Introduction to Transients in Electrical Circuits: Analytical and Digital Solution Using an EMTP-based Software (Power Systems) [1st ed. 2021] 303068248X, 9783030682484

Table of contents :
Preface
Contents
List of Figures
List of Tables
1 Introduction to Fundamental Concepts in Electric Circuit Analysis
1.1 Introduction
1.2 Preliminary Concepts
1.3 Electrical Quantities
1.4 Power and Energy
1.5 Circuit Elements
1.6 Kirchhoff’s Laws
1.7 Analytical and Digital Circuit Solution
1.8 Conclusions and Motivations for Electromagnetic Transients Studies
References
2 Singular Functions
2.1 Single Step Function
2.2 Unitary Impulse Function
2.3 The Family of Singular Functions
2.4 Causal Functions
2.5 Derivative of an Ordinary Sectionally Continuous Function
2.6 Decomposition of a Function into Singular Functions
2.7 The Distribution Concept
2.8 The Unitary Impulse Function as Distribution
2.9 Initial Condition at t = 0+
2.10 Initial Decomposition of Functions in Steps
2.11 Decomposition of Functions into Impulses
2.12 Convolution Integral
2.13 Properties of the Convolution Integral
2.14 Proposed Problems
2.15 Conclusions
References
3 Differential Equations
3.1 Introduction
3.2 Ordinary Differential Equations
3.2.1 Solution of the Homogeneous Differential Equation
3.2.2 Considerations on Algebraic Equations
3.2.3 Solution of the Non-homogeneous Differential Equation
3.3 Differential Equations with Singular and Causal Forcing Functions
3.3.1 Step Type Forcing Function {\varvec U}_{ - 1} \left( {\varvec t} \right)
3.3.2 Forcing Function Type Impulse {\varvec U}_{0} \left( {\varvec t} \right)
3.3.3 Systematic Method for Calculating Initial Conditions at {\varvec t} = 0_{ {\,+\,} }
3.4 Conclusions
References
4 Digital Solution of Transients in Basic Electrical Circuits
4.1 Introduction
4.2 Basic Algorithm for Computational Solution of EMTP-Based Programs
4.3 Solution of Differential Equations via Digital Integration
4.4 Digital Computational Model of the Elements with Concentrated Parameters
4.4.1 Resistance {\varvec R}
4.4.2 Inductance {\varvec L}
4.4.3 Capacitance {\varvec C}
4.5 Numerical Oscillations in EMTP Due to the Trapezoidal Integration Method
4.6 Conclusions
References
5 Transients in First Order Circuits
5.1 Introduction
5.2 Continuity Theorem
5.3 Response to Impulse and Response to Step
5.4 Sequential Switching
5.5 Magnetically Coupled Circuits
5.6 DuHamel Integrals
5.7 Proposed Problems
5.8 Conclusions
References
6 Transients in Circuits of Any Order
6.1 Introduction
6.2 Circuits Initially Deenergized
6.3 Thévenin and Norton Equivalents for Initially Energized Capacitances and Inductances
6.4 Circuits Initially Energized
6.5 Switching Transients
6.6 Proposed Problems
6.7 Conclusions
References
7 Switching Transients Using Injection of Sources
7.1 Introduction
7.2 Method of Voltage Source Injection
7.3 Current Source Injection Method
7.4 Displacement of Current Source Method
7.5 Proposed Problems
7.6 Conclusions
References
Appendix A Processing in the ATP
Appendix B Main Relations Involving Singular Functions
Appendix C Laplace Transform Properties
Appendix D Laplace Transform Pairs
Appendix E Heaviside Expansion Theorem

Citation preview

Power Systems

José Carlos Goulart de Siqueira Benedito Donizeti Bonatto

Introduction to Transients in Electrical Circuits Analytical and Digital Solution Using an EMTP-based Software

Power Systems

Electrical power has been the technological foundation of industrial societies for many years. Although the systems designed to provide and apply electrical energy have reached a high degree of maturity, unforeseen problems are constantly encountered, necessitating the design of more efficient and reliable systems based on novel technologies. The book series Power Systems is aimed at providing detailed, accurate and sound technical information about these new developments in electrical power engineering. It includes topics on power generation, storage and transmission as well as electrical machines. The monographs and advanced textbooks in this series address researchers, lecturers, industrial engineers and senior students in electrical engineering. **Power Systems is indexed in Scopus**

More information about this series at http://www.springer.com/series/4622

José Carlos Goulart de Siqueira Benedito Donizeti Bonatto

•

Introduction to Transients in Electrical Circuits Analytical and Digital Solution Using an EMTP-based Software

123

José Carlos Goulart de Siqueira Institute of Electrical Systems and Energy Federal University of Itajubá Itajubá, Brazil

Benedito Donizeti Bonatto Institute of Electrical Systems and Energy Federal University of Itajubá Itajubá, Brazil

ISSN 1612-1287 ISSN 1860-4676 (electronic) Power Systems ISBN 978-3-030-68248-4 ISBN 978-3-030-68249-1 (eBook) https://doi.org/10.1007/978-3-030-68249-1 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The book Introduction to Transients in Electrical Circuits: Analytical and Digital Solution Using an EMTP-based Software integrates the analytical solution with the digital solution through the ATP—Alternative Transients Program software (recognized for its use all over the world in academia and in the electric power industry) in a didactic approach and mathematical and scientific rigour suitable for undergraduate and graduate programs, as well as for industry professionals. It is clear that many other classical and advanced books indeed cover more extensive subjects in electromagnetic transients. This book fills a gap between classic works in the field of electrical circuits and more advanced works in the field of transients in electrical power systems, facilitating a full understanding of analytical and digital modelling and solution of transients in basic circuits. Therefore, it fulfils its role as an introductory book, but completes in its learning proposal in undergraduate and graduate courses. Finally, the book presents a significant number of examples and proposed problems, all with answers, without exception. The book fills an important gap between the analytical solution of traditional electrical circuits with the digital solution using a well-established tool (the ATP program) in the power engineering community. It may be particularly important when teaching some basic power system transients that can be simulated with linear networks and lumped parameters. This book presents an approach to solving singular function differential equations representing the transient and steady-state dynamics of a circuit in a structured manner and without the need for physical reasoning to set initial conditions to zero plus (0+). The method indeed facilitates the mathematical solution. That does not mean that physical reasoning and proper modelling are unimportant in engineering education. Real applications may come after a strong background in these concepts. It also presents, for each presented problem, the exact analytical solution as well as the corresponding digital solution through a computer program based on the EMTP —Electromagnetics Transients Program. The andragogical approach (science and art in adult education) integrates existing knowledge and starts from the simple to the complex, with rich exposure to a variety of problems solved (all with tested and validated solutions). v

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Preface

The teacher and the apprentice are encouraged to use modern computational tools used in the electrical power industry for the calculation of electromagnetic transients, but are warned about their limitations and modelling constraints. This integrates the “know-how” and “know why” necessary for the safe analysis of simulated results. The teacher can direct students in solving specific problems in the textbook or even propose that students present real day-to-day problems for solution, as the introductory methodology for modelling and calculation of digital and analytical solution of circuit transients in electrical systems are fully presented in the textbook. Technical–scientific articles should be consulted, as well as textbooks for more advanced studies on transients in electrical power systems can be consulted, depending on the interest and need of the problem to be solved. It is important to clarify that the motivation and intent primacy of the main author, Prof. José Carlos Goulart de Siqueira, are to leave a legacy to the teachers and students of UNIFEI, where he magistrate with excellence for decades, having been the first rector of UNIFEI-Federal University of Itajubá, which has been completed, on November 23rd 2020, 107 years of history of contributions to higher education in Brazil. We think that the book is intended for engineering students who already have been successfully approved in introductory circuit analysis, calculus and ordinary differential equations courses. Our experience as educators is that, when it comes to truly understand and apply the mathematical knowledge in electrical circuits transients, the majority of students have much more difficulties. Therefore, this book supplies, in an integrated way, all the necessary knowledge and practice through extensive examples. This book is organized as follows: Chap. 1 presents an introduction to transients in electrical circuits with a discussion about fundamentals of circuit analysis, physical phenomena and the need for mathematical modelling and simulation. The student is encouraged from the beginning to become familiar with Appendix A—processing at the ATP. Chapter 2 presents singular functions for the analytical solution, and Appendix B shows the main relations involving singular functions. Chapter 3 presents the solution of differential equations. It emphasizes the solution using the classical method in the time domain. For the operational method in the complex frequency domain using the Laplace transform, Appendix C—Laplace transform properties, Appendix D—Laplace transform pairs and Appendix E—Heaviside expansion theorem can be helpful. Chapter 4 presents the digital solution of transients in basic electrical circuits. The fundamental algorithm of EMTP-based program is introduced, and the problem of numerical oscillations due to the trapezoidal method is briefly discussed. Chapter 5 presents transients in first-order circuits. Extensive number of examples are provided with their analytical and digital solution using the ATPDraw. Chapter 6

Preface

vii

presents transients in circuits of any order, exploring the solution methods consolidated so far. Finally, Chap. 7 introduces switching transients using the injection of sources method. All chapters provide useful references to enhance the learning process, as well as to instigate further investigation about advanced topics. Enjoy it! Itajubá, Brazil

José Carlos Goulart de Siqueira Benedito Donizeti Bonatto

Acknowledgments We would like to express our gratitude to many people who have helped to bring this book project to its conclusion. We deeply thank all members of our families, friends (Antonio Eduardo Hermeto…), colleagues who by listening and encouraging have provided the fundamental support and personal care. We give a special thanks to Alexa Bonelli Bonatto and Aline Bonelli Bonatto for their work in the initial translation and editing. Professional editing was provided by Springer’s team to whom we are deeply gratefull. Finally, we thank many students who were challenged in their studies during their courses at UNIFEI by the many proposed problems.

Contents

1 Introduction to Fundamental Concepts in Electric Circuit Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Preliminary Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Electrical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Power and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Kirchhoff’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Analytical and Digital Circuit Solution . . . . . . . . . . . . . . . . . 1.8 Conclusions and Motivations for Electromagnetic Transients Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Singular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Single Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Unitary Impulse Function . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Family of Singular Functions . . . . . . . . . . . . . . . . . . . 2.4 Causal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Derivative of an Ordinary Sectionally Continuous Function 2.6 Decomposition of a Function into Singular Functions . . . . . 2.7 The Distribution Concept . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 The Unitary Impulse Function as Distribution . . . . . . . . . . 2.9 Initial Condition at t = 0+ . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 Initial Decomposition of Functions in Steps . . . . . . . . . . . . 2.11 Decomposition of Functions into Impulses . . . . . . . . . . . . . 2.12 Convolution Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 Properties of the Convolution Integral . . . . . . . . . . . . . . . . 2.14 Proposed Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Solution of the Homogeneous Differential Equation . . 3.2.2 Considerations on Algebraic Equations . . . . . . . . . . . 3.2.3 Solution of the Non-homogeneous Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Differential Equations with Singular and Causal Forcing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Step Type Forcing Function U
1 ðtÞ . . . . . . . . . . . . . 3.3.2 Forcing Function Type Impulse U0 ðtÞ . . . . . . . . . . . . 3.3.3 Systematic Method for Calculating Initial Conditions at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Digital Solution of Transients in Basic Electrical Circuits . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Basic Algorithm for Computational Solution of EMTP-Based Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Solution of Differential Equations via Digital Integration . . . 4.4 Digital Computational Model of the Elements with Concentrated Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Resistance R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Inductance L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Capacitance C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Numerical Oscillations in EMTP Due to the Trapezoidal Integration Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5 Transients in First Order Circuits . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 5.2 Continuity Theorem . . . . . . . . . . . . . . . . . 5.3 Response to Impulse and Response to Step 5.4 Sequential Switching . . . . . . . . . . . . . . . . . 5.5 Magnetically Coupled Circuits . . . . . . . . . . 5.6 DuHamel Integrals . . . . . . . . . . . . . . . . . . 5.7 Proposed Problems . . . . . . . . . . . . . . . . . . 5.8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6 Transients in Circuits of Any Order . . . . . . . . . . . . . . . . . . . . . . . . . 367 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 6.2 Circuits Initially Deenergized . . . . . . . . . . . . . . . . . . . . . . . . . . 368

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6.3

Thévenin and Norton Equivalents for Initially Energized Capacitances and Inductances . . . . . . . . . . . . . . . . . . . . 6.4 Circuits Initially Energized . . . . . . . . . . . . . . . . . . . . . . 6.5 Switching Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Proposed Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 Switching Transients Using Injection of Sources . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Method of Voltage Source Injection . . . . . . 7.3 Current Source Injection Method . . . . . . . . . 7.4 Displacement of Current Source Method . . . 7.5 Proposed Problems . . . . . . . . . . . . . . . . . . . 7.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Appendix A: Processing in the ATP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653 Appendix B: Main Relations Involving Singular Functions . . . . . . . . . . . 663 Appendix C: Laplace Transform Properties . . . . . . . . . . . . . . . . . . . . . . . 667 Appendix D: Laplace Transform Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . 669 Appendix E: Heaviside Expansion Theorem . . . . . . . . . . . . . . . . . . . . . . . 671

List of Figures

Fig. 1.1 Fig. Fig. Fig. Fig. Fig. Fig. Fig.

1.2 1.3 1.4 1.5 1.6 1.7 1.8

Fig. 1.9 Fig. 1.10 Fig. 1.11 Fig. 1.12 Fig. 1.13 Fig. 1.14 Fig. 1.15 Fig. 1.16 Fig. 1.17 Fig. 1.18 Fig. 1.19

a System input and output; b Mechanical system composed of mass, spring and friction . . . . . . . . . . . . . . . . . . . . . . . . . Force F acting on q according to Coulomb’s law . . . . . . . . . Force lines of the point charge þ Q . . . . . . . . . . . . . . . . . . . Electrostatic field in an electric dipole . . . . . . . . . . . . . . . . . Infinitesimal work dW associated with displacement dr . . . . Body moviment along the ABC path in the xy plane . . . . . . The electrostatic potential of point P . . . . . . . . . . . . . . . . . . Two charges þ q being displaced against the electrostatic field E, from the reference, one to position A and the other one to position B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric current produced by the potential difference of the battery applied to both ends of the conductor . . . . . . . . . . . . Reference arrow representing the direction assumed to be positive for current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphic of current versus time, called the current waveform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Waveform of the charge qðtÞ . . . . . . . . . . . . . . . . . . . . . . . . Power absorbed by the circuit element . . . . . . . . . . . . . . . . . Voltage and current reference arrows and power and energy being a consumed or b supplied by the circuit element . . . . Reference arrows for voltage between the terminal nodes and current across a resistor . . . . . . . . . . . . . . . . . . . . . . . . . Resistance of most good conductors as a function of temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Capacitor formed by two metallic plates, separated by a thin layer of dielectric . . . . . . . . . . . . . . . . . . . . . . . . . . Effect of dielectric on the electric field and the capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Capacitor formed by two hollow conductive cylinders, coaxial and long . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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List of Figures

Fig. 1.20

Fig. 1.21 Fig. 1.22 Fig. Fig. Fig. Fig. Fig.

1.23 1.24 1.25 1.26 1.27

Fig. 1.28 Fig. 1.29

Fig. 1.30

Fig. 1.31

Fig. 1.32 Fig. 1.33 Fig. 1.34

Fig. 1.35 Fig. 1.36 Fig. 1.37 Fig. 1.38 Fig. 1.39 Fig. 1.40 Fig. 1.41

Capacitance per linear meter in a isolated conductive wire, suspended on a metal chassis, or another grounded plane and b between two parallel conductors . . . . . . . . . . . . . . . . . Reference arrows for voltage between the terminal nodes and current across a Capacitance . . . . . . . . . . . . . . . . . . . . . Waveforms for the capacitance C of Fig. 1.21, for a voltage, b current, c power, and d energy . . . . . . . . . . . . . . . . . . . . . Repulsion or attraction between poles of magnets . . . . . . . . Magnetic dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Magnetic induction field B . . . . . . . . . . . . . . . . . . . . . . . . . . Magnetic induction field at point P . . . . . . . . . . . . . . . . . . . . Magnetic force in a conductor carrying current—charge q moving with speed v, immersed in an external magnetic field B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Magnetic flux through a circular surface . . . . . . . . . . . . . . . . a Electric current flowing through a conductor producing a magnetic field around it; b Field of magnetic forces in a plane perpendicular to the conductor [9] . . . . . . . . . . . . Right hand rule determining the relationship between the direction of current flow in a wire and the direction of the lines of force in the magnetic field around the conductor [9] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Adjacent fingers indicating the direction of the current and the thumb, the direction of the field, for conductive wire in a closed path. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Magnetic flux produced by a current across a coil with N turns and with ferromagnetic core . . . . . . . . . . . . . . . . . . . . . Permanent magnet moving to the right, with its north pole approaching a coil with N turns . . . . . . . . . . . . . . . . . . . . . . Permanent magnet moving to the right, with its north pole approaching a coil with N turns, with the winding direction of the coil inverted . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Faraday and Lenz’s laws applied to a coil in an iron core . . Relationship between B and H depending on the material from which the core is made . . . . . . . . . . . . . . . . . . . . . . . . Reference arrows for voltage between the terminal nodes and current across an indutor . . . . . . . . . . . . . . . . . . . . . . . . a Current across and b Voltage between the terminals of a coil (inductor) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Independent voltage source; b DC voltage source; c Independent current source . . . . . . . . . . . . . . . . . . . . . . . . a Common emitter NPN transistor; b Equivalent circuit model for a common emitter NPN transistor . . . . . . . . . . . . RC series circuit, powered by a constant source . . . . . . . . . .

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List of Figures

Fig. 1.42 Fig. 1.43

Fig. 2.1

Fig. 2.2 Fig. 2.3 Fig. 2.4 Fig. 2.5 Fig. 2.6 Fig. 2.7 Fig. 2.8 Fig. 2.9 Fig. 2.10 Fig. 2.11

Fig. 2.12

Fig. 2.13 Fig. 2.14 Fig. 2.15 Fig. 2.16 Fig. 2.17 Fig. 2.18 Fig. 2.19 Fig. 2.20 Fig. 2.21

xv

RC series circuit, powered by a constant source, built for the digital solution using the ATPDraw program . . . . . . RC series circuit transient response for voltage, current, power and energy with digital solution using the ATPDraw program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Unitary step function U
1 ðtÞ; b Unitary step function delayed of a: U
1 ðt
aÞ; c Amplitude k step function kU
1 ðt
aÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Pulse function; b Unitary step functions used to represent the pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Triangular pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Switching at time t ¼ 0; b Equivalent circuit with unit step function U
1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Unitary ramp function U
2 ðtÞ; b Unitary ramp function delayed of a: U
2 ðt
aÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . a Unitary parabola function U
3 ðtÞ; b Unitary parabola function delayed of a: U
3 ðt
aÞ. . . . . . . . . . . . . . . . . . . . . Unitary impulse function U 0 ðtÞ defined as a rectangular pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Unitary impulse function U 0 ðtÞ defined as a triangular pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Unitary impulse function U 0 ðtÞ defined as a gaussian function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Unitary impulse function U0 ðtÞ; b Unitary Amplitude A impulse function delayed of a: U 0 ðt
aÞ . . . . . . . . . . . . . . . a Approximate representation of a unitary step function; b Approximate representation of a unitary impulse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Triangular pulse function; b Derivative of pulse of (a); c Unitary double function, U 1 ðtÞ; d graphic representation that will be used here for a unitary double function, U 1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Synthetic representation of the family of singular functions, having the unitary impulse function as a generator . . . . . . . . Function of Example 2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . a Periodic function f 1 ðtÞ; b Alternating rectangular pulse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Function gðtÞ of Example 2.11 . . . . . . . . . . . . . . . . . . . . . . . Singular functions that make up gðtÞ . . . . . . . . . . . . . . . . . . a Unit step function U
1 ð
tÞ; b Function f ðtÞ; c Composition of functions . . . . . . . . . . . . . . . . . . . . . . . . . . 0 Graphical representation of g ðtÞ . . . . . . . . . . . . . . . . . . . . . . Graphical representation of f ðtÞ . . . . . . . . . . . . . . . . . . . . . . a Current iðtÞ applied to a capacitor; b Charge qðtÞ . . . . . . .

..

75

..

76

..

80

.. ..

80 81

..

81

..

82

..

83

..

86

..

87

..

87

..

87

..

89

..

91

.. ..

92 99

.. 99 . . 100 . . 101 . . . .

. . . .

102 103 105 106

xvi

Fig. Fig. Fig. Fig. Fig.

List of Figures

2.22 2.23 2.24 2.25 2.26

Fig. 2.27 Fig. 2.28 Fig. 2.29 Fig. Fig. Fig. Fig. Fig.

2.30 2.31 2.32 2.33 2.34

Fig. Fig. Fig. Fig.

2.35 2.36 2.37 2.38

Fig. 2.39 Fig. 2.40 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 4.1

Fig. Fig. Fig. Fig. Fig.

4.2 4.3 4.4 4.5 4.6

Fig. 4.7 Fig. 4.8

Graphical representation of current iðtÞ . . . . . . . . . . . . . . . . . a–c Alternative ways of conceiving Eq. (2.67) . . . . . . . . . . . Function f ðtÞ sectionally continuous . . . . . . . . . . . . . . . . . . . Function f ðtÞ for Example 2.23 . . . . . . . . . . . . . . . . . . . . . . a first order derivative of f ðtÞ; b second order derivative of f ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Function gðtÞ for Example 2.24; b first order derivative of gðtÞ; c second order derivative of gðtÞ . . . . . . . . . . . . . . . Function f ðtÞ for Example 2.25 . . . . . . . . . . . . . . . . . . . . . . a first order derivative of f ðtÞ; b second order derivative of f ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Functions ðtÞ; f c ðtÞ, and AU
1 ðtÞ . . . . . . . . . . . . . . . . . . . . . Functions f ðtÞ and £ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuous function f ðtÞ approximate by “steps” . . . . . . . . . Function xðtÞ for decomposition into singular functions . . . . a First order derivative of function xðtÞ; b second order derivative of function xðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Function xðtÞ expressed as a sum of singular functions . . . . Function xðtÞ decomposition in singular functions . . . . . . . . Truncated ramp function . . . . . . . . . . . . . . . . . . . . . . . . . . . . Decomposition of function xðtÞ into sequential rectangular pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Function f 1 ðtÞ; b function f 2 ðtÞ, to be convoluted . . . . . . . a Function f 2 ðt
kÞ; b function f 2 ðt
kÞ shifted to the right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Curves of f 1 ðkÞ and f 2 ðt
kÞ . . . . . . . . . . . . . . . . . . . . . . . . Functions f 1 ðtÞ and f 2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Convolution function f ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Functions f 1 ðtÞ and f 2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Convolution function f ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ in a capacitor initially de-energized . . . . . . . . . . Function f 1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Function f 2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simplified block diagram of the basic algorithm for solving programs based on the EMTP [5] . . . . . . . . . . . . . . . . . . . . . Trapezoidal integration method . . . . . . . . . . . . . . . . . . . . . . . Backward Euler integration method . . . . . . . . . . . . . . . . . . . Linear and time-invariant resistor . . . . . . . . . . . . . . . . . . . . . Linear and time-invariant inductor . . . . . . . . . . . . . . . . . . . . Thévenin equivalent digital model of the inductor, discretized using the trapezoidal integration method . . . . . . . . . . . . . . . . Norton equivalent digital model of the inductor, discretized using the trapezoidal integration method . . . . . . . . . . . . . . . . Linear and time-invariant capacitor . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

107 110 112 113

. . 114 . . 115 . . 116 . . . . .

. . . . .

116 122 122 124 127

. . . .

. . . .

128 128 129 130

. . 133 . . 135 . . . . . . . . .

. . . . . . . . .

136 137 141 141 142 143 148 149 149

. . . . .

. . . . .

210 211 211 213 213

. . 215 . . 216 . . 217

List of Figures

Fig. 4.9 Fig. 4.10 Fig. 4.11 Fig. 4.12

Fig. 4.13

Fig. 4.14

Fig. Fig. Fig. Fig.

5.1 5.2 5.3 5.4

Fig. 5.5 Fig. 5.6 Fig. 5.7 Fig. 5.8 Fig. 5.9 Fig. 5.10 Fig. 5.11 Fig. 5.12 Fig. 5.13 Fig. 5.14 Fig. 5.15 Fig. 5.16 Fig. 5.17 Fig. 5.18 Fig. 5.19

xvii

Thévenin equivalent digital model of the capacitor, discretized using the trapezoidal integration method . . . . . . . Norton equivalent digital model of the capacitor, discretized using the trapezoidal integration method . . . . . . . . . . . . . . . . Application of a current step source through an ideal inductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the trapezoidal integration method . . . . . . . . . . . . . . . . . . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the Backward Euler integration method . . . . . . . . . . . . . . . . . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the CDA—Critical Damping Adjustment technique . . . . . . . Electric circuit of Example 5.1 . . . . . . . . . . . . . . . . . . . . . . . Electric Circuit of Example 5.1 for the calculation of RTH . . Electric circuit of Example 5.2 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.2 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.2 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.3 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.3 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.3 for digital solution . . . . . . . . Transient voltage and current digital solution for circuit of Example 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.4 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.4 for digital solution . . . . . . . . Transient voltage e(t) and current i(t) digital solution for circuit of Example 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.4 for digital solution . . . . . . . . Transient power p(t) and energy W(t) digital solution for circuit of Example 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.5 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.5 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Source conversion applied to electric circuit of Example 5.5 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.6 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.6 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 219 . . 220 . . 221

. . 222

. . 223

. . . .

. . . .

224 230 230 231

. . 231 . . 231 . . 232 . . 232 . . 233 . . 233 . . 234 . . 235 . . 235 . . 236 . . 236 . . 237 . . 237 . . 237 . . 238 . . 238

xviii

Fig. 5.20 Fig. 5.21 Fig. 5.22 Fig. 5.23 Fig. 5.24 Fig. 5.25 Fig. 5.26 Fig. 5.27 Fig. 5.28 Fig. 5.29 Fig. 5.30 Fig. 5.31 Fig. 5.32 Fig. 5.33 Fig. 5.34 Fig. 5.35 Fig. 5.36 Fig. 5.37 Fig. 5.38 Fig. 5.39 Fig. 5.40 Fig. 5.41 Fig. 5.42 Fig. 5.43 Fig. 5.44 Fig. 5.45 Fig. 5.46 Fig. 5.47

List of Figures

Electric circuit of Example 5.6 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.7 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.7 for analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.7 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.7 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Source conversion at electric circuit of Example 5.7 for analytical solution at t ! 1 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.7 for digital solution . . . . . . . . Transient current i2 ðtÞ digital solution for circuit of Example 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in transient current i2 ðtÞ digital solution for circuit of Example 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current i1 ðtÞ digital solution for circuit of Example 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.8 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.8 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.8 for digital solution . . . . . . . . Transient voltage and current digital solution for circuit of Example 5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.9 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.9 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.9 for digital solution . . . . . . . . Transient voltage and current digital solution for circuit of Example 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.10 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.10 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.11 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.11 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.12 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.12 for analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.12 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.13 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.14 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.15 . . . . . . . . . . . . . . . . . . . . . .

. . 238 . . 239 . . 240 . . 240 . . 240 . . 240 . . 241 . . 242 . . 242 . . 243 . . 243 . . 243 . . 244 . . 245 . . 245 . . 246 . . 247 . . 247 . . 248 . . 248 . . 252 . . 252 . . 254 . . 254 . . . .

. . . .

254 257 259 262

List of Figures

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

5.48 5.49 5.50 5.51 5.52 5.53 5.54

Fig. 5.55 Fig. Fig. Fig. Fig.

5.56 5.57 5.58 5.59

Fig. 5.60 Fig. 5.61 Fig. 5.62 Fig. 5.63 Fig. 5.64 Fig. 5.65 Fig. 5.66 Fig. 5.67 Fig. 5.68 Fig. 5.69 Fig. 5.70 Fig. 5.71 Fig. 5.72 Fig. 5.73 Fig. 5.74 Fig. 5.75 Fig. 5.76 Fig. 5.77

xix

Equivalent electric circuit of Example 5.15 . . . . . . . . . . . . . Electric circuit of Example 5.16 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 5.16 . . . . . . . . . . . . . Electric circuit of Example 5.17 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.19 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.19 for digital solution . . . . . . . Transient voltage and current digital solution for circuit of Example 5.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage and current digital solution for circuit of Example 5.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.21 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.21 for analytical solution . . . . Electric circuit of Example 5.21 for digital solution . . . . . . . Transient voltage and current digital solution for circuit of Example 5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage and current digital solution for circuit of Example 5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Numerical oscillation in the digital solution for circuit of Example 5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at numerical oscillation in the digital solution for circuit of Example 5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.22 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.22 for analytical solution . . . . Equivalent electric circuit of Example 5.22 for analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.22 for digital solution . . . . . . . Transient voltage digital solution for circuit of Example 5.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.23 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.23 for digital solution . . . . . . . Transient current digital solution for circuit of Example 5.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient current digital solution for circuit of Example 5.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage digital solution for circuit of Example 5.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.24 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.24 for digital solution . . . . . . . Transient voltage and current digital solution for circuit of Example 5.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.25 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.25 for analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

. . . . . .

263 263 264 265 266 268

. . 269 . . . .

. . . .

269 270 271 272

. . 272 . . 273 . . 273 . . 274 . . 274 . . 275 . . 275 . . 276 . . 276 . . 276 . . 278 . . 278 . . 279 . . 279 . . 280 . . 281 . . 282 . . 282 . . 283

xx

List of Figures

Fig. 5.78 Fig. 5.79 Fig. Fig. Fig. Fig.

5.80 5.81 5.82 5.83

Fig. 5.84 Fig. 5.85 Fig. 5.86 Fig. 5.87 Fig. 5.88 Fig. 5.89 Fig. Fig. Fig. Fig.

5.90 5.91 5.92 5.93

Fig. 5.94 Fig. 5.95 Fig. 5.96 Fig. 5.97 Fig. 5.98 Fig. 5.99 Fig. 5.100 Fig. 5.101 Fig. 5.102 Fig. 5.103 Fig. 5.104 Fig. 5.105

Equivalent electric circuit of Example 5.25 for analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.25 for analytical solution at t ¼ 0 þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.25 for analytical solution . . . . Electric circuit of Example 5.26 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.26 for digital solution . . . . . . . Transient voltage and current digital solution for circuit of Example 5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current digital solution for circuit of Example 5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage digital solution for circuit of Example 5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage digital solution for circuit of Example 5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.27 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.27 for DC analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.27 for AC analytical solution at t ¼ 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.28 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.28 for analytical solution . . . . Electric circuit of Example 5.28 for digital solution . . . . . . . Transient voltage digital solution for circuit of Example 5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current digital solution for circuit of Example 5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 5.29 . . . . . . . . . . . . . Transient total current source digital solution for circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current digital solution for circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient capacitor voltage digital solution for circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient capacitor current digital solution for circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient capacitor current digital solution for circuit of Example 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuits of Example 5.30: a additive polarity; b subtractive polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.31 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.31 for analytical solution . . . . Electric circuit of Example 5.32 . . . . . . . . . . . . . . . . . . . . . .

. . 283 . . . .

. . . .

283 284 286 286

. . 286 . . 287 . . 287 . . 288 . . 288 . . 289 . . . .

. . . .

289 290 291 292

. . 293 . . 293 . . 294 . . 296 . . 297 . . 297 . . 298 . . 298 . . 299 . . . .

. . . .

300 301 302 302

List of Figures

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

5.106 5.107 5.108 5.109 5.110 5.111 5.112 5.113

Fig. 5.114 Fig. 5.115 Fig. 5.116 Fig. 5.117 Fig. 5.118 Fig. 5.119 Fig. 5.120 Fig. 5.121 Fig. 5.122 Fig. 5.123 Fig. Fig. Fig. Fig. Fig.

5.124 5.125 5.126 5.127 5.128

Fig. 5.129 Fig. 5.130 Fig. 5.131 Fig. 5.132 Fig. 5.133

xxi

Electric circuit of Example 5.32 for analytical solution . . . . Electric circuit of Example 5.33 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.34 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 5.35 . . . . . . . . . . . . . Electric circuit of Example 5.35 for digital solution . . . . . . . Electric circuit of Example 5.35 for digital solution . . . . . . . Transient current i1 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current i2 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient current i2 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current iL ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient current iL ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . Transient current i1 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations . . . . . . Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations . . . . . . Transient current i2 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations . . . . . . Zoom at transient current i2 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations . . . . . . Electric circuit of Example 5.35 for digital solution . . . . . . . Transient digital voltage in the inductance of 60½mH . . . . . . Transient digital voltage in the inductance of 120½mH  . . . . . Transient digital voltage in the inductance of
40½mH  . . . . Zoom at transient digital voltage in the inductance of 60½mH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient digital voltage in the inductance of 120½mH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transient digital voltage in the inductance of
40½mH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Steady-state digital voltage in the inductance of 60½mH . . . Steady-state digital voltage in the inductance of 120½mH . . Steady-state digital voltage in the inductance of
40½mH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

303 304 306 309 312 313 313

. . 314 . . 314 . . 315 . . 315 . . 316 . . 316 . . 317 . . 318 . . 318 . . 319 . . . . .

. . . . .

319 320 321 321 322

. . 323 . . 323 . . 324 . . 324 . . 325 . . 325

xxii

List of Figures

Fig. 5.134 Fig. 5.135 Fig. 5.136 Fig. 5.137 Fig. 5.138 Fig. 5.139

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

5.140 5.141 5.142 5.143 5.144 5.145 5.146 5.147 5.148 5.149 5.150 5.151 5.152 5.153 5.154 5.155 5.156 5.157 5.158 5.159 5.160 5.161 5.162 5.163 5.164 5.165 5.166 5.167 5.168 5.169 5.170

Transient voltage eL1 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transient voltage eL2 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transientvoltage eL1 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at transientvoltage eL2 ðtÞ digital solution for circuit of Example 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear and time-invariant (or fixed) circuit—CLI . . . . . . . . . a Invariance with time; b homogeneity property; c principle of superposition; d DuHamel, Carson, Superposition Integral or, simply, Convolution Integral . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 5.37 . . . . . . . . . . . . . . . . . . . . . . Cascade or latter circuit of Example 5.38 . . . . . . . . . . . . . . . Electric circuit of Problem P.5-1 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-2 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-3 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-4 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-6 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-7 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-8 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-9 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-10 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-11 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-12 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-14 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-15 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-16 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-17 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-18 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-19 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-20 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-21 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-22 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-23 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-24 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-25 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-26 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-27 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-28 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-29 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-30 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-32 . . . . . . . . . . . . . . . . . . . . .

. . 326 . . 326 . . 327 . . 327 . . 328

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

329 332 337 338 338 339 339 340 340 341 341 342 342 343 343 344 344 344 345 345 345 346 346 346 346 347 347 348 348 348 349 350

List of Figures

Fig. 5.171 Fig. 5.172 Fig. 5.173 Fig. 5.174 Fig. 5.175 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

5.176 5.177 5.178 5.179 5.180 5.181 5.182 5.183 5.184 5.185 5.186

Fig. 5.187 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

5.188 5.189 5.190 5.191 5.192 5.193 5.194 5.195 5.196 5.197 5.198 6.1 6.2 6.3

Fig. 6.4 Fig. 6.5 Fig. 6.6 Fig. 6.7 Fig. 6.8 Fig. 6.9

xxiii

Capacitor voltage eC ðtÞ and current iðtÞ of Problem P.5-32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Source eðtÞ, resistor eR ðtÞ and capacitor eC ðtÞ voltages of Problem P.5-32. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-33 . . . . . . . . . . . . . . . . . . . . . Capacitor current iðtÞ of Problem P.5-33 . . . . . . . . . . . . . . . Source voltage eðtÞ and Voltage at resistor eR ðtÞof Problem P.5-33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Capacitor voltage eC ðtÞ of Problem P.5-33 . . . . . . . . . . . . . . Electric circuit of Problem P.5-34 . . . . . . . . . . . . . . . . . . . . . Inductor current iðtÞ of Problem P.5-34 . . . . . . . . . . . . . . . . Source, resistor, inductor voltages of Problem P.5-34 . . . . . . Electric circuit of Problem P.5-35 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-36 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-36 for digital solution . . . . . . Inductor voltage eL ðtÞ of Problem P.5-36 . . . . . . . . . . . . . . . Resistor voltage eR ðt) of Problem P.5-36 . . . . . . . . . . . . . . . Current iðtÞ of Problem P.5-36 . . . . . . . . . . . . . . . . . . . . . . . Current iT ðtÞ, which flows through the voltage eðtÞ source of Problem P.5-36. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iC ðtÞ, which flows through the voltage source of Problem P.5-36. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Power pL ðtÞ in the inductorof Problem P.5-36 . . . . . . . . . . . Energy W L ðtÞ in the inductorof Problem P.5-36 . . . . . . . . . . Electric circuit of Problem P.5-37 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-37 for digital solution . . . . . . Voltage e1 ðtÞ and Voltage e2 ðtÞ of Problem P.5-37 . . . . . . . Zoom at current in switch 3 of Problem P.5-37 . . . . . . . . . . Electric circuit of Problem P.5-38 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-39 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-40 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-41 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.5-42 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.1 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.4 . . . . . . . . . . . . . . . . . . . . . . . Voltage e0 ðtÞ, voltage at left capacitance, source voltage eðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current at right capacitance, current at left capacitance . . . . Electric circuit of Example 6.5 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.6 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage e0 ðtÞ, source voltage . . . . . . . . . . . . . . . . . . . . . . . . Current i0 ðtÞ, Current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.7 . . . . . . . . . . . . . . . . . . . . . . .

. . 350 . . 351 . . 351 . . 352 . . . . . . . . . . .

. . . . . . . . . . .

352 353 353 354 354 355 355 356 356 357 357

. . 358 . . . . . . . . . . . . . .

. . . . . . . . . . . . . .

358 359 359 360 360 361 361 362 362 362 363 364 368 372

. . 372 . . 373 . . 373 . . . .

. . . .

377 377 378 378

xxiv

List of Figures

Fig. Fig. Fig. Fig. Fig.

6.10 6.11 6.12 6.13 6.14

Fig. Fig. Fig. Fig. Fig.

6.15 6.16 6.17 6.18 6.19

Fig. 6.20 Fig. 6.21 Fig. 6.22 Fig. 6.23 Fig. 6.24 Fig. 6.25 Fig. Fig. Fig. Fig.

6.26 6.27 6.28 6.29

Fig. Fig. Fig. Fig. Fig.

6.30 6.31 6.32 6.33 6.34

Fig. Fig. Fig. Fig. Fig.

6.35 6.36 6.37 6.38 6.39

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.40 6.41 6.42 6.43 6.44 6.45 6.46

Electric circuit of Example 6.9 . . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.9 . . . . . . . . . . . . . . Electric circuit of Example 6.10 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.11 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.11 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eC ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.12 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.12 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Supply voltage, voltage eR ðtÞ, voltage eL ðtÞ, voltage eC ðtÞ . . Electric circuit of Example 6.13 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage in the elements, current in the resistance . . . . . . . . . Current at inductance, current at capacitance . . . . . . . . . . . . Electric circuit of Example 6.14 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Inductance power, capacitance power . . . . . . . . . . . . . . . . . . Energy in inductance, energy in capacitance . . . . . . . . . . . . . Electric circuit of Example 6.15 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.15 for the numerical solution by the ATPDraw program . . . . . . . . . . . . . . . . . . . . Rectangular pulse of the voltage source eðtÞ . . . . . . . . . . . . . Zoom on the rectangular pulse of the voltage source eðtÞ . . . Response to the impulse “hðtÞ” . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.16 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.16 for the numerical solution by the ATPDraw program . . . . . . . . . . . . . . . . . . . . Current at inductance iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Current injected by sources . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage e1 ðtÞ in R, L and C . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.17 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.17 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage at capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the capacitance voltage . . . . . . . . . . . . . . . . . . . . . . A closer zoom at the capacitance voltage . . . . . . . . . . . . . . . Current at capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage and current at the 400 ½mH inductance . . . . . . . . . . Current in the resistors of 20 [X] and 10 [X] . . . . . . . . . . . . Source voltage e1 ðtÞ and e2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

382 382 383 384

. . . . .

. . . . .

388 389 389 390 390

. . 392 . . 393 . . 393 . . 394 . . 394 . . 395 . . . .

. . . .

396 396 397 397

. . . . .

. . . . .

399 400 400 401 401

. . . . .

. . . . .

403 404 404 405 405

. . . . . . . .

. . . . . . . .

406 407 408 408 409 409 410 410

List of Figures

Fig. 6.47

Fig. 6.48

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.49 6.50 6.51 6.52 6.53 6.54 6.55

Fig. Fig. Fig. Fig.

6.56 6.57 6.58 6.59

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.60 6.61 6.62 6.63 6.64 6.65 6.66 6.67 6.68 6.69 6.70 6.71 6.72 6.73 6.74 6.75 6.76 6.77 6.78

Fig. 6.79 Fig. 6.80 Fig. 6.81

xxv

Electric circuit with a a capacitance C with an initial voltage eð0
Þ; b Thévenin equivalent with capacitance C, now initially de-energized; c Norton equivalent with capacitance C, now initially de-energized . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit with a an inductance L with an initial current ið0
Þ; b Norton equivalent with inductance L, now initially de-energized; c Thévenin equivalent with inductance L, now initially de-energized . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.18 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.18 . . . . . . . . . . . . . Electric circuit of Example 6.19 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.19 . . . . . . . . . . . . . Electric circuit of Example 6.20 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.20 . . . . . . . . . . . . . Electric circuit of Example 6.20 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage e1 ðtÞ, Voltage e2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.21 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.21 . . . . . . . . . . . . . Electric circuit of Example 6.21 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ ffi i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Currents i1 ðtÞ and i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current zooms i1 ðtÞ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eðtÞ and current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ with manually altered scale . . . . . . . . . . . . . . . . Current i1 ðtÞ with manually altered scale . . . . . . . . . . . . . . . Zoom at current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.23 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.23 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage e1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in voltage e1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage e4 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 411

. . . . . . .

. . . . . . .

413 414 415 415 416 417 418

. . . .

. . . .

419 419 420 420

. . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . .

424 425 425 426 426 427 427 428 429 429 430 430 431 431 432 432 433 433 435

. . . .

. . . .

435 436 436 437

xxvi

List of Figures

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.82 6.83 6.84 6.85 6.86 6.87 6.88 6.89

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.90 6.91 6.92 6.93 6.94 6.95 6.96 6.97 6.98 6.99 6.100

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.101 6.102 6.103 6.104 6.105 6.106 6.107 6.108

Fig. 6.109 Fig. 6.110 Fig. 6.111 Fig. Fig. Fig. Fig. Fig.

6.112 6.113 6.114 6.115 6.116

Fig. 6.117 Fig. 6.118 Fig. 6.119

Zoom in voltage e4 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i3 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current i3 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.24 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.24 . . . . . . . . . . . . . Electric circuit of Example 6.24 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ in the range 50 ½ms  t  60 ½ms . . . . . . . . . . . . Voltage eR ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in voltage eR ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eC ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in voltage eC ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eC ðtÞ in the range 90 ½ms  t  100 ½ms . . . . . . . . . Electric circuit of Example 6.25 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.25 . . . . . . . . . . . . . Electric circuit of Example 6.25 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltages e1 ðtÞ ¼ e2 ðtÞ, eR ðtÞ and the source E . . . . . . . . . . . Current iR ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current ik ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current i1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in current ik ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.26 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.26 for the numerical solution by ATPDraw program . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.27 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.27 . . . . . . . . . . . . . Electric circuit of Example 6.27 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Source voltage, voltage e1 ðtÞ voltage eL ðtÞ . . . . . . . . . . . . . . Voltage ek ðtÞ and voltage eLD ðtÞ . . . . . . . . . . . . . . . . . . . . . . Zoom in voltage ek ðtÞ and voltage eLD ðtÞ . . . . . . . . . . . . . . . Current iL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.28 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage in R1 ; voltage in C; voltage in C=2, source voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage in R2 , voltage in C . . . . . . . . . . . . . . . . . . . . . . . . . Current in R1 -C Current in C2 Current in R2 -C . . . . . . . . . .

. . . . . . .

. . . . . . .

437 438 438 439 439 440 440

. . . . . . . . . . .

. . . . . . . . . . .

442 442 443 443 444 444 445 445 446 446 447

. . . . . . . .

. . . . . . . .

452 453 453 454 454 455 455 456

. . 456 . . 458 . . 458 . . . . .

. . . . .

460 461 461 462 463

. . 463 . . 464 . . 464 . . 465

List of Figures

Fig. 6.120 Fig. 6.121 Fig. 6.122 Fig. Fig. Fig. Fig. Fig.

6.123 6.124 6.125 6.126 6.127

Fig. 6.128 Fig. 6.129 Fig. Fig. Fig. Fig. Fig. Fig.

6.130 6.131 6.132 6.133 6.134 6.135

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.136 6.137 6.138 6.139 6.140 6.141 6.142

Fig. Fig. Fig. Fig. Fig.

6.143 6.144 6.145 6.146 6.147

Fig. 6.148 Fig. 6.149

Fig. 6.150 Fig. 6.151 Fig. 6.152

xxvii

Electric circuit of Example 6.29 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.29 . . . . . . . . . . . . . Electric circuit of Example 6.29 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Current i1 ðtÞ, Current i2 ðtÞ; Current iC ðtÞ . . . . . . . . . . . . . . . Source voltage voltage eR ðtÞ Voltage eL ðtÞ Volltage eC ðtÞ . . Voltage eR1 ðtÞ Voltage ek ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.30 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.30 for DC steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.30 . . . . . . . . . . . . . Electric circuit of Example 6.30 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Source voltage, Voltage eC ðtÞ, Voltage ek ðtÞ . . . . . . . . . . . . Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage in the open switch eCh ðtÞ . . . . . . . . . . . . . . . . . . . . . Zoom at voltage in the open switch eCh ðtÞ . . . . . . . . . . . . . . Electric circuit of Example 6.31 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.31 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage E, Voltage eC ðtÞ, Voltage ek ðtÞ . . . . . . . . . . . . . . . . Voltage eCh ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in voltage eCh ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current i1 ðtÞ, Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.32 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.32 . . . . . . . . . . . . . Electric circuit of Example 6.32 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eR ðtÞ, Voltage eL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eC ðtÞ, Voltage iC ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ, Current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.33 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.33 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ, Source Voltage eðtÞ . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.33 for the numerical solution by the ATPDraw program, when switch k is closed at t ¼ 0, in the worst condition . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ, with the most intense transient phenomenon . . . Zoom at current iðtÞ, with the most intense transient phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at current iðtÞ, with the most intense transient phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 465 . . 466 . . . . .

. . . . .

472 473 473 474 474

. . 475 . . 475 . . . . . .

. . . . . .

478 478 479 480 480 481

. . . . . . .

. . . . . . .

481 482 482 483 483 484 485

. . . . .

. . . . .

487 488 488 489 489

. . 492 . . 493

. . 494 . . 494 . . 495 . . 495

xxviii

Fig. 6.153

Fig. Fig. Fig. Fig.

6.154 6.155 6.156 6.157

Fig. Fig. Fig. Fig.

6.158 6.159 6.160 6.161

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.162 6.163 6.164 6.165 6.166 6.167 6.168

Fig. Fig. Fig. Fig.

6.169 6.170 6.171 6.172

Fig. Fig. Fig. Fig.

6.173 6.174 6.175 6.176

Fig. 6.177 Fig. 6.178 Fig. 6.179 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.180 6.181 6.182 6.183 6.184 6.185 6.186 6.187

List of Figures

Current iðtÞ for switch k closing at t ¼ 3:979 ms; current iðtÞ for switch k closing at t ¼ 8:146 ms; voltage eðtÞ of the inusoidal source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in Fig. 6.153 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.34 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.35 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.35 for AC steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.35 . . . . . . . . . . . . . Electric circuit of Example 6.36 . . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 6.36 . . . . . . . . . . . . . Electric circuit of Example 6.36 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Source voltage eðtÞ and Voltage in switch ek ðtÞ . . . . . . . . . . Current in the left loop iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eðtÞ, Voltage ek ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . Current in the left loop iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.37 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.37 for t [ 0 solution . . . . . . . . Electric circuit of Example 6.37 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . e1 ðtÞ, e2 ðtÞ, eðtÞ; ik ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.38 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.38 for t [ 0 solution . . . . . . . . Electric circuit of Example 6.38 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . eðtÞ, ek ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ik ðtÞ, iL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.39 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.39 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Load energizing current . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the start of the load energizing current . . . . . . . . . . Load energizing current, but now for tmax ¼ 250 ms ¼ 0:25 ½s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the end of Fig. 6.179 . . . . . . . . . . . . . . . . . . . . . . . Voltages eR ðtÞ and eL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the beginning of Fig. 6.181 . . . . . . . . . . . . . . . . . . Result of new zooms in Fig. 6.181 . . . . . . . . . . . . . . . . . . . . Voltages eR ðtÞ e eL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the end of Fig. 6.184 . . . . . . . . . . . . . . . . . . . . . . . Industrial load voltage eA ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at the beginning of Voltage eA ðtÞ . . . . . . . . . . . . . . . .

. . . .

. . . .

496 497 498 500

. . . .

. . . .

500 501 502 503

. . . . . . .

. . . . . . .

505 506 506 507 507 508 509

. . . .

. . . .

510 511 511 512

. . . .

. . . .

515 515 516 516

. . 518 . . 518 . . 519 . . . . . . . . .

. . . . . . . . .

520 520 521 521 522 523 523 524 524

List of Figures

Fig. Fig. Fig. Fig. Fig. Fig.

6.188 6.189 6.190 6.191 6.192 6.193

Fig. 6.194 Fig. 6.195 Fig. 6.196 Fig. 6.197 Fig. Fig. Fig. Fig. Fig.

6.198 6.199 6.200 6.201 6.202

Fig. Fig. Fig. Fig. Fig.

6.203 6.204 6.205 6.206 6.207

Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.208 6.209 6.210 6.211 6.212 6.213 6.214

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.215 6.216 6.217 6.218 6.219 6.220 6.221 6.222 6.223 6.224

xxix

Result of new zooms in Fig. 6.187 . . . . . . . . . . . . . . . . . . . . Zoom at the end of Fig. 6.186 . . . . . . . . . . . . . . . . . . . . . . . Voltage and current in industrial load . . . . . . . . . . . . . . . . . . Zoom at the Fig. 6.190 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Instantaneous power, pR ðtÞ, in the load resistance . . . . . . . . . Zoom in the transient phase of the instantaneous power pR ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in the steady-state phase of the instantaneous power pR ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Instantaneous power, pL ðtÞ, at load inductance . . . . . . . . . . . Zoom in the transient phase of the instantaneous power pL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom in the steady-state phase of the instantaneous power pL ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . System supply current iðtÞ with switches k1 and k2 closed . . Zoom in the supply current iðtÞ . . . . . . . . . . . . . . . . . . . . . . New zoom in the supply current iðtÞ . . . . . . . . . . . . . . . . . . Steady-state supply current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . Voltage eC ðtÞ and current iC ðtÞ of energization of the capacitor bank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at Fig. 6.203 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iF ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom at current iF ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Load voltage eA ðtÞ and supply current iðtÞ . . . . . . . . . . . . . . Zoom in the steady-state regime of the curves of Fig. 6.206 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Instantaneous power pR ðtÞ in the load resistor . . . . . . . . . . . Zoom on the load resistance power curve, pR ðtÞ . . . . . . . . . . Energy WL ðtÞ and WC ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom applied at the end of Fig. 6.210 . . . . . . . . . . . . . . . . . Power triangles pre and post power factor correction . . . . . . Electric circuit of Example 6.40 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 6.40, when the two switches are closed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-1 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-3 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-4 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-5 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-6 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-7 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-8 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-9 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-10 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-11 . . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

525 525 526 527 528

. . 528 . . 529 . . 529 . . 530 . . . . .

. . . . .

530 532 532 533 533

. . . . .

. . . . .

534 534 535 535 536

. . . . . . .

. . . . . . .

537 538 538 540 540 542 542

. . . . . . . . . . .

. . . . . . . . . . .

548 560 561 561 561 562 562 563 563 563 563

xxx

List of Figures

Fig. 6.225 Fig. 6.226 Fig. 6.227 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

6.228 6.229 6.230 6.231 6.232 6.233 6.234 6.235 6.236 6.237 6.238 6.239 6.240 6.241 6.242 6.243 6.244 6.245 6.246 6.247 6.248 6.249 6.250 6.251 6.252 6.253 7.1 7.2

Fig. 7.3 Fig. 7.4

Fig. 7.5 Fig. 7.6 Fig. 7.7

Electric circuit of Problem P.6-12 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem P.6-12 for the numerical solution by the ATPDraw program . . . . . . . . . . . . . . . . . . . . Voltage in the left capacitor and current across the switch of Fig. 6.226 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage and current in the left inductor of Fig. 6.226 . . . . . . Voltage and current in the right inductor of Fig. 6.226. . . . . Voltage at the right capacitor of Fig. 6.226 . . . . . . . . . . . . . Electric circuit of Problem 6-13 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-14 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-15 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-16 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-17 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-18 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-19 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-20 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-21 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-22 . . . . . . . . . . . . . . . . . . . . . . Voltages at capacitances and resistance of Fig. 6.240 . . . . . . Power at capacitances and resistance of Fig. 6.240 . . . . . . . . Electric circuit of Problem 6-23 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-24 . . . . . . . . . . . . . . . . . . . . . . Voltages at inductance for C ¼ 10½lF of Fig. 6.244 . . . . . . Voltages at inductance for C ¼ 40½lF of Fig. 6.244 . . . . . . Voltages at inductance for C ¼ 100½lF of Fig. 6.244 . . . . . Electric circuit of Problem 6-25 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-26 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-27 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-28 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-29 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Problem 6-30 . . . . . . . . . . . . . . . . . . . . . . Circuit with switch k open . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit in Fig. 7.1 with switch k open, replaced by the ideal voltage source F 1 . . . . . . . . . . . . . . . . . . . . . . . Circuit after the closing of switch k, simulated with the injection of the voltage source F 2 . . . . . . . . . . . . . . Circuit reduced to passive for the calculation of voltages and currents produced exclusively by the source F 2 , for t [ t0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.1 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.1 with the exclusive performance of the F2 source . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.2 . . . . . . . . . . . . . . . . . . . . . .

. . 563 . . 564 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

564 565 565 566 566 567 567 567 568 568 568 569 569 570 570 571 571 571 572 572 573 574 574 574 575 575 575 578

. . 579 . . 579

. . 580 . . 581 . . 582 . . 582

List of Figures

Fig. 7.8 Fig. 7.9 Fig. 7.10 Fig. 7.11 Fig. 7.12 Fig. 7.13 Fig. 7.14 Fig. 7.15 Fig. 7.16 Fig. 7.17 Fig. 7.18 Fig. 7.19 Fig. 7.20 Fig. 7.21 Fig. 7.22 Fig. 7.23 Fig. 7.24 Fig. 7.25 Fig. Fig. Fig. Fig. Fig. Fig.

7.26 7.27 7.28 7.29 7.30 7.31

Fig. 7.32 Fig. 7.33 Fig. 7.34 Fig. 7.35

xxxi

Electric circuit of Example 7.2 with the steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.2 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Transformed circuit of from that of Fig. 7.9, in the domain of the complex frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.3 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.3 for steady-state solution . . . . Electric circuit of Example 7.3 with the injection of the source F 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.4 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.4 with the injection of the source F 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.5 . . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.5, after the insertion of the F 2 source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.5, combining the operational impedances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.6 . . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.6, in the domain of the complex frequency with the injected voltage source F 2 . . . . Electric circuit for Example 7.7 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.7 for AC steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit of Example 7.7, after the insertion of the F 2 source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.7, after the insertion of the F 2 source, to the domain of the complex frequency . . Electric circuit of Example 7.7 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage eðtÞ voltage ek ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . Current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current ik ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.8 . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.8, for steady-state solution . . . Electric circuit of Example 7.8 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.8, in the domain of the complex frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.8 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Source voltage F 2 and current ik ðtÞ in the switch . . . . . . . . . Electric circuit for the calculation of steady-state digital solution from Example 7.8 . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 583 . . 583 . . 583 . . 584 . . 585 . . 585 . . 587 . . 588 . . 588 . . 589 . . 589 . . 590 . . 591 . . 593 . . 593 . . 595 . . 596 . . . . . .

. . . . . .

