Circuit Theory and Networks Fourth Edition WBUT-2014 9789339214265, 9339214269

Table of contents :
Title
Contents
1 Introduction to Different Types of Systems
2 Introduction to Circuit-Theory Concepts
3 Magnetically Coupled Circuits
4 Network Toplogy
5 Network Theorems
6 Laplace Transform and its Application
7 Two -Port Network
8 Fourier Series and Fourier Transform
9 Filter Circuits
10 Reasonance
Solutions to Question Paper (WBUT)

Citation preview

Circuit Theory and Networks Fourth Edition WBUT-2014

About the Authors S P Ghosh did his BE (Hon.) in Electrical Engineering from Regional Engineering College (presently, National Institute of Technology), Durgapur, and received a Master of Electrical Engineering degree from Jadavpur University, Kolkata, with specialization in High Voltage Engineering. He has over 12 years of teaching experience. Presently, he is Head of the Department of Electrical Engineering at College of Engineering and Management, Kolaghat. He is also pursuing a PhD at NIT, Agartala. Prof Ghosh has authored six different books on Electromagnetic Field Theory and Circuit Theory. He has also presented several papers in national and international conferences. His areas of interest include Power Systems, Electrical Machines, Artificial Neural Networks, and related topics. Ajoy Kumar Chakraborty received his BEE (Hon.) from Jadavpur University, Kolkata, MTech in Power System Engineering from IIT Kharagpur and PhD (Engineering) again from Jadavpur University. Presently, he is the Dean (Planning and Development) at NIT, Agartala. He taught at College of Engineering and Management, Kolaghat, for 12 years. He also worked as a Lecturer in NIT Silchar for five years. He has served in a number of industries, namely, CESC Ltd. and the Tinplate Company of India Ltd. (a Tata Enterprise) for over fourteen years before joining this institute. He is a Fellow of the Institute of Engineers (India), Chartered Engineer, Member IET (UK) and a Life Member of ISTE. He has presented several technical papers in national and international conferences, has published his essays in reputed journals, and has visited three foreign countries to deliver technical lectures. As an enthusiastic teacher, he is actively involved in guiding MTech and PhD (Engineering) scholars. His research interests are in the field of Power System Protection, Economic Operation of Power Systems, Deregulated Power System, HVDC, and related topics.

Circuit Theory and Networks Fourth Edition WBUT-2014 S P Ghosh Head Department of Electrical Engineering College of Engineering and Management Kolaghat, West Bengal

A K Chakraborty Dean (Planning and Development) and Associate Professor Department of Electrical Engineering National Institute of Technology (NIT) Agartala, Tripura

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Circuit Theory and Networks, 4e Copyright © 2015, 2012, 2011, 2010 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-392-1426-5 ISBN (10): 93-392-1426-9 Managing Director: Kaushik Bellani Head—Higher Education (Publishing and Marketing): Vibha Mahajan Senior Publishing Manager (SEM & Tech. Ed.): Shalini Jha Assistant Sponsoring Editor: Koyel Ghosh Editorial Executive: Piyali Chatterjee Manager—Production Systems: Satinder S Baveja Assistant Manager—Editorial Services: Sohini Mukherjee Assistant Manager—Production: Anjali Razdan Assistant General Manager (Marketing)—Higher Education: Vijay Sarathi Assistant Product Manager (SEM & Tech. Ed.): Tina Jajoriya Senior Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Text-o-Graphics, B1/56, Arawali Appartment, Sector 34, Noida 201 301, and printed at Avon Printers, Plot No. 16, Main Loni Road, Jawahar Nagar, Industrial Area, Shahdara, Delhi - 110 094 Cover: A. P. Offset

Dedicated to our families Wife, Lipika and My Parents S P Ghosh Wife, Indira Daughters, Amrita and Ananya A K Chakraborty

Contents

Preface Roadmaps to the Syllabus

xi xiv

1. Introduction to Different Types of Systems 1.1 Introduction 1.1 1.2 Concepts of Signals and Systems 1.1 1.3 Different Types of Systems 1.2 1.4 Different Types of Signals 1.7 1.5 Some Satandards Signals 1.10 1.6 Singularities Signals 1.12 Solved Problems 1.14 Multiple-Choice Questions 1.18 Exercises 1.20 Short-Answer Type Questions 1.20 Answers to Multiple-Choice Questions 1.21

1.1–1.21

2. Introduction to Circuit-Theory Concepts 2.1 Introduction 2.1 2.2 Some Basic Terminologies of Electric Circuits 2.3 Different Notations 2.5 2.4 Basic Circuit Elements 2.5 2.5 Passive Circuit Elements 2.6 2.6 Types of Electrical Energy Sources 2.14 2.7 Fundamental Laws 2.16 2.8 Source Transformation 2.18 2.9 Network Analysis Techniques 2.20 2.10 Duality 2.23 2.11 Star-Delta Conversion Technique 2.26 Solved Problems 2.27 Multiple-Choice Questions 2.61 Exercises 2.78 Short-Answer Type Questions 2.83 Answers to Multiple-Choice Questions 2.84

2.1–2.84 2.1

Contents

vii

3. Magnetically Coupled Circuits 3.1 Introduction 3.1 3.2 Self-Inductance 3.1 3.3 Coupled Inductor 3.2 3.4 Mutual Inductance 3.3 3.5 Mutual Inductance between Two Coupled Inductors 3.5 3.6 Dot Convention 3.6 3.7 Coefficient of Coupling 3.7 3.8 Inductive Coupling 3.10 3.9 Linear Transformer 3.12 3.10 Determination of Equivalent T and p Circuit of Linear Transformer 3.13 3.11 Ideal Transformer 3.15 3.12 Tuned Coupled Circuits 3.18 Solved Problems 3.25 Summary 3.49 Exercises 3.50 Short-Answer Type Questions 3.53

3.1–3.54

4. Network Topology (Graph Theory) 4.1 Introduction 4.1 4.2 Graph of a Network 4.1 4.3 Terminology 4.2 4.4 Concept of Tree 4.3 4.5 Incidence Matrix [Aa ] 4.4 4.6 Number of Possible Trees of a Graph 4.6 4.7 Tie-Set Matrix and Loop Currents 4.7 4.8 Cut-Set Matrix and Node-Pair Potential 4.9 4.9 Formulation of Network Equilibrium Equations 4.13 4.10 Solution of Equilibrium Equations 4.15 Solved Problems 4.16 Multiple-Choice Questions 4.47 Exercises 4.51 Short-Answer Type Questions 4.53 Answers to Multiple-Choice Questions 4.54

4.1–4.54

5. Network Theorems 5.1 Introduction 5.1 5.2 Network Theorems 5.1 Solved Problems 5.14 Multiple-Choice Questions 5.70 Exercises 5.79 Short-Answer Type Questions 5.86 Answers to Multiple-Choice Questions

5.1–5.87

5.87

viii

Contents

6. Laplace Transform and its Applications 6.1 Introduction 6.1 6.2 Advantages of Laplace Transform Method 6.1 6.3 Definition of Laplace Transform 6.2 6.4 Basic Theorems of Laplace Transform 6.2 6.5 Laplace Transform of Some Basic Functions 6.4 6.6 Laplace Transform Table 6.8 6.7 Laplace Transform of Periodic Functions 6.10 6.8 Singularity Functions and Waveform Synthesis 6.11 6.9 Inverse Laplace Transform 6.14 6.10 Applications of Laplace Transform 6.18 6.11 Application of Laplace Transform Method to Circuit Analysis 6.20 6.12 Transient Analysis of Electric Circuits using Laplace Transform 6.21 6.13 Steps for Circuit Analysis using Laplace Transform Method 6.44 6.14 Concept of Convolution Theorem 6.44 Solved Problems 6.46 Multiple-Choice Questions 6.90 Exercises 6.110 Short-Answer Type Questions 6.116 Answers to Multiple-Choice Questions 6.117

6.1–6.117

7. Two-Port Network 7.1 Introduction 7.1 7.2 Relationships of Two-Port Variables 7.1 7.3 Conditions for Reciprocity and Symmetry 7.4 7.4 Interrelationships between Two-Port Parameters 7.7 7.5 Interconnection of Two-Port Networks 7.10 7.6 Two-Port Network Functions 7.12 Solved Problems 7.13 Multiple-Choice Questions 7.58 Exercises 7.65 Short-Answer Type Questions 7.70 Answers to Multiple-Choice Questions 7.71

7.1–7.71

8. Fourier Series and Fourier Transform Part I: Fourier Series 8.1 Introduction 8.1 8.2 Definition of Fourier Series 8.2 8.3 Dirichlet’s Conditions 8.2 8.4 Fourier Analysis 8.2 8.5 Steady-State Response of Network to Periodic Signals 8.12 Part II: Fourier Transform 8.6 Introduction 8.16 8.7 Definition of Fourier Transform 8.16

8.1–8.72

Contents

ix

8.8 Convergence of Fourier Transform 8.17 8.9 Fourier Transform of Some Functions 8.18 8.10 Properties of Fourier Transforms 8.21 8.11 Energy Density and Parseval’s Theorem 8.24 8.12 Comparison between Fourier Transform and Laplace Transform 8.26 8.13 Steps for Application of Fourier Transform to Circuit Analysis 8.26 Solved Problems 8.27 Multiple-Choice Questions 8.64 Exercises 8.67 Short-Answer Type Questions 8.70 Answers to Multiple-Choice Questions 8.72 9. Filter Circuits 9.1 Introduction 9.1 9.2 Filter 9.2 9.3 Classification of Filters 9.2 9.4 Advantages of Active Filters over Passive Filters 9.5 9.5 Application of Active Filters 9.5 9.6 Low-pass Active Filter 9.5 9.7 High-pass Active Filter 9.10 9.8 Band-Pass Active Filter 9.13 9.9 Band-Reject (Notch) Active Filter 9.21 9.10 Filter Approximation 9.24 9.11 All-Pass Active Filter 9.34 Multiple-Choice Questions 9.35 Exercises 9.38 Short-Answer Type Questions 9.39 Answers to Multiple-Choice Questions 9.39 10. Resonance 10.1 Introduction 10.1 10.2 Series Resonance or Voltage Resonance 10.1 10.3 Parallel Resonance or Current Resonance or Anti-Resonance 10.11 10.4 Relation Between Damping Ratio and Quality Factor 10.16 10.5 A More Realistic Parallel Resonant Circuit 10.17 10.6 Comparison of Series and Parallel Resonance 10.27 10.7 Universal Resonance Curve 10.27 10.8 Applications of Resonance 10.28 Solved Problems 10.30 Summary 10.53 Exercises 10.54 Questions 10.57

9.1–9.39

10.1–10.59

x

Contents

Solution of 2011 WBUT Paper

SQP.1–SQP.22

Solution of 2012 WBUT Paper

SQP.1–SQP.18

Solution of 2013 WBUT Paper

SQP.1–SQP.18

Preface Brief Introduction to the Subject Circuit Theory and Networks is a gateway course to all engineering subjects; Electrical Engineering, Electronics and Communication Engineering, Computer Science and Engineering, Information Technology, and Instrumentation Engineering in particular. Almost all engineering systems use electric circuits as components. To understand the operation of these systems, the knowledge of Circuit Theory is very essential. Also, the subject of Circuit Theory provides the background for understanding the behaviour of many other electrical and electronic devices. The present book has been written keeping this in view and aims to provide requisite information of circuit theory and networks, starting from the fundamentals. It is an acceptable fact that students do not automatically acquire conceptual understanding; the concepts must be explained and the students given a chance to grapple with them. Our presentation, based on years of teaching this course, blends conceptual understanding with analytical skills.

Objective This book has been written as per the syllabus of Circuit Theory and Networks (EE 301) of the West Bengal University of Technology (WBUT). Although there are several books on Circuit Theory, there is hardly any book meeting the entire syllabi of WBUT. This text works well in our self-paced course, where students can rely on it as their primary learning resource. Nonetheless, completeness and clarity are equally advantageous when the book is used in a more traditional classroom setting. Cognizance of the present standard of the students and the difficulties of the teachers have been given due thought. The conceptual examples and practice problems and a variety of conceptual and multiple-choice questions at the end of each chapter give students a chance to check and enhance their conceptual understanding.

Scope This book is mainly written for the third-semester students under WBUT. However, as this course mitigates a definite percentage in every competitive examination of engineering professionals, namely, IES, UPSC, GATE, etc., we have written this book to help students see that a relatively small number of basic concepts are applied to a wide variety of situations.

xii

Preface

Salient Features �

� �

�

�

Chapter organization and terminology in agreement with syllabus structure: EE-301 for EEE/EE/ PE/ICE undergraduate engineering students, Sem. 3; EC 301 for ECE undergraduate engineering students, Sem. 3; EE (EI) 301 for AEIE undergraduate engineering students, Sem. 3; and BME (EC) 301 for BME undergraduate engineering students, Sem. 3 Detailed coverage of different types of systems and networks Every theorem has been explained in the following sequence: Statement > Proof > Points to be Noted Varied pedagogy including MCQs, Solved Examples, Exercises, Short-Answer-Type Questions, and Solved Exercises to explain involved concepts: � Illustrations: 700 � MCQs: 400 � Exercises: 150 � Solved Exercises: 90 � Short-Answer-Type Questions: 105 � Solved Examples: 180 Use the APP to � Study Important topics on-the-go � Revise Quick pointers to ace the examination � Test Questions with answers that test concepts learnt

Organization This book has a total of ten chapters. Chapter 1 provides information about the basic characteristics of different types of systems. Chapter 2 deals with the basic circuit theory concepts, laws and techniques for circuit analysis. Chapter 3 introduces the new topic on magnetically coupled circuits. It deals with different models in coupled circuits and analysis of the same. Chapter 4 discusses the application of graph-theory concepts in circuit analysis. In this chapter, the application of a mathematical tool like graph theory has been presented with the help of a large number of practical examples. Chapter 5 is devoted to various network theorems necessary for simplified analysis of electrical problems. This chapter is very important for examination purposes as several questions from this chapter are asked in exams. Chapter 6 introduces a new method of circuit analysis—Laplace Transform method. Starting from the very fundamental concept of Laplace transform, its applications in various complicated circuit problems have been discussed in detail in this chapter. Chapter 7 deals with the concepts of the twoport network which has vast applications in many fields like transmission lines, filters and attenuators. Chapter 8 is divided into two parts. Part I presents the fundamentals of Fourier series and its applications for circuit analysis. Part II discusses Fourier transforms and their applications. Chapter 9 is devoted to operational amplifiers and active filters. Chapter 10 talks about analysis of various resonances and their combinations in detail.

xiii

Preface

Acknowledgements The authors are indebted to Prof. Nirmal Chatterjee, Prof. Kalyan Dutta, Prof. C K Roy, Prof. A N Sanyal of Jadavpur University, Kolkata, for their encouragement. We are grateful to Prof. J K Das, founder Director of College of Engineering and Management, Kolaghat; Prof. P B Duttagupta, IIT Kharagpur; Prof. S N Bhadra, Department of Electrical Engineering, College of Engineering and Management, Kolaghat and Ex-Professor of IIT Kharagpur, Prof. H P Bhowmik, Ex-Principal, Institute of Leather Technology; and Dr Abhinandan De, Bengal Engineering and Science University, Shibpur; for their constant inspiration and encouragement in the field of academics. We would also like to thank the following reviewers of the book who sent us their suggestions and feedback. Amitava Biswas

Academy of Technology, Adi Saptagram

Jayita Sarkar

Dr Sudhir Chandra Sur Degree Engineering College, Kolkata

Animesh Bhattacharya

Netaji Subhash Engineering College, Kolkata

Debasis Banerjee

Dr B C Roy Engineering College, Durgapur

We are also thankful to the editorial and production team of McGraw Hill Education (India) for bringing out this edition. Last but not the least, we acknowledge the support offered by our respective wives and children without whom this work would not have been successful.

Feedback Criticism and suggestions for improvement shall be gratefully acknowledged. Readers may contact A K Chakraborty at [email protected] and S P Ghosh at [email protected] S P GHOSH A K CHAKRABORTY

Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and author’s name as the subject. In case you spot piracy of this book, please do let us know.

ROADMAPS TO THE SYLLABUS

Circuit Theory and Networks The text is suitable for the following paper codes and disciplines: • EE 301 Electric Circuit Theory (for EE, EEE, PE & ICE) • EC 301 Circuit Theory and Networks (for ECE) • EE (EI) 301 Circuit Theory and Networks (for AEIE) • BME (EC) 301 Circuit Theory and Networks (for Biomedical Engineering) ROADMAP TO THE SYLLABUS (EE 301) Module 1 Introduction: Continuous and discrete; Fixed and time-varying; Linear and nonlinear; Lumped and distributed; Passive and active networks and systems; Independent and dependent sources; Step, ramp, impulse, sinusoidal, square and sawtooth signals. GO TO

Chapter 1: Introduction to Different Types of Systems

Module 2 Coupled Circuits: Magnetic coupling; Polarity of coils; Polarity of induced voltage; Concept of Self- and mutual inductances; Coefficient of coupling; Modeling of coupled circuits; Solutions of problems. GO TO

Chapter 3: Magnetically Coupled Circuits

Module 3 Laplace Transforms: Impulse; Step and sinusoidal response of RL, RC and RLC circuits; Transient analysis of different electrical circuits with and without initial conditions; Concept of convolution theorem and its applications; Solutions of problems with dc and ac sources. GO TO

Chapter 6: Laplace Transform and its Applications

Module 4 Fourier Method of Waveform Analysis: Fourier series and Fourier transform (in continuous domain only); Application in circuit analysis; Solutions of problems. GO TO

Chapter 8: Fourier Series and Fourier Transform

Roadmaps to the Syllabus

xv

Module 5 Network Equations and Theorems: Formulation of network equations; Source transformation; Loopvariable analysis; Node-variable analysis; Superposition, Thevenin’s; Norton’s and maximum power transfer theorems; Millman’s theorem and its application in three-phase unbalanced circuit analysis; Solution of problems with dc and ac sources. GO TO

Chapter 4: Network Topology (Graph Theory) Chapter 5: Network Theorems

Module 6 Graph Theory and Network Equations: Concept of tree, branch, tree link; Incidence matrix; Tie-set matrix and loop currents; Cut-set matrix and node pair potentials; Duality; Solutions of problems. GO TO

Chapter 4: Network Topology (Graph Theory)

Module 7 Two-port Network Analysis: Open-circuit impedance and short-circuit admittance parameter; Transmission parameters; Hybrid parameters and their inter-relations; Driving-point impedance and admittance; Solutions of problems. GO TO

Chapter 7: Two-Port Network

Module 8 Filter Circuits: Analysis and synthesis of low-pass, high-pass, band-pass, band-reject, all-pass filters (first- and second-order only) using operational amplifier; Solutions of problems. GO TO

Chapter 9: Filter Circuits

(EC 301) APPLIED ELECTRONICS AND INSTRUMENTATION ENGINEERING Module 1 Resonant Circuits: Series and parallel resonance; Impedance and admittance characteristics; Quality factor; Half-power points; bandwidth; Phasor diagrams; Transform diagrams; Practical resonant and series circuits; Solutions of problems. Mesh Current Network Analysis: Kirchhoff’s voltage law; Formulation of mesh equations; Solutions of mesh equations by Cramer’s rule and matrix method; Driving-point impedance; Transfer impedance; Solutions of problems with dc and ac sources. GO TO

Chapter 10: Resonance Chapter 2: Introduction to Circuit Theory Concepts

xvi

Roadmaps to the Syllabus

Module 2 Node Voltage Network Analysis: Kirchhoff’s current law; Formulation of node equations and solutions; Driving-point admittance; Transfer admittance; Solutions of problems with dc and ac sources. Network Theorems: Definition and implication of superposition theorem; Thevenin’s theorem; Norton’s theorem; Reciprocity theorem; Compensation theorem; Maximum power transfer theorem; Millman’s theorem; Star-delta transformations; Solutions and problems with dc and ac sources. GO TO

Chapter 5: Network Theorems Chapter 2: Introduction to Circuit Theory Concepts

Module 3 Graph of Network: Concept of tree and branch; Tree link; Junctions; Incident matrix; Tie-set matrix; Determination of loop current and node voltages. Coupled Circuits: Magnetic coupling; Polarity of coils; Polarity of induced voltage; Concept of selfand mutual inductance; Coefficient of coupling; Solutions of problems. Circuit Transients: dc transients in RL and RC circuits with and without initial charge; RLC circuits; ac transients in sinusoidal RL; RC and RLC circuits; Solutions of problems. GO TO

Chapter 4: Network Topology (Graph Theory) Chapter 3: Magnetically Coupled Circuits Chapter 6: Laplace Transform and its Applications

Module 4 Laplace Transform: Concept of complex frequency; Transform of f(t) into F(s); Transform of step; Exponential surge, overdamped surge, critically damped surge; Damped and un-damped sine functions; Properties of Laplace transform; Linearity; Real differentiation; Real integration; Initial value and final value theorems; Inverse Laplace transform, application in circuit analysis; Partial fraction expansion; Heaviside’s expansion theorem; Laplace transform and Inverse Laplace transform; Solutions of problems. Two-port Networks: Relationship of two-port network variables; Short-circuit admittance parameters; Open-circuit impedance parameters; Transmission parameters; Relationship between parameter sets; Network functions for ladder and general networks. GO TO

Chapter 6: Laplace Transform and its Applications Chapter 7: Two-Port Network

Roadmaps to the Syllabus

xvii

ROADMAP TO THE SYLLABUS [EE (EI) 301] Module 1 Introduction: Continuous and discrete; Fixed and time-varying; Linear and nonlinear; Lumped and distributed; Passive and active networks and systems; Independent and dependent sources; Step, ramp, impulse, sinusoidal, square; sawtooth signals. Coupled Circuits: Magnetic coupling; Polarity of coils; Polarity of induced voltage; Concept of Self- and mutual inductance; Coefficient of coupling; Modeling of coupled circuits; Solutions of problems. Resonant Circuits: Series and parallel resonance; Impedance and admittance characteristics; Quality factor; Half-power points; Bandwidth; Resonant voltage rise; Transform diagrams; Solutions of problems. GO TO

Chapter 1: Introduction to Different Types of Systems Chapter 3: Magnetically Coupled Circuits Chapter 10: Resonance

Module 2 Laplace Transforms: Concept of complex frequency; Transformation of step, exponential, overdamped surges; Critically damped surge; Damped sine and undamped sine functions; Properties of Laplace transforms; Linearity; Real differentiation; Real integration; Initial value and final value theorems; Inverse Laplace transforms and applications in circuit analysis; Partial fraction expansion; Heaviside’s expansion theorem; Impulse; Step and sinusoidal response of RL, RC and RLC circuits; Transient analysis of different electrical circuits with and without initial conditions; Concept of convolution theorem and its application; Solutions of problems with dc and ac sources. GO TO

Chapter 6: Laplace Transform and its Applications

Module 3 Network Equations: Kirchhoff’s voltage law and current law; Formulation of network equations; Source transformation; Loop-variable analysis; Node-variable analysis. Network Theorems: Superposition, Thevenin’s, Norton’s and maximum power transfer theorems; Millman’s theorem and its application in three-phase unbalanced circuit analysis; Solutions of problems with dc and ac sources. Graph of Network: Concept of tree, branch, tree link, junctions; Incident matrix; Tie-set matrix and loop currents; Cut-set matrix and node pair potentials; Duality; Solutions of problems. GO TO

Chapter 2: Introduction to Circuit Theory Concepts Chapter 5: Network Theorems Chapter 4: Network Topology (Graph Theory)

xviii

Roadmaps to the Syllabus

Module 4 Two-port Network Analysis: Open-circuit impedance and short-circuit admittance parameter; Transmission parameters; Hybrid parameters and their inter-relations; Driving-point impedance and admittance; Solutions of problems with dc and ac sources. Circuit Transients: dc transient in RL and RC circuits with and without initial charge; RLC circuits; ac transients in sinusoidal RL, RC and RLC circuits; Solutions of problems. Filter Circuits: Analysis of low-pass, high-pass, band-pass, band-reject, all-pass filters (first- and secondorder only) using operational amplifier; Solutions of problems. GO TO

Chapter 7: Two-Port Network Chapter 6: Laplace Transform and its Applications Chapter 9: Filter Circuits

ROADMAP TO THE SYLLABUS

BME (EC) 301 Module 1 Resonant Circuits: Series and parallel resonance; Impedance and admittance characteristics; Quality factor; Half-power points; Bandwidth; Resonant voltage rise; Transform diagrams; Solutions of problems. GO TO

Chapter 10: Resonance

Module 2 Mesh Current Network Analysis: Kirchhoff’s voltage laws; Formulation of mesh equations; Solution of mesh equations by Cramer’s rule and matrix method; Driving-point admittance; Transfer impedance; Solutions of problems with dc and ac sources. GO TO

Chapter 2: Introduction to Circuit Theory Concepts

Module 3 Node Voltage Network Analysis: Kirchhoff’s current law; Formulation of node equations and solutions; Driving-point admittance; Transfer admittance; Solutions of problems with dc and ac sources. GO TO

Chapter 2: Introduction to Circuit Theory Concepts

Module 4 Network Theorems: Definition and implications of Superposition, Thevenin’s, Norton’s, Reciprocity, Compensation, maximum power transfer, and Millman’s theorems; Star-Delta transformations; Solutions and problems with dc and ac sources. GO TO

Chapter 5: Network Theorems

Roadmaps to the Syllabus

xix

Module 5 Graph of Networks: Concept of tree, branch, tree link, junctions, incidence matrix, tie-set matrix; cutset matrix; Determination of loop current and node voltages. GO TO

Chapter 4: Network Topology (Graph Theory)

Module 6 Coupled Circuits: Magnetic coupling; Polarity of coils; Polarity of induced voltage; Concept of selfand mutual inductances; Coefficient of coupling; Solutions of problems. GO TO

Chapter 3: Magnetically Coupled Circuits

Module 7 Circuit Transients: dc transients in RL and RC circuits with and without initial charge; RLC circuits; ac transients in sinusoidal RL, RC and RLC circuits; Solutions of problems. GO TO

Chapter 3: Magnetically Coupled Circuits

Module 8 Laplace Transforms: Concept of complex frequency; Transformation of f(t) into f(s); Transformation of step, exponential, overdamped surge, critically damped surge; Damped and undamped sine functions; Properties of Laplace transforms; Linearity; Real differentiation; Real integration; Initial value and final value theorems; Inverse Laplace transform; Applications of circuit analysis; Partial fractions expansion; Heaviside’s expansion theorem; Solutions of problems. GO TO

Chapter 6: Laplace Transform and its Applications

Module 9 SPICE: Introduction; Model statement; Elementary dc and small-signal analysis.

CHAPTER

1 Introduction to Different Types of Systems 1.1

INTRODUCTION

An electrical network is one of the many important physical systems. In order to understand the basic characteristics of an electric network, we must first know the different concepts of systems. In this chapter, the different types of systems have been discussed.

1.2

CONCEPTS OF SIGNALS AND SYSTEMS

1.2.1 Signals A signal is defined as a function of one or more variables, which provides information on the nature of a physical phenomenon. When the function depends on a single variable, the signal is said to be one-dimensional, for example, a speech signal whose amplitude varies with time, depending on the spoken word and who speaks it. When the function depends on two or more variables, the signal is said to be multidimensional, for example, an image (2-D signal).

1.2.2 Systems A system is an entity that takes an input signal and produces an output signal. It is a combination and interconnection of several components to perform a desired task. The system responds to one or more input quantities, called input signals or excitation, to produce one or more output quantities, called output signals or response.

1.2

Circuit Theory and Networks

Figure 1.1 Block diagram representation of a system

1.3 1. 2. 3. 4. 5. 6. 7. 8. 9.

DIFFERENT TYPES OF SYSTEMS Continuous and Discrete Time Systems Fixed and Time-varying Systems Linear and Non-linear Systems Lumped and Distributed Systems Instantaneous and Dynamic Systems Active and Passive Systems Causal and Non-causal Systems Stable and Unstable Systems Invertible and Non-invertible Systems

1.3.1 Continuous and Discrete Time Systems Signals are represented mathematically as functions of one or more independent variables. We classify signals as being either continuous-time (functions of a real-valued variable) or discrete-time (functions of an integer-valued variable). In other words, a continuous-time signal has a value defined for each point in time and a discretetime signal is defined only at discrete points in time. To signify the difference, we (usually) use round parenthesis around the argument for continuous time signals, e.g., x(t) and square brackets for discrete-time signals, e.g., x[n]. We will also use the notation xn for discrete-time signals.

Figure 1.2(a) Continuous-time signal

Figure 1.2(b) Discrete-time signal

Introduction to Different Types of Systems

1.3

A continuous-time system is a system which accepts only continuous-time signal to produce continuous-time internal and output signals. On the other hand, a discrete-time system is a system that transforms discrete-time input(s) into discrete-time output(s). The examples given below are common in our daily life.

Continuous-time systems (i) Atmospheric pressure as a function of altitude (ii) Electric circuits composed of resistors, inductors, capacitors driven by continuous-time sources. Discrete-time systems (i) Weekly stock market index (ii) Balance in a bank account from month to month. The sequence of values of the discrete-time signal shown in Fig. 1.2(b) defined at discrete points in time are called samples and the spacing between them is called the sample spacing. For equal sample spacing, the sequence of values are expressed as a function of the signed integer n as x[n], where n is termed as a sequence of samples or sequence, in short.

1.3.2 Time-Invariant (Fixed) and Time-Varying Systems A system is time-invariant or fixed if the behaviour and characteristics of the system do not change with time. Otherwise, the system is time-varying. Mathematically, if the input x(t) gives the output y(t), then the system is time-invariant if the input x(t – T) gives the output y(t – T) for any delay T. Hence, a time-shift of the input gives the same time-shift of the output.

Figure 1.3 Time-invariant system

Whether a system is time-invariant or time-varying can be seen in the differential equation (or difference equation) describing it. Time-invariant systems are modeled with constant coefficient equations. A constant coefficient differential (or difference) equation means that the parameters of the system are not changing over time and an input now will give the same result as the input later.

1.4

Example 1.1 Solution

Circuit Theory and Networks

A continuous system is modeled by the equation y(t) = tx(t) + 4 , and a discretetime system is modeled by y[n] = x2[n]. Are these systems time-invariant? For continuous-time system: (i) For input x(t) = x1(t), output y1(t) = tx1(t) + 4 (ii) For input x(t) = x1(t – T), output, y2(t) = tx1(t – T) + 4 From the condition of time-invariance, the output should be, y1(t – T) = (t – T)x1(t – T) + 4 (iii) From Eqs (ii) and (iii), y2(t) ¹ y1(t – T) Hence, the system is not time-invariant. For discrete-time system: For input x1[n], output y1[n] = x12[n] For input x1[n – n0], output = x12[n – n0] From the condition of time-invariance, the shifted output y1[n – n0] = x12[n – n0] Hence, the system is time-invariant.

1.3.3 Linear and Non-Linear Systems A system, in continuous-time or discrete-time, is said to be linear, if it obeys the properties of superposition, i.e., additivity and homogeneity (or scaling), while a system is non-linear if it does not obey at least any one of these properties. The superposition principle says that the output to a linear combination of input signals is the same linear combination of the corresponding output signals. Mathematically, the linearity condition is based on two properties.

1. Additivity If the input signals x1(t) and x2(t) correspond to the output signals y1(t) and y2(t), respectively, then the input signal {x1(t) + x2(t)} should correspond to the output signal {y1(t) + y2(t)}. 2. Homogeneity If the input signal x1(t) corresponds to the output signal y1(t), then the input signal a1x1(t) should correspond to the output signal a1y1(t) for any constants a1. Combining these two properties, the condition for a linear system can be written as, if the input signals x1(t) and x2(t) correspond to the output signals y1(t) and y2(t), respectively, then the input signal a1x1(t) + a2x2(t) should correspond to the output signal a1y1(t) + a2y2(t) for any constants a1 and a2. Example 1.2

Check whether the systems with the input-output relationship given below are linear. (a) y(t) = mx(t) + c, (b) y(t) = tx(t)

Solution

(a) For an input x1(t), output, y1(t) = mx1(t) + c For an input x2(t), output, y2(t) = mx2(t) + c For an input {x1(t) + x2(t)}, output, y3(t) = m{x1(t) + x2(t)} + c From the condition of linearity, the output should be {y1(t) + y2(t)} = m{x1(t) + x2(t)} + 2c From Eqs (i) and (ii), we conclude that the system is non-linear. (b) For an input x1(t), output, y1(t) = tx1(t)

(i) (ii)

Introduction to Different Types of Systems

For an input x2(t), output, y2(t) = tx2(t) For an input {k1x1(t) + k2x2(t)}, output, y3(t) = t{k1x1(t) + k2x2(t)} where, k1 and k2 are any arbitrary constants. From the condition of linearity, the output should be {k1y1(t) + k2y2(t)} = k1tx1(t) + k2tx2(t) = t{k1x1(t) + k2x2(t)} From Eqs (i) and (ii), we conclude that the system is linear.

1.5

(i)

(ii)

1.3.4 Lumped and Distributed Systems All physical systems contain distributed parameters because of the physical size of the system components. For example, the resistance of a resistor is distributed throughout its volume. However, if the size of the system components is very small with respect to the wavelength of the highest frequency present in the signals associated with it, then the system components behave as if it all were occurring at a point. This system is said to be lumped-parameter system. Distributed parameter systems are modeled as given below. 1. By partial differential equations if they are continuous-time systems 2. By partial difference equations if they are discrete-time systems. Lumped parameter systems are modeled with ordinary differential or difference equations. Example 1.3

Consider an electric power system of frequency 50 Hz. The wavelength of the signal is obtained as, C 3 ´ 105 = 6000 km = n 50 Thus, the electrical system inside a room can be treated as a lumped-parameter system, but will be treated as distributed system for a long-distance transmission line.

nl = C Þ l =

1.3.5 Instantaneous (Static or Memoryless) and Dynamic Systems An instantaneous or static or memoryless system is a system where the output at any specific time depends on the input at that time only. On the other hand, a dynamic system is one whose output depends on the past or future values of the input in addition to the present time. A static system has no memory. Physically, it contains no energy-storage elements, whereas a dynamic system has one or more energy-storage element(s). Example 1.4

An electrical circuit containing resistance R has the v–i relationship as, v(t) = Ri(t), and so the system is static. But an electrical circuit containing capacitor C has the v–i relationship as, v(t) =

1t i (t )dt , and so the system is dynamic system. C 0ò

1.6

Circuit Theory and Networks

1.3.6 Active and Passive Systems A system having no source of energy is known as a passive system, for example, electric circuits containing resistance, capacitance, inductance, diodes, etc. A system having source of energy together with other passive elements is known as an active system, for example, electric circuits containing voltage source or current source or op-amp, etc.

1.3.7 Causal and Non-causal Systems A system is said to be causal if the output of the system depends only on the input at the present time and/or in the past, but not the future value of the input. Thus, a causal system is nonanticipative, i.e., output cannot come before the input. On the other hand, the output of a non-causal system depends on the future values of the input. Example 1.5

The moving-average system described by 1 {x[n] + x[n – 1] + x[n – 2]} 3 is causal, but the moving-average system described by

y[n] =

1 {x[n + 1] + x[n] + x[n – 1]} 3 is non-causal, since the output depends on the future value of the input x[n + 1]. It is obvious that the idea of future inputs does not have any physical meaning if we take time as our independent variable and for that reason all real-time systems are causal. However, for the case of image processing, the independent variable may be the pixels to the left and right (the “future”) of the current position on the image, and thus, we can have a non-causal system.

y[n] =

Figure 1.4(a) Causal systems

Figure 1.4(b) Non-causal systems

Introduction to Different Types of Systems

1.7

1.3.8 Stable and Unstable Systems A stable system is one where the output does not diverge as long as the input does not diverge. A bounded input produces a bounded output. For this reason, this type of system is known as bounded input-bounded output (BIBO) stable system. Mathematically, a stable system must have the following property: If x(t) be the input and y(t) be the output, then the output must satisfy the condition. | y(t) | £ My < µ; for all t whenever the input satisfy the condition | x(t) | £ Mx < µ; for all t where, Mx and My both represent a set of finite positive numbers. If these conditions are not met, i.e., the output of the system grows without limit (diverges) from a bounded input, then the system is unstable.

1.3.9 Invertible and Non-invertible Systems A system is referred to as an invertible system if (i) distinct inputs lead to distinct output, and (ii) the input can be recovered from the output.

Figure. 1.5 Invertible system

The property of invariability is important in the design of communication systems. When a transmitted signal propagates through a communication channel, it becomes distorted due to the physical characteristics of the channel. An equalizer is connected in cascade with the channel in the receiver to compensate this distortion. By designing the equalizer to be inverse of the channel, the transmitted signal is restored.

1.4

DIFFERENT TYPES OF SIGNALS

Signals can be classified into different categories, as given below. 1. Continuous-time and discrete-time signals 2. Periodic and non-periodic signals 3. Odd and even signals

1.4.1 Continuous-time and Discrete-time Signals Signals are represented mathematically as function of one or more independent variables. We classify signals as being either continuous-time (functions of a real-valued variable) or discrete-time (functions of an integer-valued variable). In other words, a continuous-time signal has a value defined for each point in time and a discretetime signal is defined only at discrete points in time.

1.8

Circuit Theory and Networks x(t)

X [n] X [3] X [1]

X [ 2] X [ 4]

X [2]

X [0] 3

4 X [ 3]

1 2 0 X [ 1]

1

2

3

time [n]

time (t)

Figure 1.6(b) Discrete-time signal

Figure 1.6(a) Continuous-time signal

To signify the difference, we (usually ) use round parenthesis around the argument for continuous time signals, e.g., x(t) and square brackets for discrete-time signals, e.g., x[n]. We will also use the notation xn for discrete-time signals. The sequences of values of the discrete-time signal shown in Fig. 1.6(b) defined at discrete points in time are called samples, and the spacing between them is called the sample spacing. For equal sample spacing, the sequences of values are expressed as a function of the signed integer n as x[n], where n is termed as a sequence of samples or sequence, in short.

1.4.2 Periodic and Non-Periodic Signals A signal f(t) is said to be periodic if f(t) = f(t ± nT) (1.1) where n is a positive integer and 'T ' is the period. Thus, a periodic signal repeats itself every T seconds. Some periodic signals are shown in Fig. 1.7. v(t ) V

2T (a)

T

0

T

2T

3T

4T

t

(b)

Figure 1.7 Periodic signals

A signal not satisfying the above condition of Eq. (1.1) is called a non-periodic signal. Examples of some non-periodic signals are e¢, t, etc.

1.4.3 Odd and Even Signals A signal f(t) is said to be odd if f(t) = –f(–t)

(1.2)

1.9

Introduction to Different Types of Systems

Some examples of odd signals are sine functions, triangular functions and square function, as shown in Fig. 1.8. v(t) f(t )

V T/2 −T

0 T/4 −V

− T/2

t

T

0

ωt

Figure 1.8 Odd signals A signal f(t) is said to be even if f(t) = f(–t) Some examples of even signals are shown in Fig. 1.5.

(1.3)

f (t ) f (t ) V

− T/2

0

T/2

t

0

ωt

−V Figure 1.9 Even signals

Decomposition of a Signal into Odd and Even Components For any function f(t), let the odd component be denoted by f0(t) and the even component by fe(t), so that, f(t) = f0(t) + fe(t) (1.4) \

f(–t) = f0(–t) + fe(–t) = f0(t) + fe(t)

(1.5) [By Eq. (1.2 and (1.3)]

By addition and subtraction of Eqs (1.4) and (1.5), we get, fe(t) =

1 [ f (t ) + f (-t )] 2

1 [ f (t ) - f (-t )] 2 By these two equations, we can decompose a signal into its odd and even components.

f0(t) =

Example 1.6

(1.6) (1.7)

Decompose the following signal into its odd and even components.

Solution To find the even and odd components we need the folded signal, i.e., f(–t), as shown in Fig. 1.10(b). By point-by-point addition and subtraction, we get the even and odd components as shown in Fig. 1.10 (c) and Fig. 1.10 (d).

1.10

Circuit Theory and Networks f(t) 1

f( t ) 1

0

1

−1

t

Figure 1.10(a) Signal of Ex. 1.1

0

t

Fig. 1.10(b) Folded Signal of Fig. 1.10(a) f0(t )

fe(t )

1/2 1

1/2

1

0 1

0

1

t

1/2

t

Figure 1.10(c) Even component Figure 1.10(d) Odd component of signal of signal of Fig. 1.10(a) of signal of Fig. 1.10(a)

1.5

SOME STANDARD SIGNALS

There are some standard signals which can be generated easily in the laboratory. Some of these standard signals are discussed below. f (t )

f (t )

t

Figure 1.11(a) sin wt

t

Figure 1.11(b) cos wt

Sinusoidal Signal A sinusoid is a signal that has the form of a sine or cosine function. We consider a sinusoidal voltage, v(t) = Vm sin wt where Vm is the amplitude, wt is the argument of the sinusoid, w is the angular frequency of the sinusoid in rad/s = 2pf = and T is the time period of the sinusoid. As the sinusoid is periodic, it repeats itself; such that 2p ö æ v(t) = v(t + T) = Vm sin w ç t + ÷ = Vm sin (w t + p ) = Vm sin w t è wø

2p , T

1.11

Introduction to Different Types of Systems

A shifted sinusoid can be written as, v(t) = Vm sin (wt + f) where f is the phase of the sinusoid. Thus, we see that, –sin wt = sin (wt ± 180°) –cos wt = cos (wt ± 180°)

f (t )

Ke

at

K 0.37 K

±cos wt = sin (wt ± 90°) m sin wt = cos(wt ± 90°) Exponential Signal An exponential signal is a function of time defined as f(t) = 0, t < 0

0

τ

1/a

t

Figure 1.12 Exponential signal

Ke–at, t ³ 0 where K and a are some real constants. The reciprocal of a has the dimension of time and is known 1ö æ as time constant, ç t = ÷ . This is the time to reach 63.2% of the total change from the initial to final è aø value. Square Wave Signal A square wave is a type of waveform where the signal has only two levels. The signal switches between these levels at regular intervals and the switch is instant. An ideal square wave signal is shown in the Fig. 1.13 f(t) V

0

T

2T

3T

time

- V

Figure 1.13 Square Wave Signal

Thus, square wave is a special kind of non-sinusoidal periodic signal with a time period T. Square waves are universally encountered in digital switching circuits and are naturally generated by binary logic devices. Saw Tooth Wave The saw-tooth wave is a kind of non-sinusoidal waveform. Since the wave has some resemblance to the teeth on the blade of a saw, it is named so. In general, a saw-tooth wave rises upward and then drops sharply. However, a saw-tooth wave may also ramp downward and then rise sharply. This type of wave is known as a reverse saw-tooth wave or inverse saw-tooth wave.

1.12

Circuit Theory and Networks

A typical saw-tooth wave is shown in Fig. 1.14 f(t) 1

0

T

2T

4T

3T

t

Figure 1.14 Saw-tooth Wave

Saw-tooth signal is used in many applications, such as in PMW modulator or oscilloscope sweep circuitry.

1.6

SINGULARITY SIGNALS

(a) Step signal, (b) Ramp signal, and (c) Impulse signal, Ku(t) u(t )

K

1 0 0

t

t

Figure 1.15 (a) Unit step; (b) Step function of magnitude K u (t − T )

K 1

0

T

t

0

a

b

Figure 1.15 (c) Shifted unit step function; (d) Gate function

1.6.1 Step Signal This function is also known as Heaviside unit function. It is defined as given below, f(t) = u(t) = 1 for t > 0 = 0 for t < 0

1.13

Introduction to Different Types of Systems

and is undefined at t = 0. A step function of magnitude K is defined as f(t) = Ku(t) = K for t > 0 = 0 for t < 0 and in undefined at t = 0. A shifted or delayed unit step function is defined as f(t) = u(t – T) = 1 for t > T = 0 for t < T and is undefined at t = T. Another function, called gate function, can be obtained from step function as follows. Therefore, g(t) = Ku(t – a) – Ku(t – b)

1.6.2 Ramp Signal A unit ramp function is defined as f(t) = r(t) = t for t ³ 0 = 0 for t < 0 A ramp function of any slope K is defined as f(t) = Kr(t) = Kt for t ³ 0 = 0 for t < 0 A shifted unit ramp function is defined as f(t) = r(t – T) = t for t ³ T = 0 for t < T

1.6.3 Impulse Signal This function is also known as Dirac Delta function, denoted by d(t). This is a function of a real variable t, such that the function is zero everywhere except at the instant t = 0. Physically, it is a very sharp pulse of infinitesimally small width and very large magnitude, the area under the curve being unity. Kr (t )

r(t )

r (t − T )

0

1

K

1

1

1

1 t

0

t

0

T

Figure 1.16 (a) Unit ramp function; (b) Ramp function; (c) Shifted unit ramp function

t

1.14

Circuit Theory and Networks

Consider a gate function as shown in Fig. 1.15. The function is compressed along the time-axis and stretched a long the y-axis, keeping the area 1 under the pulse as unity. As a ® 0, the value of ® µ and the resulting function is known as a impulse. f (t ) δ (t ) It is defined as d(t) = 0 for t ¹ 0 3/a ¥

ò d (t ) dt = 1

and

∞

2/a 1/a

-¥

0 t 0 a/3 a/2 a t 1 d Also, d(t) = Lim [u (t ) - u (t - a)] = [u (t )] a® a dt Figure 1.17 (a) Generation of impulse function from gate function; (b) Impulse signal

SOLVED PROBLEMS 1.1 Check whether the system defined by,

af

af

af

af

y t = sin x t is time-invariant. Solution: For input x(t) = x1(t), output y1 t = sin x1 t

…(i)

af

a f

For input x(t) = x1(t – T), output, y2 t = sin x1 t - T From the condition of time-invariance, the output should be,

a f

a f

y1 t - T = sin x1 t - T

…(ii)

…(iii)

From equations (ii) and (iii), y2 (t) = y1(t – T ) Hence, the system is time-invariant. 1.2 Consider a system S with input x[n] and output y[n] related by, y[n] = x[n]{g[n] + g[n – 1]} (a) If g[n] = 1, for all n, show that S is time-invariant. (b) If g[n] = n, show that S is not time-invariant. (c) If g[n] = 1 + (–1) n, show that S is time-invariant. Solution:

k p

(a) If g[n] = 1, for all n, then y n = x n 1 + 1 = 2 x n For input x[n] = x1[n], output y1 n = 2 x1 n

…(i)

For input x[n] = x1[n – n0], output, y2 n = 2 x1 n - n0

…(ii)

From the condition of time-invariance, the output should be,

1.15

Introduction to Different Types of Systems

y1 n - n0 = 2 x1 n - n0

…(iii)

From equations (ii) and (iii), y2[n] = y1[n – n0] Hence, the system is time-invariant. (b) If g[n] = n, then y n = x n n + n - 1 = 2 n - 1 x n

p a

k

f

a

f n = a2 n - 1f x

For input x[n] = x1[n], output y1 n = 2 n - 1 x1 n For input x[n] = x1[n – n0], output, y2

1

…(i)

n - n0

…(ii)

From the condition of time-invariance, the output should be,

nb

g s

y1 n - n0 = 2 n - n0 - 1 x1 n - n0

…(iii)

From equations (ii) and (iii), y2[n] ¹ y1[n – n0] Hence, the system is not time-invariant.

o a f

a f t = 2x n

(c) If g[n] = 1+ (–1)n , then y n = x n 1 + -1 n + 1 + -1

n -1

This relation is same as that of part (a). Hence the system is time-invariant. 1.3 Consider the system S whose input and output are related by, y(t) = x2 (t) Check whether S is linear. Solution: For an input x1(t), output, y1(t) = x12 (t) For an input x2(t), output, y2(t) = x22 (t)

af

af

For an input {k 1 x1(t) + k2 x2(t)}, output, y3(t) = k1 x1 t + k2 x 2 t

2

…(i)

where, k1 and k2 are any arbitrary constants. From the condition of linearity, the output should be {k1 y1(t) + k2 y2(t)} = k1 x12 (t) + k2 x22(t)

…(ii)

From equations (i) and (ii), we conclude that the system is not linear. 1.4 Consider the following discrete-time system with input-output relationships as given, y[n] = Re{x[n]} Check whether the system is linear. Solution: Let, the input be, x1 n = r n + js n

l q

l

q

Therefore, the output is, y1 n = Re x1 n = Re r n + js n = r n

a

f

Now we consider scaling of the input x1[n] by a complex number, say, a + jb , i.e. the input is,

a

a

f

fl

q l

q l

x 2 n = a + jb x1 n = a + jb r n + js n = ar n - bs n + j br n + as n Corresponding output is,

l

q

l

q l

q

y2 n = Re x 2 n = Re ar n - bs n + j br n + as n = ar n - bs n

q

1.16

Circuit Theory and Networks

But the scaled output for linear system is,

aa + jbfy

1

n = ar n + jbr n

As he two outputs are not same, the system is not linear. 1.5 Consider a discrete-time system whose output y[n] is the average of the three most recent values of the input signal, x[n], given as, 1 yn = x n + x n -1 + x n - 2 3 Show that the system is BIBO stable.

l

q

Solution: Let us assume that, x n < M x < ¥ for all n, \

1 x n + x n -1 + x n - 2 3 1 £ x n + x n -1 + x n - 2 3 1 £ Mx + Mx + Mx 3 £ Mx

yn =

n

s

Hence, the absolute value of the output signal y[n] is always less than the maximum absolute value of the input signal x[n] for all n; which shows that the system is stable. 1.6 Determine whether the following continuous-time systems are stable: (a) y t = tx t (b) y t = x t sin100pt

af af

af af

Solution: Here, let the input be bounded.

af af

(a) y t = tx t As t ® ¥, y(t) ® ¥ [since x(t) is multiplied by t] Hence, the system is an unstable system. (b) y t = x t sin100pt Here x(t) is multiplied by sin100pt . We know that the value of sine varies between –1 and 1. Hence y(t) is bounded as long as x(t) is bounded. Hence the system is stable.

af af

1.7 Determine whether the following continuous-time systems are causal or non-causal: (a) y t = x t cos t + 1 (b) y t = x 2t (c)

af af a f yat f = x a - t f yat f = z x at fdt t

(e)

-¥

Solution:

af af a f

(a) y t = x t cos t + 1

(d)

af a f dyat f + 10 yat f + 5 = xat f dt

Introduction to Different Types of Systems

1.17

Here, y(t) depends on the present input x(t). A cosine function can be evaluated at (t + 1). Therefore, the system is causal. (b) y t = x 2t

af a f

af a f

Here, if t = 5, then y 5 = x 10 Thus, the output y(t) depends on the future input. Therefore, the system is non-causal. (c) y t = x - t

af a f

a f af

Here, if t = -3 , then y -3 = x 3 Thus, the output y(t) depends on the future input. Therefore, the system is non-causal. (d)

af

dy t + 10 y t + 5 = x t dt Here, y(t) depends upon the present value of x(t). Therefore, the system is causal.

af

af

t

a f z xatfdt

(e) y t =

-¥

Here, y(t) depends upon the present and the past values of x(t), but not on the future value. Therefore, the system is causal. 1.8 Determine whether the following systems are invertible:

af af yat f = x at - n f

af af yat f = x a 2 t f

(a) y t = 10 x t

(b) y t = x 2 t

(c)

(d)

Solution: (a) y t = 10 x t For this system, the inverse system will be, 1 w t = yt 10

af

af

af

af w ( t ) = 1 y( t ) = x(t) 10

y ( t ) =10 x( t )

x( t ) System

Inverse System

Therefore, the system is an invertible system.

af

af

(b) y t = x 2 t The inverse system would be,

af

af

af

af

w t = y t = x2 t = ±x t

Here, two outputs are possible: = x(t) or – x(t). This implies that there is no unique output for unique input. Therefore, the system is a non-invertible system. (c) y t = x t - n Here, the output is the delayed input, by ‘n’ samples. Clearly, the system is invertible. These can be another system for which the output is the advanced input by ‘n’ samples. The inverse system is, w t = y t+n

af a f

af a f

1.18

Circuit Theory and Networks

af a f

(d) y t = x 2t Here, the input is compressed by a factor 2. Hence, there can be another system which will expand the input by the same factor. Hence the system is invertible. The inverse system is,

a f FH 2t IK

wt =y

1.9 Determine whether the following systems are static or dynamic: d (b) y t = x t (a) y t = e x at f dt Solution:

af

af

af

Figure 1.17

(a) y t = e x at f Here, the output depends on present input only. Hence the system is a static system. d (b) y t = x t dt Here, the output depends on differentiation of the input. Calculation of differentiation depends on the present as well as past values. Therefore, the system is a dynamic system.

af

af

af

MULTIPLE-CHOICE QUESTIONS 1.1 The output y(t) and the input x(t) of a system are related by the equation y(t) = mx(t) + c, where m and c are constants. The system is (a) linear (b) non-linear (c) may be linear or non-linear depending on y(t) and x(t) (d) none of the above 1.2 If the impulse response is realizable by delaying it appropriately and is bounded for bounded excitation, then the system is said to be (a) causal and stable (b) causal but not stable (c) non-causal but stable (d) non-causal, not stable 1.3 In a linear circuit, when the ac input is doubled, the ac output becomes (a) one fourth (b) half (c) two times (d) four times 1.4 A circuit having an e.m.f. source or any energy source is (a) active circuit (b) passive circuit (c) unilateral circuit (d) bilateral circuit 1.5 A network is said to be linear if and only if (a) a response is proportional to the excitation function x(t) (b) the principle of superposition applies (c) the principle of homogeneity applies 1 (d) both the principles (b) and (c). 0 1.6 Consider the following data. t 2. Input 1. Input applied for t < t0 –1 applied for t ³ t0 Figure 1.18

1.19

Introduction to Different Types of Systems

1.7

1.8

1.9

1.10

4. State of the network at t < t0 3. State of the network at t = t0 Among these, those needed for determining the response of a linear network for t > t0 would include (a) 1, 3 and 4 (b) 2, 3 and 4 (c) 2 and 3 (d) 2 and 4. An excitation is applied to a system at t = T and its response is zero for –¥ < t < T. Such a system is (a) non-causal system (b) stable system (c) causal system (d) unstable system. The elements which are not capable of delivering energy by its own are known as (a) unilateral elements (b) non-linear elements (c) passive elements (d) active elements. The v–i characteristic of an element is shown in the given figure. The element is (a) non-linear, active, non-bilateral (b) linear, active, non-bilateral (c) non-linear, passive, non-bilateral (d) non-linear, active, bilateral What is the input-output relation of the causal moving-average system (discrete time)? 1 1 x n + x n -1 + x n - 2 (b) y n = x n -1 + x n + x n +1 (a) y n = 3 3 2 1/ 2 1 1 (c) y n = (d) y n = x n + x n +1 + x n + 2 xn + xn + xn 3 3 Which one of the following is a linear system?

l {

1.11

af

q b g b g }

af

af

af

l l

af

q q

af

(a) y t = 2u t (b) y t = 2u t + 5 (c) y t = 2u 2 t 1.12 A function f ( . ) is linear under the conditions (s)

af

af

(d) y t = 2u 2 t + 5

b g b g b g f a kx f = kf a x f only. f b x + x g = f b x g + f b x g and f a kx f = kf a x f . f b x + x g = f b x g + f b x g or f a kx f = kf a x f .

(a) f x1 + x 2 = f x1 + f x 2 only. (b) (c) (d)

1

2

1

2

1

2

1

2

1.13 The v-i characteristic of a resistor is i = 2 v 2 . The resistor is (a) linear, passive, bilateral (b) non-linear, passive, bilateral (c) non-linear, active, bilateral (d) non-linear, active, unilateral 1.14 The system y t = tx t + 4 is (a) non-linear, time-varying and unstable. (b) linear, time-varying and unstable. (c) non-linear, time-invariant and unstable. (d) non-linear, time-varying and stable. 1.15 The following is true (a) A finite signal is always bounded. (b) A bonded signal always possesses finite energy. (c) A bounded signal is always zero outside the interval [-t0, t0] for some t0. (d) A bounded signal is always finite. 1.16 The function x(t) is shown in the figure. Even and odd parts of a unit-step function u(t) are respectively, 1 1 1 1 (a) , xt (b) - , x t 2 2 2 2

af af

af

af

1.20

Circuit Theory and Networks

af

af

1 1 1 1 ,- xt (d) - , - x t 2 2 2 2 1.17 The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following description corresponds to a causal system? (b) y t = t - 4 x t + 1 (a) y t = x t - 2 + x t - 4 (c)

1.18

af a f a f af a f a f (c) yat f = at + 4 f x at - 1f (d) yat f = at + 5f x at + 5f The impulse response h(t) of a linear time-invariant continuous time system is described by hat f = exp aat f uat f + exp b bt g ua -t f , where, u(t) denotes the unit step function, and a and b are real

constants. The system is stable if (a) a is positive and b is positive (b) a is negative and b is negative (c) a is positive and b is negative (d) a is negative and b is positive 1.19 Which of the following represent a stable system? 1. Impulse response of the system decreases exponentially. 2. Area within the impulse response if finite. 3. Eigen values of the system are positive and real. 4. Roots of the characteristic equation of the system are real and negative. Select the correct answer using the codes given below. (a) 1 and 4 (b) 1 and 3 (c) 2, 3 and 4 (d) 1, 2 and 4

EXERCISES 1.1 A discrete-time system is modeled by, y[n] = x2[n] Is this system time-invariant? 1.2 Consider the systems S whose input and output are related by, (a) y(t) = t x(t) (b) y(t) = x(t) x(t – 1) (c) y(t) = x2(t) (d) y = mx + c Check whether S is linear. 1.3 Consider the following discrete-time systems with input-output relationships as given (a) y[n] = 2x[n] + 3 (b) y[n] = n x[n] Check whether the systems are linear.

SHORT-ANSWER TYPE QUESTIONS 1.1 What is a system? What are the different types of systems? Give their definitions. 1.2 Define the following and give examples of each. (a) Continuous and discrete system. (b) Time-invariant and time-varying system. (c) Lumped and distributed system.

1.21

Introduction to Different Types of Systems

(d) (e) (f) 1.5 (a) (b)

Instantaneous (Static or Memoryless) and dynamic system. Causal and non-causal system. Active and passive system. What are the conditions for a system to be a linear system? Give the conditions for a BIBO stability of a system.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.1 (b) 1.8 (c) 1.15 (b)

1.2 (a) 1.9 (b) 1.16 (a)

1.3 (c) 1.10 (a) 1.17 (a)

1.4 (a) 1.11 (a) 1.18 (d)

1.5 (d) 1.12 (c) 1.19 (b)

1.6 (c) 1.13 (b)

1.7 (c) 1.14 (a)

CHAPTER

2 Introduction to Circuit-Theory Concepts 2.1

INTRODUCTION

The fundamental theory on which many branches of electrical engineering, such as electric power, electric machines, control, electronics, computers, communications and instrumentation are built is the electric circuit theory. Thus, it is essential to have a proper grounding with electric circuit theory as the base. An electric circuit is the interconnection of electrical elements.

2.2

SOME BASIC TERMINOLOGIES OF ELECTRIC CIRCUITS

2.2.1 Concept of Electric Charge The most basic quantity in an electric circuit is the electric charge q. Electric charge is a fundamental conserved property of some subatomic particles, which determines their electromagnetic interaction. Electrically charged matter is influenced by, and produces, electromagnetic fields. It is known that an atom consists of a positively charged nucleus surrounded by negatively charged electrons. In a neutral atom, the total charge of the nucleus is equal to the total charge of the electrons. When electrons are removed from a substance, the substance becomes positively charged and if excess electrons are given to a substance, it becomes negatively charged. The SI unit of charge is Coulomb (C). The charge of an electron is 1.602 ´ 10–19C. Thus, one Coulomb charge is defined as the charge possessed by \

F 1 I electrons. H 1602 ´ 10 K . -19

1 Coulomb charge = charge of 6.24 ´ 1018 electrons

2.2

Circuit Theory and Networks

The total electric charge of an isolated system remains constant regardless of changes within the system itself. This is known as the law of conservation of charge. The law of conservation of charge states that charge can neither be created nor destroyed. The electric charge of a macroscopic object is the sum of the electric charges of its constituent particles. Often, the net electric charge is zero, because it is favourable for the number of electrons in every atom to equal the number of protons (or, more generally, for the number of anions, or negatively charged atoms, in every molecule to equal the number of cations, or positively charged atoms). When the net electric charge is non-zero and motionless, the phenomenon is known as static electricity. Even when the net charge is zero, it can be distributed non-uniformly due to an external electric field, or due to molecular motion; in such cases the material is said to be polarised. The charge due to the polarisation is known as bound charge, while the excess charge brought from outside is called free charge. The motion of charged particles (e.g., of electrons in metals) in a particular direction is said to constitute an electric current

2.2.2 Conductors, Insulators and Semiconductors In some materials, there is a large number of free electrons or loosely bound valence-band electrons present. These electrons are easily knocked out of their orbit and easily constitute a large current. Such materials are known as conductors. Almost all metals and some liquids are good conductors. In some materials, no free electrons are available; the valence-band electrons are tightly bound to the nucleus. Such materials are known as insulators. Examples of some insulators include glass, mica, plastics, etc. In between the limits of these two major categories is a third general class of materials called semiconductors; where there are no such free electrons present, but free electrons can easily be created by adding some impurities. Examples of some insulators include germanium, silicon, etc. For example, germanium, a semiconductor, has approximately one trillion times (1 × 1012) the conductivity of glass, an insulator, but has only about one thirty-millionth (3 × 10 – 8) part of the conductivity of copper, a conductor.

2.2.3 Concept of Electric Current The phenomenon of transferring electric charge from one point in a circuit to another is described by the term electric current. Electric current is defined as the rate of flow of electric charges or electrons through a cross-sectional area. By convention, the electric current flows in the opposite direction to the electrons. If Q amount of charges flow through an area in time t, then the current is given as, Q I= (2.1) t or in differential form, dq i= (2.2) dt and the charge transferred between time t0 and t is given by t

q = idt

z

t0

(2.3)

Introduction to Circuit-Theory Concepts

2.3

As Q is expressed in Coulomb, the unit of electric current is Coulomb per second and it is given the name Ampere (A). Thus, 1 A current = flow of 6.24 ´ 1018 electrons per second through an area

2.2.4 Current Density Current density at any point is a vector whose magnitude is the electric current per unit crossr I sectional area and whose direction is normal to the cross-sectional are, i.e., J = n$ . Its unit is A 2 Ampere per square meter (A/m ).

2.2.5 Concept of Electric Potential and Potential Difference To move an electron in a conductor in a particular direction, or to create a current, requires some work or energy. This work is done by the potential or the potential difference. This is also known as voltage difference or voltage (with reference to a selected point such as earth). The unit of potential is volt. The potential of a point is 1Volt if 1Joule work is done in bringing 1Coulomb charge from infinity to that point. The voltage Vab between two points a and b is the energy (or work) w required to move a unit positive charge from a to b. [Unit of voltage is volt (V).] dw Vab = (2.4) dq The potential difference between two points is 1Volt if 1Joule work is done to displace 1 Coulomb charge from one point to the other.

2.2.6 Drift Velocity Electric current is the number of coulombs of charge which pass a point in the circuit per unit time. Because of its definition, it is often confused with the quantity drift velocity. Drift velocity refers to the average distance traveled by a charge carrier per unit time. Like the velocity of any object, the drift velocity of an electron is the distance to time ratio. The path of a typical electron through a wire could be described as a rather chaotic, zigzag path characterised by collisions with fixed atoms. Each collision results in a change in direction of the electron.

A high current results from many charge carries passing through a given cross section of wine on a circuit

Figure 2.1 (a) Typical path of an electron

Figure 2.1(b) Current is constituted by flow of many charge carriers through a cross section

2.4

Circuit Theory and Networks

The net effect of these collisions results in slow drifting of the electrons with a constant average drift velocity. The drift velocity is defined as the vector average velocity of the charge carriers moving under the influence of electric field. Mathematically, if n number of charge carriers (electrons) with charge Q each passes through an area A with drift velocity v, then the current is given by, I = nQvA .

