5.13. The input is formed of the pump pressure p, the atmospheric pressure pa, and the input volume flow rate q, whereas the output consists of the two tanks pressures p1 and p2. Known are the system component properties: Cp1, Cp2, Rp1, Rp2, and Rp3. 7.18 Use the complex impedance approach and determine the transfer matrix for the pneumatic system sketched in Figure 7.35. The system input is the pressure pi, and the output is represented by the containers pressures p1, p2, and p3. Known are: Cg1, Cg2, Cg3, Rg1, Rg2, and Rg3.

pi

Rg1

Rg2 p1

q1

Rg3 p3

p2 Cg1

q2

Cg2

FIGURE 7.35 Pneumatic System With Three Containers and Resistive Duct.

q3

Cg3

371

372

CHAPTER 7 Transfer Function Approach

7.19 Consider the torsional mechanical microsystem shown in Figure 7.36, formed of three identical flexible bars and two identical rigid plates. Use a lumped-parameter model to derive the time-domain mathematical model of this system and determine the transfer function matrix connecting the input, which is formed of the moments mt1 and mt2, to the output, which consists of the rotation angle q1 and q2. Confirm the obtained result by deriving the impedance-based transfer function matrix. Consider that known are the bars’ torsional stiffness k and the plates mass moment of inertia J. θ1

θ2 Flexible bar

Fixed end

mt1

mt2

Rigid plate

Fixed end

FIGURE 7.36 Torsional Mechanical Microsystem.

7.20 Use element impedances and an impedance schematic to determine the transfer function G(s) ¼ Y(s)/U(s) for the mechanical system of Figure 7.37. Y(s) and U(s) are the Laplace transforms of the displacements y(t) and u(t). Consider known all element constant properties. y

c1 k2

k3

c2

u

m

k1

FIGURE 7.37 Lumped-Parameter Model of Translatory Mechanical System.

7.21 For the mechanical system of Figure 7.38 the parameters m, c, and k are unknown. It is determined experimentally that the logarithmic decrement is 0.05, and that the period of the free damped oscillations is 0.1 s. When

y k

k m

c

FIGURE 7.38 Single-DOF Mechanical System.

Problems

applying a unit force of 1 N to the mass, the steady-state displacement is 0.01 m. Identify the mechanical system’s parameters m, c, and k by using the transfer function approach. 7.22 For the mechanical system of Figure 7.39, use complex impedances to derive the transfer function matrix connecting the input forces u1 and u2 to the output displacements y1 and y2 of the two point masses. Known are m, c, and k. y1

k

y2

k

k u1

m

u2

m

c

c

FIGURE 7.39 Mechanical System With Two Inputs and Two Outputs.

7.23 Study the stability of the inverted pendulum of Example 7.10 and shown in Figure 7.18 when the rigid rod has a mass mr. 7.24 The transfer function of a dynamic system is G(s) ¼ 1/(s6 þ 4s5 þ 10s4 þ 22s3 þ 25s2 þ 10s). Analyze the stability of this system. 7.25 The transfer function of a dynamic system is G(s) ¼ 1/(10s3 þ p$s2 þ 2s þ 49). Analyze the stability of this system in terms of the real parameter p. 7.26 A single-mesh electrical system is formed of a capacitor C ¼ 10 mF, an inductor L, and a source voltage v ¼ b$t volts. Use the transfer function approach, with G(s) ¼ I(s)/V(s) where I(s) and V(s) are the Laplace transforms of the current and source voltage to determine the constant b that will produce a steady-state current of 5 mA. 7.27 A first-order pneumatic system, as the one of Figure 7.19, is formed of a container of capacitance C connected to a pressure source pi through a pipe segment of resistance R. The input pressure is pi(t) ¼ b$t, where b is a constant. After a time Dt ¼ 3 s, it is determined that the output (container) pressure po is 1.8 times larger than the input pressure constant b. Use the transfer function approach to determine the system’s time constant. 7.28 Consider the heat transfer between the conditioned space of a room with four walls and the outdoors. The room capacitance is C, and the total wall resistance is R. The indoor temperature qi is smaller than the outdoors temperature qo. (i) Find the transfer function corresponding to the indoor and outdoor temperatures qi and qo considering that qi is the output.

373

374

CHAPTER 7 Transfer Function Approach

(ii) Determine qi and plot it as a function of time when qo ¼ 38 C. Known also are the following data for the wall: height h ¼ 3.6 m, width w ¼ 12 m, thickness l ¼ 0.18 m, thermal conductivity k ¼ 0.035 W/(m deg), and the air in the room space: mass density r ¼ 1.12 kg/m3 and constant-pressure specific heat cp ¼ 1150 J/(kg deg). The initial room temperature is qi(0) ¼ 21 C. 7.29 Consider the mechanical system shown in Figure 7.40 with c1 ¼ 0.6 N s/m, c2 ¼ 0.7 N s/m, c3 ¼ 053 N s/m, k1 ¼ 110 N/m, k2 ¼ 160 N/m, and k3 ¼ 120 N/m. Use the complex impedance approach to find the equivalent complex impedance of all the elements between the fixed left wall and the free right end. Assuming a force f is applied at the free end that produces a displacement y at the same point, determine the transfer function Y(s)/ F(s). Plot the chain total displacement when the force is f ¼ 120$sin(65t) N. k1

k2

k3

c2

y f

c3

c1

FIGURE 7.40 Spring-Damper Mechanical System.

7.30 For the electrical circuit of Figure 7.41, determine the following: (i) The transfer matrix connecting the input currents i1, i2 to the meaningful voltages; (ii) The expressions of the (output) meaningful voltages as functions of time; also plot those voltages in terms of time. Known are i1 ¼ 15 mA, i2 ¼ 12 mA, R1 ¼ 300 kU, R2 ¼ 400 kU, and C ¼ 2 mF. R2

i1

R1

C

i2

FIGURE 7.41 Electrical Circuit With Two Current Sources.

7.31 A rotor system such as the one shown in Figure 7.42 consists of a disk with the mass moment of inertia J ¼ 0.012 kg m2, a damper with viscous damping coefficient c ¼ 8 N m s, and a rotary spring of unknown stiffness k. When applying a step input m ¼ 30 N m to the disk, it is determined experimentally that the steady-state response is qo(N) ¼ 5 . Evaluate the

Problems

m, θ

k J

c

FIGURE 7.42 Rotor System.

spring constant using the transfer function approach and plot the system response for a rotor initial velocity u(0) ¼ 50 rad/s and the step input of MATLAB. 7.32 An elastic shaft that connects two rotor disks and is supported on two bearings at its ends is sketched in Figure 7.43. Use the complex impedance approach to determine qo(0), qo(N)dthe initial and final value of the output angle. Consider the input angle is qi, and use the transfer function Qo(s)/Qi(s). θi

θo

k

J1

J2

c1

c2

FIGURE 7.43 Elastic Shaft With Inertia and Damping in Schematic Representation.

7.33 Apply the complex impedance approach to determine the transfer matrix connecting the Laplace transforms of the input moment m to the meaningful rotation angle outputs for the mechanical system of Figure 7.44. Find those angles explicitly for m ¼ 50 N m, c1 ¼ 50 N m s, c2 ¼ 65 N m s, k1 ¼ 200 N m, and k2 ¼ 260 N m. The initial conditions are zero. c1

m

k1

k2

c2

FIGURE 7.44 Serial Spring-Damper Rotary Mechanical System.

7.34 Use the complex impedance approach to find the transfer function of the op-amp circuit of Figure 7.45 and use it to plot the output voltage vo as a function of time for the following numerical values: vi ¼ 60 V, R ¼ 1.1 kU, C ¼ 10 mF, and L ¼ 200 mH. Also calculate the initial and final values of the output voltage vo.

375

376

CHAPTER 7 Transfer Function Approach

C R

R

L

– +

vi

vo

FIGURE 7.45 Operational Amplifier Circuit.

7.35 The MEMS of Figure 7.46 is formed of two pairs of beams and a shuttle mass and is subjected to capacitive actuation. Use a lumped-parameter model to formulate the system’s mathematical model for the motion about the y direction by also taking into account the damping between the fixed and mobile capacitor plates. All beams have a circular cross section of diameter d ¼ 1.8 mm. Known also are the shuttle mass m ¼ 8$1011 kg, l1 ¼ 180 mm, l2 ¼ 90 mm, Young’s modulus E ¼ 160 GPa, and the damping coefficient c ¼ 0.4 N s/m. Plot the system’s response using its transfer function and MATLAB’s impulse function when an impulsive electrostatic force f ¼ 1.2$106 d(t) N is applied by the capacitive unit superimposed to an initial displacement y(0) ¼ 1.4 mm. The transfer function is G(s) ¼ Y(s)/ F(s). Ignore the mass contribution from the springs and verify whether the initial capacitive gap g0 ¼ 10 mm is sufficient for the mobile plate motion. y

Anchor

Capacitive actuation Mobile plate

g0 l1

l2

m

l1

Anchor

l2

FIGURE 7.46 Physical Model of Translatory Mechanical Microsystem With Capacitive Actuation.

7.36 By using the transfer function Q1(s)/M(s), determined in Problem 7.5 for the rotary shaft-gear mechanical system shown in Figure 7.29, calculate and plot q1(t) using the step input command of MATLAB. Known are N1 ¼ 48, N2 ¼ 36, J1 ¼ 0.0018 kg m2, J2 ¼ 0.0012 kg m2, c ¼ 70 N m s, k1 ¼ 165 N m, k2 ¼ 100 N m, m ¼ 18 N m, and q2(0) ¼ 4 .

Problems

7.37 The liquid system of Problem 7.15 and shown in Figure 7.33 carries water with r ¼ 1000 kg/m3 and m ¼ 0.001 N s/m2. The pipe diameter is di ¼ 0.08 m, its total length is l ¼ 26 m; the tank has a diameter d ¼ 4 m and an initial liquid head h(0) ¼ 1.3 m. The input pressure is a function of time, pi ¼ (1 þ e10t)$P, where P ¼ 3.4$105 N/m2. Use the transfer function Po(s)/Pi(s) with MATLAB and its lsim input command to plot the tank pressure po in terms of time. Also calculate the steady-state value of the tank pressure. 7.38 Consider the following transfer function matrix: 2 3 sþ2 1 6 s2 þ 2s þ 16 s2 þ 2s þ 16 7 6 7 ½GðsÞ ¼ 6 7. 4 5 2 sþ1 2 2 s þ 2s þ 16 s þ 2s þ 16 Determine the system response under the following inputs: u1(t) ¼ 2 and u2(t) ¼ 3et using MATLAB and its impulse command. 7.39 For the mechanical system of Figure 7.47, calculate the components of a forcing vector that will render this free system under the initial conditions _ ¼ f 0:8 1 gT ; fyð0Þg ¼ f 0:5 0:3 gT into a forced system with fyð0Þg zero initial conditions; plot y1(t) and y2(t) for m ¼ 0.4 kg, c ¼ 1.8 N s/m, and k ¼ 50 N/m. Also determine the final values of the two displacements. y2

y1

k

k

k m

m

c

c

FIGURE 7.47 Free Mechanical System With Nonzero Initial Conditions.

7.40 Use complex impedances and the transfer function approach for the circuit of Figure 7.48 to determine and plot the output voltage for R1 ¼ 620 U, R2 ¼ 540 U, L ¼ 280 mH, C ¼ 50 mF, v1 ¼ 65 V, and v2 ¼ 30 V. L v1

R1

– +

v2 R2

C

FIGURE 7.48 Operational Amplifier Circuit With Two Input Voltages.

vo

377

378

CHAPTER 7 Transfer Function Approach

7.41 Plot the output voltage of the op-amp circuit of Problem 7.10 and shown in Figure 7.31(b) using the transfer function Vo(s)/Vi(s) and Simulink for C ¼ 1.4 mF, L ¼ 0.06 H, R1 ¼ 110 U, R2 ¼ 90 U, and vi ¼ 120 V. The resistors are nonlinear with their saturation limits being þ80 V and 60 V. 7.42 Utilize the transfer function approach and Simulink to plot the output flow rate of the liquid system shown in Figure 7.34 of Problem 7.16. Known are pa ¼ 105 N/m2, p1 ¼ 2.2$105 N/m2, qi ¼ 0.03 m3/s, d ¼ 1.2 m (diameter of the tank), r ¼ 1000 kg/m3, di ¼ 0.022 m (diameter of the pipe), l ¼ 18 m (length of pipe), m ¼ 0.001 N s/m2. 7.43 Use Simulink to plot the displacements y1(t) and y2(t) of the mechanical system shown in Figure 7.39 of Problem 7.22 based on the transfer function matrix derived in that problem. Known are m ¼ 1.2 kg, c ¼ 14 N s/m, and k ¼ 180 N/m. The input forces are u1 ¼ 50$e3t$sin(16t) N and u2 ¼ 60/(t þ 3) N. The initial conditions are zero.

Suggested Reading H. Klee and R. Allen, Simulation of Dynamic Systems with MATLAB and Simulink, 2nd Ed., CRC Press, Boca Raton, FL, 2011. N. S. Nise, Control System Engineering, 7th Ed. John Wiley & Sons, New York, 2015. D. G. Alciatore and M. B. Histand, Introduction to Mechatronics and Measurement Systems, 4th Ed. McGraw-Hill, New York, 2011. S. B. Hammond and D. K. Gehmlich, Electrical Engineering, McGraw-Hill, New York, 1970. W. T. Thomson, Laplace Transformation, 2nd Ed. New York, Prentice-Hall, 1960. I. Cochin, Analysis and Design of Dynamic Systems, Harper & Row Publishers, New York, 1980. K. Ogata, System Dynamics, 4th Ed. Prentice Hall, Upper Saddle River, 2004. W. J. Palm III, System Dynamics, 3rd Ed. McGraw-Hill, New York, 2013. J. F. Lindsay and S. Katz, Dynamics of Physical Circuits and Systems, Matrix Publishers, Champaign, IL, 1978. W. C. Sperry, Analysis of Dynamic Systems Using the Mechanical Impedance Concept, Noise Control, March/April 1961, pp. 13e21. J. A. Gatscher and G. Kawiecki, Comparison of Mechanical Impedance Methods for Vibration Simulation, Shock and Vibration, Vol. 3 (3), 1995, pp. 223e232.

CHAPTER

State-Space Modeling

8

OBJECTIVES This chapter focuses on the state-space approach to modeling dynamic systems in the time domain and determining the corresponding solution. The subject matter of the chapter is related to the modeling of Chapters 2e5, as well as with the Laplace transform of Chapter 6 and the transfer function of Chapter 7. The following topics are studied: • The state-space concept as a matrix procedure for rendering the time-domain dynamic models of SISO (single-input, single-output) and MIMO (multiple-input, multiple-output) systems into first-order differential equations and for obtaining solutions to the corresponding models. • Deriving state-space models directly in the time domain or via transfer functions in the Laplace domain when the input includes time derivatives in it. • Analytical and MATLAB methods of converting between transfer function or zero-pole-gain models and state-space models. • Nonlinear state-space models and procedures for linearizing such models. • The use of state-space models to solve for the free response with nonzero initial conditions and for the forced response of dynamic systems. • Specialized MATLAB commands designed to model and solve state-space problems. • Application of Simulink to graphically model and solve state-space system dynamics problems.

INTRODUCTION This chapter introduces the state-space modeling method for SISO and MIMO dynamic systems into the time domain. The approach is a matrix method of converting large-order differential equations into an equivalent number of first-order differential equations. The state-space methodology is able to model systems with a large number of degrees of freedom, as well as systems with nonlinearities. Similar to the transfer function approach, which is the subject of Chapter 7, the material presented here focuses on deriving state-space models of dynamic systems and solving these models to determine the time response using analytical methods, MATLAB custom commands, or Simulink. While the transfer function model belongs to the System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00008-7 Copyright © 2018 Elsevier Inc. All rights reserved.

379

380

CHAPTER 8 State-Space Modeling

Laplace domain, an state-space model operates in the time domain. The state-space approach utilizes the same matrix model for both SISO and MIMO dynamic systems. Conversions between state-space and transfer function or zero-pole-gain models are also studied here.

8.1 THE CONCEPT AND MODEL OF THE STATE-SPACE APPROACH The state-space approach employs a unique matrix formulation to model a dynamic system and to evaluate the forced response of SISO and MIMO systems. This model is derived from the particular mathematical method, which converts generally higherorder differential equations into an equivalent number of first-order differential equations, which are easier to solve. Consider, for instance, the following linear third-order differential equation with constant coefficients, the unknown y(t), and the input (forcing) u(t): € þ a1 $ yðtÞ _ þ a0 $ yðtÞ ¼ uðtÞ a3 $ 0 yðtÞ þ a2 $ yðtÞ a0 a1 a2 1 _  $ yðtÞ € þ $ uðtÞ or 0yðtÞ ¼  $ yðtÞ  $ yðtÞ a3 a3 a3 a3

(8.1)

This equation can be converted into three first-order differential equations by using the following new variables: _ € x1 ðtÞ ¼ yðtÞ; x2 ðtÞ ¼ yðtÞ; x3 ðtÞ ¼ yðtÞ

(8.2)

The following three equations are obtained from Eqs. (8.1) and (8.2): x_1 ðtÞ ¼ x2 ðtÞ x_2 ðtÞ ¼ x3 ðtÞ a0 a1 a2 1 x_3 ðtÞ ¼  $ x1 ðtÞ  $ x2 ðtÞ  $ x3 ðtÞ þ $ uðtÞ a3 a3 a3 a3

(8.3)

The three Eq. (8.3) can be written in matrix form as: 9 9 8 3 8 8 9 2 0 1 0 > > = = >

< x_1 ðtÞ = 6 7 $ uð t Þ x_2 ðtÞ ¼ 4 0 0 1 5 $ x2 ðt Þ þ 0 > > > : ; ; ; > : : x_3 ðtÞ 1=a3 x3 ðt Þ a0 =a3 a1 =a3 a2 =a3 (8.4) Eq. (8.4) allows solving for the new time-dependent variables of Eq. (8.2). The unknown of the original differential Eq. (8.1) is simply found from the first Eq. (8.2) as y(t) ¼ x1(t), but it can also be expressed in a form similar to Eq. (8.4) as: 8 9 > < x1 ðtÞ > = yðtÞ ¼ f 1 0 0 g $ x2 ðtÞ þ 0 $ uðtÞ (8.5) > > : ; x3 ðtÞ

8.1 The Concept and Model of the State-Space Approach

As we shall see shortly, Eq. (8.4) that allows solving for the new variables x1(t), x2(t), x3(t) in terms of the input u(t) together with Eq. (8.5) that yields the original output y(t) in terms of x1(t), x2(t), x3(t), and u(t) form the state-space model of the original mathematical model given in Eq. (8.1). While the example previously analyzed looked at a SISO system, it is also possible to use the same conceptual model represented by Eqs. (8.4) and (8.5) to describe the behavior of MIMO systems with a large number of outputs (or degrees of freedom (DOF)). These systems result in an equally large number of differential equations that need to be solved for these unknown outputs. Eqs. (8.4) and (8.5) are generalized as: _ ¼ ½A $ fxðtÞg þ ½B $ fuðtÞg fxðtÞg fyðtÞg ¼ ½C $ fxðtÞg þ ½D $ fuðtÞg

(8.6)

The first Eq. (8.6), which is a generalization of Eq. (8.4), is known as the state equation, and by solving it, the state vector {x(t)} is determined. The output vector {y(t)} is subsequently calculated from the output equation, which is the second Eq. (8.6)da generalization of the particular Eq. (8.5). The standard state-space model consists of these two equations. The four matrices in Eqs. (8.6) contain parameters of the dynamic system being studied; they are [A], the state matrix, [B], the input matrix, [C], the output matrix, and [D], the direct transmission matrix. Proper selection of the state variables (which are collected in state vectors forming an statespace) ensures that Eq. (8.6) can always be applied, irrespective of the number of DOF and of the system order. This topic is analyzed in the next section of this chapter. As Eq. (8.6) indicates, the input and output vectors need not have the same dimensions. Assuming the input vector has the dimension m and the output vector has the dimension p, the dimensions of the other matrices can be determined based also on the dimension of the state vector, which is n. The first Eq. (8.6) shows that [A] is square with n rows and n columns. Similarly, the matrix [B] is of n  m dimension, as it results from the same Eq. (8.6). From the second Eq. (8.6), it follows that [C] has p  n dimension, whereas the matrix [D] is of p  m dimension. A visual representation of the four matrices’ dimensions is given in Figure 8.1. State → n

Input → m

State → n

[A]

[B]

Output → p

[C]

[D]

FIGURE 8.1 Dimensions of State-Space Matrices.

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Nonuniqueness of a State-Space Model Figure 8.1 visually suggests that for any given dynamic system (defined by m inputs and p outputs) one can use different values for ndthe state variablesdand still obtain a valid state-space model. The direct inference is that there are more than one state-space model for any given dynamic system model. Moreover, two statespace models of the same system may have the same number of state variables n as argued next, and the conclusion is that state-space models are not unique. Let us assume a vector {x(t)} has been found to be a state vector, based on which a state-space model has been derived for a specific dynamic system, and consider the following transformation: fxðtÞg ¼ ½H $ fxðtÞg

(8.7)

where [H ] is a nonsingular square matrix, and fxðtÞg is another vector. Substitution of {x(t)} of Eq. (8.7) into the standard state equationdfirst Eq. (8.6)dresults in:   _ ½H $ xðtÞ ¼ ½A $ ½H $ fxðtÞg þ ½B $ fuðtÞg (8.8) Left multiplying in Eq. (8.8) by [H ]1 produces:   _ xðtÞ ¼ ½H1 $ ½A $ ½H $ fxðtÞg þ ½H1 $ ½B $ fuðtÞg

(8.9)

Similar substitution of {x(t)} from Eq. (8.7) into the generic output equation of the state-space modeldthe second Eq. (8.6)dyields: fyðtÞg ¼ ½C $ ½H $ fxðtÞg þ ½D $ fuðtÞg

(8.10)

Eqs. (8.9) and (8.10) can be rewritten as follows:       _ xðtÞ ¼ A $ fxðtÞg þ B $ fuðtÞg (8.11)   fyðtÞg ¼ C $ fxðtÞg þ ½D $ fuðtÞg       with: A ¼ ½H1 $ ½A $ ½H; B ¼ ½H1 $ ½B; C ¼ ½C $ ½H. Eq. (8.11) is of the type given in Eq. (8.6), which represents an state-space model. As a consequence, Eq. (8.11) also represents a state-space model, and since there are an infinite number of nonsingular square matrices [H], there are also an infinite number of state vectors that can be obtained from the original state vector {x(t)} according to Eq. (8.7). Therefore, an infinite number of state-space models can be derived from any specific state-space model.

Solution of the State-Space Equations There are various solution methods for solving the state equationdthe first Eq. (8.6), which is a set of first-order differential equationsdincluding the eigenvalue method (for homogeneous equations), the direct and inverse Laplace transform method, and matrix methods (such as the matrix exponential or the state-transition matrix). Section 8.3 utilizes the state-transition matrix method in conjunction with Laplace transforms to solve for the state vector {x(t)}. Laplace transforming Eq. (8.6) for zero initial conditions results in

8.1 The Concept and Model of the State-Space Approach



s $ fXðsÞg ¼ ½A $ fXðsÞg þ ½B $ fUðsÞg fYðsÞg ¼ ½C $ fXðsÞg þ ½D $ fUðsÞg

(8.12)

where {X(s)}, {U(s)}, and {Y(s)} are the Laplace transforms of {x(t)}, {u(t)}, and {y(t)}, respectively. The first Eq. (8.12) yields {X(s)} as fXðsÞg ¼ ðs $ ½I  ½AÞ1 $ ½B $ fUðsÞg

(8.13)

which, substituted into the second Eq. (8.12) gives fYðsÞg ¼ ½GðsÞ $ fUðsÞg with ½GðsÞ ¼ ½C $ ðs $ ½I  ½AÞ1 $ ½B þ ½D (8.14) [G(s)] is the transfer function matrix that is introduced in Chapter 7. As we shall see in this chapter, Eq. (8.14) also serves at converting a state-space model into a transfer function model. Once the Laplace-domain output vector is determined from Eq. (8.14), the time-domain output is simply fyðtÞg ¼ L1 ½fYðsÞg ¼ L1 ½½GðsÞ $ fUðsÞg. Let us consider a couple of examples illustrating how to derive the state-space models of dynamic systems from mathematical models. nn

Example 8.1

Derive a state-space model for a single-mesh RLC series electrical circuit formed of a resistor R, an inductor L, a capacitor C, and a voltage source v. The input is the source voltage v, and the output is the charge q.

Solution

As shown in Chapter 4, the following differential equation is the mathematical model of the RLC electrical system:

L $ q€ þ R $ q_ þ

1 1 R 1 $ q ¼ v or q€ ¼  $ q  $ q_ þ $ v C L$C L L

(8.15)

where L is the inductance, R is the resistance, C is the capacitance, q is the charge, and v is the source voltage. Let us select the following state variables to form the state vector {x}:

x1 ¼ q; x2 ¼ q;_ fxg ¼ f x1

x 2 gT

(8.16)

In Eqs. (8.15) and (8.16) the independent variable t (time) has been dropped to simplify notation, but all variables in this example (q, v, x1, and x2) are functions of time. The second form of Eq. (8.15) and the particular choice of the state variables of Eq. (8.16) indicate that

8 > < x_1 ¼ x2

1 R 1 > $ x1  $ x2 þ $ v : x_2 ¼  L$C L L

(8.17)

The source voltage is the input to the system and therefore can be denoted by u, so v ¼ u. Eq. (8.17) is collected in vector-matrix form as



x_1 x_2



2

0 ¼4 1  L$C

8 9 3   1 ; :0 > 0 C 2

(8.25)

Considering the output vector is defined by the components

y1 ¼ i; y2 ¼ vC the output equation can be written as



y1 y2



¼



1 0

    0 x1 $ þ $u x 1 0 2 0

(8.26)

(8.27)

Let us compare the state vectors of Examples 8.1 and 8.2, which are

fxg ¼ f x1 fxg ¼ f x1

x2 gT ¼ f q q_ gT ; x 2 gT ¼ f i v C gT

(8.28)

The known relationships between the electrical variables, q ¼ C $ vC ; q_ ¼ i, lead to the following transformation between the two sets of state variables:

   0 q ¼ q_ 1

  C i $ 0 vC

(8.29)

Eq. (8.29) indicates the two state vectors are linearly related, as illustrated in the generic Eq. (8.7), which showed how a state vector can be obtained by multiplying an existing state vector and a nonsingular square matrix. nn

Similar to the transfer function approach of Chapter 7, the subsequent material of this chapter is structured mainly in two segments: one is dedicated to deriving statespace models and the other one to calculating the forced response from a state-space model.

8.2 STATE-SPACE MODEL FORMULATION The state-space models can be derived from existing time-domain models or from other models, such as the zero-pole-gain and the transfer function models. Obtaining state-space models can be pursued either analytically or by means of MATLAB, as discussed in the following sections.

8.2.1 State-Space Model From the Time-Domain Mathematical Model The analytical procedures for state-space model derivation presented here apply to systems without input time derivatives, systems with input time derivatives, and nonlinear systems.

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Dynamic Systems Without Input Time Derivative When the differential equations that constitute the mathematical model of a dynamic system contain no time derivatives of the defined input, the procedure of selecting the state variables is rather straightforward, as will be discussed in the following. The cases of SISO and MIMO systems are analyzed separately.

SISO Systems SISO linear, time-invariant dynamic systems that do not include time derivatives of their input are described by the differential equation ð nÞ

ðn1Þ

_ þ a0 $ yðtÞ ¼ uðtÞ an $ y ðtÞ þ an1 $ y ðtÞ þ . þ a1 $ yðtÞ or

(8.30)

a0 a1 an1 ðn1Þ 1 _ . y ðtÞ ¼  $ yðtÞ  $ yðtÞ $ y ðtÞ þ $ uðtÞ an an an an

ð nÞ

A rule of thumb is that the number of state variables is equal to the order of the differential equation representing the mathematical model of a dynamic system, provided a0 and an of Eq. (8.30) are nonzero (the reader is encouraged to check the rationale for this assertion). This system is defined by a differential equation of order n and, therefore, requires n state variables, which can be selected as x1 ¼ y; x2 ¼ y;_ .; xn1 ¼

ðn2Þ

y ; xn ¼

ðn1Þ

y

(8.31)

The following relationships are therefore valid: x_1 ¼ x2 ; x_2 ¼ x3 ; .; x_n1 ¼ xn Eqs. (8.30) and (8.32) are collected 2 1 0 8 9 6 0 6 > > > > _ x > 1 > > 6 0 0 1 > > > 6 > < x_2 > = 6 6 . . ¼6 . . > > > x_n1 > > 6 > 6 0 0 > > > > 6 0 > : x_n > ; 6 a0 a a2 1 4   an an an

(8.32)

into the matrix equation 8 9 3 > 0 > > . 0 7 8 9 > > > > > > > x 7 > > > > 1 > > > > > > > > 0 . 0 7 > > > > > > > 7 > < > < x2 > = = > 7 > 7 . . . 7$ $ u (8.33) þ . > > > 7 > > > > > > > > > 7 xn1 > . 1 7 > > > 0 > > > > > > > > : ; > an1 7 > > 1 > > > x 5 n > > .  > : an > ; an

which represents the state equation. [A] is the square matrix multiplying the state vector, and [B] is actually a column vector multiplying the scalar u. The output equation is determined by taking into account that the differential Eq. (8.30) has one unknown, y, which can be considered the output function, but y ¼ x1. As a consequence, the first Eq. (8.31) is written as y ¼ f1

0

. 0

0 g $ f x1

x2

.

xn1

x n gT þ 0 $ u

(8.34)

and, therefore [C] is the row vector, and [D] is the zero scalar. Examples 8.1 and 8.2 actually derive state-space models of dynamic systems without input time derivatives. Let us analyze another example from the same category.

8.2 State-Space Model Formulation

nn

Example 8.3 The microresonator sketched in Figure 8.2(a) is actuated electrostatically by a comb drive. The shuttle mass is supported by two elastic beams, and there is viscous damping between the shuttle mass and the substrate. Use the lumped-parameter model sketched in Figure 8.2(b) to determine the mathematical model of this mechanical system and derive a state-space model from it. The damping coefficient is c, one beam’s spring constant is k, the shuttle mass is m, and the electrostatic actuation force is f.

f Anchor

y m

Comb drive Beam spring Beam spring

Shuttle mass

k

k c

Anchor

(a)

(b)

FIGURE 8.2 Microresonator Supported on Beam Springs, With Viscous Damping and Electrostatic Actuation: (a) Physical Model; (b) Lumped-Parameter Model.

Solution Without including the free-body diagram, it can simply be shown using Newton’s second law of motion that the equation of motion for the lumped-parameter model of Figure 8.2(b) is

m $ y€ þ c $ y_ þ 2k $ y ¼ f

(8.35)

which is a second-order differential equation with the following coefficients and forcing: a2 ¼ m, a1 ¼ c, a0 ¼ 2k, and u ¼ f. Based on these parameter assignments and according to Eqs. (8.33) and (8.34), the state-space matrices are

2

0 ½A ¼ 4 a0  a2

8 9 3 > 1 =

a1 5; ½B ¼ 1 ; ½C ¼ f 1  > ; :a > a2 2

0 g; ½D ¼ 0

(8.36)

As a consequence, the generic state Eq. (8.33) and output Eq. (8.34) become



x_1 x_2



2

0 6 ¼ 4 2k  m

3 8 9   < m1 $ y€1 ¼ f1  c1 $ y_1 y_2  k1 $ ðy1  y2 Þ 

m2 $ y€2 ¼ c1 $ y_2  y_1  k1 $ ðy2  y1 Þ  c2 $ y_2  y_3  k2 $ ðy2  y3 Þ > : m3 $ y€ ¼ f3  c2 $ y_3  y_2  k2 $ ðy3  y2 Þ 3

(8.38) The mathematical model of this mechanical system consists of three second-order differential equations. As a consequence, we need 3  2 ¼ 6 state variables, which are selected as

x1 ¼ y1 ; x2 ¼ y_1 ; x3 ¼ y2 ; x4 ¼ y_2 ; x5 ¼ y3 ; x6 ¼ y_3

(8.39)

8.2 State-Space Model Formulation

Eq. (8.39) indicates the following state variable connections:

x_1 ¼ x2 ; x_3 ¼ x4 ; x_5 ¼ x6 T

(8.40)

T

Defining the input vector as {u} ¼ {u1 u2} ¼ {f1 f3} , Eq. (8.38) can be written as

8 > k1 c1 k1 c1 1 > > y€1 ¼  $ y1  > $ y_1 þ $ y2 þ $ y_2 þ $ u1 > > m1 m1 m1 m1 m1 > > > < k1 c1 k1 þ k2 c1 þ c2 k2 c2 $ y1 þ $ y_1  $ y2  $ y_2 þ $ y3 þ $ y_3 (8.41) y€2 ¼ > m m m m m m 2 2 2 2 2 2 > > > > > k2 c2 k2 c2 1 > > > : y€3 ¼ m3 $ y2 þ m3 $ y_2  m3 $ y3  m3 $ y_3  m3 $ u2

Collecting Eqs. (8.40) and (8.41) into matrix form results in the state equation

2

6 0 6 6 k1 6 6 m 6 1 6 6 6 0 6 3 ¼6 6 k1 > x_4 > > > 6 > > > > 6 m2 > > > > 6 > > x_5 > > 6 > > > > 6 > : x_6 > ; 6 0 6 6 4 0 8 9 > > > > > x_1 > > > > > > > > > > > _ x 2 > > > > > < x_ > =

1

0

0

0

c1  m1

k1 m1

c1 m1

0

0

0

1

0

c1 m2

k1 þ k2  m2

0

0

0

0

k2 m3

c2 m3

2 6 0 6 6 1 6 6 m1 6 6 6 0 6 þ6 $ 6 > x4 > > > 6 0 > > > > 6 > > > > 6 > > > > x 6 5 > > > > 6 0 > : > ; 6 x6 6 4 0 8 9 x1 > > > > > > > > > > > > > > x2 > > > > > > > < x3 > =



c1 þ c2 m2

3 0 7 7 7 0 7 7 7 7 ( ) 0 7 u1 7 7$ 7 u2 0 7 7 7 0 7 7 7 1 7  5 m3

k2 m2 0 

k2 m3

3

0 7 7 7 0 7 7 7 7 7 0 7 7 c2 7 7 7 m2 7 7 7 1 7 7 7 c2 7  5 m3

(8.42)

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CHAPTER 8 State-Space Modeling

The matrix multiplying the state vector in Eq. (8.42) is [A] and the one multiplying the input vector is [B]. The output vector {y} ¼ {y1 y2 y3}T is connected to the state and input vectors as in the first, third, and fifth Eq. (8.39)dthey are collected in the output matrix equation:

8 9 2 1 0 > < y1 > = 6 y2 ¼ 4 0 0 > : > ; 0 0 y3

0 1

0 0

0

0

8 9 x1 > > > > > > 2 >x > > 3 > 2 > > > > 0 0 0 > =

3 6 7 þ 40 0 05$ > > x4 > > > > 0 1 0 > > > > > x5 > > > > ; : > x6

3 0   u1 7 05$ u2 0

(8.43)

The matrix multiplying the state vector in Eq. (8.43) is [C ], and the zero matrix multiplying the input vector in the same equation is [D]. nn

Dynamic Systems With Input Time Derivative In situations where the differential equations corresponding to a dynamical model include time derivatives of the defined input, the procedure used when there were no input time derivatives cannot be applied directly because a state-space model does not accommodate time derivatives of the input in its standard state and output equations. It is, however, possible to determine state-space models using an additional function or vector, as shown in the following. Only coverage of SISO systems is provided here, but the companion website Chapter 8 includes the study of MIMO systems. Let us consider a SISO dynamic system whose mathematical model consists of the following differential equation: ð nÞ

ðn1Þ

_ þ a0 $ yðtÞ an $ y ðtÞ þ an1 $ y ðtÞ þ . þ a1 $ yðtÞ ð qÞ

ðq1Þ

_ þ b0 $ uðtÞ ¼ bq $ u ðtÞ þ bq1 $ u ðtÞ þ . þ b1 $ uðtÞ

(8.44)

The state-space variable choice made for SISO systems without input time derivatives cannot be applied here, simply because, with u being defined as (or being required to be) the input, the standard state-space model cannot have additional terms with time derivatives of u on the right-hand side of Eq. (8.44). Without loss of generality, let us assume that n > q. Application of the Laplace transform with zero initial conditions to Eq. (8.44) results in the following transfer function: GðsÞ ¼

YðsÞ bq $ sq þ bq1 $ sq1 þ . þ b1 $ s þ b0 ¼ UðsÞ an $ sn þ an1 $ sn1 þ . þ a1 $ s þ a0

(8.45)

A function Z(s) is introduced in G(s) of Eq. (8.45) as GðsÞ ¼

ZðsÞ YðsÞ 1 $ ¼ UðsÞ ZðsÞ an $ sn þ an1 $ sn1 þ . þ a1 $ s þ a0 

$ bq $ sq þ bq1 $ sq1 þ . þ b1 $ s þ b0

(8.46)

8.2 State-Space Model Formulation

The following relationships can be written from Eq. (8.46): 8 ZðsÞ 1 > > > n1 < UðsÞ ¼ a $ sn þ a þ . þ a1 $ s þ a0 n n1 $ s > YðsÞ > > : ¼ bq $ sq þ bq1 $ sq1 þ . þ b1 $ s þ b0 ZðsÞ Cross multiplication in Eq. (8.47) leads to (

 an $ sn þ an1 $ sn1 þ . þ a1 $ s þ a0 $ ZðsÞ ¼ UðsÞ

 YðsÞ ¼ bq $ sq þ bq1 $ sq1 þ . þ b1 $ s þ b0 $ ZðsÞ The inverse Laplace transform is applied to Eq. (8.48), which results in 8 ð nÞ ðn1Þ < an $ z ðtÞ þ an1 $ z ðtÞ þ . þ a1 $ zðtÞ _ þ a0 $ zðtÞ ¼ uðtÞ ð qÞ ðq1Þ : yðtÞ ¼ bq $ z ðtÞ þ bq1 $ z ðtÞ þ . þ b1 zðtÞ _ þ b0 $ zðtÞ

(8.47)

(8.48)

(8.49)

The first of the two Eq. (8.49) is now an equation in z(t) that contains no time derivatives of the input u(t) and, therefore, can be modeled using the standard procedure, which has been detailed in the previous section. This system needs n state variables that are x1 ¼ z; x2 ¼ z;_ .; xn1 ¼

ðn2Þ

z ; xn ¼

ðn1Þ

z

(8.50)

Finding the state equation is done exactly as shown previously and is not detailed here. To determine the output equation, the second Eq. (8.49) is written as y ¼ b0 $ x1 þ b1 $ x2 þ . þ bq1 $ xq þ bq $ xqþ1

(8.51)

which is written in vector-matrix form as y ¼ f b0

b1

.

bq

.

0

0 g $ f x1

x2

.

xm

.

xn1

x n gT þ 0 $ u (8.52)

Note that for a SISO system with time derivatives of the input the numerator of the transfer function always contains terms in s; this problem is oftentimes referred to as numerator dynamics. nn

Example 8.5 Find a state-space representation for the electrical circuit of Figure 8.4 by using one state variable only. Consider that the input is the voltage vi and the output is the voltage vo.

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CHAPTER 8 State-Space Modeling

C vi

R

vo

FIGURE 8.4 CapacitiveeResistive Electrical System.

Solution The time-domain model can be obtained using KVL to express the input and output voltage. The reader is encouraged to check that the following differential equation describes the electrical system of Figure 8.4:

v_o þ

1 $ vo ¼ v_i or R $ C $ v_o þ vo ¼ R $ C $ v_i R$C

(8.53)

which indicates that the time derivative of vi is the input. As a consequence, the procedure described in this section needs to be utilized to derive a state-space model. The transfer function can be obtained applying the Laplace transform to Eq. (8.53) with zero initial conditions:

GðsÞ ¼

Vo ðsÞ R$C$s ¼ Vi ðsÞ R $ C $ s þ 1

(8.54)

The ratio of Eq. (8.54) is written by means of the intermediate function Z(s) as

Vo ðsÞ ZðsÞ Vo ðsÞ 1 ¼ $ ðR $ C $ sÞ ¼ $ Vi ðsÞ Vi ðsÞ ZðsÞ R$C$s þ 1

(8.55)

ZðsÞ 1 Vo ðsÞ and ¼ R$C$s ¼ Vi ðsÞ R $ C $ s þ 1 ZðsÞ

(8.56)

with

Cross multiplication in the first Eq. (8.56), followed by application of the inverse Laplace transform, yields

1 1 $z þ $ vi R $ C $ z_ þ z ¼ vi or z_ ¼  R$C R$C

(8.57)

If the state variable x ¼ z is chosen, Eq. (8.57) becomes the state equation

x_ ¼ 

1 1 $xþ $u R$C R$C

(8.58)

where the input voltage is the state-space input, u ¼ vi and A ¼ 1/(R$C), B ¼ 1/(R$C). Cross multiplication followed by inverse Laplace transformation are applied now to the second Eq. (8.56), which, combined with Eq. (8.57), yields the output equation

vo ¼ R $ C $ z_ ¼ z þ vi or y ¼ 1 $ x þ 1 $ u

(8.59)

where y ¼ vo is the output, and C ¼ 1, D ¼ 1. Eqs. (8.58) and (8.59) form a state-space model for the electrical system of Figure 8.4. nn

8.2 State-Space Model Formulation

nn

Example 8.6 Derive a state-space model for the mechanical system shown in Figure 8.5(a), where the input is the displacement u, and the output is the displacement y.

y y

u fe

k

c

fd

Displacement

(a)

(b)

FIGURE 8.5 (a) Translatory Mechanical System With Displacement Input; (b) Free-Body Diagram.

Solution Figure 8.5(b) shows the free-body diagram corresponding to the massless point defined by the coordinate y. Newton’s second law of motion corresponding to this free-body diagram results in

_ fe ¼ k $ y 0 ¼ fd  fe with fd ¼ c $ ðy_  uÞ;

(8.60)

where fd and fe are the damping and elastic forces. The following equation is obtained from Eq. (8.60):

c $ y_ þ k $ y ¼ c $ u_

(8.61)

The mechanical system is SISO, and its mathematical model consists of a first-order differential equation. As a result, a single state variable is needed. Because the time derivative of the input is present, it is necessary to apply the Laplace transform to Eq. (8.61), which results in

YðsÞ ¼ UðsÞ

s sþ

(8.62)

k c

Using the intermediate function Z(s) changes Eq. (8.62) to

YðsÞ ZðsÞ YðsÞ ¼ $ ¼ UðsÞ UðsÞ ZðsÞ

1 sþ

k c

$s

(8.63)

which indicates the following selection needs to be made

ZðsÞ ¼ UðsÞ

1 k sþ c

;

YðsÞ ¼s ZðsÞ

(8.64)

Cross multiplication in the first Eq. (8.64) and application of the inverse Laplace transform to the resulting equation yields

c $ z_ þ k $ z ¼ c $ u

(8.65)

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CHAPTER 8 State-Space Modeling

Eq. (8.65) no longer contains input derivatives; therefore, we can choose x ¼ z as the state variable. As a consequence, Eq. (8.65) becomes

k x_ ¼  $ x þ u c

(8.66)

which is the state equation with A ¼ ek/c and B ¼ 1. Cross multiplication in the second Eq. (8.64) followed by inverse Laplace transformation results in

y ¼ z_ ¼ x_

(8.67)

k y ¼  $x þ 1$u c

(8.68)

Combining Eqs. (8.66) and (8.67) yields

which is the output equation with C ¼ ek/c and D ¼ 1.

nn

Nonlinear Systems While it is almost always possible to derive a state-space model for a nonlinear dynamic system, finding a solution to it can be problematic. In such cases, the linearization technique can be utilized to derive a linear stateespace model from the original, nonlinear one when the variables vary in small amounts around an equilibrium position. The companion website Chapter 8 analyzes the linearization of nonlinear nonhomogeneous state-space models, that is, models that contain a nonzero input vector, but here we analyze only homogeneous nonlinear systems. Consider a two-DOF dynamic system defined by one second-order differential equation and one first-order differential equation. As previously discussed in this chapter, we use three state variables: two corresponding to the second-order differential equation, x1 and x2, and one for the first-order differential equation, x3. Because the system has two DOF, the DOF variables are selected to be the output variables, y1 and y2. As a consequence, the following nonlinear state equations can be formulated: 8 > < x_1 ðtÞ ¼ f1 ðx1 ; x2 ; x3 Þ x_2 ðtÞ ¼ f2 ðx1 ; x2 ; x3 Þ (8.69) > : x_3 ðtÞ ¼ f3 ðx1 ; x2 ; x3 Þ where f1, f2, and f3 are nonlinear functions of x1, x2, and x3. Similarly, the output variables can be expressed as  y1 ðtÞ ¼ h1 ðx1 ; x2 ; x3 Þ (8.70) y2 ðtÞ ¼ h2 ðx1 ; x2 ; x3 Þ with h1 and h2 being nonlinear functions of x1, x2, and x3. If small (linear) variations of the time-dependent variables and functions of Eq. (8.69) are considered (which is

8.2 State-Space Model Formulation

equivalent to using Taylor series expansions of the functions on the left-hand sides of Eq. (8.69) about the equilibrium point), the following equations result: 8 > vf1 vf1 vf1 > > > _ d x ðtÞ ¼ $ dx ðtÞ þ $ dx ðtÞ þ $ dx3 ðtÞ 1 1 2 > > vx vx vx > 1 2 3 e e e > > > < vf2 vf2 vf2 dx_2 ðtÞ ¼ $ dx1 ðtÞ þ $ dx2 ðtÞ þ $ dx3 ðtÞ (8.71) > vx1 e vx2 e vx3 e > > > > > > vf3 vf3 vf3 > > > : dx_3 ðtÞ ¼ vx1 $ dx1 ðtÞ þ vx2 $ dx2 ðtÞ þ vx3 $ dx3 ðtÞ e

e

e

where the variables preceded by the symbol d indicate small variations of those variables, and the subscript e indicates the equilibrium point. Eq. (8.71) can be arranged into the following matrix equation: 2 3 vf1 vf1 vf1 6 vx 7 vx vx 9 6 8 9 1 e 2 e 3 e7 8 6 7 > < dx1 ðtÞ > = 6 vf vf vf 7 > < dx_1 ðtÞ > = 2 2 2 6 7 (8.72) dx_2 ðtÞ ¼ 6 7 $ dx2 ðtÞ > > 7 > vx vx vx ; 6 : dx_3 ðtÞ > 6 1 e 2 e 3 e 7 : dx3 ðtÞ ; 6 7 vf3 vf3 5 4 vf3 vx1 e vx2 e vx3 e Eq. (8.72) is the linearized version of the original, nonlinear state equation. The matrix [A] is formed of the partial derivatives of f1, f2, f3 in terms of x1, x2, x3, the evaluation being made at the equilibrium point. The state vector is formed of small variations of the original state vector’s components. If small variations are now considered for the nonlinear output Eq. (8.70), the following equations are obtained: 8 > vh1 vh1 vh1 > > > < dy1 ðtÞ ¼ vx1 $ dx1 ðtÞ þ vx2 $ dx2 ðtÞ þ vx3 $ dx3 ðtÞ e e e (8.73) > vh2 vh2 vh2 > > > dy2 ðtÞ ¼ $ dx1 ðtÞ þ $ dx2 ðtÞ þ $ dx3 ðtÞ : vx1 e vx2 e vx3 e which can be written in vector-matrix form as 2 3 8 9 vh1 vh1 vh1 dx1 ðtÞ > 7 >  6  < = vx vx vx 6 dy1 ðtÞ 1 e 2 e 3 e7 7 $ dx2 ðtÞ ¼6





7 6 > dy2 ðtÞ vh2 vh2 5 > : ; 4 vh2 dx3 ðtÞ vx1 e vx2 e vx3 e

(8.74)

Eq. (8.74) is the linearized form of the original, nonlinear output Eq. (8.70). The output vector is connected to the state vector through the matrix [C].

395

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CHAPTER 8 State-Space Modeling

nn

Example 8.7 The lumped-parameter model of a microcantilever that vibrates in a vacuum is the mass-spring system shown in Figure 8.6. Considering that the spring constant is nonlinear such that the elastic force is fe ¼ (1/3)$y3, where y is the vertical displacement measured from the equilibrium position, find a state-space representation of the system and derive the corresponding linearized state-space model. Known are the lumped mass m and gravitational acceleration g.

k

Nonlinear spring

m

y

FIGURE 8.6 Lumped-Parameter Model of Microcantilever With Nonlinear Spring.

Solution At static equilibrium, the gravity force is equal to the elastic force; therefore,

pffiffiffiffiffiffiffiffiffiffiffiffi 1 m $ g ¼ $ y3e or ye ¼ 3 3m $ g 3

(8.75)

where ye is the deformation of the spring at static equilibrium. By relocating the reference frame to the static equilibrium position defined by ye of Eq. (8.75), the gravity action is removed, as known from dynamics theory, and the equation of motion is

1 m $ y€ ¼  $ y3 3

(8.76)

where y is the distance measured from the equilibrium position to an arbitrary position. The following state variables are selected:

x1 ¼ y; x2 ¼ y_

(8.77)

which enables formulating the equations

1 x_1 ¼ x2 ; x_2 ¼  $ x31 3m

(8.78)

These are state equations, and the second one is nonlinear. The output equation is linear:

y ¼ x1

(8.79)

Let us now linearize the state equation. If differentials in functions and in variables are considered, then the first Eq. (8.78), which is linear, can be written as

dx_1 ¼ dx2

(8.80)

The second Eq. (8.78) is reformulated based on the previous theory discussion as

 1 v 3  1 1 x1  dx_2 ¼  $ $ dx1 ¼  $ x21;e $ dx1 ¼  $ y2e $ dx1 3m vx1 m m x1 ¼x1;e

(8.81)

8.2 State-Space Model Formulation

Because x1 ¼ y and ye is provided in Eq. (8.75), Eq. (8.81) becomes

qffiffiffiffiffiffiffiffiffiffiffiffiffi 3 dx_2 ¼  9g2 =m $ dx1

(8.82)

Eqs. (8.80) and (8.82) can now be collected into the linearized state equation:



dx_1 dx_2



 ¼

  0 dx1 ffi 1 qffiffiffiffiffiffiffiffiffiffiffiffiffi  $ 3  9g2 m 0 dx2

(8.83)

where the state variables are dx1 and dx2. The output equation is simply obtained by differentiating Eq. (8.79) as

dy ¼ dx1 which is written in standard form as



dy ¼ f 1 0 g $

(8.84) dx1 dx2

 (8.85)

The state Eq. (8.83) and output Eq. (8.85) form the linearized state-space model of the mechanical system of Figure 8.6. nn

8.2.2 State-Space Model From Other Models The state-space models can be obtained from other models, such as transfer function or zero-pole-gain models, both analytically and by using specialized MATLAB commands. It is also possible to convert state-space models into other models, as discussed in this section.

Conversions Between Transfer Function and State-Space Models Transformation of a Transfer Function Model Into a State-Space Model The known transfer function of a SISO system can be converted analytically into a state-space model by following the steps indicated in the previous subsection, where the state-space model of a SISO system with input time derivatives is obtained using the transfer function and an intermediate function Z(s). The companion website Chapter 8 covers the topic of analytic conversion for MIMO systems. MATLAB also enables converting between transfer function and state-space models for both SISO and MIMO systems, as discussed next. A MATLAB state-space model or LTI (linear time invariant) object is defined by using the command sys = ss (A, B, C, D), where A, B, C, and D are the state-space model defining matrices. For instance, if the command is issued >> sys = ss ([1, 3; 2, 5], [1; 3], [-1, 0; 0, 5], [0; 1])

the following result will be generated:

.................................................................................................................... a =

b =

c =

x1

x2

x1

1

3

x1

1

x2

2

5

x2

3

d =

u1

x1

x2

u1

y1

-1

0

y1

0

y2

0

5

y2

1

....................................................................................................................

397

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CHAPTER 8 State-Space Modeling

which is an explicit way to mention the vectors that connect to any of the system’s matrices. Combining this command with the tf MATLAB command it is possible to realize conversions between transfer function and state-space models. Note that to be able to realize MATLAB conversion, the transfer function should not be algebraicdit needs numerical values for its coefficients. nn

Example 8.8 Determine a state-space model by converting the transfer function model of the operational amplifier circuit sketched in Figure 7.9(a) of Example 7.6 by the analytical approach. Also utilize MATLAB to obtain a state-space model. Consider the following actual elements: a resistor R1 ¼ 100 U for Z1, a resistor R2 ¼ 120 U for Z2, a resistor R3 ¼ 75 U for Z3, and a capacitor C ¼ 2$104 F for Z4.

Solution Eq. (7.43) gives the system’s transfer function, which can be written as

Vo ðsÞ Z2 $ Z4 ZðsÞ Vo ðsÞ ¼ $ ¼ Vi ðsÞ Z1 $ Z3 Vi ðsÞ ZðsÞ 8 ZðsÞ 1 > > ¼ ; > < V R ðsÞ $ R i 1 3$C$s 1 $ R2 or ¼ > R1 $ R 3 $ C $ s > > Vo ðsÞ ¼ R2 : ZðsÞ

GðsÞ ¼

(8.86)

The four impedances of Eq. (8.86)dnot to be confounded with the function Zdare Z1 ¼ R1, Z2 ¼ R2, Z3 ¼ R3, and Z4 ¼ 1/(C$s). Cross multiplication in the first Eq. (8.86) and application of the inverse Laplace transformation to the resulting equation (with zero initial conditions) results in the following equation:

z_ ¼

1 $ vi R1 $ R3 $ C

(8.87)

By selecting the state variable as x ¼ z and considering that the input is u ¼ vi changes Eq. (8.87) to

x_ ¼

1 $u R1 $ R3 $ C

(8.88)

which is the scalar form of the state equation with A ¼ 0 and B ¼ 1/(R1$R3$C) ¼ 0.667. Cross multiplication and inverse Laplace transformation with zero initial conditions are also applied to the second Eq. (8.86), which results in

y ¼ R2 $ x

(8.89)

where y ¼ vo. Eq. (8.89) is the output equation with C ¼ R2 ¼ 120 (C is the 1  1 [C] matrix of the output equation) and D ¼ 0. The following MATLAB code: >> R1 = 100; R2 = 120; R3 = 75; C = 2ee4; >> G = tf (R2, [R1*R3*C, 0]); >> ss(G) returns

8.2 State-Space Model Formulation

a¼ x1 x1 0 b = u1 x1 8 c = x1 y1 10 d = u1 y1 0 which indicates that B ¼ 8 and C ¼ 10. While individually, the B and C values obtained by MATLAB differ from the analytical results, it should be taken into account that the transfer function of the analyzed system is calculated with Eq. (8.14) as:

GðsÞ ¼ ½C $ ðs $ ½IÞ1 $ ½B ¼

C $ B 80 ¼ s s

(8.90)

since A ¼ 0 and D ¼ 0. It can also be seen that the C$B ¼ 80 for both the analytical and the MATLAB values of B and C. Moreover, any two values of B and C whose product equals 80 would be a satisfactory solution to this problem, and this illustrates again that a state-space model is not unique. nn

nn

Example 8.9 Use MATLAB to convert the transfer function matrix

2

3 6s þ s þ 1 ½GðsÞ ¼ 6 4 s 2 s þsþ1 2

3 1 s þ s þ 17 7 5 1 s2 þ s þ 1 2

into a state-space model.

Solution The following MATLAB sequence solves this example: >> s = tf (’s’); >> g = [3/(s^2 + s + 1), 1/ (s^2 + s + 1); s/( s^2 + s + 1), 1/( s^2 + s + 1)]; >> f = ss(g); To identify (and store) the four state-space matrices, the following command can be added to this sequence: >> [a,b,c,d] = ssdata(f) which returns.

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CHAPTER 8 State-Space Modeling

a = -1 1 0 0

-1 0 0 0

2 0 0 0

0 0 2 0

0 0.5

1.5 0

0 0

0 0

0 0 -1 1

0 0 -1 0

0 0

0.5 0.5

b =

c =

d =

As it can be seen from the result, the state-space generated through conversion from the given transfer function uses four state variables (see the dimensions of [A]), two inputs, and two outputs (see the dimension of [D]), which is, obviously, just one possible solution from an infinite number of potential state-space solutions. nn

Transformation of a State-Space Model Into a Transfer Function Model This subsection studies the topic of transforming a state-space model into the corresponding transfer function model. {Y(s)} and {U(s)}, the Laplace transforms of the output and input vectors, are connected by means of transfer function matrix [G(s)], as shown in Eq. (8.14). Since [G(s)] is calculated based on the state-space model matrices [A], [B], [C], and [D], it follows that the respective calculation procedure enables transformation of a state-space model into a transfer function one for both SISO and MIMO systems. Let us study the following example. nn

Example 8.10 The following matrices define a state-space model:



½A ¼

1

0

0

2



; ½B ¼

  0 1

; ½C ¼ f 2 1 g; ½D ¼ 0

Find the corresponding transfer function model using analytical calculation and also applying MATLAB conversion.

Solution The dimensions of the state-space model matrices indicate this is a SISO system and, as a consequence, the transfer function is a scalar function G(s). The definition Eq. (8.14) becomes

GðsÞ ¼ ½C $ ðs $ ½I  ½AÞ1 $ ½B

(8.91)

8.2 State-Space Model Formulation

due to the fact that D ¼ 0. The following MATLAB code: >> syms s >> A = [1, 0; 0, 2]; B = [0; 1]; C=[2, 1]; >> G = simplify(C*inv(s*eye(2)-A)*B) returns the transfer function

GðsÞ ¼

1 s2

(8.92)

The following MATLAB code >> A = [1, 0; 0, 2]; B = [0; 1]; C=[2, 1]; D=0; >> S = ss(A,B,C,D); >> tf(S) returns the transfer function of Eq. (8.92).

nn

Conversion Between Zero-Pole-Gain and State-Space Models Conversion between transfer function and zero-pole-gain models is presented in Chapter 7. Similarly, MATLAB enables conversion between state-space and zeropole-gain models. Assume that an LTI object sys1 has been defined as a zeropole-gain model; the MATLAB command >> sys2 = ss(sys1)

converts the zero-pole-gain sys1 model into another state-space model, labeled Conversely, when a state-space model, named sys2, is available, it can be transformed into a zero-pole-gain model, named sys1, by means of the MATLAB command as follows:

sys2.

>> sys1 = zpk(sys2)

nn

Example 8.11 A dynamic system’s zero-pole-gain model is defined by a gain of 2, its zeroes are 1 (double) and 3, and its poles are 0 and 1 (triple). Determine a state-space model corresponding to the given zero-polegain model then confirm that the state-space model can be converted back to the original zero-polegain model.

Solution The following MATLAB sequence solves this example: >> sys1 = zpk([1,1,3],[-1,-1,-1,0],2) Zero/pole/gain: 2 (s-1)^2 (s-3) --------------s (s+1)^3 >> sys2 = ss(sys1)

a = x1 x1

x2 0

x3 -1.414

x4 -2

1

401

402

CHAPTER 8 State-Space Modeling

x2 x3 x4

0 0 0

u1 x1 x2 x3 x4

0 0 0 4

y1

x1 -0.5

u1 y1

0

-1 0 0

-2.828 -1 0

1.414 2 -1

x2 -0.7071

x3 -1

x4 0.5

b =

c =

d =

Continuous-time model. >> sys3=zpk(sys2) Zero/pole/gain: 2 (s-3) (s-1)^2 --------------s (s+1)^3 It can be seen that the state-space model rendered by MATLAB (one of the many possible models) is the one of a SISO system (as indicated by the dimensions of the matrix [D]), which uses four state variables (see the dimensions of matrix [A]). By applying the zero-pole-gain command to the state-space model, the original zero-pole-gain object is retrieved. nn

8.3 STATE-SPACE MODEL AND THE TIME-DOMAIN RESPONSE Analytical, MATLAB, and Simulink methods can be used to determine the time (forced) response of dynamic systems by means of state-space models, which is the subject of this section.

8.3.1 Analytical Approach: The State-Transition Matrix Method As briefly mentioned at the beginning of this chapter, the time response of statespace-modeled dynamic systems can be evaluated by an algorithm that consists of combined calculations in the Laplace and time domains. In essence, solving for the output {y} based on a given input {u} and a selected state vector {x} can be performed as indicated in Figure 8.7. The state vector {x} is first determined through integration from the state equation, then substituted into the output equation, which yields the output vector {y} as a function of the input vector {u}, the state vector {x}, and the state-space model matrices [A], [B], [C], and [D]. Two subcases are studied next: The first one analyzes the homogeneous state-space model (with no forcing

8.3 State-Space Model and the Time-Domain Response

[B]{u}

{u} [B]

[D]

{x}

+

[C]{x}

{x}

∫

+

[C]

+

{y}

+

[A]

[A]{x}

[D]{u}

FIGURE 8.7 Block Diagram of Operations in a State-Space Model.

{u(t)}), and the other one focuses on the nonhomogeneous state-space model where {u(t)} s 0. Both categories consider nonzero initial conditions. The state-transition matrix method is utilized, as explained next.

Homogeneous State-Space Model The homogeneous case corresponds to the absence of an input vector, whereby the system response is caused by initial conditions only. The state and output equations are given in Eq. (8.6). The aim of this method is to express the state vector at any time instant in terms of the initial state vector as fxðtÞg ¼ ½4ðtÞ $ fxð0Þg

(8.93)

where [4(t)] is the state- transition matrix. There are two methods of calculating the state-transition matrix: the Laplace transform method, which is discussed next, and the matrix exponential method that is presented in the companion website Chapter 8. By applying the Laplace transform to the state equation, the first Eq. (8.6), and considering a nonzero initial state vector {x(0)}, the following relationship is obtained: fXðsÞg ¼ ðs $ ½I  ½AÞ1 $ fxð0Þg whose inverse Laplace transform yields the time-domain state vector: h i fxðtÞg ¼ L1 ðs $ ½I  ½AÞ1 $ fxð0Þg

(8.94)

(8.95)

By comparing Eqs. (8.93) and (8.95), it follows that the matrix connecting {x(t)} to {x(0)} is the state-transition matrix, which is therefore calculated as h i (8.96) ½4ðtÞ ¼ L1 ðs $ ½I  ½AÞ1 The output equation, which is the second Eq. (8.6), becomes h i fyðtÞg ¼ ½C $ ½4ðtÞ $ fxð0Þg ¼ ½C $ L1 ðs $ ½I  ½AÞ1 $ fxð0Þg

(8.97)

403

CHAPTER 8 State-Space Modeling

nn

Example 8.12 The

homogeneous state-space model 2 3 0 2 0 6 7 ½A ¼ 4 8 1 8 5; ½C ¼ f 1 0 1 g. 0 0 1

of

a

dynamic

system

has

the

matrices

Determine the system’s response y (t) and plot it with respect to time if the following initial condition is used: {x(0)} ¼ { 0 1 0}T.

Solution The following MATLAB commands >> >> >> >>

a = [0,2,0;-8,-1,8;0,0,1]; c = [1,0,1]; syms s c*ilaplace(inv(s*eye(3)-a))*[0;1;0]

which are based on the analytical Eq. (8.97), yield the following approximate output:

yðtÞ ¼ 0:5 $ e0:5t $ sinð4tÞ The response curve is plotted in Figure 8.8.

(8.98) nn

Output, y

404

Time (sec)

FIGURE 8.8 Time Response of a Homogeneous State-Space Model. nn

Example 8.13 Use the state-space approach to find the response of the liquid system shown in Figure 7.11 of the solved Example 7.8 by considering that there is no input flow rate. Instead, there is an initial value of the fluid head in the left tank, h1(0), which acts as an initial condition. Considering that the heads h1 and h2 are the output variables, plot h1(t) and h2(t) for R1 ¼ 20 s/m2, R2 ¼ 25 s/m2, C1 ¼ 1 m2, C2 ¼ 2 m2, and h1(0) ¼ 0.05 m.

8.3 State-Space Model and the Time-Domain Response

Solution For the case of the free response, the differential equations are derived using the definitions of hydraulic capacitances, Eq. (5.12), and resistances, Eq. (5.23):

C1 $

dh1 dh2 ¼ q; C2 $ ¼ q  qo ; R1 $ q ¼ h1  h 2 ; R2 $ qo ¼ h 2 dt dt

(8.99)

The flow rates q and qo are substituted from the last two Eq. (8.99) into the first two Eq. (8.99), which yields:

8 dh1 1 1 > > > < C1 $ dt þ R1 $ h1  R1 $ h2 ¼ 0

> dh2 1 1 1 > > C  $ $ h þ þ $ h2 ¼ 0 : 2 1 R1 R1 R2 dt

(8.100)

Two state variables are needed for this two-DOF first-order system, and they can be selected as

x 1 ¼ h1 ; x 2 ¼ h2

(8.101)

By using these state variables in conjunction with Eq. (8.100), the following state equation is obtained:

2



x_1 x_2



1 6 R1 $ C1 6 ¼6 4 1 R1 $ C2

3 1 7 x  R1 $ C 1 1 7

7$ 5 x2 1 1 1  $ þ C2 R1 R2

(8.102)

Matrix [A] of the state-space model is the one connecting the two vectors in Eq. (8.102). The output equation is obtained based on Eq. (8.101):



h1 h2





¼

y1



y2



¼

1 0

  x1 $ 1 x2 0

(8.103)

Matrix [C] of the state-space model is the identity matrix [I ]; therefore, {x(0)} ¼ {y(0)} ¼ {h1(0) 0}T. Using the parameters of this example, Eq. (8.97) yields the following time-domain output:

(

h1 ¼ 0:05 $ ½cos hð0:035 $ tÞ  0:07 sin hð0:035 $ tÞ $ e0:047 $ t h2 ¼ 0:035 $ sinhð0:035 $ tÞe0:047 $ t

(8.104)

h2 (m)

h1 (m)

Figure 8.9 contains the plots of the two response components. For t / N, the heads become h1(N) ¼ 0 and h2(N) ¼ 0. nn

Time (sec)

FIGURE 8.9 Time-Response Curves of Hydraulic Heads.

405

406

CHAPTER 8 State-Space Modeling

Nonhomogeneous State-Space Model The nonhomogeneous case implies intervention of the input (forcing) vector {u(t)}; therefore, the state and output equations are the ones originally expressed in Eq. (8.6). Application of the Laplace transform to the state Eq. (8.6), considering there is a nonzero initial state vector {x(0)}, results in the following Laplace-domain equation: fXðsÞg ¼ ðs $ ½I  ½AÞ1 $ fxð0Þg þ ðs $ ½I  ½AÞ1 $ ½B $ fUðsÞg

(8.105)

The inverse Laplace transform is now applied to Eq. (8.105), which leads to i h (8.106) fxðtÞg ¼ ½4ðtÞ $ fxð0Þg þ L1 ðs $ ½I  ½AÞ1 $ ½B $ fUðsÞg According to the convolution theorem, which is studied in Chapter 6, the second term on the right-hand side of Eq. (8.106) can be calculated as i h h i   L1 ðs $ ½I  ½AÞ1 $ ½B $ fUðsÞg ¼ L1 ðs $ ½I  ½AÞ1  L1 ½B $ fUðsÞg Z th i ¼ 4ðt  sÞ $ ½B $ fuðsÞgds 0

(8.107) As a consequence, Eq. (8.106) becomes Z th i 4ðt  sÞ $ ½B $ fuðsÞgds fxðtÞg ¼ ½4ðtÞ $ fxð0Þg þ

(8.108)

0

where [4(t)] is calculated by means of Eq. (8.96). The output vector is found as ! Z th i   4ðt  sÞ $ ½B $ fuðsÞgds þ ½D $ fuðtÞg fyðtÞg ¼ ½C $ 4ðtÞ $ fxð0Þg þ 0

(8.109)

nn

Example 8.14 A dynamic system is described in state-space form by the matrices



0 ½A ¼ 60

  1 0 ; ½B ¼ ; ½C ¼ f 1 5 1

0 g; ½D ¼ 0

A step input u ¼ 3 is applied to the system with zero initial conditions. Find the system response using the state-transition matrix approach and plot it against time.

8.3 State-Space Model and the Time-Domain Response

Solution The matrices dimensions show that the system is SISO and there are two state variables. For the particular case of this example, the output is expressed from Eq. (8.109) as

Z

fyðtÞg ¼ ½C $

t

4ðt  sÞ $ ½B $ uðsÞds

(8.110)

0

where u(s) ¼ 3. The state-transition matrix [4(t)] is first computed by using the definition Eq. (8.96) as



½4ðtÞx

½cosð7:3 $ tÞ þ 0:34 $ sinð7:3 $ tÞ 8:18 $ sinð7:3 $ tÞ

0:14 $ sinð7:3 $ tÞ $ e2:5 $ t ½cosð7:3 $ tÞ  0:34 $ sinð7:3 $ tÞ (8.111)

The variable tds is used instead of the variable t in Eq. (8.111), then the necessary operations are performed in Eq. (8.110), which yields the following time-domain response:

yðtÞ ¼ 0:005  0:017 $ ½sinð7:3 $ tÞ þ 2:933 $ cosð7:3 $ tÞ $ e2:5 $ t

nn

Output, y

Figure 8.10 displays the output as a function of time.

(8.112)

Time (sec)

FIGURE 8.10 Time Response of Nonhomogeneous State-Space Model.

8.3.2 MATLAB Approach This section presents the application of built-in MATLAB functions that use statespace modeling to calculate the forced response (including nonzero initial conditions) of dynamic systems and to plot it in terms of time. The free response with nonzero initial conditions and the forced response are studied next.

Free Response With Nonzero Initial Conditions With MATLAB, it is possible to directly model the free response of a state-space dynamic system when the initial conditions are different from zero. The basic command is as follows: >> initial (sys, x0) where x0 is the initial state vector, and sys is a state-space model. This command

directly plots the state-space output y in terms of time.

407

408

CHAPTER 8 State-Space Modeling

nn

Example 8.15 The free-response state-space model of a dynamic system is represented by the matrices [A] and [C] and the initial state vector x0, which are defined as



1 ½A ¼ 1:5

1:5 ; ½C ¼ f 4 0:1



10 g; fx0 g ¼

10 1



Use MATLAB to determine the time response and plot it in terms of time.

Solution The following MATLAB command sequence >> >> >> >> >>

a = [-1, -1.5; 1.5, 0.1]; c = [4, 10]; x0 = [10; 1]; sys = ss (a, [], c, []); initial (sys, x0)

is used to model the free response of this specific state-space model, and the result is plotted in Figure 8.11.

FIGURE 8.11 Example Plot of Free Response With Nonzero Initial Conditions. nn

Several previously defined state-space models can be plotted on the same graph and time interval and specified with the command >> initial (sys1, sys2,.,sysn, x0, t)

where t is a time interval. When using the command. >> [y,t,x] = initial (sys1, sys2,.,sysn, x0, t)

no plot is returned but the output matrix and state matrix are formed, each having as many rows as time increments. The number of columns in y is equal to the number of inputs, and the number of columns in x is equal to the number of state variables.

8.3 State-Space Model and the Time-Domain Response

Subsequent plotting is possible with a plot command. Instead of initial, can be used with the same specifications.

initialplot

Forced Response To find and plot the forced time response, the state-space models in MATLAB employ the same predefined functions as the transfer function models, step, impulse, and lsim. For instance, if a state-space model is defined as >> sys = ss ([-1, -1; 1, 0], [1; 0], [1, 0; 0, 2], [0; 1]);

the command step(sys) produces the plots shown in Figure 8.12. Because the matrices of this particular example define a one-input, two-output model, two plots result, each corresponding to a unit step input.

FIGURE 8.12 Response of a Two-Output State-Space Model to Unit-Step Input.

nn

Example 8.16 Consider the mechanical microsystem shown in Figure 8.13, which is formed of three identical serpentine springs, each having a spring constant of k ¼ 10 N/m and two identical shuttle masses with m ¼ 30 mg. A force f ¼ 5 sin(2t) mN acts on the body on the right, as shown in the figure. In addition, an initial displacement y1(0) ¼ 1.5$1011 m is applied to the body on the left of the figure. Consider that damping acts on both masses, with a damping coefficient c ¼ 0.1 N s/m. Find a state-space model of the system and use MATLAB to plot its time response.

Solution To use MATLAB for the time-domain solution, the problem is divided into two subproblems: it is considered first that the system is under the action of the force alone, then that only the initial conditions are applied. Since the system is linear, the two individual solutions are then added to obtain the total (actual) solution. The lumped-parameter model corresponding to the micromechanical system is shown in Figure 8.14, being formed of two masses m, three dampers c, and three springs k. The force f is also indicated in the figure.

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CHAPTER 8 State-Space Modeling

Spring

Spring

Spring m

m f

Anchor

Shuttle mass

Anchor

Shuttle mass

FIGURE 8.13 Two-Mass, Three-Spring Mechanical Microsystem.

y1

k

y2

k

f

m c

k

m c

c

FIGURE 8.14 Lumped-Parameter Model of Two-DOF Translatory Mechanical System.

The following dynamic equations of this system can be derived using known methods, such as Newton’s second law of motion or Lagrange’s equations:

(



m $ y€1 ¼ c $ y_1  k $ y1  c $ y_1  y_2  k $ ðy1  y2 Þ

 m $ y€2 ¼ c $ y_2  k $ y2  c $ y_2  y_1  k $ ðy2  y1 Þ þ f 8 2k 2c k c > > > < y€1 ¼  m $ y1  m $ y_1 þ m $ y2 þ m $ y_2 or > > > y€ ¼ k $ y þ c $ y_  2k $ y  2c $ y_ þ 1 $ f : 1 2 2 m m 1 m m 2 m

(8.113)

The following state variables are selected:

x1 ¼ y1 ; x2 ¼ y_1 ; x3 ¼ y2 ; x4 ¼ y_2

(8.114)

the input is the force, that is u ¼ f, whereas the output is formed of the two displacements y1 and y2. Combination of Eqs. (8.113) and (8.114) generates the state equation:

2

3

0

8 9 6 > > 6 k > > x_1 > > 2 $ < = 6 6 x_2 m ¼6 6 0 _ x > > 3 > > > 6 > : x_4 ; 6 4 k m

1 2 $ 0 c m

c m

0 k m 0 2 $

8 9 > 8 9 >0> > 7 > > > x1 > > > > 7 > > > > > > > > > 7 < 0 < = = 7 x2 7$ þ 7 > x > > 0 > $ u (8.115) > > > 3> > > 7 > ; > 7 : > > > 1> > > c5 x4 > : > ; 2 $ m m 0 c m 1

k m

8.3 State-Space Model and the Time-Domain Response

which defines the [A] and [B] matrices. The output equation is determined from the first and third Eq. (8.114):



y1 y2



 ¼

1

0 0

0

0 1

8 9 > x1 > >   > > > 0 0 < x2 = þ $u $ > > 0 0 > x3 > > > : ; x4

(8.116)

y2 (m)

y1 (m)

which identifies the [C] and [D] matrices. Eq. (8.116) relates the initial values of the output and state vectors as follows: {y1(0) y2(0)}T ¼ [C]${x1(0) x2(0) x3(0) x4(0)}T. Because y1(0) s 0 and y2(0) ¼ 0, it follows that x1(0) ¼ y1(0) s 0 and x3(0) ¼ y2(0) ¼ 0. As a consequence, the initial-condition vector can be selected as {x(0)} ¼ {y1(0) 0 0 0}T.

Time (sec)

FIGURE 8.15 Plot of Displacement Outputs for the Two-DOF Mechanical System. Figure 8.15 shows the response curves of the mechanical microsystem; the plots were obtained by the MATLAB code: >> >> >> >> >> >>

m = 30e-9; c = 0.1; k = 10; t = 0:0.001:10; u = 5e-6*sin(2*t); A = [0,1,0,0;-2*k/m,-2*c/m, k/m, c/m;0,0,0,1;k/m, c/m,-2*k/m,-2*c/m]; B = [0;0;0;1/m]; C = [1,0,0,0;0,0,1,0]; D =[0;0]; sys = ss(A,B,C,D);

>> % individual response ‘yf’ to forcing >> [yf,t,x] = lsim(sys,u,t); >> % individual response ‘yic’ to initial conditions >> y01 = -1.5e-11; >> x0 = [y01;0;0;0]; >> [yic,t,x] = initial(sys,x0,t);

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CHAPTER 8 State-Space Modeling

>> % total response ‘y’ as superposition of individual ‘yf’ and ‘yic’ >> >> >> >> >> >> >>

y = yf + yic; subplot(2 1 1); % plot of y_1 e first column of y plot(t,y1(:,1)) subplot(2 1 2); % plot of y_2 e second column of y plot(t, y1(:,2))

nn

8.3.3 Simulink Approach Simulink offers an elegant environment for creating and simulating state-space models. The minimum configuration of a Simulink state-space model is formed of an input block, the state-space block, and an output (visualization/plot) block. Let us analyze a couple of examples of formulating and solving state-space problems by means of Simulink. nn

Example 8.17 A state-space model is defined by the matrices



A¼

0:001 80 1

0





; B¼

1 1 0

0

; C ¼ f0

1 g; D ¼ f 0

0g

The input to this system has two components: u1 ¼ 20e18$e0.5 t, and u2 is a pulse with an amplitude of 20, period of 2.5 s, and pulse width of 1.25 s. The initial conditions of the problem are {x(0)} ¼ {1 2}T. Use Simulink to plot the system response.

Solution A new model window is opened and the following blocks are copied to it from the Library window: a Ramp block (with a Slope of 1 to model the time vector) and a Pulse Generator block (with an Amplitude of 20, Period of 2.5, and Pulse Width of 50%) from the Sources category. The Fcn (Function) block is taken from the User-Defined Functions library with the following Expression: 20-18*exp(-0.5*u(1)). The State-Space block is found in the Continuous category, and three To Workspace blocks are copied from the Sinks category. The system is a MISO (multiple-input/single-output) system, as the dimensions of the matrices indicate. We therefore need to use two signals combined into a single input, since the Simulink state-space operator works as a SISO one. The Simulink Mathematical Operations library possesses the Vector Concatenate operator that accepts several vectors (or matrices) as the input, which it combines into a single vector (or matrix) at the output, without modifying the original components. Configuration of the state-space block (which you have to double click) is done by filling in the following data. Function Block Parameters: State-Space Window A: [e0.001, e80; 1, 0] B: [1, e1; 0, 0] C: [0, 1] D: [0, 0] Initial conditions: [1; 2] The Simulink block diagram is shown in Figure 8.16.

8.3 State-Space Model and the Time-Domain Response

FIGURE 8.16

Output,y

Input

Simulink Block Diagram of the Two-Input, One-Output State-Space Model.

u1 u2

Time (sec)

(a)

Time (sec)

(b)

FIGURE 8.17 Simulink Plots of (a) The Pulse and Exponential Input Signals; (b) Output Signal. Figure 8.17 contains the plots of the two inputs and the one of the output e all these signals were exported to MATLAB’s workspace to be plotted as functions of time. Note: An LTI System block (as discussed in Example 7.18 in the context of transfer function) can be used instead of the State-Space block and a Mux block instead of the Vector Concatenate block e the interested reader can test that variant. nn

nn

Example 8.18

A body of mass m ¼ 1 kg slides on a horizontal surface with friction. The body is connected to a spring of stiffness k ¼ 50 N/m whose other end is fixed. The coefficient of static friction is ms ¼ 0.3, the coefficient of kinematic friction is mk ¼ 0.2, and the gravitational acceleration is g ¼ 9.81 m/s2. An initial displacement y0 ¼ 0.9 m is applied to the body. Find a state-space model of this mechanical system and plot its output (time-dependent displacement y(t)) using Simulink.

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CHAPTER 8 State-Space Modeling

Solution For the mass to vibrate, the initial displacement of 0.9 m should be larger than a threshold value determined from the static equilibrium between the spring force and the static friction force, which is ys ¼ ms$m$g/k ¼ 0.0589 mdthis condition is definitely complied with. The free-body diagram of the body is shown in Figure 8.18(a) for the situation where the displacement and velocity have identical directions, whereas Figure 8.18(b) sketches the free-body diagram for the case where the velocity direction switches. v

v y

y

fe

fe m

m

ff

ff

(a)

(b)

FIGURE 8.18 Free-Body Diagram for Body Under the Action of Elastic and Friction Forces for Velocity and Displacement Having (a) Identical Directions; (b) Opposite Directions. Applying Newton’s second law of motion for the two cases results in the following equations:



m $ y€ ¼ k $ y  mk $ m $ g m $ y€ ¼ k $ y þ mk $ m $ g

k for y $ y_ > 0 _ $ mk $ g or y€ ¼  $ y  sgnðyÞ for y $ y_ < 0 m (8.117)

Eq. (8.117) took into account that the elastic (spring) force is fe ¼ k$y and the friction force is ff ¼ mk$m$g. The new function signum, denoted as sgnðy_Þ in Eq. (8.117), assumes a value of þ1 when the velocity is positive (has the same direction as y) and a value of 1 when the velocity becomes negative (has a direction opposing the y direction). As a consequence, the last form of Eq. (8.117) captures both situations described by the two individual Eq. (8.117). Note also that Eq. (8.117) is nonlinear due to sgn function that is applied to the dependent variable y.Considering the friction action as the input, which is u ¼ sgnðy_Þ $ mk $ g , and the state variables x1 ¼ y ; x2 ¼ y_, the following state equation results via Eq. (8.117)



x_1 x_2



2

0 6 ¼4 k  m

1 0

3

7 5$

Because y ¼ x1, the output equation is





y ¼ f1 0g$ which also indicates that

 yð0Þ ¼ f 1

0g$

x1 x2

x1



 þ

0 1

 $u

(8.118)



x2 x1 ð0Þ x2 ð0Þ

þ 0$u

(8.119)

 ¼ x1 ð0Þ

The initial state variable vector is therefore as follows: {x(0)} ¼ {x1(0) x2(0)}T ¼ {y(0) 0}T.

(8.120)

8.3 State-Space Model and the Time-Domain Response

sgn( y )⋅ μ k ⋅ g

Output

Offset

Gain

μk ⋅ g

0

0

Input

−μ k ⋅ g

–Offset

–Gain

y

(a)

(b)

FIGURE 8.19 Simulink Nonlinear Coulomb and Viscous Friction: (a) Generic Structure; (b) Particular Structure. MATLAB possesses the Coulomb and Viscous Friction block in the Discontinuities library. This block, which is shown in Figure 8.19(a), can model the sgn function in a more generic fashion by means of an offset and a gain. The inputeoutput relationship of this Simulink block is expressed as follows:

Output ¼ sgnðInputÞ $ ðGain $ jInputj þ OffsetÞ

(8.121)

In our particular application, the output from the block is u ¼ sgnðy_Þ $ mk $ g , which indicates that the gain is zero, whereas the offset is mk $ g , as shown in Figure 8.19(b). Figure 8.20 shows the block diagram of the state-space model described by Eqs. (8.118) and (8.119). The State-Space block has the matrices A, B, C, and D that were defined in the same Eqs. (8.118) and (8.119). In addition, the Initial Conditions of this block are [0.9; 0]. In the Coulomb and Viscous Friction block, the Coulomb friction value (offset) is mk$g ¼ 1.962, whereas the Gain is 0. Note that the input to this block is the velocity, which results from the Derivative block taken from the Continuous library. The output y is exported to MATLAB, where the plot y(t) illustrated in Figure 8.21 is obtained. As it is noticed, the body motion stops at a distance from the equilibrium (zero) position, and this offset has the magnitude

y

u

Dy/dt

FIGURE 8.20 Simulink Block Diagram of the State-Space Model Without and With Coulomb-Friction Nonlinear Effect.

415

CHAPTER 8 State-Space Modeling

y (m)

416

Time (sec)

FIGURE 8.21 Simulink Plot of State-Space Model Time Response for Nonlinear Mechanical System With Coulomb Friction.

of the static equilibrium spring deformation, which was calculated at the beginning of this example as ys ¼ 0.0589 m. nn

SUMMARY This chapter introduces the concept of state-space as an approach to model and determine the response of dynamic systems in the time domain. Using a matrix formulation, the state-space procedure is an alternate method of characterizing mostly MIMO systems defined by a large number of coordinates (DOF). Different state-space algorithms are applied, depending on whether the input has time derivatives or no time derivatives. Methods of calculating the free response with nonzero initial conditions and the forced response using the state-space approach are presented as well. Conversion between state-space and transfer function or zero-pole-gain models is also studied, both analytically and by means of MATLAB-specialized commands. The chapter also discusses the principles of linearizing nonlinear state-space models. The material includes the application of specialized MATLAB commands to solve state-space formulated problems and examples of using Simulink to model and solve linear/nonlinear system dynamics problems by the state-space approach.

PROBLEMS 8.1 Find a state-space model for the translatory mechanical system of Figure 8.22, considering that the input is the displacement u of the chain’s free end and the output is the displacement y. Derive another state-space model when the output is formed of y and of the velocity dy/dt.

Problems

y k1

u

k2 c

FIGURE 8.22 Translatory Mechanical System With Springs, Damper, and Displacement Input.

8.2 A single-mesh electrical system comprises an inductor L, a capacitor C, and a source voltage v. Derive a state-space model where the input is v, the output is the current i, and the state variables are the voltage across the capacitor vC and the current i. 8.3 The electric system of Figure 8.23 comprises the resistor R and the capacitor C. Find a state-space model for this system by starting from the time-domain model of this system and using one state variable only. R vi

C

vo

FIGURE 8.23 Single-Stage CapacitiveeResistive Electrical System.

8.4 Find a state-space representation for the electrical circuit shown in Figure 8.24 using only one state-space variable. Known are the resistors R1, R2, R3 and the capacitor C. The input is the source current i, and the output is the voltage difference across any of the four electrical elements.

+ R1

i

R2

C

R3

– FIGURE 8.24 Electrical Circuit With Current Source.

8.5 Starting from the time-domain model of the operational-amplifier electrical circuit shown in Figure 8.25, which contains the resistor R and the inductor L, derive a state-space model using one state variable when vi is the input and vo is the output.

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CHAPTER 8 State-Space Modeling

R L

– +

vi

vo

FIGURE 8.25 Operational-Amplifier Electrical System With Resistor and Inductor.

8.6 The pneumatic system of Figure 8.26 consists of a pneumatic resistance Rg and a gas container whose capacity is Cg. The input (container) pressure is pi, and the output pressure is po. Derive a state-space model for this system.

pi po

Cg

Rg

po

qm

FIGURE 8.26 Pneumatic System With Resistance and Capacitance.

8.7 Derive a state-space model for the one-room thermal system of Figure 8.27. Consider that the input is formed of the outside temperature qo and the source heat flow rate qi, whereas the output is the inside temperature qi. The enclosed space has a thermal capacity Cth, and each of the four identical walls has a thermal resistance Rth. θo

Rth Rth

θi

Cth qi

Rth

FIGURE 8.27 Four-Wall, One-Room Thermal System.

Rth

Problems

8.8 Formulate a state-space model for the liquid system of Figure 8.28, which comprises a tank of capacitances Cl and two valves of resistances Rl1 and Rl2. The input to the system is the pressure pi, and the output is the pressure p at the bottom of the tank.

Cl Rl1

Rl2 p

pi

po = 0

FIGURE 8.28 Liquid-Level System With One Tank and Two Valves.

8.9 The MEMS of Figure 8.29 is formed of two shuttle masses m1 and m2 coupled by a serpentine spring of stiffness k and supported separately by two pairs of identical beam springsdeach beam has a stiffness k1. The shuttle masses are subjected to viscous damping individually through substrate interactiondthe damping coefficient is c, and an electrostatic force f acts on m1. Use a lumped-parameter model of this MEMS device and obtain a state-space model for it by considering that the input is f, and the output vector comprises the two masses’ displacements and velocities.

Spring Beam spring f m1

m2

Shuttle mass

Anchor

FIGURE 8.29 Translatory-Motion MEMS Device.

8.10 Find a state-space model for the translatory mechanical system of Figure 8.30 considering that the input is the force f applied to the mass m and, the output vector contains all relevant displacements. Known are also the spring stiffnesses k1, k2, and the damper’s viscous damping coefficient c.

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CHAPTER 8 State-Space Modeling

k1

f

m c

k2

FIGURE 8.30 Translatory Mechanical System With Springs, Damper, Mass, and Force Input.

8.11 Derive a state-space model for the electrical system of Figure 8.31, where the input is the voltage vi, and the output is the voltage vo. Consider known the resistance R, inductance L, and capacitance C. C vi

C L

vo

R

FIGURE 8.31 Two-Stage Electrical System.

8.12 The liquid system shown in Figure 8.32 is formed of two tanks of capacitances Cl1 and Cl2 and two valves of resistances Rl1 and Rl2. Formulate a state-space model for it by considering the input to the system is the volume flow rate qi and the pressure p3; the output consists of the pressures at the tanks’ bottoms, p1 and p2. qi

Cl1

Cl2 Rl1

p1

Rl2

p3

p2

FIGURE 8.32 Two-Tank Liquid-Level System.

8.13 Find state-space models for the SISO dynamic systems represented by the following differential equations, where y is the output, and u is the input: (i) 0y þ 5y€ þ 3y_ þ 2y ¼ 3u_ (ii) 2y€  y_ þ 4y ¼ 6u€ (iii) 0y þ y€  7y_ þ 3y ¼ u€ þ 2u_ þ 4u

Problems

8.14 Derive a state-space model for the rotary mechanical system of Figure 8.33, which is formed of a cylinder with mass moment of inertia J, a spring of stiffness k, and two dampers defined by the damping coefficients c1 and c2. The input is the rotation angle qu, and the output is the cylinder rotation angle. θu J k

c1

c2

FIGURE 8.33 Rotary Mechanical System.

8.15 Obtain a state-space model for the translatory mechanical system of Figure 8.34 assuming the input is the displacement u, and the output is the displacement y. Known are the mass m, the damper viscous damping coefficient c, and the spring stiffnesses k1 and k2. y

u

c m k2

k1

FIGURE 8.34 Translatory Mechanical System With Displacement Input.

8.16 Obtain an state-space model for the electrical system of Figure 8.35, where the applied voltage vi is the input, and the voltage vo is the output. Known are the resistances R1, R2, and inductance L.

R1 vi

R2

L

vo

FIGURE 8.35 One-Stage Electrical System With Resistors and Inductor.

8.17 Consider the operational-amplifier system shown in Figure 8.36. Knowing the resistances R1, R2, the inductance L, and the capacitance C: (i) Find a state-space model for this circuit for the voltage vi being the input, and the voltage vo being the output (ii) Convert the state-space model into the corresponding transfer function.

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CHAPTER 8 State-Space Modeling

C

R1

– + vi

vo

R2 L

FIGURE 8.36 Operational-Amplifier Electrical System.

8.18 Convert the transfer function of the electrical system sketched in Figure 7.2 of Example 7.1 into a state-space model using the analytical approach. 8.19 Same question as in Problem 8.18 for the mechanical system of Example 7.9 and shown in Figure 7.14. 8.20 Convert the zero-pole-gain model of Example 7.4 into a state-space model using both the analytical approach and MATLAB. 8.21 A unity-gain dynamic system has the following zeroes: 0 (simple) and 1 (double), and the poles: 2 þ j (double) and 4  j (simple). Determine a state-space model corresponding to the given zero-pole-gain model by using the analytical approach. Use MATLAB to confirm that the state-space model can be converted back into the original zero-pole-gain model. 8.22 Use analytical calculation to obtain a state-space model from the transfer function: GðsÞ ¼ s32sþ1 þsþ1. 8.23 Transform the state-space model of the electrical system of Problem 8.16 and pictured in Figure 8.35 into a transfer function for R1 ¼ 250 U, R2 ¼ 220 U, and L ¼ 0.5 H. 8.24 Convert the transfer function matrix 2 s sþ1 2s þ 3 3 2 2 2 6s þ s þ 1 s þ s þ 1 s þ s þ 17 7 ½GðsÞ ¼ 6 4 5 1 2 4s s2 þ s þ 1 s2 þ s þ 1 s2 þ s þ 1 into a state-space model by using MATLAB. 8.25 Convert the state-space model of the mechanical system illustrated in Figure 8.2 of Example 8.3 into a transfer function model using the analytical transformation equation.

Problems

8.26 A state-space model is defined by the following matrices:   2 1 1 0 ½A ¼ ; ½B ¼ ; ½C ¼ f 3 1 g; ½D ¼ f 1 5 3 0 2

0g

Convert this model into a transfer function model using analytical calculation and also applying MATLAB conversion. 8.27 Same question as in Problem 8.26 for the system defined by the matrices: 8 9 2 3 1 0 0 > =

6 7 1 ; ½C ¼ f 1 0 3 g; ½D ¼ 0 ½A ¼ 4 0 1 0 5; ½B ¼ > ; : > 0 1 2 1

8.28 The transfer function of a SISO dynamic system is G(s) ¼ (s þ 6)/ (s2 þ 10s þ 81). Use MATLAB to determine the corresponding zero-polegain model, and then utilize the resulting zero-pole-gain model to derive a state-space model. Also obtain a state-space model directly from the transfer function model. If the two state-space models are different, explain the discrepancy. 8.29 A point mass m and a nonlinear spring are attached to the lever sketched in Figure 8.37, which vibrates in a horizontal plane. Assume the elastic (spring) force is defined as fe ¼ 0.9 y þ 0.1$y3 (where y represents the displacement of the spring end attached to and perpendicular to the lever) and the rod is massless. For small rotations of the rod around its pin axis, obtain a linearized state-space model by considering small vibrations about the equilibrium position. l

l

m k

FIGURE 8.37 Mass-Spring Lever System.