598 598 599 599 600 600

. . 601 . . 602 . . 604 . . 605 . . 605

xxxii

Fig. 7.36 Fig. 7.37 Fig. 7.38 Fig. 7.39 Fig. 7.40

Fig. 7.41 Fig. 7.42 Fig. 7.43 Fig. 7.44 Fig. 7.45 Fig. 7.46 Fig. 7.47 Fig. 7.48 Fig. 7.49 Fig. 7.50 Fig. 7.51 Fig. 7.52 Fig. 7.53 Fig. 7.54 Fig. 7.55 Fig. 7.56 Fig. 7.57 Fig. 7.58 Fig. 7.59 Fig. 7.60 Fig. 7.61

List of Figures

Steady-state digital solution for switch voltage ek ðtÞ and inductor current iðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit before opening the switch at time t ¼ t0 . . . . . . . . . . Circuit in Fig. 7.37 with switch k closed, replaced by the current source F 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit after opening the switch k, simulated by injecting the current source F 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit reduced to passive for the calculation of voltages and currents produced exclusively by the current source F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.9 . . . . . . . . . . . . . . . . . . . . . . Adapted electric circuit for Example 7.9 . . . . . . . . . . . . . . . . Electric circuit of Example 7.9 with switch k, closed, replaced by an ideal current source F 1 . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.9 with an ideal current source F 2 injected in parallel with F 1 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.9 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.10 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.10 for steady-state solution . . . Electric circuit of Example 7.10 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.11 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.11 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Transformed circuit of Example 7.11, in the domain of the complex frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.12 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.12 for AC steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.12 with the injection of the source F 2 , for t [ 0 (new origin) . . . . . . . . . . . . . . . . Electric circuit of Example 7.12 for the numerical solution by the ATPDraw program . . . . . . . . . . . . . . . . . . . . Source voltage eðtÞ, current in switch ik ðtÞ, and voltage in switch ek ðtÞ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Alternative circuit for the numerical solution with injection of the current source F 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage ek ðtÞ, injected current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . Electric circuit for Example 7.13 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.13 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.13 for the numerical solution by the ATPDraw program . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 606 . . 606 . . 607 . . 608

. . 608 . . 609 . . 610 . . 610 . . 611 . . 611 . . 612 . . 613 . . 613 . . 614 . . 614 . . 615 . . 616 . . 616 . . 618 . . 620 . . 621 . . 622 . . 622 . . 622 . . 624 . . 626

List of Figures

Fig. 7.62 Fig. 7.63 Fig. 7.64 Fig. 7.65 Fig. 7.66 Fig. 7.67 Fig. 7.68 Fig. 7.69 Fig. 7.70 Fig. 7.71 Fig. 7.72 Fig. 7.73 Fig. 7.74 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

7.75 7.76 7.77 7.78 7.79 7.80 7.81 7.82 7.83 7.84 7.85 7.86 7.87 7.88 7.89 7.90 7.91 7.92 7.93 7.94 7.95

xxxiii

Voltage ek ðtÞ, switch current ik ðtÞ, source voltage eðtÞ . . . . . Alternative circuit for the numerical solution with injection of the current source F 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltage ek ðtÞ, injected current i2 ðtÞ . . . . . . . . . . . . . . . . . . . . Illustration of the displacement of current source method—original electric circuit . . . . . . . . . . . . . . . . . . . . . . Illustration of the displacement of current source method—equivalent electric circuit . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.14 . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.14 for AC steady-state solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.14 for calculating the exclusive effects of the (injected) source F 2 . . . . . . . . . . . . . . . . . . . . . Equivalent electric circuit of Example 7.14 with displacement of the (injected) current source F 2 . . . . . . . . . . . . . . . . . . . . Electric circuit of Example 7.14 for the numerical solution by the ATPDraw program. . . . . . . . . . . . . . . . . . . . . . . . . . . Voltages eðtÞ, ek ðtÞ and current ik ðtÞ accross the switch for this case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltages eðtÞ, ek ðtÞ and current ik ðtÞ accross the switch for this case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voltages eðtÞ, ek ðtÞ and current ik ðtÞ accross the switch for this case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-1 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-2 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-3 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-4 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-5 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-6 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-7 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-8 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-9 . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-10 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-11 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-12 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-13 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-14 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-15 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-16 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-17 . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-18 . . . . . . . . . . . . . . . . . . . . ik ðtÞ; iL1 ðtÞ; iC1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . eL1 ðtÞ; eC1 ðtÞ ¼ eL2 ðtÞ; ek ðtÞ; eðtÞ . . . . . . . . . . . . . . . . . . . . . . eL2 ðtÞ; ek ðtÞ; eC1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 627 . . 628 . . 628 . . 629 . . 629 . . 630 . . 631 . . 632 . . 632 . . 636 . . 636 . . 637 . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

638 639 639 639 640 640 640 641 641 641 642 642 643 643 643 644 644 644 645 646 647 647

xxxiv

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

7.96 7.97 7.98 7.99 7.100 7.101 7.102 7.103 A.1 A.2 A.3 A.4 A.5 A.6

Fig. A.7 Fig. A.8 Fig. A.9 Fig. A.10 Fig. A.11 Fig. A.12 Fig. A.13 Fig. B.1

List of Figures

Zoom in iL2 ðtÞ and eL2 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . eL2 ðtÞ; ek ðtÞ ¼ eC1 ðtÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . eL2 ðtÞ; ek ðtÞ; eC1 ðtÞ, for k p ¼ 0 . . . . . . . . . . . . . . . . . . . . . . . . eL2 ðtÞ; ek ðtÞ; eC1 ðtÞ, for k p ¼ 10 . . . . . . . . . . . . . . . . . . . . . . . Electric circuit for Problem P.7-19 . . . . . . . . . . . . . . . . . . . . ek ðtÞ, for k p ¼ 0 in L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ek ðtÞ, for k p ¼ 10 in L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ek ðtÞ, for k p ¼ 10 in L2 , after changing the scale . . . . . . . . . ATPDraw workspace window . . . . . . . . . . . . . . . . . . . . . . . ATPDraw menu window . . . . . . . . . . . . . . . . . . . . . . . . . . . Placing circuit components in ATPDraw window . . . . . . . . Moving circuit components in ATPDraw window . . . . . . . . Connecting circuit elements in ATPDraw window . . . . . . . Circuit components and node name identification in ATPDraw window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Label identification in ATPDraw window . . . . . . . . . . . . . . Picture resulting from the operation Run Plot . . . . . . . . . . . Selecting the variables for plotting and running the Plot program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plotting results from the transient simulation with ATPDraw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Detailed results from the transient simulation with ATPDraw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plotted curves from the transient simulation with ATPDraw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zoom on plotted curves from the transient simulation with ATPDraw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ilustrative representation of singular functions . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

648 648 649 649 650 650 651 651 654 654 655 656 656

. . 657 . . 658 . . 659 . . 660 . . 660 . . 661 . . 662 . . 662 . . 663

List of Tables

Table 1.1 Table 1.2 Table 1.3 Table 1.4 Table 3.1 Table 3.2 Table 3.3 Table 3.4 Table 4.1 Table 4.2

Table 4.3

Table 4.4

Table 6.1

Resistivity of some materials . . . . . . . . . . . . . . . . . . . . . . . . . Inferred absolute temperature for zero resistance for some common conductive materials . . . . . . . . . . . . . . . . . . . . . . . . Approximate dielectric constants of some materials . . . . . . . . Duality of electrical elements . . . . . . . . . . . . . . . . . . . . . . . . . Number of positive, negative real roots and complex conjugated roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Number of positive real roots and complex conjugated roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Number of positive, negative or complex conjugated roots . . Location of the range containing at least one real root . . . . . . Comparative synthesis of precision and stability of integration methods for discretizing integral-differential equations [5] . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the trapezoidal integration method . . . . . . . . . . . . . . . . . . . . . . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the Backward Euler integration method . . . . . . . . . . . . . . . . . Digital solutions at each discrete time point for voltage and current through the discretized inductor using the CDA—Critical Damping Adjustment technique . . . . . . . . Values for D and F ðDÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

..

30

.. .. ..

31 36 64

. . 163 . . 163 . . 166 . . 167 . . 222

. . 222

. . 223

. . 224 . . 467

xxxv

Chapter 1

Introduction to Fundamental Concepts in Electric Circuit Analysis

1.1

Introduction

This opening chapter has two main objectives. The first of them, which has a strictly revisionary character, is to approach fundamental concepts of physics, as well as laws, equations and formulas of electricity and electromagnetism. The second is to anticipate, through an example, the real raison d'être of this book, which is the solution, both analytical and digital, of transient phenomena in linear, invariant and concentrated circuits. This chapter presents then the main elements of an electrical circuit with concentrated parameters that do not vary with the frequency, as well as the dynamic response of basic circuits when disturbances occur, such as energizing or de-energizing switches, enabling a physical understanding of the phenomena and the need for mathematical modelling and computer simulations. However, it is assumed that the reader already has fundamental knowledge of analysis of electrical circuits, laws, theorems and methods of solution, as well as of solving ordinary differential equations. Thus, this chapter aims to create the motivations for the study of mathematical and computational techniques and solutions presented in the following chapters. The challenges associated with overvoltages and overcurrents caused by external and internal disturbances to the electrical networks require an understanding of the phenomena, their adequate modelling, simulation and interpretation of the results in the search for solutions that minimize the damage to equipment and that in certain more critical cases, are causing major interruptions (blackouts), with catastrophic social and economic impacts. Fundamental knowledge on circuit analysis [1–9] is essential for a better understanding of the growing complexity of modern power systems. In the emerging scenario of greater penetration of renewable and distributed energy sources (DER—Distributed Energy Resources) in current electrical systems, which tend to become more intelligent electrical networks (Smart Grids), the challenges are even greater and more complex, as they also require considering the © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. C. Goulart de Siqueira and B. D. Bonatto, Introduction to Transients in Electrical Circuits, Power Systems, https://doi.org/10.1007/978-3-030-68249-1_1

1

2

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

phenomena and effects of integrated power electronics, telecommunications and control systems on the performance of the interconnected power grid. The complete understanding of the modelling and simulation of transients in basic electrical circuits is fundamental to the approach aiming at understanding phenomena in more complex circuits, such as, for example, power electronics converter circuits (DC/ DC, DC/AC, AC/AC) used at the interfaces between renewable sources and electrical systems, or transient in HVDC or AC transmission lines in high or extra-high voltage, or transients in electrical machines (transformers, generators and motors) including effects of couplings as well as effects of non-linearity (due to lightning arresters, surge arresters, saturation in transformers, etc.), and also the consideration of the variation of parameters (resistance, inductance) with frequency. For these and several complex studies, there are very advanced technical-scientific books and articles available in the world literature. There are indeed classical books and new ones covering more detailed and extensive subjects about electromagnetic transients in electric power systems. This book fills a gap between classic works in the field of electrical circuits and more advanced works in the field of transients in electrical power systems, facilitating a full understanding of digital and analytical modelling and solution of transients in basic circuits. Therefore, it fulfils its role as an introductory book, but complete in its learning proposal in undergraduate and graduate courses. In Sect. 1.2, the concepts of linear, analogous, dual, time-invariant and concentrated systems are reviewed. In Sect. 1.3, under the generic title of electrical quantities, fundamental ideas associated with the electric charge are reviewed—Coulomb’s law and the concept of electrostatic field are discussed. The important concepts of work and electrostatic potential are remembered, defining, next, a difference in potential or voltage. Then, this section ends dealing with the electric current in all its aspects. In Sect. 1.4, the important concepts of power and energy are reviewed, emphasizing the conventions used to distinguish the supply and absorption of power and energy. In Sect. 1.5, under the title of circuit elements, the electrical resistance is initially addressed—in addition to Ohm’s law, the issue of increasing resistance with increasing temperature in conductive materials is addressed, as well as decreasing it in semiconductor materials. Then a detailed survey is made on capacitors and capacitance, including by discussing, throughout an example, the issue of the between voltage and current for sinusoidal signals, in addition to the issue of polarization of the capacitor dielectric. Before focusing on inductance, a wide approach is made to permanent magnets, taking advantage of the opportunity to introduce the concept of magnetic field and define the magnetic induction vector B and the magnetic flux U. Then, Oersted’s experience and Faraday’s law of induction are discussed, as well as Lenz’s law, used to determine the polarity of the induced electromotive force. Only then is the inductance defined, initially on a coil and then generalizing. Closing the section, comments are made on the voltage and current sources, both independent, controlled or linked. Section 1.6 is dedicated to Kirchhoff’s laws, currents and voltages, discussing their limitations on the applicability to concentrated for elements, as well as on the issue of obtaining linearly independent equations, taking advantage of the definition of branch, node, loop and mesh. In Sect. 1.7, a complete example of energizing a

1.1 Introduction

3

capacitance, linear and invariant over time, it is presented by a constant voltage source, that is, by a battery. The analytical solution is initially presented in great detail using the nodal method to obtain voltage, current, power and energy in all elements of the circuit. Then, the corresponding digital solution of the same circuit is made using the software ATPDraw, for the given numerical values. Finally, Sect. 1.8 presents the conclusions with the main motivations for electromagnetic transients’ studies in modern power systems analysis.

1.2

Preliminary Concepts

System A system is a collection of different interconnected components subject to one or more inputs, also called stimuli or excitations, and which produce one or more outputs, also called responses. In Fig. 1.1a is represented, symbolically, a system subject to an input x(t) and that produces an output y(t). In Fig. 1.1b, as an example, a mechanical system composed of mass, spring and friction is shown, in which the only input is the applied force, and whose output may be the speed resulting from the mass or the potential energy stored in the spring. Note, therefore, that the term output refers to the magnitude to be observed in the dynamics of the system. Model In the development of mathematical methods of analysis of physical systems two steps are fundamental. The first is the mathematical description of the individual components or elements of the system. The second is the establishment of a mathematical description of the effect of the interconnection of the different components, formulating the so-called interconnection laws—in the case of electrical circuits, these are Kirchhoff’s laws. Thus, a given physical system can then be described, approximately, by an idealized model or system. After that, the input– output relationships can then be established, using known mathematical methods. Linear System Systems can be classified by placing restrictions on the input–output relationship. A system is called linear when it satisfies two principles. The first is

Spring Mass

x(t)

System (a)

y(t)

Force

xxxxxxxxxxxx friction (b)

Fig. 1.1 a System input and output; b Mechanical system composed of mass, spring and friction

4

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

that of homogeneity, and the second is that of superposition. The homogeneity principle states that: When the input x(t) of a system is multiplied by a factor k, then the output y(t) is multiplied by the same factor k.

And the principle of superposition establishes the following: If two entries are applied, simultaneously or not, to the same system, then the total response will be the sum of the individual responses to each of the entries.

If y1 ðtÞ and y2 ðtÞ denote the responses corresponding to the independent entries x1 ðtÞ and x2 ðtÞ, then a system will be linear if, and only if, the response to xðtÞ ¼ k1 x1 ðtÞ þ k2 x2 ðtÞ

ð1:1Þ

yðtÞ ¼ k1 y1 ðtÞ þ k2 y2 ðtÞ;

ð1:2Þ

is

for all entries and for all values of the constants k1 and k2 . Note, for example, that for k2 ¼ 0 Eqs. 1.1 and 1.2 speak of homogeneity. And that when k1 ¼ k2 ¼ 1 they speak of superposition. Note, too, that linearity does not necessarily imply that the output time function must have the same waveform as the input. In another way, it can be said that a linear system is one whose behavior is described by linear equations, whether algebraic equations, or differential equations. Therefore, a non-linear system will always be described by non-linear equations, which are much more difficult to solve. The system described by yðtÞ ¼ xðtÞ:½dxðtÞ=dt, for example, is non-linear, because when doubling the input xðtÞ the output yðtÞ is quadrupled, contrary to the property of homogeneity. Analog and Dual Systems Sometimes the same equation describes a number of different physical situations. In such cases, the different physical systems are called analogues. Therefore, two or more systems are analogous when they are described by the same equation. When, however, the analogy occurs between physical systems of the same nature, such as two electrical circuits for example, although they have different physical functioning, they are called dual. Therefore, duality means analogy between two physical systems of the same nature. The systems described by Eqs. 1.3 and 1.4 are analogous, because although the letters are different, mathematically they are saying the same thing, namely: that one quantity is proportional to the derivative of the other. But the systems described by Eqs. 1.4 and 1.5 are dual, since in both the quantities inserted are electric. f ðtÞ ¼ M

d vð t Þ dt

ð1:3Þ

eðtÞ ¼ L

d iðtÞ dt

ð1:4Þ

1.2 Preliminary Concepts

5

iðtÞ ¼ C

d eð t Þ dt

ð1:5Þ

Time Invariant or Fixed System Delaying a function f ðtÞ of a value t0 corresponds to shifting its waveform (curve) to the right of a value equal to t0 and replacing the independent variable t, in the expression that describes it, by t  t0 . An input system xðtÞ and output yðtÞ is called a time-invariant or fixed if its input–output relationship is independent of time. That is, if your input xðtÞ is delayed by t0 , your output yðtÞ will simply be delayed by the same t0 , but maintaining its waveform and amplitude. Thus, for a time-invariant system, the response to input xðt  t0 Þ must be yðt  t0 Þ, as long as the response to xðtÞ is yðtÞ. The system described by the input–output ratio yðtÞ ¼ t dtd xðtÞ is not invariant, since the output corresponding to input xðt  t0 Þ is given by d dxðt  t0 Þ d ðt  t0 Þ dxðt  t0 Þ xð t  t 0 Þ ¼ t ¼ ðt  t0 þ t0 Þ : : ð 1  0Þ dt d ðt  t0 Þ dt d ðt  t0 Þ dxðt  t0 Þ dxðt  t0 Þ dxðt  t0 Þ ¼ ðt  t0 Þ þ t0 : ¼ yðt  t0 Þ þ t0 : 6¼ yðt  t0 Þ: d ðt  t 0 Þ d ðt  t0 Þ d ðt  t 0 Þ

y1 ðtÞ ¼ t

In time-varying systems, the form and amplitude of the response depends on the instant in which the input is applied. Thus, the response to xðt  t0 Þ is not yðt  t0 Þ. Concentrated Systems and Distributed Systems Obviously, systems with concentrated parameters are those consisting of concentrated elements; the same can be said of systems with distributed parameters. The fundamental property associated with the concentrated elements is their small size, compared to the wavelength k of their natural frequency of operation f  with v ¼ kf , where v is the wave propagation speed. As T ¼ 1=f , it can also be written that T ¼ k=v, with the period T corresponding to the time necessary for the wave to propagate a distance exactly equal to the wavelength k. Thus, for a circuit element of length l the propagation time will be s, so that s ¼ l=v. Therefore, the condition for a circuit element of length l to be considered as concentrated is that l  k, or that s  T. In other words, a circuit element is considered to be concentrated when the propagation time of the electrical signal through it can be considered negligible. This means that the actions are instantaneous; there is simultaneity. At any moment, whatever is happening at the beginning of a concentrated circuit element will be happening at the end. In a circuit with concentrated parameters, therefore, the electrical quantities are functions exclusively of time t, and the equations obtained from it are ordinary differential equations. In the case of systems with distributed parameters, as in electrical transmission lines, when propagation time cannot be neglected, and unidirectional propagation is assumed, electrical quantities are functions of distance and time, respectively x and t. Therefore, the equations obtained are partial differential equations.

6

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

In summary, what defines whether a circuit element is concentrated or distributed is its size and the frequency of the signal to which it is subjected, for the same propagation speed. A small cable l ¼ 1½m in length transmitting a sinusoidal signal at frequency f ¼ 500 ½MHz, with a propagation speed close to that of light, v ¼ 300; 000 ½km=s ¼ 3  108 ½m=s, corresponding to a wavelength k ¼ v=f ¼ 3  108 =500  106 ¼ 0:6 ½m, cannot be treated as a concentrated element, since l > k. But, itself, with a signal at frequency f ¼ 60 ½Hz and wavelength k ¼ v=f ¼ 3  108 =60 ¼ 5; 000 ½km must be treated as a concentrated component, because l  k.

1.3

Electrical Quantities

Electric Charge The existence of the electric charge dates from approximately 600 years B.C., when the Greeks observed that amber rubbed with wool could attract small light objects, such as pieces of dry straw, for example. Matter, which is something that has volume and mass, is made up of very small particles called atoms. Every atom is made up of a nucleus and its surroundings, called the electrosphere. Protons and neutrons are located in the nucleus. In the electrosphere, electrons. The electron is the fundamental negative charge of electricity, since it is indivisible. The proton is the fundamental positive charge of electricity. And the neutron, as the name implies, is the fundamental neutral charge of electricity. In their natural state, the atom of each chemical element contains an equal number of electrons and protons, called an atomic number. Since the negative charge of each electron has the same absolute value as the positive charge of each proton, the totals of positive and negative charges are canceled. And an atom in this condition is called electrically neutral or in equilibrium. Electrons orbit the nucleus of the atom in what are called quantized layers of energy, for a total of seven. They are the layers K; L; M; N; O; P; Q. Each energy layer of an atom can contain only a certain maximum number of electrons. In a copper atom, for example, with atomic number 29, 29 protons are electrically neutralized by 29 electrons, which fill the K layer with 2 electrons and the L layer with 8 electrons. The remaining 19 electrons fill the M layer with 18, leaving 1 electron in the N layer. In an equilibrium or stable atom, the amount of energy is equal to the sum of the energies of its electrons. And the energy level of an electron is proportional to its distance from the nucleus. Therefore, the energy levels of electrons in layers farther from the atom’s nucleus are higher than those of electrons in layers closer to it. The electrons located in the outermost energy layer of an atom are called valence electrons. And the outermost energy layer of an atom can contain a maximum of 8 electrons. When external energy is applied to certain materials in the form of heat, light or electrical energy, electrons acquire energy, and this can cause them to move to a

1.3 Electrical Quantities

7

higher energy layer. It is said, then, that an atom in which this happened is in an excited state. An atom in this condition is unstable. When the electron is displaced to the outermost energy layer of an atom, it is minimally attracted by the positive charges of the protons in its nucleus. Therefore, if sufficient energy is applied to the atom, some valence electrons will abandon the atom. These electrons are called free electrons. It is the movement of these free electrons that produces the electric current in the metallic conductors. Atomic electrons are attached to their nuclei by forces that, while reasonably intense, are not insuperable. Therefore, they can be transferred from one body to another when their substances come into contact. Thus, in the process of friction of two bodies, many electrons can be transferred from one to another. When this happens, one of the bodies is left with an excess of electrons, while the other is left with a deficiency of them. The one with excess electrons is negatively charged, and the one with lack of electrons is positively charged. Therefore, the amount of electrical charge that a body has is determined by the difference between the number of protons and the number of electrons that the body contains. It is worth remembering that the electric charge of a body is quantized, that is, that it appears as a whole number of electronic charges. In his experiments, Benjamin Franklin (1706–1790) showed that: 1. The rubbing of cat skin with hard rubber causes the cat skin to transfer electrons to the rubber. Thus, the cat’s skin is left with excess of protons and the hard rubber with excess of electrons. It is said, then, that the cat’s fur is positively charged ð þ Þ and that the rubber is negatively charged ðÞ. 2. Rubbing silk with glass causes the glass to transfer electrons to the silk. Thus, silk has an excess of negative charges and the glass has an excess of positive charges. It is said, then, that the glass is positively charged ð þ Þ and that the silk is negatively charged ðÞ. As described above, Benjamin Franklin defined, in a completely arbitrary way, the charge of the electrified hard rubber as negative and the charge of the electrified glass as positive. These conventions persist to this day, although perhaps it would have been more appropriate to have adopted the inverse convention, as this would avoid the need to work with the conventional direction of electric current, assuming that it is the positive charges that move in the metallic conductors. When a pair of bodies contains the same type of charge, that is, both positive ð þ Þ or both negative ðÞ, the two bodies are said to have equal charges. When a pair of bodies contain different charges, that is, one is positively charged ð þ Þ and the other is negatively charged ðÞ, they are said to have opposite or unequal charges. A fundamental law of electrical charge, called the law of charges, is as follows: Equal charges repel, opposite charges attract.

In addition, in the electrical circuits there is the principle of charge conservation, which means that the total electrical charge remains constant—the electrical charge cannot be generated or destroyed.

8

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

The charge of an electron or proton, while fundamental to electricity, is too small to be the basic unit of charge. Thus, the charge unit in SI (acronym SI, from the French Système International d'unités), the International System of Units, is Coulomb, with the symbol of unit C. Thus, the combined charge of 6:242  1018 electrons is equal to 1 ½C. And that of 6:242  1018 protons is equal to þ 1 ½C. Conversely, the charge of an electron is 1:602  1019 ½C and that of a proton is þ 1:602  1019 ½C. Coulomb’s Law The French physicist Charles Augustin de Coulomb (1736–1806), in 1784, established the law that quantitatively expresses the force between two point charges, that is, charges of infinitesimally small bodies, which contain mass, but with negligible volumes. He found that the force between the charges q1 and q2 is directly proportional to the modulus of each of the charges and inversely proportional to the square of the distance between them. He also noted that the force is directed along the straight line that passes through the point charges. Logically, the meaning is given by the law of charges. Mathematically speaking, Coulomb’s law is expressed by Eq. 1.6. F¼k

q1 q2 d2

ð1:6Þ

If q1 ¼ q2 ¼ q ¼ 1 ½C and d ¼ 1 ½m, then F ¼ 8:98742 ½N. Therefore, the proportionality constant is k ¼ 8:98742 ½N m2 =C2 . Thus, one can define, indirectly, 1 ½C with that total punctual charge that when separated by 1 ½m of a similar charge generates a force of 8:98742 ½N. It must be emphasized that Coulomb’s law applies strictly only to point charges. They are, however, necessary to deduce the expressions of forces between large distributions of charges or in large bodies. Furthermore, Coulomb’s law is universal and also applies to the force between two electrons or between an electron and a proton. Electrostatic Field The electric charge has a very important property, which is the ability to create a vector field of forces in the space that surrounds it, and that field, in turn, transmits forces to other bodies also electrically charged immersed in it and, therefore, affect their movements. Therefore, in the case of electric fields, electric charges constitute the sources of these fields. Thus, the utility of the electric field comes from its action as an intermediate agent in the transmission of forces from one or more charged bodies to others. The electrostatic field refers to the electric field created by charges at rest. Moving electric charges can also establish magnetic fields in space and these, in turn, can exert forces on charged objects in motion. At any spatial point, characterized by its orthogonal Cartesian coordinates ðx; y; zÞ, the electric field vector E is defined as the quotient of the force F acting on a positive test charge q and itself. Like this,

1.3 Electrical Quantities

9

Eðx; y; zÞ ¼

Fðx; y; zÞ q

ð q ! 0Þ

ð1:7Þ

The observation that the test charge q is infinitesimally small ðq ! 0Þ is intended to alert that, in a measurement process, it cannot influence the charge (s) that created the force field. In other words, you need to ensure that E is always independent of q. An alternative way of expressing this requirement is to define the electrostatic field as E¼

lim F q!0 q

ð1:8Þ

Evidently, in SI, the electric field unit is ½N=C. Example 1.1 A point charge þ Q is located at the origin of an orthogonal Cartesian coordinate system, as shown in Fig. 1.2. Find the electrostatic field it produces.

Fig. 1.2 Force F acting on q according to Coulomb’s law

z q

F=

KqQ r2

ir

r

ir Q

y

x Solution First, it is necessary to obtain the force that a test charge q located at a distance r from the origin experiences due to the charge þ Q. According to Coulomb’s law, Eq. 1.6, the force F acting on q is given by F¼k

qQ qQ r ir ¼ k 2 : 2 r r r

ð1:9Þ

10

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

In Eq. 1.9, ir ¼ rr is a unit vector pointing outwards, from the origin, where the charge þ Q is located, towards the observation point, where the test charge q is located. So, according to Eq. 1.7, E¼

F Q ¼ k 2 ir q r

ð1:10Þ

Therefore, the electrostatic field E also points radially outward, and is spherically symmetrical, since its modulus depends only on the distance r from the origin, for a given generator charge þ Q. And that its modulus varies in inverse proportion to the square of the distance r. It was the British physicist Michael Faraday (1791–1867) who proposed the adoption of the so-called field lines or lines of force, as a geometric means of representing the electric field, and which allow to illustrate how the electric fields behave. The construction of these field lines follows the rules: 1. The lines of force are drawn and oriented in the direction of the force that a positive test charge would experience at each spatial point in the field. 2. The density of the field lines, expressed by the number of lines crossing a unit of area perpendicular to their direction, is a measure of the intensity of the electric field. Thus, higher density of lines of force means more intense electric field. 3. The field lines can never intercept. As the force at any point can have only one direction, it is evident that the field lines can never intersect. Figure 1.3 shows the field lines of the point charge þ Q in Example 1.1, which extend radially out of the point load. It is easy to intuit that the number of lines per unit area varies inversely with the square of the distance from the charge (origin). Knowing the electric field produced in the whole space by a single charge, as in Example 1.1, or due to a distribution of charges, the Coulomb force exerted on any point charge q can always be calculated using the expression F ¼ qE. When two electrically charged bodies, with opposite polarities, are placed close to each other, constituting what is called an electric dipole, the electrostatic field is concentrated in the region between them, as indicated by the lines of force in the illustration shown in Fig. 1.4. It can be demonstrated that the electric field produced outside a hollow spherical shell, of very thin thickness, and of radius R, with a uniform charge density, totaling a charge þ Q, presents an electric field outside the sphere, at a distant point r from its center, given exactly by Eq. 1.10. On the other hand, the electric field at any point within the spherical shell charged with uniform charge density is identically zero. It is also possible to demonstrate that any spherically symmetric charge distribution produces an electric field outside the charge distribution identical to the field that would be produced if the entire charge were placed in the center of the charge distribution.

1.3 Electrical Quantities

11

Q

Fig. 1.3 Force lines of the point charge þ Q

+

+

-

Fig. 1.4 Electrostatic field in an electric dipole

Regarding the electrostatic field, it can be added that: (a) The electric field within any conductor in equilibrium is zero. (b) Any total charge produced by a conductor resides on its surface. (c) The electric field exactly outside a conductor is normal to its surface at each point. Voltage or Potential Difference Work Mechanical work can be performed by any moving force. Therefore, if there is force but there is no movement caused by it, there is no work. In this way, a person

12

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

standing with his arms on a concrete wall is not doing work. In other words, when there is force exerted, but it does not produce displacement, there is no work done. The total work done by a constant force F that moves through a straight displacement r is a scalar quantity W defined by W ¼ F  r  cos h;

ð1:11Þ

where h is the angle between the directions of F and r. So, if F and r have the same vector direction, h ¼ 0 and W ¼ F:r. Obviously, in SI the unit of work is [N.m], also called joule, [J]. As the term energy, in general, refers to the ability to do work, its unit is also joule. Equation 1.11 can also be written in vector language as W ¼Fr

ð1:12Þ

In the most general case, when the force is not constant during the displacement and the displacement itself is not straight but rather a trajectory C, which extends from a point P1 ðx1 ; y1 ; z1 Þ to a point P2 ðx2 ; y2 ; z2 Þ, as illustrated in Fig. 1.5, the total displacement r can be considered to be the vectorial sum of infinitesimal displacements dr along the path C, in each of which the force F can be considered as constant. At any point P, for example, the infinitesimal work dW associated with displacement dr, according to Eqs. 1.11 and 1.12, is dW ¼ F  cos h  dr ¼ F  dr The number of infinitesimal working elements dW can then be added along the entire path C giving the following result: Z W¼

F  dr

ð1:13Þ

C

Given the vectors A ¼ Ax ix þ Ay iy þ Az iz and B ¼ Bx ix þ By iy þ Bz iz , where ix ; iy ; iz are the unit vectors of the coordinate axes of the orthogonal Cartesian system, their scalar product is given by Fig. 1.5 Infinitesimal work dW associated with displacement dr

F P C

P1

dr

P2 r

1.3 Electrical Quantities

13

A  B ¼ Ax Bx þ Ay By þ Az Bz So, if the scalar product F  dr is expressed in terms of the components of vectors F and dr, the integral of Eq. 1.13 can be separated into three, as follows: Z W¼

ZP2 F  dr ¼



Fx dx þ Fy dy þ Fz dz

P1 Zx2

C

¼

Zy2 Fx dx þ

x1

 ð1:14Þ

Zz2 Fy dy þ

y1

Fz dz z1

Example 1.2 A body moves along the ABC path in the xy plane shown in Fig. 1.6.  The force F ¼ k xix þ 2yiy þ 0  iz ½N, where k ¼ 2 ½N=m, acts on the body during its displacement. Find the work done on it by moving it from A to C. Fig. 1.6 Body moviment along the ABC path in the xy plane

y [m] 2

C

B

1

A x [m]

0

1

2

4

Solution Applying Eq. 1.14 to segment AB, it follows that: Z2 WAB ¼

Z2 Fx dx þ

1

Z2 Fy dy ¼

1

Z2 2xdx þ

1

2 2 4ydy ¼ x2 1 þ 2y2 1 ¼ 3 þ 6 ¼ 9 ½J:

1

For the BC segment: Z4 WBC ¼

Z2 Fx dx þ

2

Z4 Fy dy ¼

2

4 2x ¼ x2 2 ¼ 12 ½J:

2

The total work done on the body when it is moved by the force F along the path from A to C is the sum of the two contributions:

14

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

W ¼ WAB þ WBC ¼ 9 þ 12 ¼ 21 ½J: The potential energy of a system is the energy that it can have due to its position or displacement. In other words, the total work that a body can do on the environment by virtue of its position is called potential energy. Thus, the potential energy of a mass m that is raised to a height h above the earth’s surface, considered as a reference, is m:g:h, since it is capable of carrying out this work when it descends to it, where g ¼ 9:81 ½m=s2  is the acceleration of Earth’s gravity. Forces such as frictional forces, for which it is not possible to specify a defined potential energy, are called non-conservative forces. Therefore, there are forces that do not accumulate potential energy and from which work cannot be recovered by releasing the accumulated potential energy. So, it is impossible to associate a potential energy with a non-conservative force. Conservative systems are those in which only conservative forces operate. Conservative forces have the following properties: 1. The work performed by a conservative force when moving from point A to point B along a given trajectory is the negative of the work performed by the same force on the return through the same trajectory, that is, from B to A. 2. The work carried out by a conservative force when moving from point A to point B is the same, regardless of the path taken to go from A to B. 3. The work performed by a conservative force when moving along a closed path is zero. 4. The work performed by a conservative force when moving from A to B is UP ðBÞ  UP ð AÞ, where UP ðx; y; zÞ is the potential energy associated with the force at any point of the orthogonal Cartesian coordinates x; y; z. An infinitesimal element of potential energy associated with a conservative force F ¼ Fx ix þ Fy iy þ Fz iz is given by dUP ðx; y; zÞ ¼ Fx dx  Fy dy  Fz dz So, Zx UP ðx; y; zÞ ¼ 

Zy Fx dx 

0

Zz Fy dy 

0

Fz dz;

ð1:15Þ

0

where the lower limit of the integrals was taken equal to zero due to the origin of the coordinate system being taken as a reference, that is, that UP ð0; 0; 0Þ ¼ 0. Electrostatic Potential Suppose that a fixed flat plate, of small thickness and large horizontal dimensions, with a uniform distribution of positive electrical charges—such as a glass plate rubbed with a silk scarf—presents, on one of its faces, an electrostatic field E,

1.3 Electrical Quantities

15

Fig. 1.7 The electrostatic potential of point P

+ + + + + + + +

0

p + q

F

r1

E

(∞)

q +

F=0 Reference

represented by their lines of force parallel and directed downwards, as illustrated in Fig. 1.7. Obviously, a test load þ q positioned at any point P within this field will experience a Coulomb force that will move it vertically downwards to a point where the force acting on it is so weak that it can be considered to be zero. This point— actually, plane—where the force no longer acts on the charge þ q and, therefore, no longer performs work on it, is called a reference. Mathematically, it is customary to consider that this reference is located at infinity. Therefore, the work done by the field to move the charge þ q from the initial point P to the reference, according to Eq. 1.13, is given by Z WP ¼

Z1 F  dr1 ¼

F  dr1

ð1:16Þ

P

C

This Eq. 1.16 also represents the energy that was expended by the electric field to carry out the work of moving the charge þ q from point P to the reference. As the electrostatic field is a conservative field, it is therefore a potential energy. The electrostatic potential of point P, in Fig. 1.7, is defined as the quotient of the potential energy of point P by the charge q, and called eP . Then, WP 1 eP ¼ ¼ q q

Z C

1 F  dr1 ¼ q

Z1 F  dr1 P

ð1:17Þ

16

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

As F ¼ qE; then Eq. 1.17 can be rewritten as Z1 eP ¼

E  dr1

ð1:18Þ

P

Consider the same situation described earlier, but now with the inversion of the direction of displacement, from r1 to r, and imagine two charges þ q being displaced against the electrostatic field E, from the reference, one to position A and one to position B, as shown in Fig. 1.8. So, the potential energy of each of these points is Z eA ¼ 

A

Z eB ¼ 

B

1

1

E  dr

ð1:19Þ

E  dr

ð1:20Þ

By the positions assumed for points A and B, eA [ eB and eA  eB [ 0. The difference between the electrostatic potentials at points A and B is what is called the voltage between A and B, designed as eAB . Therefore, 2 A 3 Z ZB eAB ¼ eA  eB ¼ 4 E  dr  E  dr5 2 ¼ 4

1

ZB

1

Fig. 1.8 Two charges þ q being displaced against the electrostatic field E, from the reference, one to position A and the other one to position B

1

ZA E  dr þ

ZB E  dr 

B

3 E  dr5

1

r + + + + + + + + 0 A + q

F

B q+

F

E q + Reference

q +

(- ∞)

1.3 Electrical Quantities

17

So, ZA eAB ¼ eA  eB ¼ 

E  dr ¼

WA WB  q q

ð1:21Þ

B

As the potential energy is expressed in joules and the electric charge in coulombs, then the voltage, like the electrostatic potential, has the unit of joules per Coulomb, also called volts. Therefore, eAB ½J=C ¼ eAB ½V. It is evident, then, that þ 1 ½C of charge produces 1 ½J of work (energy) in its environment, when suffering a potential decrease of 1 ½V. Swapping A with B in Eq. 1.21, it follows that RB RA eBA ¼ eB  eA ¼  E  dr ¼ E  dr ¼ eAB . A

Therefore,

if

eAB ¼ 6 ½V,

B

eBA ¼ 6 ½V. This is a consequence of the electrostatic field being a field of conservative forces. Once the voltage unit in volts is established, it is possible to establish an equivalent unit for the electric field, since         V J=C Nm N ¼ ¼ ¼ m m Cm C It is more common to use the unit volts per meter than newtons per coulomb for the electric field. Since the voltage is defined as the difference between electrostatic potentials, according to Eq. 1.21, and not in absolute value, one can choose the reference, where it is assumed that the electrostatic potential is zero, anywhere, and not necessarily at infinity. It is for this reason that in the solution of electrical circuits by the nodal method it is common to adopt a node of the circuit itself as a reference, to which the grounding symbol is assigned, with zero potential, and to measure the potentials of the other nodes in relation to it, calling them as nodal voltages. In this book, the voltage will always be represented by an open arrow at the tip ð Þ, where the tip refers to the highest potential ð þ Þ and the tail refers to the lowest potential ðÞ. An alternative and indirect way to define voltage or potential difference in any circuit element is: The elementary energy dW absorbed by a two-terminal circuit element, when a differential amount of charge ð þ Þ dq moves through it, from the tip to the tail of the voltage reference arrow e, is given by

18

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

dW ¼ e  dq

ð1:22Þ

or e¼

dW dq

ð1:23Þ

Therefore, a negative dW value indicates that power is being supplied by the circuit element. Then, the sign of eðtÞ is positive if, and only if, the circuit element is absorbing energy when positive charges move from the tip to the tail of the voltage reference arrow. Example 1.3 Work that consumed 110 ½J was performed to move 4:575  1018 electrons from one point to another in an element of an electrical circuit. What is the magnitude of the voltage that was established between these two points? Solution For this case, Eq. 1.23 can be written as e ¼ W=Q: So, j ej ¼

  J h i ¼ 150:09 ffi 150½V: C 4:575  1018 [e]  1:602  1019 C e 110 ½J]

Electric Current It was the invention of voltaic batteries by Alessandro Volta (1745–1827), now called batteries, that made it possible to chemically generate differences in electrical potentials, capable of producing greater fluxes of charges continuously through metallic conductors. As a consequence, the discoveries that were made since the beginning of the nineteenth century were notable. When a charge flow occurs within conductors, the conditions within the conductive material are no longer those of electrostatic equilibrium. Therefore, it is no longer valid to consider the electric field to be zero everywhere in the conductor. On the contrary, the electric field must be different from zero to establish and maintain the flow of charges. The term electrical current means charges in motion. In a metallic conductor, as already mentioned, the current is formed by the movement of so-called free electrons. Whenever an external force acts on these free electrons of a conductive substance, as occurs when there is an imposition of an electric field, these negative free charges will move, producing, then, the electric (electronic) current. The positive charges of the conductor are trapped in the nuclei of the conductor atoms and will never move freely. In a copper conductor, for example, approximately 8:5  1028 electrons per cubic meter are free to move. This is equivalent to 1:36  1010 ½C of charge that can be moved in each cubic meter.

1.3 Electrical Quantities

19

Conventional flow Electrons flow Conductive wire

-

Free electrons in movement

-

-

-

-

-

-

-

-

-

-

+

- + Battery Fig. 1.9 Electric current produced by the potential difference of the battery applied to both ends of the conductor

Figure 1.9 shows the situation where a lead wire is connected to a battery. The potential difference of the battery applied to both ends of the conductor produces the electric current. As shown in Fig. 1.9, the direction of movement of the electrons is from the negative battery terminal, passing through the conductive wire, and back to the positive battery terminal. In other words, electrons are repelled by the battery’s negative terminal and attracted to its positive terminal—and thus the battery is discharged. The direction of electron flow is from a negative potential point to a positive potential point. The continuous arrow in Fig. 1.9 indicates the direction of the current as a function of the flow of electrons. The direction of the supposed movement of the positive charges, opposite to that of the electron flow, is considered as the conventional current flow and is indicated by the dashed arrow in Fig. 1.9. Any circuit can be analyzed either through the flow of electrons or the conventional flow in the opposite direction. In this book, the current will always be considered in the conventional direction. Had the electron charge been considered positive and the proton charge negative, there would be no need to define a conventional current direction. Let Dq be the amount of positive moving charge passing through a given point P of a conductor wire over a period of time Dt. So, the electric current I is defined by I¼

Dq Dt

ð1:24Þ

For currents that vary with time, the instantaneous current iðtÞ is defined as the rate (speed) at which the charge moves passing through point P, and whose unit in SI is expressed in coulombs per second, called Ampère. Then, mathematically,

20

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

i ðt Þ ¼

lim Dq dq ¼ Dt ! 0 Dt dt

ð1:25Þ

Equations 1.24 and 1.25, alone, are not sufficient to completely define an electric current. This is because it is necessary to clarify whether positive electrical charges are moving, in the conductor, from end A to end B, or from end B to end A. To add this information, then a reference arrow is used together with the current designation iðtÞ. The reference arrow then represents the direction assumed to be positive for current iðtÞ. In this book, the reference will always be represented by an arrow with a small triangle at the end, as shown in Fig. 1.10. Therefore, the current iðtÞ will be positive if the positive charges are moving in the direction of the arrow, from A to B. If, however, the positive charges are moving in the opposite direction to the reference arrow, that is, from B to A, then the current i(t) is negative. Thus, a current of 10 ½A, from A to B, is equivalent to a current of 10 ½A, from B to A. A current is generally positive during some time intervals and negative during others. And the outline of a current versus time, as shown in Fig. 1.11, is called the current waveform. The current reference arrow can be fixed in any direction, as long as its waveform is consistent with the direction chosen for the reference arrow. Therefore, if the waveform of current iðtÞ in Fig. 1.11 refers to the current in Fig. 1.10, then the positive charges move to the right at a constant rate of 2 ½C=s ¼ 2 ½A, during the time interval 0\t\1 ½s, and to the left, at a constant rate of 1 ½C=s ¼ 1 ½A, during the time interval 1 ½s\t\3 ½s. According to Eq. 1.25, if the charge is qðtÞ ¼ Q ¼ constant, then the current iðtÞ ¼ 0½A. This means that there is no electric charge movement in the conductor, but only a body charged with Q charge. A permanently constant current is called direct current. And a current whose module varies continuously with time, and whose direction is periodically inverted is called alternating current. If the current is the derivative of the charge, as expressed by Eq. 1.25, then, inversely, the charge is the integral of the current, i.e. Z qð t Þ ¼

iðtÞdt

ð1:26Þ

But an indefinite integral requires the determination of an integration constant, which is somewhat abstract and is not very practical in determining the waveform of the charge from the current waveform. Fortunately, an indefinite integral can be replaced by a well-defined definite integral, according to the following rule:

B

i(t) A

+ i(t)>0

B

i(t) A

+ i(t)0 p=+e.i >0 (a)

Energy supply dW < t eðtÞ ¼ ðt þ 3Þ=2 > > : 0

for t 0 for 0 t 1 for 1 t 3 for t 3:

So, the current, obtained from iðtÞ ¼ C dedtðtÞ, is 8 0 > > < C iðtÞ ¼ C=2 > > : 0

for t\0 for 0\t\1 for 1\t\3 for t [ 3:

The waveform of current iðtÞ is shown in Fig. 1.22b. The power, obtained from pðtÞ ¼ eðtÞ  iðtÞ, is

1.5 Circuit Elements

45

8 0 > > < Ct pð t Þ ¼ Cðt  3Þ=4 > > : 0

for t 0 for 0 t\1 for 1\t 3 for t 3:

The waveform of the power pðtÞ is recorded in Fig. 1.22c. And the energy, calculated using Eq. 1.72, is given by 8 0 > > < 2 Ct =2 W ðt Þ ¼ Cðt þ 3Þ2 =8 > > : 0

for t 0 for 0 t 1 for 1 t 3 for t 3:

The energy at capacitance C is recorded in the curve of Fig. 1.22d. It is interesting to note that: 1. In the interval 0\t\1 the current iðtÞ, with the reference indicated in Fig. 1.21, moves to the right; however, in the interval 1\t\3 it is worth half the previous value and is directed to the left, because it has changed signs. 2. In the interval 0 t\1, the power pðtÞ is positive, meaning that the capacitance is absorbing power; in interval 1\t 3 the power is negative, meaning that the capacitance is supplying power—obviously, for the rest of the circuit of which it is part. 3. In the range 0 t 1, the capacitance is storing energy; however, in the range 1 t 3 it is returning all the energy it has accumulated at the end of the first interval. When N capacitors are connected in parallel, intuitively, one can imagine that another capacitor is formed, in such a way that the area of their plates is the sum of the individual areas of each plate and that, thus, an equivalent capacitance can be obtained through the sum of the individual capacitances. Therefore, for N capacitances in parallel, the equivalent capacitance is Ceq ¼ C1 þ C2 þ    þ CN

ð1:74Þ

So, the equivalent capacitance of N capacitances in series is obtained from 1 1 1 1 ¼ þ þ  þ Ceq C1 C2 CN

ð1:75Þ

For only two capacitances in series, Eq. 1.75 provides that Ceq ¼

C1  C2 ; C1 þ C2

ð1:76Þ

which is the famous rule of the product by the sum, also used to determine the equivalent resistance of two resistances in parallel.

46

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

Inductance Magnets The phenomena related to magnetism were discovered by the Chinese surely more than 4000 years ago. At least 3000 years ago, the magnets used in Chinese primitive compasses were pieces of an iron ore known as magnetite, or black iron oxide, Fe3 O4 , which has the property of attracting small pieces of iron. And since magnetite has such magnetic properties in its natural state, these pieces of ore are called natural magnets. In addition to these, the only other natural magnet that exists is the Earth itself. All other magnets are man-made and are, therefore, called artificial magnets or electromagnets. The regions of a body where magnetism seems to be more concentrated are called magnetic poles. Every magnet has exactly two opposite poles, which are designated the North pole ðN Þ and the South pole ðSÞ. Just as with electrical charges, which repel each other when they are equal and attract each other when they are opposite, equal magnetic poles repel each other and different poles attract each other, as shown in Fig. 1.23. When two north poles are placed next to each other, the magnetic force lines that come out of the north poles have opposite directions and, consequently, repel each other. This repelling force tends to drive the two magnets apart. On the other hand, when the north pole of a magnet is approached with the south pole of another, the magnetic force lines of the respective fields have concordant directions and come together to form loops of longer force lines. These long continuous lines tend to contract, and this force of attraction brings the two magnets together. In electrostatics there is no difficulty in producing charged bodies, either positively or negatively, as demonstrated by the experiences of Benjamin Franklin. It is also possible to break an electric dipole in two and then obtain a positively charged and a negatively charged end. The magnets, on the other hand, invariably have north and south poles of exactly the same intensity. Therefore, they are essentially dipoles. In addition, when a bar of a magnet is cut in half, an isolated north pole or an isolated south pole is not obtained, but two halves, each with their respective north and south poles of exactly equal intensities. In other words, cutting a magnet in half gives rise to two other magnets, as shown in Fig. 1.24, and not two halves, each with its respective pole. Therefore, there are no free magnetic charges; only in pairs.

Fig. 1.23 Repulsion or attraction between poles of magnets

1.5 Circuit Elements

47 N

N

N

N S

......

...... S

S N S

S N

S

Fig. 1.24 Magnetic dipoles

The right arrow in Fig. 1.24 represents an elementary magnetic dipole, that is, at the atomic level. And the diamond on its left is usually used to represent the needle of a compass. The magnetic forces exerted by permanently magnetized substances, such as magnetite, on objects made of iron constitute what is called a magnetic field. A vector field of forces that, although invisible, can be seen through the effects of its forces, spreading iron filings on a glass plate or sheet of paper placed on a bar-shaped magnet. After that, when the glass plate or paper sheet vibrates lightly and repeatedly, the grains of the filings will be distributed in a well-defined configuration, which describes the magnetic field of forces around the magnet. Another way of delineating the magnetic induction field B is to observe the direction that a small, freely suspended magnetic dipole, such as a compass needle, takes when moving at different points in the vicinity of a bar-shaped magnet, as shown in Fig. 1.25.

Fig. 1.25 Magnetic induction field B

N

B

S

48

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

The stronger the magnet, the greater the number of magnetic force lines and the greater the space covered by the magnetic field. The lines of force in the magnetic field are always closed, that is, continuous. They have no point of origin or termination, like the lines of force in the electric field. Magnetic induction B is defined as being a force field that has the same properties in relation to “the magnetic charge” that the electric field E has in relation to the electric charge—or would have, if the magnetic charge, as such, existed because, unlike the electric charge, there is no isolated magnetic pole, as can happen with the electric charge. The only source of any magnetic field is electrical current, even in natural magnets. It is defined in SI by the unit called Tesla, with the unit symbol ½T. Magnetic materials are those that are attracted or repelled by a magnet and that can be magnetized by it. Iron and steel are the most common magnetic materials. Magnetic materials are also nickel, cobalt and the alloy with 78:5% nickel and 21:5% iron, called permalloy. Permanent magnets are formed by hard magnetic materials, such as cobalt steel, which keeps its magnetism even when the magnetizing field is moved away. Magnetic Flux Not very strictly, the magnetic flux / is defined as the total number of lines of force of the magnetic induction field B that crosses a given surface S. (It is not very accurate because each line “pierces” the surface at a different angle with it, at the point where it crosses, in addition to the fact that it is not necessarily flat). Thus, at the flat top of the north pole of the magnet in Fig. 1.25, where each line is perpendicular to it, the magnetic flux is the set, or bundle, of all magnetic force lines that emerge from the area of the top of the north pole of the magnet, and enter it by the south pole. His SI unit is Weber, with the unit symbol ½Wb , in honor of the German physicist Wilhelm Weber (1804–1891). The precise definition of magnetic flux /, through a given surface S, is the integral of the component of B normal to the surface over the area established by S. Figure 1.26 shows any surface S and, over it, at any point P, an infinitesimal element of area da, in addition to a unit vector n perpendicular to the element of the surface at that point P. In general, the magnetic induction B is variable in modulus, direction and sense over the entire surface S, but, for point P, it is possible to consider the area element da as being flat and the vector B representing the local value of the magnetic field at that point P. The component of B perpendicular to the surface at this point P is nothing more than the component of B along the direction of the normal vector n, whose value is B  cos h. However, since n has a unitary module, the scalar product B  n also has the value B  cos h. Then, the surface-normal component of B at point P can be written as Bn ¼ B  cos h ¼ B  n

ð1:77Þ

1.5 Circuit Elements

49

Fig. 1.26 Magnetic induction field at point P

Bn B n

0

P da

S Therefore, the infinitesimal element of magnetic flux through the smallest surface area da is defined as the perpendicular component of the magnetic induction field at the surface times the area da, that is, d/ ¼ Bn  da ¼ B  n da

ð1:78Þ

To find the total magnetic flux through the total surface S of Fig. 1.26, it is only necessary to integrate all the infinitesimal elements of area that make up the surface. That way, Z / ¼ B  n da ð1:79Þ S

Usually the task of performing the integral in Eq. 1.79 is extremely difficult and time-consuming. The work, however, is much easier if the magnetic induction B is constant in modulus, direction and sense at all points on the surface S and if, in addition, it is planar, so that the normal vector n is also constant in all points. Under these conditions, the quantity B:n is the same for all surface area elements and can therefore be written outside the integral of Eq. 1.79, that is, Z /¼Bn

da ¼ ðB  nÞA ¼ ðB  1  coshÞA S

or / ¼ B  A  cos h;

ð1:80Þ

where A represents the total surface area S. If, in addition, the direction of B is normal to the surface S, the angle h between the vectors B and n will be zero and Eq. 1.80 will become

50

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

/¼BA

ð1:81Þ

Based on Eq. 1.81, the magnetic induction unit is more commonly called webers per square meter, with the unit symbol ½Wb=m2 , instead of Tesla. Another unit of magnetic induction is ½N=Am, resulting from F ¼ qðv  BÞ ¼ q  v  B  senh, which is the expression of the magnetic force in a conductor carrying current— charge q moving with speed v, immersed in an external magnetic field B, as shown in Fig. 1.27. Fig. 1.27 Magnetic force in a conductor carrying current— charge q moving with speed v, immersed in an external magnetic field B

F

B

0

q +

0 Right hand rule

v

Example 1.15 Find the magnetic flux through a circular surface of 32 ½cm in diameter, whose normal makes an angle of 20 with a constant magnetic induction of 0:6 ½Wb=m2 , as illustrated in Fig. 1.28.

Fig. 1.28 Magnetic flux through a circular surface

n

B 20º

0,16 m

S

1.5 Circuit Elements

51

Solution From Eq. 1.80, / ¼ B  A  cos h ¼ ð0:6Þ  pð0:16Þ2 ð0:940Þ ¼ 0:045½Wb If the magnetic field were perpendicular to the plane of the circle then h ¼ 0 and / ¼ ð0:6Þ  pð0:16Þ2 ¼ 0:048½Wb: The Oersted Experiment In 1819 the Danish scientist Hans Christian Oersted (1777–1851), using the voltaic cell capable of establishing a direct current, discovered that the needle of a compass was deflected in the vicinity of a conductor in which an electric current was flowing. And the immediate conclusion was that an electric current flowing through a conductor produces a magnetic field around it. More than that, Oersted proved that magnetism and electricity are related phenomena. In Fig. 1.29a the iron filings, when forming a well-defined configuration of concentric circumferences around the conductor, through which a current circulates, show the existence of the magnetic field created by it, so that each section of the wire has around it this field of magnetic forces in a plane perpendicular to the conductor, as shown in Fig. 1.29b. The intensity of the magnetic field around a conductor carrying an electric current depends directly on the intensity of that current. The right hand rule is a very convenient way of determining the relationship between the direction of current flow in a wire and the direction of the lines of force in the magnetic field around the conductor. It is as follows: suppose you are holding the wire that conducts the current with your right hand, with the four adjacent fingers around the wire and with your thumb extended. If the thumb extended along the wire indicates the direction of the current flow, then the other four fingers will indicate the direction of the magnetic force lines of the field around the conductor. This rule is illustrated in Fig. 1.30. If the conductive wire forms a closed path, such as the circumference conducting the current i, shown in Fig. 1.31, the right hand rule should be used inverting the previous description, that is, the adjacent fingers indicating the direction of the current and the thumb, the direction of the field, in Fig. 1.31, represented by flow /. If the conductive wire is wound with N circumferences, called turns, forming what is called a coil, as shown in Fig. 1.32, and if, in addition, an iron core, called ferromagnetic material, is placed inside it, it increases the concentration of flux within it, that is, the lines of force are approached in the the magnetic field inside the coil. Permeability, with the quantity symbol l, is a measure of this flow-intensifying property; l creates a channel for flux /. It has the unit of Henry per meter, with the symbol ½H=m in the SI. The permeability of the vacuum, designated l0 , is worth 4p  107 ½Wb=A m ¼ 0:4 ½lH=m. The permeability of

52

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

Fig. 1.29 a Electric current flowing through a conductor producing a magnetic field around it; b Field of magnetic forces in a plane perpendicular to the conductor [9]

Fig. 1.30 Right hand rule determining the relationship between the direction of current flow in a wire and the direction of the lines of force in the magnetic field around the conductor [9]

1.5 Circuit Elements

53

φ i

Fig. 1.31 Adjacent fingers indicating the direction of the current and the thumb, the direction of the field, for conductive wire in a closed path

1

φ

2

3

4

S

N

i

N

l

A

i

Fig. 1.32 Magnetic flux produced by a current across a coil with N turns and with ferromagnetic core

other materials is related to that of vacuum through a factor called relative permeability, with the symbol lr . Such a relationship is l ¼ lr :l0 , that is, lr ¼ l=l0 . Most materials have a relative permeability close to 1, but pure iron has a range of 6; 000–8; 000 and nickel has a range of 400–1; 000. The so-called permalloy has a relative permeability of around 80; 000. The relative permeability can also be expressed by lr ¼ 1 þ vm , where vm is the magnetic susceptibility. An interesting fact that is worth remembering is that hmi hmi 1 1 ffi ¼ 299:863:380; 4 ffi 3  108 pffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e0 l 0 s s 8:85  1012  4p  107   km ¼ 300; 000 ; s which is the speed of propagation of electromagnetic phenomena in free space. When all the N turns of a coil are associated with the same amount of flux /, it is said that the coil has an interlaced or concatenated flow, or an inductive coupling N/.

54

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

Magnetomotriz force— fmm is the product of the current by the number of turns of a coil. It is the “pressure” required to establish a magnetic flux in a ferromagnetic material. Its unit is ½At, which means ampere-turns. In fact, only amperes, because N is a dimensionless number. fmm ¼ N  i

ð1:82Þ

The intensity of the magnetic field is defined by the fmm quotient by the distance between the poles of a coil. Its unit symbol is H and its unit is ½At=m. So, H¼

Ni l

ð1:83Þ

H is the field at the center of a solenoid air core. With an iron core, H is the magnetic field through the entire core. Electromagnetic Induction Shortly after the Oersted experiment, the French André Marie Ampère (1775–1836) demonstrated that there is a linear relationship, that is, a line passing through the origin, between the magnetic flux / and the current i that produced it, valid for space free. In an iron core coil, however, the relationship between the magnetic flow and the current that produced it is initially almost a straight line, but as the current grows, it curves to the right, presenting the so-called saturation phenomenon, thus characterizing a non-linear relationship. As H depends on i, according to Eq. 1.83, and B is related to /, according to Eq. 1.80, then, for the linear case, it can also be said that the graphical relationship between B and H is a straight line passing through the origin. And the slope of this line is exactly l. So that, for the linear case, B¼lH

ð1:84Þ

The next innovation step occurred around 1831, when Michael Faraday (1791– 1867) and Joseph Henry (1774–1836) discovered, almost simultaneously, that a variable magnetic field could produce a voltage in a closed path - circuit, that is, that an electric current could be generated magnetically, but that such an effect was observed only when the magnetic flux through the circuit varied with time. This effect is referred to as electromagnetic induction, and the currents and electromotive forces—e.m.fs.—generated in this way are called induced currents and induced e.m.fs. Both observed, also, that when a current that varies with time flows in a given circuit, the magnetic field produced by it acts to produce an e.m.f. in this same circuit, but whose effects are opposite to e.m.f. external that makes the current circulate, and vary, first. This effect is called self-induction. Therefore, when the voltage across a coil is due to the variable flow produced by the current variation in the coil itself, it is said that the voltage that appears is self-induced or counter-electromotive force—c.e.m.f.