2.2.7 Concept of Electromotive Force (EMF) The phenomenon of electric current depends on the presence of free electrons. If a material is having a large number of free electrons, these electrons will always move in random directions as shown in Fig. 2.1 (a). If an external effort is applied to the material, it is possible to drift all the electrons in a definite direction as shown in Fig. 2.1 (b). Such an external factor is known as electromotive force (emf). In other words, the voltage or potential of an electrical energy source is known as emf. When we say something as electrical energy source, we mean that the energy is converted from non-electrical form (such as, mechanical, chemical, tidal, etc) into electrical form. Please note that emf is not a force, but it is the energy or work done.

2.2.8 Electric Circuits and Networks Any combination and interconnection of network elements like resistor or inductor or capacitor or electrical energy sources are known as ‘networks’. However, a closed energised network is known as ‘circuit’. A network need not contain an energy source; but a circuit must contain an energy source. Therefore, it can be stated that all circuits are networks, but all networks are not circuits. R1

a

b

+ –

V1

R5 R2

d

R3 e

c + –

V2

R7

I

R5 R4

f

g

Figure 2.2 Circuit illustrating terminologies

2.2.9 Loop and Mesh A loop or mesh denote a closed path obtained by starting at a node and returning back to the same node through a set of connected circuit elements without passing through any intermediate node more than once. However, the difference between mesh and loop is that a mesh does not contain any other loop within it, i.e., mesh is the smallest loop. In Fig. 2.2, some loops are a-b-e-d-c-a, a-b-e-gf-c-a, c-d-e-b-g-f-c, etc; and some meshes are: a-b-e-d-c-a, c-d-e-g-f-c, g-e-b-g (through R7) and ge-b-g (through I).

Introduction to Circuit-Theory Concepts

2.5

2.2.10 Node and Branch A node is a point in a circuit where two or more circuit elements join. A node is said to be an essential node if it joins three or more elements. Examples of nodes for Fig. 2.2 are a, b, c, d, e, f and g and examples of some essential node of Fig. 2.2 are b, c, e and g. A branch is a path that connects two nodes. Those paths that connect essential nodes without passing through an essential node are known as essential branches. Examples of branches of Fig. 2.2 are: V1, R1, R2, R3, V2, R4, R5, R6, R7 and I and some essential branches of Fig. 2.2 are c-a-b, c-de, c-f-g, b-e, e-g, b-g (through R7), and b-g (through I).

2.3

DIFFERENT NOTATIONS C E e G I i k L M N P Q q R t t V v W f Y j

2.4

Capacitance Voltage source Instantaneous value of E Conductance Current Instantaneous current Coefficient Inductance Mutual inductance Number of turns Power Charge Instantaneous charge Resistance Time constant Instantaneous time Voltage drop Instantaneous V Energy Magnetic flux Magnetic linkage Instantaneous Y

Farad, F Volt, V Volt, V Siemens, S Ampere, A Ampere, A Unit less Henry, H Henry, H Unit less Watt, W Coulomb, C Coulomb, C Ohm, W Second Second Volt, V Volt, V Joule, J Weber, Wb Weber, Wb Weber, Wb

BASIC CIRCUIT ELEMENTS

(i) Active and Passive Elements Electric Circuits consist of two basic types of elements. These are the active elements and the passive elements. An active element is capable of generating electrical energy. (In electrical engineering, generating or producing electrical energy actually refers to conversion of electrical energy from a non-electrical form to electrical form. Similarly, energy loss would mean that electrical energy is converted to a non-useful form of energy and not actually lost. Principle of Conservation of Mass and Energy).

2.6

Circuit Theory and Networks

Examples of active elements are voltage source (such as a battery or generator) and current source. Most sources are independent of other circuit variables, but some elements are dependent (modeling elements such as transistors and operational amplifiers would require dependent sources). Active elements may be ideal voltage sources or current sources. In such cases, the particular generated voltage (or current) would be independent of the connected circuit. A passive element is one which does not generate electricity but either consumes it or stores it. Resistors, Inductors and Capacitors are simple passive elements. Diodes, transistors etc. are also passive elements. Passive elements may either be linear or non-linear. Linear elements obey a straight-line law. For example, a linear resistor has a linear voltage vs current relationship which passes through the origin (V = R.I). A linear inductor has a linear flux vs current relationship which passes through the origin (f = k I) and a linear capacitor has a linear charge vs voltage relationship which passes through the origin (q = CV). [R, k and C are constants]. Resistors, inductors and capacitors may be linear or non-linear, while diodes and transistors are always nonlinear.

(ii) Linear Element A circuit/network element is linear if the relation between current and voltage involves a constant coefficient. Examples

Voltage-current relationship of resistor, inductor and capacitor (both with zero initial

(

)

di 1 , v = ò idt Hence, the elements are linear. dt c Diode and transistors are non-linear devices having non-linear characteristics.

conditions) are linear v = ri, v = L

(iii) Bilateral System In a bilateral system, the same relationship between current and voltage exists for current flowing in either direction. On the other hand, a unilateral system has different current-voltage relationships for the two possible directions of current, as in diode.

2.5

PASSIVE CIRCUIT ELEMENTS

2.5.1 Electrical Resistance Electrical resistance is a measure of the degree to which an object opposes an electric current through it. The SI unit of electrical resistance is ohm (W). Its reciprocal quantity is electrical conductance measured in Siemens. Electrical resistance shares some conceptual parallels with the mechanical notion of friction. The resistance of an object determines the amount of current through the object for a given voltage across the object. V R where, R is the resistance of the object, measured in ohm equivalent to J.s/C2

I=

(2.5)

Introduction to Circuit-Theory Concepts

2.7

V is the voltage across the object, measured in volt I is the current through the object, measured in ampere For a wide variety of materials and conditions, the electrical resistance does not depend on the amount of current through or the amount of voltage across the object, meaning that the resistance R is constant.

Resistance of a Conductor DC Resistance As long as the current density is totally uniform in the conductor, the DC resistance R of a conductor of regular cross section can be computed as l (2.6) A where, l is the length of the conductor, measured in meter, A is the cross-sectional area, measured in square meter, r (Greek: rho) is the electrical resistivity (also called specific electrical resistance) of the material, measured in ohm metre. Resistivity is a measure of the material’s ability to oppose the flow of electric current. For practical reasons, almost any connections to a real conductor will almost certainly mean the current density is not totally uniform. However, this formula still provides a good approximation for long thin conductors such as wires.

R=r

AC Resistance If a wire conducts high-frequency alternating current then the effective crosssectional area of the wire is reduced. This is because of the skin effect. This formula applies to isolated conductors. In a conductor close to others, the actual resistance is higher because of the proximity effect. Resistor A resistor is a two-terminal electrical or electronic component that resists an electric current by producing a voltage drop between its terminals in accordance with Ohm’s law: V I The electrical resistance is equal to the voltage drop across the resistor divided by the current through the resistor. Resistors are used as part of electrical networks and electronic circuits.

R=

Energy in a Resistor Instantaneous power absorbed in the resistor, (2.8) p = vi = iR ´ i = i2 R (in Watt) Therefore, the energy converted into heat energy is given by, t

t

0

0

W = ò pdt = ò i 2 Rdt = i2Rt (in Joule)

(2.7)

Figure 2.3 Resistor symbols

(2.9)

Series and Parallel Arrangements of Resistors Resistors in a parallel configuration each have the same potential difference (voltage). To find their total equivalent resistance (Req): 1 1 1 1 = + +¼+ Req R1 R2 Rn

(2.10)

2.8

Circuit Theory and Networks

The parallel, property can be represented in equations by two vertical lines “| |” (as in geometry) to simplify equations. For two resistors, Req = R1 | | R2 =

R1R2 R1 + R2

(2.11)

The current through resistors in series stays the same, but the voltage across each resistor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total resistance: Req = R1 + R2 + … + Rn

Figure 2.4 Parallel arrangement of resistors

(2.12)

Figure 2.5 Series arrangement of resistors

A resistor network that is a combination of parallel and series can sometimes be broken up into smaller parts. For instance, Req = (R1 | | R2) + R3 =

R1R2 + R3 R1 + R2

(2.13)

Current Division by Parallel Resistances When a total current IP is passed through parallel connected resistances R1 and R2, the voltage VP which appears across the parallel circuit is: VP = IPRP = IPR1R2/(R1 + R2)

Figure 2.6

Series-parallel arrangement of resistors

The currents I1 and I2 which pass through the respective resistances R1 and R2 are: I1 = VP/R1 = IPRP/R1 = IPR2/(R1 + R2) I2 = VP/R2 = IPRP /R2 = IPR1/(R1 + R2) In general terms, for resistances R1, R2, R3, …, Rn (with conductances G1, G2, G3, …, Gn) connected in parallel: VP = IPRP = IP/GP = IP/(G1 + G2 + G3 + ...) In = VP/Rn = VPGn = IPGn/GP = IPGn/(G1 + G2 + G3 + ...) where Gn = 1/Rn and In is the current through nth resistance Rn Note that the highest current passes through the highest conductance (with the lowest resistance).

2.5.2 Capacitance Capacitance is a measure of the amount of electric charge stored (or separated) for a given electric potential. The most common form of charge storage device is a two-plate capacitor. If the charges on the plates are +Q and –Q, and V gives the voltage difference between the plates, then the capacitance is given by

Introduction to Circuit-Theory Concepts

2.9

O (2.14) V The SI unit of capacitance is Farad; 1 Farad = 1 Coulomb per volt. The capacitance of the majority of capacitors used in electronic circuits is several orders of magnitude smaller than the farad. The most common units of capacitance in use today are milli-farad (mF), microfarad (mF), the nano-farad (nF) and the pico-farad (pF) The capacitance can be calculated if the geometry of the conductors and the dielectric properties of the insulator between the conductors are known. For example, the capacitance of a parallel-plate capacitor constructed of two parallel plates of area A separated by a distance d is approximately equal to the following:

C=

C=e

A d

(2.15)

where C is the capacitance in farad, F e is the permittivity of the insulator used (or e0 for a vacuum) A is the area of each plate, measured in square meter d is the separation between the plates, measured in meter The equation is a good approximation if d is small compared to the other dimensions of the plates.

Capacitor A capacitor is an electrical device that can store energy in the electric field between a pair of closely-spaced conductors. When current is applied to the capacitor, electric charges of equal magnitude, but opposite polarity, build up on each conductor. Capacitors are used in electrical circuits as energy-storage devices. They can also be used to differentiate between high-frequency and low-frequency signals and this makes them useful in electronic filters. Capacitors are occasionally referred to as condensers. This is now considered an antiquated term. Properties of Capacitance The relation between charge and voltage in a capacitor is written as, Q = CV (2.16) dQ dV dC =C +V dt dt dt In most physical cases, the capacitance is constant with time. The current,

i=

\

i=C

dW dt

(2.17)

1 idt C Taking integration on both sides,

\

dV =

vc

t

1 ò dV = C ò idt 0 0

or

vc(t) =

1t i (t )dt + vc(0) C 0ò

2.10

Circuit Theory and Networks

where, vc(0) is the initial voltage across the capacitor. For zero initial voltage, vc =

1t idt C 0ò

(2.18)

From equation (2.17), it is clear that for an abrupt change of voltage across the capacitor, the current becomes infinite. Also, from equation (2.18), it is observed that for a finite change of current in zero time the integral must be zero. Therefore, the voltage acorss a capacitor cannot change instantaneously.

Explanation of Initial Voltage vc(0) It is possible that this capacitor might have been used in some other circuit earlier, where it absorbed some energy and then it was disconnected. Because of its non-dissipative nature, the energy was stored within the capacitor. Now, as this capacitor is connected to a circuit, it gets some path to release its stored energy. Here, this stored energy is represented by the initial voltage vc (0). Energy Stored in Capacitors The energy (measured in Joule) stored in a capacitor is equal to the work done to charge it. Consider a capacitance C, holding a charge +q on one plate and –q on the other. Moving a small element of charge dq from one plate to the other against the potential difference V = q/C requires the work dW.

q dq (2.19) C where, W is the work measured in Joule q is the charge measured in Coulomb C is the capacitance, measured in Farad We can find the energy stored in a capacitance by integrating this equation. Starting with an uncharged capacitance (q = 0) and moving charge from one plate to the other until the plates have charge +Q and –Q requires the work W. dW =

Q q 1 Q2 1 = CV 2 = Wstored Wcharging = ò dq = 2 2 C C 0

(2.20)

Combining this with the Eq. (2.15) for the capacitance of a flat-plate capacitor, we get 1 1 A CV 2 = e V 2 2 2 d where W is the energy measured in Joule, C is the capacitance, measured in Farad, V is the voltage measured in Volt.

Wstored =

Series or Parallel Arrangements of Capacitors Capacitors in a parallel configuration each have the same potential difference (voltage). Their total capacitance (Ceq) is given by Ceq = C1 + C2 + … + Cn The reason for putting capacitors in parallel is to increase the total amount of charge stored. In other words, increasing the capacitance also increases the amount of energy that can be stored. Its expression is

(2.21)

Figure 2.7

Parallel arrangement of capacitors

Introduction to Circuit-Theory Concepts

Estored =

1 CV 2 2

2.11

(2.22)

Figure 2.8 Series arrangement of capacitors

The current through capacitors in series stays the same, but the voltage across each capacitor can be different. The sum of the potential differences (voltage) is equal to the total voltage. Their total capacitance is given by

1 1 1 1 = + +¼+ Ceq C1 C2 Cn

(2.23)

In parallel, the effective area of the combined capacitor has increased, increasing the overall capacitance. In series, the distance between the plates has effectively been increased, reducing the overall capacitance.

Voltage Division by Capacitances In series connection When a total voltage ES is applied to series connected capacitances C1 and C2, the charge QS which accumulates in the series circuit is: QS = iS dt = ESCS = ESC1C2/(C1 + C2) The voltages V1 and V2 which appear across the respective capacitances C1 and C2 are V1 = iS dt/C1 = ESCS/C1 = ESC2/(C1 + C2) V2 = iS dt/C2 = ESCS/C2 = ESC1/(C1 + C2) In general terms, for capacitances C1, C2, C3, … connected in series QS = iS dt = ESCS = ES/(1/CS) = ES/(1/C1 + 1/C2 + 1/C3 + …) Vn = iS dt/Cn = ESCS/Cn = ES/Cn(1/CS) = ES/Cn(1/C1 + 1/C2 + 1/C3 + …) Note that the highest voltage appears across the lowest capacitance. In parallel connection When a voltage EP is applied to parallel connected capacitances C1 and C2, the charge QP which accumulates in the parallel circuit is QP = iPdt = EPCP = EP(C1 + C2) The charges Q1 and Q2 which accumulate in the respective capacitances C1 and C2 are: Q1 = i1dt = EPC1 = QPC1/CP = QPC1/(C1 + C2) Q2 = i2dt = EPC2 = QPC2/CP = QPC2/(C1 + C2) In general terms, for capacitances C1, C2, C3, … connected in parallel: QP = iPdt = EPCP = EP(C1 + C2 + C3 + …) Qn = indt = EPCn = QPCn/CP = QPCn/(C1 + C2 + C3 + …) Note that the highest charge accumulates in the highest capacitance.

2.12

Circuit Theory and Networks

2.5.3 Inductance Inductance is the property by virtue of which a circuit opposes the changes in the value of a timevarying current flowing through it. Inductance causes opposition only to varying currents and does not cause any opposition to steady or direct current. An electric current i flowing around a circuit produces a magnetic field and hence a magnetic flux F through the circuit. The ratio of the magnetic flux to the current is called the inductance, or more accurately self-inductance of the circuit. It is customary to use the symbol L for inductance, possibly in honour of the physicist Heinrich Lenz. The quantitative definition of the inductance is, therefore,

f (2.24) i It follows that the SI unit for inductance is Webbers per ampere. In honour of Joseph Henry, the unit of inductance has been given the name Henry (H): 1 H = 1 Wb/A. In the above definition, the magnetic flux j is that caused by the current flowing through the circuit concerned. There may, however, be contributions from other circuits. Consider, for example, two circuits C1, C2, carrying the currents i1, i2. The magnetic fluxes F1 and F2 in C1 and C2, respectively, are given by F1 = L11i1 + L12i2, F2 = L21i1 + L22i2 According to the above definition, L11 an L22 are the self-inductances of C1 and C2, respectively. It can be shown (see below) that the other two coefficients are equal: L12 = L21 = M, where M is called the mutual inductance of the pair of circuits. L=

Inductor An inductor is a passive electrical device employed in electrical circuits for its property of inductance. Properties of Inductance The equation relating inductance and flux linkages can be rearranged as follows. l = Li (2.25) Taking the time derivative of both sides of the equation yields dl di dL = L +i dt dt dt

In most physical cases, the inductance is constant with time and so dl di =L dt dt

By Faraday’s Law of Induction,we have dl = –E = v dt

where E is the Electromotive force (emf) and v is the induced voltage. Note that the emf is opposite to the induced voltage. Thus

Introduction to Circuit-Theory Concepts

or

v= L

di dt

i(t) =

1t v(t )dt + i(0) L 0ò

2.13

(2.26)

where i(0) is the initial current. When initial current is zero, i(t) =

1t v(t ) dt L 0ò

(2.27)

These equations together state that, for a steady applied voltage v, the current changes in a linear manner, at a rate proportional to the applied voltage, but inversely proportional to the inductance. Conversely, if the current through the inductor is changing at a constant rate, the induced voltage is constant. From equation (2.26), it is clear that for an abrupt change in current, the voltage across the inductor becomes infinite. Also, from equation (2.27), it is observed that for a finite change in voltage in zero time the integral must be zero. Therefore, the current through an inductor cannot change instantaneously. Explanation of Initial Current i(0) It is possible that this inductor might have been used in some other circuit earlier, where it absorbed some energy and then it was disconnected. Because of its non-dissipative nature, the energy was stored within the inductor core. Now, as this inductor is connected to a circuit, it gets some path to release its stored energy. Here, this stored energy is represented by the initial current i(0). Series and Parallel Arrangement of Inductors Inductors in a parallel configuration each have the same potential difference (voltage). To find their total equivalent inductance (Leq):

1 1 1 1 = + +¼+ Leq L1 L2 Ln

(2.28)

The current through inductors in series stays the same, but the voltage across each inductor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total inductance:

Figure 2.9

Parallel arrangement of inductors

Figure 2.10 Series arrangement of inductors

Leq = L1 + L2 + … + Ln (2.29) These simple relationships hold true only when there is no mutual coupling of magnetic fields between individual inductors.

2.14

2.6

Circuit Theory and Networks

TYPES OF ELECTRICAL ENERGY SOURCES

Energy source is defined as the device that generates electrical energy. They are classified according to the current voltage characteristics. The classification is given below.

Independent Voltage Source An ideal voltage source has the following features. (i) It is a voltage generator whose output voltage remains absolutely constant whatever be the value of the output current. (ii) It has zero internal resistance so that voltage drop in the source is zero. (iii) The power drawn by the source is zero.

Figure 2.11 Independent voltage sources and their characteristics

In practical, the voltage does not remain constant, but falls slightly. This is taken care of by connecting a small resistance (r) in series with the ideal source. In this case, the terminal voltage will be, v1(t) = v(t) – ir i.e., it will decrease with increase in current i. An ideal voltage source is not practically possible. No voltage source can maintain its terminal voltage constant even when its terminals are short-circuited. The terminal voltage of a practical voltage source decreases as the load current increases. The v-i characteristics of an ideal and practical voltage source are shown in Fig. 2.14. A dc or ac generator or batteries are some examples of independent voltage sources. A lead-acid battery and a dry-cell are some examples of constant

Introduction to Circuit-Theory Concepts

2.15

voltage source which can produce constant terminal voltage within a specified range of output current.

Independent Current Source An ideal current source has the following features. (i) It produces a constant current irrespective of the value of the voltage across it. (ii) It has infinity resistance. (iii) It is capable of supplying infinity power.

Figure 2.12 Independent current sources and their characteristics

In practice, the output current does not remain constant but decreases with increase in voltage. So, a practical current source is represented by an ideal current source in parallel with a high resistance (R) and the output current becomes,

v (t ) R Similar to voltage sources, an ideal current source is not practically possible. No current source can maintain constant current even when its terminals are open-circuited. The output current of a practical current source decreases as the output voltage increases. The v-i characteristics of an ideal and practical current source are shown in Fig. 2.15. A solar cell, which can produce constant current within a specified range of output voltage, is an example of independent current source. A natural lightning can be considered to be an ideal current source. When a natural lightning strikes the top of a conductor, the resistance to the ground path is ideally zero. But, when the lightning strikes a nonconducting element (like the top of a tree) a large voltage is developed across the element which is flashed out immediately. i1(t) = i(t) –

Dependent Sources In dependent sources (also referred as controlled sources), the source voltage or current is not fixed, but is dependent on a voltage or current at some other location in the circuit. Thus, there are four types of dependent sources. (a) Voltage Controlled Voltage Source (VCVS) (b) Current Controlled Voltage Source (CCVS) (c) Voltage Controlled Current Source (VCCS) (d) Current Controlled Current Source (CCCS)

Figure 2.13 Symbols of dependent sources

2.16

Circuit Theory and Networks

Dependent sources are unilateral, because for a voltage controlled voltage source, say, v2 = kv1, the output voltage v2 is controlled by the input voltage v1, but the output current i2 has no influence on the input v1. Application in electronic systems that uses either the transistors or vacuum tubes needs dependent sources.

2.7

FUNDAMENTAL LAWS

The fundamental laws that govern electric circuits are the Ohm’s law and Kirchhoff’s laws.

2.7.1 Ohm’s Law Ohm’s law states that the voltage v(t) across a resistor R is directly proportional to the current i(t) flowing through it. v(t) µ i(t) or v (t) = R × i(t) Figure 2.14(a) An Electric Circuit Definition of Ohm’s Law Physical states (temperature, material, etc.) of a conductor remaining constant, the current flowing through a conductor is directly proportional to the potential difference across the two ends of the conductor. This general statement of Ohm’s law can be extended to cover inductances and capacitors as well under alternating current conditions and transient conditions. This is then known as the Generalized Ohm’s Law This may be stated as v(t) = Z(p) × i(t) , where p = d/dt = differential operator Z(p) is known as the impedance function of the circuit, and the above equation is the differential equation governing the behaviour of the circuit. For a resistor, Z(p) = R For an inductor Z(p) = L p For a capacitor, Z(p) =

1 Cp

Figure 2.14(b) Circuit Showing Impedance functin

In the particular case of alternating current, p = jw, so that the equation governing circuit behaviour may be written as V = Z(jw) . I, and For a resistor, For an inductor,

Z(jw )) = R Z(jw ) = jw L

For a capacitor,

Z(jw ) =

1 jw C

Introduction to Circuit-Theory Concepts

2.17

2.7.2 Kirchhoff’s Current Law (KCL) Kirchhoff’s current law is based on the principle of conservation of charge. This requires that the algebraic sum of the charges within a system cannot change. Thus, the total rate of change of charge must add up to zero. Rate of change of charge is current.

Figure 2.15 Illustration of KCL

This gives us our basic Kirchhoff’s current law as the algebraic sum of the currents meeting at a point is zero,. i.e., at a node, å In = 0, where In are the currents in the branches meeting at the node. This is also sometimes stated as the sum of the currents entering a node is equal to the sum of the currents leaving the node. The theorem is applicable not only to a node, but to a closed system. i1 + i2 – i3 + i4 – i5 = 0. Also for the closed boundary, ia – ib + ic – id – ie = 0.

2.7.3 Kirchhoff’s Voltage Law (KVL) Kirchhoff’s voltage law is based on the principle of conservation of energy. This requires that the total work done in taking a unit positive charge around a closed path and ending up at the original point is zero. This gives us our basic Kirchhoff’s voltage law as the algebraic sum of the potential differences taken round a closed loop is zero. i. e., around a loop, S Vn = 0, where Vn are the voltages across the branches in the loop. va + vb + vc + vd – ve = 0 This is also sometimes stated as the sum of the emfs taken around a closed loop is equal to the sum of the voltage drops around the loop. Although all circuits could be solved using Figure 2.16 Illustration of KVL only Ohm’s law and Kirchhoff’s laws, the calculations would be tedious. Various network theorems have been formulated to simplify these calculations.

2.18

Circuit Theory and Networks l

2.8

Sign Conventions for applying Kirchhoff’s Laws 1. When tracing through a voltage source from positive to negative terminal, the voltage should be given a positive sign. 2. When tracing through a voltage source from negative to positive terminal, the voltage should be given a negative sign. 3. When tracing through a resistance in the direction of current flow, the voltage should be given a positive sign. 4. When tracing through a resistance in a direction opposite to the direction of current flow, the voltage should be given a negative sign.

SOURCE TRANSFORMATION

Transformation of several voltage (or current) sources into a single voltage (or current) source and a voltage source into a current source or vice-versa is known as source transformation. This makes circuit analysis easier. There are some rules of source transformation. Rule (1) Several voltage sources {V1(t), V2(t), …, Vn(t)} connected in series will be replaced by a single voltage source of value V = V1(t) + V2(t) + …+ Vn(t). Similarly, a number of current sources {I1(t), I2(t), …, In(t)} connected in parallel is replaced by a single current source of value I(t) = I1(t) + I2(t) + …+ In(t).

Figure 2.17 Source transformation technique: Rule (1)

Rule (2) A number of voltage sources V1(t), V2(t), …, Vn(t) in parallel will result in a single voltage source, V(t) = V1(t) = V2(t) = … =Vn(t). Therefore, voltage sources should not be connected in parallel unless they have identical potential, as paralleling of sources with non-similar potential waveforms will result in heavy current, which may damage the equipment. Similarly, a number of current sources I1(t), I2(t), …, In(t) in series will result in a single current source of value I(t) = I1(t) = I2(t) = …= In(t) and thus, current sources cannot be connected in series if they are not identical.

2.19

Introduction to Circuit-Theory Concepts

V1

~+

V2

~+

. . .Vn

~+

º

~+

V = V1 = V2 = . . . = Vn

I 1 (t )

~

I 2 (t )

~

I n (t )

º

~

I (t ) = I 1 (t ) = I 2 (t ) = ... = I n (t )}

~

Figure 2.18 Source transformation technique: Rule (2)

Rule (3) As far as the computations in the remainder of the network are concerned, a resistor in parallel with an ideal voltage source and a resistor in series with an ideal current source may be ignored.

Figure 2.19 Source transformation technique: Rule (3)

Rule (4) A voltage source V(t) in series with a resistor R can be converted into a current source

V (t ) . R Similarly, a voltage source V(t) in series with a capacitor C may be converted into a current dV (t ) source I(t) in parallel with C, where, I(t) = C ; and a voltage source V(t) in series with an dt 1 inductor L may be converted into a current source I(t) in parallel with L, where, I(t) = ò V (t )dt L

I(t) in parallel with the same resistor R, where, I(t) =

Figure 2.20

Source transformation technique: Rule (4)

2.20

Circuit Theory and Networks

2.9

NETWORK ANALYSIS TECHNIQUES

Network analysis is the determination of the response output of a network when the input excitation is given. There are two techniques of network analysis. 1. Nodal Analysis 2. Loop or Mesh Analysis

Nodal Analysis It is based on Kirchhoff’s current law (KCL). In this method, the unknown variables are the node voltages. It is generally used when the circuit contains several current sources. Steps l If there is N number of nodes in a network, all nodes are labeled. One node is treated as datum or reference node (zero potential) and the other node voltages are treated as unknowns to be determined with respect to this reference. l KCL is written at each node in terms of node voltages. n KCL is applied at N – 1 of the N nodes of the circuit using assumed current directions, as necessary. This will create N – 1 linearly independent equations, known as node equations. n In a circuit with independent voltage sources, if two nodes of interest are separated by a voltage source instead of a resistor or current source, then the concept of supernode is used that creates constraint equations. n The current is computed based on voltage difference between two nodes. The current in any branch is obtained via ohm’s law as, i=

Vmm Vm - Vm , for D.C. = R R

Vmm Vm - Vm , for A.C. = Z Z where, Vm > Vn and current flows from node m to n. Solution of the N – 1 simultaneous equations (by Gaussian elimination or matrix method) gives the unknown node voltages. l=

l

For the network shown in Figure 2.21, apply Kirchhoff’s current law and write the node equations. Example 2.1

Let node voltages are E1, E2 and E3 at nodes 1, 2 and 3 respectively. At node 1, I1 = I3 + I4 I1 =

E1 ( E1 - E2 ) + R1 R2

E 1 1 I1 = æç + ö÷ E1 - 2 R R R2 è 1 2ø At node-2, I4 = I5 + I6

(i) Figure 2.21

Network explaining node analysis technique

Introduction to Circuit-Theory Concepts

2.21

( E1 - E2 ) ( E2 - E3 ) E2 E 1 1 1 ö E3 = + 0 = - 1 + E2 æç + + R2 R4 R3 R2 è R2 R3 R4 ÷ø R4

(ii)

At node-3, I6 = I7 + I8 – I2

( E2 - E3 ) E3 E3 = + - I2 R4 R5 R6 I2 = -

E2 1 1 1 + E3 æç + + ö R4 è R4 R5 R6 ÷ø

(iii)

Given the other values, solution of Equations (i), (ii), and (iii) gives the values of E1, E2 and E3.

Concept of Supernode This concept is used when a circuit contains voltage sources. A supernode is formed by enclosing a dependent or independent voltage source connected between two non-reference nodes and any elements connected in parallel with it. This concept is necessary for nodal analysis with voltage source, because the current through a voltage source is unknown. We consider the following two cases. Case 1 When a voltage source is connected between the reference node and a non-reference node: In this case, the voltage of the non-reference node is taken equal to the voltage of the voltage source. For the circuit shown in Fig. 2.22(a), (i) V1 = 5 V Case 2 When a voltage source is connected between two non-reference nodes: In this case, a supernode is considered enclosing the non-reference nodes. Both KCL and KVL is written for the supernode.

Figure 2.22(a) Circuit with supernode

Figure 2.22(b) KVL with supernode

For this example, nodes 2 and 3 are forming the supernode. By KCL at the supernode, i1 = i2 + i3

V1 - V2 V2 - 0 V3 - 0 (ii) = + 5 10 20 To apply KVL to the supernode, the circuit is drawn as shown in Fig. 2.22(b). By KVL, (iii) 10 + V3 – V2 = 0 Solving equations (i), (ii) and (iii), the node voltages are obtained, V1 = 5 V, V2 = 4.2857 V, V3 = –5.7143 V. or

2.22

Circuit Theory and Networks

Properties of Supernode (i) It provides the constraint equations. (ii) Both KCL and KVL are written for supernode. (iii) A supernode does not have any voltage of its own. Loop or Mesh Analysis It is based on Kirchhoff’s voltage law (KVL). In this method, the unknown variables are the loop currents. It is generally used when the circuit contains several voltage sources. Steps l If there is ‘N ’ number of loops/meshes in a network, all loops are labeled. l KVL is written at each loop/mesh in terms of loop/mesh currents. Loop currents are those currents flowing in a loop; they are used to define branch currents. n For N independent loops, total N equations are written using KVL around each loop. These equations are known as loop/mesh equations. n The concept of supermesh is used in case a circuit contains current source that provides the constraint equations. l Solution of the N simultaneous equations gives the required loop/mesh currents. Write the mesh equations for the circuit shown in Figure 2.23. Two meshes are labeled as mesh-1 and mesh-2. Applying KVL for mesh-1, Vs = R1I1 + R2(I1 – I2) By constraint equation, I2 = – Is Solving the equations, we get I1 and I2. Example 2.2

Concept of Supermesh This concept is used when a circuit contains current sources. A supermesh is formed by excluding the branch containing a dependent or independent current source connected in common to two meshes and any elements connected in series with it. This concept is necessary for loop analysis with current source, because the voltage drop across a current source is unknown. We consider the following two cases. Case-1 When a current source is in one mesh: In this case, the mesh current is taken equal to the current of the current source. For example, for the circuit shown in Fig. 2.24, i2 = –10 A Case-2 When a current source is connected between two meshes: In this case, a supermesh is considered excluding the branch with the current source and any elements connected in series with it. Both KCL and KVL is

Figure 2.23

(i) (ii)

Circuit explaining loop analysis technique

Figure 2.24 Current source in one mesh

Introduction to Circuit-Theory Concepts

2.23

written for the supermesh. For example, consider the circuit shown in Fig. 2.25. Supermesh is formed by excluding the branch with 3A current source. By KVL for the supermesh, 2(i1 – i2) + 4(i3 – i2) + 8i3 = 6

(i)

By KCL at any one node of the omitted branch (say, X), i1 = 3 + i3 (ii) Also by KVL for second mesh, 2i2 + 4(i2 – i3) + 2(i2 – i1) = 0 (iii) Solving equations (i), (ii) and (iii), the mesh currents are obtained, i1 = 3.437A, Figure 2.25 Current source connected between two i2 = 1.1052A, i3 = 0.4737A. meswhes Properties of Supermesh (i) It provides the constraint equations. (ii) Both KCL and KVL are written for supermesh. (iii) A supermesh does not have any current of its own.

Comparison of Loop and Node Analysis In any network having N nodes and B branches, there are 2B unknowns, i.e., B–branch currents and B–branch voltages. These unknowns can be determined either by loop analysis or nodal analysis. The choice of the method depends on two factors given below. 1. Nature of the network The mesh-method is generally used for circuits having many seriesconnected elements, voltage sources, or supermeshes. On the other hand, nodal analysis is more suitable for circuits having many parallel-connected elements, current sources, or supernodes. The main factor for selecting any one method is the minimum number of equations. If a circuit is having fewer nodes than meshes, then nodal analysis is used, while if a circuit with fewer meshes than nodes, then loop method is used. 2. Requirement of the problem If node voltages are required, nodal analysis is used. If branch/ mesh currents are required, loop analysis is used. However, there are some particular circuits, where only one method can be applied. For example, in analyzing transistor circuits, mesh method is the only possible method; while for op-amp circuits and for non-planar networks, node method is the only possible method.

2.10

DUALITY

Duality is a transformation in which currents and voltages are interchanged. Two phenomena are said to be dual if they are described by equations of the same mathematical form. There are a number of similarities and analogies between the two circuit analysis techniques based on loop-current method and node voltage method. The principal quantities and concepts involved in

2.24

Circuit Theory and Networks

these two methods based on KVL and KCL are dual of each other with voltage variables substituted by current variables, independent loop by independent node-pair, etc. This similarity is termed as ‘principle of duality’. Some dual relations are: v = Ri i = Gv v= L v=

di dt

i= C

1 idt Cò

i=

dv dt

1 vdt Lò

Thus, the circuit elements (R, L, C) have some dual relationship. Duality also appears as relation between two networks. For example, an RLC series circuit with voltage excitation is dual of an RLC parallel circuit with current excitation.

 Figure 26(a) Series RLC Circuit

For series circuit, v = Ri + L

Figure 2.26(b) Parallel RLC Circuit

di 1 + idt dt C ò

For parallel circuit, i = Gv + C

dv 1 + vdt dt L ò

Dual Quantities and Concepts Sl No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Quantity/Concept

Dual

Current Voltage Resistance Conductance Inductance Capacitance Impedance Admittance Reactance Susceptance Branch current Branch voltage Mesh or Loop Node or Node-pair Mesh Current or Loop Current Node Voltage or Node-pair Voltage Link Tree Branch Link Current Tree Branch Voltage Tree Branch Current Link Voltage Tie-set Cut-set Short-circuit Open-circuit Parallel Paths Series Paths

Introduction to Circuit-Theory Concepts

2.25

Construction of Dual of a Network 1. A dot is placed inside each independent loop of the given network. These dots correspond to the non-reference nodes of the dual network. 2. A dot is placed outside the network. This dot corresponds to the datum node. 3. All internal dots are connected by dashed lines crossing the common branches and placing the elements which are duals of the elements the original network. 4. All internal dots are connected to the external dot by dashed lines crossing all external branches and placing dual elements of the external branch. Conventions for Reference Polarities of Voltage Source and Reference Directions of Current Source (i) A clockwise current in a loop corresponds to a positive polarity (with respect to reference node) at the dual independent node. (ii) A voltage rise in the direction of a clockwise loop current corresponds to a current flowing towards the dual independent nodes. Finally, the dual construction can be checked by writing mesh equations and node equations of two networks. Example 2.3

Draw the dual of the network shown in figure.

Solution

Following the steps, dual network is drawn. Therefore, the dual network becomes as shown in Fig. 2.27. By KVL to the original network, I1(3 + 4) – I2(4) = 100

Figure 2.27 Circuit of example

Figure 2.28 Figure explaining drawing dual of network of Fig. 2.27

2.26

Circuit Theory and Networks

–I1(4) + I2(4 + 5 + 6) – 5I3 = 0 I3 = 5 The dual equations will be, V1(3 + 4) – V2(4) = 100 –V1I(4) + V2(4 + 5 + 6) – 5V3 = 0 V3 = 5 These equations satisfy the dual network.

2.11

Figure 2.29 Dual of network of Fig. 2.26

STAR-DELTA CONVERSION TECHNIQUE

The Y-D transform, also written Y-delta, Wye-delta, Kennelly’s delta-star transformation, starmesh transformation, T-P or T-pi transform, is a mathematical technique to simplify the analysis of an electrical network. The name derives from the shapes of the circuit diagrams, which look respectively like the letter Y and the Greek capital letter D.

Figure 2.30 (a) Star connection (b) Delta connection

The transformation is used to establish equivalence for networks with three terminals. For equivalence, the impedance between any pair of terminals must be the same for both networks. For the star connection, the impedance between terminals 1 and 2 is Z1+ Z2. For delta connection, the the impedance between terminals 1 and 2 is

Z12 | | ( Z 23 + Z31 ) =

Z12 ( Z 23 + Z31 ) Z12 + Z 23 + Z31

As the impedance between terminals 1 and 2 should be same, therefore,

Z1 + Z 2 =

Z12 ( Z 23 + Z31 ) Z12 + Z 23 + Z31

(i)

Similarly, for terminals 2 and 3 we get,

Z 2 + Z3 =

Z 23 ( Z 31 + Z12 ) Z 23 + Z31 + Z12

(ii)

Introduction to Circuit-Theory Concepts

Z3 + Z1 =

Z 31 ( Z12 + Z 23 ) Z31 + Z12 + Z 23

2.27

(iii)

Delta to Star Conversion In this case, Z1, Z2, and Z3 are to be written in terms of Z12, Z23, and Z31. By (i) – (ii) + (iii), we get

Z1 =

Similarly we get,

Z2 =

and

Z3 =

Z12

Z12 Z 31 + Z 23 + Z 31

(iv)

Z12

Z 23 Z12 + Z 23 + Z31

(v)

Z12

Z31Z 23 + Z 23 + Z 31

(vi)

Star to Delta Conversion In this case, Z12, Z23, and Z31 are to be written in terms of Z1, Z2, and Z3. Let Z = Z1Z2 + Z2Z3 + Z3Z1. Then from Eq. (iv) to Eq. (vi), we get Z=

2 2 2 Z12 Z 23 Z31 Z12 Z 23 Z31 Z12 Z 23 Z31 Z12 Z 23 Z31 + + = 2 2 2 Z12 + Z 23 + Z31 ( Z12 + Z 23 + Z31 ) ( Z12 + Z 23 + Z31 ) ( Z12 + Z 23 + Z31 )

(vii) From Eq. (vii) and Eq. (iv), we get Z = Z12Z3

Þ Z12 =

Therefore,

Z12 =

Z1Z 2 + Z 2 Z3 + Z3 Z1 ZZ = Z1 + Z 2 + 1 2 Z3 Z3

Similarly,

Z 23 =

Z1Z 2 + Z 2 Z 3 + Z 3 Z1 Z Z = Z 2 + Z3 + 2 3 Z1 Z1

and

Z31 =

Z1Z 2 + Z 2 Z3 + Z3 Z1 Z Z = Z3 + Z1 + 3 1 Z2 Z2

Z Z3

SOLVED PROBLEMS 2.1 Find the values of V, Vab and the power delivered by the 5V source. All values of resistances are in ohm.

Solution Current,

i=

2 1 = A 60 30

2.28

Circuit Theory and Networks

By KVL, 20i + 2 + 5 + v + 70i = 0 v = - 7 - 90i = - 7 - 90 ´

\

1 = - 10 V 30

vab = 20i + v + 30i = 50i - 10

1 - 10 = - 8.33 V 30 Power drawn by the 5V source = – (Power taken source) = 1 = –0.166 W –5 ´ 30 2.2 Find the equivalent resistance between the terminals A and B of the circuit shown below. = 50 ´

Solution Converting star into delta, rr ö æ 15 = 9.875 W r12 = ç r1 + r2 + 1 2 ÷ = 8 + r 8 è 3 ø

rr ö æ 40 r23 = ç r2 + r3 + 2 3 ÷ = 13 + = 26.33 W r1 ø 3 è rrö æ 24 r31 = ç r3 + r1 + 3 1 ÷ = 11 + = 15.8 W r2 ø 5 è

2.29

Introduction to Circuit-Theory Concepts

Combining the parallel connections of 5 W and 15.8 W and 4 W and 26.33 W, we have the reduced circuit. Again, converting the delta made of 6 W, 4 W and 9.875 W into equivalent star,

r1 = = r2 =

r12 r31 r1 + r2 + r3 6´4 = 1.2075 W 19.875 4 ´ 9.875 = 1.987 W 19.875

6 ´ 9.875 = 2.981 W 19.875 So, the given circuit becomes as shown in figure. r3 =

\

RAB = 1.2075 +

6.779 ´ 5.459 = 4.23 W 6.779 + 5.459

2.3 Find the equivalent resistance between (i) A and B, (ii) B and C, (iii) C and A, and (iv) A and N of the circuit shown. Solution Converting the star into delta,

rr ö 5´4 æ = 12.33 W r12 = ç r1 + r2 + 1 2 ÷ = 4 + 5 + r3 ø 6 è

rr ö 5´6 æ = 18.5 W r23 = ç r2 + r3 + 2 3 ÷ = 5 + 6 + r1 ø 4 è rrö 6´4 æ = 14.8 W r31 = ç r3 + r1 + 3 1 ÷ = 6 + 4 + r2 ø 5 è

Ans.

2.30

Circuit Theory and Networks

The circuit becomes, as shown in below figure. (i) Equivalent resistance between A and B, RAB =

3.73 ´ (10.06 + 5.52) = 3.035 W . 3.70 + 10.06 + 5.52

Ans.

(ii) RBC =

10.06 ´ (3.73 + 5.52) = 4.82 W 10.06 + 3.73 + 5.52

Ans.

(iii) RCA =

5.52 ´ (10.06 + 3.73) = 3.94 W 6.52 + 10.06 + 3.73

Ans.

(iv) Converting the delta into equivalent star, r1 =

5´6 = 0.83 W 5 + 6 + 25

r2 =

25 ´ 6 = 4.167 W 5 + 6 + 25

r3 =

25 ´ 5 = 3.472 W 5 + 6 + 25

The circuit becomes:

\

RAN =

4 ´ 6.288 = 2.4448 W 4 + 6.288

Ans.

Introduction to Circuit-Theory Concepts

2.31

2.4 Find the current through the galvanometer using delta-star conversion.

Solution Converting the delta consisting of 20 W, 30 W and 50 W, we get,

\

r1 =

20 ´ 30 =6W 20 + 30 + 50

r2 =

20 ´ 50 = 10 W 20 + 30 + 50

r3 =

30 ´ 50 = 15 W 20 + 30 + 50

RAC = 16 W

8 = 0.5 A 16 Now, to calculate potential difference between the points B and D ;

Main current i =

VXC = 10 ´ 0.5 = 5 V \ VBD = (10 ´ 0.25 – 5 ´ 0.25) = 1.25 V \ Currant through the galvanometer, (50 (W) 1.25 Ans. = 0.025 A 50 2.5 Twelve similar conductors each of R resistance form a cubical frame. Find the resistance across two opposite corners of the cube. iG =

2.32

Circuit Theory and Networks

Solution The configuration is shown in the figure.

The current distribution is shown. So, the total voltage drop between two opposite corners A and B for a total current of I is, 5 I I I VAB = R. + R. + R. = R ´ I 3 6 3 6 V 5 Equivalent resistance, RAC = AB = R Ans. 6 I 2.6 A regular hexagon is formed from 6 wires of R ohm each. The corners are joined to the centre by six more wires of 2R ohms each. Calculate the resistance of the hexagon between any two nodes diametrically opposite.

Solution The hexagon can be redrawn as shown.

2.33

Introduction to Circuit-Theory Concepts

The hexagon is symmetrical about XX¢ Equivalent resistance of the second quadrant, R1 = (2 R | | R / 2 + R ) || 4 R =

R1 A

28 R 27

B R1

So, the figure is modified as,

28 Ans. R 27 2.7 Find the input resistance of the infinite section resistive network shown below.

\

RAB = ( R1 || R1 ) + ( R1 || R1 ) = R1 =

Solution Let the equivalent resistance be Rin. The network can be terminated at A¢ B¢ instead of AB.

RA¢B ' = [ R + ( Rin ) || ( R)] By assumption,

Rin = R +

Rin R 2R R + R2 = in R + Rin R + Rin

Þ

RRin + Rin2 = 2 RRin + R 2

Þ

Rin2 - RRin - R 2 = 0

Þ

Rin =

R±

R2 + 4R2 R = [1 ± 2 2

5]

æ 5 + 1ö Taking positive sign, Rin = èç ÷R 2 ø 2.8 In the network shown, calculate the power input to each of the following elements when it is connected across A and B. (a) a resistance RAB of 59 W. (b) a voltage source of –160 V. Solution (a) Converting the two deltas into star, r1 =

18 ´ 6 =3W, 18 + 12 + 6

r3 =

18 ´ 12 =6W 36

and r11 =

r2 =

R1

6 ´ 12 = 2 W, 36

14 ´ 7 28 ´ 14 = 2 W, r21 = = 8 W, r31 = 4 W 49 49

R1

2.34

Circuit Theory and Networks

\

69 ´ 20 Req = æç 3 + + 2ö÷ = 20.5 W è ø 69 + 20

Main current, i =

2100 = 102.41 A 20.5

Current in 59W resistor, i59 = 102.41 ´

20 = 23.01 A 89

Power input, Pi = (i59 )2 ´ 59 = (23 × 01) 2 ´ 59 = 31248 ×189 W = 31 kW (b)

By KVL,

15i1 - 10i2 + 2260 = 0 and

30i2 - 10i1 - 160 = 0

Solving,

i1 = -206.285 A i2 = -63.43 A

\ Power input, Pi = v ´ i = -160 ´ (i1 - i2 ) = -160 ´ (-206 × 285 + 63 × 43) = 17.37 kW 2.9 The two-dimensional network of the figure consists of an infinite number of square meshes, each side of which has a resistance of R. Find the effective resistance between two adjacent nodes such as X and Y. Solution Let the current flowing into the circuit at node X be I. Since the infinite network is symmetrical about X, the current I in going from X to infinity, is divided equally along the branches XQ, XT, XP and XY.

Ans.

2.35

Introduction to Circuit-Theory Concepts

The current I then returns from infinity and is taken from the network at node Y. Again, by symmetry, the currents flowing along RY, XY, SY and TY are each I/4. Hence, the total current flowing along XY is VXY =

I I I + = . So, the voltage between X and Y, 4 4 2

I ´R 2

VXY R = Ans. 2 I 2.10 Use loop current analysis to find currents in all branches of the network of figure. Also, find the power delivered by 5A current source. All resistances are in ohm. Solution By KVL, So, the effective resistance between X and Y, RXY =

5i1 + 10i2 + 5(i2 - i4 ) + 15(i1 - i3 ) = 50 or,

20i1 + 15i2 - 15i3 - 5i4 = 50

and,

5(i4 - i2 ) + 30 + 10i4 + 20(i4 - i3 ) = 0

(i)

-5i2 - 20i3 + 35i4 = -30 or, By constraint equations, (i2 - i1 ) = 5 and i3 = 10 From Equation (i) and Equation (ii), 20(i2 - 5) + 15i2 - 15 ´ 10 - 5i4 = 50 or, 35i2 - 5i4 = 300 Þ 7i2 - i4 = 60 -5i2 + 35i4 = 170 Þ -i2 + 7i4 = 34 and, Solving i4 = 6.02083 A i2 = 4.4583 A, i2 = 9.4583 A i3 = 10 A and

(ii) (iii) (iv)

Power delivered by 5A current source = v ´ i = 110.83 ´ 5 = 554.16 W [To calculate the voltage across the 5A current source, v, writing KVL for Mesh (1),

5i1 + v + 15(i1 - i3 ) = 50 Þ v = 50 - 20i1 + 15i3 = 200 - 20 ´ 4.4583 = 110.83 V] 2.11 For the circuit, find the voltage Vx using nodal analysis. 40 W

ly 0.6 A

100 W 25 l y

50 W

+ Vx –

0.2 V x

+ –

2.36

Circuit Theory and Networks

Solution

By KCL at node (1),

- 0.6 + I y +

Vx v -v - 25I y + 1 2 = 0 50 40

(i)

By KCL at node (2),

v2 = 0.2 Vx and other constraint equation, V I y = x and v1 = Vx 100 From Equation (i), - 0 × 6 +

(ii)

(iii)

Vx V V v -v + x - 25 x + 1 2 = 0 100 50 100 40

Þ

-120 + 2Vx + 4Vx - 50Vx + 5Vx - 5 ´ 0.2Vx = 0

Þ

Vx =

120 = -3 V -40

Ans.

2.12 Use nodal analysis to find the voltages VA, VB and Vx in the circuit, in which I1 = 0.4 A.

Solution By KCL at node (A),

- 0.4 +

VA VA - VB + + 0.03Vx = 0 100 20

(i)

By KCL at node (B),

VB - VA VB VB - VC + + =0 20 40 40 Constraint equations,

(ii)

VB 40

(iii)

Iy =

Introduction to Circuit-Theory Concepts

VC = 80 I y

and

(iv)

(VA - VB ) = Vx and From Equation (i),

- 0.4 +

(v)

VA VA VB + + 0.03VA - 0.03VB = 0 100 20 20

Þ (9VA - 8VB ) = 40 From Equation (ii),

VA = VB

(vi)

[by Eq. (iii) and Eq. (iv)]

Thus, solving Equation (vi) VA = VB = 40 V \ Ans. Vx = (VA - VB ) = 0 2.13 For the circuit, use loop analysis to find I1 and the power absorbed by the 500 W resistor.

Solution Converting the dependent current source into dependent voltage source, By KVL, 50 = 0.083 A 600 Power absorbed by the 500 resistor 800 I1 - 200 I1 = 50 Þ I1 =

Ans.

2

50 ö = I12 R = æç ´ 500 è 600 ø÷ 500 Ans. = 3.47 W 144 2.14 Determine the currents in all the branches of the network. =

2.37

2.38

Circuit Theory and Networks

Solution By KVL, for Mesh (1)

5I1 + 10 I 2 + 10( I1 - I 2 ) + 5 I1 = 5 Þ 20 I1 = 5 Þ I1 = 0.25 A By KVL for Mesh (2),

5I 2 + 10 - 5 I1 + ( I 2 - I1 ) ´ 10 = 0 15 I 2 = 15 I1 - 10 = (3.75 - 10) = - 6.25 I 2 = -0.416 A \ Ans. 2.15 Obtain the current I in the network shown.

Solution By KVL for the second mesh

-3VR + 5 I + 4 + VR = 0 or,

-2VR + 5I + 4 = 0

Also,

VR = 2 ´ ( I - 2), putting this in Equation (i),

(i)

-2 ´ 2( I - 2) + 5I + 4 = 0 Þ

- 4 I + 8 + 5I + 4 = 0

I = -12 A Þ Ans. 2.16 The current and voltage profile of an element vs. time has been shown in figure. Determine the element and find its value. Voltage (V)

Current (A) 1A

0

5

5

Time (ms)

0

5

Time (ms)

Solution Here, the voltage is not proportional to the current; therefore, the element is not a resistance. Also, at t = 5ms, i ¹ 0, but the voltage suddenly drops to zero value, i.e., the element acts as short circuit. As the voltage across a capacitor cannot change instantaneously, the element is not a capacitor. Now, the current is zero at t = 0 and the voltage is zero at t = 5ms. Therefore, we conclude that the element is an inductor.

2.39

Introduction to Circuit-Theory Concepts

di 1 = = 200 A / s dt 5 ´ 10 -3

From the figure,

v=L

Q

di dt

Þ

L=

v 5 = = 25mH di 200 dt

2.17 The voltage across a capacitor of value C = 1 mF is shown in the figure. Find the current waveform. Solution Here, the voltage can be expressed as,

bg

v t =0 = 10t = 20 - 10t =0

and v = 5 V.

Ans.

v ( t ) (V) 10

t£0 0£ t £1 1£ t £ 2 t³2

0

1

2

t (s)

i ( t ) (mA)

10

0

1

t (s)

2

–10

Current waveform of the capacitor

bg

Since, i t = C

bg

dv t , we have the current given as, dt

bg

i t =0 = 10 -2 = -10 -2 =0

t 0 in Laplace domain is shown in the figure below. s

2

s

2

3 s

1 s

The equivalent network reduces to one as shown below. s+2

A

3 s

1 s

s+2

B

To find the current in L2, we have to find Thevenin’s equivalent circuit across the terminals A and B. The impedance between terminals A and B is given as,

Z Th = Z AB

1 s + 2g ´ b s = b s + 2g =

1 s + 2s + 1 s Short circuit current flowing from A to B is given as, s+2+

I sc =

2

=

b s + 2g bs + 1g

2

3/ s 3 = s+2 s s+2

b g

Therefore, the Norton’s equivalent circuit is shown in the figure below. Hence the current, s+2 1 3 ´ ´ 2 + 2 s s+2 s s +1 + s+2 2 s +1 3 = s s + 2 s2 + 2s + 2

IL =

b g b g b g b g b g b g b gd i

A IL 3 s ( s + 2)

( s + 2) ( s + 1) 2

B

s+2

5.46

Circuit Theory and Networks

By partial fraction expansion,

IL =

k k k3 k3 * 3 = 1+ 2 + + 2 2 1 1 1 - j1 s s s j s + + + + s s + 2 s + 2s + 2

b gd

i

Where,

k1 =

k2 =

k3 =

3

b s + 2g d s

IL =

+ 2s + 2

3 s s + 2s + 2

d

2

i

i

b gb

=

3 4

s=0

=-

3 4

s = -2

3 s s + 2 s + 1 - j1

k3* = - j

\

2

g

=j s = -1- j1

3 4

3 4

j3/4 j3/4 3/4 3/4 3 / 4 3/4 3/ 2 + = s s + 2 s +1 2 +1 s s + 2 s + 1 + j1 s + 1 - j1

b g

Taking inverse Laplace transform, we get the required current as, 3 3 3 i t = - e -2 t - e - t sin t Ans. 4 4 2 5.29 The following circuit has a dependent current source and an independent voltage source. Find the Thevenin equivalent network of the circuit across the terminals a and b.

bg

100 V +–

v1 100

20 W

a v1 b

Solution 100 V +–

v1 100

20 W

100 V +–

+

v1 –

20 W

I sc

v1 = 0

5.47

Network Theorems

With open circuit, v1 = voc . By KCL, vo c 100 + voc + = 0 Þ - vo c + 500 + 5 vo c = 0 Þ vo c = - 125 Volt 100 20 With short-circuit, v1 = 0 and the dependent current source is open, so that, I sc = -5 A -

v oc -125 = = 25 W I sc -5

125 V

a

+ –

Thus, Thevenin impedance, R T h =

25 W

b So, the Thevenin’s equivalent circuit is shown in the figure below. 5.30 In the network of figure, the switch K is closed at time t = 0, a steady state having previously existed. Obtain the current in the resistor R using Thevenin’s theorem.

K R1

R2

L1

L2

10 W

10 W

1H

1H

R3

100 V

10 W

R

10 W

Solution When the switch K is opened, under steady state condition, two inductors behave as short circuits. Therefore, the initial currents flowing through the inductors can be found out by writing the KVL equations for the circuit at t = 0–. 10 W

100 V

10 W

I2

10 W

I1

10 W

(a) Circuit at t = 0 – By KVL for the two meshes, 30 I 1 - 10 I 2 = 100 - 10 I 1 + 20 I 2 = 0

Solving, I 1 = 4 A, I 2 = 2 A Hence, the transform network for t > 0 is shown in Fig (b). L1I1= 4 V

L2I2= 2 V

+– 100 s

s

+–

10 W

s a 10 W

10 W

b

(b) Transform network for t > 0

ZL

5.48

Circuit Theory and Networks

Thevenin equivalent impedance with respect to the terminals a and b is given as, Z Th =

bs + 10g ´ 10 = 10bs + 10g s + 10 + 10 bs + 20g

To find the open circuit voltage across the terminals a and b, we have the current flowing in the left mesh,

I s =

bg

\

100 / s + 4 4 s + 100 = s + 10 + 10 s s + 20

b

VOC s = I s ´ 10 + 2 =

bg bg

g

4 s + 100 2 s 2 + 80 s + 1000 ´ 10 + 2 = s s + 20 s s + 20

b

b

g

g

Therefore, the Thevenin’s equivalent circuit is shown in Fig. (c).

2 V OC = 2 s + 80 s + 1000 s ( s + 20)

Z TH = 10( s + 10) ( s + 20) a

s

+ –

10 W

b

(c) Thevenin's equivalent circuit Hence, the current through the resistor R = 10 W is given as,

IL s =

bg

bg

VOC s = Z Th + R

2 s 2 + 80 s + 1000

=

2 s 2 + 80s + 1000 s s + 10 s + 30

gLMM10bsbs++2010g g + bs + 10gOPP b N Q

s s + 20

b

By partial fraction expansion, let,

I L ( s) = \

2 s 2 + 80s + 1000 K1 K2 K3 = + + s ( s + 10) ( s + 30) s s + 10 s + 30

LM 2s + 80s + 1000 OP = 10 N sbs + 10gbs + 30g Q 3 L2 s + 80s + 1000 OP = b s + 10gM N s bs + 10gbs + 30g Q L 2s + 80s + 1000 OP = b s + 30gM N sbs + 10gbs + 30g Q

K1 = s

2

s=0

\

2

K2

= -2

s = -10

\

2

K3

s = - 30

I L ( s) =

10 / 3 2 2 /3 + 3 s + 10 s + 30

=

2 3

gb

g

5.49

Network Theorems

Taking inverse Laplace transform, we get, 2 10 - 2e -10t + e -30t = 3.33 - 2e -10t + 0.67e -30t iL t = Ans. 3 3 5.31 For the network shown in the figure, show that the Thevenin equivalent at the terminals a-b is represented by,

bg

VTh =

V1 1 + a + b - ab 2

b

g

Z Th =

and

3-b 2

1W +–

1W

1W

I1 + –

V1

+ –

a

bI 1

aV 1 b

Solution When the terminals a-b are open-circuited no current will flow through the right side 1 W resistor. By KVL for the left mesh,

2 I1 + aV1 = V1 Þ

I1 =

V1 1- a 2

b g

\

VTh = 1 ´ I 1 + aV1 + bI 1 = 1 ´

\

VTh

V1 V V 1 - a + aV1 + +b 1 1 - a = 1 1 - a + 2a + b - ab 2 2 2

b g V = a1 + a + b - ab f a Proved f 2

b g

b

g

1

To find the Thevenin impedance we have to find the short-circuit current flowing through the terminals a-b. By KVL for the two meshes, we get, 1W

a

+–

1W

V1

1W

I1

I SC + –

+ –

bI 1

aV 1 b

2 I 1 - I SC = V1 1 - a

b g

and,

(i)

1 ´ I SC - I1 - bI1 + 1 ´ I SC = aV1 Þ - 1 + b I1 + 2 I SC = aV1

b

g

b g

(ii)

5.50

Circuit Theory and Networks

Solving (i) and (ii), we get,

I SC =

2

V1 1 - a

- 1+ b

b g

aV1

2

-1 2

b g

- 1+ b

b g

2 aV1 + V1 1 - a + b - ab V1 1 + a + b - ab = 4 -1- b 3-b

b

=

g

b

g

Therefore, the Thevenin impedance is,

Z Th

V1 1 + a + b - ab VTh 3-b = = 2 = I SC V1 1 + a + b - ab 2 3-b

b b

g g

bProved g

5.32 Find the Thevenin equivalent circuit for the network shown in figure at terminals A-B. 100 W

I = 5 –3 0° A

A

50 W

50 W

j50 W

j50 W B

Solution When the terminals A and B are open-circuited, the current flowing through the right branch (50 + j50) W is,

50 + j50 100 + 50 + j50 + 50 + j50 50 + j50 = 5Ð30o ´ 200 + j100 1+ j = 5Ð30o ´ 4 + j2

I = 5Ð30o ´

FG H FG H

IJ K

IJ K

Therefore, Thevenin voltage is,

VTh = I ´ 50 + j 50 = 5Ð30 o ´

b

g

FG 1 + j IJ ´ b50 + j50g = 1118. Ð93.43 V H 4 + j 2K o

Thevenin impedance is given as,

Z Th = 150 + j 50 50 + j 50 150 + j 50 ´ 50 + j50 = 150 + j50 + 50 + j50

b b b

gb g b g b

= 50Ð36.87 o W

g g g

5.51

Network Theorems 100 W A

50 W

50 W Z Th

j50 W

j50 W B

Thus, Thevenin equivalent circuit is shown in the figure below. 50–36.87° V a + –

111.8–93.43° V

V Th b

5.33 Find Thevenin’s equivalent circuit across the terminals A and B for the network shown in the figure. i0

12 V

10 W

A

+ –

2i0

5W

B

Solution The circuit has both dependent and independent sources. We find VTh and ISC and then taking the ratio we get ZTh. To find VTh: By KVL for the supermesh shown, i0

10 W

A +

12 V

+ –

2i0

5W

V TH –

B

(a) 10i0 + VTh - 12 = 0

Þ By KCL at node A,

VTh = 12 - 10i0

- i0 - 2i0 + Þ

VTh =0 5 VTh = 15i0

(i)

(ii)

5.52

Circuit Theory and Networks

From (i) and (ii) we get, VTh = 7.2 V 10 W

i0

12 V

Ans. A

+ –

I SC

2i0

B

(b) To find ISC When the terminals A and B are shorted, no current flows through the 5 W resistance. The circuit is shown in Fig. (c). 2W A + –

7.2 V

B

(c) By KVL for the supermesh, 10i0 = 12 Þ i0 = 1.2 A

By KCL at node A, I SC = 3i0 = 3.6 A

Ans.

Therefore, the Thevenin impedance is given as,

VTh 7.2 Ans. = =2W I SC 3.6 5.34 Find Thevenin’s equivalent circuit for the network shown in the figure. Z Th =

10 W

20 W

+ v0

0.5v 0

–

Solution This circuit does not have any independent source; it has only a dependent current source. Therefore, the Thevenin equivalent voltage will be zero. \

VTh = 0

Ans.

To find Thevenin equivalent impedance, we connect a test current source of value I. Let the voltage across this test source be V.

5.53

Network Theorems

By KCL at node x,

V - v0 - 0.5v 0 - I = 0 10 V = 10 I + 6v0

Þ

10 W

(i)

x +

20 W

+ v0

0.5v 0

I V

– –

(a) Also,

v0 = 20 ´ 0.5v0 + I = 10v0 + 20 I

Þ

v0 = -

b

g

20 I 9

(ii)

Putting the value of v0 in (i) from (ii),

3.33 W

F H

V = 10 I + 6 ´ -

I K

10 V == -3.33 W 3 I The Thevenin equivalent circuit is shown in Fig. (b). \

A

20 10 I =- I 9 3

ZTh =

Ans.