8.30 Solve Problem 8.29 when the system vibrates in a vertical plane and the lever has a mass ml. Consider small motions around the static-equilibrium position of the lever and take into consideration the gravitational acceleration g. 8.31 Use a lumped-parameter model for the single-DOF mechanical microaccelerometer of Figure 8.38. Find a transfer function model of this system and convert it into a state-space model. The five identical massless beams have

423

424

CHAPTER 8 State-Space Modeling

Flexible beam f

l

l

Rigid plate l/2 l

FIGURE 8.38 Microaccelerometer With Plate and Beam Springs.

a length l ¼ 30 mm, their cross-section is square with a side of a ¼ 1.2 mm, Young’s modulus is E ¼ 160 GPa, and the mass of the central plate is 180 mg. Use the state-space model and MATLAB to calculate the mass displacement when a force f ¼ 2$108d(t) N (where d(t) is the unit impulse) is applied to the plate. Ignore damping and inertia contributions from the beams. 8.32 Use complex impedances to determine the transfer function of the electrical system shown in Figure 8.39, and then convert the transfer function model R L

vi

C

– +

vo

FIGURE 8.39 Single-Stage Operational-Amplifier Electrical System.

into a state-space model by analytical calculation. Use MATLAB to plot the output voltage as a function of time with the aid of the state-space model. Known are R ¼ 360 U, L ¼ 0.6 H, C ¼ 50 mF, and vi ¼ 100$sin(12t) V. 8.33 (i) Determine the transfer function of the electrical system of Figure 8.40 for an input voltage vi and an output voltage vo. (ii) Convert the transfer function into a state-space model. Use the statespace model to plot vo for R1 ¼ 120 U, R2 ¼ 160 U, C1 ¼ 200 mF, C2 ¼ 300 mF, and vi ¼ 50 $ ð1 þ 10et Þ V.

Problems

C1

C2

vi

R1 vo R2

FIGURE 8.40 ResistiveeCapacitive Electrical System.

8.34 Consider an initial rotation angle of 6 degrees is applied to the pivoting rod of the lever system studied in Problem 8.29 and shown in Figure 8.37. Solve the linearized state-space model using the state-transition matrix method; also use Simulink to solve both the original nonlinear and the linearized state-space models. Known is m ¼ 0.7 kg. 8.35 Apply the state-space approach and the state-transition matrix to find the currents in the circuit of Figure 8.41 when an initial charge q0 ¼ 0.5 C is applied to the capacitor. Plot these currents in terms of time when known are R1 ¼ 150 U, R2 ¼ 180 U, L ¼ 0.6 H, and C ¼ 160 mF. L

R1

C

R2

FIGURE 8.41 Two-Mesh Electrical System With Initial Charge on Capacitor.

8.36 The homogeneous state-space model of a dynamic system is defined by the matrices  0 1 ½A ¼ ; ½C ¼ f 2 0 g 110 0:3 Use MATLAB to plot the system response y(t) if the following initial condition is applied: y(0) ¼ 0.1. 8.37 A homogeneous state-space model has the nonzero matrices: 2 3 1 0 0 6 7 ½A ¼ 4 0 1 0 5; ½C ¼ f 1 1 2 g 1 2 1 Using the state-transition method, determine the system’s response y(t) and plot it with respect to time for the initial condition: fxð0Þg ¼ f 2 1 1 gT .

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CHAPTER 8 State-Space Modeling

8.38 Use the state-transition matrix method to express and plot the output of the dynamic system described by the state-space model of Example 8.10. Consider zero initial conditions and an input is u(t) ¼ 10$e2t$sin(15t). 8.39 Consider the two-room thermal system of Figure 8.42. Derive a state-space model for the system and use MATLAB to plot the two room temperatures q1 and q2 when the outside temperature is qo ¼ 36 C and the initial room temperatures are q1(0) ¼ 28 C and q2(0) ¼ 15 C. All walls are identical and of dimensions 9 m $ 5 m $ 0.2 m; the wall material thermal conductivity is k ¼ 0.055 W/(m deg). The rooms’ thermal capacitance is Cth ¼ 130,000 J/deg. qo2

qo1

θo θ1

θ2 q

qo2

qo1 Rth Cth

Cth

Rth

qo1

qo2

FIGURE 8.42 Two-Room Thermal System.

8.40 Find a state-space model for the pneumatic system of Figure 8.43 knowing the input pressures pi,1 ¼ 40 atm and pi,2 ¼ 30 atm. Known also are the pneumatic resistances Rg1 ¼ 2100 m1 s1, Rg2 ¼ 3200 m1 s1, 1 1 Rg3 ¼ 1500 m s , and tank capacitances Cg1 ¼ 0.0006 m s2, 2 Cg2 ¼ 0.0005 m s . Use the state-transition matrix method, as well as Simulink, to find and to plot the pressures in the two containers. pi,2 Rg3

pi,1

Rg1

Rg2 p1

p2 Cg1

FIGURE 8.43 Pneumatic System With Two Containers and Ductwork.

Cg2

Suggested Reading

8.41 A dynamic system is defined by the state-space matrices    1 0 2 ½A ¼ ; ½B ¼ ; ½C ¼ f 0 1 g; ½D ¼ 3: 4 65 0 Find the system response and plot it against time when an input u ¼ 4$sin(30t) is applied to the system with zero initial conditions; use both MATLAB and Simulink for that. 8.42 A parallel RLC circuit is connected to a voltage source that provides a sinusoidal voltage v ¼ 90$sin(16t) V. It is also known that L ¼ 0.3 H and C ¼ 400 mF. Plot the currents in the three components by using Simulink and its state-space capability, when an initial charge q0 ¼ 0.2 C is applied to the capacitor and (i) The resistor is linear with R ¼ 160 U. (ii) The resistor defined at (i) has a saturation nonlinearity (discontinuity) defined by the limit voltages of 60 and 70 V. 8.43 Two bodies of masses m1 ¼ 0.8 kg and m2 ¼ 0.5 kg are connected by a spring and a damper connected in parallel defined by c ¼ 60 N s/m and k ¼ 140 N/m. The bodies slide on a horizontal surface with friction. The coefficient of static friction is ms ¼ 0.3, and the coefficient of kinematic friction is mk ¼ 0.2. A sinusoidal force f ¼ 50$sin(12t) N acts on the body of mass m2. Find a state-space model of this mechanical system and determine its time response using Simulink when nonlinear effects of Coulomb friction are considered.

Suggested Reading H. Klee and R. Allen, Simulation of Dynamic Systems with MATLAB and Simulink, 2nd Ed. CRC Press, Boca Raton, FL, 2011. N. S. Nise, Control System Engineering, 7th Ed. John Wiley & Sons, New York, 2015. B. C. Kuo and F. Golnaraghi, Automatic Control Systems, 10th Ed. John Wiley & Sons, New York, 2017. P. Marchand and O. T. Holland, Graphics and GUIs with MATLABÒ, 3rd Ed. Chapman & Hall/CRC, Boca Raton, FL, 2003. K. Ogata, System Dynamics, 4th Ed. Pearson Prentice Hall, Upper Saddle River, NJ, 2004. D. G. Luenberger, Introduction to Dynamics Systems, John Wiley & Sons, New York, 1979.

427

CHAPTER

Frequency-Domain Approach

9

OBJECTIVES This chapter studies the frequency-domain behavior of dynamic systems by introducing and using the concept of complex transfer function, which is based on the transfer function studied in Chapter 7. The frequency-domain model is also utilized in system controls, as discussed in Chapters 12 and 13. The following topics are analyzed: • Evaluation of the steady-state response of stable dynamic systems for harmonic input. • Frequency response Bode plots and asymptote (simplified) Bode plots. • Actuation, transmission, reduction, and sensing of mechanical vibrations for harmonic input. • Cascading nonloading systems with harmonic input. • Filtering in electrical and mechanical systems. • Conversion between the frequency response model and other linear, time-invariant models. • Utilization of specialized MATLAB commands to model and solve frequency-domain problems.

INTRODUCTION This chapter is centered on a particular form of the transfer function, which is the complex transfer function, and which is obtained from the transfer function using the substitution s ¼ u$j. Specifically, the main topic analyzes the steady-state response when the input (or forcing) to a dynamic system is sinusoidal (harmonic). The response, in that particular case, can be determined from the input and the complex transfer function and is obtained from the regular transfer function G(s) for single-input and singleoutput (SISO) systems or transfer function matrix [G(s)] for multiple-input and multiple-output (MIMO) systems. The frequency-domain response consists of the output amplitude and phase angle defined in terms of frequency, which becomes the variable. The two resulting graphical representations are known as Bode diagrams. MATLAB enables plotting the magnitude (ratio of the output to the input amplitudes) and the phase angle between output and input by built-in functions. System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00009-9 Copyright © 2018 Elsevier Inc. All rights reserved.

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CHAPTER 9 Frequency-Domain Approach

To help analysis and design in the frequency domain, Bode plots are simplified by using a logarithmic scale and asymptote lines instead of the actual curves. Mechanical vibration actuation, transmission, reduction, and sensing, as well as cascading nonloading dynamic systems and electrical/mechanical filters are analyzed in the frequency domain. MATLAB conversion between frequency response data and zero-pole-gain, transfer function, and state space models is also examined.

9.1 THE CONCEPT OF COMPLEX TRANSFER FUNCTION IN STEADY-STATE RESPONSE AND FREQUENCY-DOMAIN ANALYSIS The total solution (or time response) y(t) of a dynamic system to an input u(t) is the sum of the complementary solution (the solution of the homogeneous equation) yc(t)dalso known as natural solution and denoted by yn(t) as in Chapter 7dand a particular solution yp(t)dalso named the forced solution, yf (t), as presented in Chapter 7; that is, y(t) ¼ yc(t) þ yp(t). The complementary solution is related to the inherent properties of a system, whereas the particular solution is connected to the specific input. Our focus in this chapter falls on the steady-state solution, which is yss ðtÞ ¼ yðNÞ ¼ lim yðtÞ, under the assumption that the input is harmonic t/N

(sinusoidal, in particular). We demonstrate that the complementary solution is zero when time / N for stable systems; therefore, the total solution consists of only the particular solution, the latter one having a harmonic form. Before generalizing, let us solve the following example. nn

Example 9.1

Determine the steady-state response of a first-order system for a sinusoidal input u(t) ¼ U$sin(u$t) where U is the amplitude and u is the (circular) frequency.

Solution A first-order system, as introduced in Chapter 1, is defined by the differential equation

_ þ yðtÞ ¼ K $ uðtÞ s $ yðtÞ

(9.1)

with s being the time constant and K the static sensitivity. The complementary solution to the homogeneous equation attached to Eq. (9.1) has the form

yc ðtÞ ¼ c1 $ el $ t

(9.2)

where l is the root of the characteristic equation:

s$l þ 1 ¼ 0

(9.3)

Eq. (9.3) results from substituting yc(t) of Eq. (9.2) into the homogeneous equation that corresponds to u(t) ¼ 0 in Eq. (9.1). The solution to Eq. (9.3) is l ¼ 1/s; therefore, the complementary solution of Eq. (9.2) becomes

yc ðtÞ ¼ c1 $ et=s

(9.4)

9.1 The Concept of Complex Transfer Function in Steady-State

As the time constant is a positive quantity and c1 is a constant, it can be seen that

yc ðNÞ ¼ lim yc ðtÞ ¼ 0 t/N

(9.5)

therefore, the total solution is identical to the particular solution when time goes to infinity:

yðNÞ ¼ lim yðtÞ ¼ yp ðNÞ ¼ lim yp ðtÞ t/N

t/N

(9.6)

The particular solution and its time derivative are

yp ðtÞ ¼ Y $ sinðu $ t þ 4Þ; y_p ðtÞ ¼ u $ Y $ cosðu $ t þ 4Þ

(9.7)

They are substituted into the differential Eq. (9.1) where u ¼ U$sin(u$t). After expanding the sine and cosine in yp(t),y_p ðt Þ and identifying the sin(u$t) and cos(u$t) factors in the resulting equation, the unknown amplitude Y and phase angle 4 are found as

K Y ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ U; 4 ¼ tan1 ðs $ uÞ ¼ tan1 ðs $ uÞ 1 þ ðs $ uÞ2

(9.8)

and the particular solution of Eq. (9.7) becomes

  K yp ðtÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ U $ sin u $ t þ tan1 ðs $ uÞ 1 þ ðs $ uÞ2

(9.9)

The particular solution is a sinusoidal function that has the same frequency as the input function, a phase angle with respect to the input, and amplitude that is proportional to the input amplitude. Let us check that the response amplitude and phase angle can be obtained from a specific form of the system’s transfer function. The transfer function corresponding to Eq. (9.1) is

GðsÞ ¼

YðsÞ K ¼ UðsÞ s $ s þ 1

(9.10)

When the substitution s ¼ u$j is used, Eq. (9.10) changes to

Gðu $ jÞ ¼

Yðu $ jÞ K K K $s$u ¼ ¼  $j 2 Uðu $ jÞ 1 þ s $ u $ j 1 þ ðs $ uÞ 1 þ ðs $ uÞ2

¼ ReðGðu $ jÞÞ þ ImðGðu $ jÞÞ $ j

(9.11)

G(u$j) of Eq. (9.11) is a complex number defined by a real part, denoted by Re, and an imaginary part, expressed as Im. The corresponding modulus jGðu $ jÞj and phase angle ; Gðu $ jÞ are

jGð u $ jÞj ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K ½ReðGð u $ jÞÞ2 þ ½ImðGð u $ jÞÞ2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðs $ uÞ2

(9.12)

ImðGð u $ jÞÞ ¼ tan1 ðs $ uÞ ¼ 4 ; Gð u $ jÞ ¼ tan1 ReðGð u $ jÞÞ If we now compare Eqs. (9.9) and (9.12), it follows that

yp ðtÞ ¼ jGðu $ jÞj $ U $ sin½u $ t þ ; Gðu $ jÞ

(9.13)

In other words, the steady-state output characteristics can be obtained from the amplitude and phase angle of G(u$j), which is known as the complex (or sinusoidal) transfer function. nn

431

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CHAPTER 9 Frequency-Domain Approach

It is noteworthy that evaluating the behavior at infinity of a time-domain response to harmonic input shifts the analysis in the frequency domain because the amplitude and the phase angle of the sinusoidal steady-state response depend on the input frequency solely.

9.2 STEADY-STATE RESPONSE OF DYNAMIC SYSTEMS TO HARMONIC INPUT Analytical methods and MATLAB can be used to evaluate the steady-state response of dynamic systems under sinusoidal input (forcing), as shown in the following sections of this chapter.

9.2.1 Analytical Approach The steady-state response of SISO and MIMO dynamic systems are formulated and studied analytically here.

SISO Systems This section studies SISO systems in relation to the problem of harmonic excitation and the corresponding steady-state solution. It also introduces the main frequency response parameters for first-order and second-order systems.

Steady-State Solution Under Harmonic (Sinusoidal) Input The results obtained for the particular case of a first-order system (see Example 9.1) can be extended to systems of higher order, as shown in the following. When the input is of harmonic form, u ¼ U$sin(u$t), its relationship to the output and the transfer function into the s-domain is YðsÞ ¼ GðsÞ $

U $u s2 þ u2

(9.14)

Eq. (9.14) indicates that the total response adds contributions from the system itself, represented by G(s), and from the forcing (input), identified by the Laplace transform of U$sin(u$t). The partial fraction expansion on the right-hand side of Eq. (9.14) leads to   X c1 c2 þ YðsÞ ¼ Gi ðsÞ þ (9.15) ¼ Yc ðsÞ þ Yp ðsÞ s  u$j s þ u$j i where the sum denoted by Yc(s) results from expanding the Laplace transform of the complementary solution and contains partial fractions resulting from G(s). The twofraction sum denoted by Yp(s) is the Laplace transform of the particular solution and results from expanding the fraction on the right-hand side of Eq. (9.14). For a stable system (more on this topic is discussed in Chapters 7 and 12), the inverse Laplace transform of Yc(s) is zero when time goes to infinity; therefore, the steady-state

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

response consists of only the particular solution, as demonstrated for first-order systems. Consequently, we are interested in working with only Yp ðsÞ ¼ GðsÞ $

U $u c1 c2 þ ¼ 2 þu s  u$j s þ u$j

s2

(9.16)

The constants c1 and c2 are determined using the cover-up method (see Chapter 6) as 8   U $ u U > > c ¼ GðsÞ $ ¼ $ Gðu $ jÞ > 1 > < s þ u $ j s¼ju 2j (9.17)   > U $ u U > > > ¼  $ Gð u $ jÞ : c2 ¼ GðsÞ $ s  u $ j 2j s¼ju A complex number can be written in phasor form (see Appendix A) as Gðu $ jÞ ¼ jGðu $ jÞj $ e j $ ;

Gðu,jÞ

(9.18)

At the same time, the numbers G(u$j) and G(eu$j) in Eq. (9.17) are complex conjugates; therefore, they have the same modulus and their angles are symmetric with respect to the real axis. As a result, Eqs. (9.14)e(9.18) result in   1 e j $ ; Gðu $ jÞ 1 ej $ ; Gðu $ jÞ Yp ðsÞ ¼ U $ jGðu $ jÞj $ $  $ (9.19) 2j 2j s  u$j s þ u$j Application of the inverse Laplace transform to Eq. (9.19) results in yp ðtÞ ¼ U $ jGðu $ jÞj $

e j $ ½u $ tþ;

Gðu $ jÞ

 ej $ ½u $ tþ; 2j

¼ U $ jGðu $ jÞj $ sin½u $ t þ ; Gðu $ jÞ

Gðu $ jÞ

(9.20)

Eq. (9.20), which is used the exponential formulation of the sine function, shows that the steady-state response can be expressed as a sinusoidal function of time: yp ðtÞ ¼ Yp $ sin½u $ t þ 4

(9.21)

where, by comparison to Eq. (9.20), Yp ¼ U $ jGðu $ jÞj; 4 ¼ ; Gðj $ uÞ

(9.22)

In other words, • •

The steady-state amplitude results from the input amplitude U multiplied by the modulus of the complex transfer function jG(u$j)j. The phase angle between the output and the input is the angle of the complex transfer function, ; Gðu $ jÞ.

The frequency response therefore consists of the two functions, the modulus of the complex transfer function and the phase of the same function expressed in terms of the frequency u. Alternatively, these amounts are called the frequency response magnitude denoted by M(u) and the frequency response phase, denoted by 4(u).

433

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CHAPTER 9 Frequency-Domain Approach

Performing a frequency-domain analysis means studying the two functions M(u) and 4(u). The plots of M and 4 in terms of u are known as Bode plots (or diagrams) in recognition of the scientist H.W. Bode’s work in the area of frequency-domain controls of the 1920s, 1930s. As it results from G(s), the complex transfer function G(u$j) is the ratio of two complex-valued polynomials. Because a complex number can be formulated in phasor form, as in Eq. (9.18), the complex transfer function modulus is the ratio of the numerator to the denominator moduli, whereas the complex transfer function phase angle is the difference between the numerator and denominator phase angles.

Stability and Steady-State Response For an absolutely stable system, with all the poles of the transfer function situated in the left part of the complex plane, the particular solution yp(t), as well as the total solution y(t) ¼ yc(t) þ yp(t) are plotted qualitatively in Figure 9.1. The figure identifies the transient (initial) time-domain, where the total solution differs from the particular solution. The plot also shows the steady-state domain, where the total and particular solutions are identical as time grows to infinity. A different situation arises in the case of marginally stable systems, whose transfer functions possess simple imaginary poles in addition to, possibly, poles situated in the left-hand complex plane. Consider, for instance, the transfer function: GðsÞ ¼

NðsÞ 

ðs þ aÞ $ s2 þ u2n

(9.23)

where N(s) is the numerator polynomial and a > 0. The poles of this transfer function are ea (a negative real pole) and un$j (which is a simple imaginary pole). The total time response of this system to a sinusoidal input u(t) ¼ U$sin(u$t) is plotted in Figure 9.2(a). The presence of the negative pole ea generates a transient response, as indicated in the same figure, where the total response for the initial time interval Transient domain

Steady-state domain t→∞

y = yc + yp

yp Time

FIGURE 9.1 Particular and Total Time Response of Absolutely Stable System Under Sinusoidal Input.

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

Transient domain

Transient domain ≡ Steady-state domain

Steady-state domain

t→∞

y = yc + yp

y = yc + yp

t→∞

Time

Time

(a)

(b)

FIGURE 9.2 Total Time Response of Marginally Stable System Under Sinusoidal Input With: (a) Simple Imaginary Poles and Poles in the Left-Hand Plane; (b) Simple Imaginary Poles.

(slightly) differs from the periodic response (the steady-state response) that follows after that. It should be observed that the steady-state response in this case is the sum of two sine curves, one resulting from the transfer function imaginary poles un$j and the other produced by the sinusoidal input, which indicates that YðsÞ ¼ GðsÞ $ UðsÞ ¼ with

U $ NðsÞ  ¼ Ytr ðsÞ þ Yss ðsÞ ðs þ aÞ $ s þ u2n $ ðs2 þ u2 Þ

2

c N1 ðsÞ N2 ðsÞ ; Yss ðsÞ ¼ 2 þ 2 Ytr ðsÞ ¼ 2 sþa s þ un s þ u2

(9.24)

where Ytr(s) is the Laplace transform of the transient response and Yss(s) is the Laplace transform of the steady-state response, c is a constant, while N1(s) and N2(s) are numerator polynomials. The latter portion comprises now the system’s contribution through the natural imaginary poles un$j in addition to the imaginary poles corresponding to the input and which are u$j. As a consequence, the time-domain, steady-state response of such a system is the superposition of two sine partial responses: yss ðtÞ ¼ yðNÞ ¼ L1 ½Yss ðsÞ ¼ Yn $ sinðun $ t þ 4n Þ þ Yp $ sinðu $ t þ 4Þ (9.25) A purely marginally stable system has only simple imaginary poles such as the system defined by the following transfer function: GðsÞ ¼

NðsÞ þ u2n

s2

(9.26)

The Laplace-domain response of this system to the sinusoidal input u ¼ U$sin(u$t) is Y ðsÞ ¼ GðsÞ $ U ðsÞ ¼

U $ N ðsÞ 

 ¼ Yss ðsÞ s þ u2n $ s2 þ u2 2

(9.27)

435

CHAPTER 9 Frequency-Domain Approach

with Yss(s) given in Eq. (9.24). The corresponding time-domain response is yðtÞ ¼ yss ðtÞ ¼ L1 ½YðsÞ ¼ L1 ½Yss ðsÞ ¼ Yn $ sinðun $ t þ 4n Þ þ Yp $ sinðu $ t þ 4Þ

(9.28)

In other words, the steady-state response of purely marginally stable systems to sinusoidal input is identical to the total time response (since there is no transient response) and is the sum of two sinusoidal functions. A plot of y(t) from Eq. (9.28) is qualitatively illustrated in Figure 9.2(b). This chapter confines its coverage to the steady-state response of absolutely stable systems.

Frequency Response Parameters of First-Order Systems The transfer function of a first-order system whose mathematical model is given in Chapter 1 and its complex counterpart are GðsÞ ¼

K K þ 0$j ; Gðu $ jÞ ¼ s$s þ 1 1 þ ðs $ uÞ $ j

(9.29)

The magnitude and phase angle corresponding to the frequency response are obtained from Eq. (9.29) as K jGðu $ jÞj ¼ MðuÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 4ðuÞ ¼ tan1 ð0=KÞ  tan1 ðs $ uÞ 1 þ ðs $ uÞ2 ¼ tan1 ðs $ uÞ

(9.30)

Phase angle (deg)

which were also obtained in Eq. (9.12). The corresponding Bode plots are displayed in Figure 9.3 for K ¼ 1 and two values of the time constant: s ¼ 1 and s ¼ 0.2. As shown in Figure 9.3, both the magnitude and phase angle decay to zero when u / N. Larger values of the time constant produce a faster reduction in both frequency response variables.

Magnitude (abs)

436

0.707

Bandwidth

ωb

τ = 0.2

τ=1

Frequency (rad/s)

(a)

τ = 0.2 τ=1

Frequency (rad/s)

(b)

FIGURE 9.3 Linear-Scale Frequency Response Plots of First-Order System: (a) Magnitude; (b) Phase Angle.

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

MATLAB has the bode(sys) command to obtain the two plots once a linear time-invariant model, defined as sys, is available (as a zero-pole-gain, transfer function, or state-space model). Instead of a linear frequency scale (as in Figure 9.3), Bode plots utilize a logarithmic scale for frequency because using log10 u instead of simply u provides a wider dynamic range (for instance, when u ¼ 1000 rad/s, the logarithmic counterpart is only log10 u ¼ log10 1000 ¼ log10 103 ¼ 3). Oftentimes, a logarithmic scale is utilized for magnitudes as well instead of absolute values, which enables transforming products of transfer functions in linear (absolute) representation into sums of the logarithms of those transfer functions. For nonloading cascading systems (serially connected systems without interstage loading), as is shown later in this chapter and in Chapter 13, where the total magnitude is the product of individual systems’ magnitudes, using logarithmic scales for both frequency and magnitude permits plotting M(u) and 4(u) in a simplified manner by means of line segments.

–3 dB

Bandwidth ωb

FIGURE 9.4 Logarithmic-Scale Frequency Response Plots of First-Order System: (a) Magnitude; (b) Phase Angle.

The magnitude M(u) in a Bode plot can be expressed decibels, and the number of decibels is calculated as 20$log10 M(u). The magnitude is the ratio of two signal amplitudes, and power is proportional to the square of the amplitude (such as in electrical systems where power ¼ resistance  current2); in other words, jG(u$j)j2 ¼ M(u)2 is proportional to the power ratio. A factor of 10 in a power ratio is named bel (in honor of the telephone’s inventor, Alexander Graham Bell), and the number of bels corresponding to the magnitude is # bels ¼ log10jG(u$j)j2. Considering that 1 B has 10 dB, the number of decibels (dB) is # decibels ¼ 10$log10jG(u$j)j2 ¼ 20$log10jG(u$j)j. Figure 9.4 shows the MATLAB-produced magnitude and phase plots in logarithmic scale for the transfer function of Eq. (9.30) and s ¼ 0.2s.

437

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CHAPTER 9 Frequency-Domain Approach

One important characteristic of the first-order system frequency response is the bandwidth and its corresponding bandwidth frequency, ub. The bandwidth frequency is defined as the frequency where the input power amplitude is reduced by half, and since power is proportional to the square of amplitude, this frequency is given by the equation rffiffiffi .pffiffiffi 1 2 M ðub Þ ¼ ¼ 0:707 or M ðub Þ ¼ 20 $ log10 1 2 pffiffiffi ¼ 20 $ log10 2 x  3 dB; ub ¼ 1=s (9.31) The bandwidth frequency is identified in both Figure 9.3 (where M is in absolute units) and in Figure 9.4 (where M is plotted in dB). The bandwidth is the plot area defined by the 0-to-ub frequency range. nn

Example 9.2

The input to the mechanical system of Figure 9.5 is the force u ¼ U$sin(u$t). Considering that the output is the mass velocity v, design this mechanical system such that its static sensitivity is K ¼ 0.8, and the frequency response magnitude is 0.5 at a frequency of 60 rad/s. Calculate the system’s bandwidth. y c m

u

FIGURE 9.5 Mass-Damper Mechanical System With Harmonic Input.

Solution The mathematical model of this mechanical system is

m 1 $ v_ þ v ¼ $ u c c where v ¼ y_ is the mass velocity. Laplace transforming the last Eq. (9.32) results in m $ y€ þ c $ y_ ¼ u or GðsÞ ¼

VðsÞ 1=c ¼ UðsÞ ðm=cÞ $ s þ 1

(9.32)

(9.33)

Comparing Eqs. (9.29) and (9.33) it follows that the static sensitivity is K ¼ 1/c, and the time constant is s ¼ m/c. Because K ¼ 0.8, it follows that the damping coefficient is c ¼ 1/K ¼ 1.25 N s/m. Eq. (9.33) is combined with Eq. (9.30) to express the particular magnitude of this example as

0:8 0:5 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðs $ 60Þ2

(9.34)

which is solved for the time constant resulting in s ¼ 0.0208 s. As a consequence, the mass is m ¼ s$c ¼ 0.026 kg. The bandwidth is found from Eq. (9.31) as ub ¼ 1/s ¼ 48.077 rad/s. nn

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

Frequency Response Parameters of Second-Order Systems For a second-order system with unitary gain (K ¼ 1), whose mathematical model is introduced in Chapter 1, the transfer function and its corresponding complex counterpart are GðsÞ ¼

u2n u2n þ 0 $ j

 ; Gðu $ jÞ ¼ s2 þ 2x $ un $ s þ u2n u2n  u2 þ ð2x $ un $ uÞ $ j

(9.35)

The magnitude and phase angle are obtained from Eq. (9.35): u2n 1 8 jGð u $ jÞj ¼ M ðuÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ M ðbÞ;



 > 2 2 2 2 2 2 < þ ð2x $ un $ uÞ þ ð2x $ bÞ2 un  u 1b ! !   > : 0 1 2x $ un $ u 1 2x $ b  tan ¼ tan ¼ 4ðbÞ 4ðuÞ ¼ tan1 u2n u2n  u2 1  b2 (9.36)

where b ¼ u/un is the nondimensional frequency ratio. Figure 9.6 plots the magnitude and phase angle curves in terms of the frequency ratio for two different damping ratios, x1 ¼ 0.1 and x2 ¼ 0.3. As the second Eq. (9.36) shows, the phase angle becomes infinity for b ¼ 1; that is when the input frequency u is equal to the natural frequency un. The magnitude curve displays a maximum in magnitude, which is known as resonant peak, denoted as P in Figure 9.6(a), and occurs at the resonant frequency ur. The resonant frequency is found by equating the derivative of M(b) of Eq. (9.36) to zero: 

4b $ b2  1 þ 2x2 dMðbÞ ¼  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 ¼ 0 (9.37) db

2 1  b2 þ ð2x $ bÞ2 2

ξ1 = 0.1 P2 (βr2, Mr2)

Phase angle (deg)

Magnitude

P1 (βr1, Mr1) ξ2 = 0.3 ξ1 = 0.1

ξ2 = 0.3

β = ω/ωn

β = ω/ωn

(a)

(b)

FIGURE 9.6 Linear-Scale Frequency Response Plots of Second-Order System: (a) Magnitude; (b) Phase Angle.

439

440

CHAPTER 9 Frequency-Domain Approach

The valid solution of Eq. (9.37) is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi br ¼ 1  2x2 or ur ¼ un $ 1  2x2

(9.38)

P2 –3 dB

Bandwidth ωb

FIGURE 9.7 Logarithmic-Scale Frequency Response Plots of Second-Order System: (a) Magnitude; (b) Phase Angle.

Real values of the frequency are possible only for 1 e 2x2  0, which ffiffiffi presonant means that x  1 2 ¼ 0:707. For damping ratios that are larger than 0.707, there is no resonance at all. Eq. (9.38) also indicates that the resonant peak moves to the left in the magnitude plot since br decreases as the damping ratio increases. For all x  0.707, the resonant magnitude is found by combining the first Eqs. (9.36) and (9.38): Mr ¼

1 pffiffiffiffiffiffiffiffiffiffiffiffiffi 2x $ 1  x2

(9.39)

It can be seen that the resonant frequency depends on both the natural frequency and damping ratio, while the resonant magnitude depends only on the damping ratio. The bandwidth frequency is determined by solving the equation pffiffiffi Mðub Þ ¼ 1 2x0:707 with M(u) of Eq. (9.36), which results in rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ub ¼ un $

1  2x2 þ

4x4  4x2 þ 2

(9.40)

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

Figure 9.7 illustrates the logarithmic Bode plots of G(s) given in Eq. (9.35) for x ¼ 0.3 and un ¼ 100 rad/s. In addition to indicating the resonant peak P2, the magnitude plot also identifies the bandwidth and the bandwidth frequency ub that corresponds to a magnitude of M ¼ 3 dB (or to M ¼ 0.707 when M is expressed in absolute units). nn

Example 9.3 Determine the frequency response of the electrical system shown in Figure 9.8 for sinusoidal input voltage vi ¼ Vi$sin(u$t) V, when R ¼ 10 U, L ¼ 0.5 H, and C ¼ 500 mF. Determine the resonant peak (ur and Mr) and calculate the bandwidth frequency ub. Also determine the resistance value that reduces the resonant magnitude to half its original value.

R

vi

L

C

vo

FIGURE 9.8 Electrical System With Resistor, Inductor, and Capacitor.

Solution As seen in Chapter 7, the transfer function of this system is calculated based on impedances as

Vo ðsÞ Z2 1=ðC $ sÞ 1 ¼ ¼ ¼ Vi ðsÞ Z1 þ Z2 R þ L $ s þ 1=ðC $ sÞ L $ C $ s2 þ R $ C $ s þ 1 1=ðL $ CÞ ¼ 2 s þ ðR=LÞ $ s þ 1=ðL $ CÞ (9.41)

GðsÞ ¼

Z1 is the impedance representing the serially connected resistor and inductor, Z1 ¼ R þ L$s, whereas Z2 represents the impedance corresponding to the capacitor, Z2 ¼ 1/(C$s). Comparison of Eq. (9.41) with the generic Eq. (9.35) shows that the natural frequency and damping ratio are

un ¼ 1

pffiffiffiffiffiffiffiffiffiffi L $ C ; 2x $ un ¼ R=L or x ¼ R=ð2un $ LÞ

(9.42)

namely un ¼ 63.2456 rad/s and x ¼ 0.1581. Based on Eqs. (9.38) and (9.39), the resonant peak coordinates are ur ¼ 61.6441 rad/s and Mr ¼ 3.2026. The bandwidth frequency is found from Eq. (9.40) as ub ¼ 96.526 rad/s.

441

442

CHAPTER 9 Frequency-Domain Approach

Eq. (9.39) allows formulating the following equation:

4ð0:5Mr Þ2 $ ðx Þ4  4ð0:5Mr Þ2 $ ðx Þ2 þ 1 ¼ 0

(9.43)

where x* is the modified damping ratio. Solving Eq. (9.43) results in one value that is equal to 0.9437, which is not valid as it is larger than the admissible 0.707 value. The other root is x* ¼ 0.3309, which is an acceptable value. The second Eq. (9.42) allows calculating the modified resistance as

R ¼ 2x $ un $ L with a value of R* ¼ 20.928 U.

(9.44) nn

nn

Example 9.4

A microcantilever of constant rectangular cross section has a thickness h ¼ 200 nm, a width w ¼ 5 mm, and a length l ¼ 50 mm. The microcantilever undergoes harmonic out-of-plane bending vibrations. Use lumped-parameter modeling to determine the cantilever mass density r, as well as the damping coefficient c of the gaseous environment the cantilever vibrates in. Consider that known are the following parameters: un ¼ 5.65$105 rad/s, ur ¼ 5.1$105 rad/s, and E ¼ 160 GPa.

Solution The damping ratio is determined from Eq. (9.38) as

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   1 u2r x¼ 1 2 2 un

(9.45)

which yields x ¼ 0.3043. As shown in Chapter 2, the lumped-parameter model of a cantilever consists of a spring of equivalent stiffness keq, given in Table 2.2, and a point equivalent mass meq, provided in Table 2.6 as

8 > 3E $ Iz 3E w $ h3 E $ w $ h3 > > ¼ ¼ 3 $ < keq ¼ 12 l l3 4l3 > > 33 33 > : meq ¼ $m ¼ $r$w$h$l 140 140

The two Eq. (9.46) are substituted into the expression of the natural frequency un ¼ latter equation yields

r¼

35E $ h2 33u2n $ l4

(9.46) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi keq =meq. This

(9.47)

thus, r ¼ 3402.179 kg/m3. The damping coefficient is calculated as

c ¼ 2x $ un $ meq 8

whose numerical value is c ¼ 1.38$10

N s/m.

(9.48) nn

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

Asymptote Representation of Bode Plots Bode plots can be represented in a simplified manner as piecewise-continuous line segments when using the logarithmic frequency scale, as well as the logarithmic (dB) magnitude. Since a transfer function G(s) and the related function G(u$j) can be expressed in factor form, the logarithmic magnitude 20$log10 M(u)dexpressed in dBdis an algebraic sum of logarithmic terms, each being a linear function of the logarithmic frequency log10 u. The phase 4 is the algebraic sum of the angles defining the factors of the complex transfer function and can further be represented as a piecewise-continuous linear function in terms of frequency. A few simple transfer functions are analyzed next by this asymptote approach, which is also utilized in Chapter 13. 1. G(s) ¼ K Real constants appear often as gains in complex transfer functions. Since G(s) ¼ K, it follows that G(u$j) ¼ K, which means that M ¼ K and the dB magnitude is constant: MdB ¼ 20$log10 M ¼ 20$log10 K. The phase angle is 4 ¼ tan1(0/K) ¼ 0. 2. G(s) ¼ sn and G(s) ¼ 1/sn For G(s) ¼ sn, the sinusoidal transfer function is G(u$j) ¼ (u$j)n, whose magnitude is M(u) ¼ un. Using the dB definition, the magnitude is written as MdB ¼ 20$log10 M(u) ¼ (20$n)$log10 u. This equation represents a line segment of slope (20$n) when the variable is log10 u and Figure 9.9 plots this variation for several values of n. It is common practice to also indicate the actual frequency u that corresponds to log10 u, as indicated in the abscissa of Figure 9.9 Note that all segments pass through the point defined by M ¼ 0 dB and log10 u ¼ 0 (or u ¼ 1). The distance between two consecutive points on the frequency axis (in absolute values) is known as decade (short-form notation is dec) and as such, the slope of the line segments are 20$1 ¼ 20 dB/dec for n ¼ 1, 20$2 ¼ 40 dB/dec for n ¼ 2, 20$3 ¼ 60 dB/dec for n ¼ 3, and so on. The intercepts of those segments can simply be evaluated at the abscissa log10 0.1 ¼ 1: when n ¼ 1, for instance, MdB ¼ (20$1)$(e1) ¼ 20 dB, as illustrated in Figure 9.9. M(dB) 60

n = 3, 60 dB/dec n = 2, 40 dB/dec

40 20

n = 1, 20 dB/dec

0

n = −1, −20 dB/dec

−20 −40

n = −2, −40 dB/dec n = −3, −60 dB/dec

−60 −1

0

1

2

log10(ω)

0.1

1

10

100

ω

FIGURE 9.9 Log-Scale Asymptote Representation of Magnitude and Phase of G(s) ¼ sn and G(s) ¼ 1/sn.

443

444

CHAPTER 9 Frequency-Domain Approach

The angle of G(u$j) can be evaluated using De Moivre’s identity, according to which the n-th power of a complex number z ¼ jzj $ ðcosð£zÞ þ sinð£zÞ $ jÞ is calculated as zn ¼ jzjn $ ðcosðn $ £ zÞ þ sinðn $ £ zÞ $ jÞ. The angle of u$j ¼ 0 þ u$j being tan1(u/0) ¼ 90 , it follows that the angle of G(u$j) is £Gð u $ jÞ ¼ £ð u $ jÞn ¼ n $ 90+ . The sinusoidal transfer function of G(s) ¼ 1/sn is G(u$j) ¼ 1/(u$j)n ¼ (u$j)en. Its magnitude is M(u) ¼ 1/un ¼ uen and therefore, MdB ¼ 20$log10 M(u) ¼ (20$n)$log10 u. As shown in Figure 9.9, the magnitude plots are symmetric to the plots resulting from G(s) ¼ sn with respect to the zero dB line. It can be shown that the phase angles are calculated as £ Gðu $ jÞ ¼ £ðu $ jÞn ¼ n $ 90 . 3. G(s) ¼ (s$s þ 1)n and G(s) ¼ 1/(s$s þ 1)n M(dB) n = 2, 40 dB/dec 40 n = 1, 20 dB/dec

20 0

n = −1, −20 dB/dec

−20 −40

n = −2, −40 dB/dec 0.01/τ

0.1/τ

ωc = 1/τ

10/τ

100/τ

ω

(a) ϕ(deg) 90 45

ωc

n = 1, 45°/dec

0

–90 0.01/τ

(c)

n = –1, –45°/dec

(b)

–45

0.1/τ

ωc = 1/τ

10/τ

ω

FIGURE 9.10 Frequency Response for G(s) ¼ (s$s þ 1)n and G(s) ¼ 1/(s$s þ 1)n: (a) Log-Scale Asymptote Representation of Magnitude; (b) Asymptote Representation of Phase; (c) Actual Bode Plots of G(s) ¼ (0.1$s þ 1).

The sinusoidal form of G(s) ¼ (s$s þ 1)n is G(u$j) ¼ (1 þ s$u$j)n whose h i1=2 magnitude is MdB ðuÞ ¼ ð20nÞ $ log10 1 þ ðs $ uÞ2 . For u 1/s, which means u$s 1, the magnitude is approximately zero, whereas for u [ 1/s, which means u$s [ 1, the magnitude becomes MdB ðuÞ ¼ ð20nÞ $ h i1=2 log10 ðs $ uÞ2 ¼ ð20nÞ $ log10 ðs $ uÞ ¼ ð20nÞ $ log10 ðsÞ þ ð20nÞ $ log10 u: The value uc ¼ 1/s is known as corner (or break) frequency and it can be seen

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

that it leads to MdB (uc) ¼ 0. Figure 9.10(a), which shows only the absolute frequency on the abscissa, plots the magnitude for n ¼ 1 and 2. The angle of G(u$j) when u is small (u / 0) is zero. At the corner frequency (and using De

n Moivre’s identity), £Gðu $ jÞ ¼ £ 1 þ s $ 1s $ j ¼ n $ 45 , whereas for u assuming large values (u / N), the phase angle is n$90 . Figure 9.10(b) is the asymptotic plot of 4 in terms of u for n ¼ 1. In this representation, the angle values of 0, 1$45 ¼ 45 and 1$90 ¼ 90 are linearly interpolated over two decades symmetrically around the corner frequency uc ¼ 1/s; the line segment has a slope of 45 /dec. There are obviously errors in representing the actual magnitude and phase by means of asymptotes and Figure 9.10(c) shows the actual Bode plots corresponding to the transfer function G(s) ¼ 0.1$s þ 1. The maximum error in magnitude occurs at the corner frequency uc ¼ 1/s ¼ 1/ 0.1 ¼ 10, and this is 20$log10 (2)1/2 ¼ 10$log10 2 ¼ 3.03 dB. Figure 9.10(a) and 9.10(b) also show the first asymptote plots of G(s) ¼ 1/ (s$s þ 1)n ¼ (s$s þ 1)en. Note that both the magnitude and phase plots are obtained following a reasoning similar to the one described for the plots of G(s) ¼ (s$s þ 1)n. The minus sign of n instead of the plus sign generates the symmetry with respect to the 0 dB line. As such, the magnitude is simplified to h i1=2 MdB ðuÞ ¼ ð20nÞ $ log10 ðs $ uÞ2 ¼ ð20nÞ $ log10 ðs $ uÞ ¼ ð20nÞ $ log10 ðsÞ  ð20nÞ $ log10 ðuÞ when u [ 1/s, or u$s [ 1. Similarly, the phase

n ¼ n $ 45 at the corner frequency becomes £Gðu $ jÞ ¼ £ 1 þ s $ 1s $ j and is en$90 when u / N. nn

Example 9.5

Identify the transfer function G(s) whose asymptote magnitude plot is shown in Figure 9.11(a). M(dB)

M(dB)

60

60

40

40

20

20

0 –20

–1

1

2 x = log10ω

0 –20 –40

M –40 dB/dec

–1

0

M1 1

M2

2

x = log10ω M3 –20 dB/dec

–40 dB/dec –80

–80

–60 dB/dec

(a)

(b)

FIGURE 9.11 Asymptote Bode Magnitude Diagrams of (a) Transfer Function G(s); (b) Component Transfer Functions G1(s), G2(s), and G3(s).

445

446

CHAPTER 9 Frequency-Domain Approach

Solution

Figure 9.11(a) indicates that the dB magnitude corresponding to the function G(s) is expressed as



M¼

20  40x; 40  60x;

1  x  1 x1

(9.49)

where x ¼ log10 u. This function can be obtained by combining three basic transfer functions G1(s), G2(s), and G3(s), whose magnitudes are added up as follows:

(

M ¼ M1 þ M2 þ M3 ¼

20 þ ð 40xÞ þ 0

1x1

20 þ ð 40xÞ þ ð20  20xÞ; x  1 ( 0; 1  x  1 with M1 ¼ 20; M2 ¼ 40x; M3 ¼ 20  20x; x  1

(9.50)

These magnitudes point out that their corresponding transfer functions are G1(s) ¼ 10 (because 20$log10 10 ¼ 20), G2(s) ¼ 1/s2 (because the magnitude of this function has a 40 dB/dec slope and starts from a 40 dB intercept on the M axis), and G3(s) ¼ 1/(0.1$s þ 1) ¼ 10/(s þ 10); The corner frequency for G3 is uc ¼ 1/s ¼ 1/0.1 ¼ 10 rad/s or x ¼ 1. This point is also separating the two frequency subintervals, as seen in Figure 9.12(a). Also note that M3 results from its general definition for n ¼ 1: M3 ¼ 20$1$log10 (0.1) 20$1$log10 (u) ¼ 20  20$log10 (u) ¼ 20  20$x. Figure 9.11(b) illustrates the magnitudes of the three component transfer function as well as the magnitude of the overall transfer function. As a consequence, G(s) is expressed as the product of G1(s), G2(s), and G3(s):

GðsÞ ¼ G1 ðsÞ $ G2 ðsÞ $ G3 ðsÞ ¼ 10 $ ¼

1 1 1 10 ¼ 10 $ 2 $ $ s2 0:1s þ 1 s s þ 10

100 100 ¼ 3 2 s $ ðs þ 10Þ s þ 10s2

(9.51)

M1 M2

ϕ1 (deg)

Frequency (deg/sec)

ϕ2 (deg)

Frequency (deg/sec)

FIGURE 9.12 Magnitude and Phase Angle Plots for a Pneumatic System With One Input and Two Outputs. nn

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

MIMO Systems MIMO systems can also be rendered into frequency-domain models using the transfer function matrix, as shown in the following. Also discussed here is the modality of superimposing multiple sinusoidal inputs and calculating the resulting steady-state response of MIMO systems.

Complex Transfer Function Matrix Approach For MIMO systems under the action of harmonic input, it is necessary to first determine the transfer function matrix [G(s)], which creates a relationship between the Laplace-transformed input {U(s)} and the output {Y(s)} vectors. The next step is to use the substitution s ¼ u$j to obtain the complex transfer function matrix [G(u$j)]. Let us analyze an example, and then generalize the frequency response of MIMO systems. nn

Example 9.6 Determine and plot the frequency response of the pneumatic system shown in Figure 5.14 of Example 5.10, evaluating the pressures in the two vessels when the input pressure pi is sinusoidal. Known are the following pneumatic system properties: Rg1 ¼ 500 m1 s1, Rg 2 ¼ 600 m1 s1, Cg1 ¼ 0.004 m s2, Cg 2 ¼ 0.005 m s2.

Solution By combining the second and fourth of Eq. (5.102), the following equation is produced:

Rg2 $ Cg2 $

dpo ðtÞ þ po ðtÞ  pðtÞ ¼ 0 dt

(9.52)

A similar combination of the first, second, and third of Eq. (5.102) results in

Rg1 $ Cg1 $

  Rg1 Rg1 dpðtÞ þ 1þ $ po ðtÞ ¼ pi ðtÞ $ pðtÞ  dt Rg2 Rg2

(9.53)

Eqs. (9.52) and (9.53) can be assembled in vectorematrix form:



Rg1 $ Cg1 0  ¼

pi ðtÞ

9 8 2 > dpðtÞ > > Rg1 > >  > = < 0 dt 6 1 þ Rg2 þ4 $ > Rg2 $ Cg2 dpo ðtÞ > > > > > 1 ; : dt 

3 Rg1    pðtÞ Rg2 7 5$ po ðtÞ 1 (9.54)

0

The input is pi and the output components are the two pressures in the vessels, p and po. The matrix form of this system model is

_ ½a1  $ fyðtÞg þ ½a0  $ fyðtÞg ¼ fuðtÞg

(9.55)

By applying the Laplace transform with zero initial conditions to Eq. (9.55), a relationship {Y(s)} ¼ [G(s)]${U(s)} is obtained; the transfer function matrix [G(s)] is

½GðsÞ ¼ ðs $ ½a1  þ ½a0 Þ1

(9.56)

447

448

CHAPTER 9 Frequency-Domain Approach

If G11(s), G12(s) are the elements of the transfer function matrix on the first row, and G21(s), G22(s) are the elements on the second row, the following equation is formulated:



PðsÞ Po ðsÞ





¼

   Pi ðsÞ $ G21 ðsÞ G22 ðsÞ 0 G11 ðsÞ G12 ðsÞ

(9.57)

which can be reduced to the following scalar equations

8 6s þ 2 > > < PðsÞ ¼ G11 ðsÞ $ Pi ðsÞ ¼ 12s2 þ 15s þ 2 $ Pi ðsÞ > > : Po ðsÞ ¼ G21 ðsÞ $ Pi ðsÞ ¼

2 $ Pi ðsÞ 2 12s þ 15s þ 2

(9.58)

where G11(s) and G21(s) result from the numerical values of this example and MATLAB. Each equation in Eq. (9.58) represents a SISO system, which can be treated separately. The complex transfer functions corresponding to G11(s) and G21(s) are needed first. They are

G11 ð u $ jÞ ¼

2 þ 6u $ j 2 ; G21 ð u $ jÞ ¼ (9.59) 2 2 ð2  12u Þ þ 15u $ j ð2  12u Þ þ 15u $ j

The following moduli and phase angles are obtained from Eq. (9.59):

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 4 þ 36u2 15u > > > ; ; G11 ðu $ jÞ ¼ 41 ¼ tan1 ð3uÞ  tan1 < jG11 ðu $ jÞj ¼ M1 ¼ 4 2 2  12u2 144u þ 177u þ 4 > > 2 15u > : jG21 ðu $ jÞj ¼ M2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ; ; G21 ðu $ jÞ ¼ 42 ¼ tan1 4 2 2  12u2 144u þ 177u þ 4 (9.60) and these amounts are plotted in Figure 9.12.

nn

The previous example with one input and two outputs needs only two transfer functions of the 2  2 transfer function matrix to obtain the two outputs. For the general MIMO case, where there might be m inputs and p outputs, the general relationship between the input vector and output vector through the transfer function matrix is written directly with s ¼ u$j as fYðu $ jÞg ¼ ½Gðu $ jÞ $ fUðu $ jÞg

(9.61)

The complex transfer matrix [G(u$j)] contains m  p transfer functions connecting every component of the input vector to every component of the output vector. To graphically qualify the frequency response of such a system, m  p magnitude plots and an equal number of phase angle plots are needed.

Linear Superposition for Steady-State Time Response The steady-state time response of a MIMO system under multiple sinusoidal input can be found through the linear superposition of individual responses since any component of the output vector results as the sum of all input components multiplied by their corresponding transfer functions. The output component i, for instance, is computed as

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

yi ðtÞ ¼

m X jGik ð uk $ jÞj $ Uk $ sinðuk $ t þ 4ik Þ;

for i ¼ 1; 2; .; p

449

(9.62)

k¼1

with jGik ðuk $ jÞj ¼

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ImGik ðuk $ jÞ ½ReGik ðuk $ jÞ2 þ ½ImGik ðuk $ jÞ2 ; 4ik ¼ tan1 ReGik ðuk $ jÞ

(9.63)

where Re and Im are the real and imaginary parts of the complex transfer function Gik, which connects the output i to the input k. In Eq. (9.62), Uk is the amplitude of the sinusoidal input k, and 4ik is the phase angle between the output i and the input k. nn

Example 9.7

Consider the mechanical system of Figure 9.13 with m1 ¼ m2 ¼ m3 ¼ 1 kg, c1 ¼ c2 ¼ 2 N s/m, k1 ¼ k2 ¼ k3 ¼ 1 N/m is acted on by the forces u1 ¼ 5$sin(t) N; u2 ¼ 2$sin(3t) N. Express the steady-state displacements of the three bodies (time responses) of this system under the given force inputs.

y1 u1

m1

y2

c1

y3 c2

m2

u2 k1

m3

k3

k2

FIGURE 9.13 Lumped-Parameter Mechanical System With Harmonic Input Forces.

Solution The mechanical system has two inputs and three outputs; its equations of motion are

8 : > < m1 $ y€1 ¼ u1  c1 $ ðy_1  y2 Þ  k1 $ ðy1  y2 Þ m2 $ y€2 ¼ u2  c1 $ ðy_2  y_1 Þ  k1 $ ðy2  y1 Þ  c2 $ ðy_2  y_3 Þ  k2 $ ðy2  y3 Þ > : m3 $ y€ ¼ c2 $ ðy_  y_ Þ  k2 $ ðy3  y2 Þ  k3 $ y3 3 2 3 (9.64)

Eq. (9.64) are obtained either using free-body diagrams in conjunction with Newton’s second law or Lagrange’s equations. Application of the Laplace transform to Eq. (9.64) with zero initial conditions generates the equation

2 6 4

m 1 $ s2 þ c 1 $ s þ k 1

ðc1 $ s þ k1 Þ

m2 $ s2 þ ðc1 þ c2 Þ $ s þ k1 þ k2 ðc1 $ s þ k1 Þ 0 ðc2 $ s þ k2 Þ 8 9 9 8 > < Y1 ðsÞ > = = > < U1 ðsÞ > $ Y2 ðsÞ ¼ U2 ðsÞ > > > : ; ; > : Y3 ðsÞ 0

0

3

7 5 ðc2 $ s þ k2 Þ m3 $ s2 þ c2 $ s þ k2 þ k3

450

CHAPTER 9 Frequency-Domain Approach

Eq. (9.65) enables solving for Y1(s), Y2(s), and Y3(s) in terms of U1(s) and U2(s) by inverting the matrix in the left-hand side of this equation (with numerical values for the element parameters) and multiplying the inverse by the input vector in the right-hand side of the same equation. Formally, the result can be written as

8 9 2 G11 ðsÞ > < Y1 ðsÞ > = 6 Y2 ðsÞ ¼ 4 G21 ðsÞ > > : ; Y3 ðsÞ G31 ðsÞ

G12 ðsÞ G22 ðsÞ G32 ðsÞ

9 3 8 > = < U1 ðsÞ > 7 G23 ðsÞ 5 $ U2 ðsÞ > > ; : 0 G33 ðsÞ G13 ðsÞ

(9.66)

The meaningful transfer functions of the matrix in Eq. (9.66)dthe ones related to the two nonzero inputsdare

8 > s4 þ 6s3 þ 8s2 þ 8s þ 3 > > > G11 ðsÞ ¼ 6 5 > > s þ 8s þ 17s4 þ 18s3 þ 10s2 þ 4s þ 1 > > > > > > 2s3 þ 5s2 þ 6s þ 2 > > ðsÞ ¼ G ðsÞ ¼ G 12 21 > > > s6 þ 8s5 þ 17s4 þ 18s3 þ 10s2 þ 4s þ 1 > > > < s4 þ 4s3 þ 7s2 þ 6s þ 2 G22 ðsÞ ¼ 6 > s þ 8s5 þ 17s4 þ 18s3 þ 10s2 þ 4s þ 1 > > > > > > 4s2 þ 4s þ 1 > > > ðsÞ ¼ G 31 > > s6 þ 8s5 þ 17s4 þ 18s3 þ 10s2 þ 4s þ 1 > > > > > > 2s3 þ 5s2 þ 4s þ 1 > > : G32 ðsÞ ¼ 6 s þ 8s5 þ 17s4 þ 18s3 þ 10s2 þ 4s þ 1

(9.67)

where the numerical values of this example have been used. The magnitudes and phase angles of the transfer functions in Eq. (9.67) are determined in the regular manner using s ¼ u$j. The magnitude of each is then the ratio of the numerator modulus to the denominator modulus, and the phase angle is the difference between the numerator’s phase angle and the denominator’s phase angle. Using the generic Eqs. (9.62), (9.63) and (9.66), the three steady-state outputs are expressed as

8 > < y1 ¼ U1 $ jG11 ð jÞj $ sinðt þ ; G11 ð jÞÞ þ U2 $ jG12 ð3jÞj $ sinð3t þ ;G12 ð3jÞÞ y2 ¼ U1 $ jG21 ð jÞj $ sinðt þ ; G21 ð jÞÞ þ U2 $ jG22 ð3jÞj $ sinð3t þ ;G22 ð3jÞÞ > : y3 ¼ U1 $ jG31 ð jÞj $ sinðt þ ; G31 ð jÞÞ þ U2 $ jG32 ð3jÞj $ sinð3t þ ;G32 ð3jÞÞ

(9.68) with U1 ¼ 5 and U2 ¼ 2 being the amplitudes of the inputs. The six complex transfer functions G11(u$j) through G32(u$j) are obtained from the corresponding ones listed in the transfer function matrix of Eq. (9.72) for G11, G21, and G31 when u ¼ 1 rad/s, as well as for G12, G22, and G32 when u ¼ 3 rad/s. The magnitudes and phase angles of all the complex transfer functions that are needed in Eq. (9.68) can be obtained using MATLAB. For instance, to determine the magnitude and phase angle corresponding to G11(j), the following MATLAB script is used: >> >> >> >>

syms s g11s=(s^4+6*s^3+8*s^2+8*s+3)/(s^6+8*s^5+17*s^4+18*s^3+10*s^2+4*s+1); g11=subs(g11s,j);%substitutes s by j to get g11(j) from g11(s) mag11=abs(g11);%finds the magnitude of g11(j)

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

y1 (m)

y2 (m)

Time (sec)

FIGURE 9.14 Steady-State Response of Two-Input, Three-Output Dynamic System. >> ang11=angle(g11); %finds the angle of g11(j) Similar commands can be used to find the other magnitudes and phase angles to assemble the time-domain Eq. (9.68). The steady-state time responses are plotted in Figure 9.14. The response curve y3(t) is practically identical to y2(t) and its plot was not included. nn

9.2.2 MATLAB Approach MATLAB has built-in functions that perform frequency-domain calculations and generate plots quite simply. Specifically, it is possible to determine the frequency response of dynamic systems through Bode plots using either zero-pole-gain, transfer function or state space models. Discrete frequency data can also be used to obtain frequency plots. Conversions are also possible in MATLAB from zero-pole-gain, transfer function, or state space models to a frequency response data model.

Bode Plots MATLAB’s bode command has already been introduced in this chapter as a means of obtaining Bode plots for SISO systems. MIMO systems can also be handled by means of the transfer function matrix, and the frequency response can readily be determined, as illustrated in the following example. nn

Example 9.8 Use MATLAB to plot the Bode diagrams corresponding to the transfer function matrix

2

1 6 s þ 2s þ 10 ½GðsÞ ¼ 6 4 2 2 s þ 2s þ 10 2

3 sþ2 s þ 2s þ 10 7 7 5 2s s2 þ 2s þ 10 2

451

452

CHAPTER 9 Frequency-Domain Approach

Solution The following MATLAB script can be used >> >> >> >> >> >>

g11 = tf(1,[1,2,10]); g12 = tf([1,2],[1,2,10]); g21 = tf(-2,[1,2,10]); g22 = tf([2,0],[1,2,10]); m = [g11,g12;g21,g22]; bode(m)

to generate the Bode diagrams of Figure 9.15. The 2  2 transfer function matrix indicates that the system is a two-input, two-output one; therefore, four plot pairs resulted. Under Edit/Figure Properties in the generated Bode plots, the Property Editor can be configured in terms of labels, limits, units for frequency, magnitude (dB or absolute), phase (degrees or radians), scale (logarithmic or linear), and style. The plots in Figure 9.15 used linear scale and absolute units for magnitude.

FIGURE 9.15 Bode Plots for a Two-Input, Two-Output Dynamic System.

nn

Frequency Response Data Handling MATLAB possesses the capability of handling frequency response data by creating a linear timeeinvariant (LTI) model. The command frd(r,f)dwhich stands for frequency response data, with r being a vector that contains the frequency response values (the values of G(u$j)) and f being the vector comprising the corresponding frequency valuesdcan be used when discrete experimental or numerically generated frequency-domain data are available and need to be further analyzed. For instance, Bode plots can be drawn for such data, as shown in the following example.

9.2 Steady-State Response of Dynamic Systems to Harmonic Input

nn

Example 9.9 Frequency response data are furnished by a frequency analyzer corresponding to the frequency range from 0 to 100 rad/s with an increment of 10 rad/s as follows: 0.1000, -0.0246-0.0031i, -0.0052-0.0003i, -0.0023-0.0001i, -0.0013, -8.0021e-004-1.6123e-005i, -5.5038e-004-9.3211e-006i, -4.0675e-0045.8120e-006i, -3.1004e-004-3.9342e-006i, -2.5089e-004-2.7127e-006i, -2.0038e-004-2.0078e-006i pffiffiffiffiffiffiffi where i ¼ j ¼ 1. Plot the magnitude of these data in terms of frequency and determine G(u$j) for u ¼ 88 rad/s.

Solution

We need to define the frequency response vector r as well as the frequency vector f before utilizing the frd function. Once the frd LTI model has been created, the command bodemag can be used to plot just the magnitude of the complex transfer function. Retrieving data from the frequency response, other than the discrete values that have been provided, is done by means of the freqresp command. The following MATLAB code is used to solve this problem: >> >> >> >> >>

f = 0:10:100; r = []; % input-given values of G(uj)dnot included here g = frd(r, f); bodemag (g) freqresp (g,88)

The last command returns the value of -2.6123e-04-2.9181e-06$i, whereas the bodemag command generates the plot of Figure 9.16.

FIGURE 9.16 Bode Plot of Magnitude From Frequency Response Data. nn

453

454

CHAPTER 9 Frequency-Domain Approach

Frequency Response Model Conversion MATLAB allows conversion to an frd model from any of the other three (previously defined) LTI models: zpk, tf, and ss models. Conversion to a frequency response model is possible by means of the command frd (sys, f), where sys is an LTI object (zpk, tf, or ss), and f is a frequency vector. The frd vector returns frequency response data for the specified frequencies. It should be noted that conversion from an existing frd model into any of the zpk, tf, or ss models is not possible in MATLAB. nn

Example 9.10 Obtain the frequency response data model when we know: (i) A zero-pole-gain model for which: the zeroes are 1 and 2; the poles are 0, 3 (of the second order of multiplicity), and 5; the gain is 1.5. (ii) A transfer function model defined by

GðsÞ ¼

1:5s2  4:5s þ 3  11s2 þ 39s  45Þ

s $ ðs3

Plot the magnitude and phase angle corresponding to the [0; 1000] rad/s frequency range for each model.

Solution (i) The following MATLAB script generates the frequency response data model g and plots the corresponding Bode diagrams of Figure 9.17: >> z = [1,2]; >> p = [0,3,3,5];

FIGURE 9.17 Bode Plots Resulting From a Zero-Pole-Gain Model Data.

9.3 Frequency-Domain Applications

>> f = zpk(z,p,1.5); >> om = 0:1:1000; >> g = frd(f,om); >> bode(g) (ii) The following MATLAB commands >> g = tf([1.5,-4.5,3],[1,-11,39,-45,0]); >> om = 0:1:1000; >> f = frd(g,om); >> bode(g) produce a frequency response LTI model f, whose Bode plots are the ones of Figure 9.17; the reader is encouraged to check the reason for that. nn

9.3 FREQUENCY-DOMAIN APPLICATIONS This section studies several applications where frequency-domain analysis is utilized including vibration transmission in mechanical systems, serially connected (cascading) systems, and electrical/mechanical filters.

9.3.1 Mechanical Vibration Transmission Under harmonic input, mechanical vibration can be transmitted among adjacent elements/systems by either displacement or forcing. Using the concept of transmissibility, this section studies vibration absorption, isolation, and measurement.

Transmissibility for Motion Input; Mass Detection by the Frequency Shift Method in MEMS The particular type of vibration transmission where the input is a motion (displacement) is known as transmissibility for motion input, as detailed in this section. In Chapter 3, the frequency shift method is discussed in connection with the possibility of detecting small amounts of matter that attach to microresonators through the change (shift) in the natural frequency of the original device. Variations in either the bending or the torsional natural frequencies are generally used by monitoring the natural frequencies of micro- and nano-cantilevers or bridges. In either of the cases, the flexible microstructure is excited over its frequency range, and the natural frequencies of interest are measured. After mass deposition takes place, frequency excitation is applied again, and the shifts in the original natural frequencies are detected, which enables evaluating the quantity of deposited mass and/or its position. Lumped-parameter models can be used to study such problems in the frequency domain. Figure 9.18(a) shows the side view of a cantilever whose root vibrates perpendicularly to its longitudinal axis along the u direction.

455

456

CHAPTER 9 Frequency-Domain Approach

y

m u y

h l

(a)

u = U·sin(ω·t) c

k

Massless vibrating support

(b)

FIGURE 9.18 Microcantilever With Harmonic Input: (a) Side View Schematic of Physical Model; (b) Lumped-Parameter Model.

As discussed in Chapter 2, the cantilever distributed inertia and stiffness can be lumped at its free end. Lumped damping (or equivalent damping), which accounts for all system losses is also considered here. Figure 9.18(b) is the lumpedparameter model of the moving cantilever with the input displacement u and the cantilever tip (absolute) displacement y aligned. The input is a sinusoidal motion applied by the massless support to the mass-spring-damper system. The equation of motion of the mass is _  k $ ðy  uÞ or m $ y€ þ c $ y_ þ k $ y ¼ c $ u_ þ k $ u (9.69) m $ y€ ¼ c $ ðy_  uÞ The following transfer function results after applying the Laplace transform to Eq. (9.69): GðsÞ ¼ where un ¼ Eq. (9.70) to

YðsÞ c$s þ k 2x $ un $ s þ u2n ¼ ¼ 2 2 UðsÞ m $ s þ c $ s þ k s þ 2x $ un $ s þ u2n

(9.70)

pffiffiffiffiffiffiffiffiffi k=m; 2x $ un ¼ c=m. Using s ¼ u$j, as well as b ¼ u/un, changes

G ð u $ jÞ ¼

Y ð u $ jÞ u2 þ ð2x $ un $ uÞ $ j 1 þ 2x $ b $ j ¼ 2 n 2 ¼ U ð u $ jÞ un  u þ ð2x $ un $ uÞ $ j 1  b2 þ 2x $ b $ j

(9.71)

The modulus of the ratio of two complex numbers is equal to the ratio of the moduli of the same numbers. At the same time, the moduli of Y(u$j) and U(u$j) are also their amplitudes, which isjUðu $ jÞj ¼ U; jYðu $ jÞj ¼ Y. It follows from Eq. (9.71) that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 $ b2 jYðu $ jÞj Y ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi TR ¼ jGðu $ jÞj ¼ (9.72)

2 jUðu $ jÞj U 1  b2 þ 4x2 $ b2 The modulus of the transfer function that was shown to be equal to the ratio of the output amplitude Y to the input amplitude U, is known as transmissibility and denoted by TR. Figure 9.19 plots the transmissibility based on Eq. (9.72) in terms of the frequency ratio for several values of the damping ratio. As such, the underdamped (x < 1), overdamped (x > 1), and critically damped (x ¼ 1) cases are illustrated.

9.3 Frequency-Domain Applications

ξ = 0.2

TR

ξ = 0.6

ξ=1

ξ=5

β = ω/ωn

FIGURE 9.19 Transmissibility as a Function of the Frequency Ratio for Various Damping Ratios.

As Figure 9.19 shows, all curves pass through the points b ¼ 0, TR ¼ 1 and b ¼ 21/2, TR ¼ 1 (these values of b are found by solving the equation TR ¼ 1). For these two frequency ratios, the amplitude transmitted to the body of mass m is equal to the input amplitude. The spike in transmissibility is visible for the small (underdamped) damping ratio of x ¼ 0.2. It can be checked that when x / 0, the transmissibility / N. For larger values of the damping ratio, including the critically damped and overdamped cases, the transmissibility curve flattens almost to superposition over TR ¼ 1. It can also be seen in Figure 9.19 that for b > 21/2, TR < 1 irrespective of the damping ratio, which indicates that the output amplitude is less than the input amplitude. nn

Example 9.11

An external mass equal to Dm ¼ 5$1015 kg deposits at the free end of a rectangular cross-section microcantilever, which is supported on a vibratory table, as shown in Figure 9.18. Analyze the frequency spectrum before and after mass deposition by plotting the two transmissibility curves and determine the value of the frequency shift. The cantilever has a length l ¼ 90 mm, a width w ¼ 10 mm (cross-sectional out-of-plane dimension in Figure 9.18(a)), and a thickness h ¼ 120 nm. It is also known that E ¼ 160 GPa and r ¼ 2500 kg/m3. The equivalent damping coefficient is c ¼ 1.55$109 N s/m.

Solution Similar to the generic derivation of the transfer function of Eq. (9.70), when the total mass that vibrates is m (the equivalent mass of the cantilever) plus Dm, the new transfer function is

G ðsÞ ¼

Y  ðsÞ c$s þ k ¼ UðsÞ ðm þ DmÞ $ s2 þ c $ s þ k

(9.73)

457

CHAPTER 9 Frequency-Domain Approach

Using the substitutions utilized in the generic transmissibility derivation, the following modified transmissibility is obtained:

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 $ b2 jY  ðu $ jÞj Y  ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi TR ¼ jG ðu $ jÞj ¼ (9.74)  2 U jUðu $ jÞj 1  ð1 þ rm Þ $ b2 þ 4x2 $ b2 



where rm ¼ Dm/m is the ratio of the deposited mass to the cantilever’s equivalent mass. The equivalent mass m and stiffness of the cantilever are calculated according to Tables 2.2 and 2.6 as m ¼ (33/144)r$w$h$l ¼ 6.364$1014 kg, k ¼ 3E$(w$h3)/(12$l 3) ¼ 9.48$104 N/m. It follows pffiffiffiffiffiffiffiffiffiffi that the natural bending-related frequency of the cantilever is un ¼ k=m ¼ 122; 057:162 rad=s, p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi while the modified natural frequency is un ¼ k=ðm þ DmÞ ¼ 117; 527:751 rad=s. The original damping ratio is x ¼ c/(2m$un) ¼ 0.0998 and the one after mass deposition is   x ¼ c 2ðm þ DmÞ $ un ¼ 0:096065. The mass ratio is rm ¼ Dm/m ¼ 0.07856. With these values, the two transmissibility functions of Eqs. (9.72) and (9.74) are

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 0:04b2 ffi; TR ¼ jGðu $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 1  b2 þ 0:04b2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 0:04b2   TR ¼ jG ðu $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 1  1:079b2 þ 0:04b2

(9.75)

Figure 9.20 is the plot of the transmissibilities in terms of the frequency ratio. The precise frequency shift is computed by evaluating the difference between the original and altered resonant frequencies as

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 þ 8x 1 þ 8x2  1  1  un $ Dur ¼ ur  ur ¼ un $ 2x 2x

(9.76)

The resonant frequencies of Eq. (9.76) correspond to the maximum values of TR and TR*; these frequencies are obtained by annulling the  derivatives of TR and TR*in terms of b e Eqs. (9.75) e and solving for br ¼ ur /un and br ¼ ur un , respectively.

Shifted response

Original response

TR

458

β = ω/ωn

FIGURE 9.20 Transmissibility and Frequency Ratio Change as a Result of Mass Addition on a Vibrating Nanocantilever.

9.3 Frequency-Domain Applications

For the numerical values of this problem, the difference of Eq. (9.76) is Dur ¼ 4405.23 rad/s, while the difference between the natural frequencies is Dun ¼ 4529.41 rad/s. Approximate frequency values can also be obtained by right clicking the peaks of the two MATLAB plots of Figure 9.20. nn

Transmissibility for Force Input In some situations, the input to a mechanical system is a harmonic force or torque instead of displacement, and the interest lies on how that force (or moment) is transmitted to other parts of the mechanical system as either force (moment) or displacement; this connection is known as transmissibility for force input. Consider the mechanical system of Figure 9.21(a), where a sinusoidal force acts on a body of mass m connected to a fixed base, through a damper of damping coefficient c and a spring of stiffness k. We want to compare the amplitude of the input force F to the amplitude Fb, which is transmitted to the base.

f f = F·sin(ω·t) y

m

y

m

fe

fd k

c

fb Fixed base

(a)

(b)

FIGURE 9.21 Mechanical System With Input Harmonic Force: (a) Lumped-Parameter Model; (b) FreeBody Diagrams.

Based on the free-body diagram of the mass and base, Figure 9.21(b), the equation of motion of the mass is m $ y€ ¼ f  fd  fe

(9.77)

where fd is the damping force and fe is the elastic force. Using known relationships, Eq. (9.77) becomes m $ y€ þ c $ y_ þ k $ y ¼ f

(9.78)

The force transmitted to the base, fb, is equal to the sum of the damping and elastic forces, as the damper and the spring connect the moving body to the fixed base; therefore, fb ¼ c $ y_ þ k $ y

(9.79)

459

460

CHAPTER 9 Frequency-Domain Approach

Laplace-transforming Eqs. (9.78) and (9.79) with zero initial conditions, results in the following transfer function, after substituting Y(s) from one equation into the other: GðsÞ ¼

Fb ðsÞ c$s þ k ¼ FðsÞ m $ s2 þ c $ s þ k

(9.80)

Eq. (9.80) is identical to Eq. (9.70), which expressed the transfer function between the input displacement of the base U(s) and the output displacement of the body Y(s). Since the input to this system is sinusoidal, the output is sinusoidal, as well; therefore, the transmissibility can be expressed as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 $ b2 jFb ðu $ jÞj Fb TR ¼ jGðu $ jÞj ¼ ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (9.81)

2 jFðu $ jÞj F 1  b2 þ 4x2 $ b2 where Fb is the amplitude of the force transmitted to the base, this force being expressed as fb ¼ Fb $ sinðu $ t þ 4Þ

(9.82)

nn

Example 9.12 Two identical electric motors that are synchronized and rotate in opposite directions are placed symmetrically on a rigid platform connected to the ground by means of two identical dashpot supports, as shown in Figure 9.22(a). The mass center of each rotor has an eccentricity e with respect to its rotation axis, as sketched in Figure 9.22(b). Determine the amplitude of the force transmitted to the ground when the rotation frequency is one half the system’s natural frequency. Known are the mass of the platform and the motor stators, which is mp ¼ 80 kg, the mass of one rotor mr ¼ 0.5 kg, the damping coefficient c ¼ 80 N s/m, the spring stiffness k ¼ 200 N/m, and the eccentricity e ¼ 4 mm.

ω

Electric motors

y fc

y

ω

mp

mr

Rigid c

k

k

c

e ω

Rotor ω·t x

Fixed support

(a)

(b)

FIGURE 9.22 Rigid Platform With Eccentric Motors and Damper-Spring Supports: (a) Schematic of the System; (b) Enlarged Sketch of the Right Motor Showing Eccentricity and Centrifugal Force.

9.3 Frequency-Domain Applications

Solution Figure 9.22(b) is an enlarged schematic view of the rotor’s center of mass; it rotates eccentrically around the bearings axis, which coincides with the geometric center axis of the motor placed on the right of Figure 9.22(a). The eccentricity produces a centrifugal force, whose vertical projection is

fy ¼ fc $ sinðu $ tÞ ¼ mr $ u2 $ e $ sinðu $ tÞ ¼ Fy $ sinðu $ tÞ

(9.83)

Due to synchronization between the two motors, a force identical to the one of Eq. (9.83) acts on the rotor on the left in Figure 9.22(a). The horizontal projections of the centrifugal forces on the two rotors cancel out because of opposite rotation directions, whereas the vertical projections are identical. As a consequence, Figure 9.23 shows the simplified, lumped-parameter physical model of the mechanical system. f = F·sin(ω·t) y

m

2c

2k

Fixed support

FIGURE 9.23 Lumped-Parameter Model of the Platform-Motors Mechanical System. The amplitude of the force acting on the mass m (which is m ¼ mp þ 2mr) results from Eq. (9.83):

F ¼ 2Fy ¼ 2mr $ u2 $ e

(9.84)

The following equations can be written:



m $ y€ þ 2c $ y_ þ 2k $ y ¼ f fb ¼ 2c $ y_ þ 2k $ y

(9.85)

where fb is the force transmitted to the base (ground) through the two damper-spring pairs. Application of Laplace transform with zero initial conditions to Eq. (9.85) produces the following transfer function and complex counterpart:

GðsÞ ¼

Fb ð s Þ 2ð c $ s þ k Þ 2k þ ð2c $ uÞ $ j ¼ ; Gð u $ jÞ ¼ F ðsÞ m $ s2 þ 2c $ s þ 2k ð2k  m $ u2 Þ þ ð2c $ uÞ $ j (9.86)

which results in a transmissibility:

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 k2 þ ðc $ uÞ2 jFb ðu $ jÞj Fb ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi TR ¼ jGðu $ jÞj ¼ F jFðu $ jÞj 2 ð2k  m $ u2 Þ þ ð2c $ uÞ2

(9.87)

Combining Eqs. (9.84) and (9.87) yields

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4mr $ u2 $ e $ k2 þ ðc $ uÞ2 Fb ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2k  m $ u2 Þ2 þ ð2c $ uÞ2

(9.88)

461

CHAPTER 9 Frequency-Domain Approach

pffiffiffiffiffiffiffiffiffiffiffiffi The natural frequency of the mechanical system is un ¼ 2k=m ¼ 2.22 rad/s. The excitation frequency is therefore u ¼ un/2 ¼ 1.11 rad/s. With the numerical values of this example, the amplitude of the force transmitted to the base is Fb ¼ 0.0062 N. nn

Vibration Absorption and Vibration Isolation When a harmonic force acts on a body that is placed on a fixed base through a damper-spring connection, it is possible to reduce the amplitude of the force transmitted to the support, but it is impossible to completely eliminate it, as can be seen in Eq. (9.87) because the numerator, which is proportional to the force amplitude, is never zero: k2 þ (c$u)2 s 0. One way of isolating (eliminating) the base force is adding a massespring system to the existing system, as shown in the next example. nn

Example 9.13 A sinusoidal force acts on the mass-damper-spring mechanical system of Figure 9.24(a). Study the possibility of reducing to zero the force transmitted to the fixed base by adding a massespring pair to the existing system, as illustrated in Figure 9.24(b). y1

y2 k

f = F sin(ωt)

Base

462

m c

y1 k

ka ma

m f = F sin(ωt)

(a)

c

(b)

FIGURE 9.24 Mechanical System With Sinusoidal Input: (a) Original System; (b) System With Added Mass and Spring.

Solution The equations of motion for the two-degree of freedom (DOF) mechanical system of Figure 9.24(b) are

(

ma $ y€2 ¼ ka $ ðy2  y1 Þ m $ y€1 ¼ f  ka $ ðy1  y2 Þ  c $ y_1  k $ y1

(9.89)

The force transmitted to the base results from the damper and spring contact and is expressed as

fb ¼ c $ y_1 þ k $ y1

(9.90)

9.3 Frequency-Domain Applications

Laplace transforming Eqs. (9.89) and (9.90) results in

8

 2 > < ma $ s þ ka $ Y2 ðsÞ ¼ ka $ Y1 ðsÞ m $ s2 þ c $ s þ ka þ k $ Y1 ðsÞ ¼ FðsÞ þ ka $ Y2 ðsÞ > : Fb ðsÞ ¼ ðc $ s þ kÞ $ Y1 ðsÞ

(9.91)

Successive substitution of Y2(s) and Y1(s) from the first two Eq. (9.91) into the third of Eq. (9.91) results in

 ðc $ s þ kÞ $ ma $ s2 þ ka Fb ðsÞ ¼ FðsÞ ðma $ s2 þ ka Þ $ ðm $ s2 þ c $ s þ ka þ kÞ  ka2

(9.92)

The modulus of the base force can be now determined by means of the transmissibility as

Fb ¼ jFb ðu $ jÞj ¼ TR $ jFðu $ jÞj 

jk þ c $ u $ jj $ ka  ma $ u2 $F ¼ ðma $ s2 þ ka Þ $ ðm $ s2 þ c $ s þ ka þ kÞ  ka2

(9.93)

Eq. (9.93) indicates that when

ka  ma $ u2 ¼ 0 or

ka ¼ u2 ma

(9.94)

the amplitude of the force transmitted to the base is zero. In other words, when the natural frequency of the added massespring subsystem becomes equal to the input frequency of the sinusoidal force, the transmitted vibration is annihilated. This additional mass-spring system behaves as a vibration isolator. For constant additional mass and stiffness, the isolator is effective solely at the excitation frequency of Eq. (9.94)dthis type of device is known as passive isolator. Active isolators, on the other hand, can alter either of the two additional element properties and, as a result, can counteract several excitation frequencies. nn

Measuring Vibration Displacement and Acceleration Amplitudes The transmissibility principle can be used to measure the amplitude of either the displacement or the acceleration of a vibratory input, as studied in the following example. nn

Example 9.14 A massless platform undergoes a sinusoidal motion, as shown in Figure 9.25. Using a mass-damperspring system as a sensor, analyze the situations when displacement or acceleration amplitudes can be detected, using the relative motion of the body with respect to the platform.

Solution The equation of motion for the mass of Figure 9.25 is

_  k $ ðx  uÞ m $ x€ ¼ c $ ðx_  uÞ

(9.95)

€ ¼ m $ u€  c $ ðx_  uÞ _  k $ ðx  uÞ m $ ðx€  uÞ

(9.96)

which is rewritten as

463

CHAPTER 9 Frequency-Domain Approach

u x k m c

Vibrating platform

464

Scale

FIGURE 9.25 Mechanical Instrument to Measure Vibration.

The motion of interest is the one of the mass relative to the vibrating platform; therefore, the following relative motion coordinate is introduced:

y¼xu

(9.97)

m $ y€ þ c $ y_ þ k $ y ¼ m $ u€

(9.98)

Eq. (9.96) becomes

The Laplace transform is applied to Eq. (9.98), which results in the transfer function

GðsÞ ¼

YðsÞ m $ s2 ¼ UðsÞ m $ s2 þ c $ s þ k

(9.99)

The magnitude corresponding to G(u$j), which derives from G(s), is

jGðu $ jÞj ¼

m $ u2 jYðu $ jÞj Y ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jUðu $ jÞj U 2 ðk  m $ u2 Þ þ ðc $ uÞ2

u2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2

u2n  u2 þ ð2x $ un $ uÞ2

(9.100)

where the definitions of the natural frequency un and damping ratio x have been used. Depending on the relationship between u and un, two cases are possible, as discussed next. System With Small Natural Frequency and High Excitation Frequency: u [ un Eq. (9.100) can be changed to

1 Y ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi $U 2  2 un un 2  1 þ 2x $ u2 u

(9.101)

after dividing the numerator and denominator by u2. Because u [ un, the square of the natural-toexcitation frequency ratio is a very small positive quantity, which can be neglected. As a consequence, Eq. (9.101) simplifies to

YxU

(9.102)

9.3 Frequency-Domain Applications

Eq. (9.102) indicates that the instrument measurement y is approximately equal to the amplitude of the input motion. As a result, the system can be used as a displacement sensor (also known as seismometer). System With Large Natural Frequency and Small Excitation Frequency: u un In this situation, Eq. (9.100) can be written as

u2n 1 $ Y ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    ffi$U 2 u2 u2 u 2 1  2 þ 2x $ un un

(9.103)

after dividing the numerator and denominator by u2n.The square of the excitation-to-natural frequency ratio is a very small positive quantity this time, and as a consequence, the denominator of the fraction on the right-hand side of Eq. (9.103) is equal to 1 approximately. Therefore, Eq. (9.103) simplifies to

Yx

 1 2 $ u $U 2 un

(9.104)

Because the input motion is sinusoidal, the acceleration is sinusoidal:

a ¼ u2 $ U $ sinðu $ tÞ ¼ A $ sinðu $ tÞ

(9.105)

where A ¼ u $U is the acceleration amplitude. As a consequence, and as per Eq. (9.104), the amplitude of the relative motion of the mass is proportional to the amplitude of the input acceleration; therefore, the instrument can be used as an acceleration sensor. nn 2

9.3.2 Steady-State Response of Nonloading Cascading Systems There are instances, such as in measuring instrumentation, where several dynamic systems interact in a cascading (series) manner, such that the output from one system becomes the input to the next system. As introduced in Chapter 7, such systems where the output from one subsystem converts fully into the input to the following subsystem are called nonloading systems. We study here the steady-state response of nonloading cascading systems with a single harmonic input and with multiple harmonic inputs.

Single-Input Systems Figure 9.26 shows schematically a nonloading cascading system formed of two components and a single input. Due to the nonloading condition and as discussed in Chapter 7, a single transfer function G(s) ¼ G1(s)$G2(s) can be used to connect the input U(s) to the output Y(s) because YðsÞ ¼ G2 ðsÞ $ Y1 ðsÞ ¼ G1 ðsÞ $ G2 ðsÞ $ UðsÞ or YðsÞ ¼ GðsÞ $ UðsÞ with GðsÞ ¼ G1 ðsÞ $ G2 ðsÞ U(s)

Y1(s) G1(s)

Y(s) G2(s)

FIGURE 9.26 Two-Stage Nonloading Cascading Dynamic System With a Single Input.

(9.106)

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CHAPTER 9 Frequency-Domain Approach

When the input to the first system is sinusoidal with an amplitude U and a frequency u, the output from the second system is yðtÞ ¼ U $ jGðu $ jÞj $ sinðu $ t þ 4Þ where

(9.107)

 jGðu $ jÞj ¼ jG1 ðu $ jÞj $ jG2 ðu $ jÞj; 4 ¼ 41 þ 42 ¼ £ G1 ðu $ jÞ þ £ G2 ðu $ jÞ

(9.108)

This result can be generalized to a system made up of n cascading subsystems so that the output from one subsystem is input to the next one: The output from a serially connected system under sinusoidal input is of sinusoidal form; its magnitude is the product of individual magnitudes and its phase angle is the sum of all individual phase angles. nn

Example 9.15 A nonloading cascading dynamic system is formed of two subsystems: one is first order and the next one is second order, such that the output from the first-order system is the output to the second-order system. The two systems are defined by the following differential equations: 3y_1 þ y1 ¼ 20$sin(3t); y€ þ 2y_ þ 0.01 y ¼ 5y1 (see Figure 9.26). Determine the steady-state response of the system.

Solution The transfer functions corresponding to the differential equations of the two coupled systems are found by applying Laplace transforms to the differential equations:

G1 ðsÞ ¼

1 5 ; G2 ðsÞ ¼ 2 3s þ 1 s þ 2s þ 0:01

(9.109)

The magnitudes and phase angles are found from the G1(u$j) and G2(u$j) functions as

8 1 > jG1 ð u $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi; 41 ¼ tan1 ð3uÞ > > > > < 1 þ ð3uÞ2

  > 5 2u > 1 > ; 42 ¼ tan jG ð u $ jÞj ¼ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > 2 : 2 0:01  u2 0:01  u2 þ ð2uÞ2

(9.110)

Numerically, the magnitudes and phase angles of Eq. (9.110) are M1(u) ¼ 0.1104, M2(u) ¼ 0.4626, 41 ¼ 83.66 , 42 ¼ 33.72 for u ¼ 3 rad/s. The steady-state response is therefore

yðtÞ ¼ jG1 ð u $ jÞj $ jG2 ð u $ jÞj $ A $ sinðu $ t þ 41 þ 42 Þ ¼ 1:022 $ sinð3t  49:94 $ p=180Þ with A ¼ 20 and where the resulting phase angle of 49.94 is transformed in radians.

(9.111) nn

Multiple-Input Systems Consider now the nonloading cascading system of Figure 9.27, which consists of two subsystems and two inputs.

9.3 Frequency-Domain Applications

U1(s)

U2(s) Y(s)

G1(s)

G2(s)

FIGURE 9.27 Two-Stage Nonloading Cascading Dynamic System With Two Inputs.

Using the principle of linear superposition, according to which each of the two inputs produces its own output based on its flow trajectory through the cascading chain, in the end the total output is the sum of individual outputs, namely:  YU1 ðsÞ ¼ G1 ðsÞ $ G2 ðsÞ $ U1 ðsÞ (9.112) YðsÞ ¼ YU1 ðsÞ þ YU2 ðsÞ with YU2 ðsÞ ¼ G2 ðsÞ $ U2 ðsÞ or YðsÞ ¼ G1 ðsÞ $ G2 ðsÞ $ U1 ðsÞ þ G2 ðsÞ $ U2 ðsÞ

(9.113)

Considering the two inputs are sinusoidal of the form: u1 ðtÞ ¼ U1 $ sinðu1 $ tÞ; u2 ðtÞ ¼ U1 $ sinðu2 $ tÞ

(9.114)

the steady-state response y(t) is found based on Eq. (9.113) as yðtÞ ¼ U1 $ jG1 ðu1 $ jÞj $ jG2 ðu1 $ jÞj $ sinðu1 $ t þ ; G1 ðu1 $ jÞ þ ; G2 ðu1 $ jÞÞ þ U2 $ jG2 ðu2 $ jÞj $ sinðu2 $ t þ ; G2 ðj $ u2 Þ (9.115) Eq. (9.115) can definitely be generalized when more than two subsystems (transfer functions) are connected in series, and each caries its own input. nn

Example 9.16

A measuring system is formed of the nonloading cascading system shown in Figure 9.28 where G1(s) represents the system to be measuredda first-order system defined by K1 ¼ 10 and s1 ¼ 0.01 s. G2(s) is a first-order transducer characterized by K2 ¼ 12 and s2 ¼ 0.02 s, whereas G3(s) is a second-order filter (see details on filters in Section 9.3.3), which is defined by K3 ¼ 20 and un ¼ 110 rad/s and x ¼ 0.2. Known are also the two inputs: u1 ¼ 20$sin(30$t) and u3 ¼ 50$sin(60$t). Determine the steady-state response y(t) of this system and plot it with respect to time.

U3(s)

U1(s) G1(s)

G2(s)

G3(s)

Y(s)

FIGURE 9.28 Three-Stage Nonloading Cascading Measuring System With Two Inputs.

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CHAPTER 9 Frequency-Domain Approach

Solution Based on the diagram of Figure 9.28, the Laplace transform of the output is calculated as

YðsÞ ¼ G1 ðsÞ $ G2 ðsÞ $ G3 ðsÞ $ U1 ðsÞ þ G3 ðsÞ $ U3 ðsÞ

(9.116)

The transfer functions of Eq. (9.116) are

G1 ðsÞ ¼

K1 K2 K3 $ u2n ; G2 ðsÞ ¼ ; G3 ðsÞ ¼ 2 s1 $ s þ 1 s2 $ s þ 1 s þ 2x $ un $ s þ u2n

(9.117)

The steady-state response corresponding to the two harmonic inputs is calculated using Eq. (9.116) as

yðtÞ ¼ U1 $ jG1 ðu1 $ jÞj $ jG2 ðu1 $ jÞj $ jG3 ðu1 $ jÞj $ sinðu1 $ t þ ; G1 ðu1 $ jÞ þ; G2 ðu1 $ jÞ þ ; G3 ðu1 $ jÞÞ þ U3 $ jG3 ðu3 $ jÞj $ sinðu3 $ t þ; G3 ðu3 $ jÞÞ

(9.118)

Output

468

Time (sec)

FIGURE 9.29 Output of Three-Stage Cascading Nonloading System With Two Harmonic Inputs.

The magnitudes and phase angles of Eq. (9.118) are calculated as

8 K1 > > > jG1 ðu $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi; ; G1 ðu $ jÞ ¼ tan1 ðs1 $ uÞ > > > > 1 þ ðs1 $ uÞ2 > > > > > > K2 < jG2 ðu $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi; ; G2 ðu $ jÞ ¼ tan1 ðs2 $ uÞ > 1 þ ðs2 $ uÞ2 > > > >   > > > K3 $ u2n > 1 2 $ x $ un $ u > q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ðu $ jÞj ¼ ðu $ jÞ ¼ tan ; ; G jG > 3 3 >

2 2 > u2n  u2 : un  u2 þ ð2 $ x $ un $ uÞ2 (9.119)

9.3 Frequency-Domain Applications

with the particular frequencies and amplitudes needed in Eq. (9.118): u1 ¼ 30 rad/s, u3 ¼ 60 rad/s, U1 ¼ 20 and U2 ¼ 50. The steady-state time response is

yðtÞ ¼ 42; 299 $ sinð30t  0:9492Þ þ 1359:5 $ sinð60t  0:3011Þ

(9.120)

and is plotted in Figure 9.29. The phase angles in Eq. (9.120) are expressed in radians.

nn

9.3.3 Filters Electrical or mechanical filter systems are utilized to remove the input signal over certain frequency ranges and transmit it in its original or modified magnitude over subsequent frequency intervals. The most common filter types are sketched in Figure 9.30: they are low pass, high pass, band pass, or notch. A simple way to analyze filter behavior, as shown in Figure 9.30, is to have their absolute-value magnitude plotted against frequency. Ideally, the magnitude is equal to 1 over the passband and equal to 0 over the stopband. The frequency that separates the passband and the stopband is the corner frequency (also known as cutoff frequency), denoted by uc. The plots of Figure 9.30(a)e9.30(d) are ideal, highly simplified diagrams with the change between the pass and stop regions occurring at well-defined corner frequencies. M(ω)

M(ω)

1

1

0

ωc

ω

0

ωc1

(a)

ωc2

ω

(c)

M(ω)

M(ω)

1

1

Passband Transition band Stopband M(ω) 1 0.707 0

ωc

ω Roll-off slope

0

ωc

ω

(b)

0

ωc1

ωc2

(d)

ω

(e)

FIGURE 9.30 Filter Magnitude (Absolute Values) Versus Frequency: (a) Ideal Low Pass; (b) Ideal High Pass; (c) Ideal Band Pass; (d) Ideal Notch; (e) More Realistic Low Pass.

However, the more realistic behavior of a low-pass filter is illustrated in Figure 9.30(e), which uses simplified (asymptote) representation and highlights a transition band (with a roll-off slope) between the passband and the stopband. The same Figure 9.30(e) indicates pffiffiffithat the cutoff frequency is the solution to the equation Mðuc Þ ¼ Mðub Þ ¼ 1 2 ¼ 0:707, which is identical to the bandwidth frequency (as introduced in Section 9.2.1). A couple of examples of electrical and mechanical filters are presented next.