1.5 Circuit Elements

55

The experimental results of Faraday and Henry, regarding the production of f.e. ms. and induced currents, can be summarized in the following statement, which constitutes the so-called Faraday Law of induction. Whenever there is a magnetic flux that varies with time through a circuit, an e.m.f. is induced in it, the modulus of which is directly proportional to the rate of change of the magnetic flux in relation to time. Therefore, from Eq. 1.79, which defines magnetic flux, Faraday’s law can be mathematically written as eð t Þ ¼ k

d/ d ¼k dt dt

Z B  n da

ð1:85Þ

S

The integral of Eq. 1.85 is performed over the area enclosed by the circuit and, in the SI, the proportionality constant is 1. Therefore, Faraday’s law becomes eð t Þ ¼ 

d/ d ¼ dt dt

Z B  n da

ð1:86Þ

S

Under the conditions that resulted in Eq. 1.80, then d/ d ¼  ðBA cos hÞ dt dt dB dA dh  B cos h  þ ABsenh eðtÞ ¼ A cos h  dt dt dt eðtÞ ¼ 

Considering that:

dh dt

¼ x, angular speed with which the circuit can rotate, it comes

eðtÞ ¼ A cos h 

dB dA  B cos h  þ xABsenh dt dt

ð1:87Þ

Equation 1.87 is only valid if the surface S is planar and the magnetic induction B is the same at all points in S. Equation 1.87 shows that there are several ways to induce an e.m.f., according to Faraday’s law. They are: C31 þ C32 þ C33 ¼

3! 3! 3! þ þ ¼ 3þ3þ1 ¼ 7 1!ð3  1Þ! 2!ð3  2Þ! 3!ð3  3Þ!

Of these seven possibilities, three deserve more emphasis. They are: 1. The circuit does not rotate, x ¼ 0, the area does not vary, A ¼ constant and B varies with time. So,

56

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

eðtÞ ¼ A cos h 

dB dt

ð1:88Þ

2. A and B are constant and the circuit rotates with angular velocity x. In this case, eðtÞ ¼ xABsenh

ð1:89Þ

3. B is constant, the circuit does not rotate, x ¼ 0, and area A varies with time. So, eðtÞ ¼ B cos h 

dA dt

ð1:90Þ

When it comes to a coil with N turns and a concatenated flow N/, Faraday’s law is described mathematically by eð t Þ ¼ N

d/ dt

ð1:91Þ

The polarity of this voltage eðtÞ is such that any current resulting from it produces a flow that is opposed to altering the original flow. This rule for determining the polarity of the induced voltage is known as Lenz’s Law. As flow / is produced by current i, it can also be said that the polarity of the induced voltage eðtÞ is such that it tends to oppose the change in the current that was originally responsible for it. It can be summed up by saying that: an induced effect is always produced in order to oppose the cause that produced it. Therefore, in the case of a coil, Eq. 1.91 provides the magnitude of the voltage, and Lenz’s law, its polarity. And to elucidate how well to apply Lenz’s law, some examples are shown as follows. Figure 1.33 shows a permanent magnet moving to the right, with its north pole approaching a coil with N turns, wound in one of the two possible directions. The flow / of the permanent magnet is constant. But, as it approaches the coil, it becomes influenced by a flow growing upwards and, therefore, variable in it. According to Faraday’s law, then an e.m.f. is induced in the coils, given by Eq. 1.91, which appears between its two terminals. Then Lenz’s law comes in to establish the polarity of this voltage. To try to prevent the growth of the flow that is reaching it, the coil must produce another flow, /o , that is able to oppose the growth of the flow coming from the magnet. According to the right hand rule, with the adjacent fingers agreeing with the winding direction, this requires a current i entering the left terminal of the coil and exiting its right terminal. But this is only possible if the voltage arrow is directed to the left, that is, if the left terminal of the coil is positive and, of course, the right terminal, negative. Figure 1.34 shows the same description as in Fig. 1.33, but with the winding direction of the coil inverted. Similar reasoning shows that polarity has now also been reversed.

1.5 Circuit Elements

57 1

S

N

o

2

3

4

...

N

S

N

i

Displacement

i e

Fig. 1.33 Permanent magnet moving to the right, with its north pole approaching a coil with N turns

1

S

N

Displacement

o

2

3

4

...

N

S

N

i

i e

Fig. 1.34 Permanent magnet moving to the right, with its north pole approaching a coil with N turns, with the winding direction of the coil inverted

Figure 1.35 shows a ferromagnetic core of l permeability over which N turns are wound. Assuming that the magnetic flux /ðtÞ, clockwise, produced by an external source, not shown, is at a time of growth, then, according to Faraday’s law, an e.m.f. eðtÞ is induced and appears between the two winding terminals. Its magnitude is given by Eq. 1.91. Its polarity comes from the application of Lenz’s law. To try to prevent the growth of /ðtÞ it is necessary to have an opposite flow /o ðtÞ, in a counterclockwise direction. But, according to the right hand rule for coils, this requires an electric current iðtÞ entering the lower terminal and exiting the

Fig. 1.35 Faraday and Lenz’s laws applied to a coil in an iron core

(t) o

i(t)

-

1

2

e(t) :

N

+ µ

i(t)

58

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

upper terminal of the winding, which determines the voltage polarity of the voltage eðtÞ. Then the voltage reference arrow should be directed downwards, as shown. It is evident that if /ðtÞ is decreasing, /o ðtÞ should reinforce it, also clockwise. That is, there will be an inversion of the current iðtÞ and, consequently, the polarity of eðtÞ. Since flow /ðtÞ is produced by current iðtÞ Eq. 1.91 can be rewritten as eð t Þ ¼ N



d/ di di lim Di  ¼L ¼L ; Dt ! 0 Dt di dt dt

ð1:92Þ

where L¼N

d/ di

ð1:93Þ

L is called self-inductance or, simply, the coil inductance shown in Fig. 1.32. Its SI unit is Henry, with the unit symbol ½H]. According to Eq. 1.92, one Henry is the amount of inductance that allows the induction of one Volt when the current varies in the ratio of one Amp per second. Typical inductance values are in the order of ½mH. A coil’s inductance depends on its shape, how it is wound with one or more layers, the material of the core around which it is wound, the number of turns and the distance between them, and other factors. Considering that / ¼ BA and that i ¼ Hl=N, expressions obtained from Eqs. 1.81 and 1.83, we have that d/ ¼ A  dB and di ¼ ðl=N ÞdH. So, from Eq. 1.93, L¼N

A  dB ðl=N ÞdH

That is, L¼

AN 2 dB  l dH

ð1:94Þ

B is the magnetic induction in ½Wb=m2  and H is the magnetic intensity in ½At=m. The relationship between B and H depends on the material from which the core is made, and two typical curves are shown in Figs. 1.36a, b. For the air core, dB dH

¼ tgh ¼ l, where l is the core permeability. So, Eq. 1.94 becomes

1.5 Circuit Elements

59 B

B

0 H Air core

Iron core

(a)

(b)

H

Fig. 1.36 Relationship between B and H depending on the material from which the core is made

L ¼ lN 2 A=l

ð1:95Þ

As the inductance has a constant value, independent of H and, therefore, of the current throught it, it is a linear circuit element. Furthermore, Eq. 1.95 shows that the inductance L always corresponds to the product of permeability l by a geometric factor with the dimension of length. For the linear case, d/=di can be replaced by /=i. Therefore, from Eq. 1.93, L ¼ N/=i or, N/ ¼ Li

ð1:96Þ

For the iron core, the dB=dH value depends on the H value and, therefore, on the current, so that the inductance is a non-linear element. Even for the iron core, however, an equivalent linear inductance can be used to produce good results. In this book, the inductance L will always be considered constant. Physically, any extension of a conductor has some inductance associated with it, when a variable current over time flows throught it. That is, it is not necessary to build an inductor—a coil to obtain inductance. An inductor is a device built with the purpose that inductance is its main characteristic, which is the ability to store and supply energy. One can therefore speak of a conductor’s inductance or a coil’s inductance. Example 1.16 An inductance of 500 ½mH undergoes a current variation of 15 ½mA in 0:5 ½ls. What is the voltage generated at its terminals? Solution As the current curve waveform was not given, it can be considered that De ¼ LDi=Dt ¼ 500  103  15  103 ð::::Þ=0:5  106 ¼ 15; 000 ½V ¼ 15½kV Inductance can also be seen as the property of an electrical circuit to counteract sudden changes in current. According to Eq. 1.92, a sudden change in the current, that is, a finite discontinuity in the current curve iðtÞ, would require an infinite

60

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

self-induced voltage, because in the discontinuity Dt tends to zero, but Di does not. (These issues will be better understood with the help of the singular functions covered in Chap. 2 of this book). Of course, there is no similar restriction for the inductance voltage. It can suffer discontinuities or even change its polarity instantly, that is, change its signal. The currents in the inductances are not discontinued means that at t ¼ 0 þ , that is, immediately after a switching operation in the circuit, at t ¼ 0, they are the same as immediately before switching, at t ¼ 0 . In other words, that ið0 þ Þ ¼ ið0 Þ ¼ ið0Þ. This is an important data in the transient analysis of circuits containing inductances. If an attempt is made to open a branch of a circuit containing an inductance, through which an electric current is flowing and storing an energy given by ð1=2ÞLi2 , an electric arc may appear between the switch terminals, and the energy stored in the inductance will be spent on ionizing the air in the region where the arc exists. It is necessary to pay attention to the fact that Eq. 1.92 implies associated references for the voltage and current arrows in an inductance, that is, arrows in opposition; otherwise, a minus sign is required, as shown in Fig. 1.37. Equation 1.92 also reveals that if the current in an inductance is continuous, that is, constant, then the voltage at its terminals is zero, because di=dt ¼ 0. With a constant current flowing through it, but with zero voltage at its terminals, an inductance behaves like a short-circuit in relation to the direct current. From Eq. 1.92, conversely, iðtÞ ¼

1 L

Z eðtÞdt

ð1:97Þ

eðkÞdk

ð1:98Þ

Or, according to Eq. 1.27, 1 i ðt Þ ¼ L

Zt 1

Or yet, 1 iðtÞ ¼ L

e(t)

i(t)

L e(t)=Ldi(t)/dt

Zt0 1

1 eðkÞdk þ L

Zt eðkÞdk

ð1:99Þ

t0

e(t) L i(t) e(t)=—Ldi(t)/dt

Fig. 1.37 Reference arrows for voltage between the terminal nodes and current across an indutor

1.5 Circuit Elements

61

But, according to Eq. 1.98, 1 L

Zt0 eðkÞdk ¼ iðt0 Þ 1

So, 1 iðtÞ ¼ iðt0 Þ þ L

Zt eðkÞdk

ð1:100Þ

t0

Equation 1.91 can be rewritten as d ðN/Þ ¼ edt Or that

Z

Z d ðN/Þ ¼ N/ ¼

edt

ð1:101Þ

R Thus, the magnitude edt represents the concatenated magnetic flow of a coil of N turns. And the comparison of Eqs. 1.97 and 1.101 allows us to conclude that 1 iðtÞ ¼ N/ðtÞ L Or that N/ðtÞ ¼ LiðtÞ; confirming Eq. 1.96 Obviously, for N ¼ 1, /ðtÞ ¼ LiðtÞ

ð1:102Þ

The power absorbed at the inductance in Fig. 1.37 is given by pðtÞ ¼ eðtÞ  iðtÞ ¼ LiðtÞ

diðtÞ dW ðtÞ ¼ dt dt

Then the energy stored in the inductance until any moment t is Z W ðt Þ ¼

Zt pðtÞdt ¼

pðkÞdk 1

ð1:103Þ

62

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

On the other hand, Eq. 1.103 can be rewritten as Zt0 W ðtÞ ¼

Zt pðkÞdk þ

1

pðkÞdk

ð1:104Þ

t0

But, according to Eq. 1.103, Zt0

Zt0 pðkÞdk ¼ W ðt0 Þ ¼

1

1

o t 0 L Ln LiðkÞdiðkÞ ¼ ½iðkÞ2 1 ¼ ½iðt0 Þ2 ½ið1Þ2 2 2

It turns out that at 1 the inductance didn’t even exist yet, so that ið1Þ ¼ 0. Then, Zt0 pðkÞdk ¼ 1

L ½iðt0 Þ2 2

Therefore, Eq. 1.104 becomes L W ðtÞ ¼ ½iðt0 Þ2 þ 2

t L L LiðkÞdiðkÞ ¼ ½iðt0 Þ2 þ ½iðkÞ2 t0 2 2 t0

Zt

L L L ¼ ½iðt0 Þ2 þ ½iðtÞ2  ½iðt0 Þ2 2 2 2 Finally, L W ðtÞ ¼ ½iðtÞ2 2

ð1:105Þ

Equation 1.105 shows that the energy in an inductance L, at any instant t, depends only on the value of the current at that exact moment, and the value of the inductance, obviously. Being L 0, then always W ðtÞ 0, confirming that the inductance is a passive circuit element. Example 1.17 Find the inductance of a single layer coil with 50 turns wound in a cylindrical ferromagnetic core 1:5 ½cm long and 1:5 ½mm in diameter, if the relative permeability of the material is lr ¼ 7; 000. What is, approximately, the resistance of the copper wire that was wound to build the coil? Solution According to Eq. 1.95,

1.5 Circuit Elements

63

L¼

lN 2 A lr l0 N 2 A lr l0 N 2  prn2 ¼ ¼ l l l

l0 ¼ 4p  107 ½H=m; lr ¼ 7; 000; N ¼ 50; l ¼ 1:5½cm ¼ 1:5  102 ½m; dn ¼ 1:5½mm ¼ 1:5  103 ½m ) rn ¼ 0:75  103 ½m: Therefore, 2

L¼

7  103  4p  107  502  pð0:75  103 Þ ¼ 2:59½mH 1:5  102

Disregarding the thickness of the copper wire insulation film, its diameter is approximately equal to the length of the coil divided by the number of turns, as they are juxtaposed. So, df ¼

l 1:5  102 ¼ ¼ 3  104 ½m N 50

The total length of the wire is the result of the circumference of a loop multiplied by the number of turns. So, lf ffi 2prn  N ¼ 2p  0:75  103  50 ¼ 236  103 ½m The cross-sectional area of the copper wire is given by Af ¼ p

2 df2 p  ¼ 3  104 ¼ 7:069  108 ½m2  4 4

From Eq. 1.41, R¼q

lf Af

From Table 1.1, q ¼ 1:72  108 ½X m. Finally, 3

23610 R ¼ 1:72  108 7:06910 8 ¼ 0:057 ½X ffi 60 ½mX. If an inductance varies over time, it will be denoted by LðtÞ. In this case, Eq. 1.102 should be replaced by

/ðtÞ ¼ LðtÞiðtÞ Then the voltage induced in such an inductance will be

64

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

eð t Þ ¼

d/ðtÞ diðtÞ dLðtÞ ¼ LðtÞ þ iðtÞ dt dt dt

ð1:106Þ

Equation 1.106 shows that, even for direct current, there is a voltage induced in the inductance and the component cannot be treated as a short-circuit. That is, for a direct current I, the voltage at a time-varying inductance is eð t Þ ¼ I

dLðtÞ dt

ð1:107Þ

Inductance and capacitance are two dual electrical elements, as their definition equations, eðtÞ ¼ LdiðtÞ=dt and iðtÞ ¼ CdeðtÞ=dt, are essentially the same, except that the roles of eðtÞ and iðtÞ are switched. Thus, to obtain all the equations of one of them, just make the appropriate changes in all the equations of the other. For example, if in Eq. 1.49, qðtÞ ¼ CeðtÞ, C is replaced by L and eðtÞ by iðtÞ Eq. 1.102 is obtained, /ðtÞ ¼ LiðtÞ, showing that the dual of q, the charge at capacitance, is /, the flux at inductance. It is important to remember that dual is not the same as equivalent. As mentioned at the beginning of this chapter, two electrical circuits are dual when they are described by the same equations, although they have different physical functioning. The only restriction is that the duality rules are only applicable to planar circuits, that is, those where their different branches only cross at the nodes. Table 1.4 shows a list of dual element pairs. According to Table 1.4, the dual circuit of N capacitances in parallel is a circuit with N inductances in series. Thus, series inductances combine in the same way as capacitances in parallel. So for N series inductances, Leq ¼ L1 þ L2 þ L3 þ    þ LN

ð1:108Þ

Therefore, for N inductances in parallel, 1 1 1 1 1 ¼ þ þ þ  þ Leq L1 L2 L3 LN

Table 1.4 Duality of electrical elements Series connection

Parallel connection

Electric charge qðtÞ Voltage eðtÞ Voltage source Short-circuit Resistance R Capacitance C Mesh Mesh current Switch that closes Mesh with N branches in series

Magnetic flux /ðtÞ Current iðtÞ Current source Open circuit Conductance G ¼ 1=R Inductance L Node Nodal voltage Switch that opens Node with N branches in parallel

ð1:109Þ

1.5 Circuit Elements

65

In particular, for two inductances in parallel, the rule of the product by the sum applies: Leq ¼

L1  L2 L1 þ L2

ð1:110Þ

Example 1.18 Make an analysis of the phase difference between voltage and current in an inductance subjected to sinusoidal quantities. Solution If the current at the inductance in Fig. 1.37 is iðtÞ ¼ I  senxt for t 0;

ð1:111Þ

then the voltage at its terminals will be eð t Þ ¼ L

 h  d p p i iðtÞ ¼ xLI cos xt ¼ xLIsen xt þ ¼ xLIsen x t þ dt 2 2x ð1:112Þ

Comparing the two equations, we can then conclude that: for sine waveforms, in pure inductance, the voltage is ahead of the current of p=2 radians ð90 Þ or, equivalently, that the current iðtÞ is lagging by p=2½rd, (90º), in relation to the voltage eðtÞ. This conclusion is corroborated by Eq. 1.98, because if the current depends on the area under the voltage curve, obviously, it cannot anticipate it. Therefore, the current in an inductance is always behind the voltage for any waveforms. This last statement is best understood after studying the singular functions, covered in the second chapter of this book. But, you see, the lag is p=2 ½rd, ð90 Þ, when the reference is xt ½rd. When the reference is t ½s], the current delay is /=ð2xÞ ½s]. Thus, if the sinusoidal voltage goes through its maximum value at a given time t ¼ s, the corresponding sinusoidal current will pass through its peak value a time /=ð2xÞ ½s] after s, although both exist at that time t ¼ s and in all others antecedent and subsequent moments. The following is a summary of the dual correspondence between the main equations related to capacitance and inductance. Capacitance

Inductance

qðtÞ ¼ C  eðtÞ

/ðtÞ ¼ L  iðtÞ

d dt qðtÞ

d dt /ðtÞ

¼ iðtÞ ¼ C dtd eðtÞ

qðtÞ ¼

Rt

1

t

iðkÞdk ¼ qðt0 Þ þ R iðkÞdk

eðtÞ ¼ eðt0 Þ þ

t0

1 Rt C t0

iðkÞdk

¼ eðtÞ ¼ L dtd iðtÞ

/ðtÞ ¼

Rt

1

t

eðkÞdk ¼ /ðt0 Þ þ R eðkÞdk

iðtÞ ¼ iðt0 Þ þ

t0

1 Rt L t0

eðkÞdk (continued)

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1 Introduction to Fundamental Concepts in Electric Circuit Analysis

(continued) Capacitance

Inductance

pðtÞ ¼ dtd W ðtÞ ¼ eðtÞiðtÞ ¼ CeðtÞ dtd eðtÞ

pðtÞ ¼ dtd W ðtÞ ¼ iðtÞeðtÞ ¼ LiðtÞ dtd iðtÞ

W ðtÞ ¼

Rt

1

pðkÞdk ¼ W ðt0 Þ þ

Rt

t0

pðkÞdk

W ðtÞ ¼

Rt

1

pðkÞdk ¼ W ðt0 Þ þ

Rt

t0

pðkÞdk

W ðtÞ ¼ 12 C  e2 ðtÞ

W ðtÞ ¼ 12 L  i2 ðtÞ

Capacitances in parallel C ¼ C1 þ C2

Inductances in series L ¼ L1 þ L2

Capacitances in series C ¼ CC1 1þCC2 2

Inductances in parallel L ¼ LL1 1þLL2 2

Example 1.19 Find the waveform of the voltage induced in a 400 ½mH coil, when the current shown in Fig. 1.38a passes through it.

e(t) [V]

i(t) [mA] 20

8

2 0

1

2

3

4 5

6 7

t [ms]

8

0

1

2

3 4

5

6

7

8

t [ms]

-4

-10

(a)

(b)

Fig. 1.38 a Current across and b Voltage between the terminals of a coil (inductor)

Solution The induced voltage is given by eð t Þ ¼ L

d d iðtÞ ¼ 400  103 iðtÞ ¼ 0:4i0 ðtÞ dt dt

A good way to proceed is to find the slope of the current curve in each stretch, which, according to the geometric interpretation of the derivative concept, is i0 ðtÞ, and then multiply by 0:4. For 0\t\1 ½ms, the slope is 20 ½mA=1½ms ¼ 20 ½A=s. So, eðtÞ ¼ 0:4 20 ¼ 8 ½V. For 1 ½ms\t\4 ½ms, the slope is 30 ½mA=3 ½ms ¼ 10 ½A=s. So, eðtÞ ¼ 0:4  ð10Þ ¼ 4½V. For 4 ½ms\t\6 ½ms, i0 ðtÞ ¼ 0. Therefore, eðtÞ ¼ 0 ½V. For 6 ½ms\t\8 ½ms, the slope is 10 ½mA=2 ½ms ¼ 5 ½A=s. So, eðtÞ ¼ 0:4  5 ¼ 2 ½V. The waveform of the voltage induced in the coil is shown in Fig. 1.38b.

1.5 Circuit Elements

67

Sources As mentioned earlier, sources are the active elements in a circuit. They are elements of two terminals for which there is no pre-established relationship between voltage and current, as is the case with passive elements, that is, resistors, capacitors and inductors. Therefore, when one of the two quantities is given, the other cannot be calculated without knowing the rest of the circuit. They are classified into two types: independent sources and controlled sources, also called linked or dependent. Independent sources are those for which either the voltage eðtÞ or the current iðtÞ is always given, either through a constant value or through a waveform that varies with time. The independent voltage source shown in Fig. 1.39a has a voltage eðtÞ which is a specified function of time and which is independent of any external connections at its two terminals. On the other hand, the current iðtÞ depends on the external connection that is made, and can pass through the voltage source in any of the two possible directions. Thus, with concordant reference arrows, if at some point eðtÞ and iðtÞ are both positive, the voltage source will be supplying and not absorbing power, that is, pðtÞ\0. The current iðtÞ can assume any value, so that the voltage source is theoretically capable of supplying an unlimited amount of power and energy to the rest of the circuit. As ideal elements, which only approximate the real behavior of physical devices, the sources are considered to be capable of supplying energy indefinitely, that is, without limits. An independent voltage source that maintains a constant voltage at its terminals is called a continuous (DC) voltage source. And one that has a voltage that varies sinusoidally over time is called an alternating voltage source. A battery is an independent voltage source that provides an almost constant voltage, regardless of the current being requested from it. When eðtÞ is a constant, that is, when the voltage source is continuous, it is represented as in Fig. 1.39b, where the small straight lines suggest the plates of a battery. The greater sign must be associated with the positive sign. Thus, although the þ and  signs represent a redundancy of notation, they are usually included. It is worth remembering that a voltage source of zero value is equivalent to a short circuit. A generator is a machine on which electromagnetic induction is used to produce a voltage by rotating coils of conductive wire through a stationary magnetic field or by rotating a magnetic field produced by a source of direct voltage, called a

i(t)

e(t)

i(t)

e(t)=E (a)

+

i(t)

(b)

e(t)

(c)

Fig. 1.39 a Independent voltage source; b DC voltage source; c Independent current source

68

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

rotating field of the rotor through the stationary stator coils. A generator that produces a sinusoidal alternating voltage is called an alternator. When the stator has three sets of coils: a  a0 ; b  b0 ; c  c0 ; 120 spatially out of phase, sinusoidal voltages of the same amplitude are also induced, also out of phase with 120 , but in the independent variable xt ½rd, where x ¼ 2pf ½rd=s, with f ½cycles=second  ¼ f ½Hz the frequency, given by f ¼ p  n=60, with n representing the number of rotor turns in [rpm] and p, the number of pairs of magnetic poles of the rotor coils. Such a generator is called a three-phase alternator. The independent current source shown in Fig. 1.39c produces a specific current iðtÞ independent of the external connections. As the voltage eðtÞ at its terminals depends on the rest of the circuit and, theoretically, can take on any value, the independent current source can also provide an unlimited amount of power and energy. Current source causes some surprise in those who are more used to voltage sources, but many devices can be represented by a current source in conjunction with some passive elements. In addition, inductors initially energized, that is, traversed by an initial current ið0Þ ¼ I, can be represented by an equivalent consisting of a de-energized inductance L in parallel with a current source of value equal to I. It is evident that the dual situation also exists, that is, a capacitor with an initial voltage eð0Þ ¼ E can be replaced by a de-energized capacitance C in series with an independent voltage source of value E. A current source of zero value is equivalent to an open circuit. The term source at rest is often used to denote a zero value source. So, a source of voltage at rest is a short-circuit, and a source of current at rest is an open circuit. A controlled source is one whose value is not independent of the rest of the circuit of which it is a part, but is a known function of some other voltage or current in the same circuit. They are particularly important in building models of electronic components. A controlled source is not really a component of two terminals, as the voltage or current on which it depends must also be shown so that the element is fully characterized. A symbol for a common emitter NPN transistor is shown in Fig. 1.40a, where b, e and c denote the base, emitter, common to the input and output, and collector terminals. This three-terminal component can be operated in approximately linear or non-linear fashion. In the linear range, the collector current ic ðtÞ is approximately proportional to the base current ib ðtÞ, but it also depends on the voltage between the collector and the emitter ece ðtÞ. The small voltage between the base and the emitter ebe ðtÞ depends on the base current ib ðtÞ and, to a lesser extent, on the collector voltage to the emitter ece ðtÞ. A model that includes these effects is shown in Fig. 1.40b, and contains two controlled sources. A network that does not contain any active elements, that is, an independent source, is called a passive network.

1.6 Kirchhoff’s Laws

c ib(t)

69 ic(t)

ib(t)

b

c

ic(t)

rn

b ece(t)

ebe(t)

βib(t)

r0

ece(t)

µece(t)

ebe(t) (a)

e

(b)

e

Fig. 1.40 a Common emitter NPN transistor; b Equivalent circuit model for a common emitter NPN transistor

1.6

Kirchhoff’s Laws

The two laws dealing with the effects of the interconnection of different circuit elements were proposed by Gustav Robert Kirchhoff (1824–1887), a German physicist. Kirchhoff's laws are strictly valid only when the signal propagation time is neglected, that is, when the elements of the circuit are concentrated. In this book, any connection of elements in such a way that the dimensions of them and the circuit itself are negligible compared to the wavelength of the highest frequency of interest will be called a lumped circuit. The restriction on the size of the elements is a consequence of Kirchhoff’s laws being approximations of Maxwell’s famous equations, which are the general laws of electromagnetism. The approach is analogous to the fact that Newton’s laws of classical physics are approximations to the laws of relativistic mechanics. A branch of a circuit is a single path containing a single element that connects a node to any other node. Remembering that all interconnecting wires of circuit elements are considered as short-circuits, a node is defined as the meeting of all points that are subject to the same potential. Kirchhoff formulated the law of currents, also called the first law, assuming that no charge can accumulate in an element or in a node and that the charge must be conserved, that is, that a node cannot store, generate or destroy electrical charge. Therefore, the total current entering a node of any concentrated circuit, at any time, must be equal to the total current leaving it at the same time. In summary, the first law is Kirchhoff’s Currents Law The algebraic sum of currents leaving any node in a lumped circuit, at any time, is equal to zero Symbolically,

70

1 Introduction to Fundamental Concepts in Electric Circuit Analysis k X

in ðtÞ ¼ 0

ð1:113Þ

n¼1

When applying Eq. 1.113, it is necessary to remember that a current iðtÞ entering a node is equivalent to a current iðtÞ leaving it. Furthermore, in a circuit with N nodes, the application of the law of currents will only produce linearly independent equations if it is applied at most to any N  1 nodes in the circuit. Therefore, the law of currents should not be applied to all nodes in the circuit; one node must always be excluded. With linearly dependent equations, the number of unknowns will be greater than the number of truly useful equations. In other words, applying the law to the total number of nodes N of the circuit will result in N equations, but so that any one of them will be the negative of the sum of the other N  1 and, therefore, will not add any new information. Usually, this node that is excluded in the application of the law of currents is the one taken as a reference node in the application of the nodal circuit solution method, to which the grounding symbol is assigned. The Kirchhoff’s currents law is independent of the nature of the circuit elements, that is, it applies to any circuit with lumped parameters, whether linear, non-linear, active, passive, variant or invariant over time. A loop of a circuit is a set of branches that form a single closed path. So, each node in a loop must have exactly two branches of the loop attached to it. A mesh is a particular case of loop, which does not involve any other branch. Kirchhoff’s Voltages Law Kirchhoff’s voltages law is based on the principle of energy conservation. Based on the fact that voltage is energy per unit of charge, it establishes that The sum of the voltages along any closed path is equal to zero Symbolically, k X

en ð t Þ ¼ 0

ð1:114Þ

n¼1

A closed path is understood as a loop or mesh. In the application of Eq. 1.114, when the route adopted, clockwise or counter-clockwise, coincides with the voltage reference arrow, entering the tail and exiting the arrowhead, the corresponding voltage must be included positively in the equation. Otherwise, it enters the equation with a negative sign. The number of times you can apply Eq. 1.114 to a circuit also has a limit. It matches the number of loops that can be counted in a circuit, even if the equation is being done by loops and not by meshes. And the objective is always to guarantee a set of linearly independent equations, that is, one in which no equation can be obtained by linearly combining the others. Alternatively, you can calculate the maximum number of times you can apply Eq. 1.114 using the formula

1.6 Kirchhoff’s Laws

71

L ¼ B  N þ 1;

ð1:115Þ

where N is the total number of nodes in the circuit, B is the total number of branches and L is the maximum number of loops—or meshes—to which Eq. 1.114 can be applied. Note that it is the path that needs to be closed. This means that if it includes an open switch, the voltage at the switch terminals must also be included in the sum.

1.7

Analytical and Digital Circuit Solution

The following example is a preview of the main objective of this book, which is the search for a complete solution of concentrated parameter circuits, either analytically discrete-time, that is, in the domain of continuous-time, or digitally, in the domain of discrete-time. It is, therefore, expected that the reader, at the end of this chapter, will feel sufficiently motivated to continue until the last of its pages. Example 1.20 Find the current, voltage, power and energy waveforms in the elements of the RC series circuit, powered by a constant source, shown in Fig. 1.41, knowing that the capacitance is de-energized at t ¼ 0 . Then, perform the digital simulation, using the ATPDraw program, for the following numerical values E ¼ 10 ½V; R ¼ 10 ½X; C ¼ 1; 000 ½lF.

Fig. 1.41 RC series circuit, powered by a constant source

eR(t) eC(t)

E t=0

E

+

i(t)

R C

eC(t)

Solution As the digital solution uses the nodal method, it will also be used in the analytical solution. With the switch closed, there are three nodes in the circuit, namely: the lower node, consisting of all points of the connection between the negative battery terminal and the lower capacitance plate; the upper left node, consisting of all the points that connect the positive terminal of the battery and the left terminal of the resistor; and the upper right node, which gathers all the connection points from the right terminal of the resistor to the upper plate of the capacitance.

72

1 Introduction to Fundamental Concepts in Electric Circuit Analysis

As the essence of the nodal method consists of the application of the law of currents to nodes of unknown voltages in the circuit, always ignoring one of them, the lower node was taken as a reference. However, when doing this the upper left node has the source voltage, E, known, and the upper right node, with the capacitance voltage. Since there is no equation for a known voltage node, it remains, therefore, to equate only the node of unknown voltage, eC ðtÞ. Therefore, adding the two currents that supposedly leave this node and equaling to zero, it comes that eC ðtÞ  E d þ C eC ð t Þ ¼ 0 R dt Or 0

eC ð t Þ þ

1 E eC ðtÞ ¼ ; with eC ð0 Þ ¼ 0: RC RC

ð1:116Þ

The solution of the corresponding homogeneous differential equation 

Dþ

1 RC



eC ðtÞ ¼ 0 is

eCH ðtÞ ¼ kt=RC

The particular integral can be obtained using the so-called abbreviated method (Chap. 3), namely:





1 E 1 E 0:t 1 E 0:t   e CP ð t Þ ¼ ¼ ¼ ¼E 1 1 1 RC RC RC D þ RC D þ RC 0 þ RC Therefore, the general solution is eC ðtÞ ¼ eCP ðtÞ þ eCH ðtÞ ¼ E þ kt=RC

ð1:117Þ

To determine the constant k, it is necessary to know the initial condition eC ð0 þ Þ. Since the voltage at the capacitance cannot vary instantly, it is evident that eC ð0 þ Þ ¼ eC ð0 Þ ¼ 0: Taking t ¼ 0 þ in Eq. 1.117 comes that 0 ¼ E þ k then k ¼ E. So,   eC ðtÞ ¼ E  Et=RC ¼ E 1  t=RC ;

for t 0:

ð1:118Þ

1.7 Analytical and Digital Circuit Solution

73

The product RC ¼ s is called the RC circuit time-constant. Theoretically, the term containing the exponential function only tends to zero when time t tends to infinity. It turns out that for t ¼ 5s ¼ 5RC; t=RC ¼ 5 ¼ 0:007, that is, the exponential has a value equal to just 0.7% of its initial value, which is equal to 1. Thus, it is customary to consider that for t 5s the exponential function is already given as null. With this practice, it can then be said that the expression of the voltage in the capacitance, Eq. 1.118, contains a transient component, the exponential, and a component of steady-state regime, the constant E. In other words, that in a time approximately equal to 5s the capacitance has already been charged and reached the source voltage, E. It is possible to predict, then, that after a time greater than 5s, there is no longer a need to circulate current to put more positive electrical charge on the upper plate of the capacitance, with the consequent equally positive repellency of the bottom plate. The voltage between the terminals of the resistance is eR ð t Þ ¼ E  eC ð t Þ So, eR ðtÞ ¼ Et=RC ;

for t [ 0;

ð1:119Þ

The current flowing through the circuit elements is i ðt Þ ¼

eR ðtÞ E t=RC ¼  ; R R

for t [ 0:

ð1:120Þ

The powers in the passive elements are: E 2 2t=RC  ; for t [ 0 R   E pC ðtÞ ¼ eC ðtÞ  iðtÞ ¼ E 1  t=RC  t=RC R pR ðtÞ ¼ R½iðtÞ2 ¼

ð1:121Þ

Therefore, pC ð t Þ ¼

 E 2  t=RC   2t=RC ; R

for t 0:

ð1:122Þ

Note that the power in the capacitance has two time constants: s1 ¼ s ¼ RC and s2 ¼ RC=2 ¼ s1 =2. The energy dissipated in the resistance is obtained by applying Eq. 1.36, with t0 ¼ 0 þ :

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1 Introduction to Fundamental Concepts in Electric Circuit Analysis

Zt WR ðtÞ ¼ WR ð0 þ Þ þ

E2 pR ðkÞdk ¼ 0 þ R

0þ

¼

Zt 

2k=RC

t E 2 2k=RC   2  dk ¼  R RC

0þ

0þ

CE 2 2k=RC t  0þ 2

Then, W R ðt Þ ¼

 CE 2  1  2t=RC ; 2

for t 0:

ð1:123Þ

The energy stored in the capacitance is obtained by applying Eq. 1.72: W C ðt Þ ¼

C ½eC ðtÞ2 2

So, W C ðt Þ ¼

i2 CE 2   Ch  E 1  t=RC 1  2t=RC þ 2t=RC ; ¼ 2 2

for t 0: ð1:124Þ

Note that   WR ðtÞ þ WC ðtÞ ¼ CE 2 1  t=RC :

ð1:125Þ

The power at the voltage source is pF ðtÞ ¼ E  iðtÞ ¼ 

E2 t=RC  R

for t [ 0:

Note that pF ðtÞ þ pR ðtÞ þ pC ðtÞ ¼ 0: The energy at the voltage source is Zt W F ðt Þ ¼ W F ð0 þ Þ þ 0þ

E2 pF ðkÞdk ¼ 0  R

Zt

k=RC

 0þ

t E 2 k=RC  dk ¼   1   R RC

0þ

Therefore,   WF ðtÞ ¼ CE 2 t=RC  1 ; for t 0. Note that WF ðtÞ þ WR ðtÞ þ WC ðtÞ ¼ 0: Substituting the given numeric values, the following expressions result:

1.7 Analytical and Digital Circuit Solution

75

  eC ðtÞ ¼ 10 1  100t ½V; t 0 ½s; s ¼ 1=100 ¼ 10 ½ms: eR ðtÞ ¼ 10100t ½V; t [ 0 ½s: Note that eC ðtÞ þ eR ðtÞ ¼ E; t [ 0 ½s], satisfying Kirchhoff’s voltages law. iðtÞ ¼ 100t ½A; t [ 0 ½s: pR ðtÞ ¼ 10200t ½W; t [ 0 ½s:   pC ðtÞ ¼ 10 100t  200t ; t 0 ½s:

ð1:126Þ

  WR ðtÞ ¼ 50 1  200t ½mJ; t 0 ½s:   WC ðtÞ ¼ 50 1  2100t þ 200t ½mJ; t 0 ½s: The circuit built for the digital solution using the ATPDraw program is shown in Fig. 1.42. (See Appendix A) As s ¼ 10 ½ms, tmax ¼ 70½ms was adopted, enough to capture the transient and steady state regimes. In the absence of a formula for determining the value of Dt, the standard ATP value was adopted, that is, Dt ¼ 1½ls. A common practice is to adopt a value in the order of one hundredth of the time constant, that is, Dt ¼ s=100. This practice, however, produces satisfactory results for first order circuits, in which it is already known, previously, that the transient components will be simple exponentials. The waveforms obtained digitally are shown in Fig. 1.43. Deriving pC ðtÞ, given by Eq. 1.124, and equaling zero, we obtain that the instant when the power in the capacitance passes through the maximum corresponds to t ¼ s  ln2 ¼ 10  0:6931 ¼ 6:931 ½ms. At this moment, the power pCmax ¼ 2:5½W, in agreement with the value shown in the curve of pC ðtÞ shown in Fig. 1.43.

Fig. 1.42 RC series circuit, powered by a constant source, built for the digital solution using the ATPDraw program

V

UI

t=0

10 V

B 1000 uF

U

F

A 10 Ohm

76

1 Introduction to Fundamental Concepts in Electric Circuit Analysis Source Voltage-Red. Voltage in R-Green. Voltage in C-Blue

10

Current in the circuit

1,0

[V]

[A]

8

0,8

6

0,6

4

0,4

2

0,2

0

0,0 0

10

20

(f ile Exemplo1.20.pl4; x-v ar t) v :F

30 v :A

-B

v :B

40

50

60

[ms]

70

-

Power in R-Red. Power in C-Green

10

0

10

20

(f ile Exemplo1.20.pl4; x-v ar t) c:A

50

W

[mJ]

8

40

6

30

4

20

30

40

50

60

[ms]

70

-B

Energy in R-Red. Energy in C-Green

10

2

0

0 0

10

20

(f ile Exemplo1.20.pl4; x-v ar t) v :A

30 -B

v :B

-

40

50

60

[ms]

70

0

10

(f ile Exemplo1.20.pl4; x-v ar t) c:A

20 -B

30 c:B

40

50

60

[ms]

70

-

Fig. 1.43 RC series circuit transient response for voltage, current, power and energy with digital solution using the ATPDraw program

1.8

Conclusions and Motivations for Electromagnetic Transients Studies

From this brief introduction to the theme of transients in basic electrical circuits, it is evident that the understanding of physical phenomena requires adequate mathematical modelling and the use of computer simulation tools through digital models capable of representing accurately and with stability all components of the system under study. We think that the book is intended for engineering students who already have been successfully approved in introductory circuit analysis, calculus and ordinary differential equations courses. Our experience as educators is that, when it comes to truly understand and apply the mathematical knowledge in electrical circuits transients, the majority of students have much more difficulties. Therefore, this book supplies, in an integrated way, all the necessary knowledge and practice through extensive examples. In the next chapters of this book we aim to present an introduction to the modelling and mathematical calculation of transients in electrical circuits through an analytical solution and digital computational solution for a wide variety of basic cases (at least for circuits with concentrated parameters), using an EMTP-based program, the ATP software—Alternative Transients Program. At the end of the studies, it is expected that students have developed essential skills to understand the facilities, limitations and potential of these engineering tools applicable to the industry in the electrical sector.

References

77

References 1. Alexander CK, Sadiku MNO (2008) Fundamentals of electrical circuits, 3rd edn. McGraw-Hill do Brasil, Ltda, São Paulo 2. Boylestad RL (2012) Introduction to circuit analysis, 12th edn. Pearson Education of Brazil, São Paulo 3. Close CM (1966) The analysis of linear circuits. Harcourt, Brace & World Inc., New York 4. McKelvey JP, Grotch H (1979) Physics, vol 1 and 3. Harper & Row of Brazil Ltda, São Paulo 5. Nilson JW, Riedel SA (2009) Electrical circuits, 8th edn. Pearson Education of Brazil, São Paulo 6. Nussenzveig HM (1997) Basic physics course, vol 1 and 2, 3rd edn. Edgard Blücher, 7. O’Malley JR (1992) Circuit analysis. In: Schaum’s outline, 2nd edn. McGraw-Hill, New York 8. Resnick R, Halliday D, Walker J (2010) Fundamentals of physics 3: electromagnetism; LTC, 8th edn. 9. Young HD, Freedman RA (2009) Physics III: electromagnetism, vol 3, 12th edn. Pearson Education

Chapter 2

Singular Functions

2.1

Single Step Function

This chapter introduces the singular functions, also called generalized functions or distributions, that constitute a special class of functions, because they do not fit perfectly into the conventional mathematical concept of function. Here they will be designated by U n ðtÞ, n ¼ 0; 1; 2;     ; in such a way that for each value of n a specific singular function will correspond. They can be considered as a family of functions that are generated from the unit impulse function, U0 ðtÞ, by successive derivations or integrations. However, for simplicity and also for strictly didactic reasons, the study of them will be done starting from the unit step function, U1 ðtÞ. An important reason to study them is that the response of any circuit linear to an arbitrary entry can be determined, indirectly, as long as its response to a unit step or unit impulse is known. This is done through the integral convolution call. Thus, the response of a linear circuit to a unitary impulse, denoted by h(t), or to a unitary step, called r(t), can be used as a characteristic property of such a circuit. Although the step and impulse functions do not actually appear as signals on any physical circuit, that is, on any realizable circuit, they can approximate many physical signals. Often, a circuit’s excitation is a pulse whose duration is short compared to the circuit’s time constants. Under this condition, it is then possible to approximate the desired response of the circuit, to the pulse, by responding to an impulse of the same area, acting from the same moment in which the original pulse would act. Also, the energy initially stored in elements of electrical circuits can be conveniently represented by hypothetical sources of the type of step or impulse. Furthermore, in mathematical manipulations involving the differential equations that describe a given circuit, impulses may appear, even if the source that excites it is not an impulse. The unit step function, U 1 ðtÞ, shown in Fig. 2.1a, also known as the Heaviside Unit Function, is defined as being null for negative values of t, unit for positive values of t, and discontinuous for t ¼ 0. Then, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. C. Goulart de Siqueira and B. D. Bonatto, Introduction to Transients in Electrical Circuits, Power Systems, https://doi.org/10.1007/978-3-030-68249-1_2

79

80

2 Singular Functions

U-1(t)

kU-1(t-a)

U-1(t-a) k

1

1 t

0

0

t

a

0

(b)

(a)

t

a

(c)

Fig. 2.1 a Unitary step function U 1 ðtÞ; b Unitary step function delayed of a: U 1 ðt  aÞ; c Amplitude k step function kU1 ðt  aÞ


U1 ðtÞ ¼

0 for t\0 1 for t [ 0:

ð2:1Þ

Therefore, the unit step function is not defined for t ¼ 0, but is known for all values arbitrarily close to t ¼ 0. This behavior is indicated by writing that U 1 ð0 Þ ¼ 0 and U 1 ð0 þ Þ ¼ 1. If, however, the discontinuity occurs at time t ¼ a [ 0, and not at t ¼ 0, then the unit step will be denoted by U 1 ðt  aÞ, as
shown in Fig. 2.1b, for a [ 0. Therefore, 0 for t\a U1 ðt  aÞ ¼ ð2:2Þ 1 for t [ a: It is said, then, that this is a delayed unit step of a value a. It should also be noted that U 1 ða Þ ¼ 0 and that U 1 ða þ Þ ¼ 1. When the discontinuity has a value k 6¼ 1 and occurs at time t ¼ a, that is, when the step function has an amplitude equal to k and is delayed by a time a, then the function is designated by kU 1 ðt  aÞ and represented as shown in Fig. 2.1c. The possibility of making delays and amplitude changes in the unit step function allows to express analytically a large number of different pulses. For example, the rectangular pulse shown in Fig. 2.2a can be expressed by adding two step functions, namely: pðtÞ ¼ AU 1 ðt  aÞ  AU 1 ðt  bÞ;

ð2:3Þ

as explained in Fig. 2.2b. p(t) A

0

AU-1(t-a)

A

a

b

(a)

t

0

a

t

b

-A

(b)

-AU-1(t-b)

Fig. 2.2 a Pulse function; b Unitary step functions used to represent the pulse

2.1 Single Step Function

81

Fig. 2.3 Triangular pulse

p(t) 2 1

0

1

2

3

t

The multiplication of the rectangular, dashed, unitary pulse equation, between t ¼ 1 and t ¼ 3, that is, U 1 ðt  1Þ  U 1 ðt  3Þ, by the line defined by equation ðt  1Þ produces the triangular pulse pðtÞ ¼ ðt  1Þ½U 1 ðt  1Þ  U 1 ðt  3Þ;

ð2:4Þ

shown in full line in Fig. 2.3. In the theory of electrical circuits, it is common to convene the instant of application of the excitation as the instant t ¼ 0. This means that if an ideal switch is closed at time t ¼ 0, in series with a source of direct voltage E, for example, by exciting the circuit from t ¼ 0, as shown in Fig. 2.4a, then it is possible to eliminate the switch and take the voltage source as eðtÞ ¼ EU 1 ðtÞ, as shown in Fig. 2.4b. In
both Fig. 2.4a and Fig. 2.4b, we have that: 0 for t\0 eð t Þ ¼ ¼ EU1 ðtÞ. E for t [ 0 Therefore, if the switch is closed in series with a sinusoidal voltage source E cos xt, at time t ¼ 0, then the switch can be eliminated by taking the voltage source como eðtÞ ¼ E cosðxt þ hÞU 1 ðtÞ, where the phase angle h was introduced to express the value that the voltage source eðtÞ presents exactly at the moment of closing the switch, t ¼ 0, that is, eð0Þ ¼ E cos ht. (The term sinusoidal voltage source will be applied here for both E cos xt and E cos xt, as the only difference between them is a 90º lag).

Fig. 2.4 a Switching at time t ¼ 0; b Equivalent circuit with unit step function U 1 ðtÞ

82

2 Singular Functions

For example, if the switch is closed at the instant when the source voltage is maximum, then h ¼ 00 and eðtÞ ¼ E cos xt:U 1 ðtÞ. If the switch is closed when the source voltage is passing exactly zero, with a positive derivative, then h ¼ 900   and eðtÞ ¼ E cos xt  900 U 1 ðtÞ ¼ Esenxt:U 1 ðtÞ. If, however, the switch is and façade when the source voltage is minimal, h ¼ 1800   0 eðtÞ ¼ E cos xt þ 180 U 1 ðtÞ ¼ E cos xt:U 1 ðtÞ. Many other unique functions can be obtained by successive integrations of the unit step function. The unit ramp function, U 2 ðtÞ, for example, is the integral of the unit step function U 1 ðtÞ, that is, it represents the area involved by the unit step function from 1 to the instant t. Then, Z U2 ðtÞ ¼

Z U1 ðtÞdt ¼ ¼

Z

t

1 Z 0 1

U1 ðkÞdk ¼ Z 0:dt þ

t

Z

0

U1 ðkÞdk þ

1




1:dk ¼

0þ

t

U1 ðkÞdk

0þ

0

for t  0

t

for t  0: ð2:5Þ

So Eq. (2.5) can also be written as
U2 ðtÞ ¼ t:U1 ðtÞ ¼

0 t

for t  0 for t  0:

ð2:6Þ

Figure 2.5a shows the unitary ramp function U2 ðtÞ, which is continuous at t ¼ 0, and in Fig. 2.5b, the unitary ramp function delayed of a, namely:
U2 ðt  aÞ ¼ ðt  aÞU1 ðt  aÞ ¼

0 for t  a t  a for t  a:

ð2:7Þ

From the last term in Eq. (2.5), the one containing the switch, and also from Fig. 2.5a, there is that

U-2(t-a)

U-2(t)

t-a

t 45∞

0

t

(a)

45∞ 0

a

(b)

Fig. 2.5 a Unitary ramp function U 2 ðtÞ; b Unitary ramp function delayed of a: U 2 ðt  aÞ

t

2.1 Single Step Function

83

U 1 ðtÞ ¼

d U 2 ðtÞ: dt

ð2:8Þ

In turn, the integration of the unit ramp function U 2 ðtÞ, that is, the area comprised by the unit ramp function from 1 to time t, generates the singular second degree unit parabola function, U 3 ðtÞ. So, Z U3 ðtÞ ¼

Z U2 ðtÞdt ¼

Z

t

Z

t

U2 ðkÞdk ¼ U2 ðkÞdk þ 1 1
Z t Z 0 0 for t  0 0:dt þ kdk ¼ t2 ¼ for  0: 1 0 2

t

U2 ðkÞdk

0

ð2:9Þ Alternatively, Eq. (2.9) can be written as t2 U3 ðtÞ ¼ U1 ðtÞ ¼ 2




for t  0 for  0:

0 t2 2

ð2:10Þ

2

Þ The functions U 3 ðtÞ and U 3 ðt  aÞ ¼ ðta 2 U 1 ðt  aÞ are represented in Figs. 2.6a and 2.6b, respectively. From the last term of Eq. (2.9), the one containing the switch, and also from Fig. 2.6a, we have

U 2 ðtÞ ¼

d U 3 ðtÞ: dt

ð2:11Þ

Continuing with the successive integrations, other singular functions are being obtained, namely: Z U4 ðtÞ ¼

Z U3 ðtÞdt ¼ U3 ðtÞ ¼

t 1

U3 ðkÞdk ¼

t3 U1 ðtÞ 3!

ð2:12Þ

d U4 ðtÞ dt

ð2:13Þ

U-3(t-a)

U-3(t)

(t-a)2/2

t2/2

t

0

(a)

0

t a

(b)

Fig. 2.6 a Unitary parabola function U 3 ðtÞ; b Unitary parabola function delayed of a: U 3 ðt  aÞ

84

2 Singular Functions

Z U5 ðtÞ ¼

Z U4 ðtÞdt ¼

t 1

Z

Z Um ðtÞdt ¼

Um ðtÞ ¼

t4 U1 ðtÞ 4!

d U5 ðtÞ dt

U4 ðtÞ ¼ Uðm þ 1Þ ðtÞ ¼

U4 ðkÞdk ¼

t 1

ð2:14Þ ð2:15Þ

Um ðkÞdk ¼

tm U1 ðtÞ m!

d Uðm þ 1Þ ðtÞ; m ¼ 1; 2; 3; . . . dt

ð2:16Þ ð2:17Þ

Generalizing, we have that: Z U k1 ðtÞ ¼

Z U k ðtÞdt ¼

U k ðt Þ ¼

t

1

U k ðkÞdk

d U k1 ðtÞ dt

ð2:18Þ ð2:19Þ

In the process of integrating a function that is multiplied by a unit step, the unit step factor should be used to define the lower limit of the integral, but it is neither necessary nor desirable to include it in the integration process, that is, not integrating. After the integration has been carried out, the corresponding unit step factor is introduced again. This can be described mathematically as: Z

t

1

Z

t

Z

ð2:20Þ

 f ðkÞdk U 1 ðt  aÞ:

ð2:21Þ

0

Z

t

f ðkÞU1 ðk  aÞdk ¼ a

0

2.2

 f ðkÞdk U 1 ðtÞ:

t

f ðkÞU 1 ðkÞdk ¼

Unitary Impulse Function

Taking k ¼ 0 in Eqs. (2.18) and (2.19) give the expressions Z U 1 ðtÞ ¼

Z U 0 ðtÞdt ¼

U 0 ðtÞ ¼

t

1

d U1 ðtÞ dt

U0 ðkÞdk

ð2:22Þ ð2:23Þ

But, as they were obtained, Eqs. (2.18) and (2.19) are only valid for integer and negative values of k, ðk ¼ 1; 2; 3;    Þ. And you can’t just assign zero to the k index and extrapolate from there. However, and despite the inconsistency pointed

2.2 Unitary Impulse Function

85

out when adopting k ¼ 0, Eqs. (2.18) and (2.19) are absolutely correct, as we intend to show, with a wealth of details, below. Equation (2.22) suggests the existence of a singular function U 0 ðtÞ, of unit area for t [ 0 and null for t\0, according to the definition of the unit step function U 1 ðtÞ, Eq. (2.1). On the other hand, Eq. (2.23) establishes that the function U 0 ðtÞ is null for t 6¼ 0, since the unit step derivative is null for t 6¼ 0, that is, derived from zero for t\0 and from 1 to t [ 0. Thus, the existence of the function U 0 ðtÞ is restricted only to the time t ¼ 0, which, in turn, implies that the function U 0 ðtÞ is the derivative of the unit step function only at time t ¼ 0, this instant in which it is discontinuous. And this, from the point of view of conventional classical mathematics, is inadmissible (in classical mathematics there is no derivative of discontinuous function). So the function U0 ðtÞ; in addition to having to exist only at time t ¼ 0, must involve a unit area, because, Rt 1

Rt 1

U0 ðkÞdk ¼ 0, when t\0, given that U 1 ðtÞ ¼ 0 for t\0, and U0 ðkÞdk ¼ 0, when t [ 0, given that U 1 ðtÞ ¼ 1 for t [ 0.

The impossibility of proposing a rigorous definition for the function U 0 ðtÞ, based on classical mathematics of the 1930s, led physicist Paul AM Dirac, who employed it for the first time in his work in Quantum Mechanics, to call it improper function. Currently, the U 0 ðtÞ function is called the unitary impulse function or the Dirac delta function. Many authors denote it as dðtÞ. In 1950, Laurent Schwartz published a treatise entitled “The Theory of Distributions”, within which he proposed a sufficiently rigorous and satisfactory basis for Dirac’s delta function. Unfortunately, Schwartz’s theory proved to be too abstract for the mathematicians and physicists of the day. In 1953, George Temple developed a more elementary theory, although no less rigorous, under the title “Theories and Applications of Generalized Functions”, within which the impulse function was rigorously defined; thus, Eqs. (2.22) and (2.23) are duly proved. In Engineering courses, the unit impulse function is usually presented in three different ways, namely: as a limit, as an equation and through a property. ∎ Definition as Limit In this case, the unit impulse function is defined as the limit of a sequence of functions f n ðtÞ, that is: lim

n ! 1 fn ðtÞ ¼ dðtÞ ¼ U0 ðtÞ;

ð2:24Þ

provided that the functions f n ðtÞ satisfy the following conditions: Z

þ1 1

fn ðtÞdt ¼ 1; and

lim

n ! 1 fn ðtÞ ¼ 0; for t 6¼ 0: (The condition imposed by Eq. (2.26) is not always necessary).