B

(b)

Maximum Power Transfer Theorem 5.35 In the network shown, the power dissipated in R when E1, E2 or E3 acting alone is (a) 20 W, 80 W, and 5 W respectively. (b) 30 W, 270 W, and 120 W respectively. Calculate the maximum power that R can dissipate due to the simultaneous action of all the sources. Calculate both for (a) and (b). What will be the minimum power dissipated in R when all the sources are acting simultaneously? Solution Current for E1 at R, i1 = ± Current for E2 at R, i2 = ±

P1 R P2 R

P3 R \ Total current flow for simultaneous action of all the three sources is, Current for E3 at R, i3 = ±

i = ± i1 ± i2 ± i3 = ±

P1 ± R

P2 ± R

P3 R

5.54

Circuit Theory and Networks

é P Power, P = i R = ê ± 1 ± \ R ëê l For maximum power, 2

Pmax = [ P1 + (a) Pmax = [ 20 +

l

P2 +

80 +

P2 ± R

2

P3 ù ú R = [ ± P1 ± Rú û

P2 ±

P3 ]2

5 ]2 = [2 5 + 4 5 +

5 ]2 = 49 ´ 5 = 245 W

(b) Pmax = [ 30 + 270 + 120 ] 2 = [4 5 - 3 5] 2 = 1080 W For minimum power, (a) Pmin = [ - 20 + 80 -

P3 ]2

5 ]2 = [4 5 - 3 5 ]2 = 5 W

Ans.

Ans.

Ans.

(b) Pmin = [ - 30 + 270 - 120 ]2 = [ - 30 + 3 30 - 2 3]2 = 0 W Ans. 5.36 Find the value of R in the circuit of the figure such that maximum power transfer takes place. What is the amount of this power? (a)

(b)

(c)

Solution (a) Removing the résistance R, \

3i1 - 2i2 = 4

and

-2i1 + 8i2 = 0

Solving,

i2 =

\

1 ´ i2 + 6 = Voc

Þ

2 32 Voc = æç 6 + ö÷ = V è 5ø 5

2 A 5

5.55

Network Theorems

Also, to find the Rth,

17 ´1 é 1´ 2 ù 2 17 æ ö + 5ú | | [1] = ç + 5÷ | |1 = 3 = W Rth = ê è ø + 1 2 3 17 20 ë û +1 3

( )

\ For maximum power transfer, R = Rth =

17 = 0.85 W 20

Ans.

Voc2 Ans. = 12 W 4R (b) In the network, 2 W resistor is connected in parallel with an ideal voltage source of 5 V; hence this resistance can be removed without affecting the current flows in the other branches.

\ Maximum power Pmax =

Converting the voltage source into current source,

( )

5 11 +2 A = A 3 3

For maximum power transfer, R =

7 W 4

2

Maximum Power, Pmax

æ 11ö è 4ø = = 1.08 W 4 ´ 7/4

Ans.

5.56

Circuit Theory and Networks

(c) To find Rth Rth =

10 ´ 5 + 2 = 5.33 W 10 + 5

Ans.

To find Voc i=-

24 = - 1.6 A 15

\

Voc = 5i + 10 = - 8 + 10 = 2 V

\

Pmax =

4 = 0.188 W 4 ´ 5.33

Ans.

5.37 In the network shown, find the value of ZL to which the maximum power can be delivered. Hence, find the value of the maximum power.

Solution With respect to terminals A and B, the Thevenin voltage is,

Vth =

j3 æ ö 45Ð0° 5Ð0° ´ = = 2.236Ð - 26.56° (V) j 3(3 - j 3) çè 3 + j 3 - j 3 ÷ø 18 + j 9 3+ 3 - j3 + j3

and Thevenin impedance,

Z th

3 ´ j3 ö æ çè 3 + 3 + j 3 ÷ø ´ (- j 3) = = 3Ð- 53.12° W = (1.8 - j 2.4) W 3 ´ j3 - j3 3+ 3 + j3

For maximum power transfer, Z L = Z th* = (1.8 + j 2.4) W \

Current, I =

Ans.

2.236Ð - 26.56° = 0.621Ð - 26.56° A 1.8 ´ 2

The value of the maximum power is, Pmax =

(Vth ) 2 (2.236) 2 = = 0.694 W 4R 4 ´ 1.8

Ans.

5.38 A loudspeaker is connected across terminals A and B of the network. What should its impedance be to obtain maximum power dissipation in it?

Network Theorems

5.57

(a)

(b)

Solution (a) Equivalent impedance with respect to the terminals A and B is, (3 + j 4)( - j 5) = 7.9Ð - 18.43° W = (7.5 - j 2.5) W 3 + j 4 - j5

Z th =

For maximum power transfer, Z L = Z th* = (7.5 + j 2.4) W Ans. (b) Equivalent impedance with respect to the terminals A and B is,

æ -40 + j 50 + 40 + j 52 + j 60 - 78 ö é (10 + j8) j 5 ù Z th = ê + 4 + j 6 ú | | 10 = ç | |10 j j + + 10 8 5 10 + j13 è ø÷ ë û = 6.14Ð30° W = (5.316 + j 3.07) W For maximum power transfer, Z L = Z th* = 6.14Ð - 30° W = (5.316 - j 3.07) W 5.39 Two inductors each of 1 W reactance and negligible resistance are connected in series across a 2 V a. c. source. Find the value of resistance which should be connected across one of the inductors for maximum power dissipation. Also, find the maximum power. Solution Here, Z = \

R ´ j1 -1 + j 2 R + j1 = R + j1 R + j1

Current I =

2Ð0° 2Ð0° ´ ( R + j1) = Z -1 + j 2 R

\ Current through the resistance, I R = I ´ \ Power, P = | I |2 R =

4R 1 + 4R2

j1 j2 = R + j1 -1 + j 2 R

5.58

Circuit Theory and Networks

For maximum power, Þ

(1 + 4 R 2 ) ´ 4 - 4 R ´ 8 R dP =0 Þ =0 dR (1 + 4 R 2 ) 2

R = 0.5 W

Ans.

\ Maximum Power, Pmax =

4 ´ 0.4 =1W 1 + 4 ´ (0.5) 2

Ans.

5.40

In the network shown, calculate the maximum power that may be dissipated in the external resistor R. Solution Transforming the current source into voltage source, By KVL, 6i1 + 4i1 - 40 - 2i1 = 0 Þ

i1 = 5 A

eoc = 6i1 = 30 V \ For maximum power, R = Req Shorting the terminals a–b and solving by loop method, I sc = 5 A 30 =6W 5

\

Rth =

\

Pmax =

(30)2 900 = = 37.5 W 4´6 24

5.41 Find the Thevenin’s equivalent between the points a and b for the circuit given in the figure. What should be the value of impedance connected between a and b for maximum power to be transferred from the sources? Obtain the amount of the maximum power. 2W

j6 W a

100 V

+ –

(3 + j 5) W

b

Solution Here the current I is,

I=

100 100 = = 10 - j10 A 2 + 3 + j5 5 + j5

b

g

5.59

Network Theorems

\

VTh = I ´ 3 + j5 = 10 - j 10 ´ 3 + j 5

b

g b

g b

g

= 80 + j 20

b

g

= 82.46Ð14 o Volt

Ans.

2 ´ 3 + j5 . + j 6.4 W = 16 2 + 3 + j5 Thevenin’s equivalent circuit is shown.

b

Z Th = j 6 +

\

g b

g

Ans.

(1.6 + j6.4) W a + –

82.46–14° V

V Th b

For maximum power transfer, the impedance should be complex conjugate of Thevenin impedance.

. - j 6.4 W Z L = 16

\

b

g

Ans. 2

82.46 VTh 2 Ans. = = 1062.5 Watt 4R 4 ´ 16 . 5.42 In the network in the figure, two voltage sources act on the load impedance connected to the terminal A and B. If the load is variable in both reactance and resistance, for what load, ZL will receive maximum power? What is the value of maximum power?

b

Amount of the maximum power is, Pmax =

5W

50 –0° (V)

j5 W

3W

A

+ –

g

–j4 W

+ –

ZL

25–90° (V)

B

Solution

Z Th

V1 = 50Ð0 = 50 V; and V2 = 25Ð90 = j 25 V o

Here, Current in the circuit,

I=

a

o

50 - j 25 50 - j 25 = 5 + j5 + 3 - j 4 8 + j1

A

V Th

Thevenin voltage,

+ – b

VTh = 50 - I ´ 5 + j 5 = 50 -

b

g

= 9.8Ð - 78.7 o . = 1923 - j 9.615

b

g

Volt

FG 50 - j 25IJ ´ b5 + j5g = 25 - j 75 8 + j1 H 8 + j1 K Ans.

5.60

Circuit Theory and Networks

Thevenin impedance, Z Th =

b5 + j5g ´ b3 - j4g = 35 - j5 = b4.23 - j 1154 . g b5 + j5g + b3 - j 4g 8 + j 1

W

Ans.

Thus, the Thevenin’s equivalent circuit is shown.

. W For maximum power transfer to the load, Z L = Z m * = 4.23 + j 1154

b

g

Ans.

VTh 2 9.82 Ans. = = 5.676 Watt 4R 4 ´ 4.23 5.43 A network has two output terminals. The open circuit voltage at these terminals is 260 V. The current flowing through the terminals is 20 A when the terminals are short circuited. Also, the current is 13 A when a coil of 11 ohm reactance and negligible resistance is connected across the terminals. Find the impedance components of the equivalent circuit feeding the terminals. What value of load impedance will give maximum power transfer and what is the value of this power? Solution Here, VTh = 260 V; I SC = 20 A The value of the maximum power is, Pmax =

Let the Thevenin impedance across the terminals is, Z = R + j X

b

g

VTh 260 = = 13 W I SC 20

\

Z=

\

R 2 + X 2 = 169

(i)

When 11 ohm reactance is connected across the terminals, the current is 13 A. \ \

260 = 13 Þ R + j X + 11

R + j X + 11 =

b

b

R2

g + b X + 11g = 400

g

260 = 20 13

2

(ii)

Solving (i) and (ii), we get,

R = 12 W X = 5 W Therefore, Thevenin impedance, Z T h = 12 + j5 W

b

g

Ans.

For maximum power transfer, Z L = Z Th * = 12 - j5 W

b

g

Ans.

2

260 V 2 Value of maximum power, Pmax = Th = Ans. = 1408.33 W 4R 4 ´ 12 5.44 What should be the value of ZL for maximum power to be delivered in the circuit shown in the figure?

b g

3 + j2

5 cos( wt + 30°)

+ –

4 – j3

ZL

+ –

2 cos wt

Solution In this circuit, when the voltage sources are replaced by their internal impedances; i.e., when they are short-circuited, the equivalent Thevenin impedance with respect to the load terminals is given by,

5.61

Network Theorems

11 + j IW b g b4 - j3g = bb33 ++ jj22gg ´+ bb44 -- jj33gg = 187 -- jj11 = FH 127 50 50 K = b2.54 + j 0.22g W = 2.55Ð4.95 bWg

Z Th = 3 + j 2

o

For maximum power to be delivered, the load impedance should be complex conjugate of the Thevenin impedance, so that, Z L = Z Th * = 2.54 - j 0.22 W = 2.55Ð - 4.95o W

b

g

b g

Ans.

Reciprocity Theorem 5.45 Verify the Reciprocity Theorem for the network shown in the figure using current source and a voltmeter. All the values are in ohm.

Solution Using a current source and a voltmeter, Let, e1, e2 be node voltages, v1 be the voltmeter reading.

By KCL, At node (1)

Þ

3e1 - e2 - 2i1 = 0

(i)

At node (2)

Þ

-6e1 + 13e2 - 3v1 = 0

(ii)

9v1 = 5e2

(iii)

At node (3) From (ii)

From (i)

9 v - 3v1 = 0 5 1

Þ

-6e1 + 13 ´

Þ

-6e1 +

Þ

6e1 +

Þ

3´

Þ

æ i1 ö æ 21ö =ç ÷ èç v1 ø÷ è 5 ø

(

)

117 - 3 v1 = 0 5

102 17 v Þ e1 = v 5 1 5 1

17 9 v - v = 2i 5 1 5 1

(A)

5.62

Circuit Theory and Networks

Interchanging the positions of the current source and the voltmeter, Now, let v2 be the voltmeter reading

By KCL, At node (1)

Þ

3v2 = e2

At node (2)

Þ

-6v2 + 13e2 - 3e3 = 0

Þ

-6v2 + 13 ´ 3v2 - 3e3 = 0

Þ

e3 = 11v2

Þ

5e3 - 5e2 + 4e3 - 20i2 = 0

Þ

20i2 = 9e3 - 5e2 = 9 ´ 11v2 - 5 ´ 3v2 = 84v2

Þ

æ i2 ö æ 21ö =ç ÷ èç v2 ø÷ è 5 ø

At node (3)

(iv)

(v)

(B)

From equations (A) and (B), Reciprocity theorem is proved. 5.46 Solve the network shown in Figure (a) and hence find the current in the 2 W resistor in Figure (b) when an emf of 36 V is added in the branch BD as shown in Figure 7(b). All values are in ohm.

]

(a) Solution l Solve by any method of network analysis. l We consider the 36 V source acting alone. When 72 V sourer is acting alone, by network analysis, The current in 2 W resistor = 6 A and in 18 W resistor = 1 A

(b)

5.63

Network Theorems

(a) By Reciprocity theorem, 72 36 = Þ I = 0 ×5 A 1 I

(b)

[Here, I = Current in 2 W resistor when 36 V source is acting alone] \ Current in 2W resistor for simultaneous action of two sources

I = (6 - 0.5) = 5.5 A 5.47 An e.m.f. source E, having negligible internal impedance is connected in series with an impedance Z1 to the input terminals 1–2 of a linear, bilateral four terminal network. It produces a current I2 in impedance ZL connected across the output terminals 3–4. The emf source is now transferred so as to act, in series with Z2, between terminal 3–4. Z1 is disconnected and the input terminals 1–2 are short circuited. The short-circuited current traversing terminals 1–2 is then I1. Prove that the impedance looking into terminals 1–2 under the first condition is, Z12 =

Z1 I 2 I1 - I 2

Solution Let the impedance looking into terminals 1–2 be Z12. Thus the network becomes: \

I=

E Z1 + Z12

\ Voltage across 1–2, V12 =

E ´ Z12 Z1 + Z12

So, the circuit becomes as shown. The given network is linear and bilateral and according to the reciprocity theorem, if the source E is put across terminals 1–2, the response current flowing through Z2 will be I1 as shown. Now, if a voltage equal to V12 is applied instead of E, the current flowing through Z2 will be,

5.64

Circuit Theory and Networks

I1 I E ´ Z12 Z12 ´ V12 = 1 ´ = I1 ´ E E Z1 + Z12 Z1 + Z12 But, this current is equal to I2. Z12 \ I 2 = I1 Z1 + Z12

æ ZI ö Z12 = ç 1 2 ÷ (Proved) è I1 - I 2 ø 5.48 Verify the reciprocity theorem for the ladder network shown in figure. Þ

Solution Let, the three loop currents be I1, I2, and I3. By KVL for the three loops,

(20 + j10) I1 - j10 I 2 = 200Ð45° - j10 I1 + 20 I 2 + j10 I 3 = 0 j10 I 2 + (10 - j10) I 3 = 0 Solving for I3,

(20 + j10) - j10 200Ð45° - j10 20 0 j10 0 0 200Ð45° ´ 100 = I3 = + j j 200 + 100) - j10( j100 + 100) (20 10)(200 (20 + j10) - j10 0 - j10 j10 20 j10 (10 - j10) 0 = 2.169Ð57.53° (A) Now by interchanging the positions of the voltage source and the response current, we get, By KVL,

(20 + j10) I1 - j10 I 2 = 0

Network Theorems

5.65

- j10 I1 + 20 I 2 + j10 I 3 = 0 j10 I 2 + (10 - j10) I 3 = 200Ð45° Solving for I1,

- j10 0 0 0 20 0 Ð ° j 200 45 10 0 = 2.169Ð57.53° (A) I1 = (20 + j10) - j10 0 - j10 j10 20 j10 (10 - j10) 0 Since the currents in both the cases are the same, reciprocity theorem is verified. 5.49 In this circuit, find voltage V. Interchange the current source and resulting voltage V and show that the reciprocity theorem is verified.

Solution Here, the current I 2 = 5Ð90° ´

5 + j5 = 4.64Ð111.8° (A) 5 + j5 + 2 - j 2

\ The voltage, V = I 2 ´ Z C = 4.64Ð111.8° ´ (- j 2) = 9.28Ð21.8° (V) Now, interchanging the positions of the current source and the finding the resulting voltage, we get, I1 = 5Ð90° ´

- j2 - j 2 + 5 + 2 + j5

= 1.31Ð- 23.2° (A) \ The voltage, V = 1.31Ð- 23.2° ´ (5 + j 5) = 1.31Ð- 23.2° ´ 7.075Ð45° = 9.28Ð21.8° (V) As V is same as obtained before interchanging the position of the current source, reciprocity theorem is verified.

5.66

Circuit Theory and Networks

5.50 In the given circuit of the figure, find the reading of the voltmeter V. Interchange the current source and voltmeter and verify the reciprocity theorem.

1 j W

j1 W 1 –0° (A)

1W

Solution Here, the current I 2 = 1Ð0 o ´

1W

V

1 + j1 = 0.707Ð45o A 1 + j1 + 1 - j1

bg

bg

b g

\ the voltage, V = I 2 ´ Z C = 0.707Ð45o ´ 1 = 0.707Ð45o V I1

I2

j1 W

I1 1 j W

I2

j1 W

1 –0° (A)

1 j W

V

1W

1W

1W

V

1W

1 – 0° (A)

Now, interchanging the positions of the current source and the finding the resulting voltage, we get,

I 1 = 1 Ð0o ´

1 = 0.5 Ð0o A 1 - j1 + j1 + 1

bg

\ the voltage, V = 0.5Ð0o ´ 1 + j 1 = 0.5Ð - 23.2 o ´ 2 Ð45o = 0.707 Ð 45o V As ‘V’ is same as obtained before interchanging the position of the current source, reciprocity theorem is verified.

b

g

bg

Millman’s Theorem 5.51 Find the load current using Millman’s theorem. All values are in ohm. Solution Here, E1 =1 V, E2 = 2 V, E3 = 3 V Z1 = 1 J, Z2 = 2 J, Z3 = 3 J 1 J 3 By Millman’s theorem, the equivalent circuit is shown.

\

Y1 = 1 J, Y2 = 0.5 J, Y3 =

3

å EiYi

\

E=

i =1 3

å Yi

i =1

=

1 ´ 1 + 2 ´ 0.5 + 3 ´ 1 + 0.5 +

1 3

1 3 = 3 = 18 V 11 11 6

Network Theorems

and

Z=

1

=

3

å Yi

5.67

6 W 11

i =1

18 E 18 9 = 11 = = I= A \ Ans. Z + 10 6 116 58 + 10 11 5.52 Obtain the potential of node F with respect to node G in the circuit of the figure. All values are in ohm.

Solution By Millman’s theorem, equivalent voltage is, 5

V =

å Ei Yi

i =1 5

=

å Yi

1 ´ 1 - 2 ´ 1 / 2 + 3 ´ 1/ 3 - 4 ´ 1/ 4 + 5 ´ 1/ 5 60 = V 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 /5 137

i =1

Equivalent impedance, Z =

1 5

å Yi

=

1 60 W = 1 + 1 / 2 + 1/ 3 + 1/ 4 + 1/ 5 137

i =1

Therefore, the current through the 6 W resistance is, V 60 /137 60 = = A I= Z + 6 60 /137 + 6 882 Hence, the voltage between the points F and G is, 60 60 = Volt V FG = 6 ´ I = 6 ´ 882 147 5.53 In the network, two voltage sources act on the load impedance connected to terminals a, b. If the load is variable in both reactance and resistance, what load ZL will receive the maximum power? What is the value of the maximum power? Use Millman’s theorem.

5.68

Circuit Theory and Networks

Solution

V1 = 50Ð0° = 50 V; Z1 = (5 + j 5) W; Y1 =

1 1 = = (0.1 - j 0.1) J Z1 (5 + j 5)

V2 = 25Ð90° = j 25 V; Z 2 = (3 - j 4) W; Y2 =

1 1 = = (0.12 + j 0.16) J Z 2 (3 - j 4)

\ Millman voltage source, Vm =

V1Y1 + V2Y2 50(0.1 - j 0.1) + j 25(0.12 + j 0.16) = = 9.807Ð- 78.65° (V) Y1 + Y2 (0.1 - j 0.1) + (0.12 + j 0.16)

\ Millman impedance,

Zm =

1 1 = = 4.385Ð-15.25° = (4.23 - j1.15) W Y1 + Y2 0.22 - j 0.06

For maximum power transfer to the load, Z L = Z m* = (4.23 + j1.15) W \ Maximum power, Pmax =

Vm2 (9.807) 2 = = 5.68 W 4 RL 4 ´ 4.23

Ans.

Ans.

5.54 Calculate the load current I in the circuit in the figure by Millman’s theorem. I

2W

5W

2W

15 W + –

2V

3V

+ –

+ –

5V

Solution By Millman’s theorem, equivalent voltage,

2

3

5

2

2

5

å EY = 2 + 2 + 5 = 35 = 2.91667 Volt V= å Y 1 + 1 + 1 12 and equivalent impedance,

1 10 = = 0.833 W 1 1 1 12 + + 2 2 5 Therefore the current through the load resistance, Z=

I=

1

åY

=

V 2.91667 = = 0.184 A Z + 15 0.833 + 15

Ans.

5.55 Use Millman’s theorem to obtain an equivalent current source for the circuit shown in the figure. Also obtain the equivalent voltage source.

5.69

Network Theorems

20 W

10 W 100 –0° mA

15 W 20 –0° mA

j30 W

j20 W

j20 W + –

4–30° V

Solution We convert the voltage source into equivalent current source as,

V 4Ð30o . = = 01108 Ð - 26.3o A Z 20 + j 30 The modified circuit is shown in figure. I=

100 –0° mA

j20 W

110.8 ––26.3° mA

10 W

bg

20 W

15 W 20 –0° mA

j30 W

j20 W

Total equivalent current source is,

b

g

b g

I = 100Ð 0o + 110.8Ð - 26.3o - 20Ð0o = 89.33 - j 49.1 = 101.93Ð - 28.8o mA

Ans.

Total equivalent impedance is obtained as,

1 1 1 1 = + + = 0.059 - j 0.095 Z 10 + j 20 20 + j 30 15 + j 20

b

Þ

Z = 4.73 + j 7.57 W

b

g

g

Ans.

Equivalent voltage source is obtained as,

b

g

b g

V = 101.93Ð - 28.8o ´ 10 -3 ´ 4.73 + j 7.57 = 0.9Ð29.2 o V

Ans.

5.56 A symmetrical 440V, 3-phase system supplies a star-connected load. The branch impedances are

Z R = 10Ð30o W, ZY = 12Ð45o W, Z B = 15Ð40o W . Assuming the neutral of the supply to be earthed, calculate the voltage to earth of the star point. Assume the phase sequence RYB. Solution Here, line voltages are, VRY = 400 Ð 0° ; VYB = 400 Ð - 120° ; VBR = 400 Ð + 120° \ Phase voltages are 400 Ð - 30o = 254Ð - 30o V; 3 VY = 254Ð - 120o - 30o = 254Ð - 150o V; VB = 254Ð + 120o - 30o = 254Ð90o V VR =

5.70

Circuit Theory and Networks

Phase admittances are,

1 1 = = 01 . Ð - 30o S Z R 10Ð30o 1 1 = = 0.0833Ð - 45o S YY = ZY 12Ð45o 1 1 = = 0.0667Ð - 40o S YB = Z B 15Ð40o YR =

bg

bg bg

By Millman’s theorem, the voltage of the load star point with respect to earth is, VR YR + VY YY + VB YB YR + YY + YB . Ð - 30o + 254Ð - 150o ´ 0.0833Ð - 45o + 254Ð90o ´ 0.0667Ð - 40o 254Ð - 30o ´ 01 = . Ð - 30o + 0.0833Ð - 45o + 0.0667Ð - 40o 01 o = 18.59Ð - 119 Ans. . V

VN ¢N =

bg

MULTIPLE-CHOICE QUESTIONS 5.1 Millman’s theorem yields (a) equivalent voltage source. (b) equivalent voltage or current source. (c) equivalent resistance. (d) equivalent impedance. 5.2 The superposition theorem is applicable to (a) current only. (b) voltage only. (c) both current and voltage. (d)current, voltage and power. 5.3 Superposition theorem is not applicable for (a) voltage calculations. (b) bilateral elements (c) power calculations. (d) passive elements. 5.4 Thevenin’s theorem can be applied to calculate the current in (a) any load. (b) a passive load only. (c) a linear load only. (d) a bilateral load only. 5.5 Norton’s equivalent circuit consists of (a) voltage source in parallel with impedance. (b) voltage source in series with impedance. (c) current source in parallel with impedance. (d) current source in series with impedance. 5.6 The superposition theorem is applicable to (a) linear responses only. (b) linear and non-linear responses. (c) linear, non-linear and time-variant responses. 5.7 When a source is delivering maximum power to a load, the efficiency of the circuit (a) is always 50%. (b) depends on the circuit parameters. (c) is always 75%. (d) none of these.

Network Theorems

5.71

5.8 Maximum power transfer occurs at a (a) 100% efficiency. (b) 50% efficiency. (c) 25% efficiency. (d) 75% efficiency. 5.9 Which of the following statements is true? (a) A Norton’s equivalent is a series circuit. (b) A Thevenin’s equivalent circuit is a parallel circuit. (c) R-L circuit is dual pair. (d) L-C circuit is a dual pair. 5.10 For a linear network containing generators and impedances, the ratio of the voltage to the current produced in other loop is the same as the ratio of voltage and current obtained if the position of the voltage source and the ammeter measuring the current are interchanged. This network theorem is known as (a) Millman’s theorem. (b) Norton’s theorem. (c) Tellegen’s theorem. (d) Reciprocity theorem. 5.11 Under conditions of maximum power transfer from an ac source to a variable load (a) the load impedance must also be inductive, if the generator impedance is inductive. (b) the sum of the source and load impedance is zero. (c) the sum of the source reactance and load reactance is zero. (d) the load impedance has the same phase angle as the generator impedance. 5.12 Consider the following statements The transfer impedances and admittances of a network remain constant when the position of excitation and response are interchanged if the network 1. is linear 2. consists of bilateral elements 3. has high impedance or admittance as the case may be. 4. is resonant. Out of above these statements (a) 1 and 2 are correct. (b) 1, 3 and 4 are correct. (c) 2 and 4 are correct. (d) 1, 2, 3 and 4 are correct. 5.13 In a linear network, the ratio of voltage excitation to current response is unaltered when the position of excitation and response are interchanged. This assumption stems from the (a) principle of duality. (b) reciprocity theorem. (c) principle of superposition. (d) equivalence theorem. 5.14 If all the elements in a particular network are linear, then the superposition theorem hold when the excitation is (a) dc only (b) ac only (c) either ac or dc (d) an impulse. 5.15 An a.c source of voltage Es and an internal impedance of Zs =(Rs + jXs) is connected to a load of impedanceZL = (RL + jXL). Consider the following conditions in this regard 1. XL = Xs, if only XL is varied. 2. XL = Xs, if only XS is varied. 3. RL =

RS2 + ( X S + X L )2 , if only RL is varied.

4. | Z L | = | Z S | if the magnitude of ZL is varied, keeping the phase angle fixed.

5.72

5.16

5.17

5.18

5.19

5.20 5.21

5.22

5.23

Circuit Theory and Networks

Among these conditions, those which are to be satisfied for maximum power transfer from the source to the load would include (a) 2 and 3 (b) 1 and 3 (c) 1, 2 and 4 (d) 2, 3 and 4 Reciprocity theorem is applicable to a network 1. which contains R, L and C as elements. 2. which is initially relaxed system. 3. which has both independent and dependent sources. Tick out the correct combination from the combination given above (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3. Reciprocity theorem is applicable to (a) circuits with one independent source (b) circuits with only one independent source and no dependent source (c) circuits with any number of independent sources (d) circuits with any number of sources. Substitution theorem is applicable for a network which has 1. unique solution. 2. one or two non-linear elements. 3. one non-linear or time-varying element. Choose the correct combination from the combination given above (c) 2 and 3 (d) 1, 2 and 3. (a) 1 and 2 (b) 1 and 3 Substitution theorem applies to (a) linear networks. (b) non-linear networks. (c) linear time-invariant networks. (d) any networks. Which of the following theorems is applicable for both linear and non-linear circuits? (a) Superposition (b) Thevenin (c) Norton (d) none of these. A network is composed of two sub-networks N1 and N2 as shown in the given figure. If the sub-network N1 contains only linear, bilateral, time-invariant elements, then it can be replaced by its Thevenin equivalent even if the sub-network N2 contains (a) a two-terminal element which is non-linear (b) a non-linear inductance mutually coupled to an element in N1 (c) an element which is linear, but mutually coupled to some element in N1 (d) a dependent source the value of which depends upon the voltage or current in some element in N1. A certain network consists of two ideal identical voltage sources and a large number of ideal resistors. The power consumed in one of the resistors is 4W when either of the two sources is active and the other is replaced by a short-circuit. The power consumed by the same resistor when both the sources are active would be (a) zero or 16 W (b) 4 W or 8 W (c) zero or 8 W (d) 8 W or 16 W. If a network has all linear elements except for a few non-linear ones, then superposition theorem (a) cannot hold at all. (b) always holds. (c) may hold on careful selection of element values, source waveform and response. (d) holds in case of direct current excitations.

5.73

Network Theorems

5.24 The maximum power that can be dissipated in the load in the circuit shown in figure is

(a) 3 W (b) 6 W (c) 6.75 W 5.25 If Rg in the circuit shown in figure is variable between 20 W and 80 W, then the maximum power transferred to the load RL will be (a) 15 W (b) 13.33 W (c) 6.67 W (d) 2.4 W 5.26 Thevenin impedance across the terminals AB of the given network is

(d) 13.5 W

10 20 13 11 W (b) W (c) W (d) W 3 9 4 5 5.27 The V–I relation for the network shown in the given box is V = 4I – 9. If now a resistor R = 2 W is connected across it, then the value of I will be

(a)

(a) –4.5 A (b) –1.5 A (c) 1.5 A (d) 4.5 A 5.28 In the network shown in the figure, the effective resistance faced by the voltage source is

5.74

Circuit Theory and Networks

(a) 4 W

(b) 3 W

(c) 2 W

(d) 1 W

5.29 For the network shown in the figure, if Vs = V1 and V = 0, then I = –5 A and if Vs = 0 then I =

1 A. 2

The values of ISC and R1 of the Norton’s equivalent across AB would be respectively

(a) –5 A and 2 W (b) 10 A and 0.5 W (c) 5 A and 2 W (d) 2.5 A and 5 W 5.30 In the network shown in the given figure, the Thevenin source and the impedance across terminals A–B will be respectively

(a) 15 V and 13.33 W (b) 50 V and 15 W (c) 115 V and 20 W (d) 100 V and 25 W 5.31 Which one of the following combination of open-circuit voltage and Thevenin’s equivalent resistance represents the Thevenin’s equivalent of the circuit shown in the given figure?

(a) 1 V, 10 W (b) 1 V, 1 kW (c) 1 mV, 1 kW (d) 1 mV, 10 W 5.32 For the circuit shown in the given figure, the current through R, when VA = 0 and VB = 15 V is I ampere. Now, if both VA and VA are increased by 15 V, then the current through R will be

(a) I ampere

(b)

I ampere 2

(c) 3I ampere

(d)

I ampere 3

5.75

Network Theorems

5.33 Thevenin’s equivalent circuit of the network shown in the given figure, between terminals T1 and T2 is

(a)

(b)

(c)

(d)

5.34 The Thevenin equivalent of the network shown in Figure (a) is 10 V in series with a resistance of 2 W. If now, resistance of 3 W is connected across AB in Figure (b), the Thevenin equivalent of the modified network across AB will be

(a)

(b)

(a) 10 V in series with 1.2 W resistance (b) 6 V in series with 1.2 W resistance (c) 10 V in series with 5 W resistance (d) 6 V in series with 5 W resistance 5.35 A d.c. current source is connected as shown in Figure below.

The Thevenin’s equivalent of the network at terminals a–b will be

(a)

5.76

Circuit Theory and Networks

(b)

(c)

(d) is NOT feasible 5.36 Which one of the following impedance values of load will cause maximum power to be transferred to the load for the network shown in the given figure?

(a) (2 + j2) (b) (2 – j2) (c) – j2 5.37 The Thevenin’s equivalent resistance Rth for the given network is

(d) 2

(a) 1 W (b) 2 W (c) 4 W (d) infinity 5.38 The Norton’s equivalent of circuit shown in Figure (a) is drawn in the circuit shown in Figure (b). The values of ISC and Req in Figure (b) are respectively

5.77

Network Theorems

5 2 4 12 2 A and 2 W (b) A and 1 W (c) A and W (d) A and 2 W 2 5 5 5 5 5.39 For the circuit shown in the figure, the current flowing through 1 W resistor is adjusted to zero by varying the value of R. What is the value of R? (a) 2W (b) 3W (c) 4W (d) 6W 5.40 What is the Thevenin’s equivalent between A and B for the circuit shown in the figure?

(a)

A 1W

1W 8A

1W

4V

B

(a) 12V, 3 / 2 W (b) 4V, 3 / 2 W (c) 16 / 3V, 2 / 3 W (d) 16 / 3V, 3 W 5.41 If Thevenin’s equivalent resistance of the circuit shown in the figure seen from the open terminals is 2 W, then the value of ‘R’ will be 2W

5V

(a) 4W L1

+ –

2W

1A

(b) 2W

5.42 240 V, 50 Hz

+ –

R

(c) 1W

(d) zero

R2

R1

L2

C1

C2 Figure I

IL R L V th

+ –

Z th

I¢L RL

Figure II

Thevenin’s equivalent of the network shown in Figure-I would correspond to the network shown in Figure-II, if one or more of the following conditions are met: 1. I L¢ = I L 2. The equivalence is valid only if the frequency of Vth is maintained at 50 Hz 3. I L¢ = 2 I L , if the voltage Vth is doubled. The correct set of conditions would include (a) 1, 2 and 3 (b) 1 and 2 (c) 2 and 3 (d) 1 and 3 5.43 Thevenin’s theorem is not applicable for circuits with (a) passive load (b) active load (c) bilateral load (d) none of these

5.78

Circuit Theory and Networks

5.44 In the figure, Z1 = 10Ж 60°, Z2 = 10Ð60°, Z3 = 50Ð53.13°. Thevenin impedance seen from X-Y is X Z3

Z1 + –

100 –0°

Z2 Y

(a) 56Ð45° (b) 60Ð30° (c) 70Ð30° (d) 34.4Ð65° 5.45 Two ac sources feed a common variable resistive load as shown in the figure. Under the maximum power transfer condition, the power absorbed by the load resistance RL is 6W 11 0 – 0°

j8 W

+ –

6W

j8 W + –

RL

(a) 2200W (b) 1250W 5.46 In the figure, the value of R is

(c) 1000W

90 – 0°

(d) 625W

RW 14 W

1W

10 A 100 V

5A + –

+ –

2W

(a) 10 W (b) 18 W (c) 24 W 5.47 In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P-Q, is given by

40 V

(d) 12 W 10 W

(a) (2 V, 5W) (b) (2V, 7.5 W) 20 W (c) (4V, 5 W) (d) (4V, 7.5W) 5.48 In the figure, the current source is 1Ð0° A, R = 1 W, the impedances are ZC = –j W, and ZL = 2j W. The Thevenin equivalent looking into the circuit across X-Y is X

Y

4V

10 W

Unknown network

5.79

Network Theorems

(a)

2 Ð0 V, 1 + 2 j W

(b) 2Ð45°V, 1 - 2 j W

b g 2Ð45° V, b1+ j g W

b

g

(c) (d) 2 Ð45°V, 1+ j W 5.49 A source of angular frequency 1 rad/s has a source impedance consisting of 1W resistance in series with 1H inductance. The load that will obtain the maximum power transfer is (a) 1 W resistance (b) 1 W resistance in parallel with 1H inductance (c) 1 W resistance in series with 1F capacitor (d) 1 W resistance in parallel with 1F capacitor

b g

EXERCISES Reciprocity Theorem 5.1 In the network shown in figure below, verify the Reciprocity Theorem using a voltage source and an ammeter. What are the methods of verifying the Reciprocity Theorem? All values are in ohm.

5.2 Find the current in the 6 W resistor and the source current in Figure (a). Hence, determine the current in the 3 W resistor when an emf of 72 V is added in series with the 6 W resistor as shown in Figure (b). [ 0.5 A, 6 A]

(a)

(b)

5.3 In this circuit, find the voltage V. Interchange the current source and resulting voltage V and show that the reciprocity theorem is verified. [9.28Ð21.8° (V)]

5.80

Circuit Theory and Networks

5.4 Two sets of measurements are made on a linear passive resistive network in Figure (a) and (b). Find the current through the 2W resistor. [2 A]

(a)

(b)

Millman’s Theorem 5.5 Find the load current using Millman’s theorem. All values are in ohm.

[1.176 A]

5.6 Using Millman’s theorem, find the current in the load impedance, ZL =(2 + j4) W [1.06Ж58.46° (A)]

5.7 Determine the current through the branch AB using Millman’s theorem.

é 36 A ù êë 67 úû

Network Theorems

5.81

Thevenin’s and Norton’s Theorems 5.8 Determine the Thevenin equivalent circuit with respect to the terminals A and B for the circuit shown in the figure and hence the current flowing through 10 W resistor. [0.75 A]

5.9 Find the Thevenin equivalent circuit for the following networks (i)

(ii)

(iii)

[(i) 0; – 0.33 W (ii) 8 V; 10 kW (iii) 25 V; 350 W] 5.10 Determine the current in the branch AB for the circuit shown in figure by using Thevenin’s theorem. [1.818 A]

5.82

Circuit Theory and Networks

[0 A; 10.64 W]

5.11 Find Norton’s equivalent at terminals a–b.

5.12 Use Thevenin’s theorem to find the current supplied by the battery. [RTh = 33.34 W; VTh = 10 V; i = 0.3 A] 10 W

10 V

+ –

10 W

50 W

50 W

30 W

10 W

5.13 Find the Thevenin equivalent circuit with respect to the terminals A and B. (a)

50 W

A + v1 –

100 W

0.1 v 1

200 W

B 4W

(b)

1W

A ix 10 i x

+ –

2W

B 5W

(c)

j 10 W A

20 –90° ( V )

+ –

3W –j4 W

B

5.83

Network Theorems 10 W

(d)

A 5W

5W

5 – 30°(A)

j5 W

j5 W B

[(a) VTh = 0; RTh = 10.64 W; (b) VTh = 0; RTh = 1 W; (c) VTh = 10Ð0° (V); ZTh = 5.59Ж 26.6° (W); (d) VTh = 11.18Ð93.44° (V); ZTh = 5Ð36.87° (W)] 5.14 Compute I0 using Norton’s theorem. 5 cos 2 t (V)

2W

+ 1 F 4

I0

4H

– 1 F 2

d

ib g

I 0 = 0.542 cos 2t - 77.47 o A

Maximum Power Transfer Theorem 5.15 Determine the value of the resistor RL that will draw maximum power from the rest of the circuit. What is the maximum power? [4.22 W, 2.901 W]

5.16 The circuit operates in the sinusoidal steady state with w = 1000 rad/s and I s = 1Ð0° A(rms) . Find the value of the load impedance for maximum average power transfer. Also, find the average power absorbed by the load under this condition. [ (1500 + j1000) W; 83.33 W]

5.84

Circuit Theory and Networks

[40Ð0° V; (8 – j20) W, 50 W]

5.17 Determine ZL so that the maximum power is absorbed by it.

5.18 Determine the value of R such that the 6W resistor consumes the maximum power.

[R = –18 W]

5.19 Find the value of the resistance R for maximum power to be transferred to it. Also, find the maximum power. (a)

50 V

60 W

30 W

60 V

60 W

1W

(b) 10 V

40 W

R

+ –

100 V

2W

R

1W

i1

(c) 20 W

40 W R

10 i 1

+ –

+ –

50 V

[(a) 44 W, 0.568 W; (b) 4.5 W, 1.39 W; (c) 16 W]

5.85

Network Theorems

Superposition Theorem 5.20 Apply superposition theorem to the circuit to find i3.

[–0.75 A]

5.21 Find the current i0 using superposition theorem.

[–0.4706 A]

5.22 Use superposition theorem to find the voltage Vx. Vx

20 W

10 V

+ –

[12.5V]

0.1 V x

4W

2A

5.23 Determine the voltage vx in the circuit using superposition theorem. 2W

[–38.5V]

4W + vx –

50 V

+ –

+ –

0.1 v x

100 V

d

5 + 2.56 sin 500t - 39.8o

5.24 Use superposition theorem to find the voltage vx. 5W

1W +

20 sin 50 t (V)

+ –

2 mF

vx –

+ –

6V

i b Vg

5.86

Circuit Theory and Networks

5.25 Find the current through the capacitor using superposition theorem.

[4.86Ð80.8° (A)]

j1 W

5W + –

20 –0° (V)

–j5 W 10 –0° (V)

+ –

5.26 Find the current ix by the superposition theorem. ix

5W

1W

+ –

20 V

[5A]

+ –

30 A

4 ix

5.27 Find v0 using superposition theorem. 2H

+

v0

–

4W

1W 10 cos 2 t (V)

+ –

0.1 F

+ –

5V

2 sin 5 t (A)

c

h

c

h

v0 = -1 + 2.498 cos 2t - 30.79 o + 2.328 sin 5t + 10 o V

SHORT-ANSWER TYPE QUESTIONS 5.1 State and explain substitution theorem. 5.2 State and explain superposition theorem. Give a proof for a general n-mesh network indicating the conditions under which it is applicable. 5.3 State reciprocity theorem as applied to a network and give a proof of the same for a general network. Mention two networks where this theorem is not applicable. 5.4 State Thevenin’s theorem and give a proof of the same. Mention one example of a network where this network is not applicable. 5.5 (a) State Norton’s theorem as applied to a network and give a proof of the same. (b) What is ‘Dual Network’? Mention the procedure for drawing the dual of a given network. 5.6 State and prove maximum power transfer theorem. or In the circuit, the source emf ES, resistance RS and reactance jXS are fixed but both the load resistance RL and reactance jXL are variable. Show that maximum power is consumed in the load when XL = –XS and RL = RS.

5.87

Network Theorems

5.7

5.8 5.9 5.10 5.11 5.12 5.13

Prove that the load impedance which absorbs the maximum power from a source is the conjugate of the impedance of the source. State and prove the following theorem (a) Tellegen’s theorem. (b) Millman’s theorem. (c) Compensation theorem. State and explain clearly Thevenin’s theorem as applied in ac circuits. State and explain Thevenin’s theorem, specify the types of network to which it is applicable. Also, state the theorem which is the dual of the above theorem. State maximum power transfer theorem for all the various kinds of networks and loads. State maximum power transfer theorem. Derive conditions for maximum power transfer for a resistive network and resistive load. Prove the condition for maximum power transfer for an ac circuit. A source with internal impedance RS + jXS delivers power to a variable load impedance RL + j0. Show

that the condition for maximum power in the load is R L 2 = RS 2 + X S 2 . 5.14 State maximum power transfer theorem and verify that only 50% of the total power supplied by the source can be transferred to load. Or, State and explain maximum power transfer theorem. Derive the expression for efficiency for maximum power transfer. 5.15 Derive the condition for maximum power transfer for (a) Load impedance with variable resistance and variable reactance (b) Load impedance with variable resistance and fixed reactance 5.16 State and clearly prove, with the help of a suitable example, the maximum power transfer theorem as applicable to RLC circuits excited from the sinusoidal energy source. Hence explain clearly the concept and its significance in impedance matching.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 5.1 5.8 5.15 5.22 5.29 5.36 5.43

(b) (b) (d) (a) (c) (d) (b)

5.2 5.9 5.16 5.23 5.30 5.37 5.44

(c) (d) (a) (a) (c) (b) (a)

5.3 5.10 5.17 5.24 5.31 5.38 5.45

(c) (d) (b) (a) (b) (d) (d)

5.4 5.11 5.18 5.25 5.32 5.39 5.46

(a) (c) (b) (c) (a) (b) (d)

5.5 5.12 5.19 5.26 5.33 5.40 5.47

(c) (a) (d) (d) (a) (c) (a)

5.6 5.13 5.20 5.27 5.34 5.41 5.48

(a) (b) (d) (b) (b) (c) (d)

5.7 5.14 5.21 5.28 5.35 5.42 5.49

(a) (c) (a) (d) (d) (a) (c)

CHAPTER

6 Laplace Transform and its Applications 6.1

INTRODUCTION

Classical methods of solving differential equations become quite cumbersome when used for networks involving higher order differential equations. In such cases, Laplace Transform method is used. The classical methods consist of three steps: (i) determination of complementary function, (ii) determination of particular integral, and (iii) determination of arbitrary constants. But, these methods become difficult for the equations containing derivatives; and transform methods prove to be superior. The Laplace transform is an integral that transforms a time function into a new function of a complex variable. The term Laplace comes from the name of the French mathematician Pierre Simon Laplace (1749–1827). The transformation method is a very effective tool for solving integro-differential equations. Laplace transformation is also a very powerful tool for network analysis. Any linear circuit consisting of linear circuit elements can be solved by the knowledge of Laplace transformation. In this chapter, we will first discuss the basics of Laplace transformation and then apply this transform method to study the transient behaviour of electric circuits.

6.2

ADVANTAGES OF LAPLACE TRANSFORM METHOD

Laplace transforms methods offer the following advantages over the classical methods. 1. It gives complete solution.

6.2

Circuit Theory and Networks

2. 3. 4.

6.3

Initial conditions are automatically considered in the transformed equations. Much less time is involved in solving differential equations. It gives systematic and routine solutions for differential equations.

DEFINITION OF LAPLACE TRANSFORM

Let f (t) be a function of time which is zero for t < 0 and which is arbitrarily defined for t > 0, subject to some mild conditions. Then the Laplace Transform of the function f (t), denoted by F (s) is defined as, ¥

L [ f (t )] = F ( s ) = ò f (t )e - st dt 0_

Thus, the operator L[ ] transforms f(t), which is in time domain, into F(s), which is in the complex frequency domain, or simply the s-domain, where, s = Complex frequency (unit is in Hz) = (s + jw) where, s = Real part of s = neper frequency and w = Imaginary part of s = radian frequency. NB: The lower limit of the integration should be 0– instead of 0+ or simple 0. If f (t) is continuous at t = 0, then the value of f (0) is well-defined. But, if f (t) is not continuous at t = 0, then the meaning of f(0) becomes ambiguous. To consider the effect of “instantaneous energy transfer” we must use 0– as the lower limit to include the impulses at t = 0. The use of 0 will exclude the existence of any impulses at the origin. So, we use 0– as the lower limit.

6.4

BASIC THEOREMS OF LAPLACE TRANSFORM

1. Linearity Theorem If Laplace transform of the functions f1(t) and f2(t) are F1(s) and F2(s) respectively, then Laplace transform of the functions [K1 f1(t) + K2 f2(t)] will be [K1 F1(s) + K2 F2(s)]. L[K1 f1(t) + K2 f2(t)] = [K1 F1(s) + K2 F2(s)] where, K1 and K2 are constants. 2. Scaling Theorem If Laplace transform of f (t) is F(s), then

( )

1 s , where K is a constant and K > 0. F K K 3. Time Differentiation Theorem If Laplace transform of f (t) is F(s), then,

L[f(Kt)] =

é df (t ) ù = sF ( s ) - f (0- ) Lê ë dt úû 4. Frequency Differentiation Theorem If Laplace transform of f (t) is F(s), then,

L[tf (t )] = -

dF ( s) ds

Laplace Transform and its Applications

6.3

5. Time Integration Theorem If Laplace transform of f (t) is F(s), then, ét ù F (s) L ê ò f (t )dt ú = s ë0 û In general, for nth order integration,

ét1 t2 tn ù F (s) L ê ò ò ... ò f (t ) dt1 dt2 ... dtn ú = n s ë0 0 0 û 6. Shifting Theorem The shifting may be done with respect to time or frequency. (a) Time Shifting Theorem If Laplace transform of f (t) is F (s), then

L [ f (t ± a)] = e± as F ( s) (b) Frequency Shifting Theorem If Laplace transform of f(t) is F(s), then L [e m at f (t )] = F ( s ± a) 7. Initial Value Theorem If the Laplace Transform of f (t) is F (s) and the first derivative of f (t) is Laplace transformable, then, the initial value of f (t) is, f (0+ ) = Lt f (t ) = Lt [ sF ( s )] t ®0

s ®¥

Proof

¥ é df (t ) ù - st d e dt L éê f (t ) ùú = ò ê dt ë û 0- ë dt úû

or

¥ é df (t ) ù - st sF ( s ) - f (0- ) = ò ê e dt dt úû 0- ë

[by time differentiation theorem]

Taking limit s ® ¥, ¥ é df (t ) ù - st Lt [ sF ( s ) - f (0- )] = Lt ò ê e dt s®¥ s ® ¥ 0 ë dt ú û -

or

¥ é0+ df (t ) df (t ) ù Lt [ sF ( s )] - f (0- ) = Lt ê ò e0 dt + ò e- st dt ú dt dt s ®¥ s ®¥ ê 0 + úû 0 ë

or

é0+ df (t ) ù dt ú Lt [ sF ( s )] - f (0- ) = Lt ê ò e0 dt s ®¥ s ®¥ ê 0 úû ë0+

or or

[as s is not a function of time t]

Lt [ sF ( s )] - f (0- ) = Lt ò df (t ) = f (0+ ) - f (0- )

s ®¥

s ®¥ 0

f (0 + ) = Lt [ sF ( s )] s ®¥

-

6.4

Circuit Theory and Networks

8. Final Value Theorem If a function f (t) and its derivatives are Laplace transformable, then the final value of f (t) is,

f (¥) = Lt f (t ) = Lt [ sF ( s )] t ®¥

s ®0

Proof

¥ é df (t ) ù - st d e dt L éê f (t ) ùú = ò ê dt ë û 0- ë dt úû

or

¥ é df (t ) ù - st sF ( s ) - f (0- ) = ò ê e dt dt úû 0- ë

[by time differentiation theorem]

Taking limit s ® 0, ¥ ¥ t é df (t ) ù - st é df (t ) ù æ df (t ) ö Lt [ sF ( s ) - f (0- ) ] = Lt ò ê e dt dt Lt = = ò ò ç ÷ dt ú ê ú dt û s ®0 s ® 0 0 ë dt û t ® ¥ 0 è dt ø 0- ë -

Lt [ sF ( s ) - f (0- )] = Lt [ f (t ) - f (0- )]

or

s ®0

t ®¥

Lt [ sF ( s ) ] - f (0- ) = Lt [ f (t )] - f (0- )

or

s ®0

t ®¥

Lt [ f (t ) ] = Lt [ sF ( s )]

or

t ®¥

s ®0

This theorem is only applicable if the value of the function f (t) is finite as t becomes infinity, i.e., F (s) has all poles lying in the left half of s-plane or at most one simple pole at the origin.

6.5

LAPLACE TRANSFORM OF SOME BASIC FUNCTIONS

1. Exponential Function

f (t ) = eat By definition of Laplace transform, ¥

¥ ¥ é e( a - s )t ù æ 1 ö 1 F ( s ) = L[ f (t )] = ò eat × e- st dt = ò e( a - s )t dt = ê ú = èç 0 - (a - s ) ø÷ = (s - a ) ( a s ) ë û0 00-

Similarly, for f (t ) = e- at , F ( s ) = 2. Unit Step Function or, f (t ) = u (t ) = 1 for = 0 for and is undefined for

1 s+a

Heaviside Unit Function t>0 t 0), so that f(t + T) = f(t), then the Laplace transform of the function is equal to æ 1 ö times the Laplace transform of the first cycle. çè 1 - e - Ts ÷ø 1 ù L [ f (t )] = F ( s ) = F1 ( s ) éê - Ts ú ë1 - e û

\

Proof Let f(t) be the periodic function, and T be the time period, Let f1(t), f2(t), … , fn(t) be the functions representing the first, second, …, nth cycle, respectively \ f (t) = f1(t) + f2(t) + … + fn(t) + … = f1(t) + f1(t – T) + f1(t – 2T) + … Taking Laplace transform, L[f(t)] = F(s) = L[ f1 (t )] + L[ f1 (t - T )] + L[ f1 (t - 2T )] + ... = F1 ( s) + e- Ts F1 ( s) + e- 2Ts F1 ( s) + ¼ = F1 ( s)[1 + e- Ts + e- 2Ts + e- 3Ts + ¼] Therefore,

Example 6.1

1 ù F ( s ) = F1 ( s ) éê - Ts ú 1 ë e û

Find the Laplace transform of the square wave.

Figure 6.5(a) Square wave of Example 6.1

Solution

The first cycle is shown below. It can be written as, f1(t) = u(t) – 2u(t – T) + u(t – 2T)

6.11

Laplace Transform and its Applications

Taking Laplace transform of the first cycle, 1 2e- Ts e- 2Ts 1 + = (1 - e-Ts )2 s s s s By the theory of time periodicity, the Laplace transform of the square wave is given as,

F1(s) =

1 F (s) = 1 (1 - e- Ts )2 ´ s 1 - e- 2Ts (Since time period of the square wave is 2T)

Figure 6.5(b) First cycle of the square wave of Example 5.1

æ 1 - e - Ts ö 1 Ts = 1ç = tanh æç ö÷ è 2ø s è 1 + e - Ts ø÷ s

6.8

SINGULARITY FUNCTIONS AND WAVEFORM SYNTHESIS

In order to synthesise any signal, there are some standard singularity functions which can be realised in the laboratory. Other signals can be written in terms of these singularity functions. Those singularity functions are 1. 2. 3. 4.

Step Function, Ramp Function, Impulse Function, and Unit Doublet Function.

1. Step Function This function is also known as Heaviside unit function. It is defined as given below. f ( t ) = u( t ) = 1 =0

u(t) 1

for t > 0 for t < 0

f ( t ) = Ku(t ) = K for t > 0 = 0 for t < 0

and is undefined at t = 0. A shifted or delayed unit step function is defined as, f ( t ) = u( t - T ) = 1 =0

and is undefined at t = T.

for t > T for t < T

t

0

and is undefined at t = 0. A step function of magnitude K is defined as,

Figure 6.6(a) Unit Step Function Ku ( t ) K

0

t

Figure 6.6(b) Step function of magnitude K

6.12

Circuit Theory and Networks

The Laplace transform of a unit step function is given as, ¥

¥

0-

0-

z

z

F ( s) = L f (t ) = u ( t ). e - st dt = 1. e - s t dt =

LM e OP N -s Q - st

u ( t –T)

¥

1

0-

1 1 = 0= -s s

0

Figure 6.6(c) Shifted Unit Step Function

Also, the Laplace transform of step function of magnitude K is L[ Ku (t )] =

t

T

K s

Similarly, the Laplace transform of the shifted unit step function u (t – T) is, e - sT {by differentiation theorem} s Another function, called gate function can be obtained from step function as follows. Therefore, g(t) = Ku (t - a ) - Ku (t - b) L u (t - T ) =

L g (t ) =

K - a s - bs e -e s

d

i

g(t) K

0

a

b

Figure 6.7 Gate Function

2. Ramp Function A unit ramp function is defined as,

r(t)

bg bg

f t = r t = t for t ³ 0 = 0 for t < 0 A ramp function of any slope K is defined as,

bg

1 1

bg

f t = Kr t = Kt for t ³ 0 = 0 for t < 0 A shifted unit ramp function is defined as,

Figure 6.8(a) Unit Ramp Function

Kr ( t )

bg b g

f t = r t - T = t for t ³ T = 0 for t < T The Laplace transform of a unit ramp function is, ¥

¥

0-

0-

bg z

K 1 0

z

L r t = r (t ). e- s tdt = te - s t dt

t

Figure 6.8(b) Ramp Function

Integrating by parts, let, then

t

0

u=t

and

dv = e– s tdt

du = dt

and

v = e - stdt = -

z

e - st s

6.13

Laplace Transform and its Applications

Now,

¥

¥

0-

0-

z

LM t de iOP N s Q

z

L r (t ) = udv = uv|0¥- - vdu = 1 ¥ - st e dt s 01 = 2 s =

- st

¥

+

0-

z

1 ¥ - st e dt s 0-

z

r ( t –T) 1 1 0

Similarly, Laplace transform of a ramp of slope K is, K L Kr (t ) = 2 s and Laplace transform of a shifted ramp function is, L Kr (t - T ) =

t

Figure 6.8(c) Shifted Unit Ramp Function

Ke- Ts s2

3. Impulse Function This function is also known as Dirac Delta function, denoted by d (t). This is a function of a real variable t, such that the function is zero everywhere except at the instant t = 0. Physically, it is a very sharp pulse of infinitesimally small width and very large magnitude, the area under the curve being unity. Consider a gate function as shown in Fig. 6.9. f(t) 3/a 2/a 1/a 0

a/3

a/2

a

t

Figure 6.9 Generation of impulse function from gate function

The function is compressed along the time-axis and stretched along the y-axis, keeping area under the pulse unity. As a ® 0, the value of

1 ® ¥ and the resulting function is known as impulse. a

It is defined as,

bg

d t =0

for t ¹ 0

¥

and

z d (t ) dt = 1

-¥

Also,

af

d t = Lim a

0

1 u (t ) - u (t - a ) a

6.14

Circuit Theory and Networks

The Laplace transform of the impulse function is obtained as,

bg

RS 1 u (t ) - u(t - a) UV Ta W 1 L1 e O a MN s s PQ

L d t = Lim L a

= Lim a

o

- as

o

1 - e - as a o as se - a s = Lim a o s =1 = Lim

[by L' Hospital ' s rule]

4. Unit Doublet Function The derivative of unit impulse function with respect to time at any instant of time is known as unit doublet function. It is defined as, d d t -T =d t -T = 0 dt = + ¥ and

b g b g

for t ¹ 0 -¥

for t = T

The name of the function is given as doublet because it can be obtained from the function shown in Fig.6.10 (a) with a ® 0. f(t)

f(t) 1 a2 a

•

t

t

–a –•

– 12 a

Figure 6.10(a) Generation of Unit Doublet Function with a ® 0

Figure 6.10(b) Unit Doublet Function

The Laplace transform of a unit doublet function is obtained as,

a f LMN dtd d at - T fOPQ = sL d at - T f = se

L d¢ t -T =L

6.9

INVERSE LAPLACE TRANSFORM

Let, F(s) have the general form of

F (s) =

N ( s) D( s)

-T s

6.15

Laplace Transform and its Applications

where, N(s) is the numerator polynomial and D(s) is the denominator polynomial. The roots of N(s) = 0 are called the zeros of F(s) while the roots of D(s) = 0 are the poles of F(s). s -1 , the zero is at s = 1 and the poles are at s = For example, for the function F ( s ) = s ( s - 2) ( s - 3) 0, 2 and 3. We use Partial Fraction Expansion to break F(s) down into simple terms. Thus, there are two steps to find inverse Laplace transform as given below. I. Decomposition of F(s) into simple terms using Partial Fraction Expansion. II. Evaluation of the inverse of each term comparing with the standard forms of Laplace transforms. We consider the following three cases given below. I. Simple Poles Let

F(s) =

N (s) ( s + p1) ( s + p2 ) ( s + p3) ¼ ( s + pn )

where, s = –p1, –p2 –p3, …, –pn are the simple poles, and pi ¹ pj for all i ¹ j (i.e. poles are distinct) Assuming that the degree of N(s) is less than the degree of D(s), F (s) =

k3 kn k1 k2 + + +¼+ s + p1 s + p2 s + p3 s + pn

(1)

where, expansion co-efficients k1, k2, k3, …, kn are known as the residues of F(s). These can be found out by Residue method explained below. Multiplying both sides of Eq. (1), by (s + p1), ( s + p1) F ( s ) = k1 +

Putting

s = - p1 Þ

( s + p1)kn ( s + p1)k2 ( s + p1)k3 + +¼+ s + p2 s + p3 s + pn

( s + p1) F (s )|s = - p1 = k1

In general, ki = ( s + pi ) F ( s )|s = - pi . This is known as Heaviside’s Theorem. Once, the values of ki are known, the inverse Laplace is obtained as,

f (t ) = (k1e- p1t + k2 e- p2t + k3e- p3t + ¼ + kn e- pnt ) u (t ) Example 6.2

Find the inverse Laplace transform of the function,

F (s) =

Solution

2s + 1 . ( s + 1) ( s + 2) ( s + 3)

Let F (s) = \

k k k 2s + 1 = 1 + 2 + 3 ( s + 1) ( s + 2) ( s + 3) s + 1 s + 2 s + 3

k 1 = ( s + 1) F ( s )|s = - 1 =

2s + 1 ( s + 2) ( s + 3)

=s = -1

1 2

6.16

Circuit Theory and Networks

\

\ \

k 2 = ( s + 2) F ( s )|s = - 2 =

2s + 1 ( s + 1) ( s + 3)

k 3 = ( s + 3) F ( s )|s = - 3 =

2s + 1 5 =( s + 1) ( s + 2) s = -3 2

F(s) = -

=3 s = -2

1 3 5 + 2( s + 1) s + 2 2( s + 3)

Thus, the inverse Laplace transform is given as, f (t) = -

1 -t 5 e + 3e - 2t - e –3t 2 2

II. Repeated Poles Suppose, F(s) has n repeated poles at s = – p. \

F(s) =

kn - 1 kn - 2 kn k2 k1 + + +¼+ + + F1 ( s) n n -1 n-2 2 ( s + p) ( s + p) ( s + p) ( s + p) ( s + p)

where, F1(s) is the remaining part of F(s) that does not have a pole at s = –p. We find, \

k n = ( s + p ) n F ( s )|s = - p

To find kn – 1, kn – 2,…, kn – m, the procedure is, kn–1 =

d [( s + p ) n F ( s )] ds s=-p

kn–1 =

1 d2 [( s + p )n F ( s )] 2! ds 2 s=-p

In general, kn - m =

1 dm , where, m = 1, 2, …, (n – 1). [( s + p ) n F ( s )] m ! ds m s=- p

Once, the values of k1, k2, …, kn are known, the inverse Laplace is obtained as,

k kn æ ö t n -1e- pt ÷ u (t ) + f1 (t ) f(t) = ç k1e- pt + k2te- pt + 3 t 2 e- pt + ¼ + 3! (n - 1)! è ø Example 6.3 Solution

Find the inverse Laplace transform of the function F(s) = Let F ( s ) =

k k1 k 12 = + 2 + 3 2 2 s+2 s+4 ( s + 2) ( s + 4) ( s + 2)

12 . ( s + 2)2 ( s + 4)

6.17

Laplace Transform and its Applications

By residue method, k 1 = ( s + 2)2 F ( s )|s = - 2 =

\

k2 =

d d é 12 ù [( s + 2) 2 F ( s )]|s = - 2 = = -3 ds ds êë ( s + 4) úû s=-2

k 3 = ( s + 4) F ( s )|s = - 4 =

Thus, F(s) =

12 =6 ( s + 4) s = - 2

12 ( s + 2) 2

=3 s=-4

6 3 3 + 2 s s + + 2 4 ( s + 2)

Taking inverse Laplace transform, f (t ) = 3e- 4t - 3e- 2t + 6te- 2t III. Complex Poles Since N(s) and D(s) always have real co-efficients and as the complex roots of polynomials with real co-efficients occur in conjugate form, F(s) may have the general form, F (s) =

A1s + A2 k1 k2 + F1 ( s ) = + + F1 ( s ) s j s + a b + a + jb s + as + b 2

where, F1(s) is the remaining part of F(s) that does not have this pair of complex poles. Let \

( s 2 + as + b) = ( s 2 + 2a s + a 2 + b 2 ) = ( s + a )2 + b 2 s1, 2 = (- a ± j b ) = -

a a2 ± j b2 4

Thus, the coefficients are,

k1 = ( s - s1) F ( s )|s = s1 and k2 = k1* = Complex conjugate of k1 Example 6.4 Solution

Find the inverse Laplace transform of the function F ( s ) = Let \

F(s) =

2s + 1 . ( s + 1) ( s 2 + 2 s + 5)

k1 k2 2s + 1 A = + + 2 ( s + 1) ( s + 2 s + 5) s + 1 s + 1 - j 2 s + 1 + j 2

A = ( s + 1) F ( s )|s = - 1 =

2s + 1 1 =4 s 2 + 2s + 5 s = -1

k 1 = ( s + 1 - j 2) F ( s )|s = ( - 1 + j 2) =

2s + 1 ( s + 1) ( s + 1 + j 2)

1 1 = æç - j ö÷ è 8 2ø s = (- 1 + j 2)

6.18

Circuit Theory and Networks

\

\

1ö æ1 k 2 = k1* = ç + j ÷ è8 2ø 1 1 1 1 - j + j 1æ 1 ö 8 2 8 2 F(s) = - ç + + 4 è s + 1÷ø s + 1 - j 2 s + 1 + j 2

Taking inverse Laplace transform, f(t) = -

6.10

1 -t 1 [e - e - t cos 2t ] + e - t sin 2t = - e - t sin 2 t + e - t sin 2t 4 2

APPLICATIONS OF LAPLACE TRANSFORM

1. Solving Integro-Differential Equations and Simultaneous Differential Equations 2. Transient Analysis of Electrical Circuits.