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CHAPTER 9 Frequency-Domain Approach

Electrical Filter Systems Electrical filters can be passive when they contain only passive elements such as resistors, capacitors, or inductors, or active, when they incorporate elements that need external energy sources, such as actuators, sensors, or operational amplifiers. nn

Example 9.17 Analyze the frequency response of the low-pass (Butterworth) filter of Figure 9.31 using the following sets of values: R1 ¼ 100 U, C1 ¼ 0.001 F, and R2 ¼ 1000 U, C2 ¼ 0.01 F. Determine the cutoff frequency for each case. R vi

C

vo

FIGURE 9.31 Electrical Low-Pass RC (Butterworth) Filter.

Solution The transfer function of the electrical circuit of Figure 9.31 is

GðsÞ ¼

Vo ðsÞ Z2 1 ¼ ¼ Vi ðsÞ Z1 þ Z2 R $ C $ s þ 1

(9.121)

which considered the complex impedances Z1 ¼ R and Z2 ¼ 1/(C$s). The two plots of Figure 9.32 have been obtained for the combinations shown in the legend using the tf and bodemag MATLAB commands. The conclusion is that the transition band (and the roll-off slope) cannot be changed by changing the values of either R or C. However, a shift to the left is achieved by increasing the values of either R or C. The magnitude is derived from Eq. (9.121) as

R1 = 100 Ω C 1 = 0.001 F

R2 = 1000 Ω C 2 = 0.01 F

FIGURE 9.32 Magnitude Ratio Versus Frequency for an RC Low-Pass (Butterworth) Filter.

9.3 Frequency-Domain Applications

1 MðuÞ ¼ jGðu $ jÞj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðR $ C $ uÞ2 therefore, the cutoff frequency is obtained based on Eq. (9.31), which isMðuc Þ ¼ 1

uc ¼

(9.122) .pffiffiffi 2, as

1 R$C

(9.123)

Numerically, the cutoff frequency values are uc1 ¼ 10 rad/s and uc2 ¼ 0.1 rad/s, for the two cases studied here. The companion website Chapter 9 shows that, by adding two or more stages identical to the one of Figure 9.31, it is possible to reduce the transition band. nn

Mechanical Filters Mechanical systems, too, can function as filters with respect to their input and output signals, as illustrated in the following example. nn

Example 9.18 Verify whether the mechanical system of Figure 9.33 produces any filtering effects when the input is the sinusoidal displacement u, and the output is the displacement y. Consider the following numerical values: m ¼ 1 kg, c ¼ 2 N s/m, and k ¼ 200 N/m. If applicable, calculate the related cut-off frequency/frequencies. y u m k

c

FIGURE 9.33 Translatory Mechanical System as a Filter.

Solution

The dynamic equation of motion for the body of mass m is

_  k$y m $ y€ ¼ c $ ðy_  uÞ

(9.124)

which results in the transfer function

GðsÞ ¼

YðsÞ c$s ¼ 2 UðsÞ m $ s þ c $ s þ k

(9.125)

The magnitude and phase angle corresponding to G(u$j) are

8 > c$u jY ð u $ jÞj > > ffi > M ðuÞ ¼ jGð u $ jÞj ¼ ¼ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > 2 < jU ð u $ jÞj k  m $ u 2 þ c2 $ u 2 > > p c$u > 1 > > : 4ðuÞ ¼ 2  tan k  m $ u2

(9.126)

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CHAPTER 9 Frequency-Domain Approach

FIGURE 9.34 Magnitude and Phase Plots of a Mechanical Filter. and Figure 9.34 shows the Bode plots associated with Eq. (9.126). The mechanical system works as a band-pass filter, as defined in Figure 9.30(c). The cutoff frequencies are obtained by solving the equation M(uc) ¼ 0.707, which results in uc1 ¼ 13.18 rad/s and uc2 ¼ 15.18 rad/s. nn

SUMMARY The concepts of complex transfer function and complex transfer function matrix are introduced and applied in this chapter as modeling tools of the steady-state response of SISO and MIMO dynamic systems under sinusoidal input. The frequency response of dynamic systems under sinusoidal excitation consists of the output amplitude and the phase angle that depend on the input frequency. They are graphically illustrated with the aid of Bode diagrams that can be represented by asymptotes. The topics of transmission, absorption, and elimination of mechanical vibrations generated by sinusoidal input are also studied, as well as the principles of mechanical vibration sensing, the nonloading cascading systems, and electrical/mechanical filter systems. Specialized MATLAB commands are utilized to model and solve several applications of these topics.

PROBLEMS 9.1 A sinusoidal input voltage vi ¼ Vi$sin(u$t) is applied to the electrical system of Figure 9.35 where the output is the voltage vo. What is the time constant s of this electrical system such that the frequency response magnitude is 0.8 at a frequency of 80 rad/s.

Problems

vi

L R

vo

FIGURE 9.35 InductiveeResistive Electrical System.

9.2 A point mass m is connected to a dashpot (a spring k in parallel to a damper c) to a massless moving platform, as shown in Figure 9.18(b). Under sinusoidal displacement excitation applied to the massless platform, the resonant peak is Mr ¼ 3.2, and the bandwidth is ub ¼ 310 rad/s. Determine the values of the natural frequency un and of the damping ratio x. 9.3 The electrical system studied in Example 9.3 and illustrated in Figure 9.8 where L ¼ 0.1 H, needs to realize a resonant frequency ur ¼ 80 rad/s and a bandwidth ub ¼ 125 rad/s. What values are necessary for R and C to achieve these performance parameters? 9.4 Consider the mechanical system studied in Example 9.2 and illustrated in Figure 9.5. Find the values of the system’s element parameters m and c that result in a static sensitivity K ¼ 1.6 and a frequency response magnitude of 0.6 at a frequency of 70 rad/s. 9.5 A unit-gain second-order system is defined by a natural frequency un ¼ 240 rad/s and a damping ratio x ¼ 0.3. What are the modified natural frequency and damping ratio values that will reduce the resonant peak Mr by 10% and the bandwidth frequency ub by 12%? 9.6 Calculate the resonant frequency and resonant peak of a dynamic system whose transfer function is G(s) ¼ (2s þ 3)/(s2 þ 2s þ 100). Determine and plot the magnitude and the phase angle corresponding to a sinusoidal input being applied to this system by means of the analytical approach and also using MATLAB. 9.7 Use a lumped-parameter model and the analytical transfer function approach to calculate the frequency-domain response of the mechanical microsystem sketched in Figure 9.36 when a sinusoidal input force is applied to the shuttle mass in the motion direction. Use MATLAB to verify the analytical results. Each of the two identical end springs has three flexible beams of circular cross section with d ¼ 0.8 mm, l1 ¼ 9 mm, l2 ¼ 25 mm, and E ¼ 160 GPa. The mass of the shuttle mass is m ¼ 240 mg and the coefficient of viscous damping between the mass and the surrounding gas is c ¼ 0.003 N s/m. 9.8 A fixed-gap, longitudinal motion, variable-capacitor microactuator (see Figure 4.12) is connected in an electrical circuit, as sketched in Figure 9.37. Knowing the gap g ¼ 3 mm, the capacitor plate width w ¼ 40 mm, the air

473

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CHAPTER 9 Frequency-Domain Approach

Spring

l1 m

Anchor

l2 l1

Shuttle mass

FIGURE 9.36 Translatory Mechanical Microsystem.

v L

R C

FIGURE 9.37 Electrical Microcircuit With Resistor, Inductor, Voltage Source, and Variable Capacitor.

permittivity ε ¼ 8.8$1012 F/m, the resistance R ¼ 0.22 U, and the inductance L ¼ 13 mH, plot the frequency response using MATLAB and applying the complex transfer function approach for two different superposition lengths of the capacitor armatures: x1 ¼ 12 mm and x2 ¼ 18 mm. Calculate analytically and evaluate graphically the peak magnitude response for the two cases. 9.9 Apply the complex impedance approach to the fan-container-duct pneumatic system of Figure 9.38 and plot the frequency response both analytically and with MATLAB considering the HagenePoiseuille pneumatic losses in the duct. Known are the following parameters: diameter and length of duct are di ¼ 0.014 m and l ¼ 15 m, container diameter d ¼ 0.6 m, gas density r ¼ 1.2 kg/m3, and gas dynamic viscosity m ¼ 2$105 N s/m2. Do not include inertia effects and consider that the input pressure pi is sinusoidal. pi Fan Gas container po

FIGURE 9.38 Pneumatic System With Container, Duct, and Supply Fan.

Duct

Problems

9.10 A constant rectangular cross-section nanocantilever vibrates in out-of-plane bending driven by a sinusoidal input displacement applied to the anchor point, as shown in Figure 9.18. Find the equivalent viscous damping ratio of the environment when the cantilever has a length of 80 mm, a width of 6 mm, and a thickness of 140 nm. Young’s modulus is 155 GPa and the mass density is 6200 kg/m3. The experimentally determined bandwidth is 32,000 Hz. 9.11 Use the complex transfer function approach to determine the values of L and C in the electrical system of Figure 9.39 knowing the resonant frequency is 24 kHz, the resonant peak is 1.5, and R ¼ 120 U. L

vi

R

C

vo

FIGURE 9.39 Electrical System With Inductor, Resistor, and Capacitor.

9.12 Use a lumped-parameter model for the mechanical microsystem of Figure 9.40, where the three serpentine springs are identical; they are illustrated in Figure 2.5(b) with l ¼ 30 mm and a square cross section of 320 nm per side. The material Young’s modulus is 165 GPa. The masses of the shuttles are m1 ¼ 230 mg and m2 ¼ 300 mg. Determine all the magnitudes and phase angles corresponding to a sinusoidal force applied on m1 in the motion direction. Consider an overall damping coefficient c ¼ 0.04 applying to both shuttle masses’ absolute motions. Serpentine spring Shuttle mass

Anchor

m1

m2

Anchor

Motion direction

FIGURE 9.40 Mechanical Microsystem With Shuttle Masses and Serpentine Springs.

9.13 A simplified, lumped-parameter car model, as shown in Figure 9.41, is defined by the chassis mass m and moment of inertia J (both defined with

475

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CHAPTER 9 Frequency-Domain Approach

l1

l2 m, J

c1

k1

Motion

c2

k2

u1

u2

FIGURE 9.41 Simplified Car Model With Sinusoidal Input on the Wheels.

respect to the chassis center of mass), as well as by two damper-spring suspensions c1-k1 and c2-k2. Determine the system’s number of DOF and plot its steady-state frequency response. Considering small motions, calculate and plot the steady-state time variation of the chassis center of mass vertical translation and the chassis rotation when u1 ¼ 0.03$sin(5t) m, u2 ¼ 0.03$sin(5t e p/10) m, m ¼ 200 kg, J ¼ 0.03 kg m2, c1 ¼ c2 ¼ 0.5 N s/m, k1 ¼ 60 N/m, k2 ¼ 70 N/m, l1 ¼ 1 m, and l2 ¼ 1.3 m. 9.14 Consider the dynamic system of Figure 9.42 with m1 ¼ 0.1 kg, m2 ¼ 0.3 kg, m3 ¼ 0.2 kg, c1 ¼ 12 N s/m, c2 ¼ 10 N s/m, c3 ¼ 8 N s/m, k1 ¼ 80 N/m, k2 ¼ 110 N/m, and k3 ¼ 60 N/m, which is being acted upon by the force f ¼ 80 $ sinð24tÞ N. (i) Determine and plot the magnitudes and phase angles that define the frequency response of this system. (ii) Express the steady-state outputs (time responses) of this system under the given force input. y1

y2

c1

y3 c2

c3

f m1

m2 k1

m3 k2

k3

FIGURE 9.42 Lumped-Parameter Mechanical System With Harmonic Input Forces.

9.15 Use the transfer function modeling approach to obtain the Bode plots for the electrical system represented in Figure 9.43. Consider that the inputs are the sine voltages v1 and v2 and the outputs are the meaningful currents. The electrical components are L ¼ 0.8 H, R ¼ 140 U, C ¼ 100 mF.

Problems

L

R

C

v1

v2

FIGURE 9.43 Electrical System With Voltage Input.

9.16 The liquid system of Figure 9.44 is formed of two tanks and two identical pipe segments. The system input consists of the flow rate qi and the external pressure po, both of which are assumed to vary sinusoidally. Derive the transfer function matrix of this system, whose outputs are the tank pressures p1 and p2. Use MATLAB to plot the frequency response when known are d1 ¼ d2 ¼ 4 m (tank diameters), di ¼ 0.025 m (diameter of the pipe), l ¼ 50 m (length of pipe between tanks and also length of pipe at the second tank outflow), m ¼ 0.001 N s/m2, and r ¼ 1000 kg/m3. qi

Cl C2 p1

R

p2

q

R

po

qo

FIGURE 9.44 Tank-Pipe Liquid System Under Sinusoidal Input.

9.17 Derive a transfer function model and plot the Bode diagrams for the mechanical system of Figure 9.45, where f1, f2 are the input force components and y1, y2 are the output displacements. Known are m1 ¼ 1.5 kg, m2 ¼ 2 kg, c ¼ 18 N s/m, k1 ¼ 100 N/m, k2 ¼ 120 N/m, k3 ¼ 200 N/m. Also plot the steady-state time-domain solution of this system for f1 ¼ 100$sin(10t) N and f2 ¼ 120 sin(15t) N. 9.18 Consider the inputs are the voltages v1, v2, and the output is the voltage vo for the operational amplifier electrical system of Figure 9.46. Use the complex transfer function approach to calculate and plot the steady-state time-domain response of this system for R1 ¼ 220 U, R2 ¼ 190 U, L ¼ 0.25 H, C ¼ 0.5 mF, v1 ¼ 50$sin(50t) V, and v2 ¼ 60$sin(40t) V.

477

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CHAPTER 9 Frequency-Domain Approach

y2

y1 f1

c

f2 m1

m2 k2

k1

k3

FIGURE 9.45 Translatory Mechanical System. R1 L v1

– +

v2 C

vo

R2

FIGURE 9.46 Operational Amplifier System With Two Input Voltages and One Output Voltage.

9.19 The transfer function of a dynamic system has the gain K ¼ 15; the zeroes: 6, 5; and the poles: 3 (double), 2, and 1  j. Draw the asymptote (simplified) Bode magnitude diagram corresponding to the transfer function of this model. 9.20 Identify the transfer function G(s) whose asymptote magnitude plot is shown in Figure 9.47. M(dB)

60

–20 dB/dec –40 dB/dec

0

–1

0

2

x = log10ω

FIGURE 9.47 Asymptote Bode Magnitude Diagram of Transfer Function G(s).

9.21 A frequency analyzer provided the following experimental frequency response data data: 0.2, 0.2 þ 0.01i, 0.3 þ 0.1i, 0.02 þ 0.03i, 0.1 þ 0.04i, 0.1i, 0.015i, 0.002e0.6i

for the 10e40 rad/s frequency range. Plot the magnitude and phase angle corresponding to this data set and calculate G(u$j) for u ¼ 30 rad/s and u ¼ 250 rad/s.

Problems

9.22 A zero-pole-gain model has the zeroes: 1, 3, 3; the poles: 2, 4, 4, 8; and a gain of 20. Plot the Bode diagrams corresponding to this system for the 0e8000 rad/s frequency range using the conversion from the zeropole-gain model to the frequency response data model. 9.23 A dynamic system is defined by the differential equation: 30y þ y€ þ 2y_ þ 4y ¼ 110 $ sinðu $ tÞ. Use the conversion from a transfer function model to the frequency response data model to plot the Bode diagrams corresponding to this system for the 0e3000 rad/s frequency range. 9.24 A dynamic system is described by the following state space matrices:     3 1 2 ½A ¼ ; ½B ¼ ; ½C ¼ f 1 3 g; D ¼ 2: 2 5 1 Use MATLAB to convert the model to a frequency response data model corresponding to a 0e10,000 rad/s frequency range and plot the Bode diagrams of the system response. 9.25 What is the minimum mass that can be detected by a particle deposition at the free end of a microcantilever with constant rectangular cross section (dimensions w and h) if the detection equipment has a resolution of 40 Hz? For that minimum mass, plot the original and shifted transmissibilities in terms of the frequency ratio. Known are the dimensions of the cantilever, l ¼ 100 mm, w ¼ 20 mm, h ¼ 2 mm, also the material properties, E ¼ 165 GPa, r ¼ 6300 kg/m3, and the damping coefficient c ¼ 4$1010 N s/m. 9.26 A nanobridge is formed of a prismatic rigid platform having the dimensions of 250 mm$250 mm$3 mm and two identical nanowires of diameter 780 nm and length 35 mm made of a material with E ¼ 150 GPa, G ¼ 95 GPa, and r ¼ 5600 kg/m3dsee Figure 3.15. Assume that both the plate and nanowires are made of the same material. Decide which of the following two resonance shift methods is more sensitive in detecting a mass quantity of 2$1014 kg: (i) Bending mode with mass deposited at the center of the plate; the damping coefficient in translation is ct ¼ 5$109 N s/m; (ii) Torsional mode with mass deposited off the plate center at a distance of 110 mm; the damping coefficient in rotation is cr ¼ 4$106 N m s; 9.27 For the rotary mechanical system of Figure 9.48, evaluate the frequency of the sinusoidal input torque mi that generates a transmitted torque amplitude on the right base equal to 70% of the input torque amplitude at steady-state. Known are J1 ¼ 0.01 kg m2, J2 ¼ 0.014 kg m2, c ¼ 30 N m s, k1 ¼ 300 N m, k2 ¼ 260 N m.

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CHAPTER 9 Frequency-Domain Approach

c mi

k1

k2

J1

J2

FIGURE 9.48 Rotary Mechanical System With Torque Transmission.

9.28 A sinusoidal force acts on the mechanical system illustrated in Figure 9.49. Considering that m ¼ 0.6 kg and k ¼ 220 N/m, analyze the transmissibility of the input force to the force on the wall for c1 ¼ 10 N s/m and then for c2 ¼ 15 N s/m during the steady-state regime. y1 f = F sin(ωt)

y2 k

k m

m

c

FIGURE 9.49 Mechanical System With Sinusoidal Input.

9.29 The massless platform in Figure 9.50(a) is subjected to a displacement u ¼ U$sin(u$t). An additional subsystem formed of a mass ma, spring ka, and damper ca is added to the existing mass m and spring k as in Figure 9.50(b). Express the amplitude Y of the steady-state displacement y of the mass m in terms of m, k, ma, ka, ca, U, and u. Is it possible for Y to become zero?

u

u

k m

ya

y ka

ma

k m

Massless platform

y Massless platform

480

ca

(a)

(b)

FIGURE 9.50 (a) MasseSpring System With Input Displacement; (b) System With Additional MasseSpringeDamper.

Problems

9.30 The two blocks in Figure 9.51 can move along horizontal directions; as a result, the two springs and the damper have also horizontal motions. A sinusoidal horizontal force f ¼ F$sin(u$t) N is applied to the block at the left. Known are: m1 ¼ 0.5 kg, m2 ¼ 0.4 kg, c ¼ 12 N s/m, k1 ¼ 200 N/m, and k2 ¼ 160 N/m. Calculate the frequency u of the force f that generates a transmitted force amplitude on the right fixed wall equal to 80% of the input force amplitude at steady state.

f = F·sin(ω·t)

k1

m1

k2 m2

Fixed

c Fixed wall

FIGURE 9.51 Translatory Mechanical System With Sinusoidal Force Input.

9.31 Find the amplitude of the force that is transmitted to the fixed wall on the right when an input displacement u ¼ 0.4$sin(30$t) m is applied to the mechanical system shown in Figure 9.52. Known are also m ¼ 0.5 kg, c ¼ 30 N s/m, k1 ¼ 140 N/m, k2 ¼ 200 N/m. Assume steady-state conditions. u

k2 m

c

Fixed wall

y k1

FIGURE 9.52 Translatory Mechanical System With Sinusoidal Displacement Input.

9.32 The mechanical system of Figure 9.53 is formed of a massless disk that rotates eccentrically with constant angular velocity u about a fixed shaft (rotation center) that is at a distance e from the disk geometric center. The disk motion is imparted to a massless platform, which can move vertically against a spring of stiffness k1. This spring is also connected to a block of mass m, which further attaches to another spring of stiffness k2 that connects to the fixed base. Find a simple (approximate) sinusoidal input motion, which is transmitted to the platform by the rotating disk, and which depends on the eccentricity e and the angular velocity u. Based on that input and by using the transmissibility concept, calculate the force amplitude which is transmitted to the base at steady state. Is it possible that the amplitude of the force transmitted to the ground is zero?

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CHAPTER 9 Frequency-Domain Approach

Rotating disk

Disk center Rotation center

e

ω

Massless platform

k1 m

k2 Base

FIGURE 9.53 Translatory Mechanical System With Eccentric Input Displacement.

9.33 A constant rectangular cross-section microcantilever needs to be designed to sense an input base acceleration of 100g (g ¼ 9.8 m/s2 is the gravitational acceleration constant). Known are the thickness h ¼ 3 mm, the mass density r ¼ 5400 kg/m3, and Young’s modulus E ¼ 150 GPa. Calculate the microcantilever length if the tip deflection cannot exceed 1 mm during the steady-state regime. 9.34 The microbridge of Figure 9.54 operates as a sensor for the sinusoidal input rotary displacement of its base. To be effective for this purpose, the natural frequency needs to be least 120 times smaller than the input frequency. It is also expected that the actual input frequency be larger than 2300 rad/s. Find the flexible rods’ diameter d for J ¼ 2.5$1019 kg m2, l ¼ 300 mm, and G ¼ 120 GPa. Calculate the damping coefficient c corresponding to the viscous interaction between the microbridge plate and its environment knowing that the maximum value of this system’s magnitude is 1.0001. Hint: Use the transfer function connecting the output rotation angle of the plate to the base input rotation angle.

Rotating base

Flexible rod

Plate

θi

d J

θi

Rotating base l

FIGURE 9.54 Microbridge With Base Rotary Input.

l

Problems

9.35 Two dynamic systems form a nonloading cascade defined by the following differential equations: y€1 þ 3y_1 þ 2y1 ¼ 20$sin(15t); 3y€ þ 2y_ ¼ 4y_1 þ 6y1. Evaluate the steady-state response of this system and plot the corresponding Bode diagrams. 9.36 A cascading dynamic system is modeled by the differential equations: ( y€1 þ 2y_1 þ 10y1 ¼ u y_2 þ 5y2 ¼ 3y_1 þ y1 where u is the input and y2 is the output. For an input u ¼ 200$sin(50$t), express the steady-state output y2. 9.37 A system is formed of the cascading transfer functions G1(s), G2(s), ., Gm(s) that are acted upon by the inputs U1(s), U2(s), ., Um(s), as shown in Figure 9.55. Formulate the output Y(s) in terms of the corresponding transfer functions and inputs. Also, express the equation of the steady-state, time-domain output y(t) considering that the input signals are all sinusoidal: u1(t) ¼ U1$sin(u1$t), u2(t) ¼ U2$sin(u2$t), ., um(t) ¼ Um$sin(um$t). U1(s) G1(s)

U2(s)

U3(s)

G2(s)

Um(s)

G3(s)

Gm(s)

Y(s)

FIGURE 9.55 Multiple-Component Cascading System With Multiple Inputs.

9.38 Determine the equation of steady-state, time-domain response y(t) of the system shown in Figure 9.56 where G1(s) is the transfer function of a secondorder system defined by K1 ¼ 8, un1 ¼ 240 rad/s, x1 ¼ 0.5 and G2(s) is the transfer function of a first-order system with K2 ¼ 5, s2 ¼ 0.01 s. The two inputs are sinusoidal: u1(t) ¼ U1$sin(u1$t), u2(t) ¼ U2$sin(u2$t) with U1 ¼ 0.1, u1 ¼ 50 rad/s, U2 ¼ 0.01, u2 ¼ 100 rad/s. U1(s) G1(s)

U2(s) G2(s)

Y(s)

FIGURE 9.56 Two-Component Cascading System With Two Inputs.

9.39 A sinusoidal input u ¼ 10$sin(100t) is provided to an unknown first-order system, which serially connects to a first-order transducer system defined by a sensitivity of 2.4 and a time constant of 0.03 s. The output from the transducer is further fed into a filter-amplifier device, which acts as a secondorder system with a sensitivity of 5, a natural frequency of 260 rad/s, and a

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CHAPTER 9 Frequency-Domain Approach

damping ratio of 0.45. Knowing that the signal output from the filteramplifier system has an amplitude of 180 and a phase angle ofd65 with respect to the sinusoidal input, evaluate the unknown first-order system parameters. 9.40 Study the frequency response of the electrical systems of Figure 9.57(a) and 9.57(b), for which R ¼ 640 U and L ¼ 0.7 H. Using the Bode magnitude plot, demonstrate that both systems behave as high-pass filters. Calculate the two filters cutoff frequencies. R

R vi

vo

L

vi

R L

L

vo

(b)

(a)

FIGURE 9.57 Electrical Filters: (a) Single-Stage; (b) Two-Stage.

9.41 Establish the filter response of the electrical system of Example 7.1 and shown in Figure 7.2 Known are R1 ¼ 110 U, R2 ¼ 160 U, C1 ¼ 45 mF, and C2 ¼ 30 mF. If applicable, calculate the corresponding cutoff frequency/ frequencies. 9.42 Same questions as in Problem 9.41 for the mechanical system depicted in Figure 7.14 of Example 7.9. Consider that m ¼ 0.4 kg, c ¼ 2 N s/m and k ¼ 110 N/m. 9.43 Determine whether the electrical circuit of Figure 9.58 can function as a filter and establish its type in case it is a filter by plotting the Bode magnitude diagram. The electrical components are defined by R ¼ 65 U, L1 ¼ 0.6 H, L2 ¼ 0.8 H, and C ¼ 0.025 mF. Calculate the corresponding cutoff frequency/frequencies if the system behaves as a filter. R R

vi

L1

L2

C

C

– + vo

FIGURE 9.58 Operational-Amplifier Filter Circuit With Resistors, Inductors, and Capacitors.

Suggested Reading

9.44 Check whether the mechanical system of Figure 9.59 operates as a filter when connecting the input displacement u to the output displacement y. Draw the Bode magnitude plot for the following pairs of parameter values: k1 ¼ 120 N/m, c1 ¼ 11 N s/m and k2 ¼ 150 N/m, c2 ¼ 23 N s/m. If applicable, determine the system’s cutoff frequency/frequencies. y

k

u

c

FIGURE 9.59 Single-DOF Mechanical System as a Filter.

9.45 Use the complex transfer function that connects the input displacement u to the output displacement y, and the Bode magnitude plot to predict the filter behavior of the translatory mechanical system of Figure 9.60 when the following parameters are known: m ¼ 0.8 kg, c1 ¼ 60 N s/m, c2 ¼ 80 N s/m, k1 ¼ 160 N/m, k2 ¼ 190 N/m. Also calculate the cutoff frequency/frequencies if the system operates as a filter. y k2

c2

u c1 m

k1

FIGURE 9.60 Translatory Mechanical System as a Filter.

Suggested Reading J.F. Lindsay and S. Katz, Dynamics of Physical Circuits and Systems, Matrix Publishers, Champaign, IL, 1978. A.V. Oppenheim and A.S. Willsky, Signal and Circuits, 2nd Ed. Prentice Hall, Upper Saddle River, NJ, 1997. N.S. Nise, Control Systems Engineering, 7th Ed. Wiley, New York, 2015. K. Ogata, System Dynamics, 4th Ed. Pearson Prentice Hall, Upper Saddle River, NJ, 2004. B.J. Kuo and F. Golnaraghi, Automatic Control Systems, 10th Ed. Wiley, New York, 2017. N. Lobontiu, Dynamics of Microelectromechanical Systems, Springer, New York, 2007. R.S. Figliola and D.E. Beasley, Theory and Design for Mechanical Measurements, 6th Ed. Wiley, New York, 2014.

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CHAPTER

Coupled-Field Systems

10

OBJECTIVES This chapter introduces the notion of dynamic system field coupling, which studies the interaction between systems belonging to different domains (fields) and generates mathematical models reflecting that interaction. Specifically, you learn about • Actuation/sensing principles and related coupled-field system applications. • Thermomechanical coupling by means of bimetallic strips. • Electrothermomechanical coupling. • Electromechanical and electromagnetomechanical coupled-field examples, including strain-gauge sensing and rotary/translatory direct-current (dc) motors with loads. • Principles of piezoelectricity and applications of piezoelectric coupling.

INTRODUCTION Dynamic systems seldom operate in isolation but commonly interact with other systems. Interaction between dynamic systems of different physical nature is known as domain or field coupling. Some examples of the many coupled-field possibilities are presented in this chapter, such as thermomechanical or electrothermomechanical, electromechanical (including strain-gauge macro-scale and microelectromechanical systems (MEMS) applications), electromagnetomechanical (with or without optical detection), and piezoelectric. Various examples illustrate linear and nonlinear coupled-field models that apply MATLAB and Simulink to determine the solutions.

10.1 CONCEPT OF SYSTEM COUPLING Chapters 1e9 include examples of system coupling implicitly, but the coupling between coordinates was restricted to a system of a single physical type (mechanical, electrical, fluid, or thermal). Consider, for instance, the two separate mass-spring mechanical systems of Figure 10.1(a); they are physically isolated and therefore uncoupled. Once a spring of stiffness k is added between the masses m1 and m2, System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00010-5 Copyright © 2018 Elsevier Inc. All rights reserved.

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CHAPTER 10 Coupled-Field Systems

x1

x2

k1

x1 k2

m2

m1

k1

x2 k2

k m1

(a)

m2

(b)

FIGURE 10.1 (a) Two Uncoupled Mechanical Systems; (b) Spring Coupled Mechanical System.

as in Figure 10.1(b), the two systems become coupled, and the resulting mathematical model is represented by the equations

⎧m1 ⋅ x1 + ( k1 + k ) ⋅ x1 − k ⋅ x2 = 0 ⎪ ⎨ ⎪⎩m2 ⋅ x2 + ( k2 + k ) ⋅ x2 − k ⋅ x1 = 0

(10.1)

The presence of the boxed term with x2 in the first Eq. (10.1) indicates that this is a coupled system because this equation describes the motion of m1, which is defined by coordinate x1. Similarly, the boxed x1 term is included in the second Eq. (10.1), which describes the x2 motion. Because the elements of the system sketched in Figure 10.1(b) are from one field, this system is a single-field coupled system. Single-field coupled systems, such as the one just discussed, are not within the scope of this chapter but Chapters 3e5 study mechanical, electrical, fluid, and thermal single-field coupled examples. In basic terms, a multiple-field coupled system (or, simply, a coupled-field system) is defined by at least one (differential) equation with variables (coordinates or DOFs) from two or more domains of different physical nature. A basic coupled-field example is the one of a bar of original length l subjected to a temperature increase Dq; this bar deforms axially by the quantity: Dl ¼ ða $ lÞ $ Dq

(10.2)

Evaluated as an inputeoutput relationship, Eq. (10.2) shows that the input (the temperature increase Dq) is from the thermal domain, whereas the output (the bar elongation Dl ) is from the mechanical domain.

Sensing and Actuation The interaction between physical properties pertaining to (two) different physical fields is probably best exemplified through sensing and actuation. Sensors convert a physical property variation, which is difficult to detect and monitor directly (such as force, displacement, acceleration, temperature, pressure) into another physical property variation from a different field (such as electrical voltage) that can easily be monitored. As illustrated in Figure 10.2, a sensor generates ideally a linear relationship between the property to be measured (the input, denoted

o, Output (converted property that measures)

10.1 Concept of System Coupling

o(t) = K·i(t)

K, Static sensitivity i, Input (property variation to be measured)

FIGURE 10.2 Sensing Characteristics in a Linear Measurement Process.

generically by i) and the property that measures (the output, denoted generically by o) by means of a proportionality constant (K, the static sensitivity). Sensors use existing physical principles to connect the input to the output. A thermal sensor operation is based on Eq. (10.2) to predict the temperature change Dq (the input) through the corresponding change in mechanical deformation Dl (the output) by means of a $ ldthe product of thermal expansion coefficient a to the length l (the static sensitivity, K ¼ a $ l). The mechanical deformation can further be converted into electrical voltage by means of voltage dividers (potentiometers) or variable capacitorsdas discussed in Chapter 4, and as detailed later in this chapter. The raw converted (electrical) signal usually undergoes further conditioning, such as amplifying and/or filtering, before being transferred to a measuring/recording device, such as a voltmeter or oscilloscope. Conversely, an actuator produces mechanical work (as the output) from another form of (input) energy, such as thermal or electrical. As studied in Chapter 4, the voltage applied to a capacitor with one mobile armature can generate a mechanical force that attracts the mobile armature toward the fixed armature. The input in this example is electrical voltage and the output is the resulting mechanical force. This chapter analyzes several other mechanisms of producing mechanical work from other energy forms, such as thermal, electrical, electromagnetic, or piezoelectric. Eq. (10.2) described a zero-order coupled system because it is a zero-order differential equation, but higher-order coupled systems (with time derivatives of the output variable) also exist, as we see in this chapter. One convenient way of representing the mathematical model of a coupled multiple-input, multiple-output (MIMO) system is by using the Laplace domain and the transfer function matrix. Consider two systems, denoted by 1 and 2, each being formulated in a different field, see Figure 10.3, where {U(s)} ¼ {U1(s) U2(s)}T is the input vector and {Y(s)} ¼ {Y1(s) Y2(s)}T is the output vector. When at least one of the output components is a function of both inputs, the systems are coupled. Coupled systems, as shown in Figure 10.4, have nonzero terms on the secondary diagonal of the transfer function matrix. When these terms are zero, the two systems are uncoupled because the two outputs Y1(s) and Y2(s) depend on the input of their own systems solely.

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CHAPTER 10 Coupled-Field Systems

Field # 1 U1(s)

Y1(s)

+ G11(s) + G21(s) G12(s)

+ + G22(s) U2(s)

Y2(s) Field # 2

FIGURE 10.3 MIMO Coupled-Field System Studied in the Laplace Domain.

Coupled systems

Uncoupled systems

Y1(s)

G11(s)

0

U1(s)

Y1(s)

G11(s)

G12(s)

U1(s)

Y2(s)

0

G22(s)

U2(s)

Y2(s)

G21(s)

G22(s)

U2(s)

FIGURE 10.4 MIMO Uncoupled and Coupled-Field Systems.

10.2 THERMOMECHANICAL AND ELECTROTHERMOMECHANICAL COUPLING Conversion between thermal and mechanical energy can be used for either sensing or actuation purposes and the respective systems are thermomechanically coupled. In electrothermomechanical coupled systems, the electrical energy converts into thermal energy by Joule effect. Thermal energy can subsequently be transmitted to a mechanical part (or system), which will expand as a result.

10.2.1 Thermomechanical Coupling: The Bimetallic Strip The bimetallic strip (or bimaterial strip, also known as thermal bimorph) is a cantilever beam consisting of two layers of different materials, as shown in Figure 10.5(a). Exposed to a temperature increase, the two materials expand differently along the length, but because they are attached, the compound beam bends, as illustrated in Figure 10.5(b). The top layer material has a linear coefficient of thermal expansion a1 that is larger than the coefficient a2 of the bottom layer and tends to expand more in the dimension l.

10.2 Thermomechanical and Electrothermomechanical Coupling

x

h1

α1 , E 1 , ρ 1 α2 , E 2 , ρ 2

zmax

h2 R

l

(a)

z(x)

(b)

FIGURE 10.5 Bimetallic Strip (Thermal Bimorph) in: (a) Undeformed (Original) State; (b) Deformed State.

The radius of curvature R, shown in Figure 10.5(b), is related to the temperature increase Dq as R¼

h1 þ h2 h ¼ ða1  a2 Þ $ Dq ða1  a2 Þ $ Dq

(10.3)

where h ¼ h1 þ h2 is the total thickness of the bimetallic strip. Eq. (10.3) is a simplified relationship that does not account for the different elastic properties of the two materials, such as the Young’s moduli E1 and E2dthe exact equation of the curvature radius (accounting for E1 and E2) was originally derived by Timoshenko (1925). Mechanics of materials theory shows that the curvature (the reciprocal of the radius of curvature) depends on the second derivative of the beam deflection z as 1 d2 zðxÞ ¼ (10.4) R dx2 where x is measured from the fixed end of the cantilever of Figure 10.5(b). Combining Eqs. (10.3) and (10.4) followed by two integrations yields the slope dz(x)/dx and the deflection z(x): 8 > dzðxÞ ða1  a2 Þ $ Dq > > ¼ $ x þ C1 < dx h (10.5) > ða1  a2 Þ $ Dq 2 > > : zðxÞ ¼ $ x þ C1 $ x þ C2 2h Both the slope and deflection are zero at the root, for x ¼ 0, which results in the two integration constants being zero, C1 ¼ C2 ¼ 0. The maximum static deflection is at the free end and is found from the second Eq. (10.5) for x ¼ l: ða1  a2 Þ $ l2 (10.6) 2h Eq. (10.6) indicates that the bimetallic strip functions as a thermal sensor that measures a temperature variation Dq by means of its free-end deflection zs. The static sensitivity K connecting the input Dq to the output zs is identified in Eq. (10.6). zs ¼ K $ Dq with K ¼

491

492

CHAPTER 10 Coupled-Field Systems

nn

Example 10.1 A two-material microcantilever, such as the one shown in Figure 10.5, has a silicon nitride substrate (component 2) and an aluminum layer (component 1). The cantilever is exposed to a temperature increase of Dq ¼ 20 , which results in transitory vibrations for a short period of time before the cantilever reaches the final deformed shape. Known are the following parameters: a1 ¼ 23.6$106 deg1, a2 ¼ 0.8$106 deg1, E1 ¼ 80 GPa, E2 ¼ 180 GPa, r1 ¼ 2700 kg/m3, r2 ¼ 2400 kg/m3, common width w ¼ 30 mm, common length l ¼ 180 mm, h1 ¼ 125 nm, and h2 ¼ 0.5 mm. Assume a tip force feq ¼ keq$zs acts at the beam’s free end, with keq being the stiffness of the sandwich beam that results from the parallel combination of the two layers stiffnesses k1 and k2. Plot the free-tip displacement z using a lumped-parameter model, and considering that the total viscous damping coefficient is c ¼ 8$108 N s/m. Confirm that the steady-state lumped-parameter dynamic deformation is identical to the static deflection at the free end.

Solution

The static deflection at the free end is obtained by means of Eq. (10.6) as zs ¼ 11.82 mm with the static sensitivity being K ¼ 5.9098$107 m/deg. The lumped-parameter dynamic model of the twomaterial cantilever consists of two collocated masses, m1 and m2, two parallel-connection springs (of stiffnesses k1 and k2), a damper of damping coefficient c, and an equivalent point force feq that produces the same tip displacement as the quasi-static thermal effectdFigure 10.6. feq z

m2 m1 k1

k2

c

FIGURE 10.6 Lumped-Parameter Model of a Two-Material Cantilever Under Thermal Deformation. The total mass is calculated by adding the equivalent masses of the two strips, as given in Table 2.6, namely

33 33 $ ðr1 $ w $ h1 $ lÞ þ $ ðr2 $ w $ h2 $ lÞ 140 140 33 $ w $ l $ ðr1 $ h1 þ r2 $ h2 Þ ¼ 140

m ¼ m1 þ m2 ¼

(10.7)

While the equivalent stiffness of a cantilever beam formed of two different layers can be calculated precisely by mechanics of materials methods, a simplified approach is applied here, assuming that there are two beams combined in parallel. Therefore, the resulting stiffness is the sum of individuallayer stiffnesses and is calculated based on the stiffness of a cantilever beam as provided in Table 2.2:

w $ h31 w $ h32 3E1 $ Iz1 3E2 $ Iz2 3 þ E2 $ k ¼ k1 þ k2 ¼ þ ¼ 3 $ E1 $ 3 3 l l l 12 12   w ¼ 3 $ E1 $ h31 þ E2 $ h32 4l

!

(10.8)

10.2 Thermomechanical and Electrothermomechanical Coupling

With the numerical values of this example, m ¼ 1.957$1012 kg and k ¼ 0.0291 N/m. The equivalent point force is calculated by means of Eqs. (10.6) and (10.8):

feq

  w $ E1 $ h31 þ E2 $ h32 $ ða1  a2 Þ $ Dq ¼ a $ Dq ¼ k $ zs ¼ 8l $ ðh1 þ h2 Þ

(10.9)

which shows the force is proportional to the temperature variation. The proportionality constant is a ¼ 1.7219$108 N/deg and the equivalent force is feq ¼ 3.4438$107 N. Based on Figure 10.6, the equation of motion of the mass is

m $ z€ ¼ feq  c $ z_  k $ z

(10.10)

Combining Eqs. (10.9) and (10.10) results in

m $ z€þ c $ z_ þ k $ z ¼ a $ Dq

(10.11)

The transfer function corresponding to Eq. (10.11) is

GðsÞ ¼

ZðsÞ a ¼ 2 DQðsÞ m $ s þ c $ s þ k

(10.12)

The plot of Figure 10.7 results by using the MATLAB tf and step commandsdthe latter command was used as step(20*tf[ ]) to account for the 20-degree step input. It can be seen that the maximum tip displacement is approximately 18 mm. The steady-state tip displacement is calculated based on the final-value theorem as

lim zðtÞ ¼ lim s $ ZðsÞ ¼ lim s $ GðsÞ $ DQðsÞ

t/N

s/0

s/0

Deflection (m)

s$a Dq a $ Dq ¼ ¼ zs $ ¼ lim s/0 m $ s2 þ c $ s þ k s k

Time (sec)

FIGURE 10.7 Free-End Displacement of Two-Material Cantilever Under Thermal Deformation.

(10.13)

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CHAPTER 10 Coupled-Field Systems

The last equality in Eq. (10.13) is obtained by using Eq. (10.9). Eq. (10.13) proves that, indeed, the steady-state displacement, as it results from the lumped-parameter model, is equal to the static freeend deflection of the two-material cantilever. nn

10.2.2 Electrothermomechanical Coupling Consider the metallic bar of Figure 10.8 that is part of an electric circuit powered by a voltage source v. In a very simplified manner, the electrothermomechanical interaction in this system considers the bar receiving the thermal energy produced through Joule heating from the electrical energy, which is Eel ¼ (v2/R)$t over t seconds. This energy is converted, either fully or just partially, into conductive heat energy expressed as Q ¼ m$c$Dq, where m is the bar mass, c the bar material’s specific heat, and Dq is the bar temperature increase. Lastly, the bar will expand due to heating by Dl ¼ x ¼ a$l$Dq, with a being the linear coefficient of thermal expansion and l the bar’s original length. x

l

v

– Electrical circuit

+ Actuation bar

494

Fixed end

FIGURE 10.8 Metallic Bar in Electric Circuit With Joule Effect, Thermal Heating, and Mechanical Expansion.

One additional equation to be considered connects the temperature variation Dq to the electrical resistance at the final temperature R as R ¼ R0 $ ð1 þ aR $ DqÞ

(10.14)

where R0 is the electrical resistance at the original (initial) temperature and aR is the thermal coefficient of resistance. The following example derives the nonlinear model of a dynamic system with coupling between the electrical, thermal, and mechanical fields. nn

Example 10.2

A voltage v is applied to a deformable bar by means of an external electrical circuit, as sketched in Figure 10.8. The voltage generates heating of the bar through the Joule effect and its subsequent axial

10.2 Thermomechanical and Electrothermomechanical Coupling

deformation in the x direction. Derive a mathematical model of this coupled-field dynamic system with the output being the axial deformation x of the bar’s guided end. Also, evaluate the bar surface temperature qs. Known are the linear coefficient of thermal expansion a, the specific heat c, the thermal coefficient of resistance aR, as well as the electrical resistivity of the bar rel, its original length at the air temperature l0, length of square cross-sectional side a, and mass density r. The bar loses heat through convection in the air whose constant temperature is qN, while the average convection coefficient is h. Assume that the variations of the cross-sectional dimensions and mass density with the temperature are negligible. Assume that the electrical-to-mechanical energy conversion is instantaneous.

Solution

As seen in Chapter 5, the thermal capacity of the bar Cth is connected to its temperature and heat flow rates

Cth $ q_ s ¼ qi  q0 or m $ c $ q_ s ¼ qi  q0

(10.15)

where m is the bar mass, qi is the heat flow rate (power) gained through Joule heating defined as

qi ¼

v2 v2 ¼ R R0 $ ½1 þ aR $ ðqs  qN Þ

(10.16)

and qo is the heat flow rate lost through convection

qo ¼ h $ Al $ ðqs  qN Þ

(10.17)

The lateral area of the bar Al is

Al ¼ 4a $ l ¼ 4a $ l0 $ ½1 þ a $ ðqs  qN Þ

(10.18)

while the constant mass is m ¼ r$a $l0 and the original resistance is R0 ¼ rel$l0/a . Using these parameters and substituting Eqs. (10.16)e(10.18) in Eq. (10.15) results in 2

r $ a2 $ l0 $ c $ q_ s ¼

2

a2 $ v 2  4h rel $ l0 ½1 þ aR $ ðqs  qN Þ

(10.19)

$ a $ l0 $ ½1 þ a $ ðqs  qN Þ $ ðqs  qN Þ Considering that the air temperature qN is constant, the notation q ¼ qs e qN is used, which allows reformulating Eq. (10.19) as

r $ rel $ aR $ c $ a $ l20 $ q_ $ q þ r $ rel $ c $ a $ l20 $ q_ þ 4rel $ a $ aR $ h $ l20 $ q3 þ 4rel $ ða þ aR Þ $ h $ l20 $ q2 þ 4rel $ h $ l20 $ q ¼ a $ v2 (10.20) The displacement x of the bar’s movable end is expressed as

x ¼ a $ l0 $ q

(10.21)

Eqs. (10.20) and (10.21) form the mathematical model of the electrothermomechanical system of Figure 10.8. The nonlinear Eq. (10.20), which expresses the electrothermal system coupling, can be solved for q, whereas Eq. (10.21), which illustrates the thermomechanical system connection, is used to determine x, as shown in the following example. nn

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CHAPTER 10 Coupled-Field Systems

nn

Example 10.3

Solve Example 10.2 using Simulink for a copper bar and the following numerical values: l0 ¼ 0.1 m, a ¼ 1 mm, r ¼ 8900 kg/m3, a ¼ 17$106 deg1, rel ¼ 17$109 U$m, aR ¼ 0.0039 deg1, c ¼ 385 J/(kg$deg), h ¼ 30 W/(m2$deg), and v ¼ 0.017 V. Specifically, plot the time variations of the temperature change q and the axial deformation of the bar x.

Solution Eqs. (10.20) and (10.21) can be rewritten as

(

q_ ¼ a1 $ q  a2 $ q_ $ q  a3 $ q2  a4 $ q3 þ a5 x ¼ a6 $ q

(10.22)

with the coefficients a1 through a6 easily identifiable in Eqs. (10.20) and (10.21) as

4h 4h $ ða þ aR Þ 4h $ a $ aR ; a2 ¼ a R ; a3 ¼ ; a4 ¼ ; r$c$a r$c$a r$c$a v2 a5 ¼ ; a6 ¼ a $ l0 r $ rel $ c $ l20

a1 ¼

(10.23)

a4 • θ(t )3 a4

a3• θ(t )2

a2 •

d θ( t ) dt

d θ(t ) θ(t ) • dt

a2

a1 • θ(t )

θ(t )3

a3

θ(t )2

a1

θ (t )

d θ( t ) dt

a5

θ(t )

a6

x (t )

FIGURE 10.9 Simulink Diagram for Solving the Mathematical Model of the ElectroThermoMechanical System of Example 10.2. The numerical values of the constants of Eq. (10.23) are a1 ¼ 0.035, a2 ¼ 0.0039, a3 ¼ 1.3718$10 4, a4 ¼ 2.3219$10 9, a5 ¼ 0.4961, and a6 ¼ 1.7$10 6. Figure 10.9 illustrates the Simulink block diagram for solving the nonlinear Eq. (10.22). The second term in the right-hand side of the first Eq. (10.22) is generated as the product between q and q_ by means of the Product block from the Math Operations library. Figure 10.10 shows the rising and stabilizing portions of the system response over a time period of 150 s. The maximum bar temperature change is less than 14 degrees and the maximum bar deformation is approximately 23 mm.

Displacement (m)

Temperature Variation (deg)

10.3 Electromechanical Coupling

Time (sec)

Time (sec)

(a)

(b)

FIGURE 10.10 Time Plots of Bar: (a) Temperature Variation; (b) Free End Displacement. nn

10.3 ELECTROMECHANICAL COUPLING Systems defined by at least one equation comprising electrical and mechanical variables are known as electromechanical systems. Usually, differential equations can be formulated for the mechanical and electrical subsystems separately complemented by one or several coupling equations connecting amounts from the two fields. Several examples of electromechanically coupled systems are studied next, including electrostaticmechanical, electromagnetomechanical, and piezoelectric ones.

10.3.1 Electrostatic-Mechanical Coupling This section analyzes sensing and actuation of mechanical systems motion by means of voltage dividers (potentiometers) and capacitive circuits. nn

Example 10.4 A sinusoidal force is applied to a point mass, which is connected through a spring-damper to a fixed wall, as indicated in Figure 10.11. The mass motion is sensed by means of a potentiometer (voltage divider) in an electric circuit that contains another resistor and a voltage source. Known are m ¼ 0.1 kg, c ¼ 0.2 N s/m, k ¼ 10 N/m, the total resistance of the potentiometer Rp ¼ 60 U, the cross-section area of the variable resistor A ¼ 2$107 m2, the electric resistivity rel ¼ 1.8$107 U$m, R ¼ 100 U, v ¼ 80 V, and the voltmeter voltage amplitude Vx ¼ 1 V at a frequency u ¼ 40 rad/s. Calculate the force amplitude F corresponding to the steady-state regime.

Solution The time-domain and Laplace-domain mathematical models of the mechanical system are

  € þ c $ xðtÞ _ þ k $ xðtÞ ¼ f ðtÞ and m $ s2 þ c $ s þ k $ XðsÞ ¼ FðsÞ m $ xðtÞ (10.24)

497

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CHAPTER 10 Coupled-Field Systems

x k f

m

Wiper

c

Rp

Voltmeter x

l

– R

+ v

FIGURE 10.11 Coupled System for Sensing the Mechanical Motion by a Potentiometer Electric Circuit. As per Eq. (4.4), the voltage measured by the voltmeter is

vx ðtÞ ¼ i $ Rx ðtÞ ¼ Vx ðsÞ ¼

v r $ xðtÞ rel $ v or vx ðtÞ ¼ $ el $ xðtÞ and Rp þ R A AðRp þ RÞ

rel $ v $ XðsÞ AðRp þ RÞ

(10.25)

where it has been taken into account that the source voltage is v ¼ (Rp þ R)$i and the variable resistance Rx corresponding to the resistor of length x that is in the voltmeter circuit is Rx ¼ rel$x/ A. The last Eq. (10.25) is the Laplace transform of the previous equation under the assumption that all properties in the fraction that multiplies the mass displacement x(t) in Eq. (10.25) are constant. Combining the Laplace-domain Eqs. (10.24) and (10.25) results in

GðsÞ ¼

  A $ ðRp þ RÞ FðsÞ ¼ a $ m $ s2 þ c $ s þ k ; a ¼ rel $ v Vx ðsÞ

(10.26)

The steady-state amplitude of the force F is expressed in terms of the sensed voltage amplitude Vx is

F ¼ jGðu $ jÞj $ Vx ¼ a $

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðk  m $ u2 Þ2 þ ðc $ uÞ2 $ Vx

which, for the given numerical values, results in F ¼ 333.81 N.

(10.27) nn

nn

Example 10.5

A MEMS is formed of a shuttle mass m ¼ 4$1012 kg and two identical beam springs, as illustrated in Figure 10.12(a). The shuttle is actuated by two identical capacitive longitudinal units, each defined by an actuation voltage va ¼ Va$cos(u$t) mV with the amplitude Va ¼ 15 mV and the frequency u ¼ 2p$200,000 rad/s; the shuttle motion is sensed by a capacitive transverse unit. The actuation is defined by a gap ga ¼ 3 mm, superposition plate width w ¼ 200 mm, and electric permittivity ε ¼ 8.5$1012 F/m, whereas the sensing is characterized by a bias voltage vb ¼ 10 mV and an initial gap gs ¼ 10 mm. (i) Assuming no mechanical damping, find the length l of the two identical beam springs that will ensure the shuttle mass motion is less than gs. The beams are of constant cross section with a moment of inertia Iz ¼ 5$1025 m4 and the material Young’s modulus is E ¼ 170 GPa.

10.3 Electromechanical Coupling

ga Beam spring

l

Capacitive transverse sensing

x Shuttle mass

m

ks f

gs

l

x

Fixed capacitive plate

m

ks c

Capacitive longitudinal actuation

Anchor

(a)

(b)

FIGURE 10.12 MEMS With Capacitive Longitudinal Actuation and Transverse Sensing: (a) System Schematic; (b) Lumped-Parameter Model of Mechanical System.

(ii) Evaluate the viscous damping coefficient c between the shuttle and the substrate, knowing that the steady-state amplitude of the sensed voltage is Vs ¼ 10 mV when the excitation frequency is pffiffiffiffiffiffiffiffiffi ffi equal to the natural frequency un ¼ k=m of the mechanical system.

Solution

(i) The mechanical force f(t) is produced by the two capacitive longitudinal actuation units and is calculated as per Eq. (4.33):

  ε$w ε $ w $ Va2 1 þ cosð2u $ tÞ 2 f ðtÞ ¼ 2 $ $ ½Va $ cosðu $ tÞ ¼ $ 2ga 2 ga ¼

ε $ w $ Va2 ε $ w $ Va2 þ $ cosð2u $ tÞ 2ga 2ga

 ε $ w $ Va2 ε $ w $ Va2 p ¼  $ sin 2u $ t  2 2ga 2ga   p ¼ f1 þ f2 ðtÞ ¼ f1  f1 $ sin 2u $ t  2

(10.28)

The input force is therefore formed of a step component f1 ¼ ε$w$(Va)2/(2ga) and a sinusoidal part f2(t), as shown in Eq. (10.28). For no damping, the time-domain and Laplace-domain mathematical models of the mechanical system of Figure 10.12(b) are

€ þ k $ xðtÞ ¼ f ðtÞ ¼ f1 þ f2 ðtÞ; m $ xðtÞ



 m $ s2 þ k $ XðsÞ ¼ FðsÞ ¼ F1 ðsÞ þ F2 ðsÞ

(10.29) with k ¼ 2ks and ks being the stiffness of one beam spring since the two identical beams act as springs in parallel. Eq. (10.29) indicates that the transfer function is

GðsÞ ¼

XðsÞ 1 ¼ FðsÞ m $ s2 þ k

(10.30)

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CHAPTER 10 Coupled-Field Systems

Because the system is linear, superposition can be applied, and the steady-state response is calculated by adding the partial steady-state responses resulting from the two forces (inputs) of Eq. (10.29). As a consequence, combination of Eqs. (10.29) and (10.30) results in

XðsÞ ¼ X1 ðsÞ þ X2 ðsÞ with X1 ðsÞ ¼ GðsÞ $ F1 ðsÞ; X2 ðsÞ ¼ GðsÞ $ F2 ðsÞ (10.31) The steady-state response resulting from the step (constant) f1 is calculated by means of the finalvalue theorem:

X1 ¼ x1 ðNÞ ¼ lim x1 ðtÞ ¼ lim s $ X1 ðsÞ ¼ lim s $ GðsÞ $ F1 ðsÞ t/N

¼ lim

s/0

s/0

s/0

s $ f1 f1 ¼ s $ ðm $ s2 þ kÞ k

(10.32)

The amplitude of the steady-state response corresponding to the harmonic force f2(t) is

X2 ¼ jG2 ðu $ jÞj $ f1 ¼

f1 k  m $ u2

(10.33)

The total steady-state amplitude is therefore

X ¼ X1 þ X 2 ¼



f1 f1 1 1 þ þ ¼ f $ 1 k k  m $ u2 k k  m $ u2

(10.34)

The amplitude X cannot exceed the sensing original gap gs, which means X < gs. With X of Eq. (10.34) and assuming that k > m$u2 (which is un > u), this condition yields



2f1 m $ u 2 $ f1 k2  m $ u2 þ >0 $k þ gs gs

(10.35)

The two roots of Eq. (10.35) are both positive, namely k1 ¼ 6.375$1015 N/m and k2 ¼ 6.3165 N/ m. Also, the polynomial in the left-hand side of Eq. (10.35) has a minimum. Eq. (10.35) is satisfied for p 0< k < k1 and k > k2. Taking into account that k ¼ 2ks ¼ 2$12E$Iz /l 3, it follows that ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l ¼ 3 24E $ Iz =k , which yields l1 ¼ 6.84 m for k1 and l2 ¼ 68.609 mm for k2. Obviously, the only feasible dimension at the microscale of this device is l2 and so, any value larger than l2 will lead to k > k2. Let us use an actual length l ¼ 70 mm. (ii) The mathematical model of the mechanical system is in this case:

€ þ c $ xðtÞ _ þ k $ xðtÞ ¼ f ðtÞ ¼ f1 þ f2 ðtÞ; m $ xðtÞ 

 m $ s2 þ c $ s þ k $ XðsÞ ¼ FðsÞ ¼ F1 ðsÞ þ F2 ðsÞ

(10.36)

and the transfer function is

GðsÞ ¼

XðsÞ 1 ¼ 2 FðsÞ m $ s þ c $ s þ k

(10.37)

The Laplace-domain displacement X(s) is still the one given in Eq. (10.31) and the steady-state response corresponding to the step portion of the input is still f1/k as in Eq. (10.32). The amplitude of the steady-state response corresponding to the harmonic force f2(t) is

f1 f1 ffi X2 ¼ jGðu $ jÞj $ f1 ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffi 2ffi ¼ c $ pffiffiffiffiffiffiffiffi k=m 2 ðk  m $ k=mÞ þ c $ k=m

(10.38)

10.3 Electromechanical Coupling

where the natural frequency un ¼ amplitude is therefore

pffiffiffiffiffiffiffiffiffiffi k=m was used as the input frequency. The total steady-state

! f1 f1 1 1 þ pffiffiffiffiffiffiffiffiffi X ¼ X1 þ X2 ¼ þ pffiffiffiffiffiffiffiffiffi ¼ f1 $ k c $ k=m k c $ k=m

(10.39)

The time-domain relationship between the sensed voltage vs (or Dv, as in Eq. 4.31) and the mass displacement x (or Dx, as used in the same Eq. (4.31), as well as the similar Laplace-domain relationship between Vs(s) and X(s) are

vs ðtÞ ¼ 

vb vb vb $ xðtÞ and Vs ðsÞ ¼  $ XðsÞ or Vs ¼ $ X gs gs gs

(10.40)

The last form of Eq. (10.40) indicates that the steady-state amplitudes Vs and X are proportional. Eliminating now X between Eqs. (10.39) and (10.40) results in the damping coefficient equation:

c¼

pffiffiffiffiffiffiffiffiffiffi f 1 $ k $ m $ vb gs $ k $ V s  f 1 $ v b

(10.41)

whose numerical value is c ¼ 5.228$1021 N s/m. The stiffness has been recalculated with l ¼ 70 mm. nn

10.3.2 Mechanical Strain and Electrical Voltage Coupling Mechanical strain can be related to electrical voltage through sensors. Consider the electrically conductive block of Figure 10.13, which has a rectangular cross section (width is w and thickness is h) and a length l being acted upon by two axial forces f. The block has an electric resistance R that is defined along its length. It is demonstrated in the companion website Chapter 10 that the following relationship exists between the electrical resistance R variation and the mechanical axial strain εa: dR DR ¼ ¼ ð1 þ 2nÞ $ εa ¼ K $ εa R R

w

(10.42)

f

f

l h

FIGURE 10.13 Mechanical Block Under Axial Load.

501

502

CHAPTER 10 Coupled-Field Systems

where the symbol D indicates small but finite resistance variation, whereas d points at infinitesimal variations; n is Poisson’s ratio, a material constant (e.g., 0.3 for steel) and K is the material sensitivity. Eq. (10.42) is the base for a lot of sensors that measure mechanical deformation (strain) through electrical resistance variation. As known from mechanics of materials (see also Appendix F), the ratio of a length variation to the initial length is mechanical strain: Dl Dw Dh ; εt ¼ ¼ (10.43) l w h where εa is the axial strain (the strain aligned with the external force direction), and εt is the transverse strain (the strain perpendicular to the force direction). It is also known that the relationship between the two strains is εa ¼

εt ¼ n $ εa

(10.44)

Eq. (10.44) indicates that, when the axial strain is positive (the length l increases), the transverse strain is negative (the dimensions w and h decrease under the action of f ). Mechanical member St f

Sa

f

Strain gauges

FIGURE 10.14 Wire Strain-Gauge Sensing for Mechanical Deformations of Member Under Load.

The strain gauge (or gage) is a practical way of implementing the use of resistance pickup for mechanical strain. Figure 10.14 is the top view of a mechanical member under the action of two axial loads f. Two strain gauges, Sa and St, are affixed (glued) to the member. The strain gauge that has its long wire segments parallel to the forces f picks up the axial strain (this is the reason of being denoted by Sa), whereas the strain gauge identified as St, which has its longitudinal dimension perpendicular to the force line of action, senses transverse deformations. For strain gauges, an equation similar to Eq. (10.42) connects the resistance variation to the mechanical strain so that the two strain gauges of Figure 10.14 define the following resistanceestrain relationships:



DR DR ¼ Kg $ εa ; ¼ Kg $ εt ¼ Kg $ n $ εa (10.45) R a R t where Kg is the strain gauge sensitivity. Strain gauges are usually connected in a Wheatstone-bridge electrical system, such as the one studied in Chapter 4 and shown in Figure 10.15.

10.3 Electromechanical Coupling

vi

R1

R2 i1

i1

i2

i2

R4

vo R3

FIGURE 10.15 Wheatstone Full Bridge With Four Resistors.

If vi is the input voltage to the bridge, then assuming all four strain gauges are identical and affixed to a mechanical member that deforms, the following relationship between the input voltage vi and the variation of the output voltage Dvo is demonstrated in the companion website Chapter 10:

1 DR1 DR2 DR3 DR4  þ  (10.46) Dvo ¼ $ $ vi 4 R1 R2 R3 R4 where the subscripts 1, 2, 3, and 4 indicate the positions of the four strain gauges in the Wheatstone bridge. For a strain-gauge Wheatstone bridge, which is used to monitor the variation of a mechanical amount, say y, the sensor sensitivity K relates to the bridge output voltage variation Dvo in the form: Dvo ¼ K $ y

(10.47)

as discussed in the following examples. nn

Example 10.6 Design a displacement sensor to detect a sinusoidal motion using a microcantilever beam, such as the one shown in Figure 10.16, and four identical strain gauges connected in a Wheatstone bridge. Indicate the location and position of the strain gauges on the cantilever as well as their connection in the Wheatstone bridge. Express the bridge output voltage variation as a function of the mechanical strain.

Solution The sinusoidal motion of the support generates out-of-plane oscillatory bending of the microcantilever. It is known from mechanics of materials theory that the strains are maximum on the two faces (face 1 is shown in Figure 10.16 and face 2 is the opposite face). During vibration, one face shortens (and the strains on it are negative), while the opposite face extends (and its strains are positive). The maximum strain values are recorded at the microcantilever’s root. They are calculated by means of Hooke’s law (stress ¼ Young’s modulus  strain or s ¼ E$ε) as

f $l$h s mb $ h=ð2Iz Þ ¼ ε¼ ¼ E E



  2 $ w $ h3 12 6f $ l ¼ (10.48) E $ w $ h2 E

In Eq. (10.48), mb is the bending moment at a length l from the free end, mb ¼ f$l (f being a fictitious force applied at the free end of the beam that would bend the microcantilever identically to the actual support motion), and Iz is the moment of inertia of the cross section (Iz ¼ w$h3/12). Two strain gauges need to be placed on one face (such as on face 1 in Figure 10.16) and the other two on the opposite

503

504

CHAPTER 10 Coupled-Field Systems

h f (Fictitious force)

Face 1

w u = U sin(ωt)

l

R1 R3

FIGURE 10.16 Cantilever Beam Under Sinusoidal Motion. face, as close as possible to the root (fixed end). Eq. (10.48) took into account that the magnitudes of the strains on opposite faces are equal when measured at the same location along the cantilever. Let us assume that face 1 is the one that extends under bending and the opposite face 2 compresses. The strain gauges denoted by R1 and R3 are connected in the place of resistances R1 and R3 in the Wheatstone bridge of Figure 10.15. The other two strain gauges, R2 and R4, which are not shown in Figure 10.15, are placed on the face opposite to face 1 of the cantilever’s root and are connected as R2 and R4 in the Wheatstone bridge. The relative resistance variations are

DR1 DR3 DR2 DR4 ¼ ¼ Kg $ ε and ¼ ¼ Kg $ ðεÞ ¼ Kg $ ε R1 R3 R2 R4

(10.49)

As a consequence, Eq. (10.46) simplifies to

1 Dvo ¼ $ ð4Kg $ εÞ $ vi ¼ Kg $ vi $ ε or Dvo ¼ K $ ε 4 where K ¼ Kg$vi is the device sensitivity as per the definition of Eq. (10.47).

(10.50) nn

nn

Example 10.7

For the microsystem of Example 10.6 and sketched in Figure 10.16 analyze the sensor sensitivity K connecting the bridge output voltage amplitude to the input motion amplitude as a function of the subresonant excitation frequency. The output voltage amplitude is sensitivity times the cantilever’s support displacement amplitude, as defined in Eq. (10.47). Known are l ¼ 200 mm, h ¼ 100 nm, w ¼ 40 mm, r ¼ 5200 kg/m3, E ¼ 170 GPa, Kg ¼ 2.2, and vi ¼ 20 V. Consider that the system is lossless.

10.3 Electromechanical Coupling

Solution

The lumped-parameter model of the microcantilever, where m and k are effective mass and stiffness (the subscript e has been dropped), is shown in Figure 10.17. A similar application is studied in Chapter 9 when discussing transmissibility. z

m

u = U sin(ωt)

k

Vibrating support

FIGURE 10.17 Lumped-Parameter Model of a Microcantilever With Harmonic Input Motion. The dynamic equation of motion of the mass is

m $ z€ ¼ k $ ðz  uÞ

(10.51)

m $ y€ þ k $ y ¼ m $ u€ where y ¼ z  u

(10.52)

This equation can also be written as

is the relative displacement of the microcantilever free end with respect to the moving support. The transfer function corresponding to Eq. (10.52) is

GðsÞ ¼

YðsÞ m $ s2 ¼ UðsÞ m $ s2 þ k

(10.53)

and the corresponding G(u$j) function is

Gðu $ jÞ ¼

m $ u2 k  m $ u2

(10.54)

which is a real number, so its modulus is equal to the function itself. The amplitude of the relative motion, Y, can therefore be expressed in terms of the input amplitude U:

Y ¼ Gðu $ jÞ $ U ¼

m $ u2 $U k  m $ u2

(10.55)

Working with amplitudes, it is known that a force F applied at the end of the microcantilever produces a deflection Y according to the mechanics of materials equation (see also Appendix F):

Y¼

F $ l3 ¼ 3E $ Iz

F $ l3 4F $ l3 ¼ 3 E $ w $ h3 w$h 3E $ 12

(10.56)

By using Eq. (10.55), Eq. (10.56) can be written as

F¼

E $ w $ h3 E $ w $ h3 $ m $ u 2 $U $ Y ¼ 4l3 4l3 $ ðk  m $ u2 Þ

(10.57)

At the same time, the maximum strain (amplitude), X, which is recorded at the length l from the free end, can be related to the force amplitude F according to Eq. (10.48):

X¼

6l $F E $ w $ h2

(10.58)

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CHAPTER 10 Coupled-Field Systems

Eqs. (10.57) and (10.58) are combined and result into

3h $ m $ u2 $U 2l2 $ ðk  m $ u2 Þ

X¼

(10.59)

In terms of amplitudes, Eq. (10.50) can be written as

DVo ¼ Kg $ vi $ X

(10.60)

Combining Eqs. (10.59) and (10.60) results in

DVo ¼

3h $ m $ u2 $ Kg $ vi $U 2l2 $ ðk  m $ u2 Þ

(10.61)

Since DVo ¼ K$U, the system’s sensitivity K is therefore

K¼

3h $ m $ u2 $ Kg $ vi 2l2 $ ðk  m $ u2 Þ

(10.62)

Numerically, the equivalent mass is determined according to Table 2.6 as m ¼ (33/140)$ r$w$h$l ¼ 9.81$1013 kg, the equivalent stiffness is calculated based on Table 2.2 as k ¼ (3E$Iz)/l 3 ¼ (E$w$h3)/(4 l3) ¼ 2.125$104 N/m; therefore, the natural frequency is un ¼ 14,721 rad/s. Figure 10.18 shows the variation of K with u when u is subresonant, u < un. As indicated in Eq. (10.62), the microdevice sensitivity increases to infinity when the input frequency approaches the natural frequency.

Sensitivity,K (V/N)

506

Frequency, ω (rad/sec)

FIGURE 10.18 Variation of Microdevice Sensitivity With Input Frequency. nn

10.3.3 Electromagnetomechanical Coupling Systems containing magnetic components in addition to mechanical and electrical elements result in coupled electromagnetomechanical systems, as illustrated by the following examples.

10.3 Electromechanical Coupling

nn

Example 10.8 A magnet is attached to a bridge to measure velocity and can slide inside a fixed coil, which is electrically connected to a voltmeter, as sketched in Figure 10.19(a). The bridge is bent out of its equilibrium position and allowed to undergo free vibrations. Determine the overall (equivalent) viscous-type damping (loss) coefficient of this mechanical system knowing the bridge characteristics: length l ¼ 0.4 m, width w ¼ 0.02 m, thickness h ¼ 0.001 m, Young’s modulus E ¼ 210 GPa, mass density r ¼ 7800 kg/m3, the mass of the magnet mm ¼ 0.02 kg, as well as two consecutive voltage amplitude readings of 1 mV and 0.98 mV. Consider that the only electrical resistance is the one of the voltmeter.

Voltmeter

m z Fixed coil Magnet Fixed

Fixed

k eq

ceq

h Bridge

z l

(a)

(b)

FIGURE 10.19 Electromechanical System With Fixed Coil and Magnet Attached to a Bending Bridge: (a) Physical Model; (b) Lumped-Parameter Mechanical Model.

Solution

By Faraday’s law, a motional electromotive force (voltage) vm is induced in the electrical circuit containing the voltmeter and its magnitude is

vm ðtÞ ¼ B $ lc $

dzðtÞ dt

(10.63)

where lc is the coil length and B is the magnetic field. The center bridge velocity is dz(t)/dt. The actual mechanical system is equivalent to a lumped-parameter system that is formed of a mass m, a viscous damper with coefficient ceq, and a linear-motion spring of stiffness keq, all placed at the bridge midpointdsee Figure 10.19(b). The mass is the sum of the magnet mass mm and the equivalent bridge mass meq ¼ (13/35)$r$w$h$l (as given in Table 2.6), m ¼ mm þ meq. The midpoint stiffness of the bridge is provided in Table 2.2 as keq ¼ 192E$Iz /l 3, with the moment of inertia being Iz ¼ w$h3/12. As shown in Chapter 3, the equivalent damping coefficient is calculated as

ceq ¼ 2m $ un $ xeq ¼ 2 $ xeq $

pffiffiffiffiffiffiffiffiffiffiffiffiffi m $ keq

(10.64)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi where un is the natural frequency, un ¼ keq =m. Eq. (3.38) of Chapter 3 shows that the logarithmic decrement (involving two consecutive motion amplitudes Zke1 and Zk) is

d ¼ ln

2p $ xeq Zk1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffi Zk 1  x2 eq

(10.65)

507

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CHAPTER 10 Coupled-Field Systems

As shown in Eq. (3.32), the free-damped vibrations of the lumped-parameter mechanical system h qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i shown in Figure 10.19(b) are described by z ¼ Z $ e xeq $ un $ t $ sin 1  x2eq $ un $ t þ 4 . The time derivative of z is therefore

qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi  dzðtÞ 1  x2eq $ un $ t þ 4 1  x2eq $ cos ¼ Z $ un $ exeq $ un $ t $ dt qffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 1  xeq $ un $ t þ 4  xeq $ sin

(10.66)

which can be written as

 qffiffiffiffiffiffiffiffiffiffiffiffiffiffi  dzðtÞ 1  x2eq $ un $ t þ 41 ¼ Z1 $ Z $ un $ exeq $ un $ t $ sin dt

(10.67)

where the new amplitude factor Z1 is a constant amount depending only on xeq and un. Combining Eqs. (10.63) and (10.67) indicates that the voltage vm is a sinusoidal function, whose amplitude is

Vm ¼ B $ lc $ Z1 $ Z $ un $ exeq $ un $ t

(10.68)

As a consequence, the ratio of two consecutive voltage amplitudes is equal to the ratio of the consecutive deflection amplitudes at the bridge midpoint; therefore,

ln

2p $ xeq Vm;k1 Zk1 ¼ ln ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffi Vm;k Zk 1  x2

(10.69)

eq

The equivalent damping ratio is found from Eq. (10.69):

xeq

Vm;k1 ln Vm;k ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffi Vm;k1 2 2 4p þ ln Vm;k

(10.70)

For the numerical values of this example, Eq. (10.70) yields an equivalent damping ratio xeq ¼ 0.0032. The mass, equivalent midpoint bridge stiffness, and natural frequency are determined to be m ¼ 0.043 kg, keq ¼ 1050 N/m, and un ¼ 156.265 rad/s. Eq. (10.64) gives an equivalent damping coefficient ceq ¼ 0.0433 N s/m. nn

nn

Example 10.9 Derive the mathematical model of the electromechanical system of Figure 10.20. The mechanical subsystem is formed of an arm (rotor) of mass moment of inertia J, which can rotate with small motions in two end bearings (assumed to be ideal/frictionless), and of a rotary spring of stiffness k that is anchored at one end. This mobile arm is metallic and closes an electrical circuit that has a voltage source vs, a resistance R, and an inductance L. An external constant magnetic field B is applied perpendicularly to the electrical circuit plane.

10.3 Electromechanical Coupling

Anchor Electrical system Bearing Rotary spring

i

k r

Mechanical system

+ Arm

ω

l

vs B

–

Bearing

FIGURE 10.20 Schematic of an Electromagnetomechanical System With Rotary Mechanical Elements.

i r B v f

θ il

FIGURE 10.21 Top View of Mechanical Rotary Arm.

Solution As shown next, a differential equation governs the dynamics of the mechanical system, another one defines the dynamic behavior of the electrical circuit, and two coupling equations connect mechanical to electrical variables. Let us discuss first the coupling equations based on Figure 10.21, which shows the top view of the mechanical rotary arm. The interaction between the magnetic field B and the current i passing through the element of length l (which is part of both the rotary mechanical system and the electrical system) generates the following force f (which is shown in Figure 10.21):

f ¼ l $ i  B; f ¼ l $ i $ B

(10.71)

Because there is a right angle between the two product vectors of Eq. (10.71), the magnitude f is the product of the magnitudes of the two vectors. The force f generates a torque about the pivot point that is

mt ¼ f $ r $ cos q ¼ l $ r $ B $ i $ cos q ¼ ðA $ B $ cos qÞ $ i

(10.72)

where A ¼ l$r is the area of the rotating surface (see Figure 10.20). Eq. (10.72) is a coupling equation as it connects the (mechanical) torque mt to the (electrical) current i. It is also known from electromagnetics that, due to the Hall effect, a back electromotive force (actually voltage) vb is generated by the motion of a current-carrying conductor in a magnetic field

509

510

CHAPTER 10 Coupled-Field Systems

R

i

L

+ vs

+ vb

–

–

FIGURE 10.22 Electrical Subsystem of the Electromechanical System. B, which is perpendicular to the conductor; this voltage, which opposes the source voltage vs, is calculated as

dq vb ¼ v  B $ l ¼ v $ B $ l $ sin ð90  qÞ ¼ v $ B $ l $ cos q ¼ r $ $ B $ l $ cos q dt dq ¼ ðA $ B $ cos qÞ $ dt (10.73) where v denotes velocity and v ¼ r$u ¼ r$dq/dt. Eq. (10.73) relates the voltage vb (which is an electrical amount) to the angular velocity of the rotor u ¼ dq/dt (which is a mechanical amount); therefore, this is another coupling equation. The time-domain differential equation of the mechanical subsystem motion is

J$

d 2 qðtÞ ¼ mt ðtÞ  k $ qðtÞ dt2

(10.74)

where k$q is the opposing elastic torque produced by the spring. Based on Figure 10.22, the differential equation defining the dynamic response of the electrical system is

vs ðtÞ  vb ðtÞ ¼ R $ iðtÞ þ L $

diðtÞ dt

(10.75)

By combining Eqs. (10.72) and (10.74), as well as Eqs. (10.73) and (10.75), the following equations are obtained:

8 > d2 qðtÞ > > þ k $ qðtÞ ¼ A $ B $ iðtÞ $ cos qðtÞ

> diðtÞ dqðtÞ > :L$ þ R $ iðtÞ ¼ vs ðtÞ  A $ B $ $ cos qðtÞ dt dt

(10.76)

where the input is the voltage vs, for instance, and the output is formed of the unknown rotation angle q and the current i. The two Eq. (10.76) can be combined into a single differential equation to express either i or q in terms of vs. nn

nn

Example 10.10

Use Simulink to plot the time variation of the arm rotation angle q(t) and the current i(t) corresponding to the electromagnetomechanical system of Example 10.9. Known are r ¼ 0.01 m, l ¼ 0.04 m, k ¼ 0.1 N m, J ¼ 2$106 kg m2, vs ¼ 60 V, R ¼ 80 U, L ¼ 0.04 H, B ¼ 1 T, as well as the nonzero initial condition q(0) ¼ 0.05 rad.

10.3 Electromechanical Coupling

Solution The differential Eq. (10.76) can be written as

8 > d2 qðtÞ k A$B > > $ iðtÞ $ cos qðtÞ ¼  $ qðtÞ þ < 2 J J dt > > diðtÞ 1 R A $ B dqðtÞ > : ¼ $ vs  $ iðtÞ  $ $ cos qðtÞ dt L L L dt 8 > d2 qðtÞ > > ¼ a1 $ qðtÞ þ a2 $ iðtÞ $ cos qðtÞ < dt2 or > > diðtÞ dqðtÞ > : ¼ a3 $ vs  a4 $ iðtÞ  a5 $ $ cos qðtÞ dt dt

(10.77)

where A ¼ r$l ¼ 4$104 m2 is the rotating loop area and the parameters in the right-hand side of Eq. (10.77) are assuming the following numerical values: a1 ¼ 50,000, a2 ¼ 200, a3 ¼ 25, a4 ¼ 2000, a5 ¼ 0.01. It can be seen that the differential equations are nonlinear because the variable q enters as a trigonometric function’s argument and also because it multiplies the other variable i. Figure 10.23 is the Simulink block diagram allowing to solve the two differential Eq. (10.77). The upper part of the diagram solves the first Eq. (10.77), while the lower part corresponds to the second Eq. (10.77). Blocks used in previous Simulink-solved examples are seen in Figure 10.23; a new block here is the Trigonometric Function, which enables calculation of cos q; this block is taken from the Math Operations library. The nonzero initial condition of this example is specified by clicking the second integration block on the upper part of the block diagram (Integrator 2 that which gives the angle q), and specifying the value of q(0). Figure 10.24 illustrates the plots of q and i as functions of time. It can be seen that the current reaches a constant value of less than 0.8 A very fast, whereas the arm has an oscillatory motion with an amplitude of approximately 3 degrees.

d θ(t ) dt

a1

θ(t ) (rad)

d 2θ(t ) dt 2 cos θ(t )

d θ(t ) dt

cos θ(t )

cos θ(t ) i (t ) a5 a2

θ (t ) (deg) di (t ) dt

vs

i (t )

a3

a4

FIGURE 10.23 Simulink Diagram for Solving the Mathematical Model of the Electromagnetomechanical System of Example 10.9.

511

CHAPTER 10 Coupled-Field Systems

i (A)

θ (deg)

512

Time (sec)

Time (sec)

(a)

(b)

FIGURE 10.24 Plot of Solution in Terms of Time: (a) Rotation Angle q (deg); (b) Current i (A), Time Is Measured in Seconds. nn

Rotary Direct-Current Electric Motor With Mechanical Load The electromechanical system of Figure 10.25 consists of a rotary dc (direct-current) electric motor and load shaft. The dynamic model of this system consists of equations that describe the mechanical, electrical, and mechanicaleelectrical (coupledfield) behavior. Essentially, the electromechanical system sketched in Figure 10.25 is formed of a mobile part (the rotor armature), which rotates under the action of a magnetic field produced by the electrical circuit of an armature (the stator). Ra

+

Rotor armature Damper

ia vb

va

– La

Jl θ, ma Stator

c Load

FIGURE 10.25 Schematic of a Rotary-Motion DC Motor With Load as an Electromagnetomechanical System.

The electrical circuit is governed by Kirchhoff’s second law, according to which dia ðtÞ ¼ va ðtÞ  vb ðtÞ (10.78) dt where the subscript a indicates the armature and vb is the back electromotive force (voltage). The mechanical part of the system is governed by the equation Ra $ ia ðtÞ þ La $

Jl $

d2 qðtÞ dqðtÞ ¼ ma ðtÞ  c $ dt2 dt

(10.79)

10.3 Electromechanical Coupling

where ma is the torque developed due to the stator-rotor interaction. It is also known that the following equations couple the mechanical and electrical fields: dqðtÞ (10.80) dt with Km (measured in N$m/A) and Ke (measured in V$s/rad) being constants. The first Eq. (10.80) is similar to Eq. (10.72), which expressed the mechanical moment resulting from the interaction of an external magnetic field with the mobile arm carrying an electric current. The second Eq. (10.80) is similar to Eq. (10.73) that exemplified the back electromotive force being generated in a circuit with a current-carrying mobile arm in an external magnetic field. When neglecting the armature inductance, which is La ¼ 0 in Eq. (10.78), the following equation can be written by combining Eqs. (10.78) and (10.80) ma ðtÞ ¼ Km $ ia ðtÞ; vb ðtÞ ¼ Ke $

ma ¼

ma;b Km Ke $ Km $ va  $ u or ma ¼ ma;b  $u Ra Ra u0

(10.81)

and using the angular velocity u(t) ¼ dq(t)/dt instead of the rotation angle q(t). Eq. (10.81) indicates a linear relationship between the actuation torque ma and the angular velocity u for a specified armature voltagedthis relationship is plotted with solid line in Figure 10.26. The actuation torqueeangular velocity curve generated by an armature voltage va1 intersects the coordinate axes at u0dthe free (no-load) angular velocity, when basically the dc motor shaft spins freely with no external load acting on it, and at ma,bdthe block (stall) torque, where an external load torque equal to ma,b stalls the shaft rotation altogether. These two parameters are obtained from Eq. (10.81) as u0 ¼

va Km ; ma;b ¼ $ va Ke Ra

(10.82)

Any operation point defined by an angular velocity and a delivered torque is located on this characteristic curve and is at the intersection of the actuation torque and the load torque curves. For the particular case of a constant load torque ml illustrated in Figure 10.26, the operation point has the coordinates: uop ¼

va Ra  $ ml ; mop ¼ ml Ke Ke $ Km

m Actuation torque va,1

Operation point

m a,b

Load torque

m op va O

ωop

ω0

ω

FIGURE 10.26 Actuation and Load Torques Versus Shaft Angular Velocity in a Rotary DC Motor.

(10.83)

513

514

CHAPTER 10 Coupled-Field Systems

For a different armature voltage, say va1 > va, the torqueeangular velocity characteristic shifts up but preserves its slope for constant values of Kt, Ke, and Ra, as indicated in Eq. (10.83). nn

Example 10.11 Consider the electromechanical system represented by the dc motor of Figure 10.25. Assume that, in addition to the given parameters, a load torque ml acts on the mechanical cylinder. (i) Discuss the coupling of this system based on the transfer function matrix that connects the input vector {va, ml} to the output vector {ia, u}. (ii) Calculate the free angular velocity u0, the block torque ma,b, as well as the operation point (uop, mop) for the following numerical parameters: Jl ¼ 0.001 kg m2, c ¼ 0.00005 N m s, ml ¼ 0.3 N m, Ra ¼ 3 U, La ¼ 0.003 H, va ¼ 80 V, Ke ¼ 0.105 N m/A (1 N m/A ¼ 1 V s/rad), and Km ¼ 0.105 N m/A. Also plot the system response u(t).

Solution (i) The electrical circuit is governed by Kirchhoff’s voltage law, according to Eq. (10.78). Eq. (10.79) changes due to the presence of the load torque:

Jl $

d 2 qðtÞ dqðtÞ duðtÞ or Jl $ þ c $ uðtÞ  ma ðtÞ ¼ ml ðtÞ ¼ ma ðtÞ  ml ðtÞ  c $ 2 dt dt dt (10.84)

where u is the angular velocity. Combining Eqs. (10.78), (10.80) and (10.84) results in

8 > dia ðtÞ > > þ Ra $ ia ðtÞ þ Ke $ uðtÞ ¼ va ðtÞ < La $ dt > duðtÞ > > : Jl $ þ c $ uðtÞ  Km $ ia ðtÞ ¼ ml ðtÞ dt

(10.85)

This system is field coupled as the two Eq. (10.85) comprise variables from both the electrical and the mechanical fields. Laplace transforming Eq. (10.85) results in



L a $ s þ Ra Km

     1 0 Va ðsÞ Ia ðsÞ ¼ $ UðsÞ 0 1 Ml ðsÞ Jl $ s þ c Ke

(10.86)

where U(s) is the Laplace transform of uðt Þ ¼ dqðt Þ=dt, the angular velocity. Eq. (10.86) can be written as

Ia ðsÞ



UðsÞ

¼ ½GðsÞ $

Va ðsÞ



Ml ðsÞ

(10.87)

which, by comparison to the equation of Figure 10.4, indicates that the transfer function matrix [G(s)] is

2



La $ s þ Ra ½GðsÞ ¼ Km 

Ke Jl $ s þ c

G11 ðsÞ G12 ðsÞ ¼ G21 ðsÞ G22 ðsÞ



1  1 $ 0

Jl $ s þ c  6 0 6 DðsÞ ¼6 4 Km 1 DðsÞ

Ke DðsÞ

3

7 7 7 La $ s þ Ra 5  DðsÞ

(10.88)

10.3 Electromechanical Coupling

with the denominator D(s) being

DðsÞ ¼ Jl $ La $ s2 þ ðJl $ Ra þ c $ La Þ $ s þ Ke $ Km þ c $ Ra

(10.89)

and G11(s), G12(s), G21(s), G22(s) representing individual transfer functions. Field coupling is also indicated by the nondiagonal transfer function matrix of Eq. (10.88). (ii) The free angular velocity and block torque are determined from Eq. (10.82) as: u0 ¼ 761.905 rad/s and ma,b ¼ 2.8 N m. Similarly, the operation point coordinates are found using Eq. (10.83) as uop ¼ 680.272 rad/s and mop ¼ ml ¼ 0.3 N m. Combining Eq. (10.87) e(10.89), the Laplace-transformed angular velocity is

UðsÞ ¼ G21 ðsÞ $ Va ðsÞ þ G22 ðsÞ $ Ml ðsÞ

(10.90)

with Va(s) and Ml(s) being the Laplace transforms of va(t) and ml(t), respectively. The time-domain solution u(t) (which is not given here) is obtained based on Eqs. (10.87)e(10.90) and plotted using the following MATLAB code: >> % introduce numerical values of all physical parameters: jl, c,. La, Ra, ml, va, km, ke. >>s=tf(‘s’); >>g=inv([la*s+ra,ke;-km,jl*s+c])*[1,0;0,-1]; >> t = 0:0.0001:3; >>om=step(va*g(2,1),t)+ step(ml*g(2,2),t);

ω (rad/sec)

>>plot(t,om)

Time (sec)

FIGURE 10.27 Time-Domain Plot of the Shaft Angular Velocity for a Coupled-Field DC Motor System. The shaft angular velocity is plotted in Figure 10.27 for the numerical values of this example.

nn

Translatory Direct-Current Electric Motor With Mechanical Load Electromagnetic translatory (also called linear) dc motors operate similarly with the rotary motor previously studied here. Figure 10.28 is the schematic representation of a dc linear-motion motor with its actuation circuit and mechanical load subsystems.

515

516

CHAPTER 10 Coupled-Field Systems

y, fa c R

+

ml

ia

va

–

Fixed coil

L

vb

k

Mobile Load magnetic rod

FIGURE 10.28 Schematic of a Translatory-Motion DC Motor With Load as an Electromagnetomechanical System.

The interaction between the current passing through a fixed cylindrical coil and a magnetic rod that can slide inside the coil generates an actuation force fa, which is used as input to the massedamperespring mechanical load system. A back electromotive force (voltage) vb is also generated in the actuation electrical circuit, which opposes the actuation voltage va. These two amounts, namely fa and vb are expressed similarly to the ones of the rotary dc motor in Eq. (10.80): dyðtÞ (10.91) dt Eq. (10.91) are the relationships that realize coupling between the electromagnetic and mechanical fields. The equations governing the behavior of the electrical and of the mechanical systems illustrated in Figure 10.28 are 8 diðtÞ > > > < R $ iðtÞ þ L $ dt ¼ va ðtÞ  vb ðtÞ (10.92) > d 2 yðtÞ dyðtÞ > > : ml $  k $ yðtÞ ¼ f ðtÞ  c $ a dt dt2 fa ðtÞ ¼ Km $ iðtÞ; vb ðtÞ ¼ Ke $

Substituting the actuation force and back voltage of Eq. (10.91) into Eq. (10.92), yields: 8 diðtÞ dyðtÞ > > >L$ þ R $ iðtÞ ¼ va ðtÞ  Ke $ < dt dt (10.93) 2 > d yðtÞ dyðtÞ > > : ml $ þ k $ yðtÞ ¼ Km $ iðtÞ þ c$ dt dt2 Eq. (10.93) form the mathematical model of the coupled-field system of Figure 10.28. All the other features connected to the rotary dc motor with load can similarly be derived for the translatory dc motor, such as block force, free velocity, or operation point.

10.3 Electromechanical Coupling

10.3.4 Electromagnetomechanical Coupling With Optical Detection in MEMS Optical capturing of the reflected beam from a tilting surface provides information about the amount of tilt by using a photodiode that can detect the deflection of a reflected ray. This principle is implemented in MEMS applications to measure angular position. The distance d characterizing the deflection of a reflected ray after the reflecting surface has tilted an angle q can be calculated simply based on Figure 10.29: d ¼ h $ ½tana0  tanða0  2qÞ

(10.94)

which allows determining the tilt angle q as q¼

a0  tan1 ðtana0  d=hÞ 2

Tilted reflected ray n

Incoming ray α h

θ

Original reflected ray

n0 Photodiode screen

α α0

(10.95)

θ

α0

d

Tilted position Reflecting surface Original position

FIGURE 10.29 Optical Detection by Means of a Tilting Surface and Photodiode.

nn

Example 10.12

A sinusoidal current i ¼ I$sin(u$t) passes through the circular-loop electrical circuit of radius r, which is printed on a rigid thin plate of thickness hp and mass density r. The plate attached to a flexible microrod of length l, diameter dr, and shear modulus G. The microdevice is placed under the action of an unknown constant magnetic field B, as shown in Figure 10.30(a). Determine the magnetic field B when I ¼ 0.002 A, u ¼ 40 rad/s, r ¼ 100 mm, dr ¼ hp ¼ 1 mm, l ¼ 50 mm, r ¼ 5500 kg/m3, G ¼ 500 GPa. The tilt of the rigid plate is monitored optically by a photodiode, which is placed at a distance h ¼ 2 mm from the plate, the incident laser beam angle being a0 ¼ 30 (see Figure 10.29). It is also known that the maximum deviation of the reflected ray is d ¼ 4 mm.

517

518

CHAPTER 10 Coupled-Field Systems

Current-carrying loop

Flexible rod

mt , θ

r

l

B

J

mt, θ

kt

Anchor Rigid plate

(a)

(b)

FIGURE 10.30 (a) Top View of Electromagnetomechanical Sensor; (b) Lumped-Parameter Model of Mechanical System.

Solution It can be shown using Faraday’s law that the interaction between a current passing through a circular loop and an external magnetic field parallel to the loop plane results in a moment (torque) applied to the loop plane; the moment acts along an axis that passes through the circular-loop center and perpendicular to the magnetic field direction (see Lobontiu (2007) or Bueche and Jerde (1995) for more details). Moreover, the torque value is equal to the product of the loop area, magnetic field, and current:

mt ¼ A $ B $ i ¼ pr 2 $ B $ I $ sinðu $ tÞ ¼ Mt $ sinðu $ tÞ

(10.96)

where the torque amplitude is Mt ¼ pr $B$I. The direction of the torque mt being that of the flexible rod axis, its effect is alternating torsion. For small tilt angles, the lumped-parameter mathematical model of the rigid plate mechanical motion is derived using Newton’s second law of motion based on Figure 10.30(b): 2

€ ¼ m t  kt $ q J $q

(10.97)

where J is the plate mass moment of inertia, kt is the torsional stiffness of the rod, and q is the tilt angle of the plate (identical to the rotation angle of the rod end). The mass moment of inertia for thin plates (see Table 2.4) and torsional stiffness (see Table 2.3) are calculated as

J¼

G $ Ip pG $ dr4 m $ ð2rÞ2 4r $ r 2 $ dr $ ð2rÞ2 4r $ r 4 $ dr ¼ ¼ ; kt ¼ ¼ (10.98) 12 12 3 l 32 $ l

Because the plate is thin, the square of its thickness dr was neglected in J of Eq. (10.98). With the numerical data of this example, J ¼ 73.33$1020 kg m2 and kt ¼ 9.82$1010 N m. By applying the Laplace transform to Eq. (10.97), the following transfer function is obtained:

GðsÞ ¼

QðsÞ 1 ¼ Mt ðsÞ J $ s2 þ kt

(10.99)

The complex transfer function is therefore

Gðu $ jÞ ¼

1 Q ¼ jGðu $ jj ¼ kt  J $ u2 Mt

(10.100)

where Q and Mt are the amplitudes of the tilt angle q and torsional moment mt, respectively. Combination of Eqs. (10.96) and (10.100) results in

Q ¼ jGðu $ jÞj $ Mt ¼

p $ r2 $ B $ I kt  J $ u 2

(10.101)

10.3 Electromechanical Coupling

Considering now that the angle Q is q of Eq. (10.95), Eq. (10.101) yields

   kt  J $ u2 $ a0  tan1 ðtan a0  d=hÞ B¼ 2p $ r 2 $ I 

(10.102)

and the numerical value of B is 0.0117 T.

nn

10.3.5 Piezoelectric Coupling Another type of interaction between the mechanical and electrical fields with applications in actuation and sensing is through piezoelectric coupling, which is studied in this section.