ð2:25Þ ð2:26Þ

86

2 Singular Functions

As can be seen, there are a large number of functions that satisfy Eqs. (2.25) and (2.26). Figure 2.7 shows the rectangular pulse pD ðtÞ, with base D, between D=2 and þ D=2, and height 1=D. The pulse sequence formed when D ! 0 satisfies Eq. (2.26). It also satisfies Eq. (2.25), because Z

þ1

1

pD ðtÞdt ¼ 1;

regardless of the value assigned to D. In other words, when D tends to zero, the limit results in “a zero-based pulse, infinite amplitude and unit area”, that is, lim

U0 ðtÞ ¼ D ! 0 pD ðtÞ:

ð2:27Þ

The triangular pulse shown in Fig. 2.8 also becomes a unitary pulse when a tends to zero. Like this, lim

U0 ðtÞ ¼ a ! 0 pa ðtÞ: Another function that generates the unit impulse is shown in Fig. 2.9. So, t =e U0 ðtÞ ¼ lim e!0 pffiffiffiffiffi pe 2

The unit impulse function will be represented as shown in Fig. 2.10a. Figure 2.10b shows an area A pulse, delayed by time a. Logically, to obtain a pulse of area A, starting from Fig. 2.7, for example, the amplitude of the rectangular pulse must be A=D. It is interesting to note, then, that the factor A that multiplies an impulse does not represent its amplitude, which is infinite, but the area it contains. Positive and negative impulses must be represented by arrows of the same size, drawn above and below the time axis, respectively.

p∆(t)

pa(t)

1/∆

-∆/2 0

1/a

∆/2

t

-a

a

Fig. 2.7 Unitary impulse function U 0 ðtÞ defined as a rectangular pulse

t

2.2 Unitary Impulse Function

1/∆

87

p∆(t)

-∆/2 0

pa(t) 1/a

t

∆/2

t

a

-a

Fig. 2.8 Unitary impulse function U 0 ðtÞ defined as a triangular pulse

Fig. 2.9 Unitary impulse function U 0 ðtÞ defined as a gaussian function

Fig. 2.10 a Unitary impulse function U 0 ðtÞ; b Unitary Amplitude A impulse function delayed of a: U 0 ðt  aÞ

a

b AU0(t-a)

U0(t)

0

t

0

t

a

∎ Definition by Equation The definition by the equation is simply as follows:


tÞ ¼ 0 for t 6¼ 0 RU0þð1 1 U0 ðtÞdt ¼ 1:

ð2:28Þ

As the unit impulse is equal to zero for t 6¼ 0, one can also write that Z

0þ

U0 ðtÞdt ¼ 1;

0

because the total area of the unit impulse is “concentrated’’ at t ¼ 0. In consequence,

ð2:29Þ

88

2 Singular Functions

Z

Z

0

þ1

U0 ðtÞdt ¼

1

U0 ðtÞdt ¼ 0:

ð2:30Þ

0þ

∎ Definition by Property Also known as the pulse sampling property, it is the following: Z

þ1 1

f ðtÞU 0 ðt  aÞdt ¼ f ðaÞ;

ð2:31Þ

provided that f ðaÞ exists. Equation (2.31) is easily justified, since U 0 ðt  aÞ ¼ 0 for t 6¼ a. So, f ðtÞU0 ðt  aÞ ¼ 0 for all t 6¼ a: For t ¼ a, where f ðtÞ is defined at this point, f ðtÞU 0 ðt  aÞ becomes f ðaÞU 0 ðt  aÞ. Then, f ðtÞU0 ðt  aÞ ¼ f ðaÞU 0 ðt  aÞ

ð2:32Þ

So, Z

þ1

1

Z f ðtÞU0 ðt  aÞdt ¼

þ1

1

Z f ðaÞU0 ðt  aÞdt ¼ f ðaÞ

þ1

1

U0 ðt  aÞdt

¼ f ðaÞ:1 ¼ f ðaÞ: As a particular case of Eq. (2.31), it is obvious that: Z

þ1 1

f ðtÞU 0 ðtÞdt ¼ f ð0Þ:

ð2:33Þ

One way of defining the unit impulse function also emerges from the Inverse Fourier Transform: U 0 ðt Þ ¼

1 2p

Z

þ1

1

jxt dx ¼

1 2p

Z

þ1 1

cos xt dx ¼

1 p

Z

þ1

cos xt dx:

0

Using Eq. (2.33), we can now present a first justification for the very important Eq. (2.23). R R Integrating by parts ð udv ¼ uv  vduÞ the expression f ðtÞ dtd U 1 ðtÞ; between the limits 1 and þ 1, comes that:

2.2 Unitary Impulse Function

Z

þ1

1

89

Z þ1 þ1 d d  f ðtÞ U1 ðtÞdt ¼ U1 ðtÞf ðtÞj U1 ðtÞ f ðtÞdt dt dt 1 1 Z þ1 d ½ f ðt Þ ¼ f ð þ 1Þ 

ð2:34Þ

0

¼ f ð þ 1Þ  f ðtÞj

þ1 ¼ f ð0Þ: 0

Comparing Eq. (2.34) with Eq. (2.33), or considering the operators that produce the same result to be the same, it is concluded that U 0 ðt Þ ¼

d U 1 ðtÞ; dt

Equation (2.23) is justified. A second way of justifying Eq. (2.23), and also (2.22), which is more intuitive and less abstract, is done with the help of Figs. 2.11a, b. It is easy to see that: d f ðt Þ dt 1

f 0 ðtÞ ¼ Z f1 ðtÞ ¼

t 1

ð2:35Þ

f0 ðkÞdk

ð2:36Þ

When a tends to zero, the function f 1 ðtÞ converges to the function U 1 ðtÞ, and the function f 0 ðtÞ, to the function U 0 ðtÞ: Thus, Eqs. (2.36) and (2.35) are transformed, respectively, into Eqs. (2.22) and (2.23), namely: Z U1 ðtÞ ¼

t 1

Z U0 ðkÞdk ¼

U 0 ðt Þ ¼

U0 ðtÞdt:

d U 1 ðtÞ: dt

It is justified, therefore, that Eqs. (2.18) and (2.19) are also valid for k ¼ 0. f0(t) f-1(t)

1/a

1

0

a

t

(a)

0

a

t

(b)

Fig. 2.11 a Approximate representation of a unitary step function; b Approximate representation of a unitary impulse function

90

2 Singular Functions

It is also observed, from the last expression, that the unit of the unitary impulse function is s1 , when the unit step function is dimensionless. Otherwise, the unit is second−1 times the unit of the step function. For example, if the step function is the electrical charge. q(t) ¼ QU1 (t)[C], then, q(0 ) ¼ 0 [C], q(0 þ ) ¼ Q[C], and   d d C iðtÞ ¼ qðtÞ ¼ ½QU1 ðtÞ ¼ QU0 ðtÞ ¼ QU0 ðtÞ½A: dt dt S It should be noted that the area of a current impulse, Q, represents the electrical charge that the current impulse transfers instantly at t ¼ 0, causing qð0 þ Þ ¼ Q½C 6¼ qð0 Þ ¼ 0½C. If the step has amplitude A and its discontinuity occurs at time t ¼ a, that is, if the step is AU 1 ðt  aÞ, then, according to the chain derivation rule, d d dU 1 ðt  aÞ d ðt  aÞ : ¼ AU 0 ðt  aÞ ½ AU 1 ðt  aÞ ¼ A U 1 ðt  aÞ ¼ A dt dt d ðt  aÞ dt ð2:37Þ Equation (2.37) highlights a most relevant fact, namely: The derivative of a finite discontinuity A, present in any step function, is an impulse function of area A, located exactly at the moment when the step function discontinuity occurs. this conclusion can be expanded to any function f ðtÞ, which presents a finite discontinuity at any time t ¼ a. Its derivation should be done normally for the intervals 1\t  a e a þ  t\ þ 1: Then, the impulse ½ f ða þ Þ  f ða ÞU 0 ðt  aÞ must be added, exactly at time t ¼ a:

2.3

The Family of Singular Functions

When k ¼ 1 is assigned to Eqs. (2.19) and (2.18), it follows that: d U 0 ðtÞ: dt Z U1 ðkÞdk ¼ U1 ðtÞdt:

U 1 ðt Þ ¼ Z U 0 ðt Þ ¼

t

1

ð2:38Þ ð2:39Þ

Although again there are difficulties when taking k ¼ 1 in Eqs. (2.18) and (2.19), which have been deducted for integer and negative values of k, we will no longer enter into the merits of this question, simply accepting that Eqs. (2.38) and (2.39) are correct. The singular function U1 ðtÞ is called a unit double.

2.3 The Family of Singular Functions

91

One way to try to visualize the unit stunt function uses Fig. 2.12. In part (a) of the figure, a triangular pulse f 0 ðtÞ is shown, which, as seen above, generates a unit impulse when it tends to zero. In part (b) the function f 1 ðtÞ is shown, which is the derivative of the pulse of part (a). When it tends to zero, the function f 1 ðtÞ will converge to the unitary double function, U 1 ðtÞ. Part (c) of Fig. 2.12 shows the graphical representation used by some authors. But the graphic representation that will be used here is the one shown in part (d) of the same Fig. 2.12. That is, a single arrow will be used, not two, but writing next to it that it is the stunt function, U 1 ðtÞ: And the main reason for the adoption of this representation is justified in the next paragraph. Assigning the integer and positive values 2; 3; 4; :::; to the k index in Eq. (2.19), other singular functions are being obtained, namely: U 2 ðtÞ ¼ dtd U 1 ðtÞ;U 3 ðtÞ ¼ dtd U 2 ðtÞ;U 4 ðtÞ ¼ dtd U 3 ðtÞ, etc., which, of course, also satisfy Eq. (2.18). They will also always be represented here by a single arrow, next to which the function referred to will be noted. Otherwise, using the notation of other authors, the function U 2 ðtÞ would have to be represented by three arrows, U 3 ðtÞ by four, and so on, creating a lot of confusion when two or more of them are occurring at the same point. Obviously, the impulsive functions U 0 ðtÞ, U 1 ðtÞ, U 2 ðtÞ, … must be understood in the sense of generalized, symbolic functions or distributions, and not in the classical sense of functions.

1/a

-a

f1(t)

f0(t)

0

-1/a2

t

a

(a)

-a

0

t

a

(b) -1/a2

U 1(t)

∞

U 1(t)

1 0

t 1

-∞

t

0

(c)

(d)

Fig. 2.12 a Triangular pulse function; b Derivative of pulse of (a); c Unitary double function, U 1 ðtÞ; d graphic representation that will be used here for a unitary double function, U 1 ðtÞ

92

2 Singular Functions

It remains evident, then, that the unit impulse function, U 0 ðtÞ, can be considered as the starting point for obtaining the entire family of singular functions, through successive integrations or derivations. Seen from this point of view, the reason for choosing negative and positive indices used in the notation of singular functions is perfectly understood. Thus, negative indices mean that the singular function in question was obtained by successively integrating the unit impulse function, U 0 ðtÞ. And positive indexes, by successive derivation of it. Therefore, to derive or integrate a singular function just add or subtract one to its respective index. There is no simpler way to derive or integrate a function! The notation U n ðtÞ used here for singular functions also has the advantage of bringing together the Laplace Transform of all of them in a single formula, namely: L½U n ðtÞ ¼ sn ; n ¼ 0; 1; 2;    

ð2:40Þ

Then, 1 1 L½U0 ðtÞ ¼ s0 ¼ 1; L½U1 ðtÞ ¼ s; L½U2 ðtÞ ¼ s2 ; L½U1 ðtÞ ¼ ; L½U2 ðtÞ ¼ 2 : s s Figure 2.13 presents a synthetic representation of the singular functions, with the main objective of reaffirming that they constitute a family of functions, having the unitary impulse function as a generator. The singular functions obtained by deriving the unit impulse function, on the right in Fig. 2.13, include the following properties, demonstrated through the process of integration by parts: Z

þ1 1

U1 ðtÞf ðtÞdt ¼ U0 ðtÞf ðtÞj ¼ U0 ðtÞf ð0Þj

þ1 1 þ1 1

Z 

þ1 1 0

 f ð 0Þ

U0 ðtÞf 0 ðtÞdt

Z

þ1 1

ð2:41Þ 0

U0 ðtÞdt ¼ f ð0Þ:

Fig. 2.13 Synthetic representation of the family of singular functions, having the unitary impulse function as a generator

2.3 The Family of Singular Functions

Z

þ1 1

93

Z þ1 þ1  U2 ðtÞf ðtÞdt ¼ U1 ðtÞf ðtÞj U1 ðtÞf 0 ðtÞdt 1 1 Z þ1 ¼ U1 ðtÞf 0 ðtÞdt ¼ ½f 00 ð0Þ ¼ f 00 ð0Þ:

ð2:42Þ

1

Generally, Z

þ1

1

Z Un ðtÞf ðtÞdt ¼ 

þ1

1

U n1 ðtÞf 0 ðtÞdt ¼ ð1Þn f ðnÞ ð0Þ; n ¼ 1; 2; 3; . . . ð2:43Þ

where, f ðnÞ ð0Þ ¼

dn f ðtÞjt¼0: dtn

And also, Z

þ1

1

Un ðt  aÞf ðtÞdt ¼ ð1Þn f ðnÞ ðaÞ;

ð2:44Þ

for n ¼ 0; 1; 2; 3; . . .. Note that Eqs. (2.43) and (2.44) encompass Eqs. (2.33) and (2.31), respectively. Example 2.1 Prove the following properties: Z

þ1

(aÞ 1

Z

þ1

(bÞ 1

Z U0 ðt  aÞf ðtÞdt ¼

U 0 ðatÞf ðtÞdt ¼

1 jaj

Z

þ1 1

þ1

1

U0 ðtÞf ðt þ aÞdt ¼ f ðaÞ:

U0 ðtÞf

 t 1 dt ¼ f ð0Þ; a 6¼ 0: a jaj

ð2:45Þ ð2:46Þ

Solution (a) Considering t  a ¼ s; t ¼ s þ a; dt ¼ ds; Z

þ1

1

Z U0 ðt  aÞf ðtÞdt ¼

Because of Eq. (2.33),

þ1 1

Z U0 ðsÞf ðs þ aÞds ¼

þ1 1

U0 ðtÞf ðt þ aÞdt

94

2 Singular Functions

Z

þ1

1

U0 ðtÞf ðt þ aÞdt ¼ f ðt þ aÞjt¼0 ¼ f ðaÞ:

(b) Considering, now, at ¼ s; t ¼ s=a; dt ¼ ð1=aÞds, we have for a [ 0, that: Z

þ1

1

Z Z 1 þ1 1 þ1 U0 ðatÞf ðtÞdt ¼ U0 ðsÞf ðs=aÞds ¼ U0 ðtÞf ðt=aÞdt a 1 a 1 1 1 ¼ f ð0Þ ¼ f ð0Þ: a j aj

When a\0, t and s you have opposite signs and Z

þ1

1

Z Z 1 1 1 þ1 U0 ðsÞf ðs=aÞds ¼  U0 ðtÞf ðt=aÞdt a þ1 a 1 1 ¼ f ð0Þ: jaj

U0 ðatÞf ðtÞdt ¼

Example 2.2 Prove that the unit impulse function is even. Solution From Eqs. (2.46) and (2.33), it follows that Z

þ1

1

U0 ðatÞf ðtÞdt ¼

1 1 f ð 0Þ ¼ j aj j aj

Z

þ1

1

Z U0 ðtÞf ðtÞdt ¼

þ1 1

½U0 ðtÞ=jajf ðtÞdt

From the extreme integrals, considering that f ðtÞ is any function, we conclude that U 0 ðatÞ ¼

1 U 0 ðt Þ j aj

ð2:47Þ

Taking a ¼ 1 in the above relationship, it follows that U 0 ðtÞ ¼ U0 ðtÞ;

ð2:48Þ

showing that U0 ðtÞ is an even function. Example 2.3 Demonstrate that Z

þ1 1

U1 ðatÞf ðtÞdt ¼

1 aj aj

Z

þ1

1

U1 ðtÞf ðtÞdt ¼

1 0 f ð0Þ; a 6¼ 0: aj aj

ð2:49Þ

2.3 The Family of Singular Functions

95

Solution For a [ 0, considering s ¼ at; t ¼ s=a; dt ¼ ð1=aÞds, it comes that Z

þ1

1

Z Z 1 þ1 1 þ1 U1 ðatÞf ðtÞdt ¼ U1 ðsÞf ðs=aÞds ¼ U1 ðtÞf ðt=aÞdt a 1 a 1     1 d t 1 df ðt=aÞ ðt=aÞ : ¼  f jt¼0 ¼  j a dt a a d ðt=aÞ dt t¼0   Z þ1 1 0 1 1 0 1 ¼  f ðt=aÞ: jt¼0 ¼  2 f ð0Þ ¼ 2 U1 ðtÞf ðtÞdt; a a a a 1 ð2:50Þ

according to Eq. (2.41). For a\0, with the same variable change, considering that now when t ! 1; s ! þ 1 and vice versa, it comes that Z

þ1

1

Z Z 1 1 1 þ1 U1 ðatÞf ðtÞdt ¼ U1 ðsÞf ðs=aÞds ¼  U1 ðsÞf ðs=aÞds a þ1 a 1   Z þ1 1 1 d t  f U1 ðtÞf ðt=aÞdt ¼ ¼ j a 1 a dt a t¼0   1 df ðt=aÞ ðt=aÞ 1  : ¼ j ¼ ½f 0 ðt=ajt¼0 a d ðt=aÞ dt t¼0 a2 Z þ1 1 0 1 ¼ 2 f ð 0Þ ¼  2 U1 ðtÞf ðtÞdt: a a 1 ð2:51Þ


With ajaj ¼

a2 a2

para a [ 0 , Equations (2.50) and (2.51) can be combined para a\0

into one: Z

þ1

1

1 U1 ðatÞf ðtÞdt ¼ aj aj

Z

þ1

1

U1 ðtÞf ðtÞdt ¼

1 0 f ð0Þ; a 6¼ 0: ajaj

Example 2.4 Prove that the unit double function is odd. Solution From the two integrals of Eq. (2.49) it appears that: U 1 ðatÞ ¼

1 U 1 ðt Þ aj aj

ð2:52Þ

96

2 Singular Functions

For a ¼ 1, U 1 ðtÞ ¼ U1 ðtÞ; confirming that the unit stunt function U 1 ðtÞ is odd. Example 2.5 Prove that f ðtÞU 1 ðtÞ ¼ f ð0ÞU 1 ðtÞ  f 0 ð0ÞU 0 ðtÞ:

ð2:53Þ

Then, calculate tU 1 ðtÞ. Solution From Eq. (2.32), for a ¼ 0, we have that: f ðtÞU 0 ðtÞ ¼ f ð0ÞU 0 ðtÞ:

ð2:54Þ

Deriving the two members of Eq. (2.54), it results 0

f ðtÞU 0 ðtÞ þ f ðtÞU 1 ðtÞ ¼ f ð0ÞU 1 ðtÞ 0

0

But, according to Eq. (2.54),f ðtÞU 0 ðtÞ ¼ f ð0ÞU 0 ðtÞ. So, 0

f ðtÞU 1 ðtÞ ¼ f ð0ÞU 1 ðtÞ  f ð0ÞU 0 ðtÞ: For f ðtÞ ¼ t, tU 1 ðtÞ ¼ U 0 ðtÞ:

ð2:55Þ

f ðtÞU 2 ðtÞ ¼ f ð0ÞU 2 ðtÞ  2f 0 ð0ÞU 1 ðtÞ þ f 00 ð0ÞU 0 ðtÞ:

ð2:56Þ

Example 2.6 Establish that

Then, calculate t2 U 2 ðtÞ. Solution Deriving Eq. (2.53), it follows that: 0

0

f ðtÞU 1 ðtÞ þ f ðtÞU 2 ðtÞ ¼ f ð0ÞU 2 ðtÞ  f ð0ÞU 1 ðtÞ where, 0

0

f ðtÞU 2 ðtÞ ¼ f ð0ÞU 2 ðtÞ  f ð0ÞU 1 ðtÞ  f ðtÞU 1 ðtÞ

2.3 The Family of Singular Functions

97

But, because of Eq. (2.53) itself, f 0 ðtÞU1 ðtÞ ¼ f 0 ð0ÞU1 ðtÞ  f 00 ð0ÞU0 ðtÞ So, after replacing the last equation on the penultimate, f ðtÞU2 ðtÞ ¼ f ð0ÞU2 ðtÞ  2f 0 ð0ÞU1 ðtÞ þ f 00 ð0ÞU0 ðtÞ: For f ðtÞ ¼ t2 , Eq. (2.56) provides that: t2 U 2 ðtÞ ¼ 2U 0 ðtÞ:

ð2:57Þ

The generalization of the identities of examples 2.5 and 2.6 is f ðtÞUn ðtÞ ¼

n X

ð1Þk

k¼0

n! f ðkÞ ð0ÞUnk ðtÞ; for n ¼ 1; 2; 3; . . . k!ðn  kÞ!

ð2:58Þ

It can also be shown that f ðtÞUn ðt  aÞ ¼

n X

ð1Þk

k¼0

n! f ðkÞ ðaÞUnk ðt  aÞ; for n ¼ 1; 2; 3; . . . k!ðn  k Þ! ð2:59Þ

Example 2.7 Prove that
U1 ðatÞ ¼

for a [ 0 U1 ðtÞ U1 ðtÞ for a\0:

Solution The definition of unitary step, given by Eq. (2.1), allows to write that
U1 ðatÞ ¼

0 for at\0 1 for at [ 0:

Or,
U1 ðatÞ ¼

0 1

for t\0 ¼ U1 ðtÞ for a [ 0: for t [ 0

Now let a ¼ b; b [ 0. From Eq. (2.1),

ð2:60Þ

98

2 Singular Functions


U1 ðatÞ ¼ U1 ðbtÞ ¼

0

if bt\0

1

if bt [ 0


¼

0 if bt [ 0 1 if bt\0


¼

0

if t [ 0

1

if t\0

¼ U1 ðtÞif a\0; Thus, Eq. (2.60) is proved. Example 2.8 Prove that U n ðtÞ ¼ ð1Þn U n ðtÞ; for n ¼ 0; 1; 2; 3; . . .

ð2:61Þ

Solution For n ¼ 0, the identity has already been demonstrated, Eq. (2.48), showing that the impulse function is even, namely: U0 ðtÞ ¼ þ U0 ðtÞ: For n ¼ 1, the Eq. (2.61) has already been demonstrated in example 2.4, showing that the unit double function is odd. But it's easy to prove it. From Eq. (2.19), for k ¼ 1, U 1 ðt Þ ¼

d U 0 ðt Þ dt

So, U1 ðtÞ ¼

d d d U0 ðtÞ ¼ U0 ðtÞ ¼  U0 ðtÞ d ðtÞ dt dt

¼ U1 ðtÞ: Also from Eq. (2.19), for k ¼ 2, U 2 ðt Þ ¼

d U 1 ðt Þ dt

So, U2 ðtÞ ¼

d d d d U1 ðtÞ ¼ U1 ðtÞ ¼  ½U1 ðtÞ ¼ U1 ðtÞ ¼ þ U2 ðtÞ: d ðtÞ dt dt dt

Still from Eq. (2.19), U 3 ðt Þ ¼

d U 2 ðt Þ dt

2.3 The Family of Singular Functions

99

So, U3 ðtÞ ¼

d d d U2 ðtÞ ¼ U2 ðtÞ ¼  ½ þ U2 ðtÞ ¼ U3 ðtÞ: d ðtÞ dt dt

Therefore, generalizing, in fact, Un ðtÞ ¼ ð1Þn Un ðtÞ; for n ¼ 0; 1; 2; 3; . . . Example 2.9 Express the function f ðtÞ of Fig. 2.14 in a series of singular functions.

f(t)

Fig. 2.14 Function of Example 2.9

4 3 2 1 0

T

2T

3T

t

4T

Solution By simple inspection, it is easy to conclude that f ðtÞ ¼ U1 ðtÞ þ U1 ðt  T Þ þ U1 ðt  2T Þ þ U1 ðt  3T Þ þ    Example 2.10 Perform the step expansion of the periodic function f 1 ðtÞ shown in Fig. 2.15a.

2K

f1(t) K

0

K

A

2A

-K

3A

4A

5A

t

0

A

2A

3A 4A

-K

(a) -2K (b)

Fig. 2.15 a Periodic function f 1 ðtÞ; b Alternating rectangular pulse functions

5A

t

100

2 Singular Functions

Solution According to Fig. 2.2 and Eq. (2.3), f 1 ðtÞ can be written as the sum of alternating rectangular pulse equations, namely: f1 ðtÞ ¼ K ½U1 ðtÞ  U1 ðt  AÞ  K ½U1 ðt  AÞ  U1 ðt  2AÞ þ K ½U1 ðt  2AÞ  U1 ðt  3AÞ  K ½U1 ðt  3AÞ  U1 ðt  4AÞ þ . . . ¼ K ½U1 ðtÞ  2U1 ðt  AÞ þ 2U1 ðt  2AÞ  2U1 ðt  3AÞ þ . . . Or, f 1 ðtÞ ¼ KU 1 ðtÞ þ 2K

þ1 X

ð1Þn U 1 ðt  nAÞ

n¼1

Figure 2.15b shows the first steps contained in the solution obtained. Example 2.11 Decompose the function gðtÞ shown in Fig. 2.16 into singular functions. g(t)

Fig. 2.16 Function gðtÞ of Example 2.11

2

A B

1 0

1

2

3

t

Solution The function gðtÞ can be considered as the juxtaposition of the next three pulses. The rectangular pulse of amplitude 2, between t ¼ 0 and t ¼ 1, and which is defined by 2½U 1 ðtÞ  U 1 ðt  1Þ The equation of the line passing through points A and B is given by –t + 3. Therefore, the trapezoidal pulse between t ¼ 1 e t ¼ 2 is given by ðt þ 3Þ½U1 ðt  1Þ  U1 ðt  2Þ

2.3 The Family of Singular Functions

101

Finally, the amplitude 1 rectangular pulse, between t ¼ 2 and t ¼ 3, is defined by U 1 ðt  2Þ  U 1 ðt  3Þ Considering, then, that the function gðtÞ is the juxtaposition, in sequence, of the three pulses mentioned, it can be written that: gðtÞ ¼ 2½U1 ðtÞ  U1 ðt  1Þ þ ðt þ 3Þ½U1 ðt  1Þ  U1 ðt  2Þ þ U1 ðt  2Þ  U1 ðt  3Þ ¼ 2U1 ðtÞ  2U1 ðt  1Þ  ðt  1ÞU1 ðt  1Þ þ 2U1 ðt  1Þ þ ðt  2ÞU1 ðt  2Þ  U1 ðt  2Þ þ U1 ðt  2Þ  U1 ðt  3Þ ¼ 2U1 ðtÞ  ðt  1ÞU1 ðt  1Þ þ ðt  2ÞU1 ðt  2Þ  U1 ðt  3Þ: Referring to Eq. (2.7), we can finally write that: gðtÞ ¼ 2U 1 ðtÞ  U 2 ðt  1Þ þ U 2 ðt  2Þ  U 1 ðt  3Þ:

ð2:62Þ

The singular functions that make up gðtÞ are shown, in sequence, in Fig. 2.17. Fig. 2.17 Singular functions that make up gðtÞ 2 1 0

1

2

3

4

5

t

-1 -2

Example 2.12 Find the derivative of the functions U 1 ðtÞ e f ðtÞ shown in Figs. 2.18a, b. Solution As can be seen from Fig. 2.18c, U1 ðtÞ ¼ U 1 ðt þ 1Þ  U1 ðtÞ ¼ 1  U1 ðtÞ: So, d d d U 1 ðtÞ ¼ ½1  U 1 ðtÞ ¼  U 1 ðtÞ: dt dt dt

102

2 Singular Functions U-1(-t)

f(t) 1

1 t

0

t

0

(a)

-2

(b)

1 t

0 -1

-U-1(t)

(c) Fig. 2.18 a Unit step function U 1 ðtÞ; b Function f ðtÞ; c Composition of functions

Therefore, d U 1 ðtÞ ¼ U 0 ðtÞ: dt That is, the derivative of the function U 1 ðtÞ is a pulse with area equal to 1, which is exactly the value of the discontinuity present in the function U 1 ðtÞ, at time t ¼ 0, that is: 0  1 ¼ 1. Similar reasoning shows that the function f ðtÞ, in Fig. 2.18b, can be expressed as f ðtÞ ¼ 2U 1 ðt þ 1Þ þ 3U1 ðtÞ ¼ 2 þ 3U1 ðtÞ: So, 0

f ðtÞ ¼ 3U 0 ðtÞ: Therefore, the derivative of f ðtÞ is an impulse of area equal to 3, which is the discontinuity value of f ðtÞ exactly at time t ¼ 0, that is: 1  ð2Þ ¼ 3. Example 2.13 Derive the function gðtÞ from example 2.11 and graph it.

2.3 The Family of Singular Functions

103

Solution From Eq. (2.62), we have that: d d d U2 ðt  1Þ þ U2 ðt  2Þ  U1 ðt  3Þ dt dt dt d d ð t  1Þ d d ð t  2Þ U2 ðt  1Þ: þ U2 ðt  2Þ: ¼ 2U0 ðtÞ  d ð t  1Þ dt d ð t  2Þ dt d d ð t  3Þ  U1 ðt  3Þ: d ð t  3Þ dt

g0 ðtÞ ¼ 2U0 ðtÞ 

¼ 2U0 ðtÞ  U1 ðt  1Þ þ U1 ðt  2Þ  U0 ðt  3Þ: Or, 0

g ðtÞ ¼ 2U 0 ðtÞ  ½U 1 ðt  1Þ  U 1 ðt  2Þ  U 0 ðt  3Þ: 0

The graphical representation of g ðtÞ is shown in Fig. 2.19. Note that the two finite discontinuities present in gðtÞ, Fig. 2.16, gave rise to the impulses contained 0 in g ðtÞ. Furthermore, that the areas of the two pulses correspond to the values of gðtÞ, discontinuities. g’(t)

Fig. 2.19 Graphical 0 representation of g ðtÞ

2U0(t)

0 -1

1

2

t

3 -U0(t-3)

Example 2.14 Calculate the following integrals: R þ1 (a) 1 ejxt :U0 ðt  T Þdt R þ1 (b) 1 cost:U0 ðt  p=3Þdt R þ1 (c) 1 U0 ðt  aÞ:U1 ðt  bÞdt; a [ b R þ1 (d) 1 sent:½U0 ðt  p=4Þ þ U1 ðt  p=2Þ þ U2 ðt  pÞdt R 2t (e) I ¼ 0 f ð xÞdx; sendo que f ð xÞ¼ U1 ð xÞ:U1 ðt  xÞ; t [ 0 e parameter:

104

2 Singular Functions

Solution By Eq. (2.31), R þ1 (a) 1 ejxt :U0 ðt  T Þdt ¼ ejxt jt¼T ¼ ejxT R þ1   (b) 1 cost:U0 ðt  p=3Þdt ¼ cos p=3 ¼ cos 60 ¼ sen 30 ¼ 1=2 R þ1 (c) 1 U0 ðt  aÞ:U1 ðt  bÞdt ¼ U1 ðt  bÞjt¼a ¼ U1 ða  bÞ With a [ b, U 1 ða  bÞ ¼ U 1 ðsÞ for s [ 0, that is, U 1 ða  bÞ ¼ 1. So, Z

þ1

U0 ðt  aÞ:U1 ðt  bÞdt ¼ 1; so a [ b

1

(d) Under Eq. (2.44), Z

þ1

1

sent:½U0 ðt  p=4Þ þ U1 ðt  p=2Þ þ U2 ðt  pÞdt Z þ1 sent:U0 ðt  p=4Þdt ¼ 1 Z þ1 Z þ1 sent:U1 ðt  p=2Þdt þ sent:U2 ðt  pÞdt þ 1 pffiffiffi 1 ¼ senp=4  cos p=2  senp ¼ 2=2


U1 ð xÞ ¼

0

1
1 ¼ 0
1 ¼ 0

(e)

if x\0

; U1 ðxÞ x[0 x\0 )U1 ðt  xÞ ¼ U1 ½ðx  tÞ x[0
x  t\0 1 if x\t : if x  t [ 0 0 if x [ t

if if if if

So,
U1 ðt  xÞ ¼

1 for  1\x\t eU ð xÞ ¼ 0 for t\x\ þ 1 1




0 for  1\x\0 1 for 0\x\ þ 1:

Then, Z I¼ 0

2t

Z f ð xÞdx ¼ 0

t

Z 1:dx þ t

2t

0:dx ¼ t:

2.3 The Family of Singular Functions

105

Example 2.15 Synthesize the function f ðtÞ given in Eq. (2.63) and graph the result obtained. f ðtÞ ¼ U 2 ðt þ 1Þ  2U 1 ðtÞ  U 2 ðt  1Þ

ð2:63Þ

Since U 2 ðtÞ ¼ tU 1 ðtÞ, Eq. (2.6), then U 2 ðt  1Þ ¼ ðt  1ÞU 1 ðt  1Þ, and Eq. (2.63) can be rewritten as f ðtÞ ¼ ðt þ 1ÞU 1 ðt þ 1Þ  2U 1 ðtÞ  ðt  1ÞU 1 ðt  1Þ Adding these three functions from the moments when they become non-zero, that is, synthesizing them, it results that: 8 0 for  1\t   1 > > < t þ 1 for  1  t\0 f ðtÞ ¼ t >  1 for 0\t  1 > : 0 for 1  t\ þ 1:

ð2:64Þ

The graphical representation of f ðtÞ is made with the information contained in Eq. (2.64), and is shown in Fig. 2.20.

f(t)

Fig. 2.20 Graphical representation of f ðtÞ

1 0

-1

t

1

-1

Example 2.16 To a capacitance C, initially de-energized, that is, at rest, the current iðtÞ½A shown in Fig. 2.21-a is applied. (a) Calculate the charge q½C stored in it at time t ¼ 5½s, that is, qð5Þ. (b) Find the waveform of q (t), for t > 0. Solution R Rt (a) For iðtÞ ¼ dtd qðtÞ, then qðtÞ ¼ iðtÞdt ¼ 1 iðkÞdk and, from Fig. 2.21a, Z qð 5Þ ¼

Z

5 1

Z

þ

iðkÞdk ¼ 4

3þ

0:dk þ

1

1 Z 4þ 4

Z 0:dk þ

1þ

Z 3U0 ðk  1Þdk þ

1

U0 ðk  4Þdk þ

Z

2

1þ 5

0:dk: 4þ

Z 0:dk þ

3

2þ

ð1Þdk

106

2 Singular Functions q(t)

i(t)

3 3U0(t-1)

0

1

2

2 3

-1

(a)

t

4 -U0(t-4)

1 0

1

2

3

4

5

t

(b)

Fig. 2.21 a Current iðtÞ applied to a capacitor; b Charge qðtÞ

Therefore, qð5Þ ¼ 0 þ 3 þ 0  kj32 þ 0  1 þ 0 ¼ 3  1  1 ¼ 1½C: Considering that the load is the net area under the current curve, alternatively, we have that: qð5Þ ¼ area of impulse 3U 0 ðt  1Þ þ area of square pulse þ area of impulse U 0 ðt  4Þ ¼ 3 þ ð1Þ þ ð1Þ ¼ 1 [C]. (b) The decomposition of i(t) into singular functions, from Fig. 2.21a, gives that iðtÞ ¼ 3U0 ðt  1Þ  U1 ðt  2Þ þ U1 ðt  3Þ  U0 ðt  4Þ: Remembering that to integrate any single function, just subtract 1 from its index, we have to Z

qðtÞ ¼ iðtÞdt ¼ 3U1 ðt  1Þ  U2 ðt  2Þ þ U2 ðt  3Þ  U1 ðt  4Þ ¼ 3U1 ðt  1Þ  ðt  2ÞU1 ðt  2Þ þ ðt  3ÞU1 ðt  3Þ  U1 ðt  4Þ ¼ 3U1 ðt  1Þ þ ðt þ 2ÞU1 ðt  2Þ þ ðt  3ÞU1 ðt  3Þ  U1 ðt  4Þ: The synthesis of these unique functions gives 8 0 for t\1 > > > > < 3 for 1\t  2 qðtÞ ¼ t þ 5 for 2  t  3 > > > 2 for 3  t\4 > : 1 for t [ 4: The waveform of qðtÞ is shown in Fig. 2.21b. R Example 2.17 Calculate I ¼ f ðtÞdt, where f ðtÞ is the function shown in Fig. 2.20, and which is expressed in terms of singular functions by Eq. (2.63) of example 2.15. Graph the result obtained.

2.3 The Family of Singular Functions

107

Solution As mentioned earlier, and noted in Fig. 2.13, to integrate any singular function, just subtract 1 from its index. Therefore, from Eq. (2.63), Z ð2:65Þ I ¼ f ðtÞdt ¼ U 3 ðt þ 1Þ  2U 2 ðtÞ  U 3 ðt  1Þ 2

Considering now that U 3 ðtÞ ¼ t2 U 1 ðtÞ, Eq. (2.10), then. 2

Þ U 3 ðt  1Þ ¼ ðt1 2 U 1 ðt  1Þ: Then,

ð t þ 1Þ 2 ð t  1Þ 2 U1 ðt þ 1Þ  2tU1 ðtÞ  U1 ðt  1Þ 2 2

2 

2  t 1 t 1 þtþ ¼ U1 ðt þ 1Þ  2tU1 ðtÞ þ  þ t  U1 ðt  1Þ 2 2 2 2 8 0 for  1\t   1 > > > < t2 þ t þ 1 for  1  t  0 2 2 ¼ t2 >  t þ 12 for 0  t  1 > > : 2 0 for 1  t  þ 1:

I¼

ð2:66Þ

The graphical representation of I is shown in Fig. 2.22. It is interesting to note that integration, in general, tends to “smooth out” the curve that has been integrated. More than that: it eliminates the finite discontinuities present in the curve that has been integrated. Fig. 2.22 Graphical representation of current iðtÞ

(t+1)2/2

1

I

(t-1)2/2

0,5

-1

-0,5

0

0,5

1

t

Example 2.18 Transform f ðtÞ ¼ 10senð200t þ p=6Þ:U 1 ðtÞ into a sum of singular functions. Solution According to Eq. (2.53), d ½10senð200t þ p=6Þjt¼0 :U0 ðtÞ dt pffiffiffi ¼ 10senp=6:U1 ðtÞ  2:000cos p=6 ¼ 5U1 ðtÞ  1:000 3U0 ðtÞ:

f ðtÞ ¼ 10senð200t þ p=6Þjt¼0 :U1 ðtÞ 

108

2 Singular Functions

Example 2.19 Eliminate the step function of the function f ðtÞ ¼ U 1 ðt2 þ 5t þ 6Þ. Solution t2 þ 5t þ 6 ¼ 0 has roots of 2 and 3.That is, t2 þ 5t þ 6 ¼ ðt þ 2Þðt þ 3Þ: So,
f ðtÞ ¼ U1 ½ðt þ 2Þðt þ 3Þ ¼

0 1

for ðt þ 2Þðt þ 3Þ\0ð AÞ for ðt þ 2Þðt þ 3Þ [ 0ðBÞ

Study of (A) A-1: ðt þ 2Þ\0 and ðt þ 3Þ [ 0 or t\  2 and t [  3; or, yet, 3\t\  2: A-2: ðt þ 2Þ [ 0 and ðt þ 3Þ\0 or t [  2 and t\  3; inconsistent conditions. Study of (B) B-1: ðt þ 2Þ [ 0 and ðt þ 3Þ [ 0 or t [  2 and t [  3; then t [  2. B-2: ðt þ 2Þ\0 and ðt þ 3Þ\0 or t\  2 and t\  3; then t\  3. Therefore,
f ðt Þ ¼

for  3\t\  2 for th3 and for ti  2

0 1

Finally, f ðt Þ ¼

8
> : 1 for 1  t\ þ 1: The convolution function f ðtÞ is shown in Fig. 2.43.

f(t)

Fig. 2.43 Convolution function f ðtÞ

1 -2

-1

0

1

2

3

t

-1

Example 2.35 Using properties, calculate f ðtÞ ¼ f 1 ðtÞ f 2 ðtÞ, where f 1 ðtÞ and f 2 ðtÞ are the functions shown in Fig. 2.44.

142

2 Singular Functions f1(t)

f2(t) 1

1 0

-1

t

1

-1

-1

0

1

t

-1

Fig. 2.44 Functions f 1 ðtÞ and f 2 ðtÞ

Solution According to property 6, f 000 ðtÞ ¼ f100 ðtÞ f20 ðtÞ: On the other hand, f100 ðtÞ ¼ U1 ðt þ 1Þ þ U0 ðt þ 1Þ  U1 ðt  1Þ  U0 ðt  1Þ: 0

f 2 ðtÞ ¼ U 0 ðt þ 1Þ  2U 0 ðtÞ þ U 0 ðt  1Þ: For properties 2, 16 and 15, f 000 ðtÞ ¼ f100 ðtÞ ½U0 ðt þ 1Þ  2U0 ðtÞ þ U0 ðt  1Þ ¼ f100 ðt þ 1Þ  2f100 ðtÞ þ f100 ðt  1Þ

¼ U1 ðt þ 2Þ þ U0 ðt þ 2Þ  U1 ðtÞ  U0 ðtÞ þ 2U1 ðt þ 1Þ  2U0 ðt þ 1Þ þ 2U1 ðt  1Þ þ 2U0 ðt  1Þ  U1 ðtÞ þ U0 ðtÞ  U1 ðt  2Þ  U0 ðt  2Þ: Simply put, it results f 000 ðtÞ ¼ U1 ðt þ 2Þ þ U0 ðt þ 2Þ þ 2U1 ðt þ 1Þ  2U0 ðt þ 1Þ  2U1 ðtÞ þ 2U1 ðt  1Þ þ 2U0 ðt  1Þ  U1 ðt  2Þ  U0 ðt  2Þ: Integrating, successively, three times, it comes that f ðtÞ ¼ U2 ðt þ 2Þ þ U3 ðt þ 2Þ þ 2U2 ðt þ 1Þ  2U3 ðt þ 1Þ  2U2 ðtÞ þ 2U2 ðt  1Þ þ 2U3 ðt  1Þ  U2 ðt  2Þ  U3 ðt  2Þ:

2.13

Properties of the Convolution Integral

143

Or,

2  t þ 2t þ 2 U1 ðt þ 2Þ þ ð2t þ 2ÞU1 ðt þ 1Þ f ðtÞ ¼ ðt  2ÞU1 ðt þ 2Þ þ 2  2  þ t  2t  1 U1 ðt þ 1Þ þ ð2tÞU1 ðtÞ þ ð2t  2ÞU1 ðt  1Þ   þ t2  2t þ 1 U1 ðt  1Þ þ ðt þ 2ÞU1 ðt  2Þ

2  t þ  þ 2t  2 U1 ðt  2Þ: 2 Grouping terms, we have

2    t þ t U1 ðt þ 2Þ þ t2 þ 1 U1 ðt þ 1Þ þ ð2tÞU1 ðtÞ f ðt Þ ¼ 2

2  2  t þ t  1 U1 ðt  1Þ þ  þ t U1 ðt  2Þ: 2 Finally,

f ðt Þ ¼

8 0 > > > t2 > > 2 þt > > <  t2 þ t þ 1 2

>  t2  t þ 1 > > > t2 > t > > :2 0 2

for  1\t   2 for  2  t   1 for  1  t  0 for 0  t  1 for 1  t  2 for 2  t\ þ 1:

The f ðtÞ waveform is shown in Fig. 2.45. Note that f ðtÞ satisfies properties 21 and 23. Appendix B presents a summary of the main relationships and properties of the singular functions covered in this chapter.

f(t) 1

-2

-1

0 -0,5

Fig. 2.45 Convolution function f ðtÞ

1

2

t

144

2.14

2 Singular Functions

Proposed Problems

P.2-1 Given the function f ð xÞ ¼ U 1 ð xÞ:U 1 ðt  xÞ, where t is a parameter, 0 determine f ðxÞ. Answer: 0

f ð xÞ ¼ U 0 ð xÞ  U 0 ð x  tÞ: P.2-2 Eliminate the step function at f ðtÞ ¼ U 1 ðt2 þ 3t þ 2Þ: Answer:

8 < 1 for t\  2 f ðtÞ ¼ 0 for  2\t\  1 : 1 for t [  1:

P.2-3 Eliminate the ramp functions of f ðtÞ ¼ U2 ðt þ 1Þ:U2 ð3t  1Þ: Answer:

8 for t   1

> > > t < (a) f1 ðtÞ ¼ 2t þ 3 > > t3 > > : 0

for for for for for

t0 0t1 1t2 2t3 t  3:

146

2 Singular Functions

8 0 for t  0 > > > > for 0  t\1

> t þ 4 for 2t4 > > : 0 for t  4: 8 for t  0 >0 > < 2 for 0  t  1 t þt (c) f3 ðtÞ ¼ > t2 =2  3t þ 9=2 for 1  t  3 > : 0 for t  3: P.2-11 Synthesize the function f ðtÞ, for 0  t  6, if f ðtÞ ¼ 0 for t  0 and f ðt Þ ¼

þ1 X ð1Þn :ð2n þ 1Þ:U2 ðt  nÞ; for t  0: n¼0

Answer:

8 t; > > > > 2t þ 3; > > < 3t  7; f ðt Þ ¼ 4t þ 14; > > > > 5t  22; > > : 6t þ 33;

0t1 1t2 2t3 3t4 4t5 5  t  6:

P.2-12 Starting from f ðtÞU 0 ðt  aÞ ¼ f ðaÞU 0 ðt  aÞ prove that: 0

(a) f ðtÞU 1 ðt  aÞ ¼ f ðaÞU 1 ðt  aÞ  f ðaÞU 0 ðt  aÞ: (b) f ðtÞU2 ðt  aÞ ¼ f ðaÞU2 ðt  aÞ  2f 0 ðaÞU1 ðt  aÞ þ f 00 ðaÞU0 ðt  aÞ:

P.2-13 Make the decomposition into singular functions of the next function, with x being a positive parameter. f ðtÞ ¼ U 2 ðt  xÞ:U 2 ð2x  tÞ: Answer: f ðtÞ ¼ xU 2 ðt  xÞ  2U 3 ðt  xÞ þ xU 2 ðt  2xÞ þ 2U 3 ðt  2xÞ:

2.14

Proposed Problems

147

P.2-14 Find the value of the following integrals: R2 (a) 1 ej2t U0 ðt  1; 8Þdt: R3 (b) 2 ej2t U0 ðt  1; 8Þdt: R3 (c) 2 ð2t þ 3Þ½U1 ðt  2Þ  U1 ðt  3ÞU0 ðt  2; 5Þdt:   R3 (d) 2 ðt3  1ÞU2 2t  1 dt:   R4  (e) 3 100e500t þ 220 cos 300t þ p8 U2 ðt  1Þdt: R 4t (f) 2t U3 ð4t  xÞdx: Answer: (a) (b) (c) (d) (e) (f)

ej3:6 ; 0, 8, 96, 0, 4 3 3t :

P.2-15 Make the following integrations: R þ1 (a) 1 U 0 ðatÞdt: Rt (b) 1 U 0 ðatÞdt: R þ1 (c) 1 f ðkÞU 1 ðt  kÞdk; with f ðtÞ ¼ 1a U 1 ðtÞ  1a U 1 ðt  aÞ  U 0 ðt  aÞ and a a constant.   R þ1 (d) 1 ðt2 þ 2ÞU 0 3t  1 dt: Answer: (a) (b)

1 jaj : 1 jaj U 1 ðt Þ:

8

> > > 6 > > < t þ 7 b) qðtÞ ¼ tþ1 > > > > 5 > > : 2

for for for for for for

t\0 0\t  1 1t3 3t4 4  t\5 t [ 5:

P.2-19 Develop functions f 1 ðtÞ e f 2 ðtÞ in singular functions shown in Figs. 2.47 and 2.48. Answer: f 1 ðtÞ ¼ 2  4U 2 ðtÞ þ 8U 2 ðt  1Þ  8U 2 ðt  2Þ þ 4U 2 ðt  3Þ: f 2 ðtÞ ¼ U 1 ðtÞ þ U 2 ðtÞ  4U 1 ðt  2Þ  U 1 ðt  4Þ  U 2 ðt  4Þ:

2.14

Proposed Problems

149

f1(t)

f2(t)

2

3

1

2

0

1

2

4

3

t

-1

1 0

-2

1

2

3

4

1

2

3

4

t

-1

Fig. 2.47 Function f 1 ðtÞ

f2(t)

f1(t) 2

3

1 0

2 1

2

4

3

t

-1

1 0

-2

t

-1

Fig. 2.48 Function f 2 ðtÞ

P.2-20 Prove that if the function uðtÞ is continuous and successively differentiable at t ¼ a, then the decomposition into singular functions of the function f ðtÞ ¼ uðtÞU 1 ðt  aÞ is

f ðt Þ ¼

þ1 X

uðn1Þ ðaÞU n ðt  aÞ;

n¼1

where, uðn1Þ ðaÞ ¼

P.2-21 Knowing that

d n1 uðtÞjt¼a : dtn1

150

2 Singular Functions

U 0 ½ f ðtÞ ¼

X k

1  0  U ðt  tk Þ; f ðtk Þ 0

tk being the roots of f ðtÞ ¼ 0; calculate: (a) U 0 ðt2  a2 Þ; a ¼ constante: (b) U 0 ðsentÞ: (c) 2U 0 ðcos 4tÞ: Answer: (a) (b) (c)

1 2jaj :½U 0 ðt þ aÞ þ U 0 ðt  aÞ: Pþ1 k¼1 U 0 ðt  kpÞ: P 1 p k¼1;2;3;... U 0 t  ð2k þ 1Þ 8 : 2

P.2-22 Demonstrate the formula contained in Eq. (2.57). R þ1 1 jtx P.2-23 Determine f ðtÞ if f ðtÞ ¼ F 1 ½F ðxÞ ¼ 2p 1 F ðxÞe dx and

F ðxÞ ¼ F ½ f ðtÞ ¼

2 þ jx pU 0 ðxÞ: 3 þ jx

Answer: 1 f ðt Þ ¼ : 3

P.2-24 Show that: (a) If tf1 ðtÞ ¼ tf2 ðtÞ, then f1 ðtÞ ¼ f2 ðtÞ þ kU0 ðtÞ (b) If t2 f1 ðtÞ ¼ t2 f2 ðtÞ, then f1 ðtÞ ¼ f2 ðtÞ þ k1 U0 ðtÞ þ k2 U1 ðtÞ:

P.2-25 Perform the decomposition in (a) steps and (b) in ramps of the function defined by

2.14

Proposed Problems

151

8 < 0 for t  0 f ðtÞ ¼ t2 þ 2at for 0  t  2a : 0 for t  2a: Confirm the results by making the integrals obtained. Answer: (a) f ðtÞ ¼

R 2a

ð2k þ 2aÞU 1 ðt  kÞdk: R 0ð2aÞ þ (b) f ðtÞ ¼ 0 ½2aU 0 ðkÞ  2 þ 2aU 0 ðk  2aÞU 2 ðt  kÞdk: P.2-26 Convolute the following functions, using the convolution definition directly. 8 < 0 for  1\t\0 f2 ðtÞ ¼ f1 ðtÞ: (a) f1 ðtÞ ¼ 1 for 0\t\1 : 0 for 1\t\ þ 1: 8 8 0 for  1\t\0 > > < 0 for  1\t\0 < 1 for 0\t\1 f2 ðtÞ ¼ 1 for 0\t\2 (b) f1 ðtÞ ¼ 1 for 1\t\2 : > > 0 for 2\t\ þ 1: : 0 for 2\t\ þ 1: 8 for  1\t   1 >0 > < tþ1 for  1  t  0 (c) f1 ðtÞ ¼ > t þ 1 for 0  t  1 > : 0 for 1  t\ þ 1:

f 2 ðt Þ ¼

X

U0 ðt  kÞ:

k¼0;1;2...

8 8 < 0 for  1\t\0

> < t (a) f ðtÞ ¼ t þ 2 > > : 0 8 >0 > > >

> t4 > > : 0

for  1\t  0 for 0  t  1 for 1  t  2 for 2  t\ þ 1: for  1\t  0 for 0  t  1 for 1  t  3 for 3  t  4 for 4  t\ þ 1:

for  1\t  0 for 0  t  1 for 1  t\ þ 1:

152

2 Singular Functions

(c) f ðtÞ ¼ 8 1 for  1\t\ þ 1: for  1\t  0 > 02 >

 þ 2t  1 for 1  t  2 > : 2 1 for 2  t\ þ 1: P.2-27 If f ðtÞ ¼ f 1 ðtÞ f 2 ðtÞ, calculate f ð2Þ,being

8 0 > > > >

> t þ 3 > > : 0

8 for  1\t  0 0 > > for 0  t  1 < 2t f2 ðtÞ ¼ t3 for 1  t  2 > > for 2  t  3 : 2 0 for 3  t\ þ 1:

for  1\t  0 for 0  t\1 for 1\t  3 for 3  t\ þ 1:

Answer: f ð 2Þ ¼

7 : 12

P.2-28 Using only properties, convert the following functions f1 ðtÞ and f2 ðtÞ 8 0 for  1\t\  1 > > < 1 for  1\t\0 f2 ðtÞ ¼ U1 ðt þ 1Þ: (a) f1 ðtÞ ¼ 1 for 0\t\1 > > : 0 for 1\t\ þ 1: 8 < 1 for  1\t\  1 for  1\t\1 f2 ðtÞ ¼ U0 ðt þ 1Þ  U0 ðt  1Þ: (b) f1 ðtÞ ¼ 1 : 1 for1\t\ þ 1: 8 8 0 for  1\t  0 > > < < 0 for  1\t\0 2t for 0  t  1 f 2 ðt Þ ¼ (c) f1 ðtÞ ¼ 1 for 0\t\1 2t þ 4 for 1  t  2 > : > 0 for 1\t\ þ 1: : 0 for 2  t\ þ 1: 8 8 < 2 for  1\t   1 < 0 for  1\t\  1 (d) f1 ðtÞ ¼ 2t for  1  t  1 f2 ðtÞ ¼ 1 for  1\t\1 : : 2 for 1  t\ þ 1: 0 for 1\t\ þ 1: 8
< 0 for  1\t\  1 1 for  1\t\0 (e) f1 ðtÞ ¼ f ðtÞ ¼ 1 for  1\t\1 1 for 0\t\ þ 1: 2 : 0 for 1\t\ þ 1:

2.14

Proposed Problems

8 1 > > < 2 (f) f1 ðtÞ ¼ 2 > > : 1

for  1\t\  1 for  1\t\0 for 0\t\1 for 1\t\ þ 1:

153

8 < 0 for  1\t\  1 f2 ðtÞ ¼ 1 for  1\t\1 : 0 for 1\t\ þ 1:

Answer:

(a)

(b)

(c)

(d)

(e)

(f)

8 0 for  1\t   2 > > < t þ 2 for  2  t   1 f ðt Þ ¼ for  1  t  0 > 1 > : 0 for 0  t\ þ 1: 8 0 for  1\t\  2 > > < 2 for  2\t\0 f ðt Þ ¼ 2 for 0\t\2 > > : 0 for 2\t\ þ 1: 8 0 for  1\t  0 > > > > for 0  t  1 < t2 f ðtÞ ¼ 2t2 þ 6t  3 for 1  t  2 > > t2  6t þ 9 for 2  t  3 > > : 0 for 3  t\ þ 1: 8 2 for  1\t   2 > > < t2 þ 2t for  2t0 f ðt Þ ¼ 2 t 2 >  2 þ 2t for 0  t  2 > : for 2  t\ þ 1: 82 < 2 for  1\t   1 f ðtÞ ¼ 2t for  1  t  1 : 2 for 1  t\ þ 1: 8 2 for  1\t   2 > > > > < t þ 4 for  2  t   1 f ðtÞ ¼ 3t for  1  t  1 > > t  4 for 1  t  2 > > : 2 for 2  t\ þ 1:

2.15

Conclusions

In this chapter it was presented the mathematical topic of singular functions, also called generalized functions or distributions, that constitute a special class of functions, because they do not fit perfectly into the conventional mathematical concept of function. Here they were designated by U n ðtÞ, n ¼ 0; 1; 2; :::; in such a way that for each value of n a specific singular function will correspond. They can be considered as a family of functions that are generated from the unit impulse function,U 0 ðtÞ, by successive derivations or integrations. However, for

154

2 Singular Functions

simplicity and also for strictly didactic reasons, the study of them was done starting from the unit step function, U 1 ðtÞ. An important reason to study them is that the response of any circuit linear to an arbitrary entry can be determined, indirectly, as long as its response to a unit step or unit impulse is known. This is done through the integral convolution call. Thus, the response of a linear circuit to a unitary impulse, denoted by hðtÞ, or to a unitary step, called rðtÞ, can be used as a characteristic property of such a circuit. This is relevant for developing in the next chapter a systematic method to solve ordinary differential equations in which xðtÞ is a singular and / or causal function. This method allows to determine the initial conditions at t ¼ 0 þ , from the respective initial conditions at t ¼ 0 and the differential equation itself, without any physical reasoning about the given circuit. Therefore, a good understanding of this chapter is an indispensable condition for the good follow-up of the following chapters.

References 1. Adkins WA, Davidson MG (2012) Ordinary differential equations, undergraduate texts in mathematics. Springer 2. Boyce WE, DiPrima RC (2002) Elementary differential equations and boundary value problems, 7th ed. LTC 3. Chicone C (2006) Ordinary differential equations with applications, 22nd ed. Springer 4. Close CM (1975) Linear circuits analysis, LTC 5. Doering CI, Lopes AO (2008) Ordinary differential equations, 3rd ed. IMPA 6. Irwin JD (1996) Basic engineering circuit analysis. Prentice Hall 7. Kreider DL, Kluler RG, Ostberg DR (2002) Differential equations. Edgard Blucher Ltda 8. Leithold L (1982) The calculus with analytical geometry, vol 1, 2nd ed. Harper & How of Brasil 9. Perko L (2001) Differential equations and dynamical systems, 3rd edn. Springer, New York 10. Stewart J (2010) Calculus, vol 1, 6th ed. Thomson 11. Zill DG, Cullen MR (2003) Differential equations, 3rd ed. Makron Books

Chapter 3

Differential Equations

3.1

Introduction

When a linear system, with concentrated parameters and invariant over time, is excited with an input signal xðtÞ, the classic mathematical model for determining the output (or response) yðtÞ is an n order ordinary linear differential equation with constant coefficients. In case the system is an electrical circuit, the order n is generally at most equal to the sum of the number of capacitances and inductances of the circuit, computing these numbers, however, only after capacitances and inductances in series and / or parallel have been replaced by their respective equivalents. This chapter presents a fundamental review about the theory and solution of ordinary differential equations. Moreover, it presents a systematic method to solve ordinary differential equations in which xðtÞ is a singular and / or causal function. This method allows to determine the initial conditions at t ¼ 0 þ , from the respective initial conditions at t ¼ 0 and the differential equation itself, without any physical reasoning about the given circuit. Therefore, a good understanding of this chapter is an indispensable condition for the good follow-up of the following chapters. When a linear system, with concentrated parameters and invariant over time, is excited with an input signal xðtÞ, the classic mathematical model for determining the output (or response) yðtÞ is an ordinary linear differential equation with constant coefficients [1–11], as indicated in Eq. 3.1. dn d n1 d n2 d n3 d y þ a1 n1 y þ a2 n2 y þ a3 n3 y þ . . . þ an1 y þ an y n dt dt dt dt dt dm d m1 d m2 d ¼ bm m x þ bm1 m1 x þ bm2 m2 x þ . . . þ b1 x dt dt dt dt þ b0 x; with n  m:

ð3:1Þ

The ai and bj are constant and n is the order of the differential equation. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. C. Goulart de Siqueira and B. D. Bonatto, Introduction to Transients in Electrical Circuits, Power Systems, https://doi.org/10.1007/978-3-030-68249-1_3

155

156

3 Differential Equations

In case the system is an electrical circuit, the order n is generally at most equal to the sum of the number of capacitances and inductances of the circuit, computing these numbers, however, only after capacitances and inductances in series and / or parallel have been replaced by their respective equivalents. In Sect. 3.2 there is a wide review of the most important and necessary topics in the theory of ordinary differential equations, which take the form of Eq. 3.1. The study of transient phenomena in linear electrical circuits, with lumped or concentrated parameters, can be done both in the time domain, by the so-called classical method, and in the domain of the complex frequency, by the well-known operational method of the Laplace transforms. The classical method consists of the direct resolution, that is, in the time domain itself, of the system of ordinary differential equations that are obtained by equating the circuit. And it requires determining the arbitrary constants that appear in the general solution, either of the current, or of the voltage sought, from the initial conditions at t ¼ 0 þ , which must be obtained from the respective initial conditions at t ¼ 0 , a circumstance that constitutes the fundamental difficulty of this method, since it requires physical reasoning, sometimes very elaborate and quite difficult, about the given circuit. The operational method, on the other hand, transforms the obtained ordinary differential equations into algebraic equations of the complex variable s ¼ r þ jx, incorporating, inherently to this transformation process, the initial conditions at t ¼ 0 . This eliminates the need to determine arbitrary constants, based on the initial conditions at t ¼ 0 þ . However, the process of inverse transformation, that is, the return to the time domain, can become excessively laborious. As in the following chapters it is intended to work almost exclusively with the classical method, then, in Sect. 3.2 a systematic method is developed to solve ordinary differential equations in which xðtÞ is a singular and / or causal function. This method allows to determine the initial conditions at t ¼ 0 þ , from the respective initial conditions at t ¼ 0 and the differential equation itself, without any physical reasoning about the given circuit. Therefore, a good understanding of Sect. 3.2 is an indispensable condition for the good comprehension of the following chapters.