6.10.1 Solving Integro Differential Equations and Simultaneous Differential Equations An integro-differential equation is an integral equation in which various derivatives of the unknown function can also be present. Using the Laplace transform of integrals and derivatives, an integrodifferential equation can be solved. Similarly, it is easier with the Laplace transform method to solve simultaneous differential equations by transforming both equations and then solving the two equations in the s-domain and finally obtaining the inverse to get the solution in the time domain. Example 6.5

(Integro-Differential Equation) Solve the equation for the response i(t), given that t di + 2i + 5ò idt = u (t ) and i(0) = 0. dt 0

Solution

Let L[i(t)] = I(s).

di L éê ùú = sI ( s) - i (0) = sI ( s ) - 0 = sI ( s ) ë dt û Taking Laplace transform on both sides of the given equation, \

I (s) 1 = s s 1 1 2 I (s) = 2 = 2 2 s + 2s + 5 ( s + 1) + (2)2

sI(s) + 2I(s) + 5 or

Laplace Transform and its Applications

6.19

Taking inverse Laplace transform, we get i (t ) =

Example 6.6

1 -t e sin 2t ( A), t > 0 2

(Integro-Differential Equation) Solve the initial value problem for y(t) when d2y + y (t ) = 3 sin 2t and y (0) = 1, y ¢(0) = - 2 . dt 2

Solution

Let L[y(t)] = Y(s). \

éd2y ù L ê 2 ú = s 2Y ( s ) - sy (0) - y ¢(0) = s 2Y ( s ) - s + 2 ë dt û

or Y(s) =

s 2 - 2 s +1 s + 4 2

Taking inverse Laplace transform, we get, y (t ) = (cos t - sin 2t ) Example 6.7

(Simultaneous Differential Equations) Find the solution of the system:

dy dx - 2 x - y = 4et with initial conditions x(0) = –1, y(0) = 0. - 6 x + 3 y = 8et and dt dt Solution

Taking Laplace transform,

( s - 6) X + 3Y =

-s + 9 s -1

(i)

4 s -1

(ii)

- 2 X + ( s - 1)Y =

Solving for X and Y, X=

-s + 7 2 1 =+ ( s - 1) ( s - 4) s -1 s - 4

Y=

- 2 /3 2/3 2 = + ( s - 1) ( s - 4) s - 1 s - 4

Taking inverse Laplace transform, 2 2 x(t) = - 2et + e 4t and y (t ) = - et + e 4t 3 3

Example 6.8

(Simultaneous Differential Equations) Solve for x(t) and y(t), given that x(0) = 4, y(0) = 3 and dy dx + x + 4 y = 10 and x - y=0 dt dt

6.20

Circuit Theory and Networks

Solution

Following the same procedures, as in Ex (6.7), we get,

4s 2 + 2s + 10 3s 2 + s + 10 Y = and s ( s 2 + 3) s ( s 2 + 3) Taking inverse Laplace transform, we get the desired results. X=

6.11

APPLICATION OF LAPLACE TRANSFORM METHOD TO CIRCUIT ANALYSIS

We now apply the mathematical tool for the analysis of electric circuits.

6.11.1 Transform Impedance of Network Elements Element

Time Domain

s-Domain

1. Resistor (R)

v(t) = Ri(t)

V(s) = RI(s)

2. Inductor (L)

v(t) = L

di(t ) dt

V(s) = L[sI(s) – i(0–)]

t

i(t) =

3. Capacitor (C)

1 v(t )dt L -ò¥

i(t) = C

dv(t ) dt

I(s) =

I(s) = sCV(s) – Cv(0–)

t

v(t) =

1 i (t ) dt C -ò¥

1 é V ( s ) i (0-) ù + L êë s s úû

V(s) =

I ( s ) v(0- ) + Cs s

Laplace Transform and its Applications

6.12

6.21

TRANSIENT ANALYSIS OF ELECTRIC CIRCUITS USING LAPLACE TRANSFORM

In electrical engineering, a transient response or natural response is the electrical response of a system to a change from equilibrium. The condition prevailing in an electric circuit between two steady-state conditions is known as the transient state; it lasts for a very short time. The currents and voltages during the transient state are called transients. In general, transient phenomena occur whenever (i) a circuit is suddenly connected or disconnected to/from the supply, (ii) there is a sudden change in the applied voltage from one finite value to another, (iii) a circuit is short-circuited. A simple example would be the output of a 5 volt DC power supply when it is turned on: the transient response is from the time the switch is turned on and the output is a steady 5 volt. At this point the power supply reaches its steady-state response of a constant 5 volt. The transient response is not necessarily tied to “on/off” events but to any event that affects the equilibrium of the system. If in an RC circuit the resistor or capacitor is replaced with a variable resistor or variable capacitor (or both) then the transient response is the response to a change in the resistor or capacitor. The transient currents are not caused by any part of the supply voltage, but are entirely associated with the changes in the stored energy in capacitor and inductors. As there is no energy stored in resistors, there are no transients in purely resistive circuits. Although transients last for a very short time, their study is very important because. (i) They indicate what dangerous rises in voltage or current may happen in individual sections of a circuit. (ii) They indicate how signals are distored in waveform or amplitude as they pass through amplifiers, filters, or other circuit elements. We consider the transient analysis for the following circuits subject to step input, impulse input and sinusoidal input: 1. RL Series Circuit, 2. RC Series Circuit, 3. RLC Series Circuit, and 4. RLC Parallel Circuit.

6.12.1 RL Series Circuit 1. RL Series Circuit with Step Input We consider an RL series circuit as shown in the figure.

Figure 6.11 R-L series circuit

6.22

Circuit Theory and Networks

If the switch is closed at time t = 0, the voltage across the RL combination would be v(t) which is a step of magnitude V [or Vu(t)] and not a constant as is the supply voltage V. v(t) = 0, for t £ 0 = V, for t ³ 0 Thus the differential equation governing the behaviour of the circuit would be

di (t ) = Vu (t ) dt Taking Laplace transform, we get Ri (t ) + L

RI ( s ) + L[ sI ( s ) - i (0 -)] =

or,

I (s) =

V L

( RL )

+

s s+

V s

i (0 -) V æ 1 1 ö i (0 -) = + R Rçs R÷ R s+ s + ÷ø s + çè L L L

Taking inverse Laplace transform,

i (t ) =

R R R - æç ÷ö t ö -çæ ö÷ t - æç ÷ö t ö Væ Væ è1 - e è L ø ø + i (0 -) e è L ø = è1 - e è L ø ø with i(0–) = 0. R R

The transient part of the current response, itr = [i (t ) - is ] = -

R

V - Lt e R

L V V , i = (1 - e - 1 ) = 0.63 = 0.63is R R R When the switch is first closed, the voltage across the inductor will immediately jump to battery voltage (acting as though it were an open-circuit) and decay down to zero over time (eventually acting as though it were a short-circuit). Voltage across the inductor is determined by calculating how much voltage is being dropped across R, given the current through the inductor, and subtracting that voltage value from the battery. When the switch is first closed, the current is zero, then it increases over time until it is equal to the battery voltage divided by the series resistance. This behavior is precisely opposite that of the series resistor-capacitor circuit, where current started at a maximum and capacitor voltage at zero.

From the current equation at t = t =

V R The variation of the current is shown in Figure 6.12.

The steady state part of the current response, is =

The quantity t =

L is known as the Time-constant of the circuit and it is defined as follows. R

6.23

Laplace Transform and its Applications

Figure 6.12 Variation of current with time RL series circuit with step input

Definitions of Time-constant (t) 1. It is the time taken for the current to reach 63% of its final value. Thus, it is a measure of the rapidity with which the steady state is reached. Also, at t = 5t, i = 0.993is; the transient is therefore, said to be practically disappeared in five time constants. 2. The tangent to the equation i =

(

R

- t V 1- e L R

) at t = 0, intersects the straight line, i = VR

at

L . Thus, time-constant is the time in which steady state would be reached if the R current increases at the initial rate. Physically, time-constant represents the speed of the response of a circuit. A low value of timeconstant represents a fast response and a high value of time-constant represents a sluggish response. t =t =

Calculations of the Voltage Across Elements

(

) ù di (t ) d éV ) Voltage across the inductor, V = L = L ê (1 - e ú = Ve dt dt ë R û Voltage across the resistor, VR = Ri (t ) = V 1 - e

-

R t L

-

L

R t L

-

R t L

2. RL Series Circuit with Impulse Input By KVL, the mesh equation becomes, di (t ) = V d (t ) dt Taking Laplace transform, RI ( s) + sLI ( s ) = V with i (0 -) = 0 Ri (t ) + L

6.24

or

Circuit Theory and Networks

I (s) =

Væ 1 ö L çè s + R / L ÷ø

Taking inverse Laplace transform, R

V - Lt e L Here, the plot of the current is shown in Figure 6.13. i (t ) =

Figure 6.13 Variation of voltages with time in RL series circuit with impulse input

Voltage across the resistor, VR = Ri (t ) = Voltage across the inductor, VL = L

R

VR - L t e L

R R di (t ) d æV - tö VR - L t e = L çè e L ÷ø = dt dt L L

3. RL Series Circuit with Sinusoidal Input Here, the input voltage is given as, v(t ) = V sin w t By KVL, di (t ) = V sin wt , with i(0–) = 0 dt Vw I ( s)[ R + sL] = 2 s + w2

Ri (t ) + L or

or

ì ü ï 1 Vw ï = I(s) = í ý L R R ï ( s + jw ) ( s - jw ) s + ï (s 2 + w 2 ) s + L L þ î Vw L

( )

( )

6.25

Laplace Transform and its Applications

=

where,

A3 ù A2 V w é A1 + + ê L s - jw s + jw Rú s+ ú ê Lû ë

ì ü ïï ïï 1 L = A 1 = í( s - jw ) ý 2 ( w + jw L) j R R ï ( s + jw ) ( s - jw ) æ s + ö ï è ø L ïþs = jw îï ì ü ï ï 1 L A 2 = í( s + jw ) =ý 2 jw ( R - jw L ) R ï ï ( s + jw ) ( s - jw ) s + L þs = - jw î

( )

and

ì ü ï ï R 1 L2 A3 = í s + = 2 ý L R ï ( R + w 2 L2 ) ï ( s + jw ) ( s - jw ) s + L þs = - R î

( )

( )

L

ù Vw é L L L2 + ê ú L 2 jw ( R + jw L) ( s - jw ) 2 jw ( R - jw L) ( s + jw ) R ê ú ( R 2 + w 2 L2 ) s + L û ë Taking inverse Laplace transform, \

I(s) =

( )

R é ù jw t 2 - Lt - jw t w V Le Le L e ê ú i(t) = + 2 L êë 2 jw ( R + jw L ) 2 jw ( R - jw L ) R + w 2 L2 úû

-

R

t

jwt é e- jwt ù e L = V ê e + VwL 2 ú 2 j ë R + jw L R - jw L û R + w 2 L2

Let, ( R + jw L) = Ze jq and ( R - jw L) = Ze- jq so that, Z = Putting these values, - jwt é jwt i(t) = V ê e jq - e - jq 2 j ë Ze Ze

=

V Z

-

R

ù e L ú + VwL 2 Z û

t

é e j (wt - q ) - e - j (wt - q ) ù V w L - R t ê ú+ 2 e L 2j Z ë û

æ wLö ( R 2 + w 2 L2 ) and q = tan -1 ç è R ÷ø

6.26

Circuit Theory and Networks

or, finally, the current is,

i (t ) =

R

V VwL - t sin (w t - q ) + 2 e L Z Z

From this result, it is clear that the current in RL series circuit lags behind the voltage by an angle, wLö q = tan -1æç . If the resistance R = 0, then q = 90° as is the case for a perfect inductor. è R ÷ø

6.12.2 RC Series Circuit 1. RC Series Circuit with Step Input We consider an RC series circuit as shown in Figure 6.14. By KVL, Ri (t ) +

1t i (t ) dt = Vu (t ) C 0ò

Taking Laplace transform, RI ( s ) +

or

1 é I ( s ) q (0 -) ù V + = C êë s s úû s

1 ù V q(0 -) I ( s ) éê R + = Cs Cs ë ûú s

Figure 6.14 RC series circuit

q (0 -) q (0 -) 1 V- C C or I (s) = = s ( R + 1/ Cs ) R ( s + 1/ RC ) Taking inverse Laplace transform, V-

t éV q(0 -) ù - RC e ; for t ³ 0 i(t) = ê RC úû ëR t

V - RC ; if q (0–) = 0 e R The steady state part of the current response, is = 0

=

t

The transient part of the current response, itr = [i(t) – is] =

V - RC e R

V -1 V e = 0.37 R R When the switch is first closed, the voltage across the capacitor (which was fully discharged) is zero volt; thus, it first behaves as though it were a short-circuit. Over time, the capacitor voltage will rise to equal battery voltage, ending in a condition where the capacitor behaves as an open-circuit. Current through the circuit is determined by the difference in voltage between the battery and the capacitor, divided by the resistance. As the capacitor voltage approaches the battery voltage (V), the current approaches zero. Once the capacitor voltage has reached V, the current will be exactly zero.

From the current equation at t = t = RC, i =

Laplace Transform and its Applications

6.27

The variation of current in the circuit is shown in Figure 6.15.

Figure 6.15 Variation of current with time in RC series circuit with step input

The quantity t = RC is known as the Time-constant of the circuit and it is defined as follows.

Definitions of Time-constant (t) 1. It is the time in which the current decays to 37% of its initial value. Also, at t = 5t, i = 0.07

V ; the transient is therefore, said to be practically disappeared in five R

time constants. t

V - RC at t = 0, intersects the time axis at t = t = RC. e R Thus, time-constant is the time in which the current would reach the steady state zero value if the current decays at the initial rate. Physically, time-constant represents the speed of the response of a circuit. A low value of timeconstant represents a fast response and a high value of time-constant represents a sluggish response.

2. The tangent to the equation i =

Calculations of the Voltage Across Elements Voltage across the resistor, VR = Ri (t ) = Ve Voltage across the capacitor, VC =

-

t RC

-t -t ö æ 1t 1 t V RC RC i ( t ) dt e dt V 1 e = = ç ÷ C 0ò C 0ò R è ø

2. RC Series Circuit with Impulse Input With zero initial condition, q(0–) = 0, KVL equation becomes, Ri (t ) +

RI ( s ) +

1t i (t )dt = V d (t ) C 0ò

I (s) =V Cs

6.28

Circuit Theory and Networks

1 ù é ú Væ s ö Vê RC I (s) = 1= = or, Rç 1 1 ÷ Rê 1 ú R+ çè s + ÷ ê s + RC ú Cs RC ø ë û Taking inverse Laplace transform, V

i (t ) =

-t Vé 1 RC ù e ú ; for t ³ 0 êd (t ) Rë RC û

-t é 1 RC ù Voltage across the resistor, VR = Ri (t ) = V êd (t ) e ú RC ë û

-t

V RC e RC These variations of the voltages are shown in Figure 6.16. Voltage across the capacitor, VC = {V d (t ) - VR } =

Figure 6.16 Variation of voltages with time in RC series circuit with impulse input

3. RC Series Circuit with Sinusoidal Input Here, the input voltage is given as, v(t) = V sin wt By KVL, Ri (t ) +

or

or

1 t i (t )dt = V sin w t , with q(0–) = 0 C 0ò

1 ù Vw I ( s ) éê R + = 2 ú Cs ë û s + w2 ì V w Cs Vw ï s = I(s) = 2 í 2 R ( s + w ) (1 + sRC ) ï ( s + jw ) ( s - jw ) s + 1 RC î A3 ù A2 V w é A1 + + = L ê s - jw s + jw 1 ú s+ ê ú RC û ë

(

)

ü ï ý ï þ

6.29

Laplace Transform and its Applications

where,

ì ï 1 A 1 = í( s - jw ) 1 ï ( s + jw ) ( s - jw ) s + RC î

(

ì ï 1 A 2 = í( s + jw ) 1 ï ( s + jw ) ( s - jw ) s + RC î

(

and

)

ü ï RC = ý 2(1 jw RC ) + ï þs = jw

)

ü ï RC =ý 2(1 - jw RC ) ï þs = - jw

ì ü ï ï 1 1 A3 = í s + ý RC 1 ï ï ( s + jw ) ( s - jw ) s + RC þs = î

(

)

(

)

1 RC = æw 2 + 1 ö ÷ èç R 2C 2 ø -

1 RC

1 é ù ú Vw ê RC RC RC + \ I(s) = R ê 2(1 + jw RC ) ( s - jw ) 2(1 - jw RC ) ( s + jw ) æ 2 1 öæ 1 öú ê çè w + 2 2 ÷ø è s + RC ø ú êë úû R C Taking inverse Laplace transform, -t

é ù e jwt e - jw t Ve RC i(t) = V w C ê ú ë 2(1 + jw RC ) 2(1 - jw RC ) û w RC æ R 2 + 1 ö çè ÷ w 2C 2 ø -t

V é e jwt e - jwt ù Ve RC = ê ú 2j 1 1 1 ö æ 2 RêR + ú jw C jw C û w RC èç R + w 2 C 2 ø÷ ë j ö j ö æ æ 1 ö æ 1 ö æ = R+ = Ze jq = R= Ze - jq and ç R Let, ç R + jw C ÷ø çè jw C ÷ø çè w C ÷ø w C ÷ø è è

so that, Z =

1 ö æ 2 æ 1 ö R + 2 2 ÷ and q = tan - 1 ç è w RC ÷ø èç ø w C

Putting these values, -t é jwt e- jwt ù V RC i(t) = V ê e e 2 j ë Ze - jq Ze jq úû w CZ 2 -t é e j (wt + q ) - e - j (wt + q ) ù V RC = Vê e ú2 Zê 2j ë ûú w CZ

6.30

Circuit Theory and Networks

or, finally, the current is, i (t ) =

-t

V V sin (w t + q ) e RC Z w CZ 2

From this result, it is clear that the current in RC series circuit leads the voltage by an angle, 1 ö q = tan -1 æç . If the resistance R = 0, then q = 90° as is the case for a perfect capacitor. è w RC ÷ø

6.12.3 RLC Series Circuit 1. RLC Series Circuit with Step Input With zero initial conditions, the Kirchhoff’s voltage law equation becomes, Ri(t) + L or

di (t ) 1 t + ò i(t )dt = Vu (t ) dt C0

RI(s) + sLI ( s ) +

1 V I (s) = Cs s Figure 6.17 RLC series circuit

V L or I(s) = R 1 2 s + s+ L LC The roots of the denominator polynomial of equation are,

s2 + or Let Then, So,

\

R 1 =0 s+ L LC

s1 = w0 =

(6.1)

R + 2L

1 LC

R2 1 2 LC 4L

and

xw 0 =

s1 = - xw 0 + w 0 x 2 - 1

and, s2 = -

R R i.e. x = 2 2L

R 2L

R2 1 2 LC 4L

C = Damping Ratio L

and s2 = - xw 0 - w 0 x 2 - 1

V A B L I(s) = = + ( s - s1) ( s - s2 ) s - s1 s - s2

V L A = ( s - s1) ( s - s1) ( s - s2 ) s = s1

V V L = = ( s1 - s2 ) 2w L x 2 - 1 0

V L and, therefore B = ( s - s2 ) ( s - s1) ( s - s2 ) s = s2

V V L = =( s2 - s1) 2w 0 L x 2 - 1

6.31

Laplace Transform and its Applications

Putting these values of A and B, we get, I(s) =

é 1 1 ù ês - s - s - s ú 1 2û 2w 0 L x - 1 ë V

2

Taking inverse Laplace transform, i(t) =

V 2w 0 L x 2 - 1

[e s1t - e s2t ] =

V 2w 0 L x 2 - 1

e- xw0t [e(w0

x 2 -1)t

- e- (w0

x 2 -1)t

]

Depending upon the values of R, L and C, three cases may appear: R 1 > (Overdamped condition) (a) 2L LC (b)

R < 2L

(c)

R = 2L

1 (Underdamped condition) LC 1 (Critically Damped condition) LC

A. Overdamped Condition The condition is,

R > 2L

1 1 or, x > 1 or Q < 2 LC

w0 L æ çè Since, Quality Factor, Q = R and w 0 =

Under this condition, the current becomes, 2 2 V V i (t ) = e- xw0t [e(w0 x -1)t - e- (w0 x -1)t ] = e- xw0t sinh (w 0 x 2 - 1)t 2 2 2w 0 L x - 1 w0 L x - 1 The graphical plot for the current is shown in Figure 6.18. 0.6 Underdamped condition

0.5

Amplitude

0.4 Critically damped condition 0.3 0.2

Overdamped condition

0.1 0 –0.1

0

1

2

3

4

5 6 Time (sec)

7

8

9

10

Figure 6.18 Current response in RLC series circuit for three different damping conditions

1 ö LC ø÷

6.32

Circuit Theory and Networks

R = 2L

B. Critically Damped Condition The condition is,

1 1 or, x = 1 or Q = 2 LC

From equation (6.1),

V V 1 ö L I (s) = 2 = æç 2 2÷ L s + 2w 0 s + w 0 è (s + w0) ø Taking inverse Laplace transform, V - w 0t te L The graphical plot for the current is shown in Figure 6.13. i (t ) =

C. Underdamped Condition The condition is,

R < 2L

1 1 or, x < 1 or Q > 2 LC

So, the current becomes, V

i(t) =

2w 0 L x 2 - 1

é æç jw0 ê eè V = e- xw 0t ê ë w0 L 1 - x 2 =

V w0 L 1 - x

2

x 2 - 1)t

e- xw0t [e(w0

1- x2

ö ÷ø t

- e- (w0

-e 2j

æ -ç è

x 2 - 1)t

jw 0 1 - x 2

ö t ø÷

]

ù ú ú û

e- xw0t sin (w 0 1 - x 2 )t

So, the circuit is oscillatory. When R = 0, x = 0, the oscillations are undamped or sustained. The frequency of the undamped oscillation (w0) is known as undamped natural frequency.

2. RLC Series Circuit with Impulse Input With zero initial conditions, the Kirchhoff’s voltage law equation becomes, Ri (t ) + L or

or

di (t ) 1 t + ò i (t )dt = V d (t ) dt C0

RI ( s ) + sLI ( s ) +

1 I (s) = V Cs

æV ö s è Lø I(s) = 1 R s2 + s + L LC

(6.2)

6.33

Laplace Transform and its Applications

The roots of the denominator polynomial of equation are, s2 +

or

Let Then,

So,

\

R 1 s+ =0 L LC

s1 = -

w0 =

R2 1 2 LC 4L

R + 2L

1 LC

(VL ) ( s - s1) ( s - s2 )

=

R 2L

R R i.e. x = 2 2L

xw 0 =

and

s1 = - xw 0 + w 0 x 2 - 1

I(s) =

and, s2 = -

R2 1 2 LC 4L

C = Damping Ratio L

and s2 = - xw 0 - w 0 x 2 - 1

A B + s - s1 s - s2

æV ö s è Lø A = ( s - s1) ( s - s1) ( s - s2 )

æV ö s è Lø 1 Vs1 = = ( s1 - s2 ) 2w L x 2 - 1 0 s = s1

æV ö s è Lø 2 and, therefore B = ( s - s2 ) ( s - s1) ( s - s2 )

æV ö s è Lø 2 Vs2 = =( s2 - s1) 2w 0 L x 2 - 1 s = s2

Putting these values of A and B, we get, I(s) =

s2 ù é s1 ês - s - s - s ú 1 2û 2w 0 L x - 1 ë V

2

Taking inverse Laplace transform, i(t) =

=

V 2w 0 L x 2 - 1 V 2w 0 L x 2 - 1

[ s1e s1t - s2 e s2t ] =

V 2w 0 L x 2 - 1

e- xw0t [ s1e(w0

e- xw0t [(- xw 0 + w 0 x 2 - 1)e(w0

x 2 -1)t

- s2e - (w0

x 2 - 1)t

- (- xw 0 - w 0 x 2 - 1)e - (w 0

x 2 - 1)t

]

x 2 -1)t

]

6.34

Circuit Theory and Networks

Three cases are considered: (A)

R > 2L

1 (Overdamped condition) LC

(B)

R < 2L

1 (Underdamped condition) LC

(C)

R = 2L

1 (Critically Damped condition) LC

A. Overdamped Condition Here, x > 1 The current becomes, i(t) =

V 2w 0 L x - 1 2

e- xw0t [(- xw 0 + w 0 x 2 - 1)e(w0

x 2 - 1)t

– (- xw 0 - w 0 x 2 - 1)e - (w 0 =

V L x -1 2

[ x 2 - 1 cosh (w 0 x 2 - 1)t - x sinh (w 0 x 2 - 1)t ]

B. Critically Damped Condition The condition is, x = 1 From equation (6.2), I(s) =

where

and

(V / L) s V s A B ù ö =V é = æç + 2 2 2 ÷ ê s + w 0 ûú s + 2w 0 s + w 0 L è ( s + w 0 ) ø L ë ( s + w 0 ) 2

A = (s + w 0 )2

B=

s ( s + w 0 )2

= - w0 s = - w0

d é s ù (s + w 0 )2 =1 ds êë ( s + w 0 ) 2 úû s = - w 0

So,

I(s) =

V L

w0 é 1 ù ê s + w - (s + w )2 ú 0 0 û ë

Taking inverse Laplace transform, i(t) =

V [1 - w 0 t ]e - w 0t L

x 2 - 1)t

]

6.35

Laplace Transform and its Applications

C. Underdamped Condition The condition is, x < 1 So, the current becomes, i (t ) =

V 2w 0 L x 2 - 1

e- xw0t [(w 0 x 2 -1) {e(w0 + e- (w0

=

V 2w 0 Lj 1 - x

2

x 2 -1) t

=

V L 1 - x2

x 2 -1)t

} - xw 0 {e(w0

e- xw0t [( jw 0 1 - x 2 ) {e( jw0

- xw 0 {e( jw0 i (t ) =

x 2 -1)t

1 - x 2 )t

1 - x 2 )t

- e - (w0

+ e- ( jw0

- e - ( jw0

1 - x 2 )t

1 - x2 ) t

x 2 -1) t

}]

}

}]

e- xw0t [ 1 - x 2 cos (w 0 1 - x 2 )t - x sin (w 0 1 - x 2 )t ]

æ 1 - x2 ö e - xw 0t cos {(w 0 1 - x 2 )t + q}, where q = tan - 1 ç ÷ x ø è L 1 - x2 V

3. RLC Series Circuit with Sinusoidal Input Sinusoidal voltage v(t) = Vm sin (wt + q) is applied to a series RLC circuit at time t = 0. We want to find the complete solution for the current i(t) using Laplace transform method. v (t ) = Vm sin (w t + q ) By KVL, di (t ) 1 t + Ri (t ) + L ò i(t )dt = Vm sin (w t + q ) dt C -¥ Taking Laplace transform with zero initial conditions, ( s sin q + w cos q ) 1 ù I ( s ) éê R + sL + = Vm ú Cs ë û s2 + w 2

or

I(s) =

=

Vm s ( s sin q + w cos q ) R 1 ö L( s 2 + w 2 ) æ s 2 + s + è L LC ø

Vm s ( s sin q + w cos q ) L ( s + jw ) ( s + jw ) ( s - s1) ( s - s2 )

where, s1, s2 are the roots of the quadratic equation:

æ s2 + R s + 1 ö = 0 çè L LC ÷ø

v(t)

Figure 6.19

RLC series circuit with sinusoidal input

6.36

Thus,

Now, let

Circuit Theory and Networks

s1 = -

R + 2L

R2 1 2 LC 4L

and,

s2 = -

R 2L

R2 1 2 LC 4L

K3 K1 K2 K4 s ( s sin q + w cos q ) = + + + ( s + jw ) ( s - jw ) ( s - s1) ( s - s2 ) s - s1 s - s2 s + jw s - jw

So, by residue method, multiplying by (s – s1) and putting s = s1, K1 =

s1 ( s1 sin q + w cos q ) s2 ( s2 sin q + w cos q ) and K 2 = ( s1 + jw ) ( s1 - jw ) ( s1 - s2 ) ( s2 + jw ) ( s2 - jw ) ( s2 - s1)

Similarly, multiplying by (s + jw) and putting s = – jw,

and,

K3 =

- jw (- jw sin q + w cos q ) w (cos q - j sin q ) = (- jw - jw ) (- jw - s1) (- jw - s2 ) 2( s1 + jw ) ( s2 + jw )

K4 =

jw (- w sin q + w cos q ) w (cos q + j sin q ) = ( jw + jw ) ( jw - s1) ( jw - s2 ) 2( s1 - jw ) ( s2 - jw )

Hence the current response becomes,

Vm V [ K1e s1t + K 2 e s2t ] + [ K 3e- jwt + K 4 e jwt ] = I tr + I ss L L Thus, the transient part of the total current is, i(t) =

é ù ê ú Vm ê s1 ( s1 sin q + w cos q ) s1t s2 ( s2 sin q + w cos q ) s2t ú I tr = e e L ê ú R2 R2 4 4 2 2 2 2 s w ( ) + ê ( s2 + w ) ú 2 L2 LC L2 LC ëê ûú

The steady-state part of the total current is obtained as follows. Iss =

=

=

=

=

Vm é ù w e- jq e- jwt w e jq e jwt + ê 2 L ë ( s1 + jw ) ( s2 + jw ) ( s1 - jw ) ( s2 - jw ) úû ù Vmw é e- j (wt + q ) e j (wt + q ) + ê 2 L ëê ( s1 + jw ) ( s2 + jw ) ( s1 - jw ) ( s2 - jw ) úûú 2 L( s12

Vmw [e- j (wt + q ) ( s1s2 - w 2 - jw s1 - jw s2 )] + w 2 ) ( s22 + w 2 )

2 L( s12

Vmw [( s1s2 - w 2 ) 2 cos(w t + q ) - (w s1 + w s2 ) 2 sin (w t + q )] + w 2 ) ( s12 + w 2 )

Vmw 1 L ( s12 + w 2 ) ( s22 + w 2 )

éæ 1 ù æ wRö 2ö êçè LC - w ÷ø cos (w t + q ) - çè - L ÷ø sin (w t + q ) ú ë û

6.37

Laplace Transform and its Applications

or

éw R ù 1 ö sin (w t + q ) - æ w 2 cos (w t + q ) ú è LC ø Vmw êë L û Iss = L ( s12 + w 2 ) ( s22 + w 2 )

ì æ wL - 1 öü Vmw ï 1 wC ÷ ï sin íw t + q - tan - 1 çè = øý L ( s12 + w 2 ) ( s22 + w 2 ) R ï ï î þ

´

or,

Vm

I ss =

1 ö R2 + æw L è wC ø

2

w L

1 ö R 2 + æç w L è w C ø÷

2

ìï æ w L - 1 ö üï wC ÷ ý sin íw t + q - tan - 1 èç ø ïþ ïî R

This gives the steady-state current of the series RLC circuit to a sinusoidal voltage.

6.12.4 RLC Parallel Circuit 1. RLC Parallel Circuit with Step Current Input With zero initial conditions, the Kirchhoff’s current law equation becomes, v(t ) dv(t ) 1 t +C + ò v(t )dt = Iu (t ) R dt L0 or

or

V (s) 1 I + sCV ( s ) + V ( s ) = R sL s V(s) =

Figure 6.20 RLC parallel circuit

I /C 1 1 s2 + s+ RC LC

(6.3)

The roots of the denominator polynomial of equation are, s2 +

1 1 s+ =0 RC LC

or

s1 = -

1 + 2 RC

Let

w0 =

1 LC

1 1 4 R 2 C 2 LC

and

xw 0 =

and, s2 = -

1 1 i.e. x = 2R 2RC

1 2 RC

1 1 4 R 2 C 2 LC

L = Damping Ratio C

6.38

Circuit Theory and Networks

s1 = - xw 0 + w 0 x 2 - 1

Then, So,

V(s) =

\

A=

and s2 = - xw 0 - w 0 x 2 - 1

I /C A B = + ( s - s1) ( s - s2 ) s - s1 s - s2 ( s - s1)

and, therefore, B = ( s - s2 )

I /C I /C I = = ( s - s1) ( s - s2 ) s = s ( s1 - s2 ) 2w C x 2 - 1 1 0 I /C I /C I = =( s - s2 ) ( s - s2 ) s = s ( s2 - s1) 2w 0C x 2 - 1 2

Putting these values of A and B, we get, V(s) =

é 1 1 ù ês - s - s - s ú 1 2û 2w 0C x - 1 ë I

2

Taking inverse Laplace transform, v(t) =

I 2w 0C x - 1 2

[e s1t - e s2t ] =

I 2w 0C x - 1 2

e- xw0t [e(w0

x 2 -1)t

- e- (w0

x 2 -1)t

]

Depending upon the values of R, L and C, three cases may appear: (a)

1 > 2RC

1 (Overdamped condition) LC

(b)

1 < 2RC

1 (Underdamped condition) LC

(c)

1 = 2RC

1 (Critically Damped condition) LC

A. Overdamped Condition The condition is,

1 > 2RC

1 1 or, x > 1 or Q < 2 LC

æ Since, Quality Factor, Q = 1 and w = 0 çè w 0 RC

1 ö ÷ LC ø

Under this condition, the current becomes, v(t ) =

I 2w 0C x - 1 2

e- xw0t [e(w0

x 2 - 1)t

- e- (w0

x 2 - 1)t

]=

I w 0C x - 1 2

(

)

e- xw0t sinh w 0 x 2 - 1 t

6.39

Laplace Transform and its Applications

The graphical plot for the voltage is shown in Figure 6.21. 0.6 Underdamped condition

0.5 0.4 Amplitude

Critically damped condition 0.3 0.2

Overdamped condition

0.1 0 –0.1

0

1

2

3

4

5 6 Time (sec)

7

8

9

10

Figure 6.21 Voltage response in RLC parallel circuit for three different damping conditions

B. Critically Damped Condition The condition is,

1 = 2RC

1 1 or, x = 1 or Q = 2 LC

From equation (6.3), V(s) =

I /C

s + 2w 0 s + w 02 Taking inverse Laplace transform, 2

=

I æ 1 ö C èç ( s + w 0 )2 ø÷

I - w 0t te C The graphical plot for the voltage is shown in Fig. 6.21.

v(t) =

C. Underdamped Condition The condition is,

1 < 2RC

1 1 or, x < 1 or Q > 2 LC

So, the voltage becomes, v(t) =

=

=

I 2w 0C x 2 - 1

e- xw0t [e(w0

é e( jw0 e- xw0t ê êë w 0C 1 - x 2 I

I w 0C 1 - x 2

x 2 - 1)t

1 - x 2 )t

- e- (w0

- e- ( jw0 2j

e- xw0t sin (w 0 1 - x 2 )t

x 2 - 1)t

]

1 - x 2 )t

ù ú úû

6.40

Circuit Theory and Networks

Similarly we can find out the impulse response and sinusoidal response of a parallel RLC circuit using Laplace transform method as for the series RLC circuit.

6.12.5 Response with Pulse Input Voltage 1. RC Series Circuit circuit, then by KVL,

If a voltage pulse of width as shown in Fig. 6.22 is applied to an RC series

1 i (t )dt = v (t ) Cò Taking Laplace transform with zero initial condition, Ri (t ) +

1 V Ve - sT I (s) = Cs s s - sT V 1- e or I(s) = R s + 1/ RC Taking inverse Laplace transform, V - t / RC i(t) = [e - e - (t - T ) / RC ] R Hence the voltage across the resistance is given as, RI ( s ) +

Figure 6.22 Pulse voltage

vR(t) = Ri (t ) = V [e- t / RC - e- (t - T )/ RC ] and the voltage across the capacitor is given as, vc(t) = V - vR (t ) = V [e- t / RC + e- (t - T )/ RC ] To plot the two voltages with varying time, we have the following observations: (i) At t = 0, all the voltage appears across the resistance R and thus, vR = V and vC = 0

Figure 6.23 Voltage response of RC series circuit with pulse input

Laplace Transform and its Applications

6.41

(ii) As the time increases, the voltage vC grows and the voltage vR decays exponentially, with time-constant t = RC. (iii) At t = T, voltage across the network drops abruptly to zero from V. Again this entire drop is instantaneously felt across the resistance R. (iv) For time t > T, total voltage across the circuit is zero. So, at any instant of time t, vR(t) + vc(t) = 0 and both vR and vC asymptotically approach zero.

Case (1) If Time-constant (t = RC) Pulse-width (T) In this case, the voltage across the capacitor varies with time almost linearly and the value is far from the steady state value V; i.e. vR = V.

Figure 6.25 Voltage response of RC series circuit (RC >> T) with pulse input

6.42

\

Circuit Theory and Networks

vC =

1t 1 t vR 1 t 1 t idt = dt » Vdt or, \ vC » Vdt ò ò ò C0 C0 R RC 0 RC 0ò

Thus, the voltage vC is the integration of the input voltage and hence the circuit acts as an Integrator.

2. RL Series Circuit If a similar pulse voltage is applied to an RL series circuit, then the KVL equation will be, di = v (t ) dt Taking Laplace transform with zero initial condition, Ri (t ) + L

RI ( s ) + sLI ( s ) = or

V Ve- sT s s

é ù 1 e- sT I(s) = V ê L ë s ( s + R / L) s( s + R / L) úû

Taking inverse Laplace transform, R R ù - t - (t - T ) V é u (t - T ) úû êë 1 - e L u (t ) - 1 - e L R The variation of the two voltages is shown in Figure 6.26.

i(t) =

(

)

(

)

Figure 6.26 Voltage response of RL series circuit with pulse input

Case (1) If Time-constant (t = L/R) Pulse-width (T) In this case, the voltage across the resistor varies with time almost linearly and the value is far from the steady state value V; i.e. vL = V. 1.0V 0.8V Voltages voltage across L

0.6V

voltage across R

0.4V 0.2V 0V –0.2V

0s

2s

4s

6s

8s

10s 12s Time

14s

16s

18s

20s

Figure 6.28 Voltage response of RL series circuit ( L / R >> T ) with pulse input

\

vR = Ri = R

1t Rt Rt vL dt » ò Vdt or, \ vR » ò Vdt ò L0 L0 L0

Thus, the voltage vR is the integration of the input voltage and hence the circuit acts as an Integrator.

6.44

6.13

Circuit Theory and Networks

STEPS FOR CIRCUIT ANALYSIS USING LAPLACE TRANSFORM METHOD

1. All circuit elements are transformed from time-domain to Laplace domain with initial conditions. 2. Excitation function is transformed into Laplace domain. 3. The circuit is solved using different circuit analysis techniques, such as, mesh analysis, node analysis, etc. 4. Time domain solution is obtained by taking inverse Laplace transform of the solution.

6.14

CONCEPT OF CONVOLUTION THEOREM

6.14.1 Convolution Integral If h(t) is the impulse response of a linear network, then the response of the same network y(t) subject to any arbitrary input w(t) is given by the convolution integral as, ¥

¥

-¥

-¥

y (t ) = ò h(t ) w(t - t )d t = ò w(t )h(t - t )d t Thus, if the impulse response of any linear time-invariant system is known, we can obtain the zerostate response of the system to any other type of input.

6.14.2 Convolution Theorem If f1(t) and f2(t) are two functions of time which are zero for t < 0, and if their Laplace transforms are F1(s) and F2(s), respectively, then the convolution theorem states that the Laplace transform of the convolution of f1(t) and f2(t) is given by the product F1(s) F2(s). Mathematically, if the convolution of f1(t) and f2(t) is written as, t

t

0

0

f1 (t ) * f 2 (t ) = ò f1 (t ) f 2 (t - t ) d t = ò f1 (t - t ) f 2 (t ) d t = f 2 (t ) * f1 (t ) where, t is a dummy variable for time t, then the convolution theorem is written as, Proof

L[ f1 (t ) * f 2 (t )] = F1 ( s ) F2 ( s ) By the definition of convolution,

ét ù ¥ ét ù L[ f1 (t ) * f 2 (t )] = L ê ò f1 (t ) f 2 (t - t )d t ú = ò ê ò f1 (t - t ) f 2 (t )d t ú e - st dt ë0 û 0 ë0 û Also, by the definition of a shifted unit step function, using dummy variable, u (t - t ) = 1; for t £ t = 0; for t > t

\

t

¥

0

0

ò f1 (t - t ) f 2 (t ) d t = ò f1(t - t ) u (t - t ) f 2 (t ) d t

(i)

6.45

Laplace Transform and its Applications

Putting this in (i), we get, ¥ é¥ ù L[ f1 (t ) * f 2 (t )] = ò ê ò f1 (t - t ) u (t - t ) f 2 (t )d t úe - st dt 0 ë0 û Now, let (t – t ) = x \ dt = dx,

t x

(ii)

µ µ

0 –t

From equation (ii), we get, ¥é¥ ù L[ f1 (t ) * f 2 (t )] = ò ê ò f1 ( x) u ( x) f 2 (t ) d t ú e - s ( x + t ) dx 0 ë -t û ¥

¥

-t

0

= ò f1 ( x) u ( x) e - sx dx ò f 2 (t ) e - st d t ¥

¥

0

0

= ò f1 ( x) e- sx dx ò f 2 (t ) e- st d t

{Q u( x) = 0

for x < 0}

\ L [[ f1 (t ) * f 2 (t )]] = F1 ( s ) F2 ( s ) Thus, the convolution in time domain becomes multiplication in the frequency domain, and viseversa.

6.14.3 Application of Convolution Theorem The convolution theorem is used to find the response of a linear system to any arbitrary excitation if the impulse response of the system is known. We know that the transfer function is defined as the ratio of response transform to excitation transform with zero initial conditions. Thus, Transfer Function =

or Thus,

H(s) =

Y (s) W (s)

Laplace transform of Response Laplace transform of Excitation all initial conditions reduced to zero

IC = 0

Y ( s ) = H ( s )W ( s )

Here, W(s) = L[w(t)], is the input Laplace transform and Y(s) = L[y(t)], is the output Laplace transform. Now, if the input is an impulse function, then w(t) = d (t) or W(s) = 1 \ Y(s) = H(s)W(s) = H(s) Taking inverse Laplace transform, y(t) = h(t)

6.46

Circuit Theory and Networks

Thus, h(t) is the impulse response of the system. If this impulse response of the system is known, we can find out the response of the system due to any arbitrary input w(t) from the following relation: Y(s) = H(s)W(s) t

t

0

0

y (t ) = h(t ) * w(t ) = ò h(t )w(t - t )d t = ò h(t - t )w(t )d t

or

Example 6.9 Solution

Find the convolution integral when f1 (t ) = e- at and f 2 (t ) = t . Here, the convolution integral is given as, t

t

0

0

f1 (t ) * f 2 (t ) = ò e- a(t - t ) t d t = e- at ò t eat d t t

= e

- at

é t eat ù eat êë a - ò 1 × a d t úû 0 t

= e

- at

é t eat t eat ù ê a - 2 ú a û0 ë

at é at ù = e- at ê te - e 2 + 12 ú a a a û ë

=

1 [at - 1 + e - at ] a2

Ans.

SOLVED PROBLEMS 6.1 (a) Find the initial value of the function whose Laplace Transform is,

( s + a)sin q + b cos q ( s + a)2 + b2 Check the result by solving it for v(t). V(s) = A ×

(b) Find the final value of the function whose Laplace Transform is, I(s) = Solution (a) By initial value theorem, V(0+) = lim sV ( s ) s ®¥

= lim SA s ®¥

( s + a ) sin q + b cos q (s + a)2 + b2

s+6 s ( s + 3)

6.47

Laplace Transform and its Applications

= lim A s ®¥

æ1 + a ö sin q + b cos q è sø s 2

æ1 + a ö + æ b ö è è sø sø

2

= A sin q Ans. In order to check this result, we find v(t) and then put t = 0.

é ( s + a ) sin q + b cos q ù v(t) = L-1 ê A ú ( s + a)2 + b2 ë û é ( s + a) sin q b cos q ù + = AL-1 ê ú 2 2 ( s + a)2 + b2 û ë (s + a) + b = A[sin q e- at cos bt + cos q e- at sin bt ] = Ae- at sin (bt + q ) At t = 0, v(0 +) = Ae0 sin (0 + q ) = A sin q [Checked] (b) By final value theorem, I (¥) = lim sI ( s ) = lim s s ®0

s ®0

s+6 s+6 = lim =2 s ( s + 3) s ® 0 ( s + 3)

Ans.

é s+6 ù 1 ù -1 é 2 - 3t For checking it, i (t ) = L-1 ê ú = L ê s - s + 3ú = 2 - e + s s ( 3) ë û ë û At t = µ, i (¥) = 2 - e-¥ = 2 [Checked] 6.2 (a) Obtain the Laplace Transform of square wave of unit amplitude and periodic time 2T, as shown. (b) Find the Laplace Transform of the following function:

Solution (a) The equation of the square wave is,

f (t ) = u (t ) - u (t - T ) - u (t - T ) + u (t - 2T ) + u (t - 2T ) - u (t - 3T ) - ¼ = u (t ) - 2u (t - T ) + 2u (t - 2T ) - 2u (t - 3T ) + ...

6.48

Circuit Theory and Networks

Taking Laplace transform, F(s) =

1 2e- Ts 2e- 2Ts 2e- 3Ts + +¼ s s s s

=

1 [1 - 2e - Ts (1 - e - Ts + e - 2Ts - e - 3Ts + ¼)] s

=

1 s

é 2e- Ts ù ê1 - Ts ú ë 1+ e û

=

1 s

é1 - e - Ts ù ê - Ts ú ë1 + e û

F(s) =

1 Ts tanh æç ö÷ è 2ø s

1 ü ìQ sum of G.P. series = í ý 1 e- Ts þ î

Ans.

(b) The equation can be written as,

( )

f (t) = 2r(t) – 4r t -

1 + 2r (t - 1) 2

Taking Laplace transform, 1 - s

1 4e 2 2e- s 2 2 + 2 = 2 [1 - 2e- s / 2 + e – s ] = 2 [1 - 2e- s / 2 ]2 F (s) = 2 2 2 s s s s s 6.3 A sinusoidal voltage 25 sin 10t is applied at time t = 0 to a circuit as shown in the figure. Find the current i(t), by Laplace transform method. R = 5W and L = 1H. Solution

By KVL, RI ( s) + sLI ( s) = 25

10 s 2 + 100

with zero initial condition. I(s) =

250 250 = ( s + 5) ( s 2 + 100) ( s + 5) ( s + j10) ( s - j10)

A3 ù A2 é A + = 250 ê 1 + + + s s j s j10 úû 5 10 ë

A1 = ( s + 5)

where,

1 + s ( 5) ( s 2 + 100)

= s = -5

1 125

A2 = ( s + j10)

1 ( s + 5) ( s + j10) ( s - j10)

A3 = ( s - j10)

1 ( s + 5) ( s + j10) ( s - j10)

=s = - j10

= s = j10

1 1 =j 20(5 - j10) 100(2 + j )

1 100( - 2 + j )

Laplace Transform and its Applications

6.49

Substituting these, A3 ù A2 é A1 + + I(s) = 250 ê + + s s j s j10 úû 5 10 ë Taking inverse Laplace transform,

i(t) = 250[ A1e- 5t + A2 e- j10t + A3e j10t ]

ì ü 1 1 = 2e - 5t + 250 íe - j10t + e j10t ý 100( - 2 + j ) î 100(2 + j ) þ = 2e - 5t -

- j10t (- 2 - j )e j10t ïü 5 ïì (2 - j )e í ý 2 îï 5 5 þï

= 2e - 5t -

1 {2e - j10t - je - j10t + 2e j10t + je j10t } 2

or i(t) = 2e- 5t - 2 cos 10t + sin 10t ( A) 6.4 The circuit of the figure is initially in the steady state. The switch S is closed at t = 0. (a) Find Vc(t) (b) Determine the final value of Vc(t) and verify it from the final value theorem of Laplace Transform.

Solution At steady-state before closing the switch, the capacitor becomes open-circuited. So, the circuit becomes as shown above. v(0 +) =

2 V 3

For t > 0, by KVL, RI1 + R ( I1 - I 2 ) =

and

V V Þ 2 RI1 - RI 2 = s s

(i)

1 2V 1ö 2V I + R ( I 2 - I1) = Þ - RI1 + æç R + I =è Cs 2 Cs ÷ø 2 3s 3s

Solving equations (i) and (ii),

V /s 2R - R - 2V /3s I2 = -R 2R - R ( R + 1/ Cs)

4VR VR + V æ Cs ö s 3s = =- ç 2 s è 2 + RCs ÷ø 3 2 R( R + 1/ Cs) - R -

(ii)

6.50

Circuit Theory and Networks

ù V 1 2V 2V V é 1 + =+ = 2Cs 3s RCs + 2 úû 3s (2 + RCs ) 3s 3s êë V V æ 1 ö + = 2s 6 çè s + 2/ RC ÷ø Taking inverse Laplace transform, \

VC (s) = I 2 ´

V V - 2t / RC + e (Volt), t > 0 2 6 Thus, the final value of the voltage,

vC (s) =

vC (¥) = lim vC (t ) = t ®¥

V 2

Ans.

Ans.

By final value theorem,

æV ö V Vs = vC (¥) = lim SVC ( s ) = lim ç + (Proved) ( s + 2/ RC ) ø÷ 2 s ®0 s ®0 è 2 6.5 In the network shown in the figure, the switch S is closed and a steady state is attained. At t = 0, the switch is opened. Determine the current through the inductor for t > 0.

Solution When the switch S is closed and the steady-state exists, the current through the inductor is, V 5 =2A = R 2.5 The voltage across the capacitor, VC (t) = 0 as it is shorted. For t > 0, the switch is opened. By KVL,

i(0–) =

t

L

di 1 idt = 0 + dt C ò0

Taking Laplace transform, L[ sI ( s ) - i (0-)] +

I ( s) =0 Cs

1 ù I ( s ) éê sL + úû = Li (0-) Cs ë Putting the values,

or

s s 2 + 104 Taking inverse Laplace transform, I (s) = 2

i (t ) = 2 cos 100t ( A); t ³ 0

Ans.

Laplace Transform and its Applications

6.51

6.6 The circuit shown in the figure is initially in the steady state with the switch S open. At t = 0, the switch S is closed. Obtain the current through the inductor for t > 0. Take R1 = R2 = R4 =1W and R3 = 2W and L = 1H. Solution When the switch S is open and steady state exists, the current through the inductor is, i2 (0-) =

1 =1A R3 ( R1 + R2 )/ R3 + R1 + R2

After S is closed, for t > 0, by KVL, 2i1 – i2 – i3 = 1

di2 – i3 = 0 dt –i1 – i2 + 4i3 = 0 Taking Laplace transform, –i1 + 2i2 +

1 s –I1(s) + I2(s)[s + 2] – I3(s) = i2(0–) = 1

2I1(s) – I2(s) – I3(s) =

–I1(s) – I2(s) + 4I3(s) = 0 By Cramer’s Rule,

2 1/ s -1 -1 1 -1 5 1 -1 0 4 6 = + 6 I2(s) = -1 -1 s s + 6 2 7 -1 ( s + 2) -1 -1 -1 4 Taking inverse Laplace transform, 5 1 - 6t / 7 Ans. + e ( A); t > 0 6 6 6.7 A series R-L-C circuit with R = 3W, L = 1H and C = 0.5 F is excited with a unit step voltage. Obtain an expression for the current, using Laplace transform. Assume that the circuit is relaxed initially. Solution By KVL,

i2(t) =

RI ( s) + sLI ( s) - Li(0 -) + Since the circuit is initially relaxed,

i (0-) = 0 and Q(0-) = 0 \ Putting the values, 2 1 I ( s ) éê3 + s + ùú = s ë û s

Q(0 -) 1 1 I ( s) + = sC sC s

6.52

Circuit Theory and Networks

or

I(s) =

where,

A1 =

A A 1 1 = = 1 + 2 s + 3s + 2 ( s + 1) ( s + 2) s + 1 s + 2 2

1 s+2

I (s) =

\

= 1 and A2 = s = -1

1 = -1 s + 1 s=-2

1 1 s +1 s + 2

Taking inverse Laplace transform, i(t) = e–t + e–2t (A) t = 2e 3t / 2 sinh æç ö÷ (A) è 2ø

Ans.

6.8 The switch S in the figure is opened at t = 0. Determine the voltage v(t), for t > 0. What is the nature of the response?

(a)

(b)

Solution (a) By KVL, t v(t ) dv 1 + i (0 -) + ò vdt + C =I R L0 dt Taking Laplace transform,

1 1 I V ( s ) éê + + sC ùú = R sL ë û s Putting the values, 2 s 2 V ( s ) éê 2 + + ùú = s 2û s ë

4 4 = s + 4s + 4 ( s + 2)2 Taking inverse Laplace transform, or

V (s) =

2

v(t ) = 4te- 2t (V ), t > 0 Ans. The response is critically damped (Q x = 1)

Ans.

Laplace Transform and its Applications

(b) Proceeding in the same way as Prob. 6.8(a),

( 3 /2) 1 2 = ´ 2 2 s2 + s + 1 æ 3 æ 3ö 1 s+ ö +ç è 2ø è 2 ø÷

V ( s) =

Þ

æ 3 ö 2 -t / 2 e t (V); t > 0 sin ç è 2 ÷ø 3

v(t ) =

Ans.

The response is under-damped (Q x < 1) Ans. 6.9 In the R-C series circuit of figure, the capacitor has an initial charge of 2.5 mC. At t = 0, the switch is closed and a constant voltage source of V = 100 V is applied. Use the Laplace transform method to find the current i(t) in the circuit. Solution By KVL, after the switch is closed, t é ù êQ(0-) + ò i (t )dt ú = V 0 ë û Taking Laplace transform,

1 C

Ri (t ) +

10 I ( s ) +

or

I ( s) =

2.5 ´ 10- 3 100 I ( s) = s 50 ´ 10- 6 s 50 ´ 10- 6 s

15 s + 2 ´ 103

Taking inverse Laplace transform,

i (t ) = 15e - 2 ´10 t (A); t > 0 Ans. 6.10 In the R-L circuit as shown, the switch is in position-1 long enough to establish steady state condition and at t = 0 it is switched to position-2. Find the resulting current, i(t). Solution When the switch is in position 1, steady-state exists and the initial current through the inductor is, 3

50 =2A 25 After the switch is moved to position 2, the KVL gives, in Laplace transform, i (0 -) =

25 I ( s ) + 0.01sI ( s ) - 0.01 ´ 2 =

or

I (s) =

where,

A1 =

100 s

A A2 10 4 2 2 = 1 + s ( s + 2500) s + 2500 s s + 2500 s + 2500

104 ( s + 2500)

= 4 and A2 = s =0

104 s

= -4 s = - 2500

6.53

6.54

Circuit Theory and Networks

\

I (s) =

4 4 2 4 6 = s s + 2500 s + 2500 s s + 2500

Taking inverse Laplace transform,

i (t ) = 4 - 6e- 2500 t (A); t > 0 Ans. 6.11 In the series RLC circuit as shown, there is no initial charge on the capacitor. If the switch is closed at t = 0, determine the resulting current at i(t).

Solution By KVL, for t > 0, Ri + L

t

di 1 idt = V [Q i (0 -) = 0] + dt C ò0

Taking Laplace transform, RI ( s ) + sLI ( s ) +

I (s) V = Cs s

Putting the values, 2 I ( s ) + sI ( s ) + 2

I ( s ) 50 = s s

50 50 50 = = s 2 + 2s + s ( s + 1 + j ) ( s + 1 - j ) ( s + 1) 2 + 1 By Partial Fraction Expansion, j 25 j 25 I ( s) = s +1+ j s +1- j Taking inverse Laplace transform, or

I (s) =

i(t ) = j 25 [e( -1- j )t - e( -1+ j )t ] = 50e- t sin t (A); t > 0 Ans. 6.12 In the two-mesh network shown in the figure, there is no initial charge on the capacitor. Find the loop currents i1(t) and i2(t) which result when the switch is closed at t = 0.

Solution Writing two mesh equations, 10i1 (t ) +

and

t

1 i (t )dt + 10i2 (t ) = 50 0.2 0ò- 1

50i2(t) + 10i1(t) = 50

Laplace Transform and its Applications

6.55

Taking Laplace transform, 10 I1( s ) +

I1( s ) 50 5 50 + 10 I 2 ( s ) = Þ I1 ( s ) éê10 + ùú + 10 I 2 ( s ) = s s s 0.2 s ë û

and

10 I1( s ) + 50 I 2 ( s ) =

Solving,

I1( s ) =

50 s

5 1 1 and I 2 ( s ) = s + 0.625 s s + 0.625

Taking inverse Laplace transform,

i1(t ) = 5e- 0.625t (A) and i2 (t ) = 1 - e- 0.625t (A), t > 0 6.13 Find using Final value theorem, the steady state value of I2(t) in the circuit shown in figure below. Switch S is closed at t = 0. The inductor is initially de-energized.

Solution Circuit for t > 0 is, By KVL, in Laplace transform, I1( s )[2 + 2 + 0.5s ] - [2 + 0.5s ]I 3 ( s ) = 48 s

or

I1( s )[ s + 8] - [ s + 4]I 3 ( s ) =

and

- I1 ( s )[2 + 0.5s ] + [4 + 0.5s ]I 3 ( s ) = 0

- I1 ( s )[ s + 4] + [ s + 8]I 3 ( s ) = 0 or Solving equations (i) and (ii),

I1 ( s ) =

24 s

(i)

(ii)

48/ s - ( s + 4) s +8 0 6( s + 8) = s ( s + 8) s +8 - ( s + 4) - ( s + 4)

s +8

and

I3 (s) =

s +8 48/ s - ( s + 4) 0 6( s + 4) = s +8 - ( s + 4) s ( s + 6) - ( s + 4)

s +8

6.56

Circuit Theory and Networks

\

I 2 ( s ) = I1( s ) - I 3 ( s ) =

6( s + 8) 6( s + 4) 24 = s ( s + 6) s ( s + 6) s ( s + 6)

24 =4A Ans. s+6 6.14 In a series LC circuit, the supply voltage being v = Vm cos (t), find i(t) with zero initial conditions. Assume L = 1H, C = 1F. Solution By KVL, for t > 0, Final value of the current, i2 (¥) = lim sI 2 ( s ) = lim

\

s ®0

s ®0

1 ù sVm = I ( s) éê sL + Cs úû s 2 + 1 ë or

I(s) =

é s2 ù sVm é ù s2 = Vm ê 2 ú = Vm ê ( s + j ) ( s - j ) ( s + j ) ( s - j ) ú 2 1 ë û ë ( s + 1) û ( s 2 + 1) æ s + ö è sø

é ù s2 = Vm ê 2 2ú ë (s + j ) (s - j ) û é K1 K1* K2 K 2* ù + + + = Vm ê 2 2 ( s - j ) ( s + j ) úû (s + j) ë (s - j) K1 = I ( s ) ´ ( s - j ) 2

s= j

=

1 4

where,

K2 =

( s + j )2 2s - s 2 ´ 2( s + j ) j d 1 = =( s - j )2 I ( s) 4 (2 - 1)! ds 4 (s + j) s= j

\

K1* =

j 1 ; and K 2* = 4 4

Thus

I(s) =

Vm 4

é 1 j j ù 1 + + ê 2 2 ( s - j ) ( s + j ) ûú (s + j) ë (s - j)

Taking inverse Laplace transform,

Vm V [te jt + te- jt - je jt + je – jt ] = m [t cos t + sin t ] (A); t > 0 4 4 6.15 The series RC circuit of figure has a sinusoidal voltage source, v = 180 sin (2000t + f) (V) and an initial charge on the capacitor Q0 = 1.25 mC with polarity as shown. Determine the current if the switch is closed at a time corresponding to f = 90o. What is the current at time t = 0? Solution By KVL, for t > 0, i(t) =

40i (t ) +

1 25 ´ 10- 6

t é ù -3 ê1.25 ´ 10 + ò i (t ) dt ú = 180 cos 2000t 0 ë û

Ans.

6.57

Laplace Transform and its Applications

Taking Laplace transform, 40 I ( s ) +

Þ

I (s) =

1.25 ´ 10- 3 4 ´ 104 180 s + I (s) = 2 s s + 4 ´ 106 25 ´ 10- 6 s

4.5s 2 1.25 ( s + 4 ´ 106 ) ( s + 103) s + 103 2

Applying Heaviside expansion formula to find the first term on the right hand side, we have, P (s) = 4.5s2, Q (s) = s3 ´ 103 s2 + 4 ´ 106 s + 4 ´ 109, Q¢ (s) = 3s2 + 2 ´ 103 s + 4 ´ 106, a1 = - j 2 ´ 103 ; a2 = j 2 ´ 103 and a31 = -103 Then,

i(t) =

3 P ( - j 2 ´ 103) - j 2 ´103 t P ( j 2 ´ 10 3) j 2 ´103 t P ( -10 3 ) -103 t + + - 1.25e -10 t e e e Q ¢( - j 2 ´ 103) Q ¢( j 2 ´ 10 3 ) Q ¢( -10 3 )

= (1.8 - j 0.9) e - j 2 ´10 t + (1.8 + j 0.9) e j 2 ´10 t - 0.35 e -10 3

3

3

= -1.8 sin 2000t + 3.6 cos 2000t - 0.35 e -10

3

t

= 4.02 sin (2000t + 116.6°) - 0.35 e -10 t (A); t > 0 3

6.16 In the RL circuit of Figure, the source is v = 100 sin (500t + f ) . Determine the resulting current if the switch is closed at a time corresponding to f = 0. Solution By KVL,

RI ( s ) + sLI ( s ) - Li (0-) = V ( s ) or

5I ( s ) + 0.01sI ( s ) =

or

I ( s) =

100 ´ 500 [Q i (0-)] = 0 s 2 + 25 ´ 104

5 ´ 106 ( s + 25 ´ 104 ) ( s + 500) 2

By Partial Fraction Expansion,

æ -1 + j ö æ -1 - j ö 10 + 5ç + I(s) = 5 ç è s + j 500 ÷ø è s - j 500 ÷ø s + 500 Taking inverse Laplace transform, i(t) = 10 sin 500t - 10 cos 500t + 10e- 500t = 14.14 sin (500t - 45°) + 10 e- 500t (A); t > 0

t

6.58

Circuit Theory and Networks

6.17 Determine the Laplace transform of the following periodic waveform. f( t ) 1 t(second) 0

p

2p

3p

4p

Solution Let, for the first half sine wave, the transform is F1(s). Now, f1(t) = sin tu(t) + sin(t – p) u(t – p) Taking Laplace transform, F1(s) =

e- p s 1 + e- p s 1 + 2 = 2 s +1 s +1 s +1 2

By the theory of periodicity of Laplace transform, the Laplace transform of the full periodic waveform will be, F(s) = F1 ( s) ´

1 + e- p s 1 1 = ´ - Ts 2 s +1 1- e 1 - e- p s

[Q T = p for the waveform given]

æ 1 + e- p s ö 1 = ç è 1 - e- p s ÷ø s 2 + 1 =

1 æ psö coth ç ÷ è 2ø s +1 2

Ans.