Brief Introduction to Piezoelectricity In piezoelectric actuation or sensing coupling between the mechanical and electrical fields is realized by means of the direct and converse piezoelectric effects, which are constitutive features of piezoelectric materials. In the direct piezoelectric effect, external application of a mechanical strain (deformation) to a piezoelectric material generates a voltage, which is proportional to the mechanical action. A device based on the direct piezoelectric effect can be used as a sensor or generator. The converse piezoelectric effect generates mechanical strain (deformation) from a piezoelectric material connected to an external voltage source (with the strain being proportional to the voltage); this principle can be used for actuation resulting in motors. Some natural materials, such as quartz crystals, the Rochelle salt or tourmaline, as well as ceramics, such as PZTs (based on lead [plumbum], zirconium, and titanium or barium and titanium) display piezoelectric behavior. Piezoelectric materials are formed of electric dipoles oriented randomly in the material structure, see Figure 10.31(a). Through a poling process, a strong external voltage (approximately

+ Electrode

Electric dipole

– – + –

Poling direction

+ – +

+ – + – +

– Electrode

(a)

FIGURE 10.31 Piezoelectric Block: (a) Unpoled; (b) Poled.

(b)

+ vp –

519

520

CHAPTER 10 Coupled-Field Systems

2000 V/m) is applied to the material, which has been heated to an elevated temperature that aligns the dipoles, as sketched in Figure 10.31(b); this structure is permanently impressed on the material by a fast cooling process. Through poling, the final dimension about the poling direction becomes larger, as illustrated in Figure 10.31(b).

Longitudinal Actuation and Sensing With Piezoelectric Block We briefly present the actuator (motor) and sensor (generator) behaviors of a piezoelectric block with longitudinal action. The block’s deformation and charge variation pertaining to the longitudinal direction result from superimposing the actions produced by the mechanical and electrical fields.

Actuation Consider an actuation voltage va is applied to a piezoelectric block along the poling direction (produced by the poling voltage vp), as illustrated in Figure 10.32(a) and with va and vp having the same polarity. The result is the extension of the block dimension parallel to poling direction and compression of the cross section. Conversely, when an external voltage is applied that has a polarity inverse to the one of the poling voltage, as indicated in Figure 10.32(b), the dimension parallel with the poling direction shrinks, whereas the cross-sectional dimensions extend due to volume conservation. In either of the two situations, linear translatory actuation is mainly possible along the poling axis, although deformations and associated forces are also generated in the section perpendicular to the poling axis. Actuation based on deformation about the poling axis is known as longitudinal actuation, whereas transverse actuation generates mechanical motion about either of the cross-sectional axes. A third operation mode, shearing, is presented in the companion website Chapter 10, together with detailed presentation of the constitutive piezoelectric laws.

Actuation direction + Electrode

– Electrode y –

+ Poling direction

–

– Electrode

f

A

va

+ va

l –

+ + Electrode Fixed surface

(a)

(b)

(c)

FIGURE 10.32 Piezoelectric Block as a Longitudinal Actuator Realizing: (a) Expansion; (b) Compression (Original Block Is Shown With Solid Lines and Deformed Block Is Indicated With Dashed Lines); (c) Boundary Conditions, Force, and Displacement.

va

10.3 Electromechanical Coupling

The aim is to evaluate the mechanical deformation (strain) resulting from the actuator (motor) behavior of a piezoelectric block under the action of an external (actuation) voltage va that has the same direction as the poling direction. A block of original length l and cross-sectional area A with one end fixed is acted on by an external voltage va, as shown in Figure 10.32(c). According to the converse piezoelectric effect principle, the deformation is proportional to the applied voltage; therefore, Dl ¼ d $ va

(10.103)

with d being known as the piezoelectric charge (or strain) constant. The constant is measured in m/V or, alternatively, in C/N. Eq. (10.103) illustrates the coupling between the mechanical field (represented by Dl) and the electrical field (va).

Piezoelectric Block Actuator and Load Spring One common design is the one sketched in Figure 10.33(a) and consists of a piezoelectric block actuator working axially against a spring that represents the external load. A lumped-parameter model of this actuator-spring model is illustrated in Figure 10.33(b), where the actual distributed properties of the piezoelectric actuator have been substituted by equivalent lumped-parameter ones: mass meq, damping coefficient ceq, and stiffness kPZT, as well as the piezoelectrically generated force denoted by fa,b. A quasi-static model can be obtained from the dynamic one of Figure 10.33(b) considering meq ¼ 0 and ceq ¼ 0. The equation that yields the actuation force in terms of the end displacement is f ¼ fa;b 

fa;b $y y0

(10.104)

where fa,b is the block force (the minimum external force that results in no piezoelectric deformation, so y ¼ 0) and y0 is the free displacement of the piezoelectric (when there is no external load and that is y0 ¼ Dl of Eq. 10.103). This actuation characteristic expresses f in terms of y and is illustrated in Figure 10.32(c). Note that Eq. (10.104) is similar to Eq. (10.81), which is the torqueevelocity equation of a rotary dc motor. f Anchor

k

k PZT

Spring

fa,b

k

va,2

c eq

fop 0

(a)

(b)

Operation point

fa,b

meq

PZT actuator l

Actuation force

y

y

Spring force

va,1

yop

y0

(c)

FIGURE 10.33 Linear-Motion Piezoelectric Actuation: (a) Against Spring; (b) Lumped-Parameter Mechanical Model; (c) ForceeDisplacement Characteristic.

y

521

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CHAPTER 10 Coupled-Field Systems

The block force balances out the actuation force that produces a free displacement y0: fa,b ¼ kPZT$y0. Taking into account that kPZT ¼ E$A/l (see Table 2.2), with A and l being the piezoelectric cross-sectional area and length, while E is the material Young’s modulus, the two parameters of Eq. (10.104) become E$A $ d $ va (10.105) l Note, again, similarities of y0 and fa,b of Eq. (10.105) with u0 and ma,b of Eq. (10.82), which describe the rotary dc motor properties. When the piezoelectric motor works against an external load, such as the spring of Figure 10.33(a) and 10.33(b), an operation point, defined by the coordinates yop and fop, results at the intersection of the motor and spring characteristics, as illustrated in Figure 10.33(c). Because the spring force can be expressed as fop ¼ k$yop, Eq. (10.104) becomes y0 ¼ d $ va ; fa;b ¼ kPZT $ y0 ¼

k $ yop ¼ fa;b 

fa;b $ yop or k $ yop ¼ fa;b  kPZT $ yop or fa;b ¼ ðk þ kPZT Þ $ yop y0 (10.106)

which indicates that the piezoelectric block force is balanced quasi-statically by the elastic reaction of the two parallel-connection springs. The operation point coordinates are obtained from Eqs. (10.105) and (10.106) using f ¼ fop ¼ k$yop: fa;b ðE $ A $ d $ va Þ=l E $ A $ d $ va ¼ ; fop ¼ k $ yop ¼ k þ ðE $ AÞ=l k þ kPZT k$l þ E$A k $ E $ A $ d $ va ¼ k$l þ E$A

yop ¼

(10.107)

For a voltage va2 > va1, the actuation characteristic changes into one that is parallel to the original one (for va1)das shown by the dotted line in Figure 10.33(c). When inertia and damping are taken into account, as indicated in Figure 10.32(b), the balance Eq. (10.106) transforms into meq $ y€ þ ceq $ y_ þ keq $ y ¼ fa;b with keq ¼ k þ kPZT ; fa;b ¼

E $ A $ d $ va l (10.108)

Eq. (10.108) can be used to study the transient or steady-state dynamics of a system that is actuated by a piezoelectric block. nn

Example 10.13

A longitudinal piezoelectric actuator of length l ¼ 0.05 m, cross-sectional area A ¼ 25$106 m2, mass density r ¼ 7500 kg/m3, Young’s modulus E ¼ 70 GPa, and piezoelectric charge constant d ¼ 107 m/V is attached to a block of mass m ¼ 0.20 kg, which further connects to a spring of stiffness k ¼ 200 N/m, as illustrated in Figure 10.34.

10.3 Electromechanical Coupling

Piezoelectric actuator Block

k

m l

y

FIGURE 10.34 Longitudinal Piezoelectric Actuator With Block and Spring.

(i) Calculate the actuation voltage va,b that would result in a block force fa,b ¼ 185 N. For va ¼ 0.6$va,b, identify the operation point of the device. (ii) Evaluate the total system mechanical losses as an equivalent damping coefficient ceq if experimental frequency-domain testing indicates the system’s resonant peak is Mr ¼ 3.5. Consider the mass and stiffness contributions of the piezoelectric actuator. (iii) Plot the system’s response y as a function of time when an impulse voltage va ¼ d(t)$103 Vdwhere d(t) is the unit impulsedis applied to the piezoelectric actuator. Note: the electrical circuit powering the actuator is not shown in the figure.

Solution (i) The maximum actuation voltage that results in the block force is obtained from Eq. (10.105) as

va;b ¼

fa;b $ l E$A$d

(10.109)

which is va,b ¼ 52.857 V for the given parameter numerical values. The operation point’s coordinates are calculated from Eq. (10.107) for va ¼ 0.6$va,b ¼ 31.714 V as yop ¼ 3.171 mm and fop ¼ 6.343$104 N. (ii) The dynamics of the system shown in Figure 10.34 are governed by an equation similar to Eq. (10.108), which depicts a second-order system. For sinusoidal excitation, the frequency response of this system exhibits resonance, as discussed in Chapter 9. Eq. (9.39) gives the relationship between the resonant peak Mr and in our case, the equivalent damping ratio xeq. Eq. (9.39) can be written as

 4x4eq  4x2eq þ 1 Mr2 ¼ 0

(10.110)

Eq. (10.110) is solved for the equivalent damping ratio, which results in xeq,1 ¼ 0.144 and xeq,2 ¼ 0.989. The second solution is not valid because the maximum value of the damping ratio for a resonant response of a second-order system, as this one, is 0.707. The mechanical lumped-parameter model of the piezoelectrically actuated system of Figure 10.34 is shown in Figure 10.33(b). The total mass meq is formed of the mass m and the lumped mass fraction of the piezoelectric actuator, meq,PZT (which is provided in Table 2.6); as a consequence,

1 1 meq ¼ m þ meq; PZT ¼ m þ $ mPZT ¼ m þ $ rPZT $ l $ A 3 3

(10.111)

Numerically, the total mass is meq ¼ 0.2031 kg. The equivalent stiffness is calculated by adding the piezoelectric block stiffness kPZT and the spring stiffness k:

keq ¼ k þ kPZT ¼ k þ

E$A l

(10.112)

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CHAPTER 10 Coupled-Field Systems

with a numeric value of keq ¼ 35,000,200 N/m. The equivalent damping coefficient can be determined from the equivalent damping ratio, the equivalent mass and equivalent stiffness as

ceq ¼ 2xeq $

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi meq $ keq

(10.113)

and its numerical value is ceq ¼ 767.9089 N s/m. (iii) The mathematical model of Eq. (10.108) is written as

meq $ y€ þ ceq $ y_ þ keq $ y ¼ C $ va with C ¼

E$A$d l

(10.114)

where the constant multiplying the actuation voltage is of C ¼ 3.5 N/V. Laplace transforming Eq. (10.114) yields the transfer function

GðsÞ ¼

YðsÞ C ¼ Va ðsÞ meq $ s2 þ ceq $ s þ keq

(10.115)

FIGURE 10.35 Displacement Response of Piezoelectrically Actuated Mechanical System Under Unit Impulse Voltage Input. By using the tf and impulse MATLAB commands, the plot of Figure 10.35 is obtained. The transfer function numerator is actually C$103, to conform to the va ¼ d(t)$103 V actual input. nn

Sensing We now evaluate the sensor (generator) response of the block of Figure 10.36 under an external mechanical force f that produces an elongation Dl of the block whose original length is l and cross-sectional area is A. Assume that the fixed surface and the movable surface of the piezoelectric block are electrodes incorporated into an electric circuit, with a voltmeter that can measure the voltage variation. This approach is simplified as it does not take into account other electrical components that are typically present in a piezoelectric sensor, and we attempt to describe

10.3 Electromechanical Coupling

External force,f Δl A l

Piezoelectric block

Voltmeter Fixed surface

FIGURE 10.36 Piezoelectric Block as a Longitudinal Force Sensor.

the basic sensing action only. A more comprehensive discussion on the complex nature of coupling between mechanically generated charge and electrical charge is beyond the scope of the current coverage here but can be found in (more advanced) instrumentation texts. The voltage variation Dv picked up by the voltmeter is proportional to both the block deformation Dl and the external force f, as follows: 1 l $f (10.116) Dv ¼ $ Dl or Dv ¼ d E$A$d which means that the piezoelectric block can be used as either a displacement sensor or as force sensor (load cell). Eq. (10.116) used the equation of the axial deformation of a bar produced by an axial force: Dl ¼ f$l/(E$A). nn

Example 10.14 An accelerometer uses a piezoelectric plate to sense the acceleration of an input sinusoidal displacement in a horizontal plane, using the device shown in Figure 10.37(a). Use a lumped-parameter model to express the acceleration amplitude by means of the sensing voltage amplitude. Known are the

Moving platform z k

ceq

k

ceq

meq,PZT

z

meq

m ceq

m u = U·sin(ωt)

kPZT

keq

u = U·sin(ωt)

u = U·sin(ωt)

Piezoelectric sensor

(a)

(b)

(c)

FIGURE 10.37 (a) Piezoelectric Sensing of Acceleration; (b) Lumped-Parameter Model of Original System; (c) Lumped-Parameter Model of Equivalent System.

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CHAPTER 10 Coupled-Field Systems

dimensions of the piezoelectric plate: thickness h ¼ 0.003 m, side of square cross-sectional area a ¼ 0.01 m, as well as Young’s modulus E ¼ 50 GPa, piezoelectric mass density r ¼ 7500 kg/m3, and piezoelectric charge constant d ¼ 2.7$1010 m/V. We also know the mass m ¼ 0.05 kg, and spring stiffness k ¼ 5$108 N/m. Assume that the total system losses are lumped into the equivalent viscous damping coefficient ceq ¼ 1670 N s/m. The voltage amplitude read by the sensor is V ¼ 0.05 V. Consider that the spring and damper are precompressed to maintain permanent contact between the vibrating mass m and the piezoelectric sensor. Also, draw the Bode plot magnitude ratio to verify the results.

Solution By taking into account that the piezoelectric plate can be represented as a massespring system, Figure 10.37(b) shows the lumped-parameter model of the original system of Figure 10.37(a), which includes the equivalent mass and spring stiffness of the piezoelectric sensor. Because the actual mass m and the piezoelectric plate equivalent mass move together, the lumped-parameter model can be simplified further to the one of Figure 10.37(c). The equation of motion of the equivalent mass is obtained by applying Newton’s second law of motion, and using the lumped-parameter model of Figure 10.37(c):

_  keq $ ðz  uÞ meq $ z€ ¼ ceq $ ðz_  uÞ

(10.117)

where z monitors the absolute motion of the mass m about the equilibrium position (the one obtained by the precompression of the dashpot). The equivalent mass and equivalent spring rate are

8 1 1 > 2 > > < meq ¼ m þ meq;PZT ¼ m þ 3 $ mPZT ¼ m þ 3 $ r $ h $ a > E $ a2 > > : keq ¼ k þ kPZT ¼ k þ h

(10.118)

A similar accelerometer problem is studied in Chapter 9, and it is demonstrated that Eq. (10.117) can be brought to the form

meq $ y€ þ ceq $ y_ þ keq $ y ¼ meq $ u€

(10.119)

where y ¼ z e u is the coordinate indicating the relative motion of the mass with respect to the moving platform. The piezoelectric sensor generates a voltage v proportional to the relative displacement y, as shown in Eq. (10.116) when eliminating the variation symbol D:

y ¼ d$v

(10.120)

The Laplace transform is applied to Eqs. (10.119) and (10.120) and that results in the following transfer function:

GðsÞ ¼

meq $ s2 VðsÞ ¼ UðsÞ d $ ðmeq $ s2 þ ceq $ s þ keq Þ

(10.121)

Figure 10.38 is the MATLAB Bode plot of the magnitude ratio corresponding to the transfer function of Eq. (10.121), obtained for the following parameter values: meq ¼ 0.0508 kg and keq ¼ 2,166,666,666.7 N/m. Our interest lies on transmissibility, the amplitude ratio of voltage to input displacement, which is the modulus of the complex function G(u$j), as discussed in Chapter 9; that is,

jGð j $ uÞj ¼

jVðu $ jÞj V ¼ ¼ jUðu $ jj U

meq $ u2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d $ ðkeq  meq $ u2 Þ2 þ ðceq $ uÞ2

(10.122)

10.3 Electromechanical Coupling

FIGURE 10.38 MagnitudeeRatio Bode Plot of Piezoelectric System’s Transfer Function.

Dividing the numerator and denominator of Eq. (10.122) by keq, yields:

u2 $U u2n sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V¼

2

u2 u 2 d$ 1  2 þ 2xeq $ un un

(10.123)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   where the natural frequency and damping ratio are un ¼ keq =meq ; xeq ¼ ceq $ un 2keq . The numerical value of the natural frequency is un ¼ 206,622.7 rad/s; this value can also be retrieved by right clicking on the peak of the plot obtained using MATLAB’s bodemag command, as illustrated in Figure 10.38. The damping ratio is xeq ¼ 0.0796. An accelerometer is defined by a high natural frequency un; therefore, the square root in Eq. (10.123) is approximately equal to 1. As a consequence,

Vx

u2 1 A $U $ ¼ d d $ u2n u2n

(10.124)

where A ¼ u2$U is the acceleration amplitude of the external motion. The acceleration amplitude is therefore expressed as A ¼ d $ u2n $ V . By using the numerical values of the example, the acceleration amplitude is found to be A ¼ 0.576 m/s2. nn

Piezoelectric and Strain Gauge Sensory-Actuation A piezoelectric stack can be equipped with two strain gauges, as shown in Figure 10.39, which is the photograph of an actuator realized by Thorlabs. When the strain gauges are connected in a half Wheatstone bridge, the deformation of the piezoelectric stack can be monitored, and the block functions as both actuator and sensor.

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CHAPTER 10 Coupled-Field Systems

FIGURE 10.39 Piezoelectric Stack With Two Strain Gauges in a Half Wheatstone Bridge Connection.

nn

Example 10.15

An axial force f acts on the block shown in Figure 10.40(a). Two identical strain gauges, each of resistance R and gauge factor Kg, are glued to the block and are connected in a Wheatstone bridge, as shown in Figure 10.40(b) together with two identical resistors of constant resistances Rc. Determine the sensitivity of this device by relating the bridge output voltage to the applied force. Known are the block’s constant cross-section area A and the material’s Young’s modulus E.

f vi R1 = R R1

R2

R2 = Rc i1

i1

i2

i2

R4 = Rc

(a)

vo R3 = R

(b)

FIGURE 10.40 (a) Piezoelectric Block With Two Strain Gauges; (b) Half Wheatstone Bridge With Two Variable Resistors and Two Constant Resistors.

10.3 Electromechanical Coupling

Solution Eq. (10.46) simplifies to





1 DR1 DR3 1 DR DR 1 DR þ $ vi Dvo ¼ $ þ $ vi ¼ $ $ vi ¼ $ 4 4 R R 2 R R1 R3

(10.125)

Because the constant resistances variations are zero: DR2/R2 ¼ DR4/R4 ¼ DRc/Rc ¼ 0. Eq. (10.125) took into account that the two resistances vary by the same quantity when the piezoelectric block deforms axially. According to Eq. (10.45) their relative variation is

DR ¼ Kg $ εa R

(10.126)

As a consequence, Eq. (10.125) becomes

K g $ vi 1 1 f $ vi ¼ $f Dvo ¼ $ Kg $ εa $ vi ¼ $ Kg $ 2 2 E$A 2E $ A

(10.127)

Eq. (10.127) used εa ¼ f/(E$A). The device sensitivity (resulting from Dvo ¼ K$f ) is therefore

K¼

Kg $ vi 2E $ A

(10.128) nn

nn

Example 10.16

A two strain-gauge piezoelectric block is used to eliminate the effects of the force f ¼ 5$sin(4t) N, which is applied at the end of the rigid pinned rod shown in Figure 10.41(a). A spiral (rotary) spring connects the rod to the pin. Known are L ¼ 20 mm, the rod mass m ¼ 0.05 kg, and the spring stiffness k ¼ 5.833 N m. For the piezoelectric sensory-actuator known are l ¼ 20 mm, the cross-sectional area A ¼ 6.25 mm2, E ¼ 520 GPa, d ¼ 390$1011 m/V, vi ¼ 15 V, and Kg ¼ 2. (i) What actuation voltage is necessary for the piezoelectric block to statically counteract the external force f ? (ii) The readout indicates an output voltage amplitude of DVo ¼ 420 mV, which suggests there is an error in the actuator placement. What is the actual position of the actuator?

L/2 Spiral spring

L/2

f

f

Rigid rod l

Piezoelectric actuator

k fa

(a)

(b)

FIGURE 10.41 (a) Lever-Type Mechanical System With External Force and Counteracting Piezoelectric Block; (b) Equivalent Lumped-Parameter Model.

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CHAPTER 10 Coupled-Field Systems

Solution (i) The lumped-parameter model of the system sketched in Figure 10.41(a) is shown in Figure 10.41(b), where the piezoelectric actuator has been replaced by a point force fa. Static rotary balancing of the horizontal rod with respect to the pivot point requires that

L fa $ ¼ f $ L or fa ¼ 2f 2

(10.129)

In this case, there is no rotation of the rod about the pivot point and therefore no elastic torque generated by the torsional spring. By taking into account that the axial deformation of the piezoelectric block is Dl ¼ fa$l/(E$A) ¼ d$va, Eq. (10.129) allows expressing the actuation voltage as

va ¼

2l 2l $f ¼ $ 5 $ sinð4tÞ E$A$d E$A$d

(10.130)

Numerically, the actuation voltage is va ¼ 157.79$sin(4t) V. (ii) For a perfectly balanced mechanical system (with no rotation about the pivot point), the tip displacement of the piezoelectric actuator should be zero and this should translate into an output voltage of zero. The output voltage not being zero means that one possible cause for this discrepancy is the position of the actuator with respect to the pivot point. Assume the actual distance of the actuator with respect to the pivot point is a instead of L/2. In this case, the dynamic equation of motion of the rod around the pivot point is

€ ¼ fa $ a  f $ L  k $ q J $q

(10.131)

According to Eq. (10.127), the following relationship can be written between the output voltage variation Dvo and the actuation force fa:

fa ¼ c1 $ Dvo ; c1 ¼

2E $ A K g $ vi

(10.132)

Assuming that the rod of Figure 10.41(b) rotates by an angle q in the direction of action of fa, this angle can be expressed for small deformations as

q¼

Dl fa $ l ¼ a E$A$a

(10.133)

where Dl ¼ fa$l/(E$A) is the axial deformation of the actuator. Substituting fa of Eq. (10.132) into Eq. (10.133) results in

q¼

c2 2l $ Dvo ; c2 ¼ K g $ vi a

(10.134)

Combining Eqs. (10.131), (10.132), and (10.134) produces the following differential equation:



J $ c2 d2 k $ c2 $ 2 ðDvo Þ þ  c1 $ a $ Dvo ¼ f $ L a dt a

(10.135)

The Laplace transform applied to Eq. (10.135) yields the transfer function that connects DVo(s), the Laplace transform of Dvo(t), to F(s), the Laplace transform of f(t):

GðsÞ ¼

DVo ðsÞ L ¼ c2   2 FðsÞ $ J $ s þ k  c1 $ a a

(10.136)

Problems

Using s ¼ u$j in Eq. (10.136), the ratio of the steady-state amplitudes of the two signals Vo and F is expressed as

DVo L ¼ c2   2 F $ J $ u  k þ c1 $ a a

(10.137)

As a consequence, the following equation in a is produced:

c1 $ a 2 

  L$F $ a þ c2 $ J $ u2  k ¼ 0 DVo

(10.138)

The mass moment of inertia of the rigid rod is determined as J ¼ m$L2/3. By using the numerical values of this example, the value of a ¼ 11 mm is obtained. Compared to the required position of the actuator at L/2 ¼ 10 mm from the pivot point, the value of a indicates that, for the given sensing voltage, the piezoelectric actuator needs to be moved 1 mm to the right to cancel the error. nn

SUMMARY This chapter studies the interaction among systems containing elements of different physical fields by formulating mathematical models of these coupled-field systems. Several applications of coupled-field systems are analyzed, including bimetallic strips as thermomechanically coupled problems and electrothermomechanical systems, electromechanical and electromagnetomechanical ones, and piezoelectrically coupled systems. Examples of using Simulink to model the dynamics of linear and nonlinear coupled-field systems are utilized.

PROBLEMS 10.1 A dynamic system is modeled in the Laplace domain by the matrix equation       a$s þ b b Y1 ðsÞ a b U1 ðsÞ $ ¼ $ Y2 ðsÞ U2 ðsÞ b sþ1 b a The outputs Y1(s), Y2(s) are from different fields. Find the values of the real parameters a and b that will render the system uncoupled. 10.2 The two outputs y1(t), y2(t) of the dynamic systems defined below through state-space models are from different fields. Decide whether these systems are field-coupled or not.         1 2 2 1 1 0 0 0 (i) ½A ¼ ; ½B ¼ ; ½C ¼ ; ½D ¼ 0 1 1 0 0 1 0 0     3 1 0 (ii) ½A ¼ 1; ½B ¼ ½ 2 1 ; ½C ¼ ; ½D ¼ 2 0 1

531

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CHAPTER 10 Coupled-Field Systems

10.3 Solve Example 10.1 considering the strip is formed of three layers instead of two; the third layer (component 3) is deposited atop the aluminum layer, and is defined by h3 ¼ 100 nm, a3 ¼ 30$106 deg1, E3 ¼ 70 GPa, r3 ¼ 2500 kg/m3. 10.4 The bimetallic strip and end-point mass of Figure 10.42, in conjunction with two identical strain gauges applied at the cantilever root on the top layer, are used as a temperature sensor. Formulate the transfer function between the input temperature and the output voltage of the strain-gauge Wheatstone bridge. Plot the output voltage using Simulink when the sensor is exposed to a temperature increase of Dq ¼ 50 . Known are l ¼ 50 mm, h1 ¼ 1 mm, h2 ¼ 2 mm, w ¼ 8 mm (the strips width), a1 ¼ 30$106 deg1, a2 ¼ 4$106 deg1, E1 ¼ 80 GPa, E2 ¼ 200 GPa, and m ¼ 0.05 kg. Do not include the inertia contributions from the two beam layers. Strain gauge α1

y

h1

h2 m

Anchor α2 l

FIGURE 10.42 Bimetallic Strip With End-Point Mass and Strain Gauge.

10.5 Solve Problem 10.4 considering there are two metallic strips, identical to the one of Problem 10.4, and situated on each side of the lumped mass. 10.6 Solve Example 10.1 when the two-layer strip is clamped at both ends. Consider that the equivalent force feq ¼ keq$zs acts at the bridge’s midpoint whose static deflection is zs and equivalent stiffness is keq. 10.7 Use a lumped-parameter model for the system of Figure 10.43, which is formed of two bimetallic strips and a connecting spring, to derive a l1 α1

h1

h2

α2 k Anchor

α3

h4

α4 l3

FIGURE 10.43 Two Bimetallic Strips With Connecting Spring.

h3

Problems

time-domain mathematical model describing the mechanical motion in the spring direction when a temperature increase of Dq is applied. Consider the inertia of the four thin cantilevers. 10.8 Use Simulink and the state lumped-parameter model of Problem 10.7 to plot the time response of the system sketched in Figure 10.43. Known are l1 ¼ 70 mm, l3 ¼ 90 mm, h1 ¼ h3 ¼ 1 mm, h2 ¼ h4 ¼ 2.3 mm, w ¼ 10 mm (all strips’ width), a1 ¼ 35$106 deg1, a2 ¼ 3$106 deg1, a3 ¼ 42$106 deg1, a4 ¼ 5$106 deg1, E1 ¼ 85 GPa, E2 ¼ 200 GPa, E3 ¼ 80 GPa, E4 ¼ 190 GPa, r1 ¼ r3 ¼ 5600 kg/m3, r2 ¼ r4 ¼ 7200 kg/m3, k ¼ 35 N/m, and Dq ¼ 50 . 10.9 Solve Example 10.2 by including the equivalent inertia and stiffness of the bar actuator. Consider that known is the material Young’s modulus E. 10.10 Solve Example 10.3 with all its numerical data by using the additional assumptions of Problem 10.9. The material Young’s modulus is E ¼ 180 GPa. 10.11 Four identical strain gauges Kg ¼ 2.3 are glued to the bar of Example 10.3; they are connected in a Wheatstone bridge with a voltage source vi ¼ 50 V. Use Simulink to plot the voltage sensed by the bridge, vo. 10.12 Obtain a mathematical model for the MEMS sketched in Figure 10.44 to connect the input voltage v to the output mechanical motion x. Consider electrical, thermal, and mechanical field interaction in the model. The thermal actuator bar has a square cross section of side h, its length is la, the electrical resistivity is rel, thermal coefficient of resistance aR, coefficient of thermal expansion a, specific heat c, mass density ra, and Young’s modulus Ea. The two identical beams have a length l, crosssectional diameter d, Young’s modulus E, and mass density r. Known are also the circuit resistance R1 and source voltage v.

Flexible beam l

la

x Thermal actuator l

– Electrical circuit

R1

+ v

FIGURE 10.44 MEMS With Electrothermal (Joule Effect) Actuation.

533

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CHAPTER 10 Coupled-Field Systems

10.13 Use Simulink to plot x as a function of time and the input voltage v for the MEMS of Problem 10.12. Known are h ¼ 12 mm, la ¼ 220 mm, rel ¼ 18$109 U m, aR ¼ 0.004 deg1, c ¼ 370 J/(kg$deg), a ¼ 18$106 deg1, ra ¼ 8500 kg/m3, Ea ¼ 160 GPa, l ¼ 150 mm, d ¼ 3 mm, r ¼ 7800 kg/m3, E ¼ 210 GPa, R1 ¼ 0.12 U, and v ¼ 6 mV. 10.14 Calculate the amplitude of the voltage in Example 10.4 when the amplitude of the sinusoidal force applied to the point mass is F ¼ 400 N. All other parameters assume the values provided in Example 10.4. 10.15 Consider the MEMS device of Example 10.5 and illustrated in Figure 10.12. Find the viscous damping coefficient c for all provided numerical data. It is also known that the beams have a length l ¼ 60 mm and the sensed voltage amplitude is Vs ¼ 9 mV, at a frequency u ¼ 1.3$un. 10.16 The sensor of Example 10.5 is moved to another environment where the sensed voltage is Vs ¼ 8 mV and the excitation frequency is u ¼ 0.8$un. What is the value of the damping coefficient of the new environment, all other parameters of Example 10.5 remaining unchanged? 10.17 The shuttle mass m ¼ 1.8$1012 kg of Figure 10.45 is supported by two identical beam springs, each of stiffness k1 ¼ 2.6 N/m. Two identical longitudinal electrostatic actuators create a capacitive force on the shuttle mass in the y direction, which is 1.5 times larger than the force required to close the gap d ¼ 10 mm between the mass and a mobile, massless plate supported by a serpentine spring of stiffness k2 ¼ 4 N/m. The superposition width of the capacitor plates is w ¼ 280 mm, the constant capacitive gap is g ¼ 6 mm, the electric permittivity is ε ¼ 8.6$1012 F/ m, and the viscous damping coefficient between the mass and substrate is c ¼ 0.0012 N s/m. Determine and plot the mass displacement before and after contact with the mobile stop plate assuming that the capacitive actuation is discontinued after contact.

g

Mobile stop Spring

Beam spring y Shuttle mass

m

d

Anchor

FIGURE 10.45 MEMS With Longitudinal Capacitive Actuation.

Anchor

Capacitive actuation

Problems

10.18 A voltage v ¼ 0.2 sin(130t) V is supplied to the rotary capacitive MEMS actuator of Figure 10.46. Find the steady-state response of the mobile armature knowing its moment of inertia is J ¼ 1.2$1018 kg m2, the viscous damping coefficient (not symbolized in the Figure) is c ¼ 20 N m s, and the spiral spring stiffness is k ¼ 2.2 N m. Known also are the fixed gap g ¼ 4 mm, the radius R ¼ 220 mm and electric permittivity ε ¼ 8.4$1012 F/m. k Spiral spring

g

Mobile armature

R Fixed armature

FIGURE 10.46 MEMS With Rotary Capacitive Actuation.

10.19 The microbridge of Figure 10.47 is used to evaluate the dynamic viscosity of a fluid by transverse capacitive sensing of the midpoint displacement relative to the input displacement u ¼ 6$sin(240t) mm of a massless cage. Use a lumped-parameter model to determine the fluid dynamic viscosity knowing that the bridge has a length of 300 mm, width of 40 mm, thickness of 1.1 mm, Young’s modulus of 180 GPa, and mass density of 6200 kg/m3. The maximum voltage sensed by the capacitor is DV ¼ 3 mV and the mass of capacitor plate attached to the microbridge is mp ¼ 5$1012 kg. Capacitive sensing Microbridge Anchor

u

Massless cage

FIGURE 10.47 Microbridge With Displacement Input and Transverse Capacitive Sensing.

10.20 Solve Example 10.6 when using three identical strain gauges instead of four. What is the location/position of the gauges on the cantilever and what is their connection in the Wheatstone bridge? 10.21 Calculate the input displacement amplitude U of the microcantilever of Example 10.7 knowing that the sensed voltage amplitude is Vo ¼ 2.5 mV at a frequency u ¼ 0.2$un. All other parameters of Example 10.7 remain unchanged.

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CHAPTER 10 Coupled-Field Systems

10.22 A microcantilever with a length l ¼ 210 mm, width w ¼ 30 mm, thickness h ¼ 1.9 mm, Young’s modulus E ¼ 170 GPa, and mass density r ¼ 6000 kg/m3 has two strain gauges (Kg ¼ 2) attached at its root; the gauges are connected in a Wheatstone bridge (with vi ¼ 15 mV). The cantilever is attached to a massless platform (base), which has a sinusoidal motion u ¼ 12$sin(100,000$t) mm. Determine the point mass that attaches at the cantilever free end knowing the output voltage amplitude of the Wheatstone bridge, which is DVo ¼ 8 mV. Use a lumped-parameter model without damping. 10.23 A microcantilever is placed on a shaker that provides a variable-frequency input displacement motion u ¼ 8$sin(u$t) mm. The microcantilever has a length l ¼ 300 mm, width w ¼ 40 mm, thickness h ¼ 5 mm, Young’s modulus E ¼ 150 GPa, mass density r ¼ 5500 kg/m3 and is used in conjunction with a strain gauge (Kg ¼ 2) attached to the cantilever root and connected in a Wheatstone bridge (with vi ¼ 25 mV). Knowing that the maximum voltage readout is DVo ¼ 6 mV, determine the overall coefficient of viscous damping c. 10.24 The microcantilever of Problem 10.23 is used to sense the mass of a particle mp that attaches to its free end. Calculate mp knowing that the lumpedparameter viscous damping coefficient is c ¼ 4$109 N s/m, the maximum voltage readout is DVo ¼ 4.6 mV, while all other parameters are the ones specified in Problem 10.23. 10.25 A straight deformable bar of cross-sectional area A ¼ 30 mm2, length l ¼ 85 cm, mass density r ¼ 7800 kg/m3, Young’s modulus E ¼ 210 GPa, and Poisson’s ratio n ¼ 0.3 is anchored at one end and subjected to a sinusoidal axial load of unknown amplitude and frequency u ¼ 30,000 rad/s at the other end. Disposing of three identical strain gauges (Kg ¼ 2.1) that can be attached to the bar and connected in a Wheatstone bridge (vi ¼ 32 V), determine the force amplitude when the maximum output voltage of the bridge is Dvo ¼ 2.8$106 V. 10.26 The electromagnetomechanical system of Example 10.8 and shown in Figure 10.19 is used as a sensor for a sinusoidal force of frequency u ¼ 100 rad/s and unknown amplitude. Use a lumped-parameter model together with the parameters of Example 10.8 and the amplitude of the sensed voltage of 0.05 V to determine the input force amplitude. Also viscous damping affects the mechanical motion with a coefficient c ¼ 30 N s/m. 10.27 The support of the microbridge described in Example 10.8 receives an external sinusoidal motion along the direction perpendicular to the bridge longitudinal axis in the plane of the figure. If the sensed voltage amplitude is Vm ¼ 5.8 mV at a frequency u ¼ 310 rad/s, what is the

Problems

amplitude U of the input motion? All other parameters assume the values given in Example 10.8. 10.28 Formulate a state-space model for the dc motor of Example 10.11, shown in Figure 10.48, using two inputs (the armature voltage va and load moment ml) and one output (the shaft angular velocity u), then plot the system’s time response (output) by means of MATLAB. Use the numerical values of Example 10.11 except for va ¼ 90 e 10/(t þ 1) V and ml ¼ 0.25 sin(5t) N m. Ra

Rotor armature ml

+ va

Jl

–

La

θ Stator

c Load

FIGURE 10.48 Schematic of a DC Motor as an Electromechanical System.

10.29 Consider that the mechanical system of Figure 3.40(a), is actuated by a dc motor and its corresponding electrical circuit, which provides the moment ma. Demonstrate the electromechanical coupling of this system by deriving the transfer function matrix connecting the armature voltage va and load moment ml as input variables to the rotation angle q of the middisk of inertia J and armature current ia as output variables. 10.30 Evaluate the steady-state angular response of the midshaft cylinder studied in Problem 10.29. All mechanical parameters furnished in Example 3.19 are applicable here except for the actuation moment ma. The other parameters are Ra ¼ 5 U, La ¼ 0.004 H, va ¼ 80$sin(80t) V, Ke ¼ 0.120 N m/A, and Km ¼ 0.125 N m/A. 10.31 Use Simulink to plot the time displacement y(t) of the translatory dc motor and load system of Figure 10.28. Known are: m ¼ 0.05 kg, c ¼ 3.1 N s/m, k ¼ 20 N/m R ¼ 10 U, L ¼ 0.002 H, va ¼ 90$e0.1t$sin(110t) V, Ke ¼ 0.1 N m/A and Km ¼ 10 N m/A. 10.32 Calculate the external magnetic field B of the sensor described in Example 10.12 of Figure 10.30 for the following parameters. The current is i ¼ 0.005$sin(60t) A, the loop radius is r ¼ 160 mm, the rigid plate thickness is hp ¼ 10 mm and its mass density is r ¼ 6550 kg/m3. The flexible rod is of length l ¼ 100 mm, diameter dr ¼ 2 mm, and Young’s modulus E ¼ 1.5 GPa. The photodiode is placed at a distance h ¼ 3 mm from the plate; the incident laser beam angle is a0 ¼ 25 , and the maximum deviation of the reflected beam is d ¼ 6 mm.

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CHAPTER 10 Coupled-Field Systems

10.33 An external magnetic field B ¼ 1.3 T interacts with a sinusoidal current of unknown amplitude and rotates the rigid plate shown in Figure 10.49 by an angle q. Optical detection with an incident beam at a0 ¼ 25 and a capture screen at a distance h ¼ 220 mm from the microdevice plane indicates a maximum deflection of the reflected beam of d ¼ 60 mm (see Figure 10.29). The two identical flexible bars have a circular cross section of diameter dr ¼ 3 mm, their length is l ¼ 260 mm, while their material modulus is G ¼ 110 GPa. The rigid plate is defined by r ¼ 100 mm, a thickness hp ¼ 8 mm, and a mass density r ¼ 6000 kg/m3. Knowing that the current’s frequency is one half the natural rotary frequency of the mechanical microsystem in the q direction, determine the current amplitude. Current-carrying loop Bar Anchor l

θ

B r

l Rigid plate

FIGURE 10.49 Top View of Compliant Microsystem for External Magnetic Field Detection.

10.34 Use Simulink to plot the time response of the system described in Example 10.12 and shown in Figure 10.30 when B ¼ 0.3/(1 þ 0.2 t) T by using the numerical data provided in that example. The optical detection system is not used here. 10.35 An unknown actuation voltage is applied to the piezoelectric block of Example 10.15 and shown in Figure 10.40 and generates an output (sensing) voltage variation of Dvo ¼ 0.0001$sin(20t) V. Determine the characteristics of the actuation voltage, as well as the maximum free-end displacement and the maximum force that can be generated by the piezoelectric block. Known are Kg ¼ 2, l ¼ 20 mm, A ¼ 42.25 mm2, vi ¼ 150 V, E ¼ 52 GPa, d ¼ 390$1012 m/V. 10.36 Use Simulink to plot the displacement y of the block of Example 10.13 when an actuation voltage va ¼ 80$e10t$sin(120t) V is applied. All other parameters assume the values provided in Example 10.13. 10.37 A longitudinal piezoelectric actuator of length l ¼ 0.04 m, cross-sectional area A ¼ 0.8 cm2, mass density rPZT ¼ 7600 kg/m3, Young’s modulus Em ¼ 60 GPa, and piezoelectric charge constant d ¼ 3.8$1010 m/V is attached to a mass m ¼ 0.18 kg, which further connects to a spring k ¼ 190 N/m, as illustrated in Figure 10.34.

Suggested Reading

(i) Evaluate the total system mechanical losses as an equivalent damping coefficient ceq if experimental frequency-domain testing indicates the system’s bandwidth (cutoff frequency) is ub ¼ 1.2$un (where un is the mechanical system’s natural frequency). Consider the mass and stiffness contributions of the piezoelectric actuator. (ii) Plot the system’s response y as a function of time when an input voltage va ¼ 80$e2t V is applied to the piezoelectric actuator. 10.38 Consider four strain gauges (Kg ¼ 2.2) are attached on the piezoelectric actuator of Problem 10.37; they are connected in a full Wheatstone bridge with vi ¼ 18 V. Express the relationship between the actuation voltage of the piezoelectric block, va ¼ 50$sin(120t) V and the readout voltage Dvo for this device, then evaluate the viscous damping coefficient corresponding to the total (equivalent) system losses. 10.39 Solve Example 10.16 when the external force is f ¼ 30$sin(110t) N and the sensed voltage amplitude is DVo ¼ 85 mV. Use four strain gauges attached to the piezoelectric block and connected in a Wheatstone bridge. All other necessary parameters assume the values provided in Example 10.16. 10.40 A circular piezoelectric plate with diameter D ¼ 20 mm, thickness h ¼ 5 mm, Young’s modulus of 150 GPa, mass density of 6800 kg/m3, and d ¼ 3$10 11 m/V is attached at the midpoint of a bridge, as shown in Figure 10.50. Determine the amplitude of external the sinusoidal motion provided to the bridge knowing that the maximum voltage produced by the piezoelectric sensor is 0.001 V. Known are the bridge data: length of 180 mm, width of 20 mm, thickness of 1 mm, Young’s modulus of 210 GPa, mass density of 7800 kg/m3. The system losses are equivalent to a viscous damping coefficient c ¼ 14 N s/m.

Microbridge

u

Anchor

Piezoelectric sensor

FIGURE 10.50 Microbridge With Piezoelectric Sensor.

Suggested Reading R.S. Figliola and D.E. Beasley, Theory and Design for Mechanical Measurements, 6th Ed. Wiley, New York, 2014. P.L. Reece (Editor), Smart Materials and Structures, Nova Science Publishers, New York, 2006.

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R.H. Bishop (Editor), The Mechatronics Handbook, 2nd Ed. CRC Press, Boca Raton, FL, 2008. N. Lobontiu and E. Garcia, Mechanics of Microelectromechanical Systems, Kluwer, New York, 2004. MORGAN Electro Ceramic Piezoelectric Tutorials, available at: www. morganmatroc.commec_tutorials. K. Ogata, System Dynamics, 4th Ed. Pearson Prentice Hall, New York, 2004. W.C. Young and R.G. Budynas, Roark’s Formulas for Stress and Strain, 8th Ed. McGraw-Hill, New York, 2012. N. Lobontiu, Dynamics of Microelectromechanical Systems, Springer, New York, 2007. F.J. Bueche and D.A. Jerde, Principles of Physics, 6th Ed. McGraw-Hill, New York, 1995. O. Buecher and M. Weeks, Introduction to MATLABÒ & SimulinkÒdA Project Approach, 3rd Ed. Infinity Science Press, Boston, 2008. S. Timoshenko, Analysis of bi-metal thermostats, Journal of the Optical Society of America and Review of Scientific Instruments, 1925, Vol. 11, pp. 233e255.

CHAPTER

Block Diagrams and Feedback Control System Modeling

11

OBJECTIVES This chapter utilizes the material presented in Chapters 1e10 to introduce the main concepts, terminology, elements, and analytical tools of feedback control dynamic systems. Chapters 12 and 13 incorporate the basic notions introduced in this chapter to investigate the behavior of control systems in more detail. The following topics are studied here: • Modeling of single-input, single-output (SISO) control systems by means of negativefeedback block diagrams and typical transfer functions such as feed-forward, open-loop, and closed-loop transfer functions. • Basic controllers achieving proportional, integral, derivative or mixed functions; lag, lead, and lagelead compensators. • Sensitivity analysis of feedback systems. • Modeling of multiple-input, multiple-output (MIMO) feedback systems by means of transfer function matrices; study of feedback systems with disturbances. • Modeling of SISO and MIMO feedback systems by means of state-space models.

INTRODUCTION As mentioned in the Preface of this book, while many of its ingredients are included in system dynamics, controls is a standalone discipline, and it is thus studied thoroughly in dedicated textbooks and monographs; this chapter, as well as Chapters 12 and 13, introduces the controls fundamentals. Single-field and coupled-field dynamic systems have been studied in previous chapters under the assumption that once an accurate mathematical model has been derived and a given input has been specified, the output (or response) of the system of interest can be determined, and then conclusions can be formulated with respect to the system behavior. In all instances, the output of a dynamic system theoretical model was never interrogated or checked against experimental results. However, in real-life applications, errors due to modeling (such as those induced by severe simplifying assumptions) or to the presence of unwanted factors System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00011-7 Copyright © 2018 Elsevier Inc. All rights reserved.

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(such as disturbances or interference) affect the theoretical model predictions. Controlling the dynamics of a regular system (also known as open-loop system) becomes thus necessary to correct the deviation of an actual system response from the theoretically predicted response. One practical control method is picking up the output signal and comparing it to the input (or reference) signal, which is the desired response of the dynamic system; in case differences are noted, a controller is set into action, which modifies the level and quality of the input provided to the originally uncontrolled system. The main tool in modeling such feedback control systems is the block diagram that consists of the transfer functions defining the various subsystems and of the Laplace-domain signals connecting them. It becomes thus possible to model, analyze, and design control systems from the viewpoint of stability, transient response, and steady-state response.

11.1 CONCEPT OF FEEDBACK CONTROL OF DYNAMIC SYSTEMS Consider the piezoelectric (PZT) actuator of Figure 11.1, which needs to provide an output force that is proportional to the applied voltage va and the block free displacement, as discussed in Chapter 10.

+ –

Actuation circuit va

R Mechanical deformation Strain gauge

+ vs

Guide

Mechanical deformation PZT actuator

– Sensing circuit

FIGURE 11.1 Piezoelectric (PZT) Actuator With Strain Gauge Sensing.

The free displacement of the PZT block can be evaluated by an electrical strain gauge circuit through the sensed voltage vs. Due to modeling approximations, fabrication/assembly imprecision, as well as material constants that may differ from specifications, the actual free displacement is not identical to the one predicted through modeling, and, as a consequence, the delivered force is other than the intended one. It is therefore necessary to somehow modify the actuation voltage to obtain the required force. The necessity of controls arises naturally when the output of a SISO system, for instance, is not the one expected through the mathematical model that represents the system, as discussed in the previous example and sketched generically in Figure 11.2.

11.1 Concept of Feedback Control of Dynamic Systems

Input

Output System Model

FIGURE 11.2 Uncontrolled System.

This mismatch between actual and model responses can be generated by several factors such as accuracy level of the mathematical model or disturbance (noise) inputdwhich was not accounted for through modeling. The objective is therefore reducing (or eliminating, ideally) the difference between actual and desired outputs without modifying the existing system. Actual control procedures operate as illustrated in Figure 11.3 where the system output is back fed into a unit that compares it to the reference signal (or set pointd the desired, model output) and then an error results, which is supplied to a control unit whose role is of minimizing (or eliminating) the error supplied to the actual system (or plant) through an actuator. This type of control is known as negativefeedback control because of the output signal being directed to the minus terminal of the summing point (summer or comparator). A sensor, which is placed on the feedback branch, converts the actual output into a physical parameter of the same nature with the reference signal, such that comparison between the two signals is possible. In many applications the actuator and system are designed or considered as a single unit making up the plant. Error signal Summing point

Control signal Actuation signal

+

Controlled output signal Controller

Reference signal

Actuator

System (Plant)

_ Feedback signal Sensor

FIGURE 11.3 Diagram of Control System With Negative Feedback.

The essence of any control system that follows the diagram of Figure 11.3 is that the controlled output signal should be as close as possible to the reference signal (or the feedback/transformed signal, in case the output and input signals are not of the same nature)dideally, they should be identical and this important trait is known as trackability. Figure 11.4 shows the time response of a typical (second-order) control system under a unit-step input (reference). As it can be seen, the response reaches a uniform level after an initial period where it shows some oscillations. The initial period

543

CHAPTER 11 Block Diagrams and Feedback Control System Modeling

Unit-step input Transient response

Steady-state response

Steady-state error

544

Controlled output

FIGURE 11.4 Typical Time Response of a Second-Order Control System With Negative Feedback.

illustrates the transient response, whereas the final-stage, known as steady-state response, might be set off from the unit-step input by an error, known as steadystate error. Both stages, the transient response and the steady-state response, are studied in detail in Chapter 13. Controls are also utilized to ensure that the output of a feedback system is bounded for any bounded input, which means the control system is stabledthis important aspect is covered in Chapter 12. Other important control objectives are rejection of disturbances (such as external noise) and sensitivity minimization of system performance to parameter variation. It should be noted that while the diagram of Figure 11.3 presupposes that the mathematical model of each component is known, there are situations where the system to be controlled is too complex and cannot be described adequately by mathematical means in terms of identifying an equation (algebraic or differential) to connect the plant input and output. Although more time-consuming, satisfactory control is still possible in these cases by feedback adjustments to the controllerdthe feedback system is called adaptive in these situations. A particularly simple and inexpensive feedback system category is the on-off control, also known as two-position control. In on-off control there is a switching action between the actuated state, where the feedback control is applied (the on state), and a state where actuation is discontinued (the off state). The best known example is that of a thermostat regulating the temperature in a room. When the room temperature is below the set-point temperature, the thermostat is on and the heating system operates; as a result, the room temperature increases. Once the set-point temperature has been reached, the thermostat stops the heating system and then this cycle repeats itself. Another example of on-off controls is the liquid system illustrated in Figure 11.5(a), which consists of a float rigidly connected to the switch of an electromagnetic system with a plunger that opens or closes a valve through which liquid

11.1 Concept of Feedback Control of Dynamic Systems

Solenoid with moving plunger

Switch Voltage source

qi Control valve Float h

Tank

qo

Liquid

Relief valve

(a) Head, h Off Δh, differential gap (neutral zone)

hn On

Time, t

(b)

FIGURE 11.5 On-Off Control of Liquid-Level System: (a) Physical System Schematic; (b) Head Versus Time With Differential Gap Control. Modified from K. Ogata, Modern Control Engineering, 4th Ed. Prentice Hall, Pearson, Upper Saddle River, NJ, 2009.

flows into the tank. As long as the liquid level does not reach a nominal head hn, the float does not open the switch, which keeps the electromagnetic plunger in its upper position, the valve open and so more liquid enters the tank. Once the liquid level reached the nominal head, the float goes up and opens the switch, which powers off the solenoid; consequently the plunger descends and closes the valve discontinuing the liquid flow into the tank. A practical solution of implementing on-off control provides a differential gap (or neutral zone) instead of the single value hn for the two actions. As depicted in Figure 11.5(b), there is an offset Dh defining the off and on positions, which allows the system to respond over a small interval of time and eliminate the oversensitivity to the on and off actions occurring at a single set value, such as hn. Due to the differential gap, the on action starts the filling for a head slightly smaller than hn. This action continues to a head value just above hn when the off action is applied and emptying of the tank starts, and this cycle is repeated continuously. Note that Figure 11.5(b) only depicts the small head variations around the nominal head, not the full filling/emptying of the tank. Another type of control, named open-loop control, is schematically illustrated in Figure 11.6. As indicated in Figure 11.6, the output is not compared to the input and the output does not affect the control action at alldthese two features are present in

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

Control signal Actuation signal Output signal

Input signal Controller

Actuator

System (Plant)

FIGURE 11.6 Diagram of Open-Loop Control Process.

feedback control. Open-loop control is used when the inputeoutput relationship is well defined and when the levels of disturbance and error are reduced so as not to adversely affect the output. Examples of open-loop control systems include applications that are conceived on a time-predefined sequence of events such as: automatic washing machines, car washing systems, hair driers, coffee makers, or traffic control with stop lights. All these examples cannot correct errors that are due to external disturbances (such as noise or temperature change) or to internal or model errors. The model-based approach depicted in the diagram of Figure 11.3 is applied in this chapter and Chapters 12 and 13 to model, design, or analyze feedback control systems.

11.2 BLOCK DIAGRAMS AND SISO FEEDBACK SYSTEMS The structure and element connectivity of control systems can be modeled graphically by means of flow graphs and transfer function block diagrams. This section introduces the block diagram graphical representation of feedback control systems; the remainder of this chapter and Chapters 12 and 13 utilize this functional tool. The state-space approach is also utilized to model SISO feedback control systems. The section discusses the main types of control functions and related controller systems, the sensitivity of control systems to parameter variations, and includes examples of actual (physical) control systems together with their mathematical models.

11.2.1 Transfer Functions and Basic Block Diagrams Connection charts such as the one of Figure 11.3, where linkage between the various subsystems is assumed to take place in the time domain, are usually displayed by using the Laplace (s) domain transfer functions inside the blocks and signals connecting the blocksdthis pictorial structure, which is illustrated in Figure 11.7, is known as a block diagram. It should be mentioned that the block interconnections are assumed ideal, which means there is no interstage loading.

Components of Block Diagrams The main components of a block diagram corresponding to SISO feedback control systems, such as the one of Figure 11.7, are signals, blocks, summing points (or summers), and junctions (or pick off points)dthey are represented graphically in Figure 11.8. The transfer function block of Figure 11.8(a) has an input (incoming) signal I(s) and an output (outgoing) signal O(s) that connect to the transfer function G(s) as: OðsÞ ¼ GðsÞ $ IðsÞ

(11.1)

11.2 Block Diagrams and SISO Feedback Systems

+

M(s)

E(s)

R(s)

C(s)

A(s) Ga(s)

Gc(s)

Gs(s) = Gp(s)

_ B(s)

C(s) H(s)

FIGURE 11.7 Block Diagram of a Basic SISO Negative-Feedback Control System. I(s) I1(s) I(s)

O(s) G(s)

+

I2(s)

I(s)

I(s)

_ I3(s)

(a)

O(s)

Junction

+ (b)

I(s)

(c)

FIGURE 11.8 Components of a Block Diagram: (a) Transfer Function Block With Connecting Signals; (b) Summing Point (Summer); (c) Junction (Pick Off) Point.

A summing point is another block diagram component, which combines several incoming signalsdsuch as I1(s), I2(s), and I3(s) in Figure 11.8(b)dinto an outgoing signal O(s), according to the signal conservation rule: OðsÞ ¼ I1 ðsÞ  I2 ðsÞ þ I3 ðsÞ

(11.2)

The junction (or pick off) point sketched in Figure 11.8(c) is used to branch the same signal across several connection lines in a block diagram. The signals are shown as directed lines connecting to any of the three components depicted in Figure 11.8. The two basic mathematical operations given in Eqs. (11.1) and (11.2), together with the signal preservation at a junction point, form the algebra of block diagram signals, which enables connecting various signals of interest by means of combined transfer functions that are derived from the basic (subsystem) ones. It should also be noted that two transfer functions that are connected in series (also called cascading), such as the ones illustrated in Figure 11.9(a), result in a transfer function, which is the product of the two component transfer functions; two transfer functions combined in parallel result in a single transfer function that is the sum of the two transfer functions, as indicated in Figure 11.9(b). Eq. (11.1) has been used to obtain the transfer function of the two cascading blocks of Figure 11.9(a); similarly, Eqs. (11.1) and (11.2), as well as the identity of signals connected to a pick off point, were used to derive the parallel blocks transfer function.

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

G2(s)

G1(s)

G2(s) O(s)

I(s)

G1(s)

I(s)

G2(s)·G2(s) I(s)

O(s)

+

+

O(s)

G2(s) + G2(s)

I(s)

(a)

O(s)

(b)

FIGURE 11.9 Blocks Combined in: (a) Series (Cascade); (b) Parallel. nn

Example 11.1 (i) The mechanical system of Figure 11.10(a) is formed of six massless rods. Rods 1, 2, 4, and 6 can translate horizontally; rod 3 has a planar motion, while rod 5 rotates around the fixed point O. The two input translations x1 and x2 move rod 3, which displaces horizontally rod 4 by a quantity x; this is the input to the pivoting rod 5, which displaces horizontally the output rod 6 by x3. Demonstrate that this mechanical device can operate as a summing point with the inputs being x1 (positive), x2 (negative), and the output x3. x1

Rod 1

A

A x3

l1

x

C l1 x2

E

Rod 4

D Rod 5 O

Guide

Rod 2 B Rod 3

x1

A’

Rod 6

l2

C x

C’

l2 B’ x2

B

(a) R R

R v1 –v2 R

– +

R v

– +

v3

(b)

FIGURE 11.10 Summing Point Function Resulting From: (a) Mechanical System With Massless Rods; (b) Electrical System With Resistances and Two Operational-Amplifier Stages.

11.2 Block Diagrams and SISO Feedback Systems

(ii) Demonstrate that the two-stage, resistor-based operational amplifier of Figure 11.10(b) can function as a summing point where the inputs are the voltages v1 (positive), v2 (negative), and the output is the voltage v3.

Solution (i) As demonstrated in Example 4.24, which analyzed a mechanical device (see Figure 4.43) similar to the one studied here, and as shown in Figure 11.10(a), the horizontal displacement of point C (which is also the displacement of the translating rod 4) is calculated from the horizontal input displacements of the top rod 1 and bottom rod 2 as:

x¼

x1  x2 2

(11.3)

Rod 4 pushes the pivoting rod 5 by x at point D and, as a result, point E displaces by twice this distance, which is 2x. This displacement is also the displacement of the output rod; as a consequence, the output displacement is:

x3 ¼ 2x ¼ 2 $

x1  x2 ¼ x1  x2 2

(11.4)

Laplace transforming Eq. (11.4) yields:

X3 ðsÞ ¼ X1 ðsÞ  X2 ðsÞ

(11.5)

and, therefore, the device of Figure 11.10(a) behaves as a mechanical summing point with the positive input X1(s), the negative input X2(s), and the output X3(s). (ii) The following current relationship can be written for the first-stage operational-amplifier unit:

v1  0 v2  0 0v þ ¼ or v1  v2 ¼ v R R R

(11.6)

The current relationship corresponding to the second stage of the circuit is:

v0 0  v3 ¼ or v ¼ v3 R R

(11.7)

Eqs. (11.6) and (11.7) took into account that the input voltages to both operational amplifiers are zero because of negative feedback and ground connection of the positive input terminals. Substituting v of Eq. (11.7) into Eq. (11.6) results in:

v3 ¼ v1  v2

(11.8)

In the Laplace domain, Eq. (11.8) becomes:

V3 ðsÞ ¼ V1 ðsÞ  V2 ðsÞ

(11.9)

which proves that the circuit of Figure 11.10(b) operates as an electrical summing point with the positive input being V1(s), the negative input V2(s), and the output V3(s). nn

Returning to the basic block diagram of Figure 11.7, its component transfer functions are Gc(s) for the controller, Ga(s) for the actuator, Gs(s) ¼ Gp(s) for the system (plant), and H(s) for the sensor/conditioner. The signals connecting the blocks are R(s)dfor reference, E(s)dfor error, M(s)dfor modified (manipulated), A(s)dfor actuation, C(s)dfor controlled output, and B(s)dfor back signal. The summing point (or block) connects two input signals (one negative, the other positive) and an output signal, according to the algebraic relationship: RðsÞ  BðsÞ ¼ EðsÞ

(11.10)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

The controller, actuator, and plant can be grouped serially into a single transfer function G(s), which is named the feed-forward transfer function and is calculated as: GðsÞ ¼

CðsÞ CðsÞ AðsÞ MðsÞ ¼ $ $ ¼ Gp ðsÞ $ Ga ðsÞ $ Gc ðsÞ EðsÞ AðsÞ MðsÞ EðsÞ

(11.11)

such that the block diagram of Figure 11.7 is simplified to the one of Figure 11.11(a).

+

E(s)

C(s) G(s)

R(s)

R(s)

_

C(s) GCL(s)

B(s) H(s)

(a)

(b)

FIGURE 11.11 (a) Simplified Block Diagram of Negative-Feedback Control System; (b) Equivalent Closed-Loop Transfer Function.

Another transfer function defined based on the block diagram of Figure 11.11(a) is the open-loop transfer function GOL(s): GOL ðsÞ ¼

BðsÞ CðsÞ BðsÞ ¼ $ ¼ GðsÞ $ HðsÞ EðsÞ EðsÞ CðsÞ

(11.12)

The third important qualifier of a negative-feedback control system is the closedloop transfer function, which is defined as: GCL ðsÞ ¼

CðsÞ RðsÞ

(11.13)

Using operations of the algebra of block diagrams, it can be shown that: EðsÞ ¼ RðsÞ  BðsÞ ¼ RðsÞ  HðsÞ $ CðsÞ

(11.14)

which, in turn, enables expressing the reference signal as: RðsÞ ¼ EðsÞ þ HðsÞ $ CðsÞ ¼

CðsÞ 1 þ GðsÞ $ HðsÞ þ HðsÞ $ CðsÞ ¼ $ CðsÞ GðsÞ GðsÞ (11.15)

The closed-loop transfer function is obtained by substituting R(s) of Eq. (11.15) into Eq. (11.13): GCL ðsÞ ¼

CðsÞ GðsÞ ¼ RðsÞ 1 þ GðsÞ $ HðsÞ

(11.16)

11.2 Block Diagrams and SISO Feedback Systems

The closed-loop transfer function relationship of Eq. (11.16), which can be written as C(s) ¼ GCL(s)$R(s) and is illustrated in Figure 11.11(b), is very important as it indicates that a feedback control system can formally be transformed into an uncontrolled system defined by a unique transfer function GCL(s) whose input is R(s) and output is C(s). However, GCL(s) contains all information pertaining to a feedback control system with controller, actuator, plant, feedback loop, and summing point action incorporated.

+

E(s)

C(s) G(s)

R(s)

_

FIGURE 11.12 Block Diagram of a Unity-Feedback Control System.

When there is no sensor on the feedback branch (or rather, there is a sensor with a transfer function H(s) ¼ 1), the system becomes a unity-feedback system, as sketched in Figure 11.12. From Eqs. (11.12) and (11.16), the open-loop and closed-loop transfer functions of such a system become: GOL ðsÞ ¼ GðsÞ; GCL ðsÞ ¼

GðsÞ 1 þ GðsÞ

(11.17)

nn

Example 11.2

Convert the control system depicted by the block diagram of Figure 11.13dwhere G1(s) is the transfer function of a prefilterdinto a unity-feedback control system as the one of Figure 11.12. Determine the feed-forward transfer function G(s) of the equivalent unity-feedback system.

R(s) G1(s)

R1(s)

+

C(s)

E(s) G2(s)

_ B(s)

G3(s)

FIGURE 11.13 Block Diagram of a Control System With Prefilter Transfer Function.

Solution The inputeoutput relationship pertaining to the summing point is:

EðsÞ ¼ R1 ðsÞ  BðsÞ

(11.18)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

The three blocks of the diagram permit formulating the following equations:

R1 ðsÞ ¼ G1 ðsÞ $ RðsÞ; EðsÞ ¼ CðsÞ=G2 ðsÞ; BðsÞ ¼ G3 ðsÞ $ CðsÞ

(11.19)

Combining Eq. (11.19) and Eq. (11.18) results in:

CðsÞ G1 ðsÞ $ G2 ðsÞ ¼ GCL ðsÞ ¼ RðsÞ 1 þ G2 ðsÞ $ G3 ðsÞ

(11.20)

This closed-loop transfer function has to be identical to the closed-loop transfer function of the unity-feedback system given in Eq. (11.17), which is: G(s)/[1 þ G(s)] ¼ G1(s)$G2(s)/[1 þ G2(s)$ G3(s)]. As a consequence, the feed-forward transfer function of the unity-feedback system that is equivalent to the original one is found as:

GðsÞ ¼

G1 ðsÞ $ G2 ðsÞ 1 þ G2 ðsÞ $ G3 ðsÞ  G1 ðsÞ $ G2 ðsÞ

(11.21) nn

nn

Example 11.3 Convert the control system described by the block diagram of Figure 11.14(a) into a basic feedback control system as the one represented in Figure 11.11(a).

R(s) +

E1(s)+

E2(s)

_

A(s) G1(s)

C(s) G2(s)

_ B(s) G3(s)

(a) R(s) +

E1(s)

A(s) GCL,1(s)

C(s) G2(s)

_

(b)

FIGURE 11.14 (a) Block Diagram of a Control System With Two Feedback Loops; (b) Block Diagram of the Control System With Inner Feedback Loop Collapsed.

Solution The following signal equation can be written for the summing point on the left in Figure 11.14(a):

E1 ðsÞ ¼ RðsÞ  CðsÞ

(11.22)

Similarly, the equation for the next summing point is:

E2 ðsÞ ¼ E1 ðsÞ  BðsÞ

(11.23)

11.2 Block Diagrams and SISO Feedback Systems

Based on the transfer functions G3(s), the signal B(s) is expressed as:

BðsÞ ¼ AðsÞ $ G3 ðsÞ

(11.24)

A(s) depends on G1(s) according to the equation:

AðsÞ ¼ E2 ðsÞ $ G1 ðsÞ

(11.25)

Combination of Eqs. (11.23)e(11.25) yields:

E2 ðsÞ $ ½1 þ G1 ðsÞ $ G3 ðsÞ ¼ E1 ðsÞ

(11.26)

At the same time, the transfer function G2(s) and Eq. (11.25) result in:

CðsÞ ¼ AðsÞ $ G2 ðsÞ ¼ E2 ðsÞ $ G1 ðsÞ $ G2 ðsÞ

(11.27)

which yields:

E2 ðsÞ ¼

CðsÞ G1 ðsÞ $ G2 ðsÞ

(11.28)

E2(s) of Eq. (11.28) is substituted in Eq. (11.26), which produces:

E1 ðsÞ ¼

½1 þ G1 ðsÞ $ G3 ðsÞ $ CðsÞ G1 ðsÞ $ G2 ðsÞ

(11.29)

Similarly, E1(s) of Eq. (11.29) is introduced in Eq. (11.22), which results in:

CðsÞ ¼

G1 ðsÞ $ G2 ðsÞ $ RðsÞ 1 þ G1 ðsÞ $ G2 ðsÞ þ G1 ðsÞ $ G3 ðsÞ

(11.30)

Because C(s) ¼ GCL(s)$R(s), Eq. (11.30) indicates that the closed-loop transfer function of the control system is:

GCL ðsÞ ¼

G1 ðsÞ $ G2 ðsÞ 1 þ G1 ðsÞ $ G2 ðsÞ þ G1 ðsÞ $ G3 ðsÞ

(11.31)

Note that the closed-loop transfer function of Eq. (11.31) can also be obtained by collapsing the inner loop of Figure 11.14(a), as illustrated in Figure 11.14(b), and substituting it by the closed-loop transfer function GCL,1(s) ¼ G1(s)/[1 þ G1(s)$G3(s)]. As a consequence, the closed-loop transfer function of the system shown in Figure 11.14(b) is GCL(s) ¼ GCL,1(s)$G2(s)/[1 þ GCL,1(s)$G2(s)], which yields the result of Eq. (11.31). Comparison of Eq. (11.31) to the definition Eq. (11.16) shows that one condition is:

GðsÞ ¼ G1 ðsÞ $ G2 ðsÞ

(11.32)

The other condition necessary for Eqs. (11.16) and (11.31) to be identical is:

1 þ G1 ðsÞ $ G2 ðsÞ þ G1 ðsÞ $ G3 ðsÞ ¼ 1 þ HðsÞ $ GðsÞ

(11.33)

By combining now Eqs. (11.32) and (11.33) results in the following transfer function of the feedback function:

HðsÞ ¼

G1 ðsÞ $ G2 ðsÞ þ G1 ðsÞ $ G3 ðsÞ G3 ðsÞ ¼1þ G1 ðsÞ $ G2 ðsÞ G2 ðsÞ

(11.34)

Eqs. (11.32) and (11.34) thus demonstrate that the control system of Figure 11.14 can be rendered into the basic control system of Figure 11.11(a). nn

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

11.2.2 Controllers This section introduces the basic controllers, which are proportional, integral or derivative, and combinations thereof such as proportional þ integral, proportional þ derivative, and proportional þ integral þ derivative. Also introduced are some particular combinations of proportional, integral, and derivative controllers that are known as compensators; this class includes lag, lead, and lagelead compensators. The purpose of this section is to simply introduce these basic controllers through their transfer functions and to provide examples of physical systems that can act as controllers. The time- and frequency-domain features of all these controllers and the way they affect the response of feedback control systems are investigated in more depth in Chapters 12 and 13.

Proportional, Integral, Derivative, and Combined Controllers The main types of control functions are proportional (P), integral (I), and derivative (D). For a proportional controller and based on the block diagram of Figure 11.7, the transfer function is: Gc ðsÞ ¼

MðsÞ ¼ KP EðsÞ

(11.35)

where the constant KP is the proportional gain. Proportional control is therefore utilized to mainly change the overall gain and increase/decrease the controlled output. As discussed in Chapter 13, P control can also alter the time constant of first-order systems and the natural frequency of second-order systems. Integral controllers apply integration to their input and are defined by a transfer function: Gc ðsÞ ¼

Kp MðsÞ KI ¼ ¼ EðsÞ s TI $ s

(11.36)

where KI is the integral gain (also related to the proportional gain KP by means of the integral time TI, as shown in Eq. 11.36). It is known that time-domain integration is transformed into division by the variable s in the Laplace domain, which accounts for the integral name of the controller introduced in Eq. (11.36). I control is usually applied to increase the system order and reduce the steady-state errors, but increasing the number of integrations can decrease the control system stability. Derivative controllers are defined by a transfer function: Gc ðsÞ ¼

MðsÞ ¼ KD $ s ¼ KP $ TD $ s EðsÞ

(11.37)

where the constant KD is the derivative gain, and the constant TD is known as derivative time. The derivative name comes from the effect produced by such a controller, which is derivationdindicated by the multiplication by s in the Laplace domain. Derivative control contributes to increasing the overall damping and to improve stabilitydas is detailed in Chapters 12 and 13.

11.2 Block Diagrams and SISO Feedback Systems

There are multiple examples of components or systems realizing the basic controller transfer functions that have been just mentioned. Assuming zA is the displacement input and zB is the displacement output, the transfer function for the small-rotation mechanical lever of Figure 11.15(a) is:

zA A'

O lA

qv

B' zB

A

B

i Cl

lB p

(a)

L

–

+

(b)

v

(c)

FIGURE 11.15 Basic Controllers: (a) Mechanical Lever as a P Controller; (b) Tank With Fluid as an I Controller; (c) Inductor as a D Controller.

Gc ðsÞ ¼

ZB ðsÞ zB lB ¼ ¼ ¼ KP ZA ðsÞ zA lA

(11.38)

and therefore this mechanical device can be used as a proportional controller. The transfer function of the tank with liquid of Figure 11.15(b) is: Gc ðsÞ ¼

PðsÞ 1 KI 1 ¼ ¼ ; KI ¼ Qv ðsÞ Cl $ s Cl s

(11.39)

which indicates the tank acts as an integral controller. In Figure 11.15(b), qv is the volume flow rate, Cl is the tank liquid capacitance, and p is the pressure at the tank’s bottom. The transfer function: Gc ðsÞ ¼

VðsÞ ¼ L $ s ¼ KD $ s; KD ¼ L IðsÞ

(11.40)

corresponds to the inductor of Figure 11.15(c), which shows this electrical element can be used as a derivative controller. Various combinations between these basic functions, such as proportionaleintegral (P þ I), proportionalederivative (P þ D), and proportionaleintegralederivative (P þ I þ D) are also possible, as discussed in the following. These compound controllers have transfer functions that are obtained by adding basic controller transfer functions. A P þ I controller, for instance, is defined as:
 KI 1 TI $ s þ 1 s þ 1=TI Gc ðsÞ ¼ KP þ ¼ KP $ 1 þ ¼ KP $ ; ¼ KP $ TI $ s TI $ s s s KP KI ¼ TI (11.41)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

Similarly, a P þ D controller is characterized by the transfer function: Gc ðsÞ ¼ KP þ KD $ s ¼ KP $ ð1 þ TD $ sÞ; KD ¼ KP $ TD

(11.42)

whereas the P þ I þ D controller has the transfer function:
 KI 1 þ TD $ s Gc ðsÞ ¼ KP þ þ KD $ s ¼ KP $ 1 þ TI $ s s ¼ KP $

TI $ TD $ s2 þ TI $ s þ 1 TI $ s

(11.43)

The mixed controllers mentioned above can be modeled graphically by combining individual proportional, derivative, integral blocks with the aid of summing points and pick off points, as depicted in Figure 11.16.

KP

I(s)

KP

+

1 s

KI

I(s)

KP +

KI s

+ s

KD

I(s)

O(s)

I(s)

O(s)

KP + K D ⋅ s

+

+ O(s)

O(s)

(b)

(a) 1 s

KI

+

KP

I (s) KD

I(s)

O(s)

+

s

KP +

+

KI + KD ⋅ s s

O(s)

(c)

FIGURE 11.16 Combinations of Individual P, I, and D Controllers With Summing and Pick off Points to Realize: (a) P þ I Controllers; (b) P þ D Controllers; P þ I þ D Controllers.

11.2 Block Diagrams and SISO Feedback Systems

–

+

+ I(s)

E(s)

I(s)

1 a

1 a ⋅ s+ 1

(a)

1 s

O(s)

+ O(s)

I(s)

E(s)

I(s)

1 a

1 a ⋅ s− 1

1 s

O(s)

O(s)

(b)

FIGURE 11.17 Shifted First-Order Transfer Functions From Elementary Block Algebra Operators: (a) G(s) ¼ 1/(a$s þ 1); (b) G(s) ¼ 1/(a$s  1).

In a similar manner, elementary block algebra operations and rules can be used to generate the shifted first-order transfer functions illustrated at the bottom of Figure 11.17(a) and 11.17(b) where a is a constant parameter. The elementary blocks and operations at the top of Figure 11.17(a) result in: 9 8 > = < IðsÞ  OðsÞ ¼ EðsÞ > 1 /OðsÞ ¼ $ IðsÞ (11.44) 1 1 > > a $ s þ1 : OðsÞ ¼ $ $ EðsÞ ; a s which shows that the transfer function connecting the input I(s) to the output O(s) is the one shown in the block at the bottom of Figure 11.17(a). The transfer function in the block at the bottom of Figure 11.17(b) is obtained through similar algebra operations from the elementary blocks shown in the diagram at the top of Figure 11.17(b) and using Eq. (11.44) where a plus sign is used instead of the minus sign to connect I(s) and O(s) in the first Eq. (11.44) to account for the positive input terminal in Figure 11.17(b). nn

Example 11.4

A unity-feedback control system is defined by the closed-loop transfer function GCL ðsÞ ¼ s3 þ3ssþ3 2 þ3sþ3, which relates an input R(s) to the output C(s). Propose a block diagram formed of only proportional gains, integrators, derivation operators, summers, and pick off points to represent this transfer function.

Solution The open-loop transfer function of the unity feedback is calculated based on Eq. (11.17) in terms of the closed-loop transfer function as:

  ðs þ 3Þ s3 þ 3s2 þ 3s þ 3 GCL ðsÞ sþ3 ¼ ¼ GðsÞ ¼ GOL ðsÞ ¼ 1  GCL ðsÞ 1  ðs þ 3Þ=ðs3 þ 3s2 þ 3s þ 3Þ s3 þ 3s2 þ 2s (11.45)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

1 s+ 1

C(s)

1 s

– + 3

– +

1 2

R(s)

1 s

+

s

+

+

–

1 s

2

s+ 3

1

(1 / 2) ⋅ s + 1 FIGURE 11.18 Unity-Feedback Control System Represented by Elementary Blocks and Operations.

This function can further be written as:

sþ3 sþ3 ¼ s $ ðs2 þ 3s þ 2Þ s $ ðs þ 1Þ $ ðs þ 2Þ 0 1


 1 1 1 B 1 C ¼ $ $ ð s þ 3Þ $ @ A$ 1 2 s sþ1 $s þ 1 2

GOL ðsÞ ¼

(11.46)

The factored form of Eq. (11.46) suggests that GOL(s) is a cascade combination of a proportional gain 1/2, an integrator 1/s, a P þ D transfer function s þ 3, and two shifted first-order transfer functions, 1/ (s/2 þ 1) and 1/(s þ 1). This series-form GOL(s) is added to a summing block in a unity-feedback diagram as depicted in Figure 11.18. The figure identifies the structure of the P þ D transfer function s þ 3, as well as the shifted first-order transfer functions 1/(s þ 1) and 1/(s/2 þ 1)dnote that for the latter shifted function the generic parameter a of Eq. (11.44) is actually 1/2, which explains the presence of 2 in the corresponding elementary-operation block. nn

nn

Example 11.5

(i) Design a P þ I controller with two operational amplifiers that use only resistors and inductors. (ii) Identify a mechanical microsystem formed of a mass m, a damper c, and a spring k to act as a P þ I þ D controller by using displacements at the input and output ports.

11.2 Block Diagrams and SISO Feedback Systems

Solution (i) Two operational-amplifier cascading stages are needed to generate input and output voltages having the same sign. It is demonstrated in Chapter 7 that the transfer function of the impedance system of Figure 11.19(a) is:

GðsÞ ¼

Vo ðsÞ ¼ Vi ðsÞ







 Z2 Z4 Z2 $ Z4 $  ¼ Z1 Z3 Z1 $ Z3

(11.47)

For the electrical system of Figure 11.19(b), the four impedances of Eq. (11.47) are:

Z1 ¼

R1 $ L $ s ; Z 2 ¼ R 2 ; Z3 ¼ R 3 ; Z4 ¼ R 4 R1 þ L $ s

(11.48)

Combining Eqs. (11.47) and (11.48) yields the transfer function:

GðsÞ ¼ Gc ðsÞ ¼

R2 R4 R2 $ R4 $ ðR1 þ L $ sÞ KI ¼ KP þ $ ¼ R1 $ L $ s R3 R1 $ R3 $ L $ s s R1 þ L $ s

(11.49)

with

R2 $ R4 R2 $ R4 ; KI ¼ R1 $ R3 R3 $ L

KP ¼

(11.50)

and, therefore, the op-amp system of Figure 11.19(b) behaves as a P þ I controller.

Z2 Z1 Vi

Z4

– +

Z3

– +

Vo

(a) R2

R4

R1

– +

vi

R3

– +

v

L

vo

(b)

FIGURE 11.19 Two-Stage Operational-Amplifier System as a P þ I Controller With: (a) Impedances; (b) Actual Electrical Elements.

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

Shuttle mass

Anchor Beam spring

u c

k

y

Anchor Actuation

m

Plate

Serpentine spring

(a)

(b)

FIGURE 11.20 Mechanical System as a P þ I þ D Controller: (a) Lumped-Parameter Model; (b) Actual Microsystem. (ii) Consider the cascade arrangement of the mechanical elements shown in Figure 11.20(a). The two displacements u and y are the relevant parameters defining the mechanical systems. Newton’s second law of motion is applied to mass m as:

_ or m $ u€ þ c $ u_ þ k $ u ¼ c $ y_ m $ u€ ¼ k $ u  c $ ðu_  yÞ

(11.51)

Applying the Laplace transform to Eq. (11.51) with zero initial conditions results in the following transfer function:

Gc ðsÞ ¼

YðsÞ m $ s2 þ c $ s þ k k 1 m ¼ ¼ 1 þ $ þ $s UðsÞ c$s c s c

(11.52)

Eq. (11.52), where Y(s) is the output and U(s) is the input, indicates the system acts as a P þ I þ D controller with KP ¼ 1, KI ¼ k/c, and KD ¼ m/c. A mechanical microsystem whose lumped-parameter model is the one of Figure 11.20(a) is the device of Figure 11.20(b). The shuttle mass plays the role of m and the serpentine spring is designed with a stiffness k about the motion direction. There is damping between the mobile plate to the right of Figure 11.20(b) and the shuttle mass, which can be modeled by the damper c with two mobile ends of Figure 11.20(a). nn

Hydraulic Controllers In addition to directly generating actuation forces through hydraulic cylinders, the hydraulic energy can result in various control actions, such as those produced by dashpots and hydraulic servomotors.

Dashpot A serial connection between a hydraulic viscous-damping cylinder and a spring (this combination is known as dashpot, although the same name is also being used for the parallel connection of these two elements) is analyzed heredsee the schematic representation of Figure 11.21(a). The piston can slide inside the hydraulic cylinder and the liquid circulates through an external circuit of liquid resistance Rl. The input motion u of the piston displaces the liquid from the right cylinder chamber into the left chamber externally due to the pressure difference Dp ¼ p1 e p2. The same pressure difference generates

11.2 Block Diagrams and SISO Feedback Systems

1 s

– +

qv u

1

τ

O(s)

I(s)

Rl p2

p1 A y

k I(s)

(a)

τ ⋅s τ ⋅ s+1

O(s)

(b)

FIGURE 11.21 Dashpot as a Controller: (a) Lumped-Parameter Model; (b) Transfer Functions From Elementary Block Algebra Operators.

a force A$Dp that compresses the spring k a distance y until the pressure and the spring force balance out. At the same time, the liquid volume that exits the right cylinder chamber is equal to the volume circulating through the external duct. The following equations model this behavior: 8 > < Dp ¼ qv $ Rl YðsÞ c$s _ ¼ k $ y/Gc ðsÞ ¼ A $ Dp ¼ k $ y /A2 $ Rl $ ðu_  yÞ ¼ > UðsÞ c $ sþk : qv ¼ A $ ðu_  yÞ _ ¼

ðc=kÞ $ s s$s 1 ¼ ¼ 1 þ ðc=kÞ $ s 1 þ s $ s 1 þ 1=ðs $ sÞ (11.53)

where c ¼ A2$Rl is the viscous-damping coefficient The controller’s transfer function Gc(s) was obtained by Laplace transforming the time-domain equation that resulted from the three equations to the left of Eq. (11.53). Figure 11.21(b) shows the elementary operations and their block diagram that realize the transfer function of Eq. (11.53) for I(s) ¼ U(s) and O(s) ¼ Y(s).

Hydraulic Servomotor Figure 11.22 illustrates a hydraulic servomotor. It is formed of a spool valve, which receives the input displacement u and directs the fluid in two possible directions in one of the two chambers of the power cylinder, which can move to the right or left. There are three pressure ducts connecting the spool valve: two of them are the output ports that are at low pressure po, whereas the third (middle) port is the input one providing high pressure pi. There are also two ducts connecting the spool valve and the power cylinder. When the piston of the spool valve moves to the right in Figure 11.22 a distance u, the right input orifice of the cylinder is open and highpressure fluid (of pressure pi) enters the right cylinder chamber, pushing the cylinder

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

po

pi

po

u Spool valve

y

qm

Fixed

qm

p1 Power cylinder p2 Fixed

FIGURE 11.22 Schematic Representation of a Hydraulic Servomotor.

piston to the left a distance y and producing the necessary external work. The following equations apply here:  K YðsÞ K=ðr $ AÞ KI qm ¼ K $ u $ u/Gc ðsÞ ¼ ¼ ¼ (11.54) / y_ ¼ qm ¼ r $ A $ y_ r$A UðsÞ s s While the second equation in the curly bracket of Eq. (11.54) models the mass flow rate conservation, the first Eq. (11.54) is the linearized flow rate equation through the spool valve’s orifice, with K being a constant that depends on geometrydOgata (2009). As per Eq. (11.54), the hydraulic servomotor of Figure 11.22 acts as an integral controller with a gain KI ¼ K/(r$A) where A is the power piston area and r is the liquid mass density.

Pneumatic Controllers Pneumatic controllers are used in many industrial processes due to their constructive simplicity, compactness, reliability, ease of maintenance, reduced cost, and explosion/fire proof properties. They can produce P, P þ D, P þ I, or P þ I þ D functions. This section discusses a P þ D pneumatic controller with its own feedback loop, which uses a mechanical displacement as input and generates a relatively large gas pressure output. Figure 11.23(a) sketches the structure of this controller. Its main component is a gas conduit that receives compressed air at supply pressure ps. The air passes through an orifice/restriction (not shown in the figure) before exiting the conduit through a nozzle. The air is ejected toward a flapper and the small interstice po

ps

R

Bellows ub Cp B pb

A ⋅ l1 k ⋅ ( l1 + l 2 )

l1 l2

x Flapper

1 R ⋅C ⋅ s + 1

–

O

Nozzle

P(s)

U(s)

l2 l1 + l2

+

K X(s)

A

Po (s)

u

(a)

(b)

FIGURE 11.23 Nozzle-Flapper Pneumatic System as P þ D Controller: (a) Schematic Physical Diagram; (b) Block Diagram With Feedback.

11.2 Block Diagrams and SISO Feedback Systems

(gap) between the nozzle and flapper generates a back pressure pb. This pressure is approximately equal to the output pressure po, which can be used to control a power cylinder or a control valve. One end of the flapper is connected to the moving end B of a bellows (an elastic cylinder that can extend axially under the action of compressed air at a pressure p), whose axial displacement is ub. Before entering the bellows, whose pneumatic capacitance is C, the air passes through another restriction of resistance R. An input displacement u at the other end A of the flapper generates a displacement x of the flapper toward the nozzle, which decreases the gap and increases (controls) the back and output pressure. The back pressure (equal to the output pressure) is considered to be proportional to the displacement x of the flapper: po ¼ pb ¼ K$x, with K assuming large values. The flapper’s inner point’s displacement x can be expressed in a manner similar to the mechanisms discussed in Example 4.24 and in Example 11.1 in terms of the opposite-direction displacements u and ub as: x¼

l2 l1 1 l2 l1 $ po ¼ $u  $ ub or $u  $ ub K l1 þ l2 l1 þ l2 l1 þ l2 l1 þ l2

(11.55)

where the previous relationship between po and x has been used. Note that for l1 ¼ l2, the first Eq. (11.55) reduces to Eq. (11.3) by taking u ¼ x1 and ub ¼ x2. The bellows also functions as a mechanical spring of stiffness k along its axial direction; the force generated by the inner pressure p over a cross-sectional bellows area A is balanced by the spring elastic force for an axial deformation ub, namely: A $p (11.56) k Combining Eqs. (11.55) and (11.56) results in the following time- and Laplacedomain equations: p $ A ¼ k $ ub or ub ¼

1 l2 A $ l1 A $ l1 $ po ¼ $ p/ $ Pð s Þ $u  K l1 þ l2 k $ ðl1 þ l2 Þ k $ ðl1 þ l2 Þ l2 1 ¼ $ U ðsÞ  $ P o ðsÞ K l1 þ l2

(11.57)

The pneumatic subsystem formed of the resistance R and the capacitance C can be modeled as detailed in Chapter 5 as: R $ C $ p_ þ p ¼ po /ðR $ C $ s þ 1Þ $ PðsÞ ¼ Po ðsÞ

(11.58)

The second Eq. (11.58) is the Laplace transform of the first, time domain Eq. (11.58). Substituting P(s) from Eq. (11.58) into the Laplace domain Eq. (11.57) yields the following actuator transfer function: A $ l1 l2 1 $ Po ðsÞ ¼ $ UðsÞ  $ Po ðsÞ/ K k $ ðl1 þ l2 Þ $ ðR $ C $ s þ 1Þ l1 þ l2 Po ðsÞ K $ l2 =ðl1 þ l2 Þ ¼ GC ðsÞ ¼ UðsÞ 1 þ K $ l1 $ A=k=ðl1 þ l2 Þ=ðR $ C $ s þ 1Þ K $ l2 =ðl1 þ l2 Þ k $ l2 $ R $ C k $ l2 x ¼ $s þ ¼ KP þ KD $ s K $ l1 $ A=k=ðl1 þ l2 Þ=ðR $ C $ s þ 1Þ A $ l1 A $ l1 (11.59)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

Eq. (11.59) considered that the term added to 1 in the denominator is significantly larger than 1 due to large K; as a consequence, the final, approximate Eq. (11.59) indicates that the pneumatic device of Figure 11.1(a) is a P þ D controller. Figure 11.23(b) is the block diagram corresponding to the transfer function of Eq. (11.59); it shows that the controller itself is actually a feedback control subsystem.

Compensators: Lag, Lead, and LageLead Controllers are sometimes called compensators as an acknowledgement of their adjusting functions. This section introduces three types of compensators: the lag, lead, and lagelead designs, which are generalized versions of P þ I, P þ D, and P þ I þ D controllers, respectively. More details of these compensators’ time- and frequencydomain action in feedback control systems are provided in Chapter 13. The general equation of a basic compensator transfer function is: Gc ðsÞ ¼ K $

s þ 1=s s þ 1=ða $ sÞ

(11.60)

where K is a gain, s is a time constant and a is a positive parameter. When a > 1 (which is equivalent to a assuming large values), the first-order characteristic polynomial of Eq. (11.60) has its negative root (or the pole of the transfer function) very close to the origin of the complex plane, and the corresponding controller is known as a phase-lag (or simply, lag) compensator. On the contrary, when 0 < a < 1 (which means a is small), the negative pole of the transfer function given in Eq. (11.60) is farther from the origin of the complex plane and the resulting controller is known as a phase-lead (or simply, lead) compensator. Figure 11.24 shows the relative zero-pole positions in the complex plane for lag and lead compensators. ω·j

Zero –1/τ

ω·j

Pole –1/(α·τ) O

(a)

Pole σ

–1/(α·τ)

Zero –1/τ

O

σ

(b)

FIGURE 11.24 Pole-Zero Position for: (a) Lag Compensator and Large a; (b) Lead Compensator and Small a.

Lag Compensators Consider a P þ I controller as the one defined in Eq. (11.41) whose transfer function has a pole at zero (because the root of the characteristic equation is p ¼ 0). The original P þ I controller can be altered through slight

11.2 Block Diagrams and SISO Feedback Systems

shifting of the original zero pole in the left-hand semiplane (LHP) of the complex plane by a small offset ε > 0 to become p ¼ eε; as such, the transfer function becomes:
 1 s þ 1=TI s þ 1=TI s þ 1=TI xKP $ ¼ KP $ Gc ðsÞ ¼ KP $ 1 þ ¼ KP $ TI $ s s sþε s þ 1=ða $ TI Þ (11.61) where a is a large positive number. Note that Eq. (11.61) expresses the small quantity ε in terms of the integral time TI and the large parameter a as ε ¼ 1/(a$TI). This equation is of the general form of Eq. (11.60) with K ¼ KP (the proportional gain) and s ¼ TI; therefore, it represents a compensator, specifically a lag compensator due to the large parameter a. The complex number Gc(u$j) together with its modulus and angle are: 1=s þ u $ j ; 1=ða $ sÞ þ u $ j sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1=sÞ2 þ u2 1 þ s2 $ u2 ; ¼ K $ a $ jGc ðj $ uÞj ¼ K $ 2 1 þ a 2 $ s2 $ u 2 ½1=ða $ sÞ þ u2

Gc ðj $ uÞ ¼ K $

(11.62)

;Gc ðj $ uÞ ¼ tan1 ½u=ð1=sÞ  tan1 ½u=ð1=ða $ sÞÞ ¼ tan1 ðs $ uÞ  tan1 ða $ s $ uÞ The angle is negative because a > 1 and the inverse tangent is a monotonically increasing functiondthe lag qualifier is due to this negative angle. For large a and if we ignore the K$a multiplier, the modulus of Eq. (11.62) indicates a lowpass filter response, as will be illustrated shortly in an example. Lead Compensators Consider now a P þ D controller, such as the one whose transfer function is defined in Eq. (11.42). If we add a pole that is farther away from the imaginary axis in the LHP, the modified transfer function becomes:
 KD 1 s þ 1=TD s þ 1=TD ¼ KD $ þ KD $ s ¼ KD $ s þ Gc ðsÞ ¼ xKD $ TD TD s þ 1=ε s þ 1=ða $ TD Þ (11.63) where ε is a small positive number setting the pole p ¼ 1/ε far in the LHP. Eq. (11.63) is of the general form of Eq. (11.60) with the derivative time TD being the generic time constant s and so it defines a compensator. For small positive values of the parameter a, the small offset ε is expressed in Eq. (11.63) as ε ¼ a$TD and, so, this particular design is a lead compensator as per the previous definition. The name of lead is related to the positive angle corresponding to Gc(u$j) that results from Gc(s) of Eq. (11.63) since 0 < a < 1. For small a and if we ignore the K$a

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multiplier, the modulus of Gc(u$j) resulting from Eq. (11.63) displays a high-pass filter frequency response, as illustrated in the following example. nn

Example 11.6 Establish whether the passive circuit of Figure 11.25 can be utilized as a lag or as a lead compensator by analyzing its transfer function Vo(s)/Vi(s). Also study its frequency response to determine whether the system behaves as a filter.