3.2

Ordinary Differential Equations

Since xðtÞ is a known function of time, the second member of Eq. 3.1, as a whole, is also a known function of time, called the Forcing Function, and denoted by f ðtÞ. Thus, Eq. 3.1 can be rewritten as in Eq. 3.2. dn d n1 d n2 d n3 d y þ a1 n1 y þ a2 n2 y þ a3 n3 y þ . . . þ an1 y þ an y ¼ f ðtÞ n dt dt dt dt dt

ð3:2Þ

If f ðtÞ is identically null, that is, if f ðtÞ ¼ 0, then Eq. 3.2 becomes Eq. 3.3, known as the Homogeneous Differential Equation.

3.2 Ordinary Differential Equations

dn d n1 d n2 d n3 d y þ a y þ a y þ a y þ . . . þ an1 y þ an y ¼ 0 1 2 3 n n1 n2 n3 dt dt dt dt dt

157

ð3:3Þ

When f ðtÞ 6¼ 0, then Eq. 3.2 is called the Non-homogeneous Differential Equation. Homogeneous Eq. 3.3, which is of order n, always presents n solutions. In the vast majority of cases these n solutions are linearly independent, which means that none of them is the linear combination of the others n1, and that they are denoted by: y1 ðtÞ; y2 ðtÞ; y3 ðtÞ; . . .; yn ðtÞ: Thus, the most general solution of Eq. 3.3 is given by: yH ðtÞ ¼ k1 y1 ðtÞ þ k2 y2 ðtÞ þ k3 y3 ðtÞ þ . . . þ kn yn ðtÞ;

ð3:4Þ

where k1 ; k2 ; k3 ; . . .; kn are arbitrary constants. The general solution of Eq. 3.2, which is non-homogeneous, is given by: yð t Þ ¼ yH ð t Þ þ yP ð t Þ

ð3:5Þ

The term yH ðtÞ is the solution of the corresponding homogeneous equation, as given by Eq. (3.4), and is called the Complementary Solution. The term yP ðtÞ is any other solution, free of arbitrary constants, that satisfies Eq. 3.2, that is, that makes its two members identical, and is called a Solution or Particular Integral. As yH ðtÞ represents the solution of Eq. 3.2 for f ðtÞ ¼ 0, that is, the system response without external excitation, when input xðtÞ ¼ 0, the complementary solution is also called the Free Response of the system. And considering that this free response depends on the general nature of the system, that is to say: the types of components present, their numerical values and the way they are interconnected, in addition to the energy stored in its non-dissipative elements, it is also called the Natural Response of the system. In other words, the natural response is due only to the energy initially stored in the system’s own non-dissipative interior elements. A system is considered dissipative when it transfers energy, in the form of heat, to its exterior. In an electrical circuit, for example, this role fits its resistances. And as in a dissipative system, yH ðtÞ tends to zero after some time of operation of the system, it is still called the Transient Response of the system. Therefore, in a non-dissipative system, the free response cannot be called a transient response, because it does not disappear after a certain time; it has a permanent character. On the other hand, since yP ðtÞ depends on f ðtÞ, that is, on the external source (s) that excites the system, it is called the Forced Response or Permanent Regime (Steady-State) Response. It is necessary to emphasize, therefore, that in a non-dissipative system, such as, for example, an electrical circuit without resistances, there is no damping of yH ðtÞ, which does not tend to zero over time. So, the natural response is also permanent, just like the forced response. Every time a dissipative system undergoes an intervention, which forces it to move from one condition of stability to another, either through a change in the

158

3 Differential Equations

applied energy source, either through a change in its elements or even in its structure, there is a transition period, during which the variables associated with its elements change from their primitive values to new ones. This period is called the system’s Transient Regime. In it, we have that yðtÞ ¼ yH ðtÞ þ yP ðtÞ. After the transitional regime has been extinguished, when yH ðtÞ can already be considered null, it is said that the system is in Permanent Regime or Stationary State, when yðtÞ ¼ yP ðtÞ: In summary, it can be said that: the complete response of a stable and dissipative system is made up of the sum of the transient response yH ðtÞ and the forced or permanent response yP ðtÞ. And that, after sufficient time has elapsed for the transitional regime to be extinguished, generally of the order of approximately five times the largest time constant of the system, only the permanent regime or steady-state will remain. As a consequence, one can study the forced behavior, or the permanent regime of the systems, regardless of the transient behavior. This is what is done when studying the permanent regime of electrical circuits subjected to sinusoidal sources using phasors and impedances. Forced oscillations, in response, occur whenever a system is subjected to excitations that vary periodically over time.

3.2.1

Solution of the Homogeneous Differential Equation

Suppose that a solution of Eq. 3.3 is of the form yðtÞ ¼ rt , where r is a constant to be determined and “e” is the Neperian number. Substituting this supposed solution in Eq. 3.3, it is obtained that:


 r n þ a1 r n1 þ a2 r n2 þ a3 r n3 þ . . . þ an1 r þ an rt ¼ 0

For this equation to be satisfied for all values of t it is necessary that F ðr Þ ¼ r n þ a1 r n1 þ a2 r n2 þ a3 r n3 þ . . . þ an1 r þ an ¼ 0

ð3:6Þ

Equation 3.6 is called the Characteristic Equation and is an algebraic equation of order n, with constant coefficients, which contains n roots: r1 ; r2 ; r3 ; . . .; rn , called Characteristic Roots. Thus, by factoring the Polynomial Operator F ðr Þ, we have that: F ðr Þ ¼ ðr  r1 Þðr  r2 Þðr  r3 Þ. . .ðr  rn Þ So, the solutions of the homogeneous differential equation are: y1 ðtÞ ¼ r1 t ; y2 ðtÞ ¼ r2 t ; y3 ðtÞ ¼ r3 t ; . . .; yn ðtÞ ¼ rn t : And the general solution will fit into one of the following five cases.

ð3:7Þ

3.2 Ordinary Differential Equations

159

(a) If all the roots of the characteristic equation FðrÞ ¼ 0 are real and distinct, the solutions mentioned are independent and the general solution is the linear combination of them, namely: yH ðtÞ ¼ k1 r1 t þ k2 r2 t þ k3 r3 t þ . . . þ kn rn t

ð3:8Þ

(b) If some roots of FðrÞ ¼ 0 are complex and distinct, yH ðtÞ can be written in a different form. When the homogeneous differential equation, 3.3, has coefficients a1 , a2 , a3 , …, an real, the complex roots of F ðr Þ ¼ 0 always occur in conjugated pairs. If, for example, r1 ¼ a þ jb, where a and b are real numbers, then r2 ¼ a  jb. Then, k1 r1 t þ k2 r2 t ¼ at ðk1 jbt þ k2 jbt Þ ¼ at ½k1 ðcos bt þ j sin btÞ þ k2 ðcos bt  j sin btÞ ¼ at ½ðk1 þ k2 Þ cos bt þ jðk1  k2 Þ sin bt ¼ at ðk10 cos bt þ k20 sin btÞ ¼ kat cosðbt þ /Þ Therefore, if r1 ¼ a þ jb, r2 ¼ a  jb and r3; r4 ; . . .; rn are real and distinct roots, then, yH ðtÞ ¼ at ðk1 cos bt þ k2 sin btÞ þ k3 r3 t þ . . . þ kn rn t

ð3:9Þ

yH ðtÞ ¼ kat cosðbt þ /Þ þ k3 r3 t þ . . . þ kn rn t ;

ð3:10Þ

or

where k and / are arbitrary constants; being that: k¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 k12 þ k22 and / ¼ arctg k1

(c) If there are repeated real roots, the terms of Eq. 3.8 will not all be independent and the equation will no longer represent the general solution of the homogeneous Eq. 3.3, since the number of arbitrary constants will be less than the order n of the differential equation. If r1 ¼ r2 , for example, k1 r1 t þ k2 r2 t ¼ ðk1 þ k2 Þr1 t ¼ k10 r1 t

160

3 Differential Equations

In this case, it can be shown that y1 ðtÞ ¼ r1 t and y2 ðtÞ ¼ t r1 t are independent solutions, so that the general solution of the homogeneous Eq. 3.3 is: yH ðtÞ ¼ ðk1 þ k2 tÞr1 t þ k3 r3 t þ . . . þ kn rn t

ð3:11Þ

(d) If the root r1 is repeated ðm  1Þ times, that is, if r1 ¼ r2 ¼ r3 ¼ . . . ¼ rm , then the general solution is:
 yH ðtÞ ¼ k1 þ k2 t þ k3 t2 þ . . . þ km tm1 r1 t þ . . . þ kn rn t

ð3:12Þ

(e) If there are repeated complex roots, the terms of Eqs. 3.9, or 3.10, will also not be independent and the solution, in the case of r1 ¼ r2 ¼ a þ jb and r3 ¼ r4 ¼ a  jb, for example, must be taken like: yH ðtÞ ¼ at ½ðk1 þ k2 tÞ cos bt þ ðk3 þ k4 tÞsin bt þ . . . þ kn rn t

ð3:13Þ

Example 3.1 Find the solution of the following homogeneous equations: 2

(a) ddt2y þ dy dt  6y ¼ 0 (b) ðr þ 1Þðr2Þðr þ 3Þy ¼ 0 (c) ðr  1Þ3 ðr þ 2Þðr þ 3Þy ¼ 0 2 (d) ddt2y  2 dy dt þ 10y ¼ 0 (e) DðD2 þ 4ÞðD  1Þ2 y ¼ 0 (f) ðD4 þ 2D2 þ 1Þy ¼ 0 (g) DðD4 þ 6D2 þ 9Þy ¼ 0 D is the Heaviside Differential Operator, defined by: d d2 d3 D ¼ ; D2 ¼ 2 ; D3 ¼ 3 ; . . . dt dt dt 2

5

Thus, D2 z ¼ ddt2z ; D5 W ¼ ddtW5 . Therefore, conversely: R D1 ¼ D1 ¼ . . . dt is the Integral Operador. Thus, the voltage-current relationship in an inductance, eðtÞ ¼ L dtd iðtÞ, can be written as eðtÞ ¼ LDiðtÞ and, conversely, R 1 iðtÞ ¼ L1 eðtÞdt can be written as iðtÞ ¼ LD eðtÞ. Likewise, the current-voltage relationship in a capacitance, iðtÞ ¼ C dtd eðtÞ R 1 ¼ CDeðtÞ, and eðtÞ ¼ C1 iðtÞdt ¼ CD iðtÞ

3.2 Ordinary Differential Equations

161

Solution (a) Using the operator D, the differential equation can be written as:


 D2 þ D  6 y ¼ 0

Therefore, the polynomial operator is F ðDÞ ¼ D2 þ D  6, and the characteristic equation is D2 þ D  6 ¼ 0. The characteristic roots are r1 ¼ 2 and r2 ¼ 3, and the general solution, according to Eq. 3.8, is yH ðtÞ ¼ k1 2t þ k2 3t (b) The characteristic roots are r1 ¼ 1, r2 ¼ 2 e r3 ¼ 3. So, yH ðtÞ ¼ k1 t þ k2 2t þ k3 3t (c) The roots of the characteristic equation ðr  1Þ3 ðr þ 2Þðr þ 3Þ ¼ 0 are: r1 ¼ r2 ¼ r3 ¼ 1, r4 ¼ 2 and r5 ¼ 3. So, according to Eq. 3.12, the general solution of the homogeneous equation is
 yH ðtÞ ¼ k1 þ k2 t þ k3 t2 t þ k4 2t þ k5 3t (d) The characteristic equation is D2  2D þ 10 ¼ 0. And its roots are r1 ¼ 1 þ j3 and r2 ¼ 1  j3. According to Eq. 3.9, yH ðtÞ ¼ t ðk1 cos 3t þ k2 sin 3tÞ or yH ðtÞ ¼ kt cosð3t þ /Þ (e) The roots are r1 ¼ 0, r2 ¼ j2, r3 ¼ j2, r4 ¼ r5 ¼ 1. So, by Eqs. 3.9 and 3.11, yH ðtÞ ¼ k1 þ k2 cos 2t þ k3 sin 2t þ ðk4 þ k5 tÞt (f) The roots of the characteristic equation are obtained first by doing D2 ¼ z, D4 ¼ z2 , which generates the following characteristic equation: z2 þ 2z þ 1 ¼ pffiffiffiffiffiffiffi pffiffiffiffi ðz þ 1Þ2 ¼ 0. Its roots are z1 ¼ z2 ¼ 1. Therefore, D ¼  z1 ¼  1 ¼ j, pffiffiffiffi that is, r1 ¼ þ j and r2 ¼ j. On the other hand, also D ¼  z2 ¼ pffiffiffiffiffiffiffi  1 ¼ j, that is,r3 ¼ þ j and r4 ¼ j. Therefore, in view of Eq. 3.13, with a ¼ 0,

162

3 Differential Equations

yH ðtÞ ¼ ðk1 þ k2 tÞcost þ ðk3 þ k4 tÞ sin t pffiffiffi pffiffiffi (g) The characteristic roots are r1 ¼ 0, r2 ¼ r3 ¼ j 3, r4 ¼ r5 ¼ j 3. The general solution, therefore, is pffiffiffi pffiffiffi yH ðtÞ ¼ k1 þ ðk2 þ k3 tÞcos 3t þ ðk4 þ k5 tÞ sin 3t

3.2.2

Considerations on Algebraic Equations

Obtaining the solution of the homogeneous differential equation, as seen in Sect. 3.2.1, depends essentially on determining the roots of the characteristic equation F ðDÞ ¼ 0, which is always an algebraic equation. For this reason, it is opportune to make some considerations about them. Descartes rule “If the coefficients of an algebraic equation f ð xÞ ¼ 0 are real, then the number of positive real roots (NPRR) that it has is equal to, or less than, an even number, to the number of sign inversions that the succession of its coefficients presents”. Applying the same rule in the algebraic equation that is obtained by exchanging x for −y in the original equation f ð xÞ ¼ 0, the number of real negative roots (NNRR) of the given equation is estimated. And the difference establishes the number of pairs of complex conjugated roots (NCCR). Example 3.2 Apply the Descartes rule to the algebraic equation f ð xÞ ¼ x5  1:7x4 þ 2x2 þ 3:1x  12 ¼ 0 Solution Since the equation is of fifth order, there are a total of 5 roots, and the number of sign inversions is equal to 3. Therefore, the equation must have either 1 or 3 positive real roots. Replacing x by −y, we get:y5  1:7y4 þ 2y2  3:1y  12 ¼ 0. Or that: y5 þ 1:7y4  2y2 þ 3:1y þ 12 ¼ 0: The y-equation now has 2 sign inversions. So, the y-equation must have either zero or 2 positive real roots (PRR). But, as x ¼ y, positive real roots in y mean negative real roots (NRR) in x. Therefore, the original equation f ð xÞ ¼ 0 must have zero or 2 negative real roots. It is then possible to put together the following possibilities (combinations) in Table 3.1.

3.2 Ordinary Differential Equations Table 3.1 Number of positive, negative real roots and complex conjugated roots

163

Cases

NPRR

NNRR

NCCR

Total

1st 2nd 3rd 4th

1 1 3 3

0 2 0 2

4 2 2 0

5 5 5 5

Example 3.3 Research the number of positive and negative real roots of the equation f ð xÞ ¼ x5 þ 25x4 þ 230x3 þ 950x2 þ 1689x þ 945 ¼ 0: Solution Since there is no sign inversion, the equation has no positive real roots. Taking x ¼ y, it comes that: y5 þ 25y4  230y3 þ 950y2  1689y þ 945 ¼ 0: There are 5 signal inversions. So, the given equation has either 1 or 3 or 5 real negative roots. Therefore, the following table of possibilities can be put together in Table 3.2. In fact, the given equation was assembled from the five roots, making the product (x + 1)(x + 3)(x + 5)(x + 7)(x + 9). That is, the roots are: 1; 3; 5; 7; and 9, corresponding to the 3rd case in Table 3.2. Bolzano’s theorem “If f ðaÞ and f ðbÞ have the same sign, then f ð xÞ ¼ 0 has an even number of roots, or has none, in the range ½a; b. If f ðaÞ and f ðbÞ have opposite signs, then f ð xÞ ¼ 0 admits an odd number of roots in the range ½a; b”. Upper limit of positive real roots of an algebraic equation of degree n If f ð xÞ ¼ xn þ an1 xn1 þ . . . þ a1 x þ a0 ¼ 0, then, x\1 þ

pffiffiffiffi A;

nm

where m is the highest degree of the negative coefficient terms. A is the absolute value (modulus) of the lowest negative coefficient in the equation.

Table 3.2 Number of positive real roots and complex conjugated roots

Cases

NNRR

NCCR

Total

1st 2nd 3rd

1 3 5

4 2 0

5 5 5

164

3 Differential Equations

Example 3.4 Establish the interval within which all the real roots of the equation f ð xÞ ¼ x5  1:2x4 þ x2 þ 3:4x  8 ¼ 0 must be contained. Solution A ¼ j8j ¼ 8; m ¼ 4; n ¼ 5: x\1 þ

pffiffiffi 8¼9

54

Taking x ¼ y, results that: y5 þ 1:2y4  y2 þ 3:4y þ 8 ¼ 0. So, now, A ¼ j1j ¼ 1; m ¼ 2; n ¼ 5: y\1 þ

pffiffiffi 1 ¼ 2: Then  x\2; or x [  2

52

Therefore, all the real roots of f ð xÞ ¼ 0 are contained in the range 2\x\9. Example 3.5 Find the range that contains all the real roots of the algebraic equation f ð xÞ ¼ 3x6 þ 2x3  2x2  1 ¼ 0 Solution Initially, the coefficient of the highest degree term in x must be reduced to 1, according to the standard initially established for f ð xÞ ¼ 0. It follows, then, that: 2 3 2 2 1 x  x  ¼0 3 3 3    2 2 A ¼   ¼ ; n ¼ 6; m ¼ 2 3 3 x6 þ

rffiffiffi 1 x\1 þ ¼ 1:904 3 62

For x ¼ y, 2 2 1 y6  y3  y2  ¼ 0 3 3 3 2 A ¼ ; n ¼ 6; m ¼ 3 3 rffiffiffi 63 2 y\1 þ ¼ 1:874 3

3.2 Ordinary Differential Equations

165

Therefore, the real roots of the equation are contained in the range 1:874\x\1:904. Newton-Raphson method It is a powerful recurrence process, which allows starting from an initial estimate x0 , from a root of a given algebraic equation f ð xÞ ¼ 0, or any non-linear equation, and progressing, iteratively, until reaching the desired precision for the true root sought x. Its general form is: xi þ 1 ¼ xi 

f ð xi Þ ; for i ¼ 0; 1; 2; . . . f 0 ð xi Þ

Therefore, x is the limit of the succession x0 ; x1 ; x2 ; x3 ; . . . The second part of Bolzano’s theorem is fundamental in the search for the referred initial estimate, x0 , since it guarantees that when f ðaÞ and f ðbÞ have opposite signs, there is at least one real root between a and b. (It is interesting to note that, digitally speaking, the signs are opposite when the product of f ðaÞ by f ðbÞ is negative). On the other hand, the knowledge of the interval where all the real roots of the equation must be located facilitates the search for the values of a and b, in which the ordinates have opposite signs, since it avoids the search without any criteria of the initial estimate. (It is worth mentioning that the Newton-Raphson method also applies to complex roots). Depending on the initial estimate adopted, however, the Newton-Raphson method may not converge to the desired solution, that is, it will not converge to the sought x root. Studies on the convergence of the method, however, establish that: “If f 0 ð xÞ does not change sign in the interval a\x\b, the Newton-Raphson method always converges to the root x of f ð xÞ ¼ 0, included in this interval, as long as the process is started at the end where f ð xÞ:f 00 ð xÞ [ 0. In other cases, the process may or may not converge”. In the search for two values of x with ordinates of opposite signs, especially when it comes to algebraic equations, it is necessary to build tables for different values of x and their corresponding f ð xÞ values. This is a tedious and boring job, as it requires, repeatedly, the calculation of different powers of x. However, this work can be shortened by adopting the following system, based on successive factors. Let the following polynomial be: f ð xÞ ¼ Ax5 þ Bx4 þ Cx3 þ Dx2 þ Ex þ F Factoring x4 : f ð xÞ ¼ ðAx þ BÞx4 þ Cx3 þ Dx2 þ Ex þ F Factoring x3 : f ð xÞ ¼ ððAx þ BÞx þ CÞx3 þ Dx2 þ Ex þ F Factoring x2 : f ð xÞ ¼ ðððAx þ BÞx þ CÞx þ DÞx2 þ Ex þ F Factoring x: f ð xÞ ¼ ððððAx þ BÞx þ C Þx þ DÞx þ EÞx þ F

166

3 Differential Equations

Knowing that the number of parentheses that open must always be equal to the number of those that close, there is no greater problem in leaving the parentheses that open, and then write, simply, that: f ð xÞ ¼ Ax þ BÞx þ CÞx þ DÞx þ EÞx þ F Proceeding in this way, the only necessary algebraic operations are multiplication and addition, alternately, without the need to repeatedly calculate increasing powers of x. Furthermore, if a given numerical value of x is stored, for example, in the memory ∎ of an electronic calculator, then it follows that: f ð xÞ ¼ A
þ BÞ
þ CÞ
þ DÞ
þ EÞ
þ F Example 3.6 Given the algebraic equation f ð xÞ ¼ x3 þ 2x2 þ 10x  20 ¼ 0, which had one of its roots calculated in 1225, with ten correct significant numbers, by Leonardo de Pisa, (a) apply the Descartes rule; (b) establish the range that contains its real roots; (c) apply Bolzano’s theorem and make an initial estimate, x0 , for one of these roots; (d) from x0 , refine by Newton-Raphson up to six correct significant numbers; (e) by deflating the polynomial, find the other two roots of the equation. Solution (a) There is a sign inversion in the given equation. So, there is a real positive root. Replacing x by y, we get: y3 þ 2y2  10y  20 ¼ 0: Or, y3  2y2 þ 10y þ 20 ¼ 0: There are now two signal inversions. So, there are zero or two negative real roots. The possibilities are presented in Table 3.3 (b) A ¼ 20; n ¼ 3; m ¼ 0 p ffiffiffiffiffi 3 x\1 þ 20 ¼ 3:714 From the y equation, from item (a), A ¼ 2; n ¼ 3; m ¼ 2

Table 3.3 Number of positive, negative or complex conjugated roots

Cases

NPRR

NNRR

NCCR

Total

1st 2nd

1 1

0 2

2 0

3 3

3.2 Ordinary Differential Equations Table 3.4 Location of the range containing at least one real root

167

x

−3

−2

−1

0

1

2

3

f ðxÞ

−59

−40

−29

−20

−7

16

55

y\1 þ

pffiffiffi 2 ¼ 3:or; x [  3

32

Therefore, the real roots are contained in the range 3\x\3:714. (c) The function f ð xÞ written in the factored form is (Table 3.4) f ð xÞ ¼ x þ 2Þx þ 10Þx  20: Since f ð1Þ ¼ 7 and f ð2Þ ¼ þ 16 have opposite signs, there is certainly a root between x ¼ 1 and x ¼ 2. Initial estimate: f 0 ð xÞ ¼ 3x2 þ 4x þ 10 and f 00 ð xÞ ¼ 6x þ 4: Since the derivative f 0 ð xÞ does not change sign in the interval 1\x\2, as it always remains positive in it, the iterative process must start at the end where f ð xÞ:f 00 ð xÞ [ 0, to ensure convergence to the sought root., Thus, f ð1Þ:f 00 ð1Þ ¼ 7:10 ¼ 70\0 f ð2Þ:f 00 ð2Þ ¼ 16:16 ¼ 256 [ 0 Therefore, the iterative process must start at the right end of the range, that is, the initial estimate must be x0 ¼ 2. (d) The Newton-Raphson refinement is as follows: xi þ 1 ¼ xi  ¼

For For For

2x3i þ 2x2i þ 20 2xi þ 2Þxi Þxi þ 20 ¼ 3xi þ 4Þxi þ 10 3x2i þ 4xi þ 10

2x0 þ 2Þx0 Þx0 þ 20 2:2 þ 2Þ2Þ2 þ 20 ¼ 3x0 þ 4Þx0 þ 10 3:2 þ 4Þ2 þ 10 44 ¼ 1:46667 ¼ 30 30:61215 i ¼ 1: x2 ¼ 22:32000 ¼ 1:37151 i ¼ 2: x3 ¼ 28:92181 21:12916 ¼ 1:36881 i ¼ 3: x4 ¼ 28:87660 21:09616 ¼ 1:36881 i ¼ 0 : x1 ¼

For

f ð xi Þ x3 þ 2x2i þ 10xi  20 ¼ xi  i 0 f ð xi Þ 3x2i þ 4xi þ 10

168

3 Differential Equations

Therefore, x1 ¼ 1:36881, with six correct significant numbers, since x4 ¼ x3 : (e) Polynomial deflation The division of the polynomial x3 þ 2x2 þ 10x  20 by x  1:36881 is done next. x3 þ 2x2 þ 10x  20 =x  1:368810000000000 x3 þ 1:36881x2 x2 þ 3:36881x þ 14:61126 2 3:36881x þ 10x 3:36881x2 þ 4:61126x 14:61126x  20 14:61126x þ 20:00004 ffi0 The new equation, then, is: x2 þ 3:36881x þ 14:61126 ¼ 0 And the other two roots are: x2 ¼ 1:68441 þ j 3:43133 x3 ¼ 1:68441  j 3:43133:

3.2.3

Solution of the Non-homogeneous Differential Equation

As highlighted by Eq. 3.5, the general solution of a non-homogeneous differential equation is composed of two parts, namely: the complementary solution yH ðtÞ, which involves a number of arbitrary constants equal to the order of the differential equation, and the particular solution yP ðtÞ, free of arbitrary constants. The purpose of this section is to recall one of the methods employed in determining yP ðtÞ. The two main methods used to determine the particular solution yP ðtÞ are the Method of Variation of Parameters and the Method of Coefficients to be Determined. It will be remembered here a method called Abbreviated Method, which is a slight variation of the method of the coefficients to be determined. The Fasorial Method, used in the case of sinusoidal forcing functions, will also be considered as an abbreviated method. A particular solution of a linear differential equation with constant coefficients F ðDÞyðtÞ ¼ f ðtÞ is given by yP ðtÞ ¼ F ð1DÞ f ðtÞ, where the right member must be understood as “Fð1DÞ operating on f ðtÞ”, which produces the following result:

3.2 Ordinary Differential Equations

Z yP ð t Þ ¼ 



r1 t

ðr2 r1 Þt

169

Z 

ðr3 r2 Þt

Z ...



ðrn rn1 Þt

Z

f ðtÞrn t dtn ;

ð3:14Þ

where r1 , r2 , r3 ,…, rn are the n roots of the characteristic equation F ðDÞ ¼ 0. For n ¼ 1, Eq. 3.14 is reduced to yP ð t Þ ¼

1 f ðtÞ ¼ r1 t D  r1

Z

f ðtÞr1 t dt

ð3:15Þ

For certain special forms of f ðtÞ, however, the work required to calculate can be greatly reduced. The shortened method will be shown for some of these more frequent special forms. 1º Case: f ðtÞ ¼ Aat , where A and a are any constants. In this case,

1 F ðDÞ f ðt Þ

yP ð t Þ ¼

1 Aat ðAat Þ ¼ ; provided that FðaÞ 6¼ 0: FðDÞ FðaÞ

ð3:16Þ

Example 3.7 Solve the following differential equations: (a) (b) (c) (d)

ðD3  2D2  5D þ 6ÞyðtÞ ¼ ðD  1ÞðD þ 2ÞðD  3ÞyðtÞ ¼ 4t 2 ðD  1ÞðD þ 2ÞðD  3ÞyðtÞ ¼ ð2t þ 3Þ 3t ðD  1ÞðD þ 2ÞðD  3ÞyðtÞ ¼  ðD þ 1ÞðD þ 2ÞðD þ 3ÞyðtÞ ¼ 30:

Solution (a) The characteristic roots are r1 ¼ 1, r2 ¼ 2 and r3 ¼ 3. So, the complementary solution is: yH ðtÞ ¼ k1 t þ k2 2t þ k3 3t The particular solution is: 1 4t ðD  1ÞðD þ 2ÞðD  3Þ 1 1 1 4t ¼ 4t ¼ 4t ¼ ð4  1Þð4 þ 2Þð4  3Þ ð3Þð6Þð1Þ 18

yP ð t Þ ¼

And the general solution is: yð t Þ ¼ yP ð t Þ þ yH ð t Þ ¼

1 4t  þ k1 t þ k2 2t þ k3 3t 18

170

3 Differential Equations

(b) The particular solution is: 1 ð2t þ 3Þ2 ðD  1ÞðD þ 2ÞðD  3Þ 1 4t ð4t þ 62t þ 90:t Þ ¼ ¼ ðD  1ÞðD þ 2ÞðD  3Þ ð3Þð6Þð1Þ 2t 4t 2t 6 9  3 3 ¼ þ  þ þ 2 ð1Þð4Þð1Þ ð1Þð2Þð3Þ 18 2

yP ð t Þ ¼

So, the general solution is: yð t Þ ¼

4t 32t 3  þ þ k1 t þ k2 2t þ k3 3t 2 18 2

(c) A particular solution is: yP ð t Þ ¼

1 3t ðD  1ÞðD þ 2ÞðD  3Þ

In this case, F ðaÞ ¼ F ð3Þ ¼ ð2Þ:ð5Þ:ð0Þ ¼ 0 and the method cannot be applied directly. One way to get around this difficulty is to proceed as follows, with the help of Eq. 3.15:    3t  1 1 1  3t  ¼ y P ðt Þ ¼ ð D  3Þ ð D  1Þ ð D þ 2Þ ð D  3Þ ð 2Þ ð 5Þ  Z  1 1 1 3t ¼ 3t 3t 3t dt ¼ 10 D  3 10 Z 3t 1 t ¼ 3t dt ¼ 10 10 Therefore, yð t Þ ¼

ðdÞ

yP ð t Þ ¼

t3t þ k1 t þ k2 2t þ k3 3t 10

1 30 ð300:t Þ ¼ ¼5 ð D þ 1Þ ð D þ 2Þ ð D þ 3Þ ð1Þð2Þð3Þ

3.2 Ordinary Differential Equations

171

yðtÞ ¼ 5 þ k1 t þ k2 2t þ k3 3t Whenever the complementary solution yH ðtÞ contains exponential terms that participate in the forcing function f ðtÞ, then F ðaÞ ¼ 0 and the method fails. To get around this obstacle, proceed as in item c) of example 3.7, using the formula 1 f ðtÞ ¼ at Dþa

Z f ðtÞat dt

ð3:17Þ

2º Case: f ðtÞ ¼ tn , where n is a positive integer number. In this case, the inverse of the polynomial operator F ðDÞ must be expanded in a series of increasing powers of D, by direct division, truncating the series in the term Dn , because from Dn þ 1 tn derivatives will always be null, as indicated below: yP ð t Þ ¼

 1 n
t ¼ a0 þ a1 D þ a2 D2 þ a3 D3 þ . . . þ an Dn tn ; a0 6¼ 0 FðDÞ

ð3:18Þ

Example 3.8 Solve the differential equations: (a) ðD2  4D þ 3ÞyðtÞ ¼ t2 þ 2t þ 1 (b) ðD3  4D2 þ 3DÞyðtÞ ¼ t2 Solution (a) As r1 ¼ 1 and r2 ¼ 3, the complementary solution is:

yH ðtÞ ¼ k1 t þ k2 3t The particular integral is: yP ð t Þ ¼

1 ðt2 þ 2t þ 1Þ 3  4D þ D2

The obtaining of the inverse of the expanded polynomial operator in a series of increasing powers of D, truncated in D2 , is shown below.

172

3 Differential Equations

So,   1 4 13 2
2 þ Dþ D yP ð t Þ ¼ t þ 2t þ 1 3 9 27  4d
2  1
2 t þ 2t þ 1 þ ¼ t þ 2t þ 1 þ 3 9 dt 2   13 d 2
2 t 2 1 8 8 t þ t þ þ t þ 2t þ 1 ¼ þ þ 27 dt2 3 3 9 9 3  2 26 t 14 59 tþ þ ¼ þ 27 9 27 3 Finally, yð t Þ ¼

t2 14 59 tþ þ k1 t þ k2 3t þ 9 27 3

Obviously, the method of the coefficients to be determined can be applied directly, as indicated below. Assuming that yP ðtÞ ¼ At2 þ Bt þ C, because the forcing function is of the second degree in t, then y0P ðtÞ ¼ 2At þ B and y00P ðtÞ ¼ 2A. The given differential equation is: y00 ðtÞ  4y0 ðtÞ þ 3yðtÞ ¼ t2 þ 2t þ 1 The substitution of the assumed values for the particular integral, and its derivatives, in the differential equation results in:

3.2 Ordinary Differential Equations

173


 2A  4ð2At þ BÞ þ 3 At2 þ Bt þ C ¼ t2 þ 2t þ 1 3At2 þ ð8A þ 3BÞt þ ð2A  4B þ 3C Þ ¼ t2 þ 2t þ 1 Identifying similar terms in the two members of this algebraic equation results in that 3A ¼ 1 ) A ¼ 1=3 

8 þ 3B ¼ 2 ) B ¼ 14=9 3

2 56  þ 3C ¼ 1 ) C ¼ 59=27 3 9 2

Therefore, yP ðtÞ ¼ t3 þ

14 59 9 t þ 27

(b) The complementary solution is: yH ðtÞ ¼ k1 þ k2 t þ k3 3t The particular solution is:  1 1 1 4 13 2 2 2 þ D þ D t ¼ t DðD2  4D þ 3Þ D 3 9 27  1 t2 8 26 þ tþ ¼ D 3 9 27

yP ð t Þ ¼

Since

1 D f ðt Þ

¼

R

f ðtÞdt,

Z 2 t 8 26 t3 4t2 26t yP ð t Þ ¼ þ tþ þ dt ¼ þ 9 27 27 3 9 9 So, yð t Þ ¼

t3 4t2 26t þ k1 þ k2 t þ k3 3t þ þ 27 9 9

3º Case: f ðtÞ ¼ A cosðxt þ hÞ
 Initially, it is worth remembering that sinðxt þ hÞ ¼ cos xt þ h  90 .

174

3 Differential Equations

In the case of a sinusoidal forcing function, the abbreviated method is the Phasor Method, namely: the forcing function f ðtÞ is replaced by its Fasor F_ ¼ Ajh ¼ A=h; the operator D is replaced by jx; D2 , by ðjxÞ2 ¼ x2 ; D3 , by ðjxÞ3 ¼ jx3 ; D4 , by ðjxÞ4 ¼ x4 ; … Logically, the particular solution sought, yP ðtÞ, is replaced by its corresponding phasor Y_ P . So, for example, the differential equation


 D4 þ aD3 þ bD2 þ cD þ d yðtÞ ¼ A cosðxt þ hÞ

is converted to the simple frequency domain as: ðx4 jax3 bx2 þ jcx þ dÞY_ P ¼ F_ ¼ A=h Therefore, Y_ P ¼

ðx 4



bx2

F_ ¼ M=u þ d Þ þ jðcx  ax3 Þ

To return to the time domain, that is, to make explicit yP ðtÞ, the phasor definition is used: yP ðtÞ ¼ Real Part of½Y_ P :jxt  So, h i

 yP ðtÞ ¼ RP Mju :jxt ¼ RP Mjðxt þ uÞ ¼ RPfM ½cosðxt þ uÞ þ j sinðxt þ uÞg Therefore, yP ðtÞ ¼ M cosðxt þ uÞ Example 3.9 Solve the following differential equations: (a) ðD2 þ 4ÞyðtÞ ¼ cos 3t
 (b) ðD2 þ 3D  4ÞyðtÞ ¼ sin 2t ¼ cos 2t  90 Solution (a) Since r1 ¼ j2 and r2 ¼ j2, the complementary function is:

yH ðtÞ ¼ k1 cos 2t þ k2 sin 2t

3.2 Ordinary Differential Equations

175

Converting the differential equation to the simple frequency domain, with

 x ¼ 3 rds , it follows that: 1   ð9 þ 4ÞY_ P ¼ 1=0 ) Y_ P ¼  =0 5 Therefore, yP ðtÞ ¼ 0; 2 cos 3t So, yðtÞ ¼ 0; 2 cos 3t þ k1 cos 2t þ k2 sin 2t (b) The characteristic roots are r1 ¼ 1, r2 ¼ 4, and the complementary solution is: yH ðtÞ ¼ k1 t þ k2 4t In the frequency domain, we have that:  ð22 þ j64ÞY_ P ¼ 1=90 

Y_ P ¼

1 1\  90    :1=90 ¼  ¼ 0; 1=233:13 ¼ 0; 1=53:13 8 þ j6 10\143:13

So, in the time domain,

   yP ðtÞ ¼ 0:1 cos 2t  53:13 ¼ 0:1 sin 2t  53:13 þ 90

   ¼ 0; 1 sin 2t þ 36:87 ¼ 0; 1 sin 2t: cos 36:87 þ sin 36:87 : cos 2t ¼ 0:1ð0:8 sin 2t þ 0:6 cos 2tÞ ¼ 0:02ð4 sin 2t þ 3 cos 2tÞ Therefore,
 yðtÞ ¼ 0; 1 sin 2t þ 36:87 þ k1 t þ k2 4t In the sinusoidal case, the abbreviated method itself is as follows:
 1 1 cosðxt þ /Þ; for F x2 6¼ 0; yP ðtÞ ¼
2  cos ðxt þ /Þ ¼ 2 Fðx Þ F D

ð3:19Þ


 1 1 sinðxt þ /Þ; for F x2 6¼ 0: yP ðtÞ ¼
2  sinðxt þ /Þ ¼ 2 Fðx Þ F D

ð3:20Þ

176

3 Differential Equations

Example 3.10 Find the particular solution of the following differential equations: (a) ðD2 þ 4ÞyðtÞ ¼ 10 cos 3t
 (b) ðD4 þ 20D2 ÞyðtÞ ¼ 128 sin 4t  45
 (c) ðD2 þ 3D  4ÞyðtÞ ¼ 100 cos 2t þ 25 Solution (a) The particular solution is yP ð t Þ ¼

D2

1 1 10 cos 3t ¼ 10 cos 3t ¼ 2 cos 3t: þ4 ð3Þ2 þ 4


1  128 sin 4t  45 þ 20Þ

1   h i 128 sin 2t  45 ¼ 2 sin 2t  45 : ¼ 2 2 ð4Þ ð4Þ þ 20

yP ð t Þ ¼

(b)

D2 ðD2

(c) The operator is not of the standard 1/F(D2) and the abbreviated method does not apply ready. Fortunately, there are two ways to get around this difficulty. The first is
1  100 cos 2t þ 25 ðD  1ÞðD þ 4Þ

ðD þ 1ÞðD  4Þ ðD2  3D  4Þ   100 cos 2t þ 25 100 cos 2t þ 25 ¼ ¼ 2 100 ðD  1ÞðD2  16Þ

  ¼ 8 cos 2t þ 25 þ 6 sin 2t þ 25 :

yP ðtÞ ¼

The second is

1 1   100 cos 2t þ 25 ¼ 100 cos 2t þ 25 2 þ 3D  4 ð2Þ þ 3D  4

1 3D þ 8   100 cos 2t þ 25 ¼ ¼ 100 cos 2t þ 25 2 3D  8 9D 
64

   ¼ ð3D þ 8Þ cos 2t þ 25 ¼ 8 cos 2t þ 25 þ 6 sin 2t þ 25 :

yP ðtÞ ¼

D2

4º Case: f ðtÞ ¼ kt uðtÞ, with k being a constant and uðtÞ any function. In this case, yP ðtÞ ¼

1 kt 1  uðtÞ ¼ kt uð t Þ FðDÞ FðD þ kÞ

ð3:21Þ

3.2 Ordinary Differential Equations

177

Example 3.11 Find the general solution of the following differential equations: (a) y0 ðtÞ þ 4yðtÞ ¼ t2 2t (b) ðD  2Þ2 yðtÞ ¼ t þ t2t : Solution (a) ðD þ 4ÞyðtÞ ¼ t2 2t

yH ðtÞ ¼ k1 4t 1 1 1 2 ðt2 2t Þ ¼ 2t  t2 ¼ 2t  t Dþ4 ð D  2Þ þ 4 Dþ2 2  1 1 1 t t 1  D þ D2 t2 ¼ 2t  þ ¼ 2t 2 4 8 2 2 4

yP ðtÞ ¼

Therefore, yðtÞ ¼ k1 4t þ

ðbÞ

2 t t 1 2t  þ  2 2 4

yH ðtÞ ¼ ðk1 þ k2 tÞ2t yP ð t Þ ¼




1

 t þ t2t ¼

1

t þ

1

t2t ð D  2Þ 2 1 ¼ t þ 2t  t ¼ t þ 2t : 2 ðtÞ 2 2 D ð 1  2Þ ½ðD þ 2Þ  2 Z Z 2 1 t t3 t dt ¼ t þ 2t : dt ¼ t þ 2t : ¼ t þ 2t : D 2 6 ð D  2Þ 1

2

ð D  2Þ 1

2

Finally, yð t Þ ¼

t3 2t  þ t þ ðk1 þ k2 tÞ2t 6

178

3.3

3 Differential Equations

Differential Equations with Singular and Causal Forcing Functions

When the forcing function f ðtÞ of a differential equation is singular and / or causal, it may happen that the solution yðtÞ and its derivatives are discontinuous at t ¼ 0. That is, for a given differential equation of order n, it may happen that yð0 þ Þ 6¼ yð0 Þ y0 ð0 þ Þ 6¼ y0 ð0 Þ y00 ð0 þ Þ 6¼ y00 ð0 Þ . . .. . .. . .. . .. . .. . .. . . yn1 ð0 þ Þ 6¼ yn1 ð0 Þ

ð3:22Þ

In electrical circuit problems, for example, the initial conditions are generally known at t ¼ 0 , but the determination of the arbitrary n constants in the general solution, the circuit response, requires knowledge of the initial conditions at t ¼ 0 þ . Therefore, one of the problems to be solved is the determination of the initial conditions at t ¼ 0 þ , based on their knowledge at t ¼ 0 , as they may or may not be the same. To facilitate the treatment of such by questions, a second order differential equation will be used as a reference, which, facilitating the exposure that will be made, will not cause any loss of generality. In addition, some essentially heuristic reasoning will be made. Suppose that the differential equation that relates the input xðtÞ and the output yðtÞ of a linear and time-invariant system, with concentrated parameters, is d2 y dy þ a1 þ a2 y ¼ b0 x ð t Þ dt2 dt or y00 ðtÞ þ a1 y0 ðtÞ þ a2 yðtÞ ¼ b0 xðtÞ

ð3:23Þ

The general solution of a differential equation of order n, as shown in Sect. 3.2, involves n essential arbitrary constants, derived from the complementary solution, and the determination of these constants is only possible when n boundary conditions, as they are called, are known. In the case of electrical circuits, for example, n initial conditions at t ¼ 0 þ . Consequently, the solution of Eq. 3.23 requires knowledge of yð0 þ Þ and y0 ð0 þ Þ.

3.3 Differential Equations with Singular and Causal Forcing Functions

3.3.1

179

Step Type Forcing Function U 1 ðtÞ

When xðtÞ ¼ U1 ðtÞ, then, for the two members of Eq. 3.23 to be identical, at any time t, it is necessary that in some part of the first member of the differential equation the step b0 U1 ðtÞ also appear. And it is evident that this portion has to be y00 ðtÞ. This is because if y0 ðtÞ contains the step b0 U1 ðtÞ, then y00 ðtÞ will contain its derivative, which is the impulse b0 U0 ðtÞ, and this will break the balance of the two members of that differential equation, as there is no impulse occurring at t ¼ 0 in the second member of the same. For similar reasoning, it can be concluded that yðtÞ cannot contain the step b0 U1 ðtÞ. For if yðtÞ contains the step b0 U1 ðtÞ, y0 ðtÞ will contain its derivative, which is the impulse b0 U0 ðtÞ and, even worse, y00 ðtÞ will contain the double function b0 U1 ðtÞ, with the latter two occurring exactly at t ¼ 0. It turns out that in the second member of Eq. 3.23 there is neither impulse nor double at t ¼ 0. However, if y00 ðtÞ contains the step b0 U1 ðtÞ, then y0 ðtÞ must contain its integral, which is the ramp function b0 U2 ðtÞ, and yðtÞ must contain the ramp integral, which is the second degree parabola function b0 U3 ðtÞ. And since the ramp and parabola functions are continuous at any time t, containing no step function, then, according to Eq. 1.92 and subsequent conclusion, it can be concluded, finally, that yð0 þ Þ ¼ yð0 Þ and that y0 ð0 þ Þ ¼ y0 ð0 Þ. But the presence of the step b0 U1 ðtÞ in y00 ðtÞ causes y00 ð0 þ Þ 6¼ y00 ð0 Þ. It turns out, however, that the initial condition y00 ð0 þ Þ is not necessary, since the differential equation being second order, only the initial conditions yð0 þ Þ and y0 ð0 þ Þ are needed to determine the constants k1 and k2 that arise from the solution of the corresponding homogeneous differential equation. Therefore, there is no difficulty in calculating the solution of Eq. 3.23, when xðtÞ is a unit step U1 ðtÞ. The particular solution is found using the abbreviated method to the equation


 D2 þ a1 D þ a2 yðtÞ ¼ b0 ; since for t [ 0; U1 ðtÞ ¼ 1:

With the values of yð0 þ Þ, which is equal to yð0 Þ, and of y0 ð0 þ Þ, which is equal to y0 ð0 Þ, the arbitrary constants k1 and k2 of the general solution are determined. Although it is not necessary to know the value of y00 ð0 þ Þ for the complete solution of the differential equation, it can be calculated by making t ¼ 0 þ in Eq. 3.23 itself. Thus, y00 ð0 þ Þ þ a1 y0 ð0 þ Þ þ a2 yð0 þ Þ ¼ b0 So, y00 ð0 þ Þ ¼ b0  ½a1 y0 ð0 þ Þ þ a2 yð0 þ Þ

180

3 Differential Equations

Example 3.12 Solve the differential equation y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 5U1 ðtÞ, with yð0 Þ ¼ 0 and y0 ð0 Þ ¼ 0. Solution The characteristic roots are r1 ¼ 1 and r2 ¼ 2, and the complementary solution is yH ðtÞ ¼ k1 t þ k2 2t The particular solution is obtained considering the forcing function equal to 5 and using the Eq. 3.16, with a ¼ 0. So, yP ð t Þ ¼

1 5 ð50:t Þ ¼ D2 þ 3D þ 2 2

Therefore, the general solution is yð t Þ ¼

5 þ k1 t þ k2 2t ; and 2

y0 ðtÞ ¼ k1 t  2k2 2t Since the yðtÞ and y0 ðtÞ functions must be continuous at t ¼ 0, yð0 þ Þ ¼ yð0 Þ ¼ 0 and y0 ð0 þ Þ ¼ y0 ð0 Þ ¼ 0: Substituting these initial conditions in yðtÞ and y0 ðtÞ, for t ¼ 0 þ , it turns out that k1 þ k2 ¼ 

5 2

k1  2k2 ¼ 0 Solving this system of algebraic equations, it turns out that k1 ¼ 5 and k2 ¼ 5=2. Finally,    5
yðtÞ ¼ 5t þ 1 þ 2t :U1 ðtÞ 2 The multiplication by U1 ðtÞ is to point out that the expression found for yðtÞ, enclosed in brackets, is only valid for t [ 0, since the forcing function 5U1 ðtÞ only acts for t [ 0 and, in addition, the initial conditions at t ¼ 0 are null. At the beginning of this section, it was stated that when a second order differential equation has a step forcing function, then the solution yðtÞ must contain a parabola U3 ðtÞ ¼ ðt2 =2!ÞU1 ðtÞ. To confirm this statement, just expand the solution obtained using the Mac-Laurin Series of the exponential function:

3.3 Differential Equations with Singular and Causal Forcing Functions

at ¼ 1 þ Thus,

at 1!

þ

a2 t 2 2!

þ

a3 t 3 3!

181

þ . . ., converging to 1\at\ þ 1:

    t t2 t3 5 2t 4t2 8t3 1þ1  þ yðtÞ ¼ 5 1  þ  þ . . . þ  þ . . . :U1 ðtÞ 1! 2! 3! 2 1! 2! 3!  2 3 4 t t t ¼ 5:  15: þ 35:  . . . :U1 ðtÞ 2! 3! 4! or, yðtÞ ¼ 5U3 ðtÞ  15U4 ðtÞ þ 35U5 ðtÞ  . . . y0 ðtÞ ¼ 5U2 ðtÞ  15U3 ðtÞ þ 35U4 ðtÞ  . . . y00 ðtÞ ¼ 5U 1 ðtÞ  15U2 ðtÞ þ 35U3 ðtÞ  . . . The series expansions of singular functions above clearly show that yðtÞ contains the parable 5U3 ðtÞ; y0 ðtÞ, the ramp 5U2 ðtÞ; and y00 ðtÞ, the step 5U1 ðtÞ. Furthermore, that the solution in the expanded form of yðtÞ; together with its two derivatives, satisfies the given differential equation. And also, that y00 ð0 þ Þ ¼ 5, a result that can be proved by doing t ¼ 0 þ in the given differential equation itself: y00 ð0 þ Þ þ 3y0 ð0 þ Þ þ 2yð0 þ Þ ¼ 5 Since yð0 þ Þ ¼ y0 ð0 þ Þ ¼ 0, it follows that y00 ð0 þ Þ ¼ 5. Note, too, that the singular functions present in the expansions of yðtÞ and y0 ðtÞ are all null at t ¼ 0 þ , exactly because they do not have the step function U1 ðtÞ and, for this very reason, they make yðtÞ and y0 ðtÞ continuous at t ¼ 0. When xðtÞ in Eq. 3.23 is a singular forcing function of the type obtained by successive integration of the unit step, that is, ramps and parabolas, then all initial conditions are equal at t ¼ 0 and at t ¼ 0 þ , i.e., yð0 þ Þ ¼ yð0 Þ; y0 ð0 þ Þ ¼ y0 ð0 Þ; y00 ð0 þ Þ ¼ y00 ð0 Þ;

ð3:24Þ

and the particular solution yP ðtÞ can be obtained by using the abbreviated method directly, or the method of the coefficients to be determined, without any difficulty. Example 3.13 A linear system is described by the differential equation. y00 ðtÞ þ 5y0 ðtÞ þ 6yðtÞ ¼ 6xðtÞ Determine your answer yðtÞ for an input xðtÞ equals to a unit ramp U2 ðtÞ, given the initial conditions yð0 Þ ¼ 0 and y0 ð0 Þ ¼ 1.

182

3 Differential Equations

Solution The characteristic roots are r1 ¼ 2 and r2 ¼ 3. So, the complementary solution is: yH ðtÞ ¼ k1 2t þ k2 3t ; regardless of what the forcing function is. When it is a unitary ramp, the equation becomes:


 D2 þ 5D þ 6 yðtÞ ¼ 6U2 ðtÞ ¼ 6tU1 ðtÞ

Therefore, a particular solution is determined by considering the forcing function equal to 6t, for 0 þ t\1, and using the second case of the abbreviated method, that is, Eq. 3.18. Thus, 1 yP ð t Þ ¼ ð6tÞ ¼ 6 þ 5D þ D2

 1 5 5  D 6t ¼ t  6 36 6

So, yð t Þ ¼ t 

5 þ k1 2t þ k2 3t 6

y0 ðtÞ ¼ 1  2k1 2t  3k2 3t Since the initial conditions at t ¼ 0 and at t ¼ 0 þ are the same, yð0 þ Þ ¼ yð0 Þ ¼ 0 and y0 ð0 þ Þ ¼ y0 ð0 Þ ¼ 1. Taking t ¼ 0 þ in yðtÞ and y0 ðtÞ results in the following system of algebraic equations: 5 0 ¼  þ k1 þ k2 6 1 ¼ 1  2k1  3k2 or k1 þ k2 ¼ 5=6 2k1 þ 3k2 ¼ 0 The solution of this system results in k1 ¼ 5=2 and k2 ¼ 5=3: Finally, yð t Þ ¼ t 

5 5 2t 5 3t þ    ; for t  0: 6 2 3

3.3 Differential Equations with Singular and Causal Forcing Functions

183

Note that now the result obtained cannot be multiplied by U1(t). Because if yðtÞ ¼ ðt 

5 5 2t 5 3t þ    ÞU1 ðtÞ; 6 2 3

it turns out that yð0 Þ ¼ 0, due to the presence of the unitary step, confirming the initial condition given in the example statement. But on the other hand, y0 ð t Þ ¼

 
 5 5 5 t  þ 2t  3t U1 ðtÞ ¼ 1  52t þ 53t U1 ðtÞ 6 2 3   
 5 5 5 þ t  þ 2t  3t  :U0 ðtÞ ¼ 1  52t þ 53t U 1 ðtÞ; 6 2 3 t¼0 d dt

causing y0 ð0 Þ ¼ 0, in a clear contradiction with the statement of the example, which establishes that y0 ð0 Þ ¼ 1.

3.3.2

Forcing Function Type Impulse U 0 ðtÞ

If xðtÞ ¼ U0 ðtÞ, the differential Eq. 3.23 becomes: y00 ðtÞ þ a1 y0 ðtÞ þ a2 yðtÞ ¼ b0 U0 ðtÞ

ð3:25Þ

To ensure that the left side of Eq. 3.25 is identical to its right side, at any time t, y00 ðtÞ must contain the impulse b0 U0 ðtÞ, y0 ðtÞ, in turn, contain the step b0 U1 ðtÞ and yðtÞ, the ramp b0 U2 ðtÞ, as first terms in their respective expansions in infinite series of singular functions, with decreasing indices. Therefore, for the case of Eq. 3.25, which is second order, yðtÞ does not contain the step function, causing yð0 þ Þ ¼ yð0 Þ. But both y0 ðtÞ and y00 ðtÞ contain the step b0 U1 ðtÞ and, as a consequence, y0 ðtÞ and y00 ðtÞ are discontinuous in t ¼ 0, that is: y0 ð0 þ Þ 6¼ y0 ð0 Þ and y00 ð0 þ Þ 6¼ y00 ð0 Þ. It is evident that if the differential equation is of third order, y000 ðtÞ will contain impulse, y00 ðtÞ will contain step, y0 ðtÞ will contain ramp and yðtÞ will contain second degree parabola, in a way that y000 ð0 þ Þ 6¼ y000 ð0 Þ, y00 ð0 þ Þ 6¼ y00 ð0 Þ, y0 ð0 þ Þ ¼ y0 ð0 Þ and yð0 þ Þ ¼ yð0 Þ. Conclusion: for a differential equation of any order with a forcing function involving unit impulse, U 0 ðtÞ, the two derivatives of the highest order of yðtÞ are discontinuous at t ¼ 0. Therefore, in the case of a differential equation of second order, it is necessary to calculate only y0 ð0 þ Þ, from the knowledge of the value of y0 ð0 Þ, since yð0 þ Þ ¼ yð0 Þ, and it is not necessary to have knowledge of y00 ð0 þ Þ for the calculation of the two essential arbitrary constants that appear in the general solution of the equation.

184

3 Differential Equations

And one way to calculate y00 ð0 þ Þ is to integrate Eq. 3.25 between the limits 0 and 0 þ , namely: Z0 þ

Z0 þ

00

y ðtÞdt þ a1 0

0

0

Z0 þ

y ðtÞdt þ a2 0

Z0 þ yðtÞdt ¼ b0

0

U0 ðtÞdt 0

Z0 þ

0

½y ð0 þ Þ  y ð0 Þ þ a1 ½yð0 þ Þ  yð0 Þ þ a2

yðtÞdt ¼ b0 0

Since yðtÞ is continuous at t ¼ 0, with greater strength its integral will be continuous, resulting, therefore, that: y 0 ð 0 þ Þ  y 0 ð 0 Þ ¼ b0 thus y 0 ð 0 þ Þ ¼ b0 þ y 0 ð 0 Þ

ð3:26Þ

Note that b0 is the area of the impulsive forcing function, and that it is also the step amplitude in y0 ðtÞ. If it were necessary to find the value of y00 ð0 þ Þ, it would be enough to substitute t for 0 þ in Eq. 3.25, obtaining that y00 ð0 þ Þ ¼ ½a1 y0 ð0 þ Þ þ a2 yð0 þ Þ

ð3:27Þ

Example 3.14 Solve the differential equation y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 5U0 ðtÞ; with the initial conditions yð0 Þ ¼ 0 and y0 ð0 Þ ¼ 0. Solution Since the forcing function is an impulse, the series expansion of singular functions of y00 ðtÞ must start with an impulse, that of y0 ðtÞ with a step and that of yðtÞ with a ramp. Therefore, yð0 þ Þ ¼ yð0 Þ ¼ 0 and, according to Eq. 3.26, y0 ð0 þ Þ ¼ 5 þ 0 ¼ 5. On the other hand, for t [ 0 the equation is:
And its general solution is

 D2 þ 3D þ 2 yðtÞ ¼ 0

3.3 Differential Equations with Singular and Causal Forcing Functions

185


 yðtÞ ¼ k1 t þ k2 2t U1 ðtÞ Then,
 y0 ðtÞ ¼  k1 t þ 2k2 2t U1 ðtÞ þ ðk1 þ k2 ÞU0 ðtÞ Substituting the initial conditions yð0 þ Þ ¼ 0 and y0 ð0 þ Þ ¼ 5 in yðtÞ and y0 ðtÞ, and remembering that U0 ð0 þ Þ ¼ 0, results the system of equations 0 ¼ k1 þ k2 5 ¼ k1  2k2 ; from which it follows that que k1 ¼ 5 and k2 ¼ 5. Finally,
 yðtÞ ¼ 5t  52t U1 ðtÞ
 y0 ðtÞ ¼ 5t þ 102t U1 ðtÞ
 y00 ðtÞ ¼ 5U0 ðtÞ þ 5t  202t U1 ðtÞ So, checking,
 y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 5U0 ðtÞ þ 5t  202t U1 ðtÞ
 þ 15t þ 302t U1 ðtÞ
 þ 10t  102t U1 ðtÞ ¼ 5U0 ðtÞ Also, from the last expressions, it is evident that yð0 þ Þ ¼ 0, y0 ð0 þ Þ ¼ 5 and that y ð0 þ Þ ¼ 15. From the given differential equation itself, for t ¼ 0 þ , one has to 00

y00 ð0 þ Þ þ 3y0 ð0 þ Þ þ 2yð0 þ Þ ¼ 0; since U0 ðtÞ ¼ 0 for t 6¼ 0: Hence, in accordance with Eq. 3.27, y00 ð0 þ Þ ¼ ½3y0 ð0 þ Þ þ 2yð0 þ Þ ¼ 15 Therefore,
 yðtÞ ¼ 5t  52t U1 ðtÞ it is really the solution sought.