6.18 Determine the Laplace transform of the sawtooth waveform as shown below.

Solution For the first cycle, 1 1 r (t ) - u (t - T ) - r (t - T ) T T Taking Laplace transform,

f1(t) =

1 1 1 - Ts 1 1 - Ts 1 - e e = 2 [1 - (1 + Ts )e - Ts ] T s2 s T s2 Ts By Scalling Theorem (the theory of periodicity), Laplace transform of the given periodic function is, F1(s) =

F(s) = F1 ( s) ´ =

1 1 1 = [1 - (1 + Ts)e- Ts ] ´ 1 - e- Ts Ts 2 1 - e- Ts

e- Ts 1 2 Ts s (1 - e - Ts )

Ans.

Laplace Transform and its Applications

6.59

6.19 Find the Laplace transform of the waveform shown in figure.

( )

a 2 4 2 r (t ) - r t + r (t - a) a a a 2 Taking Laplace transform, Solution Here, v1 (t ) =

V1(s) = =

2 1 4 e - as / 2 2 e - as + 2 a s a s2 a s2

2 (1 - 2e - as / 2 + e - as ) as 2

2 (1 - e- as / 2 )2 as 2 By Scalling Theorem (the theory of periodicity), Laplace transform of the given periodic function is, =

V(s) = V1 ( s) ´

=

2 as 2

1 2 1 = (1 - e- as / 2 ) 2 ´ 1 - e- Ts as 2 1 - e- as

æ 1 - e - as / 2 ö çè 1 + e - as / 2 ÷ø

as 2 Ans. tanh æç ö÷ è 4ø as 2 6.20 The unit step response of a network is given by (1 – e–bt) . Determine the unit impulse response h(t) of this network. 1 Solution Here, the input is, w(t ) = u (t ) Þ W ( s ) = s 1 1 b = and the output is y (t ) = (1 - e - bt ) Þ Y ( s ) = s s + b s ( s + b) By convolution theorem,

=

Y ( s) = H ( s)W ( s) Þ

b 1 = H (s) s ( s + b) s

Þ

H (s) =

b ( s + b)

Taking inverse Laplace transform, the impulse response is,

h(t ) = be- bt

Ans.

6.60

Circuit Theory and Networks

6.21 The unit impulse response of current of a circuit having R =1 ohm and C = 1 F in series is given by

[d (t ) - exp( - t )u (t )] . Find the current expression when the circuit is driven by the voltage given as, [1 - exp(- 2t )]u (t ) . Solution Here, the impulse response is, h(t ) = [d (t ) - exp(- t ) u (t )] Þ H ( s ) = 1 The input is, w(t ) = [1 - exp(- 2t )]u (t ) Þ W ( s ) =

1 s . = s +1 s +1

1 1 2 = s s + 2 s ( s + 2)

By convolution theorem, the output is given by,

Y ( s ) = H ( s )W ( s ) =

s 2 2 2 2 ´ = = s + 1 s ( s + 2) ( s + 1) ( s + 2) s + 1 s + 2

Taking inverse Laplace transform,

y (t ) = (2e- t - 2e- 2t )

Ans.

6.22 The response of a network to an impulse is h(t ) = 0.18(e- 0.32t - e- 2.1t ) . Find the response of the network to a step function using convolution theorem. Solution By convolution theorem,

é 1 1 ù 1 Y(s) = H ( s )W ( s ) = 0.18 ê ú´ ë s + 0.32 s + 2.1 û s =

0.32 s ( s + 0.32) ( s + 2.1)

=

A3 A1 A2 + + s s + 0.32 s + 2.1

A1 =

0.32 ( s + 0.32) ( s + 2.1)

\

A2 =

0.32 s( s + 2.1)

\

A3 =

0.32 s ( s + 0.32)

\

= 0.477 s =0

= - 0.562 s = - 0.32

= 0.0856 s = - 2.1

Putting these values,

0.477 0.562 0.0856 + s s + 0.32 s + 2.1 Taking inverse Laplace transform, Y(s) =

y(t) = 0.477 - 0.562e- 0.32t + 0.0856e- 2.1t

Ans.

6.61

Laplace Transform and its Applications

6.23 Find the initial and final value of the functions given as,

bg

(a) F s =

b g

4 s+1

b g sds

(b) F s =

s2 + 4s + 6

5s 3 - 1600 3

i

+ 18s 2 + 90s + 800

Solution (a) By initial value theorem, the initial value of the functions is given as,

LM 4 + 4 OP L 4b s + 1g O s f b0 + g = Lim s F b sg = Lim M s ´ MN s + 4s + 6 PPQ = Lim MM1 + 4s + 6 PP = 4 s Q N s®¥

2

s®¥

s®¥

2

By final value theorem, the final value of the function is given as,

bg

bg

LM MN

f ¥ = Lim s F s = Lim s ´ s®0

s®0

b g OP = 0 + 4 s + 6 PQ

4 s+1 s

2

Ans.

(b) By initial value theorem, the initial value of the functions is given as,

b g

af

LM MN scs

f 0 + = Lim sF s = Lim s ´ s®¥

s®¥

5s 3 - 1600 + 18 s 2 + 90 s + 800

3

OP h PQ

LM 5s - 1600 OP MN cs + 18 s + 90 s + 800h PQ LM 5 - 1600 OP s = Lim M 18 90 800 P 1 + + MN s s + s PQ = Lim

s®¥

3

3

2

3

s® ¥

2

=5

3

Ans.

By final value theorem, the final value of the function is given as,

af

af

LM MN s cs

f ¥ = Lim sF s = Lim s ´ s ®0

= Lim s®0

=

LM MN cs

3

s®0

3

5s 3 - 1600 + 18 s 2 + 90 s + 800

5s 3 - 1600 + 18 s 2 + 90 s + 800

-1600 = -2 800

Ans.

OP h PQ

OP h PQ

Ans.

6.62

Circuit Theory and Networks

6.24 Express the function in terms of the standard signals and find its Laplace transform. f(t) 1

0

t

3

2

1

Solution The function can be written as the summation of some ramp functions as given below.

bg bg b g b g b g e e 1 e F b sg = + s s s s f t = r t - r t -1 - r t - 2 + r t - 3

\

2

-s

-2 s

-3s

2

2

2

6.25 Find the current i(t) flowing through the circuit if the circuit is initially relaxed. Find the voltage across the capacitor vc (t) also. What is the value of the steady state current? 5W 1 F 2

10 V

Vc(t)

Solution By KVL,

\

F 5 + 1 I I bsg = 10 Þ H s/2 K s 10 2 I b sg = = 5s + 2 s + 2 /5

b gb

g

I s 5s + 2 = 10

Taking inverse Laplace transform, the current in the circuit,

bg

i t = 2 e- 2t /5

bA g

Ans.

Voltage across the capacitor is,

bg bg

FG H

2 2 4 1 1 1 = ´ = = 10 1 s s + 2 / 5 s s + 2 /5 s s + 2 /5 s 2 Taking inverse Laplace transform, VC s = I s ´

b

bg

VC t = 10 1 - e- 2 t /5

g

b Vg

IJ K

Ans.

From the current expression, as t ® µ, i(t) ® 0. So, the steady state value of the current is, I ss = 0

Ans.

6.26 The series RL circuit shown in the figure is excited by a dc voltage of 50 V. Assume the initial current flowing through the inductor to be 5 A and find the current i(t) for t > o. Use Laplace transform method.

6.63

Laplace Transform and its Applications Switch

5W

50 V

1H

i( t )

Solution Applying KVL for the loop, we get,

bg

Ri t + L

bg

di t = 50 dt

Taking Laplace transform,

af

a f b g 50s 50 +5 5 I asf + sI asf = s bs + 5g I asf = 50s + 5 1 1 O 5 50 5 I asf = + = 50 LM + P s + 5 s + s b s + 5g b s + 5g s b N Q 5g

RI s + L sI s - i 0 - = Þ Þ Þ

Taking inverse Laplace transform, we get,

bg

d

i

d

i t = 50 1 - e -5 t + 5e-5 t = 50 - 45e - 5t

i bA g

t>0

Ans.

bg

bg

6.27 Find the current i (t) for the circuit shown in the figure, if the voltage source is v t = 5e- 2 t u t and

b g

vC 0 - = 0 . Switch

v(t)

1W

i( t )

1F

Solution Applying KVL for the loop,

Ri (t ) +

t

z

bg

1 i (t ) dt = v (t ) = 5e-2t u t C0

Taking Laplace transform,

RI ( s) +

LM N

OP Q 1 5 I ( s) LM1 + OP = N sQ s + 2

1 I ( s) q ( 0 - ) 5 + = C s s s+2

b g

(since vC 0 - = 0 )

6.64

Circuit Theory and Networks

I ( s) =

5s

10

5

bs + 1gbs + 2g = s + 2 - s + 1

Taking inverse Laplace transform,

i (t ) = 10 e- 2 t - 5e - t ; for t ³ 0 Ans. 6.28 Determine the current i (t) in a series RLC circuit consisting of R = 5 W, L = 1 H and C = ¼ F when the source voltage is given as: (a) ramp voltage 12r (t – 2) and (b) step voltage 3u (t – 3). Assume that the circuit is initially relaxed. Solution Applying KVL for the series RLC circuit we get,

b g didtbt g + C1 z ibt gdt = vbt g dibt g 1 Þ + 5ibt g + ibt gdt = vbt g 1 /4 z dt (a) When v bt g = 12 r bt - 2g di bt g 1 5i b t g + + i bt g dt = v bt g dt 1 /4 z Ri t + L

Taking Laplace transform,

F 5 + s + 4 I I bsg = 12 e H sK s

-2 s

2

Þ

bg

I s =

12 e - 2 s

d

s s 2 + 5s + 4

=

i

= 12 e - 2 s

LM 1 OP = 12e L 1/ 4 - 1/ 3 + 1/12 O MN s s + 1 s + 4 PQ MNsbs + 1gbs + 4g PQ -2 s

3e -2 s 4e -2 s e -2 s + s s +1 s + 4

Taking inverse Laplace transform, we, get,

i t = 3u t - 2 - 4 e- bt - 2 g + e- 4 bt - 2 g

bg b g (b) When v bt g = 3u bt - 3g di bt g 1 5i bt g + + i bt g dt = v bt g dt 1/ 4 z

Ans.

Taking Laplace transform,

F 5 + s + 4 I I b sg = 3 e H sK s Þ

bg

I s =

=

-3s

3e - 2 s

ds

2

+ 5s + 4

i

e -2s e -2s s+ 4 s+1

= 3e - 2 s

LM 1 OP = 3e L 1 / 3 - 1/3 O MN s + 4 s + 1PQ MN bs + 1gbs + 4g PQ -2 s

6.65

Laplace Transform and its Applications

Taking inverse Laplace transform, we, get,

i t = e- bt - 3g + e- 4 bt - 3g Ans. 6.29 For the RC parallel circuit shown in the figure, determine the voltage across the capacitor using Laplace transform method. Assume the capacitor to be initially relaxed. Solution Applying KCL at the upper node,

bg

bg

5W

10 A

1F

bg bg

v t dv t +C = i t = 10 R dt Taking Laplace transform and putting the values of R and C,

bg

V s 10 + sV s = 5 s

Þ

bg

LM 1 1 OP = 50 M V b sg = 1P s 1 sF s + I H 5K MN s + 5 PQ 10

Taking inverse Laplace transform, we get,

bg

d

v t = 50 1 - e - t /5

i b Vg

Ans.

6.30 The circuit was in steady state with the switch in position 1. Find the current i(t) for t > 0 if the switch is moved from position 1 to 2 at t = 0. 10 W 1

2

10 V

50 V

0.5 H

Solution When the switch is in position 1, steady-state exists and the initial current through the inductor is, 10 = 1A i (0 -) = 10 After the switch is moved to position 2, the KVL gives, in Laplace transform, 50 10 I ( s) + 0.5sI ( s) - 0.5 ´ 1 = s

I ( s) =

or,

LM N

b

g

Taking inverse Laplace transform,

i (t ) = 5 - 4 e- 20t

OP Q

100 1 1 1 1 + =5 + s + 20 s s + 20 s + 20 s s + 20 ( A); t > 0;

Ans.

6.66

Circuit Theory and Networks

6.31 (a) In the circuit shown in the figure, The switch S has been thrown to position 1 for a long period of time. Find the complete expression for the current after throwing the switch S to 2 which removes R1 from the circuit. R1

1 R2

S

t=0

+ 2

V

L

–

(b) If the values of V, R1, R2 and L be 10 V, 1 ohm, 2 ohms and 1 H respectively, calculate (i) steady state current (ii) the energy stored in the inductance at steady state period (iii) time constant of the circuit for both the positions of the switch S Also calculate the voltage across the resistor R2 and inductor L, at 0.05 second after the switch S has been thrown to position 2. Solution (a) For t < 0, as the circuit was in steady state with the switch in position 1, the circuit becomes as shown below.

b g

i 0- =

\

V R1 + R2

For t > 0, the circuit becomes as shown. R1

R2

R2

+

+

V

V

i(0–)

L

i(t)

–

– Circuit fort < 0

Circuit fort > 0

By KVL,

bg

bg

b g Vs V VL + sL I b sg = + s R +R OP + V F 1 I V L 1 I b sg = M R MN s b s + R /Lg PQ R + R GH b s + R /Lg JK

R 2 I s + sLI s - Li 0 - =

Þ

R2

1

Þ

2

2

2

1

2

2

Taking inverse Laplace transform,

bg

i t =

V V e - b R2 /L g t 1 - e- b R2 /L g t + R2 R1 + R2

e

j

bAg, t > 0

Ans.

6.67

Laplace Transform and its Applications

(b) V = 10 V, R1 = 1 ohm, R2 = 2 ohm and L = 1H

V 10 Ans. = =5A R2 2 (ii) Energy stored in the inductance at steady state period, 1 1 W = L I 2 = ´ 1 ´ 52 = 12.5 watts 2 2 (iii) Time constant of the circuit for switch in position 1 is, (i) Steady state current, Iss =

Ans.

L 1 Ans. = = 0.33 s R1 + R2 1 + 2 Time constant of the circuit for switch in position 2 is, t1 =

L 1 = = 0.5 s R2 2 For t = 0.05, voltage across the resistor, t2 =

LM d N

i

VR2 = i ´ R2 = 5 1 - e - 2 t +

20 - 2 t e 3

b

Ans.

OP Q

´2=7V

Ans.

t = 0.05

g

Ans. and voltage across the inductor, V L = 10 - 7 = 3 V 6.32 The circuit given in figure is initially at steady state with the switch ‘K’ open. If the switch is closed at time t = 0, find the voltage ‘VC(t)’ across the capacitor. 1 kW

1 kW 1 m F V c(t)

6V

1 kW

K

Solution At steady-state before closing the switch, the capacitor becomes open-circuited. So, the circuit becomes as shown above. 2 v(0 - ) = ´ 6 = 4 V 3 1 kW 1 kW 6V

+ –

v(0+)

1 kW

6.68

Circuit Theory and Networks

For t > 0, by KVL,

b

g

1 ´ 10 3 ´ I 1 + 1 ´ 10 3 ´ I 1 - I 2 =

and

6 s

Þ

FG H

10 6 4 I + 1 ´ 10 3 ´ I 2 - I 1 = s 2 s

b

2000 I 1 - 1000 I 2 =

g

Þ - 1000 I 1 + 1000 +

6 s

10 6 s

(i)

IJ I K

2

=-

4 s

(ii)

Solving equations (i) and (ii),

2000 6 /s -1000 - 4 /s I2 = -1000 2000 -1000 1000 + 106 / Cs

d

\

Vc ( s) = I 2 ´ =

=-

FG H

2 1 1000 s + 2000

i

IJ K

10 6 4 2000 4 + =+ s s s s + 2000 s

b

g

3 1 + s s + 2000

Taking inverse Laplace transform,

vc (t ) = 3 + e- 2000 t (Volt ), t > 0

Ans.

6.33 In the circuit in the figure, the steady state exists when the switch S is in position a for a considerable period of time. Find the current response after throwing the switch from position a to b. What will be the steady state value of the current? 10 W a b 20 V 100 m F

1H

Solution When the switch is in position a, steady-state exists and the initial current through the inductor is, 20 = 2A i (0 -) = 10 After the switch is moved to position b, the KVL gives, in Laplace transform,

1 I ( s) + 1sI ( s) - 1 ´ 2 = 0 100 ´ 10 - 6 s or,

I ( s) =

2 2s = 4 2 10 s + 10 4 s+ s

d

i

10 W

20 V

6.69

Laplace Transform and its Applications

Ans.

i

• The steady state current will oscillate sinusoidally following the relation i (t ) = 2 cos100t with peak magnitude of 2 A and frequency of 100 rad/s or 16.9 Hz. 6.34 A dc voltage applied to a coil of inductance L and resistance R is suddenly changed from V1 to V2. (a) Find an expression for current in the circuit. (b) If R = 10 W, L = 1 H, V1 = 100 V, and V2 = 200 V, find current at t = 0.5 s. (c) If R = 10 W, L = 1H, V1 = 200 V, and V2 = 100 V, find current at t = 0.5 s.

10 s

s

4

2 Volt

+ –

Taking inverse Laplace transform, i (t ) = 2 cos100 t ( A); t > 0;

V1 R (a) After changing the voltage, the KVL equations is,

b g

Solution Here, initial current in the circuit, i 0 - =

bg

Ri t + L

bg

di t = V2 u t dt

bg

Taking Laplace transform,

b g b g Vs V VL I b sg R + sL = + s R V VLF 1 I V /L F V /R I I asf = + = + R H R + sL K s b s + R /Lg H s + R /L K s b R + s Lg

bg

RI s + L sI s - i 0 - = Þ

2

2

Þ

1

2

1

2

1

Taking inverse Laplace transform,

bg

i t =

FG H

IJ K

V2 V V V V 1 - e- b R / L gt + 1 e- b R / L gt = 2 + 1 - 2 e- b R/ L gt R R R R R

Ans.

(b) If R = 10 W, L = 1 H, V1 = 100 V, and V2 = 200 V, we get the current at t = 0.5 s as,

bg

i t =

F H

I K

200 100 200 - b10 /1g ´ 0.5 + = 20 - 10e -5 = 19.93 A e 10 10 10

Ans.

(c) If R = 10 W, L = 1 H, V1 = 200 V, and V2 = 100 V, we get the current at t = 0.5s as,

bg

i t =

F H

I K

100 200 100 - b10 /1g ´ 0.5 + = 10 + 10 e - 5 = 10.07 A e 10 10 10

Ans.

6.35 A 50 mF capacitor and 20000 W resistor are connected in series across a 100 V battery at t = 0. At t = 0.5 s, the battery voltage is suddenly increased to 150 V. Find the charge on capacitor at t = 0.75 s. Solution When the circuit is connected to 100 V supply, the equation of voltage across the capacitor is,

d

i

e

vC = E 1 - e - t / RC = 100 1 - e - t / 20000 ´ 50 ´ 10

-6

j = 100 d1 - e i -t

6.70

Circuit Theory and Networks

d

i

At t = 0.5s, the voltage across the capacitor is, vC = 100 1 - e- 0.5 = 39.347 V \ At t = 0.5 s, charge on the capacitor is, q = CvC = 50 ´ 10 -6 ´ 39.347 = 1967.35 ´ 10 - 6 C This charge is the initial charge q0 when the battery voltage is suddenly increased to 150V. When the circuit is connected to 150 V, the KVL equation becomes,

bg

Ri t +

t

zbg

bg

1 i t dt = Vu t C0

Taking Laplace transform,

b g C1 LM I bssg + qs OP = Vs N Q

RI s +

0

V q0 I s = R RC 1 s+ RC

bg

Þ Taking inverse Laplace transform, i (t ) =

LMV - q OP e N R RC Q 0

- t / RC

Therefore, the voltage across the capacitor,

VC =

z FGH

IJ K

t t q q 1 1 V i (t ) dt = - 0 e- t /RC dt = V 1 - e- t /RC + 0 e- t /RC C0 C 0 R RC RC

z

d

i

Substituting the values, the voltage across the capacitor at t = 0.75 s i.e., 0.25 second after changing the battery voltage,

d

i

VC = V 1 - e - t /RC +

q 0 - t / RC 1967.35 ´ 10 - 6 - 0.25 = 150 1 - e- 0.25 + = 63.82 V e e RC 1

d

i

. = 319 . ´ 10 - 3 C Ans. \ charge on the capacitor is, q = CvC = 50 ´ 10 - 6 ´ 6382 6.36 For the circuit shown in the figure, find an expression for the current supplied by the source. How much time will it take for the current to reach 25 mA? Assume the circuit to be initially relaxed.

10 V

500 W

700 W i2

100 m F

i1

Solution Applying KVL for the two meshes, we get,

500i1 - 500i2 = 10 -500i1 + 1200i2 +

t

1 i2 dt = 0 100 ´ 10 -6 0

z

6.71

Laplace Transform and its Applications

Taking Laplace transform,

bg b g 10s F 10 I I b sg = 0 - 500 I b sg + G 1200 + s JK H 500 I1 s - 500 I 2 s =

4

1

2

Solving for I1(s), we get,

- 500

10 / s

FG1200 + 10 IJ 0 s K H 24 s + 200 24 200 I b sg = = = + - 500 500 s d700s + 10 i 700 s + 10 s d700 s + 10 i F 10 IJ - 500 G 1200 + s K H IJ + 2 FG 1 IJ = 1 F 1I + 1 F 1 I 24 F 1 = G 700 H s + 100 / 7 K 7 H s b s + 100 / 7 g K 50 H s K 70 H s + 100 / 7 K 4

1

4

4

4

4

Taking inverse Laplace transform,

bg

i1 t =

1 1 -100t / 7 + e 50 70

For the current to be 25 mA, we get, 1 1 - 100 t /7 + 25 ´ 10 - 3 = e 50 70

b Ag

Þ t = 0.0735 second

Ans.

6.37 The figure shows a parallel RLC circuit fed from a dc current source through a switch. The circuit elements are R = 400 W, L = 25 mH, C = 25 nF. The source current is 24 mA. The switch which has been in closed position for a long time is opened at t = 0.

I

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

S

L

R

C

v(t)

What is the initial value of current iL (i.e., at t = 0)? What is the initial value of voltage across L at t = 0? What is the expression for current through inductance, capacitance and resistance? What is the final value of iL? What happens to iL(t) if R is increased from 400 to 625 W? Assume that initial energy is zero.

Solution Applying KCL for the node, we get,

v (t ) 1 t dv vdt + C + =I dt R L0

z

6.72

Circuit Theory and Networks

Taking Laplace transform, V ( s)

Þ

LM 1 + 1 + sCOP = I N R sL Q s V b sg = CF s H

2

I s 1 + + RC LC

I K

Substituting the values,

af

V s =

24 ´ 10 - 3

FG H

25 ´ 10 -9 s 2 + =

s 1 + 400 ´ 25 ´ 10 - 9 25 ´ 10 - 3 ´ 25 ´ 10 -9

IJ K

=

24 ´ 10 6 25 s 2 + 10 5 s + 16 ´ 10 8

d

i

24 ´ 10 6 16 16 = 4 4 4 s + 2 ´ 10 s + 8 ´ 10 4 25 s + 2 ´ 10 s + 8 ´ 10

c

hc

h

Taking inverse Laplace transform, v ( t ) = 16 e- 2 ´ 10 t - 16e -8 ´ 10 Also, the current through the inductor, 4

a f VsLasf = 25 ´V a10sf

IL s =

-3

s

=

25 ´ 10

-3

4

t

( V)

24 ´ 10 6 ´ 25 s s + 2 ´ 10 4 s + 8 ´ 10 4

c

hc

=

384 ´ 10 5 s s + 2 ´ 10 4 s + 8 ´ 10 4

=

24 32 8 + 1000 s 1000 s + 2 ´ 10 4 1000 s + 8 ´ 10 4

c

hc c

h

h

h

c

h

Taking inverse Laplace transform,

i L t = 24 - 32 e- 2 ´ 10 t + 8e-8 ´ 10

bg

bg (b) At t = 0, we get, vb0g = 0

(a) At t = 0, we get, i L 0 = 0

4

4

t

bmAg

Ans.

(c) Current through inductance

i L t = 24 - 32 e- 2 ´ 10 t + 8e-8 ´ 10 Current through capacitance,

bg

bg

iC t = C

bg

4

4

t

bmAg Ans.

dv t 4 4 4 4 d = 25 ´ 10 - 9 16e -2 ´ 10 t - 16e -8 ´ 10 t = 32e -8 ´ 10 t - 8e -2 ´ 10 t dt dt

bmAg

Ans.

6.73

Laplace Transform and its Applications

Current through resistance, 1 ´ 16e b g v Rbt g = 400

- 2 ´ 10 4 t

iR t =

- 16e - 8 ´ 10

4

t

=

4 1 - 2 ´ 10 4 t - 16e- 8 ´ 10 t e 25

bg

b g

(d) At t = ¥, the final value of iL is, i L ¥ = 24 - 32e - ¥ + 8e - ¥ = 24 mA (e) Putting the value of resistance R = 625 W in the expression of iL, we get,

b g VsLbsg =

IL s =

sLC F s H

I 2

+

s 1 + RC LC

=

Ans.

Ans.

I K 24 ´ 10 - 3

F GH

s ´ 25 ´ 10 - 3 ´ 25 ´ 10 - 9 s 2 + =

bA g

s 1 + -9 -3 625 ´ 25 ´ 10 25 ´ 10 ´ 25 ´ 10 - 9

I JK

384 ´ 10 5

d

s s 2 + 64 ´ 10 3 s + 16 ´ 10 8

i

Taking inverse Laplace transform and simplifying, we get,

bg

i bA g

d

i L t = 106.67 ´ 10 9 e - 32 ´ 10 t sin 14.4 ´ 10 3 t 3

Ans.

Here, with R = 400 W, the circuit was in overdamped condition. As the value of resistance is increased to 625 W, the circuit becomes underdamped. 6.38 In the network of the figure, the switch S has been closed for a long time. The switch is suddenly opened at t = 0 and reclosed at t = 20 ms. Find expression for voltage V0 for t £ 20 ms and t > 20 ms. 1000 W

S 120 V

3000 W

+ –

2000 W

V0

0.001 mF

Solution With the switch closed, the initial voltage across the capacitor is,

b g

vC 0 - =

120 ´ 2000 = 80 V 1000 + 2000 1000 W

120 V

+ –

2000 W

v C(0–)

6.74

Circuit Theory and Networks

After the switch is opened, the transformed circuit is shown in the figure below. 1000 W

3000 W 120 s

X

+ –

1

V0(S)

10 –8 s + –

2000 W

80 s

Applying KCL at node X, we get,

bg

120 80 V s + VX + X s =0 4000 2000 1 10 -8 s

VX s -

Þ

VX

LM 1 + 1 + 18 sOP = 120 + 80 ´ 100 N 4000 2000 Q 4000s -8

Þ

VX =

=

-8

=

0.03 + 80 ´ 100 - 8 s

80 ´ 100 - 8 0.03 + 0.00075 + 10 - 8 s s 0.00075 + 10 - 8 s

d

i

40 40 80 40 40 + = + s s s + 0.075 ´ 10 6 s + 0.075 ´ 10 6 s + 0.075 ´ 10 6

Taking inverse Laplace transform,

V X t = 40 + 40e- 0.075 ´10 Therefore, the desired voltage is,

bg

6

t

bg

120 - V X t 6 ´ 3000 = 100 + 10e- 0.075 ´ 10 t for 0 £ t £ 20 ms 4000 At t = 20 ms, the voltage of node X is,

bg

bg

V0 t = V X t +

-6

V X = 40 + 40e- 0.075 ´ 10 ´ 20 ´ 10 = 48.925 V When the switch is reclosed at t = 20 ms, the voltage across the capacitor will be 48.925 V. After reclosing the switch, the transformed circuit is shown in the figure below. Now let the voltage of node X be VX’. Applying KCL at node X, we get, 6

120 48.925 V¢ s s + VX¢ s + X s =0 1000 2000 1 10 -8 s

af

VX¢ s -

af

af

Ans.

120 s

1000 W

V¢x X

+ –

1

V0(S)

10 –8 s

2000 W

+ –

48.925 s

6.75

Laplace Transform and its Applications

1 1 120 + + 10 s O = a fLMN1000 PQ 1000s + 48.925 ´ 10 2000 0.12 + 48.925 ´ 10 V ¢ asf 0.0015 + 10 s = s -8

VX¢ s

Þ Þ

-8

X

-8

-8

.925 ´ 10 a f sc0.00150.12+ 10 sh + 048.0015 + 10 0.12 48.925 = + scs + 0.15 ´ 10 h s + 0.15 ´ 10 80 31.075 V ¢ asf = s s + 0.15 ´ 10

Þ

VX¢ s =

-8

-8

X

s

6

6

Þ

-8

6

Taking inverse Laplace transform, we get,

bg

V X¢ t = 80 - 31075 . e- 0.15´10

6

t

In this case, the output voltage V0 is equal to VX’(t). Since time t is to be counted from the instant the

dt - 20 ´ 10 i . -6

switch is reclosed, t is replaced by

- 0.15 ´ 10 6 t - 20 ´ 10 - 6

bg

d

V0 t = 80 - 31075 . e

\

i for t > 20 ms

Ans.

6.39 In the circuit of the figure the switch S is closed at t = 0 and opened again at t = p second. Prior to closing the switch at t = 0, vC = 10 V while L and C2 do not have any stored energy. Find the voltages vC1 and vC2 at t = p second. C1 = C2 = 1 F; L = 2 H Solution After closing the switch, applying KVL in the circuit, we get,

L

bg

di t 1 + dt C1

t

S

t

z ibt gdt + C1 z ibt gdt = 0

+

2 -¥

-¥

C1

Since initial voltage across C1 is 10 V, we get,

vc

1

–

b g I bssg - 10s + I bssg = 0 2 10 I b sgLM2 s + OP = N sQ s 5 I b sg = s +1

2 sI s +

Þ Þ

2

Taking inverse Laplace transform, we get,

bg bt g = - 10 + C1

i t = 5sin t \

vC1

p

z 5sin tdt = - 10 + - 5csot

1 0

p 0=

0

Ans.

C2

+ vc –

2

6.76

Circuit Theory and Networks

\

bg

vC2 t =

p

1 5 sin t dt = - 5csot C2 0

z

p 0

= 10 V

Ans.

6.40 The network shown in the figure is in steady state with S1 closed and S2 open. At t = t1, S1 is opened and S2 is closed. Find current through capacitor for t ³ t1. 2W

2H S1

10 V

S2 1 mF

3H

Solution When switch S1 is closed and S2 is opened, the initial current through the 3 H inductor is, 10 = 5A . Initial voltage across the capacitor is zero. i 0- = 2 When switch S1 is opened and S2 is closed, the current through the capacitor is given by the KVL equation as,

b g

3

bg

t di t 1 i t dt = 0 + 6 dt 1 ´ 10 0

zbg

Taking Laplace transform,

bg

1 I s =0 10 - 6 s

bg

3sI s - 3 ´ 5 +

bg

15s = 3s + 106

I s =

Þ

2

5s 106 s2 + 3

Taking inverse Laplace transform we get,

bg

i t = 5 cos

LMF MNGH

I OP JK P Q

10 6 t = 5 cos 577.35t 3

b

g

b g

Since the switch is closed at t = t1, the time will be shifted by t - t1 so that the current through the capacitor is given as,

bg

b g

i t = 5 cos 577.35 t - t1

for t ³ t1

Ans.

6.41 The switch in figure has been in position A for a long time. At t = 0 it is moved to B and at t = 1 second, it is moved to A again. Find voltage across capacitor after a further lapse of 1 millisecond. 500 ¥ 10 3 W

B A

10 V

1500 W

1 mF

vc

6.77

Laplace Transform and its Applications

Solution As the switch is in position A for a long time, initial charge across the capacitor is zero. When the switch is moved to position B, the current in the circuit is obtained from the KVL equation as,

b g FGH

I s 500 ´ 10 3 +

I JK 10 1 F 1 I I b sg = = G J 500 ´ 10 b s + 2g 5 ´ 10 H s + 2 K

1 10 = -6 s 1 ´ 10 s

Þ

3

4

Taking inverse Laplace transform,

bg

bA g

1 e-2 t 5 ´ 104 Therefore, voltage across the capacitor is, i t =

bg

ibg

d

d

vC t = 10 - 500 ´ 103 i t = 10 - 500 ´ 10 3

i 5 ´110

4

d

e-2 t = 10 1 - e -2 t

i b Vg

Therefore, voltage across the capacitor at t = 1second is,

bg

d

i

v C t = 10 1 - e -2 = 8.65

bV g

At t = 1second, the switch is moved to position A, so that the KVL equation becomes, t

zbg

bg

1 i t dt + 1500i t = 0 1 ´ 10 - 6 0 Taking Laplace transform,

b g LMN

I s 1500 +

OP Q 8.65 F 1 I I b sg = 1500 H s + 666.67 K

1 8.65 = since initial voltage across the capacitor is 8.65 V. s 1 ´ 10 -6 s

Þ

Taking inverse Laplace transform,

bg

8.65 - 666.67 t e 1500 Hence, the voltage across the capacitor is, i t =

bg

bg

vC t = 1500i t = 1500 ´

bA g

8.65 -666.67 t = 8.65e -666.67 t e 1500

At t = 1ms, the voltage is, vC = 8.65e -666.67 ´ 10 ´ - 3 = 4.44 V

bg

6.42 Determine the Laplace transform of f t =

2 - 2e t

Ans.

-t

.

Solution

bg

f t =

FG H

IJ K

2 - 2e- t 2 2 2 t2 t3 t4 t5 = 1 - e- t = t e t - 1 = t 1 + t + + + + + ... - 1 (Expanding et) 2! 3! 4! 5! t t te te

d

i

d

i

6.78

Circuit Theory and Networks

bg

\ f t =

FG H

IJ K

t2 t3 t4 t5 2 t + + + + + ... 2! 3! 4! 5! te t

F H

I K

1 -t 1 2 -t 1 1 te + t e + t 3e - t + t 4 e - t + ... 2! 3! 4! 5! Taking Laplace transform of each term, we get,

= 2 e -t +

L O b g MM s +1 1 + 21! bs +1!1g + 31! bs +2!1g + 41! bs +3!1g + ...PP N Q L 1 F 1 I F 1 I F 1I F 1 I F 1 I F 1 = 2M MN s + 1 + H 2 K GGH bs + 1g JJK + H 3K GGH bs + 1g JJK + H 4 K GGH bs + 1g

F s =2

2

3

4

2

3

4

I OP JJ + ...P K Q

Ans.

6.43 Express the following functions in terms of singularity functions and find their Laplace transform: (a) f(t)

(b) f(t)

Vm

K t

0

(c)

0

p

t

1

f(t)

f(t)

(d)

1

K

e –t/2 K (2 – t) 2

Kt 2

t 0

1

t

2

0

1

2

3

4

5

Solution (a) Here, the signal can be expressed in terms of step signal as,

bg

bg sin w t ubt g + V

6

f(t) Vm

F T I uFt - T I H 2K H 2K sin w bt - p g ubt - p g

f t = Vm sin w t u t + Vm sin w t = Vm

m

t 0

Taking Laplace transform of individual terms, we get the Laplace transform of the functions as,

bg

F s =

Vm V e -p s V + m2 = m 1 + e-p s 2 s + 1 s + 1 s2 + 1

d

i

Ans.

p

6.79

Laplace Transform and its Applications

(b) Here, the signal starts with a straight line of slope K passing through origin and then comes to zero at t = 1.Hence the signal can be expressed in terms of ramp and step signals as,

bg

bg

b g

b g

f t = Kr t - Kr t - 1 - Ku t - 1

f(t) K

Ans. 0

1

t

Taking Laplace transform of individual terms, we get the Laplace transform of the functions as,

K Ke - s Ke - s K - 2 = 2 1 - 1 + s e- s s s2 s s (c) The function can be written as,

bg

b g

F s =

bg

Ans.

bg b g b g b g b g b g = Kt u bt g - K t - b2 - t g u bt - 1g - K b2 - t g u bt - 2g = Kt u bt g - K 4 bt - 1g u bt - 1g - K b2 - t g u bt - 2g 2

2

f t = Kt 2 u t - Kt 2u t - 1 + K 2 - t u t - 1 - K 2 - t u t - 2 2

2

2

2

2

2

f(t) K K (2 – t) 2

Kt 2

t 0

1

2

Taking Laplace transform of individual terms, we get the Laplace transform of the functions as,

bg

F s =K (d)

LM 2 - 4e Ns s 3

-s

2

-

OP Q

2e - 2 s 2 = K 3 1 - 2 se- s - e - 2 s 3 s s

Ans.

f(t) 1

e –t/2

t 0

1

2

3

4

5

6

The function can be written as,

bg

bg b g b g b g b g b g ubt g - u bt - 1g + u bt - 2g - u bt - 3g + u bt - 4g - u bt - 5g + ...

f t = e - t /2 u t - u t - 1 + e - t /2 u t - 2 - u t - 3 + e - t /2 u t - 4 - u t - 5 + ... =e

- t /2

6.80

Circuit Theory and Networks

Taking Laplace transform of individual terms, we get the Laplace transform of the functions as,

bg

1 e- b s + 1/2 g e -2 b s +1/2 g e-3 b s + 1/2 g + + ... s + 1/ 2 s + 1 /2 s + 1 /2 s + 1/ 2 1 = 1 - e - b s + 1/2 g + e - 2b s + 1/2 g - e- 3b s + 1/2 g + ... s + 1/ 2

F s =

FG IJ H K F 1 IJ FG 1 =G H s + 1/ 2 K H 1 + e b

I

- s + 1/2

g JK

Ans.

6.44 A pulse voltage of width a and magnitude 10 V is applied at time t = 0 to a series RL circuit consisting of a resistance R = 4 W and an inductor L = 2 H. Find the current i(t). Assume zero current through the inductor L before application of the voltage pulse. Solution The pulse voltage can be written as,.

bg

bg

b g

v t = 10u t - 10u t - a Applying KVL for the RL series circuit with the pulse voltage,

bg bg

di t =v t dt

bg

Ri t + L

Taking Laplace transform,

bg

bg b g bg

RI s + L sI s - i 0 - = V s With zero initial current, substituting the values we get, 10 4 I s + 2 sI s = 1 - e - as s

bg

Þ

bg

d i 10 F 1 - e I 5d1 - e i 5 I b sg = G = = d1 - e s H 2 s + 4 JK 2 sb s + 2g 5 L1 1 e e O = M + 2 Ns s+ 2 s s + 2 PQ - as

- as

- as

- as

iLMN 1s - s +1 2 OPQ

- as

Taking inverse Laplace transform,

b g 25 LNMd1 - e i ubt g - e1 - e

i t =

b g j u bt - a gO

-2 t - a

-2 t

QP

Ans.

6.45 A voltage pulse of width b and magnitude 10 V is applied at time t to a series RC circuit consisting 1 of a resistor R = 1 W and a capacitor C = F. Find the current i(t). Assume zero charge across the 4 capacitor C before application of the voltage pulse. Solution The pulse voltage can be written as,.

bg

bg

b g

v t = 10u t - 10u t - b Applying KVL for the RC series circuit with the pulse voltage,

bg

Ri t +

t

zbg bg

1 i t dt = v t C -¥

6.81

Laplace Transform and its Applications

Taking Laplace transform,

LM b g b g OP = V bsg N Q

v 01 I s + C s s

bg

RI s +

With zero initial voltage, substituting the values we get,

b g sI1b/s4g = 10s d1 - e i F 1 - e IJ = 10d1 - e i = LM 10 - 10e OP I b sg = 10G H s+ 4 K s+ 4 Ns+ 4 s+ 4 Q - bs

4I s +

- bs

- bs

Þ

-bs

Taking inverse Laplace transform,

i t = 10 e -4 t u t - e -4b t -b g u t - b

bg

bg

b g

Ans.

6.46 Find the response current of a series RL circuit consisting of a resistor R = 3 W and an inductor L = 1 H when each of the following driving force voltage is applied:

b g (b) unit impulse voltage d bt - 2g , (c) unit step voltage u bt - 2g , (d) unit doublet voltage d ¢bt - 2g , and (a) unit ramp voltage r t - 2 ,

(e) pulse of width a and magnitude 1 V beginning at time t = 2 seconds. Solution

b g

(a) Unit ramp voltage r t - 2 Applying KVL to RL series circuit, Ri + L

bg b g

di =v t = r t - 2 dt

Taking Laplace transform,

b R + sLgI bsg = s1 e e I b sg = s bsL + Rg -2 s

2

-2 s

2

Substituting the values,

bg

I s = \

\

K1 =

K2 =

LM N

K K K e -2 s = e -2 s 21 + 2 + 3 s s+3 s s s+3 2

b g

1 s+3

LM N

= s=0

d 1 ds s + 3

OP Q

OP Q

1 3

= s=0

1

b g s+3

=-

2 s=0

1 9

6.82

Circuit Theory and Networks

\ \

K3 =

1 s2

bg

= s=-3

I s = e -2 s

1 9

LM 1 /3 + -1 /9 + 1 /9 OP N s s s + 3Q 2

Taking inverse Laplace transform, 1 1 1 i t = - u t - 2 + r t - 2 + e -3b t - 2 g u t - 2 9 3 9

bg

b g

b g

b g

Ans.

b g

(b) Unit impulse voltage d t - 2 : In this case, Ri + L

bg b g

di = v t =d t - 2 dt

Taking Laplace transform,

a R + sLfIasf = e e e I asf = asL + Rf = as + 3f -2 s

-2 s

-2 s

Taking inverse Laplace transform,

i t = e -3bt - 2 g u t - 2

bg

b g

Ans.

b g

(c) Unit step voltage u t - 2 In this case,

Ri + L

bg b g

di = v t = u t -2 dt

Taking Laplace transform, -2 s

a R + sLfIasf = e s L1 1 O e e 1 I asf = asL + Rf = sas + 3f = 3 e MN s - as + 3f PQ -2 s

-2 s

-2 s

Taking inverse Laplace transform, 1 1 i t = u t - 2 - e- 3b t - 2 gu t - 2 3 3 (d) Unit doublet voltage d ¢ t - 2 : In this case, di Ri + L = v t = d ¢ t - 2 dt Taking Laplace transform,

bg

b g

b g

Ans.

b g

bg b g

-2 s

a R + sLf Iasf = se L 3 O e se I asf = bsL + Rg = bs + 3g = e MN1 - bs + 3g PQ -2 s

-2 s

-2 s

6.83

Laplace Transform and its Applications

Taking inverse Laplace transform,

i t = d t - 2 u t - 2 - 3e-3bt - 2gu t - 2 Ans. (e) Pulse of width a and magnitude 1V beginning at time t = 2 seconds In this case, di Ri + L = v t = u t + 2 - u t - 2 - a dt Taking Laplace transform, 1 1 R + sL I s = e - 2 s - e - b 2 + a g s s s

bg b g b g

b g

bg b g b

b

g

g bg

bg

I s =

=

e -2 s - e -b2 + a g s 1 1 = e -2 s - e - b 2 + a g s = e -2 s - e - b 2 + a g s 3 s sL + R s s+3

b

g

LM MN

1 e -2 s e - b2 + a g s 3 s s

b g e e b g O + s+3 s + 3 PPQ - 2+a s

-2 s

Taking inverse Laplace transform, 1 i t = u t - 2 - u t - 2 - a - e- 3b t - 2 gu t - 2 + e- 3b t - 2 - a g u t - 2 - a Ans. 3 6.47 Find the response current of a series RC circuit consisting of a resistor R = 2 W and a capacitor C = 1 F when each of the following driving force voltage is applied: 4

bg

b g b

g

b g

b

g

b g (b) impulse voltage 2d bt - 3g , (c) step voltage 2u bt - 3g , and (d) doublet voltage 2d ¢ bt - 3g . (a) ramp voltage 2r t - 3 ,

Solution

b g

(a) ramp voltage 2r t - 3 : Applying KVL to RC series circuit,

Ri +

t

z

bg b g

1 idt = v t = 2r t - 3 C -¥

Taking Laplace transform,

F R + 1 I I b sg = 2 e H Cs K s 2e 2e 1 1 1 O e I b sg = = = = e LM s s + 2 PQ 1I 4 I sb s + 2g 2 N F F s R+ H Cs K s H 2 + s K -3 s

2

-3s

2

-3 s

2

-3s

-3s

6.84

Circuit Theory and Networks

Taking inverse Laplace transform, 1 1 i t = u t - 3 - e -2b t - 3g u t - 3 2 2

bg

b g

b g

Ans.

b g

(b) Impulse voltage 2d t - 3 In this case,

1 C

Ri +

t

zid t = vbt g = 2d bt - 3g

-¥

Taking Laplace transform,

FH R + Cs1 IK Iasf = 2e 2e 2e se L 2O I asf = FH R + Cs1 IK = FH 2 + 4s IK = as + 2f = e MN1 - s + 2 PQ -3s

-3s

-3s

-3s

-3s

Taking inverse Laplace transform,

i t = d t - 3 u t - 3 - 2e- 2bt - 3g u t - 3

bg b g b g

b g

Ans.

b g

(c) Step voltage 2u t - 3 In this case,

Ri +

1 C

t

zidt = vbt g = 2u bt - 3g

-¥

Taking Laplace transform,

FH R + Cs1 IK I asf = 2e 1s 2e 2e e I asf = = = 1 4 s sF R + H Cs IK sFH 2 + s IK b + 2g -3s

-3 s

-3 s

-3s

Taking inverse Laplace transform,

i t = e -2bt - 3gu t - 3

bg

b g

Ans.

b g

(d) Doublet voltage 2d ¢ t - 3 In this case,

Ri +

t

bg

b g

1 idt = v t = 2d ¢ t - 3 C -¥

z

Taking Laplace transform,

FH R + Cs1 IK Iasf = 2se 2 se 2 se s e 4 O L I asf = FH R + Cs1 IK = FH 2 + 4s IK = as + 2f = e NMs - 2 + s + 2 QP -3 s

-3s

-3s

2 -3 s

-3s

6.85

Laplace Transform and its Applications

Taking inverse Laplace transform,

i t = d ¢ t - 3 - 2d t - 3 + 4e -2bt - 3g u t - 3 t ³ 3 Ans. 6.48 Find the response current of a series RLC circuit consisting of a resistor R = 2 W, an inductor L = 1 H 1 and a capacitor C = F when each of the following driving force voltage is applied: 4

bg b g b g

b g

b g (b) step voltage 3u bt - 3g , (c) impulse voltage 3d bt - 1g , and (d) doublet voltage 2d ¢ bt - 3g . (a) ramp voltage 12r t - 2 ,

Solution Applying KVL for the series RLC circuit we get,

af

Þ

af

di t 1 i t dt = v t + dt C

z af af diat f 1 + iat fdt = vat f 5iat f + dt 1/4 z

Ri t + L

b g When v bt g = 12r bt - 2g

(a) Ramp voltage 12r t - 2

b g didtbt g + 11/ 4 z i bt gdt = v bt g

5i t +

Taking Laplace transform,

F 5 + s + 4 I I bsg = 12 e H sK s

-2 s

2

Þ

a f scs 12+e5s + 4h = 12e LM s bs + 11gbs + 4g OP = 12e LMN 1 /s4 - s1+/ 31 + s1/+124 OPQ N Q -2 s

I s =

=

-2 s

-2s

2

e -2 s 3e -2 s 4e -2 s + s s +1 s + 4

Taking inverse Laplace transform, we, get,

i t = 3u t - 2 - 4 e-bt - 2g + e- 4bt - 2g

bg b g

b g When v bt g = 3u bt - 3g

t³2

(b) Step voltage 3u t - 3

b g didtbt g + 11/4 z i bt g dt = v bt g

5i t +

Taking Laplace transform,

F 5 + s + 4 I I b sg = 3 e H sK s

- 3s

Ans.

6.86

Circuit Theory and Networks

Þ

a f cs

I s =

=

2

3e -2 s = 3e -2 s + 5s + 4

h

LM 1 OP = 3e L 1/3 - 1/3 O MN s + 4 s + 1PQ N bs + 1gbs + 4g Q -2 s

e -2 s e -2 s s + 4 s +1

Taking inverse Laplace transform, we, get,

i t = e -bt - 3g + e -4bt - 3g

bg

t ³ 3 Ans.

b g When v bt g = 3d bt - 1g

(c) Impulse voltage 3d t - 1

b g didtbt g + 11/4 z ibt gdt = v bt g

5i t +

Taking Laplace transform,

F 5 + s + 4 I I bsg = 3e H sK

b g ds 3+se5s + 4i = 3e LMM bs + 1gbs s + 4g OPP = 3e LMN sK+ 1 + sK+ 4 OPQ N Q s O -1 K = LM N s + 4 QP = 3 s O 4 K = LM N s + 1PQ = 3 LM - 1 4 OP 4e e I b sg = 3e M 3 + 3 P = s+1 s+ 4 s+ 4 s+1 MN PQ

Þ

I s =

\

1

\

\

-s

-s

-s

-s

2

s =-1

2

s =-4

-s

-s

-s

Taking inverse Laplace transform, we, get,

i t = 4e- 4 bt -1g - e -bt -1g

bg

t ³ 1 Ans.

b g When v bt g = 2d ¢bt - 3g

(d) Doublet voltage 2d ¢ t - 3

b g didtbt g + 11/4 z ibt g dt = vbt g

5i t +

Taking Laplace transform,

F 5 + s + 4 I I bsg = 2se H sK

-3 s

1

2

6.87

Laplace Transform and its Applications

b g ds 2+s 5es + 4i = e LM2 - s 10+s5+s +8 4 OP N Q

Þ

I s =

2 -3 s

-3 s

2

2

10s + 8 K K2 = 1 + s 1 s + +4 s + 5s + 4

Let,

2

LM10s + 8 OP N s+4 Q 10s + 8 O = LM N s + 1 PQ

K1 =

\ \

K2

bg

\

=

-2 3

=

32 3

s =-1

I s = 2e

-3 s

s =-4

-e

-3 s

LM - 2 32 OP MM s +31 + s +3 4 PP = 2e N Q

-3 s

+

2 e -3s 32 e -3s 3 s+1 3 s+ 4

Taking inverse Laplace transform, we, get, 2 32 i t = 2d t - 3 + e - b t - 3g - e -4b t - 3g t ³ 3 Ans. 3 3 6.49 A voltage pulse of magnitude 6 V and duration 3 seconds to 6 seconds is applied to a series RL circuit consisting of R = 6 W and L = 2 H. Obtain the current i(t). Also, calculate the voltage across L and R. Solution Applying KVL for the series RL circuit, di Ri + L = v t = 6 u t - 3 - u t - 6 dt Taking Laplace transform, 6 R + sL I s = e -3s - e -6 s s

bg

b g

bg

b

Þ

b g b g

gbg

b g 6s LMN e R +- sLe OPQ = 6s LMN e 6 +-2es OPQ = 3s LMN e -3s

I s =

-6 s

-3s

-6 s

OP Q

-3s

1 1 - e -6 s = e -3s - e -6 s s+3 s s+3

LM N

Taking inverse Laplace transform, i t = 1 - e -3b t - 3g u t - 3 - 1 - e-3 b t - 6g u t - 6

bg e

jb g e

j b g

Ans.

Voltage across inductor,

LM{ b g N b g

} b gOQP

di = 2 - - 3 e -3b t - 3g u t - 3 - - -3 e -3b t - 6g u t - 6 dt = 6e -3b t - 3g u t - 3 - 6e -3b t - 6g u t - 6 Ans.

vL = L

}b g {b g b g

Voltage across resistor,

e

jb g e

jb g

v R = Ri = 6i = 6 1 - e -3bt - 3g u t - 3 - 1 - e -3bt - 6g u t - 6

Ans.

OP Q

6.88

Circuit Theory and Networks

6.50 Voltage having waveform of truncated ramp as shown in the figure is applied to an RL series circuit consisting of a resistor R = 3 W and inductor L = 1 H. The rise time t0 = 2 ms. Find the current i(t). v(t)

1

t

t0

0

Solution The applied voltage can be synthesised in terms of two ramp functions as,

bg

vt =

1 1 r t - r t - t0 t0 t0

bg

b

g

bg

b

Applying KVL for the series RL circuit,

Ri + L

bg

1 1 di = v t = r t - r t - t0 dt t0 t0

g

Taking Laplace transform,

b R + sLgI bsg = t1 LMN s1 - s1 e 1L1 1 I b sg = M - e t Ns s 0

Þ

0

2

2

2

2

- t0 s

- t0 s

OP Q OP F 1 I = 1 d1 - e i 1 s b s + 3g Q H R + sL K t - t0 s

2

0

K K K 1 = 21 + 2 + 3 s s +3 s s+3 s

Let

2

b g

LM 1 OP = 1 N s + 3Q 3 d 1 O 1 K = LM ds N s + 3PQ = - 9 L1O =1 K =M P Ns Q 9 L 1 1 1 1 1 1 I OP 1 I asf = d1 - e i M F I - F I + F t N 3 H s K 9 H s K 9 H s + 3K Q L 1 1 1 1 1 1 I OP 1 = d1 - e i M- F I + F I + F t N 9 H s K 3 H s K 9 H s + 3K Q K1 =

\ \

s=0

2

\

s=0

3

\

2

s =-3

- t0 s

2

0

- t0 s

2

0

Taking inverse Laplace transform,

LM N

OP b g Q

LM N

OP b Q

1 1 1 1 1 1 1 1 - + r t + e -3t u t - + r t - t 0 + e -3b t - t0 g u t - t 0 9 3 9 9 3 9 t0 t0 where, t0 = 2 ms.

bg

i t =

bg

b

g

g

Ans.

6.89

Laplace Transform and its Applications

6.51 The figure shows a staircase voltage waveform. Assuming that the staircase is not repeated, express its equation in terms of step functions. If this voltage is applied to a series RL circuit with R = 2 ohms and L = 1 H, find an expression for the resulting current i (t); i(0+) = 0.

Voltage in volt

5 4 3 2 1 0

2

4

6 8 Time t in second

10

12

Solution Here, the applied voltage is a combination of several shifted step functions and can be written as,

bg b g b g b g b g b

g b

v t = u t - 2 + u t - 4 + u t - 6 + u t - 8 + u t - 10 - 5u t - 12

g

Taking Laplace transform,

bg

V s =

1 -2 s e + e -4 s + e -6 s + e -8 s + e -10 s - 5e -12 s   s

If this voltage is applied to RL series circuit, applying KVL we get,

bg

Ri t + L

bg bg

di t =v t dt

Taking Laplace transform,

b R + sLgI bsg = V bsg = 1s e + e + e + e + e - 5e 1 I b sg = e +e +e +e +e - 5e sb s + 2g 1 1 1 O = LM - 5e e +e +e +e +e 2 N s s + 2 PQ -2 s

Þ

-2 s

-4 s

-6 s

-4 s

-2 s

-8 s

-6 s

-8 s

-4 s

-6 s

-10 s

-12 s

-10 s

-8 s

-12 s

-10 s

-12 s

Taking inverse Laplace transform,

af

it =

1 1 1 1 1 - e -2b t - 2 g u t - 2 + 1 - e -2b t - 4 g u t - 4 + 1 - e -2b t - 6 g u t - 6 + 2 2 2 2

1 - e -2bt - 8g

a f 1 uat - 8f + 1 - e 2

a

-2 t -10

f

a f 5 u at - 10 f - 1 - e 2

b

-2 t - 12

g

b g u bt - 12 g

Ans.

6.90

Circuit Theory and Networks

MULTIPLE-CHOICE QUESTIONS 6.1 The condition for over-damped response of an RLC series circuit is R2 R2 R2 1 1 1 (b) (c) = > < 2 2 LC LC 4L 4L 4 L2 LC 6.2 Transient current in an RLC circuit is oscillatory when

(d)

(a)

L L L (b) R > 2 (c) R < 2 C C C Laplace transform analysis gives (a) time domain response only (b) frequency domain (c) both (a) and (b) (d) None of these. A function f(t) is shifted by a then it is correctly represented as (a) f(t – a)u(t) (b) f(t)u(t – a) (c) f(t – a)u(t – a) Laplace transform of a delayed unit impulse function ds(t) = d (t – 1) is (a) unity. (b) zero. (c) e–s. The condition for under damped response of an RLC series circuit is

(a) R = 2

6.3

6.4 6.5 6.6

6.7 6.8 6.9 6.10

6.11

-

6.12 6.13 6.14 6.15

6.16

(d) R = 0. response only

(d) f(t – a)(t – a) (d) s.

R2 R2 R2 1 1 1 (b) (c) (d) = > < 2 2 2 LC LC LC 4L 4L 4L The value of the impulse function d (t) at t = 0 is (a) 0 (b) µ (c) 1 (d) The value of the ramp function at t = + ¥ is (a) infinity (b) unity (c) zero (d) The value of the ramp function at t = –¥ is (a) 0 (b) ¥ (c) –¥ (d) The value of the impulse function d (t) for t > 0 is (a) zero (b) unity (c) k, where k is a constant (d) infinity. The free response of RL and RC series networks having a time constant t is

(a)

t

-

t

-

t

-

t

R2 1 £ 4 L2 LC

R2 1 £ 2 LC 4L

indeterminate indeterminate 1

of the form -

t

(a) A + Be t (b) Ae t (c) Ae t + Be t (d) ( A + Bt )e t In the complex frequency s = s + jw, w has the units of rad/s and s has the units of (a) Hz (b) neper/s (c) rad/s (d) rad Time constant of a series RC circuit is (a) C/R (b) R/C (c) RC (d) 1/RC Time constant of a series RL circuit is (a) L/R (b) R/L (c) LR (d) 1/LR A coil with a certain number of turns has a specified time constant. If the number of turns is doubled, its time constant would (a) remain unaffected (b) become doubled (c) become four-fold (d) get halved. An RLC series circuit has R = 1W, L = 1 H and C = 1F. Damping ratio of the circuit will be (a) more than unity (b) unity (c) 0.5 (d) zero

6.91

Laplace Transform and its Applications

6.17 A step function voltage is applied to an RLC series circuit having R = 2W, L = 1H and C = 1F. The transient current response of the circuit would be (a) over-damped (b) critically damped (c) under damped (d) over, under or critically damped depending upon magnitude of the step voltage. 6.18 For an RC circuit comprising a capacitor C = 2 mF in series with a resistance R = 1 MW period 6 seconds will be equal to (a) one time constant (b) two time constants (c) three time constants (d) four time constants 6.19 A series RL circuit with R = 100 ohm; L = 50H, is supplied to a d.c. source of 100V. The time taken for the current to rise 70% of its steady state value is (a) 0.3s (b) 0.6s (c) 2.4s (d) 70% of time required to reach steady state. 6.20 If f(t) and its first derivative are Laplace transformable then the initial value of f(t) is given by (a) (c)

Lt f (t ) = Lt sF ( s )

(b)

F (s) s

(d)

t ®0

s ®0

Lt f (t ) = Lt

t ®0

s ®0

Lt f (t ) = Lt

t ®0

s ®¥

F (s) s

Lt f (t ) = Lt sF ( s )

t ®0

s ®¥

6.21 If f(t) and its first derivative are Laplace transformable then the final value of f(t) is given by (a) (c)

Lt f (t ) = Lt sF ( s )

(b)

F (s) s ®0 s

(d)

t ®¥

s ®0

Lt f (t ) = Lt

t ®¥ +

Lt f (t ) = Lt

t ®¥

s ®¥

F (s) s

Lt f (t ) = Lim sF ( s )

t ®¥

s ®¥

6.22 At t = 0 with zero initial condition which of the following will act as short circuit? (a) Inductor (b) Capacitor (c) Resistor (d) None of these 6.23 At t = 0+ with zero initial condition which of the following will act as open circuit? (a) Inductor (b) Capacitor (c) Resistor (d) None of these 6.24 A capacitor at time t = 0+ with zero initial charge acts as a (a) short circuit (b) open circuit (c) current source (d) voltage source. 6.25 A series RC circuit is suddenly connected to a dc voltage of V volt. The current in the series circuit just after the switch is closed is equal to V VC V (c) (d) RC R R 6.26 A series LC circuit is suddenly connected to a dc voltage of V volt. The current in the series circuit just after the switch is closed is equal to

(a) zero

(b)

V V V (b) (c) zero (d) L C LC 6.27 The steady state current in the RC series circuit, on the application of step voltage of magnitude E will be

(a)

(a) zero

(b)

E R

(c)

E - t / CR e R

(d)

E -t e RC

6.92

Circuit Theory and Networks

6.28 A 10 W resistor, a 1H inductor and 1F capacitor are connected in parallel. The combination is driven by a unit step current. Under steady state conditions, the source current flows through (a) resistor (b) inductor (c) capacitor only (d) All of three elements. 6.29 When a unit impulse voltage is applied to an inductor of 1H, the energy supplied by the source is 1 Joule (d) 0. 2 6.30 Which of the following conditions are necessary for validity of Initial Value Theorem:

(a) ¥

(b) 1 Joule

(c)

Lim sF ( s ) = Lim f (t ) ? s ®¥

(a) (b) (c) (d)

t ®0

f (t) and its derivative f ¢(t)must have Laplace transform. If the Laplace transform of f (t) is F(s), then Lim sF(s) must exist. Only f (t) must have Laplace transform. (a) and (b) both.

6.31 Inverse Laplace transform of

1 is s-a

(d) e – at (a) sin at (b) cos at (c) e at 6.32 The impulse response of an RL circuit is a (a) rising exponential function. (b) decaying exponential function. (c) step function. (d) parabolic function. 6.33. Laplace transform of the output response of a linear system is the system transfer function when the input is (a) a step signal. (b) a ramp signal. (c) an impulse signal. (d) a sinusoidal signal. 6.34 An initially relaxed RC series network with R = 2MW and C = 1mF is switched on to a 10V step input. The voltage across the capacitor after 2 seconds will be (a) zero (b) 3.68 V (c) 6.32 V (d) 10 V 6.35 For V(s) =

( s + 2) , the initial and final values of v(t) will be respectively s ( s + 1)

(a) 1 and 1

(b) 2 and 2

6.36 The Laplace transform of the function i(t) is: I(s)

(c) 2 and 1

(d) 1 and 2.

10 s + 4 . Its final value will be s ( s + 1) ( s 2 + 4 s + 5)

(a) 4/5 (b) 5/4 (c) 4 (d) 5 6.37 An initially relaxed 100 mH inductor is switched ‘ON’ at t = 1 second to an ideal 2A dc current source. The voltage across the inductor would be (a) zero (b) 0.2d (t) V (c) 0.2d (t – 1) V (d) 0.2tu (t – 1) V 6.38 If the unit step response of a network is (1 – e–a t ), then its unit impulse response will be 1 -t / a 1 -t / a (c) (d) (1 – a) e–at e e a a 6.39 The response of an initially relaxed system to a unit ramp excitation is (1 + e–t). Its step response will be -a t (a) a e

(a)

1 2 -t t -e 2

(b)

(b) 1 – e–t

(c) –e –t

(d) t.

6.93

Laplace Transform and its Applications

6.40 A series circuit containing R, L and C is excited by a step voltage input. The voltage across the capacitance exhibits oscillations. Damping coefficient (ratio) of this circuit is given by (a) x =

R 2 LC

(b) x =

R LC

(c) x =

R 2 C/L

(d) x =

R 2 L /C

6.41 Consider the following statements: A unit impulse d (t) is mathematically defined as 1. d (t) = 0, t ¹ 0

¥

2. ò d (t ) dt = 1 0+

¥

3.