L1

R1

L2

vi

vo R2

FIGURE 11.25 Passive Electrical Network as a Lag or Lead Compensator.

Solution Considering that the two series-connection inductoreresistance groups in Figure 11.25 result each in one impedance (L1 and R1 produce Z1 while L2 and R2 form Z2), as demonstrated in Chapter 7 (Eq. 7.35 and Figure 7.7(a)), the transfer function of the circuit is:

Z2 L 2 $ s þ R2 ¼ Z1 þ Z2 L1 $ s þ R1 þ L2 $ s þ R2 1 sþ L2 ðL2 =R2 Þ ¼ $ 1 L1 þ L2

sþ R2 $ ðL1 þL2 Þ ðL2 =R2 Þ $ L2 $ ðR1 þR2 Þ

Gc ðsÞ ¼

(11.64)

The particular Eq. (11.64) is of the general type of Eq. (11.60), which demonstrates that the electrical system of Figure 11.25 is a compensator with:

K¼

L2 L2 R2 $ ðL1 þ L2 Þ 1 þ L1 =L2 ¼ ; s¼ ; a¼ L2 $ ðR1 þ R2 Þ 1 þ R1 =R2 L1 þ L2 R2

(11.65)

The compensator is a lag compensator for a > 1; this condition is satisfied when L1 is (very) large and R1 is (very) small for a given set of values of L2 and R2das Eq. (11.65) indicates. Consider L1/L2 ¼ 50 and R1/R2 ¼ 0.05, which results in a ¼ 48.5714. When L2 ¼ 1 H (which yields L1 ¼ 50 H) and R2 ¼ 200 U (which results in R1 ¼ 10 U), the transfer function of Eq. (11.64) gives the Bode plot (obtained with a bodemag command in MATLAB) of Figure 11.26(a), which indicates a low-pass filter response. Conversely, the electrical system is a lead compensator for 0 < a < 1, which is equivalent to small values of adcondition that requires small ratios L1/L2 and large ratios R1/R2. For L1/L2 ¼ 0.05 and R1/R2 ¼ 50, a value of a ¼ 0.0206 is obtained. Particularly, for L1 ¼ 0.05 H, L2 ¼ 1 H, R1 ¼ 250 U, and R2 ¼ 5 U, the frequency-response magnitude plot of the

11.2 Block Diagrams and SISO Feedback Systems

(a)

(b)

FIGURE 11.26 Frequency-Response Magnitude Plots for the Electrical System of Figure 11.25: (a) Low-Pass Filter (Lag) Response for L1 ¼ 50 H, L2 ¼ 1 H, R1 ¼ 10 U, R2 ¼ 200 U; (b) High-Pass Filter (Lead) Response for L1 ¼ 0.05 H, L2 ¼ 1 H, R1 ¼ 250 U, R2 ¼ 5 U. transfer function given in Eq. (11.64) is shown in Figure 11.26(b), which represents a typical highpass filter behavior. nn

LageLead Compensators A lagelead compensator, as the name suggests, combines a lag component with a lead one in a cascade manner. Following a reasoning similar to the ones used to derive the transfer function of a lag compensator from that of P þ I controller and the transfer function of a lead compensator from that of a P þ D controller, it can be shown (without being demonstrated here) that the transfer function of a lag-lead compensator results from modifying the transfer function of a P þ I þ D controller to result in:



s þ 1=s1 s þ 1=s2 Gc ðsÞ ¼ Gc1 ðsÞ $ Gc2 ðsÞ ¼ K1 $ $ K2 $ s þ 1=ða1 $ s1 Þ s þ 1=ða2 $ s2 Þ ¼K$

s þ 1=s1 s þ 1=s2 $ s þ 1=ða1 $ s1 Þ s þ 1=ða2 $ s2 Þ

(11.66)

The subscript 1 denotes the lag component and the subscript 2 represents the lead component; evidently, a1 > 1 and 0 < a2 < 1. The lagelead compensator behaves as a lag design at small frequencies (and therefore, as a low-pass filter) and as a lead configuration for high frequencies (which is a high-pass filter) resulting in an overall notch-filter response. While not studied here, leadelag compensators can be designed when reversing the sequence and using a P þ I unit first that connects to a subsequent P þ D unit. nn

Example 11.7 Verify that the operational-amplifier electrical network of Figure 11.27 can function as a lagelead compensator and determine the necessary conditions applicable to the electrical element parameters

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

R2 R1

C2

R4

C4

C1 R3

– +

vi

C3

– +

vo

FIGURE 11.27 Two-Stage Operational-Amplifier Electrical System as LageLead Compensator.

that would enable this condition. Check whether the following set of element parameter values satisfies this requirement: R1 ¼ R4 ¼ 200 U, C1 ¼ 100 mF, R2 ¼ 100 U, C2 ¼ 20 mF, R3 ¼ 10 U, C3 ¼ 2 mF, C4 ¼ 1 mF, and plot the magnitude frequency response.

Solution The electrical system is a two-stage operational network as the one pictured in Figure 11.19(a). Its transfer function is given in Eq. (11.47) where the impedances are:

Zi ¼ Ri þ

1 Ri $ Ci $ s þ 1 ¼ Ci $ s Ci $ s

with i ¼ 1; 2; 3; 4

(11.67)

The impedance ratio of the first op-amp stage is written as:

Z 2 R2 $ C 2 $ s þ 1 C1 $ s C1 R2 $ C2 $ ¼ ¼ $ $ C2 $ s R1 $ C1 $ s þ 1 C2 R1 $ C1 Z1

R2 ¼ $ R1

1 ð R2 $ C 2 Þ 1  sþ ðR2 $ C2 Þ $ RR12 $$ CC12

1 R2 $ C2 1 sþ R1 $ C1 sþ

sþ

(11.68)

Similarly, the impedance ratio corresponding to the second stage is:

Z4 R4 ¼ $ Z3 R3

1 ðR4 $ C4 Þ 1
 sþ ðR4 $ C4 Þ $ RR34 $$ CC34 sþ

(11.69)

11.2 Block Diagrams and SISO Feedback Systems

The system’s transfer function is therefore:


Gc ðsÞ ¼


 Z2 Z4  $  Z1 Z3


 R2 $ R 4 ¼ $ R1 $ R 3

1 1 sþ ðR2 $ C2 Þ ðR4 $ C4 Þ $ (11.70) 1 1

 sþ  sþ ðR2 $ C2 Þ $ RR12 $$ CC12 ðR4 $ C4 Þ $ RR34 $$ CC34 sþ

Eq. (11.70) is of the generic form given in Eq. (11.66) with:

K¼


 R2 $ R4 R1 $ C1 R3 $ C 3 ; s2 ¼ R4 $ C4 ; a2 ¼ ; s1 ¼ R2 $ C2 ; a1 ¼ R1 $ R3 R2 $ C2 R4 $ C 4 (11.71)

and therefore, the electrical system of Figure 11.27 is a lagelead compensator provided that:

a1 ¼

R1 $ C1 R3 $ C3 > 1 and 0 < a2 ¼

dia ðtÞ > > þ Ra $ ia ðtÞ þ Ke $ uðtÞ ¼ va ðtÞ < La $ dt > duðtÞ > > : Jl $ þ c $ uðtÞ ¼ Km $ ia ðtÞ dt

(11.73)

Laplace transforming Eq. (11.73) with zero initial conditions yields:

(

ðLa $ s þ Ra Þ $ Ia ðsÞ þ Ke $ UðsÞ ¼ Va ðsÞ

(11.74)

ðJl $ s þ cÞ $ UðsÞ ¼ Km $ Ia ðsÞ

Substituting Ia(s) from the second Eq. (11.74) into the first Eq. (11.74) yields Gap(s), the transfer function that characterizes both the actuator (defined by Ga(s)) and the plant defined by Gp(s):

Gap ðsÞ ¼ Ga ðsÞ $ Gp ðsÞ ¼ ¼

La $ Jl

$ s2

UðsÞ Va ðsÞ

Km þ ðLa $ c þ Ra $ Jl Þ $ s þ Ra $ c þ Ke $ Km

(11.75)

Tachometers are sensors that convert the input (mechanical) angular velocity into an output that can be electrical voltagedin this case the tachometer can be a dc generator connected to the rotating shaft and whose back electromotive force is related to the angular velocity as given in the second Eq. (10.80), which is reformulated here in time domain and in Laplace domain as:

vðtÞ ¼ Ks $ uðtÞ; VðsÞ ¼ Ks $ UðsÞ or Gt ðsÞ ¼

VðsÞ ¼ Ks UðsÞ

(11.76)

where Gt(s) is the generic tachometer transfer function. The two tachometers of Figure 11.29(a) are defined by the transfer functions:

Gr ðsÞ ¼

Vr ðsÞ Vs ðsÞ ¼ Ks ¼ Ks ; HðsÞ ¼ Ur ðsÞ UðsÞ

(11.77)

With these transfer functions, the block diagram of Figure 11.30 illustrates the structure of the full velocity-control system. The feed-forward transfer function G(s) is:

Ωr(s)

Vr(s) +

Ve(s)

Gr(s) = Ks

Ω(s)

Va(s) Gc(s) = KP

Gap(s)

_ Vs(s)

H(s) = Ks

FIGURE 11.30 Block Diagram of the Velocity-Control of a DC Motor and Rotary Mechanical Plant With Prefilter and P Control.

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GðsÞ ¼ Gc ðsÞ $ Gap ðsÞ ¼

La $ Jl

$ s2

KP $ Km þ ðLa $ c þ Ra $ Jl Þ $ s þ Ra $ c þ Ke $ Km (11.78)

Eq. (11.78) took into account the controller transfer function, which is Gc(s) ¼ KP. Example 11.2 analyzed the equivalent conversion of a nonunity feedback system with prefilter into a regular unity-feedbacks system. As per Eq. (11.21), the feed-forward transfer function of the equivalent, unity-feedback system is:

Gr ðsÞ $ GðsÞ ¼ Gr ðsÞ $ GðsÞ 1 þ GðsÞ $ HðsÞ  Gr ðsÞ $ GðsÞ Ks $ KP $ Km ¼ 2 La $ Jl $ s þ ðLa $ c þ Ra $ Jl Þ $ s þ Ra $ c þ Ke $ Km

G ðsÞ ¼

(11.79)

when taking into account that H(s) ¼ Gr (s). The closed-loop transfer function GCL(s) of the feedback system shown in Figure 11.30 is calculated based on Eqs. (11.17) and (11.79) as:

GCL ðsÞ ¼ ¼

G ðsÞ 1 þ G ðsÞ Ks $ KP $ Km La $ Jl $ s2 þ ðLa $ c þ Ra $ Jl Þ $ s þ Ra $ c þ Ke $ Km þ Ks $ KP $ Km (11.80) nn

nn

Example 11.9

The mechanical system of Figure 11.31(a) consists of a shuttle mass m rigidly connected to a piezoelectric (PZT) actuator of negligible inertia having an axial stiffness kPZT, a PZT charge constant d and PZT circuit actuation voltage va. The shuttle mass is supported by two identical beam springs, each of stiffness k. A transverse capacitive unit defined by an original gap g0 and a bias voltage vb senses the

Beam spring

y

Capacitive sensing g0

y k PZT

2k

m

m PZT actuator

fb

Ground va

vs

Anchor

(a)

Shuttle mass

(b)

FIGURE 11.31 (a) System With Piezoelectric (PZT) Actuation and Transverse Capacitive Sensing; (b) Lumped-Parameter Model of Mechanical Plant.

11.2 Block Diagrams and SISO Feedback Systems

mass motion by means of the sensed voltage vs. The system is incorporated into a feedback position control system (such as the generic one of Figure 11.13) whose reference displacement is transformed into a voltage by a capacitive unit identical to the sensing one. Design an amplified, resistor-based, electrical summing point that would also operate as a proportional controller. Draw the block diagram, determine the individual transfer function of each component, and calculate the closed-loop transfer function.

Solution As detailed in Chapter 10 in the section dedicated to PZT actuation, the actual PZT block can be modeled as a spring of stiffness kPZT fixed at one end and acted upon at the other end by the block force fb (shortened notation for fa,b of Chapter 10). As a consequence, the mechanical system of Figure 11.31(a)dthe plantdcan be represented in lumped-parameter form as in Figure 11.31(b). Applying Newton’s second law of motion to the lumped-parameter mechanical unit and ignoring the losses yields

€ ¼ ð2k þ kPZT Þ $ yðtÞ þ fb ðtÞ m $ yðtÞ

(11.81)

Eq. (11.81) took into account that two original springs and the spring corresponding to the PZT act in parallel. Laplace transforming Eq. (11.81) results in the following transfer function of the mechanical system

Gp ðsÞ ¼

YðsÞ 1 ¼ Fb ðsÞ m $ s2 þ 2k þ kPZT

(11.82)

As also discussed in the section dedicated to PZT transduction in Chapter 10, the PZT block force is calculated as:

fb ðtÞ ¼ kPZT $ y0 ðtÞ ¼ kPZT $ d $ va ðtÞ

(11.83)

where y0 is the PZT block free displacement, va is the actuation voltage, and d is the PZT charge constant. The actuation transfer function is found after applying the Laplace transform to Eq. (11.83):

Ga ðsÞ ¼

Fb ðsÞ ¼ kPZT $ d Va ðsÞ

(11.84)

The transverse capacitive sensing unit converts the shuttle mass displacement y into voltage variation vs. Eq. (4.31) is reformulated here as:

vb Dv ¼ vs ¼  $ y g0 R2

R

R2

R1 vr

R vs

(11.85)

– +

(a)

– vs

–vs

R1

– +

R2 v

– +

(b)

FIGURE 11.32 (a) Single-Stage, Unit-Gain Operational Amplifier; (b) Two-Stage Operational Amplifier With Resistors as a Summing Point and P Controller for the Device of Figure 11.31.

ve

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

The Laplace transform is applied to Eq. (11.85), and this generates the transfer function of the feedback sensor:

HðsÞ ¼

Vs ðsÞ vb ¼  ¼ Ks YðsÞ g0

(11.86)

which is a constant negative gain denoted by Ks. The prefilter has the same transfer function of Eq. (11.86). The negative sign of the feedback sensor constant can be transformed into a positive sign since the output signal is connected to the minus input terminal of the summing block. However, for the prefilter transfer function, we need an additional operational-amplifier inverting circuit with two identical resistors, as the one indicated in Figure 11.32(a) to serially connect to the capacitive unit. The transfer function of this circuit is 1, which, multiplied by H(s) of Eq. (11.86)dbecause the two transfer functions are connected in seriesdresults in þKs. The summing point will be designed now as a two-stage operational-amplifier systemdsee Figure 11.32(b), which is similar to the one studied in Example 11.1dassuming that R2 ¼ n$R1 with n > 1. The following relationship exists between the voltages corresponding to the first- and second-stage operational-amplifier units:

v ¼

R2 $ ðvr  vs Þ; ve ¼ v R1

(11.87)

Combining the two Eq. (11.87) results in:

ve ¼

R2 $ ðvr  vs Þ ¼ n $ ðvr  vs Þ R1

(11.88)

In the Laplace domain, Eq. (11.88) becomes:

Ve ðsÞ ¼ n $ ½Vr ðsÞ  Vs ðsÞ

(11.89)

which indicates the summing point has also an amplifying effect, and therefore the physical system of Figure 11.32(b) can operate as both summing point and proportional controller.

Yr(s)

Vr(s) +

–Vr(s)

H(s)

Va(s) Ga(s)

–1

Fb(s)

Y(s) Gp(s)

− Vs(s)

H(s)

FIGURE 11.33 Block Diagram of the Position-Control Feedback System With Mechanical Plant, Piezoelectric Actuation, Capacitive Sensing, P Control, and Capacitive Prefilter. Figure 11.33 is the block diagram of the feedback system described in the example statement. The system of Figure 11.33 is identical to the generic system of Example 11.2 and illustrated in Figure 11.13, which is defined by three arbitrary transfer functions: G1(s)dthe one before the summing point, G2(s)dthe feed-forward transfer function, and G3(s)dthe feedback transfer function, if:

8 G1 ðsÞ ¼ HðsÞ $ ð1Þ ¼ HðsÞ ¼ Ks > > > > < kPZT $ d G2 ðsÞ ¼ Ga ðsÞ $ Gp ðsÞ ¼ 2 > m $ s þ 2k þ kPZT > > > : G3 ðsÞ ¼ HðsÞ ¼ Ks

(11.90)

11.2 Block Diagrams and SISO Feedback Systems

The closed-loop transfer function can now be calculated with Eq. (11.20) as:

GCL ðsÞ ¼

G1 ðsÞ $ G2 ðsÞ d $ kPZT $ Ks ¼ 1 þ G2 ðsÞ $ G3 ðsÞ m $ s2 þ d $ kPZT $ Ks þ 2k þ kPZT

(11.91) nn

11.2.4 Sensitivity Analysis System characteristics and performance depend, in general, on the parameters defining the system. One fundamental-related requirement is that system characteristics (or response/output) vary as little as possible even when parameter values change substantially. In other words, a system should be minimally sensitive to parameter variation. In addition to ensuring system trackability, disturbance rejection, and satisfactory time/frequency-domain performance, feedback controls is also tasked with reducing the dynamic system sensitivity to parameter alterations. To better grasp the concept of sensitivity from a mathematical standpoint, consider a function f depending on two parameters p1 and p2 (also variables here) as: f ðp1 ; p2 Þ ¼ p1 þ p22

(11.92)

It is clear that a variation of p2 will produce a variation in f that is larger than the one generated by the same variation of p1. Common-sense language would state that the function f is more sensitive to p2 than to p1. One measure of sensitivity of the function f to a parameter p is the ratio of the change in function Df to the change in parameter Dp. A large Df/Dp value signals that f is (very) sensitive to changes in p, while small Df/Dp values indicate reduced sensitivity of f to p. Instead of using the ratio of absolute variations Df and Dp, one can utilize relative (percent) variations Df/f and Dp/p. The mathematical definition of sensitivity evaluates this latter ratio in the limit when Dp approaches zero; as such, the sensitivity of a function f to a parameter p, denoted as Sf,p, is: Sf ;p

Df p Df p vf f ¼ $ ¼ $ lim ¼ lim Dp/0 Dp f Dp/0 Dp f vp p

(11.93)

where vf/vp is the partial derivative of f with respect to p. In feedback control, sensitivity analysis studies open-loop and closed-loop transfer functions or time-response qualifiers as functions f in terms of various parameters. nn

Example 11.10

A two-stage operational amplifier containing only resistors acts as a proportional controller of gain K to an electrical system (plant) formed of an inductor, a resistor, and a capacitor whose transfer function is Gp(s) ¼ Q(s)/V(s) ¼ 1/(s2 þ 2s þ 5)dwith Q(s) being the charge in the plant circuit and V(s) being the voltage provided by the controller. These two subsystems are connected in a unity-feedback control

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system. Relate the sensitivity of the open-loop transfer function to the sensitivity of the closed-loop transfer function, both calculated in terms of the gain K.

Solution

The open-loop transfer function is G(s) ¼ Gc(s)$Gp(s) ¼ K/(s2 þ 2s þ 5). The sensitivity of the open-loop and of the closed-loop transfer functions to K is calculated according to Eq. (11.93):

SG;K ¼

K vGðsÞ K vGCL ðsÞ $ ; SGCL ;K ¼ $ GðsÞ vK GCL ðsÞ vK

(11.94)

Taking in consideration that GCL(s) ¼ G(s)/[1 þ G(s)] for a unity-feedback system, the partial derivative of GCL(s) in terms of K is calculated as:

vGCL ðsÞ vGCL ðsÞ vGðsÞ 1 vGðsÞ ¼ $ ¼ $ vK vGðsÞ vK ½1 þ GðsÞ2 vK

(11.95)

Combining now Eqs. (11.94) and (11.95) results in the following sensitivity connection:

SGCL ;K ¼

1 $ SG;K 1 þ GðsÞ

(11.96)

For the particular expression of the open-loop transfer function of this example it follows that:

8 > < SG;K ¼ 1 > : SGCL ;K ¼

s2 þ 2s þ 5 s2 þ 2s þ 5 þ K

(11.97)

Eq. (11.97) shows that the sensitivity of the open-loop transfer function does not vary with K, whereas the sensitivity of the closed-loop transfer function decreases with the increase of the gain K, which means that GCL(s) becomes less sensitive to larger gains. nn

11.2.5 State-Space Modeling of SISO Feedback Systems For a SISO feedback control system, the closed-loop transfer function GCL(s) that connects the reference (input) signal R(s) to the controlled (output) signal C(s) should be used to derive a state-space model using the standard procedure presented in Chapter 8. As such, the closed-loop transfer function is expressed in terms of an intermediate transfer function Z(s) as: GCL ðsÞ ¼

CðsÞ ZðsÞ CðsÞ ¼ $ RðsÞ RðsÞ ZðsÞ

(11.98)

After inverse Laplace transformation, the fraction Z(s)/R(s) will provide the state equation while the fraction C(s)/Z(s) will result in the output equationdthe state and output equations forming the state-space model. The following example illustrates the procedure of converting the closed-loop transfer function of a feedback system into a state-space model. nn

Example 11.11 Find a state-space model for the position-control system of Example 11.9 whose closed-loop transfer function is given in Eq. (11.91).

11.3 Block Diagrams and MIMO Feedback Systems

Solution The closed-loop transfer function of Eq. (11.91) is written in short form as:

GCL ðsÞ ¼

CðsÞ ZðsÞ CðsÞ 1 ¼ $ ¼ $b RðsÞ RðsÞ ZðsÞ a2 $ s2 þ a0

(11.99)

with:

a2 ¼ m; a0 ¼ d $ kPZT $ Ks þ 2k þ kPZT ; b ¼ d $ kPZT $ Ks

(11.100)

Taking:

ZðsÞ 1 ¼ RðsÞ a2 $ s2 þ a0

(11.101)

results in the following time-domain equation after inverse Laplace transformation:

a2 $ zðtÞ € þ a0 $ zðtÞ ¼ rðtÞ

(11.102)

The following choices are made for the two state variables x1 and x2, the input u, and the output y:

x1 ðtÞ ¼ zðtÞ; x2 ðtÞ ¼ zðtÞ; _ uðtÞ ¼ rðtÞ; yðtÞ ¼ cðtÞ

(11.103)

Eqs. (11.102) and (11.103) allow formulating the state equation:



x_1 x_2



2

0 ¼ 4 a0  a2

2 3

0 x1 6 7 5$ þ 4 1 5$u 0 x2 a2 1

3



(11.104)

The second fraction of Eq. (11.99) is:

CðsÞ ¼b ZðsÞ

(11.105)

Cross-multiplication in Eq. (11.105) and application of the Laplace transform to the result yield:



cðtÞ ¼ b $ zðtÞ or yðtÞ ¼ ½ b 0  $

x1 þ 0$u x2

(11.106)

The last form of Eq. (11.106) is the output equation and together with the state equationdEq. (11.104)dthey form the state-space model. nn

11.3 BLOCK DIAGRAMS AND MIMO FEEDBACK SYSTEMS Complex systems, such as rockets, with multiple inputs and multiple outputs (MIMO or multivariable systems) might need simultaneous control of several variables. This section discusses a few notions related to the feedback control of MIMO dynamic systems. Systems with disturbances are first studied here, which are often times systems with multiple inputs and one output. Actual MIMO systems are then modeled, by the matrix method that connects the controlled output vector {C(s)} to the reference (input) vector {R(s)} through the closed-loop transfer function matrix [GCL(s)].

577

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

11.3.1 MISO Feedback Systems With Disturbances As mentioned in the introduction of this chapter, one important function of controls is disturbance rejection/minimization (or regulation). In addition to the normal input to feedback controlled systems (which is the reference signal R(s) for SISO systems), disturbances can also act being produced by unwanted external effects and/or imprecision in the mathematical models describing various transfer functions, as well as by the interstage loading between various components of the control system. In such situations, the original SISO system becomes a MISO (multiple input, single output) system actually. One common situation is when the disturbance D ( s) R(s)

+

+

+

Gc(s)

C(s) Gp(s)

_ H (s)

FIGURE 11.34 Block Diagram of a Nonunity Control System With Disturbance Input.

signal D(s) acts between the controller and the plant, as illustrated in Figure 11.34. The following signal balance is valid for the plant transfer function Gp(s): f½RðsÞ  HðsÞ $ CðsÞ $ Gc ðsÞ þ DðsÞg $ Gp ðsÞ ¼ CðsÞ

(11.107)

After some algebraic calculation, Eq. (11.107) can be written as: CðsÞ ¼ GCL;R $ RðsÞ þ GCL;D $ DðsÞ where the two partial closed-loop transfer functions are: 8 > Gc ðsÞ $ Gp ðsÞ > > > GCL;R ðsÞ ¼ < 1 þ Gc ðsÞ $ Gp ðsÞ $ HðsÞ > Gp ðsÞ > > > : GCL;D ðsÞ ¼ 1 þ G ðsÞ $ G ðsÞ $ HðsÞ c p

(11.108)

(11.109)

Comparing the two transfer functions of Eq. (11.109) and their contributions to the total response C(s), it can be seen that for large (or very large) values of Gc(s), such as for cases where a large proportional gain KP is used in the controller, the weight of GCL,D in C(s) is relatively small, and therefore the effect of disturbances is substantially reduced. nn

Example 11.12 Consider the dc motor of Figure 1.18 is the plant of the feedback control system sketched in the block diagram of Figure 11.34 with the input being the armature voltage va and the output being the shaft angular velocity u. The reference signal is a voltage. The controller is proportional with a gain KP, and

11.3 Block Diagrams and MIMO Feedback Systems

the sensor converts the controlled output u into a voltage and is proportional with a gain Ks. Express the closed-loop transfer functions GCL,R(s) and GCL,D(s).

Solution Eq. (1.16) described the equation governing the time-domain behavior of the dc motor with the armature voltage va(t) being the input and the shaft rotation angle q(t) being the output. Eq. (1.16) can be reformulated using the shaft angular velocity u(t) as the plant output, namely:

€ þ a1 $ uðtÞ _ a2 $ uðtÞ þ a0 $ uðtÞ ¼ va ðtÞ

(11.110)

with:

a2 ¼

La $ Jl La $ cr þ Ra $ Jl R a $ cr ; a1 ¼ ; a0 ¼ þ Ke Km Km Km

(11.111)

The transfer function of the plant is obtained by applying the Laplace transform to Eq. (11.110), namely:

Gp ðsÞ ¼

UðsÞ 1 ¼ Va ðsÞ a2 $ s2 þ a1 $ s þ a0

(11.112)

The transfer functions of the controller and feedback sensor are:

Gc ðsÞ ¼ KP ; HðsÞ ¼

Vs ðsÞ ¼ Ks UðsÞ

(11.113)

The closed-loop transfer functions of Eq. (11.109) are:

KP ; a2 $ s 2 þ a1 $ s þ a0 þ K P $ K s 1 GCL;D ðsÞ ¼ 2 a2 $ s þ a1 $ s þ a0 þ KP $ Ks

GCL;R ðsÞ ¼

(11.114)

Eq. (11.114) show that GCL,D(s) ¼ GCL,R(s)/KP, which confirms that increasing the controller gain KP reduces the weight of disturbance influence on the control system response. nn

11.3.2 MIMO Feedback Systems Transfer-Function Matrix Modeling Similar to SISO feedback systems, basic MIMO feedback systems are governed by the block diagram of Figure 11.35(a) where signal vectors are used instead of signal scalars and transfer function matrices for regular transfer functions. The significance of the vectors and matrices involved in Figure 11.35(a) is the same as the one of their SISO counterparts. The controller, actuator, and plant are coupled in cascade and thus, the feedforward transfer function matrix denoted by [G(s)] results when connecting the output vector {C(s)} to the error vector {E(s)}: fCðsÞg ¼ ½Gp ðsÞ $ fAðsÞg ¼ ½Gp ðsÞ $ ½Ga ðsÞ $ fMðsÞg ¼ ½Gp ðsÞ $ ½Ga ðsÞ $ ½Gc ðsÞ $ fEðsÞg ¼ ½GðsÞ $ fEðsÞg or ½GðsÞ ¼ ½Gp ðsÞ $ ½Ga ðsÞ $ ½Gc ðsÞ

(11.115)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

{R(s)}

+

{E(s)}

{M(s)}

{C(s)}

{A(s)}

[Gc(s)]

[Ga(s)]

[Gp(s)]

_

{C(s)}

{B(s)} [H(s)]

(a) {R(s)}

+

{C(s)}

{E(s)} [G(s)]

_

{B(s)} [H(s)]

(b)

FIGURE 11.35 (a) Block Diagram of a Basic Negative-Feedback MIMO Control System; (b) Simplified Block Diagram of a Basic Negative-Feedback MIMO Control System.

The block diagram of Figure 11.35(b) uses [G(s)] instead of the original transfer function matrices of Figure 11.35(a). Assuming that {C(s)} and {R(s)} have the same dimension, it follows that [G(s)] and [H(s)] are square and of the same dimension as {C(s)} and {R(s)}. The coupling between controller, actuator, plant (all combined into [G(s)]), and feedback is realized by means of the open-loop transfer function matrix [GOL(s)], which is derived as: fBðsÞg ¼ ½HðsÞ $ fCðsÞg ¼ ½HðsÞ $ ½GðsÞ $ fEðsÞg ¼ ½GOL ðsÞ $ fEðsÞg or ½GOL ðsÞ ¼ ½HðsÞ $ ½GðsÞ ¼ ½HðsÞ $ ½Gp ðsÞ $ ½Ga ðsÞ $ ½Gc ðsÞ

(11.116)

where Eq. (11.115) has been used. Eventually, the closed-loop transfer function matrix [GCL(s)] connects the controlled output signal vector {C(s)} to the reference signal vector {R(s)}. The summing point relates its three signal vectors as follows: fEðsÞg ¼ fRðsÞg  fBðsÞg ¼ fRðsÞg  ½HðsÞ $ fCðsÞg

(11.117)

11.3 Block Diagrams and MIMO Feedback Systems

Substituting {E(s)} of Eq. (11.117) into {C(s)} ¼ [G(s)]${E(s)} results in the closed-loop transfer function matrix: fCðsÞg ¼ ð½I þ ½GðsÞ $ ½HðsÞÞ1 $ ½GðsÞ $ fRðsÞg ¼ ½GCL ðsÞ $ fRðsÞg or ½GCL ðsÞ ¼ ð½I þ ½GðsÞ $ ½HðsÞÞ1 $ ½GðsÞ (11.118) where [I ] is the identity matrix that has the dimension of the matrix resulting from the product [G(s)]$[H(s)]. Because [G(s)] and [H(s)] are square and of the same dimension with {C(s)} and {R(s)}, as discussed here, the product [G(s)]$[H(s)] is also a square matrix of the same dimension with its factor matrices. Using {E(s)} ¼ [G(s)]1${C(s)} from Eq. (11.115) in conjunction with Eq. (11.117) results in the alternative closed-loop transfer function:  1 ½GCL ðsÞ ¼ ½GðsÞ1 þ ½HðsÞ (11.119)

{R(s)} [P(s)]

{R1(s)}

+

{C(s)}

{E(s)} [G(s)]

_

{B(s)}

[H(s)]

FIGURE 11.36 Block Diagram of a Negative-Feedback MIMO Control System With Prefilter.

A variant of the MIMO simplified control system of Figure 11.35(b) is the one whose block diagram is shown in Figure 11.36, and which adds a prefilter transfer function matrix [P] before the summing point to convert the reference signal vector {R(s)} into a vector {R1(s)} that has the same physical nature with the back signal vector {B(s)}. Applying a reasoning similar to the one used to derive [GCL(s)] of Eq. (11.118) and corresponding to the feedback system of Figure 11.35(b), the following closed-loop transfer function matrix is obtained for the system of Figure 11.36: fCðsÞg ¼ ð½I þ ½GðsÞ $ ½HðsÞÞ1 $ ½GðsÞ $ ½PðsÞ $ fRðsÞg ¼ ½GCL ðsÞ $ fRðsÞg or ½GCL ðsÞ ¼ ð½I þ ½GðsÞ $ ½HðsÞÞ1 $ ½GðsÞ $ ½PðsÞ (11.120) When Eq. (11.119) is used instead of Eq. (11.118), the closed-loop transfer function matrix of the control system shown in Figure 11.36 is:  1 ½GCL ðsÞ ¼ ½GðsÞ1 þ ½HðsÞ $ ½PðsÞ (11.121)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

nn

Example 11.13 (i) Derive the mathematical model for a feedback MIMO system whose actuation-plant section is sketched in Figure 11.37(a). The plant is formed of two point masses m1 and m2 and three springs of stiffnesses k1, k2, and k. Each mass is actuated by a unit formed of a translatory-motion dc servomotor similar to the one studied in Chapter 10. The translation damping coefficient related to the motion of the magnetic rod inside its cylinder is c. The coil inductance is L, the resistance is R, the force factor is Km, and the back electromotive force factor is Ke. Each actuation circuit is driven by its own voltage source (v1 and v2). The displacement of each mass is sensed by a Hall-effect sensor of constant KH and the two sensed voltages are vs1(t) and vs2(t). Similar sensors are used as prefilters to convert the reference displacements r1(t) to the mass m1 into the voltage vr1(t) and r2(t) to the mass m2 into the voltage vr2(t). Consider the magnetic rods have negligible mass. (ii) Draw the block diagram of this control system and formulate the closed-loop transfer function matrix connecting the reference signal vector {r1 r2}T to the mass displacement vector {y1 y2}T ¼ {c1 c2}T. Consider that each mass is controlled individually by one controller.

y1

y2

k1 c L

m2 c

L

R v1

k2

k

m1

R v2

(a) y2

y1 k1·y1 ⋅ c ⋅ y1

m1

k·(y1 – y2)

k2·y2

k·(y2 – y1) m2 fa2

fa1

c ⋅ y⋅ 2

(b)

FIGURE 11.37 (a) Physical Diagram Showing the Mechanical Plant, Actuation, and Sensing of a MIMO Feedback System; (b) Free-Body Diagrams of the Two Point Bodies.

Solution (i) Based on the free-body diagrams of Figure 11.37(b), Newton’s second law is used to formulate the differential equations of motion for the two point masses:

(

(

m1 $ y€1 ¼ fa1  c $ y_1  k1 $ y1  k $ ðy1  y2 Þ or m2 $ y€2 ¼ fa2  c $ y_2  k2 $ y2  k $ ðy2  y1 Þ m1 $ y€1 ¼ Km $ i1  c $ y_1  k1 $ y1  k $ ðy1  y2 Þ m2 $ y€2 ¼ Km $ i2  c $ y_2  k2 $ y2  k $ ðy2  y1 Þ

(11.122)

11.3 Block Diagrams and MIMO Feedback Systems

583

The actuation forces acting on the two masses are:

fa1 ¼ Km $ i1 ; fa2 ¼ Km $ i2

(11.123)

with i1 being the current through the actuation electrical circuit producing fa1 and i2 being the current through the actuation electrical circuit producing fa2. Kirchhoff’s voltage law is used to obtain the equations corresponding to the two actuation electrical circuits of Figure 11.37(a):

8 8 > > di1 di1 > > > > þ R $ i1 ¼ v1  vb1 þ R $ i1 ¼ v1  Ke $ y_1

di2 di2 > > > > þ R $ i2 ¼ v2  vb2 þ R $ i2 ¼ v2  Ke $ y_2 :L$ :L$ dt dt

(11.124)

where the back electromotive forces generated in the two circuits were:

vb1 ¼ Ke $ y_1 ; vb2 ¼ Ke $ y_2

(11.125)

Applying the Laplace transform to the last-form Eqs. (11.122) and (11.124) yields:

8  > m1 $ s2 þ c $ s þ k1 þ k $ Y1 ðsÞ  k $ Y2 ðsÞ ¼ Km $ I1 ðsÞ > > > > >   > < k $ Y1 ðsÞ þ m2 $ s2 þ c $ s þ k2 þ k $ Y2 ðsÞ ¼ Km $ I2 ðsÞ > > ðL $ s þ RÞ $ I1 ðsÞ þ Ke $ s $ Y1 ðsÞ ¼ V1 ðsÞ > > > > > : ðL $ s þ RÞ $ I2 ðsÞ þ Ke $ s $ Y2 ðsÞ ¼ V2 ðsÞ

(11.126)

The currents I1(s) and I2(s) are expressed from the first two Eq. (11.126) and substituted into the last two Eq. (11.126), which results in:

2

 m1 $ s2 þ c $ s þ k1 þ k $ ðL $ s þ RÞ þ Ke $ s 6 Km 6 6 6 4 k $ ðL $ s þ RÞ  Km ( ) ( ) Y1 ðsÞ V1 ðsÞ $ ¼ Y2 ðsÞ V2 ðsÞ

3 k $ ðL $ s þ RÞ  7 Km 7 7   7 5 m2 $ s2 þ c $ s þ k2 þ k $ ðL $ s þ RÞ þ Ke $ s Km

(11.127)

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

The solution to Eq. (11.127) is {Y(s)} ¼ [Gap(s)]${V(s)} with {Y(s)} ¼ {Y1(s) Y2(s)}T, {V(s)} ¼ {V1(s) V2(s)}T, and the actuation þ plant transfer function matrix [Gap(s)] being the inverse of the matrix in the left-hand side of Eq. (11.127):

31  m1 $ s2 þ c $ s þ k1 þ k $ ðL $ s þ RÞ k $ ðL $ s þ RÞ þ Ke $ s  7 6 7 6 Km Km 7 6 ½Gap ðsÞ ¼6 7   7 6 2 5 4 m2 $ s þ c $ s þ k2 þ k $ ðL $ s þ RÞ k $ ðL $ s þ RÞ  þ Ke $ s Km Km " # Gap11 ðsÞ Gap12 ðsÞ ¼ Gap21 ðsÞ Gap22 ðsÞ (11.128) 2

The symbolic (algebraic) equation of [Gap(s)] is too complex and is not given explicitly here. The prefilters and sensors connect their inputs and outputs as:

vr1 ¼ KH $ r1 ; vr2 ¼ KH $ r2 ; vs1 ¼ KH $ y1 ; vs2 ¼ KH $ y2

(11.129)

Because the sensor/prefilter gain KH is constant, the four transfer functions resulting from Laplace transforming Eq. (11.129) are identical and equal to KH. (ii) Figure 11.38 is the block diagram of the two-input, two-output feedback system modeled at (i). With individual signals identified and connected to scalar transfer functions, this block diagram is the detailed variant of the generic block diagram shown Figure 11.36, which used signal vectors and transfer function matrices. The actuation voltage vector {V(s)} is connected to the error signal vector {E(s)} as:



fVðsÞg ¼ ½Gc ðsÞ $ fEðsÞg or

V1 ðsÞ V2 ðsÞ



¼

Gc1 ðsÞ 0

V (s) s1



0 E1 ðsÞ $ Gc2 ðsÞ E2 ðsÞ (11.130) Y1(s)

KH

_ R1(s)

KH

+

E1(s)

V1(s) Gc1(s)

Gap11(s)

Vr1(s) R2(s)

KH

Y11(s)

+ Vr 2(s)

E2(s)

V2(s)

Gc2(s)

Gap12(s)

_ V1(s) Gap21(s)

Gap22(s) Vs2(s)

KH

+

Y12(s) Y2(s)

+ Y21(s)

V2(s)

Y1(s)

+

+

Y22(s) Y2(s)

FIGURE 11.38 Block Diagram of a Negative-Feedback MIMO Control System With Prefilter.

11.3 Block Diagrams and MIMO Feedback Systems

by means of the diagonal control transfer function matrix [Gc(s)]. The feed-forward transfer function matrix is therefore:

" ½GðsÞ ¼ ½Gap ðsÞ $ ½Gc ðsÞ ¼ " ¼

Gap11 ðsÞ Gap12 ðsÞ

# "

0

0

Gc2 ðsÞ

$

Gap21 ðsÞ Gap22 ðsÞ

#

Gc1 ðsÞ

Gap11 ðsÞ $ Gc1 ðsÞ

Gap12 ðsÞ $ Gc2 ðsÞ

Gap21 ðsÞ $ Gc1 ðsÞ

Gap22 ðsÞ $ Gc2 ðsÞ

#

(11.131) as per Eq. (11.115). The sensed voltage vector {Vs(s)}dwhich is the generic {B(s)} in Figure 11.36dis connected to the output controlled signal vector {Y(s)} as:

 fVs ðsÞg ¼ ½HðsÞ $ fYðsÞg or

Vs1 ðsÞ

¼

Vs2 ðsÞ

KH 0



Y1 ðsÞ $ KH Y2 ðsÞ 0

(11.132)

by means of the diagonal feedback transfer function matrix [H(s)]. Consequently and as per Eq. (11.116), the open-loop transfer function matrix becomes:

½GOL ðsÞ ¼ ½HðsÞ $ ½GðsÞ ¼ KH $



Gap11 ðsÞ $ Gc1 ðsÞ

Gap12 ðsÞ $ Gc2 ðsÞ

Gap21 ðsÞ $ Gc1 ðsÞ

Gap22 ðsÞ $ Gc2 ðsÞ (11.133)

It can be shown that the prefilter transfer function matrix [P(s)] connecting the reference (input) signal vector {R(s)} to the corresponding reference voltage signal vector {Vr(s)} is identical to the feedback transfer function [H(s)] because the individual prefilters are identical to the individual feedback sensors; as a consequence:

 fVr ðsÞg ¼ ½PðsÞ $ fRðsÞg or

Vr1 ðsÞ Vr2 ðsÞ

¼

KH 0



0 R1 ðsÞ $ KH R2 ðsÞ

(11.134)

The closed-loop transfer function matrix of the feedback system analyzed in this Example can now be calculated by means of the definition Eq. (11.120), the feed-forward transfer function matrix of Eq. (11.131), and the prefilter transfer function matrix of Eq. (11.134), which is identical to the feedback transfer function matrix. nn

nn

Example 11.14 Determine the closed-loop transfer function matrix [GCL(s)] for the unity-feedback MIMO system whose block diagram is shown in Figure 11.39. Consider the two proportional controllers are also actuators and are defined as: Gc1(s) ¼ 1, Gc2(s) ¼ 2; the plant transfer functions are Gp11(s) ¼ 1/ (s þ 2), Gp12(s) ¼ Gp21(s) ¼ 3/(s þ 2), and Gp22(s) ¼ 2/(s þ 2).

Solution The block diagram of Figure 11.39 identifies the following transfer function matrices:

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

_ R1(s) +

E1(s)

M1(s) Gc1(s)

C11(s) R2(s) +

E2(s)

M2(s)

Gc2(s)

Gp12(s)

_

+

C12(s) C 2( s)

+

Gp21(s)

C21(s) Gp22(s)

C 1( s)

+

Gp11(s)

+

C22(s)

FIGURE 11.39 Block Diagram of a Negative-Feedback MIMO Control System.

3 1 3 6s þ 2 s þ 27 Gp11 ðsÞ Gp12 ðsÞ 7 6 ½Gp ðsÞ ¼ ¼6 7; 4 Gp12 ðsÞ Gp22 ðsÞ 3 2 5 sþ2 sþ2 " # " # " Gc1 ðsÞ 0 1 0 H1 ðsÞ ½Gc ðsÞ ¼ ; ½HðsÞ ¼ ¼ 0 Gc2 ðsÞ 0 0 2 "

#

2

0 H2 ðsÞ

#

" ¼

1

0

#

0 1 (11.135)

Because [H(s)] is the identity matrix, the feed-forward and the closed-loop transfer function matrices are calculated as per Eqs. (11.115) and (11.118):

3 6 s þ 27 7 7; 7 4 5

2

1 6sþ 2 6 ½GðsÞ ¼ ½Gp ðsÞ $ ½Gc ðsÞ ¼ 6 6 4 3 sþ2

2

sþ2

3 s  12 6s þ 12 6 s2 þ 9s s2 þ 9s 7 6 7 1 7 ½GCL ðsÞ ¼ ð½I þ ½GðsÞÞ $ ½GðsÞ ¼ 6 6 7 4 3s þ 6 4s  6 5 s2 þ 9s s2 þ 9s 3 2 GCL11 ðsÞ GCL12 ðsÞ 5; ¼4 GCL21 ðsÞ GCL22 ðsÞ

(11.136)

nn

Problems

SUMMARY This chapter introduces the main notions, concepts, and modeling tools of feedback control dynamic systems. Using transfer functions interconnected in negativefeedback block diagrams, models are presented for SISO systems. The closedloop transfer function synthesizes the added actions of the controller and feedback and directly connects the input (reference) signal to the output (controlled) signal. Studied is also the sensitivity of SISO systems to parameter variation. MIMO feedback systems are also studied in this chapter and are modeled by means of transfer function matrices connected in block diagrams. Feedback systems with disturbances are analyzed as a specific case of MIMO systems. SISO systems are also modeled by using the state-space approach. The chapter illustrates and models numerous examples of actual physical systems that can operate as individual components in or full feedback control systems.

PROBLEMS 11.1 Convert the control system of Example 11.3 and shown in the block diagram of Figure 11.14(a) into a basic unity-feedback control system as the one shown in Figure 11.12. 11.2 Convert the control system of block diagram of Example 11.2 and Figure 11.13 into a basic nonunity-feedback control system. 11.3 A MIMO feedback control system is connected as in Figure 11.13 of Example 11.2 where [G1(s)], [G2(s)], [G3(s)] are matrices, while {R(s)}, {C(s)} are vectors. Convert this system into the regular feedback system of Figure 11.35(b). 11.4 Convert the system of Problem 11.3 into a unity-feedback system where [H(s)] is the unity matrix in Figure 11.35(b). 11.5 Convert the control system described by the block diagram of Figure 11.40 into a unity-feedback control system as the one illustrated in Figure 11.12. 11.6 Convert the control system of Figure 11.40 into a basic nonunity control system as the generic one of Figure 11.11(a). 11.7 Convert the control system described by the block diagram of Figure 11.41 into a unity-feedback control system as the generic one shown in Figure 11.12. 11.8 Convert the control system of Figure 11.41 into a basic nonunity control system as the generic one of Figure 11.11(a).

587

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

+

R(s)

+

C(s)

G1(s)

G2(s)

_

_

G3(s)

FIGURE 11.40 Block Diagram of a Control System With Embedded Feedback Loop and Prefilter

_ +

R(s)

+

G1(s)

G2(s) C ( s)

_

G3(s)

FIGURE 11.41 Block Diagram of a Control System With Two Feedback Loops and Prefilter.

11.9 Calculate the controlled output C(s) resulting from the feedback system of Figure 11.42 when R(s) ¼ 1/s. R(s) +

+

1 s ⋅ ( s+ 1)

1 s ⋅ ( s+ 3)

2

_

_

C ( s)

1 s s +1 s+ 2

FIGURE 11.42 Block Diagram With Two Feedback Loops.

11.10 Using only basic P, D, and I controllers, as well as summing and pick off points, design the block diagrams to model the controllers defined by Gc1(s) ¼ 1/(a  b$s) and Gc2(s) ¼ (c d$s)/(a  b$s) where a, b, c, and d are real numbers.

Problems

11.11 A unity-feedback control system is defined by the closed-loop transfer function GCL ðsÞ ¼ ðs þ 1Þ=ðs þ 4Þ. Design a block diagram formed of only proportional gains, integrators, derivation operators, summers, and pick off points to represent this transfer function. 11.12 Solve Problem 11.11   for the closed-loop transfer function: GCL ðsÞ ¼ ðs þ 2Þ s2 þ s  3 . 11.13 Solve Problem 11.11 for  the  GCL ðsÞ ¼ s s3 þ 4s2 þ 5s þ 2 .

closed-loop

transfer

function:

11.14 Solve Problem 11.11   for the  closed-loop GCL ðsÞ ¼ s2 þ s þ 1 s3 þ 2s2 þ 3s .

transfer

function:

11.15 Demonstrate that electrical circuit of Figure 11.43 with three cascading operational-amplifier stages and seven identical electrical elements can operate as a summing point in a basic control system when considering the three signals are the voltages Vs(s), Vr(s), and Ve(s). Vr Z Z Vs

– +

Z

Z Z −Vs

– +

Z Z V

– +

Ve

FIGURE 11.43 Electrical System as a Summing Point in a Feedback Control System.

11.16 Disposing of three identical pulleys, two identical translatory-motion springs, and some rigid rods, design a rotary mechanical system to operate as a summing point when considering small rotations. Hint: Use pulley rotation angles as the signals of the summing point. 11.17 Design an active electrical system to operate as P þ I controller. The system comprises two cascading operational-amplifier stages, one inductor, and four identical resistors. Knowing the resistance ranges in the 200e300 U interval and the inductance is within the 1e5 H range, find the electrical components that minimize the controller’s integral time TI. 11.18 Design an active electrical system to operate as a P þ D controller. The system comprises two cascading operational-amplifier stages, one capacitor, and four identical resistors. Knowing the resistance ranges in the 150e250 U interval and the capacitance is within the 3e5 mF range, find the electrical components that minimize the controller’s derivative time Td.

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

11.19 Derive the transfer function of the fluid system of Example 7.3 and shown in Figure 7.4 when considering that the flow rate qi is the input and the flow rate qo is the output. Using basic P, D, I controllers together with summing and junction points, propose a block diagram to model the system’s transfer function. 11.20 Design a P þ I þ D controller formed of the following rotary-motion mechanical elements: a disk of mass moment of inertia J, a damper of damping coefficient cr, and a spring of stiffness kr by defining a transfer function that connects two rotation angles. 11.21 Design a P þ I þ D controller with two operational amplifiers, three identical resistors, and three identical inductors. Determine the values of R and L that will result in a derivative time TD ¼ 0.05 s and an integral time TI ¼ 0.1 s. 11.22 Disposing of two operational amplifiers, four resistors (two of resistance R1 and two of resistance R2), and four inductors (two of inductance L1 and two of inductance L2), design an active lagelead compensator. Express the gain K, the time constants s1, s2 and the constants a1, a2, as well as the relationships between the four electrical elements properties resulting in a lagelead compensator. 11.23 Design a unity-feedback control system entirely composed of mechanical elements where the controller is a proportional one. As shown in Figure 11.44, the plant contains a rotary (torsional) spring of stiffness kr and a cylinder of mass moment of inertia J. The plant’s input and output are the angles qi and qo.

θi kr

J

θo

FIGURE 11.44 Rotary Mechanical Plant With Angular Input and Output.

11.24 Consider the plant is the circuit of Figure 7.7(a). Design a unity-feedback control system entirely composed of electrical elements resulting in a P þ I controller. 11.25 Demonstrate that the passive circuit of Figure 11.45 can be utilized as a lag compensator.

Problems

R1

C

vi

vo R2

FIGURE 11.45 Passive Electrical Network as a Lag Compensator.

11.26 Decide whether the electrical circuit of Figure 7.2 in Example 7.1 can operate as a lead or a lag compensator in terms of its four electrical elements. 11.27 Can the subsystem defined by the transfer function Gc(s) ¼ (s2 þ a$s þ b)/ (s2 þ c$s þ d) operate as a lagelead compensator where a, b, c, d are real parameters? 11.28 A mechanical second-order microplant is controlled by a P þ D unit, resulting in the feed-forward transfer function shown in Figure 11.46. Study the sensitivity of the open-loop transfer function and of the closed-loop transfer function to the gain K. R(s) +

s+3 s + 3s + 2

C(s)

2

_

K s

FIGURE 11.46 Block Diagram of Nonunity-Feedback Control System.

11.29 Calculate and analyze the sensitivity of the closed-loop transfer function of the control system shown in Figure 11.13 of Example 11.2 in terms of K when G1(s) ¼ G3(s) ¼ K and G2(s) ¼ 1/(s2 þ s þ 1). 11.30 Calculate and analyze the sensitivity of the closed-loop transfer function of the control system sketched in Figure 11.40 of Problem 11.5 in terms of K1, K2, and K3 for G1(s) ¼ K1, G2(s) ¼ K2/[s$(s þ 1)$(s þ 2)2] and G3(s) ¼ K3/s. 11.31 Obtain a state-space model for the control system of Problem 11.28 for K ¼ 0.5 and plot its time response c(t) when r(t) ¼ 5. 11.32 Considering K1 ¼ 10, K2 ¼ 2, and K3 ¼ 120 in Problem 11.30, find a statespace model for the control system and plot its time response c(t) for r(t) ¼ 40$sin(110t). 11.33 Compare the time responses of the control systems sketched in Figures 11.34 and 11.47. Use Simulink to plot the time response c(t) for both systems considering that r(t) ¼ 1 and d(t) ¼ 1/(t2 þ 1). Known are Gp(s) ¼ 100/(s2 þ 5s þ 100), Gc(s) ¼ 0.02/s, and H(s) ¼ 1.

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CHAPTER 11 Block Diagrams and Feedback Control System Modeling

D(s) R(s)

+

+ Gc(s)

+

C(s)

Gp(s)

_ H(s)

FIGURE 11.47 Block Diagram of a Nonunity-Feedback Control System With Disturbance.

11.34 A two-input, two-output unity-feedback control system similar to the one depicted in Figure 11.39 is defined by the following control and plant transfer function matrices: 3 2 3 2 s 2 100 6 s2 þ 2s þ 10 s2 þ 2s þ 10 7 0 7 6 20 þ s 7; ½Gc ðsÞ ¼ 4 5; ½Gp ðsÞ ¼ 6 5 4 2 sþ1 0 10 þ 0:02s 2 2 s þ 2s þ 10 s þ 2s þ 10 Calculate the closed-loop transfer function matrix and plot the system response for a reference input {r(t)} ¼ {1 1}T. 11.35 The control and plant transfer functions of Problem 11.34 are combined in a nonunity-feedback control system as the general one sketched in Figure 11.36. For identical prefilter and feedback transfer function matrices

1200 0 ½PðsÞ ¼ ½HðsÞ ¼ , calculate the closed-loop transfer 0 1200 function matrix of this system and plot the time response for {r(t)} ¼ {1 1/ (t þ 1)}T.

Suggested Reading W.R. Evans, Control-System Dynamics, McGraw-Hill, New York, 1954. N.S. Nise, Control Systems Engineering, 7th Ed. Wiley, Hoboken, NJ, 2015. K. Ogata, Modern Control Engineering, 4th Ed. Prentice Hall, Pearson, Upper Saddle River, NJ, 2009. G.F. Franklin, J.D. Powell, and A. Emami-Naeini, Feedback Control of Dynamic Systems, 6th Ed. Pearson, Upper Saddle River, NJ, 2010. R.T. Stefani, B. Shahian, C.J. Savant, and G.H. Hostetter, Design of Feedback Control Systems, 4th Ed. Oxford University Press, New York, 2002. C.H. Phillips and J.M. Parr, Feedback Control Systems, 5th Ed. Prentice Hall, Upper Saddle River, 2011. R.C. Dorf and R.H. Bishop, Modern Control Systems, 13th Ed. Prentice Hall, Upper Saddle River, 2016.

CHAPTER

Stability of Feedback Control Systems

12

OBJECTIVES Based on the introductory control’s material of Chapter 11, this chapter studies the stability of feedback control systems and discusses the design of stable control dynamic systems. The following topics are presented here: • Definition of stable, marginally stable, and unstable feedback control systems through transfer function and state-space models. • Stability study of single-input, single-output (SISO) and multiple-input, multipleoutput (MIMO) control systems by analytical or MATLAB study of the closed-loop poles location in the complex plane. • Analysis of stability of SISO and MIMO control systems by the RoutheHurwitz criterion. • Design of stable SISO and MIMO control systems with variable parameters by means of the RoutheHurwitz criterion. • Analysis of stability of feedback systems with variable gain using the root locus method, the Nyquist criterion, Bode plots, and the phase/gain margin.

INTRODUCTION This chapter focuses on the stability of feedback control systems. It utilizes the transfer function modeling and an approach similar to the one of Chapter 7, which introduces the concept of stability applied to open-loop (uncontrolled) dynamic systems. The closed-loop transfer function is employed for SISO systems, and the closedloop transfer function matrix is used for MIMO (multivariable) systems to determine the pole position in the complex plane and to establish stable, marginally stable, or unstable behavior of feedback systems. The state-space model is also applied to determine the stability behavior of dynamic control systems. Analytical or MATLAB methods can be used to solve the characteristic equation related to the closed-loop transfer function or the closed-loop transfer function matrix when such solutions are available. The RoutheHurwitz test or criterion is the analytical tool of choice for designing control systems with parameters, such as gains, that are originally undetermined. The method is illustrated by several examples. The System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00012-9 Copyright © 2018 Elsevier Inc. All rights reserved.

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root locus method and the Nyquist criterion are introduced here to monitor the closed-loop pole migration between the left-hand complex plane (LHP) and the right-hand complex plane (RHP) for feedback system with a variable gain by analyzing the poles and zeros of the open-loop transfer function. For the same purpose, the Bode plots are utilized, as well as the phase and/or gain margins.

12.1 CONCEPT OF STABILITY APPLIED TO FEEDBACK CONTROL SYSTEMS Together with transient response characteristics and steady-state errors (which are studied in Chapter 13), stability is a fundamental topic (probably the most important) in control system’s analysis and design. If a system is unstable, it is pointless to analyze the other two categoriesdthe transient response and the steady-state errors. The concept of stability is introduced in Chapter 7 in connection with the bounded/ unbounded time-domain response y(t) of SISO uncontrolled (open-loop) dynamic systems. Similarly, the stability of both SISO- and MIMO-controlled (feedback) dynamic systems is discussed in this chapter by analyzing the controlled time-domain response c(t) for SISO systems and {c(t)} for MIMO systems. Therefore, a stable feedback control system will display a response that does not grow unboundedly for any specific bounded input r(t) or {r(t)}.

12.1.1 SISO Systems The total response of a SISO feedback control system is formed of the natural and forced responses; this is expressed in the time domain as c(t) ¼ cn(t) þ cf (t) and in the Laplace domain as C(s) ¼ Cn(s) þ Cf (s). The forced response cf (t) results from the action of the reference (input) signal r(t) on the system, whereas the natural response cn(t) is determined by the system’s closed-loop transfer function GCL(s), which reflects features inherent to the feedback control system. For a SISO feedback system with the closed-loop transfer function GCL(s) ¼ NGL(s)/DGL(s) and the Laplace-domain input signal R(s) ¼ NR(s)/DR(s)dwhere the numerators NGL(s), NR(s) and denominators DGL(s), DR(s) are s-polynomialsdthe controlled output C(s) is calculated as: 0 1 CðsÞ ¼ GCL ðsÞ $ RðsÞ ¼ Cn ðsÞ þ Cf ðsÞ ¼ ¼

X

g X i¼1

GCL;ij ðsÞ þ Cf ðsÞ

@

gi X

aij

j¼1

ðs  pi Þj

A þ Cf ðsÞ (12.1)

i;j

The closed-loop transfer function of Eq. (12.1) is assumed to have g different poles pi (roots of the denominator DGL(s)) of order of multiplicity gi and is expanded in partial-fraction form with GCL,ij(s) being one such generic (arbitrary) partial fraction.

12.1 Concept of Stability Applied to Feedback Control Systems

Consider that a unit step r(t) ¼ 1 is applied to this control system; the Laplace transform of the response is: 1 X k (12.2) CðsÞ ¼ Cn ðsÞ þ Cf ðsÞ ¼ GCL ðsÞ $ ¼ GCL;ij ðsÞ þ s s i;j where k is a constant, and k/s represents the forced response in the Laplace domain. By inverse Laplace transforming Cf(s) of Eq. (12.2) it follows that the forced response is cf(t) ¼ k, which is a step function, and therefore the forced response is finite (bounded). It depends on the Laplace transforms of the components GCL,ij(s), whether the natural (and therefore total) response is finite (bounded) or grows to infinity with time, which will generate an unbounded response that corresponds to an unstable control system. As discussed in Chapter 7 and illustrated in Figure 7.17, SISO uncontrolled (open-loop) dynamic systems can be stable, marginally stable, or unstable depending on the transfer function poles position in the complex plane and the pole order or multiplicity. Similarly, the stability of controlled systems is established by analyzing the position of the poles of the closed-loop transfer function GCL(s)dnamed closedloop polesdin the complex plane. In terms of stability behavior, the following feedback control systems are encountered: • •

•

Stable (also known as absolutely stable) systems: they have all the closed-loop poles, either simple or multiple, situated in the LHP. Marginally stable systems: they have simple closed-loop poles that are placed on the imaginary axis of the complex plane. In addition, these systems might have closed-loop poles in the LHP. Unstable systems: they have one or more closed-loop poles: • Either simple or multiple and located in the RHP. • Multiple and situated on the imaginary axis of the complex plane.

Closed-Loop Poles at the Origin Consider a closed-loop transfer function defined as follows: NCL ðsÞ n s $ DCL ðsÞ and DCL ðsÞ is

GCL ðsÞ ¼

(12.3)

where n is an integer larger than zero, a factor polynomial of the denominator. The right-hand side of Eq. (12.3) is expanded in partial fractions. For n ¼ 1, the fraction k/s (k is a constant) results from the factor s1 in the denominator. Its inverse Laplace transform is k, so the effect of a simple closed-loop pole at the origin is a constant, which indicates the corresponding time response is constant. For n  2, the fractions k1/s, k2/s2, ., kn/sn are obtained from the sn factor. While the time-domain counterpart of k1/s is the constant k1 (resulting in a bounded time response), the subsequent fractions result in the time-domain terms k2$t, k2 $ t; k3 $ t2 ; . ; kn $ tn1 , which are all unbounded.

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CHAPTER 12 Stability of Feedback Control Systems

As a conclusion: • •

Simple closed-loop poles at the origin lead to (bounded) stable time-domain response. Multiple closed-loop poles at the origin generate (unbounded) unstable timedomain response.

MATLAB can be used to plot the closed-loop pole position in the complex by using either pzmap or pzplot commands applied to the closed-loop transfer function. These maps locate the closed-loop poles in the complex plane, which establishes the stability character of a feedback system. However, these two MATLAB commands generate results that can be misleading when multiple poles are located on the imaginary axis, as there is no direct graphical way of indicating the order of multiplicity through these commands, and therefore such roots can be either simple or multiple, and the system can be either marginally stable or unstable. To avoid this drawback, MATLAB has the command pole, which, as mentioned in Chapter 7, identifies all the closed-loop poles. nn

Example 12.1 Use MATLAB commands to decide the stability behavior of the controlled systems that are defined by the following closed-loop transfer functions: (i) GCL ðsÞ ¼ s6 þ 2s 5 þ 8s4 þ 8s13 þ 20s2 þ 8s þ 16 þ 5s  3s  1 (ii) GCL ðsÞ ¼ s7 þ 12s 6 þ 51s5 þ2s100s 4 þ 146s 3 þ 216s 2 þ 96s þ 128 3

2

Solution (i) The following MATLAB code: >> g = tf(1, [1, 2, 8, 8, 20, 8, 16]); >> pzmap(g) yields the plot of Figure 12.1, where the closed-loop poles are graphically identified by the symbol “x.” The sixth-degree characteristic polynomial has six roots (closed-loop poles), but the plot of Figure 12.1 only comprises four, which indicates that two poles are multiple. To identify the multiple poles, as well as the simple ones, the command pole(g) is added to the sequence above and MATLAB returns: ans = -1.0000 + 1.7321i -1.0000 - 1.7321i 0.0000 + 1.4142i 0.0000 - 1.4142i -0.0000 + 1.4142i -0.0000 - 1.4142i

12.1 Concept of Stability Applied to Feedback Control Systems

FIGURE 12.1 Plot of Closed-Loop Poles in the Complex Plane. which shows the imaginary pole is of the order 2 of multiplicity, and therefore this system is unstable. (ii) There are instances where a polynomial, such as the one in the denominator of a closed-loop transfer function, can be factored outdthat will allow rapid identification of the nature of that system’s stability. The function factor(sys) can be used as follows in this example: >> syms s >> factor(s^7+12*s^6+51*s^5+100*s^4+146*s^3+216*s^2+96*s+128) which returns: ans = [s^2 + 2, s^2 + 1, s + 4, s + 4, s + 4] The result above shows that there is a real negative closed-loop pole of the order 3 of multiplicity (p ¼ 4) and four simple imaginary poles (p ¼ ej, p ¼ jdthey result from solving the equation s2 þ 1 ¼ 0 and p ¼ 1.41$j, p ¼ 1.41$jdthe latter two poles are the solution to the equation s2 þ 2 ¼ 0); this system is therefore marginally stable. nn

MATLAB enables finding and/or plotting the closed-loop poles by specifying the feed-forward transfer function G(s) and feedback transfer function H(s)dbased on a typical control system such as the one illustrated in Figure 11.11(a). The command feedback(sys1, sys2) defines a linear time invariant (LTI) object, where sys1 is G(s) and sys2 is H(s); this LTI object is actually the closed-loop transfer function GCL(s) corresponding to G(s) and H(s). Let us analyze an example.

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CHAPTER 12 Stability of Feedback Control Systems

nn

Example 12.2 A feedback control system is formed of a unit-gain integral controller, a mechanical filter microsystem (plant), which is formed of two shuttle masses, and a connecting micro spring, with one mass being subjected to viscous damping and connected to another micro spring to the substrate. When a force is applied to the free shuttle mass, the plant transfer function connecting the displacement of the other mass to the force is

Gp ðsÞ ¼

2s4

þ

3s3

1 þ 15s2 þ 200s þ 9

This control system is also provided with a feedback sensor whose transfer function is HðsÞ ¼ s þ 5. Establish the stability character of the system using MATLAB.

Solution The feed-forward transfer function is the product of the unit-gain integral controller transfer function Gc(s) ¼ 1/s and the plant transfer function Gp(s). MATLAB can operate on transfer function LTI objects by multiplying them, so that G(s) does not need to be obtained separately. The following MATLAB code can be used: >> sys1=tf(1,[1,0])*tf(1,[2,3,15,200,9]); % G(s) is obtained as the . product of G_c(s) and G_p(s) >> sys2=tf([1,5],1); >> pole(feedback(sys1,sys2)) which returns: ans = 1.5648 + 4.3924i 1.5648 - 4.3924i -4.5812 + 0.0000i -0.0242 + 0.1566i -0.0242 - 0.1566i The system is unstable as it has a pair of complex conjugated closed-loop poles in the RHP.

nn

Feedback control can render an originally unstable plant into a stable control system, as illustrated by the following example. nn

Example 12.3

A plant is defined by the transfer function Gp(s) ¼ 1/(s2 þ s  2). Study the possibility of obtaining a stable system by connecting a P (proportional) controller to the plant in a unity-feedback control system.

Solution The plant is an unstable system because it has a positive pole (þ1) in addition to a negative pole (2). Adding a proportional controller defined by Gc(s) ¼ KP to the plant in a unity-feedback control structure results in the following closed-loop transfer function:

  K P s2 þ s  2 Gc ðsÞ $ Gp ðsÞ GðsÞ ¼ ¼ GCL ðsÞ ¼ 1 þ GðsÞ 1 þ Gc ðsÞ $ Gp ðsÞ 1 þ KP =ðs2 þ s  2Þ KP ¼ 2 s þ s þ KP  2

(12.4)

12.1 Concept of Stability Applied to Feedback Control Systems

The closed-loop poles (roots of the characteristic equation) resulting from Eq. (12.4) are:

p1;2 ¼ 0:5  0:5 $

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  4ðKP  2Þ

(12.5)

When the quantity under the square root is a positive number, i.e., 1 e 4(KP  2)  0 or KP  2.25, the two closed-loop poles of Eq. (12.5) are real numbers. They are both in the LHP (they are both negative) when 1  4(KP  2) < 1, which means KP > 2. Therefore, the control system is stable, and the closed-loop transfer function has two real poles in the LHP when 2 < KP < 2.25. Conversely, when 1  4(KP  2) < 0 or KP > 2.25, the two closed-loop poles of Eq. (12.5) are complex conjugated. Because their real part is negative, they are both in the LHP, and the control system is, again, stable. Combining the two possible situations described above, we conclude that the system is stable for proportional gains that are larger than 2, which is KP > 2. nn

Note: It is also possible that an originally stable system becomes unstable when incorporated in a feedback control loop in conjunction with certain controllers. For instance, the plant defined by Gp(s) ¼ 1/(s þ 1) is stable. Including the plant in a unity-feedback control system alongside the controller Gc(s) ¼ 1/s3, results in the closed-loop transfer function GCL(s) ¼ 1/(s4 þ s3 þ 1). Two of the closed-loop poles, namely: 0.5189  0.6666j, are in the RHP, and, therefore, the feedback system is unstable.

12.1.2 MIMO Systems The closed-loop transfer function matrix of a MIMO control system is given in Eqs. (11.119) and (11.115) for a nonunity feedback, namely:  1  1 1 ½GCL ðsÞ ¼ ½GðsÞ1 þ ½H ðsÞ ¼ Gp ðsÞ $ ½Ga ðsÞ $ ½Gc ðsÞ þ ½ H ðsÞ (12.6) For a unity-feedback system [GCL(s)] is obtained from Eq. (12.6) taking into account that the feedback transfer function matrix is the identity matrix: [H(s)] ¼ [I]. The closed-loop transfer function matrix of Eq. (12.6) is calculated based on the following adjoint matrix and determinant:   adj ½GðsÞ1 þ ½HðsÞ   ½GCL ðsÞ ¼ (12.7) det ½GðsÞ1 þ ½HðsÞ As a consequence, the closed-loop poles are found by solving the characteristic equation of Eq. (12.7), which is:   (12.8) det ½GðsÞ1 þ ½HðsÞ ¼ 0

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CHAPTER 12 Stability of Feedback Control Systems

nn

Example 12.4 A unity-feedback MIMO control system is formed of a proportional controller and a plant whose trans2 3 1 2s þ 1

 6 s 2 þ 3s þ 2 s 2 þ 3s þ 2 7 1 0 6 7 fer function matrices are ½Gc ðsÞ ¼ ; ½Gp ðsÞ ¼ 6 7. Analyze the 4 2s þ 1 5 0 5 s1 s 2 þ 3s þ 2 s 2 þ 3s þ 2 stability of this system.

Solution The following MATLAB code is used here: >> >> >> >> >> >>

syms s gc=[1,0;0,5]; den=s^2+3*s+2; gp=[1/den,(2*s+1)/den;(2*s+1)/den,(s-1)/den]; g=gp*gc; det(inv(g)+eye(2))% eye(2) is the 2x2 [H]=[I] identity matrix

The last command returns the determinant expression of Eq. (12.8), which is also the characteristic expression:

  s4  11s3  4s2 þ 5s þ 14 det ½GðsÞ1 þ ½I ¼ 5ð4s2 þ 3s þ 2Þ

(12.9)

Solving the characteristic equation corresponding to Eq. (12.9)dnumerator equals zerodresults in the following closed-loop poles (roots of the characteristic equation): - 10.5895 + 0.0000i 1.0697 + 0.0000i - 0.7401 + 0.8296i - 0.7401 - 0.8296i which are obtained using, for instance, the MATLAB command: >> roots([-1,-11,-4,5,14]) Because of the real positive root, the system is unstable.

nn

State-Space Stability As discussed in Chapter 8, a transfer function matrix can be obtained from a statespace model. The closed-loop transfer function matrix is expressed in terms of the state-space matrices [A], [B], [C], and [D] as in Eq. (8.14): ½GCL ðsÞ ¼ ½C  $ ðs $ ½I   ½ AÞ1 $ ½B þ ½D

(12.10)

12.1 Concept of Stability Applied to Feedback Control Systems

The closed-loop transfer-function matrix [GCL(s)] of Eq. (12.10) can be written as: ½GCL ðsÞ ¼ ½C $ ¼

adjðs $ ½I  ½AÞ $ ½B þ ½D detðs $ ½I  ½AÞ

1 $ ð½C $ adjðs $ ½I  ½AÞ $ ½B þ detðs $ ½I  ½AÞ $ ½DÞ detðs $ ½I  ½AÞ (12.11)

which indicates that the closed-loop poles are the roots of the characteristic equation: detðs $ ½I  ½AÞ ¼ 0

(12.12)

At the same time, the eigenvalues l of a square matrix [A] are connected to the corresponding eigenvectors {y} as ðl $ ½I  ½AÞ $ fyg ¼ 0

(12.13)

Eq. (12.13) is the concise form of a system of homogeneous equations with the unknown being the components of the eigenvector {y}dits solution is nontrivial when the system determinant is zero: detðl $ ½I  ½AÞ ¼ 0

(12.14)

Comparing Eqs. (12.12) and (12.14) shows that the closed-loop poles of the feedback system are identical to the eigenvalues of the state matrix [A] that corresponds to the state-space model of the same system. As a consequence, the stability of a feedback system, which is formulated in state space, can be studied by solving for or discussing the nature of matrix [A] eigenvalues as per Eq. (12.14). nn

Example 12.5 Analyze the stability of a feedback control system modeled in state-space form whose state matrix is 2 3 1 0 2 6 7 ½A ¼ 4 3 5 1 5. 1 4 1

Solution According to Eq. (12.14), the eigenvalue equation is:

detðl $ ½I  ½AÞ ¼ l3 þ 3l2  15l þ 23

(12.15)

whose roots are 6.086, 1.543  1.183$j. These eigenvalues are also obtained using the MATLAB code: >> a=[1,0,2;-3,-5,1;1,4,1]; >> eig(a) Because the eigenvalues are also the closed-loop poles of a state-space system and because two eigenvalues are positive, it follows that two closed-loop poles are located in the RHP and one in the nn LHPdas a consequence, this system is unstable.

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CHAPTER 12 Stability of Feedback Control Systems

12.2 THE ROUTHeHURWITZ STABILITY TEST The RoutheHurwitz stability test (or array) offers the means of verifying whether a system is stable, marginally stable, or unstable by predicting the number of closedloop poles in the LHP, on the imaginary axis, and in the RHP. The method, which does not calculate the closed-loop poles, is particularly useful in designing stable feedback systems whose closed-loop transfer function contains at least one unknown parameter, such as a gain K. A unity-feedback control system with a feed-forward transfer function: GðsÞ ¼

a4

$ s4

þ a3

$ s3

K þ a2 $ s 2 þ a1 $ s þ a 0

(12.16)

has a closed-loop transfer function: GCL ðsÞ ¼

GðsÞ K ¼ 1 þ GðsÞ a4 $ s4 þ a3 $ s3 þ a2 $ s2 þ a1 $ s þ a0 þ K

(12.17)

The poles of the closed-loop transfer function of Eq. (12.17) are the roots of the characteristic equation a4$s4 þ a3$s3 þ a2$s2 þ a1$s þ a0 þ K ¼ 0. Normally, they cannot be found analytically, and the stability features cannot be established because the parameter K is unknown. The RoutheHurwitz test (also known as criterion or theorem) is a method capable of determining the stability of such a system. The RoutheHurwitz theorem, as well as several of its consequences, are not demonstrated here, that being beyond the introductory scope of the current text. The interested reader can find more on the RoutheHurwitz in Chen (1984), Matsumoto (2001), or Ho et al., (1998). The RoutheHurwitz test comprises two steps: (1) construction of the RoutheHurwitz array and (2) interpretation of the array and deriving conclusions on system stability. Step (1) oftentimes needs some additional calculations, particularly in two special cases where normal construction of the array becomes impossible. Essential in the interpretation of the RoutheHurwitz array step (2) are the elements in the first column, as the number of closed-loop poles that lie in the RHP is equal to the number of changes in the sign of coefficients of the first column. When there are no sign changes in the first column, there is no closed-loop pole in the RHP, and the system is stable or marginally stable.

12.2.1 Construction of the RoutheHurwitz Array Let us consider a particular characteristic polynomial corresponding to the closedloop transfer function, although the method is applicable to any characteristic polynomial: DCL ðsÞ ¼ a5 $ s5 þ a4 $ s4 þ a3 $ s3 þ a2 $ s2 þ a1 $ s þ a0

(12.18)

12.2 The RoutheHurwitz Stability Test

The following Table 12.1 is the related RoutheHurwitz array: Table 12.1 RoutheHurwitz Array of a Fifth-Degree Characteristic Polynomial Column 1 Row 1

s5

Row 2

s4 s3

s2

a5

A1= −

B1 = −

s1 C1 = −

s0

Zero Column

D1= −

a3

a4

a2

a5 a 3 a4 a2

a5 a1 a4 a0

a4 a4 a2 A1 A2 A1 A1 A2 B1 B2 B1

A2 = −

B2 = −

C2 = −

a4 a4 a0 A1 0 A1

A1 0 B1 0 B1

A3= −

B3 = −

a1

0

a0

0

a5 0 a4 0 a4 a4 0 A1 0 A1

=0 0

=0

=0

B1 B2 C1 0 B1

As can be seen, the RoutheHurwitz array is built by following some simple rules, namely: • • •

•

The first row contains every other coefficient of the characteristic polynomial starting with the one of the largest exponent (power) of s. The second row contains every other coefficient of the characteristic polynomial starting with the one of the second largest exponent of s. All other rows are formulated by using data from the two rows immediately above them as follows: they are all negative ratios of an order-2 determinant to the first-column number in the row immediately above; the first column of the determinant is formed of the first-column elements in the rows immediately above, whereas the second column is formed of the elements on the column to the right that belong to the rows immediately above. Note that all determinants in a row have the same first column, and the divisor (pivot) in any given row is also the same. An additional last column with zero elements in the first three rows is used to enable calculation of the last determinant of the third and fourth rows.

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CHAPTER 12 Stability of Feedback Control Systems

•

•

Inspection of Table 12.1 indicates that the last row contains one calculated element in its first column, the first to the last row has two calculated elements, of which the one in the second column is zero, whereas the row above this last row has three calculated elements, of which the third one (in the third column) is zero. If a positive factor is found in any given row, that factor can be eliminated and the resulting (smaller) quotient numbers can be used instead of the original numbers in that row. Let us study a concrete example.

nn

Example 12.6 Analyze the stability of a unity-feedback control system, which is formed of a proportional controller, a second-order electrical system, and a second-order electrical filter whose aggregate feed-forward transfer function is G(s) ¼ 1/(s4 þ 4s3 þ 10s2 þ 12s þ 4). Use the RoutheHurwitz test and MATLAB.

Solution For a unity-feedback control system with the feed-forward transfer function of this example, the

GðsÞ 1 ¼ s4 þ4s3 þ10s closed-loop transfer function is GCL ðsÞ ¼ 1þGðsÞ 2 þ12sþ5. The RoutheHurwitz array corre-

sponding to the given closed-loop transfer function is calculated below:

s4 s3 s2

s1

s0

1 4

10 12

5 0

0 0

1 10 1 5 1 0 4 12 4 0 4 0 = 0 = 5 − =7 − − 4 4 4 4 12 4 0 7 5 7 0 = 0 = 9.1 4 − − 7 7 7 5 9.1 4 0 =5 − 9.14

The elements in the first column are all positive, so there is no sign change. According to the RoutheHurwitz criterion, this means there are no closed-loop poles in the RHP, and therefore the system is stable. The MATLAB command roots is used to directly obtain the closed-loop poles as: >> roots([1,4,10,12,5]) ans = -1.0000 + 2.0000i -1.0000 - 2.0000i

12.2 The RoutheHurwitz Stability Test

-1.0000 + 0.0000i -1.0000 - 0.0000i It can be seen that all roots of the characteristic polynomial have negative real parts, and therefore the system is confirmed to be stable. nn

12.2.2 Excepted Cases There are two particular cases where it is impossible to continue past a certain point with the RoutheHurwitz array normal construction. Obviously when a zero element results in the first column, the formulation cannot continue because the elements on the next row cannot be computed (due to division by a zero pivot). A second special case is when all elements of a row are zero. Both cases are discussed next.

Zero Element in First Column Two methods are analyzed here enabling continuation of the RoutheHurwitz array formation in case a zero results in the first column (but not all elements in the respective row are zero)dwhich would normally halt further generation of the array. One is the method of the polynomial with reciprocal roots and the other one is the socalled method of epsilon.

Method of the Polynomial With Reciprocal Roots This method is based on two elementary features of the polynomial with reciprocal roots, namely: 1. The roots of the original polynomial and their reciprocals (which are roots of another polynomial, named polynomial with reciprocal roots) are distributed identically in the complex plane, i.e., if a root of the original polynomial is in the LHP, the corresponding reciprocal root is also in the LHP. Consider the root of the characteristic polynomial (or closed-loop pole) p ¼ s þ u$j has a positive real part s > 0. The reciprocal of this root, which is 1/p ¼ s/(s2 þ u2)  u/ (s2 þ u2)$j, also has a positive real part, which is s/(s2 þ u2), and so both p and 1/p are placed in the RHP. A similar reasoning shows that if p is in the LHP, 1/p is also in the LHP. 2. The polynomial with reciprocal roots is obtained from the original polynomial by using the coefficients in reversed order. Consider that p is a nonzero root of the characteristic polynomial: DCL(s) ¼ an$sn þ ane1$sn1 þ . þ a1$s þ a0. The polynomial that results from DCL(s) and has its coefficients in reversed order is DCL ðsÞ ¼ a0$sn þ a1$sn1 þ . þ ane1$sþ an. The value of this polynomial for s ¼ 1/p is DCL ð1=pÞ ¼ a0/pn þ a1/pn1 þ . þ ane1/p þ an ¼ (an$pn þ ane1$pn1 þ . þ a1$p þ a0)/pn ¼ DCL(p)/pn ¼ 0 because p is a root of DCL(s). This proves the statement (2). When a zero element is produced in the first column (but not all elements in the respective row are zero), the polynomial with reversed coefficients can be used

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CHAPTER 12 Stability of Feedback Control Systems

instead of the original characteristic polynomial from the RoutheHurwitz array. Let us consider the following example to illustrate the method. nn

Example 12.7 Study the stability of a feedback control system whose closed-loop transfer function has the following characteristic polynomial: DCL ðsÞ ¼ s 4 þ 2s 3 þ s 2 þ 2s þ 3. Verify the results with MATLAB.

Solution The RoutheHurwitz array corresponding to the given closed-loop transfer function is calculated down to the third row where a zero element results in the first column:

s4

1

1

3

s

3

2

2

0

s

2

0

3

0

At this stage, the array generation is stopped because the elements of the fourth row cannot be calculated due to the pivot (the first element of the previous row) being zero. The polynomial with reversed coefficients is formed from the original DCL(s) as:

DCL ðsÞ ¼ 3s4 þ 2s3 þ s2 þ 2s þ 1

(12.19)

and the corresponding RoutheHurwitz array becomes

s4 s3

3 2

1 2

1 0

s2 s1

2 3

1 0

0

s0

1

There are two sign changes in the first column of the array (from 2 to 2 and from 2 to 3), and therefore there are two closed-loop poles in the RHP, which indicates the control system is unstable. The same conclusion is reached using MATLAB’s command: >> roots([3,2,1,2,1]) which renders: ans = 0.3245 + 0.7913i 0.3245 - 0.7913i -0.6578 + 0.1515i -0.6578 - 0.1515i showing there are two roots with positive real parts that are in the RHP.

nn

12.2 The RoutheHurwitz Stability Test

Method of Epsilon The method of epsilon (ε) uses a small real number ε instead of the 0 in the first column and continues building the RoutheHurwitz array with ε. At the end, limits are calculated of the elements in the first column of the array for ε / 0, and conclusions are derived with respect to system stability. Let us study the previous example by means of the epsilon method. nn

Example 12.8 Study the stability of a feedback control system of Example 12.7 by using the method of epsilon.

Solution

Instead of 0 in the third line and first column of the array developed in Example 12.7, ε is used; and the element calculation is completed as shown below:

s4

1

1

3

s

3

2

2

0

s2

ε 2ε  6 ε

3

0

s1 s0

0

3

The following limit calculations are performed for the elements in the first column of the array:

ε > 0; ε/0

ε < 0; ε/0

1

þ

þ

2

þ

þ

ε 2ε  6 ε 3

þ





þ

þ

þ

As the array above indicates, for small ε, either positive or negative, there are two sign changes in the first column, which indicates there are two closed-loop poles in the RHP, so the system is unstable. This was also the conclusion of Example 12.7, where a different solution method has been applied. Note: It is customary to consider only the case with ε > 0 to evaluate the signs of the elements in the array’s first column. nn

All Elements in a Row Are Zero When all elements in a row are zero, the method of the auxiliary polynomial and the method of epsilon (an extension of the method with the same name that was used to circumvent the excepted case with a zero element in the first column) can be utilized. We will cover the auxiliary polynomial method only in this section.

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CHAPTER 12 Stability of Feedback Control Systems

Method of the Auxiliary Polynomial This method is presented and described hereda demonstration of it can be found in Chen (1984), for instance. According to this method, an auxiliary polynomial P(s) is selected that is formed with the coefficients of the previous (nonzero) row, which are made to correspond to the respective powers of s. The derivative in terms of s is taken of the auxiliary polynomial, and the resulting coefficients are used instead of the original zero-row coefficients, after which the formation of the Routhe Hurwitz array continues in the normal way. Let us consider the following example: nn

Example 12.9 Establish the stability character of the feedback control system whose closed-loop denominator is the polynomial DCL ðsÞ ¼ s 4 þ 4s 3  s 2  16s  12. Use the RoutheHurwitz test and check the result with MATLAB.

Solution The following is the RoutheHurwitz array for the given closed-loop characteristic polynomial:

s4

1

1

12

s

3

1

4

0

s s1

1 0

4 0

0

2

In the row of s3, the actual coefficients are 4, e16, and 0, so that 4 has been factored from that row and eliminated. A similar operation has been applied to the coefficients resulting in the row corresponding to s2, where the factor 3 has been eliminated. It can be seen that the row corresponding to s1 comprises only zeros. The auxiliary polynomial P(s) is formed with the coefficients of the row that is located above the zero row. The polynomial P(s) and its s-derivative are:

PðsÞ ¼ s2  4; dPðsÞ=ds ¼ 2 $ s þ 0

(12.20)

As a consequence, the coefficients that are used in the row corresponding to s instead of the zeros are 2 (corresponding to the largest coefficient in the derivative of P(s)) and 0, and the array is now completed as: 1

s4 s3

1 1

1 4

s2

1

4

s s0

2 4

0

1

12 0

There is one sign change in the first column of the array, and therefore the system is unstable, having one closed-loop pole in the RHP (on the real axis) and three closed-loop poles in the LHP, this can easily be checked with MATLAB, which yields the following roots of the original characteristic polynomial: 3, 2, 1, and 2. nn

12.2 The RoutheHurwitz Stability Test

nn

Example 12.10 Using the RoutheHurwitz test, determine the stability behavior of the unity-feedback control system whose feed-forward transfer function is GðsÞ ¼ s 41þ3.

Solution Generally, the feed-forward transfer function can be written as the ratio of two polynomials, i.e., G(s) ¼ NG(s)/DG(s). As a consequence, the closed transfer function of a unity-feedback transfer function becomes

GCL ðsÞ ¼

GðsÞ NG ðsÞ ¼ 1 þ GðsÞ DG ðsÞ þ NG ðsÞ

(12.21)

which indicates that the closed-loop poles are the roots of the characteristic equation:

DCL ðsÞ ¼ DG ðsÞ þ NG ðsÞ ¼ 0

(12.22)

In this example, the characteristic polynomial is

DCL ðsÞ ¼ DG ðsÞ þ NG ðsÞ ¼ s4 þ 3 þ 1 ¼ s4 þ 4

(12.23)

The roots of this polynomial are 1  j and 1  j, so the system has two closed-loop poles in the RHP and is unstable. The other closed-loop poles, as shown above, are in the LHP. The corresponding RoutheHurwitz array is shown below

s4 s3

1 0

0 4 0 0

The row of zeroes in this array requires using the auxiliary polynomial P(s) formed with the elements of the row situated immediately above the zero row. P(s) and its s-derivative are:

PðsÞ ¼ 1 $ s4 þ 0 $ s2 þ 4 ¼ s4 þ 4; dPðsÞ=ds ¼ 4 $ s3 þ 0 $ s

(12.24)

The characteristic polynomial and the auxiliary polynomial are identical and, as a consequence, the closed-loop poles are identical to the roots of the auxiliary polynomial. The coefficients 4 and 0 are therefore used in the modified second row instead of the original zeros, and the resulting RoutheHurwitz array becomes

s4 s3

1 4

0 0

4 0

s2 s1

ε 16=ε

4 0

0

s0

4

Note that the element of the first column in the row corresponding to s2 is originally zero. Using the method of ε, the RoutheHurwitz array shown above is obtained. For either ε > 0 or ε < 0, there are two sign changes in the first column of the array. As a result, there are two closed-loop poles in the RHP and two closed-loop poles in the LHP. The system is unstable, as also shown through direct calculation. nn

A row with zero elements in the RoutheHurwitz array indicates that the original polynomial is divisible by the polynomial formed of the coefficients in the row immediately above the row of zeroes, which is the auxiliary polynomial. As a

609

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CHAPTER 12 Stability of Feedback Control Systems

consequence, the roots that are involved with the rows including and below the first row above the original row of zeros are the roots of both the auxiliary polynomial P(s) and the original characteristic polynomial DCL(s). The rows above the row that is situated immediately above the row of zeros are related to the quotient polynomial resulting from the division of DCL(s) by P(s). In Example 12.9, for instance, the roots of the auxiliary polynomial of Eq. (12.20) are e 2 and 2; MATLAB verified that, indeed, these are also roots of the original characteristic polynomial. In the same example, dividing DCL(s) by P(s) results in a polynomial whose roots are 3 and 1. These are also the other two roots of DCL(s) identified by MATLAB. A related feature of the characteristic polynomial is that it is always an even polynomial, which means it only contains even powers of sdthis can also be seen in Eq. (12.20). A direct result is that the roots of the auxiliary polynomial are paired and are symmetric with respect to the origin (as e2 and 2 of the previous Example 12.9). Therefore, should a complex number, say s þ u$j, be a root of the auxiliary polynomial (with both the real part s and imaginary part u being positive, for instance), three other complex numbers are also roots of that polynomial, namely: s  u$j, s þ u$j, and s  u$j. On the other hand, if the polynomial has an imaginary root u$j, it also has its conjugate u$j, as a root. However, simple poles of the type þu$j or u$j are located on the imaginary axis, and they indicate marginal stability, according to the definition. In other words, a necessary condition for marginal stability is a row of zero in the RoutheHurwitz array. Let us analyze the next example involving several of these aspects. nn

Example 12.11 The closed-loop characteristic polynomial of a feedback control system is DCL ðsÞ ¼ s 6 þ s 5  3s 4 þ 3s 3  22s 2  4s þ 24. Determine the number of closed-loop poles that are located in the LHP, RHP, and on the imaginary axis.

Solution The RoutheHurwitz array for this system, after factoring out and eliminating 6 on the row corresponding to s4, generates zeroes on the row corresponding to s3, as shown below:

s6 s5 s4 s3

1

3

1 3 1 3 0

0

22 24 4 4

0

0

The auxiliary polynomial, which is formed of the elements of the row above the zeroes, and its s-derivative are:

PðsÞ ¼ s4  3s2 þ 4;

dPðsÞ ¼ 4s3  6s þ 0 ds

(12.25)

12.2 The RoutheHurwitz Stability Test

The coefficients 4, 6, and 0 (or rather 2, 3, and 0, after factoring out and eliminating the factor 2) are used instead of the original zeros in the row of s3, as shown in the new array:

− 3 − 22 24 s6 1 5 s 1 3 −4 0 −1 − 3 4 0 s4 3 − − 2 3 0 s 0 P(s) s2 −1.5 4 1 − 8.33 0 s 4 s0

DCL(s) ÷ P(s)

The poles corresponding to the auxiliary polynomial P(s) (which divides the original polynomial DCL(s)) can be located by analyzing the coefficients on the first column from the row corresponding to s4 down. There is one sign change in that region, which indicates there is one pole in the RHP. Due to symmetry, there is also a pole in the LHPdthey are both on the real axis. The remaining two poles of the auxiliary polynomial have to be placed on the imaginary axis, due to origin symmetry requirements, as well. By analyzing the remaining elements on the first column, a sign change is spotted between 1 and 1, which shows there is another pole in the RHPdpole that belongs to the second-degree polynomial corresponding to the first two rows of the array. As a consequence, the system is unstable, and the poles are located as follows: 2 poles in the RHP (both on the real axis) 2 poles in the LHP (both on the real axis) 2 poles on the imaginary axis This qualitative prediction of the closed-loop poles is confirmed by the following actual roots of the characteristic polynomial DCL(s): 3, 1, 1, 2, 2$j. nn

12.2.3 Design Problems While Sections 12.2.1 and 12.2.2 have introduced the method of RoutheHurwitz to establish the stability behavior of control systems with specified numerical parameters, the same method is utilized in this section as a design tool enabling to identify the values (or value ranges) of certain control parameters (such as gains) that would render the control system stable, marginally stable, or unstable. nn

Example 12.12 The feed-forward transfer function of a unity-feedback control system, which contains an integral controller and a second-order mechanical microsystem, is GðsÞ ¼ s3 þ 15sK2 þ 50s . Determine the values of the positive gain K that will render this system stable, unstable, or marginally stable.

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CHAPTER 12 Stability of Feedback Control Systems

Solution The closed-loop transfer function is:

GCL ðsÞ ¼

GðsÞ K ¼ 1 þ GðsÞ s3 þ 15s2 þ 50s þ K

(12.26)

and the corresponding RoutheHurwitz array is:

s3

1

50

s2

15 750  K 15

K

s1 s0

0

K

The system is (absolutely) stable when there is no sign change in the first column. Because K > 0, it follows that for K < 750, all elements in the first column are positive, and therefore the system is absolutely stable for 0 < K < 750. When, on the contrary, K > 750, there are two sign changes in the first column, the system is unstable, and two closed-loop poles are located in the RHP. For K ¼ 750, the row corresponding to s1 contains only zeros. The auxiliary polynomial is generated with the elements of the row immediately above; this polynomial and its s-derivative are:

PðsÞ ¼ 15s2 þ 750;

dPðsÞ ¼ 30s þ 0 ds

(12.27)

The new array is constructed below:

s3

1

50

s2 15 750 0 s1 30 0 s 750 There are no sign changes in the first column underneath the line separating P(s), and therefore there are no poles of the second-degree auxiliary polynomial P(s) in the RHP, which means there are no symmetric poles on the left plane either. The only possibility that remains is that the two poles are on the imaginary axis. Because there are no sign changes in the full first column, there are no closed-loop poles at all in the RHP, which means the third closed-loop pole is in the LHP on the real axis. As a consequence, the system is marginally stable for K ¼ 750. nn

nn

Example 12.13 The block diagram of a feedback control system is shown in Figure 12.2. Determine the values of the real gain K that make this system stable, unstable, or marginally stable.