186

3 Differential Equations

By also replacing the exponential functions present in yðtÞ, y0 ðtÞ and y00 ðtÞ by their respective Mac-Laurin series, the infinite series expansions of singular functions are obtained: yðtÞ ¼ 5U2 ðtÞ  15U3 ðtÞ þ 35U4 ðtÞ  . . . y0 ðtÞ ¼ 5U1 ðtÞ  15U2 ðtÞ þ 35U3 ðtÞ  . . . y00 ðtÞ ¼ 5U 0 ðtÞ  15U1 ðtÞ þ 35U2 ðtÞ  . . . And you can confirm the presence of the ramp 5U2 ðtÞ, the step 5U1 ðtÞ and the impulse 5U0 ðtÞ as first terms in the expansions in singular functions of yðtÞ, y0 ðtÞ and y00 ðtÞ, respectively. And also, that yð0 þ Þ ¼ 0, y0 ð0 þ Þ ¼ 5 and y00 ð0 þ Þ ¼ 15, which are the amplitudes of the steps in yðtÞ, y0 ðtÞ and y00 ðtÞ, respectively. Note, too, that the ramp function that is present in yðtÞ, is also present in y0 ðtÞ and 00 y ðtÞ, with different coefficients, of course. And also that the step function U1 ðtÞ that is contained in y0 ðtÞ, with amplitude 5, is also present in y00 ðtÞ, but with amplitude 15. In addition, since the initial conditions at t ¼ 0 are all null, the initial conditions at t ¼ 0 þ of y, y0 e y00 are the amplitudes of the steps in the respective expansions thereof. Logically, the expansion of y (t) in singular functions must be understood as: yðtÞ ¼ 0:U 1 ðtÞ þ 5U 2 ðtÞ  15U3 ðtÞ þ 35U4 ðtÞ  . . . Example 3.15 Solve example 3.14 with the initial conditions yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2. Solution The initial conditions at t ¼ 0 þ are: yð0 þ Þ ¼ yð0 Þ ¼ 1 and y0 ð0 þ Þ ¼ y0 ð0 Þ þ b0 ¼ 2 þ 5 ¼ 7, according to Eq. 3.26. And the general solution is: yðtÞ ¼ k1 t þ k2 2t ; for t [ 0: In the present case, in which the initial conditions at t ¼ 0 are different from zero, the answer cannot be multiplied by U1 ðtÞ, because if this were done it would be obtained that:
 yðtÞ ¼ k1 t þ k2 2t U1 ðtÞ

3.3 Differential Equations with Singular and Causal Forcing Functions

187

and that
 y0 ðtÞ ¼  k1 t þ 2k2 2t U1 ðtÞ þ ðk1 þ k2 ÞU0 ðtÞ Since yð0 þ Þ ¼ k1 þ k2 ¼ yð0 Þ ¼ 1 6¼ 0, then there would be the impulse function ðk1 þ k2 ÞU0 ðtÞ ¼ U0 ðtÞ in y0 ðtÞ and, consequently, the double function U1 ðtÞ in y00 ðtÞ, which is completely unacceptable, because it would destroy the balance of the two members of the differential equation. In other words, as yð0 þ Þ ¼ yð0 Þ 6¼ 0, multiplying the solution by U1 ðtÞ would imply the inclusion of the step yð0 þ ÞU1 ðtÞ in yðtÞ, which would go against the requirement that yðtÞ contain a U2 ðtÞ ramp, and its successive integrals, when expanded into singular functions, but not a step function, as highlighted at the end from example 3.14. Therefore, the solution to the equation is really yðtÞ ¼ k1 t þ k2 2t ; for t [ 0; and y0 ðtÞ ¼ k1 t  2k2 2t ; for t [ 0: Entering the initial conditions at t ¼ 0 þ , it results that: 1 ¼ k1 þ k2 7 ¼ k1  2k2 ; whose solution is k1 ¼ 9 and k2 ¼ 8. Finally, yðtÞ ¼ 9t  82t ; for t [ 0: When the forcing function contains one or more impulsive functions occurring at t ¼ 0, the differential equation is equal to zero for t [ 0, since impulsive functions are null for t 6¼ 0. Therefore, the solution of the differential equation is always reduced to the solution of a homogeneous differential equation. From the study carried out in Sect. 3.2, it appears that whatever the roots of the characteristic equation F ðDÞ ¼ 0, are the solution of the homogeneous equation can always be expressed by exponential functions, that is, decomposed into a series of powers of t by means of Mac-Laurin series. This fact forms the basis for the development of a systematic method for determining the initial conditions at t ¼ 0 þ , which is presented below.

188

3.3.3

3 Differential Equations

Systematic Method for Calculating Initial Conditions at t ¼ 0 þ

To facilitate the understanding, example 3.15 is taken up, that is, the solution of the differential equation y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 5U0 ðtÞ; with yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2: The need for y00 ðtÞ to contain the impulse function, y0 ðtÞ the step function and yðtÞ the ramp function, together with the possibility of expressing the solution in the form of a series of powers of t, let write that yðtÞ ¼ 0:U1 ðtÞ þ C0 U 2 ðtÞ þ C1 U3 ðtÞ þ C2 U4 ðtÞ þ C3 U5 ðtÞ þ . . .

ð3:28Þ

y0 ðtÞ ¼ C0 U 1 ðtÞ þ C1 U2 ðtÞ þ C2 U3 ðtÞ þ C3 U4 ðtÞ þ . . .

ð3:29Þ

y00 ðtÞ ¼ C0 U 0 ðtÞ þ C1 U1 ðtÞ þ C2 U2 ðtÞ þ C3 U3 ðtÞ þ . . .

ð3:30Þ

where Ci are coefficients to be determined. (Obviously, Eqs. 3.28 and 3.29 contain two inconsistencies. Namely: they imply that yð0 Þ ¼ 0 and that y0 ð0 Þ ¼ 0, since all the singular functions contained in them are null at t ¼ 0 , which is in complete disagreement with the statement of the problem, which claims to be yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2. In the meantime, the comments that follow will put things in their proper terms). Substituting Eqs. 3.28 to 3.30 in the given differential equation, it follows that: ½C0 U0 ðtÞ þ C1 U1 ðtÞ þ C2 U2 ðtÞ þ C3 U3 ðtÞ þ . . . þ 3½C0 U1 ðtÞ þ C1 U2 ðtÞ þ C2 U3 ðtÞ þ C3 U4 ðtÞ þ . . . þ 2½C0 U2 ðtÞ þ C1 U3 ðtÞ þ C2 U4 ðtÞ þ C3 U5 ðtÞ þ . . . ¼ 5U0 ðtÞ Grouping similar terms together, it follows that: C0 U0 ðtÞ þ ðC1 þ 3C0 ÞU1 ðtÞ þ ðC2 þ 3C1 þ 2C0 ÞU2 ðtÞ þ ðC3 þ 3C2 þ 2C1 ÞU3 ðtÞ þ . . . ¼ 5U0 ðtÞ þ 0U1 ðtÞ þ 0:U2 ðtÞ þ 0:U3 ðtÞ þ . . . Identifying the two members of this equation, it is obtained that:

3.3 Differential Equations with Singular and Causal Forcing Functions

0

= 5………………………

0

=5

1

+3

0

= 0………………

1

= −15

2

+3

1

+2

0

= 0………

2

= 35

3

+3

2

+2

1

= 0………

3

= −75

189

ð3:31Þ

… … … … … … … … … … … … … … … ….

If the initial conditions at t ¼ 0 were all null, evidently the initial conditions at t ¼ 0 þ would be the coefficients of the step functions of each of the Eqs. 3.28 to 3.30, namely: yð0 þ Þ ¼ 0, y0 ð0 þ Þ ¼ C0 ¼ 5 and y00 ð0 þ Þ ¼ C1 ¼ 15, as in example 3.14. However, as the initial conditions at t ¼ 0 are not null, it is necessary to add to the initial conditions at t ¼ 0 the step coefficients existing in yðtÞ, y0 ðtÞ and y00 ðtÞ, to obtain the respective initial conditions at t ¼ 0 þ , as established in Eq. 3.32. Therefore, yð0 þ Þ ¼ yð0 Þ þ 0 ¼ yð0 Þ ¼ 1 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C0 ¼ 2 þ 5 ¼ 7 y00 ð0 þ Þ ¼ y00 ð0 Þ þ C1 ¼ y00 ð0 Þ ¼ 15

ð3:32Þ

Obviously, the values found, yð0 þ Þ ¼ 1 and y0 ð0 þ Þ ¼ 7, agree with those obtained in example 3.15. An additional observation, however, allows to simplify the exposed procedure considerably. The initial conditions at t ¼ 0 þ , listed in expressions 3.32, show that it is necessary to calculate only the coefficients of the steps present in yðtÞ and its derivatives. Then, from the series expansions of singular functions of yðtÞ, y0 ðtÞ and y00 ðtÞ, contained in Eqs. 3.28 to 3.30, it is sufficient to compute only the first terms of the series, truncating them in steps, as the subsequent functions are all continuous, and null at t ¼ 0 þ . Thus, looking at Eqs. 3.28 to 3.30, it is possible to write only that : yðtÞ ¼ 0:U 1 ðtÞ: : y0 ðtÞ ¼ C0 U1 ðtÞ: : y00 ðtÞ ¼ C0 U0 ðtÞ þ C1 U 1 ðtÞ:

ð3:33Þ

The notation ≐ is used only to remember that the function yðtÞ and its derivatives are not only made up of the singular functions that are indicated in each one, that is,

190

3 Differential Equations

that the expansions in singular functions, which contain an infinite number of terms, were truncated on the steps. Therefore, replacing the expressions contained in Eq. 3.33 in the given differential equation, results: : C0 U0 ðtÞ þ ðC1 þ 3C0 ÞU1 ðtÞ¼5U0 ðtÞ þ 0:U1 ðtÞ So, C0 ¼ 5 C1 þ 3C0 ¼ 0; C1 ¼ 15 Example 3.16 Solve the differential equation y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 5U1 ðtÞ; with yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2: Solution For t [ 0, the equation is y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 0; And its general solution is: yðtÞ ¼ k1 t þ k2 2t y0 ðtÞ ¼ k1 t  2k2 2t Since the forcing function is a double function, the balance of the two members of the given equation requires that y00 ðtÞ contain a double, y0 ðtÞ an impulse and yðtÞ a step. Therefore, for the determination of the initial conditions at t ¼ 0 þ , one must consider that : yðtÞ ¼ C0 U 1 ðtÞ: : y0 ðtÞ ¼ C0 U0 ðtÞ þ C1 U 1 ðtÞ: : y00 ðtÞ ¼ C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U1 ðtÞ: Substituting in the differential equation, results in : C0 U 1 ðtÞ þ ðC1 þ 3C0 ÞU 0 ðtÞ þ ðC2 þ 3C1 þ 2C0 ÞU 1 ðtÞ¼5U 1 ðtÞ þ 0:U0 ðtÞ þ 0:U 1 ðtÞ:

So,

3.3 Differential Equations with Singular and Causal Forcing Functions

0

= 5………………………

0

=5

1

+3

0

= 0………………

1

= −15

2

+3

1

+2

2

= 35

0

= 0………

191

Therefore, yð0 þ Þ ¼ yð0 Þ þ C0 ¼ 1 þ 5 ¼ 6 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C1 ¼ 2  15 ¼ 13 y00 ð0 þ Þ ¼ y00 ð0 Þ þ C2 ¼ y00 ð0 Þ þ 35 So, from yðtÞ and y0 ðtÞ, to t ¼ 0 þ , it comes that 6 ¼ k1 þ k2 13 ¼ k1  2k2 Solving this system results that k1 ¼ 1 and k2 ¼ 7. Finally, yðtÞ ¼ 72t  t ; for t [ 0: When the given differential equation is integrated between the limits 0 and 0 þ it results that 0

Z0 þ

0

½y ð0 þ Þ  y ð0 Þ þ 3½yð0 þ Þ  yð0 Þ þ 2

Z0 þ yðtÞdt ¼ 5

0

U1 ðtÞdt 0

The function yðtÞ presents a finite discontinuity of 5 units at t ¼ 0, since yð0 þ Þ ¼ 6 and yð0 Þ ¼ 1. However, the process of integrating a function with finite discontinuity(s) always produces an absolutely continuous function, which allows you to write that Z0 þ yðtÞdt ¼ 0 0

Since the double function is odd,

0Rþ 0

U1 ðtÞdt ¼ 0.

192

3 Differential Equations

Therefore, y0 ð0 þ Þ  y0 ð0 Þ þ 3yð0 þ Þ  3yð0 Þ ¼ 0: This equation is satisfied by the initial conditions given and calculated. Example 3.17 An input system xðtÞ and output yðtÞ is described by the differential equation y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 4x0 ðtÞ þ 2xðtÞ. Find the output corresponding to the input xðtÞ ¼ U0 ðtÞ, considering that yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 3. Solution When xðtÞ ¼ U0 ðtÞ, the equation becomes y00 ðtÞ þ 3y0 ðtÞ þ 2yðtÞ ¼ 4U1 ðtÞ þ 2U0 ðtÞ Its solution, for t [ 0, is yðtÞ ¼ k1 t þ k2 2t y0 ðtÞ ¼ k1 t  2k2 2t As a consequence of the second member of the differential equation, it follows that y00 ðtÞ must contain a double, y0 ðtÞ an impulse and yðtÞ a step, as first terms in their respective expansions in series of singular functions. So, : yðtÞ ¼ C0 U 1 ðtÞ: : 0 y ðtÞ ¼ C0 U0 ðtÞ þ C1 U 1 ðtÞ: : 00 y ðtÞ ¼ C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U1 ðtÞ: Substituting in the given equation, it results that: : C0 U 1 ðtÞ þ ðC1 þ 3C0 ÞU 0 ðtÞ þ ðC2 þ 3C1 þ 2C0 ÞU 1 ðtÞ ¼ 4U 1 ðtÞ þ 2U 0 ðtÞ þ 0:U 1 ðtÞ:

So, 0

= 4………………………

0

=4

1

+3

0

= 2………………

1

= −10

2

+3

1

+2

2

= 22

0

= 0………

3.3 Differential Equations with Singular and Causal Forcing Functions

193

Thus, yð0 þ Þ ¼ yð0 Þ þ C0 ¼ 1 þ 4 ¼ 5 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C1 ¼ 3  10 ¼ 7 Therefore, k1 þ k2 ¼ 5 k1 þ 2k2 ¼ 7 So, k1 ¼ 3 and k2 ¼ 2. Finally, yðtÞ ¼ 3t þ 22t ; for t [ 0: Example 3.18 Solve the integral-differential equation 00

0

Zt

y ðtÞ þ 8y ðtÞ7

yðkÞdk ¼ 3U1 ðtÞ; 0

with the initial conditions yð0 Þ ¼ 2 and y0 ð0 Þ ¼ 4. Solution Considering that yðtÞ and y0 ðtÞ are continuous at t ¼ 0, since there will only be a step in y00 ðtÞ, then, yð0 þ Þ ¼ yð0 Þ ¼ 2 and y0 ð0 þ Þ ¼ y0 ð0 Þ ¼ 4. As the differential equation is of the third order, there will be three arbitrary constants to be determined in its general solution and, as a result, it is also necessary to calculate y00 ð0 þ Þ. And this is done by replacing t with 0 þ in the integral-differential equation itself: 00

0

Z0 þ

y ð0 þ Þ þ 8y ð0 þ Þ  7

yðkÞdk ¼ 3U1 ð0 þ Þ ¼ 3 0

Since yðtÞ is continuous at t = 0, with greater strength it will be its integral, so 0Rþ yðkÞdk ¼ 0. that 0

194

3 Differential Equations

So, y00 ð0 þ Þ ¼ 3  8y0 ð0 þ Þ ¼ 3  8 4 ¼ 29 Deriving the given integral-differential equation, we have y000 ðtÞ þ 8y00 ðtÞ  7yðtÞ ¼ 3U0 ðtÞ For t [ 0, it is


 D3 þ 8D2  7 yðtÞ ¼ 0

The characteristic equation is F ðDÞ ¼ D3 þ 8D2  7 ¼ 0 If there is only one signal inversion in the sequence of coefficients, there is the possibility of a single positive real root, according to Descartes’ rule. Replacing D with D, it becomes: D3 þ 8D2  7 ¼ 0. If there are two sign inversions, then the possibility of having negative real roots is none or two. On the other hand, it is reasonably easy to see that a root is r1 ¼ 1. (If a root is not noticed to be -1, the alternative is to look for an initial estimate for r1 and apply the Newton-Raphson method). After deflation of the polynomial, it follows that D2 þ 7D  7 ¼ 0, which gives the roots r2 ¼ 0:89 and r3 ¼ 7:89. So, the general solution of the equation is: yðtÞ ¼ k1 t þ k2 0:89t þ k3 7:89t y0 ðtÞ ¼ k1 t þ 0:89k2 0:89t  7:89k3 7:89t y00 ðtÞ ¼ k1 t þ 0:79k2 0:89t þ 62:25k 3 7:89t Considering the instant t ¼ 0 þ , the following system results: k1 þ k2 þ k3 ¼ 2 k1 þ 0:89k 2  7:89k3 ¼ 4 k1 þ 0:79k 2 þ 62:25k 3 ¼ 29 And the solution for this system is: k1 ¼ 1:00; k2 ¼ 0:56; k3 ¼ 0:44. Finally, yðtÞ ¼ t  0:560:89t  0:447:89t , for t [ 0. Example 3.19 Given the differential equation y00 ðtÞ þ 12y0 ðtÞ þ 4yðtÞ ¼ U2 ðtÞ þ þ 2U0 ðtÞ, with yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2, determine yð0 þ Þ and y0 ð0 þ Þ.

3.3 Differential Equations with Singular and Causal Forcing Functions

195

Solution As the most positive singular forcing function, U2 ðtÞ, has to be in y00 ðtÞ, it comes that: : yðtÞ ¼ C0 U 0 ðtÞ þ C1 U 1 ðtÞ: : y0 ðtÞ ¼ C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U 1 ðtÞ: : y00 ðtÞ ¼ C0 U2 ðtÞ þ C1 U1 ðtÞ þ C2 U0 ðtÞ þ C3 U 1 ðtÞ: Substituting in the differential equation, it comes that: C0 U2 ðtÞ þ ðC1 þ 12C0 ÞU1 ðtÞ þ ðC2 þ 12C1 þ 4C0 ÞU0 ðtÞ þ ðC3 þ 12C2 þ 4C1 ÞU1 ðtÞ : ¼U2 ðtÞ þ 0:U1 ðtÞ þ 2U0 ðtÞ þ 0:U1 ðtÞ

0

= 1………………………

0

=1

1

+ 12

0

= 0…………….

1

= −12

2

+ 12

1

+4

2

= 142

0

= 2…….

Since the initial condition at t ¼ 0 þ is the sum of the initial condition at t ¼ 0 with the amplitude of the step present in the respective function, yð0 þ Þ ¼ yð0 Þ þ C1 ¼ 1  12 ¼ 11 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C2 ¼ 2 þ 142 ¼ 140 It is interesting to note that the solution of the differential equation will contain the impulse function, that is, yð t Þ ¼

U0 ðtÞ; yH ðtÞ;

for t ¼ 0 for t [ 0:

Example 3.20 Solve the differential equation y000 ðtÞ þ 4y0 ðtÞ ¼ 6U1 ðtÞ þ 10 t U1 ðtÞ ¼ 6U1 ðtÞ þ 10½U1 ðtÞ  U2 ðtÞ þ U3 ðtÞ  U4 ðtÞ þ . . .; with yð0 Þ ¼ y0 ð0 Þ ¼ y00 ð0 Þ ¼ 0:

196

3 Differential Equations

Solution For t [ 0: DðD2 þ 4ÞyðtÞ ¼ 10t The characteristic roots are: r1 ¼ 0; r2 ¼ j2; r3 ¼ j2. So, yH ðtÞ ¼ k1 þ k2 cos 2t þ k3 sin 2t yP ðtÞ ¼

D3

1 10t ¼ 2t þ 4D

And the general solution is yðtÞ ¼ 2t þ k1 þ k2 cos 2t þ k3 sin 2t y0 ðtÞ ¼ 2t  2k2 sin 2t þ 2k3 cos 2t y00 ðtÞ ¼ 2t  4k2 cos 2t  4k3 sin 2t From the differential equation, with its second member fully expressed in singular functions, one can then write that : yðtÞ ¼ 0:U 1 ðtÞ: : y0 ðtÞ ¼ C1 U1 ðtÞ: : y00 ðtÞ ¼ C1 U0 ðtÞ þ C2 U 1 ðtÞ: : y000 ðtÞ ¼ C1 U1 ðtÞ þ C2 U0 ðtÞ þ C3 U 1 ðtÞ: Substituting y0 ðtÞ and y000 ðtÞ in the given differential equation, it comes that : C1 U1 ðtÞ þ C2 U0 ðtÞ þ ðC3 þ 4C1 ÞU1 ðtÞ ¼ 6U1 ðtÞ þ 0:U0 ðtÞ þ 10U1 ðtÞ: Therefore, C1 ¼ 6 C2 ¼ 0 C3 þ 4C1 ¼ 10; C3 ¼ 14 Thus, y ð 0 þ Þ ¼ y ð 0 Þ þ 0 ¼ 0 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C1 ¼ 6

3.3 Differential Equations with Singular and Causal Forcing Functions

197

y00 ð0 þ Þ ¼ y00 ð0 Þ þ C2 ¼ 0 Taking t ¼ 0 þ in yðtÞ, y0 ðtÞ and y00 ðtÞ, it follows that: 0 ¼ 2 þ k1 þ k2 6 ¼ 2 þ 2k3 0 ¼ 2  4k2 Solving the algebraic system results in k1 ¼ 5=2; k2 ¼ 1=2; k3 ¼ 2. Finally,   5 1 yðtÞ ¼ 2t þ þ cos 2t þ 2 sin 2t U1 ðtÞ: 2 2 Example 3.21 Solve the differential equation y00 ðtÞ þ 25yðtÞ ¼ U1 ðtÞ þ 40 cos 3tU1 ðtÞ ¼ U1 ðtÞ þ 40U1 ðtÞ  360U3 ðtÞ þ . . ., with

the initial conditions yð0 Þ ¼ 0 and y0 ð0 Þ ¼ 1.

Solution For t [ 0, we have that


 D2 þ 25 yðtÞ ¼ 40 cos 3t

So, the characteristic equation is D2 þ 25 ¼ 0 and r1 ¼ j5; r2 ¼ j5. And the complementary solution is yH ðtÞ ¼ k1 cos 5t þ k2 sin 5t To use the phasor method in determining the particular solution of the preceding differential equation, the following changes are required, from the time domain to the simple frequency domain: 40 cos 3t       ! 40=0



yP ðtÞ       ! Y_ P y0P ðtÞ     ! ðj3ÞY_ P y00P ðtÞ     ! ðj3Þ2 Y_ P ¼ 9Y_ P

198

3 Differential Equations

Thus, the differential equation becomes the algebraic equation ð9 þ 25ÞY_ P ¼ 40=0



whence 40   Y_ P ¼ =0 ¼ 2:5=0 16 So, yP ðtÞ ¼ 2:5 cos 3t; yðtÞ ¼ 2:5 cos 3t þ k1 cos 5t þ k2 sin 5t y0 ðtÞ ¼ 7:5 sin 3t  5k1 sin 5t þ 5k2 cos 5t Observing the expansion in singular functions of the forcing function of the given differential equation, it appears that y00 ðtÞ must contain the double function U1 ðtÞ. So, : yðtÞ ¼ C0 U1 ðtÞ: : y0 ðtÞ ¼ C0 U0 ðtÞ þ C1 U1 ðtÞ: : y00 ðtÞ ¼ C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U 1 ðtÞ: Substituting in the given equation, it comes that: : C0 U1 ðtÞ þ C1 U0 ðtÞ þ ðC2 þ 25C0 ÞU1 ðtÞ ¼ U1 ðtÞ þ 0:U0 ðtÞ þ 40U1 ðtÞ: C0 ¼ 1; C1 ¼ 0; C2 þ 25C0 ¼ 40; C2 ¼ 15: So, yð0 þ Þ ¼ yð0 Þ þ C0 ¼ 0 þ 1 ¼ 1 ¼ 2:5 þ k1 ; k1 ¼ 1:5 y0 ð0 þ Þ ¼ y0 ð0 Þ þ C1 ¼ 1 þ 0 ¼ 1 ¼ 5k2 ; k2 ¼ 0:2 Finally, yðtÞ ¼ 2:5 cos 3t  1:5 cos 5t  0:2 sin 5t; for t [ 0: This expression can be simplified using the trigonometric identity

3.3 Differential Equations with Singular and Causal Forcing Functions

199

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  B 2 2 A cos xt þ B sin xt ¼ A þ B cos xt  arctg : A But it is possible to do the same using phasors. Defining
 zðtÞ ¼ 1:5 cos 5t  0:2 sin 5t ¼ 1:5 cos 5t  0:2 cos 5t  90 ; the corresponding phasor is    Z_ ¼ 1:5=0  0:2=90 ¼ 1:5 þ j0:2 ¼ 1:513=172:41   ¼ 1:513=172:41  180


  Z_ ¼ 1:513=7:59 and zðtÞ ¼ 1:513 cos 5t  7:59 Therefore, the solution can also as:
be written  yðtÞ ¼ 2:5 cos 3t  1:513 cos 5t  7:59 , for t [ 0. Example 3.22 Solve the following system of differential equations: x0 ðtÞ þ 3xðtÞ  4yðtÞ ¼ U1 ðtÞ 2xðtÞ þ y0 ðtÞ  yðtÞ ¼ U0 ðtÞ; with the initial conditions xð0 Þ ¼ x0 ð0 Þ ¼ yð0 Þ ¼ y0 ð0 Þ ¼ 0. Solution By introducing the differential operator D ¼ d=dt, the system can be written as: ðD þ 3ÞxðtÞ  4yðtÞ ¼ U1 ðtÞ 2xðtÞ þ ðD  1ÞyðtÞ ¼ U0 ðtÞ You can now obtain a differential equation only in xðtÞ and another only in yðtÞ, algebraically solving the system, as follows:    U1 ðtÞ 4    U0 ðtÞ D  1   xðtÞ ¼  4  Dþ3  2 D  1 Hence, eliminating the determinant of the denominator, one obtains that  Dþ3 F ðDÞxðtÞ ¼  2

   U ðt Þ 4  xðtÞ ¼  1  D1 U0 ðtÞ

 4  D 1

200

3 Differential Equations

Calculating the determinants results


 D2 þ 2D þ 5 xðtÞ ¼ 5U 0 ðtÞ  U 1 ðtÞ

ð3:34Þ

Similarly,  Dþ3 F ðDÞyðtÞ ¼ ðD2 þ 2D þ 5ÞyðtÞ ¼  2

 U1 ðtÞ  U 0 ðt Þ 

Whence


 D2 þ 2D þ 5 yðtÞ ¼ U 1 ðtÞ þ 3U 0 ðtÞ  2U 1 ðtÞ

ð3:35Þ

For t [ 0,
2  D þ 2D þ 5 xðtÞ ¼ 1
2  D þ 2D þ 5 yðtÞ ¼ 2 The characteristic roots are obtained from the algebraic equation F ðDÞ ¼ D2 þ 2D þ 5 ¼ 0, and are r1 ¼ 1 þ j2 and r2 ¼ 1  j2. So, the complementary solutions are: xH ðtÞ ¼ t ða1 cos 2t þ a2 sin 2tÞ yH ðtÞ ¼ t ðb1 cos 2t þ b2 sin 2tÞ And the particular solutions, xP ð t Þ ¼ yP ð t Þ ¼


 1 1 10:t ¼  D2 þ 2D þ 5 5 D2


 1 2 20:t ¼  5 þ 2D þ 5

Therefore, xð t Þ ¼ 

1 þ t ða1 cos 2t þ a2 sin 2tÞ 5

ð3:36Þ

yð t Þ ¼ 

2 þ t ðb1 cos 2t þ b2 sin 2tÞ 5

ð3:37Þ

In view of the second members of Eqs. 3.34 and 3.35, we have that:

3.3 Differential Equations with Singular and Causal Forcing Functions

201

: xðtÞ ¼ 0:U1 ðtÞ: : x0 ðtÞ ¼ C0 U1 ðtÞ: : x00 ðtÞ ¼ C0 U0 ðtÞ þ C1 U1 ðtÞ: : yðtÞ ¼ D0 U1 ðtÞ: : y0 ðtÞ ¼ D0 U0 ðtÞ þ D1 U1 ðtÞ: : y00 ðtÞ¼D0 U1 ðtÞ þ D1 U0 ðtÞ þ D2 U1 ðtÞ: Replacing in Eqs. 3.34 and 3.35, it comes that: : C0 U0 ðtÞ þ ðC1 þ 2C0 ÞU1 ðtÞ¼5U0 ðtÞ  U1 ðtÞ : D0 U1 ðtÞ þ ðD1 þ 2D0 ÞU0 ðtÞ þ ðD2 þ 2D1 þ 5D0 ÞU1 ðtÞ¼U1 ðtÞ þ 3U0 ðtÞ  2U1 ðtÞ Thus, 0

1

= 5 ..............................

0

=5

= −1 ...............

1

= −11

+2

0

0

1

= 1.........................

0

=1

= 3 ..............

1

=1

+2

0

Since all initial conditions at t ¼ 0 are null, the initial conditions at t ¼ 0 þ are given only by the coefficients of the step functions. Therefore, xð0 þ Þ ¼ 0; x0 ð0 þ Þ ¼ 5; yð0 þ Þ ¼ 1; y0 ð0 þ Þ ¼ 1 From Eqs. 3.36 and 3.37, we have that: x0 ðtÞ ¼ t ða1 cos 2t þ a2 sin 2tÞ þ t ð2a1 sin 2t þ 2a2 cos 2tÞ

ð3:38Þ

y0 ðtÞ ¼ t ðb1 cos 2t þ b2 sin 2tÞ þ t ð2b1 sin 2t þ 2b2 cos 2tÞ

ð3:39Þ

Entering the initial conditions at t ¼ 0 þ in Eqs. 3.36 to 3.39, it follows that: 0 ¼  15 þ a1 5 ¼ a1 þ 2a2

1 ¼  25 þ b1 1 ¼ b1 þ 2b2

202

3 Differential Equations

So, a1 ¼ 1=5; a2 ¼ 13=5; b1 ¼ 7=5; b2 ¼ 6=5: Finally, xð t Þ ¼

t 1 ðcos 2t þ 13 sin 2tÞ  ¼ 2:608t cosð2t  85:6 Þ  0:2 5 5

yð t Þ ¼


t 2  ð7 cos 2t þ 6 sin 2tÞ  ¼ 1:844t cos 2t  40:6  0:4 for t  0: 5 5

for t  0:

Proposed problems P.3.1. Prove the theorems: (a) FðDÞkt ¼ FðkÞkt , where k is any constant. (This theorem results in the first case of the abbreviated method). (b) If k is a constant and uðtÞ is an arbitrary function of t, then: FðDÞkt uðtÞ ¼ kt FðD þ kÞuðtÞ. (This theorem results in the fourth case of the abbreviated method). P.3.2. Using the theorems of P.3.1, calculate: (a) (b) (c) (d) (e)

ðD2Þ3t ð
D2  D  2Þ2t D5  6D3 þ 9D t ðD2  D þ 2Þt2t ðD2  13D þ 36Þt sin 10t

Answer (a) (b) (c) (d) (e)

3t 42t 4t ð8t5Þ2t t ð150 cos 10t  50 sin 10tÞ

P.3.3. Calculate: (a) (b) (c) (d)

1 2 D þ 1 ðt þ 1Þ 1 t 2 D þ 2 ð þ 3t Þ 1 2t sin tÞ D þ 2 ð 1 sin 3t Dþ2

3.3 Differential Equations with Singular and Causal Forcing Functions

203

Answer (a) t2  2t þ 3 t 2 (b) 3 þ 3t2  3t2 þ 34 (c) 2t cos t 1 ð2 sin 3t  3 cos 3tÞ (d) 13 P.3.4. Find the general solution of the following homogeneous differential equations: (a) DðD þ 1ÞðD þ 2ÞðD þ 3ÞyðtÞ ¼ 0 (b) D2 ðD2 þ 16ÞðD  5Þ3 yðtÞ ¼ 0 4 3 2 (c) ddt4h þ ddt3h  2 ddt2h ¼ 0 2

(d) ddt2z þ 2 dz dt þ 10z ¼ 0 (e)
ðD4 þ 2D2 þ 10Þz ðtÞ ¼ 0 (f) D5 þ 6D3 þ 9D yðtÞ ¼ 0 (g) ðD3 þ 2:7D2 þ 0:15D  1:771ÞxðtÞ ¼ 0 Answer (a) (b) (c) (d) (e) (f) (g)

yðtÞ ¼ k1 þ k2 t þ k3 2t þ k4 3t yðtÞ ¼ k1 þ k2 t þ k3 cos 4t þ k4 sin 4t þ ðk5 þ k6 t þ k7 t2 Þ5t hðtÞ ¼ k1 þ k2 t þ k3 t þ k4 2t zðtÞ ¼ t ðk1 cos 3t þ k2 sin 3tÞ zðtÞ ¼ 1:04t ðk1 cos 1:44t þ k2 sin 1:44tÞ þ 1:04t ðk3 cos 1:44t þ k4 sin 1:44tÞ pffiffiffi pffiffiffi yðtÞ ¼ k1 þ ðk2 þ k3 tÞ cos 3t þ ðk4 þ k5 tÞ sin 3t xðtÞ ¼ k1 0:7t þ k2 2:3t þ k3 1:1t

P.3.5. Given the algebraic equation f ð xÞ ¼ x7 þ 10x6  8x5 þ 3x3  9x þ 20 ¼ 0, (a) specify the possibilities regarding the number of real positive, real negative and complex conjugated roots; (b) establish the interval within which the real roots of the equation must necessarily be. Answer (a) There are six possibilities: 0; 1; 6; 0; 3; 4; 2; 1; 4; 2; 3; 2; 4; 1; 2; 4; 3; 0: (b) 11\x\4. P.3.6. Find the general solution of the following non-homogeneous differential equations: (a) (b) (c) (d)

DðD  1ÞðD  3ÞðD þ 2ÞyðtÞ ¼ 40 ðD  1ÞðD  3ÞðD þ 2ÞyðtÞ ¼ 202t ðD  1ÞðD  3ÞðD þ 2ÞyðtÞ ¼ 202t ðD2 þ 7D þ 10ÞyðtÞ ¼ 200U3 ðtÞ þ 50U2 ðtÞ

204

(e) (f) (g) (h) (i) (j)

3 Differential Equations

ðD3 þ 7D2 þ 10DÞyðtÞ ¼ t3 ðD2  4D þ 4ÞhðtÞ ¼ t3 2t þ t2t ðD2 þ 4D þ 4ÞiðtÞ ¼ ðD þ 1Þ100 cos 3t ðD2  2DÞwðtÞ ¼ t sin t
 y000 ðtÞ þ 20y00 ðtÞ þ 100y0 ðtÞ þ 800yðtÞ ¼ 400 cos 6t þ 35 ðD2  1ÞyðtÞ ¼ sin3 t

Answer (a) (b) (c) (d) (e)

t 3t 2t yðtÞ ¼ 20 3 t þ k1 þ k2  þ k3  þ k4  yðtÞ ¼ 52t þ k1 t þ k2 3t þ k3 2t yðtÞ ¼ 43 t2t þ k1 t þ k2 3t þ k3 2t yðtÞ ¼ 10t2  9t þ 4:3 þ k1 2t þ k2 5t ; t [ 0 4 3 2 2t 5t  yðtÞ ¼ 0:025t

 0:070t þ 0:117t þ 0:1218 þ k1 þ k2  þ k3 

(f) hðtÞ ¼ (g) (h) (i) (j)

t5 20

þ

t3 6

2t þ ðk1 þ k2 tÞ2t
 iðtÞ ¼ 24:33 cos 3t  41:1 þ ðk1 þ k2 tÞ2t t wðtÞ ¼ 10 ð2 cos t þ sin tÞ
 yðtÞ ¼ k1 16:88t þ 1:56t ðk2 cos 6:71t þ k3 sin 6:71tÞ þ 1:02 cos 6t  43:2 1 sin 3t þ k1 t þ k2 t yðtÞ ¼  38 sin t þ 40

P.3.7. Find the solution of the equation: i00 ðtÞ þ 4i0 ðtÞ þ iðtÞ ¼ 100U1 ðtÞ þ 400U0 ðtÞ, with zero initial conditions at t ¼ 0 . Answer iðtÞ ¼ 107:7370:268t  7:7373:732t ; for t [ 0: P.3.8. Solve the differential equation y00 ðtÞ þ yðtÞ ¼ U2 ðtÞ, with yð0 Þ ¼ y0 ð0 Þ ¼ 0. Answer yðtÞ ¼ U0 ðtÞ  sin t:U1 ðtÞ: P.3.9. Given the differential equation y00 ðtÞ þ 2y0 ðtÞ ¼ 2U2 ðtÞ þ U0 ðtÞ þ 2U1 ðtÞ, with yð0 Þ ¼ 1; y0 ð0 Þ ¼ 0 and y00 ð0 Þ ¼ 1, determine: (a) yð0 þ Þ, y0 ð0 þ Þ and y00 ð0 þ Þ. (b) the solution yðtÞ, for t  0.

3.3 Differential Equations with Singular and Causal Forcing Functions

205

Answer 0 00 (a) yð0 þ Þ ¼ 3, y ð0 þ Þ ¼ 9 and y ð0 þ Þ ¼ 15. 2U0 ðtÞ; for t ¼ 0 (b) yðtÞ ¼ 1 þ t  42t ; for t [ 0:

P.3.10. Given the differential equation y00 ðtÞ þ 2y0 ðtÞ ¼ 4U2 ðt  1Þ, with the initial conditions yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2, find yð1 þ Þ and y0 ð1 þ Þ. Answer yð1 þ Þ ¼ 2  2  8 ¼ 6:135; y0 ð1 þ Þ ¼ 22 þ 16 ¼ 16:271: P.3.11. Given the differential equation y00 ðtÞ þ 2y0 ðtÞ ¼ U1 ðt  2Þ, with the initial conditions yð0 Þ ¼ 1 and y0 ð0 Þ ¼ 2, find yð2 þ Þ and y0 ð2 þ Þ. Answer yð2 þ Þ ¼ 3  4 ¼ 2:982 and y0 ð2 þ Þ ¼ 2 þ 24 ¼ 1:963. P.3.12. Solve the differential equation
 y000 ðtÞ þ 9y0 ðtÞ ¼ 2U2 ðtÞ þ 400 cos 2t  30 U1 ðtÞ; with yð0 Þ ¼ y0 ð0 Þ ¼ y00 ð0 Þ ¼ 0.

Answer
 yðtÞ ¼ 11:111 þ 10:889 cos 3t  23:094 sin 3t þ 40 cos 2t  120 ; for t [ 0: P.3.13. Solve the following system of differential equations, knowing that all initial conditions are null at t ¼ 0 , and that eðtÞ ¼ 100cos300tU1 ðtÞ: ð2D þ 1Þi1 ðtÞ  i2 ðtÞ ¼ DeðtÞ
 i1 ðtÞ þ 2D2 þ 4D þ 1 i2 ðtÞ ¼ 0: Answer i1 ðtÞ ¼ 50 cos 300t  50t þ 501:5t ; for t [ 0: i2 ðtÞ ¼ 2:78 104 cos300t þ 8:34 104 t  5:56 104 1:5t ; for t  0:

206

3.4

3 Differential Equations

Conclusions

In this chapter it was made a revision of the principal methods of solution of homogeneous and non-homogeneous differential equations. It was also presented a systematic method to solve ordinary differential equations in which xðtÞ is a singular and / or causal function. This method allows to determine the initial conditions at t ¼ 0 þ , from the respective initial conditions at t ¼ 0 and the differential equation itself, without any physical reasoning about the given circuit. Therefore, a good understanding of this chapter is an indispensable condition for the good comprehension of the following chapters.

References 1. Adkins, William A, Davidson MG (2012) Ordinary Differential Equations, Undergraduate Texts in Mathematics, Springer 2. Boyce WE, Diprima RC (2002) Elementary differential equations and boundary value problems, LTC, 7th ed 3. Chicone C (2006) Ordinary differential equations with applications, Springer, 22nd ed 4. Close CM (1975) Linear circuits analysis, LTC 5. Doering CI, Lopes AO (2008) Ordinary differential equations, IMPA, 3rd ed 6. Irwin JD (1996) Basic engineering circuit analysis, Prentice Hall 7. Kreider DL, Kuller RG, Ostberg DR (2002) Differential equations, Edgard Blucher Ltda 8. LEITHOLD, Louis., The calculus with analytical geometry, vol.1, Harper & How of Brasil, 2nd. ed., (1982) 9. Perko L (2001) Differential equations and dynamical systems, New York: Springer, 3rd ed 10. Stewart J (2010) Calculus. vol 1, Thomson, 6th. ed 11. Zill DG, Cullen MR (2003) Differential equations, Makron Books, 3rd ed

Chapter 4

Digital Solution of Transients in Basic Electrical Circuits

4.1

Introduction

This chapter presents the fundamental algorithm of the digital computational solution of electromagnetic transients in electrical circuits used in programs based on the EMTP-Electromagnetics Transients Program. Since the nodal method is used in the fundamental EMTP algorithm, therefore, the discretization of circuit elements to concentrated parameters (inductances and capacitances) is implemented through the trapezoidal integration method, resulting in Norton Equivalent Digital Circuits for these elements. It is emphasized that such digital models depend on the appropriate choice of the integration step (Dt) so that the results of the computer simulation present the acceptable precision for the phenomena under study. The simplicity for computational implementation of the trapezoidal integration method when compared to other numerical integration methods, its precision and stability justified its use in EMTP and programs based on it. Because of the trapezoidal integration method's own discretization algorithm, under certain conditions numerical oscillations may arise in the computational solution. Therefore, it is essential to understand its origin to evaluate the different methods proposed for its solution in different computer programs based on EMTP. The use of computer simulation programs based on the EMTP—Electromagnetic Transients Program [1, 2] is applied in several areas of Electrical Engineering, such as, for example, studies on the calculation of energization transients of transmission lines, transformers, reactors, capacitor banks or various loads. Thus, the study of electromagnetic transients in electrical power systems is of paramount importance, as in these conditions, electrical equipment is subjected to very stressful situations, which can exceed their rating specifications. Even though electrical systems operate most of the time in a permanent or steady-state regime, it is essential that these systems can be able to withstand the transient regimes caused by short-circuits, switching, lightning strikes, among others. The electrical system constantly presents a dynamic behavior that integrates generation, loads, control and protection systems. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. C. Goulart de Siqueira and B. D. Bonatto, Introduction to Transients in Electrical Circuits, Power Systems, https://doi.org/10.1007/978-3-030-68249-1_4

207

208

4 Digital Solution of Transients in Basic Electrical Circuits

The seminal paper [1] by Prof. Dr.-Ing. Hermann W. Dommel (the creator of EMTP) presents the digital computational solution of electromagnetic transients in single-phase and multiphase networks, along with the digital computational models for basic circuit elements such as resistors, inductors and capacitors and also transmission lines. Several technical-scientific articles and other publications [3–10] followed this development, resulting in several models and studies of practical applications that enshrined different versions of programs based on the EMTP, such as Microtran [11], EMTP-RV [12], ATP (Alternative Transients Program) [13, 14], PSCAD [15], DIGSILENT [16], SABER [17], eMEGAsim [18], HYPERsim [19], PSIM [20] among others. This universal platform satisfactorily applies to transient analysis of electrical systems, usually offline. For real time simulations (Real Time Digital Simulations), especially for tests on protection and control equipment and devices (HIL—Hardware in the Loop Simulations), hardware and software such as OPAL [21] and RSCAD constituting an RTDS [22] have been increasingly used. A well-known computational platform used in academia is MatLab [23], mainly for prototyping and model testing. There is also free software with similar applications such as Scilab [24], etc. as well as software dedicated to dynamic simulations of electronic circuits such as Pspice [25], Multisim [26], etc. The modeling and study of Electromagnetic Transients is based on fundamental concepts of analysis of electrical circuits, in which concentrated elements are described by ordinary differential equations, and transmission lines with distributed parameters are described by partial differential equations. Various other elements and components of electrical systems are represented by combinations of basic element models. Mathematical models of electrical systems, such as differential equations, are often difficult to solve analytically, especially when they involve a large number of energy storage circuit elements, switching, non-linear components, among others. To overcome this difficulty, digital computational modeling of the elements is used, in which the differential equations are discretized over time, involving the adoption of numerical integration methods and thus obtaining approximate solutions, but with acceptable precision and numerical stability, for complex engineering problems.

4.2

Basic Algorithm for Computational Solution of EMTP-Based Programs

The solution algorithm for calculating electromagnetic transients in electrical power systems [1, 2] uses the Nodal Method, which allows to obtain the voltage in each of the nodes of an electrical circuit. The systematization of the resolution is presented in a simplified form below:

4.2 Basic Algorithm for Computational Solution of EMTP-Based Programs

209

½G:½eðtÞ ¼ ½I

ð4:1Þ

½I ¼ ½iðtÞ þ ½Ih ðtÞ

ð4:2Þ

where: ½G is the N  N matrix of nodal conductance; ½eðtÞ is the vector of nodal voltages; ½iðtÞ are the known current sources; and ½Ih ðtÞ are “historical” sources that depend on the initial conditions of the system and the used discretization rule. The matrix ½G is real, symmetrical and remains constant as long as there is no change in the switches status and in the time integration step ðDtÞ. The formation of the conductance matrix follows the same rules as the formation of the nodal admittance matrix in steady-state analysis. The system of equations represented before can be arranged as follows, using partitioning techniques for the nodal conductance matrix and voltage and current vectors:


½GAA  ½GAB  ½GBA  ½GBB 





 ½eA ðtÞ ½IA  ¼ ½IB  ½eB ðtÞ

ð4:3Þ

where: ½eA ðtÞ is the vector of unknown nodal voltages; ½eB ðtÞ is the vector of known nodal voltages, in nodes with connected voltage sources. To calculate the voltages at the nodes of interest, the Gaussian elimination method is used in the matrix system: ½GAA ½eA ðtÞ ¼ ½IA   ½GAB ½eB ðtÞ ¼ ½RHS:

ð4:4Þ

Thus, we have the values of nodal voltages for a given time t. Then the time t is increased with the step Dt, updating the values of the Right-Hand-Side [RHS] vector and calculating the nodal voltages, branch voltages and branch currents. The very simplified block diagram presented in Fig. 4.1 represents the basic algorithm for calculation and computational solution in the time domain of transients in electrical circuits [5]. A simplified source code for the base EMTP algorithm implemented in MATLAB is available at https://github.com/aptis.

210

4 Digital Solution of Transients in Basic Electrical Circuits Reading input data Building the [Y] Matrix for the steady-state solution Building the [G] Matrix of discretized elements for the transient solution

Calculating the steady-state solution for the initialization of history sources at t 1/3R

Fig. 5.17 Source conversion applied to electric circuit of Example 5.5 for analytical solution at t ! 1

1/R

R

R

R

^ | eo

By Eq. 5.3, e0 ðtÞ ¼ e0 ð1Þ þ ½e0 ð0 þ Þ  e0 ð1Þt=T ¼ Therefore, e0 ð t Þ ¼

  1 1 þ 0  3Rt=2L 3 3

1 3Rt 1   2L ½V; t  0½s 3

Example 5.6 The circuit shown in Fig. 5.18 was operating steadily when switch k was opened at t ¼ 0. Find the voltage expression ek ðtÞ, at the switch terminals, for t [ 0. Solution Instant t ¼ 0 : the inductor is short-circuited and switch k is still closed. So ið0 Þ ¼ E=R. Instant t ¼ 0 þ : the inductor is replaced by an independent source of current equal to a ið0 þ Þ ¼ ið0 Þ ¼ E=R, as shown in Fig. 5.19, and switch k is already open.

238

5 Transients in First Order Circuits

--> i (t) L

R +

+ R

E

t=0

ek (t) _

Fig. 5.18 Electric circuit of Example 5.6

--) E/R

R +

+ R

E

ek (t) _

Fig. 5.19 Electric circuit of Example 5.6 for analytical solution at t ¼ 0 þ

ek ð0 þ Þ ¼ R:ið0 þ Þ ¼ E: For t ! 1: there is a new permanent regime, and the inductor again behaving like a short circuit, as shown in Fig. 5.20. ið1Þ ¼ E=2R; where ek ð1Þ ¼ E=2. The resistance seen by the inductance with the k switch open and the short-circuit voltage source is RTH ¼ 2R. So, the time constant is T ¼ L=RTH ¼ L=2R. Like this, t

ek ðtÞ ¼ ek ð1Þ þ ½ek ð0 þ Þ  ek ð1Þ T ¼

  E E 2Rt þ E  L 2 2

---> i R +

+

E

R

ek _

Fig. 5.20 Electric circuit of Example 5.6 for analytical solution at t ! 1

5.2 Continuity Theorem

239

Therefore, ek ð t Þ ¼

E 2Rt 1 þ  L ; for t [ 0½s: 2

The expression of ek ðtÞ only applies to t [ 0 because ek ð0 þ Þ ¼ E and ek ð0 Þ ¼ 0. Example 5.7 The circuit shown in Fig. 5.21 was operating in a steady state when, at time t ¼ 0, switch k was closed. Find i2 ðtÞ, for t [ 0. =>> i2 (t)

=>> i1 (t) 10 Ohm

2 mH

5 Ohm t=0 k

+

2 Ohm

100 V

5 Ohm

Fig. 5.21 Electric circuit of Example 5.7

Solution Instant t ¼ 0 : switch k is still open, the circuit operates in a steady state and the inductance can be exchanged for a short circuit, as shown in Fig. 5.22. Considering the hourly mesh currents i1 ð0 Þ and i2 ð0 Þ, the equations are obtained: 12i1 ð0 Þ  2i2 ð0 Þ ¼ 100 2i1 ð0 Þ þ 12i2 ð0 Þ ¼ 0: Solving the system of equations, it turns out that i1 ð0 Þ ¼ 60=7 ¼ 8:57½A and i2 ð0 Þ ¼ 10=7 ¼ 1:43½A. Instant t ¼ 0 þ : switch k is already closed and the inductance is replaced by an independent source of current equal to i1 ð0 Þ, as in the circuit of Fig. 5.23. (Note that By current divider: i2 ð0 þ Þ ¼ 260=7 5 þ 2 ¼ 120=49 ¼ 2:45½A. i2 ð0 þ Þ 6¼ i2 ð0 Þ). When t ! 1: again the circuit is in steady state, with the configuration shown in Fig. 5.24, in which the inductance has been replaced again by a short circuit. Converting the voltage source to a current source, we have the circuit of Fig. 5.25. Replacing the resistors that are in parallel with their equivalent and applying a current divider

240

5 Transients in First Order Circuits

10 Ohm

5 Ohm =>> i1 (0-)

=>> i2 (0-)

+

k

2 Ohm

100 V

5 Ohm

Fig. 5.22 Electric circuit of Example 5.7 for analytical solution at t ¼ 0

=>> 60/7 A 10 Ohm

=>> i2 (0+) 5 Ohm

+

k 2 Ohm

100 V

Fig. 5.23 Electric circuit of Example 5.7 for analytical solution at t ¼ 0 þ

=>> i2 10 Ohm

5 Ohm

+

k 2 Ohm

100 V

Fig. 5.24 Electric circuit of Example 5.7 for analytical solution at t ! 1

=>> i2 (inf) 5 Ohm

k 10 A

10 Ohm

2 Ohm

Fig. 5.25 Source conversion at electric circuit of Example 5.7 for analytical solution at t ! 1

5.2 Continuity Theorem

241


 10  20
12 ¼ 2:5½A i2 ð1Þ ¼ 5 þ 20 12 Time constant: opening the current source in Fig. 5.25, the inductance sees the following equivalent Thevenin resistance: RTH ¼ 10 þ T¼

2  5 80 ¼ ¼ 11:43½X 2þ5 7

  L 2:103 7 ¼ 0:175½ms ¼ 175½ls and 1=T ¼ 5:714; 3 s1 ¼ ¼ RTH 40 80=7

According to Eq. 5.4, i2 ðtÞ ¼ i2 ð1Þ þ ½i2 ð0 þ Þ  i2 ð1Þt=T ¼ 2:5 þ ð2:45  2:5Þ5714;3t i2 ðtÞ ¼ 2:5  0:055714;3t ½A; for t [ 0½s: The numerical solution, using ATPDraw, performed based on the circuit shown in Fig. 5.26, in which the inductance is depicted by the model that includes the initial condition, is shown in Figs. 5.27 and 5.28. In Fig. 5.29 the current curve i1 ðtÞ is plotted. As the initial values i2 ð0 þ Þ ¼ 2:45½A and final values i2 ð1Þ ¼ 2:50½A are very close, the result shown in Fig. 5.27 was not very illustrative. For this reason, it was necessary to change the scale of values of the ordinate of the graph, and the result is what is shown in Fig. 5.28.

A

100 V

---|> i ( t ) 2 I

---|> i ( t ) i(0) 1 +

2 mH

5 Ohm B

+

10 Ohm

I

2 Ohm

Fig. 5.26 Electric circuit of Example 5.7 for digital solution

k

C 5 Ohm

t=0

242

5 Transients in First Order Circuits

2,5 [A] 2,0

1,5

1,0

0,5

0,0 0,0 0,2 (f ile Ex3.7.pl4; x-v ar t) c:B

0,4

0,6

0,8

[ms]

1,0

[ms]

1,0

-C

Fig. 5.27 Transient current i2 ðtÞ digital solution for circuit of Example 5.7

2,50 [A] 2,48

2,46

2,44

2,42

2,40 0,0

0,2

(f ile Ex3.7.pl4; x-v ar t) c:B

0,4

0,6

0,8

-C

Fig. 5.28 Zoom in transient current i2 ðtÞ digital solution for circuit of Example 5.7

5.2 Continuity Theorem

243

8,75 [A] 8,71

8,67

8,63

8,59

8,55 0,0

0,2

(f ile Ex3.7.pl4; x-v ar t) c:A

0,4

0,6

0,8

[ms]

1,0

-B

Fig. 5.29 Transient current i1 ðtÞ digital solution for circuit of Example 5.7

Example 5.8 The capacitor of the circuit shown in Fig. 5.30 has an initial charge qð0 Þ ¼ 800½lC . Obtain the expressions of the current iðtÞ circulating in the circuit and the charge qðtÞ in the capacitor, for t [ 0½s. Solution ½lC  ¼ 200½V. qðtÞ ¼ CeðtÞ. Therefore, eð0 Þ ¼ qðC0 Þ ¼ 800 4½lF 10 Ohm

+

Fig. 5.30 Electric circuit of Example 5.8

t=0

--> i (t)

100 V

+ 4 uF

_

^ | e (t)

10 Ohm

Fig. 5.31 Electric circuit of Example 5.8 for analytical solution at t ¼ 0 þ

100 V

+

+

--> i (0+) 200 V

244

5 Transients in First Order Circuits

As there is no impulse in current iðtÞ, due to the presence of the resistor of 10[X], and eð0 þ Þ ¼ eð0 Þ ¼ 200½V. Instant t ¼ 0 þ : the capacitor is replaced by a voltage source with a value equal to eð0 þ Þ, as in Fig. 5.31. ið0 þ Þ ¼

100  200 ¼ 10½A 10

When t ! 1: eð1Þ ¼ 100½V], because, due to the counterclockwise direction of the current, the capacitor discharges until it has a voltage equal to that of the source and ið1Þ ¼ 0½A , because in the steady state the capacitor behaves like an open circuit. Time constant: T ¼ RC ¼ 10  4:106 ¼ 40½ls and 1=T ¼ 25; 000½s1 . t

iðtÞ ¼ ið1Þ þ ½ið0 þ Þ  ið1Þ T ¼ 0 þ ð10  0Þ25;000t iðtÞ ¼ 1025;000t ½A; for t [ 0½s:
 eðtÞ ¼ 100 þ ð200  100Þ25;000t ¼ 100 1 þ 25;000t ½V; for t  0½s:
 qðtÞ ¼ CeðtÞ ¼ 4:106 :100 1 þ 25;000t
 qðtÞ ¼ 400 1 þ 25;000t ½lC ; for t  0½s: But the load can also be obtained by integrating the current: qð t Þ ¼ qð 0 þ Þ þ

Zt

iðkÞdk ¼ qð0 Þ þ

0þ

Zt

iðkÞdk ¼ 800:106 þ

0þ

Zt

1025;000k dk

0þ

t
 25;000k

6 þ 400:106 25;000t  1 . qðtÞ ¼ 800:106 þ 10 25;000 ¼ 800:10 0þ

Then,
 qðtÞ ¼ 400 1 þ 25;000t ½lC ; for t  0½s: The numerical solution by the ATPDraw program, performed through the circuit of Fig. 5.32, in which the capacitance is shown by the model that includes the initial voltage, is shown in Fig. 5.33. A

Fig. 5.32 Electric circuit of Example 5.8 for digital solution

B

10 Ohm U

+

t=0

I

U(0)

4 uF

+

100 V

C

5.2 Continuity Theorem

245

0

200,00 [V]

[A]

183,45

-4

166,89 -8 150,34 -12 133,78 -16

117,23

-20 0,00 0,04 (f ile Ex3.8.pl4; x-v ar t) c:A

-B

0,08 v :B

0,12

0,16

100,68 [ms] 0,20

-C

Fig. 5.33 Transient voltage and current digital solution for circuit of Example 5.8

Example 5.9 Determine the voltage expressions for e1 ðtÞ and e2 ðtÞ, for t [ 0, in the circuit shown in Fig. 5.34, knowing that e1 ð0 Þ ¼ E1 and e2 ð0 Þ ¼ E2 6¼ E1 . e2 (t) > i (t) e1 (t) ^|

t=0

C

C R

^| eR (t)

Fig. 5.34 Electric circuit of Example 5.9

Solution As there is no impulse in the hourly current, due to the presence of resistance R, then e1 ð0 þ Þ ¼ e1 ð0 Þ ¼ E1 and e2 ð0 þ Þ ¼ e2 ð0 Þ ¼ E2 . Without resistance R, necessarily e1 ð0 þ Þ ¼ e2 ð0 þ Þ, requiring current impulse at t ¼ 0; as shown in Example 5.10.