ò d (t ) dt = 1

-¥

Of these statements (a) 1, 2 and 3 are correct. (b) 1 and 2 are correct. (c) 2 and 3 are correct. (d) 1 and 3 are correct. 6.42. With symbols having their usual meanings, the Laplace transform of u(t – a) is

e- as eas 1 1 (b) (c) (d) s s s s-a Two coils having equal resistances but different inductances are connected in series. The time constant of the series combination is the (a) sum of the time constants of the individual coils. (b) average of the time constants of the individual coils. (c) geometric mean of the time constants of the individual coils. (d) product of the time constants of the individual coils. If the step response of an initially relaxed circuit is known then the ramp response can be obtained by (a) integrating the step response. (b) differentiating the step response. (c) integrating the step response twice. (d) differentiating the step response twice. If a capacitor is energized by a symmetrical square wave current source, then the steady state voltage across the capacitor will be a (a) square wave (b) triangular wave (c) step function (d) impulse function. A square wave is fed to an RC circuit, then (a) voltage across R is square and across C is not square. (b) voltage across C is not square and across R is not square. (c) voltage across both R and C is square. (d) voltage across both R and C is not square. A step voltage is applied to an under-damped series RLC circuit with variable R. Which of the following statements correctly describe the behaviour of the circuit? If R is increased, the steady state voltage across C will be reduced If R is increased, the frequency of transient oscillation across C will be reduced. If R is reduced, the transient oscillation will die down faster. If R is reduced to zero, the peak amplitude of the voltage across C will be double the input step voltage. Select the correct answer using the codes given below. Codes: (a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 1, 3 and 4. The number of turns of a coil having a time constant T is doubled. Then the new time constant will be (i ) T (b) 2T (c) 4T (d) T/2 (a)

6.43

6.44

6.45

6.46

6.47 1. 2. 3. 4.

6.48

6.94

Circuit Theory and Networks

6.49 The response of a network is of the form kest,where s = s + jw, then s is known as (a) radian frequency (b) neper frequency (c) complex frequency (d) None of these. 6.50 In Laplace transform the variable ‘s’ equals (s + jw). Which of the following represent the true nature of s ? 1. s has a damping effect. ¥

2. s is responsible for convergence of integral ò f (t )e - st dt . 0

3. s has a value less than zero. Select the correct answer using the coeds given below. Codes: (a) 1, 2 and 3 (b) 1 and 2 (c) 2 and 3 6.51 Laplace transform of tne–at is (a)

6.52

n ( s - a ) n +1

(b)

n! ( s + a ) n +1

(c)

n! ( s - a )n

(d) 1 and 3.

(d)

n! ( s - a ) n +1

s is the Laplace transform of (s + w 2 ) 2

(a) sin w t (b) cos w t (c) cosh w t (d) sinh w t 6.53 Consider the following statements regarding an RC differentiating network. 1. For an applied rectangular pulse, the output is spiky in nature for RC 0. If a signal 3u(t) + d (t) is applied to the same initially relaxed system, the response will be (a) (3 - 6e-3t )u (t ) (b) (3 - 3e-3t )u (t ) (c) 3u (t ) (d) (3 + 3e-3t ) u (t ) 6.63 A unit impulse input to a linear network has a response R(t) and a unit step input to the same network has response S(t). The response R(t)

dS (t ) dt (c) is the reciprocal of S(t)

(a) equals

(b) equals the integral of S(t) (d) has no relation with S(t)

- 2t 6.64 The response of an initially relaxed linear circuit to a signal VS is e u (t ) . If the signal is changed

dVS ö æ , the response would be to ç VS + 2 dt ÷ø è

(a) –4e–2t u(t)

(b) – 3e–2t u(t)

(c) 4e–2t u(t)

(d) 5e–2t u(t)

6.96

Circuit Theory and Networks

6.65 The impulse response of a circuit is given by h(t ) = R æ - tö (a) ç1 - e L ÷ u (t ) è ø

(b)

R

1 - Lt e u (t ) . Its step response is given as L

R R - tö - tö 1æ Læ L L 1 1 ( ) (c) e u t e ÷ ÷ u (t ) (d) None of these. R çè R çè ø ø

6.66 The time constant of the network shown in the figure is (a) CR (b) 2CR CR CR (d) (c) 4 2 6.67 Non-linear system cannot be analyzed by Laplace transform because (a) it has no zero initial conditions. (b) superposition law cannot be applied. (c) non-linearity is generally not well defined. (d) All of the above. 6.68 In the circuit shown in figure, the response i(t) is

(a)

V t ö exp æç è RC ÷ø R

(c)

V R

é 1 t öù æ êd (t ) - RC exp çè - RC ÷ø ú ë û

(b)

V d (t ) R

(d)

V R

é t öù æ êd (t ) - exp çè - RC ÷ø ú ë û

6.69 A voltage v(t ) = 6e-2t is applied at t = 0 to a series RL circuit with L = 1H. If i(t ) = 6[e-2t - e-3t ] , then R will have a value of 2 W (b) 1 W (c) 3 W 3 6.70 The Laplace transform of the signal described in figure is

(a)

(a) e –as/s

(b) e–bs/s2

(c) (e–as + e–bs)/s

(d)

1 W 3

(d) (e–as – e–bs)/s

6.97

Laplace Transform and its Applications

6.71 If a pulse voltage v(t) of 4V magnitude and 2 second duration is applied to a pure inductor of 1H, with zero initial current, the current (in A) drawn at t = 3 second, will be (a) zero (b) 2 (c) 4 (d) 8. 6.72 At certain current, the energy stored in an iron-cored coil is 1000J and its copper loss is 2000W. The time constant (in second) of the coil is (a) 0.25 (b) 0.5 (c) 1.0 (d) 2.0. 6.73 Consider the voltage waveform shown in the given figure.

The equation for v(t) is (a) u (t - 1) + u (t - 2) + u (t - 3)

(b) u (t - 1) + 2u (t - 2) + 3u (t - 3)

(c) u (t ) + u (t - 1) + u (t - 2) + u (t - 4) (d) u (t - 1) + u (t - 2) + u (t - 3) - 3u (t - 4) 6.74 For the circuit given in the figure V0 = 2 V and inductor is initially relaxed. The switch S is closed at t = 0. The value of v at t = 0+ is

(a) 3 V (b) 2 V (c) 0.5 V (d) 0.25 V 6.75 In the circuit shown in the given figure, S is open for a long time and steady state is reached. S is closed at t = 0. The current I at t = 0+ is

(a) 4 A

(b) 3 A

(c) 2 A

(d) 2 A

6.98

Circuit Theory and Networks

6.76 The circuit shown in the given figure is in steady state with switch S open. The switch is closed at t = 0. The values of VC (0+) and VC (¥) will be respectively

(a) 2 V, 0 V (b) 0 V, 2 V (c) 2 V, 2 V (d) 0 V, 0 V 6.77 In the circuit shown, the switch is opened at t = 0. Prior to that switch was closed, i(t) at t = 0+ is

(a)

2 A 3

(b)

3 A 2

(c)

1 A 3

(d) 1 A.

¥

6.78 Given the Laplace transform L[ v(t )] = ò e - st v(t )dt , the inverse transform v(t) is 0

s + j¥

(a)

s + j¥

st ò e V ( s )ds

(b)

1 e stV ( s )ds 2p j s -òj¥

(d)

1 e - stV ( s )ds 2p j s -òj¥

s - j¥

s + j¥

¥

(c)

1 e stV ( s )ds 2p j ò0

6.79 In the circuit shown in the given figure, switch S is closed at t = 0. After some time when the current in the inductor was 6A, the rate of change of current through it was 4 A/s. The value of the inductor is

(a) indeterminate

(b) 1.5 H

(c) 1.0 H

(d) 0.5 H

6.99

Laplace Transform and its Applications

6.80 A circuit consisting of a 1W resistor and a 2F capacitor in series is excited from a voltage source with the voltage expressed as 3e–t, as shown in the given figure. If the i(0–) and vc(0–) are both zero, then the values of i(0+) and i(¥) will be respectively

(a) 3 A and 1.5 A (b) 1.5 A and zero (c) 3 A and zero (d) 1.5 A and 3 A 6.81 The time constant associated with the capacitor charging in the circuit shown in the given figure is

(a) 6 ms (b) 10 ms (c) 15 ms (d) 25 ms 6.82 In the network shown in the figure, the switch S is closed and a steady state is attained. If the switch is opened at t = 0, then the current i(t) through the inductor will be

(a) cos 50t A (b) 2 A (c) 2 cos100t A (d) 2 sin 50t A 6.83 In the network shown, the switch is opened at t = 0. Prior to that, the network was in the steady state. Vs(t) at t = 0+ is

(a) 0

(b) 5 V

(c) 10 V

(d) 15 V

6.100

Circuit Theory and Networks

6.84 The steady state in the circuit, shown in the given figure is reached with S open. S is closed at t = 0. The current I at t = 0+ is

(a) 1 A (b) 2 A (c) 3 A (d) 4 A 6.85 For the circuit shown in the given figure, the current through L and the voltage across C2 are respectively

(a) zero and RI (b) I and zero (c) Zero and zero (d) I and RI 6.86 In the circuit shown in the given figure, the switch is closed at t = 0. The current through the capacitor will decrease exponentially with a time constant

(a) 0.5 s

(b) 1 s

(c) 2 s

(d) 10 s

w , the final value of f (t) is s2 + w 2 (a) infinity (b) zero (c) one (d) None of the above 6.88 The v–i characteristics as seen from the terminal-pair (A, B) of the network of Figure (a) is shown in Figure (b). If an inductance of value 6 mH is connected across the terminal-pair (A, B), the time constant of the system will be

6.87 The Laplace transformation of f (t) is F(s). Given F ( s ) =

(a) 3 ms (c) 32 s

(b) 12 s (d) unknown, unless the actual network is specified.

6.101

Laplace Transform and its Applications

6.89 In the circuit shown in figure, it is desired to have a constant direct current i(t) through the ideal inductor L. The nature of the voltage source v(t) must be (a) constant voltage (b) linearly increasing voltage (c) an ideal impulse (d) exponentially increasing voltage. ¥

6.90 The value of the integral ò e5t d (t - 5)dt is -¥

(a) 1 (b) (e5 – 1) (c) e 25 6.91 An inductor at t = 0+ with initial current I0 acts as (a) voltage source (b) current source (c) open-circuit 6.92 A capacitor at t = 0+ with initial charge Q0 acts as (a) voltage source (b) current source (c) open-circuit 6.93 Consider the following statements 1. Current through an inductor cannot change abruptly. 2. Voltage across the capacitor cannot change abruptly. 3. Initial value of a function f(t) is Lim sF ( s ) s ®0

4. Final value of a function f(t) is Lim sF ( s ) s ®¥

Of these statements (a) 3 and 4 are correct (b) 1 and 4 are correct (c) 1 and 2 are correct (d) 2 and 3 are correct. 6.94 An inductor with inductance L and initial current I0 is shown as

The correct admittance diagram for it is (a)

(b)

(c)

(d)

(d) zero. (d) short-circuit (d) short-circuit

6.102

Circuit Theory and Networks

6.95 An inductor with inductance L and initial current I0 is shown as

The correct impedance diagram for it is (a)

(b)

(c)

(d)

6.96 A capacitor with capacitance C and initial voltage vc(t) is shown here

The correct admittance diagram for this circuit is

6.97

(a)

(b)

(c)

(d)

6.103

Laplace Transform and its Applications

Laplace transform of f (t) shown in the given figure is (a) F ( s ) =

1 2 -s 3 -s - e + e s s s

(b) F ( s ) =

1 2 - s 3 -2 s 2 -3s - e + e - e s s s s

1 e- s 2 -2 s 2 -3s 1 2 3 + e - e (d) F ( s ) = + e - s - e - s s s s s s s s 6.98 The time constant of the circuit shown in the given figure is (c) F ( s ) =

(a) R C (b) R C (c) R C (d) R C 6.99 Consider the following functions for the rectangular voltage pulse shown in the given figure

(a) v(t ) = u (t - a ) - u (t - b )

(b) v(t ) = u (b - t ) - u ( a - t )

(c) v(t ) = u (b - t ).u (t - a )

(d) v(t ) = u ( a - t ).u (t - b )

bg

6.100 If F1 s =

1 2 ,F s = 2 ; what is the Laplace transform of the product F1(s) F2(s)? s+3 2 s +4

b g 15 e 1 f bt g = e 7

bg

(a) f t =

-t

(c)

-2 t

b g 131 2e 1 f bt g = e 11

+ 3 cos 2t - 2 sin t

(b) f t =

+ 2 sin 2t - cos 2t

(d)

-3t

-2 t

+ 3 sin 2t - 2 cos 2t

+ sin t - 2 sin 2t

6.101 The impulse response of a linear network is given by e -2 t . Which one of the following gives its unit step response? 1 1 -t (a) 1 - e -2 t (b) e - t - e -2 t (c) (d) 1 - e -2 t e - e -2 t 2 2

d

i

d

i

6.104

Circuit Theory and Networks

6.102 The network shown in the figure given above reaches a steady state with the switch K in position a. At t = 0, the switch is moved from a to b by a make-before-break mechanism. Assume the initial current in 2 H inductor as zero. What is the current in 1H inductor at t = 0 + and t = ¥, respectively? 2W

a

K

5V

b

(a) 1 A and 0 A (c) 1 A and 2.5 A

bg

6.103 If F s =

b g

2 s+1

s + 2s + 5 2

1 W 2

1H

2H

(b) 2.5 A and 0 A (d) 2.5 A and 2.5 A , then what are the values of f (0+) and f (¥) respectively?

(a) 0, 2 (b) 2, 0 (c) 0, 1 (d) 2/5, 0 6.104 In the circuit shown in the figure, the switch S is closed at t = 0. Which one of the following gives the expression for the voltage across the inductance as a function of time? 1W

S

1V

1H

d1 - e i -t

(a) e - t /2

d

i

(c) 1- e - t (d) e - t 2 6.105 For the circuit shown in figure, the initial capacitor voltage is 2 V and I is a unit step function. Then, what is the expression for v (t) for t > 0? (b)

1W I

1W 0.25 F

(a) 2 + e - t

(b) 2 - e -2t

v(t)

(c) 1 + e -2 t

(d) 1 - e -2 t

6.106 In the circuit shown in the figure, steady state is reached with switch S open. Switch S is then closed at t = 0. What is the value of voltage V under steady state (when t = ¥)?

6.105

Laplace Transform and its Applications 5W +

5A

10 W

S 5W

V –

(a) 50 V

(b) 12.5 V

(c) 25 V

(d) 0V

6.107 If f1 (t) and f2(t) have the widths (duration) T1 and T2 respectively, then what is the width (duration)

bg bg

of f 1 t * f 2 t where * denotes convolution)? (b) The smaller of T1 and T2 (a) The larger of T1 and T2 (c) T1 - T2 (d) T1 + T2 v(t) 6.108 The Laplace transform of v(t) shown in the figure is (a) (c)

1 1 1 - e- s - e - 2 s s s2

d i 1 1 1+ e i + e d s s -s

(b)

-2 s

(d)

2

1 1 1- es - e2s s s2

d i 1 1 1+ e i + e d s s s

1

2s

0

2

1

2

t

6.109 If f (t) and F (s) form the Laplace transform pair, then what is the Laplace transform of f ( t /t 0 ) ?

b g

(a) t 0 F t 0 s

(b)

1 F t0s t0

b g

(c) t 0 F

FG 1 sIJ Ht K

(d)

0

FG IJ H K

1 1 F s t0 t0

6.110 The switch in the circuit is closed at t = 0. The current through the battery at t = 0 + and t = ¥ is, respectively 1W

10 V

1H

1F

(a) 10 A and 10 A (b) 0 A and 10 A (c) 10 A and 0 A (d) 0 A and 0 A 6.111 The Laplace transform of the voltage across the capacitor of 0.5 F is

bg

V s =

s+1 s3 + s2 + s + 1

Then the value of the current through the capacitor at t = 0 + is given by (a) 0 A (b) 0.5 A (c) 1.0 A (d) 1 .5 A 6.112 If u (t) and d (t) are the step function and the impulse function respectively at t = 0, then the Laplace

bg b g bg

transform of the function f t = u t - 1 d t is equal to (a) 1

(b)

1 s

(c) 0

(d)

1 s+1

6.106

Circuit Theory and Networks

6.113 The step response of a system is C t = 1 - 5e- t + 10e -2 t - 6e - 3 t . The impulse response of the system is

bg

(a) 5e - t - 20e - 2 t + 18e - 3 t

(b) 5e t + 20e 2 t + 18e - 3t

(b) 5e - t + 20e - 2 t + 18e - 3t

(d) 5e - t + 20e - 2 t + 18e 3 t

6.114 The Laplace transform of ea t cos a t is (a)

(c)

bs + a g bs + a g + a bs + a g bs - a g + a 2

2

(b)

2

(d)

2

1

bs - a g bs - a g + a bs - a g bs + a g + a 2

2

2

+ 2

u(t)

2

–

2

6.115 For the circuit shown in the figure, the initial inductor current is 2A. The value of i(t) for t > 0 is (a) 0.5 - 0.75e - t

bg

6.116 Consider a system described by the transfer function G s =

bg

(d) 0.5 + 0.75e - t

(c) 0.5 - 0.25e - t

(b) 1- e - t

2s + 3 . It is subjected to an input s + 2s + 5 2

bg

f t = 10u t . The initial and final values of the response are given by (a) 0, 2 / 3 (b) 1, 4 (c) 0, 6 6.117 The impulse response of a linear time invariant system is given by

(d) 0, 4

h t = 2 e- t u t

bg

bg

The unit step response is given by

bg d i bg y bt g = 2 d1 - e i u bt g

(a) y t = 2 1 - e - t u t (c)

bg d i bg y bt g = 2 d2 - e i u bt g

(b) y t = 2 e - t - 1 u t

-2 t

(d)

-2 t

6.118 In the given circuit, if the inductor is initially relaxed, then the current in the circuit will be L (a) zero (b) d t R

bg

(c)

1 L

Rt e L

(d)

1 L

F GH

Rt 1- e L

R + d( t )

L

I JK

6.119 For the circuit shown in figure, the switch ‘K’ was closed for a long time till steady state conditions reached. At time t = 0, the switch ‘K’ is opened, then the current through inductor will be K

10 V

(a) 5 cos 10 t

(b) 5 cos 100 t

2W

1H

1 mF

(c) 5 cos 1000 t

(d) 5 cos 10000 t

6.107

Laplace Transform and its Applications

6.120 The response of a system to a unit ramp input is

1 1 1 t - u t + e - 4 t . Which one of the following 2 8 8

bg

is the unit impulse response of the system?

d

(b) 2 1 - e- 4 t

(a) 1 - e - 4 t

i

(c) e - 4 t

(d) 2e- 4 t

6.121 The Laplace transform of current in an RLC series circuit with R = 2 W, L = 1 H and C =

bg

I s =

1 F is 2

1 . The voltage across the inductor ‘L’ will be s + 2s + 2 2

(a) e - t sin tu t

(b) e- t cos tu t

bg

-t

bg

-t

(c) e sin t + cos t u t (d) e cos t - sin t u t 6.122 For the network shown in the figure, the initial position of switch ‘S’ is ‘1’. After reaching steadystate, if the position of the switch is changed over to ‘2’, the current ‘i’ for t ³ 0 will be equal to

b

g bg

1

b

g bg

L

2R

2

V

i R

(a) (c)

F H

I K V 3Rt I exp F H LK R V Rt exp 2R L

(b) (d)

F H

I K V 3Rt I exp F H LK 2R

V 2 Rt exp R L

6.123 The correct value of the current i(t) at any instant when K is switched R K on at t = 0 in the network shown in the given figure is E E b R /L gt E E b R /L g t (a) (b) + e - e L E R R R R i(t) E E - b R /L gt E E - b R /L gt (c) (d) + e - e R R R R 6.124 In the circuit shown in figure, the switch S is closed at t = 0. The voltage across the inductance at t = 0 +, is 3W S 10 V

(a) 2V

+ –

(b) 4V

4F

4W

4W

(c) –6 V

4V

(d) 8V

6.108

Circuit Theory and Networks

b g s ds

6.125 Consider the function, F s =

2

5 where F (s) is the Laplace transform of the function f (t). + 3s + 2

i

The initial value of f (t) is equal to 5 5 (c) (d) 0 (a) 5 (b) 2 3 6.126 In the figure, the capacitor initially has a charge of 10 coulombs. The current in the circuit one second after the switch S is closed will be

2W

S 100 V

+

+ –

0.5 F –

(a) 14.7 A (b) 18.5 A (c) 40.0 A (d) 50.0 A 6.127 In the figure given, the initial capacitor voltage is zero. The switch is closed at t = 0. The final steadystate voltage across the capacitor is T=0

10 W

10 m F

20 V

10 W

(a) 20 V (b) 10 V (c) 5V (d) 0V 6.128 The circuit shown in the figure is in steady state, when the switch is closed at t = 0. Assuming that the inductance is ideal, the current through the inductor at t = 0 + equals 10 W

10 V

10 mH

t=0

(a) 0 A (b) 0.5 A (c) 1 A (d) 2 A 6.129 If, at t = 0 +, the voltage across the coil is 120 V, the value of resistance R is 1 S

20 W 2

10 H

120 V R

(a) 0 W

(b) 20 W

40 W

(c) 40 W

(d) 60 W

6.109

Laplace Transform and its Applications

6.130 For the value obtained in Q. 129, the time taken for 95% of the stored energy to be dissipated is close to (a) 0.10 second (b) 0.15 second (c) 0.50 second (d) 1.0 second 6.131 An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let w 0 =

1 , the voltage across the LC

capacitor at time t > 0 is given by

b g

b g

(a) V0 (b) V0 cos w 0 t (c) V0 sin w 0 t (d) V0 e -w 0t cos w 0t 6.132 In the circuit shown in the figure, switch SW1 is initially CLOSED and SW2 is OPEN. The inductor L carries a current of 10 A and the capacitor is charged to 10V with polarities as indicated. S W2 is initially CLOSED at t = 0– and SW1 is OPENED at t = 0. The current through C and the voltage across L at t = 0+ is R 210 W

SW 1

SW 2

R 110 W

L

1F

1F

+ 10 V

10 A

(a) 55 A, 4.5 V (b) 5.5 A, 45 V 6.133 The time constant for the given circuit will be

1F

1 1 s (b) s 9 4 6.134 The Laplace transform of i(t) is given by

(a)

b g

C

(c) 45 A, 5.5 V

(d) 5.5 A, 5.5 V

3W

3W

(c) 4 s

3A

(d) 9 s

b g sb12+ sg

I s = As t ® ¥, the value of i (t) tends to (a) 0 (b) 1

(c) 2

(d) ¥

6.135 In what range should Re (s) remain so that the Laplace transform of the function e b a + 2 g t + 5 exits? (a) Re (s) > a + 2 (b) Re (s) > a + 7 (c) Re (s) < 2 (d) Re (s) > a + 5 6.136 A square pulse of 3 V amplitude is applied to C-R circuit shown in the figure. The capacitor is initially uncharged. The output voltage V0 at time t = 2 seconds is

6.110

Circuit Theory and Networks –j2 W

0.1 m F

Vi 3V

+

+

Vi 2 second

–

t

–

(a)

(a) 3V

V0

1 KW

(b)

(b) –3 V

(c) 4V

(d) –4 V

EXERCISES 6.1 (a) Find the initial values of the functions:

bg

(i) f t = e - at cosw tu t

bg

(ii) F s =

bg

b g

2 s +1

s + 2s + 5 2

[(i) 1, (ii) 2] (b) Find the final value of the functions:

b g sbs +7 3g

(ii) F s =

b g bs + 1sgb- s1+ 2g

(ii) F s =

2

LM(i) 7 , (ii) 0OP N 9 Q

6.2 Obtain the Laplace transform of the following functions: f(t) V

f(t) A

0

1

2

t 0

–V (i)

LM(i) F bsg = A (1 - e Ts N

t

T (ii)

OP Q

A (1 - e - Ts - Te - Ts ) Ts 2 6.3 In the network shown, the switch is closed and a steady state is reached in the network. At time t = 0, the switch is opened. Find an expression for the current through the inductor i2(t). - Ts

2

bg

- se - Ts ) (ii) F s =

10 W i 2(t) 100 V

1H

100 m F

[10 cos 100t (A)]

6.111

Laplace Transform and its Applications

6.4 Find for the circuit shown, the current through C using Laplace transform. The switch is closed at t = 0 and the initial charge in the capacitor, i.e., at t = 0 is zero. 2W

100 mF

10 V

bg

5e -5000t A

6.5 The circuit of a figure was initially in the steady state with the switch S in position a. At t = 0, the switch goes from a to b. Find an expression for the voltage v0(t) for t > 0. Take the initial current in

LMv (t ) = 1 e 2 N

the inductor L2 to be zero.

0

a

( V)

OP Q

b +

R1 = 2 W 2V

3 - t 2

R2 = 1 W

L1 = 2 H

L2 = 1 H VO –

6.6 In the circuit of the figure, the applied voltage is v(t) = 10sin(10t + p/6), R = 1 W, C = 1 F. Using Laplace Transformation, find complete solution for current i(t). Switch K is closed at time t = 0. Assume zero charge across the capacitor before switching. 5 100 é ù -t o êi (t ) = 101 (1 - 10 3)e + 101 cos (10t - 54 8¢ )(A)ú ë û 1 + –

V

2 1W

K

i(t)

1F

6.7 A series RLC circuit, with R = 5 W, L = 0.1 H and C = 500 µF, has a sinusoidal voltage source, v = 1000 sin 250t. Find the resulting current if the switch is closed at t = 0. i ( t ) = e -25t (5.42 cos139t + 189 . sin 139t ) + 5.65 sin( 250t - 73.6 o )( A)

6.8 The two-mesh network shown in the figure contains a sinusoidal voltage source, v = 100 sin(200t + f )( V) . The switch is closed at an instant when the voltage is increasing at its maximum rate. Find the resulting mesh currents, with directions as shown in the figure.

50 mH + V –

2

1

10 W i1

10 W i2

[i1 (t ) = 3.01e -100t + 8.96 sin(200t - 63.4 o )i2 (t ) = 1505 . e -100t + 4.48 sin(200t - 63.4 o )]

6.112

Circuit Theory and Networks

6.9 Find i2(t) for t > 0; assume the all initial conditions to be zero. 2

1H

10 W

1H 10 W

10 W 1

i 2( t )

100 V

LMi (t ) = 10 + 5 e 3 3 N 2

-30t

- 5e -10t for t > 0

OP Q

6.10 In the network shown, (a) determine Va(t), using Laplace transform method if k1 = -3. (b) determine i2(t), using Laplace transform method if k1 = 3. i1

Va

1H

k1i1 +–

1W + –

1W

1W

v 1(t ) = 5 u (t )

1W 1F

2

i (t )

[ (a ) v a (t ) = 4 - e -0.75t (15 . cos 0.25t - 0.5 sin 0.25t ) (b) i 2 (t ) = -5 + 16.3375e -0.707t - 13375 . e -0.707t (A )] 6.11 The network shown in the figure, has reached steady state when the switch S moves from a to b. (i) Determine initial values for iL(t) and Vc(t) with switch in position b. (ii) Determine Vc(t) for t > 0. Sketch Vc(t) as a function of time. (iii) Determine damping ratio, undamped and damped natural frequencies. b 10 V

S

a

1H iL(t)

5V

+ 1F

1W

Vc(t) –

LM(i) 5 A, 5 V; (ii) 15 - 10e MN

- t /2

cos

FG 3 tIJ - 20 e H2 K 3

- t /2

sin

FG 3 tIJ bV gOP H 2 K PQ

6.12 Find the source current after the switch is closed at t = 0. Take initial current to be zero.

d3 - e i bA g -25t

6.113

Laplace Transform and its Applications 100 W t=0

50 W

100 V

4H

6.13 Find an expression for the current in the inductor at time t after the switch is closed. What is the final value of the current and how long will it take for the inductor current to reach 95% of its final value? iL

t=0

50 W

100 W 100 V

4H

+ –

d

2 1 - e -50 t /6

i b Ag; 2A, 0.36 second

6.14 In the circuit, find the initial and final values of currents i1 and i2 when the switch is closed at t = 0.

bg

bg

bg

bg

[i1 0 = 7.14A , i1 ¥ = 10A; i2 0 = 7.14A , i2 ¥ = 0A]

Use initial value and final value theorems.

15 W t=0 150 V

+ –

i1

30 mH

i2

6W

6.15 In the network shown in figure, the switch is closed at t = 0, prior to which the circuit is in zero state. Using Thevenin’s theorem, transform the circuit to the left of points A and B into Thevenin equivalent in frequency domain and find the current in 30 W resistance. Convert the expression for current

. - 0.265e -13.14 t + 0.083e -41.86t (A)] [ 01818

in time domain. 10 W

10 V

+ –

1H

A

2H

t=0

20 W

30 W

B

6.16 The network shown in figure is in steady state with switch S1 and S2 open. At t = t1, S1 is opened and S2 is closed. Find the current through the capacitor for t ³ t1.

6.114

Circuit Theory and Networks 2W

10 V

2H S1

+ –

S2 1 mF

3H

[i (i ) = 5 cos {0.577 ´ 10 3 t - t1 } for t ³ t1 ]

b

g

6.17 An RC series circuit has R = 2 W, C = 0.25 F. Find the current response if the driving voltage is: (a)

b g

b g LM(a) e N

step voltage 2u t - 3 , (b) ramp voltage 2r t - 3 .

b g u bt - 3g; ( b) 1 ubt - 3g - 1 e- 2bt - 3g u bt - 3gO

-2 t -3

2

6.18 Show that the Laplace transform of the square wave is, F ( s) =

QP

2

1 s (1 + e - as )

f(t) 1 0

a

2a

3a

t (second)

4a

6.19 Determine the current response of a series RL circuit with R = 6 W and L = 3 H for each of the following driving voltages:

b g (b) a ramp voltage 2r bt - 3g (a) a step voltage 2u t - 2

Assume that the circuit is initially relaxed.

LM(a) 1 e1 - e N 3

b g j u bt - 2g; ( b) 2 L2r bt - 3g - 1 u bt - 3g + 1 e -2 bt - 3g u bt - 3g OO

-2 t -2

3

MN

4

PQPQ

4

6.20 A series RL circuit has a resistor R = 4 W and an inductor L = 2 H. A pulse of magnitude 10 V and duration 5 ms is applied to the circuit at t = 3 ms. Find i(t). Assume that the circuit was initially relaxed.

LM 5 u bt - 0.003g - u bt - 0.008g - 5 e 2 N2

b

g u bt - 0.003g - e -2bt - 0.008g u bt - 0.008g O

-2 t - 0.003

PQ

6.21 A voltage pulse of 20 V magnitude and 10 ms duration is applied to an RC circuit. Determine the current. Assume that the circuit was initially relaxed. Take R = 10 W and C = 10 mF.

LM2 Le N MN

-10 - 4 t

-e

d

-10 - 4 t - 10 - 5

i O u dt - 10 -5 iO

PQ

PQ

6.115

Laplace Transform and its Applications

b g

6.22 A unit doublet voltage d ¢ t - 5 is applied at t = 0 to a series RLC circuit consisting of a resistor R = 4 W, L = 1 H and C =

1 F. Determine i(t). Assume that the circuit was initially relaxed. 3

LMd bt - 5g + 1 e b 2 N

g u bt - 5g - 9 e-3bt - 5g u bt - 5gO

- t -5

PQ

Voltage in volt

2 6.23 Figure shows a staircase voltage waveform. Assuming that the staircase is not repeated, express its equation in terms of step functions. If this voltage is applied to a series RL circuit with R = 4 ohms and L = 2 H, find an expression for the resulting current i (t); i (0+) = 0.

4 3 2 1 0

LMi atf = 1 - e MM N

b g

-2 t - 2

2

4

6 8 Time t in second

10

b gOP P u bt - 10g P Q

u t - 2 + 1 - e- 2b t - 4 g u t - 4 + 1 - e - 2bt - 6 g u t - 6 + 1 - e-2 bt - 8g u t - 8

b g

b g

b g

- 4 1 - e- 2 bt -10 g

6.24 Find the current i(t) in a series RLC circuit comprising resistor R = 4 W, inductor L = 1 henry and capacitor C = 1 Farad when each of the following driving voltage is applied: 3

b g (d) step voltage 4u bt - 3g , and (e) impulse voltage 2d bt - 1g . LM(a) L3 - 9 e b g - 3 e 2 MM MN 2 N (c) ramp voltage 9r t - 2 ,

- t -2

b g O u bt - 2g;

-3 t -2

PQ

OP b g PP b g bt - 1g - e u bt - 1g Q f bt g = 2u bt g and f bt g = exp b- 3t g u bt g is

(b) 2 e- bt - 3g - e- 3bt - 3g u t - 3 ; c) 3e- 3 bt -1g u

6.25 Verify that the convolution between two functions

- t -1

1

2 1 - exp - 3t ; t > 0 where u (t) is the unit step function. 3

b g

2

6.116

Circuit Theory and Networks R=1W

6.26 Find the response of the network shown in figure when the input voltage is: (a) unit impulse, and (b) vi t = e - 2 t .

bg

d

( a ) e - t ; ( b) e - t - e - 2 t

i

Vi(t)

C = 1F

V0(t)

SHORT-ANSWER TYPE QUESTIONS 6.1 (a) What do you understand by Complex Frequency? Give its physical significance. (b) Define Laplace transform of a function f (t). What are the advantages of Laplace transform? or Explain clearly the advantages of Laplace transform method over classical method of solving differential equation with constant co-efficient describing electrical network. (c) State and deduce initial-value and final value theorems. (d) Write notes on: Application of Laplace transform to network analysis. 6.2 Define unit-step, unit ramp and unit impulse functions and derive their Laplace transform from first principles. 6.3 (a) Find the current i(t) if unit step voltage is applied to an RL circuit. or Derive an expression for the current response in an R–L series circuit excited with constant voltage source. (b) Define the term ‘time-constant’ of a circuit. What is the physical significance of time-constant of a circuit? Find its value for R–L series circuit. 6.4 (a) Derive an expression for the decay current in an RC circuit excited by a unit step voltage. What is the time-constant of the circuit? Also, determine the nature of the voltage response across the capacitor. (b) Under what conditions an RC series circuit will act as (i) a Differentiator? (ii) an Integrator? 6.5 (a) Explain the terms critical resistance, damping ratio and frequency as applied to the study of RLC series circuit. How they help in simplifying the analysis of the circuit? (b) Derive an expression for the current i(t) flowing through an RLC series circuit. Explain with suitable sketches the variation of current with time under three conditions: (I) Under damped, (II) Critically damped, (III) Over damped. 6.6 What do you understand by the impulse response of a network? Briefly explain its importance in network analysis. 6.7 What do you understand by transient and steady state parts of response? How can they be identified in a general solution? or Discuss the natural and steady state response of an electrical circuit with illustrative examples. or Write notes on: (a) Transient and steady state response (b) Free and forced response. 6.8 State and prove Convolution Theorem. What is the necessity of convolution theorem in circuit analysis?

6.117

Laplace Transform and its Applications

6.9 What is Laplace transformation? Give reasons for its wide use in the electric circuit analysis. 6.10 Discuss the advantages of analysing the circuits using frequency domain rather than the time domain. How can the initial conditions of a circuit is incorporated using Laplace transform?

LM z f (t ) e dtOP is taken as 0– instead of N Q ¥

6.11 Explain why the lower limit of the Laplace transform integral

- st

0-

0+. 6.12 What is the Laplace transform of a function which is nonzero for t < 0? 6.13 Does every signal f (t), such f (t) = 0 for t < 0, have a Laplace transform? 6.14 Define unit-step, unit ramp and unit impulse functions and derive their Laplace transform from first principles. 6.15 Define and sketch ramp, unit step and unit impulse functions. 6.16 Derive from the first principle the Laplace transform of a unit step function. Hence or otherwise, determine the Laplace transform of unit ramp function and unit impulse function. 6.17 Explain gate function. Obtain the equation of a gate function starting at origin and of duration T.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 6.1 6.8 6.15 6.22 6.29 6.36 6.43 6.50 6.57 6.64 6.71 6.78 6.85 6.92 6.99 6.106 6.113 6.120 6.127 6.134

(b) (a) (b) (b) (c) (a) (b) (b) (a) (b) (d) (b) (d) (a) (a) (b) (a) (d) (b) (c)

6.2 6.9 6.16 6.23 6.30 6.37 6.44 6.51 6.58 6.65 6.72 6.79 6.86 6.93 6.100 6.107 6.114 6.121 6.128 6.135

(c) (a) (c) (a) (d) (a) (a) (b) (b) (b) (b) (d) (b) (c) (b) (d) (b) (d) (c) (a)

6.3 6.10 6.17 6.24 6.31 6.38 6.45 6.52 6.59 6.66 6.73 6.80 6.87 6.94 6.101 6.108 6.115 6.122 6.129 6.136

(a) (a) (b) (a) (c) (a) (b) (b) (b) (a) (d) (c) (d) (a) (c) (a) (d) (d) (a) (b)

6.4 6.11 6.18 6.25 6.32 6.39 6.46 6.53 6.60 6.67 6.74 6.81 6.88 6.95 6.102 6.109 6.116 6.123 6.130

(c) (a) (c) (d) (b) (c) (d) (d) (c) (a) (b) (a) (a) (c) (b) (a) (c) (d) (c)

6.5 6.12 6.19 6.26 6.33 6.40 6.47 6.54 6.61 6.68 6.75 6.82 6.89 6.96 6.103 6.110 6.117 6.124 6.131

(c) (b) (b) (c) (c) (d) (c) (c) (a) (c) (a) (c) (c) (a) (b) (a) (a) (b) (b)

6.6 6.13 6.20 6.27 6.34 6.41 6.48 6.55 562 6.69 6.76 6.83 6.90 6.97 6.104 6.111 6.118 6.125 6.132

(c) (c) (d) (a) (c) (d) (b) (c) (c) (c) (b) (b) (c) (b) (d) (a) (c) (d) (d)

6.7 6.14 6.21 6.28 6.35 6.42 6.49 6.56 6.63 6.70 6.77 6.84 6.91 6.98 6.105 6.112 6.119 6.126 6.133

(c) (a) (a) (b) (d) (d) (b) (b) (a) (d) (d) (b) (b) (c) (c) (c) (c) (a) (c)

CHAPTER

7 Two-port Network

7.1

INTRODUCTION

A port is a pair of nodes across which a device can be connected. The voltage is measured across the pair of nodes and the current going into one node is the same as the current coming out of the other node in the pair. These pairs are entry (or exit) points of the network. So, a network with two input terminals and two output terminals is called a four-terminal network or a two-port network. It is convenient to develop special methods for the systematic treatment of networks. In the case of a single-port linear active network, we obtained the Thevenin’s equivalent circuit and the Norton’s equivalent circuit. When a linear passive network is considered, it is convenient to study its behaviour relative to a pair of designated nodes. In a two-port network, there are two voltage variables and two current variables. According to the choice of input and output port, these voltage and current variables can be arranged in different equations, giving rise to different port Figure 7.1 Block diagram of a two-port network parameters. In this chapter, we will discuss the behaviours of two-port networks.

7.2

RELATIONSHIPS OF TWO-PORT VARIABLES

In order to describe the relationships among the port voltages and currents of an n-port network, ‘n’ number of linear equations is required. However, the choice of two independent and two dependent variables is dependent on the particular application.

7.2

Circuit Theory and Networks

For n-port network, the number of voltage and current variables is 2n. The number of ways in 2n! 2n! . So, there will be which these 2n variables can be arranged in two groups of n each is = n! ´ n! (n!)2 2n! types of port parameters. (n!)2 For a two-port network (n = 2), there are six types of parameters as mentioned below:— 1. Open-Circuit Impedance Parameters (z-parameters), 2. Short-Circuit Admittance Parameters (y-parameters), 3. Transmission or Chain Parameters (T- parameters or ABCD – parameters), 4. Inverse Transmission Parameters (T ¢-parameters), 5. Hybrid Parameters (h-parameters), and 6. Inverse Hybrid Parameters (g-parameters). Note: Inverse parameters (T ¢ and g) are not included in WBUT syllabus.

7.2.1 Open-Circuit Impedance Parameters (z-parameters) The impedance parameters represent the relation between the voltages and the currents in the twoport network. The impedance parameter matrix may be written as, éV1 ù é z11 êV ú = ê z ë 2 û ë 21

z12 ù é I1 ù z22 úû êë I 2 úû

or

V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2

In this matrix equation, it is easily seen without even expanding the individual equations, that z11 =

V1 I1

I2 = 0

z12 =

V1 I2

I1 = 0

z21 =

V2 I1

I2 = 0

z22 =

V2 I2

I1 = 0

= Driving Point Impedance at Port-1.

= Transfer Impedance

= Transfer Impedance

= Driving Point Impedance at Port-2

It can be seen that the z-parameters correspond to the driving point and transfer impedances at each port with the other port having zero current (i.e. open circuit). Thus these parameters are also referred to as the open circuit parameters.

7.2.2 Short-Circuit Admittance Parameters (y-parameters) The admittance parameters represent the relation between the currents and the voltages in the twoport network.

Two-port Network

7.3

The admittance parameter matrix may be written as

I1 = y11V1 + y12 V2 é I1 ù é y11 y12 ù éV1 ù or êI ú = ê y ú ê ú I2 = y21V1 + y22 V2 ë 2 û ë 21 y22 û ëV2 û The parameters y11, y12, y21, y22 can be defined in a similar manner, with either V1 or V2 on short circuit. y11 =

I1 V1 V

2

= Driving Point Admittance at Port-1 =0

y12 =

I1 V2

V1 = 0

y21 =

I2 V1

V2 = 0

y22 =

I2 V2

V1 = 0

= Transfer Admittance

= Transfer Admittance

= Driving Point Admittance at Port-2

It can be seen that the y-parameters correspond to the driving point and transfer admittances at each port with the other port having zero voltage (i.e., short circuit). Thus these parameters are also referred to as the short circuit parameters.

7.2.3 Transmission Line Parameters (ABCD-parameters) The ABCD parameters represent the relation between the input quantities and the output quantities in the two-port network. They are thus voltage-current pairs. However, as the quantities are defined as an input-output relation, the output current is marked as going out rather than as coming into the port.

Figure 7.2 Two-port current and voltage variables for calculation of transmission line parameters

The transmission parameter matrix may be written as

V1 = AV2 - BI2 éV1 ù é A B ù é V2 ù ê I ú = êC D ú ê- I ú or I = CV - DI 1 2 2 û ë 2û ë 1û ë The parameters A, B, C, D can be defined in a similar manner with either port 2 on short circuit or port 2 on open circuit. A=

V1 V2

= Open Circuit Reverse Voltage Gain I2 = 0

7.4

Circuit Theory and Networks

B= -

C=

V1 I2

I1 V2

D= -

= Short Circuit Transfer Impedance V2 = 0

= Open Circuit Transfer Admittance I2 = 0

I1 I2

= Short Circuit Reverse Current Gain V2 = 0

These parameters are known as transmission parameters as in a transmission line, the currents enter at one end and leaves at the other end, and we need to know a relation between the sending end quantities and the receiving end quantities.

7.2.4 Hybrid Parameters (h-parameters) The hybrid parameters represent a mixed or hybrid relation between the voltages and the currents in the two-port network. The hybrid parameter matrix may be written as

éV1 ù é h11 ê I ú = êh ë 2 û ë 21

h12 ù é I1 ù h22 úû êëV2 úû

or

V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2

The h-parameters can be defined in a similar manner and are commonly used in some electronic circuit analysis. h11 =

V1 I1 V

2

= Short Circuit Impedance at Port-1 =0

h12 =

V1 V2

I1 = 0

h21 =

I2 I1

V2 = 0

h22 =

I2 V2

I1 = 0

= Open Circuit Reverse Voltage Gain

= Short Circuit Current Gain

= Open Circuit Output Admittance

As the h-parameters are dimensionally mixed, they are also named mixed parameters. Transistor circuit models are generally represented by these parameters as the input impedance (h11) and the short-circuit current gain (h21) can be easily measured by making the output short-circuited.

7.3

CONDITIONS FOR RECIPROCITY AND SYMMETRY

A network is said to be reciprocal if the ratio of the response transform to the excitation transform is invariant to an interchange of the positions of the excitation and response of the network.

7.5

Two-port Network

A two-port network will be reciprocal if the interchange of an ideal voltage source at one port with an ideal current source at the other port does not alter the ammeter reading. A two-port network is said to be symmetrical if the input and output ports can be interchanged without altering the port voltages and currents.

1. Conditions in terms of z-parameters Condition for Reciprocity We short circuit port 2 – 2¢ and apply a voltage source Vs at port 1 – 1¢. Therefore, V1 = Vs, V2 = 0, I2 = – I2¢ Writing the equations of z-parameters, V s = z11 I1 - z12 I 2¢

Fig. 7.3(a) Reciprocal network

0 = z21 I1 - z22 I 2¢ Solving these two equations for I2¢,

I 2¢ = Vs

z21 z11 z22 - z12 z21

(7.1)

Now, interchanging the positions of response and excitations, i.e., shorting port 1 – 1¢ and applying Vs at port 2 – 2¢; V1 = 0, V2 = Vs, I1 = I1¢ Writing the equations of z-parameters,

Fig. 7.3(b) Reciprocal network

0 = - z11 I1¢ + z12 I 2 V s = - z21 I1¢ + z22 I 2 Solving these two equations for I1¢,

I1¢ = Vs

z12 z11 z22 - z12 z21

(7.2)

For the two-port network to be reciprocal, from Eq. (7.1) and Eq. (7.2), we have the condition as,

z12 = z 21 Condition for Symmetry Applying a voltage Vs at port 1 – 1¢ with port 2 – 2¢ open, we have the equation,

Vs = z11I1 - z12 × 0 = z11 I1 Þ

Vs I1

= z11

(7.3)

I2 = 0

Now, applying a voltage Vs at port 2 – 2¢ with port 1 – 1¢ open, we have the equation,

Vs = z21 × 0 + z22 I 2 = z22 I 2 Þ

Vs I2

= z22

(7.4)

I1 = 0

For the network to be symmetrical, the voltages and currents should be same. From Eq. (7.3) and Eq. (7.4), we have the condition for symmetry as,

z11 = z 22

7.6

Circuit Theory and Networks

2. Conditions in terms of y-parameters Condition for Reciprocity From Fig. 7.3(a), writing the y-parameter equations, I1 = y11Vs I¢ Þ - 2 = y21 Vs - I 2¢ = y21Vs

(7.5)

From Fig. 7.3(b), writing the y-parameter equations,

- I1¢ = y12Vs I¢ Þ - 1 = y12 V I 2 = y22Vs s

(7.6)

From the principle of reciprocity, the condition for reciprocity is,

y12 = y 21 Condition for Symmetry As already stated, a two-port network is said to be symmetric if the ports can be interchanged without changing the port voltages and currents and thus the condition of symmetry becomes, y11 = y 22

3. Conditions in terms of ABCD-parameters Condition for Reciprocity From Fig. 7.3(a), writing the ABCD-parameter equations, Vs = A × 0 - B(- I 2¢ ) = BI 2¢ I¢ 1 Þ 2 = Vs B I1 = C × 0 - D(- I 2¢ ) = DI 2¢

(7.7)

From Fig. 7.3(b), writing the ABCD-parameter equations, 0 = AVs - BI 2 I ¢ AD - BC Þ 1 = Vs B - I1¢ = CVs - DI 2

From the principle of reciprocity, the condition for reciprocity is,

(7.8)

1 ( AD - BC ) = B B

(AD – BC) = 1 Condition for Symmetry From Eq. (7.7), I1 = DI 2¢ = D

Vs B

(7.9)

ü I1¢ + CVs 1 ì æ AD - BC ö A = íVs è ø + CVs ýþ = Vs B D Dî B From Eq. (7.9) and Eq. (7.10), we have the condition for symmetry as,

From Eq. (7.8), I 2 =

A=D

(7.10)

7.7

Two-port Network

4. Conditions in terms of h-parameters Condition for Reciprocity From Fig. 7.3(a), writing the h-parameter equations, Vs = h11 I1 + h12 × 0 = h11 I1 I¢ h Þ 2 = - 21 Vs h11 - I 2¢ = h21 I1 + h22 × 0 = h21 I1

(7.11)

From Fig. 7.3(b), writing the h-parameter equations,

0 = - h11I1¢ + h12Vs I¢ h Þ 1 = 12 V h11 ¢ I 2 = - h21I1 + h22Vs s

(7.12 )

From the principle of reciprocity, the condition for reciprocity is,

h12 = – h 21 From Eq. (7.11), I1 =

Vs h11

(7.13)

h h -h h æh ö From Eq. (7.12), I 2 = - h21 ç 12 Vs ÷ + h22Vs = Vs 11 22 12 21 h11 è h11 ø

(7.14)

From Eq. (7.13) and Eq. (7.14), we have the condition for symmetry as, (h11h22 - h12 h21 ) = 1

Table 7.1 Conditions of Reciprocity and Symmetry in terms of different Two-Port Parameters Parameter

7.4

Condition of Reciprocity

Condition of Symmetry

z

z12 = z21

z11 = z22

y

y12 = y21

y11 = y22

T (ABCD) h

(AD – BC ) = 1 h12 = –h21

A=D (h11h22 – h12h21) = 1

INTERRELATIONSHIPS BETWEEN TWO-PORT PARAMETERS

Each type of two-port parameter has its own utility and is suited for certain specific applications. However, it is sometime necessary to convert one set of parameters to another. It is possible through simple mathematical manipulations to convert one set to any of the remaining sets. It is discussed below.

1. z-parameters in Terms of Other Parameters (a) In terms of y-parameters The z-parameter equations are, V 1 = z11 I1 + z12 I 2

(7.15)

7.8

Circuit Theory and Networks

V 2 = z21 I1 + z22 I 2 The y-parameter equations are, I 1 = y11V1 + y12V2

(7.16)

I 2 = y21V1 + y 22V2

I2 y - 21 V ; substituting this in first equation, y22 y22 1 y22 y y æ I ö I - 12 I I 1 = y11V1 + y12 ç 2 - 21 V1 ÷ or V1 = Dy 1 Dy 2 è y22 y22 ø where, Dy = (y11y22 – y12y21) Substituting this value in second equation of Eq. 7.16 y21 y y æy ö I1 + 11 I 2 I 2 = y21ç 22 I1 - 12 I 2 ÷ + y22V2 or, V2 = Dy Dy Dy ø è Dy Comparing Eqs (7.15), (7.17) and (7.18), we get, From Eq. (7.16), V2 =

z11 =

(7.17)

(7.18)

y22 y y y ; z = - 12 ; z21 = - 21 ; z22 = 11 Äy 12 Äy Äy Äy

(b) In terms of transmission parameters The Transmission parameter equations are, V1 = AV2 - BI 2 I1 = CV2 - DI 2 From second equation of Eq. (7.19), æ 1ö æ Dö V 2 = ç ÷ I1 + ç ÷ I 2 èCø èCø From first equation of Eq. (7.19),

é 1 D ù æ AD - BC ö A I1 + I 2 ú - BI 2 = æç ö÷ I1 + ç V1 = A ê ÷ø I 2 è ø C C C C è ë û Comparing Eq. (7.20) and (7.21) with Eq. (7.15), we get,

() ()

(7.19)

(7.20)

(7.21)

AD - BC ÄT A D 1 ; z12 = ;z = ;z = = C C C 21 C 22 C (c) In terms of hybrid parameters The hybrid parameter equations are, V1 = h11 I1 + h12V2 z11 =

I 2 = h21 I1 + h22V2

(7.22)

æ h ö æ 1 ö From second equation, V2 = ç - 21 ÷ I1 + ç I è h22 ø è h22 ÷ø 2

(7.23)

éæ h ö æ 1 ö ù æ h11h22 - h12 h21 ö æh ö From first equation, V1 = h11 I1 + h12 êç - 21 ÷ I1 + ç I2 ú = ç I1 + ç 12 ÷ I 2 ÷ ÷ h h h è 22 ø û è ø è h22 ø 22 ëè 22 ø

(7.24)

7.9

Two-port Network

Comparing Eqs (7.23) and (7.24) with Eq. (7.15), we get, z11 =

h11h22 - h12 h21 Äh h h 1 ; z = 12 ; z = - 21 ; z22 = = h22 h22 12 h22 21 h22 h22

Similarly, the inter-relation of the other parameter in terms of the remaining parameters is obtained by writing the remaining parameter equations in the same format as those of the other parameter; and comparing the co-efficients of the two sets of equations, a relation is obtained. A summary of the relationships between impedance z-parameters, admittance y-parameters, hybrid h-parameters, and transmission ABCD-parameters is shown in Table where Dz = (z11z22 – z12z21), Dh = (h11h22 – h12h21), DT = (AD – BC), DT¢ = (A¢D¢ – B¢C¢), and Dg = (g11g22 – g12g21). Table 7.2 Interrelationships between Two-Port Parameters [z]

z11 z21

[z]

[y]

z22 Dz z - 21 Dz

[ABCD]

[A¢B¢C¢D¢]

[h]

z12 z22

y22 Dy - y21 Dy

z12 Dz z11 Dz

-

[ABCD]

y12 Dy y11 Dy

-

y11 y21

y12 y22

z11 z21

Dz z21

-

y22 y21

-

1 y21

1 z21

z22 z21

-

Dy y21

-

y11 y21

z22 z12

Dz z12

-

y11 y12

-

1 y12

1 z12

z11 z12

-

Dy y12

-

y22 y12

Dz z22 -

[g]

[y]

z21 z22

1 z11 z21 z11

-

z12 z22

1 y11

1 z22

y21 y11

z12 z11

Dy y22

y12 y22

y - 21 y22

1 y22

Dz z11

-

y12 y11 Dy y11

A C 1 C

D B 1 B

DT C D C

-

DT B A B

A B C D

[A¢ B¢ C¢D¢] D¢ C¢ DT ¢ C¢

A¢ B¢ DT ¢ B¢ D¢ DT ¢ C¢ DT ¢

D DT C DT

B DT A DT

B D 1 D

DT D C D

B¢ A¢ DT ¢ A¢

DT A B A

C¢ D¢ DT ¢ D¢

C A 1 A

-

1 C¢ A¢ C¢

[h] Dh h22 -

1 h11

1 B¢ D¢ B¢

-

B¢ DT ¢ A¢ DT ¢

A¢ B¢ C ¢ D¢

1 A¢ D¢ B¢ 1 D¢ B¢ D¢

-

h21 h22

h21 h11

-

[g] h12 h22

1 g11

1 h22

g 21 h11

Dg g11

h12 h11

Dg g 22

g12 g 22

Dh h11

-

-

g 21 g 22

g12 g11

1 g 22

-

Dh h21

-

h11 h21

1 g 21

-

h22 h21

-

1 h21

g11 g 21

1 h22

h11 h12

-

Dg g12

-

g 22 g12

h22 h12

Dh h12

-

g11 g12

-

1 g12

g 22 Dg g - 211 Dg

-

g12 Dg g11 Dg

h11 h12 h21 h22

h22 Dh h - 21 Dh

h12 Dh h11 Dh

-

g 22 g 21

-

Dg g 21

g11 g 21

g12 g 22

7.10

7.5

Circuit Theory and Networks

INTERCONNECTION OF TWO-PORT NETWORKS

In certain applications, it becomes necessary to connect the two-port networks together. The common connections are (a) series, (b) parallel and (c) cascade. (a) Series Connection of Two-port Networks As in the case of elements, a series connection is defined when the currents in the series elements are equal and the voltages add up to give the resultant voltage. In the case of two-port networks, this property must be applied individually to each of the ports. Thus, if we consider 2 networks r and s connected in series At port 1, Ir1 = Is1 = I1, and Vr1 + Vs1 = V1 Similarly, at port 2, Ir2 = Is2 = I2 and Vr2 + Vs2 = V2 The two networks, r and s can be connected in the following manner to be in series with each other.

Figure 7.4 Series connection of two-port networks

Under these conditions, V1 = (Vr1 + Vs1) = ( z11r + z11s ) I1 + ( z12 r + z12 s ) I 2

V2 = (Vr 2 + Vs 2 ) = ( z21r + z21s ) I1 + ( z22 r + z22 s ) I 2

It is seen that, the resultant impedance parameter matrix for the series combination is the addition of the two individual impedance matrices. [Z] = [Zr] + [Zs] Note: In the interconnection of series networks, there is a strong requirement of isolation, since the ground node of upper network form the non-ground node of the lower network. For the port properties to be valid, the voltages Va and Vb must be identically zero for the two networks r and s to be connected in series. If Va and Vb are not zero, then by connecting the two ports there will be a circulating current and port property of the individual networks r and s will be violated. (b) Parallel Connection of Two-port Networks As in the case of elements, a parallel connection is defined when the voltages in the parallel elements are equal and the currents add up to give the resultant current.

7.11

Two-port Network

In the case of two-port networks, this property must be applied individually to each of the ports. Thus, if we consider 2 networks r and s connected in parallel, At port 1, Ir1 + Is1 = I1, and Vr1 = Vs1 = V1 Similarly, at port 2, Ir2 + Is2 = I2 and Vr1 = Vs1 = V1 The two networks, r and s can be connected in following manner to be in parallel with each other. Under these conditions, I1 = ( I r1 + I s1) = ( y11r + y11s )V1 + ( y12 r + y12 s )V2 I 2 = ( I 2 r + I 2 s ) = ( y21r + y21s )V1 + ( y22 r + y22 s )V2 It is seen that, the resultant admittance parameter matrix for the parallel combination is the addition of Figure 7.5 Parallel connection of two-port networks the two individual admittance matrices. [Y] = [Yr] + [Ys]

Figure 7.6(a) Vb = 0

Figure 7.6(b) Va = 0

Note: As in series connection, parallel connection is also possible under the condition that Va = Vb = 0; otherwise they cannot be connected in parallel as that will violate the port properties. (c) Cascade Connection of Two-port Networks A cascade connection is defined when the output of one network becomes the input to the next network.

Figure 7.7 Cascade connection of two-port network

It can be easily seen that Ir2 = Is1 and Vr2 = Vs1. Therefore it can easily be seen that the ABCD parameters are the most suitable to be used for this connection. éVr1 ù é Ar Br ù éVr 2 ù éVs1 ù é As Bs ù éVs 2 ù ê I ú = êC D ú ê I ú , ê I ú = êC D ú ê I ú s û ë s2 û r û ë r 2 û ë s1 û ë s ë r1 û ë r

7.12

Circuit Theory and Networks

éV1 ù éVr1 ù é Ar ê I ú = ê I ú = êC ë 1 û ë r1 û ë r

Br ù éVr 2 ù é Ar = Dr úû êë I r 2 úû êëCr

Br ù éVs1 ù é Ar = Dr úû êë I s1 úû êëCr

Br ù é As Dr úû êëCs

Bs ù éVs 2 ù Ds úû êë I s 2 úû

é A Br ù é As Bs ù éV2 ù = ê r úê ú úê ëCr Dr û ëCs Ds û ë I 2 û Thus it is seen that the overall ABCD matrix is the product of the two individual ABCD matrices. This is a very useful property in practice, especially when analyzing transmission lines. é A B ù é Ar êC D ú = êC ë û ë r

7.6

Br ù é As Dr úû êëCs

Bs ù Ds úû

TWO-PORT NETWORK FUNCTIONS

Two-port network functions are broadly divided into two groups: 1. Transfer function, and 2. Driving point functions.

7.6.1 Transfer Function It is defined as the ratio of an output transform to an input transform, with zero initial condition and with no internal energy sources ecxcept the controlled sources. For a two-port network, having the variables I1(s), I2(s), V1(s) and V2(s), the transfer function can take the following four forms: V (s) V (s) Voltage Transfer Function G12 ( s ) = 1 ; G21 ( s ) = 2 V2 ( s ) V1 ( s ) I (s) I (s) Current Transfer Function a12 ( s ) = 1 ; a 21 ( s ) = 2 I 2 (s) I1 ( s ) V (s) V (s) Transfer Impedance Function Z12 ( s ) = 1 ; Z 21 ( s ) = 2 V2 ( s ) V1 ( s ) I1 ( s ) I 2 (s) Transfer Admittance Function Y12 ( s ) = ; Y21 ( s ) = I 2 (s) I1 ( s ) Note: (i) For a one-port network, Z(s) = 1/Y(s); but for a two-port network, in general Z12 ¹ 1/Y12; G12 ¹ 1/a12; (ii) Z and Y functions will becomes z and y parameters under the conditions of open-circuits or short-circuits, respectively.

7.6.2 Driving Point Function It takes two forms: Driving Point Impedance [Z(s)] For a two-port newtork in zero state with no internal energy sourceds, the driving point impedance s the ratio of transform voltage at any port to the transform current at the same port.

Two-port Network

Z11 ( s ) =

7.13

V1 ( s ) V (s) ; Z 22 ( s ) = 2 I1 ( s ) I 2 (s)

Driving Point Admittance [Y(s)] For a two-port network in zero state with no internal energy sources, the driving point admittance is the ratio of transform current at any port to the transform voltage at the same port

Y11 ( s ) =

I1 ( s ) I ( s) ; Y22 ( s ) = 2 V1 ( s ) V2 ( s )

Note: (i) Driving point impedance and admittance functions together are known as immittance function. (2) Z and Y functions will becomes z and y parameters under the conditions of open circuits or short circuits,

SOLVED PROBLEMS 7.1 Find the Z and Y parameter for the networks shown in figure. (a)

(b)

(c)

(d)

Solution (a) By KVL, ( Z a + Z c ) I1 + Z c I 2 = V1 and Z c I1 + ( Z b + Z c ) I 2 = V2 Thus, the Z-parameters are: z11 = ( Z a + Z c ), z12 = z21 = Z c , z22 = ( Z b + Z c ) (b) By KCL, V -V 1 1 I1 = 1 2 = V1 - V2 Z Z Z V -V 1 1 and I2 = 2 1 = - V1 + V2 Z Z Z Thus, the y-parameters are, 1 1 y 11 = = y22 y12 = y21 = Z Z Since, Dy = y11 y22 - y12 y21 = 0, the z-parameters do not exist for this network.

7.14

Circuit Theory and Networks

(c) By KVL, V1 =

I1 + I 2 1 1 1 1 = V2 or, V1 = æç ö÷ I1 + æç ö÷ I 2 and V2 = æç ö÷ I1 + æç ö÷ I 2 èY ø èY ø èY ø èY ø Y

Thus, the z-parameters are, z11 = z22 =

1 = z12 = z21 Y

Since, Dz = z11 z22 - z12 z21 = 0 , the y-parameters do not exist for this network. (d) By KCL, I1 = YaV1 + (V1 - V2 )Yc = V1 (Ya + Yc ) - V2Yc I 2 = YbV2 + (V2 - V1 )Yc = - V1Yc + V2 (Yb + Yc )

Thus, the y-parameters are:

y11 = Ya + Yc ; y12 = y21 = - Yc ; y22 = Yb + Yc 7.2 Obtain the z-parameters for the circuit shown in figure. (a)

(b)

Solution (a) The given circuit can be considered as the cascade connection of the following two networks:

(a)

(b)

From Prob. 7.1(a), z11a = z11b = z22 a = z22b = 3W z12 a = z21a = z12b = z21b = 2W So, the transmission parameters are, \

Aa = Ab =

z11 3 = z21 2

Two-port Network

\

Ba = Bb =

Dz 9 - 4 5 = = W z21 2 2

\

Ca = Cb =

1 1 = J z21 2

\

Da = Db =

z22 3 = z21 2

So, the transmission parameters of the resulting network are:

é3/2 5/2 ù é3/2 5/2ù é7/2 15/2 ù T = Ta ´ Tb = ê úê ú=ê ú ë1/2 3/2 û ë1/2 3/2 û ë3/2 7/2 û So, the z-parameters are: A 7 = W üï C 3 DT 2 ï = Wïï z12 = C 3 ý 1 2 z21 = = W ï C 3 ï D 7 z22 = = W ï ïþ C 3 z11 =

(b) By KVL, V1 = 2I1 + 4I3 V2 = I1 + I2 – I3 and 2(I1 – I3) + I1 + I2 – I3 – 4I3 = 0 Eliminating I3 from above equations, V1 =

26 4 I + I 7 1 7 2

4 6 I + I 7 1 7 2 Thus, the z-parameters are:

V2 =

é26/7 4/7 ù [z] = ê úW ë 4/7 6/7 û 7.3 For the network shown, find z and y-parameters.