Solution The feed forward transfer function of this system is:

GðsÞ ¼

s3

þ

K þ 5s þ 1

3s2

(12.28)

12.2 The RoutheHurwitz Stability Test

+ R(s)

1 s3 + 3s 2 + 5s+ 1

K

_

C(s)

FIGURE 12.2 Unity-Feedback System With Variable Gain.

and the closed-loop transfer function is:

GCL ðsÞ ¼

GðsÞ K ¼ 1 þ GðsÞ s3 þ 3s2 þ 5s þ ð1 þ KÞ

(12.29)

The RoutheHurwitz array corresponding to the closed-loop transfer function of Eq. (12.29) is:

s3

1

5

s

3 14  K 3

1þK

2

s1 s0

0

1þK

The system is stable for 1 < K < 14 because for this range all coefficients in the first column of the array are positive. As a consequence, all four closed-loop poles are in the LHP. For K ¼ 1, the resulting array becomes

s3

1 5

s2 s1

3 0 5 0

s0

0

The characteristic polynomial corresponding to this array is DCL(s) ¼ s3 þ 3s2 þ 5s ¼ s$(s2 þ 3s þ 5); its roots (closed-loop poles) are 0 and e1.5  1.658$j, which confirms that two closed-loop poles are in the LHP, and one closed-loop pole is at the origin. As a consequence, the system is stable. Note: A zero element in the last row of the first column identifies a closed-loop pole at the origin, as indicated in the generic partial array shown below.

s2 a 0 s1 b 0 s0

0

The second element in the row of s is always zero (as per the last simple rule of constructing the RoutheHurwitz array); this makes the second element in the row of s2 be also zero (otherwise, it would not be possible to get zero in the last row). In other words, a zero in the last row and first column will always result in an array whose last three rows are as illustrated above. Because the 1

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CHAPTER 12 Stability of Feedback Control Systems

row of s0 is a row of zeros, the characteristic polynomial is P(s) ¼ b$s whose root is s ¼ 0; this root is also a closed-loop pole. When K < 1, the last coefficient in the first column is the only one negative, which means there is one sign change in the first column resulting in an unstable condition with one closed-loop pole in the RHP; the other two poles are in the LHP. For K > 14, the coefficient corresponding to s1 is negative, this leads to two sign changes in the first column, and therefore the system is again unstable with two closed-loop poles in the RHP and the other pole in the LHP. For K ¼ 14, the resulting array is

s3 s2

1 3

5 15

s1

0

0

The row of zeroes in this table results in the following auxiliary polynomial formed with the coefficients of the row above the zero-elements row:

PðsÞ ¼ 3s2 þ 15

(12.30)

whose s-derivative is dP(s)/ds ¼ 6s þ 0. Using the coefficients of 6 and 0 instead of the row of zeroes corresponding to s1 changes the previous array into

s3

1

5

s2 3 15 s1 6 0 s0 15 Because from the row of s2 down in the first column of this array there are no sign changes, it means that the two roots of P(s), which are also closed-loop poles, need to be on the u$j axis. There is also no sign change from the coefficient of s3 to the one of s2 in the first column, which indicates that the third closed-loop pole is in the LHP. As a result, the system is marginally stable. To conclude, the system is stable for e1  K < 14, unstable when K < e1 (with one closed-loop pole in the RHP and three poles in the LHP) and K > 14 (with two closed-loop poles in both the RHP and the LHP), and marginally stable for K ¼ 14. nn

nn

Example 12.14 The RoutheHurwitz array corresponding to the closed-loop characteristic polynomial of a control system is shown below.

s3

5

?

s

? 2K  10 K

?

2

s1 s0

12.2 The RoutheHurwitz Stability Test

where K is a positive gain. (i) Express the closed-loop characteristic polynomial knowing that it has no missing terms. (ii) Complete the RoutheHurwitz table, determine the positions of the closed-loop poles in the complex plane in terms of the gain K, and discuss the corresponding conditions of stability, instability, and marginal stability.

Solution (i) The given portion of the RoutheHurwitz array can be written as:

s3

5

x

2

y

z

s1

2K  10 K

s

s0 The element in the first column that corresponds to s1 is:

2K  10 x $ y  5z ¼ K y

(12.31)

The only possibility for the relationship of Eq. (12.31) to be valid is when x ¼ 2, y ¼ K, and z ¼ 2. The closed-loop characteristic polynomial is formed of the elements in the first two rows of the RoutheHurwitz array:

DCL ðsÞ ¼ 5s3 þ y $ s2 þ x $ s þ z ¼ 5s3 þ K $ s2 þ 2s þ 2

(12.32)

(ii) With the characteristic polynomial of Eq. (12.32), the following RoutheHurwitz array results

s3 s

2

s1

5

2

K 2 2K  10 0 K

s0

2

For K > 5, all coefficients in the first column of the array are positive, so the system has no closedloop poles in the RHP and is stable, with all three closed-loop poles being in the LHP. For 0 < K < 5, there are two sign changes in the first column, which signifies there are two closedloop poles in the RHP and one closed-loop pole in the LHPdthe system being unstable. For K ¼ 5, a row of zeros is formed with the coefficients corresponding to s1. The auxiliary polynomial and its derivative are:

PðsÞ ¼ 5s2 þ 2;

dPðsÞ ¼ 10 $ s þ 0 ds

(12.33)

The coefficients 10 and 0 are used in the row of zeroes, which changes the previous RoutheHurwitz array to

s3 s2 s1 s0

5 5 10 2

2 2 0

615

616

CHAPTER 12 Stability of Feedback Control Systems

Because there are no sign changes in the first column, there are no closed-loop poles in the RHP. As a consequence, the two roots of P(s), which are also closed-loop poles, are imaginary. The third closed-loop pole is in the LHP. The roots of P(s) are found solving Eq. (12.33) and are 0:6325,j. The system is marginally stable for K ¼ 5. nn

nn

Example 12.15 A MEMS plant is modeled as a second-order system by the following differential equation: € Þ þ a1 $ cðt _ Þ þ a0 $ cðt Þ ¼ a0 $ mðt Þ, with m being the input and c the output. Its natural frea2 $ cðt quency is 50 rad/s, and damping ratio is 0.6. A proportional þ integral controller (P þ I ) defined by the positive constants KP and KI and a proportional feedback sensor defined by the outputeinput relationship b(t) ¼ 20$c(t) are connected to the plant in a feedback control system. Determine the values of KP and KI that will render the feedback system stable, marginally stable, or unstable.

Solution The block diagram of this control system is shown in Figure 12.3.

+ R(s)

E(s)

KP +

_ B(s)

KI s

C (s )

M(s) G p (s)

H(s)

FIGURE 12.3 Nonunity-Feedback System With P þ I Controller. The differential equation governing the plant dynamics can be written as

€ þ cðtÞ

a1 a0 a0 _ þ $ cðtÞ ¼ $ mðtÞ or cðtÞ € þ 2x $ un $ cðtÞ þ u2n $ cðtÞ ¼ u2n $ mðtÞ $ cðtÞ a2 a2 a2 (12.34)

The plant transfer function is found by applying the inverse Laplace transform with zero initial conditions to Eq. (12.34), namely:

Gp ðsÞ ¼

CðsÞ u2n 2500 ¼ 2 ¼ 2 2 MðsÞ s þ 2x $ un $ s þ un s þ 60s þ 2500

(12.35)

where the numerical values of the problem have been used. The P þ I controller transfer function is Gc(s) ¼ M(s)/E(s) ¼ KP þ KI/s ¼ (KP$s þ KI)/s, whereas the feedback transfer function is H(s) ¼ B(s)/C(s) ¼ 20. The feed-forward transfer function is

GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼



KI 2500 2500KP $ s þ 2500KI ¼ 3 KP þ $ 2 s þ 60s þ 2500 s s þ 60s2 þ 2500s (12.36)

The closed-loop transfer function of the feedback system is

GCL ðsÞ ¼

GðsÞ 2500KP $ s þ 2500KI ¼ 1 þ GðsÞ $ HðsÞ s3 þ 60s2 þ 2500ð1 þ 20KP Þ $ s þ 50; 000KI (12.37)

12.2 The RoutheHurwitz Stability Test

The RoutheHurwitz array corresponding to the characteristic polynomial of Eq. (12.37) is

s3

1

2500ð1 þ 20KP Þ

s

2

3

2500KI

s

1

3 þ 60Kp  KI

0

0

KI

0

s

after factoring out and eliminating 20 in the second row, as well as after eliminating the factor 2500/3 in the third row. When 3 þ 60KP  KI > 0, which is KP > (KI  3)/60, there are no sign changes in the first column of the array (since KI > 0); therefore, all closed-loop poles are in the LHP, the control system being stable. When 3 þ 60KP e KI < 0, which is 0 < KP < (KI  3)/60, there are two sign changes in the array, which indicates the system is unstable with two closed-loop poles in the RHP and one closedloop pole in the LHP. For 3 þ 60KP  KI ¼ 0, which results in KP ¼ (KI  3)/60, there is a row of zeroes in the Routh eHurwitz array corresponding to s1. The resulting auxiliary polynomial and its derivative are

PðsÞ ¼ 3s2 þ 2500 $ KI ;

dPðsÞ ¼ 6$s þ 0 ds

(12.38)

The resulting first column of the RoutheHurwitz array becomes

s3 s2 s1 s0

1 3 6 KI

There are no sign changes in the first column because KI > 0, and therefore, there are no closed-loop poles in the RHP. The system has one closed-loop pole in the LHP and two closed-loop poles (the pffiffiffiffiffiffi roots of P(s) of Eq. 12.38) on the imaginary axis, namely 50 33KI $ j, and the control system is marginally stable. nn

nn

2

5 2K1 6 Analyze the stability of a control system whose state matrix is ½A ¼ 4 3 1 constants K1 and K2. 4 K2

Example 12.16

1

3

7 2 5 in terms of real 1

Solution The closed-loop poles of the system are the solution to Eq. (12.12), whose left-hand side is:

detðs $ ½I  ½AÞ ¼ s3 þ 7s2 þ ð7  6K1  2K2 Þ $ s þ ð1 þ 10K1  7K2 Þ (12.39) where [I] is the 3  3 identity matrix. The particular expression in the right-hand side of Eq. (12.39) can be obtained using the MATLAB code: >> syms s K1 K2 >> a=[-5, 2*K1, 1; 3, -1, -2; 4, -k2, -1]; >> d=collect(det(s*eye(3)-a), s) with d being the determinant expression.

617

CHAPTER 12 Stability of Feedback Control Systems

The corresponding RoutheHurwitz array is

s3

1

7  6K1  2K2

s2

7

1 þ 10K1  7K2

s1

48  52K1  7K2 7

0

s0

1 þ 10K1  7K2

The system is stable when all coefficients in the first column of the array are positive, this results in the system of inequalities:



48  52K1  7K2 > 0 1 þ 10K1  7K2 > 0

(12.40)

The graphical solution of this system is shown in Figure 12.4; it was obtained by plotting the lines represented by the equations 48  52K1  7K2 ¼ 0 and 1 þ 10K1  7K2 ¼ 0. The common (shaded) area underneath the two lines represents the system solution domain, which indicates that

8 > < K2 < 1:226 7K2  1 48  7K2 > < K1 < : 10 52

(12.41)

The value K2 ¼ 1.226 was obtained by intersecting the two line segments, while the inequalities in the second Eq. (12.41) resulted from expressing K1 from the two Eq. (12.40). For values of K1 and K2 outside the shaded area in Figure 12.4, which is

8 > < K2 > 1:226 48  7K2 7K2  1 > < K1 < : 10 52

(12.42)

25 20

K2 = 15

48 − 52K1 7

10

(0.758, 1.226)

1 + 10K1 7

2

5

K2 =

K

618

0 -5

Solution domain

-10 -15 -20 -2

-1.5

-1

-0.5

0

0.5 K1

1

1.5

FIGURE 12.4 Solution Domain of Parameters K1 and K2 for System Stability.

2

2.5

3

12.2 The RoutheHurwitz Stability Test

the system is unstable because the expressions in the left-hand side of Eq. (12.40) are negative. This leads to one sign change in the first column of the RoutheHurwitz array; as a result, one closed-loop pole is in the RHP (on the real axis), whereas the other closed-loop poles are in the LHP. The designs with the last two elements in the first column are zero are discussed next. When K2 ¼ (48  52K1)/7, the row corresponding to s1 in the RoutheHurwitz array becomes a row of zeroes. The auxiliary polynomial

PðsÞ ¼ 7s2 þ 1 þ 10K1  7K2 ¼ 7s2 þ 62K1  47

(12.43)

is constructed from the coefficients of the row above the zero row, and its derivative dP(s)/ ds ¼ 14s þ 0 yields the new coefficients to be introduced instead of the zeros in the row of s1 such that the RoutheHurwitz array becomes

s3

1

s2 7 s1 14 s0 −47 + 62 K1

62 K1 − 47 7 −47 + 62 K1 0

When K1  47/62 ¼ 0.758, the two roots of the auxiliary polynomial (which are also closed-loop poles and are related to the signs of the elements in the first column underneath the separating line in the array) need to be on the imaginary axis because there is no sign change in the first column between the coefficients corresponding to s2, s1, and s0. The third closed-loop pole is in the LHP because there is no sign change between the coefficient of s3 and that of s2 in the array’s first column. As a consequence, the system is marginally stable when K1  47/62. When K1 < 47/62, due to the fact that last coefficient in the first column is negative, the system is unstable having one closedloop pole in the RHP and the other two closed-loop poles in the LHP. When K1 ¼ 47/62, the lastcolumn zero element indicates a zero closed-loop pole at the origindsee Note in Example 12.13. The other closed-loop poles are in the LHP because there are no sign changes in the first column of the array, and the system is stable. For K2 ¼ (1 þ 10K1)/7, the original RoutheHurwitz array transforms into

s3

1

s2

7 47  62K1 7

s1 s0

47  62K1 7 0 0

0

For K1 < 47/62, there are no sign changes in the first column of the array so there are no closed-loop poles in the RHP: two closed-loop poles are in the LHP, whereas the third is at the origin (as signaled by the zero element in the last row), and the system is stable. When K1 > 47/62, there is one change of sign in the first column, and therefore the system is unstable with one closed-loop pole in the RHP and one pole in the LHP, whereas the third closed-loop pole is at the origin. This can better be seen by noticing that the characteristic polynomial resulting becomes in this case:



47  62K1 DCL ðsÞ ¼ s s2 þ 7s þ 7

(12.44)

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CHAPTER 12 Stability of Feedback Control Systems

with one root equal to zero, one root real and negative, and the other real and positive. Eventually, when K1 ¼ 47/62, the row corresponding to s1 becomes a row of zeros, but this case has already been analyzed previously. nn

12.3 STABILITY OF FEEDBACK SYSTEMS BY THE ROOT LOCUS METHOD Using the poles and zeroes of the open-loop transfer function, the root locus method is a graphical procedure that tracks the successive positions of the closed-loop poles in the complex plane when a parameter (such as a gain K) variesdthe resulting plot is the root locus. This method is able to qualify changes in a feedback system’s stability through monitoring the closed-loop pole migration between the LHP and RHP. Vr(s)

+

Ve(s)

KI /s

Va(s)

Ka

Ma(s)

_

Gp(s)

Ω(s) m a, θ

Vs(s)

J c

Ks (a)

(b)

FIGURE 12.5 (a) Block Diagram of a Nonunity-Feedback Control System With Feedback Gain; (b) Rotary Mechanical Plant.

Consider, for instance, the control system of Figure 12.5(a), which includes an integral controller of constant KI and a gain-type actuator converting the actuation voltage Va(s) into a torque Ma(s), its transfer function is the constant Ka. The forward path comprises a plant, which is formed of a disk of rotary inertia J ¼ 0.001 kg m2 and a damper of damping coefficient c ¼ 0.1 N m s, as sketched in Figure 12.5(b). The time-domain mathematical model of the plant and the plant transfer function connecting the input torque Ma(s) to the output shaft angular velocity U(s) are: _ J $ uðtÞ þ c $ uðtÞ ¼ ma ðtÞ; Gp ðsÞ ¼

UðsÞ 1 1 ¼ (12.45) ¼ Ma ðsÞ J $ s þ c 0:001s þ 0:1

A tachometer sensor picks up the output shaft angular velocity U(s) and converts it to a proportional voltage Vs(s), which means the feedback sensor transfer function is H(s) ¼ Vs(s)/U(s) ¼ Ks. Vs(s) is further transmitted to a summing point, where it is compared to the reference voltage Vr(s); the result, the error voltage Ve(s), is applied to the integral controller. The feed-forward transfer function of this control system is: GðsÞ ¼

UðsÞ KI 1 KI $ Ka KI $ Ka ¼ ¼ $ Ka $ ¼ Ve ðsÞ J $ s þ c s $ ðJ $ s þ cÞ J $ s2 þ c $ s s

(12.46)

12.3 Stability of Feedback Systems by the Root Locus Method

and therefore, its closed-loop transfer function is GCL ðsÞ ¼

UðsÞ GðsÞ KI $ Ka ¼ ¼ Vr ðsÞ 1 þ Ks $ GðsÞ J $ s2 þ c $ s þ Ks $ KI $ Ka

(12.47)

Assume the gain Ks of the feedback sensing unit varies from 0 to N, whereas the other two gains, KI and Ka, are constant. In this case K ¼ Ks$KI$Ka also varies from 0 to N. The closed-loop poles are the roots of the characteristic polynomial J$s2 þ c$s þ K ¼ 0.001s2 þ 0.1s þ K in the denominator of the closed-loop transfer function of Eq. (12.47). When K varies from 0 to 5, in increments of 0.5, it can be seen that the two closed-loop poles migrate on the real axis from 100 and 0 to meet at 50 for a gain K ¼ 2.5, as shown in Figure 12.6. With the gain increasing past 2.5, the closed loop-poles leave the real axis and move vertically, one up and the other one down, as illustrated in the same Figure 12.6. ω·j

50 K=0 –100

K = 2.5 –50

K=0 O

σ

–50

FIGURE 12.6 Root Locus of a Second-Order Feedback System.

It can be seen that this particular system is stable as the closed-loop poles are always in the LHP. Many SISO feedback control systems can be modeled as in the block diagram of Figure 12.7, where the gain K is variable, and G(s) is the feed-forward transfer function. The closed-loop transfer function of this system is: GCL ðsÞ ¼

R(s)

+

K $ GðsÞ 1 þ K $ GðsÞ $ HðsÞ

(12.48)

C(s) K

G(s)

_ H(s)

FIGURE 12.7 SISO Feedback Control System With Variable Gain.

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CHAPTER 12 Stability of Feedback Control Systems

The closed-loop poles of this feedback system are the roots of the characteristic equation of Eq. (12.48); this can be written as K $ GðsÞ $ HðsÞ ¼ 1

(12.49)

but K$G(s)$H(s) ¼ GOL(s) is the open-loop transfer function. The open-loop transfer function can be written as a ratio of factorized polynomials in terms of zeroes and poles, namely m Y

K $ GðsÞ $ HðsÞ ¼ GOL ðsÞ ¼

j¼1 n Y

ðs  zj Þ (12.50) ðs  pi Þ

i¼1

assuming GOL(s) has m zeros and n poles. For a given s (which is a complex number), the factors s  zj and s  pi are also complex numbers, and therefore GOL(s) of Eq. (12.50) is a complex number as well. Eq. (12.49) is valid when the following angle 4 and magnitude (modulus) M conditions are satisfied:  4 ¼ ;K $ GðsÞ $ HðsÞ ¼ ;GðsÞ $ HðsÞ ¼ ð2k þ 1Þ $ p; k ¼ 0; 1; 2; . M ¼ K $ jGðsÞ $ HðsÞj ¼ 1 (12.51) and Figure 12.8 depicts the vector GOL(s) in the complex planednote that 4 ¼ p for k ¼ 0. ω·j π

–1

GOL(s)

O

σ

FIGURE 12.8 Representation of GOL(s) in the Complex Plane.

The first Eq. (12.51) took into account that the direction of a vector is not influenced by a constant (such as K) multiplying that vector. Recalling the phasor form of complex numbers, the factors of Eq. (12.50) are written as ( s  zj ¼ js  zj j $ e;ðszj Þ $ j (12.52) s  pi ¼ js  pi j $ e;ðspi Þ $ j and are graphically represented in Figure 12.9.

12.3 Stability of Feedback Systems by the Root Locus Method

ω·j s s − pi

s

s − zj

(s− z )

( s − pi ) pi

j

σ

O

zj

FIGURE 12.9 Representation of a Pole pi and Zero zj in Connection to a Number s in the Complex Plane.

As a consequence, the angle condition of Eq. (12.51) results from the particular form of the open-loop transfer function given in Eq. (12.50), namely: 4¼

m X

;ðs  zj Þ 

j¼1

n X

;ðs  pi Þ ¼ ð2k þ 1Þ $ p;

k ¼ 0; 1; 2; .

(12.53)

i¼1

Similarly, the magnitude condition of Eq. (12.51) results in m Y

K $ jGðsÞ $ HðsÞj ¼ K

j¼1 $ n Y

js  zj j ¼1

(12.54)

js  pi j

i¼1

which is also written as n Y

1 ¼ i¼1 K¼ m jGðsÞ $ HðsÞj Y

js  pi j (12.55) js  zj j

j¼1

A complex number s is a closed-loop pole if it satisfies Eqs. (12.53) and (12.55). If Eq. (12.53) is not complied with, the test point s is clearly not a closed-loop pole. If, instead, the test point s checks Eq. (12.53), it means the point is on the root locus (because it is a closed-loop pole), and Eq. (12.55) needs to be used to determine the corresponding gain K. nn

Example 12.17

The open-loop transfer function of a feedback transfer function has the poles 5 and 4 and the zeroes 1 and 3. Verify whether the point s ¼ 2 is on the root locus and calculate the corresponding gain K if the point is on the root locus.

623

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CHAPTER 12 Stability of Feedback Control Systems

Test point s = 2 p1 = –5 p2 = –4

z1 = 1

z2 = 3

σ

s – p1 s – p2 s – z1 s – z2

FIGURE 12.10 Relative Positions of Open-Loop Poles and Zeroes Connected to a Test Point.

Solution Figure 12.10 depicts the relative positions of the open-loop poles and zeros with respect to the test point and the complex numbers s  p1 ¼ s þ 5, s  p2 ¼ s þ 4, s  z1 ¼ s  1, and s  z2 ¼ s  3. The angle of Eq. (12.53) is calculated as

4 ¼ ½;ðs  z1 Þ þ ;ðs  z2 Þ  ½;ðs  p1 Þ þ ;ðs  p2 Þ ¼ ð0 þ pÞ  ð0 þ 0Þ ¼p (12.56) Because 4 is an odd multiple of p, it follows that the test point is on the root locus and therefore is a closed-loop pole. The corresponding gain is calculated based on Eq. (12.55) as

K¼

js  p1 j $ js  p2 j j2  ð5Þj $ j2  ð4Þj ¼ 42 ¼ js  z1 j $ js  z2 j j2  1j $ j2  3j

(12.57) nn

12.3.1 Basic Rules for Sketching the Root Locus There are several rules, also known as Evans rules (they have been introduced by W.R. Evans in 1948), for sketching the root locus plot. These rules are only enunciated next without a demonstrationdthe interested reader is referred to dedicated controls texts such as the ones mentioned in the suggested reading section at the end of this chapter. 1. The number of root locus branches is equal to the number of open-loop poles. 2. The root locus is symmetric with respect to the real axis. 3. The root locus starts (begins) at open-loop poles and ends at open-loop zeros. As a consequence of Rule #1, it follows that the number of open-loop poles is equal to the number of open-loop zeros. 4. Real-axis portions of the root locus are always located to the left of an odd number of open-loop poles and zeros (this is the sum of the open-loop poles and zeros numbers) that are on the real axis.

12.3 Stability of Feedback Systems by the Root Locus Method

5. The behavior at infinity of the root locus is described by asymptotes. Because usually the number of finite open-loop poles np is larger than the number of finite open-loop zeros nz (due to the fact that the degree of the polynomial in the numerator of the open-loop transfer function is normally smaller than the degree of the open-loop transfer function characteristic polynomial), there should be np  nz infinite zeros (that are situated at infinity), so that the end of root locus branches corresponding to these infinite zeros is determined by means of asymptotes. The np  nz asymptotes are centered at an abscissa: P P ðfinite open-loop polesÞ  ðfinite open-loop zeroesÞ (12.58) sa ¼ np  nz and the asymptotes’ angles with respect to the real axis are calculated as 4a;k ¼

ð2k þ 1Þ $ p np  n z

(12.59)

with k ¼ 0, 1, ., np  nz  1. As a result of Rule #2, the asymptotes are sym metric with respect to the real axis. 6. The points where two root locus branches meet on the real axis and continue on this axis as K increases are known as break-in points. The points where two realaxis root locus branches meet on the real axis and then leave this axis are named break-away points. The gain K is minimum for break-in points and is maximum for break-away points compared with the gains of all other points located on the same real-axis segment of the root locus. 7. The points where the root locus crosses the u$j (imaginary) axis can be calculated using the RoutheHurwitz criterion. These u$j-axis crossing points ultimately indicate values of the gain K that separate stable from unstable regions. Analyzing the characteristic polynomial of the closed-loop transfer function, it is enforced that a row that involves K and relates to an odd power of s in the Routh array be zerodthis results in the value of the gain. The corresponding frequency is obtained by annulling the auxiliary polynomial P(s) formed of the coefficients on the row above the zero row (see method of RoutheHurwitz). Assuming there are no sign changes in the first-column coefficients related to P(s), it follows that the roots of P(s) are located on the u$j axis. Note that these roots (also closed-loop poles) come always in even numbers and are symmetric with respect to the origin, which is in compliance with both the nature of P(s) and Rule #2 of sketching the root locus. nn

Example 12.18 Consider the feedback system having the open-loop transfer function of the previous Example 12.17 and a feedback gain K. (i) Sketch the root locus and determine the exact values of K corresponding to the break-in and break-away points. (ii) Calculate the u$j-axis crossing points of the root locus and establish the range of K for stability.

625

626

CHAPTER 12 Stability of Feedback Control Systems

Solution (i) Because there are two finite open-loop poles and two finite open-loop zeros, it follows that np ¼ nz, and therefore there are no infinite zeros and no asymptotes. In addition, only two root locus segments are located on the real axis, between 1 and 3 (to the left of one open-loop pole, which is an odd number) and between 5 and 4, which is to the left of three open-loop poles and zeroes. Figure 12.11 sketches the root locus.

ω·j ω·j – axis crossing Break-away point

–5

Break-in point Root Locus

O

–4

1

3

σ

Root Locus ω·j – axis crossing

FIGURE 12.11 Root Locus Sketch for Example 12.17.

The root locus starts from the two open-loop poles, 5 and 4, and the two branches proceed on the real axis until they reach the break-away point. After that, the two branches continue off the real axis and land again on the real axis meeting at the break-in point, between the two open-loop zeros. They continue moving on the real axis until each branch reaches one of the two zeros, 1 and 3. To calculate the break-away and break-in points, the open-loop transfer function needs to be first determined. Knowing the two open-loop poles and zeros, the open-loop transfer function is written as:

GOL ðsÞ ¼ K $

ðs  1Þ $ ðs  3Þ s2  4s þ 3 ¼K$ 2 ðs þ 4Þ $ ðs þ 5Þ s þ 9s þ 20

(12.60)

Points on the root locus satisfy the propriety GOL(s) ¼ 1, which, by using Eq. (12.60), leads to

   s2 þ 9s þ 20 K¼ s2  4s þ 3

(12.61)

According to Rule #6, at break-in and break-away the gain K assumes extreme values, which means that the derivative of the function K in terms of s should be zero: dK(s)/ds ¼ 0. Taking the s-derivative of K in Eq. (12.61) yields the equation 13s 2 þ 34s  107 ¼ 0 whose roots are s1 ¼ 4.461 and s2 ¼ 1.845. It can be seen that s1 is the break-away point (for which K ¼ 0.006 from Eq. 12.61) and s2 is the break-in point (its gain is K ¼ 40.994 from the same Eq. 12.61).

12.3 Stability of Feedback Systems by the Root Locus Method

(ii) The closed-loop transfer function resulting from the open-loop transfer function of Eq. (12.60) is

  K s2  4s þ 3 GOL ðsÞ ¼ GCL ðsÞ ¼ 1 þ GOL ðsÞ ðK þ 1Þs2 þ ð4K þ 9Þs þ 3K þ 20

(12.62)

whose characteristic polynomial is DCL(s) ¼ (K þ 1)$s2 þ (e4K þ 9)$s þ 3K þ 20. The following RoutheHurwitz coefficient array is formed related to this polynomial:

s2 s1 s0

K + 1 3K + 20 0 − 4K + 9 3K + 20

The only possibility of having a row of zeroes is when K ¼ 9/4 ¼ 2.25, which produces zeros in the entire row of s1. In this case, the auxiliary polynomial is as follows:

PðsÞ ¼ ðK þ 1Þ $ s2 þ 3K þ 20 or PðsÞ ¼ 3:25s2 þ 26:75

(12.63)

The roots of P(s) are 2.869$j, which are imaginary roots. Because the roots of the auxiliary polynomial are also closed-loop poles, this indicates that the root locus intersects the imaginary axis at 2.869j. Since 2.869j ¼ u$j, it follows that u ¼ 2.869 rad/s. Therefore the u$j-axis crossing from the LHP into the RHP (and when unstable behavior occurs) takes place for a gain K ¼ 2.25 and at a frequency u ¼ 2.869 rad/s. nn

nn

Example 12.19

An actuated electrical plant is defined by the transfer function Gp(s) ¼ Ka/(s2 þ 2s þ 3), where Ka is the actuator’s gain. A basic feedback control design is used with an integral (I) controller of unity gain (KI ¼ 1) and a proportional gain Ks acting as a sensor on the feedback branch. (i) Sketch the root locus corresponding to this transfer function considering the variable gain K ¼ Ka$Ks. (ii) Calculate the u$j-axis crossing points of the root locus. What is the range of K for stability?

Solution (i) The open-loop transfer function of the system is

GOL ðsÞ ¼ Gc ðsÞ $ Gp ðsÞ $ HðsÞ ¼

Ka $ Ks K ¼ s $ ðs2 þ 2s þ 3Þ s3 þ 2s2 þ 3s

(12.64)

This function has three finite poles, which are the roots of the characteristic polynomial s3 þ 2s2 þ 3s, and which are

s1 ¼ 0; s2 ¼ 1 

pffiffiffi pffiffiffi 2j; s3 ¼ 1 þ 2j

(12.65)

This indicates that np ¼ 3, which further shows that the root locus has three branches. At the same time, there are no finite zeros, as can be seen in Eq. (12.64) and therefore nz ¼ 0, but

627

628

CHAPTER 12 Stability of Feedback Control Systems

ω·j Asymptote

Root Locus Open-loop pole

2

–2/3 Root Locus

Open-loop pole –1

Asymptote

O

σ

− 2

Open-loop pole Root Locus

Asymptote

FIGURE 12.12 Root Locus Sketch for Example 12.19. that means there need be three infinite zeros, and, as a consequence, three asymptotes. The center point of the asymptotes is calculated using Eq. (12.58) as

pffiffiffi   pffiffiffi   0 þ 1  2 j þ 1 þ 2 j 2 ¼ sa ¼ 3 3

(12.66)

The corresponding asymptote angles are given by Eq. (12.59) as

p 3p 5p ¼ p; 4a;2 ¼ 4a;0 ¼ ; 4a;1 ¼ 3 3 3

(12.67)

Using Rule #4, according to which only one real-axis portion of the root locus exists to the left of the origin, the root locus of Figure 12.12 is sketched. (ii) The characteristic polynomial of the closed-loop transfer function

GCL ðsÞ ¼

GOL ðsÞ K ¼ 3 2 1 þ GOL ðsÞ s þ 2s þ 3s þ K

(12.68)

is s3 þ 2s2 þ 3s þ K. The following RoutheHurwitz array is formed based on this polynomial:

s3

1

3

s2 s1

2 6K

K 0

When K ¼ 6, the coefficients corresponding to the s1 row are all zero. The auxiliary polynomial is

  PðsÞ ¼ 2s2 þ K ¼ 2s2 þ 6 ¼ 2 s2 þ 3

(12.69)

pffiffiffi and its roots are  3 $ j ¼ 1.732$j. They indicate that the u$j-axis crossing from the LHP into the RHP occurs for a gain K ¼ 6 and at a frequency u ¼ 1.732 rad/s. As a consequence, the feedback system is unstable for gains larger than 6. nn

12.3 Stability of Feedback Systems by the Root Locus Method

12.3.2 Using MATLAB to Plot the Root Locus The MATLAB command rlocus(sys), which belongs to the Control System Toolbox, facilitates plotting the root locus trajectories for an open-loop transfer function previously defined as sys and the variable feedback gain K. Either of the two feedback sys-

R(s)

+

C(s) G(s)

_

R(s)

+

C(s) G(s)

_ K

K

H(s)

FIGURE 12.13 Block Diagrams of Feedback Systems Modeled by the MATLAB rlocus Command.

tems shown in Figure 12.13 are covered by this command. As such, sys represents the open-loop transfer function G(s) in the left block diagram of Figure 12.13, whereas in the right block diagram of Figure 12.13, sys stands for G(s)$H(s). It should be noticed that instead of being on the feedback branch, the gain K could be placed as well next to G(s)dthis alteration would not change the characteristic equation of the closed-loop transfer function, which gives the closed-loop poles, and which is 1 þ K$G(s) ¼ 0 or 1 þ K$G(s)$H(s) ¼ 0. Several root loci plots corresponding to the previously defined open-loop transfer functions sys1, sys2, . are obtained by means of the rlocus (sys1, sys2, .) command. A variant is the command rlocus(sys, K), which allows selecting a range for the feedback gain K by defining this range as a vector. nn

Example 12.20 Use the MATLAB rlocus command to obtain the root locus plot for the application studied analytically in Example 12.19. Determine graphically and with approximation the gain and frequency corresponding to u$j-axis crossings.

Solution The MATLAB code: >> sys=tf([1],[1,2,3,0]) >> rlocus(sys) produces the plot of Figure 12.14, which confirms the analytically obtained plot of Figure 12.12. By approximately clicking the point of intersection between the upper branch and the imaginary axis, the box shown in Figure 12.14 is obtained, which indicates an approximate gain of K ¼ 6.26 and

629

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CHAPTER 12 Stability of Feedback Control Systems

FIGURE 12.14 MATLAB Root Locus for Example 12.19. a frequency of 1.75 rad/s corresponding to the crossing into the unstable domain. These values are close to the ones obtained analytically in Example 12.19. nn

12.4 NYQUIST PLOT AND BODE PLOTS FOR STABILITY OF FEEDBACK SYSTEMS This section gives a brief introduction to the Nyquist stability criterion and the Nyquist plot, which can be used in conjunction with Bode plots to study the stability of feedback systems. A similar method, which is not covered here, uses Nichols polar plots. The Nyquist (or polar) plot is a graphical representation in the s (or frequency) domain of the open-loop transfer function through which conclusions can be derived with respect to the stability of a feedback control system. Assume such a system is formed of the feed-forward transfer function K$G(s), with K being a gain and G(s) incorporating the controller and plant and the feedback function H(s). The open-loop transfer function is therefore GOL(s) ¼ K$G(s)$H(s). As shown previously in this chapter, the closed-loop transfer function is GCL ðsÞ ¼

K $ GðsÞ GOL ðsÞ GOL ðsÞ ¼ or 1 þ GOL ðsÞ ¼ 1 þ GOL ðsÞ HðsÞ $ ½1 þ GOL ðsÞ HðsÞ $ GCL ðsÞ (12.70)

Eq. (12.70) demonstrates that 1. The poles of the closed-loop transfer function (the values of s which makes GCL(s) infinite) are the zeroes of 1 þ GOL(s)dthe values of s which make 1 þ GOL(s) zero.

12.4 Nyquist Plot and Bode Plots for Stability of Feedback Systems

2. The poles of 1 þ GOL(s) are also the poles of GOL(s). Consider now the following particular form of 1 þ GOL(s): 1 þ GOL ðsÞ ¼

ðs  zÞ ðs  p1 Þ $ ðs  p2 Þ

(12.71)

and also consider a generic complex number s traveling along a closed contour C1, which is connected to the zero z and to the poles p1 and p2 as shown in Figure 12.15(a). Im(s)

Im[ Pc(s)]

s Pc(s) s − p1

ϕp,1 p1

s− z

ϕz z

s − p2 p2

O

ϕp,2

C1

Re(s)

(a)

C2

ϕ O

Re[Pc(s)]

(b)

FIGURE 12.15 Contour Transformation (Mapping) Through the Characteristic Polynomial of the ClosedLoop Transfer Function: (a) Original Contour and Open-Loop Pole/Zero Positions; (b) Transformed Contour.

Assume the generic point s makes a full rotation around the contour C1 in a clockwise direction. The vectors s  z and s  p1 (with the zero z and pole p1 external to the closed contour C1) will only go between one limit position to another limit position and back without performing a full rotation; therefore, the net angular motion of each is zero. However, the pole vector s  p2 (with the pole p2 inside the contour C1) makes a full rotation when s realizes the same process. At the same time, the point Pc, which is the mapped image of s through 1 þ GOL(s), moves along the transformed contour C2, which is shown in Figure 12.15(b). This point’s radial (polar) position is measured by the angle 4. Beacuse the three vectors s e z, s e p1, and s e p2 are actually complex numbers, they can be written in phasor form. As a consequence, the angle 4 (or argument) is calculated based on the component angles 4z, 4p,1, and 4p,2 as discussed in the section dedicated to the root locus and with the aid of Eq. (12.53), namely:   4 ¼ 4z  4p1 þ 4p2 (12.72) Since the net variations of 4z and 4p1 are zero, as discussed, and the net variation of 4p2 is 360 , it follows from Eq. (12.72) that the net variation of 4 is 360 ; this means that the transformed point Pc travels a full revolution around the origin of its complex plane in a counterclockwise direction. If instead of a pole, a zero is inside the contour C1, it would follow that the net variation of 4 is þ360 , so Pc would

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CHAPTER 12 Stability of Feedback Control Systems

undergo a full rotation around its origin in a clockwise direction. Similarly, if an equal number of poles and zeroes are placed inside C1, the net travel by the polar radius of Pc would be zero. As a conclusion, we can state the following: 3. The number of full rotations N of the point Pc (that is mapped through 1 þ GOL(s)) in a counterclockwise direction is equal to the difference between the number of the open-loop poles Np inside contour C1 and the number of openloop zeros Nz inside the same contour: N ¼ Np  Nz or Nz ¼ Np  N

(12.73)

Let us now combine the conclusions derived at (1), (2), and (3) in this section and make an addition to the process to be able to formulate Nyquist criterion. According to conclusion (1), Nz is also the number the closed-loop poles, which are the parameters whose position in the complex plane define stability. According to (2), the poles of 1 þ GOL(s) are also the poles of GOL(s), so Np of Eq. (12.73) represents the number of open-loop poles. Eq. (12.73) can therefore be used to express the number of closed-loop poles in terms of the number of open-loop poles Np inside contour C1 and the number of counterclockwise encirclements (full rotations) of the origin by the image point Pc on the mapped contour C2. To help define the number of closedloop poles in the RHP, consider that the contour C1 is the infinite semicircle shown in Figure 12.16. Im(s)

R

∞ Re(s) C1

FIGURE 12.16 The Infinite Semicircle.

The poles and zeros of the open-loop transfer function GOL(s) are easy to determine (or known), whereas the poles and zeroes of 1 þ GOL(s) are not straightforward to evaluate. In addition, the image of GOL(s) is the image of 1 þ GOL(s) translated to the left by one unit, see Figure 12.17. As a consequence, instead of considering encirclement of the origin by the transformation function 1 þ GOL(s), one can equivalently use the encirclements of 1 by GOL(s) and count the rotations around 1 to obtain N. Now we can enunciate the Definition of the Nyquist criterion: The number of closed-loop poles Nz in the RHP is equal to the number of open-loop poles Np

12.4 Nyquist Plot and Bode Plots for Stability of Feedback Systems

that are enclosed in an infinite-radius semicircle placed in the RHP and traveled in a clockwise direction minus the number of counterclockwise encirclements N of the point 1 made by the contour that is the map of the semicircle through the openloop transfer function GOL(s). The number of encirclements can be established by counting the number of intersections of an arbitrary-position test radius stemming from the critical point and the Nyquist plot; the plot of Figure 12.17, for instance, intersects the arbitrary test radius in one point and so N ¼ 1. Note that N is negative when the encirclements of the critical point are clockwise. Im[1 + GOL(s)]

Im[ GOL(s)] Test radius

C2

C*2

Re[1 +GOL(s)]

Re[GOL(s)] −1

0

(a)

−1

0

(b)

FIGURE 12.17 Mapped Contours of (a) GOL(s) Around 1; (b) 1 þ GOL(s) Around the Origin.

When the open-loop transfer function contains a variable gain K as a factor in it, the Nyquist plot, which is drawn for a value of K ¼ 1, can be used to calculate the gain ranges for stability, as well as K values for marginal stability. Different values of K would produce Nyquist plots that are similar, values of K larger than 1 would expand the normal plot, whereas values of K smaller than 1 will shrink the original unit-gain plot. As a result of varying K, the Nyquist plot can move to the right and to the left of the critical 1 point, which would possibly change the stability condition of the system. An equivalent situation is generated when instead of using the point 1 as the origin for counting the encirclements, we scale the open-loop transfer function K$G(s)$H(s) by K, and therefore analyze the encirclements of a new critical point 1/K. As K varies, the position of this critical point changes on the negative real axis. The Nyquist diagram is obtained by calculating the magnitude and phase of the open-loop transfer functions for sequentially increasing u from 0 to N in a first phase, which covers the positive imaginary axis of the source infinite semicircle of Figure 12.16. For the source points located on the positive u$j axis, the unitgain open-loop transfer function is G(u$j)$H(u$j). Note that the Bode plots, which are introduced in Chapter 9, represent separately the variation of the modulus (magnitude) of a transfer function G(u$j) and the variation of the same function’s phase angle, both in terms of frequency u. Thus, the Bode plots can be used to obtain the polar plot of the magnitude in terms of the phase angle of any

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frequency-expressed (sinusoidal) transfer function, such as the open-loop transfer function G(u$j)$H(u$j), for instance. When the Nyquist plot passes through 1, the magnitude of the open-loop transfer function is equal to 1, which indicates that the particular value of s is a closedloop pole, according to the root locus method. That means the equation G(u$j)$ H(u$j) ¼ 1 results in a closed-loop pole that is located on the u$j axis and the system is marginally stable; in other words, the Nyquist plot intersects the real axis at 1 for the same frequency of the root locus intersecting the imaginary axis. The exact Nyquist plot is easily obtainable using any computer software with polar plotting capabilities, or, even simpler, with software such as MATLAB, which has the built-in function nyquist, as we see in the following example. The marginal stability point can be found similarly by means of the Bode plots. The marginalstability frequency is found by solving the equation £GOL ðu $ jÞ ¼ 180 . For the system to be stable, the magnitude of GOL(u$j) should be less than 1 for the marginal-stability frequency; this last condition can be used to establish the range of K, should the open-loop transfer function contain a gain factor K. nn

Example 12.21 Utilize the Nyquist plot and the Bode plots for the open-loop transfer function of Example 12.18 to find the range of the gain K for stable and unstable behavior. Also determine the value of K and the corresponding frequency for marginal stability.

Solution Based on the transfer function of Eq. (12.60), the following MATLAB code has been used to obtain the Nyquist plot of Figure 12.18: >> g = tf([1,4,3],[1,9,20]); >> nyquist(g)

C (ω = 2.869) −1/K2 −0.444

−1/K1

A (ω = 0) B (ω 0.15

FIGURE 12.18 Nyquist Plot of GOL(s) ¼ (s2  4s þ 3)/(s2 þ 9s þ 20).

∞) 1

12.4 Nyquist Plot and Bode Plots for Stability of Feedback Systems

The points on the Nyquist plot corresponding to the test point moving on the positive imaginary (u$j) axis can be determined from GOL(s) as provided in Eq. (12.60) by using s ¼ u$j and K ¼ 1, which is as follows:

GOL ð u $ jÞ ¼

  u2 þ 3  4u $ j u4  59u2 þ 60 u $ 13u2  107 ¼ þ $j u2 þ 20 þ 9u $ j u4 þ 41u2 þ 400 u4 þ 41u2 þ 400 (12.74)

 (u$j) is 60/400 ¼ 0.15 and its angle is 0 and, therefore, point A on For u ¼ 0, the magnitude of GOL the real axis is the corresponding point of the Nyquist plot. As u increases, the mapped point PC moves clockwise along the Nyquist contour, as illustrated in Figure 12.18. The clockwise rotation  (u$j), 4 ¼ tan1 can be checked if a test value of u ¼ 1 rad/s is used in the angle of GOL [u$(13u2  107)/(u4  59u2 þ 60)], which results in 4 ¼ 1.5495 rad. When u / N on the  (u$j) is 1 and its angle is 0 as per infinite semicircle’s imaginary axis, the magnitude of GOL Eq. (12.74), which define point B on the Nyquist plot. It should be noted that the remainder of the Nyquist plot (which corresponds to u assuming negative values on the imaginary axis of the infinite semicircle) is symmetric to the portion constructed already, and which corresponded to positive u, as shown in Figure 12.18.  (u$j) becomes zero. From The Nyquist plot crosses the real axis when the imaginary part of GOL 2 Eq. (12.74), this leads to 13u  107 ¼ 0 or u ¼ 2.8969 rad/s and with it, the magnitude of  (u$j) is 0.444; point C of Figure 12.18 identifies this point, which corresponds to a marginally GOL stable condition of the feedback system. The corresponding gain is calculated by solving the equation 1/K ¼ 0.444, which produces K ¼ 2.25. Example 12.18 has also identified the values u ¼ 2.8969 rad/s and K ¼ 2.25 as the ones associated with the crossing of the u$j axis by the root locus. Considering now that the critical point assumes the value 1/K1, as illustrated in Figure 12.18, there are two clockwise encirclements of this critical point by the Nyquist plot, which is N ¼ 2. Since there are no open-loop poles in the RHP, it means that Np ¼ 0 and the number of closed-loop poles in the RHP is Nz ¼ Np  N ¼ 0  (2) ¼ 2. This critical point’s position is such that 0.444 < 1/K1 or K1 > 2.25. As a consequence, for gain values that are larger than K ¼ 2.25 the system is unstable. On the contrary, when the critical point lies to the left of 0.444 in Figure 12.18, there are no encirclements of this point by the Nyquist plot, which means N ¼ 0. As a result, the number of RHP closed-loop poles is Nz ¼ Np  N ¼ 0  0 ¼ 0, and the system is stable; this occurs for 0.444 > 1/K2 or K2 < 2.25. These results were also obtained in Example 12.18. Similar conclusions are derived using information from the Bode plots. The sinusoidal open-loop transfer function of Eq. (12.74) has the following phase angle:

£GOL ð u $ jÞ

¼ tan

1



 u $ ð13u2  107 u4  59u2 þ 60

(12.75)

As just mentioned, at marginal stability, the phase angle is 180 , which results in u$(13u2  107) ¼ 0. The valid solution is u ¼ 2.8969 rad/s, which was also obtained by the Nyquist plot method. As the actual open-loop transfer function of Eq. (12.60) has a gain factor K in it, the actual transfer function corresponding to s ¼ u$j is

GOL ð u $ jÞ ¼ K $ GOL ð u $ jÞ ¼ K $

u2 þ 3  4u $ j u2 þ 20 þ 9u $ j

(12.76)

the magnitude of it is

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðu2 þ 3Þ þ ð4uÞ2 jK $ GOL ðu $ jÞj ¼ K $ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðu2 þ 20Þ þ ð9uÞ2

(12.77)

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CHAPTER 12 Stability of Feedback Control Systems

For stability, this magnitude should be less than 1 at the critical frequency u ¼ 2.8969 rad/s, which results in a gain K < 2.25, value which was also obtained by the Nyquist method. nn

Gain Margin and Phase Margin Consider a system with no open-loop poles in the RHP and the portion of the Nyquist plot illustrated in Figure 12.19, which also shows the unit-radius circle. Because point 1 is not encircled by the Nyquist plot, this system has Nz ¼ Np  N ¼ 0  0 ¼ 0 closed-loop poles in the RHP, and therefore the feedback system is a stable one.

Im[ GOL(s)]

1 ω

P PM = ϕ

−1 G

∞ Re[GOL(s)]

0 GOL (ω g ⋅ j )

Nyquist plot ω=0 1 = GOL (ω p ⋅ j ) GM

FIGURE 12.19 Nyquist Plot Portion of a Stable System.

Point P of Figure 12.19 where the Nyquist plot intersects the real axis is the phase crossover point. For this point the phase of GOL(u$j) is 180 , and the ratio of 1 (the length from the origin to 1) to the magnitude (modulus) of GOL(u$j) is known as the gain margin, GM ¼ 1/jGOL(u$j)j. When measured in decibels, the gain margin is GMdB ¼ 20$logjGOL(u$j)j. The frequency at which the gain margin is calculated is the phase crossover frequency up. A small gain margin results from a large magnitude of GOL(u$j) and indicates proximity to point 1, and therefore marginal stability; conversely, large gain margins correspond to intersection points that are closer to 0 and farther away from 1, so to more stable systems. Point G is the intersection of the Nyquist plot with the unit circle and is known as the gain crossover point. For this point the magnitude of GOL(u$j) is unity, and the difference between the phase of GOL(u$j) and 180 is known as the phase margin, 4 (or PM) which is PM ¼ 4 ¼ £GOL ðug $ jÞ  ð180 Þ and is calculated at the gain crossover frequency ug. A large phase margin is indicative of a large lag of GOL(u$j) with respect to the negative real semiaxis. Recommended ranges of the gain margin and phase margin resulting in acceptable/satisfactory performance are GMdB > 6 dB (which is GM > 2 approximately) and 30 < PM < 60 . When the Nyquist plot passes through the point 1 on the real axis in Figure 12.19, the system is marginally

12.4 Nyquist Plot and Bode Plots for Stability of Feedback Systems

stable, which yields GMdB ¼ 0 and PM ¼ 0 for one cross-over frequency ug ¼ up. For an unstable system, the Nyquist plot crosses the real axis to the left of 1. The Bode plots can also be used to visualize and assess the PM and GM of feedback control system. Consider, for instance, the open-loop transfer function of Example 12.19 and given in Eq. (12.64). The system was shown to be marginally stable for a gain K ¼ 6. Gain values smaller than 6 render the system stable, while gains K larger 6 result in unstable systems. Figure 12.20 plots the Bode diagrams for three different gains (and systems): K1 ¼ 6, K2 ¼ 1, and K3 ¼ 20.

MdB,3 > 0 K2 = 1 MdB,2 < 0

K3 = 20 K1 = 6

ωg2

ωp = ω g1

ω g3

PM2 > 0 PM3 < 0

FIGURE 12.20 Bode Diagrams of GOL(s) ¼ K/(s3 þ 2s2 þ 3s) for K1 ¼ 6 (Marginally Stable System), K2 ¼ 1 (Stable System), and K3 ¼ 20 (Unstable System).

For the marginally stable system (K ¼ 6), the magnitude diagram crosses the 0 dB line (where the absolute magnitude is equal to unity and GMdB ¼ 0) at the same frequency with the phase plot crossing the 180 line, which shows the phase and gain cross-over frequencies are identical, up ¼ ug1, as shown in Figure 12.20. When the gain is K < 6 (which is for stable systems), the magnitude curve shifts to the left and intersects the 0 dB line at a gain cross-over frequency ug2 < ug1. Because the phase plot remains unchanged through gain variations, the corresponding phase margin is PM2 ¼ 42  (180 ) > 0, as illustrated in the same Figure 12.20. The left shift of the magnitude plot results in a gain margin GMdB,2 which is positive; the reason is that the gain margin is calculated as GMdB,2 ¼ 20$log10M2(up) and 20$log10M2(up) < 0 (the dB magnitude of the system defined by K2), as seen in the same Figure 12.20. To conclude, a stable system of this type has a positive phase margin and a positive gain margin. It is straightforward to note that when K > 6 and the system is unstable, both the phase margin and the

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CHAPTER 12 Stability of Feedback Control Systems

K1 = 1 MdB,1 > 0 MdB,2 < 0

K2 = 0.1 2

FIGURE 12.21 Bode Diagrams of GOL(s) ¼ K$(s þ 4)$(s þ 5)/[(s  1)$(s  3)] for K1 ¼ 1 (Stable System) and K2 ¼ 0.1 (Unstable System).

gain margin become negative (such as PM3 ¼ 43  (180 ) < 0 and GMdB,3 < 0 because MdB,3 > 0, as in Figure 12.20). These conclusions regarding stability are valid for systems that display the (more) common behavior of being stable for relatively smaller values of the gain and becoming unstable for larger gains. In the case of feedback control systems that are unstable for smaller gains and regain stability at higher gains, the phase margin and gain margin ranges resulting in stable systems are opposite to the ones mentioned previously with reference to the system of Example 12.19. Consider the open loop-transfer function GOL(s) ¼ K$(s þ 4)$(s þ 5)/[(s  1)$(s  3)], for instance. Figure 12.21 shows the Bode plots corresponding to K1 ¼ 1, which results in a stable system, and K2 ¼ 0.1, which yields an unstable system. While MdB,1 > 0 gives a negative gain margin GMdB,1 < 0, MdB,2 < 0 relates to a positive gain margin GMdB,2 > 0. Because neither stable nor unstable systems can reach a 0 dB gain, the phase margins occur at an infinite frequency, this can be checked by using a margin(sys) MATLAB command, where sys is the studied transfer function. nn

Example 12.22

A unity-feedback control system has a plant defined by Gp(s) ¼ 1/(s2 þ 2s þ 1) and a controller Gc(s) ¼ K/s with K > 0. (i) Determine the gain value K1, which will render the system marginally stable. (ii) Calculate analytically and with MATLAB the gain margin GM, the phase margin PM, as well as the corresponding cross-over frequencies for a gain K2 ¼ 0.4$K1

12.4 Nyquist Plot and Bode Plots for Stability of Feedback Systems

Solution (i) The open-loop and closed-loop transfer functions are as follows:

K K ¼ ; s $ ðs2 þ 2s þ 1Þ s3 þ 2s2 þ s GOL ðsÞ K GCL ðsÞ ¼ ¼ 1 þ GOL ðsÞ s3 þ 2s2 þ s þ K

GOL ðsÞ ¼

(12.78)

The marginally stable condition occurs when two roots of the characteristic polynomial (the denominator DCL(s) of GCL(s) in Eq. 12.78) are imaginary; that means the third root of DCL(s) is a real root. As a consequence, DCL(s) can be written as follows:

  s3 þ 2s2 þ s þ K ¼ ðs þ aÞ $ s2 þ u2c ¼ s3 þ a $ s2 þ u2c $ s þ a $ u2c

(12.79)

where the real root is s ¼ a and the imaginary roots are s ¼ uc$j, with uc being the crossing frequency corresponding to the imaginary closed-loop poles. Identifying coefficients in Eq. (12.79), yields a ¼ 2, K ¼ 2, and uc ¼ 1 rad/s. Therefore the system is marginally stable for a gain K1 ¼ 2 at a frequency uc ¼ 1 rad/s. (ii) Figure 12.22 shows the Bode diagrams that result from using MATLAB’s bode command applied to the open-loop transfer function of Eq. (12.78) for K2 ¼ 0.4$K1 ¼ 0.5. The values of the gain and phase margin, as well as of their corresponding cross-over frequencies, are included in the same figure; these values were obtained with MATLAB’s margin command.

FIGURE 12.22 Bode Diagrams of GOL(s) ¼ 0.5/[s$(s2 þ 2s þ 1)].

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CHAPTER 12 Stability of Feedback Control Systems

For a gain K2, the complex open-loop transfer function together with its modulus and phase angle are as follows:

GOL ð u $ jÞ ¼

K2 1   ; $ u 2u þ 1  u2 $ j

jGOL ð u $ jÞj ¼

K2 1 $ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 ; u 4u2 þ 1  u2

£GOL ð u $ jÞ ¼ tan

1



(12.80)

2

 1  u2 1 u  1 ¼ tan 2u ð2uÞ

The phase margin PM is calculated for the gain cross-over frequency ug, which is evaluated by solving the equation jGOL ðu $ jÞj ¼ 1. With K2 ¼ 0.4$K1 ¼ 0.5 and the magnitude of Eq. (12.80), the following algebraic equation results:

u6g þ 2u4g þ u2g  0:25 ¼ 0

(12.81)

whose valid solution is ug ¼ 0.424 rad/s. The associated phase angle is given by Eq. (12.80) as 44.056 . However, as shown in Figure 12.22, the phase angle is actually negative. This indicates that, instead of the 44.056 value, the valid phase angle (resulting in the same tangent value) is in the third quadrant (where the tangent function is positive) and is £Gðug $ jÞ ¼ ð180  44:056 Þ. The phase margin is therefore PM ¼ £Gðug $ jÞ  ð  180 Þ ¼ 44:056 , as also indicated in Figure 12.22. The gain margin GM is calculated for the phase cross-over frequency up, which is determined as root of the equation £Gðup $ jÞ ¼ 180 . Note that 180 is used instead of þ180 for the marginal-stability threshold value because the phase angle is negative. Combining the condition above with the phase angle and magnitude of Eq. (12.80) yields:

8 u2  1 > 1 p > > tan ¼ 180 or up ¼ 1 rad=s > < 2up rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi >  2ffi > up 1 > 2 2 >   ¼ $ 4up þ 1  up ¼ 4 ¼ 12:04 dB : GM ¼ jGOL up $ j j K2 which are also the values provided in Figure 12.22.

(12.82)

nn

SUMMARY This chapter studies stable, marginally stable, and unstable feedback control dynamic systems. Using transfer function or state-space models, the stability of SISO and MIMO feedback control systems can be assessed analytically or with MATLAB by establishing the closed-pole position in the complex plane. The RoutheHurwitz array/criterion is utilized as the principal method of designing feedback control systems with variable parameters based on stability. The root locus method and the Nyquist plot (together with the Bode plots), as well at the phase and gain margins, are also introduced and applied as graphical tools of investigating the stability of SISO systems by analyzing the open-loop transfer function that comprises a variable gain.

Problems

PROBLEMS 12.1 Use MATLAB to determine the stability condition of a feedback control system whose closed-loop transfer function is GCL ðsÞ ¼ s2 þ s þ 1 by calculating and plotting 6 5 2s þ 15s þ 40s4 þ 12s3 þ 8s2 þ s þ 81 the position of the closed-loop poles in the complex plane. 12.2 A unity-feedback control system is formed of a P þ I þ D controller with KP ¼ 3, TI ¼ 0.1 s and TD ¼ 0.05 s and a plant defined as 1 . Analyze the stability of the sysGp ðsÞ ¼ 4 s þ s3 þ 10s2 þ 128s þ 16 tem using MATLAB. 12.3 A feedback control system is formed of a P þ D controller with KP ¼ 10 sþ1 , and and TD ¼ 0.01 s, a plant with Gp ðsÞ ¼ 3 2s þ 82s2 þ 10s þ 121 a feedback sensor defined by a transfer function H(s) ¼ 220/s. Analyze the stability of the system using MATLAB. 12.4 A unity-feedback MIMO control system is formed of a controller and a

 2 0 ; plant whose transfer function matrices are ½Gc ðsÞ ¼ 0 3 3 2 sþ1 1 6 s3 þ 2s2 þ s þ 1 s3 þ 2s2 þ s þ 1 7 7 6 ½Gp ðsÞ ¼ 6 7. Analyze the stability of 5 4 3 sþ2 3 2 3 2 s þ 2s þ s þ 1 s þ 2s þ s þ 1 this system using MATLAB commands. 12.5 Utilize MATLAB commands to establish the stability condition of a nonunity-feedback MIMO system defined by the following transfer functions: 2 3 3 1 0 6 s2 þ s þ 1 2 3 s2 þ s þ 1 7 6 7 sþ2 0 0 6 7 6 7 6 7 s 6 7; 6 7 0 0 ½Gc ðsÞ ¼ 4 0 3 0 5; ½Gp ðsÞ ¼ 6 2 7 s þsþ1 6 7 6 7 0 0 sþ1 4 2 sþ1 5 0 s2 þ s þ 1 s2 þ s þ 1 2 3 10 0 0 6 7 ½HðsÞ ¼ 6 0 7 4 0 30 5: 0 0 120

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CHAPTER 12 Stability of Feedback Control Systems

12.6 Establish the stability condition of a feedback control system modeled in state-space form whose state matrix corresponding to the closed-loop 2 3 2 3 1 6 7 transfer function matrix is ½A ¼ 4 8 10 5 5. 6 4 7 3 2 1 0 0 2 6 0 2 5 10 7 7 6 12.7 Solve Problem 12.6 for the state matrix ½A ¼ 6 7. 40 0 1 3 5 0

0 0

8

12.8 The feed-forward transfer function of a unity-feedback control system is G(s) ¼ (s  a)/[(s þ b)$(s þ c)] where a, b, and c are real positive parameters. Use analytical calculation to establish the stability condition of this system. 12.9 Use analytical calculation to determine the stability condition of a MIMO

 a aþb feedback control system whose state matrix is ½A ¼ with a b a and b being real parameters. 12.10 A unity-feedback control system has the following characteristic polynomial: DCL(s) ¼ s4 þ 7s3 þ 18s2 þ 20s þ 8. Use the RoutheHurwitz criterion to study the stability of this system and confirm your conclusions with MATLAB. 12.11 Solve Problem 12.10 for the characteristic polynomial: DCL(s) ¼ s5 þ s4 þ 2s3 þ 2s2 þ 5s þ 3. 12.12 Solve Problem 12.10 for the characteristic polynomial: DCL(s) ¼ s6 þ 4s5 þ 16s4 þ 42s3 þ 65s2 þ 54s þ 18. 12.13 Solve Problem 12.10 for the characteristic polynomial: DCL(s) ¼ s5 þ s4 þ 2s3 þ 2s2 þ 7s þ 7. 12.14 A unity-feedback control system is defined by a controller Gc(s) ¼ 20 þ 5/ s þ 10s and a plant Gp(s) ¼ 2/(s2 þ 15s þ 50). Use the RoutheHurwitz criterion to study the stability of this system and confirm your conclusions with MATLAB. 12.15 A nonunity-feedback control system is defined by a controller (compensator) Gc(s) ¼ (s þ 20)/(s þ 1), a plant Gp(s) ¼ 1/(s2 þ 2s þ 100), and a feedback (sensor) H(s) ¼ 52/s. Use the RoutheHurwitz criterion to study the stability of this system and confirm your conclusions with MATLAB. 12.16 Use the RoutheHurwitz criterion to establish the stability condition of the MIMO feedback control system of Problem 12.4.

Problems

12.17 Use the RoutheHurwitz criterion to establish the stability condition of the MIMO feedback control system of Problem 12.5. 12.18 Use the RoutheHurwitz criterion to establish the stability condition of the MIMO feedback control system of Problem 12.6. 12.19 Discuss the stability of the feedback control system of Problem 11.28 and shown in the block diagram of Figure 11.46 in terms of the variable gain K. 12.20 The closed-loop characteristic polynomial of a feedback control system is DCL(s) ¼ 2s6 þ s5 þ 2s4 þ s3 þ K$s2 þ 6s þ K þ 10. Analyze its stability condition in terms of the positive gain K. 12.21 The closed-loop characteristic polynomial of a feedback control system is DCL(s) ¼ s6 þ 3s5 þ s4 þ 3s3 þ K1$s2 þ 2s þ K2 þ 4. Analyze its stability condition in terms of the positive gains K1 and K2. 12.22 The closed-loop transfer function of a unity-feedback control system is GCL(s) ¼ 1/(s2  1). Evaluate whether it is possible or not to stabilize this system by adding a P þ D controller. In case stabilizing is doable, determine the constants of the additional controller. 12.23 The transfer function of a dc motor (which acts as a plant, whose input is the armature voltage and the output is the shaft rotation angle) is 1 Gp ðsÞ ¼ ðsþ1Þ $ ðsþ2Þ $ ðsþ3Þ. The plant interacts with a P þ I þ D controller

in a unity-feedback control system. Considering that KI ¼ 1, determine the proportional gain KP and the derivative gain KD such that the system has two stable poles and two imaginary poles. Specify the necessary values, ranges, or relationships between these constants.

12.24 A feedback control system similar to the one of Example 12.15 and illustrated in Figure 12.3 is studied. The plant is defined by the € þ a1 $ cðtÞ _ þ a0 $ cðtÞ ¼ a0 $ mðtÞ. Its differential equation: a2 $ cðtÞ natural frequency is 220 rad/s, and damping ratio is 0.4. A P þ D controller is used instead of the P þ I controller. This controller is defined by the positive constants KP and KD, and the feedback sensor is defined as H(s) ¼ 60/s. Determine the values of KP and KD, which will render the system stable, marginally stable, or unstable. 12.25 A unity-feedback MIMO control system is formed of a controller and a

 s þ 10 0 ; plant whose transfer function matrices are ½Gc ðsÞ ¼ 0 25 3 2 5 s 6 s2 þ 2s þ 100 s2 þ 2s þ 100 7 7 6 ½Gp ðsÞ ¼ 6 7. Use the RoutheHurwitz cri5 4 3s þ 4 K 2 2 s þ 2s þ 100 s þ 2s þ 100 terion to analyze the stability of this system for K being a positive gain.

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CHAPTER 12 Stability of Feedback Control Systems

12.26 The open-loop and feedback transfer functions of a MIMO feedback control system are ½GOL ðsÞ ¼ 2 3 1 2s 6 s3 þ 2K $ s2 þ s þ K þ 100 s3 þ 2K $ s2 þ s þ K þ 100 7 1 2 1 2 6 7 6 7; 4 5 sþ1 3 s3 þ 2K1 $ s2 þ s þ K2 þ 100 s3 þ 2K1 $ s2 þ s þ K2 þ 100

 180 0 ½HðsÞ ¼ . Use the RoutheHurwitz criterion to analyze the 0 625 stability of this system for positive gains K1 and K2. 12.27 Use the RoutheHurwitz criterion to analyze the 2 1 2K 6 system whose state matrix is ½A ¼ 4 2 5 1 1 positive gain.

stability of a control 3 3 7 8 5 with K being a 0

12.28 The open-loop transfer function of a feedback control system has the poles 4, 2, 1 and the zeroes 6, 5. Verify whether the point s ¼ 3 is on the root locus and calculate the corresponding gain K if the point is on the root locus. Is this system stable? 12.29 The open-loop transfer function of a feedback control system has the pole 4 þ j and the zero 1 þ 2j. Verify whether the point s ¼ 2 is on the root locus. If not, what happens when adding the open-loop zero 3? In case the point is on the root locus, calculate the corresponding gain K. 12.30 Sketch the root locus for a unity-feedback system whose open-loop transfer function is K$(s2 þ s)/[(s  1)$(s þ 2)$(s þ 3)] and determine the range of the gain K for stability. 12.31 The open-loop transfer function of a feedback control system is

K . s½ðsþ2Þ2 þ1

(i) Sketch the root locus corresponding to this transfer function. (ii) Calculate possible break-away, break-in, and u$j-axis crossing points of the root locus. What is the range of K for stability? 12.32 The open-loop transfer function of a feedback control system is K/ (s3 þ 4s2 þ 5s). (i) Sketch the root locus corresponding to this transfer function and confirm the sketch using MATLAB rlocus command. (ii) Calculate possible break-away, break-in, and u$j-axis crossing points of the root locus. What is the range of K for stability?

Problems

12.33 The open-loop transfer function of a feedback system with variable gain K has the poles of 1 þ j and 1; it also has a zero of 1. (i) Sketch the root locus of this system knowing that it has a total of three closed-loop poles, and that the root locus plot has only one asymptote. (ii) Determine the open-loop transfer function of this system and verify the validity of the sketched root locus by plotting the root locus with MATLAB. (iii) Calculate possible break-in, break-away, and u$j-axis crossing points, as well as the values of the gain corresponding to these points. 12.34 The open-loop transfer function of a feedback control system has the poles 3, 2, and 1, and no finite zeroes. Sketch the root locus path and confirm your sketch by using MATLAB. Should there be any, determine the break-away, break-in, and u$j-axis crossing points. 12.35 The open-loop transfer function of a unity-feedback system is

Kðs2 6sþ8Þ s2 þ6sþ25 .

(i) Sketch the root locus for this system. (ii) Determine possible break-in, break-away, and u$j-crossing points, as well as the values of the gain corresponding to those points. (iii) Find the range of the gain K for which the system is stable. 12.36 A root locus plot is shown in Figure 12.23. Is this root locus possible at all? In case it does, find the corresponding open-loop transfer function. Assuming the open-loop transfer function is connected in a unity-feedback system, check whether the system is stable for a gain K ¼ 0. 12.37 Utilize the Nyquist plot and the Bode plots for the open-loop transfer function of Problem 12.35 to find the range of the gain K for stable and stable behavior. 12.38 The open-loop transfer function of a feedback transfer function has the poles 5 (double), 3, 2 and the zeroes 10 and 8 (double). Use the Nyquist and Bode plots to study the stability of this unity-feedback system in terms of a variable gain K. 12.39 The open-loop transfer function of a feedback control system is G(s) ¼ 1/ (s3 þ 3s2 þ s þ 1). Calculate the gain margin, the phase margin, and their corresponding crossing frequencies both analytically and with MATLAB’s margin command. Is this system stable? 12.40 Solve Problem 12.39 for G(s) ¼ 1/[s$(s2 þ 4s þ 1)].

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CHAPTER 12 Stability of Feedback Control Systems

FIGURE 12.23 Root Locus of an Unknown Unity-Feedback Control System.

Suggested Reading W.R. Evans, Control-System Dynamics, McGraw-Hill, New York, 1954. N.S. Nise, Control Systems Engineering, 7th Ed. Wiley, Hoboken, NJ, 2015. K. Ogata, Modern Control Engineering, 5th Ed. Prentice Hall, Pearson, Upper Saddle River, NJ, 2009. G.F. Franklin, J.D. Powell, and A. Emami-Naeini, Feedback Control of Dynamic Systems, 6th Ed. Pearson, Upper Saddle River, NJ, 2010. C.-T. Chen, Linear System Theory and Design, Holt, Rinehart and Winston, New York, 1984. N. Matsumoto, Simple proof of the Routh stability criterion based on order reduction of polynomials and principle of argument, ISCAS 2001 e The 2001 International Symposium on Circuits and Systems, 2001, 1, 699e702. M.-T. Ho, A. Datta, and S.P. Bhattacharyya, An elementary derivation of the Routh-Hurwitz criterion, IEEE Transactions on Automatic Control, 43(3), 1998, 405e409. A.P. Sage, Linear Systems Control, Matrix Publishers, Champaign, IL, 1978. R.C. Dorf and R.H. Bishop, Modern Control Systems, 13th Ed. Prentice Hall, Upper Saddle River, NJ, 2016. C.L. Phillips and J.M. Parr, Feedback Control Systems, 5th Ed. Prentice Hall, Upper Saddle River, NJ, 2011. R.T. Stefani, B. Shahian, C.J. Savant, and G.H. Hostetter, Design of Feedback Control Systems, 4th Ed. Oxford University Press, Oxford, 2002. J.P. den Hartog, Mechanical Vibrations, Dover, New York, 1985. T.D. Roopamala and S.K. Katti, Comments on Routh Stability Criterion, International Journal of Computer Science and Information Security, 7(2), 2010, 77e78.

CHAPTER

Time- and FrequencyDomain Controls of Feedback Systems

13

OBJECTIVES Based on the material presented in previous chapters, this last chapter focuses on the time-domain and frequency-domain modeling, analysis, and design of feedback control systems. The following topics are studied here: • Transient response characteristics of first- and second-order control systems under unit-step input. • Steady-state response of control systems. • Utilization of the root locus method to characterize the time response of feedback control systems. • Control effects on the time response of single-input, single-output (SISO) and multiple-input, multiple-output (MIMO) feedback control systems. • Modeling and analysis of control systems in the frequency domain by means of Bode plots. • Design of phase-lead and phase-lag compensators for frequency-domain performance enhancement of feedback control systems.

INTRODUCTION This chapter is dedicated to the analysis and design of feedback control systems, as well as to defining, characterizing, and improving the performance of these systems in the time domain and the frequency domain. The time domain is divided into the transient range and the steady-state range. Characteristics defining both subdomains are formulated, and examples are used to illustrate these characteristics for first- and second-order systems. The root locus, as well as the time-domain solution, is utilized to model the time response of SISO and MIMO control systems. The frequencydomain response of SISO control systems is mostly modeled with the Bode plots by defining system characteristics pertaining to this domain. The connections between frequency- and time-domain performance qualifiers are formulated and studied here. The chapter also introduces notions of phase-lead and phase-lag compensator design in the frequency domain. System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.00013-0 Copyright © 2018 Elsevier Inc. All rights reserved.

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

13.1 TIME-DOMAIN RESPONSE OF SISO FEEDBACK CONTROL SYSTEMS The performance of systems in the time domain is studied by focusing on the transient (initial) response and on the steady-state (final) response of SISO systems. The control system time response is defined and characterized in terms of specific performance parameters by analyzing first-, second-, and higher-order systems. While the systems studied are mostly linear, nonlinear feedback control time-domain response is also exemplified in this section.

13.1.1 Transient Time-Domain Response and Specifications One area where the performance of dynamic systems can be improved through controls is the transient time-domain response, which describes the natural response (and is governed by the homogeneous differential equation that is the mathematical model) of a control system. Transient time response specifications are defined for first- and second-order uncontrolled and feedback-uncontrolled systems. Section 13.1.3 is looking at the specific influence of various types of controls on the transient response of feedback control systems.

First-Order Systems The differential equation of a first-order feedback control system connecting the input r(t) to the output c(t) is of the general form given in Eq. (1.4): a1 $ c_ðtÞ þ a0 $ cðtÞ ¼ b $ r ðtÞ or s $ c_ðtÞ þ cðtÞ ¼ K $ r ðtÞ

(13.1)

where s ¼ a1/a0 is the time constant and K ¼ b/a0 is the static sensitivity. The closed-loop transfer function results by applying the Laplace transform to Eq. (13.1), namely: GCL ðsÞ ¼

C ðsÞ K K=s ¼ ¼ RðsÞ s $ s þ 1 s þ 1=s

(13.2)

For a unit-step input r(t) ¼ 1, which means R(s) ¼ 1/s, and unit sensitivity (K ¼ 1) the Laplace-domain output C(s)dobtained from Eq. (13.2)dand the corresponding time-domain response c(t) are: CðsÞ ¼ GCL ðsÞ $ RðsÞ ¼

1=s ; cðtÞ ¼ 1  et=s s $ ðs þ 1=sÞ

(13.3)

Figure 13.1 shows a typical response curve that corresponds to a time constant of s ¼ 0.1 s. For t ¼ s, a response value of c(s) ¼ 1 e e1 ¼ 0.63 is obtained from Eq. (13.3), which shows that after a time interval equal to the time constant, the response (output) reaches 63% of the input, which, for this particular case, is equal to 1 (one). Two other time parameters are representative of a first-order system time response, namely the rise time sr and the settling time ss. Various definitions are

13.1 Time-Domain Response of SISO Feedback Control Systems

0.98 0.9

τ

τr

τs

FIGURE 13.1 Unit-Step Time Response of First-Order System With Time Specifications.

being used in the controls literature for sr; we shall consider here that the rise time is the time necessary for the output to reach 90% of the input. Solving the equation 1  esr =s ¼ 0:9, results in sr ¼ 2.3s, and this point is indicated in the plot of Figure 13.1. The settling time ss is defined here as the time necessary for the output to reach 98% of the input, but other definitions are also offered in the controls literature. The value of ss is determined by solving the equation: 1  ess =s ¼ 0:98 as ss ¼ 3.91s, but is generally approximated as four times the time constant, ss ¼ 4s. The point corresponding to the settling time is also indicated in Figure 13.1. In general, the settling time is associated with the start of the steady-state intervald the steady-state response and the corresponding errors (differences between reference and output signals) are studied in more detail in Section 13.1.2 of this chapter.

Second-Order Systems The differential equation of a second-order feedback control system connecting the input r(t) to the output c(t) is: a2 $ c€ðtÞ þ a1 $ c_ðtÞ þ a0 $ cðtÞ ¼ b $ r ðtÞ or

(13.4) c€ðtÞ þ 2x $ un c_ðtÞ þ u2n $ cðtÞ ¼ K $ u2n $ r ðtÞ  
with un being the natural frequency u2n ¼ a0 a2 , x the damping ratio ð2x $ un ¼ a1 =a2 Þ, and K the static sensitivity (K ¼ b/a0)dparameters that have

649

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

been introduced in Eq. (1.10). The closed-loop transfer function of the second-order control system is obtained by taking the Laplace transform of Eq. (13.4) as: GCL ðsÞ ¼

C ðsÞ K $ u2n ¼ 2 RðsÞ s þ 2x $ un $ s þ u2n

(13.5)

The time response c(t) that is produced by a unit-step input r(t) ¼ 1 for unit sensitivity (K ¼ 1) is analyzed here. Different response situations, which depend on the value of the damping ratio, can be connected to the nature and position of the closed-loop poles, which are the roots of the characteristic equation: s2 þ 2x $ un $ s þ u2n ¼ 0. For a unit-step input, the Laplace-domain output is obtained from Eq. (13.5) as: CðsÞ ¼ GCL ðsÞ $ RðsÞ ¼

u2n  s $ s2 þ 2x $ un $ s þ u2n


(13.6)

There are four situations possible, depending on the value of the damping ratio, namely: overdamping (x > 1), critical damping (x ¼ 1), underdamping (0 < x < 1), and no damping (x ¼ 0). Figure 13.2 illustrates the closed-loop pole equations, the pole positions in the complex plane, as well as the corresponding unit-step time responses for second-order control systems, with various damping ratios. p = −ξ ⋅ ω ± 1 − ξ ⋅ ω ⋅ j

ω·j 1 − ξ ⋅ ωn

σ − 1− ξ ⋅ ω

ω·j

1.8

ω

1.6

0 1

0

0.05

0.1

0.15

0.2 0.25 0.3 Time (seconds)

0.35

0.4

0.45

p = −ω

0.5

( −ξ +

(

)

ξ −1 ⋅ ω

= −ξ ± ξ −

σ

)⋅ω

FIGURE 13.2 Closed-Loop Pole Positions of Second-Order Control System and Unit-Step Time Response: Underdamped (0 < x < 1), Critically Damped (x ¼ 1), Overdamped (x > 1), and Undamped (x ¼ 0).

Underdamped Second-Order Feedback Systems Many control system applications involve values of the damping ratio that are less than 1, so they address the underdamped case. For first-order systems, two timedomain specifications, the rise time and the settling time, have been introduced to

13.1 Time-Domain Response of SISO Feedback Control Systems

characterize the response to unit input. For second-order systems, in addition to these time specifications, two other parameters are important, namely the peak time and the maximum overshoot. They are best defined based on the plot of c(t), so c(t) will be determined explicitly for underdamping. C(s) of Eq. (13.6) is expanded in partial fractions as: pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s þ x $ un x 1  x2 $ un ffiffiffiffiffiffiffiffiffiffiffiffi ffi p C ðsÞ ¼   $

 2 2 s ðs þ x $ un Þ2 þ 1  x2 $ u2n 1  x2 ðs þ x $ un Þ þ 1  x $ u2n (13.7) The inverse Laplace transform of C(s) of Eq. (13.7) is: "

# x c ðt Þ ¼ 1  e $ cosðud $ tÞ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi $ sinðud tÞ 1  x2 pffiffiffiffiffiffiffiffiffiffiffiffiffi ! 2 1 x $ un $ t 1 1  x $ sin ud $ t þ tan ¼ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffi $ e x 1  x2 x $ un $ t

(13.8)

pffiffiffiffiffiffiffiffiffiffiffiffiffi where ud ¼ 1  x2 $ un is the damped frequency, which is introduced in Chapter 3. Figure 13.3 contains several plots of c(t) that were obtained for various values of the damping ratio and for un ¼ 100 rad/s.

2 fu(t) ξ = 0.05 ξ = 0.2

c(t)

1.5

1

0.5 ξ = 0.7 fl(t)

0

0

0.1

0.2

0.3

0.4

0.5

Time (sec)

FIGURE 13.3 Response Curves c(t) for Second-Order Control System With Underdamping and UnitStep Input.

651

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

It can be seen that smaller amounts of damping generate larger and more sustained oscillations (e.g., x ¼ 0.05), whereas larger damping ratios produce vibrations that are more rapidly decaying to the input level (as the case is with x ¼ 0.7). In Figure 13.3, envelope curves are drawn that are tangent to the crests of one of the curves corresponding to x ¼ 0.05 and are defined by the equations: fl ðtÞ ¼ 1  ex $ un $ t and fu ðtÞ ¼ 1 þ ex $ un $ t , where the subscript “l” stands for lower and the subscript “u” denotes the upper curve. Both are decaying curves whose limits are 1 when time grows to infinity. A similar exponentially decaying curve is the unit-step time response of a first-order control systemdEq. (13.3), where the exponential is f ðtÞ ¼ et=s , with s being the time constant. If we compare f(t) with the exponential parts of fl(t) and fu(t), it follows that the time constant of an underdamped second-order control system under unit-step input is: s¼

1 x $ un

(13.9)

Large values of the damping ratio and the natural frequency result in small time constants and therefore in fast system responsesdthis is seen in Figure 13.3 and the response curves corresponding to x ¼ 0.2 and x ¼ 0.7. The four parameters that define the time response of second-order control systems under unit-step input are: the peak time, maximum overshoot, rise time, and settling time, and they are defined based on Figure 13.4. Because the response reaches the input and exceeds it, the rise time sr can be defined as the time the control system needs to reach the unit input for the first time, and in our case, the equation to determine it is c(sr) ¼ 1. By using this condition with c(t) of Eq. (13.8) yields: " # x x $ un $ sr $ cosðud $ sr Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi $ sinðud $ sr Þ ¼ 0 or e 1  x2 pffiffiffiffiffiffiffiffiffiffiffiffiffi ! (13.10) 1  x2 tan1 x sr ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  x2 $ u n The rise time results from annulling the bracket of Eq. (13.10) and when the minus sign is neglected. The settling time ss is the time after which the response reaches and remains within 2% of the input. Mathematically, this requirement is: jcðss Þ  1j  0:02; in conjunction with Eq. (13.8), this results in:  pffiffiffiffiffiffiffiffiffiffiffiffiffi !  2  1  x $ un $ ss  1 1  x pffiffiffiffiffiffiffiffiffiffiffiffiffi $ e $ sin ud $ ss þ tan  2   x 1x (13.11) 1 x $ un $ ss  0:02  pffiffiffiffiffiffiffiffiffiffiffiffiffi $ e 1  x2

13.1 Time-Domain Response of SISO Feedback Control Systems

cmax c(t)

τr

cmax – 1

r(t) = 1

τp

τs Time (sec)

FIGURE 13.4 Typical Response Curve c(t) for Second-Order Control System With Underdamping and Unit-Step Input.

which, at limit, means solving the equation: 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi ex $ un $ ss ¼ 0:02 or ss ¼  1  x2

 pffiffiffiffiffiffiffiffiffiffiffiffiffi  ln 0:02 1  x2 x $ un

(13.12)

An accurate approximation of the settling time given in Eq. (13.12) is ss ¼

4 ¼ 4s x $ un

(13.13)

where s is the time constant of Eq. (13.9). The typical response curve that is plotted in Figure 13.4 displays a peak value, which defines the maximum overshoot (cmax  1). The time corresponding to this maximum response is known as peak time, and it can be computed by solving the equation: dcðtÞ=dtjt¼sp ¼ 0. This condition is combined with the first form of Eq. (13.8) and the result is:
 un pffiffiffiffiffiffiffiffiffiffiffiffiffi $ ex $ un $ sp sin ud $ sp ¼ 0 or ud $ sp ¼ p or 1  x2 (13.14) p p sp ¼ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ud 1  x2 $ un

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

For an input different from unity, the maximum (peak) overshoot is defined as: Op ¼

cmax  cðNÞ cðNÞ

(13.15)

where c(N) is the steady-state response (that is equal to the input in the case we analyze). For the concrete situation where the input is a unit step and where c(N) / 1, the maximum overshoot becomes: Op ¼ cmax  1 ¼ e

$x ppffiffiffiffiffiffi 1x2

(13.16)

which resulted by substituting the peak time of Eq. (13.14) into c(t) of the first-form Eq. (13.8). While the peak time depends on both the natural frequency and the damping ratio, the maximum overshoot is only dependent on damping ratio. The time specifications defined herein can be connected by means of the damping ratio and the natural frequency to the pole position of a second-order, underdamped unity-feedback control system with unit-step input. Figure 13.5 shows the position of one of the two conjugate closed-loop poles corresponding to underdamping. ω·j P

1 − ξ 2 ⋅ ωn = ωd

ωn α −ξ ⋅ ωn = σ d

O

σ

FIGURE 13.5 Closed-Loop Complex Plane Pole Position for Underdamped Unity-Feedback Control System With Unit-Step Input.

The underdamped pole can be expressed in complex number form as p ¼ sd þ ud$j where sd is the real part and ud is the imaginary part, as shown in Figure 13.5; the pole magnitude and its direction are: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sd ¼x (13.17) jpj ¼ OP ¼ s2d þ u2d ¼ un ; jcosaj ¼ un This result indicates that if the pole moves around a circular trajectory (centered at O), the natural frequency of the system remains unchanged. It also shows that when the pole moves radially along a line of constant inclination, the damping ratio is unaltered. Clockwise rotation of the radial line OP (which increases the angle a) reduces the damping ratio and conversely, counterclockwise rotation increases the damping effect.

13.1 Time-Domain Response of SISO Feedback Control Systems

1.6

100

1.5

80

Op (%)

τr (sec)

1.4 1.3 1.2

40 20

1.1 1

60

0

0.2

0.4

0.6

0.8

1

0

0

0.2

0.4

0.6

ξ

ξ

(a)

(b)

0.8

1

FIGURE 13.6 (a) Rise Time Versus Damping Ratio; (b) Maximum Overshoot Versus Damping Ratio.

Because sd ¼ x$un, the settling time of Eq. (13.13) can be written as ss ¼ 4=sd , which shows that the settling time will remain unmodified as long as the pole moves up or down on a specified vertical line in the complex plane. Moving the vertical line to the left in Figure 13.5 increases the abscissa sd and thus the settling time decreases. Eq. (13.14) demonstrates that the peak time is inversely proportional to the damped frequency ud, which shows that when the pole moves on a specified horizontal line (to the left or to the right in Figure 13.5), the peak time remains constant. To reduce the peak time, a horizontal line has to be displaced upward in the complex plane, which increases the coordinate ud. If a value of un ¼ 1 rad/s is used in Eq. (13.10), the rise time becomes a function of only the damping ratiodthis function is plotted in Figure 13.6(a). It can be seen that the rise time sr decreases when the damping ratio increases. Also, when a radial line OP rotates counterclockwise (which reduces the angle ad see Figure 13.5), the damping ratio becomes larger. The rotation of the pole while preserving the radius keeps the natural frequency unchanged (the natural frequency is in the denominator of sr), but, overall, the rise time is reduced. There is a similar trend in the maximum overshoot Op, which decreases when the damping increases, as shown in Figure 13.6(b). Again, counterclockwise rotation of the radial line OP in Figure 13.5 generates an increase in damping, which reduces the maximum overshoot. nn

Example 13.1 The closed-loop transfer function of a unit-sensitivity feedback control system is 2500 GCL ðsÞ ¼ s 2 þ40sþ2500 , and its time response under unit-step input is being analyzed. We need reductions in the maximum overshoot of 15% and in the rise time of 20%. What changes in the damping

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

ratio and natural frequency need to be operated for that? What will be the modifications in the settling time and peak time? How will the closed-loop poles move in the complex plane as a result of these modifications?

Solution

The closed-loop transfer function of this problem shows that un ¼ 50 rad/s, and therefore 2x$un ¼ 40, which results in a damping ratio of x ¼ 0.4. The original rise time is calculated by means of Eq. (13.10) as sr ¼ 0.0253 s, and the original maximum overshoot is determined by Eq. (13.16) as Op ¼ 25.38%. According to the requirements of the problem, the modified rise time and maximum overshoot are:

(

sr ¼ sr  0:2 $ sr ¼ 0:0202

Op ¼ Op  0:15 $ Op ¼ 21:575%

(13.18)

The modified (new) damping ratio is obtained by expressing it in terms of the maximum overshoot of Eq. (13.16) as:

ln

Op

!

100 x ¼ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " !#2ffi ¼ 0:4387 u  u O p tp2 þ ln 100 

(13.19)

The modified natural frequency is calculated from Eq. (13.10) as:

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  ðx  Þ2

un

tan1 x ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 61:4 rad=s 1  ðx Þ2 $ sr

(13.20)

With these values of the damping ratio and natural frequency, the closed-loop transfer function of this example problem modifies to:

GCL ðsÞ ¼

s2

3770 þ 53:87s þ 3770

(13.21)

To plot the pole position in the complex plane we use MATLAB’s pzmap(sys), which enables plotting the poles and zeroes of an LTI object (such as a transfer function in our case) in the complex plane. The following MATLAB code generates the pole position plot of Figure 13.7: >> tf1 = tf(2500,[1,40,2500]); >> tf2 = tf(3770,[1,53.87,3770]); >> pzmap(tf1,tf2) The arrows indicate how the poles moved after the modifications of the damping ratio and natural frequency. The poles motion to the left indicates that the settling time decreased, while the poles moving away vertically reduce the peak time value.

13.1 Time-Domain Response of SISO Feedback Control Systems

Pole-Zero Map 60

Imaginary Axis (seconds-1)

40

20

0

-20

-40

-60 -30

-25

-20

-15

-10

-5

0

Real Axis (seconds-1)

FIGURE 13.7 Closed-Loop Pole Position in the Complex Plane.

nn

13.1.2 Steady-State Time-Domain Response and Errors In addition to model inaccuracies in controlled systems, the interaction between the input and the configuration of the control system can lead to differences between the controlled output and the reference (input) signals in the form of an error. In its initial period, the time response is different from the input inherently, as we have seen, for both first-order and second-order systems. As a consequence, it is important that the errors are minimal or completely eliminated after the system enters its settled phase. The error of a general nonunity-feedback system, as the one sketched in Figure 11.11(a), is defined as the difference between the input (reference) signal and the back output (control) signal: eðtÞ ¼ rðtÞ  bðtÞ

(13.22)

The steady-state error is the limit of this error when time grows to infinity and can be calculated by means of the final-value theorem as: eðNÞ ¼ lim eðtÞ ¼ lim ½s $ EðsÞ ¼ lim ½s $ ðRðsÞ  BðsÞÞ t/N s/0 s/0

s $ RðsÞ ¼ lim ½s $ ð1  HðsÞ $ GCL ðsÞÞ $ RðsÞ ¼ lim s/0 s/0 1 þ HðsÞ $ GðsÞ

(13.23)

657

658

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

Eq. (13.23), which used GCL(s) ¼ G(s)/[1 þ H(s)$G(s)], indicates that the steadystate errors are influenced by the control system through its controller, plant and feedback transfer functions, as well as by the type of input signal.

Unity-Feedback Control Systems For a unity-feedback control system, as the one illustrated in Figure 11.12, where H(s) ¼ 1, Eq. (13.23) yields the steady-state error:

s $ RðsÞ (13.24) eðNÞ ¼ lim eðtÞ ¼ lim t/N s/0 1 þ GðsÞ A more systematic approach to studying the steady-state errors of feedback plant systems takes into account possible variants of systems (as the control-plant combination) and inputs. In general, three basic input types are used to test and study unityfeedback control systems, namely: unit step, unit ramp, and unit parabola. Table 13.1 Correlation Between Basic Input, Steady-State Errors, and Constants r(t)

R(s)

unit step

r (t ) r (t ) = 1

RðsÞ ¼

e(N)

Constant

1 s

1 1 ¼ 1 þ lim GðsÞ 1 þ Kp

Position Kp ¼ lim GðsÞ

1 s2

1 1 ¼ lim s $ GðsÞ Kv

Velocity Kv ¼ lim s $ GðsÞ

1 1 ¼ lim s2 $ GðsÞ Ka

Acceleration Ka ¼ lim s2 $ GðsÞ

s/0

s/0

t

unit ramp

r (t )

RðsÞ ¼

r (t ) = t

s/0

s/0

t

unit parabola

r (t )

r (t ) =

2

t 2

RðsÞ ¼

1 s3

s/0

s/0

t

Table 13.1 illustrates these inputs, their Laplace transforms, and the steady-state errors that can be calculated with Eq. (13.24) for each particular input in terms of a related constant (Kpdposition constant for unit-step input, Kvdvelocity constant for unit ramp input, and Kadacceleration constant for unit parabola input). Consider for instance a unit-parabola input r(t) ¼ t2/2, which corresponds to R(s) ¼ 1/s3 being the input to a feedback system. According to Eq. (13.24), the steady-state error is: eðNÞ ¼ lim

s/0

s 1 1 1 $ ¼ ¼ 1 þ GðsÞ s3 lim ½s2 $ GðsÞ Ka

(13.25)

s/0

which is also the result included in the last row of Table 13.1, with  Ka ¼ lim s2 $ GðsÞ being the acceleration constant. The other errors and their cors/0

responding constants of Table 13. 1, are obtained similarly.