246

5 Transients in First Order Circuits

E2 > i (0+) ^ E1 |

R

Fig. 5.35 Electric circuit of Example 5.9 for analytical solution at t ¼ 0 þ

Instant t ¼ 0 þ : the circuit can be viewed as in Fig. 5.35. So, ið0 þ Þ ¼

E1  E2 R

When t ! 1: the two capacitances being equal, e1 ð1Þ ¼ e2 ð1Þ and ið1Þ ¼ 0. (Capacitances in series have the same charge. If the capacitances are the same, then the voltages must also be). Time constant: C Ceq ¼ CCC þ C ¼ 2 . Therefore, T ¼ R:Ceq ¼ RC=2: t

iðtÞ ¼ ið1Þ þ ½ið0 þ Þ  ið1Þ T i ðt Þ ¼ e1 ð t Þ ¼ e1 ð 0 þ Þ 

E1  E2 2t  RC ; for t [ 0: R

1 Zt E1  E2 Zt 2k iðkÞdk ¼ E1   RC dk C 0þ RC 0 þ

So, e1 ð t Þ ¼

E1 þ E2 E1  E2 2t þ  RC ; for t  0: 2 2 2t

eR ðtÞ ¼ RiðtÞ ¼ ðE1  E2 Þ RC ; for t [ 0: e2 ðtÞ ¼ e1 ðtÞ  eR ðtÞ Therefore, e2 ð t Þ ¼

E1 þ E2 E1  E2 2t   RC ; for t  0: 2 2

5.2 Continuity Theorem

247

Fig. 5.36 Electric circuit of Example 5.9 for digital solution

I

B

U

t=0

C

+

100 uF

U

A

U(0)

10 Ohm

U (0)

+

100 uF

200

10

[V]

[A]

180

8

160

6

140

4

120

2

0

100 0,0

0,5

(f ile ex100.pl4; x-v ar t) v :A

1,0 -

v :B

1,5 -C

c:A

2,0

[ms]

2,5

-B

Fig. 5.37 Transient voltage and current digital solution for circuit of Example 5.9

Alternatively, e2 ðtÞ could be obtained from: e2 ð t Þ ¼ e2 ð 0 þ Þ þ

1 Zt 1 Zt iðkÞdk ¼ E2 þ iðkÞdk: C 0þ C 0þ

The numerical solution, through ATPDraw, for e1 ð0 Þ ¼ 200½V, e2 ð0 Þ ¼ 100½V, R ¼ 10 [X] and C ¼ 100 [lF] is shown in Figs. 5.36 and 5.37. Example 5.10 If the capacitances C1 and C2 of the circuit shown in Fig. 5.38 have initial voltages e1 ð0 Þ ¼ E1 and e2 ð0 Þ ¼ E2 , respectively, find the voltage e1 ð0 þ Þ ¼ e2 ð0 þ Þ ¼ E, for all t [ 0. Also find the current iðtÞ. Consider E1 [ E2 . Solution At t ¼ 0 the voltage at the open switch terminals is E1  E2 , with positive polarity at its left terminal, since E1 [ E2 . At t ¼ 0 þ , the circuit is shown in Fig. 5.39. For linear and time-invariant capacitances, qðtÞ ¼ CeðtÞ. Therefore,

248

5 Transients in First Order Circuits

Fig. 5.38 Electric circuit of Example 5.10

t=0 e 1 (t) ^|

C1

C2

^ e2 (t) |

=|> i (t) ^ E = e1 (0+) |

C1

C2

^| E = e2 (0+)

Fig. 5.39 Electric circuit of Example 5.10 for analytical solution at t ¼ 0 þ

q1 ð0 Þ ¼ C1 e1 ð0 Þ ¼ C1 E1 q1 ð0 þ Þ ¼ C1 e1 ð0 þ Þ ¼ C1 E q2 ð0 Þ ¼ C2 e2 ð0 Þ ¼ C2 E2 q2 ð0 þ Þ ¼ C2 e2 ð0 þ Þ ¼ C2 E qtotal ð0 Þ ¼ q1 ð0 Þ þ q2 ð0 Þ qtotal ð0 þ Þ ¼ q1 ð0 þ Þ þ q2 ð0 þ Þ By the Law of Conservation of Load, qtotal ð0 þ Þ ¼ qtotal ð0 Þ. So: C1 E þ C2 E ¼ C1 E1 þ C2 E2 Then, C 2 E2 E ¼ e1 ð0 þ Þ ¼ e2 ð0 þ Þ ¼ C1 CE11 þ þ C2 , for 0 þ  t\1 and C1 þ C2 , Cp . Therefore,

q1 ð0 þ Þ ¼ C1

C1 E1 þ C2 E2 C1 E1 þ C2 E2 ¼ C1 C1 þ C2 Cp

q2 ð0 þ Þ ¼ C2

C1 E1 þ C2 E2 C1 E1 þ C2 E2 ¼ C2 C1 þ C2 Cp

As Cp ¼ C1 þ C2 and q1 ð0 þ Þ 6¼ q2 ð0 þ Þ, the two capacitances are in parallel from t ¼ 0 þ , and not in series, not least because the circuit is closed and there is no electric current flowing for t [ 0. When C1 ¼ C2 ¼ C, then E ¼ E1 þ2 E2 and q1 ð0 þ Þ ¼ q2 ð0 þ Þ ¼ C E1 þ2 E2 : Supposing that the capacitances were already charged well before t ¼ 0; we can write, for example, that: q2 ðtÞ ¼ q2 ð0 ÞU 1 ðtÞ þ q2 ð0 þ ÞU 1 ðtÞ C1 E1 þ C2 E2 ¼ C2 E2 U1 ðtÞ þ C2 U1 ðtÞ C1 þ C2 So, the clockwise current flowing through the circuit is

5.2 Continuity Theorem

iðtÞ ¼

249

d C1 E1 þ C2 E2 q2 ðtÞ ¼ C2 E2 U0 ðtÞ:ð1Þ þ C2 U 0 ðt Þ dt C1 þ C2

As the impulse function is even, U0 ðtÞ ¼ U0 ðtÞ and it can be written that 

C1 E1 þ C2 E2 iðtÞ ¼ C2  C2 E2 U0 ðtÞ C1 þ C2 Or, iðtÞ ¼

C1 C2 ðE1  E2 ÞU0 ðtÞ; at t ¼ 0: C1 þ C 2

The area of the current pulse, C1 C2 ðE1  E2 Þ=ðC1 þ C2 Þ, represents the amount of electric charge that is transferred instantly from capacitance C1 to the capacitance C2 , at time t ¼ 0, to equalize the voltages at t ¼ 0 þ . This can be confirmed by calculating q1 ð0 Þ  q1 ð0 þ Þ. q1 ð0 Þ  q1 ð0 þ Þ ¼ C1 E1  C1

C1 E1 þ C2 E2 C1 C2 ¼ ðE1  E2 Þ: C1 þ C2 C1 þ C2

Therefore, iðtÞ ¼ QU0 ðtÞ; where Q ¼ q1 ð0 Þ  q1 ð0 þ Þ: It is interesting to observe that the quotient C1 C2 =ðC1 þ C2 Þ ¼ Cs represents the equivalent of capacitances connected in series, which is absolutely consistent with the fact that at moment t ¼ 0 an impulsive current circulates in the circuit. It can be said, therefore, that at time t ¼ 0 the capacitances are in series, allowing the circulation of the impulsive current iðtÞ ¼ QU0 ðtÞ. And that for t [ 0 the capacitances are in parallel, with the same voltage E, no more current circulating in the circuit, since iðtÞ ¼ 0 for t [ 0 , and constant voltage does not add current in capacitances. It is also possible to confirm the results obtained by calculating, for example, the voltage e1 ðtÞ from the current iðtÞ: e1 ðtÞ ¼

1 Zt 1 0Z 1 Zt q ð 0 Þ 1 Zt iðkÞ dk ¼ iðkÞ dk  iðkÞdk ¼ 1  iðkÞdk C1 1 C1 1 C 1 0 C1 C1 0

Or e1 ð t Þ ¼

1 Zt 1 0Z 1 Zt q1 ð 0 Þ 1 Zt iðkÞdk ¼ iðkÞdk þ iðkÞdk ¼ þ iðkÞdk C1 1 C1 1 C1 0 C1 C1 0

250

5 Transients in First Order Circuits

It should be noted that the current that carries capacitance C1 , between 1 and 0 , has the opposite sense of the one that causes it to discharge; hence the negative sign for the current in the range between 0 and t. Moving on, e 1 ð t Þ ¼ e 1 ð 0 Þ  ¼ E1 

0Zþ 1 Zt 1 C1 C2 iðkÞdk ¼ E1  : ðE1  E2 Þ U0 ðkÞdk C1 0 C1 C1 þ C2 0

C2 C1 E1 þ C2 E2 ðE1  E2 Þ ¼ ; for all t [ 0: So, C1 þ C2 C1 þ C2 e1 ð 0 þ Þ ¼

C1 E1 þ C2 E2 ¼ E: C1 þ C2

This Example 5.10 establishes the opportune moment to state the following corollary to the Continuity Theorem: Corollary 5.1 A unitary current impulse, iðtÞ ¼ U 0 ðtÞ½A, circulating through a capacitance C½F produces an instantaneous variation in its voltage equal to 1=C½V : This means that if the capacitance voltage at the initial time t ¼ 0 is eð0 Þ½V, then the voltage at time t ¼ 0 þ will be eð0 þ Þ ¼ eð0 Þ  1=C½V  . Being that the 1/C signal will be positive or negative, depending on the direction in which the current will circulate, that is, if the current will charge the capacitor even more, or if it will act to decrease its load and, consequently, its voltage. In the present case, in which the current is a unitary impulse, the variation in the electric charge is 1 [ C]. Obviously, the impulsive current iðtÞ ¼ QU 0 ðtÞ½A transfers Q½C of electric charge, instantly. Considering, on the other hand, that [A]¼[C/s], it is evident, once again, that the unit of U 0 ðtÞ is [ s−1], being t in seconds. For a capacitance C with voltage and current arrows from opposite directions, the current is given by: iðtÞ ¼ C

d eðtÞ: dt

If eðtÞ is discontinuous, then the current iðtÞ will contain impulse, and also the power, because, pðtÞ ¼ eðtÞ:iðtÞ ¼ CeðtÞ

d eðtÞ: dt

Since pðtÞ ¼ dtd W ðtÞ, then the energy is given by

5.2 Continuity Theorem

251

deðkÞ Z Zt Zt Zt dk ¼ C W ðtÞ ¼ pðtÞdt ¼ pðkÞdk ¼ CeðkÞ eðkÞdeðkÞ dk 1 1 1

t  e2 ðkÞ

C 2 e ðtÞ  e2 ð1Þ ¼ ¼C

2 1 2 Whereas at t ! 1 the capacitor, of capacitance C; has just left the factory completely de-energized, so qð1Þ ¼ eð1Þ ¼ 0. So, W ðt Þ ¼

C 2 e ðtÞ: 2

Therefore, W ð0 Þ ¼ ðC=2Þe2 ð0 Þ, regardless of the waveform of the current that flowed through the capacitance from 1 to t ¼ 0 . On the other hand, it can also be written that: W ðtÞ ¼

Zt

1

pðkÞdk ¼

0Z

pðkÞdk þ

1

Zt

pðkÞdk

0

t deðkÞ C 2 e2 ðkÞ

dk ¼ e ð0 Þ þ C ¼ W ð0 Þ þ C eðkÞ dk 2 2 0 0

t C e2 ðkÞ

¼ e 2 ð 0 Þ þ C 2 2 0  C C C 2 e ðtÞ  e2 ð0 Þ ¼ e2 ðtÞ: ¼ e 2 ð 0 Þ þ 2 2 2 Zt

Therefore, regardless of the instant considered, when calculated through the voltage, the energy in a capacitance is always given by W ðtÞ ¼ ðC=2Þe2 ðtÞ. But, directly from the power, you can also use the formula W ðt Þ ¼ W ðt 0 Þ þ

Zt

pðkÞdk:

t0

Under no circumstances is it assumed that the energy stored in a capacitance is infinite. And since the energy is given by WC ¼ ð1=2ÞCe2C , this means that the voltage at the terminals of a capacitance can never be an impulse, as it has infinite amplitude. In an equivalent way, considering that iC ¼ CdeC =dt, the current iC ðtÞ through a capacitance can never be a double. Example 5.11 The circuit shown in Fig. 5.40 is, strictly, the dual of that of example 5.10. Considering that it is possible that the inductances L1 and L2 have initial currents i1 ð0 Þ ¼ I1 and i2 ð0 Þ ¼ I2 , find the current i1 ð0 þ Þ ¼ i2 ð0 þ Þ ¼ I; for all t [ 0. Also find the voltage eðtÞ at the open switch terminals. Consider that I1 [ I2 .

252

5 Transients in First Order Circuits

Fig. 5.40 Electric circuit of Example 5.11

i1 (t) ^|

^ e (t) |

L1

t=0

L2

| i2 (t) v

Solution At t ¼ 0 the current through the closed switch is I1  I2 , downwards, since I1 [ I2 . For t [ 0, the circuit is shown in Fig. 5.41. For any linear and time-invariant inductance, /ðtÞ ¼ LiðtÞ. So, /1 ð0 Þ ¼ L1 i1 ð0 Þ ¼ L1 I1

/1 ð0 þ Þ ¼ L1 i1 ð0 þ Þ ¼ L1 I

/2 ð0 Þ ¼ L2 i2 ð0 Þ ¼ L2 I2

/2 ð0 þ Þ ¼ L2 i2 ð0 þ Þ ¼ L2 I

/total ð0 Þ ¼ /1 ð0 Þ þ /2 ð0 Þ /total ð0 þ Þ ¼ /1 ð0 þ Þ þ /2 ð0 þ Þ By the Flow Conservation Law, /total ð0 þ Þ ¼ /total ð0 Þ. So, L1 I þ L2 I ¼ L1 I1 þ L2 I2 Then, L2 I2 I ¼ i1 ð0 þ Þ ¼ i2 ð0 þ Þ ¼ L1LI11 þ þ L2 , for 0 þ  t\1 and L1 þ L2 ¼ Ls . Therefore, /1 ð0 þ Þ ¼ L1

L1 I1 þ L2 I2 L 1 I 1 þ L2 I 2 ¼ L1 L 1 þ L2 Ls

/2 ð0 þ Þ ¼ L2

L1 I1 þ L2 I2 L 1 I 1 þ L2 I 2 ¼ L2 L 1 þ L2 Ls

As Ls ¼ L1 þ L2 and /1 ð0 þ Þ 6¼ /2 ð0 þ Þ, the two inductances are in series starting at t ¼ 0 þ . When L1 ¼ L2 ¼ L; I ¼ I1 þ2 I2 and /1 ð0 þ Þ ¼ /2 ð0 þ Þ ¼ L I1 þ2 I2 : Whereas

Fig. 5.41 Electric circuit of Example 5.11 for analytical solution at t ¼ 0 þ

L1 ^ I |

| e1 (t) v

^ e (t) |

L2 e2 (t) ^|

| I v

5.2 Continuity Theorem

253

/2 ðtÞ ¼ /2 ð0 ÞU1 ðtÞ þ /2 ð0 þ ÞU1 ðtÞ ¼ L2 I2 U1 ðtÞ þ L2

e2 ð t Þ ¼

L1 I1 þ L2 I2 U1 ðtÞ; L1 þ L2

d L1 I1 þ L2 I2 /2 ðtÞ ¼ L2 I2 U0 ðtÞ  ð1Þ þ L2 U 0 ðt Þ dt L 1 þ L2

Since U0 ðtÞ ¼ U0 ðtÞ, 

L1 I1 þ L2 I2 e2 ðtÞ ¼ L2  L2 I 2 U 0 ð t Þ L1 þ L2 or e2 ðtÞ ¼ L1 LL21ðþI1LI2 2 Þ U 0 ðtÞ ¼ eðtÞ ¼ e1 ðtÞ, only at t ¼ 0. The area of the voltage impulse, L1 L2 ðI 1  I 2Þ =ðL1 þ L2 Þ; represents the amount of magnetic flux that was transferred instantly from inductance L1 to inductance L2 , at time t ¼ 0, to equal the currents at t ¼ 0 þ : This can be proved by calculating /1 ð0 Þ  /1 ð0 þ Þ. /1 ð0 Þ  /1 ð0 þ Þ ¼ L1 I1  L1

L1 I1 þ L2 I2 L1 L2 ðI1  I2 Þ ¼ : L1 þ L 2 L1 þ L2

Therefore, eðtÞ ¼ UU0 ðtÞ, where U ¼ / 1 ð 0 Þ  / 1 ð 0 þ Þ It is interesting to note that the quotient L1 L2 =ðL1 þ L2 Þ ¼ Lp is the equivalent of parallel inductances, which agrees with the fact that at moment t ¼ 0 a voltage impulse appears in the circuit. Therefore, it can be said that at time t ¼ 0 the inductances are in parallel, allowing the presence of the voltage eðtÞ ¼ UU0 ðtÞ: And that, for t [ 0, they are in series, with the same current I, but with eðtÞ ¼ 0; because constant current does not induce electromotive force in inductances. Corollary 5.2 A unitary voltage impulse, eðtÞ ¼ U 0 ðtÞ½V , applied to the terminals of an L[ H] inductance causes an instantaneous change in its current value equal to 1=L ½A. This corollary states that if the inductance current at t ¼ 0 is ið0 Þ½A, then the current at t ¼ 0 þ will be ið0 þ Þ ¼ ið0 Þ  1=L½A. Being that the 1=L signal depends on the polarity of the applied voltage, that is, if it will increase the current, or if it will act to decrease it. In this case, where the voltage is a unitary impulse, the magnetic flux variation is 1 [Wb]. Obviously, the impulsive voltage eðtÞ ¼ UU0 ðtÞ½V transfers U [Wb] of magnetic flux, instantly. Remembering, on the other hand, that ½V ¼ ½Wb=S, it is evident, once again, that the unit of U0 ðtÞ is [s−1], being t in seconds.

254

5 Transients in First Order Circuits

Under no circumstances is it assumed that the energy stored in an inductance is infinite. And since the energy given by WL ¼ ð1=2ÞLi2L , this means that the current through an inductance can never be a impulse, which has infinite amplitude. On the other hand, considering that eL ¼ LdiL =dt, then the voltage eL at the ends of an inductance can never be a double. Example 5.12 The circuit shown in Fig. 5.42 was operating steadily when switch k was opened at time t ¼ 0 [s]. Find the current i1 ðtÞ for t [ 0 [s]. Solution At t ¼ 0 switch k is still closed and the circuit is operating steadily. So the inductances behave like short circuits, as shown in Fig. 5.43. Therefore, E E=2 E 5E i1 ð0 Þ ¼ ¼ I1 ; ið0 Þ ¼ ¼ ; i 2 ð 0 Þ ¼ i 1 ð 0 Þ þ i ð 0 Þ ¼ : R 2R 4R 4R

L1

2R

=|> i1 (t)

i (t) i1 (0+) E

L2

|| V i2 (0+)

5.2 Continuity Theorem

255

At t ¼ 0 þ the switch k is already open and i1 ð0 þ Þ ¼ i2 ð0 þ Þ, as in the circuit in Fig. 5.44. From the result of Example 5.11, I ¼ i 1 ð0 þ Þ ¼ i 2 ð0 þ Þ ¼

L1 I1 þ L2 I2 LR1 E þ 5L4R2 E 4L1 þ 5L2 E ¼ ¼  : L1 þ L2 L1 þ L2 4ðL1 þ L2Þ R

For t ! 1 the inductances are again replaced by short circuits and i1 ð1Þ ¼ ER ¼ i2 ð1Þ. The time constant is: T ¼ ðL1 þ L2 Þ=R. As the inductances are in series, the circuit is of the first order and it can be written that: t

i1 ðtÞ ¼ i1 ð1Þ þ ½i1 ð0 þ Þ  i1 ð1Þ T  E 4L1 þ 5L2 E E L Rt i 1 ðt Þ ¼ þ :   1 þ L2 R 4ðL1 þ L2Þ R R Finally,

 Rt E L2 L1 þ L2 1þ i1 ðtÞ ¼ i2 ðtÞ ¼  ; for t [ 0: R 4ðL1 þ L2 Þ

To complete, the inductance voltages are calculated, assuming associated references, that is, with opposite voltage and current arrows. As there is discontinuity in the current i1 ðtÞ, since i1 ð0 Þ ¼ ER, i1 ð0 þ Þ ¼ 4L1 þ 5L2 4ðL1 þ L2Þ

 ER and e1 ðtÞ ¼ L1 dtd i1 ðtÞ, then the voltage e1 ðtÞ presents an impulse at

t ¼ 0, of value e1 ðtÞ ¼ L1 ½i1 ð0 þ Þ  i1 ð0 ÞU0 ðtÞ ¼

L1 L2 E U0 ðtÞ; t ¼ 0: 4RðL1 þ L2 Þ

For t [ 0, e 1 ð t Þ ¼ L1

  Rt d E L2 L1 L2 E Rt 1þ L1 þ L2  L1 þ L2 ; t [ 0: ¼ 2 dt R 4 ð L1 þ L 2 Þ ðL1 þ L2 Þ 4

Therefore, it can be written, alternatively, that: e1 ðtÞ ¼

Rt L1 L2 E L1 L2 E U 0 ðt Þ  L1 þ L2 U 1 ðtÞ: 2 4RðL1 þ L2 Þ 4ð L 1 þ L 2 Þ

4L1 þ 5L2 Since i2 ð0 Þ ¼ 5E 4R and i2 ð0 þ Þ ¼ 4ðL1 þ L2 Þ.

also a discontinuity in i2 ðtÞ. Since e2 ðtÞ ¼ presents an impulse at t ¼ 0, of value

E R

 ¼ 4þ

L2 dtd i2 ðtÞ,

L2 L1 þ L2



E 4R \i2 ð0 Þ;

there is

then the voltage e2 ðtÞ also

256

5 Transients in First Order Circuits

e2 ðtÞ ¼ L2 ½i2 ð0 þ Þ  i2 ð0 ÞU0 ðtÞ ¼ 

L1 L2 E U0 ðtÞ; t ¼ 0: 4RðL1 þ L2 Þ

It is interesting to note that the impulses present in e1 ðtÞ and e2 ðtÞ satisfy Kirchhoff's voltage law, that is, that e1 ðtÞ þ e2 ðtÞ ¼ 0, at time t ¼ 0, as there is no impulse in the voltage of the resistance R at any time, although there is discontinuity in the current i1 ðtÞ circulating through it. For t [ 0,

  Rt d E L2 L22 E Rt L1 þ L2 1þ  : L1 þ L2 ; t [ 0: e 2 ð t Þ ¼ L2 ¼ 2 4 dt R 4 ð L1 þ L 2 Þ ðL1 þ L2 Þ Putting the results together for t ¼ 0 and t [ 0, e2 ð t Þ ¼ 

5.3

Rt L1 L2 E L22 E U 0 ðtÞ  L1 þ L2 U1 ðtÞ: 2 4RðL1 þ L2 Þ 4ðL1 þ L2 Þ

Response to Impulse and Response to Step

The response of a circuit, without energy initially stored at time t ¼ 0 , to a unitary impulse source U0 ðtÞ or to a unitary step source U1 ðtÞ, for its importance, is called h(t) or r(t), respectively. It is important to note that the terms Impulse Response and Response to Step always refer to singular unitary functions, that is, unitary area impulse at t ¼ 0 and unitary amplitude step for t [ 0. And more: that the circuit does not contain any energy stored in its capacitors and inductors before the application of these singular functions. Thus, if the input and output of a circuit are denoted by xðtÞ and yðtÞ, respectively, then: yðtÞ ¼ hðtÞ where xðtÞ ¼ U0 ðtÞ. yðtÞ ¼ rðtÞ where xðtÞ ¼ U 1 ðtÞ. It is also worth noting that, being U0 ðtÞ the derivative of U1 ðtÞ, then, as a consequence, hðtÞ ¼

d rðtÞ: dt

ð5:11Þ

And that, conversely, Z

rðtÞ ¼ hðtÞdt ¼

Zt

1

hðkÞdk ¼

Zt

0

hðkÞdk:

ð5:12Þ

5.3 Response to Impulse and Response to Step

257

Example 5.13 Determine r ðtÞ and hðtÞ for the circuit in Fig. 5.45. Solution (a) Where iðtÞ ¼ U1 ðtÞ, eðtÞ ¼ r ðtÞ ¼ ? According to the Continuity Theorem, the voltage eðtÞ that is at the capacitance terminals, and that is zero for t\0; cannot be changed instantly, as the current source is unitary step finite. Therefore, eð0 þ Þ ¼ 0. On the other hand, at t ¼ 0 þ , when ið0 þ Þ ¼ 1½A, there can still be no current going down through the resistance, because, if it did, the voltage at the capacitance terminals, which is the same in resistance terminals, it would jump to a non-zero value equal to R  1 ¼ R. Thus, all current from the source, which is 1½A, is initially directed to capacitance, which begins to accumulate electrical charge and voltage. But as the charge and voltage increase in capacitance, the current going down through the resistance, iðtÞ ¼ eðtÞ=R, also increases, and the current going down through the capacitance decreases, which in turn instead, it slows the growth of tension eðtÞ. When the current in the resistor reaches 1 ½A, the current in the capacitance becomes zero and there is no more voltage growth e (t), which reaches its maximum value eð1Þ ¼ R. As the circuit time constant is T ¼ RC, according to Eq. 5.3, it can be written that. r ðtÞ ¼ eðtÞ ¼ eð1Þ þ ½eð0 þ Þ  eð1Þt=RC ¼ R þ ð0  RÞt=RC . So,  t r ðtÞ ¼ R 1  RC U1 ðtÞ: This result can be confirmed by writing the nodal circuit equation: C

d 1 eð t Þ þ eð t Þ ¼ i ð t Þ dt R

Then, 1 r 0 ðtÞ þ RC r ðtÞ ¼ C1 U1 ðtÞ; with r ð0 Þ ¼ 0: For t [ 0,

  1 1 Dþ r ðt Þ ¼ RC C rH ðtÞ ¼ kt=RC   1 1 0t  rP ðt Þ ¼ ¼R 1 C D þ RC Fig. 5.45 Electric circuit of Example 5.13

^i (t) |

R

C

^ e (t) = ? |

258

5 Transients in First Order Circuits

r ðtÞ ¼ rP ðtÞ þ rH ðtÞ ¼ R þ kt=RC Since r ðtÞ does not contain a step, since the step of the second member of the differential equation[1-11] must appear in r 0 ðtÞ, so r ð0 þ Þ ¼ r ð0 Þ ¼ 0 ¼ R þ k and k ¼ R. Soon,  t r ðtÞ ¼ R 1  RC U1 ðtÞ: (b) Where iðtÞ ¼ U0 ðtÞ; eðtÞ ¼ hðtÞ ¼ ? As the energy in the capacitor, ð1=2ÞCe2 , cannot be infinite, because this would require a double in its current, which is prohibited, the current impulse of the source, which happens at t ¼ 0, cannot circulate through the circuit resistance, as this would create the voltage RU0 ðtÞ, an infinite voltage, in the resistance and also in the capacitance. Thus, at t ¼ 0 , the current impulse of the source circulates through the capacitance. But, by Corollary 5.1, the passage of this unitary current impulse through the capacitor at t ¼ 0 causes eð0 þ Þ ¼ 1=C. At t ¼ 0 þ , on the other hand, the impulsive current source is already zero, which means that the current source has become an open circuit for t [ 0. Therefore, the energy ð1=2ÞC:ð1=C Þ2 ¼ 1=2C, existing at t ¼ 0 þ in the capacitance, will be entirely dissipated in the resistance, through an anti-clockwise current in the RC loop, causing eð1Þ ¼ 0. As T ¼ RC, then, according to Eq. 5.3, hðtÞ ¼ eðtÞ ¼ eð1Þ þ ½eð0 þ Þ  eð1Þt=T ¼ 0 þ ð1=C  0Þt=RC Then, hð t Þ ¼

1 t=RC  U1 ðtÞ: C

The impulse response can also be obtained using Eq. 5.11, that is, deriving the response to the step: hð t Þ ¼

 i 

dh  1 t t t

R 1  RC U1 ðtÞ ¼ R  RC U1 ðtÞ þ 1  RC U0 ðtÞ dt RC t¼0

Whence hð t Þ ¼

1 t=RC  U1 ðtÞ: C

Another way to obtain hðtÞ is the nodal equation of the circuit:

5.3 Response to Impulse and Response to Step

C

259

d 1 eð t Þ þ eð t Þ ¼ i ð t Þ dt R

Then, h0 ð t Þ þ

1 RC hðtÞ

¼ C1 U0 ðtÞ, with hð0 Þ ¼ 0:

For t [ 0,   1 Dþ hð t Þ ¼ 0 RC hðtÞ ¼ kt=RC Since the impulse ð1=CÞU0 ðtÞ must appear in h0 ðtÞ; then hðtÞ must contain the step ð1=C ÞU1 ðtÞ: Thus, hð0 þ Þ ¼ hð0 Þ þ 1=C ¼ 1=C ¼ k. Soon, hð t Þ ¼

1 t=RC  U1 ðtÞ: C

Example 5.14 In the circuit shown in Fig. 5.46 the capacitance is de-energized at t ¼ 0 . (a) If the voltage source is a unitary impulse, find the expressions for current iðtÞ and voltages eR ðtÞ and eC ðtÞ. (b) Establish the expressions of the power pC ðtÞ and the energy WC ðtÞ in the capacitance. Solution (a) The circuit mesh equation is RiðtÞ þ

1Z iðtÞdt ¼ eðtÞ C

Or, i0 ðtÞ þ

1 1 i ð t Þ ¼ e0 ð t Þ RC R

For eðtÞ ¼ U0 ðtÞ, it comes that. R

^ e ( t ) ||

+

Fig. 5.46 Electric circuit of Example 5.14

-----/> i ( t ) 1

+

t=0 ^ E cos wt |

R

---\> i 2 1 * L 1

L2 *

Fig. 5.109 Electric circuit of Example 5.35

(t)

R2

310

5 Transients in First Order Circuits

Solution (a) Since k ¼ 1, M ¼

pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L1 L2 ¼ 0:02  0:08 ¼ 0:04½H :

The primary equation is: 0

0

R1 i1 ðtÞ þ L1 i1 ðtÞ þ Mi2 ðtÞ ¼ eðtÞ

ð5:19Þ

The secondary equation is: 0

0

Mi1 ðtÞ þ L2 i2 ðtÞ þ R2 i2 ðtÞ ¼ 0

ð5:20Þ

Or, ðL1 D þ R1 Þi1 ðtÞ þ MDi2 ðtÞ ¼ eðtÞ MDi1 ðtÞ þ ðL2 D þ R2 Þi2 ðtÞ ¼ 0 Then,

L1 D þ R1

MD




eð t Þ MD

i ðtÞ ¼

L2 D þ R2 1 0

MD

L2 D þ R2

 0 L1 L2 D2 þ L1 R2 D þ L2 R1 D  M 2 D2 þ R1 R2 i1 ðtÞ ¼ L2 e ðtÞ þ R2 eðtÞ

As M 2 ¼ L1 L2 , it follows that:  Dþ

 R1 R2 L2 R2 e0 ðtÞ þ eð t Þ i1 ðtÞ ¼ L1 R2 þ L2 R1 L1 R2 þ L2 R1 L1 R2 þ L2 R1

For eðtÞ ¼ E cos xtU1 ðtÞ; e0 ðtÞ ¼ EU0 ðtÞ  xE sin xtU1 ðtÞ. Substituting and entering numerical values, the differential equation becomes: ðD þ 35:714Þi1 ðtÞ ¼ 28:571U0 ðtÞ  1; 795:196 sin 20ptU1 ðtÞ þ 3; 571:429 cos 20ptU1 ðtÞ: Or, ðD þ 35:714Þi1 ðtÞ ¼ 28:571U0 ðtÞ þ 3; 997:228 cosð20pt þ 26:687 ÞU1 ðtÞ; i1 ð0 Þ ¼ 0: For t [ 0,

5.5 Magnetically Coupled Circuits

311

ðD þ 35:714Þi1 ðtÞ ¼ 3; 997:228 cosð20pt þ 26:687 Þ i1H ðtÞ ¼ k1 35:714t ð35:714 þ j62:832ÞI_1P ¼ 3; 997:228=26:687

3; 997:228\26:687 I_1P ¼ ¼ 55:307\  33:699 72:273\60:386 i1P ðtÞ ¼ 55:307 cosð20pt33:699 Þ i1 ðtÞ ¼ k1 35:714t þ 55:307 cosð20pt33:7 Þ From the original i1 ð0 þ Þ ¼ 28:571½A. So,

differential

equation

of

i1 ðtÞ,

we

have

that

28:571 ¼ k1 þ 55:307 cosð33:7 Þ ¼ k1 þ 46:013 and k1 ¼ 17:442: Therefore, i1 ðtÞ ¼ 17:44235:714t þ 55:307cosð20pt33:7 Þ½A; for [ 0½s: Solving the system of differential equations for i2 ðtÞ, it follows that:  Dþ

 R1 R2 M e0 ð t Þ i2 ðtÞ ¼ L1 R2 þ L2 R1 L1 R2 þ L2 R1

By making the numerical substitutions, the equation becomes: ðD þ 35:714Þi2 ðtÞ ¼ 14:286U0 ðtÞ þ 897:598 sin 20ptU1 ðtÞ; with i2 ð0 Þ ¼ 0: For t [ 0, ðD þ 35:714Þi2 ðtÞ ¼ 897:598 sin 20pt i2H ðtÞ ¼ k2 35:714t ð35:714 þ j62:832ÞI_2P ¼ 897:598\0 I_2P ¼

897:598\0 ¼ 12:420\  60; 4 72:273\60:386

i2P ðtÞ ¼ 12:420 sinð20pt  60:4 Þ ¼ 12:420 cosð20pt  60:4  90 Þ

312

5 Transients in First Order Circuits

¼ 12:420 cosð20pt  150:4 Þ ¼ 12:420cosð20pt þ 29:6 Þ i2 ðtÞ ¼ k2 35:714t  12:420cosð20pt þ 29:6 Þ i2 ð0 þ Þ ¼ 14:286 ¼ k2  12:420 cos 29:6 ¼ k2  10; 799 and k2 ¼ 3:487: Finally, i2 ðtÞ ¼ 3:48735:714t  12:420cosð20pt þ 29:6 Þ½A; for t [ 0½s: The time constant, both for the primary circuit and the secondary circuit, is T ¼ 1=35:714 ¼ 28½ms. And the period of the sinusoidal permanent regime component is Ts ¼ 2p=x ¼ 2p=20p ¼ 100½ms]. (b) In terms of the transformer T equivalent, in this case where the polarity is additive, that is, in which the two currents enter the windings through the dotted terminals, the equivalent circuit that produces the same equations, that is, Eqs. 5.19 and 5.20, is shown in Fig. 5.110. Then, the circuit for the numerical solution by ATPDraw is the one shown in Fig. 5.111. After typing RUN, with Dt ¼ 10½ls, the program apparently ran. But, when having the currents i1 ðtÞ and i2 ðtÞ plotted, it issued the following message: “Invalid floating point operation”. To explain what happened, it is necessary to remember that the initial values of the currents are: i1 ð0 þ Þ ¼ 28:571½A and i2 ð0 þ Þ ¼ 14:286½A. Therefore, the currents in the three inductances in Fig. 5.110 show discontinuities at t ¼ 0½s, that is: at the inductance of 60½mH  the current jumps from 0 [A] at t ¼ 0 ½s to 28:571½A at t ¼ 0 þ ; at the inductance of 120 [mH] it changes from 0 [A] at t ¼ 0 ½s to 14:286½A at t ¼ 0 þ ½s; and at the inductance of 40½mH  it varies from 0 [A] at t ¼ 0 ½s to i1 ð0 þ Þ  i2 ð0 þ Þ ¼ 42:857½A at t ¼ 0 þ ½s. However, according to Eq. (5.8), or in the face of Corollary 5.2, discontinuities in the waveforms of the inductance currents generate impulses in the voltages at their terminals, that is, infinite voltages. It is these infinite tensions that in the ATPDraw program, unable to represent them, generate so-called numerical oscillations. And that in the present case resulted in the message R1

L1 + M

L2 + M

i2 ( t ) ---\>

+

i1 ( t ) ---\>

^ e ( t )|

-M

Fig. 5.110 Equivalent electric circuit of Example 5.35

R2

5.5 Magnetically Coupled Circuits

313 AI

--|> i1 ( t ) t = 0

B

C 120 mH --|> i 2 ( t )

60 mH

1 Ohm

I

100 cos (20pi t) [V]

10 Ohm

- 40 mH

Fig. 5.111 Electric circuit of Example 5.35 for digital solution

“Invalid floating point operation”. In fact, in ATPDraw,
that is, in discrete time, the initial values of the currents are i1 ðDtÞ ¼ i1 105 ¼ 37:868½A and
 i2 ðDtÞ ¼ i2 105 ¼ 15:905½A, but which generate impulses in the inductance voltages in the same way and, consequently, the mentioned difficulties. To circumvent this difficulty, what the ATPDraw program provides is an “artificial damping”, through the introduction of large resistances in parallel with the pure inductances, expressed by R ¼ k:2L=Dt, allowing to choose k ¼ 5 or k ¼ 10: Thus, using k ¼ 10; for the inductance of 60½mH , the parallel resistance is 10  2  60:103 =105 ¼ 0:12½MX; for the 120½mH  inductance, the resistance is 0:24 [MX]; and for 40½mH , the resistance is 0:08 [MX]. Thus, the new circuit for the numerical solution by ATPDraw is the one shown in Fig. 5.112. It is interesting to note, in Fig. 5.112, that the representation of the inductances now differs from that of the inductances in Fig. 5.111. In Fig. 5.113 the current i1 ðtÞ is plotted and in Figs. 5.114 and 5.115, two zooms in the same current, in the vicinity of t ¼ 0[s], on the right. In Fig. 5.115, with the help of the plotter cursor, it is possible to capture that i1 ð10lsÞ ¼ i1 ðDtÞ ¼ 28:585½A, which, logically, differs from 37:868 [A], since this last value refers to to the case where the circuit still did not have the built-in parallel resistors. However, 28:585[A]ffi i1 ð0 þ Þ ¼ 28:571½A, the true, that is, the initial value obtained analytically from the original circuit of Fig. 5.109. In Fig. 5.116 the current i2 ðtÞ is plotted. And in Fig. 5.117 a very strong zoom is shown in the current i2 ðtÞ in the vicinity of t ¼ 0 [s], on the right. It is possible to extract from it, with the help of the plotter cursor, that i2 ðDtÞ ¼ i2 ð10lsÞ ¼ 14:283½A ffi i2 ð0 þ Þ ¼ 14:286½A, the true initial current, obtained analytically from the original circuit of Fig. 4.109. The current that descends through the 40½mH inductance is shown in Fig. 5.118. AI --|> i1 ( t ) t = 0

1 Ohm

B

U

U

D

C

120 mH --|> i 2 ( t )

60 mH

- 40 mH

Fig. 5.112 Electric circuit of Example 5.35 for digital solution

I

I

100 cos (20pi t) [V]

10 Ohm

314

5 Transients in First Order Circuits

60 [A] 40

20

0

-20

-40

-60 0,0 0,1 (f ile Ex34.pl4; x-v ar t) c:A

0,2

0,3

0,4

[s]

0,5

[ms]

90

-B

Fig. 5.113 Transient current i1 ðtÞ digital solution for circuit of Example 5.35

60 [A] 40

20

0

-20

-40

-60 0

15

(f ile Ex34.pl4; x-v ar t) c:A

30

45

60

75

-B

Fig. 5.114 Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35

5.5 Magnetically Coupled Circuits

315

60 [A] 40

20

0

-20

-40

-60 0

5

(f ile Ex34.pl4; x-v ar t) c:A

10 -B

15

20

25

30

[us]

35

Fig. 5.115 Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35

15 [A] 10

5

0

-5

-10

-15 0,0 0,1 (f ile Ex34.pl4; x-v ar t) c:C

0,2

0,3

0,4

-

Fig. 5.116 Transient current i2 ðtÞ digital solution for circuit of Example 5.35

[s]

0,5

316

5 Transients in First Order Circuits

15 [A] 10

5

0

-5

-10

-15 0

5

(f ile Ex34.pl4; x-v ar t) c:C

10

15

20

[us]

25

[s]

0,5

-

Fig. 5.117 Zoom at transient current i2 ðtÞ digital solution for circuit of Example 5.35

70,0 [A] 52,5 35,0 17,5 0,0 -17,5 -35,0 -52,5 -70,0 0,0

0,1

(f ile Ex34.pl4; x-v ar t) c:D

0,2

0,3

0,4

-

Fig. 5.118 Transient current iL ðtÞ digital solution for circuit of Example 5.35

5.5 Magnetically Coupled Circuits

317

70,0 [A] 52,5 35,0 17,5 0,0 -17,5 -35,0 -52,5 -70,0 0

10

20

(f ile Ex34.pl4; x-v ar t) c:D

30

40

50

60

70

[us]

80

-

Fig. 5.119 Zoom at transient current iL ðtÞ digital solution for circuit of Example 5.35

It is interesting to note that this current at the inductance of 40½mH  presents oscillation (numeric damped), as illustrated in Fig. 5.119, which shows a zoom of the same, in the vicinity of t ¼ 0 [s]. This fact, however, is not relevant, because this current does not even exist in the original circuit of Fig. 5.109. It is also worth noting that, because the inductances of the T equivalent in Fig. 5.111 have a common point (node D in Fig. 5.112), all three must receive the addition of large resistances in parallel. Otherwise, the primary and secondary currents will have damped numerical oscillations. For example, when only the inductance of 40½mH  receives the incorporation of the resistance of 0:08 [MX] in parallel, the currents i1 ðtÞ and i2 ðtÞ present damped numerical oscillations, as shown in Figs. 5.120 and 5.121, 5.122 and 5.123. Obviously, these oscillations are due to the digital process of solution of the circuit, because, as seen in the analytical solution, they are neither part of the current i1 ðtÞ, nor the current i2 ðtÞ. (c) From Eq. 5.19, the voltage at inductance L1 is given by: 0

0

0

0

eL1 ðtÞ ¼ L1 i1 ðtÞ þ Mi2 ðtÞ ¼ 0:02i1 ðtÞ þ 0:04i2 ðtÞ ¼ 0:02

  d  17:44235:714t þ 55:307 cosð20pt  33:7 Þ U1 ðtÞ dt

þ 0:04

 d  3:48735:714t  12:420 cosð20pt þ 29:6 Þ U1 ðtÞ: dt

318

5 Transients in First Order Circuits

70,0 [A] 52,5 35,0 17,5 0,0 -17,5 -35,0 -52,5 -70,0 0,0

0,1

(f ile Ex34.pl4; x-v ar t) c:A

0,2

0,3

0,4

[s]

0,5

-B

Fig. 5.120 Transient current i1 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations

70,0 [A] 52,5 35,0 17,5 0,0 -17,5 -35,0 -52,5 -70,0 0,00

0,02

(f ile Ex34.pl4; x-v ar t) c:A

0,04

0,06

0,08

0,10

[s]

0,12

-B

Fig. 5.121 Zoom at transient current i1 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations

5.5 Magnetically Coupled Circuits

319

20 [A] 10

0

-10

-20

-30 0,0

0,1

0,2

(f ile Ex34.pl4; x-v ar t) c:C

0,3

0,4

[s]

0,5

-

Fig. 5.122 Transient current i2 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations

20 [A] 10

0

-10

-20

-30 0,00

0,02

0,04

(f ile Ex34.pl4; x-v ar t) c:C

0,06

0,08

0,10

0,12

0,14

[s]

0,16

-

Fig. 5.123 Zoom at transient current i2 ðtÞ digital solution for circuit of Example 5.35 with damped numerical oscillations

320

5 Transients in First Order Circuits

  eL1 ðtÞ ¼ 12:45835:714t  69:501 sinð20pt  33:7 Þ U1 ðtÞ þ 0:571U 0 ðtÞ   þ 4:98135:714t þ 31:215 sinð20pt þ 29:6 Þ U1 ðtÞ  0:571U 0 ðtÞ:

Simply put, it comes that: eL1 ðtÞ ¼ 17:43935:714t þ 62:090 cosð20pt þ 29:613 Þ½V ; for t [ 0½s: Therefore, eL1 ð0 þ Þ ¼ 71:419½V: From Eq. 5.20, the voltage at inductance L2 is given by: 0

0

0

0

eL2 ðtÞ ¼ Mi1 ðtÞ þ L2 i2 ðtÞ ¼ 0:04i1 ðtÞ þ 0:08i2 ðtÞ   ¼ 24:91735:714t  139:002 sinð20pt  33:7 Þ U1 ðtÞ þ 1:143U 0 ðtÞ   þ 9:96335:714t þ 62:430 sinð20pt þ 29:6 Þ U1 ðtÞ  1:143U 0 ðtÞ: Simply put, it follows that: eL2 ðtÞ ¼ 34:88035:714t þ 124:180 cosð20pt þ 29:6 Þ½V ; for t [ 0½s: Therefore, eL2 ð0 þ Þ ¼ 142:854½V: The voltage at inductance L2 could also be obtained from eL2 ðtÞ ¼ eR2 ðtÞ ¼ R2 i2 ðtÞ: (d) For the numerical solution of the voltages in L1 and L2 using the ATPDraw program, the corresponding circuit is shown in Fig. 5.124. Initially, the voltages in the inductances of 60½mH ; 120½mH  and 40½mH , are represented in Figs. 5.125, 5.126 and 5.127, respectively. V

A --|> i1 ( t ) t = 0

V

B 1 Ohm

U

U

D

C

120 mH --|> i 2 ( t )

60 mH

- 40 mH

Fig. 5.124 Electric circuit of Example 5.35 for digital solution

U

U

100 cos (20pi t) [V]

10 Ohm

5.5 Magnetically Coupled Circuits

321

400 [kV] 300 200 100

0 -100 -200 -300 0,0

0,1

(f ile Ex34.pl4; x-v ar t) v :B

0,2

0,3

0,4

[s]

0,5

0,4

[s]

0,5

-D

Fig. 5.125 Transient digital voltage in the inductance of 60½mH 

300 [kV] 200 100 0

-100 -200 -300 -400 0,0 0,1 (f ile Ex34.pl4; x-v ar t) v :D

0,2

0,3

-C

Fig. 5.126 Transient digital voltage in the inductance of 120½mH 

Apparently “nothing came out”. It turns out that the ordinate scale is in [kV] and, in addition, the transient regime lasts approximately 5T ¼ 5  28½ms ¼ 140½ms: Thus, the observation time tmax ¼ 0:5½s is extremely long to be able to see the transient (exponential) component. On the other hand, along the analytical solution

322

5 Transients in First Order Circuits

300 [kV] 200 100 0

-100 -200 -300 -400 0,0

0,1

(f ile Ex34.pl4; x-v ar t) v :D

0,2

0,3

0,4

[s]

0,5

-

Fig. 5.127 Transient digital voltage in the inductance of 40½mH 

for the voltages, impulses appeared in the self and mutually induced electromotive forces, which canceled each other out. This suggests numerical oscillations dampened in the voltages in Figs. 5.125, 5.126 and 5.127. Therefore, it is better to adopt a very short observation time; for example, on the order of 20Dt ¼ 0:2½ms. The results obtained are shown in Figs. 5.128, 5.129 and 5.130. To capture the sinusoidal steady state voltages, the scale of ordinate values was changed from [kV] to [V], taking tmax ¼ 0:3½s. The results are in Figs. 5.131, 5.132 and 5.133. After detailing the waveforms of the voltages in the three inductances that make up the T equivalent of the transformer, contained in Fig. 5.124, it remains to be clarified that they do not represent the requested numerical solutions: eL1 ðtÞ and eL2 ðtÞ. This is because the voltage eL1 ðtÞ is the potential difference between node B and the reference node (Earth) in Fig. 5.124, and that voltage eL2 ðtÞ is the negative of the voltage in the resistance of 10 [X], that is, the negative of the potential difference between node C and Earth. It is for this reason that there is a Probe Volt at node B and another at node C in Fig. 5.124. Thus, the actual requested voltages are shown in Figs. 5.134 and 5.135. From Figs. 5.136 and 5.137, with the plotter cursor, it can be determined that: eL1 ðDtÞ ¼ 71:415½V and that eL2 ðDtÞ ¼ 142:83½V:

5.5 Magnetically Coupled Circuits

323

400 [kV] 300 200 100

0 -100 -200 -300 0,00

0,04

(f ile Ex34.pl4; x-v ar t) v :B

0,08

0,12

0,16

[ms] 0,20

0,16

[ms] 0,20

-D

Fig. 5.128 Zoom at transient digital voltage in the inductance of 60½mH

300 [kV] 200 100 0

-100 -200 -300 -400 0,00

0,04

(f ile Ex34.pl4; x-v ar t) v :D

0,08

0,12

-C

Fig. 5.129 Zoom at transient digital voltage in the inductance of 120½mH

324

5 Transients in First Order Circuits

300 [kV] 200 100 0

-100 -200 -300 -400 0,00 0,04 (f ile Ex34.pl4; x-v ar t) v :D

0,08

0,12

0,16

[ms] 0,20

-

Fig. 5.130 Zoom at transient digital voltage in the inductance of 40½mH

400 [V] 300 200 100 0 -100 -200 -300 -400 0,00 0,05 (f ile Ex34.pl4; x-v ar t) v :B

0,10

0,15

0,20

-D

Fig. 5.131 Steady-state digital voltage in the inductance of 60½mH

0,25

[s]

0,30

5.5 Magnetically Coupled Circuits

325

400 [V] 300 200 100 0 -100 -200 -300 -400 0,00

0,05

(f ile Ex34.pl4; x-v ar t) v :D

0,10

0,15

0,20

0,25

[s]

0,30

0,25

[s]

0,30

-C

Fig. 5.132 Steady-state digital voltage in the inductance of 120½mH

400 [V] 300 200 100 0 -100 -200 -300 -400 0,00 0,05 (f ile Ex34.pl4; x-v ar t) v :D

0,10

0,15

0,20

-

Fig. 5.133 Steady-state digital voltage in the inductance of 40½mH

326

5 Transients in First Order Circuits

80 [V] 50

20

-10

-40

-70 0,00 0,05 (f ile Ex34.pl4; x-v ar t) v :B

0,10

0,15

0,20

0,25

[s]

0,30

[s]

0,30

Fig. 5.134 Transient voltage eL1 ðtÞ digital solution for circuit of Example 5.35

150 [V] 100

50

0

-50

-100

-150 0,00 0,05 (f ile Ex34.pl4; x-v ar t) v :C factors: 1 -1 offsets: 0,00E+00 0,00E+00

0,10

0,15

0,20

0,25

Fig. 5.135 Transient voltage eL2 ðtÞ digital solution for circuit of Example 5.35

5.5 Magnetically Coupled Circuits

327

80 [V] 50

20

-10

-40

-70 0

5

10

15

20

[us]

25

(f ile Ex34.pl4; x-v ar t) v :B

Fig. 5.136 Zoom at transientvoltage eL1 ðtÞ digital solution for circuit of Example 5.35

150 [V] 100

50

0

-50

-100

-150 0 (f ile Ex34.pl4; x-v ar t) factors: 1 offsets: 0,00E+00

10 v :C -1 0,00E+00

20

30

40

[us]

Fig. 5.137 Zoom at transientvoltage eL2 ðtÞ digital solution for circuit of Example 5.35

50

328

5 Transients in First Order Circuits

Of the analytical solutions, as noted earlier, eL1 ð0 þ Þ ¼ 71:419½V and eL2 ð0 þ Þ ¼ 142:854½V:

5.6

DuHamel Integrals

Figure 5.138a symbolically represents a linear and time-invariant (or fixed) circuit —CLI, with no energy initially stored in its capacitors and inductors at t ¼ 0 , with input (excitation) xðtÞ and output (response) yðtÞ. Assuming an arbitrary and causal input xðtÞ, that is, that xðtÞ ¼ 0 for t\0, we want to determine the corresponding output yðtÞ, for t  0, from the knowledge of the response hðtÞ, when a unitary impulse U0 ðtÞ is applied at the input, at time t ¼ 0, as shown in Fig. 5.138b. The invariance with time means that when the input is delayed by a value k, the only consequence is an equal delay in the output, which, however, keeps its waveform and intensity unchanged. Thus, if the impulsive input is applied not at t ¼ 0, but at t ¼ k, the same response will exist, but only from t ¼ k. This means that the response to input U0 ðt  kÞ is simply hðt  kÞ, as illustrated in Fig. 5.139a. The homogeneity property of linear systems states that: if the input is multiplied by a constant factor, the output is simply multiplied by the same constant. So, if the input is not a unit impulse applied at time t ¼ k, but an impulse of intensity xðkÞ, applied at time t ¼ k, where xðkÞ is the value of xðtÞ for t ¼ k, that is, a constant, the answer will also be multiplied by xðkÞ. In other words, if the input is changed to xðkÞU0 ðt  kÞ, then the corresponding output is xðkÞU0 ðt  kÞ, as shown in Fig. 5.139b. The principle of superposition of linear systems, called the superposition theorem when it comes to electrical circuits, establishes that: the response of a circuit to several independent sources is the sum (superposition) of the individual responses of these sources. Thus, considering as input the sum of the infinite inputs xðkÞU0 ðt  kÞ, for all possible values of kð0  k\ þ 1Þ, the superposition theorem imposes that the output must be equal to the sum responses resulting from the use of k , equally for all possible values. In other words: the input integral produces the Fig. 5.138 Linear and time-invariant (or fixed) circuit—CLI

x ( t ) ----->

C.L.I.

------> y ( t )

(a)

U ( t ) ------> 0

C.L.I.

(b)

------>h ( t )

5.6 DuHamel Integrals

329

C.L.I.

------>

------>

(a)

C.L.I.

------>

------>

(b)

------>

C.L.I.

------>

(c)

------>

C.L.I.

------>

(d) Fig. 5.139 a Invariance with time; b homogeneity property; c principle of superposition; d DuHamel, Carson, Superposition Integral or, simply, Convolution Integral

output integral, as shown in Fig. 5.139c. However, according to Eq. 2.112, adapted for causal signs, þZ1

xðkÞU0 ðt  kÞdk ¼ xðtÞ:

0

Therefore, the corresponding output is yð t Þ ¼

þZ1

xðkÞhðt  kÞdk;

ð5:21Þ

0

as shown in Fig. 5.138d. Equation 5.21 is called the DuHamel, Carson, Superposition Integral or, simply, Convolution Integral. Considering, however, that hðtÞ ¼ 0 for t\0, then hðt  kÞ ¼ 0 for k [ t, and Eq. 5.21 can be written as Zt

yðtÞ ¼ xðkÞhðt  kÞdk; for all t  0: 0

ð5:22Þ

330

5 Transients in First Order Circuits

As established in 2.12 (convolution integrals), then yðtÞ ¼ xðtÞ hðtÞ

ð5:23Þ

In summary, Eq. 5.23 establishes that: the output yðtÞ is equal to the convolution of the input xðtÞ with the response to the impulse hðtÞ. Due to the commutability property of the convolution integral, yðtÞ ¼ hðtÞ xðtÞ ¼

Zt

hðkÞxðt  kÞdk:

ð5:24Þ

0

The lower limit of the integral in Eq. 5.24 was taken as k ¼ 0 ; predicting a possible presence of impulse, at time t ¼ 0, at hðtÞ; as in Example 5.16, where the impulse response also contains impulse. If there is no impulse at t ¼ 0, the limit must be taken as k ¼ 0. The property that deals with the convolution derivative, Eq. 2.116, establishes that to derive a convolution it is enough to derive any of the two functions that are being convoluted. So, from Eq. 5.23, y0 ðtÞ ¼ x0 ðtÞ hðtÞ:

ð5:25Þ

Equation 5.25 shows that when deriving the input from an invariant linear circuit, the output is also derived. Logically, the inverse is also true, that is: integrating the function h(t) in Eq. 5.25 yields yðtÞ. It turns out that the integral of hðtÞ is r ðtÞ, the response to the unitary step U1 ðtÞ. Therefore, yðtÞ ¼ x0 ðtÞ rðtÞ ¼

Zt

x0 ðkÞrðt  kÞdk; for all t  0:

ð5:26Þ

0

In summary, Eq. 5.26 states that: the output yðtÞ is equal to the convolution of the derivative of the input, x0 ðtÞ; with the response to the step rðtÞ. The lower limit of the integral was taken as 0 because it may happen that xðtÞ presents finite discontinuity at t ¼ 0, causing an impulse at x0 ðkÞ. Due to the property of commutativity, Zt

yðtÞ ¼ rðtÞ x0 ðtÞ ¼ rðkÞ 0

dxðt  kÞ dk: d ð t  kÞ

ð5:27Þ

In conclusion, it can then be said that Eq. 5.22 is a consequence of the fact that the sign xðtÞ has been decomposed into impulses, and that Eq. 5.26 is a consequence of it having been decomposed in steps. In general, Eqs. 5.24 and 5.27 are more difficult to apply.