7.15

7.16

Circuit Theory and Networks

Solution From the figure, we can write the KVL equations,

V1 = I3 V2 = 2I2 – 4I1 – 2I3 and,

2I1 – 2I3 + 2I2 – 4I1 – 2I3 – I3 = 0 Þ I3 =

From (i), V1 = -

2 2 I + I = - 0.4 I1 + 0.4 I 2 5 1 5 2

From (ii), V2 = 2 I 2 - 4 I1 \

4 4 I + I = - 3.2 I1 + 1.2 I 2 5 2 5 1

é- 0.4 0.4 ù [z] = ê úW ë- 3.2 1.2 û Dz = (- 0.4 ´ 1.2) - 0(0.4) ´ (- 3.2) = 0.8

é1.2/0.8 - 0.4 ù ê é1.5 - 0.5ù 0.8 ú \ [y] = ê úJ=ê úJ ë 4 - 0.5û ê3.2/0.8 - 0.4 ú 0.8 û ë 7.4 Find the y-parameters for the 2-port networks shown. (a)

(b)

(c)

(i) (ii) 2 (I - I ) 5 2 1

7.17

Two-port Network

Solution (a) We consider two cases to find out the y-parameters. Case (I) Making port- 2 shorted and applying a voltage of V1 at port- 1

By KVL, and

17I1 + 20I2 = V1 12I1 + 20I2 = 0

I1 =

20 20

V1 0

17 20

= 0.2V1 Þ y11 =

I1 V1

= 0.2 J V2 = 0

12 20 Solving,

I2 =

17 V1 12 0 17 20

= - 0.12V1 Þ y21 =

I2 V1

= - 0.12 J V2 = 0

12 20 Case (II) Making port-1 shorted and applying a voltage of V2 at port- 2

By KVL, and

17I1 + 20I2 = – 0.2V2 12I1 + 20I2 = V2

I1 =

- 0.2V2

20

V2

20

17 20 12 20

= - 0.24V2 Þ y12 =

I1 V2

= - 0.24 J V1 = 0

7.18

Circuit Theory and Networks

Solving,

17 - 0.2V2 I2 =

12

V2

17 20

= 0.194V2 Þ y22 =

I2 V2

= 0.194 J V1 = 0

12 20

- 0.24 ù é 0.2 [ y] = ê úJ ë- 0.12 0.194 û (b) We consider two cases. Case (I) V1 = 0 Case (II) V2 = 0 Thus,

By KCL,

V2 æ 0 - V2 ö 1 ü + Þ y12 = Jï 4 çè 12 ÷ø 6 ï V2 V2 5 ï + Þ y22 = I 2 = y22V2|V1 = 0 = J ï 3 12 12 ý 1ö 1 æ1 I1 = y11V1|V2 = 0 = ç + ÷ V1 Þ y11 = J ï è 12 12 ø 6 ï ï V1 1 I 2 = y21V1|V2 = 0 = - Þ y21 = - J ï 12 12 þ

I1 = y12V2|V1 = 0 =

(c) For V1 = 0, the circuit becomes as shown. \

I 2 = y22V2 = (1 + 2)V2 = 3V2 Þ y22 = 3 J

Also,

-

I1 = V2 Þ y12 = - 1 J 1

Two-port Network

7.19

For V2 = 0, the circuit becomes as shown. \

-

I2 = 3V1 1

(i)

I3 + 3V1 = V1 Þ 2V1 = - I 3 1 I1 = I3 + I4 I4 1 From (i) to (iv), and V1 =

(ii) (iii) (iv)

I1 = V1 + I 3 = V1 - 2V1 = - V1 Þ y11 = -1 J From (i), y21 = – 3 J Thus, the y-parameters are:

é -1 -1ù [y] = ê úJ ë- 3 3 û From the interrelationship, we get the z-parameters as: é-1 0 ù [z] = ê ú (W) ë-1 1/3û 7.5 Measurements were made on a two-port network shown in the figure.

(i) With port-2 open, a voltage of 100Ð0° volt is applied to port-1, resulted in, I1 = 10Ð0° amp and V2 = 25Ð0° volt. (ii) With port-1 open, a voltage of 100Ð0° volt is applied to port-2, resulted in, I2 = 20Ð0° amp and V1 = 50Ð0° volt. (a) Write the loop equations for the network and also find the driving point and transfer impedance. (b) What will be the voltage across a 10 W resistor connected across port-2 if a 100Ð0° volt source is connected across port-1. Solution (a) From the given data, we get the z-parameters as: z11 =

z21 =

z12 =

V1 I1 V2 I1 V1 I2

=

100Ð0° = 10 W 10Ð0°

=

25Ð0° = 2.5 W 10Ð0°

=

50Ð0° = 2.5 W 20Ð0°

I2 = 0

I2 = 0

I1 = 0

7.20

Circuit Theory and Networks

z22 =

V2 I2

= I1 = 0

100Ð0° =5W 20Ð0°

So, the loop equations are:

V1 = 10 I1 + 2.5 I 2 ü ý V2 = 2.5 I1 + 5I 2 þ (b) Here, V1 = 100Ð0° and V2 = - I 2 RL = -10 I 2 Putting these values in loop equations, 100 = 10I1 + 2.5I2 Þ I1 = 10 – 0.25I2 and –10I2 = 2.5I1 + 5I2 or,

-10 I 2 = 2.5(10 - 0.25 I 2 ) + 5 I 2

or,

-15I 2 = 25 - 0.625 I 2

- 25 = –1.74 A 14.375 \ Voltage across the resistor = –I2RL = 17.4 V 7.6 (a) The following equations give the voltages V1 and V2 at the two ports of a two port network, V1 = 5I1+2I2 , V2 = 2I1+I2 ; A load resistance of 3 W is connected across port-2. Calculate the input impedance. (b) The z-parameters of a two port network are z11 = 5 W, z22 = 2 W, z12 = z21 = 3 W. Load resistance of 4 W is connected across the output port. Calculate the input impedance. Solution (a) From the given equations, (i) V1 = 5I1 + 2I2 (ii) V2 = 2I1 + I2 At the output, V2 = – I2RL = – 3I2 Putting this value in (ii), –3I2 = 2I1 + I2 Þ I2 = –I1/2 or,

I2 =

æ -I ö Putting in (i), V1 = 5I1 + ç 1 ÷ = 4I1 è 2 ø \ Input impedance, Zin = (b) [Same as Prob. (a)] Zin =

V1 = 4W I1

V1 = 3.5W I1

7.7 Determine the h-parameter with the following data: (i) with the output terminals short circuited, V1 = 25 V, I1 = 1 A, I2 = 2 A (ii) with the input terminals open circuited, V1 = 10 V, V2 = 50 V, I2 = 2 A Solution The h-parameter equations are, V1 = h11I1 + h12V2 I2 = h21I1 + h22V2

Two-port Network

7.21

(a) With output short-circuited, V2 = 0, given: V1 = 25 V, I1 = 1 A and I2 = 2 A.

\ and

25 = h11 ´ 1ü Þ h11 = 25 W, and h21 = 2 2 = h21 ´ 1ýþ

(b) With input open-circuited, I1 = 0, given: V1 = 10 V, V2 = 50 V and I2 = 2 A.

\ and

10 = h12 ´ 50 ü 1 1 Þ h12 = = 0.2 and h23 = J = 0.04 J 2 = h22 ´ 50þý 5 25

Thus, the h-parameters are:

0.2 ù é25 W = ê ú 0.04 W-1 û ë 2 7.8 The y-parameters for a two-port network N are given as, [y11 = 4 J, y22 = 5 J, y12 = y21 = 4 J] If a resistor of 1 ohm is connected across port-1 of N, then find out the output impedance. Solution Output impedance is given as, [h]

Zout =

z11 z22 - z12 z21 z11 + Z L

Here,

y11 = 4 W-1 , y12 = y21 = 4 W-1 , y22 = 5 W-1

\

z11 =

y22 5 5 = = W Dy 20 - 16 4

z12 = z21 = and

z22 =

y12 4 = - = -1 W Dy 4

y11 4 = =1W Dy 4

Putting these values, Zout

5 z11 z22 - z12 z21 4 ´ 1 - ( - 1) ´ (- 1) + 1 ´ 1 5 = = = W z11 + Z L 5/4 + 1 9

7.9 (a) The h-parameters of a two-port network are h11 = 100 W, h12 = 0.0025, h21 = 20 and h22 = 1 mJ. Find V2/V1. (b) The h-parameters of a two-port network are h11 = 1 W, h12 = –h21 = 2, h22 = 1 J. The power absorbed by a load resistance of 1 W connected across port-2 is 100 W. The network is excited by a voltage source of generated voltage Vs and internal resistance 2 W. Calculate the value of Vs. Solution (a) The h-parameter equations are: (i) V1 = 100I1 + 0.0025V2

7.22

Circuit Theory and Networks

I2 = 20I1 + 0.001V2 By KVL at the output mesh, V2 = –2000I2

(ii) (iii)

æ V2 ö é I - 0.001V2 ù V1 = 100 ê 2 ú + 0.0025V2 = 5 çè - 2000 ÷ø - 0.005V2 + 0.0025V2 20 ë û From (i), or

V2 = - 200 V1

(b) The h-parameter equations are: V1 = I1 + 2V2 I2 = – 2I1 + V2 Since the load resistance of 1 W is connected across port-2, \

(i) (ii)

V22 = 100 Þ V2 = 10 V 1

By KVL, V2 = - I 2 RL = - I 2 Þ I 2 = -10 A and 2I1 + V1 = Vs From (ii), putting the values of I2 and V2,

(iii)

–10 = - 2 I1 + 10 Þ I1 = 10 A From (iii), Vs = 2 ´ 10 + V1 = 20 + I1 + 2V2 {by (i)} = 20 + 10 + 2 ´ 10 or, Vs = 50 V 7.10 The z-parameters for a network N are:

é2 1ù ê 2 5ú ë û The terminal connections for the network are shown in the adjacent figure. Calculate the voltage ratio V2/Vs, current ratio –I2/I1 and input resistance V1/I1. Solution The z-parameter equations are: V1 = 2I1 + I2 V2 = 2I1 + 5I2 By KVL at the input and output circuits, I1 + V1 = Vs Þ 3I1 + I2 = Vs and 5I2 + V2 = 0 Þ 2I1 + 10I2 = 0 Solving (iii) and (iv), Vs I1 =

0 3

1 3 Vs 10 10 2 0 2 = Vs and I 2 = = - Vs 28 28 1 3 1

2 10

2 10

(i) (ii) (iii) {by (i)} (iv) {by(ii)}

7.23

Two-port Network

I2 1 = I1 5

\

-

Now,

20 10 ö 10 V2 = (2 I1 + 5 I 2 ) = æç V = V è 28 28 ÷ø s 28 s

\

V2 5 = Vs 14

Again, 20 2 ö 18 V1 = (2I1 + I2) = æç V = V è 28 28 ÷ø s 28 s

\

V1 9 = W I1 14

7.11 For the two-port network in figure, terminated in a 1 W resistance, show that,

V2 z21 V z + Dz and 1 = 11 = I1 1 + z22 I1 1 + z22

Solution The z-parameter equations are: V1 = z11I1 + z12I2 V2 = z21I1 + z22I2

(i) (ii)

By KVL at the output, V2 = - I 2 ´ 1 Þ I 2 = - V2 V2 = z21 I1 + z22 I 2 = z21 I1 + z22 (- V2 ) From (ii), or, V2 (1 + z22 ) = z21 I1 or

(iii)

V2 z21 = I1 1 + z22

(Proved)

From (i),

é V (1 + z22 ) ù V1 = z11 ê 2 ú + z12 (- V2 ) z21 ë û

{by (iii)}

éz + z z - z z ù = V2 ê 11 11 22 12 21 ú z21 ë û é z11 + Dz ù = V2 ê z ú 21 ë û \

V1 V1 V2 z11 + Dz z21 z + Dz = ´ = ´ = 11 I1 V2 I1 z21 1 + z22 1 + z22

(Proved)

7.12 Calculate the T-parameters for the block A and B separately and then using these results, calculate the T-parameters of the whole circuit shown in the figure. Prove any formula used.

7.24

Circuit Theory and Networks

(a)

(b)

Solution (a) We consider the given network as a cascade connection of two networks as shown. For Block A: Opening the port-2, By KCL, æ 1 + 1ö V - 1 V = I çè 2 3 ÷ø 1 3 2 1

and

1 1 - V1 + çæ + s÷ö V2 = 0 è3 ø 3

Solving for V1 and V2, V1 =

\ and

2 I1 (1 + 3s ) 2 I1 and V2 = (1 + 5s ) (1 + 5s )

ü = (1 + 3s) ï I2 = 0 ï ý I1 (1 + 5s) ï = Ca = 2 ï V2 I = 0 2 þ Aa =

V1 V2

Short-circuiting port-2, \ and

and

I1 =

V1 V1 5 + = V1 2 3 6

V1 = - 3I 2 Þ Ba = -

Da = -

I1 I2

= V2 = 0

V1 I2

= 3W V2 = 0

5V1 3 5 ´ = 6 V1 2

7.25

Two-port Network

For Block B: Opening the port-2, By KCL, æ 1 + sö V - 1 V = I çè 5 ÷ø 1 5 2 1 1 1 1 - V1 + æç + ö÷ V2 = 0 è 5 4ø 5

and

Solving for V1 and V2, V1 =

9 I1 4 I1 and V2 = (1 + 9 s ) (1 + 9 s )

ü ï I2 = 0 ï ý I (1 + 9 s ) ï = Cb = 1 4 ï V2 I = 0 2 þ Ab =

\ and

V1 V2

=

9 4

Short-circuiting port-2, 1 I1 = æç + sö÷ V1 è5 ø

\

V1 = - 5 I 2 Þ Bb = -

and

Db = -

and

I1 I2

V1 I2

=5W V2 = 0

= (5s + 1) V2 = 0

Since the two networks are connected in cascade, the overall transmission parameter matrix is obtained as,

é (3s + 1) [T] = [Ta ] ´ [Tb ] = êæ 5s + 1ö ê ëêè 2 ø

3 ù é 9/4 ú ´ ê 1 + 9s æ ö 5/2 ú êç ÷ ûú êëè 4 ø

(b) [Same as Prob. (a)] Here,

\

1 ù é1 é3/2 1ù [Ta] = ê ú and [Tb ] = ê3/2 1ú 1/2 3/2 ë û ë û é3 2ù [T] = [Ta ] ´ [Tb ] = ê ú ë3 2û

ù (30s + 8) ù ú é (13.5s + 3) = ê (5s + 1) ú ë(11.25s + 1.75) (25s + 5) úû úû 5

7.26

Circuit Theory and Networks

7.13 Two identical sections of the network shown in the figure are connected in parallel. Obtain the y-parameters of the resulting network and verify by direct calculation. Solution

For the circuit,

y11 = 3 W-1 , y12 = y21 = - 2 W-1

and

y22 = 3W-1 The y-parameters for the combination will be, ü ¢ + y11 ¢¢ ) = 6 W-1 y11 = ( y11 ï 1 ¢ + y12 ¢¢ ) = - 4 W ý y12 = y21 = ( y12 ï ¢ + y22 ¢¢ ) = 6 W-1 y22 = ( y22 þ

To find the y-parameters by direct calculation, we consider the resulting network as shown. For the entire network, y11 = 4 + 2 = 6 W-1 ; y12 = y21 = - 4 W-1 ; y22 = 4 + 2 = 6 W-1

(Proved)

7.14 Two networks have general ABCD parameters as shown below: Parameter

Network-1

Network-2

A B C D

1.50 11W 0.25 siemens 2.5

5/3 4W 1 siemens 3.0

If the two networks are connected with their inputs and outputs in parallel, obtain the admittance matrix of the resulting network. Solution For network-1: y 11 =

D 2.5 5 -1 = = W B 11 22

y 12 = -

AD - BC 1.5 ´ 2.5 - 11 ´ 0.25 1 == - W-1 B 11 11

y 21 = -

1 1 = - W -1 B 11

y 22 =

A 1.5 3 -1 = = W B 11 22

Two-port Network

For network-2: y 11 =

D 3 -1 = W B 4

y 12 = -

AD - BC 1 = - W-1 B 4

y 21 = -

1 1 = - W -1 B 4

y 22 =

A 5 5 -1 = = W B 3 ´ 4 12

So, the admittance matrix of the resulting network is:

é 5/22 -1/11ù é 3/4 -1/4 ù é 43/44 -15/44 ù -1 [y] = ê ú+ê ú=ê úW ë-1/11 3/22 û ë-1/4 5/12 û ë-15/44 73/132 û 7.15 Two identical sections of figure are connected in series. Obtain the z-parameters of the resulting network and verify by direct calculation. All values are in ohm. Solution The z-parameters of each section: z11 = 3 W, z12 = z21 = 1 W, z22 = 3 W So, the z-parameters of the combined series network are: z11 = (3 + 3) = 6 W, z12 = z21 = (1 + 1) = 2 W, z22 = (3 + 3) = 6 W To find the z-parameters by direct calculation, we consider the resulting network as shown.

For the resulting network,

ü = 2 Wï ï I2 = 0 I2 = 0 ý V2 V1 = 6 W z12 = = 2 Wï z22 = I2 I =0 I2 I =0 ï 1 1 þ

z11 =

V1 I1

= 6 W z21 =

V2 I1

7.27

7.28

Circuit Theory and Networks

7.16 (a) Find out the z- and h-parameters for the circuit shown in Fig. (a). All values are in ohm. (b) Hence, obtain the hybrid parameters for the two-port network of Fig. (b).

(a) Solution (a) For Fig. (a), the z-parameters are: z11 =

\

V1 I1

= 4 W, z12 = z21 = 2 W, z11 = I2 = 0

(b)

V2 I2

=4W I1 = 0

ü Dz 16 - 4 = =3W ï z12 4 ï z 2 ï h12 = 12 = = 0.5 ï z22 4 ý z21 2 = - = - 0.5 ï h21 = z22 4 ï ï 1 1 -1 = = 0.25 W ï h22 = z12 4 þ

h11 =

(b) The connection is series-parallel connection. For this connection, the overall h-parameters will be the sum of individual h-parameters. \

h11 = (3 + 3) = 6W ü ï h12 = (0.5 + 0.5) = 1 ï ý h21 = (- 0.5 - 0.5) = -1 ï h22 = (0.25 + 0.25) = 0.5W-1 þï

7.17 (a) Find the equivalent p-network for the T-network shown in the Fig. (a). (b) Find the equivalent T -network for the p-network shown in the Fig. (b).

(a)

(b)

7.29

Two-port Network

Solution (a) Let the equivalent p-network have YC as the series admittance and YA and YB as the shunt admittances at port-1 and port-2, respectively. Now, the z-parameters are given as: z11 = ( Z A + Z C ) = 7 W, z12 = z21 = Z C = 5 W, z22 = ( Z B + Z C ) = 7.5 W \

Dz = (7 ´ 7.5 - 5 ´ 5) = 27.5 W2

\

y 11 =

z22 7.5 J = Dz 27.5

y 12 = y21 = -

zC 5 J =27.5 Dz

z11 7 J = Dz 27.5

y 22 = \

YA = ( y11 + y12 ) =

2.5 1 = J 27.5 11

\

YB = ( y22 + y12 ) =

2 J 27.5

5 2 = J 27.5 11 Thus, the impedances of the equivalent p-networks are:

and

YC = - y21 =

ü ï ï ï = 13.75 W,ý ï ï = 5.5 W ï þ

ZA =

1 = 11 W, YA

ZB =

1 YB

ZC =

1 YC

Equivalent p-network

(b)

p-network

Equivalent T-network

The y-parameters, y 11 = 1.2 J, y12 = y21 = -1 J, and y22 = 1.5 J \

Dy = (1.2 ´ 1.5 - 1) = 0.8

7.30

Circuit Theory and Networks

y22 1.5 y y 1 1.2 = W, z12 = z21 = - 12 = W, z22 = 11 = W Dy 0.8 Dy 0.8 Dy 0.8

\

z11 =

\

Z A = ( z11 - z12 ) =

0.5 = 0.625 Wüï 0.8 ïï 0.2 = 0.25 W ý Z B = ( z22 - z12 ) = 0.8 ï 1 ï = 1.25 W ZC = z12 = ïþ 0.8

7.18 The z-parameter of a 2-port network are: z11 = 10 W, z22 = 20 W, z12 = z21 = 5 W. Find the ABCD-parameters. Also find the equivalent T-network. Solution From the inter-relationship, we get the ABCD parameters as: A=

z11 10 = =2 z21 5

B=

z11 Z 22 - Z12 Z 21 10 ´ 20 - 5 ´ 5 = = 35 W z21 5

C=

1 1 = = 0.2 J z21 5

D=

z22 20 = =4 z21 5

Equivalent T-network

To find the equivalent T-network, we have the relations,

and

z11 = ( Z A + ZC ) = 10 W ü ï z12 = z21 = ZC = 5 W ý Þ Z A = 5 W, Z B = 15 W, ZC = 5 W z22 = ( Z B + ZC ) = 20 Wïþ

7.19 Z-parameters of the two-port network N in figure. are, z11 = 4s, z12 = z21 = 3s, z22 = 9s. (a) Replace N by its T-equivalent. (b) Use part (a) to find the input current I1 for Vs = cos1000t.

7.31

Two-port Network

Solution

é4s 3s ù (a) The z-parameters are: [ z ] = ê ú (W) ë3s 9s û Since the network is reciprocal, its T-equivalent exists. Its elements are: ZA = ( z11 - z12 ) = s, Z B = ( z22 - z21) = 6s, and

Equivalent T-network

ZC = z21 = z12 = 3s

So, the equivalent circuit is shown in figure.

(b) We repeatedly combine the series and parallel elements of above figure, with resistors in kW and s in Krad/s to find the input impedance, Zin in kW. \

Zin =

Vs (6 s + 12) (3s + 6) =s+ = (3s + 4) I1 (6 s + 12) + (3s + 6)

or Zin(j) = (3 j + 4) = 5Ð36.9° kW So, the current, i(t) =

vs (t ) 1 = cos (1000t - 36.9°) (mA) Z in ( j ) 5

7.20 The z-parameters of a two-port network N are given by, z11 = ( 2s + 1/s ), z12 = z21 = 2s, z22 = ( 2s + 4). (a) Find the T-equivalent of N. (b) The network N is connected to a source and a load as shown in figure. Replace N by its T-equivalent and then find I1, I2, V1, and V2.

7.32

Circuit Theory and Networks

Solution (a) To find the equivalent T-network, we have the relations,

and

ü 1 z11 = ( Z A + ZC ) = æç 2 s + ö÷ Wï è sø ï ý Þ Z = 1 W, Z = 4W, Z = 2 s W z12 = z21 = Z C = 2 s W A B C ï s z22 = ( Z B + Z C ) = (2 s + 4) W þï

Equivalent T-network (b) The equivalent circuit is shown below.

By KVL,

I1(3 + j) + I2( j2) = 12Ð0° I1( j2) + I2(5 + j3) = 0

I1 =

12Ð0° j2 0 (5 + j 3) (3 + j )

j2

j2

(5 + j 3)

12Ð0° 2Ð90° 0 5.831Ð30.96° = = 3.29Ð -10.22° (A) 16 + j14

Solving,

and

(3 + j ) 12Ð0° j2 0 = 1.13Ð -131.19° (A) I2 = (3 + j ) j2 j2

(5 + j 3)

\

V1 = 12Ð0° - I1 ´ 3 = 12 - 3.29 ´ 3Ð-10.22°= 2.28 + j1.75 = 2.88Ð37.504° (V)

and

V2 = - I 2 (1 + j ) = -1.13(1 + j )Ð -131.186° = 1.59Ð93.81°

7.33

Two-port Network

So, the currents and voltages are:

i1(t ) = 3.29 cos (t - 10.2°) (A) ü i2 (t ) = 1.13 cos (t - 131.2°) (A) ïï ý v1(t ) = 2.88 cos (t + 37.5°) (A) ï v2 (t ) = 1.6 cos (t + 93.8°) (A) ïþ 7.21 For the bridge-TRC network, find the y-parameters and its equivalent p-network.

Solution The given network is the parallel combination of the two networks:

(a) Network

(b) Network

é s /2 - s /2 ù For network (a), the y-parameters are: [ya] = ê úJ ë- s /2 s /2 û

2/ s é(1 + 2/ s) ù For network (b), the z-parameters are: [zb] = ê úW + 2/ s (1/2 2/ s ) ë û z22b s+4 (1/2 + 2/ s) = = Dzb (1 + 2/ s ) (1/2 + 2/ s) - 4/ s 2 s + 6

\

y 11b =

\

y 12b = y21b = -

\

y 22b =

z12b 2/ s 4 = = Dzb ( s + 6)/2s s + 6

z11b ( s + 2)/2 2( s + 2) = = Dzb ( s + 6)/2s s +6

és + 4 ês + 6 For network (b), the y-parameters are: [yb] = ê ê 4 êë s + 6

4 ù s+6 ú ú 2( s + 2) ú s + 6 úû

7.34

Circuit Theory and Networks

Thus, the overall y-parameters are: és + 4 é s /2 - s /2 ù ê s + 6 [y] = [ ya ] + [ yb ] = ê ú+ê ë - s /2 s /2 û ê 4 êë s + 6 é s 2 + 8s + 8 ê 2( s + 6) = ê ê s 2 + 6s + 8 ê2( s + 6) ëê

4 ù s+6 ú ú 2( s + 2) ú s + 6 úû

s 2 + 6s + 8 ù 2( s + 6) úú s 2 + 10 s + 8 ú ú 2( s + 6) ûú

-

Equivalent p network can be found out from the relations: s Ya = ( y11 + y12 ) = ; Y = ( y22 + y12 ) ( s + 6) b

s 2 + 6s + 8 2s ; Yc = - y12 = - y21 = ( s + 6) 2( s + 6) 7.22 For the notch-filter network, determine the y-parameters. =

Solution The given network is the parallel combination of the two networks:

(a) Network For network (a), \ \

(b) Network

1 + 2s 1 + 2s 1 1 z11a = çæ + 1÷ö = + 1÷ö = ; z12 a = z21a = 1; z22 a = çæ è 2s ø è 2s ø 2s 2s 1 + 4s 4s 2 z22 a 2s(1 + 2s) z = ; y12 a = y21a = - 12a y 11a = Dza Dza (1 + 4s)

Dza =

= -

z 2s(1 + 2s) 4s 2 ; y = 11a = (1 + 4s) 22a Dza (1 + 4s)

7.35

Two-port Network

For network (b), z11b = (1/ s + 2) =

1 + 2s 1 + 2s 1 ; z12b = z21b = ; z22b = (1/ s + 2) = s s s

\

Dzb =

4( s + 1) s

\

y 11b =

z22b (1 + 2s) z z (1 + 2 s) 1 = ; y = y21b = - 12b = ; y = 11b = Dzb 4( s + 1) 12b Dzb 4( s + 1) 22b Dzb 4( s + 1)

Thus, the overall y-parameters are, y 11 = y22 = (y11a + y11b) = and

2s(1 + 2s) (1 + 2s) (1 + 2 s) (8s 2 + 12 s + 1) + = 1 + 4s 4 + 4s 4( s + 1) (4s + 1)

y 12 = y21 = (y12a + y12b) = -

16s3 + 16s 2 + 4s + 1 4s 2 1 =1 + 4s 4( s + 1) 4(4s + 1) ( s + 1)

7.23 A network has two input terminals a, b and two output terminals c, d. The input impedance with c-d open-circuited is (250 + j100) ohm and with c-d short-circuited is (400 + j300) ohm. The impedance across c-d with a-b open-circuited is 200 ohm. Determine the equivalent T-network parameters. Solution For c-d Terminals opened,

( Z A + Z B ) = (250 + j100) But, for c-d terminals shorted,

ZA +

(i)

Z B ZC = (400 + j 300) Z B + ZC

(ii)

Again, with a-b terminals opened, (ZB + ZC) = 200

(iii)

From (ii) and (i), we get,

Z B ZC - Z B = 150 + j 200 Z B + ZC or

Z B ZC - Z B2 - Z B ZC = 200(150 + j 200)

or

Z B2 = 200(-150 - j 200) = 104 (1 - j 2) 2

\ \

Z B = (100 - j 200)W ü ï Z A = (150 + j 300)W ý ZC = (100 + j 200)Wïþ

and

{by (iii)

7.24 Find the driving point impedance at the terminals 1-1¢ of the ladder network shown in figure. (a)

7.36

Circuit Theory and Networks

(b)

Solution (a) The driving point impedance at 1-1¢ is

Z11 = s +

1 s+

=

1 s+

s 4 + 3s 2 + 1 s 2 + 2s

1 s

(b) The driving point impedance at 1-1¢ is,

Z11 = ( s + 1) +

1 s+

=

1 ( s + 1) +

s 6 + 3s 5 + 8s 4 + 11s 3 + 11s 2 + 6 s + 1 s 5 + 2 s 4 + 5 s 3 + 4 s 2 + 3s

1 s+

1 ( s + 1) +

1 s

7.25 For the Notch-filter (Twin-T) network, determine: (a) y-parameters, (b) the voltage ratio transfer function V2/V1 when noload impedance is present, and (c) the value of the frequency at which the output voltage is zero. Solution (a) The given network is the parallel combination of the two networks:

(a) Network

(b) Network

For network (a), 2 + RCs 2 + RCs R R R 1 1 z11a = æç + ö÷ = + ö÷ = ; z12 a = z21a = ; z22 a = æç è Cs 2 ø è Cs 2 ø 2Cs 2 2Cs

\

Dza =

1 + RCs C 2 s2

7.37

Two-port Network

\

y11a =

z22 a RCs(2 + RCs) = ; Dza 2 R(1 + RCs)

z y22 a = 11a = Dza For network (b),

y12a = y21a = -

z12a R 2C 2 s 2 =; Dza 2 R(1 + RCs)

1 Cs æ1 + Csö è 2 ø (1 + RCs)

1 + 2 RCs 1 + 2 RCs 1 1 1 z11b = æç + Rö÷ = ; z12b = z21b = ; z22b = æç + 2ö÷ = è 2Cs ø è ø s 2Cs 2Cs 2Cs

\ \

1 + RCs C 2 s2 z (1 + 2 RCs) y11b = 22b = ; Dzb 2 R( RCs + 1)

Dzb =

y22b =

y12b = y21b = -

z12b 1 =; Dzb 2 R( RCs + 1)

z11b (1 + 2 RCs) = Dzb 2 R( RCs + 1)

Thus, the overall y-parameters are, y 11 = y22 = ( y11a + y11b ) = and (b) Now,

RCs(2 + RCs) (1 + 2 RCs) ( R 2C 2 s 2 + 4 RCs + 1) + = 2 R(1 + RCs) 2 R( RCs + 1) 2 R( RCs + 1)

y 12 = y21 = ( y12 a + y12b ) = -

R 2C 2 s 2 + 1 R 2C 2 s 2 1 =2 R(1 + RCs) 2 R( RCs + 1) 2 R ( RCs + 1)

I1 = y11V1 + y12V2 I 2 = y21V1 + y22V2

When no-load impedance is present, I2 = 0, V2 y R 2C 2 s 2 +1 2 R( RCs + 1) R 2C 2 s 2 + 1 = - 21 = ´ 2 2 2 = 2 2 2 V1 y22 2 R ( RCs + 1) ( R C s + 4 RCs + 1) ( R C s + 4 RCs + 1) 2 2 2 (c) For V2 = 0 Þ 1 + R C s = 0 Putting s = jw, 1 – w2R2C2 = 0

\

\

w=

1 RC

1 2p RC 7.26 Find the open circuit impedance parameters for the two-port network shown in the figure below.

Thus, the notch frequency is given by, fN =

10 mH 2

1

5W 1¢

10 W 2¢

7.38

Circuit Theory and Networks

Solution For this p-network, the y-parameters are given as,

FH 15 + 0.011 s IK = FH 0.2 + 100s IK ;

y11 =

y12 = y21 = -

1 100 =; s 0.01s

FH 101 + 0.011 s IK = FH 0.1 + 100s IK 100 100 100 Dy = b y y - y y g = F 0.2 + H s IK ´ FH 0.1 + s IK - FH - s IK

y22 =

\

11 22

= 0.02 +

FH

2

12 21

2

FH IK - FH - 100s IK

30 100 + s s

= 0.02 +

30 s

2

IK

Thus, the z-parameters are,

U| || y - 100 /s 100 5000 W =z === = Dy 0.02 + 30 /s 0.02 s + 30 s + 1500 V || y 0.2 + 100 /s 0.2 s + 100 10 s + 5000 W = = = = |W 0.02 s + 30 s + 1500 Dy 0.02 + 30 /s

z11 = z12 z22

y22 0.1 + 100 /s 0.1s + 100 5s + 5000 = W = = s + 1500 Dy 0.02 + 30 /s 0.02 s + 30 12

Ans.

21

11

7.27 Find the open-circuit impedance parameters of the circuit given in the figure. Also, find the hparameters of the circuit. I1

I2 2

1

j10 W 5W

j15 W

2¢

1¢

Solution By KVL, (j10 + 5) I1 + 5I2 = V1 5I1 + (j15 + 5)I2 = V2

and Thus, the z-parameters are:

a

(i) (ii)

a

f

f

z11 = 5 + j10 W z12 = z21 = 5 W Z22 = 5 + j15 W The hybrid parameter matrix may be written as

LMV OP = LMh N I Q Nh 1

11

2

21

h12 h22

OP LM I OP Q NV Q 1

2

Ans.

7.39

Two-port Network

From Eq (ii), we get, I2 = – =-

V2 5 I + 5 + j15 1 5 + j15

1 1 I + V 1 + j 3 1 5 + j15 2

(iii)

Putting this value of I2 in Eq (i), we get,

LM N

(5 + j10) I1 + 5 -

OP Q

V2 5 I + = V1 5 + j15 1 5 + j15

Þ

V1 = =

a5 + j10f ´ a5 + j15f - 25 I + 5 V 5 + j15 a5 + j15f 1

2

30 + j 25 1 I + 1 + j3 1 1 + j3

(iv)

Comparing Eq (iii) and (iv) with the standard equations of h-parameters, we get,

30 + j 25 1 1 1 W; h12 = J ; h 21 = ; h22 = 1 + j3 1 + j3 1 + j3 5 + j15 7.28 Determine the z-parameters for the network shown in the figure. h11 =

I1

Ans.

I2 2

1 + V1

5W 20 W

+ V2

10 W

– 1¢

– 2¢

Solution We consider two situations: (a) When I1 = 0, i. e. port-1 is open-circuited: In this case no current will flow through the 5W resistor. I1= 0 1 + V1

I2 2

5W 20 W

+ V2

10 W

– 1¢

– 2¢

Figure(a) When I1 = 0 By KVL in the right mesh, we get,

10 I2 + 20 I2 - V2 = 0 \

z22 =

V2 I2

= 30 W I1 = 0

7.40

Circuit Theory and Networks

From Fig. (a), we get, V1 = 20I2

z12 =

\

V1 I2

= 20 W I1 = 0

(b) When I2 = 0, i.e., port-2 is open-circuited: In this case no current will flow through the 10 W resistor. I1 I2 = 0 By KVL in the left mesh, we get, 2 1 + + 5 I1 + 20 I1 - V1 = 0 5W 10 W

V1 I1

z11 =

\

V1

= 25 W I2 = 0

I1 1 + V1

20 W

– 1¢

V2 – 2¢

I2 = 0 2

5W 20 W

+ V2

10 W

– 1¢

– 2¢

Figure (b) When I2 = 0 From Fig. (b), we get, V2 = 20I1 \

z21 =

V2 I1

= 20 W I2 = 0

Therefore, the z-parameters of the network are:

z =

LM25 20OP aWf N20 30Q

Ans.

7.29 Find the y-parameters for the network shown in the figure. 50 W

20 W

10 W

Solution We consider two situations: When V1 = 0, i.e., port-1 is short-circuited In this case, no current will flow through the 20 W resistor. The modified circuit is shown in Fig. (a). By KCL at node 2, V2 - 0 V2 - 0 + = I2 10 50 \

y22 =

I2 V2

= V1 = 0

1 1 + = 0.12 J 10 50

Ans.

7.41

Two-port Network I1

50 W

I2

2

V1 = 0

10 W

V2

Figure(a) When V1 = 0 Also, from Fig. 7.5 (a) we get, I1 =

y12 =

\

0 - V2 50

I1 V2

= V1 = 0

1 = 0.02 J 50

Ans.

When V2 = 0, i.e., port-2 is short-circuited In this case, no current will flow through the 10 W resistor. The modified circuit is shown in Fig. (b). By KCL at node 1, V1 - 0 V1 - 0 + = I1 20 50

y11 =

\ I1

= V2 = 0

50 W

1

V1

I1 V1

1 1 + = 0.07 J 20 50 I2

V2 = 0

20 W

Figure(b) When V2 = 0 Also, from Fig. 7.5 (b) we get, I2 =

\

y21 =

0 - V1 50

I2 V1

= V2 = 0

1 = 0.02 J 50

Ans.

Therefore, the y-parameters of the network are

y =

LM0.07 N0.02

OP Q

0.02 J 0.12

Ans.

Ans.

7.42

Circuit Theory and Networks

7.30 For the network shown in the figure, determine the ABCD parameters. I1 1 W

2W

I2

1W

1 +

+

2W

V1

2W

2

V2 – 2¢

– 1¢

Solution The ABCD-parameter equations are, V1 = AV2 - BI 2 I1 = CV2 - DI2

For the network shown in the figure. we convert the delta consisting of the resistances of 2 W each into its equivalent star so that the circuit becomes as shown in Fig. (a) and Fig. (b). 2´2 2 r1 = r2 = r3 = = W 2+2+2 3 2 W 3

I1 1 W 1 + V1 – 1¢

2 W 3

1W

I2

I1 +

2 W 3

2

1.67 W

1 +

1.67 W

V1 V2 – – 2¢ 1¢

(b)

To find the ABCD parameters, we consider two situations: When V2 = 0, i.e., port-2 is short-circuited As shown in Fig. (c), by KVL we get, 1.67I1 + 0.67(I1 + I2) = V1 2.33I1 + 0.67I2 = V1 0.67(I1 + I2) + 1.67I2 = 0 2.33 I1 = I = -3.5 I 2 0.67 2

or,

D= -

\ I1

V1 – 1¢

I1 I2

= 3.5 V2 = 0

1.67 W I 2

1.67 W

2

1 +

V2 = 0

1.67 W

2¢ (c)

I2 +

(a) Modified network

or, and,

1.67 W

2

V2 – 2¢

7.43

Two-port Network

Putting this value in the first equations, we get,

a f

2.33 ´ -3.5 I2 + 0.67 I2 = V1 Þ B = -

V1 I2

= 7.5 W V2 = 0

When I2 = 0, i. e. port-2 is open-circuited Here, no current will flow through the right side 1.67 W resistance. By KVL, we get, and,

V1 = (1.67 + 0.67)I1 = 2.33I1 V2 = 0.67I1

\

C=

A=

\

I1 V2

V1 V2

=

1 = 1.5 J 0.67

=

2.33I1 = 3.5 0.67 I1

I2 = 0

I2 = 0

I 1 1.67 W

1.67 W I 2 = 0 2

1 +

V2

1.67 W

V1 – 1¢

2¢ (d)

Therefore, the ABCD parameters of the network are B = 7.5 W;

A = 3.5;

C = 15 J;

and

D = 3.5

Ans.

7.31 Find the hybrid parameters for the network shown in the figure. I2 = 0

I1

2

1

10 W 5W

15 W

2¢

1¢

Solution By KVL, 15I1 + 5I2 = V1 5I1 + 20I2 = V2

(i) (ii)

Thus, the z-parameters are

a

a

f

f

z11 = 5 + j10 W z12 = z21 = 5 W Z22 = 5 + j15 W The hybrid parameter equations are, V1 = h11 I1 + h12 V2 I 2 = h21 I1 + h22 V2

Ans.

7.44

Circuit Theory and Networks

From Eq (ii), we get, V 5 I + 2 20 1 20 1 1 V = - I1 + 4 20 2

I2 = -

(iii)

Putting this value of I2 in Eq (i), we get,

LM N

15 I1 + 5 -

OP Q

V 1 I + 2 = V1 4 1 20

55 1 I + V 4 1 4 2 Comparing Eq (iii) and (iv) with the standard equations of h-parameters, we get, 55 1 1 1 h11 = W; h12 = ; h21 = - ; h22 = J Ans. 4 4 4 20 7.32 Find the y parameters for the following network: V1 =

Þ

(iv)

20 W

+

5W

10 W

V1

40 W

–

+ V2 –

Solution This two-port network can be considered as the parallel connection of two two-port networks as shown below. 20 W +

5W

10 W

V1

40 W

–

+

+

+

V2

V1

V2

–

–

–

(b)

(a)

For network (a), the z-parameters are:

z11a = 50 W; z12 a = z21a = 40 W; z22 a = 45 W;

\ Dz = 50 ´ 45 - 40 2 = 650

c

Thus, the y-parameters are y11a =

z22 a 45 9 = = mho 650 130 Dz

y12 a = y21a = y22 a =

z12 40 4 ==mho Dz 650 65

z11a 50 1 = = mho Dz 650 13

h

7.45

Two-port Network

For network (b), the y-parameters are

1 1 mho; y12 b = y21b = mho 20 20 We know that for parallel connection of two two-port networks the overall y-parameters are the summation of individual y-parameters. Thus, y11b = y22 b =

U| g FH 1309 + 201 IK = 0.119 mho || 4 1I F = y = by + y g = - H 65 20 K = - 0.111 mhoV| || 1 1 = b y + y g = F + I = 0.127 mho H 13 20 K W b

y11 = y11a + y11b = y12 y22

12 a

21

22 a

Ans.

12 b

22 b

7.33 Obtain the ABCD parameters for the network shown in the figure. 10 W

20 W

Input

50 W

50 W

20 W

Output

10 W

Solution This two-port network can be considered as the cascade connection of two two-port networks as shown below. 10 W

20 W

50 W

50 W

20 W

Network (a)

10 W

Network (b)

For Network (a), as this is a T-network, the z-parameters are given as,

b

z11 = 60 W; z12 = 50 W; z22 = 70 W; \

g c

\ Dz = z11 z22 - z12 z21 = 60 ´ 70 - 50 2 = 1700

Aa =

z11 60 6 = = z 21 50 5

Ba =

Dz 1700 = = 34 W z21 50

Ca =

1 1 mho = z21 50

Da =

z 22 70 7 = = z 21 50 5

h

For Network (b), as this is a p-network, the y-parameters are given as,

FH 501 + 201 IK = 1007 mho; y = y = - 501 mho; 7 3 F 1I 1 Dy = b y y - y y g = ´ - = 100 25 H 50 K 125 y11 =

\

12

21

2

11 22

12 21

y22 =

FH 501 + 101 IK = 253 mho

7.46

Circuit Theory and Networks

Ab = -

y22 3 / 25 ==6 y21 - 1 /50

Cb = -

Dy 1 /125 2 == mho y21 - 1 / 50 5

\

Bb = -

1 1 == 50 W y21 - 1 /50

Db = -

y11 7 /100 7 == y21 - 1 / 50 2

For the entire network, the ABCD parameters are given as,

LM A BOP = LM A NC DQ NC

a a

OP LM Q N

OP LM Q N

Ba Ab ´ Da Cb

OP LM Q N

Bb 6 /5 = Db 1 / 50

OP LM Q N

20.8 179 50 = 0.68 5.9 7/ 5

34 6 ´ 7 /5 2 /5

OP Q

7.34 Calculate the ABCD parameters of the network shown in the figure below. j20 W

j20 W

1

2

30 W 1¢

2¢

Solution For this T-circuit, the z-parameters are given as,

a

f

z11 = z22 = 30 + j 20 W z12 = z21 = 30 W

\ \ \

2

a z 30 + j 20 2 U| = F1 + j I = A= 3K Dz a60 + j 20f j 20 H || Dz a60 + j 20f j 20 F 40 = = B= H 3 + j 40IK W||V z 30 1 1 || = C= mho z 30 || z 30 + j 20 F 2I = = 1+ j D= H 3K |W z 30 b

g b

Dz = z11 z22 - z12 z21 = 30 + j 20

g

f

11

21

\

a

f

- 30 2 = 60 + j 20 j 20 = -400 + j1200

Ans.

12

\

22

12

7.35 Determine the hybrid parameters for the network in the figure shown below. r2

I1 r1

V1

I2 r3

V2

Solution For this p-network, the y-parameters are given as,

y11 =

F 1 + 1 I = F r + r I; Hr r K H r r K 1

1

2

2

1 2

y12 = y21 = -

1 ; r2

y22 =

FG 1 + 1 IJ = FG r + r IJ Hr r K H r r K 2

2

3

3

2 3

Ans.

7.47

Two-port Network

By inter-relationship, the h-parameters are obtained as,

h11 =

F H

r r 1 = 1 2 r1 + r2 y11

I K

1 y12 r2 =h12 = y11 r1 + r2 r1 r2

F H

1 y21 r2 = h21 = y11 r1 + r2 r1 r2 h22

r1 r1 + r2

=

I K

=-

r1 r1 + r2

FG IJ H K Dy |R br + r gbr + r g - r r |U F r r I br + r gbr + r g - r r = =S V|W ´ H r + r K = r br + r g y r r r |T 1

2

2

3

1 3

11

1 2

1

1 2

2

1

3

2

2

2

2

3

1

2

1 3

7.36 Find the hybrid parameters of the circuit given in the figure. 2W

I1 V1

1W

I2 V2

3W

Solution For this p-network, the y-parameters are given as, y11 =

FH 11 + 12 IK = 23 ;

1 y12 = y21 = - ; 2

Dy = y11 y22 - y12 y21 =

\

y22 =

FH IK

3 5 1 ´ - 2 6 2

FH 12 + 13IK = 65

2

=1

By inter-relationship, the h-parameters are obtained as,

h11 =

1 2 = W y11 3

h12 = -

y12 - 1 /2 1 == y11 3/2 3

1 y21 - 2 Dy 1 3 3 = == 1´ = J h22 = y11 3 /2 y11 3 2 2 2V 3 7.37 For the network shown in the figure, determine the z and y parameters. Solution By KVL for the three meshes, we get, I 1 10 W V1 = 10I1 + 3I2 + 2(I1 + I2) Þ 12I1 + 5I2 = V1 (i) +– V2 = 2(I2 – 2V3) + 2(I1 + I2) Þ 2I1 + 4I2 – 4V3 = V2 (ii) + 3I 2 V3 = 2(I1 + I2) (iii) V1 2 W V3 h21 =

–

2W

I2 + V2 –

7.48

Circuit Theory and Networks

From (ii) and (iii), V2 = 2I1 + 4I2 – 4(2I1 + 2I2) Þ V2 = – 6I1 – 4I2 From (i) and (iv), we get,

z=

LM12 N-6

y= z

\

-1

OP a f Q

5 W -4 =

LM 2/9 N - 1 /3

(iv)

Ans.

OP Q

5/18 (J) - 2 /3

Ans.

7.38 The h-parameters of a two-port network shown in figure are h11 = 1000W, h12 = 0.003, h21 = 100,

and h22 = 50 ´ 10 -6 mho. Find V2 and z-parameters of the network if Vs = 10 -2 Ð0 o V .

af

500 W

I1

I2 +

+ Vs –

2000 W

V2 –

Solution The h-parameter equations are, V1 = h11I1 + h12V2 = 1000I1 + 0.003V2 I2 = h21I1 + h22V2 = 100I1 + 50 ´ 10–6V2

(i) (ii)

By KVL for the two meshes, V1 = Vs – 500I1 V2 = – 200I2

(iii) (iv)

From (i) and (iii), or, From (ii) and (iv),

Vs – 500I1 = 1000I1 + 0.003V2 10–2 – 1500I1 = 0.003V2 -

or, From (v) and (i),

(v)

V2 = 100 I1 + 50 ´ 10 -6 V2 2000

I1 = -5.5 ´ 10 -6 V2 0.003V2 = 10 -2 + 1500 -5.5 ´ 10 -6 V2

c

V2 = -1.905 V Þ The z-parameters are calculated as follows.

(vi)

h

Ans.

h h Dh = - 500W z12 = 12 = 60W z21 = - 21 = - 2 ´ 10 6 W h22 h22 h22 1 = 20 ´ 10 3 W Ans. z22 = h22 z11 =

7.49

Two-port Network

7.39 For the two-port network shown in the figure, find the z-parameters. I1 1 W

2V 1

1W

1W

I2

+–

+ V1

2W

+ V2

2W

–

–

Solution We consider two cases: When I2 = 0 Here, as the output port is open-circuited, no current will flow through the 1W resistor connected at port 2. The modified circuit is shown in Fig (a). I1 1 W

2V 1

1W

1W

I2= 0

+–

+

+

I (I 1 – I )

V1

2W

V2

2W

–

–

(a) By KVL for the middle mesh, we get, I + 2V1 + 2I – 2 ´ (I1 – I) = 0 Þ

I=

FH 25 I - 25 V IK 1

(i)

1

By KVL for the left mesh, we get, V1 = I1 + 2 ´ (I1 – I) = 3I1 – 2I = 3I1 – 2 ´ or,

V1 = 11I1

\

z11 =

V1 I1

FH 25 I - 25 V IK 1

{by equation (i)}

1

= 11 W I2 = 0

Also, by KVL for the right mesh, we get, V2 = 2 I = 2 ´

\

z21 =

V2 I1

FH 25 I - 25 V IK = 45 I - 45 V = 45 I - 45 ´ 11 ´ I 1

= -8 W I2 = 0

1

1

1

1

1

= -8 I1

7.50

Circuit Theory and Networks

When I1 = 0 Here, as the output port is open-circuited, no current will flow through the 1 W resistor connected at port 1. The modified circuit is shown in Fig (b). I1= 0

2V 1

1W

1W

+–

+

I2 +

I V1

2W

(I 2 – I )

2W

–

V2 –

(b) By KVL for the middle mesh, we get, I – 2V1 + 2I – 2 ´ (I2 – I) = 0 Þ

I=

FH 25 I

2

+

2 V 5 1

IK

(ii)

By KVL for the left mesh, we get,

FH 25 I

V1 = 2 I = 2 ´

Þ

V1 = 4 I 2

\

z12 =

V1 I2

2

+

IK

2 4 4 V = I 2 + V1 5 1 5 5

=4W I1 =

Also, by KVL for the right mesh, we get,

b

g

V2 = I 2 + 2 ´ I 2 - I = 3 I2 - 2 I = 3I2 - 2 ´ =

\

z22 =

FH 25 I

2

+

2 V 5 1

IK

lby equation (ii)q

11 4 11 4 I - V = I - ´ 4 I2 = - I2 5 2 5 1 5 2 5

V2 I2

= -1 W I1 = 0

Therefore, the z-parameters of the network are,

z =

LM11 4 OP aWf N-8 -1Q

7.40 Find the z and y parameters of the network shown in the figure. Solution We convert the dependent current source into its equivalent voltage source as shown in the figure below.

Ans. 0.9 I 1

I1 1 W

10 W

I2

+

+

V1

V2

–

–

7.51

Two-port Network 9 I1

1W

I2

10 W

+–

I1 +

+

V1

V2

1W

–

–

By KVL for the two meshes, we get, I1 + 1 ´ (I1 + I2) = V1 Þ V1 = 2I1 + I2

(i)

10I2 + 9I1 + 1 ´ (I1 + I2) = V2 Þ V2 = 10I1 + 11I2

and,

(ii)

From (i) and (ii), we get the z-parameters as,

z =

LM 2 1 OP aWf N10 11Q

Ans.

Therefore, the y-parameters are, y = z

-1

-1

=

LM 2 1 OP = LM 11/12 N10 11Q N- 10/12

- 1/12 2 /12

OP J Q

Ans.

7.41 The network shown in the figure contains both dependent current source and dependent voltage source. For this circuit, determine the y and z parameters. 2 V1

1W

I1

I2

+– 2 V2

V1 1 W

2 W V2

Solution We first find out the y parameters. To find the y parameters, we consider two situations: When V1 = 0 Here, port 1 is shorted and hence, the dependent voltage source is zero, i.e., shortcircuited. The 1 W resistance in port 1 becomes redundant. The circuit is shown in Fig (a). I1

V1= 0

I2

1W

A

I2 –

2V 2

V2 2

2W

V2

(a) By KCL at node (A), we get,

F H

– I1 – 2V2 – I 2 -

V2 2

I =0ÞI K

1

+ I2 = –

3V2 2

(i)

7.52

Circuit Theory and Networks

By KVL for the outer loop, we get,

F H

V2 = 1 ´ I2 Þ

3 V = I2 2 2

\

y22 =

I2 V2

I K

V2 V = I2 - 2 2 2

= V1 = 0

3 J 2

Substituting the value of I2 in (i), we get, 3 3 I1 + V2 = - V2 2 2 I1 = -3V2 Þ

y12 =

\

I1 V2

= -3 J V1 = 0

When V2 = 0 Here, port 2 is shorted and hence, the dependent current source is zero, i.e., opencircuited. The 2 W resistance in port 2 becomes redundant. The circuit is shown in Fig (b). I1

1W

+–

V1

2V 1

1W

I2

V2= 0

(I1 + I2)

(b) By KVL for the left loop, we get, V1 = (I1 + I2) By KVL for the outer loop, we get, 2V1 + I2 + V1 = 0 Þ I2 = - 3V1

y21 =

\

I2 V1

= -3 J V2 = 0

From (ii),

V1 = I1 - 3V1 \

y11 =

I1 V1

Þ I1 = 4V1

= 4J V2 = 0

Therefore, the y parameters of the netwrok is given as,

y =

LM 4 -33OP MN-3 2 PQ

Ans.

(ii)

7.53

Two-port Network

Hence, the z parameters are given as, z = y

-1

-3 3 2

L4 =M MN-3

OP PQ

-1

LM -21 = MM -1 N

-1 4 3

OP PP aWf Q

Ans.

7.42 The model of a transistor in CE mode is shown in the figure. Determine the h parameters of the model. rb

I1

re

I2 re

a cbI 1

+ –

mbcV 2

V1

rd

V2

Solution The equations of h parameters are,

V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 To find h parameters, we consider two cases: When I1 = 0 Here, the dependent current source is open-circuited. The modified circuit is shown in Fig (a). I1= 0

rb

re

I2 + –

mbcV 2

V1

re rd

V2

(a) \

V1 = m bc V2

h12 =

Þ

b

Also, V2 = I2 re + rd

V1 V2

I1 = 0

= m bc

g h22 =

Þ

I2 V2

= I1 = 0

1 J re + rd

When V2 = 0 Here, the dependent voltage source is short-circuited. The modified circuit is shown in Fig (b). I1

rb

re

I2 abcI 2

V1

V2

(b)

7.54

Circuit Theory and Networks

b

\

V1 = I1 rb + re

Þ

h11 =

V1 I1

g

b

g

= rb + re W

V2 = 0

Also, I 2 = a cb I1

h21 =

Þ

I2 I1

V2 = 0

= a cb

Therefore, the h parameters for the transistor model is given as,

h =

LMbr + r g MN a b

e

cb

m bc 1 re + rd

OP PQ

Ans.

7.43 Find the hybrid parameters for the network of the figure (which represents a transistor). aI 1

I1

R1

I2

+

+

R3

V1

V2

R1

–

–

Solution Case (I): When V2 = 0 The circuit is modified as shown in the figure. aI 1

I1

R1 x

I2 y

+

R3

V1

V2 = 0

R1

–

By KCL at node x, Vx Vx + + a I1 = I1 R2 R3

b

Þ

Vx = 1 - a

g RR +RR 2

3

2

3

By KVL,

b

V1 = I1 R1 + Vx = I1 R1 + 1 - a

g FGH RR +RR IJK I 2

2

3

3

1

I1

7.55

Two-port Network

h11 =

\

V1 I1

LM N

= R1 +

V2 = 0

a1 - a fR R OP R +R Q 2

2

3

Ans.

3

By KCL at node y, 0 - Vx = I 2 + a I1 R3

h21 =

\

I2 I1

b g FGH RR +RR IJK I

Þ

I 2 = - a I1 - 1 - a

2

3

2

=-

FG R + a R IJ H R +R K 2

3

2

V2 =0

1

= - I1

3

FG R + a R IJ H R +R K 2

2

3

3

Ans.

3

Case (II): When I1 = 0 Here, the dependent current source is to be opened (since I1 = 0). The circuit is modified as shown in the figure. I1= 0 +

I2 +

R3

R1

V1

R2

V2

–

–

\ and

V2 = I2(R2 + R3) V1 = I2 R2

\

h12 = h22 =

and

V1 V2 I2 V2

=

R2 R2 + R3

Ans.

=

1 R2 + R3

Ans.

I1 = 0

I1 = 0

Therefore, the hybrid parameters are

LM MN

h11 = R1 +

b1 - a gR R OP; R +R PQ 2

2

3

3

h12 =

R + a R3 R2 1 ; h21 = - 2 ; h22 = R2 + R3 R2 + R3 R2 + R3

FG H

IJ K

Ans.

7.44 Determine the y and z parameters for the network shown in the figure. 1W

2 V1 +–

+ V1 1 W –

+ 2 V2

2W

V2 –

Solution We convert the dependent current source into equivalent dependent voltage source. The modified network is shown in the figure.

7.56

Circuit Theory and Networks

+ V1 –

2 V1

1W

I1

I2

+–

+

1W 2 V2

V2

2W

I3

+ –

–

By KVL for three meshes, we get, V1 = 1 ´ (I1 – I3) + 2V2 Þ I3 = I1 + 2V2 – V1 and 1 ´ I3 – 2V1 + 2(I2 + I3) – 2V2 + 1 ´ (I3 – I1) = 0 Þ 2V1 + 2V2 = – I1 + 2I2 + 4I3 and, V2 = 2 ´ (I2 + I3)

(i) (ii) (iii)

Substituting the value of I3 from (i) into (ii) and (iii), we get, 2V1 + 2V2 = – I1 + 2I2 + 4(I1 + 2V2 – V1) Þ 6V1 – 6V2 = 3I1 + 2I2 and, V2 = 2(I2 + I1 + 2V2 – V1) Þ 2V1 – 3V2 = 2I1 + 2I2 By (iv) – (v), we get, I1 = 4V1 – 3V2 Also, from (v) and (vi), we get, 2V1 – 3V2 = 2 (4V1 – 3V2) + 2I2 3 I2 = – 3V1 + V2 2

Þ

y=

-1

-1

\

OP bW g 4P - P 3Q -1

7.45 Find the h-parameters for the two-port network shown in the figure. I1 + V1 –

0.5V 1

3W

I2 +

4W + –

3 I2

1W

(vi)

(vii)

From (vi) and (vii), we get,

LM 4 -33OP bmhog Ans. MN-3 2 PQ LM- 1 4 -3O L 2 3 = =M z= y MN-3 2 PPQ MMN -1

(iv) (v)

V2 –

Solution To find h parameters, we consider two cases: When I1 = 0 Here, no current will flow through the 3 W resistance.

Ans.

7.57

Two-port Network

By KVL at the left mesh, we get,

I1 = 0

b g

V1 = 4 ´ 0.5V1 + 3I 2

+

= 2V1 + 3I 2

0.5V 1

3W

+ –

–

I2 =

I2 +

4W

V1

Þ V1 = - 3I 2 Also, by KCL at Node (X), we get,

X

1W 3 I2

V2 –

V2 + 0.5V1 = V2 + 0.5V1 1

b g

= V2 + 0.5 ´ - 3I 2 2.5I 2 = V2

Þ \

h22 =

I2 V2

= I1 = 0

1 = 0.4 J 2.5

\

V1 = - 3I 2 = - 3 ´

\

h12 =

V1 V2

FG V IJ = - 1.2V H 2.5K 2

2

= - 12 . I1 = 0

When V2 = 0 Here, the port 2 is short circuited. The 1W resistance becomes redundant. The modified circuit is shown in Fig (b). I1

0.5V 1

3W

+

I2

4W

V1

V2 = 0

+ –

–

3 I2

(b) \

I 2 = 0.5V1

= 0.5 ´ 3I1 + 4 I1 + 4 I 2 + 3I 2 = 35 . I1 + 35 . I2 Þ

2.5I 2 = -3.5I 1

\

h21 =

I2 I1

=V2 = 0

35 . = - 14 . 2.5

Also,

b

V1 = 3I1 + 4 I1 + 4 I 2 + 3I 2 = 7 I1 + 7 I 2 = 7 I1 + 7 ´ -14 . I1 = - 2.8 I1 \

h11 =

V1 I1

= - 2.8 W V2 = 0

g

7.58

Circuit Theory and Networks

Therefore, the h parameters of the network are given as,

h =

LM-2.8 N-14.

. -12 0.4

OP Q

Ans.

MULTIPLE-CHOICE QUESTIONS 7.1 Which one of the following pairs is correctly matched? (a) Symmetrical two-port network: AD – BC = 1 (b) Reciprocal two-port network: z11 = z22. (c) Inverse hybrid parameters: A, B, C, D (d) Hybrid parameters: (V1, I2) = f (I1, V2) 7.2 What is the condition for reciprocity in terms of h-parameters? (a) h11 = h22 (b) h12h21 = h11h22 (c) h12 + h21 = 0 (d) h12 = h21 7.3 For a reciprocal network, the two-port ABCD parameters are related as follows (a) AD – BC = 1 (b) AD – BC = 0 (c) AC – BD = 0 (d) AC – BD = 1 7.4 For a symmetrical two port network 2 =0 (a) z11 = z22 (b) z12 = z21 (c) z11 z22 - z12 (d) z11 = z22 and z12 = z21 7.5 For a two port network to be reciprocal, it is necessary that (b) z11 = z22 and AD – BC = 0. (a) z11 = z22 and y12 = y21 (c) h21 = –h12 and AD – BC = 0 (d) y12 = y21 and h21 = –h12 7.6 A two port network is symmetrical if (a) z11 z22 - z12 z21 = 1 (b) AD – BC = 1 (c) h11h22 – h12h21 = 1(d) y11y22 – y12y21 = 1 7.7 A two port network is reciprocal if and only if (b) BC – AD = –1 (c) y12 = –y21 (d) h12 = h21 (a) z11 = z22 7.8 In terms of ABCD parameters, a two port network is symmetrical if and only if: (a) A = B (b) B = C (c) C = D (d) D = A 7.9 The condition for reciprocity of a two port network having different parameters are: 2. g12 = –g21 3. A = D 1. h12 = –h21 Choose the correct combination. (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3. 7.10 Two two-port networks with transmission parameters A1, B1, C1, D1 and A2, B2, C2, D2 respectively are cascaded. The transmission parameter matrix of the cascaded network will be é A1 B1 ù é A2 B2 ù é A1 B1 ù é A2 B2 ù +ê (a) ê (b) ê ú ú úê ú ëC1 D1 û ëC2 D2 û ëC1 D1 û ëC2 D2 û

é A1 A2 B1 B2 ù (c) ê ú ëC1C2 D1 D2 û 7.11 Consider the following statements. For a bilateral network, 1. A = D 2. z12 = z21 Of these statements. (a) 1, 2 and 3 are correct (c) 1 and 3 are correct

é ( A1 A2 + C1C2 ) ( A1 A2 - B1 D2 ) ù (d) ê ú ë(C1 A2 - D1C2 ) (C1C2 + D1 D2 ) û

3. h12 = –h21 (b) 1 and 2 are correct (d) 2 and 3 are correct.