13.1 Time-Domain Response of SISO Feedback Control Systems

The feed-forward transfer function can be written in factor form as: K $ ðs  z1 Þ $ ðs  z2 Þ. ðs  p1 Þ $ ðs  p2 Þ.

GðsÞ ¼

(13.26)

where K is a gain, z1, z2,. are zeroes (they are written here as distinct ones, but they can also be multiple) and p1, p2,. are the poles (some of them can also be multiple). Table 13.1 shows that the position constant Kp is constant, while the velocity constant Kv and the acceleration constant Ka are zero for the transfer function of Eq. (13.26). That results in steady-state errors that are constant for unit input, and infinite for unit ramp and unit parabola input, as can be checked in the same Table 13.1. As a consequence, to realize zero steady-state error, it is necessary that the feed-forward transfer function be of the form: GðsÞ ¼

K $ ðs  z1 Þ $ ðs  z2 Þ. sn $ ðs  p1 Þ $ ðs  p2 Þ.

(13.27)

where n is a positive integer. The value of n defines the system type, in the sense that a type-zero system has n ¼ 0, a type-one system has n ¼ 1, while a type-two system has n ¼ 2, and so forth. Having defined the feed-forward transfer function as in Eq. (13.27) allows calculation of the three types of constants, and of the corresponding steady-state errors for various system types. A type-zero system will have the following position constant: K $ ðs  z1 Þ $ ðs  z2 Þ. s/0 s0 $ ðs  p1 Þ $ ðs  p2 Þ.

Kp ¼ lim

¼

K $ z1 $ z2 . $ ð  1Þ p1 $ p2 .

(13.28)

which has a constant value; as a consequence, the steady-state error that is associated with itdsee Table 13.1dis also constant. For a type-one system, the position constant is: K $ ðs  z1 Þ $ ðs  z2 Þ. s/0 s1 $ ðs  p1 Þ $ ðs  p2 Þ.

Kp ¼ lim

¼N

(13.29)

and therefore the corresponding steady-state error is zero. It is clear that for type-one systems and higher, the steady-state error is zero for unit-step input. Similar calculations can be carried for unit ramp and unit parabola inputs. Table 13.2 synthesizes the resulting conclusions. Table 13.2 Correlation Between System Type, Basic Input, Constants, and Steady-State Errors Unit Step

Unit Ramp

Unit Parabola

System Type

Kp

e(N)

Kv

e(N)

Ka

e(N)

0

constant

constant

0

N

0

N

1

N

0

constant

constant

0

N

2

N

0

N

0

constant

constant

3

N

0

N

0

N

0

659

660

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

It can be seen from Table 13.2 that a type-three system realizes zero steady-state errors for any of the three basic inputs that have been analyzed here. However, increasing the number of integrations past three, increases the chances that the controlled system becomes unstable. nn

Example 13.2

5ðsþ1Þ A controller-and-plant is defined by the feed-forward transfer function GðsÞ ¼ s4 þ9s 3 þ20s 2 , and is

connected in a unity-feedback system. Verify the stability of the system, determine the system type, evaluate the Kp, Kv, Ka constants, and determine the steady-state error for r(t) ¼ t2/2.

Solution The stability of this system can be verified by finding the poles of the closed-loop transfer function; this can be done in MATLAB by means of the following commands: >> g = tf([5,5],[1,9,20,0,0]); >> pole(feedback(g,1)) which returns the following poles: ans = -5.4971 -3.3621 -0.0704 + 0.5153i -0.0704 - 0.5153i Because all real parts of the closed-loop poles are negative, the unity-feedback control system is stable. The feed-forward transfer function can also be written as:

GðsÞ ¼

5ðs þ 1Þ 5ðs þ 1Þ ¼ s2 $ ðs2 þ 9s þ 20Þ s2 $ ðs þ 4Þ $ ðs þ 5Þ

(13.30)

which shows that this is a type-two system because the exponent in the denominator of G(s) is n ¼ 2. Table 13.2 indicates that Kp ¼ N and Kv ¼ N, whereas the acceleration constant Ka is finite. Its value is calculated according to the definition as:

Ka ¼ lim s2 $ GðsÞ ¼ lim s/0

s/0

5ðs þ 1Þ ¼ 0:25 ðs þ 4Þ $ ðs þ 5Þ

As a consequence, the constant steady-state error is e(N) ¼ 1/Ka ¼ 4.

(13.31) nn

Nonunity-Feedback Systems The steady-state error of a SISO nonunity-feedback system is expressed in Eq. (13.23). Consider the more complex system of Figure 11.13, which uses a prefilter G1(s), a feed-forward transfer function G2(s), and a feedback transfer function G3(s). Such a system can equivalently be transformed into a standard unity-feedback system, whose feed-forward transfer function is obtained from the prefilter transfer function G1(s),

13.1 Time-Domain Response of SISO Feedback Control Systems

and the functions G(s) ¼ G2(s) and H(s) ¼ G3(s) as in Eq. (11.21) of Example 11.2, which is rewritten here: G ðsÞ ¼ ¼

G1 ðsÞ $ GðsÞ 1 þ GðsÞ $ H ðsÞ  G1 ðsÞ $ GðsÞ

G1 ðsÞ¼1



! G ðsÞ

GðsÞ 1 þ GðsÞ $ H ðsÞ  GðsÞ

(13.32)

The last form of Eq. (13.32) gives the feed-forward transfer function of a unity-feedback system that is equivalent to the nonunity-feedback control system without prefilter. The steady-state error can now be calculated with Eq. (13.32), or by using Tables 13.1 and/or 13.2. nn

Example 13.3 Find the open-loop transfer function of a unity-feedback system without prefilter that is equivalent to the feedback system whose block diagram is illustrated in Figure 13.8. Calculate the steady-state error corresponding to r (t) ¼ 1. R(s)

+

10

1/(s 2 + 2s +100)

C(s)

_ 1/(s + 200)

FIGURE 13.8 Block Diagram of a Nonunity-Feedback Control System With Prefilter.

Solution

Taking into consideration that G1(s) ¼ 10, G(s) ¼ 1/(s2 þ 2s þ 100), and H(s) ¼ 1/(s þ 200), the feed-forward transfer function of the unity-feedback system without prefilter that is equivalent to the system of Figure 13.8 is calculated by means of Eq. (13.32):

G ðsÞ ¼

s3

10s þ 2000 þ 202s2 þ 490s þ 18001

(13.33)

The closed-loop transfer function is therefore:

GCL ðsÞ ¼

G ðsÞ 10s þ 2000 ¼ 1 þ G ðsÞ s3 þ 202s2 þ 500s þ 20001

(13.34)

whose poles are: 200 and 1  9.95$j. They are all located in the LHP, and therefore the feedback system is stable. The denominator of G*(s) from Eq. (13.33) does not have any integrator, and the unity-feedback system is consequently a zero-type system. The corresponding static error constant and steady-state error are:

Kp ¼ lim G ðsÞ ¼ 0:111; eðNÞ ¼ s/0

1 ¼ 0:9 1 þ Kp

(13.35) nn

661

662

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

13.1.3 Transitory and Steady-State Time Response of Feedback Control Systems This section studies the effect of basic controls on the time response of first- and second-order plants in unity-feedback systems under unit-step input.

First-Order Plants The time response of a first-order plant Gp(s) ¼ 1/(s$s þ 1) that is connected to proportional, derivative, or integral control is analyzed here.

Proportional (P) Control Connecting a P controller Gc(s) ¼ KP to a first-order plant results in the following open-loop and closed-loop transfer functions: KP ; s$s þ 1 GðsÞ KP KP =s ¼ GCL ðsÞ ¼ ¼ 1 þ GðsÞ s $ s þ ðKP þ 1Þ s þ ðKP þ 1Þ=s

GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

(13.36)

For a unit-step input, the Laplace- and time-domain responses corresponding to the closed-loop transfer function of Eq. (13.36) are:    s= K t þ1 KP =s KP ð Þ P ; cðtÞ ¼ $ 1e C ðsÞ ¼ GCL ðsÞ $ RðsÞ ¼ s $ ½s þ ðKP þ 1Þ=s KP þ 1 (13.37) The exponent of e in Eq. (13.37) indicates that the modified time constant of this first-order system is s* ¼ s/(KP þ 1), which is smaller than the original (plant) time constant. This will result in smaller rise and settling times, and therefore the response of the P-controlled first-order plant is faster. The zero-type system (as indicated by the open-loop transfer function of Eq. (13.36)) results in a constant steadystate error produced by a unit-step input: e(N) ¼ 1/(KP þ 1), as per Table 13.1. Increasing the proportional gain results in improved transitory response characteristics (smaller time constant, rise time, and settling time) and steady-state response (smaller steady-state error e(N)).

Derivative (D) Control The following open-loop and closed-loop transfer functions result from coupling a D controller Gc(s) ¼ KD$s to a first-order plant: KD $ s ; s$s þ 1 G ðsÞ KD $ s ½KD =ðKD þ sÞ $ s GCL ðsÞ ¼ ¼ ¼ 1 þ GðsÞ ðKD þ sÞ $ s þ 1 s þ 1=ðKD þ sÞ GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

(13.38)

13.1 Time-Domain Response of SISO Feedback Control Systems

When using a reference signal R(s) ¼ 1/s, the Laplace-domain and time-domain responses are: CðsÞ ¼ GCL ðsÞ $ RðsÞ ¼ KD  t $ e KD þs cðtÞ ¼ KD þ s

½KD =ðKD þ sÞ $ s KD =ðKD þ sÞ ¼ ; s $ ½s þ 1=ðKD þ sÞ s þ 1=ðKD þ sÞ

(13.39)

The time response c(t) is a decaying exponential function, which starts from KD/(KD þ s)dthat is a value always less than the unit-step input r(t) ¼ 1dand becomes zero as time grows to infinity. This response reflects poor tracking capabilities, as well as a steady-state error equal to 1 (see Table 13.2), which indicates that use of derivative control alone for first-order plants is regularly not advisable.

Integral (I) Control An integral controller Gc(s) ¼ KI/s and a first-order plant generate the following open-loop and closed-loop transfer functions: KI ; s $ ð s $ s þ 1Þ GðsÞ KI KI =s ¼ GCL ðsÞ ¼ ¼ 1 þ GðsÞ s $ s2 þ s þ KI s2 þ ð1=sÞ $ s þ KI =s KI =s ¼ 
pffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffi 2 s þ 2 1= 2 s $ KI $ KI =s $ s þ KI =s GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

(13.40)

The first-order original plant transfer function is transformed into a second-order system in the feedback configuration with an I controller. This system’s natural frepffiffiffiffiffiffiffiffiffiffi quency and damping ratiodas illustrated in Eq. (13.40)dare: un ¼ KI =s . pffiffiffiffiffiffiffiffiffiffiffi and x ¼ 1 ð2 s $ KI Þ. Note that in this case, the new time constant of the resulting system is double the time constant of the first-order original plant, namely: s ¼ 1=ðx $ un Þ ¼ 2s, which results in a more sluggish response of the new second-order system. In addition, there are also oscillations introduced by this system. However, because of the integrator in the open-loop transfer function of Eq. (13.40), the system is a type one system and therefore the steady-state error generated by a unit-step input r(t) ¼ 1 is zero, which indicates that using I control in a feedback system eliminates the steady-state error.

Second-Order Plants This section analyzes the effect of using proportional-derivative (P þ D) and proportional-integral 
(P þ I) control of a second-order plant, defined by the transfer  function Gp ðsÞ ¼ u2n s2 þ 2x $ un $ s þ u2n in a unity-feedback system.

663

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

ProportionaleDerivative (P þ D) ControldAdditional Zero The open-loop and closed-loop transfer functions corresponding to connecting a P þ D controller defined by Gc(s) ¼ KP þ KD$s ¼ KP$(1 þ TD$s) to a secondorder plant are: GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

u2n $ ðKP þ KD $ sÞ u2n $ ð1 þ TD $ sÞ ¼ K $ ; P s2 þ 2x $ un s þ u2n s2 þ 2x $ un s þ u2n

GðsÞ KP $ u2n $ TD $ s þ KP $ u2n ¼ 2 GCL ðsÞ ¼ 1 þ GðsÞ s þ ð2x þ KP $ un $ TD Þ $ un $ s þ ðKP þ 1Þ $ u2n

Root Locus

100

(13.41)

Root Locus

150

80

100

60

-1

Imaginary Axis (seconds )

-1

Imaginary Axis (seconds )

664

40 20 0 -20 -40 -60

50 0 -50

-100

-80 -100 -200

-150

-100 -50 Real Axis (seconds-1)

(a)

0

50

-150 -100

-50

0 50 100 150 Real Axis (seconds-1)

200

250

(b)

FIGURE 13.9 Root Loci of Second-Order Plant With P þ D Control in Unity-Feedback System With: (a) TD > 0 and Stable System; (b) TD < 0 and Unstable System.

Figure 13.9 plots the root locus of the open-loop transfer function for un ¼ 100 rad/s, x ¼ 0.5 corresponding to a positive derivative constant TD ¼ 0.1dFigure 13.9(a)d and to a negative derivative constant TD ¼ 0.1, as in Figure 13.9(b). While for positive values of TD, the feedback control system is stable with the root locus in the LHP, negative derivative constants TD may render the system unstable. The characteristic polynomial of the closed-loop transfer function of Eq. (13.41) indicates that the system is defined by a new natural frequency and damping ratio as: pffiffiffiffiffiffiffiffiffiffiffiffiffiffi un ¼ KP þ 1 $ un ; (13.42) 2x þ KP $ un $ TD pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x $ un ¼ ð2x þ KP $ un $ TD Þ $ un or x ¼ 2 KP þ 1 As per Eq. (13.42), the addition of the P þ D controller increases the natural frequency and the damping ratio of the plant for KP > 0. Eq. (13.42) also indicates

13.1 Time-Domain Response of SISO Feedback Control Systems

that it is possible to change both the natural frequency (by changing the proportional gain KP) and the damping properties (by independently modifying the derivative constant TD). The time constant of this modified second-order system is:
  1 1
¼ x $ un ¼ x $ un þ KP $ TD $ u2n 2 ¼ þ KP $ TD $ u2n 2 (13.43)  s s Eq. (13.43) shows that 1/s* > 1/s, which is s* < s; therefore, the settling time of the P þ D controlled second-order system, which is ss ¼ 4s , is smaller than the settling time of the original plant, which is ss ¼ 4s. Note that the steady-state response of this system (which is a type-zero system with no integrator in the open-loop transfer function) to a unit-step input is: eðNÞ ¼

1 1 ¼ 1 þ lim GðsÞ 1 þ KP

(13.44)

s/0

Eqs. (13.42) and (13.44) show that by increasing the proportional gain KP the natural frequency increases (which leads to decreasing the settling time), while the steadystate error decreases, both effects being beneficial. Proportional control alone can alter the natural frequency but cannot affect the damping properties, see the closed-loop characteristic polynomial of Eq. (13.41), which becomes s2 þ 2x $ un $ s þ ðKP þ 1Þ $ u2n for TD ¼ 0. P þ D control is mostly implemented to improve the transient response characteristics of second- or higherorder control systems. The derivative component of P þ D control results in sharp responses to sudden input variation. Additional Zero The net result of combining P þ D control with a second-order plant (as well as with plants of any order) is the addition of a zero (s ¼ z ¼ 1/ TD) to the closed-loop transfer function, as seen in Eq. (13.41). The closed-loop transfer function of Eq. (13.41) can also be written as:
 2 un KP GCL ðsÞ ¼ $
 1 þ KP s2 þ 2x $ u $ s þ u 2 n n
 2 un KP þ $ ð a $ sÞ $ (13.45)
2 $ s  2 1 þ KP s þ 2x $ u $ s þ u n

n

In Eq. (13.45), the derivative constant TD is expressed in terms of the plant time constant s ¼ 1/(x$un) as TD ¼ a$s, where a is a constant. Multiplying Eq. (13.45) by R(s) in the left and right-hand sides, and taking into account that GCL(s)$R(s) ¼ C(s), yields: C ðsÞ ¼ Cm ðsÞ þ ða $ sÞ $ s $ Cm ðsÞ; Cm ðsÞ
 2 un KP ¼ $
 $ Rð s Þ 1 þ KP s2 þ 2x $ u $ s þ u 2 n

n

(13.46)

665

666

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

where the subscript “m” stands for modified. Inverse Laplace transforming Eq. (13.46) yields the time response of the altered control system: cðtÞ ¼ cm ðtÞ þ a $ s $ c_m ðtÞ

(13.47)

Eq. (13.47) indicates that the time response c(t) of the additional-zero system is the superposition between the time response of the modified system cm(t) and the one of its time derivative. Small values of the parameter a make the contribution of cm(t) significant compared to that of its time derivative, which means the zero introduced by the derivative action can be neglected, and the system can be considered a standard second-order system. However, if a is close to or larger than 1, the contribution of the derivative of cm(t) cannot be ignored. Figure 13.10 contains the plots of cm(t), a$s$dcm(t)/dt, and c(t) for a ¼ 0.2, a ¼ 2, and a ¼ 2. For positive values of a, the

0.18

0.7

. a . τ cm (t )

0.16 0.14

. a . τ . cm (t )

0.5

0.12

0.4

0.1 0.08

0.3

c

0.06

0.2

0.04

cm

0.1

0.02

cm

0

0 -0.02

c

0.6

0

0.05

0.1 0.15 Time (seconds)

0.2

0.25

-0.1

0

0.05

0.1 0.15 Time (seconds)

(a)

0.2

(b) 0.3

c

0.2 0.1

cm

0

. a . τ . cm (t )

-0.1 -0.2 -0.3

0

0.05

0.1

0.15 0.2 0.25 Time (seconds)

0.3

0.35

(c)

FIGURE 13.10 Unit-Step Time Response of Original Second-Order Control System, and System With Additional Pole in Terms of Parameter a: (a) a ¼ 0.2; (b) a ¼ 2; (c) a ¼ 2.

0.25

13.1 Time-Domain Response of SISO Feedback Control Systems

derivative term is positive, and is added to cm(t), resulting in a higher transient c(t) with a larger overshoot. Smaller values of a (such as a ¼ 0.2 in the plot of Figure 13.13) result in a time response c that is very close to cm; a rule of thumb is that for a positive and smaller than 0.2 ¼ 1/5, the derivative effect can be neglected (which is the same as ignoring the additional zero). Larger values of a result in larger overshootsdas it can be seen when comparing Figure 13.10(a) and 13.10(b). For negative values of a, see Figure 13.10(c), the derivative term is negative initially and is therefore subtracted from cm(t), resulting in a time response c(t) that is negative for an initial durationdthis particular type of response is known as nonminimum phasedmore details on it are provided in the frequency-domain analysis section of this chapter. Negative values of the parameter a lead to negative values of the derivative constant TD, which render the feedback system into the unstable domaindsee the root locus of Figure 13.9(b).

ProportionaleIntegral (P þ I) ControldAdditional Zero and Additional Pole The open-loop and closed-loop transfer functions that result from connecting a P þ I controller defined by Gc(s) ¼ KP þ KI/s ¼ KP$[1 þ 1/(TI$s)] to a second-order plant are: GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼ GCL ðsÞ ¼

u2n $ ðKP þ KI =sÞ u2n $ ð1 þ TI $ sÞ
; ¼ K $ P s2 þ 2x $ un s þ u2n TI $ s s2 þ 2x $ un s þ u2n

GðsÞ KP $ u2n $ TI $ s þ KP $ u2n ¼ 3 1 þ GðsÞ TI $ s þ 2x $ un $ TI $ s2 þ u2n $ TI $ ðKP þ 1Þ $ s þ KP $ u2n (13.48)

Compared to the original second-order plant transfer function, the closed-loop transfer function of Eq. (13.48) adds a zero (s ¼ z ¼ 1/TI) and a poledas indicated by the third-order characteristic polynomial. The open-loop transfer function of Eq. (13.48) enables plotting the root locus of this feedback system, as shown in Figure 13.11(a), which corresponds to TI ¼ 0.1, and Figure 13.11(b), which was plotted for TI ¼ 0.1; this latter value indicates that the zero of the open-loop transfer function is in the RHP, and the system is unstable. Expressing the integral time as TI ¼ a$s (s ¼ 1/(x$un) being the plant time constant), with x ¼ 0.5 and un ¼ 100 rad/s results in s ¼ 0.02 and a ¼ 5 for TI ¼ 0.1. For TI ¼ 0.1, the factor is a ¼ 5. The root locus of Figure 13.11(a) shows that while two closed-loop poles are always complex (and conjugated), the third (additional) pole is real. The two abscissas of this root locus are directed at 90 and 270 , and their convergence abscissa is obtained from the open-loop poles and zeros as: sa ¼

2x $ un  ð1=TI Þ 1 ¼ x $ un þ 31 2TI

(13.49)

667

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

Root Locus

300

200

Imaginary Axis (seconds-1)

100 0 -100

100 0 -100 -200

-200 -300 -60

Root Locus

300

200 Imaginary Axis (seconds-1)

-50

-40 -30 -20 -10 Real Axis (seconds-1)

0

10

-300 -60

-50

-40

-30

-20

-10

0

10

20

-1 Real Axis (seconds )

(a)

(b) 1500

Root Locus

1000 Imaginary Axis (seconds-1)

668

500 0 -500

-1000 -1500 -300 -250 -200 -150 -100 v

-50

0

50

100

(c)

FIGURE 13.11 Root Loci of Second-Order Plant With P þ I Control in Unity-Feedback System With: (a) TI ¼ 0.1 > 0 and Stable System; (b) TI ¼ 0.1 < 0 and Unstable System; (c) Positive Small TI (TI ¼ 0.004) and Unstable System.

Eq. (13.49) points to the fact that the abscissa of convergence (together with the two asymptotes) can move into the RHP and the system becomes unstable when 1/(2TI) > x$un. This situation is depicted in Figure 13.11(c) that was plotted for TI ¼ 0.004 or a ¼ 0.2 because 1/(2TI) ¼ 125 > x$un. The presence of the integrator in the open-loop transfer function of Eq. (13.48) results in a type-one system whose steady-state error for a unit-step input, r(t) ¼ 1, is zero. For unit ramp input, r(t) ¼ t, the steady-state error is constant, namely: e(N) ¼ 1/Kv ¼ Ti/KP. Figure 13.12 plots the time response resulting from a unitstep input applied to the closed-loop transfer function of Eq. (13.48) for two integral

13.1 Time-Domain Response of SISO Feedback Control Systems

1.4

Plant response 1.2

1

c (a = 0.2) 0.8

c (a = 5) 0.6

0.4

0.2

0

0

0.1

0.2

0.3 0.4 Time (seconds)

0.5

0.6

0.7

FIGURE 13.12 Unit-Step Time Response of Plant and Plant With P þ I Controller.

times, namely TI ¼ 0.1 (which corresponds to a ¼ 5) and TI ¼ 0.004 (resulting in a ¼ 0.2)dboth plotted for x ¼ 0.5 and un ¼ 100 rad/s. These responses are denoted by c in Figure 13.12. P þ I control is generally utilized to meet steady-state specifications. The integral component of this control helps rejecting disturbances, but it may also increase the transient oscillations of the system. Dominant Poles The additional real closed-loop pole of Eq. (13.48) is inversely proportional to the parameter a. Also, as illustrated in Figure 13.12, small values of the coefficient a (such as a ¼ 0.2) result in a response that is very close to that of the original second-order system (a ¼ 0 in the same figure). This indicates that the additional closed-loop pole is farther away in the LHP from the original closed-loop poles. On the contrary, larger values of a (such as a ¼ 5), which is the same as having the additional pole closer to the origin, and the original poles farther away on the LHP, lead to responses that have the character of a first-order system (see Figure 13.12). These two observations are an illustration of the notion of dominant poles, which are the closed-loop poles that are closer to the origin and determine the character of the control system time response. The effect of the other closed-loop poles can be neglected, provided they are at least five times farther than the dominant poles in the LHP away from the origindNise (2015). The order of a higherorder control system could therefore be reduced by keeping the dominant poles

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

and eliminating the other poles. The response corresponding to a ¼ 0.2 in Figure 13.12, for instance, indicates that the additional pole is 5 times farther away to the left in the LHP compared to the two original-system closed-loop poles and, as a consequence, the dominant poles are the original ones. The additional pole can be eliminated, and the third-order system can be reduced to a second-order system. Likewise, the response for a ¼ 5, which is similar to a first-order system response, is mostly due to the additional pole, which is five times closer to the origin in the LHP compared to the original-system closed-loop poles. The additional pole is the dominant pole and the third-order system can be reduced to a first-order system. nn

Example 13.4

The mechanical microsystem (plant) of Figure 13.13 is composed of a mass m ¼ 1012 kg and two beam springs, each having a stiffness k ¼ 5 N/m. An external force f acts on the mass displacing it about the y direction. The plant is originally connected in a unity-feedback control system, with a unity gain controller. Design a P þ D controller to replace the original unity-gain P controller that will generate a maximum overshoot of 10% and will reduce the steady-state error of the resulting feedback control system under a unit-step reference r(t) ¼ 1 input 100 times.

y k

k

m

Shuttle mass

f

Beam spring

FIGURE 13.13 Translatory Mechanical Microsystem (Plant).

Solution The dynamic model of the mechanical microsystem and the corresponding plant transfer function are:

m $ y€ ¼ f  2k $ y; Gp ðsÞ ¼

YðsÞ 1 ¼ FðsÞ m $ s2 þ 2k

(13.50)

The resulting closed-loop transfer function is:

GCL ðsÞ ¼

Gp ðsÞ GðsÞ 1 ¼ ¼ 1 þ GðsÞ 1 þ Gp ðsÞ m $ s2 þ 2k þ 1

(13.51)

The original control system is shown in Figure 13.14(a), but the controller’s transfer function is not included, since its gain is unity. Addition of the P þ D controller to the mechanical microplant of Figure 13.14(a) generates the block diagram of Figure 13.1(b). The open-loop transfer function corresponding to the control system of Figure 13.14(b) is:

G ðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

KP $ TD $ s þ KP m $ s2 þ 2k

(13.52)

13.1 Time-Domain Response of SISO Feedback Control Systems

R(s)

+

E(s)

C(s)

1/ (m·s + 2k)

_

R(s)

+

E(s)

K ·T ·s + K

M(s)

C(s)

1/ (m·s + 2k)

_

(a)

(b)

FIGURE 13.14 Block Diagrams of Mechanical Microplant in a Unity-Feedback Control System With: (a) No Controller; (b) P þ D Controller.

and the closed-loop transfer function of the same system is:

GCL ðsÞ ¼

G ðsÞ KP $ TD $ s þ KP ¼ 1 þ G ðsÞ m $ s2 þ KP $ TD $ s þ 2k þ KP

(13.53)

whose characteristic polynomial is:

DCL ðsÞ ¼ m $ s2 þ KP $ TD $ s þ 2k þ KP

(13.54)

 ðsÞ ¼ 0, which are the closed-loop poles, are found from The roots of the characteristic equation DCL Eq. (13.54):

p1;2 ¼ 

KP $ TD 1 $  2m 2m

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðKP $ TD Þ2  4m $ ð2k þ KP Þ

(13.55)

They are located in the LHP, and therefore this system is stable as well. Eq. (13.54) shows that the effect of the derivative control translates into damping arising in the characteristic polynomial due to the s term. Eq. (13.54) also indicates that the (total) equivalent stiffness increased through the addition of the proportional control. Division of the characteristic equation DCL (s) ¼ 0 by m leads to the following equivalent natural frequency and damping ratio:




un

2

¼

2k þ KP KP $ TD ; 2x $ un ¼ m m

(13.56)

Since the maximum overshoot only depends on the damping ratio according to Eq. (13.16), for Op ¼ 0.1 (as the example specifies) the solution of Eq. (13.16) yields x ¼ 0.5912. The steady-state errors of the original system (with unity-gain P control) e(N) and the system with the P þ D control, which is e*(N), are calculated based on Table 13.1 as:

8 s $ RðsÞ 1 2k > > eðNÞ ¼ lim ¼ ¼ > > > s/0 1 þ GðsÞ 1 þ lim 2k þ1 GðsÞ < s/0 > s $ RðsÞ 1 2k > > > ¼ ¼ e ðNÞ ¼ lim  > : s/0 1 þ G ðsÞ 1 þ lim G ðsÞ 2k þ KP s/0

(13.57)

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

This example requires that e  ðNÞ ¼ eðNÞ=100. Using this condition in Eq. (13.57) results in KP ¼ 198k þ 100 ¼ 1090. Substituting now un from the first Eq. (13.56) into the second Eq. (13.56) yields:

TD ¼

2x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ ð2k þ KP Þ $ m ¼ 3:6 $ 108 s KP

(13.58)

The P þ D controller is thus designed with the calculated values of the proportional gain KP and derivative time TD. nn

Zero-Pole Cancellation or NeareZero-Pole Cancellation When the open-loop transfer function of a feedback control system has a zero and a pole that are identical, z ¼ p, the numerator factor s  z and the denominator factor s  p cancel out and the system is reduceddthe situation is known as zero-pole cancellation. It is important to establish whether a similar procedure, named nearezero-pole cancellation, can be applied when the zero and the pole are very close to each other but are not identical. The following example discusses this aspect. nn

Example 13.5

To modify the damping characteristics of a second-order plant with I control, P þ D control is added to this unity-feedback system. As a result, the open-loop transfer function becomes Gi(s) ¼ K$(s þ ai þ ε)/ [s$(s þ ai)$(s þ bi)], with i ¼ 1, 2; a1 > b1 and a2 < b2, where ai, bi, ε are positive parameters, and ε is very small. Using root locus, compare these systems that display near-zero-pole cancellation at a, with the zero-pole cancellation system defined by the open-loop transfer function G(s) ¼ K /[s$(s þ b1)]. Also, plot the time response of the three systems resulting from a unit-step input.

Solution Figure 13.15(a) plots the root locus of the reduced second-order system (with zero-pole cancellation) for b ¼ 1, while Figure 13.15(b) and 13.15(c) show the root loci of two systems with nearezero-pole cancellation as follows: Figure 13.15(b) is plotted for a1 ¼ 10, b1 ¼ 1; Figure 13.15(c) corresponds to a2 ¼ 1, b2 ¼ 10; for both Figure 13.15(b) and 13.15(c), a value ε ¼ 0.5 has been used. Note that the addition of the zero (a1 þ ε) ¼ (10 þ 0.5) ¼ 10.5 and of the pole a1 ¼ 10 to the original second-order system results in a root locus that is almost identical to the one of the original systemd as seen in Figure 13.15(a) and 13.15(b). This is also shown in Figure 13.16, where the unit-step time responses of the two feedback control systems are almost identical. When the additional zero is (a2 þ ε) ¼ (1 þ 0.5) ¼ 1.5 and the additional pole is a2 ¼ 1, the root locus is the one of Figure 13.15(c), which is quite different from the one of the original system. This is confirmed by the time response of Figure 13.16, which has the characteristics of an overdamped system, so it is substantially different from the underdamped response of the original system. The different results produced by the two systems have to be expected: the additional zero and pole of the root locus system shown in Figure 13.15(b) are farther away to left from the original-system poles, and, therefore, it is expected that their contribution to the time response be minor. As a consequence, the nearezeropole cancellation is effective in this design. On the contrary, the system whose root locus is plotted in Figure 13.15(c) adds a zero that is very close to one of the original-system poles (s ¼ 0), which alters the system’s time response, despite the additional pole being relatively farther away in the LHP. It can therefore be concluded that nearezero-pole cancellation in this situation alters fundamentally the traits of the original second-order system. The unit-step time responses of the systems G(s), G1(s), and G2(s) are identified as c, c1 and c2, respectively in the plots of Figure 13.16.

Root Locus

Root Locus

0.6

3 2

-1

Imaginary Axis (seconds )

Imaginary Axis (seconds-1)

0.4 0.2 0 -0.2 -0.4 -0.6 -1.2

-1

-0.8

-0.6

-0.4

-0.2

0

1 0 -1 -2 -3 -12

0.2

-10

Real Axis (seconds-1)

-4 -2 -8 -6 Real Axis (seconds-1)

(a)

0

2

(b) 10 8

-1

Imaginary Axis (seconds )

6 4 2 0 -2 -4 -6 -8 -10 -12

-10

-8

-6

-4

-2

2

0

vc

(c)

FIGURE 13.15 Root Loci for: (a) Original Second-Order System G(s) ¼ 1/[s$(s þ b1)]; (b) System With G1(s) ¼ (s þ a1 þ ε)/[s$(s þ b1)$(s þ a1)]; (c) System With G2(s) ¼ (s þ a2 þ ε)/ [s$(s þ b2)$(s þ a2)]. 1.2

c1

c

1

0.8

c2 0.6

0.4

0.2

0 0

5

10 15 Time (seconds)

20

25

FIGURE 13.16 Unit-Step Time Response of Systems Defined by Open-Loop Transfer Function G(s), G1(s), and G2(s).

nn

674

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

ProportionaleIntegraleDerivative (P þ I þ D) Control

P þ I þ D control combines the capabilities of P þ D control to adjust the transient response properties and of P þ I control to improve the steady-state characteristics of a dynamic system. As discussed by Nise (2015) for instance, sequential design of the two different stages (usually P þ D design is followed by P þ I design) requires trade-offs as it is not possible to improve one time segment without somewhat deteriorating the performance of the other segment. Most often, the P þ I þ D controller design requires several iterations (redesign) until satisfactory transient and steady-state performance criteria are met. Connecting a P þ I þ D controller, whose transfer function is Gc(s) ¼ KP þ KI/ s þ KD$s ¼ KP$[1 þ 1/(TI$s) þ TD$s], to a second-order plant in a unity-feedback system, generates the following open-loop and closed-loop transfer functions:
 u2n $ ðKP þ KI =s þ KD $ sÞ u2n $ KD $ s2 þ KP $ s þ KI GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼ ; ¼ 3 s2 þ 2x $ un $ s þ u2n s þ 2x $ un $ s2 þ u2n $ s
 u2n $ KD $ s2 þ KP $ s þ KI GðsÞ ¼ 3 GCL ðsÞ ¼ 1 þ GðsÞ s þ ð2x þ un $ KD Þ $ un $ s2 þ u2n $ ðKP þ 1Þ $ s þ KI $ u2n (13.59) As Eq. (13.59) points out, the closed-loop poles can uniquely be determined through the three control parameters of the P þ I þ D controller; therefore, the nature of the dynamic response of this control system can be designed by adequate selection of KP, KI, KDdthis process is known as controller tuning.

ZieglereNichols P þ I þ D Controller Tuning Algorithm J.G Ziegler and N.B Nichols proposed in 1942e1943 two methods for tuning the P þI þ D controller for plants whose dynamic mathematical model is not precisely known, but whose time response is available from experimental testing. One method, which is not covered here, was designed for plants displaying a specific transient time response to unit-step input, resembling an S-shaped curve, also known as process reaction curveddetails of this method are provided in Ogata (2004) and Franklin et al. (2010). The second P þI þ D tuning algorithm, also named the critical (ultimate) sensitivity methoddFranklin et al. (2010)dis applicable to systems that can become marginally stable; this method is briefly discussed here. Proportional control only is

13.1 Time-Domain Response of SISO Feedback Control Systems

Tc R(s)

+ KP → Kc

Gp(s)

C(s)

c

_ t

(a)

(b)

FIGURE 13.17 (a) Unity-Feedback Plant With P Control; (b) Continuous Oscillatory Time-Domain Response of Marginally Stable Control System.

applied to a plant in a unity-feedback system, as illustrated in Figure 13.17(a). This is equivalent to TI / N and TD ¼ 0 in the original, three-constant P þ I þ D controller. Under a short impulse input, for instance, the controller gain KP is increased to a critical value Kc when the controlled time response c(t) becomes oscillatory (the system is marginally stable), see Figure 13.17(b), which identifies the critical period of oscillations Tc. Based on Kc and Tc, Ziegler and Nichols proposed values for the parameters that define P, P þ I, and P þI þ D controllers, as per Table 13.3. With the values of Table 13.3 and after a bit of algebra, the P þ I þ D controller transfer function becomes:   1 ðs þ 4=Tc Þ2 Gc ðsÞ ¼ 0:6 $ Kc $ 1 þ þ 0:125 $ Tc $ s ¼ 0:075Kc $ Tc $ 0:5 $ Tc $ s s (13.60)

Table 13.3 P þ I þ D Controller Parameters Based on ZieglereNichols Critical Sensitivity Tuning Method Control Type

Controller Parameter

P

KP ¼ 0.5$Kc

PþI

KP ¼ 0.45$Kc; TI ¼ Tc/1.2

PþIþD

KP ¼ 0.6$Kc; TI ¼ 0.5$Tc; TD ¼ 0.125$Tc

675

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

The ZieglereNichols tuning algorithm can also be applied when the plant transfer function is known, as studied in the following example. nn

Example 13.6

Design a P þ I þ D controller using the critical sensitivity method of ZieglereNichols to interact with a plant of transfer function Gp(s) ¼ 1/[s$(s þ 2)$(s þ 10)] in a unity-feedback control system. Study the time response of this control system that results from a unit-step input r (t) ¼ 1.

Solution

Considering that TI / N and TD ¼ 0, which reduces the original P þI þ D controller to a P controller of gain KP, the open-loop and closed-loop transfer functions of the resulting unity-feedback system are:

KP KP ¼ ; s $ ðs þ 2Þ $ ðs þ 10Þ s3 þ 12s2 þ 20s GðsÞ KP ¼ GCL ðsÞ ¼ 1 þ GðsÞ s3 þ 12s2 þ 20s þ KP

GðsÞ ¼ Gc ðsÞ $ Gp ðsÞ ¼

(13.61)

The following RoutheHurwitz array is formed with the coefficients of the characteristic polynomial DCL(s) ¼ s3 þ 12s2 þ 20s þ KP:

s3

1

20

s2

12

KP

s1

240  KP 12

0

s0

KP

As it can be checked, the system becomes marginally stable when the coefficients in the row of s1 are zero, which means that the critical gain is: Kc ¼ KP ¼ 240. With this gain value, the characteristic polynomial becomes DCL(s) ¼ s3 þ 12s2 þ 20s þ 240 ¼ (s2 þ 20)$(s þ 12). The imaginary root of this polynomial results from solving s2 þ 20 ¼ 0, which yields a critical frequency uc ¼ 201/2 when considering s ¼ u$j. The corresponding critical period is Tc ¼ 2p/uc ¼ 1.405 s. With these values, the P þ I þ D controller transfer function of Eq. (13.60) is Gc(s) ¼ 25.2893$(s þ 2.8471)2/s. The time response of the control system to a unit-step input is shown in Figure 13.18 as c. It has been obtained combining the MATLAB commands step(feedback(g,1)), where g is the open-loop transfer function and 1 is the feedback H(s). It can be seen that both the maximum overshoot (approximately 60%) and the settling time are relatively large. To reduce these two parameters and improve the transient characteristics, some fine-tuning needs to be performed using a few different sets of parameters Kc and Tc. For Kc1 ¼ 500 and Tc1 ¼ 2.8 s, for instance, the controller transfer function Gc1(s) ¼ 105$(s þ 1.4286)2/s is obtained, which result in the unit-step response c1, also included in Figure 13.18. It can be seen that the maximum overshoot has been reduced to approximately 25% while also increasing the response speed. Because the system is a type-1 system (with one integrator in the open-loop transfer function), the unit-step steady-state error is zero. nn

13.1 Time-Domain Response of SISO Feedback Control Systems

Step Response

1.8 1.6

c

1.4

Amplitude

1.2

c1

1 0.8 0.6 0.4 0.2 0

0

1

2

3 4 5 Time (seconds)

6

7

8

FIGURE 13.18 Unit-Step Time Response of P þ I þ D-Control Systems Defined by Control Transfer Functions Gc(s) and Gc1(s).

Nonlinear Control Systems Nonlinearity in a feedback control system can result from any of its components: the plant, the actuation, the controller, or the sensing (feedback). A sensor or an actuator can saturate, and therefore, the respective component will have a nonlinear model. Likewise, the plant can have inherent nonlinearities. In those cases, transfer functions cannot be used for the nonlinear components, but rather inputeoutput differential equations are needed. The overall nonlinear control system can therefore be modeled as a hybrid by means of both transfer functions for the linear components and differential equations for the nonlinear ones, using the Simulink environment, for instance. nn

Example 13.7 The mechanical system studied in Example 3.8 and illustrated in Figure 3.19 is the plant in a feedback control system. The input to the plant is a force f acting along the direction of the output displacement x. As shown in Figure 13.19, the physical control system also contains an actuator that produces the force f ¼ ka$va, a controller that generates the actuation voltage va, a feedback sensor that converts the mass displacement into voltage as vs ¼ ks$x, and another transducer, identical to the sensor,

677

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

xr

Transducer, ks

+ vr

e

Controller

va

Actuator, ka

f

x

System (Plant)

_ vs

x

Sensor, ks

FIGURE 13.19 Diagram of Feedback Control System With Position Control.

d x/dt

x k

k

dx/dt

x

1/m

k

2c/m

FIGURE 13.20 Block Diagram of Nonlinear-Plant Feedback Control System. which converts the reference displacement signal xr into voltage as vr ¼ ks$xr. Consider known the mass m ¼ 1 kg, the damping coefficient c ¼ 0.2 N s/m, the spring stiffness k ¼ 100 N/m, the spring original length l ¼ 0.4 m, the actuation gain ka ¼ 0.5 N/V, and the sensing constant ks ¼ 1000 V/m. Plot the time response under a step input xr ¼ 0.05 m considering P þ I control with KP ¼ 1 and KI ¼ 1.

Solution

The mathematical model of the plant is obtained from Eq. (3.46) by using the force f in the right-hand side of that equation, namely:

  l m $ x€ þ 2c $ x_ þ 2k $ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ x ¼ f or l2 þ x2   2k l 2c 1 x€ ¼  $ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi $ x  $ x_ þ $ f m m m l2 þ x2

(13.62)

Eq. (13.62) is nonlinear in the mass displacement x. The equations representing the transducer, the actuator, and the feedback sensor have proportional relationships between their respective inputs and outputs. Figure 13.20 is the Simulink block diagram modeling the physical diagram of Figure 13.19, which uses a P þ I controller transfer function. For the P þ I controller, the transfer function is Gc(s) ¼ (KP$s þ KI)/s ¼ (s þ 1)/s and is modeled by the Transfer Function (Fcn) block in the diagram of Figure 13.20. The Fcn block (from the User-Defined Functions library) had the following definition: f(u) = 2*k/m*(1 e l/sqrt(l^2 + u(1)^2))*u(1). Also note that the summing point of Figure 13.19 is the Add1 block in the Simulink diagram of Figure 13.20. Figure 13.21 shows the time responses of the control system and the reference signal. It can be seen that the P þ I controller eliminates the steady-state error altogether.

13.2 Time-Domain Response of MIMO Feedback Control Systems

0.1 0.09 0.08

x (m)

0.07 0.06 0.05 0.04 0.03 0.02 0.01 0

0

2

4

6

8

10

12

Time (s)

FIGURE 13.21 Unit-Step Time Response of Nonlinear Feedback System With P þ I Control.

nn

13.2 TIME-DOMAIN RESPONSE OF MIMO FEEDBACK CONTROL SYSTEMS MIMO feedback control systems are composed of several plants that are combined with adequate controllers, as introduced in Chapter 11. The closed-loop transfer function matrix approach, or the state space model, can be used to analyze or design MIMO feedback systems. The transient-response features of a MIMO feedback system can easily be studied based on the (unique) characteristic polynomial of the closed-loop transfer function matrix, similarly to SISO systems. Steady-state errors to various inputs can be evaluated for each open-loop transfer function or collectively as an error vector, as discussed in the following.

13.2.1 MISO Feedback Systems With Disturbances As mentioned in Chapter 11, feedback systems with disturbances are a particular case of MIMO systems that are characterized by multiple inputs of which one is the reference input r(t) and the others (one or more) represent disturbances affecting the control systems and are treated as additional inputs. Because there is a single output, this type of system is known as MISO (multiple-input, single-output). We will study here systems with nonunity feedback in the presence of one disturbance

679

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

input d(t) adding to the reference input r(t). Figure 11.34 is the block diagram of a nonunity-feedback system with two inputs: the reference signal R(s) and one disturbance signal D(s). As demonstrated in Chapter 11, the controlled output of this system is expressed as: CðsÞ ¼ GCL;R ðsÞ $ RðsÞ þ GCL;D ðsÞ $ DðsÞ ¼

Gc ðsÞ $ Gp ðsÞ Gp ðsÞ $ RðsÞ þ $ DðsÞ (13.63) 1 þ Gc ðsÞ $ Gp ðsÞ $ HðsÞ 1 þ Gc ðsÞ $ Gp ðsÞ $ HðsÞ

Eq. (13.63) results from combining Eqs. (11.108) and (11.109). One possibility to evaluate the error for MISO feedback systems is to use the definition of Eq. (13.23), namely: eðNÞ ¼ lim eðtÞ ¼ lim ½s $ EðsÞ ¼ lim ½s $ ðRðsÞ  HðsÞ $ CðsÞÞ t/N s/0 s/0



s $ Gp ðsÞ $ HðsÞ s $ RðsÞ $ DðsÞ ¼ lim þ lim  s/0 1 þ Gc ðsÞ $ Gp ðsÞ $ HðsÞ s/0 1 þ Gc ðsÞ $ Gp ðsÞ $ HðsÞ ¼ eR ðNÞ þ eD ðNÞ (13.64) Eq. (13.64) points to the fact that the two inputs combine to the overall steady-state error. The two contributions are identified as eR(N), which is the steady-state error of a SISO unity-feedback system, and eD(N) that is produced by the disturbance signal and can be reformulated as:

s eD ðNÞ ¼ lim  $ DðsÞ (13.65) s/0 1=½Gp ðsÞ $ HðsÞ þ Gc ðsÞ To reduce the weight of the disturbance in the overall steady-state error (when disregarding the minus sign in Eqs. 13.64 and 13.65), the limit of the product transfer function Gp(s)$H(s) should be as small as possible, while the limit of the controller transfer function Gc(s) should be as large as possible. nn

Example 13.8 Consider the dc motor of Example 11.12, which is the plant in a feedback control system with disturbances, where the reference input is the armature voltage, and the output is the shaft angular velocity. With the closed-loop transfer functions GCL,R(s) and GCL,D(s) of Example 11.12, and using Simulink, plot the time response c(t) when the reference signal is r (t) ¼ 1 V and the disturbance signal is d(t) ¼ 3$sin(10t) V. Known are the following parameters: KP ¼ 25, Ks ¼ 0.1 V s/rad, Jl ¼ 0.025 kg m2, c ¼ 0.14 N m s, Ra ¼ 10 U, La ¼ 1.8 H, Km ¼ 0.2 N m/A, and Ke ¼ 0.02 N m/A. Also, calculate the maximum steady-state error with respect to the unit-step input.

13.2 Time-Domain Response of MIMO Feedback Control Systems

Solution The maximum steady-state value of the controlled output is calculated by linearly superimposing the steady-state value due to the unit-step input, and the amplitude of the steady-state response produced by the sinusoidal disturbance input as:

  1 cðNÞ ¼ lim s $ GCL;R ðsÞ $ þ GCL;D ðu $ jÞ $ D s/0 s    ¼ lim GCL;R ðsÞ þ GCL;D ðu $ jÞ $ D

(13.66)

s/0

where D ¼ 3 is the amplitude and u ¼ 10 rad/s, both related to d(t). Taking into account Eq. (11.114) that express the two closed-loop transfer functions, the steady-state error corresponding to r(t) ¼ 1 is calculated as:

eðNÞ ¼ jKs $ cðNÞ  1j      K $K  Ks $ D   s P ¼ þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 a0 þ KP $ Ks  2 2   ða0 þ KP $ Ks  a2 $ u2 Þ þ ða1 $ uÞ

(13.67)

Eq. (13.67) confirms that large values of the proportional controller and sensor gains result in reduced errors due to the disturbance signal. With the numerical values of the example, the coefficients of Eq. (13.67) whose expressions are given in Eqs. (11.111) are a0 ¼ 7.02, a1 ¼ 2.51, a2 ¼ 0.225, and the resulting steady-state error of the same equation is e(N) ¼ 0.727. The block diagram of Figure 13.22 is built by means of blocks that have already been used in previous chapters in Simulink examples.

D(s) Gp(s) R(s)

E(s)

C(s)

Gc(s)

ω(t)

e(t) H(s)

FIGURE 13.22 Simulink Block Diagram of the Feedback Control System of a DC Motor Plant With Sinusoidal Disturbance Input.

Figure 13.23 plots both the angular velocity u and the error as functions of time. The shaft angular velocity has a vibratory motion around a value slightly larger than 2.5 rad/s after a transient period, while the maximum value is around 2.7. The error has a (maximum) steady-state value of approximately 0.7, which was more accurately determined from Eq. (13.67) as 0.727.

681

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

3

2.5

ω (rad/s)

2

1.5

1

error (V)

0.5

0

-0.5 0

1

2 3 Time (seconds)

4

5

FIGURE 13.23 Plot of Shaft Angular Velocity and Error by Simulink for the Feedback-Controlled DC Motor Plant With Sinusoidal Disturbance Input. nn

13.2.2 MIMO Feedback Systems by the Transfer Function Matrix and State-Space Methods The steady-state error vector of a nonunity-feedback MIMO system is the difference between the input (reference) vector and the transformed output (control) or back vector: feðtÞg ¼ frðtÞg  fbðtÞg

(13.68)

The assumption in Eq. (13.68) is that {r(t)} and {b(t)} have the same number of components. The steady-state error vector is calculated based on the final-value theorem as the limit of {e(t)} when time grows to infinity as: feðNÞg ¼ lim feðtÞg ¼ lim ½s $ fEðsÞg ¼ lim ½s $ ðfRðsÞg  fBðsÞgÞ t/N

s/0

s/0

¼ lim ½s $ ð½I   ½H ðsÞ $ ½GCL ðsÞÞ $ fRðsÞg s/0

where ½GCL ðsÞ ¼ ð½I  þ ½GðsÞ $ ½H ðsÞÞ1 $ ½GðsÞ  with ½GðsÞ ¼ Gp ðsÞ $ ½Ga ðsÞ $ ½Gc ðsÞ

(13.69)

The closed-loop and feed-forward transfer function matrices of Eq. (13.69) are derived in Eqs. (11.110) and (11.113). Eq. (13.69) shows that the steady-state errors depend on both the feedback system and the input. As seen in Eq. (13.69), the steady error vector of a MIMO feedback system requires knowledge of the closed-loop transfer function matrix [GCL(s)] and reference vector {R(s)}.

13.2 Time-Domain Response of MIMO Feedback Control Systems

For unity-feedback systems, the feedback transfer function matrix is the identity matrix, and the time-domain error is: feðtÞg ¼ frðtÞg  fcðtÞg

(13.70)

The corresponding steady-state transfer function is calculated from Eq. (13.69) by simply using [H(s)] ¼ [I]. Note that the closed-loop transfer function matrix can also be obtained from an existing state-space model in terms of the matrices [A], [B], [C], and [D], as discussed in Chapter 12, namely: fCðsÞg ¼ ½GCL ðsÞ $ fRðsÞg with ½GCL ðsÞ ¼ ½C $ ðs $ ½I  ½AÞ1 $ ½B þ ½D (13.71)

nn

Example 13.9 A two-variable feedback system is controlled as shown in the block diagram of Figure 13.24. The plant transfer functions are Gp11(s) ¼ 0.1/(s þ 0.2), Gp12(s) ¼ 0.5/(s þ 0.2), Gp21(s) ¼ 0.2/(s þ 0.2), Gp22(s) ¼ 0.3/(s þ 0.2). Find the gains of two proportional controllers Gc1(s) ¼ K1 and Gc2(s) ¼ K2 that will produce a second-degree closed-loop characteristic polynomial, with a damping ratio x ¼ 0.7 and a natural frequency un ¼ 10 rad/s. Plot the two controlled time responses c1(t) and c2(t) corresponding to r1(t) ¼ 1 and r2(t) ¼ 1 and calculate the resulting steady-state errors e1(N) and e2(N).

R1(s) +

_ E1(s)

M1(s)

Gc1(s)

+

Gp11(s)

C11(s) R2(s)

+

E2(s)

_

M2(s)

Gc2(s)

Gp12(s)

+

C12(s)

+

Gp21(s)

C21(s) Gp22(s)

C1(s)

C2(s)

+

C22(s)

FIGURE 13.24 Block Diagram of a Negative-Feedback MIMO Control System.

Solution The feedforward transfer function matrix is:



K1 ½GðsÞ ¼ ½Gp ðsÞ $ ½Gc ðsÞ ¼ $ 0:2=ðs þ 0:2Þ 0:3=ðs þ 0:2Þ 0 0:1=ðs þ 0:2Þ 0:5=ðs þ 0:2Þ

0



K2 (13.72)

683

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

2

1.8

1.8

1.6

1.6

1.4

c2

1.4 1.2

1

1

c1

0.8

0.8

c2

0.6

0.6 0.4

0.4

0.2

0.2

0

c1

1.2

0

0.5

1

2

1.5

0

0

1

0.5

2

1.5

Time (seconds)

Time (seconds)

(a)

(b)

FIGURE 13.25 Unit-Step Time Response of MIMO Feedback System: (a) K1 ¼ 161.763, K2 ¼ e8.587; (b) K1 ¼ e25.763, K2 ¼ 53.921.

as in Eq. (11.115) with [Ga(s)] ¼ [I]. The closed-loop transfer function matrix of this unity-feedback system is calculated as in Eq. (11.118) resulting in:

½GCL ðsÞ ¼ ð½I þ ½GðsÞÞ1 $ ½GðsÞ

K1 $ ð10s  7K2 þ 2Þ=DCL ðsÞ 10K2 $ ð5s þ 1Þ=DCL ðsÞ ¼ 4K1 $ ð5s þ 1Þ=DCL ðsÞ K2 $ ð30s  7K1 þ 6Þ=DCL ðsÞ (13.73) where the closed-loop characteristic polynomial is:

DCL ðsÞ ¼ 100s2 þ 10ðK1 þ 3K2 þ 4Þ $ s þ 2K1 þ 6K2  7K1 $ K2 þ 4

(13.74)

Comparison of this polynomial to the standard form s þ 2x$un$s þ (un) results in: 2

2x $ un ¼

2

10ðK1 þ 3K2 þ 4Þ K1 þ 3K2 þ 4 2K1 þ 6K2  7K1 $ K2 þ 4 ¼ ; u2n ¼ 100 10 100 (13.75)

Solving Eq. (13.75) with x ¼ 0.7, un ¼ 10 rad/s, yields two sets of values for the proportional gains: K1 ¼ 161.763, K2 ¼ 8.587 and K1 ¼  25.763, K2 ¼ 53.921. The plots of the controlled signals c1 and c2 are plotted in Figure 13.25. They resulted by applying the MATLAB step command to the corresponding transfer functions of the closed-loop transfer function matrix of Eq. (13.73). The steady-state errors are calculated based on Eq. (13.69), which simplifies to:

feðNÞg ¼ lim ð½I  ½GCL ðsÞÞ $ s/0

  1 1

(13.76)

For the first set of values of the proportional gain, the two steady-state errors are e1(N) ¼ 0.0038 and e2(N) ¼ 0.032, whereas for the second set, the errors are e1(N) ¼ 0.012 and e2(N) ¼ 0.0056. nn

13.3 Feedback Control Systems in the Frequency Domain

13.3 FEEDBACK CONTROL SYSTEMS IN THE FREQUENCY DOMAIN Feedback control systems can also be studied in the frequency domain by means of Bode plots, Nyquist plots, or the root locus. Time-domain characteristics are related with frequency-domain characteristics in the first part of this section. The design of phase-lead and phase-lag compensators is presented subsequently, as they are used to improve the frequency response of existing feedback control systems. This section utilizes the simplified (asymptote-representation) Bode plots that are introduced in Chapter 9.

13.3.1 Frequency-Domain and Time-Domain Connections The connections between time-domain and frequency-domain characteristics of second-order feedback control systems are analyzed in this section.

Frequency Response and Transient Time-Response Characteristics Consider a second-order unity-feedback control system defined by the closed-loop 
 transfer function GCL ðsÞ ¼ u2n s2 þ 2x $ un $ s þ u2n . Frequency-domain specifications, including the bandwidth frequency ub, the resonant-peak coordinates (ur, Mr), and the phase margin (PM) can be related to transient-response, time-domain specifications, such as the settling time ss and peak coordinates (sp, Op) resulting from r(t) ¼ 1. For instance, both the resonant peak Mr of Eq. (9.39) and the maximum overshoot Op of Eq. (13.16) depend on the damping ratio x solely. Eliminating x between these two equations yields:
2 p2 þ ln Op (13.77) Mr ¼  2p $ ln Op The minus sign in Eq. (13.77) ensures that Mr > 0 since ln Op < 0 for underdamping. Substituting the natural frequency un from Eq. (13.13), which gives the settling time ss, into the bandwidth frequency ub of Eq. (9.40), results in: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 $ 1  2x2 þ 4x4  4x2 þ 2 ub ¼ (13.78) ss $ x A similar substitution of un between Eq. (9.40) and Eq. (13.14), which provides the peak time sp, generates the following relationship between ub, sp, and x: p pffiffiffiffiffiffiffiffiffiffiffiffiffi $ ub ¼ sp $ 1  x 2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  2x2 þ

4x4  4x2 þ 2

(13.79)

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

The frequency ub increases when reducing the settling time and the peak timedsee Eqs. (13.78) and (13.79). Two other frequency-domain characteristics are the gain margin and the phase margin, each directly connected to a cross-over frequency, as introduced in Chapter 12, in the context of stability. These parameters can also be related to transient time response characteristics. At gain cross-over, for instance, the modulus of the openloop transfer function is equal to 1, which results in the gain cross-over frequency. For the second-order unity-feedback system, the open-loop transfer function and the related frequency-dependent function are: GOL ðsÞ ¼

GCL ðsÞ u2n ¼ ; 1  GCL ðsÞ s $ ðs þ 2x $ un Þ

GOL ð u $ jÞ ¼

u2n u2n ¼ 2 ð u $ jÞ $ ð2x $ un þ u $ jÞ u þ ð2x $ un $ uÞ $ j

jGOL ð u $ jÞj ¼

u2n qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; u $ u2 þ 4x2 $ u2n

£GOL ð u $ jÞ ¼ 0  tan

1



2x $ un  u







Actual £GOL ð u $ jÞ ¼  180  tan

¼ tan 1

1



 2x $ un u

 2x $ un ; u 

(13.80)

The phase angle of GOL(u$j) in Eq. (13.80), which is tan1(2x$un/u) and which is positive, should actually be negativedit can be checked easily with a simplified Bode plot that starts from 90o (due to the 1/s factor) and grows negative with a 45o/dec slope (due to the 1/(s þ 2x$un) factor). As a consequence (and as discussed in Chapter 12), the last form of the phase angle in Eq. (13.80) reflects the angle’s negative value and should be used. Solving the equation jGOL ðug $ jÞj ¼ 1, results in the following gain cross-over frequency: rq ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ug ¼ un $

4x4 þ 1  2x2

(13.81)

which, in conjunction with the phase angle of Eq. (13.80), yields the phase margin PM: 0 1  
 2x $ un 2x B C ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiA PM ¼ £GOL ug $ j  ð180 Þ ¼ tan1 ¼ tan1 @qp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ug 4x4 þ 1  2x2 (13.82)

13.3 Feedback Control Systems in the Frequency Domain

nn

Example 13.10 The open-loop transfer function of a second-order unity-feedback system is defined by a phase margin of 55 . The bandwidth of this system is 100 rad/s. Identify the second-order control system, i.e., find its natural frequency and damping ratio, and also determine the peak time and maximum overshoot.

Solution Eq. (13.82) changes to:

2x tan 55 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qp ¼ tan 55 / x ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 2 1 þ ðtan 55 Þ2 4x4 þ 1  2x2

(13.83)

whose solution is x ¼ 0.541. Eq. (13.78) enables calculation of the settling time:

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  2x2 þ 4x4  4x2 þ 2 ss ¼ ub $ x 4

(13.84)

which is ss ¼ 0.0905 s. The natural frequency is determined from Eq. (13.13) as un ¼ 4=ðss $ xÞ ¼ 81:71 rad=s. The peak time and maximum overshoot are calculated by means of Eqs. (13.14) and (13.16), respectively, as sp ¼ 0.0457 s and Op ¼ 13.27%. nn

Frequency Response and Steady-State Error Constants The position, velocity, and acceleration constants can be evaluated from available frequency-domain data, particularly from simplified magnitude Bode plots, as discussed here.

Position Constant Kp Consider a type-0 system defined by the open-loop transfer function and its associated complex function:   Q
Q
K$ m K$ m j¼1 szj $ s þ 1 j¼1 szj $ u $ j þ 1  ; Gð u $ jÞ ¼ Qn
 (13.85) GðsÞ ¼ Qn
i¼1 spi $ s þ 1 i¼1 spi $ u $ j þ 1 where K is a constant gain, while szj (j ¼ 1, 2,., m) and spi (i ¼ 1, 2,., n) are time constants associated with the zeroes (subscript z) and the poles (subscript p) of the transfer function, respectively. The position constant is calculated as per Table 13.1 and using the first Eq. (13.85): Kp ¼ lim GðsÞ ¼ K s/0

(13.86)

From Eq. (13.85), the magnitude of the complex transfer function G(u$j) and its limit when u / 0 are: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 Q 1 þ szj $ u K$ m j¼1 lim M ðuÞ ¼ K (13.87) M ðuÞ ¼ jGð u $ jÞj ¼ Q qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ; M0 ¼ u/0 n 1 þ s $ u pi i¼1

687

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CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

Comparing Eqs. (13.86) and (13.87), it follows that Kp ¼ M0. When the magnitude is expressed in dB, the position constant is evaluated as: 20 $ log10 Kp ¼ M0;dB .

Velocity Constant Kv A type-1 system has the following open-loop transfer function and associated harmonic function:   Q
Q
K$ m K$ m j¼1 szj $ s þ 1 j¼1 1 þ szj $ u $ j ; Gð u $ jÞ ¼  (13.88) Q
G ðsÞ ¼ 1 Qn
s $ i¼1 spi $ s þ 1 u $ j $ ni¼1 1 þ spi $ u $ j The velocity constant of this system is calculated based on Table 13.1 and G(s) of Eq. (13.88) as: Kv ¼ lim ðs $ GðsÞÞ ¼ K

(13.89)

s/0

For small frequency values u 1 because a < 1, as per Eq. (11.63). Figure 13.29 illustrates the asymptote Bode plots of the phase-lead, unity-gain (K ¼ 1) compensator transfer function of Eq. (13.100), evaluated by superimposing the magnitudes and phase angles of the component transfer functions G1(s) and G2(s). The two corner frequencies are calculated with the time constants s1 of G1(s) and s2 of G2(s) as: uc1 ¼ 1/s1 ¼ 1/(b$s) and uc2 ¼ 1/s2 ¼ 1/s. For K ¼ 1, the frequency-based compensator transfer function and its dB magnitude are expressed from Eq. (13.100) as: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðb $ s $ u Þ $ j 1 þ ðb $ s $ uÞ2 ; Mc ðuÞ ¼ Gc ð u $ jÞ ¼ 1 þ ðs $ uÞ $ j 1 þ ðs $ u Þ2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðb $ s $ u Þ2 1 þ ðb $ s $ uÞ2 Mc;dB ðuÞ ¼ 20 $ log10 ¼ 10 $ log 10 1 þ ðs $ uÞ2 1 þ ðs $ u Þ2 (13.101) 8 10 $ log10 1 ¼ 0; u  1 > > <  x 1 u2 þ ðb $ sÞ2 > > : 10 $ log10 ¼ 20 $ log10 b; u[1 1=u2 þ s2 The actual phase angle plot, which is shown in simplified asymptote-rendition in Figure 13.29, has a maximum point situated between the corner frequencies; to

693

694

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

calculate this maximum, the phase angle and its frequency derivative are expressed from the frequency transfer function of Eq. (13.101): fðuÞ ¼ tan1 ðb $ s $ uÞ  tan1 ðs $ uÞ;

n h i h io 2 2 s $ b $ 1 þ ð s $ u Þ  1 þ ð b $ s $ u Þ dfðuÞ b$s s h i h i ¼  ¼ 2 du 1 þ ðb $ s $ u Þ 1 þ ðs $ u Þ2 1 þ ðb $ s $ u Þ2 $ 1 þ ðs $ u Þ2 (13.102)

Equating the numerator of the derivative to zero results in: s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 1 pffiffiffi ¼ ¼ uc1 $ uc2 um ¼ $ b$s s s$ b

(13.103)

which is known as the geometric mean frequency; in a logarithmic frequency scale, um is placed at the midpoint between the two corner frequencies, as shown in Figure 13.29. The maximum phase angle (which is 90 only in the simplified representation of Figure 13.29) is calculated by substituting um of Eq. (13.103) into 4 of Eq. (13.102), and results in: b1 tan 4m b1 1 þ sin 4m ffi¼ which is b ¼ tan 4m ¼ pffiffiffi or sin 4m ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 b þ 1 1  sin 4m 2 b 1 þ tan 4m (13.104) Adding a phase-lead compensator to an existing open-loop transfer function can increase the phase margin, the bandwidth, and/or the gain margin. The phase-lead compensator design aims at finding the parameter b, the time constant s, and the derivative gain KD. When the main task is increasing the phase margin to a target value, the following steps need to be followed: • •

•

Evaluation of K from known/necessary steady-state error. Calculation of the gain cross-over frequency ug of the uncompensated system by solving the equation MdB(ug) ¼ 0, and evaluation of the phase margin as PM ¼ 4g  ð  180 Þ, where 4g ¼ 4(ug). Calculation of the phase to be added by the lead compensator as 4m ¼ PMd  PM þ D4, where PMd is the design (desired) phase margin, and D4 is an additional phase that covers for the regular phase decrease at higher (compensated) cross-over frequencies. Figure 13.30 shows the open-loop transfer function Bode plots of the original system and of the system with phase-lead compensation (the latter one is identified by an asterisk*), as well as the asymptote Bode plots of the compensator. As seen in Figure 13.30,  the increase from ug to ug due to the gain increase results in a phase 4 ug < 4ðug Þ

as point A descends to A* in thefigure;  the increase D4 is necessary to counteract the phase decrease 4ðug Þ  4 ug . •

Calculation of b by means of Eq. (13.104).

13.3 Feedback Control Systems in the Frequency Domain

M(dB) M*

20·log10β Mc

M

10·log10β −10·log10β

0 ωg

ω*g

ω

φ(deg)

φm

−90 A PM –180

ωg

A* φ

φc φ*

ω*g

ω

FIGURE 13.30 Bode Plots of Feedback System With Phase-Lead Compensator.

•

   u ¼ 0. It Evaluation of the shifted gain cross-over frequency ug where MdB g is known that the (maximum) gain of the phase-lead compensator is 20$log10bdsee Eq. (13.101); ifthemagnitude of the uncompensated system at

the new frequency ug is MdB ug ¼ 10 $ log10 b, then the additional gain   generated by the compensator should be Mc;dB ug ¼ 10 $ log10 b, to result in a 0 dB gain at ug dsee Figure 13.30. With b calculated at the previous step, ug is   the solution of the equation: MdB ug ¼ 10 $ log10 b. •

•

If we now choose the new gain cross-over frequency to be equal to the mean frequency, which is ug ¼ um , the maximum phase 4m will be added to the original-system phase at this frequency; therefore the .compensator pffiffiffi pffiffiffi  time constant is given by Eq. (13.103) as: s ¼ 1=ðum $ b Þ ¼ 1 ug $ b . Calculation of KD ¼ b$K.

nn

Example 13.12

The open-loop transfer function of a unity-feedback control system is G(s) ¼ K /[s$(s þ 4)]. Calculate the gain K, and design a phase-lead compensator that will result in a unit-ramp steady-state error e(N) ¼ 0.04 and in a phase margin larger than 48 .

Solution For a phase-lead compensator defined as in Eq. (13.100), the modified open-loop transfer function becomes:

G ðsÞ ¼ Gc ðsÞ $ GðsÞ ¼

K $ ð b $ s $ s þ 1Þ K $b$s$s þ K ¼ s $ ðs þ 4Þ $ ðs $ s þ 1Þ s $ s3 þ ð4s þ 1Þ $ s2 þ 4s (13.105)

695

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

The velocity error constant is

Kv ¼ lim s $ G ðsÞ ¼ K=4

(13.106)

s/0

and since e(N) ¼ 1/Kv, it follows that K ¼ 4Kv ¼ 4/e(N) ¼ 100. Using MATLAB’s bode command for the original (uncompensated) open-loop transfer function, the corresponding Bode diagrams are plotted; the Characteristics: Minimum Stability Margins option results in a phase margin PM ¼ 22.6 and a related gain cross-over frequency ug ¼ 9.61 rad/s. The difference between the target phase margin of PMd ¼ 48 and the uncompensated-system phase margin of PM ¼ 22.6 requires a minimum 25.4 phase increase. Adding an extra D4 ¼ 10 , the maximum phase difference needing to be generated by the compensator is 4m ¼ 25.4 þ 10 ¼ 35.4 . From Eq. (13.104), the parameter of the lead-phase compensator is b ¼ 3.7538. As a consequence, the new gain cross-over frequency is calculated by solving the equation MdB ¼ 10$log10(b) ¼ 5.7447. MdB is the decibel magnitude of the original system, which results from the corresponding frequency-domain uncompensated transfer function G(u$j) ¼ 100/[eu2 þ (4u)$j], and is calculated as:

100 MdB ¼ 20 log10 M ¼ 20 log10 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 5:7447 u4 þ 16u2

(13.107)

Eq. (13.107) yields the new gain cross-over frequency ug ¼ 13:635 rad=s. Identifying this frequency with the geometric mean frequency of the compensator um results in the following time constant, as . pffiffiffi  per Eq. (13.103): s ¼ 1 ug $ b ¼ 0:0379. The derivative gain is KD ¼ b$K ¼ 375.38. As a consequence, the new open-loop transfer function of Eq. (13.105) is:

G ðsÞ ¼

14:21s þ 100 0:0379s3 þ 1:151s2 þ 4s

(13.108)

Magnitude (dB)

Bode Diagram Gm = Inf dB (at Inf rad/s) , Pm = 51.7 deg (at 13.6 rad/s) 50

0

-50

-100 -90

Phase (deg)

696

-135

-180 10-1

100

101 Frequency (rad/s)

102

FIGURE 13.31 Bode Plots of Phase-LeadeCompensated Feedback System.

103

13.3 Feedback Control Systems in the Frequency Domain

The Bode plots of the modified open-loop transfer function of Eq. (13.108) is shown in Figure 13.31; they reveal a phase margin of PM* ¼ 51.7 , which is larger than the minimum target value of 48 . nn

Phase-Lag Compensation As discussed in Chapter 11, a phase-lag compensator is a generalized P þ I controller. This compensator improves the static error of the feedback system, similarly to the P þ I controller. The lag compensator limits the low-frequency amplification (gain) of the P þ I controller, which produces a minimal effect on the transient response, particularly when the zero-pole couple is close to the origin. The comparator transfer function of Eq. (11.61) is reformulated here: s þ 1=TI TI $ s þ 1 ¼ ðKP $ aÞ $ a $ TI $ s þ 1 s þ 1=ða $ TI Þ 1 1 ¼ ðKP $ aÞ $ $ ð s 2 $ s þ 1Þ ¼ ðKP $ aÞ $ ðs $ s þ 1Þ $ a$s$s þ 1 s1 $ s þ 1

Gc ðsÞ ¼ KP $

¼ K $ G1 ðsÞ $ G2 ðsÞ with K ¼ KP $ a; TI ¼ s; s1 ¼ a $ s; s2 ¼ s; G1 ðsÞ ¼ 1=ðs1 $ s þ 1Þ; G2 ðsÞ ¼ s2 $ s þ 1 (13.109) where a > 1 (a is large), as indicated in Eq. (11.61). The asymptote Bode plots of the phase-lag compensator defined by the transfer function of Eq. (13.109) are shown in Figure 13.32. These diagrams superimpose the magnitudes and phase M(dB) 20 dB/dec 0

M2

ωc1 ωc2

ω M

−20 dB/dec M1 φ(deg) 90

φ2

45 ωc1

ωc2

0

φ

ωm

–45 –90

FIGURE 13.32 Asymptote Bode Plots of Phase-Lag Compensator.

φ1

ω

697

698

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

angles of the component transfer functions G1(s) and G2(s). The corner frequencies of Figure 13.32 are found by means of the time constants s1 of G1(s) and s2 of G2(s) as: uc1 ¼ 1/s1 ¼ 1/(a$s) and uc2 ¼ 1/s2 ¼ 1/s. For unit gain (K ¼ 1), the sinusoidal transfer function and its associated magnitude result from Gc(s) of Eq. (13.109): sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðs $ uÞ $ j 1 þ ðs $ uÞ2 ; Mc ðuÞ ¼ Gc ð u $ jÞ ¼ 1 þ ða $ s $ uÞ $ j 1 þ ða $ s $ uÞ2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðs $ uÞ2 1 þ ðs $ uÞ2 Mc;dB ðuÞ ¼ 20 $ log10 ¼ 10 $ log 10 1 þ ða $ s $ uÞ2 1 þ ða $ s $ uÞ2 (13.110) 8 10 $ log10 1 ¼ 0; u  1 > > <  2 x 1 u þ s2 > > ¼ 20 $ log10 a; u[1 : 10 $ log10 1=u2 þ ða $ sÞ2 The phase angle equation is: fðuÞ ¼ tan1 ðs $ uÞ  tan1 ða $ s $ uÞ

(13.111)

The actual phase angle reaches a maximum value at a frequency um, which is the one of Eq. (13.103) where a is used instead of b; substituting this frequency into Eq. (13.111) results in: 1a 1a 1  sin fm which is a ¼ tanðfðum ÞÞ ¼ tan fm ¼ pffiffiffi or sin fm ¼ 1þa 1 þ sin fm 2 a (13.112) Phase-lag compensation can improve/realize the stability of an existing feedback system by reducing the high-frequency gain to a negative value from a positive value, at a phase angle of 180 . Other effects are reducing the steady-state error by increasing the low-frequency gain associated with the compensator, without affecting the transient response sensibly. Consider a feedback system whose openloop transfer function Bode plots, together with the simplified Bode plots of the compensator, are schematically drawn in Figure 13.33 over a limited frequency interval. This system is unstable because the dB magnitude corresponding to a 180 (occurring at the phase cross-over frequency up) is positive, MdB(up) > 0. Designing a phase-lag compensator (which means finding the parameters K, s, and a) to render such a system into a stable system, while improving its steadystate errors, requires the following steps: • •

Evaluation of the gain K that results in a suitable steady-state error and plotting of the uncompensated-system Bode plots. Given a design phase margin PMd, calculation of the phase margin of the compensated system as PM* ¼ PMd þ D4, where D4 is an additional phase that

13.3 Feedback Control Systems in the Frequency Domain

M(dB) M M 0

ωc1

M(ωg) M(ωp)

*

ωc2

ωp

ω*g

ω

–20 dB/dec

Mc –M(ωg)

φ(deg) 0 φ*g

ω*g

φc

ωp

PM *

ω

φ

–180

FIGURE 13.33 Bode Plots of Feedback System With Phase-Lag Compensator.

•

•

•

•

•

covers for the phase decrease due to the compensation; Ogata (2004) recommends D4 be in the 5e12 range. Calculation of the phase angle 4g corresponding to PM* as fg ¼ ð180  PM  Þdsee phase plot of Figure 13.33dand of the associated frequency ug from the same phase  plot.  Calculation of the uncompensated gain M ug from the uncompensated gain plot. Since ug is the compensated-system gain cross-over frequency, the   compensator gain at this point should be equal to M ug . Selection of the high-frequency corner frequency as one decade smaller than the gain cross-over frequency: uc2 ¼ 0:1 $ ug ; this particular choice ensures that the phase drop due to compensation is within the 5e12 , as recommended by Ogata (2004). The compensator time constant can now be calculated as s ¼ 1/uc2. Use a 20 dB/dec line starting from the point defined by uc2 on the highfrequency gain asymptote to the intersection with the 0 dB low-frequency gain asymptote; the intersection point is at the other corner frequency uc1. The compensator parameter a can now be calculated as a ¼ 1/(s$uc1). Plot the Bode diagrams of the compensated system (identified by an asterisk in the magnitude Bode plot of Figure 13.33). Make further parameter refinements if still necessary to alter the system frequency-response performance. Calculate the proportional gain KP ¼ K/a.

nn

Example 13.13

A unity-feedback control system is defined by the open-loop transfer function G(s) ¼ K / [s$(s2þ4s þ 1)]. Calculate the gain K that will result in a steady-state error e(N) ¼ 0.1 for unitramp input. Design a phase-lag compensator that produces a phase margin of at least 45 and a gain margin of at least 25 dB. Also plot the time response of the resulting control system for r(t) ¼ 1.

699

700

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

Solution

The velocity constant error is evaluated as Kv ¼ lim s $ GOL ðsÞ ¼ K , which means that K ¼ Kv ¼ 1/ s/0

e(N) ¼ 1/0.1 ¼ 10. The system is unstabledthis can be demonstrated by simply calculating the closed-loop poles (two of each are in the RHP). The gain cross-over frequency needs to be determined first. For a design PMd of 45 , an additional 12 (to account for the phase decrease due to the phaselag action) is added to result in an actual PM* of 57 . The angle corresponding to this phase is

4g ¼ 180 þ 57 ¼ 123 . The frequency open-loop transfer function together with its magnitude and phase angle are:

Gðu $ jÞ ¼

4u2

8 10 > > qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi MðuÞ ¼ > >
2 > < u $ ð4uÞ2 þ 1  u2

10 ;   þ u $ ð1  u2 Þ $ j > > > 1  u2 > 1 > : 4ðuÞ ¼ tan 4u

(13.113)   Solving the equation 4 ug ¼ 123 , results in ug ¼ 0:158 rad=s. The large corner frequency is therefore calculated as uc2 ¼ 0:1 ug ¼ 0:0158 rad=s. The corresponding time constant is s2 ¼ s ¼ 1/uc2 ¼ 63.1774 s. The magnitude corresponding to ug is calculated from   Eq. (13.113) as MdB ug ¼ 34:7035 dB. The compensator needs to provide a gain of   Mc;dB ug ¼ 34:7035 dB. Taking into account that the frequency is in logarithmic scale, the following relationship connects the two corner frequencies, the compensator gain at the cross-over frequency, and the 20 dB/dec slope of the segment line of the phase diagram: 34.7053 ¼ (20)$ log10(uc2/uc1). Solving this equation, results in the other corner frequency: uc1 ¼ 2.9125$104 rad/s, whose corresponding time constant is s1 ¼ a$s ¼ 1/uc1 ¼ 3433.48 s. The parameter a is therefore a ¼ s1/s2 ¼ 54.374 and the proportional gain is KP ¼ K/a ¼ 10/54.374 ¼ 0.184. With these values, the compensated open-loop transfer functions become:

G ðsÞ ¼ Gc ðsÞ $ GðsÞ ¼

631:77s þ 10 3433:48s4 þ 13734:93s3 þ 3437:48s2 þ s

(13.114)

where Gc(s) is given in Eq. (13.109). The Bode diagrams of G*(s), together with the phase margin and the gain margin, are given in Figure 13.34dboth values satisfy the requirements of this example. The phase and gain margins with their frequencies were plotted on the Bode diagrams by using the MATLAB command margin(sys)dwhere sys was the compensated transfer function. An additional MATLAB command step(feedback(sys,1)) generates the unit-step response of Figure 13.35, which illustrates the overdamped character of the feedback system (due to the original overdamped second-order plant system). As the figure indicates, the response is slow (the large time constants were an early indication of this behavior), with a peak response slightly larger than 1.2 after approximately 20 s and a steadystate response reached after 90 s.

13.3 Feedback Control Systems in the Frequency Domain

FIGURE 13.34 Bode Plots of Phase-LageCompensated Feedback System.

FIGURE 13.35 Unit-Step Time Response of Compensated Feedback System. nn

701

702

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

SUMMARY This chapter discusses the introductory notions and methods for the modeling, analysis, and design of feedback control SISO and MIMO dynamic systems in the time and frequency domains. System characteristics are presented that define the transient time response of first- and second-order control systems. The steady-state errors are defined and analyzed in terms of the system type and input. Simulink is applied to determine and plot the full time response of feedback control systems that may include nonlinearities. Bode diagrams are used to analyze and design feedback control systems in the frequency domain. MATLAB built-in commands are utilized throughout this chapter to aid in the solution of almost all the problems. Phase-lead and phase-lag compensators, and their design in the frequency domain, are studied as well.

PROBLEMS 13.1 Identify the static sensitivity K and time constant s of the first-order closedloop transfer function of a feedback control system. Under unit-step input, the response rise time is sr ¼ 0.03 s and the response at t ¼ s is 25. 13.2 Identify the natural frequency un and damping ratio x of the second-order closed-loop transfer function of a unit-sensitivity feedback electrical system whose unit-step time response display a peak time sp ¼ 0.2 s and maximum overshoot Op ¼ 0.12. What are the rise time sr and the settling time ss? Plot the position of the closed-loop poles in the complex plane. 13.3 Determine the damping ratio x and natural frequency un of a unitsensitivity second-order closed-loop transfer function corresponding to a unity-feedback control system. Known are the time constant s ¼ 0.02 s and the maximum overshoot Op ¼ 15%. What change in the natural frequency will generate a reduction in the rise time sr of 30%? 13.4 The closed-loop transfer function of a unity-feedback control MEMS is 
 GCL ðsÞ ¼ 490; 000 s2 þ 650s þ 490; 000 . (i) What are the changes needed in the equivalent damping ratio x and natural frequency un that produce a reduction of 20% in the maximum overshoot Op and a reduction of 15% in the rise time sr under unit-step input? (ii) How do the closed-loop poles move in the complex plane as a result of these changes?

Problems

13.5 Determine the equivalent damping ratio and the natural frequency of the control system sketched in Figure 13.36. Calculate the peak time sp, maximum overshoot Op, rise time sr, and settling time ss.

_ R(s) 70

+

+ _

1 s

1 s+ 8

C(s)

8000

FIGURE 13.36 Control System With Two Nonincluding Feedback Loops.

13.6 The velocity constant of a unity-feedback system is Kv ¼ 100. Determine the system type and calculate the steady-state errors resulting from unitstep, unit-ramp, and unit-parabola inputs. 13.7 The acceleration constant of a unity-feedback system is Ka ¼ 25. Determine the system type and calculate the steady-state errors resulting from unit-step, unit-ramp, and unit-parabola inputs. 13.8 A plant Gp(s) ¼ (s þ 8)/(s3 þ 5s2 þ s) is connected in a unity-feedback control system. Verify the stability of the resulting system, determine its type, calculate the error constants, and determine the steady-state errors for unit-step, unit-ramp, and unit-parabola inputs. 13.9 Solve Problem 13.8 for Gp(s) ¼ 60/(2s5 þ 3s4 þ s3). 13.10 Consider a nonunity-feedback control system defined by the feed-forward transfer function G(s) ¼ (s þ 3)/(s2 þ 3s þ 2) and the feedback transfer function H(s) ¼ 58/s. Determine the system type, calculate the position, velocity, and acceleration constants, as well as the steady-state errors corresponding to unit-step, unit-ramp, and unit-parabola inputs. 13.11 Solve Problem 13.10 for G(s) ¼ (s þ 1)/[s2$(s2 þ 3s þ 2)] and the feedback transfer function H(s) ¼ (s þ 3)/(s þ 4). 13.12 Consider the feedback control system shown in Figure 11.13 of Example 11.2. For G1(s) ¼ 60/(s þ 2), G2(s) ¼ 1/(s3 þ 2s þ s þ 3), and G3(s) ¼ (s þ 10)/ (s þ 1), establish the system type and calculate the steady-state error for r(t) ¼ 1.

703

704

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

13.13 Consider the feedback control system shown in Figure 11.14 of Example 11.3. For G1(s) ¼ 3/((s2 þ s þ 5), G2(s) ¼ (s þ 1)/(s þ 3), and G3(s) ¼ 60/(s þ 2), establish the system type and calculate the steady-state error for r(t) ¼ t. 13.14 (i) Using three identical resistors and one capacitor, design a two-stage operational amplifier circuit to function as an integral controller. (ii) The plant being the electrical circuit shown in Figure 13.37 (vi is the voltage input and vo is the voltage output), find the system type and determine the steady-state error of the unity feedback control system that adds the integral controller identified at (i) to the plant when the reference signal is a unit ramp. Numerical application: R ¼ 350 U (for both controller and plant), L ¼ 0.8 H, C ¼ 25 mF (for both controller and plant). Z1 R

Z2 vi

vo C

L Z3

FIGURE 13.37 Electrical Plant Circuit.

13.15 A first-order plant, which has a static sensitivity K ¼ 1, is connected in a unity-feedback control system. To reduce its steady-state error by 40% under unit-step input, the plant is connected to a P controller in the same unity-feedback system. Calculate the necessary proportional gain KP that will produce this change. 13.16 A first-order plant, which is defined by a time constant s ¼ 0.5s and static sensitivity K ¼ 1, is connected to a P þ D controller in a unity-feedback control system. It is necessary that the initial response is c(0) ¼ 0.01 and the steady-state error is minimum under unit-step input. Knowing that 0 < KP < 100 and KD > 0, determine the proportional gain KP and derivative time TD that will realize this performance by the feedback system.

Problems

13.17 The first-order plant of Problem 13.16 is connected to a P þ I þ D controller in a unity-feedback system. The closed-loop transfer function characteristic polynomial is a second-degree polynomial with a damping ratio x ¼ 0.6 and a natural frequency un ¼ 20 rad/s. It is also known that the steady-state error of this controlled system is e(N) ¼ 0.02 for a unitramp input. Determine the proportional gain KP, the integral time TI, and the derivative time TD that will satisfy these performance figures. 13.18 A unit-sensitivity second-order plant is defined by un ¼ 230 rad/s and x ¼ 0.45. Design a P þ D controller to connect to the plant in a unityfeedback system whose closed-loop transfer function characteristic polynomial has a damping ratio x* ¼ 0.55 and a steady-state error e(N) ¼ 0.6 under unit-step input. Plot the root locus of this system and establish whether the system is a minimum-phase system. 13.19 A feed-forward second-order transfer function is connected in a unityfeedback control system. It results in a peak time sp ¼ 0.02 s and a settling time ss ¼ 0.5 s under unit-step input. An I controller is added to the original system. Evaluate the integral gain KI that will result in an additional closed-loop pole that is six times farther away to the left from the two closed-loop poles of the original system. Plot the root locus and time response for both the original and altered control systems. 13.20 A P þ I control unit is added to the feed-forward transfer function of Problem 13.19. Evaluate the proportional and integral gains KP and KI that will result in an additional closed-loop pole that is eight times farther away to the left from the smallest closed-loop pole of the original system, as well as into an additional zero that is within 5% away from the additional closedloop pole. Plot the root locus and time response for both the original and the altered control systems. Is the modified system displaying nearezero-pole cancellation behavior? 13.21 For the mechanical microsystem (plant) shown in Figure 13.38 an actuation force displaces the shuttle mass along the y direction. The shuttle has a mass Anchor Shuttle mass

Beam spring

Actuation y

FIGURE 13.38 Translatory Mechanical Microsystem With Actuation Force.

705

706

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

m ¼ 2$1010 kg and the four beam springs supporting the shuttle mass are identical. When P control is applied in a unity feedback loop, the natural frequency of the controlled system is un ¼ 900,000 radians/s. Under a unit-step input, this controlled system records a maximum overshoot of 8% and a steady-state error e(N) ¼ 0.04. (i) Determine the beam spring stiffness k about the motion direction, the viscous damping coefficient c, and the proportional gain KP. (ii) Design a proportional-plus-derivative P þ D controller to replace the original P controller to reduce the steady-state error to e(N) ¼ 0.015 under unit-step input and to produce an effective damping ratio of x ¼ 0.6. 13.22 The microcantilever of Figure 13.39, receives an input displacement u from a rigid, massless frame, and undergoes bending vibrationsdthe cantilever free end deflection is y. This deflection is sensed optoelectronically by a photodiode whose output voltage is proportional (constant Ks) to the cantilever free end deflection. These two units are connected in a standard nonunity feedback control system with a summing point. (i) Express the steady-state error in terms of the transfer functions involved and a generic reference signal. Use a lumped-parameter model for the cantilever with mass m, viscous damping coefficient c, and stiffness kdall expressed at the free end of the cantilever. (ii) For I control, calculate the steady-state error when applying individually a unit step, unit ramp, and a unit parabola input to this control system. Rigid frame Position sensor Microcantilever

y

u

FIGURE 13.39 Microcantilever as a Plant With Position Sensing Unit.

13.23 The plant of Problem 13.22 has its original I controller combined serially to a P þ I unitdboth control units have the same integral constant KI. Considering unity feedback (Ks ¼ 1), evaluate the steady-state error sensitivity to the proportional constant KP and the integral constant KI. 13.24 Solve Example 13.6 for a plant defined by the transfer function: Gp(s) ¼ 1/ [s2$(s þ 1)$(s þ 6)].

Problems

13.25 Plot the time response of the nonlinear feedback control system of Example 13.7 using P þ I þ D controls with KP ¼ 10, TI ¼ 0.08 s, and TD ¼ 0.11 s. Consider an input xr ¼ 0.02 m and separately an input xr ¼ 0.02$sin(5p$t) m. Use all other numerical values provided in Example 13.7. 13.26 A MISO feedback with disturbances has the block diagram of Figure 11.34. For Gp(s) ¼ 1/(s3 þ s2 þ 3s þ 2), Gc(s) ¼ (s þ 1)/(s þ 5), and H(s) ¼ 100/ (s þ 2), calculate the total steady-state error resulting from r(t) ¼ d(t) ¼ 1. Also, use Simulink to plot the system’s time response c(t) for r(t) ¼ 5 and d(t) ¼ 5$e10 t$sin(20p$t). 13.27 Solve Example 13.9 for x ¼ 0.45 and un ¼ 20 rad/s. All other parameters and conditions are the ones provided in Example 13.9. 13.28 Use Simulink to plot the time response of the system modeled in Example 11.13 considering proportional control with Gc1(s) ¼ Gc2(s) ¼ 10, proportional sensing KH ¼ 170 V/m when r1(t) ¼ r2(t) ¼ 1m. We know: m1 ¼ 0.05 kg, m2 ¼ 0.08 kg, c ¼ 6 N s/m, k ¼ 120 N/m, k1 ¼ 100 N/m, k2 ¼ 160 N/m, L ¼ 0.1 H, R ¼ 210 U, Km ¼ 1000 N/A, and Ke ¼ 120 V s/m. 13.29 The feed-forward transfer function of a second-order system, which is placed in a unity-feedback control system has a damping ratio x ¼ 0.5 and a natural frequency un ¼ 180 rad/s. Calculate the corresponding phase margin, resonant peak, and bandwidth frequency. 13.30 A second-order feed-forward transfer function is connected in a unityfeedback control system. Knowing its natural frequency un ¼ 180 rad/s, and its gain crossover frequency ug ¼ 150 rad/s, calculate the damping ratio x, resonant frequency ur, resonant peak Mr, and phase margin PM. 13.31 Identify the open-loop transfer function, whose asymptote Bode magnitude is shown in Figure 13.40. Determine the steady-state error of the unity-feedback control system formed with this transfer function for unitramp and unit-parabola inputs; also, calculate the bandwidth frequency. MdB 80 −40 dB/dec 40

GOL(s) −20 dB/dec

0 1

10

100

ω

FIGURE 13.40 Asymptote Magnitude Bode Plots of Open-Loop Transfer Function for a Unity-Feedback Control System.

707

708

CHAPTER 13 Time- and Frequency-Domain Controls of Feedback Systems

13.32 The open-loop transfer function of a unity-feedback control system is G(s) ¼ K/[s$(s þ 3)]. Calculate the gain K, and design a phase-lead compensator that will result in a unit-ramp steady-state error e(N) ¼ 0.05 and in a phase margin larger than 50 . 13.33 Using the equations of gain margin, phase margin, and of their corresponding frequencies, solve Problem 13.32 that will result in precisely a phase margin of 50 . Also, plot the corresponding Bode diagrams and the controlled time response for r(t) ¼ t. 13.34 The open-loop transfer function of a unity-feedback control system is G(s) ¼ K/[s$(s2þ15s þ 2)]. Calculate the gain K that will result in a steadystate error e(N) ¼ 0.15 for unit-ramp input. Also design a phase-lag compensator that produces a phase margin of at least 55 and a gain margin of at least 45 dB. 13.35 Using the equations of gain margin, phase margin, and of their corresponding frequencies, solve Problem 13.34 that will result in precisely a phase margin of 55 and a gain margin of 45 dB. Also, plot the corresponding Bode diagrams and the controlled time response for r(t) ¼ t.

Suggested Reading K. Ogata, System Dynamics, 4th Ed. Prentice Hall, Upper Saddle River, 2004. N.S. Nise, Control Systems Engineering, 7th Ed. Wiley, Hoboken, NJ, 2015. G.F. Franklin, J.D. Powell, A. Emami-Naeini, Feedback Control of Dynamic Systems, 6th Ed. Pearson, Upper Saddle River, NJ, 2010. R.T. Stefani, B. Shahian, C.J. Savant, and G.H. Hostetter, Design of Feedback Control Systems, 4th Ed. Oxford University Press, New York, 2002. C.H. Phillips and J.M. Parr, Feedback Control Systems, 5th Ed. Prentice Hall, Upper Saddle River, 2011. B.C. Kuo and F. Golnaraghi, Automatic Control Systems, 10th Ed. McGraw-Hill, New York, 2017. K. Ogata, Modern Control Engineering, 5th Ed. Prentice Hall, Pearson, Upper Saddle River, NJ, 2009. R.C. Dorf and R.H. Bishop, Modern Control Systems, 13th Ed. Prentice Hall, Upper Saddle River, 2016. S.M. Tripathi, Modern Control Systems, Jones & Bartlett Learning, Sudbury, 2008.