5.6 DuHamel Integrals

331

Similarly, if the input signal xðtÞ is decomposed into ramps, the output yðtÞ will be given by yðtÞ ¼ x00 ðtÞ aðtÞ ¼

Zt

x00 ðkÞaðt  kÞdk;

ð5:28Þ

0

R where aðtÞ ¼ r ðtÞdt is the response to the unitary ramp U2 ðtÞ. Equation 5.23, together with property 2.11 of Sect. 2.12 of Chap. 2, shows why the response of a circuit to a causal pulse of duration Tx lasts much longer than Tx According to the aforementioned property, the duration of the output signal, Ty , is the sum of the duration of the input signal, Tx , with the duration of the impulse response, Th : T y ¼ T x þ Th :

ð5:29Þ

Therefore, when the impulse response is an exponential signal with time constant T, the output signal lasts approximately Ty ¼ Tx þ 5T. When hðtÞ contains impulse, that is, when hðtÞ ¼ AU0 ðtÞ þ h1 ðtÞ, where h1 ðtÞ does not contain impulse, according to Eq. 5.23, yðtÞ ¼ xðtÞ hðtÞ ¼ xðtÞ ½AU0 ðtÞ þ h1 ðtÞ ¼ xðtÞ AU0 ðtÞ þ xðtÞ h1 ðtÞ ¼A

tZþ

xðkÞU0 ðt  kÞdk þ

Zt

xðkÞh1 ðt  kÞdk

0

0

Therefore, yðtÞ ¼ AxðtÞ þ

Zt

xðkÞh1 ðt  kÞdk:

ð5:30Þ

0

Equation 5.30 indicates that, in this special case, the input signal must appear as part of the output signal. Example 5.36 If the impulse response of a linear and time-invariant circuit is given by hðtÞ ¼ ðt  3t ÞU1 ðtÞ; determine the your answer yðtÞ for the input xðtÞ ¼ ð2t  4t ÞU1 ðtÞ. Solution
 xðkÞ ¼ 2k  4k U1 ðkÞ
 hðt  kÞ ¼ t þ k  3t þ 3k U1 ðt  kÞ

332

5 Transients in First Order Circuits

By Eq. 5.20, y ð t Þ ¼ x ð t Þ hð t Þ ¼

Zt





 2k  4k t þ k  3t þ 3k dk

0

¼ t

Zt




  Zt
k  3k dk  3t k  k dk

0

¼ t

0

 k 


 t  3k

t þ  3t k þ k

0 0 1 3

  3t 2 þ ¼ t t þ  3t ðt þ t  2Þ 3 3 So,

 t   4t 2t 3t yðtÞ ¼ 2   þ  U1 ðtÞ: 3 3

Input xðtÞ lasts approximately 2.5 units of time; hðtÞ lasts approximately 5 units of time. Therefore, yðtÞ lasts approximately 7.5 units of time. Example 5.37 Using the convolution integral, find the current iðtÞ in the circuit of Fig. 5.140 when the input voltage is: (a) eðtÞ ¼ t U1 ðtÞ; (b) eðtÞ ¼ sin t½U1 ðtÞ  U1 ðt  pÞ: Solution The impulse response of this circuit was determined in example 5.14, and it is worth: hð t Þ ¼ i ð t Þ ¼

1 1 U0 ðtÞ  2 t=RC U1 ðtÞ; for eðtÞ ¼ U0 ðtÞ: R RC

Then, hð t  kÞ ¼

1 1 U0 ðt  kÞ  2 ðtkÞ=RC U1 ðt  kÞ: R R C

Fig. 5.140 Electric circuit of Example 5.37 +

R

^ e ( t )| |

-----------/> i(t) C

5.6 DuHamel Integrals

333

So, iðtÞ ¼ eðtÞ hðtÞ ¼

Zt

eðkÞhðt  kÞdk:

0

(a) For eðtÞ ¼ t U1 ðtÞ, iðtÞ ¼ t U1 ðtÞ ¼

¼



1 1 U0 ðtÞ  2 t=RC U1 ðtÞ R R C



1 tZþ k 1 Zt k ðtkÞ=RC  U0 ðt  kÞdk  2   dk R0 R C0

1 t t=RC Zt 1RCk t t=RC RC 1RCk

t :  2  RC

 2 :  RC dk ¼ R RC 0 R R C 1  RC 0 ¼

t t  t=RC t t=RC  1RCt   RC  1 ¼  R Rð1  RC Þ R Rð1  RC Þ

¼

t=RC  RCt :U1 ðtÞ: Rð1  RC Þ

This result can be confirmed by directly solving the circuit by the usual classic method, for eðtÞ ¼ t U1 ðtÞ: From Example 5.14, or directly from Fig. 5.140, we have that: i 0 ðt Þ þ

1 1 iðtÞ ¼ e0 ðtÞ: RC R

If eðtÞ ¼ t U1 ðtÞ; e0 ðtÞ ¼ U0 ðtÞ  t U1 ðtÞ. So, i 0 ðt Þ þ

1 1 1 1 : 1 iðtÞ ¼ U0 ðtÞ  t U1 ðtÞ¼ U0 ðtÞ  U1 ðtÞ þ . . .; ið0 Þ ¼ 0: RC R R R R

For t [ 0, 

 1 t : Dþ i ðt Þ ¼  RC R iH ðtÞ ¼ k1 t=RC :

334

5 Transients in First Order Circuits

i P ðt Þ ¼

 t  t  R 1  Ct :  ¼ ¼ 1 1 R 1 þ RC RC  1 D þ RC Ct : RC  1

iðtÞ ¼ k1 t=RC þ

From the previous differential equation, we have that ið0 þ Þ ¼ R1 . So, 1 C 1 C 1 ¼ k1 þ : Then, k1 ¼  ¼ : R RC  1 R RC  1 Rð1  RC Þ Therefore, i ðt Þ ¼

t=RC Ct t=RC  RCt  ¼ :U1 ðtÞ; Rð1  RC Þ 1  RC Rð1  RC Þ

confirming the result obtained through the convolution integral. (b) For 0\t\p, eðtÞ ¼ sint and i ðt Þ ¼

1 tZþ 1 Zt sin kU0 ðt  kÞdk  2 sin kðtkÞ=RC dk R0 R C0 ¼

1 t=RC Zt k=RC sin t  2  sin kdk R R C 0

R ax cos bxÞ From ax sin bxdx ¼  ða sina2bxb , with x ¼ k, a ¼ 1=RC and b ¼ 1, comes þ b2 that:
1  t sin k  cosk

1 t=RC k=RC RC iðtÞ ¼ sin t  2 :


1 2

R R C þ1 RC

"

1 t=RC  ¼ sin t  2 : R R C ¼

¼

Ct=RC 1 þ ðRC Þ

2

þ


t=RC 1

RC sin t  cost
1 2 RC þ 1

0



1 
2 1 RC þ 1

#

1 RCcost  sin t Ct=RC i sin t þ h R 1 þ ðRC Þ2 R 1 þ ðRC Þ2

Cðcost þ RCsentÞ 1 þ ðRC Þ

2

¼

Ct=RC 1 þ ðRC Þ

2

þ

C cos½t  arctgðRC Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; for 0\t\p: 1 þ ðRC Þ2

5.6 DuHamel Integrals

335

For t [ p the integral from 0 to t must be replaced by the integral from 0 to p, since eðtÞ ¼ 0 for t [ p. Zp

iðtÞ ¼ eðtÞ hðtÞ ¼ sin kU0 ðt  kÞdk  0

Since U0 ðt  kÞ occurs only at k ¼ t [ p,

Rp 0

1 Zp sin kðtkÞ=RC dk: R2 C 0

sin kU0 ðt  kÞdk ¼ 0. So,


1  p sin k  cosk

t=RC Zp k=RC t=RC k=RC RC i ðt Þ ¼  2  sin kdk ¼  2 :


1 2

R C 0 R C þ1 RC

0

p
 t=RC k=RC ðRC sin k  R2 C 2 coskÞ

t=RC R2 C2 p=RC þ 1 ¼ 2 :

¼ 2 :

R C R C 1 þ ðRC Þ2 1 þ ðRC Þ2 0




¼

C p=RC þ 1 1 þ ðRC Þ2

 :t=RC ; for t [ p:

To confirm the results of this item b, using the classic method, the superposition theorem will be used, considering that: eðtÞ ¼ sin t½U1 ðtÞ  U1 ðt  pÞ ¼ sin t:U1 ðtÞ  sin t:U1 ðt  pÞ ¼ sin t:U1 ðtÞ þ sinðt  pÞ:U1 ðt  pÞ: Therefore, the response iðtÞ to eðtÞ ¼ sin t:U1 ðtÞ. will be obtained first. Recalling, from Example 5.14, that the differential equation of the circuit is i 0 ðt Þ þ i 0 ðt Þ þ

1 1 iðtÞ ¼ e0 ðtÞ; then e0 ðtÞ ¼ cost:U1 ðtÞ and; RC R

1 1 1 iðtÞ ¼ cost:U1 ðtÞ ¼ U1 ðtÞ þ . . .; ið0 Þ ¼ 0: RC R R iH ðtÞ ¼ k1 t=RC :

Using the phasor method to find the particular integral, it comes that: 

I_P ¼

 1 _ IP ¼ 1\0 j1 þ RC

C C \0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \  arctgðRCÞ 1 þ jRC 1 þ ðRC Þ2

336

5 Transients in First Order Circuits

  C iP ðtÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos t  tg1 ðRC Þ : 1 þ ðRC Þ2   C iðtÞ ¼ k1 t=RC þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos t  tg1 ðRC Þ : 1 þ ðRC Þ2 Denominating S ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðRC Þ2 and / ¼ tg1 ðRC Þ, it becomes that iðtÞ ¼ k1 t=RC þ

C cosðt  /Þ: S

From the previous differential equation, ið0 þ Þ ¼ 0. So, 0 ¼ k1 þ CS cos / ) k1 ¼  CS cos /: Then, iðtÞ ¼  CS cos /et=RC þ CS cosðt  /Þ: On the other hand, tg/ ¼

sin / RC ¼ ; cos / 1

sin2 / ðRC Þ2 sin2 / þ cos2 / ðRC Þ2 þ 1 ¼ ; ¼ cos2 / cos2 / 1 1 1 cos / 1 ¼ cos / ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; : S 1 þ ðRC Þ2 1 þ ðRC Þ2 2

3 t=RC

C 6 C 7 i ðt Þ ¼ 4 þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosðt  /Þ5:U1 ðtÞ: 2 2 1 þ ðRC Þ 1 þ ðRC Þ Therefore, the answer to eðtÞ ¼ sinðt  pÞ:U1 ðt  pÞ is

ð5:31Þ

5.6 DuHamel Integrals

337

2

3 p=RC t=RC

 6C iðt  pÞ ¼ 4 1 þ ðRC Þ2 2

C 7 þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosðt  p  /Þ5:U1 ðt  pÞ 2 1 þ ðRC Þ 3 ð5:32Þ

p=RC t=RC  C 6C 7 ¼4  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosðt  /Þ5:U1 ðt  pÞ: 2 2 1 þ ðRC Þ 1 þ ðRC Þ

The answer sought is to overlap Eqs. 5.31 and 5.32, resulting in

iðtÞ ¼

8 Ct=RC C cosðt  /Þ; for0\t\p < 1 þ ðRCÞ2 þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 þ ðRC Þ

: Cðp=RC þ 1Þ :et=RC ; for t [ p; 1 þ ðRC Þ2

thus confirming the results obtained through the convolution integral. Example 5.38 In Fig. 5.141 we have what is called cascade circuits, since the output of circuit N1 is the input of circuit N2 . If the connection of N2 does not affect the output of N1 , they are called isolated. Otherwise, it is said that N2 carries N1 . If h1 ðtÞ and h2 ðtÞ are the impulse responses of two isolated circuits connected in cascade, as in Fig. 5.141, find the impulse response of the combination. Also find the answer to the step of the cascading connection. Solution From the statement, h1 ðtÞ ¼ zðtÞ when xðtÞ ¼ U0 ðtÞ and h2 ðtÞ ¼ yðtÞ when zðtÞ ¼ U0 ðtÞ: The Convolution Theorem applied to the N2 circuit states that yðtÞ ¼ zðtÞ h2 ðtÞ. If xðtÞ ¼ U0 ðtÞ, then zðtÞ ¼ h1 ðtÞ. Therefore, yðtÞ ¼ h1 ðtÞ h2 ðtÞ. On the other hand, for the connection as a whole, yðtÞ ¼ hðtÞ xðtÞ ¼ hðtÞ U0 ðtÞ ¼ hðtÞ. So by comparison, hðtÞ ¼ h1 ðtÞ h2 ðtÞ:

ð5:33Þ

r1 ðtÞ ¼ zðtÞ when xðtÞ ¼ U1 ðtÞ and r2 ðtÞ ¼ yðtÞ when zðtÞ ¼ U1 ðtÞ.

x ( t ) --------->

N1

z( t ) -------------->------------

Fig. 5.141 Cascade or latter circuit of Example 5.38

N

2

----------> y ( t )

338

5 Transients in First Order Circuits

The DuHamel Integral applied to the N2 circuit results in yðtÞ ¼ z0 ðtÞ r2 ðtÞ: If 0 0 xðtÞ ¼ U1 ðtÞ, then zðtÞ ¼ r1 ðtÞ and z0 ðtÞ ¼ r1 ðtÞ.Therefore, yðtÞ ¼ r1 ðtÞ r2 ðtÞ. However, yðtÞ ¼ x0 ðtÞ r ðtÞ ¼ r ðtÞ x0 ðtÞ ¼ r ðtÞ U0 ðtÞ ¼ r ðtÞ. By comparison, 0

r ðtÞ ¼ r1 ðtÞ r2 ðtÞ:

ð5:34Þ

Example 5.39 If the impulse response of each of the circuits in Fig. 5.141 is h1 ðtÞ ¼ h2 ðtÞ ¼ Aat for t [ 0, determine the impulse response of the cascade combination. Solution Zt

Zt

0

0

hðtÞ ¼ h1 ðtÞ h2 ðtÞ ¼ h1 ðkÞh2 ðt  kÞdk ¼ Aak AaðtkÞ dk Zt

¼ A2 at dk ¼ A2 tat :U1 ðtÞ: 0

5.7

Proposed Problems

P.5-1 Find the circuit time constant in Fig. 5.142, for t [ 0. Answer: T ¼ 0:2½s:

t=0

40 uF

25 kOhm 4 uF

6 uF

Fig. 5.142 Electric circuit of Problem P.5-1

8 Ohm

40 V

4 Ohm

+ +

t=0

60 V

24 Ohm

Fig. 5.143 Electric circuit of Problem P.5-2

16 mH

40 Ohm

5.7 Proposed Problems

339

L R

R

L

L

R

Fig. 5.144 Electric circuit of Problem P.5-3

P.5-2 Find the circuit time constant shown in Fig. 5.143, for t\0 and for t [ 0. Answer: T ¼ 0:4½ms for t\0 and T ¼ 2½ms for t [ 0. P.5-3 Find the circuit time constant in Fig. 5.144. Answer: T ¼ L=R. P.5-4 The switch shown in the circuit of Fig. 5.145 has been open for a long time, before being closed at time t ¼ 0. (a) Determine i1 ð0 Þ; e1 ð0 Þ; iC ð0 Þ; eC ð0 Þ; i2 ð0 Þ; e2 ð0 Þ; ik ð0 Þ; ek ð0 Þ; i3 ð0 Þ; e3 ð0 Þ; iL ð0 Þ; eL ð0 Þ:

(b) Determine i1 ð0 þ Þ; e1 ð0 þ Þ; iC ð0 þ Þ; eC ð0 þ Þ; i2 ð0 þ Þ; e2 ð0 þ Þ; ik ð0 þ Þ; ek ð0 þ Þ; i3 ð0 þ Þ; e3 ð0 þ Þ; iL ð0 þ Þ; eL ð0 þ Þ:

(c) Determine i1 ð1Þ; e1 ð1Þ; iC ð1Þ; eC ð1Þ; i2 ð1Þ; e2 ð1Þ; ik ð1Þ; ek ð1Þ; i3 ð1Þ; e3 ð1Þ; iL ð1Þ; eL ð1Þ:

Answer: (a) i1 ð0 Þ ¼ 1½A; e1 ð0 Þ ¼ 40½V; iC ð0 Þ ¼ 0½A; eC ð0 Þ ¼ 80½V; i2 ð0 Þ ¼ 1½A; e2 ð0 Þ ¼ 50½V; ik ð0 Þ ¼ 0½A; ek ð0 Þ ¼ 30½V; i3 ð0 Þ ¼ 1½A; e3 ð0 Þ ¼ 30½V; iL ð0 Þ ¼ 1½A; eL ð0 Þ ¼ 0½V:

i1( t ) --|>

40 Ohm

5 uF

^e ( t ) | C

Fig. 5.145 Electric circuit of Problem P.5-4

50 Ohm

|i (t) Vk ^ | e k( t )

30 Ohm i (t)

+ ^ | e (t)

2 uF

+

t=0

_

200 V

Fig. 5.146 Electric circuit of Problem P.5-6

Fig. 5.147 Electric circuit of Problem P.5-7

t=0 ^ e1( t ) |

C1

---|> i(t) C 2

|e (t) v 2

5.7 Proposed Problems

341

P.5-7 If capacitances C1 ¼ 2½lF and C2 ¼ 1½lF, shown in Fig. 5.147, have voltages e1 ð0 Þ ¼ 150½V and e2 ð0 Þ ¼ 50½V, find the voltages e1 ð0 þ Þ and e2 ð0 þ Þ and the current iðtÞ. Answer: e1 ð0 þ Þ ¼ 250=3½V, e2 ð0 þ Þ ¼ 250=3½V, iðtÞ ¼ CC1 1þCC2 2 ½e1 ð0 Þ þ e2 0 ÞU0 ðtÞ ¼ ð400=3ÞU0 ðtÞ½lA. P.5-8 In the circuit shown in Fig. 5.148, the capacitor is de-energized at t ¼ 0 ½s. Find iðtÞ and eðtÞ, for t [ 0½s:

10 Ohm

+

t=0

--|> i(t) 10 Ohm

50 V

2 uF

^ | e(t)

Fig. 5.148 Electric circuit of Problem P.5-8


 Answer: iðtÞ ¼ 2:5 1 þ t=10 ½A; for t [ 0½ls:  eðtÞ ¼ 25 1  t=10 ½V; for t  0½ls. P.5-9 Capacitors C1 and C2 in the circuit of Fig. 5.149 are de-energized at t ¼ 0 ½s. If the switch is closed at t ¼ 0½s, establish the expressions of iðtÞ, e1 ðtÞ and e2 ðtÞ, for t [ 0. Answer: iðtÞ ¼ ER ðC1 þ C2 Þt=RC1 C2 :U1 ðtÞ: e1 ð t Þ ¼

i C2 E h 1  ðC1 þ C2 Þt=RC1 C2 U1 ðtÞ: C1 þ C2

Fig. 5.149 Electric circuit of Problem P.5-9

t=0

1

^ (t) | e1

C2

^e ( t ) | 2

C

+

R

E i(t)

: 0

the voltage e0 ðtÞ in the circuit of Fig. 5.166, for t  0; knowing for t\0. for 0\t\1. or iðtÞ ¼ 2U1 ðtÞ  3U1 ðt  1Þ þ U1 ðt  2Þ: for 1\t\2. for t [ 2:

348

5 Transients in First Order Circuits

Fig. 5.166 Electric circuit of Problem P.5-27

2H 4 Ohm 2 Ohm

i(t)

8 0 > > < 4 ð1  3t Þ Answer: e0 ðtÞ ¼ 3 2
3 4 3t  þ 2   > > :
3 3 2 6 43 3t 2  3   3  P.5-28 Find e0 ðtÞ, for 1\t\1, in that: 8 I

^ |eR(t)

Fig. 5.182 Electric circuit of Problem P.5-36 for digital solution

160 [V] 120 80 40 0 -40 -80 -120 -160 0

1

(f ile Exp32.pl4; x-v ar t) v :D

2 -

Fig. 5.183 Inductor voltage eL ðtÞ of Problem P.5-36

3

4

[ms]

5

5.7 Proposed Problems

357

160 [V] 120 80 40 0 -40 -80 -120 -160 0

1

2

(f ile Exp32.pl4; x-v ar t) v :C

3

4

[ms]

5

[ms]

8

-

Fig. 5.184 Resistor voltage eR ðt) of Problem P.5-36

2,0 [A] 1,5 1,0 0,5 0,0 -0,5 -1,0 -1,5 -2,0 0

1

2

(f ile Exp32.pl4; x-v ar t) c:C

3 -D

Fig. 5.185 Current iðtÞ of Problem P.5-36

4

5

6

7

358

5 Transients in First Order Circuits

9 [A] 6

3

0

-3

-6

-9 0

1

(f ile Exp32.pl4; x-v ar t) c:B

2

3

4

5

[ms]

6

-C

Fig. 5.186 Current iT ðtÞ, which flows through the voltage eðtÞ source of Problem P.5-36

40 [A] 30 20 10 0 -10 -20 -30 -40 0,0 0,5 1,0 (f ile Exp32.pl4; x-v ar t) c:A -B

1,5

2,0

2,5

3,0

Fig. 5.187 Current iC ðtÞ, which flows through the voltage source of Problem P.5-36

[ms]

3,5

5.7 Proposed Problems

359

160 W 120 80 40

0 -40 -80 -120 0

1

(f ile Exp32.pl4; x-v ar t) v :D

2

3

4

5

[ms]

6

4

5

[ms]

6

-

Fig. 5.188 Power pL ðtÞ in the inductorof Problem P.5-36

20 [mJ] 16

12

8

4

0 0

1

(f ile Exp32.pl4; x-v ar t) c:D

2

3

-

Fig. 5.189 Energy W L ðtÞ in the inductorof Problem P.5-36

360

5 Transients in First Order Circuits

P.5-37 The capacitors of the circuit of Fig. 5.190 were de-energized when switches 1 and 2 were closed, simultaneously, at t ¼ 0. After 30 [ls] switch 3 was closed. Establish the expressions of the voltages e1 ðtÞ and e2 ðtÞ for t [ 0. Also, perform the numerical solution of the problem (Figs. 5.191, 5.192 and 5.193). (  6 16 1  10 t=20 ½V; for 0  t\30½ls Answer: e1 ðtÞ ¼ 106 t=16 ½V; for t [ 30½ls 11:008  7:655 e2 ð t Þ ¼

(  6 6 1  10 t=12 ½V; 11:008  7:65510

6

for 0  t\30½ls t=16

½V;

for t [ 30½ls

The numerical oscillation in the current of switch 3 means that, at time t ¼ 30 [ls], an impulsive current passed through it, and it circulated clockwise through the mesh formed by the capacitances and switch 3, to make e1 ð30ls þ Þ ¼ e2 ð30ls þ Þ ¼ 9:834 [V], because e1 ð30ls Þ ¼ 12:430½V and e2 ð30ls Þ ¼ 5:507½V: 5 kOhm

4 kOhm

3

1

t =30 us

2

+

t=0 ^ e1 ( t ) |

20 kOhm

t=0

^ e 2( t ) |

5 nF

3 nF

+

20 V

6V

Fig. 5.190 Electric circuit of Problem P.5-37

1 t=0

t =30 us

+

20 kOhm

^ e1 ( t ) |

B 5 nF

^ e 2( t ) |

2 U

A

U

20 V

4 kOhm

3

t=0 3 nF

+

5 kOhm

6V

Fig. 5.191 Electric circuit of Problem P.5-37 for digital solution

5.7 Proposed Problems

361

15 [V] 12

9

6

3

0 0,00 0,02 (f ile p.36.pl4; x-v ar t) v :A

-

0,04 v :B

0,06

0,08

0,10

[ms]

0,12

30,09

[us]

30,11

-

Fig. 5.192 Voltage e1 ðtÞ and Voltage e2 ðtÞ of Problem P.5-37

Z00m in the current of the switch 3

3 [A] 2

1

0

-1

-2

-3 29,99

30,01

(f ile p.36.pl4; x-v ar t) c:A

30,03

30,05

-B

Fig. 5.193 Zoom at current in switch 3 of Problem P.5-37

30,07

362

5 Transients in First Order Circuits

P.5-38 Find the equivalent inductance of the configuration shown in Fig. 5.194. Answer: Leq ¼ 84½H  Fig. 5.194 Electric circuit of Problem P.5-38

20 H

* ^ |

4H | v *

^ | | | -- 1 H -----> -----> |-- 2 H | v

30 H *

40 H

P.5-39 Find the equivalent inductance seen from terminals a-a' in the circuital structure of Fig. 5.195. Answer: Leq ¼ L1 L2 þ ðLL12þþLM2 ÞM3M

2

Fig. 5.195 Electric circuit of Problem P.5-39

a

*

--------- M ---------v| v| L

* L2

1 M

a'

P.5-40 Find the equivalent inductance of the configuration shown in Fig. 5.196. Answer: Leq ¼ 52 L  2ML

2

*

L ^ ^ | | | | M -- -- M ------> L | -- -- M------> | * | | v v L

*

Fig. 5.196 Electric circuit of Problem P.5-40

L

5.7 Proposed Problems

363

P.5-41 Find the currents i1 ðtÞ and i2 ðtÞ, for t [ 0, in the circuit of Fig. 5.197, knowing that the coupling coefficient k is unitary and that: 8 for t  0

> < Et i 1 ðt Þ ¼ R 1 þ > > :E þ R1

i 2 ðt Þ ¼

  1 ½A   t 1 ET1 T1 þ T2 T1 þ T2 ½A R1 1   ET1 R1

8 0½A  > > < qffiffiffiffiffiffiffi T1 T2



t T1 þ T2

for t  1½s: for t  0½s



t T1 þ T2

for t  0½s for 0  t  1½s

 1 E½A for 0  t  1½s R1 R2  qffiffiffiffiffiffiffi  > 1 t > : T1 T2 1  T1 þ T2 ET1 þ T2 ½A for t  1½s: R1 R2

where, T1 ¼

L1 L2 ; T2 ¼ : R1 R2

+

---|> i ( t ) [A] 1

e ( t ) [V]

R Ohm 1

---|> i 2 ( t ) [A] * L [H] 1

M [H]

L2 [H] *

Fig. 5.197 Electric circuit of Problem P.5-41

R

2 Ohm

364

5 Transients in First Order Circuits

P.5-42 The circuit shown in Fig. 5.198 is a cutter, which has the function of controlling the average current through resistor R. For this, the switch oscillates periodically, starting from t ¼ 0, between positions A and B, at intervals equal to T ¼ L=R seconds. After a large number of cycles, the current iðtÞ tends towards a periodic behavior, in which it oscillates between two levels: I1 ¼ imin and I2 ¼ imax . Determine the general expression of the current iðtÞ, for t [ 0, as well as the values of the current levels I1 and I2 . Answer: ( " # ) k X E 1 þ ð1Þk j t=T i ðt Þ ¼  ðÞ  ; kT  t  ðk þ 1ÞT; k ¼ 0; 1; 2; . . . R 2 j¼0 I1 ¼

1 E E : ¼ 0:269 : þ1 R R

I2 ¼

 E E : ¼ 0:731 : þ1 R R

Fig. 5.198 Electric circuit of Problem P.5-42 +

A

E

^ v| | k

R

B

L i(t) i ( t ) 1

10 Ohm

t=0

BI

--|> i 0 ( t )

10 Ohm U

+

C

100 V

100 mH

100 mH

^ |e ( t ) 0

Fig. 6.6 Electric circuit of Example 6.6 for the numerical solution by the ATPDraw program 100

100

[V]

[V]

78

80

56

60

34

40

12

20

-10 0,00

0,03

0,06

(f ile Ex4.6.pl4; x-v ar t) v :C

-

0,09

0,12

[s]

0,15

0

v :A

Fig. 6.7 Voltage e0 ðtÞ, source voltage

Example 6.7 Find the initial conditions e0 ð0 þ Þ, e00 ð0 þ Þ and e000 ð0 þ Þ in the circuit of Fig. 6.9, knowing that it is totally de-energized at t ¼ 0 , and that the source voltage is eðtÞ ¼ E cosðxt þ hÞU1 ðtÞ: Solution 1 eðtÞ, and the nodal equations of the circuit are: If eðtÞ ¼ L dtd iðtÞ ¼ LDiðtÞ; then iðtÞ ¼ LD e e e1 e 1 1 0 Node 1: R þ LD e1 þ R ¼ 0 1 1 Node 0: e0 e R þ LD e0 ¼ 0: Multiplying both equations by RLD and factoring, it follows that: ð2LD þ RÞe1  LDe0 ¼ LDe LDe1 þ ðLD þ RÞe0 ¼ 0:

378

6 Transients in Circuits of Any Order

10 [A] 8

6

4

2

0 0,00

0,03

0,06

(file Ex4.6.pl4; x-var t)

c:B

-C

0,09

c:A

0,12

[s]

0,15

-B

Fig. 6.8 Current i0 ðtÞ, Current i1 ðtÞ

Fig. 6.9 Electric circuit of Example 6.7

e1( t )

e(t)

R

+

R

e0 ( t )

e(t)

L

L

Therefore,

2LD þ R

LD




2LD þ R LD

e0 ðtÞ ¼


LD þ R LD


LDeðtÞ

0


So, the differential equation for e0 ðtÞ is: 

 L2 D2 þ 3RLD þ R2 e0 ðtÞ ¼ L2 e00 ðtÞ

Or, e000 ðtÞ þ

3R 0 R2 e0 ðtÞ þ 2 e0 ðtÞ ¼ e00 ðtÞ: L L

eðtÞ ¼ E cosðxt þ hÞU1 ðtÞ:

^ | e0 ( t )

6.2 Circuits Initially Deenergized

379

e0 ðtÞ ¼ E cos hU0 ðtÞ  xE sinðxt þ hÞU1 ðtÞ: e00 ðtÞ ¼ E cos hU1 ðtÞ  xE sin hU0 ðtÞ  x2 E cosðxt þ hÞU1 ðtÞ: Using only the first term of the series expansion of singular functions of the last term of the expression of e00 ðtÞ, the differential equation in e0 ðtÞ becomes: e000 ðtÞ þ

3R 0 R2 e0 ðtÞ þ 2 e0 ðtÞ¼E _ cos hU1 ðtÞ  xE sin hU0 ðtÞ  x2 E cos hU1 ðtÞ: L L

The balance of the two members of this equation allows us to write that: : e0 ðtÞ¼C0 U1 ðtÞ: : e00 ðtÞ¼C0 U0 ðtÞ þ C1 U1 ðtÞ: : e000 ðtÞ¼C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U1 ðtÞ: Therefore   3R 3R R2 C0 U0 ðtÞ þ C2 þ C1 þ 2 C0 U1 ðtÞ C0 U1 ðtÞ þ C1 þ L L L : 2 ¼E cos hU1 ðtÞ  xE sin hU0 ðtÞ  x E cos hU1 ðtÞ: So, C0 ¼ E cos h: 3R 3RE E cos h ¼ xE sin h; C1 ¼ xE sin h  cos h: L L  3R 3RE R2 xE sin h  cos h þ 2 E cos h ¼ x2 E cos h; C2 þ L L L C1 þ

C2 ¼

3xRE 8R2  x2 L2 sin h þ E cos h: L L2

Since all initial conditions are null at t ¼ 0 , finally, e0 ð0 þ Þ ¼ E cos h: e00 ð0 þ Þ ¼ xE sin h  e000 ð0 þ Þ ¼

3RE cos h: L

3xRE 8R2  x2 L2 sin h þ E cos h: L L2

380

6 Transients in Circuits of Any Order

For h ¼ 0, that is, eðtÞ ¼ E cos xtU1 ðtÞ, e0 ð0 þ Þ ¼ E: e00 ð0 þ Þ ¼ 

3RE : L

and e000 ð0 þ Þ ¼

8R2  x2 L2 E: L2

Example 6.8 Find the impulse response hðtÞ of the circuit in Fig. 6.9. Solution From Example 6.7, the differential equation at e0 ðtÞ is: e000 ðtÞ þ

3R 0 R2 e0 ðtÞ þ 2 e0 ðtÞ ¼ e00 ðtÞ: L L

When eðtÞ ¼ U0 ðtÞ, e0 ðtÞ ¼ hðtÞ and the differential equation becomes: h00 ðtÞ þ

3R 0 R2 h ðtÞ þ 2 hðtÞ ¼ U2 ðtÞ; with hð0 Þ ¼ h0 ð0 Þ ¼ 0: L L

For t > 0,  3R R2 2 D þ 2 hð t Þ ¼ 0 D þ L L The characteristic roots are:r1 ¼ 0:382R=L and r2 ¼ 2:618R=L: So, hðtÞ ¼ k1 0:382Rt=L þ k2 2:618Rt=L R R h0 ðtÞ ¼ 0:382 k1 0:382Rt=L  2:618 k2 2:618Rt=L L L The balance of Eq. (6.5), at the time t ¼ 0, allows to write that: : hðtÞ¼C0 U0 ðtÞ þ C1 U1 ðtÞ: : h0 ðtÞ¼C0 U1 ðtÞ þ C1 U0 ðtÞ þ C2 U1 ðtÞ: : h00 ðtÞ¼C0 U2 ðtÞ þ C1 U1 ðtÞ þ C2 U0 ðtÞ þ C3 U1 ðtÞ:

ð6:5Þ

6.2 Circuits Initially Deenergized

381

Substituting in Eq. (6.5), it comes that:     R 2 C0 3RC1 0 C0 U2 ðtÞ þ C1 þ 3RC ð t Þ þ C þ þ U 0 ðt Þ U 2 1 2 L L L   : 3RC2 R2 C1 þ C3 þ L þ L2 U1 ðtÞ¼U2 ðtÞ þ 0:U1 ðtÞ þ 0:U0 ðtÞ þ 0:U1 ðtÞ: C0 ¼ 1 C1 þ C2 þ

3RC0 3R ¼ 0; C1 ¼  L L

3RC1 R2 C0 8R2 þ 2 ¼ 0; C2 ¼ 2 : L L L

0 8R Therefore, hð0 þ Þ ¼ C1 ¼  3R L and h ð0 þ Þ ¼ C2 ¼ L2 . So, 2



3R ¼ k1 þ k2 L

8R2 R R ¼ 0:382 k1  2:618 k2 : L L L2 Solving this system of algebraic equations results that k1 ¼ 0:065 R=L and k2 ¼ 3:065 R=L Finally,  R 0:382Rt R 2:618Rt L L hðtÞ ¼ U0 ðtÞ þ 0:065   3:065  U1 ðtÞ: L L Example 6.9 Determine the initial conditions hð0 þ Þ, h0 ð0 þ Þ and h00 ð0 þ Þ in the circuit shown in Fig. 6.10. Solution The circuit is the same as in Fig. 6.11. Just replace the part to the left of the capacitance with the Thévenin equivalent to confirm. The loop (mesh) equations are: 1 Loop 1: i1 þ 4D ði 1  i 0 Þ ¼ e 1 Loop 0: 4D ði0  i1 Þ þ 2Di0 þ i0 ¼ 0 Or, ð4D þ 1Þi1  i0 ¼ 4De i1 þ ð8D2 þ 4D þ 1Þi0 ¼ 0 Solving the system of equations for i0 ðtÞ, the differential equation results

382

6 Transients in Circuits of Any Order

1 Ohm +

1 Ohm

0.5 Ohm

e(t)

_| 2 H V i ( t ) =? 0

4F

+

Fig. 6.10 Electric circuit of Example 6.9

e(t)

1 Ohm -----| i ( t ) __| 4 F 0, i ðt Þ ¼ 


2

1  x=x0 xL

Ek sinxt 

E
2 sinxt xL 2  x=x0


2

1  x=x0 E E ¼
2 sinxt 
2 sinxt x x xL 2  =x0 xL 2  =x0 i ðt Þ ¼ 

E sinxt; para t  0: xL

ð7:8Þ

Obviously, Eqs. 7.7 and 7.8 are referred to at time t ¼ 0 of the closing of switch k. To refer to the (original) time t ¼ 0 of the voltage source eðtÞ, it is necessary to correct them. For this, the xt phase must be replaced by xt  p, that is, t must be replaced by t  p=x. Thus, from Eqs. 7.7 and 7.8, it follows that i k ðt Þ ¼ 

CE E sinðxt  pÞ:U1 ðt  p=xÞ:
2 U0 ðt  p=xÞ  xL 2  x=x0 i ðt Þ ¼ 

E sinðxt  pÞ; xL

for t  p=x  0:

CE E sinxt:U1 ðt  p=xÞ:
2 U0 ðt  p=xÞ þ xL 2  x=x0

i k ðt Þ ¼ 

i ðt Þ ¼

E sinxt; xL

for t  p=x:

ð7:9Þ

ð7:10Þ

Summing up, 8 0; > >

> :E sinxt; xL 8
2

> x > 1 = > x0 E > <
2 sinxt; iðtÞ ¼ x > xL 2 =x0 > > > :E xL sinxt;

for 0  t\p=x for t ¼ p=x

ð7:11Þ

for t [ p=x:

for 0  t  p=x for t  p=x:

ð7:12Þ

598

7 Switching Transients Using Injection of Sources

Needless to say, Eqs. 7.11 and 7.12 refer to the closing of switch k at time t ¼ p=x; and are related to time t ¼ 0 of the voltage source eðtÞ. Entering the numerical values, Eqs. 7.11 and 7.12 become: 8 for 0  t\8:33 ½ms < 0; ik ðtÞ ¼ 5:382U 0 ðt  8:33Þ ½A; for t ¼ 8:33 ½ms ð7:13Þ : 26:526sinð376:99tÞ ½A; for t [ 8:33 ½ms:  iðtÞ ¼

12:248sinð376:99tÞ ½A; for 0  t  8:33 ½ms 26:526sinð376:99tÞ ½A; for t  8:33 ½ms:

ð7:14Þ

(b) The circuit for the numerical solution by the ATPDraw program is shown in Fig. 7.25. Figure 7.26 shows the source and switch voltages. From the source, eðtÞ ¼ E cos xt ¼ 1; 000 cos 376:99t ½V;

for t [ 0:

Fig. 7.25 Electric circuit of Example 7.7 for the numerical solution by the ATPDraw program

1000 [V] 750 500 250 0 -250 -500 -750 -1000

0

(f ile Ex5.7.pl4; x-v ar t) v :A

10

v :B

-

Fig. 7.26 Voltage eðtÞ voltage ek ðtÞ

20

30

40

[ms]

50

7.2 Method of Voltage Source Injection

599

The voltage on the capacitance, which is the same as on the switch, ek ð t Þ ¼

E
2 cos xt ¼ 538:248 cosð376:99tÞ ½V; 2  x=x0 ek ðtÞ ¼ 0;

for 0\t\8:33 ½ms:

for t [ 8:33 ½ms:

Figure 7.27 shows the waveform of current iðtÞ passing through the source, which is given analytically by Eq. 7.14. As given by the expressions in Eq. 7.13, there is an impulse in the current circulating through switch k, at the moment of its closing at t0 ¼ 8:33 ½ms, which explains the numerical oscillation in the current waveform shown in Fig. 7.28.

30 [A] 20

10

0

-10

-20

-30

0

10

(f ile Ex5.7.pl4; x-v ar t) c:A

-B

20

30

40

[ms]

50

20

30

40

[ms]

50

Fig. 7.27 Current iðtÞ

12 [kA] 8

4

0

-4

-8

-12

0

10

(f ile Ex5.7.pl4; x-v ar t) c:B

Fig. 7.28 Current ik ðtÞ

-

600

7 Switching Transients Using Injection of Sources

Example 7.8 The circuit in Fig. 7.29 was operating in a steady-state when switch k was closed at the instant taken as t ¼ 0. (a) Knowing that at the moment of switching i ¼ 0, with i0 ðtÞ\0, find the analytical expression of the current ik ðtÞ, flowing through the switch k, for t [ 0. (b) Using the ATPDraw program, obtain the numerical solution for E ¼ 100 ½V; f ¼ 60 ½Hz; L ¼ 10 ½mH; K ¼ 3 and C ¼ 100 ½lF.

L

k

ik ( t ) --------/>

t=0

i(t) e ( t ) = E coswt

C

KL

Fig. 7.29 Electric circuit for Example 7.8

Solution (a) The steady-state solution, with switch k open, comes from the circuit in the simple frequency domain shown in Fig. 7.30.

jwL

Fig. 7.30 Electric circuit for Example 7.8, for steady-state solution

k

. E

+

-------/> . I 1 / jwC

^ . | | E

k

jwKL

pffiffiffiffiffiffi E_ E_ jxCE\0 xCE\90 I_ ¼ ¼ ¼ ¼ ; where x ¼ 1= LC :
 0 2 Z jxL þ 1=jxC 1  x2 LC 1  x=x0 1 _ :I ¼ E_ k ¼ jxC

E\0
2 : 1  x=x0

7.2 Method of Voltage Source Injection

601

Therefore, iðtÞ ¼

xCE xCE 
2 cosðxt þ 90 Þ ¼ 
2 sinxt and 1  x=x0 1  x=x0 i0 ðtÞ ¼ 

ek ð t Þ ¼




E

1  x=x0

2 cos xt ¼

x2 CE
2 cos xt: 1  x=x0

x20 E x20 E cos xt ¼ E cos xt; E ¼ : k k x20  x2 x20  x2

For t ¼ 0, i ¼ 0 and i0 ðtÞ\0, assuming x0 [ x, which usually happens, meaning that the closing time of switch k coincides with the instant t ¼ 0 (reference) of the voltage source eðtÞ. Thus, there is no need for any change in the expression of voltage ek ðtÞ at the terminals of switch k, operating in open circuit, and that ek ðtÞ is in phase with eðtÞ. Then, the time domain circuit, for t [ 0, containing only the injected voltage source F2 , is the one shown in Fig. 7.31. Transposing the circuit of Fig. 7.31 to the domain of the complex frequency, through the Laplace Transform, results the equivalent circuit of Fig. 7.32. Ek ðsÞ ¼

sEk : s2 þ x2

1 sL: sC sL ¼ sKL þ 1 LCs2 þ 1 sL þ sC h 2 i Ks 2 þ ðK þ 1Þ 2 3 sL x0 KL Cs þ ðK þ 1ÞLs ¼ ¼ :  LCs2 þ 1 s2 2 þ 1 x0

Z ðsÞ ¼ sKL þ

F2

L ------/> i(t)

+

ik (t) ----------/>

------------------------> e k ( t ) = E coswt.U ( t ) k -1 C

KL

Fig. 7.31 Electric circuit of Example 7.8 for calculating the exclusive effects of the (injected) source F 2

602

7 Switching Transients Using Injection of Sources sL

F2

------/> I(s )

------------> E (s ) k

+

Ik ( s ) ----------/>

1/sC

sKL

Fig. 7.32 Transformed circuit of Example 7.8, in the domain of the complex frequency

Or,   KLs s2 þ K Kþ 1 x20 Z ðsÞ ¼ : s2 þ x20 Ik ðsÞ ¼ ¼

E k ðsÞ sEk s2 þ x20   ¼ 2 : 2 2 Z ðsÞ s þ x KLs s þ K Kþ 1 x20 Ek : KL

s2 þ x20
qffiffiffiffiffiffiffiffiffiffiffi 2 : ðs2 þ x2 Þ s2 þ 1 þ K1 x0

On the other hand, considering that uþA A1 A2 ¼ þ ; ðu þ BÞðu þ CÞ u þ B u þ C uþA u!B u þ C

where A1 ¼ lim uþA u!C u þ B

A2 ¼ lim

AB ¼ CB

¼  AC CB ;

results the identity:   uþA 1 AB AC ¼  ðu þ BÞðu þ CÞ C  B u þ B u þ C

ð7:15Þ

Considering, then, that u ¼ s2 , A ¼ x20 , B ¼ x2 and C ¼ ð1 þ 1=K Þx20 , Eq. 7.15 allows to write that 2 3   2 2 1 2 2 x x  1 þ Ek 1 x  x 6 0 0 0 7 K  Ik ðsÞ ¼ :  4
qffiffiffiffiffiffiffiffiffiffiffi 2 5: KL 1 þ K1 x20  x2 s2 þ x2 s2 þ 1 þ K1 x0

7.2 Method of Voltage Source Injection

603

Considering also that, from the steady-state solution, x2 E

0 Ek ¼ x2 x 2 , then,

2

0

Ik ðsÞ ¼

x2 E  20 KL x0 

x2

:

1þ

1 K

1  2 x0 

2 2 6x0  x : 4 x2 s2 þ x2

þ

2 ¼

x2 E  20 : L x0  x2 ðK

2 2 1 6x  x :4 20 þ 2 2 2 s þx þ 1Þx0  Kx

¼

E x h i: 2 x2 s þ x2 xL ðK þ 1Þ  K x2

x20 K

3

7
qffiffiffiffiffiffiffiffiffiffiffi 2 5 s2 þ 1 þ K1 x0 3 x20 K

7
qffiffiffiffiffiffiffiffiffiffiffi 2 5 s2 þ 1 þ K1 x0

qffiffiffiffiffiffiffiffiffiffiffi 1 þ K1 x0

þ qffiffiffiffiffiffiffiffiffiffiffi : q
ffiffiffiffiffiffiffiffiffiffiffi 2 :
2 h i 2 2þ s 1 þ K1 x0 : ð K þ 1Þ  K x 1 þ K1 x0 KL: 1  xx0 x2 0

E

0

Finally, ik ðtÞ ¼

E
2 sinðxtÞ xL ðK þ 1Þ  K x=x0 E þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2

ðK þ 1Þ  K x=x0 K ðK þ 1Þx0 L 1  x=x0 rffiffiffiffiffiffiffiffiffiffiffiffi ! 1 sin 1 þ x0 t ; K for t  0:

For the given numerical values,

1 1 rad ¼ ¼ 16:667 ½ms; x ¼ 2pf ¼ 120p ¼ 376:9911 : f 60 s



1 rad x0 rad 1 x0 ¼ pffiffiffiffiffiffi ¼ 1000 ¼ 159:1549 ; f0 ¼ ; T0 ¼ ¼ 6:2832 ½ms: s s f0 2p LC T¼

ik ðtÞ ¼ 7:4226sinð376:9911tÞ þ 0:9416sinð1; 154:7005tÞ; t  0: Then,

604

7 Switching Transients Using Injection of Sources

ik ð0Þ ¼ 0 ½A ik ð0:005Þ ¼ 7:059  0:459 ¼ 6:600 ½A ik ð0:010Þ ¼ 4:363  0:802 ¼ 5:165 ½A ik ð0:008333Þ ¼ iðT=2Þ ¼ 0:002  0:186 ¼ 0:188 ½A ik ð0:020Þ ¼ 7:059  0:840 ¼ 6:219 ½A (b) The circuit for the numerical solution is shown in Fig. 7.33. Ek ¼

x20 E 106  102 108 ¼ 116:5667 ½V: ¼ ¼ x20  x2 106  ð120pÞ2 857;877:6966

The results are shown in Fig. 7.34, with Δt = 1 [ls]. With the cursor, you can see that: Ik (0.005) = 6.598 [A] Ik (0.010) = −5.167 [A] Ik (0.008334) = −0.189 [A] Ik (0.020) = 6.218 [A]. If the analytical solution in item a had not been requested, the steady-state digital solution would be necessary to determine the instant when the current satisfies condition iðtÞ ¼ 0, with i0 ðtÞ\0, as well such as the voltage curve at the switch terminals, ek ðtÞ, in order to establish the future voltage of the source F2 , to be injected. And the circuit for this is shown in Fig. 7.35. In Fig. 7.36, the curves of ek ðtÞ and iðtÞ are plotted. And it clearly shows that the condition iðtÞ ¼ 0, with i0 ðtÞ\0 occurs exactly at the original t ¼ 0, therefore coinciding with the time

V

ik ( t ) ----------/> A I

B

30 mH 10 mH

100 uF

F2

- 116,5667 cos 376,9911t V

Fig. 7.33 Electric circuit of Example 7.8 for the numerical solution by the ATPDraw program

7.2 Method of Voltage Source Injection

605

120

8 [A]

[V]

6

80

4 40

2

0

0 -2

-40

-4 -80

-120

-6

0

10

(f ile Ex5.9.pl4; x-v ar t) v :B

c:A

20

-B

30

40

[ms]

50

-8

Fig. 7.34 Source voltage F 2 and current ik ðtÞ in the switch

t ¼ 0 of the closing of the switch k, no thus requiring any phase correction in the quantities of interest. From the same Fig. 7.36, the maximum values of permanent regime of 116:57 ½V for ek ðtÞ and 4:3945 ½A for iðtÞ are also obtained with the cursor. These being exactly the values that are provided by Ek ¼

x20 E and I ¼  x2

x20

xCE
2 : 1  x=x0

Substituting the numerical values, results in Ek ¼ 116:5667 ½V, I ¼ 4:3945 ½A.

F

I

K

-----|> i ( t ) 10 mH

U

e(t)=Ecoswt E=100 V

100 uF

^| |e ( t ) k

30 mH

f=60 Hz StartA=-1

Fig. 7.35 Electric circuit for the calculation of steady-state digital solution from Example 7.8

606

7 Switching Transients Using Injection of Sources 120

5,00 [A]

[V]

3,75

80

2,50 40

1,25

0

0,00 -1,25

-40

-2,50 -80

-120

-3,75

0

10

(f ile Ex-5.8.pl4; x-v ar t) v :K

-

c:F

-K

20

30

40

[ms]

50

-5,00

Fig. 7.36 Steady-state digital solution for switch voltage ek ðtÞ and inductor current iðtÞ

7.3

Current Source Injection Method

It is a technique that can be applied to any linear circuit and is based on the theorems of substitution and superposition. In a way, it is the dual procedure of the voltage source injection method, and its main objective is to replace the process of opening a switch, at any time t ¼ t0 , with an equivalent assembly, that is, simulating the opening of a switch at time t0 . It basically consists of the injection of a current source equal to and contrary to that which would continue to pass through the switch, after the t0 instant, if the switch was not opened at that instant. It can be applied to circuits already operating in a steady-state, as well as to those still in transitory phase. It is described in the following five steps. 1. Referring to Fig. 7.37, the current ik ðtÞ circulating through the switch k and the other voltages and currents in the elements inside the networks N1 and N2 , which are of interest, are initially calculated. Assuming that ik ðtÞ ¼ Icosðxt þ hÞ, then, ek ðtÞ ¼ 0; pk ðtÞ ¼ ek ðtÞ:ik ðtÞ ¼ 0: 2. Then, the switch k is replaced by an ideal current source F1 , whose current i1 ðtÞ has the same value as the current that circulated and that would continue to circulate through switch k, if it were not opened at time t ¼ t0 . As the switch is ideal, that is, without voltage at its terminals when closed, then the substitutive

Fig. 7.37 Circuit before opening the switch at time t ¼ t0

7.3 Current Source Injection Method

607

current source F1 always has zero voltage at its terminals, whenever acting alone between terminals a and b, as shown in Fig. 7.38. i1 ðtÞ ¼ ik ðtÞ ¼ Icosðxt þ hÞ ¼ Icosðxt þ hÞ½U 1 ðt þ t0 Þ þ U 1 ðt  t0 Þ; ek ðtÞ ¼ 0; pF1 ðtÞ ¼ pk ðtÞ ¼ 0: 3. To portray the opening of switch k at time t ¼ t0 , a second ideal current source F2 , in the opposite direction, is injected in parallel with the source F1 , from the same time t ¼ t0 . Its value is equal to that of the future current through the switch, that is, the current that would continue to circulate through the switch if it had not been opened at time t ¼ t0 , nor after it. Thus, the total current ik ðtÞ, circulating through the branch containing the switch k, will be zero, after the injection of the current source F2 , that is, for t [ t0 , which is equivalent to the opening of the switch, as illustrated in Fig. 7.39. In Fig. 7.39, i1 ðtÞ ¼ Icosðxt þ hÞ; i2 ðtÞ ¼ Icosðxt þ hÞ:U 1 ðt  t0 Þ; pk ðtÞ ¼ 0: In node b: ik ðtÞ ¼ i1 ðtÞ  i2 ðtÞ. In detail, we have, then, that: For t\t0 , switch k closed, i1 ðtÞ ¼ I cosðxt þ hÞ; i2 ðtÞ ¼ 0 (F2 is an open circuit) and ik ðtÞ ¼ I cosðxt þ hÞ. For t [ t0 , switch k open, i1 ðtÞ ¼ i2 ðtÞ ¼ I cosðxt þ hÞ and ik ðtÞ ¼ 0: Since ik ðtÞ ¼ i1 ðtÞ  i2 ðtÞ; ek ðtÞ:ik ðtÞ ¼ ek ðtÞ:i1 ðtÞ  ek ðtÞ:i2 ðtÞ. So, the instantaneous power on the switch is pF1 ðtÞ þ pF2 ¼ 0; showing that the power released by source F2 is consumed by source F1 . In detail: For t\t0 , switch k closed, ik ðtÞ ¼ i1 ðtÞ 6¼ 0; ek ðtÞ ¼ 0. Therefore, pk ðtÞ ¼ 0. For t [ t0 , switch k open, ik ðtÞ ¼ 0, ek ðtÞ 6¼ 0.

Fig. 7.38 Circuit in Fig. 7.37 with switch k closed, replaced by the current source F 1

608

7 Switching Transients Using Injection of Sources

Fig. 7.39 Circuit after opening the switch k, simulated by injecting the current source F 2

The voltage ek ðtÞ is a consequence of the exclusive action of the current source F2 , for t [ t0 , which, due to the parallelism of the two current sources, also appears at the terminals of the source F1 . Therefore, pk ðtÞ ¼ 0. 4. The calculation of voltages and currents over all elements of the circuit, on which there is interest for t [ t0 , that is, with the switch k open, is done using the superposition theorem, a property of linear systems. As the power at source F1 is zero in the absence of source F2 , that is, with source F2 at rest—open, because in this condition ek ðtÞ ¼ 0 (Fig. 7.38), it does not establish any voltage or current on the passive elements of the circuit inside the N1 and N2 networks. Therefore, the effect of the current source F1 on the circuit is always null. On the other hand, the voltages and currents produced by the sources internal to the N1 and N2 networks, with the source F2 at rest, have already been previously calculated in the initial phase of the process (Fig. 7.37). Thus, it remains to calculate only the voltages and currents produced by the current source F2 , acting alone for t [ t0 , that is, with the source F1 at rest and the networks N1 and N2 reduced to the passive form, that is, with all their independent voltage sources short-circuited and current sources opened. In this phase, the voltage ek ðtÞ at the terminals of the current source F2 is also calculated, as shown in Fig. 7.40, and that is actually the voltage itself at the terminals of switch k, after it was opened at time t ¼ t0 .

Fig. 7.40 Circuit reduced to passive for the calculation of voltages and currents produced exclusively by the current source F2

7.3 Current Source Injection Method

609

In Fig. 7.40, i2 ðtÞ ¼ Icosðxt þ hÞ:U 1 ðt  t0 Þ; ek ðtÞ 6¼ 0; ik ðtÞ ¼ i2 ðtÞ ¼ Icosðxt þ hÞ: For t [ t0 , pF2 ¼ ek ðtÞ:i2 ðtÞ 6¼ 0: 5. Finally, the resulting values of voltage and current in the circuit elements, after opening the switch k at time t ¼ t0 , are obtained by superposition of the values found with the isolated action of the current source F2 (Fig. 7.40) with those obtained before opening the switch k (Fig. 7.37). It is necessary to point out that if the opening time of the switch k; t ¼ t0 , is taken as a new time t ¼ 0, that is, if the time will be computed from the time of opening of the switch k, then the analytical expressions of all voltages and currents in the circuit, both those that were given and those that were calculated before the opening of switch k (Fig. 7.37), must undergo corrections in their initial phases, to be related to this new origin of time. And this is done simply by adding the value xt0 to their respective phases, that is, replacing t with t þ t0 in the corresponding expressions. So, for example, i2 ðtÞ would be i2 ðtÞ ¼ I cos½xðt þ t0 Þ þ hÞU1 ½ðt þ t0 Þ  t0  ¼ I cosðxt þ h þ xt0 Þ:U1 ðtÞ: Example 7.9, although it does not refer to switching transients, because the circuit contains neither capacitance nor inductance, nor switch opening, it is quite useful for a better understanding of the current source injection method in solving circuit electric problems. Example 7.9 Use the substitution and superposition theorems to determine the variation that current I undergoes in the circuit of Fig. 7.41, when the resistance R undergoes an increase DR, that is, it changes from R to R þ DR. Determine, also, the new current I and the new voltage er , in the resistance r.

er

+

r R

E Solution The problem can be replaced as shown in Fig. 7.42, where the resistive addition DR, of resistance R, is placed in parallel with a switch k, which changes from the closed to the open position, at t ¼ 0.

610

7 Switching Transients Using Injection of Sources

Fig. 7.42 Adapted electric circuit for Example 7.9

1. The current I flowing through switch k, at whose terminals the voltage is zero, is I¼

E : rþR

And the voltage at resistance r, by voltage divider, is er ¼

rE : rþR

2. Switch k, closed, can be replaced by an ideal current source F1 , of value I and zero voltage, as shown in Fig. 7.43. 3. To simulate the opening of switch k, at time t ¼ 0, an ideal current source F2 is injected in parallel with F1 , with the same value as this one, but in the opposite direction, as shown in Fig. 7.44, valid for t [ 0. 4. The solution of the circuit of Fig. 7.44 is made using the superposition theorem, calculating, initially, the simultaneous effects of the independent voltage source E and the current source F1 and, later, the effects of the source F2 . It turns out that this first calculation was already done in phase 1, as the source F1 is nothing more than the replacement of the closed k switch (Fig. 7.42). So, it

Fig. 7.43 Electric circuit of Example 7.9 with switch k, closed, replaced by an ideal current source F 1

7.3 Current Source Injection Method

611

Fig. 7.44 Electric circuit of Example 7.9 with an ideal current source F 2 injected in parallel with F 1

is only necessary to calculate the exclusive effects of the F2 source, from the circuit shown in Fig. 7.45. From the circuit of Fig. 7.45, by current divider, we have to DI ¼ 

DR: r þE R DR:I DR:E ¼ : ¼ r þ R þ DR ðr þ RÞðr þ R þ DRÞ r þ R þ DR

So, Der ¼ r:DI ¼ 

r:DR:E : ðr þ RÞðr þ R þ DRÞ

5. Thus, the new values for current I and voltage er , by superposition, are: I¼

E DR:E E  ¼ : r þ R ðr þ RÞðr þ R þ DRÞ r þ R þ DR

er ¼

rE rDR:E rE  ¼ : r þ R ðr þ RÞðr þ R þ DRÞ r þ R þ DR

Fig. 7.45 Electric circuit of Example 7.9 for calculating the exclusive effects of the (injected) source F 2

612

7 Switching Transients Using Injection of Sources

Example 7.10 Switch k in the circuit of Fig. 7.46 has long been closed. If it is opened at t ¼ 0, determine the voltage ek ðtÞ at its terminals, for t [ 0. Fig. 7.46 Electric circuit for Example 7.10

k

t=0

R

+

i k( t )

2R

E

2R

Fig. 7.47 Electric circuit of Example 7.10 for steady-state solution

e k( t ) = ?

+

F R 2

| i (t) | L 2R | |---------

2R

L

Fig. 7.48 Electric circuit of Example 7.10 for calculating the exclusive effects of the (injected) source F 2

Therefore, ek ð t Þ ¼

  E E þ 4Rt=L :U1 ðtÞ: 2

Se a fonte independente de tensão E não for considerada no cálculo da tensão ek ðtÞ, representada por um curto-circuito na Figura 7.48, então não é necessário considerar também os seus efeitos sobre as resistências R e 2R, e nem aplicar superposição, ou seja: If the voltage independent source E is not considered in the calculation of the voltage ek ðtÞ, represented by a short-circuit in Fig. 7.48, then it is not necessary to consider its effects on resistors R and 2R, nor to apply superposition, that is: 

E E E
4Rt=L þ  ek ðtÞ ¼ eR ðtÞ þ e2R ðtÞ ¼ R: :U1 ðtÞ þ 2R:  1 :U1 ðtÞ 2R 2R 4R   E 4Rt=L ¼ Eþ  :U1 ðtÞ: 2 Example 7.11 The circuit in Fig. 7.49 has been operating in a steady-state for a long time. If the switch k is opened at t ¼ 0, find the expression of the voltage ek ðtÞ at its terminals, for t [ 0.

614

7 Switching Transients Using Injection of Sources

k

t=0

ik ( t ) ---------/> R

L R

+