APPENDIX

Complex Numbers

A complex number z is expressed in standard (algebraic) form as: pffiffiffiffiffiffiffi z ¼ x þ y$j or z ¼ x þ y$i with j ¼ i ¼ 1

A (A.1)

where the real numbers x and y are the real and imaginary parts, respectively. The powers of j are j2 ¼ 1, j3 ¼ j, j4 ¼ 1, ., jnþ4 ¼ jn. When y ¼ 0, the complex number z is real, z ¼ x, whereas for x ¼ 0, z ¼ y$j is imaginary. Two complex numbers z1 ¼ x1 þ y1$j and z2 ¼ x2 þ y2$j are added/subtracted as: z1  z2 ¼ ðx1  x2 Þ þ ðy1  y2 Þ$j

(A.2)

Two complex numbers z1 and z2 are equal if and only if x1 ¼ x2 and y1 ¼ y2. The complex number z ¼ x  y$j is the conjugate of the original complex number z ¼ x þ y$j. The following property: z$z ¼ ðx þ y$jÞ$ðx  y$jÞ ¼ x2 þ y2

(A.3)

enables expressing the reciprocal of a complex number, which is 1/z, in standard form as: 1 z x  y$j x y ¼ ¼ ¼  $j z z$z x2 þ y2 x2 þ y2 x2 þ y2

(A.4)

The trigonometric (or polar) form of a complex number is: z ¼ jzj$½cos£z þ ðsin£zÞ $j ¼ r$½cos 4 þ ðsin 4Þ$j

(A.5)

where r ¼ jzj is the modulus, and 4 ¼ £z is the argument. The image of a complex number in the complex plane is illustrated in Figure A.1, which shows the two sets of parameters, algebraic and trigonometric, defining the complex number z. The following transformations apply between algebraic and trigonometric parameters: 8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( < x ¼ r$cos 4 r ¼ x2 þ y 2 and (A.6) : y ¼ r$sin 4 4 ¼ tan1 ðy=xÞ Euler’s identity: e4$j ¼ cos 4 þ ðsin 4Þ$j

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15001-8 Copyright © 2018 Elsevier Inc. All rights reserved.

(A.7)

709

APPENDIX A

Imaginary axis

710

y

z z=r

Real axis

ϕ= z

x

FIGURE A.1 Graphical Representation of Complex Number in Complex Plane (Argand Diagram).

leads to: cosð4Þ ¼

e4$j þ e4$j e4$j  e4$j ; sinð4Þ ¼ 2 2j

(A.8)

Euler’s identity permits expressing the trigonometric complex number of Eq. (A.5) in phasor (or exponential) form: z ¼ r$e4$j ¼ jzj$eð£zÞ$j

(A.9)

The phasor form simplifies division, multiplication, or raising to power of complex numbers, as shown in Eq. (A.10): z¼

z1 jz1 j$eð£z1 Þ$j jz1 j ð£z1 £z2 Þ$j $e ¼ ¼ z2 jz2 j$eð£z2 Þ$j jz2 j

z ¼ z1 $z2 ¼ jz1 j$eð£z1 Þ$j $jz2 j$eð£z2 Þ$j ¼ jz1 j$jz2 j$eð£z1 þ£z2 Þ$j zn ¼ ðjzj$eð£zÞ$j Þn ¼ jzjn $eðn$£zÞ ¼ jzjn $fcosðn$£zÞ þ ½sinðn$£zÞ $jg ¼ r n $fcosðn$4Þ þ ½sinðn$4Þ$jg

(A.10)

The last Eq. (A.10) is known as De Moivre’s theorem (or identity). The hyperbolic functions: e4 þ e4 e4  e4 ; sinhð4Þ ¼ (A.11) 2 2 are connected to exponential complex numbers by means of Euler’s identity as: coshð4Þ ¼

e4$j þ e4$j ¼ cos 4; 2 e4$j  e4$j e4$j  e4$j ¼ $j ¼ ðsin 4Þ$j sinhð4$jÞ ¼ 2 2j

coshð4$jÞ ¼

(A.12)

APPENDIX

B

Matrix Algebra

SPECIAL-FORM MATRICES A column vector {x}c and a row vector {x}r, both having n elements, are defined as: 8 9 x1 > > > > >

= 2 ; fxgr ¼ f x1 x2 . xn g (B.1) fxgc ¼ > .> > > > > : ; xn A square matrix of the n dimension (or an n  n matrix) comprises an equal number n of rows and columns: 3 2 a11 a12 . a1n 7 6a 6 21 a22 . a2n 7 (B.2) ½A ¼ 6 7 4. . . .5 an1

an2

.

ann

where a11, a22, ., ann are diagonal elements (or elements on the main diagonal). A symmetric matrix is a square matrix for which aij ¼ aji: 3 2 a11 a12 . a1n 7 6a 6 12 a22 . a2n 7 (B.3) ½A ¼ 6 7 4. . . .5 a1n

a2n

.

ann

A diagonal matrix is a square matrix with all elements zero except for the elements placed on the main diagonal: 2 3 d11 0 . 0 6 0 d 0 7 22 . 6 7 ½D ¼ 6 (B.4) 7 4. . . .5 0

0

.

dnn

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15002-X Copyright © 2018 Elsevier Inc. All rights reserved.

711

712

APPENDIX B

The identity (or unity) matrix is a diagonal matrix with the nonzero elements being equal to 1: 3 2 1 0 . 0 6 0 1 . 0 7 7 6 (B.5) ½I ¼ 6 7 4. . . .5 0

0

.

1

A zero matrix, zero column vector, or zero row vector has all elements equal to 0. The transpose of an n  m matrix [A]nm is an m  n matrix [B]mn defined as ½Bmn ¼ ½ATmn , whose elements are bij ¼ aji, where aij are the elements of the original matrix: 3 3 2 2 a11 a21 . an1 a11 a12 . a1m 7 7 6a 6a 6 12 a22 . an2 7 6 21 a22 . a2m 7 T ½A ¼ 6 7; ½B ¼ ½A ¼ 6 7 (B.6) 4. . . .5 4. . . .5 an1 an2 . anm a1m a2m . anm Transposing a column vector results in a row vector and, similarly, transposing a row vector produces a column vector, as per Eqs. (B.1) and (B.6).

BASIC MATRIX OPERATIONS The result of multiplying 2 a11 a12 6a 6 21 a22 b$½A ¼ b$6 4. . an1

an2

a scalar b by an n  m matrix [A] is: 3 2 3 b$a11 b$a12 . b$a1m . a1m 7 6 . a2m 7 7 6 b$a21 b$a22 . b$a2m 7 7¼6 7 ¼ ½b$A . .5 4 . . . . 5 .

b$an1

anm

b$an2

. b$anm (B.7)

Two matrices need the same dimensions to add algebraically; adding/subtracting the matrices [A] and [B], both of an n  m dimension, results in: 2 3 2 3 a11 a12 . a1m b11 b12 . b1m 6a 7 6 7 6 21 a22 . a2m 7 6 b21 b22 . b2m 7 ½A  ½B ¼ 6 76 7 4. . . .5 4. . . . 5 an1

an2

.

bn1

a11  b11 6a  b 21 6 21 ¼6 4 .

a12  b12 a22  b22

. .

.

.

bn2 . bnm 3 a1m  b1m a2m  b2m 7 7 7 5 .

an1  bn1

an2  bn2

.

anm  bnm

2

anm

(B.8)

Basic Matrix Operations

Two matrices [A] and [B] can be multiplied only if the number of columns of [A] is equal to the number of rows of [B]; in other words, [A] is of the n  m dimension, [B] is of the m  p dimension, and [C] is of the n  p dimension. A generic element of the product matrix [C]np ¼ [A]nm$[B]mp, say cij (located in the ith row and the jth column of [C]), is calculated as: cij ¼

m X

ðaik $bkj Þ

(B.9)

k¼1

for i ¼ 1, 2, ., n and j ¼ 1, 2, ., p. Except for special cases, the product of two matrices [A] and [B] is noncommutative, namely: ½A$½Bs½B$½A. Multiplying a square matrix or a vector of matching dimensions by the identity matrix leaves the original matrix or vector unchanged: ½A$½I ¼ ½I$½A ¼ ½A; ½I$fxg ¼ fxg

(B.10)

The associative and distributive laws are applicable to matrix addition and multiplication: ð½A$½BÞ$½C ¼ ½A$ð½B$½CÞ; ½A$ð½B þ ½CÞ ¼ ½A$½B þ ½A$½C

(B.11)

1

The inverse of a square matrix [A] is denoted by [A] , and the following property applies to the original matrix and its inverse: ½A$½A1 ¼ ½A1 $½A ¼ ½I

(B.12)

The inverse of a square matrix [A] is calculated as: ½A1 ¼

adj½AT det½A

(B.13)

The determinant of a square matrix [A], det[A], is a number that can be calculated based on the elements of a single row or column. When the first row is used, the determinant is: det½A ¼ a11 $b11 þ a12 $b12 þ . þ a1n $b1n

(B.14)

b1j ¼ ð1Þ1þj $det½A1j 

(B.15)

where

(with j ¼ 1, 2, ., n) is the 1jth cofactor of [A] and [A1j] is the 1jth minor of [A] obtained by deleting row number 1 and column number j of the original matrix [A]. The determinant of a 2  2 matrix is calculated as det[A] ¼ a11$a22  a12$a21. The adjoint matrix of [A]T, which is denoted by adj[A]T in the numerator of Eq. (B.13), is a matrix that contains the cofactors of the elements of the transposed matrix [A]T. A square matrix that can be inverted has its determinant det[A] s 0dthis matrix is known as nonsingular.

713

714

APPENDIX B

The solution to a system of n linear algebraic equations with constant coefficients of the form: 8 a11 $y1 þ a12 $y2 þ . þ a1n $yn ¼ b1 > > > < a21 $y1 þ a22 $y2 þ . þ a2n $yn ¼ b2 (B.16) > . > > : an1 .y1 þ an2 .y2 þ . þ ann $yn ¼ bn where y1, y2, ., yn are the unknowns, can be obtained using matrix operations. The system of Eq. (B.16) can be written as: ½A$fyg ¼ fbg where 2

a11

6a 6 21 ½A ¼ 6 4. an1

a1n

3

a12

.

a22 .

. .

a2n 7 7 7; .5

an2

.

ann

8 9 8 9 b1 > y1 > > > > > > > > > >

=

> .> .> > > > > > > : > ; ; : > yn bn

(B.17)

(B.18)

Left multiplying Eq. (B.17) by [A]1, results in the solution to Eq. (B.17): ½A1 $½A$fyg ¼ ½A1 $fbg / ½I$fyg ¼ ½A1 $fbg / fyg ¼ ½A1 $fbg (B.19) For {b} ¼ {0}, the Eq. (B.18) are homogeneous. Combining Eqs. (B.13) and (B.19) yields: {y}$det[A] ¼ adj[A]T${0} ¼ {0}. This condition indicates that a set of homogeneous equations has a nontrivial (nonzero) solution {y} s {0} only when det[A] ¼ 0. The eigenvalue problem is defined by the following equation: ½A$fYg ¼ l$fYg

(B.20)

where [A] is an n  n square matrix, {Y} is a vector known as eigenvector, and l is an eigenvalue. Eq. (B.20) can be written as: ð½A  l$½IÞ$fYg ¼ f0g

(B.21)

which represents a set of n homogeneous algebraic equations, the unknowns being Y1, Y2, ., Yndthe eigenvector components. For {Y} to be nontrivial (nonzero), the system’s determinant has to be zero: detð½A  l$½IÞ ¼ 0

(B.22)

which is the characteristic equation associated with the eigenvalue problem of Eq. (B.21). The roots of the characteristic equation are the eigenvalues l1, l2, ., ln. Each of the eigenvalues is substituted back into Eq. (B.21), which can be solved for a corresponding eigenvector. Only n  1 components of any eigenvector are independent; each of those components can be expressed in terms of the nth

Basic Matrix Operations

component, which can be chosen arbitrarily. Many eigenvalue applications use a value of 1 for the arbitrary component of each eigenvector. In other instances, the condition of unit norm is used, which requires that the norm of each eigenvector be 1: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Y 21 þ Y 22 þ . þ Y 2n ¼ 1 (B.23)

715

APPENDIX

Solutions to Linear Homogeneous Ordinary Differential Equations With Constant Coefficients

C

Consider the following homogeneous, linear differential equations with the unknown time-dependent function y: dn yðtÞ d n1 yðtÞ dyðtÞ þ a0 $yðtÞ ¼ 0 þ a $ þ . þ a1 $ (C.1) n1 dtn dtn1 dt where a0, a1, ., ane1, an are constant coefficients. Searching for solutions of the form: an $

yðtÞ ¼ el$t

(C.2)


el$t $ an $ln þ an1 $ln1 þ . þ a1 $l þ a0 ¼ 0

(C.3)

in Eq. (C.1) results in:

The specific function of Eq. (C.2) is a solution to the differential Eq. (C.1) only when: an $ln þ an1 $ln1 þ . þ a1 $l þ a0 ¼ 0

(C.4)

Eq. (C.4) is the characteristic equation associated with Eq. (C.1) whose roots are l. Solving Eq. (C.4) for the roots l1, l2, ., lne1, ln results in the following solution to Eq. (C.1): y1 ðtÞ ¼ el1 $t ; y2 ðtÞ ¼ el2 $t ; .; yn1 ðtÞ ¼ eln1 $t ; yn ðtÞ ¼ eln $t

(C.5)

As a consequence, the general homogeneous solution of Eq. (C.1) is a linear combination of the individual solutions of Eq. (C.5): yh ðtÞ ¼ c1 $y1 ðtÞ þ c2 $y2 ðtÞ þ . þ cn1 $yn1 ðtÞ þ cn $yn ðtÞ ¼ c1 $el1 $t þ c2 $el2 $t þ . þ cn1 $eln1 $t þ cn $eln $t

(C.6)

where c1, c2, ., cn are constant coefficients. There are four different cases, depending on the possible types of roots of the characteristic equation, as the eigenvalues can be: (1) real and simple, (2) complex System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15003-1 Copyright © 2018 Elsevier Inc. All rights reserved.

717

718

APPENDIX C

and simple, (3) real and multiple, and (4) complex and multiple. Therefore, the general solution contains contributions from all four categories and can formally be written as: yðtÞ ¼ ya ðtÞ þ yb ðtÞ þ yc ðtÞ þ yd ðtÞ

(C.7)

with ya(t), yb(t), yc(t), and yd(t) being provided in Table C.1. All coefficients in the Solution column of Table C.1 are constant. Table C.1 Solutions to Linear Homogeneous Ordinary Differential Equations With Constant Coefficients Roots l Number or Order of Multiplicity

Solution

Real and simple, li

na

ya ðtÞ ¼ ca;1 $el1 $t þ ca;2 $el2 $t þ . þ ca;na $elna $t

Complex and simple, li ¼ si  ui$j

nb

yb ðtÞ ¼ cb;1 $es1 $t $cosðu1 $tÞ þ c0b;1 $es1 $t $sinðu1 $tÞ þ .

Real and multiple, l

nc

Complex and multiple, l ¼ s  u$j

nd

Type

þcb;nb $esnb $t $cosðunb $tÞ þ c0b;nb $esnb $t $sinðunb $tÞ yc ðtÞ ¼ cc;1 $el$t þ cc;2 $t$el$t þ . þcc;nc 1$tnc 2 $el$t þ cc;nc $tnc 1 $el$t yd ðtÞ ¼ cd;1 $es$t $cosðu$tÞ þ c0d;1 $es$t $sinðu$tÞ þ cd;2 $t$es$t $cosðu$tÞ þ c0d;2 $t$es$t $sinðu$tÞ þ . þ cd;nd $tnd 1 $es$t $cosðu$tÞ þ c0d;nd $tnd 1 $es$t $sinðu$tÞ

APPENDIX

D

Basics of Simulink

This appendix introduces the program Simulink as a graphical tool for modeling, simulating, solving, and plotting the time response of system dynamic problems. While Simulink can solve complex, nonlinear dynamic problems, this introduction is limited to basic dynamic systems that are modeled by linear, constant-coefficient ordinary differential equations (ODE). Simulink is a graphical user interface module incorporated into MATLAB that employs diagrams formed of various signalconnected operators (or blocks). Figure D.1 shows a generic block diagram consisting of an input (a forcing function or/and initial conditionsdnamed source in Simulink) that is applied to a mathematical model (represented as an operation (or transformation) block in Simulink) and results in an output (the system’s time-domain responsedknown as sink in Simulink). The three blocks of Figure D.1 are connected by means of signal lines indicating the direction of the information flow. Any of the three basic blocks of Figure D.1 can be formed of other, simpler interconnected blocks. Simulink can also be used as a simple plotting tool; in those cases, transmission blocks are not needed in the schematic block diagram of Figure D.1. The Simulink input and output can be exchanged with MATLAB, either interactively (while working in MATLAB’s Workspace) or via specific MATLAB files (with the “mat” extension). The following steps are necessary to generate and run a Simulink model: •

Build the Simulink block diagram based on a given mathematical model. • Start Simulink from MATLAB, which opens the Simulink Library Browser comprising all available blocks.

SOURCE BLOCK INPUT

OPERATION BLOCK MATHEMATICAL MODEL

SINK BLOCK OUTPUT

SIGNAL LINE

FIGURE D.1 Basic Block Diagram in Simulink With Source, Operation, and Sink Blocks Connected by Signal Lines. System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15004-3 Copyright © 2018 Elsevier Inc. All rights reserved.

719

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APPENDIX D

Create a new file or open an existing file, which opens a Model Window; this is the working space for creating, modifying, and running the Simulink model. • Select and drag necessary blocks from the Simulink Library Browser into the Model Window; define the blocks’ parameters; define MATLAB import if input is available in MATLAB. • Connect blocks by signal lines to complete the block diagram and save the model file. Run the Simulink model. • Specify the simulation parameters (such as the running time, time step). • Start simulation and obtain numerical solution/results. • Save the model file. Visualize/export the results. • Plot results directly in Simulink. • Export results into MATLAB for further conditioning/plotting. •

•

•

AN EXAMPLE Let us apply this step-by-step sequence to design a Simulink block diagram for plotting the function:
 uðtÞ ¼ 20 sinð30tÞ$et when 0  t  5: The function u(t) is the product of the two terms in the bracket multiplied by a constant. The input is the independent variable t, while the output is the function u(t), which will be plotted in Simulink and also exported to MATLAB. To access/start Simulink, open first MATLAB, and either type >> simulink at the MATLAB prompt or click the Simulink icon (button) on MATLAB’s toolbar, as indicated in Figure D.2. Either action will open the Simulink Library Browser window shown to the left in Figure D.3. To create a new Simulink model, select the New Model from the top bar menu, which will open the window shown to the right of Figure D.3. Simulink

Simulink button

FIGURE D.2 Starting Simulink From MATLAB’s Toolbar Simulink Button.

An Example

Simulink Library Browser window

New Model window

New Model

FIGURE D.3 The Two Windows Needed to Create a Simulink Model.

versions newer than 2015b allow you to first create a file as a New Blank Model window, followed by opening the Simulink Library Browser, which is identified by the icon shown at the top left in Figure D.3. Save this new file as “example” using File/ Save As in the New Model Windowdnote that the extension of this file is “mdl” (from model). The blocks that are needed to generate the diagram will be selected from the Library window and dragged into the Model window. Select the Sine Wave block from the Sources library of the Simulink Library Browser (by clicking on Sources to open it, as shown to the left of Figure D.4) and copy (drag) it into the Model window, as indicated to the right of Figure D.4. Double click the Sine Wave block and type 30 under Frequency (rad/sec) into the new window that opens up, as shown in Figure D.5; click OK to close this window. Note that the window also allows to specify an amplitude (in this case the amplitude is one and is also the window’s default value), a phase angle, and a bias valuedthis is an average value around which the sinusoidal variation takes place (by default, which is also the current situation, this value is zero). The other input is the independent time variable (or rather t), which is needed to further define the exponential function et. To create an input t, select the Ramp block from the same Sources library and drag it into the Model window, as also indicated in Figure D.4. Double click the block to open its Parameters window, and use 1 for the slope, as shown in Figure D.5; the default value of the slope is þ1, which generates an input of þt. Note that this block has also options of Start Time and Initial Output other than zero (the default values); for instance, one can choose a Ramp value of t þ 2 (where 2 is the Initial Output) and can select to have it applied for t  3 (which requires a Start Time of 3).

721

722

APPENDIX D

FIGURE D.4 Select the Sine Wave and Ramp Blocks From the Sources Library of the Simulink Library Browser and Copy Them Into the Model Window (Saved as example.mdl).

FIGURE D.5 Ramp and Sine Wave Blocks Parameter Selection Window.

An Example

FIGURE D.6 All Blocks Necessary to Build the Block Diagram and the Related Libraries.

The other blocks that are necessary to build the block diagram in this example, together with their libraries, are shown in Figure D.6. The Math Function (with the option eudwhere u ¼ tdselected from the parameters options) and the Product (with the default options) blocks are taken from the Math Operations library. The Gain block, which plays the role of an amplifier of the input by a factor of 20, is dragged from the Commonly Used Blocks library (it can also be used from the Math Operations library) and uses a value of 20 in the gain parameters option. The Scope and To Workspace blocks are in the Sinks library. While the Scope block plots u(t) directly in Simulink, the To Workspace block exports the time history to MATLAB’s Workspace. Figure D.7 shows this block’s parameters that have to be configured, which are: Variable name: simout (this is the default, but a different name can be used for u), Save Format: Array. Under Configuration Parameters in the Model Window menu bar, select Data Import/ Export, Save to workspace or file: check the Time box, use a name for the time variable (such as tout, the default), and then check the Output box and use the same name for the output (simout in this case). Once the necessary blocks have been copied into the Model window, and their parameters have been set, it is necessary to connect the blocks by means of signal lines to form the block diagram of Figure D.8. Note that prior to connection, or after that, all blocks can be resized, renamed, copied, deleted, moved, rotated, flipped, or aligned in the Model window quite easily. For instance, to align several blocks, select the blocks: click one block, and then shift-click each of the other

723

724

APPENDIX D

FIGURE D.7 Configuration of a To Workspace Block to Export Data to MATLAB’s Workspace.

remaining blocks; select Format from the Model window toolbar, and select the desired alignment operation. To connect two blocks, select the starting block, and then drag a line from its output port to the input port of the connecting block. Alternatively, select the start block, and then click the destination block by holding down the Ctrl key. To create a branch line (such as the one going to the To Workspace block), select the source line and draw (drag) the branch line while pressing the Ctrl button. To run the simulation, go in the Model window toolbar and select Simulation, then Configuration Parameters, which opens another window, and use a Simulation time of 5 s, which coincides with the default Stop Time value, provided the Start time is 0. This selection dictates the time interval specified for both the sine and the ramp functions. Under Configuration Parameters/Solver/Solver options, you can use Fixed-step instead of the default Variable-step option, and then chose an integration time step small enough (such as 0.0005 s) under Additional options/Fixed-step sizedsee Figure D.9. The Simulink numerical solver can use either a variable time-step calculation scheme or a fixed-step one. The variable-step option enables the integrators (when integration is necessary) to −t

e−t

20 [ e−t ·sin (30t)] e−t ·sin (30t)

sin (30t)

FIGURE D.8 Simulink Block Diagram.

An Example

FIGURE D.9 Using the Fixed-Step Solution.

adapt to various gradients and overcome potential numerical problems in complex (nonlinear) problems, but might be large at times and lead to uneven/coarselooking plots. However, the fixed-step variant can be utilized for simpler applications, such as the one studied here, with small time steps producing smoother plots. After that, click Start under Simulation, or click the Run button on the top bar menu. Double-clicking the Scope block in the Model window will open another window, similar to the one illustrated in Figure D.10, and which is the plot of u(t) obtained by exporting the vectors t and u to the Workspace. The plot is generated by the plot(tout, simout) command in MATLAB. Plot of u(t)

20 15 10 5

n

0 -5

-10 -15 -20 0

1

2

3

4

5

Time (sec)

FIGURE D.10 Plot of u(t) ¼ 20[sin(30t)∙et] After Importing Simulink Data Into MATLAB’s Workspace.

725

726

APPENDIX D

SOLVING LINEAR ORDINARY DIFFERENTIAL EQUATIONS AND SYSTEMS Dynamic system models are mostly defined by differential equations. A linear ODE of the n-th order needs n integrations to be performed to determine the dependent variable, which is also the philosophy used in Simulink to find numerical solutions to differential equations and plot them in terms of time. The centerpiece operator is the Integrator block shown in Figure D.11 and which is in the Commonly Used Blocks library. Note that the input is a function, and the output is its integral. u(t) = dy(t)/dt

y(t)

1 s

Integrator

FIGURE D.11 Simulink’s Integration Block.

Mathematically, the operation indicated in Figure D.11 amounts to: Z t Z t yðtÞ ¼ dyðtÞ ¼ uðtÞdt t0

t0

(D.1)

It should be noted that while 1/s is an s-domain symbol used to indicate integration, both the input and output of the Integrator block are time-domain signals, as indicated in Figure D.11. In the Block Parameters window of the Integrator, the important box to fill is the Initial Condition, and according to Eq. (D.1), the value of u(t) at t ¼ t0, that is u(t0), needs to be input. The upper integral limit t is defined separately by specifying the total simulation time, as illustrated in Figure D.9. The following presentation considers linear ODE with constant coefficients, but the Simulink approach is also applicable to linear ODE with time-variable coefficients and to nonlinear differential equations.

FIRST-ORDER DIFFERENTIAL EQUATIONS Consider the first-order differential equation: a0 1 _ þ a0 $yðtÞ ¼ uðtÞ or yðtÞ _ ¼  $yðtÞ þ $uðtÞ a1 $yðtÞ a1 a1

(D.2)

with the input u(t) being a known time-domain function and the unknown dependent variable (the output) being y(t). The second form of Eq. (D.2) suggests that the time derivative of y(t) in the left-hand side can be the input to an integration operator that will provide y(t) at its output. Eq. (D.2) also indicates that the time derivative of y(t) is the algebraic sum of a term containing y(t) and one based on u(t). Simulink has the Add block operator, shown in Figure D.12, and which is in the Math Operations library.

Solving Linear Ordinary Differential Equations and Systems

u1(t)

y(t)

+ +

u2(t)

Add

FIGURE D.12 Simulink’s Addition Block.

This block, which has two inputs and one output, is equivalent to: u1(t) þ u2(t) ¼ y(t). The Main option of this block’s Parameters contains the List of Signs box that enables not only using a number of inputs larger than two but also using  signs instead of only þ. If one chooses the following sequence: “ þ þ þ ” at this option, the result will be an Add box with three positive inputs and a negative input in a top-down sequence. (1/a1) . u(t)

u(t) u(t)

1

Source

GainGain = 1/a1

dy(t)/dt Add

1 s

y(t)

Integrator

(−a0/a1) . y(t)

Sink

Splitting point (node) 1

y(t)

Gain = −a0/a1

FIGURE D.13 Generic Block Diagram for Solving First-Order Differential Equations.

With the Integrator and the Add blocks, Eq. (D.2) is used with the block diagram of Figure D.13 to solve for y(t), provided that we use a generic source for u(t) and a generic sink to plot/display y(t). Note that the factors multiplying u(t) and y(t) in Eq. (D.2) are represented by Gain blocks in Figure D.13. Also, note that a splitting point (or node) is needed to branch y(t) back through its gain and to the Add block. A splitting point can be added to the line connecting the integrator to the sink as previously discussed. The branch line can also be created by selecting the source line, then starting from any point of the selected line, and arriving at the destination point (the Gain block in this case) while right-clicking.

SECOND- AND HIGHER-ORDER DIFFERENTIAL EQUATIONS The general second-order differential equation: a1 a0 1 € þ a1 $yðtÞ _ þ a0 $yðtÞ ¼ uðtÞ or yðtÞ € ¼  $yðtÞ _  $yðtÞ þ $uðtÞ a2 $yðtÞ a2 a2 a2 (D.3)

727

728

APPENDIX D

u(t)

u(t)

Source

1

(1/a2) . u(t)

Gain = 1/a2

2

d y(t)/dt Add

(−a1/a2) . dy(t)/dt

2

dy(t)/dt

1 s Integrator

1 s

y(t)

Integrator

Sink

1

Gain = −a1/a2 (−a0/a2) . y(t)

1

Gain = −a0/a2

FIGURE D.14 Generic Block Diagram for Solving Second-Order Differential Equations.

indicates that the second derivative of the dependent function y(t) is the sum of the three terms involving the first derivative of y(t), the function y(t), and the input function u(t). Also, a second-order differential equation requires two integrations, and therefore, Figure D.14 shows the Simulink block diagram necessary to integrate a general second-order differential equation. It is straightforward to extend the principles used to build block diagrams for integrating first- and second-order differential equations to higher-order ones. The Simulink block diagram for solving a differential equation of the n-th order requires n Integrator blocks, an Add block with n þ 1 inputs, as well as n þ 1 signal lines: one corresponding to the input u(t) and the other n lines starting (branching) from y(t), dy(t)/dt, ., dn1y(t)/dtne1, and using each a Gain block.

SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS Consider the first-order differential equations system ( a11 $y_1 ðtÞ þ a01 $y1 ðtÞ þ a12 $y_2 ðtÞ þ a02 $y2 ðtÞ ¼ u1 ðtÞ a11 $y_1 ðtÞ þ a01 $y1 ðtÞ þ a12 $y_2 ðtÞ þ a02 $y2 ðtÞ ¼ u2 ðtÞ

(D.4)

where u1(t), u2(t) are the inputs, which are known time-domain functions. The output comprises the unknown dependent variables y1(t) and y2(t). Eq. (D.4) is reformulated as: 8 < y_1 ðtÞ ¼ c1 $y1 ðtÞ  c2 $y_2 ðtÞ  c3 $y2 ðtÞ þ c4 $u1 ðtÞ : y_ ðtÞ ¼ d $y_ ðtÞ  d $y ðtÞ  d $y ðtÞ þ d $u ðtÞ 1 1 2 1 3 2 4 2 2 c1 ¼

a01 a12 a02 1 a a01 ; c2 ¼ ; c3 ¼ ; c4 ¼ ; d1 ¼ 11 ; d ¼ ; 2 a11 a11 a11 a11 a12 a12

d3 ¼

a02 1 ; d4 ¼  a12 a12

(D.5)

Suggested Reading

c1. y1(t)

1

Source

u1(t)

1

c4.u1(t)

1

dy1(t)/dt

c2. dy2(t)/dt

1 s

y1(t) Sink

1 1

c3. y2(t)

d1. dy1(t)/dt

1

1

d2.y1(t) 1

u2(t)

1

d4.u2(t)

dy2(t)/dt

1 s

Source

y2(t)

Sink

d3.y2(t) 1

FIGURE D.15 Generic Block Diagram for Solving a System of Two First-Order Differential Equations.

Eq. (D.5) indicates that two integrations are needed using the time derivatives of y1(t) and y2(t), as well as two additions, each of the latter ones with four inputsd based on the right-hand sides of the two equations. A reasoning similar to the one discussed for a singular equation previously can be applied to generate the block diagram corresponding to Eq. (D.5) and which is shown in Figure D.15. Systems formed of higher-order differential equations can be modeled and solved following a procedure similar to the one described here, coupled with the algorithm used to model single higher-order differential equations.

Suggested Reading M. Nuruzzaman, Modeling and Simulation in Simulink for Engineers and Scientists, Author House, Bloomington, IN, 2004. O. Beucher and M. Weeks, MATLAB and Simulink: A Project Approach, 3rd Ed. Infinity Science Press, Hingham, 2008. A.B. Bejan, What Every Engineer Should Know About MATLAB and Simulink, CRC Press, Boca Raton, 2011. H. Klee and R. Allen, Simulation of Dynamic Systems with MATLAB and Simulink, 2nd Ed. CRC Press, Boca Raton, 2011.

729

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APPENDIX D

D. Xue and Y.Q. Chen, System Simulation Techniques with MATALB and Simulink, John Wiley and Sons, Chichester, UK, 2014. J.D. Dabney and T.L. Harman, Mastering Simulink, Pearson Prentice Hall, Upper Saddle River, 2004. A.K. Tyagi, MATLAB and Simulink for Engineers, Oxford University Press, New Delhi, 2012. N. Nise, Control Systems Engineering, 7th Ed. Wiley, New York, 2015. L. Zamboni, Getting Started with Simulink, Packt Pushing, 2013.

APPENDIX

Essentials of MATLAB and System DynamicseRelated Toolboxes

E

This appendix reviews several useful commands and options that are relevant to the material of this text. The commands can be applied either directly in MATLAB or through the Symbolic Math Toolbox and the Control System Toolboxdwhich are incorporated in MATLAB. For quick reference, they have been organized under the following functional categories: mathematical calculations, visualization and graphics, linear systems modeling, time-domain analysis, frequency-domain analysis, controls, and linear time invariant (LTI) objects. A complete function list with documentation (including Simulink information) can be found at www. mathworks.com/access/helpdesk/help/helpdesk.html. Comprehensive information on these functions and related functions can be accessed directly in MATLAB by means of the doc command. For instance, if you need more information on Bode plots, just type >> doc bode at the MATLAB prompt and you will be directed on the Help page to the pertinent information on the bode command.

MATHEMATICAL CALCULATIONS abs(z)

Returns the absolute value of a scalar or an array z. It also returns the modulus (magnitude) of a complex number or of an array of complex numbers defined as z. angle(z)

Calculates the phase angle (in radians) of a complex number or the phase angles of an array of complex numbers defined as z. collect(f)

Orders/rewrites the symbolic function f in terms of the powers of the default variable, say t. In the variant collect(f, x), it produces a similar result in terms of the powers of the variable x. conj(z)

Calculates the conjugate of the complex number z.

System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15005-5 Copyright © 2018 Elsevier Inc. All rights reserved.

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APPENDIX E

conv(v1, v2)

The command convolves two vectors v1 and v2. When v1 and v2 are the coefficients of two polynomials p1 and p2, the command multiplies the two polynomials and results in a vector listing the coefficients of the product polynomial p = p1$p2. deconv(v1, v2)

When v1 and v2 are the coefficients of two polynomials p1 and p2, the command operates the division of p1 by p2. Its most common form is [q, r] = deconv(v1, v2); its result is p1/p2 = q + r, where q is the quotient, and r is the remainder. det(m)

Calculates the determinant of a square matrix m. diag(v)

Generates a diagonal matrix from the elements of a vector v; the elements of v are placed on the main diagonal of the square matrix. When v is a square matrix, the command returns a vector comprising the main diagonal elements of the matrix. diff(f, v, n)

Calculates symbolically the nth derivative of a function f in terms of the variable v. The command diff(f, v) calculates the first derivative of f with respect to v. dirac(t)

Represents the unit-impulse (also known as Dirac delta) function. The related command dirac(n, t) generates the nth derivative of the unit-impulse function. dsolve(’eqn1,eqn2,. ’,’cond1,cond2,. ’)

Solves symbolically the differential equations eqn1, eqn2,. with the initial conditions cond1, cond2,.dall using string notationdby considering the independent variable is time t. If a differential equation contains, for instance, the second time derivative of an unknown function y, the MATLAB notations D2y or diff(y, t, 2) can be used. Similarly, the first time derivative of y is denoted as Dy or as diff(y, t). An initial condition that specifies the value of the first time derivative as being equal to a symbolic value of b at the initial time moment 0 is denoted as ’Dy(0) = b’. expand(f)

Performs symbolic expansion of a function f such as polynomial multiplication and distribution of products over sums. eval(f)

Evaluates the function f. The command is also used to convert a symbolic timedependent function f(t) into a vector f paired with a previously defined time vector t.

Mathematical Calculations

eye(n)

Returns the n
n identity (unity) matrix with aii ¼ 1, for i ¼ 1 to n. When this command is used as eye(m, n), the result is a matrix with m rows and n columns where aii ¼ 1 (for i ¼ 1 to n if n < m, and for i ¼ 1 to m if m < n). All other elements are aij ¼ 0, for i s j. factor(f)

Transforms a rational-coefficient polynomial f into a product of irreducible, rational-coefficient, lower-degree polynomials in terms of the default variable. When used as factor(f, x), with x a vector, the command returns all irreducible factors of the vector x variables/elements. fzero(f,x0)

Finds the root of a nonlinear equation. This command works in conjunction with a function file. It finds a root of the function f near the point x0 where the sign of f changes. Instead of a point x0, an interval [xmin, xmax] can be specified. Both x0 and [xmin, xmax] are part of a vector x, and f is a function handle (denoted by the symbol @), which can be called in a MATLAB session. heaviside(t)

Represents the unit-step (or Heaviside) function. ilaplace(f)

Returns the inverse Laplace transform of the symbolic function f that depends on the variable s. imag(z)

Calculates the imaginary part of the complex number z. int(f,x)

Evaluates symbolically the indefinite integral of the function f that depends on the variable x. The following command int(f,x,a,b) evaluates the definite integral of f as a function of x between the integration limits a and b. inv(m)

Calculates numerically or symbolically the inverse of a square nonsingular matrix m. An equivalent command is m\ (the backslash operator substitutes inv), which is recommended in solving linear equation systems. laplace(f)

Returns the Laplace transform of the symbolic function f that depends on the time variable t.

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APPENDIX E

limit(f,x,a)

Calculates the symbolic limit of the function f when the variable x reaches the value of a. It can also be written as limit(f,x,a,’right’), in the case where x > a, or as limit(f,x,a,’left’), when x < a. For a being infinity, a is substituted by inf. ode45

The command is used to solve first-order (or sets of) differential equations that can be moderately nonlinear. Its most usual configuration is [t, y] = ode45(eqfun, [t0, tf], [incond]); eqfun is a function file that formulates the first-order differential equations, the vector [t0, tf] defines the limits of the integration time domain, and [incond] is a vector listing the initial conditions. poly(f)

Returns the coefficients of a polynomial f, where f is vector containing the roots of that polynomial; the returned coefficients are ordered in descending order of powers in a row vector. When f was defined as a matrix, the same command returns a row vector with the coefficients (also ordered in descending powers) of the characteristic polynomial, which is det(s[I]  [f]), with [I] being the identity matrix. polyval(f, v)

Returns the value of polynomial f for the value v of the polynomial variable. pretty(f)

Writes the symbolic object f in a mathematical typeset format instead of the regular MATLAB return format. real(z)

Calculates the real part of the complex number z. roots(n)

Returns the roots of a polynomial that was previously defined as the row vector n containing the coefficients of the polynomial in descending order of powers. simplify(f)

Applies algebraic identities and functional identities (trigonometric, exponential and logarithmic) to simplify the more complex symbolic function f. solve(eq1, eq2, ., eqn, v1, v2, ., vn)

Solves the system formed of the linear algebraic equations eq1, eq2, ., eqn for the variables v1, v2, ., vn. In the form solve(eq, v), it solves the equation defined as eq for the variable v.

Visualization and Graphics

subs(f, old, new)

It is used for symbolic substitution in the function f of the old variable by the new variable. syms a real

Specifies that the amount a is treated as a symbolic (algebraic) object, so no numerical value is expected of a, and also that a is real. Instead of the “real” qualifier, “unreal,” or “positive” can be used. Also, if the nature of the amount a is not of interest, the command syms a is to be used. The alternate string command a = syms(’a’,’real’) has the same effect. zeros(m, n)

Returns an m
n zero matrix.

VISUALIZATION AND GRAPHICS axis(xmin, xmax, ymin, ymax)

Sets limits to axes to better visualize a portion of interest from a two-dimensional plot. In the configuration axis(xmin, xmax, ymin, ymax, zmin, zmax), the command operates in a similar manner with three-dimensional plots. fplot(f)

Plots the function f in terms of a variable (say t) for 5  t  5 without having to define the independent variable t as symbolic. The command is normally applied with the handle function @ as fplot(@t f(t)). When used as fplot(@t f(t), [tmin, tmax]), the plot covers the [tmin, tmax] time interval. ltiview

Generates or opens an existing LTI object plot and enables changing the type of input and visualization of the plot characteristics; see more on this function at the end of this appendix. mesh(z)

Generates a three-dimensional plot as a wireframe mesh of the function z, which depends on two variables x and y; these are vectors having the dimensions m and n, respectively. The function works in connection with the meshgrid command. Instead of the mesh(z) command, the surf(z) command can be used, which generates a three-dimensional surface of the same shape as the wireframe. meshgrid(x, y)

Creates two-dimensional arrays to enable three-dimensional plots of analytical functions of two variables. In the configuration [X,Y] = meshgrid(x, y), where x is a

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APPENDIX E

vector of dimension m, and y is another vector of dimension n, the command generates two arrays: One has the vector x as a row that repeats itself n times and the other array has the vector y as a column that repeats itself m times. With another variable, z, defined in terms of x and y, the command mesh(z) generates a three-dimensional plot if used after the meshgrid command. plot(t,y,’LineSpec’)

Returns a two-dimensional line plot of the pairs (t, y), where t is the independent variable vector, and y is the vector function that depends on t. LineSpec is a string of symbols that enable customizing the line used in the plot (its style, width, and color) as well as the marker used (its type, size, and color). Several function pairs, (t, y1), (t, y2), ., can be graphically represented on the same plot, each pair with its own line specifications, using the command plot(t,y1, ’LineSpec1’,t,y2,’LineSpec2’,.). Several spatially separated plots can be generated using the subplot command. For instance, if subplot(3,2,1) is used before an actual plot command, MATLAB positions the respective plot in the first row and first column of a matrix with three rows and two columns. If the command subplot(3,2,3) is used, MATLAB places the plot in the second row and first column, which is the third position in a sequence that runs from left to the right and from top to bottom over the matrix elements. plot3(x,y,z,’LineSpec’)

Returns a three-dimensional line plot of the pairs (x, y, z), where x, y, and z can be defined as vectors. LineSpec specifies graphic details of the plot line. Several function pairs, (x1, y1, z1), (x2, y2, z2), ., can be graphically represented on the same three-dimensional plot, each pair with its own line specifications, using the command plot3(x1,y1,z1, ’LineSpec1’,x2,y2,z2,’LineSpec2’,.). Related MATLAB functions are xlabel(’text1’), ylabel(’text2’), zlabel(’text3’), title(’text’), and legend(’text’), which enable using customized text for the reference axes, a title for the plot and a legend. surf(z)

Generates a three-dimensional surface plot; see meshgrid.

LINEAR SYSTEM MODELING [A,B,C,D] = tf2ss(n,d)

Converts a transfer function model defined by means of the numerator and denominator coefficient lists n, d into a state-space model defined by the matrices A, B, C, D. frd(r,f)

Creates a frequency response data LTI object from a vector r formed of the values of the G(u$j), the complex transfer function, and a vector f formed of the frequencies

Time-Domain Analysis

u (in rad/s) corresponding to the values of G(u$j). The adjacent command freqresp(frd(r,f),om0) where om0 is a frequency value that was, for instance, not specified in the vector f, generates the value of G(u$j) that corresponds to om0. Mention should be made that om0 can also be a vector. frd (sys, f)

Converts an LTI object (such as zero-pole-gain, transfer function, or state space) sys into a frequency response data object (model) that corresponds to a vector f containing specified frequencies. [n,d] = ss2tf(A,B,C,D)

Returns the transfer function that corresponds to the state-space model defined by the A, B, C, D matrices. If the command [n, d] = ss2tf(A,B,C,D,ui) is used, MATLAB returns the transfer function corresponding to the same state-space model and its ith input. ss(A,B,C,D)

Creates a state-space LTI model based on the previously defined matrices A, B, C, The command ss(sys) converts a zero-pole-gain (zpk) model or a transfer function (tf) models into a state space model previously defined as sys.

D.

tf(n,d)

Generates a transfer function (tf) LTI object (model) based on a numerator n and a denominator d. The two row vectors n and d contain the coefficients of the numerator and of the denominator in descending order (from the largest power of s to the constant term). The function tf(sys) converts another LTI model (such as zeropole-gain, zpk, or state space, ss) into a transfer function model. The s = tf(‘s’) command enables defining a transfer function object using regular MATLAB syntax. zpk(z,p,k)

Creates a zero-pole-gain (zpk) LTI model in the form of a factored polynomial fraction by providing the list of zeroes z, the list of poles p, and the gain k. The command zpk(sys) converts a transfer function tf model or state-space ss model denoted by sys into a zero-pole-gain model.

TIME-DOMAIN ANALYSIS impulse(sys)

Produces the two-dimensional plot of an LTI object sys subjected to a unit-impulse (Dirac delta function) input. All the features presented for the step command are also valid for the impulse command.

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APPENDIX E

initial(sys,x0)

Generates a two-dimensional plot based on a state-space model sys and an initialstate vector x0. Several state-space models, sys1, sys2,. can be used in conjunction with the same initial condition x0 with the command initial(sys1,sys2,.,x0). Time can also be specified with this command using initial(sys,x0,t) where t can be either a value or a previously defined vector. The command [y,x,t] = initial(sys,x0) stores the state-space output vector y, the state vector x, and the time vector t of a state-space model sys subjected to the initial-state vector x0 input. The same results are obtained using the function initialplot instead of initial, with the same syntax. lsim(sys,u,t)

Generates the two-dimensional plot of the time response of an LTI object sys to a time-dependent input u over a time interval defined by the vector t. step(sys)

Produces the two-dimensional plot of an LTI object sys to a unit-step input. Time is the independent variable, and the time range for the plot is chosen by MATLAB. If a time vector t has been defined, then the command step(sys,t) will generate the plot over the specified range of t. Another variant of this command is x = step(sys), whereby the time response is not plotted, but the vector x (which is the time-domain function resulting from applying a unit-step input to a system) is generated and stored for further manipulation. For a nonunity step input n, the command changes to step(n*sys).

FREQUENCY-DOMAIN ANALYSIS bode(sys)

Produces the magnitude and phase angle Bode plots of an LTI object sys, such as transfer function, state space, zero-pole-gain, or frequency response data. The command bode(sys1,sys2,.) generates Bode plots for the LTI objects sys1, sys2, .. bodemag(sys)

Generates the Bode plot of the magnitude only of the LTI object sys. damp(sys)

Returns four groups of data, pertaining to the poles of an LTI object sys, such as a zero-pole-gain, transfer function, or a state-space model. The return consists of poles, damping ratios x, natural frequencies un, and time constants s ¼ 1/(x$un).

Controls

db2mag (mdb)

Converts the decibel magnitude mdb into its absolute magnitude counterpart. mag2db(m)

Converts the absolute magnitude m into the corresponding decibel magnitude. eig(m)

Calculates the eigenvalues (the squares of the natural frequencies) of a square matrix m. The use of the command [V,D] = eig(m) returns the modal matrix V (where each column is an eigenvector) and a diagonal matrix D with the eigenvalues located on the main diagonal. pole(sys)

The command finds the poles of the LTI object sys. For sys being a transfer function, the poles are the roots of the denominator (characteristic equation). pzmap(sys)

or pzplot(sys)

Returns a plot displaying the pole and zero positions of an LTI object sys. The command [p,z] = pzmap(sys) returns the vector p containing the poles and the vector z containing the zeroes of the LTI object. The command pole(sys) plots only the pole positions, whereas the zero(sys) plots only the zero positions associated with the LTI object sys. residue(n,d)

Generates the residues (which are the numerators in a partial fraction expansion where all fractions have first-degree polynomials in their denominators) of a fraction (such as a transfer function) that uses the polynomial n as its numerator and the polynomial d as its denominator. If instead of this command, the following command is used, [r,p,k] = residue(n,d), MATLAB returns three groups of numbers: the residues r, the poles p (which are the roots of d, the denominator polynomial, with changed signs), and the direct term k (which is the quotient of dividing n by d and returns 0 when the degree of n is less than the degree of d). zero(sys)

Finds the zeroes of the LTI object previously defined as sys. For a transfer function, the zeroes are the roots of the numerator.

CONTROLS feedback(sys1, sys2)

Generates the closed-loop transfer function of a basic negative feedback control system, where sys1 is the open-loop transfer function G(s) and sys2 is the feedback transfer function H(s).

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APPENDIX E

margin(sys)

Generates the magnitude and phase plots of the LTI object sys. It also calculates the gain margin, phase margin, and their corresponding crossover frequencies. A command [gm,pm,ogm,opm] = margin(sys) calculates and stores the gain margin (gm), phase margin (pm), and their crossover frequencies (ogm and opm). nyquist(sys)

Plots the Nyquist chart on an LTI object sys. The alternate command nyquist(sys,om) returns the Nyquist plot for a set of specified frequencies om. rlocus(sys)

Calculates and plots the root locus for the open-loop transfer function of a single-input, single-output system sys. In the format rlocus(sys1, sys2, .), MATLAB plots the root loci for the open-loop transfer functions previously defined as sys1, sys2, ..

LINEAR TIME INVARIANT (LTI) OBJECTS The Control System Toolbox of MATLAB has the capability of creating single entities from zero-pole-gain, transfer function, state-space, and frequency response data models (which are defined as collections of vectors and matrices), as exemplified in Chapters 7e10. These objects are known as linear time invariant (LTI) objects (or models) and are invoqued by the MATLAB commands shown in Figure E.1. Details on creating individual LTI objects are given in the main text; a few additional details are included here on object manipulation (operations), conversion, and visualization. Simple mathematical operations, such as addition and multiplication, can be applied to LTI objects. Consider, for instance, the following transfer function object, G(s) ¼ 1/(s2 þ 2s þ 3), and assume the transformation 2G(s) þ 5 is needed. The following MATLAB code realizes this task: >> g = tf(1,[1,2,3]); >> 2*g+5

MATHEMATICAL MODEL

LTI OBJECT

Zero-pole-gain

zpk

Transfer function

tf

State space

ss

Frequency response data

frd

FIGURE E.1 Linear Time Invariant (LTI) MATLAB Objects and Their Mathematical Models.

Linear Time Invariant (LTI) Objects

which returns Transfer function: 5 s^2 + 10 s + 17 ----------------s^2 + 2 s + 3

Converting between LTI objects, as seen in the book chapters, can be performed by calling a particular LTI object as a function of a different LTI object that has already been defined. To convert the transfer function model used in the previous example into a state-space model, for instance, the command >> ss(g) realizes this conversion. Similarly, to obtain a zero-pole-gain model from the same transfer function model, the command >> zpk(g) is needed. As mentioned in Chapter 9, it is not possible to transform any of the zpk, tf, or ss models into a frd model, but an existing frd model can be converted to any of the other three LTI models. There are no restrictions on converting zpk, tf, and ss models between themselves. Another important rule, the precedence rule, frd > ss > zpk > tf, basically states that operations among various LTI objects are possible and arithmetical combinations among all four objects produce an frd model. Similarly, combination of a zpk model with a tf model results in a zpk model. Assume we want to calculate 2gd3z (g is the transfer function defined previously, and z is a zpk model described here). Using the code >> z = zpk(0,[1,2],1); >> 2*g-3*z

returns Zero/pole/gain: -3 (s-0.2473) (s^2 + 1.581s + 5.391) -----------------------------------------------(s-2) (s-1) (s^2 + 2s + 3)

LTI objects can subsequently be used to plot a system’s time-domain or frequency-domain response by means of built-in MATLAB commands, such as step, impulse, lsim, or bode. Once an LTI object plot was defined, the command ltiview (typed at the MATLAB prompt) opens the Linear System Analyzer graphical user interface, which is represented by the LTI Viewer window. You can import an existing LTI object either from the Workspace or from a MAT-file, which is achieved by selecting File/Import in the window menu bar. By right clicking the plot space, you can change the plot type by selecting another input or response type altogether. Once an input has been selected and the corresponding plot has been obtained, important system response characteristics, such as the peak response, rise time, or settling time can easily be obtained from the plot.

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APPENDIX

F

Deformations, Strains, and Stresses of Basic Line Mechanical Members

BARS UNDER AXIAL FORCE OR TORQUE A bar of length l, rectangular cross-sectional area A, and modulus of elasticity (or Young’s modulus) E, which is acted on by axial forces fx, as sketched in Figure F.1(a), elongates (or compresses if the forces have opposite directions) by a quantity: fx $l E$A By definition, the axial (or normal) strain is: ux ¼ Dl ¼

(F.1)

ux Dl fx ¼ ¼ (F.2) l l E$A The axial (normal) stress, which is perpendicular to any cross section of the bar of Figure F.1 (and therefore parallel to the forces fx), is calculated as: εa ¼ εx ¼

sa ¼ sx ¼

fx A

(F.3)

Comparison of Eqs. (F.2) and (F.3) yields: sa ¼ E$εa

(F.4)

which expresses the Hooke’s law. y

fx

ux/2

h l

mx, θx

uy /2

Original shape

d

fx

x

x ux/2

uy /2

(a)

l

(b)

FIGURE F.1 Bar Under: (a) Axial Point-Force Loading; (b) Point-Torque Loading. System Dynamics for Engineering Students. http://dx.doi.org/10.1016/B978-0-12-804559-6.15006-7 Copyright © 2018 Elsevier Inc. All rights reserved.

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APPENDIX F

Extension along the axial (x) direction is accompanied by compressions in the y and z directions of the bar shown in Figure F.1(a). These compressive deformations are perpendicular to the force direction and result in transverse (or lateral) strains, which are defined as: uy Dh uz Dw (F.5) ¼ ; εt ¼ εz ¼ ¼ h w h w where w is the bar width (dimension in the z direction, which is normal to the drawing plane of Figure F.1). The axial and transverse strains are related as: εt ¼ ε y ¼

εt ¼ y$εa

(F.6)

where y is Poisson’s ratio, a material constant. Eqs. (F.1)e(F.6) are also valid for a bar that is fixed at one end and free at the other end, where a force fx is applied. A point moment (torque) mx applied at the free end of a fixed-free bar with a circular cross section of diameter d, such as the one of Figure F.1(b), produces an angular deformation qx at the free end of the bar that is equal to: qx ¼

mx $l p$d4 ; Ip ¼ G$Ip 32

(F.7)

where G is the shear modulus of elasticity and Ip is the polar moment of inertia of the circular cross section. The moduli of elasticity E and G are related by means of Poisson’s ratio as: E (F.8) G¼ 2ð1 þ yÞ

BEAMS IN BENDING For a cantilever (or fixed-free beam), such as the one of Figure F.2(a), the maximum deflection and rotation (or slope) produced at the free end by a point force fy applied at the same point are: uy ¼

fy $l3 fy $l2 ; qz ¼ 3E$Iz 2E$Iz

(F.9)

with E being the elasticity modulus and Iz being the cross-sectional moment of inertia about the bending z axis. For a circular cross-section of diameter d, Iz ¼ p$d4/64 and for a rectangular cross section of width w and thickness h (parallel to the y axis), Iz ¼ w$h3/12. When a moment mz acts at the cantilever’s free end, the deflection and rotation at the same point are: uy ¼

mz $l2 mz $l ; qz ¼ E$Iz 2E$Iz

(F.10)

For a bridge (or fixed-fixed beam), such as the one of Figure F.2(b), the maximum deflection produced at the midpoint by a point force applied at the same point is: fy $l (F.11) uy ¼ 192E$Iz

Beams in Bending

y

y θz

uy uy Undeformed beam

fy

mz

x

x l /2

l

fy l

(a)

(b)

FIGURE F.2 Beams in Bending Under Point-Force Loading: (a) Cantilever; (b) Bridge.

y

R σmin

mz

Upper fiber

mz σ

Neutral axis Lower fiber

σmax

FIGURE F.3 Portion of a Bent Beam.

Figure F.3 shows a portion of a bent beam under the action of two end bending moments mz. The upper layer of the beam is compressed, while the lower layer is extended under the action of the two bending moments. As a consequence, a layer (or fiber) between the two external layers is undeformed, and this is the neutral axis (or fiber). Compressive stresses are set up on all the layers between the neutral axis and the upper fiber, while extensional stresses are applied to the fibers limited by the neutral axis and the lower fiber. As is the case with bars under axial loading, the bending stresses are normal to any cross section. According to Navier’s equation, the maximum and minimum stresses corresponding to the lower and upper fibers are equal to: h mz $ 2 smax ¼ jsmin j ¼ (F.12) Iz where h is the thickness of the beam cross section (dimension measured between the undeformed lower and upper fiber) along the y axis in the plane of Figure F.3. The cross section is assumed to be symmetric about the y axis.

745

Index Note: ‘Page numbers followed by “f” indicate figures and “t” indicate tables.’

A Abscissa of convergence, 275 Acceleration constant Ka, 688, 689f Actuation, 155e160, 156f Actuator, 18, 489, 543 Adaptive, 544 Additional zero, 665e667, 666f Adiabatic transformation, 237 Adjoint matrix, 713 Algebra of block diagram signals, 546e547 Analogous first-order systems, 170f, 192f, 193t Analytical partial fraction expansion, 296e297 Arbitrary input, 358e362, 361f Attraction/repulsion forces, 11e12 Auxiliary polynomial method, 608e611 Axial strain, 501e502, 743 Axial stress, 743

B Bandwidth frequency, 438 Basic matrix operations, 712e715 Bath-thermometer thermal system, 13, 13f Bimaterial strip, 490 Bimetallic strip, 490e494 Block diagrams and MIMO feedback systems MISO feedback systems, disturbances, 578e579 transfer-function matrix modeling, 579e586 and SISO feedback systems controllers, 554e569 feedback-control physical systems, 570e575 sensitivity analysis, 575e576 state-space modeling, 576e577 transfer functions and basic block diagrams, 546e553 Block (stall) torque, 513e515 Bode plots, 433e434, 630e640 asymptote representation, 443e446 MATLAB, 451e452 Branch, 162 Break frequency, 444e445 Break-in points, 625

C Cantilever beam, 6, 6f sinusoidal motion, 504f Capacitance, 216e219, 219f, 241

Capacitive-resistive electrical system, 392f Capacitor connections, 155f Capacitor elements, 154e160, 154f Cart-pendulum mechanical system, 109f Cascading systems, 339f, 547e549 Centrifugal pump, 226 Characteristic equation, 714e715 Circular pitch, 46 Closed-loop poles, 595 Closed-loop transfer function, 550e551, 595e601, 597f matrix, 580e581 Coefficient of dynamic viscosity, 40e43 Comb drive/longitudinal motion, 155e160, 156f Compensators, 564f lag compensators, 564e565 lag-lead compensator, 567e569, 568f lead compensators, 565e567, 566fe567f Complementary solution, 7, 430e431 Complex impedance, 334 Complex numbers, 709, 710f Complex transfer function, 430e432 matrix approach, 447e448 Compliant mechanisms, 8e12, 9fe11f Components, 7, 546e553, 547fe548f, 550fe552f Composite walls, 244e249, 244f Conduction, 242e243, 242f, 245e249 Conical-segment vertical tank, 217f Conservative system, 85e86 Constant gain, 12 Constant-pressure transformations, 237 Constant-volume transformations, 237 Constitutive piezoelectric laws, 520 Controllers, 555fe560f derivative control, 554 hydraulic controllers dashpot, 560e561, 561f hydraulic servomotor, 561e562, 562f integral control, 554 pneumatic controllers, 562e569 proportional control, 554 proportional-derivative, 555e556 proportional-integral, 555e556 proportional-integral-derivative, 555e556 Convection, 243, 245e249 Converse piezoelectric effect, 519, 521 Convolution theorem, 293e295, 311e312

747

748

Index

Corner frequency, 444e445 Coulomb-friction nonlinear effect, 415f Coupled-field systems electromechanical coupling, 497 electrical voltage coupling, 501e506 electromagnetomechanical coupling, 506e516 electrostatic-mechanical coupling, 497e501 mechanical strain, 501e506 piezoelectric coupling, 519e531 multiple-field coupled system, 488 overview, 487 sensing and actuation, 488e489, 489fe490f single-field coupled systems, 488 spring coupled mechanical system, 488f thermomechanical coupling bimetallic strip, 490, 491f electrothermomechanical coupling, 494e497, 494f free-end displacement, two-material cantilever, 493f lumped-parameter model, two-material cantilever, 492f radius of curvature, 491e494 two uncoupled mechanical systems, 488f Coupled-field (multiple-field) systems, 17e18, 17fe18f Critical damping, 92e93 Cutoff frequency, 469

D d’Alembert equations, 84e85 Damped frequency, 92e93 Dampers, 51e53, 51fe53f Damping ratio, 14, 655e657, 655f Dashpot, 560e561, 561f Decade, 443e444 Decibels, 437 Deformations, 743e745 Degrees of freedom (DOF), 2, 165e167, 165f, 227 coordinate connections, 81e83 cylinder rolling without slippage, 82f motion transmission, 81 movable elements, 79e80, 79f multiple-DOF, 105e130 pinion, 81, 82f potential DOF, 81e83 racks, 82f related coordinates, motion transmission, 81e83, 81f rigid connections, 80, 80f rotary motion transmission, 83f single-degrees of freedom, 78f

single-DOF, 83e105 spring and block, 82f springs and dampers, translatory and rotary motion, 78, 79f three-DOF body system, planar motion, 78f translatory mechanical system, 80f with points of different motions, 80f two-DOF point mass system, planar motion, 78f De Moivre’s theorem, 709e710 Derivative control, 554, 662e663 Derivative gain, 554 Derivative time, 554 Diagonal matrix, 711e712 Dielectric permittivity, 154 Differential gap, 544e545 Differential input voltage, 161 Direct/inverse Laplace transformations, 274e301, 274fe275f Laplace transform pairs, 275e279, 276t MATLAB’s, 277e279 unit-ramp function, 277f Direct piezoelectric effect, 519 Direct-terms polynomial, 297e301 Direct transmission matrix, 381 Displacement amplification, 54 Displacement reduction, 54 Disturbance rejection/minimization, 578 Dominant poles, 669e672, 669f, 671f Dynamic matrix, 111e118 Dynamic systems feedback control control system with negative feedback, 543, 543f on-off control of liquid-level system, 544e545, 545f open-loop control process, 546f, 545e546 piezoelectric (PZT) actuator, 542, 542f second-order control system with negative feedback, 543e544, 544f uncontrolled system, 542, 543f harmonic input analytical approach, 432e451 MATLAB approach, 451e455 with input time derivative, 390e394, 392fe393f without input time derivative multiple-input, multiple-output systems, 388e390, 388f single-input, single-output systems, 386e387, 387f

E Eigenvalue problem, 111e118, 714e715 Eigenvector, 111e118, 714e715 Elastic torque, 24e25

Index

Electrical filter systems, 470e471, 470f Electricalefluidethermal system analogy, 257e259, 258f, 258t, 259f Electrical systems, 334e340 circuits/networks, 162e191, 162f configuration, 165e167 degrees of freedom, 165e167, 165f forced response, 178e191 free response, 167e178 Kirchhoff’s laws, 163e164, 163f rule of thumb, 166 single-degrees of freedom systems, 178e185 single-mesh electrical circuit, 164, 164f electrical elements actuation, 155e160, 156f capacitor connections, 155, 155f capacitor elements, 154e160, 154f comb drive/longitudinal motion, 155e160, 156f inductor elements, 160e161, 160f joule effect, 152 mechanical displacement sensing, 151e153, 151f microelectromechanical systems (MEMS), 155e160, 156f operational amplifiers, 161e162, 161f resistor elements, 150e153, 150f sensing, 155e160, 156f voltage and current sources, 150, 150f mechanicaleelectrical analogy, 191e197 analogous first-order systems, 170f, 192f, 193t forceecurrent analogies, 193, 194t forceevoltage analogies, 193, 193t general procedure, 196 three-mass mechanical system, 194f two-mesh electrical network, 195f objectives, 149 Electrical voltage coupling, 501e506 Electromagnetic translatory, 515 Electromagnetomechanical coupling, 512f electrical subsystem, 510f fixed coil and magnet attached, bending bridge, 507f mechanical rotary arm, 509f with optical detection, microelectromechanical systems, 517e519, 517fe518f rotary direct-current electric motor, mechanical load, 512e515, 512fe513f, 515f rotary mechanical elements, 509f Simulink diagram, 511f translatory direct-current electric motor, mechanical load, 515e516, 516f

Electromechanical coupling, 497 electrical voltage coupling, 501e506 electromagnetomechanical coupling, 506e516 electrostatic-mechanical coupling, 497e501 mechanical strain, 501e506 piezoelectric coupling, 519e531 Electromechanical systems, 497 Electrostatic-mechanical coupling coupled system sensing, mechanical motion, 498f microelectromechanical systems, capacitive longitudinal actuation and transverse sensing, 499f Electrothermomechanical coupling, 494e497, 494f Elements, 7e8 Energy method, 85e86 Engineering system dynamics, 1e2 feedback control, 21e22, 22f linear and nonlinear dynamic systems, 18e21, 19fe20f modeling process, 2, 3f lumped-parameter modeling and solution, 6e7 modeling variants, 3e5 system order, 12e17 time- and frequency-domain, 21 Epsilon method, 607 Equivalent lumped-parameter model, 6 Equivalent mechanical elements levers of small rotation, 54e60 toothed gears, 45e53 Equivalent rotary spring model, 29f Euler’s identity, 709e710 Evans rules, 624e628 Exponential order, 275

F Feedback control systems, 2e3, 21e22, 22f, 542e546 Bode plots, 630e640 control system with negative feedback, 543, 543f MIMO systems, 599e601 closed-loop transfer-function, 600e601 state-space stability, 600e601 Nyquist plot, 630e640 definition, 632e633 gain margin, 636e640 open-loop transfer functions, 633e634 phase margin, 636e640 stable system, 636e637, 636f on-off control of liquid-level system, 544e545, 545f open-loop control process, 546f, 545e546

749

750

Index

Feedback control systems (Continued) physical systems, 570e575, 570fe573f piezoelectric (PZT) actuator, 542, 542f root locus method, 620e630, 620f MATLAB, 629e630, 629fe630f open-loop transfer function, 621e624 rules, 624e628 second-order feedback system, 621f sketch, 626f RoutheHurwitz stability test, 602e620 all elements in a row are zero, 607e611 auxiliary polynomial method, 608e611 constructions, 602e605, 603t design problems, 611e620, 613f epsilon method, 607 nonunity-feedback system, 616f polynomial with reciprocal roots method, 605e606 rules, 603e604 solution domain of parameters, 618f unity-feedback system, 613f zero element in first column, 605e607 second-order control system with negative feedback, 543e544, 544f SISO systems, 594e599 closed-loop transfer function, 595e599, 597f MATLAB, 596e598 stability concept, 594e601 uncontrolled system, 542, 543f Feed-forward transfer function, 550, 659 matrix, 579e580 Field coupling. See Coupled-field systems Filters cutoff frequency, 469 electrical filter systems, 470e471, 470f filter magnitude, 469, 469f mechanical filters, 471e472, 471fe472f types, 469 Final-value theorem, 289e291 Finite transforms, 274 First-order systems, 13e14, 13fe14f, 648e649, 649f Fixed-gap, 159f Flexure-based planar compliant mechanism, 10, 10f Flexure hinges, 9e10 Fluid/thermal systems, 342e344, 343f electricalefluidethermal system analogy, 257e259, 258f, 258t, 259f liquid elements, 214e226 capacitance, 216e219, 219f hydraulic energy sources, 226

inertance, 214e216 resistance, 219e226, 220f liquid systems, 226e236 forced response, 231e236 modeling, 213e236 natural response, 227e231 objectives, 213 pneumatic systems modeling, 236e240 gas laws, 236e237 pneumatic elements, 237e239 pneumatic systems, 239e240 thermal systems modeling, 240e257 multiple-degrees of freedom systems, 253e257, 254fe257f single-degrees of freedom systems, 250e253 thermal elements, 241e249 Force-amplification, 54 Forced response, 100fe101f, 178e191, 231e236 actuation moment definition, 127f damping ratio, 100e105 gear-shaft rotary mechanical system, 127f Lagrange’s equations, 123e130 linear-motion mechanical filter microsystem, 123f liquid-level systems, 231e236 multiple-degrees of freedom liquid-level systems, 232e236, 234f single-degrees of freedom liquid-level systems, 231e232, 232f lumped-parameter models and free-body diagrams, 99f MATLAB, 99, 409e412, 410fe411f multiple-degrees of freedom systems, 185e191, 185f Lagrange’s equations, 186e191 natural frequency, 100e105 Newton’s law of motion, 99 shuttle masses free-body diagrams, 124f Simulink, 99 Simulink diagram, 125f, 129f single-degrees of freedom systems, 178e185 operational amplifier circuits, 178e185 static sensitivity, 100e105 system reduced, middle shaft, 127f time variation angular velocities, mechanical system, 130f shuttle mass displacements, 126f Forced solution, 430e431 Forced/steady-state response, 7 Force-reduction device, 54 Forcing functions, 8 Fraction combination, 297

Index

Free angular velocity, 513e515 Free-body diagram, 2, 84e85, 393f Free response, 7, 167e178 conservative systems mathematical model, 84e86, 105e110 physical model of inertia-spring systems, 84f solution of mathematical model, 86e91, 87f, 90f, 92f, 110e118 free damped response, 175e178, 176f, 178f Lagrange’s equations, 169e175 mesh analysis, 169 natural (free lossless) response, 167e175 node analysis, 169e175, 170f with nonzero initial conditions, 407e409 single-degrees of freedom conservative electrical systems, 167e169, 168f systems with losses and free damped response mathematical model, 92 solution of mathematical model, 92e99, 93f, 95fe96f, 98f Free undamped systems, 83 Frequency-domain analysis, 433e434 applications filters, 469e472 mechanical vibration transmission, 455e465 steady-state response, nonloading cascading systems, 465e469 approach applications, 455e472 complex transfer function, steady-state response, 430e432 dynamic systems to harmonic input, steady-state response, 432e455 overview, 429e430 modeling, 21 Frequency-domain response, 685e701 lead- and lag-phase compensation design, 692e701 geometric mean frequency, 693e694 phase-lag compensation, 697e701, 697f, 699f, 701f phase-lead compensation, 692e697, 693f, 695fe696f time-domain connections, 685e692 steady-state error constants, 687e692 transient time-response characteristics, 685e687 Frequency ratio, 439e440 Frequency response data handling, 452e453, 453f magnitude, 433e434

model conversion, 454e455, 454f phase, 433e434 Frequency shift method, 455e459 Frequency shift theorem, 280e282

G Gear-shaft-cam rotary system, 50f Generalized coordinates, 105e110 Generator, 519 Geometric mean frequency, 693e694

H HagenePoiseuille resistance, 222e226 Heating, ventilating, and air conditioning (HVAC), 249e250 Helical spring, 24, 24f Higher-order systems, 14e17 High gain, 162 High input impedance, 162 Homogeneity, 19 Homogeneous state-space model, 403e405, 404f Hooke’s law, 743 Hydraulic actuators, 226 Hydraulic controllers dashpot, 560e561, 561f hydraulic servomotor, 561e562, 562f Hydraulic diameter, 222e223 Hydraulic energy sources, 226 Hydraulic servomotor, 561e562, 562f

I Ideal current source, 150 Ideal voltage source, 150 Identity matrix, 711e712 Impedance transfer function, 334e347, 334f, 335t electrical systems, 334e340 fluid systems, 342e344, 343f loading cascading (series) electrical systems, 338e340, 339f mechanical systems, 344e347, 345fe346f nonloading cascading (series) electrical systems, 338e340, 339f operational amplifier configuration, 337f, 339f series and parallel equivalent impedances, 334e336, 334f thermal systems, 340e342, 341f voltage divider configuration, 337f, 339f Impulse function, 284f Indefinite integrals, 288e289 Independent variable, 301e302 Individual electrical elements, 7, 8f Inductor elements, 160e161, 160f

751

752

Index

Inertance, 214e216 Inertia coupling, 110e111 Inertia elements bevel gear, 33f elastic members, equivalent mass fractions, 35e38, 36t equivalent mechanical moments, bars in torsion, 35e38, 36t flexible bar with axial inertia and bridge, 37f inertia force, 33 inertia force/moment and kinetic energy, 33, 33t inertially equivalent lumped-parameter (point) mass, 37f lumped-parameter inertia, 31, 32f, 35e38 mass moments of inertia, 32, 32t mechanical microsensor, flexible segments and central rigid plate, 38f parallel-axis theorem, 32 right circular conical solid frustum, 33f Inertia force, 33 Inertia moment, 33 Inertia torque, 33 Initial conditions, 8, 86e91 Initial-value problems, 7 Initial-value theorem, 289e291 Input, 8 functions, 8, 9f matrix, 381 signals, 8 Integral and integraledifferential equations, 309e312 Integral control, 554, 663 Integraledifferential equations, 309 Integral equations, 288e289 Integral gain, 554 Inverse Laplace transform, 7, 432e433 Inverse PZT effect, 17e18 Isothermal transformation, 237

J Joule effect, 152 Junction point, 546e547

K Kirchhoff’s laws, 1e2, 6e7, 162e164, 163f

L Lag compensators, 564e565 Lag-lead compensator, 567e569, 568f Lagrange’s equations, 105e110, 106fe108f, 233 advantage, 7

mesh analysis, 186 method, 162 node analysis, 186e191 two-mesh electrical network, 187f Laplace-domain functions, 295e296 Laplace-domain output vector, 382e385 Laplace transform, 432e433 analytical partial fraction expansion, 296e297 convolution theorem approach, 293e295, 311e312 derivatives, 287e288 direct and inverse Laplace transformations, 274e301, 274fe275f Laplace transform pairs, 275e279, 276t MATLAB’s, 277e279 unit-ramp function, 277f final-value theorem, 289e291 indefinite integrals, 288e289 initial-value theorem, 289e291 integral and integraledifferential equations, 309e312 linear ordinary differential equations, 301e309 constant coefficients, 301e306 systems, 306e309 MATLAB partial fraction expansion, 297e301 objectives, 273 partial fraction expansion, 295e301 periodic functions, 291e293, 292f piecewise continuous functions, 285e286, 285f properties, 279e301, 279t frequency shift theorem, 280e282 linearity, 279e280 time-shift theorem, 279t, 282e286, 282f, 284fe285f time-domain system identification, 312e313 Lead- and lag-phase compensation design, 692e701 Lead compensators, 565e567, 566fe567f Lever-mass-spring system, 88f Levers, of small rotation dampers, 58e60, 58fe59f displacement amplification, 54 displacement reduction, 54 force-reduction device, 54 inertia, 56e58, 56fe57f mechanical levers, 54 small-angle approximation, 54 springs, 55e56, 55fe56f Linear dynamic system, 18e21 Linear homogeneous ordinary differential equations, 717e718, 718t Linearity, 279e280 Linearization technique, 394

Index

Linear-motion mechanical filter microsystem, 123f Linear ordinary differential equations, 301e309 constant coefficients, 301e306 systems, 306e309 Linear resistance, 222e226 Linear superposition, 19, 448e451, 449f, 451f Linear time invariant (LTI) models, 333 Liquid elements, 214e226 capacitance, 216e219, 219f hydraulic energy sources, 226 inertance, 214e216 resistance, 219e226, 220f Liquid systems, 226e236 forced response, 231e236 modeling, 213e236 natural response, 227e231 Loading cascading (series) electrical systems, 338e340, 339f Logarithmic decrement, 94e99 Log-scale asymptote representation, 443f Longitudinal actuation, 520 Longitudinal-motion (comb-drive) variable capacitor, 159f Loop, 162 Low output impedance, 162 LTI models. See Linear time invariant (LTI) models Lumped-parameter inertia, 31, 32f, 35e38 Lumped-parameter modeling, 6f, 29f microcantilever with harmonic input motion, 505f microcantilever with nonlinear spring, 396f modeling methods, 6e7 platform-motors mechanical system, 461f serpentine spring and bridge, 31f solution methods, 7 system response, 7 two-degrees of freedom translatory mechanical system, 410f two-material cantilever, thermal deformation, 492f Lumped-parameter stiffness, 25e26, 25te26t

M Marginally stable systems, 348, 349f, 351, 595 Mass (inertia), 111e118 Mass-damper mechanical system, 438f Mass flow rate, 214 Material sensitivity, 501e502 MATLAB, 2, 277e279, 357e362, 361f, 397, 596e598, 629e630, 629fe630f frequency-domain approach

Bode plots, 451e452 frequency response data handling, 452e453, 453f frequency response model conversion, 454e455, 454f partial fraction expansion, 297e301 state space model forced response, 409e412 free response with nonzero initial conditions, 407e409 and system dynamics controls, 739e740 frequency-domain analysis, 738e739 linear system modeling, 736e737 linear time invariant objects, 740e741 mathematical calculations, 731e735 time-domain analysis, 737e738 visualization and graphics, 735e736 Matrix algebra, 711e715 Matrix exponential method, 403 Mechanicaleelectrical analogy, 191e197 Mechanical elements, 7 equivalent mechanical elements, 45e60 inertia elements, 31e38 spring elements, 23e31 viscous damping elements, 38e45 Mechanical filters, 471e472, 471fe472f Mechanical levers, 54 Mechanical strain, 501e506 Mechanical systems, 344e347, 345f degrees of freedom (DOF), 78e83 multiple-DOF, 105e130 single-DOF, 83e105 Mechanical vibration transmission transmissibility for force input, 459e462, 459fe461f transmissibility for motion input and mass detection, 458f damping ratios, 457e459, 457f lumped damping, 456 microcantilever with harmonic input, 455, 456f vibration absorption and vibration isolation, 462e463 vibration displacement and acceleration amplitudes measurement, 463e465, 464f Mechanics of materials, 501e502 MEMS applications, 11 Mesh analysis method, 162, 169 Microdevice sensitivity, 506f Microelectromechanical systems (MEMS), 8e12, 155e160, 156f, 455e459

753

754

Index

MIMO. See Multiple-input and multiple-output (MIMO) systems Minimum-phase and nonminimum-phase angle systems, 691e692, 691f Mixed heat flow, 244e249, 244f Model formulation, 326e347 Modeling variants, 3fe4f analysis and design (synthesis), 5, 5f center of gravity (CG), 3e4 dynamic modeling, 5 physical model, 5, 5f system parameters, 4 Mode shape, 110 Moivre’s identity, 444e445 Monolithic (single-piece) devices, 10 Moody friction factor, 222e223 Motion transmission, 81 Multilayer pipe joint, 248f Multiple closed-loop poles, 596 Multiple-degrees of freedom systems forced response, 123e130 free response conservative systems, 105e118 free damped response, 118e123, 118fe119f, 122f Multiple-field coupled systems, 488. See also Coupled-field systems Multiple-input, 8, 8f Multiple-input and multiple-output (MIMO) systems, 8, 326, 328e332, 381, 388e390, 388f complex transfer function matrix approach, 447e448 feedback systems MISO feedback systems, disturbances, 578e579, 578f transfer-function matrix modeling, 579e586, 580fe582f, 584f, 586f linear superposition, steady-state time response, 448e451, 449f, 451f

N Natural frequency, 14, 86 Natural/modal motion, 86, 110 Natural response, 227e231 multiple-degrees of freedom conservative liquid systems, 227e231, 228f single-degrees of freedom conservative liquid systems, 227 Natural solution, 430e431 Negative-feedback control, 543 Neutral zone, 544e545

Newton’s law of cooling, 243 Newton’s law of viscous flow, 40e43 Newton’s second law of motion, 1e2, 6e7, 83e85, 92, 105 Node analysis, 162, 169e175 Nonhomogeneous state-space model, 406e407, 407f Nonlinear control systems, 677e679, 678f Nonlinear dynamic system, 18e21, 394 Nonlinear resistance, 220e222, 220f Nonloading cascading (series) electrical systems, 338e340, 339f Nonloading cascading systems multiple-input systems, 466e469, 467f single-input systems, 465e466, 465f Nonsingular, 713 Nonunity-feedback systems, 660e661, 661f Nonzero initial conditions, 357e358, 360f Nozzle-flapper pneumatic system, 562e563, 562f Numerator dynamics, 391e394 Nyquist plot, 630e640 definition, 632e633 gain margin, 636e640 open-loop transfer functions, 633e634 phase margin, 636e640 stable system, 636e637, 636f

O Ohm’s law, 150e151 On-off control, 544e545 Open-loop control process, 546f, 545e546 Open-loop transfer function, 550e551, 621e624 function matrix, 580 Operational amplifiers, 161e162, 161f circuits, 178e185 inverting amplifier circuits, 178e181, 179f mathematical operations, 182e185, 182fe184f saturation-type nonlinearity, 180f two-stage operational amplifier system., 180f unsaturated and saturated output voltages, 181f configuration, 337f, 339f Output defined, 8 equation, 381 matrix, 381 Overdamping, 92e93

P Parallel-axis theorem, 32 Parallel-plate/transverse-motion, 155e160

Index

Partial fraction expansion, 295e301 Particular solution, 7, 430e431 Periodic functions, 291e293, 292f Phase-lag compensation, 697e701, 697f, 699f, 701f Phase-lead compensation, 692e697, 693f, 695fe696f Piecewise continuous functions, 285e286, 285f Piezoelectric actuation, 519 Piezoelectric (PZT) actuators, 10, 542, 542f Piezoelectric charge constant, 521 Piezoelectric coupling actuation, 520e521, 520f overview, 519e520, 519f piezoelectric and strain gauge sensory-actuation, 527e531, 528fe529f piezoelectric block actuator and load spring, 521e524, 521f, 523fe524f sensing, 524e527, 525f, 527f Piezoelectric materials, 519 Pinion, 81 Pitch circle, 46 Pneumatic controllers compensators, 564e569 nozzle-flapper pneumatic system, 562e563, 562f Pneumatic elements capacitance, 238 inertance, 237 pneumatic energy sources, 239 resistance, 238 Pneumatic systems, 239e240 forced response, 239e240, 240f modeling, 236e240 gas laws, 236e237 pneumatic elements, 237e239 pneumatic systems, 239e240 natural response, 239 Poisson’s ratio, 501e502, 744 Poling process, 519e520 Polynomial with reciprocal roots method, 605e606 Polytropic transformation, 237 Position constant Kp, 687e688 Potentiometers, 151e152 Proportional control, 554, 662 Proportional-derivative, 555e556 Proportional-integral, 555e556 Proportional-integral-derivative, 555e556 Pulleys, 45e46 Pulse function, 284f PZT block, 18

R Radiation, 243e249 Rayleigh dissipation function, 118e120 Reference signal, 543 Resistance, 219e226, 220f HagenePoiseuille resistance, 222e226 linear resistance, 222e226 lost head, 222e226 nonlinear resistance, 220e222, 220f Resistive wheatstone bridge circuit, 153f Resistor connections, 152e153, 152f Resistor elements, 150e153, 150f Resonant frequency, 439e440 Resonant peak, 439e440 Rigid bodies, 31 Rigid lever, 54f Root locus method, 620e630, 620f MATLAB, 629e630, 629fe630f open-loop transfer function, 621e624 rules, 624e628 second-order feedback system, 621f Rotary-motion damping, 42f Rotation potentiometer, 151f RoutheHurwitz stability test, 602e620 all elements in a row are zero, 607e611 auxiliary polynomial method, 608e611 constructions, 602e605, 603t design problems, 611e620, 613f epsilon method, 607 nonunity-feedback system, 616f polynomial with reciprocal roots method, 605e606 rules, 603e604 solution domain of parameters, 618f unity-feedback system, 613f zero element in first column, 605e607 Rule of thumb, 166

S Saturation, 20e21 Scalability, 19 Second-order systems, 14, 15f, 621f, 649e657, 650f Sensing, 155e160, 156f Sensing coupling, 519 Sensitivity analysis, 575e576 Sensors, 18, 488e489, 519 Series/parallel equivalent impedances, 334e336, 334f Serpentine spring and bridge, 30f Shearing, 520 Shear stresses, 40e43

755

756

Index

Signals, 546e547 Simple closed-loop poles, 596 Simulink, 2, 357e362, 362f, 719, 719fe725f linear ordinary differential equations and systems, 726e729, 726fe729f state space model, 412e416 free-body diagram, elastic and friction forces, 414f nonlinear Coulomb and viscous friction, 415f nonlinear mechanical system with Coulomb friction, 416f pulse and exponential input and output signals, 413f two-input, one-output state space model, 413f without and with Coulomb-friction nonlinear effect, 415f Single-degrees of freedom systems, 178e185 forced response, 100fe101f damping ratio, 100e105 lumped-parameter models and free-body diagrams, 99f MATLAB, 99 natural frequency, 100e105 Newton’s law of motion, 99 Simulink, 99 static sensitivity, 100e105 free response conservative systems, 84e91 systems with losses and free damped response, 92e99 free undamped systems, 83 Newton’s second law of motion, 83 Single-field coupled systems, 488 Single-input and single-output (SISO) systems, 8, 325e328, 381, 386e387, 387f asymptote representation, Bode plots, 443e446, 443fe446f feedback systems controllers, 554e569 feedback-control physical systems, 570e575 sensitivity analysis, 575e576 state-space modeling, 576e577 transfer functions and basic block diagrams, 546e553 frequency response parameters, first-order systems, 436e438, 436fe438f frequency response parameters, second-order systems, 439e442, 439fe441f stability and steady-state response, 434e436, 434fe435f steady-state solution, harmonic (sinusoidal) input, 432e434

Single-mesh electrical circuit, 164, 164f Sinusoidal pulses, 292f SISO. See Single-input and single-output (SISO) systems Small-angle approximation, 54 Special-form matrices, 711e712 Spool valve, 561 Spring elements, 23 series and parallel spring connections, 26e31 lumped-parameter model, 31f symmetric shaft-disk system, 29f translatory microspring system, 30f translatory spring combinations, 27fe28f and stiffness elastic torque, 24e25 helical spring, 24, 24f lumped-parameter stiffness, 25e26, 25te26t translatory and rotary springs, 24, 24t Square matrix, 711e712 Stability concept, 594e601 Stable systems, 348, 349f, 595 Standard state space model, 381 State equation, 381 State matrix, 381 State-space methods, 682e684 State-space model formulation conversion between zero-pole-gain and state-space models, 401e402 conversions between transfer function and state-space models, 397e401 time-domain mathematical model dynamic systems with input time derivative, 390e394 dynamic systems without input time derivative, 386e390 nonlinear systems, 394e397 State-space modeling conversions between transfer function and state-space models, 397e401 conversions between zero-pole-gain and statespace models, 401e402 MIMO system, 381 nonuniqueness, 382 SISO feedback systems, 576e577 SISO system, 381 solution of state equation, 382e385 state space matrices, 381, 381f time-domain mathematical model dynamic systems with input time derivative, 390e394

Index

dynamic systems without input time derivative, 386e390 nonlinear systems, 394e397 and time-domain response analytical approach, 402e407 MATLAB approach, 407e412 Simulink approach, 412e416 State-space stability, 600e601 State-transition matrix method, 402e407, 403f State variables, 381 State vectors, 381 Static sensitivity, 12, 488e489 Steady-state error, 543e544 constants, 687e692 acceleration constant Ka, 688, 689f minimum-phase and nonminimum-phase angle systems, 691e692, 691f position constant Kp, 687e688 velocity constant Kv, 688, 689f Steady-state response, 543e544 complex transfer function, 430e432 harmonic input analytical approach, 432e451 MATLAB approach, 451e455 nonloading cascading systems, 465e469 Steady-state solution, 430e431 Steady-state time-domain response and errors, 657e661 feed-forward transfer function, 659 nonunity-feedback systems, 660e661, 661f unity-feedback control systems, 658e660, 658te659t Steady-state time response, 448e451, 449f, 451f Stiffness coupling, 110e111 Stiffness matrices, 111e118 Strain gauge, 502 sensitivity, 502 Summing point, 543 Surface/bulk micromachining, 10 Symmetric shaft-disk system, 29f System behavior, 4 System coordinates, 165 System coupling concept, 487e489. See also Coupled-field systems System order, 7 first-order systems, 13e14, 13fe14f second-order systems, 14 third-order system model, 16 zero-order system, 12, 12f System parameters, 4 System’s response, 4 System stability, 347e353, 349fe351f

T Taylor series expansions, 394e395 Thermal bimorph, 490 Thermal coefficient of resistance, 494 Thermal elements, 241e249 capacitance, 241 heat sources, 249 resistance, 241e249 composite walls, 244e249, 244f conduction, 242e243, 242f, 245e249 convection, 243, 245e249 mixed heat flow, 244e249, 244f radiation, 243e249 Thermal systems, 340e342, 341f modeling, 240e257 multiple-degrees of freedom systems, 253e257, 254fe257f single-degrees of freedom systems, 250e253 thermal elements, 241e249 Thermomechanical coupling bimetallic strip, 490, 491f electrothermomechanical coupling, 494e497, 494f free-end displacement, two-material cantilever, 493f lumped-parameter model, two-material cantilever, 492f radius of curvature, 491e494 Third-order system model, 16 Time-domain, 21 Time-domain dynamic system, 312 Time-domain mathematical model, 326e332 dynamic systems with input time derivative, 390e394 dynamic systems without input time derivative, 386e390 MIMO system, 328e332 nonlinear systems, 394e397 SISO system, 326e328 Time-domain response analytical approach homogeneous state-space model, 403e405 nonhomogeneous state-space model, 406e407 MATLAB approach, 407e412 MIMO system, 679e684, 683fe684f MISO feedback systems, 679e682, 681fe682f state-space methods, 682e684 transfer function matrix, 682e684 Simulink approach, 412e416 SISO system, 648e679 steady-state time-domain response and errors, 657e661

757

758

Index

Time-domain response (Continued) transient time response specifications, 648e657 transitory and steady-state time response, 662e679 Time-domain response y(t), 302, 306f Time-domain system identification, 312e313 Time response MIMO systems, 362e367 analytical approach, 362e364, 364f MATLAB, 365e367 Simulink approach, 365e367 SISO systems, 353e362 analytical approach, 353e357, 355f arbitrary input, 358e362, 361f MATLAB, 357e362, 361f nonzero initial conditions, 357e358, 360f Simulink approach, 357e362, 362f unit impulse, 358e362, 360f unit step, 358e362, 360f transfer function, 353e367 Time-shift theorem, 279t, 282e286, 282f, 284fe285f Toothed gears circular pitch, 46 dampers, 51e53, 51fe53f and inertia, 49e51, 49fe50f meshing gears, 45e46, 46f pair at contact point, 45e46, 46f pitch circle, 46 pulley disks with belt connection, 45e46, 46f pulleys, 45e46 and springs, 47e49, 47fe48f Torsional micromirror, 11e12 Torsional mirror, 11e12 Total solution, 430e431 Trackability, 543 Transduction, 18 Transfer function characteristics, 324 definition, 324e326 impedance transfer function, 334e347, 334f, 335t electrical systems, 334e340 fluid systems, 342e344, 343f loading cascading (series) electrical systems, 338e340, 339f mechanical systems, 344e347, 345fe346f nonloading cascading (series) electrical systems, 338e340, 339f operational amplifier configuration, 337f, 339f series and parallel equivalent impedances, 334e336, 334f

thermal systems, 340e342, 341f voltage divider configuration, 337f, 339f marginally stable systems, 348, 349f model formulation, 326e347 multiple-input, multiple-output (MIMO) systems, 326 simple imaginary, 351 single-input, single-output (SISO) systems, 325 stable systems, 348, 349f system stability, 347e353, 349fe351f time-domain mathematical model, 326e332 MIMO system, 328e332 SISO system, 326e328 time response. See Time response unstable systems, 348, 349f zero-pole-gain mathematical model, 332e334 linear time invariant (LTI) models, 333 Transfer function matrix, 382e385 Transfer-function matrix modeling, 579e586, 580fe582f, 584f, 586f Transient response, 7, 543e544 Transient time response specifications, 648e657 damping ratio, 655e657, 655f first-order systems, 648e649, 649f peak time, 653e654 second-order systems, 649e657, 650f underdamped second-order feedback systems, 650e657, 651f Transitory/steady-state time response, 662e679 first-order plants, 662e663 derivative control, 662e663 integral control, 663 proportional control, 662 second-order plants, 663e673 additional zero, 665e667, 666f dominant poles, 669e672, 669f, 671f nonlinear control systems, 677e679, 678f proportionalederivative control, 664e667, 664f proportionaleintegral control, 667e673, 668f proportionaleintegralederivative control, 674 zero-pole cancellation, 672e673 ZieglereNichols P+I+D tuning algorithm, 674e676 Translation potentiometer, 151f Translatory mechanical system with displacement input, 393f Translatory microspring system, 30f Translatory-motion microelectromechanical systems device, 329f

Index

Translatory spring combinations series and parallel, 26, 27f three parallel branches, 28f three series portions, 28f Translatory spring stiffness, 24, 24t Transmissibility for force input, 459e462, 459fe461f Transmissibility for motion input, 455 Transverse actuation, 520 Transverse-motion (parallel-plate) variable capacitor, 156f Transverse strains, 501e502, 744 Two-port systems, 338 Two-position control, 544e545 Two shaft-gear-rack lumped-parameter mode, 89f Two-shaft gear system, 95f Two-tank liquid-level system, 343f

U Underdamped second-order feedback systems, 650e657, 651f Underdamping, 92e93 Unit impulse, 358e362, 360f Unit norm, 714e715 Unit-ramp function, 277f Unit-step function, 275, 277f, 358e362, 360f Unity-feedback control systems, 658e660, 658te659t Unity-feedback system, 551e553, 551f Unstable systems, 348, 349f, 595

V Variable cross-section vertical tank, 217f Variable-gap, 156f Vectorematrix formulation, 306e307 Velocity constant Kv, 688, 689f Viscous damping coefficient, 40e43 Viscous damping elements, 38 Couette’s flow, 40e43, 40f cylinder front face, elementary area and damping torque, 43f rotary-motion damping, 42f series and parallel damper connections damper combination, required velocity reduction, 45f massless body with two dampers, translatory motion, 44f possible damper combinations, additional dampers, 45f rotary parallel, 43f translatory parallel, 43f translatory series, 43f

translatory and rotary viscous damping, 39, 39f viscous damping coefficient, 40e43 viscous damping force/moment and energy, 39, 39t Voltage divider configuration, 337f, 339f Volume flow rate, 214

W Wetted perimeter, 222e223 Wheatstone full bridge, 503f Wire electro-discharge machining, 10 Wire strain-gauge sensing, 502f

Z Zero element in first column, 605e607 Zero-order system, 12, 12f Zero-pole-gain mathematical model, 332e334 Zero-to-infinity time interval, 291e293 ZieglereNichols P+I+D tuning algorithm, 674e676

759