7.59

Two-port Network

7.12 In a two port network containing linear bilateral passive circuit elements, which one of the following conditions for z parameters would hold? (a) z11 = z22 (b) z12z21 = z11z22 (c) z11z12 = z22z21 (d) z12 = z21 7.13 The relation AD – BC = 1, where A, B, C and D are the elements of a transmission matrix of a network, is valid for (a) any type of network. (b) passive but not reciprocal network. (c) passive and reciprocal network. (d) both active and passive network. 7.14 When a number of 2-port networks are connected in cascade, the individual: (b) Ysc matrices are added. (a) Zoc matrices are added. (c) chain matrices are multiplied. (d) H-matrices are multiplied. 7.15 The h parameters h11 and h22 are related to z and y parameters as 1 (a) h11 = z11 and h22 = (b) h11 = z11 and h22 = y22 z22 Dz 1 1 (c) h11 = and h22 = (d) h11 = and h22 = y22 y11 z22 z22 7.16 Two two-port networks a and b having A B C D parameters as Aa = 4 = Da Ab = 3 = Db Ba = 5, Ca = 3 and Bb = 4, Cb = 2 are connected in cascade in the order of a, b. The equivalent A parameters of the combination is (a) 17 (b) 22 (c) 24 (d) 31. 7.17 With the usual notation, a two-port resistive network satisfies the condition A = D =

3 4 B= C 2 3

The z11 of the network is 5 4 2 1 (b) (c) (d) 3 3 3 3 7.18 The reciprocal of a network function is (a) an immittance function, if the original function is an immittance function. (b) a transfer function, if the original function is a transfer function. (c) never an immittance function. (d) never a transfer function.

(a)

7.19 A two-port network is defined by the relations I1 = 2V1 + V2 , I 2 = 2V1 + 3V2 . Then z12 is (a) –2 W

(b) –1 W

(c) -

1 W 2

(d) -

1 W 4

7.20 Consider the following statements 1. Transfer impedance is the reciprocal of transfer admittance. 2. One can derive transfer impedance of a network if its driving-point impedance and admittance are known. 3. Driving-point impedance is the ratio of the Laplace transform of voltage and current functions at the input. Of these statements: (a) 1, 2 and 3 are correct (b) 1 and 2 are correct (c) 2 and 3 are correct (d) 3 alone is correct.

7.60

Circuit Theory and Networks

7.21 Consider the following statements 1. The two-port network shown below does NOT have an impedance matrix representation.

2. The two-port network shown below does NOT have an admittance matrix representation.

7.22

7.23

7.24

7.25

7.26

7.27

3. A two-port network is said to be reciprocal if it satisfies z12 = z21 or an equivalent relationship. Of these statements: (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 1 and 3 are correct (d) None is correct. If two two-port networks are connected in series, and if the port current requirement is satisfied, which of the following is true? (a) The z-parameter matrices add (b) The y-parameter matrices add. (c) The ABCD-parameter matrices add. (d) None of these. If two two-port networks are connected in parallel, and if the port current requirement is satisfied, which of the following is true? (a) The z-parameter matrices add (b) The y-parameter matrices add. (c) The ABCD-parameter matrices add (d) None of these. If two two-port networks are connected in cascade, and if the port current requirement is satisfied, which of the following is true? (a) The z-parameter matrices add (b) The y-parameter matrices add. (c) The ABCD-parameter matrices add (d) None of these. The z11 and z22 parameters of the given network are (a) 8 W, 7.75 W (b) 13 W, 9 W (c) 12 W, 8.5 W (d) None of the above. For the network shown, the parameters h11 and h21 are (a) 5 W and –2/3 W (b) 3.4 W and –2/5 W (c) 3.4 W and –3/5 W (d) None of the above. The maximum value of the transmission parameter A for a passive, reciprocal, linear two-port network is (a) 1 (b) 2 (c) 3 (d) none of the above.

7.61

Two-port Network

7.28. The unique feature of ABCD parameters as compared to x, y and h parameters is (a) none (b) short-circuit functions (c) open-circuit functions (d) reverse transverse functions 7.29. The driving point impedance of the infinite ladder network shown in the given figure is

(given R1 = 2 W and R2 = 1.5 W) (a) 3 W

(b) 3.5 W

(c)

3 ö (d) ln æç1 + W è 3.5 ÷ø

3 W 3.5

7.30 A 2-port network is described by the relations: V1 = 2V2 + 0.5I 2 I1 = 2V2 + I 2 What is the value of the h22 parameter of the network? (a) 1 mho (b) 2 W (c) – 2 mho (d) 4 W 7.31 What are the suitable values for Z1 and Z2, to make the input impedance, Zin, of the network equal to R? Z2 R Z in

(a) R and R I1

7.32

(b) 2R and R

rb

R

(c) 3R and 2R

re

mbcV 2

V1

Z1

(d) 4R and 4R

I2

+ –

a cbI 2

re rd

V2

Which one of the following gives the h-parameter matrix for the network shown in the figure? (a)

LM 1 MN ra+ r LMr + r MN a e

d

cb

b

(c)

e

cb

OP r + r PQ m O 1 P r + r PQ m bc

b

d

e

bc

e

bc

e

LMr + r MN m LM m MNr + r b

(b)

bc

(d)

b

e

OP PQ a O 1 P r + r PQ a cb 1 re + rd cb

e

d

7.62

Circuit Theory and Networks

7.33 In a two-port network, the output short-circuit current was measured while the source voltage at the input was 1 V; the value of the output current would provide the parameter (c) h 21 (d) y 21 (a) B (b) y12 7.34 The y-parameter ‘y21’ of the network shown in the figure I1

6W

I2

4W +–

+ V1

+

14I 1

6W

V2

–

–

(a) is 2 mho (b) is 6 mho (c) is 3 mho 7.35 The phasor current through the inductance in the circuit shown is (a)

FG 10 IJ Ð - 45 H 2K

o

(b)

FG 10 IJ Ð45 H 2K

(d) does not exist

o

2W

i = 10cos 2 t

(c) 5Ð45o (d) 5Ð - 45o 7.36 For the two-port network, the parameter y21 will be Y3

1

V1

Y1

2

gmV1

Y2

1¢

2¢

(a) Y2 + Y3 (b) g m - Y3 7.37 For the given two-port network, z21 will be

(c) Y3 - g m

2W

(d) g m + Y2 + Y3

2W

1

2 V1

2 W V2

1W

1

(a) 2/5 W

(b) 3/5 W

2

(c) 1/5 W

7.38 The h-parameters for a two-port network are defined by network shown in the figure, the value of h12 is given by (a) 0.125 (b) 0.167 (c) 0.625 (d) 0.25

I1

E1

(d) 4/5 W

LM E OP = LMh N I Q Nh 1

11

2

21

h12 h22

4W

OP LM I OP . For the two-port Q NE Q

2W

2W

1

2

2W

4W

I2

E2

7.63

Two-port Network

7.39 The z matrix of a two-port network as given by

LM0.9 N0.2

OP Q

0.2 . The element y22 of the corresponding y 0.6

matrix of the same network is given by (a) 1.2 (b) 0.4 (c) – 0.4 7.40 For the two-port network shown in the figure, the z-matrix is given by (a)

LM Z Z + Z OP NZ + Z Z Q LM Z Z OP NZ Z + Z Q 1

1

(c)

1

2

1

2

2

(b)

2

1

1

2

1

LM Z Z OP NZ + Z Z Q LMZ Z OP NZ Z + Z Q

(d)

2

i1

i2

1

2

1 1

(d) 1.8

2

Z2 v1

Z1

v2

1

1

2

7.41 The parameters of the circuit shown in the figure are Ri = 1 M W, R0 = 10 W, A = 106 V/V. If Vi = 1 mV, then output voltage, input impedance and output impedance respectively are Ri

R0 + + –

Vi

AV i

–

(a) 1 V, ¥, 10 W (b) 1 V, 0, 10 W (c) 1 V, 0, ¥ (d) 10 V, ¥, 10 W 7.42 The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are I1

LM0 N0 L0 (c) h parameters, M N0 (a) z parameters,

I2

+

+

V1

V2

–

–

OP Q 0O 0PQ

LM1 N0 L1 (d) z parameters, M N0

0 0

(b) h parameters,

OP Q 0O 1PQ 0 1

7.43 The impedance parameters z11 and z12 of the two-port network in the figure are 3W

5W

10 W

4W

5W

7.64

Circuit Theory and Networks

(a) (b) (c) (d) 7.44 For

z11 = 2.75 W and z12 = 0.25 W z11 = 3 W and z12 = 0.5 W z11 = 3 W and z12 = 0.25 W z11 = 2.25 W and z12 = 0.5 W the lattice circuit shown in the figure, Za = j2 W and Zb = 2 W. The values of the open circuit

impedance parameters z =

LMz Nz

11 21

OP Q

z12 are z22 Zb

1

Za

Za

Zb

2

(a)

(c)

LM1 - j N1 + j LM1 + j N1 - j

3

OP Q 1 + jO 1 - j PQ 1+ j 1+ j

4

(b)

(d)

LM 1 - j N-1 + j LM 1 - j N-1 - j

1+ j 1- j

OP Q

-1 + j 1- j

OP Q

7.45 The ABCD parameters of an ideal n : 1 transformer shown in the figure are will be

I1

1 (a) n (b) n 1 (c) n 2 (d) 2 n 7.46 The h-parameters of the circuit shown in the figure are

(a)

(c)

LMn 0 OP . The value of X N0 X Q

LM 01. 01. OP N- 01. 0.3Q LM30 20OP N20 20Q

(b)

(d)

I2

V1

V2

n:1

LM10 N1 LM10 N-1

OP Q 1 O 0.05PQ

I1

-1 0.05

V1

7.47 In the two-port network shown in the figure below, z12 and z21 are, respectively, I1

I2

re

bI 1

r0

10 W

I2

20 W

7.65

Two-port Network

(a) re and b r0 (b) 0 and –b r0 (c) 0 and b r0 7.48 A two-port network is represented by ABCD parameters given by

(d) re and –br0

LMV OP = LM A B OP LM V OP N I Q NC DQ N - I Q 1

2

1

2

If port-2 is terminated by RL, then the input impedance seen at port-1 given by (a)

A + BR L C + DR L

(b)

AR L + C BR L + D

(c)

DR L + A BR L + C

(d)

B + AR L D + CR L

EXERCISES 7.1 Current I1 and I2 entering at ports 1 and 2 respectively of a two-port network are given by the following equations: I1 = 0.5V1 – 0.2V2 I2 = –0.2V1 + V2 where V1 and V2 are the voltages at ports 1 and 2 respectively. Find the y, z and ABCD parameters for the network. Also find its equivalent p-network. [y11 = 0.5 J; y12 = – 0.2 J; y21 = –0.2 J; y22 = 1 J; z11 = 2.174 W; z12 = z21 = –0.435 W; z22 = 1.086 W; A = 5, B = 5 W, C = 2.3 J, D = 2.5; Y1 = 0.3 J; Y2 = 0.2 J; Y3 = 0.8 J] 7.2 Determine the z-and y-parameters of the networks shown in figure.

(a)

ì ü é - j /120 j110 ù é- j120 - j160 ù ï -1 ï ê ú = W) = W z y ( ; ( ) í ý ê ú ê j /110 - j 3 ú ë- j160 - j80 û ïî ïþ 4û ë

(b)

j 40 ù é(30 + j 40) ïì ïü íz = ê ú (W)ý + j 40 (30 j 80) ë û îï þï

(c)

ì ü 1 ù é(1 + 25) ï ê ú (W)ï = z í ý æ1 + 1 ö ú ê 1 çè ÷ø ï ï 5 ê ú ë û î þ

7.66

Circuit Theory and Networks

7.3 Obtain the z-parameters for the circuit shown in figure and hence draw the z-parameter equivalent circuit.

ì é14 ê5 ïï íz = ê ï ê2 ë5 îï

ü 2ù ú ïï 5 ú (W)ý 6ú ï 5û þï

7.4 Find the open-circuit and short-circuit impedances of the network shown in figure.

é é 31 ê ê 44 êy = ê ê ê - 19 êë ë 44

ù 19 ù ú 44 ú ; z -parameters do not exist ú 23 úú ú úû 44 û

-

7.5 Find the z-parameters for the 2-port networks shown in figure containing a controlled source.

ì ü é-2 -1ù ï ï í z = ê 1 3 ú (W) ý ê ú ïî ïþ ë2 2û

7.6 A 2-port network made up of passive linear resistors is fed at port 1 by an ideal voltage source of V volt. It is loaded at port 2 by a resistor R. (i) With V = 10 volt and R = 6 W currents at ports 1 and 2 were 1.44 A and 0.2 A respectively. (ii) With V = 15 volt and R = 8 W current at port 2 was 0.25 A. Determine the p-equivalent circuit of the 2-port network. {YA = 0.2; YB = 0.3; YC = 0.5 (mho)} 7.7 Calculate the T-parameters for the block A and B separately and then using these results calculate the T-parameters of the whole circuit shown in figure. Prove any formula used.

{

Aa = Ab = Da = Db =

3 2

1 5 ; B = Bb = ; 2 a 2 é7/2 15/2 ùüï T =ê úý ë3/2 7/2 ûþï Ca = Cb =

7.67

Two-port Network

7.8 Find out the z-parameters of the two-port network shown in the figure.

ìï üï é6 2 ù ( W) ý íz = ê ú ïî ïþ ë2 6 û

7.9 Find the z-parameters for the lattice network shown in the figure.

ì æ Zb + Z a ö æ Z - Za ö ; z12 = z21 = ç b ; í z11 = z22 = ç ÷ 2 2 ÷ø è ø è î é2( Z a + Z b ) Z b2 ùïü 1 ê úý 2Z a + Zb ë Z b2 2 Z a Z b ûïþ 7.10 Current I1 and I2 entering at port-1 and port-2 respectively of a two port network are given by the following equations: I1=0.5V1-0.2V2, I2=-0.2V1+V2, where V1 and V2 are the voltages at port-1 and port-2 respectively. Find the y, z and ABCD parameters for the network. Also find the equivalent pnetwork. z=

ìï é 0.5 - 0.2 ù -1 é2.174 0.435ù (W ); Z = ê íy = ê ú ú (W), 1 û ë- 0.2 ë0.435 1.087 û îï üï 5 Wù é 5 T =ê ; Ya = 0.3 J, Yb = 0.8 J, Yc = 0.2 Jý ú ë2.3 J 2.5 û þï 7.11 Two identical sections of the circuit shown in the figure are connected in series. Obtain the zparameters of the combination and verify by direct calculation. [ z11 = z22 = 6 W; z12 = z21 = 4 W]

7.68

Circuit Theory and Networks

7.12 Test results for a two-port network are (a) port 2 open-circuited, I 1 = 0.01Ð0o A , V1 = 1.4Ð45o V , V2 = 2.3Ð - 26.4 o V

b g = 0.01Ð 0 bA g, V

bg = 1Ð - 90 bV g, V

b g = 15 . Ð - 531 . b Vg

o o (b) port 1 open-circuited, I 2 1 2 The source frequency in both the tests was 1000 Hz. Find z-parameters.

o

LM L 140Ð45 MN MN230Ð - 26.4

OP bWgOP . Q 150Ð - 531 PQ 100Ð - 90 o

o

o

o

7.13 Find the z-parameters for the network shown in the figure. I1

I2

6W

4W

+

+

V1

9W

V2

9W

–

–

LM L10 N MN 3

OP bWgOP 6Q Q 3

7.14 For the network shown in the figure, find the y-parameters and also the equivalent T-network. 8W

+

+

4W

1W

V1

V2

2W

–

–

LM L 62 /112 N MN- 30/112

OP Q

- 30/112 , Za = 8/13 W, Zb = 32 / 13 W, Z c = 30 /13 W 38 /112

OP Q

7.15 Find the h-parameters for the network shown in the figure. I1

I2

8W

+ V1

+

16 W

16 W

–

V2 –

8W

LMh N

11

=

32 1 1 1 W; h12 = ; h21 = - ; h22 = mho 3 3 3 12

OP Q

7.69

Two-port Network

7.16 The h-parameters of a two-port network are

h11 = 35W; h12 = 2.6 ´ 10- 4 ; h21 = - 0.98; h22 = 0.3 ´ 10 -6 mho The input terminals are connected to 0.001V sinusoidal source and a 104 ohm resistance is connected across the output port. Find the output voltage. [0.26 V] 7.17 Find the y and z-parameters for the network shown in the figure. I1

1W

I2

1W

+

+

V1

1 W 2

2W

V2

–

–

LML13/ 7 NMN 2 / 7

OPa fOP Q Q

2/7 - 3/ 5 - 2 / 5 W; mho 3/ 7 -2 / 5 13 / 5

OP a Q

f LMN

7.18 Find the y-parameters for the network shown in the figure. I1

5W

0.2I 2

I2

+–

V1

V2

20 W 0.4I 2

LML 0.2 NMN-0.333

OP OP Q Q

-0.24 J 0.4833

7.19 Find the transmission parameters of the network shown in the figure. I1

0.3 V 1

5W

+–

+ V2 10

V1

I2 +

4W

–

V2 –

LML 55 MMMM 267 NMN 20

50 13 1

OPOP PPPP QQ

7.20 Determine the T-parameters for the network shown in the figure using the concept of interconnection of two two-port networks. 1H

1H

1

1H 2

LM1 + 3s + s N 2s + s 2

1F 1¢

1F

3

2¢

4

3s + 4 s 3 + s5 1 + 3s 2 + s 4

OP Q

7.70

Circuit Theory and Networks

7.21 Determine the y parameters of the overall network, considering two networks connected in parallel. 1W

+ V1

+

1W

1W

V2

1W

–

–

LM y N

11

= y22 =

5 4 J; y12 = y21 = - J 3 3

OP Q

7.22 The z-parameters of a two-port network are z11 = 50 W; z22 = 30 W; z12 = z21 = 20 W

Calculate the y-parameters and ABCD parameters of the network.

LM y = 0.0273 mho; y = 0.0454 mho; y = y N A = 2.5; B = 55W; C = 0.05 mho; D = 15. 11

22

12

21

OP Q

= -0.01818 mho;

7.23 For the symmetrical two-port network, calculate the z-parameters and ABCD parameters. 1

I1

V1 1¢

40 W

+

–

40 W

I2 +

20 W

2

V2 –

2¢

z11 = z22 = 60 W; z12 = z21 = 20 W; A = D = 3; B = 160 W; C = 0.05 mho;

SHORT-ANSWER TYPE QUESTIONS 7.1 (a) Consider a linear passive two-port network and explain what are meant by (i) open-circuit impedance parameters and (ii) short-circuit admittance parameters. (b) What are the open-circuit impedance parameters of a two-port network? How can the transmission parameters be obtained from open-circuit impedance parameters? (c) Establish, for two-port networks, the relationship between the transmission parameters and the open-circuit parameters. (d) Define z- and y-parameters of a typical four terminal network. Determine the relationship between the z and y parameters. (e) Express h-parameters in terms of z-parameters for a two-port network. (f) Derive expressions for the y-parameters in terms of ABCD parameters of a two-port network. 7.2 (a) What do you understand by a reciprocal network? What is a symmetrical network? (b) Write technical note on derivation of short-circuit admittance parameter y12 of a symmetrical and reciprocal two-port lattice network.

7.71

Two-port Network

(c) How will you find the p-equivalent of a given network when its y-parameters are known? 7.3 (a) Explain what are meant by the transmission (ABCD) parameters of a two-port network. Derive the conditions necessary to be satisfied for the network to be (i) reciprocal and (ii) symmetrical. Or, Prove that for a reciprocal two-port network, DT = (AD – BC) = 1 (b) Prove that for a symmetrical two-port network, Dh = (h11h22 – h12h21) = 1 7.4 (a) Two two-port networks are connected in parallel. Prove that the overall y-parameters are the sum of corresponding individual y-parameters. (b) Two two-port networks are connected in cascade. Prove that the overall transmission parameter matrix is the product of individual transmission parameter matrices. (c) Two two-port networks are connected in series. Prove that the overall z-parameters are the sum of corresponding individual z-parameters. 7.5 What are transmission parameters? Where are they most effectively used? Establish, for two-port networks, the relationship between the transmission parameters and the open circuit impedance parameters.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 7.1 7.8 7.15 7.22 7.29 7.36 7.43

(d) (d) (c) (a) (a) (b) (a)

7.2 7.9 7.16 7.23 7.30 7.37 7.44

(c) (a) (b) (b) (c) (a) (d)

7.3 7.10 7.17 7.24 7.31 7.38 7.45

(a) (b) (b) (d) (a) (d) (b)

7.4 7.11 7.18 7.25 7.32 7.39 7.46

(a) (d) (a) (a) (c) (d) (d)

7.5 7.12 7.19 7.26 7.33 7.40 7.47

(d) (d) (d) (b) (d) (d) (b)

7.6 7.13 7.20 7.27 7.34 7.41 7.48

(c) (c) (d) (d) (d) (a) (d)

7.7 7.14 7.21 7.28 7.35 7.42

(b) (c) (b) (d) (a) (c)

CHAPTER

8 Fourier Series and Fourier Transform

PART I: FOURIER SERIES 8.1

INTRODUCTION

In 1807, the French mathematician Joseph Fourier (1768–1830) submitted a paper to the Academy of Sciences in Paris. In it he presented a mathematical description of problems involving heat conduction. Although the paper was at first rejected, it contained ideas that would develop into an important area of mathematics named in honour, Fourier analysis. One surprising ramification of Fourier’s work was that many familiar functions can be expanded in infinite series and integrals involving trigonometric functions. The idea today is important in modeling many phenomena in physics and engineering. In this chapter, in the first part, we will discuss the basic concepts of Fourier series. Then we will apply this concept to find the steady-state response of an electric circuit subject to a periodic excitation. A function of time f (t) is said to be periodic if f ( t ) = f ( t ± nT ) ; where, n is a positive integer and ‘T ’ is the period. Thus, a periodic function repeats itself every T second. v (t ) V

–2T (a)

–T

0

T (b)

Figure 8.1 Periodic functions

2T

3T

4T

t

8.2

Circuit Theory and Networks

In the second part of this chapter, we will learn about another transform method, namely Fourier transform, which is used to find the steady-state response of a network to aperiodic excitation.

8.2

DEFINITION OF FOURIER SERIES

French mathematician J.B.J. Fourier first studied the periodic function in 1822 and published his theorem which states that, “Any arbitrary periodic function can be represented by an infinite series of sinusoids of harmonically related frequencies.” This infinite series is known as Fourier series. Thus, if f (t) is a periodic function, then the Fourier series is, f (t) = a0 + a1 cos w t + a2 cos 2w t + ¼ + an cos nw t + ¼ + b1 sin w t + b2 sin 2w t + ¼ + bn sin nw t + ¼ ¥

= a0 + å (an cos nwt + bn sin nwt ) n =1

2p T nw is the nth harmonic of fundamental frequency a0, an, bn are the Fourier Co-efficients

where, w is the fundamental frequency =

8.3

DIRICHLET’S CONDITIONS

The conditions, under which a periodic function f (t) can be expanded in a convergent Fourier series, are known as Dirichlet’s conditions. These are as follows: (i) f (t) is a single valued function. (ii) f (t) has a finite number of discontinuities in each period, T. (iii) f (t) has a finite number of maxima and minima in each period, T. T

T

(iv) The integral, ò | f (t )| dt exists and is finite or in other way, ò [ f (t )]2 dt < ¥ . 0

0 T

Note: If f (t) is current or voltage, ò [ f (t )]2 dt represents energy which would be supplied by the 0

source in one cycle. That means the energy in the waveform for each cycle must be finite. All physical waveforms would, of course, satisfy this criterion. Therefore, in practical engineering problems, it is not necessary to check whether a function satisfies Dirichlet condition.

8.4

FOURIER ANALYSIS

This involves two operations: 1. The evaluation of the co-efficient a0, an and bn.

8.3

Fourier Series and Fourier Transform

2. Truncation of the infinite series after a finite number of terms so that f (t) is represented within allowable error (-Done later).

8.4.1 Evaluation of Fourier Coefficients ¥

f (t) = a0 + å ( an cos nw t + bn sin nw t )

(8.1)

n =1

From (8.1), T

T

¥ T

0

0

n =1 0

ò f (t )dt = a0 ò dt + å ò (an cos nw t + bn sin nw t )dt = a0T

ìï t0 + T íQ ò sin mw tdt = 0 for all m; and îï t0 \ a0 =

t0 + T

üï

t0

þï

ò cos nw tdt = 0 for all n;ý

1 T f (t )dt T ò0

This shows hat a0 is the average value of f(t) over a period; therefore, called dc value of the signal. Now from equation (8.1), T

T

¥ T

0

0

n =1 0

ò f (t ) cos kw tdt = ò a0 cos kw tdt + å ò (an cos kw t cos nw t + bn cos kw t sin nw t ) dt = 0 + ak

T +0 2

ìï t0 + T íQ ò sin nw t sin mw tdt = 0 for m ¹ n and ïî t0

= \

ak =

T for n = m 2

t0 +T

ò cos nwt cos mwtdt = 0 for n ¹ m

t0

=

}

T for n = m 2

2T f (t ) cos kw tdt T ò0

Again from equation (8.1), T

T

¥ T

0

0

n =1 0

ò f (t )sin kw tdt = ò a0 sin kw tdt + å ò (an sin kw t cos nw t + bn sin kw t sin nw t ) dt = 0 + 0 + bk

\

bk =

T 2

2T f (t ) sin kw tdt T ò0

8.4

Circuit Theory and Networks

Example 8.1

For the periodic waveform shown in the figure, find the Fourier series expansion.

Figure 8.2 Periodic waveform of Example (8.1)

Solution

Here, v(t) = V, for 0 < t < T/2 = 0, for T/2 < t < T T /2

a0 =

1T 1 V v (t )dt = Vdt = T ò0 T ò0 2

an =

2T 2 2p v (t ) cos nw tdt = ò V cos æ n ö dt = 0 è T ø T ò0 T 0

bn =

2T 2 2p v (t ) sin nw tdt = V sin æ n ö dt è T ø T ò0 T ò0

T /2

T /2

and,

=

V (1 - cos np ); n = ± 1, ± 2, ± 3,¼ np

= 0; for even n

V ; for odd n np So, the Fourier series of the square wave is given as, =

1 2 2 2 v(t) = V éê + sin w t + sin 3w t + sin 5w t + ¼ùú 3p 5p ë2 p û

Exponential Form of Fourier Series We have the trigonometric Fourier series, ¥

f(t) = a0 + å ( an cos nw t + bn sin nw t ) n =1

We know that, sin nw t =

e jnwt + e - jnwt e jnwt - e - jnwt and cos nwt = 2 2j

Thus, ¥ é (e jnwt + e - jnwt ) (e jnwt - e - jnwt ) ù + bn f (t) = a0 + å êan ú 2 2j n =1 ë û ¥

1 n =1 2

= a0 + å

éæ ù bn ö jnwt æ b ö + ç an - n ÷ e - jnwt ú êèç an + j ø÷ e j è ø ë û

¥ é a - jb æ n ö jnw t æ an + jbn ö - jnw t ù = a0 + å êç n e +ç ÷ø e ú 2 2 ÷ø è è n =1 ë û

Fourier Series and Fourier Transform

8.5

æ a - jbn ö æ a + jbn ö and Cn* (or C-n ) = ç n C0 = a0, Cn = ç n 2 ø÷ 2 ø÷ è è

Let,

Thus the series becomes, ¥

f(t) = C0 + å [Cn e jnwt + C- n e - jnwt ] n =1

¥

or

f (t ) = C0 + å Cn e jnwt This is the exponential form of the Fourier series. n =-¥

Cn =

Now,

=

Thus, Cn =

1T f (t ) (cos n w t - j sin n w t ) dt T ò0

1T f (t )e - jnwt dt This equation is valid for both positive, negative and zero values of n. T ò0

Example 8.2 Solution

ù an - jbn 1 é 2 T 2T = ê ò f (t ) cos nw tdt - j ò f (t ) sin nw tdt ú T0 2 2 ëT 0 û

For the square wave shown in Example 8.1, find the exponential Fourier series. f (t) = v(t) = V, for 0 < t < T/2 = 0, for T/2 < t < T T /2

So,

Cn =

1T 1 f (t )e - jnwt dt = Ve - jnwt dt T ò0 T ò0 T /2

For n = 0, C0 = For n ¹ 0 Cn =

1 V Vdt = 2 T ò0 T /2 jV - jnp 1 V 1 [e - jnw T / 2 - 1] = [e Ve - jnwt dt = - 1] 2p n T ò0 T - jnw

(since wT = 2p) or Cn = 0 for even n jV for odd n pn Thus, the exponential Fourier series becomes,

= -

v(t) = ... + -

jV j 5wt jV j 3wt jV jwt V jV - jwt e e e + e + + 5p 3p 2 p p

jV - j 3wt jV - j 5wt e e - ... for odd n 3p 5p

8.6

Circuit Theory and Networks

Amplitude and Phase Spectrum From the trigonometric Fourier series, ¥

f (t) = a0 + å ( an cos nw t + bn sin nw t ) n =1 ¥

= A0 + å An cos ( nw t - fn ) n =1

æb ö an2 + bn2 ; fn = tan -1 ç n ÷ è an ø Also, for exponential form, Cn is complex and we may write it as, where, A0 = a0, An =

An æb ö and fn = tan - 1 ç n ÷ 2 è an ø th The quantities An and fn are called the amplitude and the phase of the n harmonic, respectively. l Variation of An with n (or nw) is known as the amplitude spectrum or Frequency – spectrum. l Variation of fn with n (or nw) is known as the phase- spectrum of the signal. As both An and fn occurs at discrete values of the frequency, i.e., n = 1, 2, 3, etc. these spectra are called Line spectra.

Cn = |Cn| e jfn and |Cn| =

1 2

an2 + bn2 =

An ; there is a scale factor of ½ for the amplitude spectrum for exponential form for 2 the Fourier series compared to the trigonometric form for all lines except the one for n = 0. Also, in the case of exponential form spectral lines are drawn for both for positive and negative values of n.

Since |Cn | =

Example 8.3 Solution

For the square wave shown in Example 8.1, draw the amplitude and phase spectra. From the results of Example 8.1, we have, 1 2 2 2 v(t) = V éê + sin w t + sin 3w t + sin 5w t + ¼ùú 3p 5p ë2 p û 2V 2V V Ð90° ; V2 = 0; V3 = Ð90° [since the cosine Magnitudes: V0 = Ð0°; V1 = 3p 2 p æb ö components are all zero, the phase angle will be tan -1 ç n ÷ = tan -1 ( ¥) = 90° ] è 0ø So, the line spectra become,

(a) Amplitude Spectrum

(b) Phase Spectrum

Figure 8.3 Amplitude and phase spectra of Example 8.3

8.7

Fourier Series and Fourier Transform

Significance for Line Spectra: The amplitude- spectrum renders valuable information as to where to truncate the infinite series and yet maintain a good approximation to the original waveform.

Effective Value of a Periodic Function The effective (or R.M.S.) value of a periodic function f (t) is defined as, Feff (Frms) =

1T [ f (t )]2 dt = T 0ò

=

1 T

¥ é 2 ù 2T ê A0 T + å An 2 ú n =1 ëê ûú

Feff ( Frms ) =

A02

æ A ö + åç n ÷ n =1 è 2 ø ¥

2

¥ ù 1Té A0 + å An cos (nw t - fn ) ú dt ê ò T 0ë n =1 û

2

This shows that the effective value of a periodic function is the square root of the effective values of the harmonic components and the square of the d. c. value.

Waveform Symmetry There are few methods by which the evaluation of Fourier co-efficients is simplified by symmetry consideration. These methods reduce the amount of labour involved in finding out the co-efficients. T /2 ù 1T 1 é 0 f (t )dt = ê ò f (t )dt + ò f (t )dt ú ò T 0 T ê- T / 2 úû 0 ë Putting t = –x in the first integrand and t = x in the second integrand, we get

Now,

a0 =

a0 =

Now,

an =

T /2 ù 1 é [ f ( x ) + f ( - x )] dx ú ê T ë ò0 û

T /2 0 ù 2T 2é f (t ) cos nw tdt = ê ò f (t ) cos nw tdt + ò f (t ) cos nw tdt ú ò T 0 Tê 0 úû -T / 2 ë

2 [I + I ] T 1 2 Since the variable t in I1 and I2 integrals is dummy variable, let x = t in I1 and x = – t in I2. =

\

an =

Thus,

an =

T /2 T /2 ù 2 é f ( x ) cos nw xdx - ò f ( - x ) cos nw x ( - dx ) ú ê ò T ë 0 0 û T /2

2 [ f ( x ) + f ( - x )]cos nw xdx T ò0

8.8

Similarly,

Circuit Theory and Networks

bn =

T /2

2 [ f ( x ) - f ( - x )] sin nw xdx T ò0

Following symmetries are considered: 1. Odd or Rotation Symmetry, 2. Even or Mirror Symmetry, 3. Half-Wave or, Alternation Symmetry, and 4. Quarter-Wave Symmetry. 1. Odd Symmetry A function f (x) is said to be odd if, f (x) = – f (–x)

Figure 8.4 Odd function T /2

Hence, for odd functions a0 = 0 and an = 0 and bn =

1 f ( x) sin nw x dx T 0ò

Thus, the Fourier series expansion of an odd function contains only the sine terms, the constant and the cosine terms being zero. 2. Even Symmetry A function f (x) is said to be even, if f (x) = f (–x) T /2

\

a0 =

2 f ( x )dx T ò0

\

an =

4 f ( x ) cos nw xdx T ò0

T /2

Figure 8.5 Even function

and bn = 0 Thus, the Fourier series expansion of an even periodic function contains only the cosine terms plus a constant, all sine terms being zero. 3. Half –Wave or Alternation Symmetry A periodic function f (t) is said to have half wave symmetry if it satisfies the condition, f (t) = – f (t ± T/2), where T – time period of the function

8.9

Fourier Series and Fourier Transform

\

a0 =

T /2 ù 1 1 é 0 ê ò f (t )dt + ò f (t )dt ú = [ I1 + I 2 ] T ê- T / 2 T 0 ë ûú

x T /2 0 t 0 -T /2

For I1, let x = (t + T/2); so, f (t) = f (x – T/2) = – f (x) and dt = dx \

I1 =

0

T /2

T /2

-T / 2

0

0

ò f (t )dt = ò - f ( x )dx = - ò f ( x )dx

\

a0 =

T /2 T /2 T /2 ù 1 éT / 2 ù 1é - ò f ( x )dx + ò f (t )dt = ú = ê ò f ( x )dx - ò f ( x )dx ú = 0 ê Të 0 0 0 û Të 0 û

\

an =

T /2 ù 2 2é ê ò f (t ) cos nw tdt ú = T ê- T / 2 úû T ë

T /2 é 0 ù 2 ê ò f (t ) cos nw tdt + ò f (t ) cos nw tdt ú = [ I1 + I 2 ] êë- T / 2 úû T 0

Again putting x = (t + T/2) and following the same procedure, I1 =

0

T /2

T /2

-T / 2

0

0

ò f (t ) cos nwtdt = ò - f ( x ) cos nw ( x - T /2) dx = ò - f ( x ) cos(nw x - np ) dx

T /2

T /2

0

0

= ò - f ( x ) cos np cos nw xdx = ò - f (t ) cos np cos nw tdt T /2

an =

2 (1 - cos np ) ò f (t ) cos nw tdt T 0

= 0; for even n, and T /2

= Similarly,

4 f (t ) cos nw tdt , for odd n. T ò0

bn = 0, for even n; and T /2

=

4 f (t )sin nw tdt , for odd n. T ò0

Thus, the Fourier series expansion of a periodic function having half-wave symmetry contains only odd harmonics, the constant term being zero. 4. Quarter–Wave Symmetry The symmetry may be regarded as a combination of first three kinds of symmetry provided that the origin is properly chosen.

8.10

Circuit Theory and Networks

Figure 8.6(a)

Sin w t: combination of half-wave and odd symmtery

Figure 8.6(b)

Cos w t: combination of half-wave and even symmetry

For Figure 8.6(a), the wave has alternation and odd symmetry; thus the Fourier series consists of odd sine terms only. T /4

8 f (t ) sin nw t dt , n being odd only. T 0ò For Figure 8.6(b), the origin, having chosen one quarter cycle away, as in Figure 8.6(a), the wave has alternation and even symmetry; thus the Fourier series consists of odd cosine terms only. \

a0 = 0, an = 0 and bn =

T /4

\

a0 = 0; bn = 0; and an =

8 f (t ) cos nw tdt , n being odd only. T ò0

Note: (i) The sum or product of two or more even functions is an even function, and with the addition of a constant, the even nature of the function is still preserved. (ii) The sum of two or more odd functions is an odd function, but the addition of a constant removes the odd nature of the function. The product of two odd functions is an even function.

8.4.2 Truncating Fourier Series When a periodic function is represented by a Fourier series, the series is truncated after a finite number of terms. So, the periodic function is approximated by a trigonometric series of (2N + 1) terms as, N

SN (t) = a0 + å ( an cos nw t + bn sin nw t )

(8.2)

n =1

such that the co-efficients a0, an and bn are chosen to give the least mean square error. The truncation error is, eN (t) = f (t) – SN (t) So, the mean square error/figure of merit/the cost criterion for optimal minimal error is, E N = eN2 (t ) =

1T [e (t )]2 dt T ò0 N

(8.3)

(8.4)

where, EN is a function of a0 , an and bn, but not of t. Example 8.4

Show that the mean square error is a minimum if the co- efficients in the approximated trigonometric series SN (t) are the Fourier co- efficients.

Fourier Series and Fourier Transform

Solution

8.11

In order to make ‘EN’ minimum, the necessary conditions are, ¶E N = 0, for n = 0, 1, 2, … (8.5a) ¶an ¶E N = 0, for n = 0, 1, 2, … (8.5b) and ¶bn These two equations give (2N + 1) equations from which (N + 1) number of an for n = 1, 2, …, N and N number of bn for n = 1, 2, …, N can be determined. From Equations 8.4 and 8.5 ¶E N 2 T ¶e (t ) 2T = ò eN (t ) N dt = ò [ f (t ) - S N (t )] cos nw tdt = 0 T 0 ¶an T 0 ¶an T

or

T

ò f (t ) cos nw tdt = ò S N (t ) cos nw tdt 0

0

N é ù = ò êa0 + å ( an cos nw t + bn sin nw t ) ú cos nw tdt n =1 ê 0ë ûú T

T T = ò an cos2 nw tdt = an 2 0

2T f (t ) cos nw tdt (n = 0, 1, 2, …, N ) T ò0 Similarly, from equation 8.5(b), we get,

or

an =

2T f (t ) sin nw tdt (n = 0, 1, 2, …, N ) T ò0 Therefore, it is proved that a Fourier series with a finite number of terms represents the best approximation for a given periodic function by any trigonometric series with the same number of terms. However, there is no analytical method for the evaluation of estimation of error due to truncation of infinite series; i.e., we can not predict the number of minimum terms to be retained in the series within a prescribed accuracy. The minimisation of error is done by trial and error method, using more terms until specifications are met.

\ bn =

Example 8.5

If f (t) is approximated by

8 sin w t , i.e., the first term in the Fourier Series, find p2

the mean square error.

Figure 8.7 Waveform of Example 8.5

8.12

Solution

Circuit Theory and Networks

Truncation Error, eN = f (t ) -

8 sin w t p2 2

T /4

Mean Square Error, EN = eN2 =

1T 2 4 é f (t ) - 8 sin w t ù dt e (t )dt = úû T ò0 N T ò0 êë p2 (from symmetry consideration) T /4

2

4 é 4t - 8 sin w t ù dt úû T ò0 êë T p 2 = 0.0047

=

8.5

STEADY- STATE RESPONSE OF NETWORK TO PERIODIC SIGNALS

The voltage (periodic) is, ¥

v(t) = A0 + å An cos (nw t - fn ) n =1

We want to find out the steady state current, i(t). Phasors corresponding to terms in right hand side are, V0 = A0 e j 0 and Vn = An e - jfn Let, Z (jw) = Impedance phasor of the network at any frequency w. So, the current phasors are, I0 =

V0 A e j0 = 0 = |I 0| e j 0 Z( j 0) Z ( j 0)

Vn A e - jfn = 0 = |I n| e - ja n Z( jw ) Z ( jw ) By superposition principle, the net current phasor is, i(t) = I0 + I1 + I2 + … So, transforming from frequency domain to time domain,

In =

¥

i(t) = I 0 + å |I n| cos ( nw t - a n ) n =1

8.5.1 Average Power Calculation ¥

v(t) = V0 + å Vn cos ( nw t - fn ) n =1 ¥

i(t) = I 0 + å I n cos ( nw t - a n ) n =1

Here,

V 0 = DC voltage component V n = the amplitude of the nth harmonic voltage fn = the phase angle of the nth harmonic voltage

Fourier Series and Fourier Transform

8.13

I0 = DC current component In = the amplitude of the nth harmonic current an = the phase angle of the nth harmonic current Instantaneous power, P(t) = v(t) i(t) Average power, Pav =

¥ ¥ öæ öù 1T 1 T éæ v (t )i (t ) = ò êçV0 + å Vn cos ( nwt - fn )÷ ç I 0 + å I n cos ( nw t - a n )÷ ú dt ò T 0 T 0 êè øè ø úû n =1 n =1 ë ¥ T

or

Pav = V0 I 0 + å ò Vn I n cos (nw t - fn ) cos (nw t - a n ) dt n =1 0 ¥

or

Vn I n cos (fn - a n ) n =1 2

Pav = V0 I 0 + å

8.5.2 Steady State Current in Series Circuits It is known that when a sinusoidal voltage is applied to a single phase series circuit, the resulting current will also be a sinusoidal. But if an alternating voltage containing various harmonics is applied to such a circuit, each harmonic voltage will produce a component current independent of the others and the resulting current will be the phasor sum of all the harmonic currents. The wave-shape of the current may altogether be different from the wave-shape of the applied voltage. We consider the following four series circuits: 1. Purely Resistive Circuit We consider a voltage as given below be applied to a pure resistor R. v = V1m sin wt + V2 m sin 2wt + V3m sin 3wt +...

Since the impedance offered by different harmonics is constant and equal to R, the resulting current will be as given. V V v V i = = 1m sin wt + 2 m sin 2wt + 3m sin 3wt +... R R R R Therefore, the waveform of the current and voltage will be the same and the percentage harmonics in the current wave is the same as that in the voltage wave. 2. Purely Inductive Circuit We consider a voltage as given below be applied to a pure inductor L. v = V1m sin wt + V2 m sin 2wt + V3m sin 3wt +...

The inductance reactances for different harmonics are as given. X L = wL; for fundamental = 2wL; for second harmonic = 3wL; for third harmonic,and so on.

Hence, the current waveform is obtained by the principle of superposition considering the different harmonic components.

8.14

Circuit Theory and Networks

V1m V V sin wt - 90 0 + 2m sin 2wt - 90 0 + 3m sin 3wt - 90 0 +... wL 2wL 3wL From v and i, it is seen that the percentage harmonics in the current wave is less than that in the 1 voltage wave. For nth harmonic, the percentage harmonic in the current wave is -times than in the n voltage wave. Respective RMS values of the voltage and current are given as,

d

i=

\

i

d

V RMS =

1 V1m 2 + V2 m 2 + V3m 2 +... 2

I RMS =

1 2

2

2

FG V wCIJ + FG V IJ + FG V IJ H wL K H 2wL K H 3wL K 1m

2m

3m

i

2

+... =

d

i

1 V 2 V 2 V1m 2 + 2 m 4 + 3m 9 +... 2

From the above discussion, we conclude that, when a complex voltage wave is applied to a pure inductor, the current wave has lesser harmonics than the applied voltage wave and thus, the current waveform will be smoother than the voltage wave. 3. Purely Capacitive Circuit We consider a voltage as given below be applied to a pure capacitor C. v = V1m sin wt + V2 m sin 2wt + V3m sin 3wt +... The capacitance reactances for different harmonics are as given.

1 ; for fundamental wC 1 ; for second harmonic = 2wC 1 = ; for third harmonic,and so on. 3wC

XC =

Hence, the current waveform is obtained by the principle of superposition considering the different harmonic components. \

b g d

i

b g d

i

b g d

i

i = V1m wC sin wt + 90 0 + V2 m 2wC sin 2wt + 90 0 + V3m 3wC sin 3w t + 90 0 +....

From v and i, it is seen that the percentage harmonics in the current wave is more than that in the voltage wave. For nth harmonic, the percentage harmonic in the current wave is n times than in the voltage wave. Respective RMS values of the voltage and current are given as, 1 V1m 2 + V2 m 2 + V3m 2 +... 2 2 2 1 = V1mwC + 2V2 mwC + 3V3mwC 2

V RMS = I RMS

b

g b

g b

g

2

+... =

1 V1m 2 + 4V2 m 2 + 9V3m 2 +... 2

From the above discussion, we conclude that, when a complex voltage wave is applied to a pure capacitor, the current wave has more harmonics than the applied voltage wave and thus, the current

8.15

Fourier Series and Fourier Transform

waveform will be more distorted than the voltage wave. 4. General RLC Series Circuit We consider a voltage as given below be applied to a general RLC series circuit. v = V1m sin wt + V2 m sin 2wt + V3m sin 3wt +... The impedance for different harmonics are as given. 2

F 1 I ; for fundamental H wC K 1 I + F 2wL H 2wC K ; for second harmonic 1 I + F 3wL H 3wC K ; for third harmonic, and so on.

Z 1 = R 2 + wL -

2

Z2 = R 2

2

Z3 = R 2

Hence, the current waveform is obtained by the principle of superposition considering the different harmonic components. \

i=

where,

V1m V V sin w t - f 1 + 2 m sin 2wt - f 2 + 3m sin 3wt - f 3 +.... Z1 Z2 Z3

f n = tan -1

b

g

b

g

b

g

LM 1 F nw L - 1 I OP . N R H nwC K Q

8.5.3 Steps for Application of Fourier Series to Circuit Analysis 1. Fourier series of the given periodic excitation function is obtained. 2. The circuit elements are transformed from time domain to frequency domain (i.e., R ® R, L 1 for nth harmonic). jw nC 3. The Fourier series of the DC and AC components of the response are calculated. 4. Using Superposition, the Fourier series of the response is obtained by summing up the individual DC and AC response components.

® jw nL, C ®

8.5.4 Power Spectrum It is the distribution of the average power over the different frequency components. Let, Pn be the average power for the nth harmonic component. Note: Pn is always positive so that only a magnitude spectrum is possible. Another form of line spectrum for power is also possible [Fig. 8.8(b)]; obtained by assuming half of Pn to the positive frequency nw and half to the negative frequency.

8.16

Circuit Theory and Networks

Figure 8.8(a) Power Spectrum for positive w

Figure 8.8(b)

Power Spectrum for both positive and negative w

PART II: FOURIER TRANSFORM 8.6

INTRODUCTION

The Fourier series representation of a period function describes the function in the frequency domain in terms of amplitude and phase spectra. The Fourier transform extends this frequency domain description to functions that are not periodic. Fourier transform is a powerful tool in the study of power spectra, correlation functions, noise and other advanced problems.

8.7

DEFINITION OF FOURIER TRANSFORM

The Fourier Transform or the Fourier integral of a function f (t) is denoted by F ( jw) and is defined by, ¥

F ( jw) = F [ f (t)] = ò f (t )e - jwt dt

(8.6)

-¥

and the inverse Fourier transform is defined by, f (t) = F

–1

[F ( jw)] =

¥ 1 ¥ jw t j 2p f ( ) F j e d df w w = ò F ( j 2p f ) e 2p -ò¥ -¥

(8.7)

Equations 8.6 and 8.7 form the Fourier transform pair. Explanation Consider the exponential Fourier Series, ¥

f (t) = å Cn e jnwt -¥

(8.8)

Fourier Series and Fourier Transform

8.17

T /2

where,

Cn =

1 f (t )e - jnwt dt T - Tò / 2

(8.9)

If the period T becomes infinite, the function does not repeat itself and becomes aperiodic or nonperiodic. So, the interval between adjacent harmonic frequencies is, Dw = (n + 1) – nw = w =

2p T

1 w Dw = = (8.10) T 2p 2T As T ® ¥, Dw ® dw and the frequency goes from a discrete variable over to a continuous variable. or

1 dw ® T 2p From 8.7 and 8.11,

and nw ® w

(8.11)

¥

CnT ® ò f (t )e - jwt dt . This is the Fourier Transform of f (t) i.e., F ( jw). -¥

¥

F ( jw ) = F [ f (t )] = ò f (t )e - jwt dt -¥

So, from equation (8.8), ¥ 1 f (t) = å (CnT )e jnwt æ ö è ø T -¥

As T ® ¥, CnT ® F ( jw), nw ® w and

(8.12)

1 dw ® and å ® ò (summation approaches integration). T 2p

Thus, from (8.12), f (t ) =

1 ¥ f ( jw )e jwt d w 2p -ò¥

Spectra Let, F( jw) = |F( jw)|e jf (w) The variation of |F( jw)| with w is referred to as the amplitude spectrum. The variation of f (w) with w is referred to as the phase–spectrum. Since F( jw) is a continuous function, the corresponding amplitude and phase spectra are continuous spectra.

8.8

CONVERGENCE OF FOURIER TRANSFORM

When f (t) is a single-valued function and is different from zero over an infinite interval of time, the behavior of f (t) as t ® ± ¥ determines the convergence of the Fourier transform.

8.18

Circuit Theory and Networks

The Fourier transform will exist, if ¥

ò | f (t )| dt < ¥

-¥

8.9

FOURIER TRANSFORM OF SOME FUNCTIONS

1. f(t) = Ae–at u(t), a > 0 Fourier transform will exist, if a > 0 \

¥ ¥ e - ( a + jw ) t F ( jw) = F [ f (t )] = ò f (t ) e - jwt dt = A ò e - at e - jwt dt = A - ( a + jw ) 0 -¥

Amplitude, |F ( jw)| =

Phase,

¥

= 0

A a + jw

A a + w2 2

w f ( jw) = - tan -1 æ ö èaø

2. f(t) = Ke–a|t|, for all values of t ¥

0

¥

-¥

-¥

0

F ( jw) = F [ Ke - a|t |] = ò Ke - a t e - jw t dt = ò Ke( a - jw )t dt + ò Ke - ( a + jw )t dt =

K K 2 Ka + = a - jw a + jw a 2 + w 2

Thus the Fourier transform of the double exponential function has zero phase for all values of w and the magnitude spectrum is shown in Fig. 8.9. | F ( jw ) f(t)

2K a K

0

0

Time, t

(a) Double exponential function

w

(b) Fourier transform

Figure 8.9 Double exponential function and its Fourier transform

NB: There are some important functions which do not have Fourier transforms in a strict sense; ¥

because they do not satisfy the Dirichlet’s condition, i.e.,

z f (t ) dt in infinite (such as, the step

-¥

8.19

Fourier Series and Fourier Transform

function and sinusoidal function). However, the Fourier transform of these functions are evaluated by approximating these functions in time domain as the limiting value of another function which possesses Fourier transform. 3. Fourier transform of some constant, K; for all values of t Here, we can approximate the constant as,

f (t ) = Lt Ke - a t 0

a

F K = Lt

\

a

\

¥ 0

z Ke

- a t - jwt

e

dt = Lt a

-¥

0

2 Ka a2 + w 2

F K = 0; for w ¹ 0 = ¥; for w = 0

[by L Hospital’s rule, i.e., differentiating both numerator and denominator with respect to ‘a’] Thus, F [K] is an impulse function at w = 0. The strength (amplitude) of the impulse function is obtained as, ¥

¥

-¥

-¥

z F K dw = z a 2+Kaw dw = 2pK 2

2

af

\

F K = 2pKd w

Hence, Fourier transform of a constant K is an impulse of magnitude 2pK as shown in Fig. 8.10. | F ( j w )|

f(t)

2pKd(w)

K

0

0

Time, t

w

(b) Magnitude spectrum of constant K

(a) Constant K

Figure 8.10 Constant K and its magnitude

4. Unit impulse function or Dirac Delta Function, d (t) Some problems involve the concept of an impulse, which may be intuitively thought of as a force of very large magnitude impacting just for an instant. \

¥

z

F d (t ) = d (t )e - jwt dt -¥

We use shifting property of impulse function as explained below. The product of any arbitrary function f (t) with unit impulse function d (t) provides the function f (t) to exist only at t = 0.

8.20

Circuit Theory and Networks

Mathematically, ¥

z f bt gd (t )dt = f bt g

-¥

t =0

This shifting property can also be applied at any instant of time, say t = t0, so that we can write, ¥

z f bt gd (t - t )dt = f bt g 0

-¥

t = t0

bg

= f t0

Using this property, we have the Fourier transform of unit impulse function as, ¥

z

F d (t ) = d (t )e - jwt dt = e 0 = 1

\

-¥

Thus, Fourier transform of an impulse function is unity as shown in Fig. 8.11. f(t) | F ( jw )|

d( t )

1 0

0

Time, t (a) Impulse function

Time, t

(b) Fourier transform of impulse function

Figure 8.11 Impulse function and its Fourier transform f (t )

5. Fourier transform of Signum Function, Sgn(t) A signum function is defined as, Sgn(t) = +1 =0 = –1 \

+1

for t > 0 for t = 0 for t < 0

0

Time, t –1

¥

z Sgn(t )dt is infinite, direct evaluation of Fourier transform

Figure 8.12(a) Sgn (t)

-¥

is not possible. Therefore, the given function has to be expressed as limiting case of some other function and then the Fourier Transform is computed. Let, the Sgn(t) be multiplied by e - a t

F Sgn(t ) = Lim a

= Lim a

or,

F Sgn(t ) =

¥ 0

0

z

LM z ea N

e - a t Sgn(t )e - jwt dt = Lim -

-¥

a

0

0

-¥

and a ® 0.

f dt + ¥ e -aa + jw ft dt OP

a - jw t

z

0

LM -1 + 1 OP N a - jw a + j w Q

2 jw

Figure 8.12 shows the magnitude and phase spectrum of the Signum function.

Q

8.21

Fourier Series and Fourier Transform –F ( j w )

| F ( jw )| 2 jw

p 2 w –

0

p 2

w

(b) Magnitude spectrum of sgn( t )

(c) Phase spectrum of sgn( t )

Figure 8.12 Signum function and its magnitude spectrum

6. Fourier transform of Unit Step Function, u(t) u(t) = 1 =0

for t > 0 for t < 0

¥

Since

z u(t )dt is infinite, direct evaluation of Fourier transform is imposable.

-¥

Let, \ or,

u( t ) =

1 1 + Sgn(t ) 2 2

LM 1 OP + F LM 1 Sgn(t )OP = 2p ´ 1 d (w ) + 1 ´ 2 2 2 jw N2Q N2 Q 1 F uat f = pd aw f + jw

F u( t ) = F

Thus, the amplitude of unit step function u(t) in frequency domain will be a combination of rectangular hyperbola and impulse function (of strength p at w = 0) as shown in Fig. 8.13. | F ( j w)| 1 jw p

0

w

Figure 8.13 Magnitude spectrum of unit step function

8.10

PROPERTIES OF FOURIER TRANSFORMS

1. Linearity If a, b, Î C, then F {a f (t ) + b g (t )} = a F { f (t )} + b F {g (t )} = a F (w ) + bG (w )

8.22

Circuit Theory and Networks

provided the Fourier transforms of f (t) and g (t) exist. 2. Scaling If F{f(t)} = F(w) and c Î R, then F {cf (t )} =

1 æwö F |c| è c ø

3. Time shifting If F { f (t )} = F (w ) and t0 Î R , then F { f (t - t0 )} = e - jw t0 F (w ) 4. Frequency shifting

If F { f (t )} = F (w ) and w Î R , then F (w - w 0 ) = F {e jw0 f (t )} 5. Symmetry

If F { f (t )} = F (w ) , then F {F (t )} = 2p f ( - w ) 6. Modulation If F { f (t )} = F (w ) and w 0 Î R , then

F { f (t ) cos (w0t )} = F { f (t )sin(w0t )} =

1 [ F (w + w0) + F (w - w0)] 2

1 [ F (w + w0) - F (w - w0)] 2

7. Differentiation in time Let n Î N and suppose that f (n) is piecewise continuous. Assume that Lim f ( k ) (t ) = 0 , then x ®¥

F { f ( n ) (t )} = ( jw )n F (w )

In particular F { f ¢(t )} = jw F (w )

and F { f ² (t )} = - w 2 F (w ) 8. Frequency differentiation Let n Î N and suppose that f is piecewise continuous. Then F {t n f (t )} = j n F ( n ) (w )

In particular F {tf (t )} = jF ¢(w )

and F {t 2 f (t )} = - F ² (w )

8.23

Fourier Series and Fourier Transform

These properties can be tabulated as follows (Table 8.1). Table 8.1 Properties of Fourier Transforms Sl No.

Time Domain f (t ) =

1 2p

¥

ò

Frequency Domain F ( jw ) =

F ( jw )e jw t dt

-¥

¥

ò

f (t )e - jw t dt

-¥

*

1 2 3

f (t) real f(t) even, f(t) = f(– t) f (t) odd, f (t) = – f (– t)

F ( jw) = F (– jw) F ( jw) = F(– jw), F( jw) is real F ( jw) = – F (– jw), F( jw) is imaginary,

4

y (t) = tn f (t)

Y ( jw) = ( j )n

5

y(t) = f (at)

F ( jw) =

6

y (t ) = f (t - t0)

7

y(t) =

8

y(t) =

Y( jw) = ( jw)n F( jw)

¥

Y( jw) =

f (t )dt

-¥

9

y(t) = f (t )e jw0t

Example 8.6

Solution

1 æ jw ö F ,a>0 a çè a ÷ø

Y ( jw ) = e - jw t0 F ( jw )

d n f (t ) dt n

ò

d ² F ( jw ) dw ²

F ( jw ) jw

Y( jw) = F [ j(w – w0)]

Show that when f (t) is an even function of t, its Fourier transform F ( jw) is a function of w and is real; while when f(t) is an odd function of t, its Fourier transform F ( jw) is an odd function of w and is imaginary. From the definition, ¥

¥

-¥

-¥

F ( jw) = ò f (t )e - jwt dt = ò f (t )(cos w t - j sin w t )dt ¥

¥

-¥

-¥

= ò f (t ) cos w tdt - j ò f (t ) sin w tdt = P(w ) + jQ (w ) ¥

where, P(w) = ò f (t ) cos w tdt = Even function of w , i.e., P (w ) = P (- w ) -¥ ¥

and

Q(w) = ò f (t ) sin w tdt = Odd function of w , i.e., Q (w ) = - Q (- w ) -¥

Now, F ( jw) = | F ( jw )| e jf (w ) |F ( jw)| =

P 2 (w ) + Q 2 (w ) = Even function of w

8.24

Circuit Theory and Networks

and l

é Q (w ) ù F ( jw) = tan -1 ê = Odd function of w ë P (w ) úû When f(t) is an even function f(t) cos w t is an even function f(t) sin w t is odd function. \

¥

P(w) = 2 ò f (t ) cos w tdt 0

l

Q(w) = 0 so, F( jw) = P(w) = Even and Real (Proved) When f(t) is an odd function f(t) cos wt is an odd function f(t) sin wt is an even function \ P(w) = 0 ¥

and \ Q(w) = - 2 ò f (t ) sin w tdt 0

so,

8.11

F( jw) = jQ(w) = Odd and Imaginary (Proved)

ENERGY DENSITY AND PARSEVAL’S THEOREM

This theorem states that the energy content (W ) of a waveform (periodic or non-periodic) over the whole frequency band is, ¥ 1 ¥ |F ( jw )|2 d w W = ò f 2 (t )dt = 2p -ò¥ -¥

Proof: We have, ¥

¥

-¥

-¥

W = ò f 2 (t )dt = ò f (t ) × [ f (t )dt ] ¥ é 1 ¥ ù = ò f (t ) ê F ( jw )e jwt dt ú dt ò 2p -¥ ëê - ¥ ûú

or

=

é¥ ù 1 ¥ F ( jw ) ê ò f (t )e jwt dt ú dt ò 2p - ¥ êë - ¥ úû

=

1 ¥ F ( jw ) × F (- jw ) d w 2p -ò¥

=

1 ¥ |F ( jw )|2 d w 2p -ò¥

¥ 1 ¥ 2 W = ò f 2 (t ) dt = ò |F ( jw )| d w p 2 -¥ -¥

Proved

8.25

Fourier Series and Fourier Transform

Note (i) Since |F( jw)| is an even function of w, ¥ 1¥ W = ò f 2 (t )dt = ò |F ( jw )|2 d w p -¥

0

(ii) Since w = 2p f, where f is the frequency, ¥

¥

¥

-¥

-¥

0

2

W = ò f 2 (t )dt = ò |F ( j 2p f )|2 df = 2 ò F ( j 2p f ) df The quantity |F ( j 2p f )|2 df is the energy in an infinitesimal band of frequency df. It represents the energy density in the frequency domain and has unit of Joule/Hertz. Total energy content within the frequency band f1 and f2 is, f2

Wb = 2 ò F ( j 2p f )|2 df f1

For the integration range –¥ to + ¥, the total energy is, - f2

f2

- f1

f1

Wb = ò |F ( j 2p f )|2 df + ò |F ( j 2p f )|2 df (iii) If f (t) is the voltage across a 1 W resistance or current through the same resistance, then Wb is known as 1 W energy. Example 8.7 Solution

The current in a 10 W resistor is i (t ) = 10e -2t u (t ) ( A). What is the energy associated with the frequency band 0 £ w £ 2 rad/s? Here, f(t) = i(t) = 10e–2tu(t) \

F( jw) =

10 2 + jw

So, the energy associated with the given frequency band is, W=

=

w ù 10 2 10 2 100 d w 103 é 1 = | F ( jw )|2 d w = ò tan -1 æç ö÷ ú ò è 2 øû p 0 p 0 4 + w2 p ëê 2

103 p

ép ù êë 8 úû

= 125 Joule

Ans.

2

0

8.26

Circuit Theory and Networks

8.12

COMPARISON BETWEEN FOURIER TRANSFORM AND LAPLACE TRANSFORM

The defining equations are, ¥

¥

0

-¥

F(s) = ò f (t )e - st dt and F ( jw ) = ò f (t )e - jwt dt Following are some differences and similarities: 1. Laplace Transform is one-sided in the interval 0 < t < ¥ and Fourier Transform is double-sided in the interval –¥ < t < ¥. Thus, Laplace Transform is applicable for positive time function, f(t), t > 0; while Fourier Transform is applicable for functions defined for all times. 2. Laplace Transform includes the initial conditions and is applicable for transient analysis; while Fourier Transform is only applicable for steady-state analysis. ¥

3. For functions f (t) = 0 for t < 0 and ò | f (t )| dt < ¥ , the two transforms are related as, 0

F ( jw ) = F ( s )|s = jw . Thus, Laplace Transform is associated with entire s-plane, while, Fourier

Transform is restricted to the imaginary ( jw) axis. 4. Laplace Transform is applicable to a wider range of functions than the Fourier Transform. On the other hand, Fourier Transforms exist for signals that are not physically realizable and have no Laplace Transform.

8.13

STEPS FOR APPLICATION OF FOURIER TRANSFORM TO CIRCUIT ANALYSIS

By Fourier Transform, we can find the response of a circuit due to non-periodic functions. The general procedure is described below. 1. Fourier Transform of the given excitation function is obtained. 1 ö 2. Fourier Transform of the circuit elements is obtained æç i.e., R ® R, L ® jw L, C ® . j C ÷ø w è

3. The transfer function in Fourier Transform Domain is defined as, H ( jw ) =

Y ( jw ) or X ( jw )

Y ( jw ) = H ( jw ) × X ( jw ) ; where, Y( jw) is the response transform and X( jw) is the excitation transform. 4. Taking the inverse Fourier Transform of the product H ( jw ) × X ( jw ) , we get the response y(t).

Fourier Series and Fourier Transform

8.27

SOLVED PROBLEMS 8.1 Determine the Fourier series for the square waveform shown below and plot the magnitude and the phase spectra.

Solution The waveform, f (t) = V; 0 < t < T/4 = –V; T/4 < t < 3T/4 = V; 3T/4 < t < T Obviously, the given function is an even function. \ bn = 0 Now,

a0 =

2 T

T /2

f (t ) dt =

ò

0

2 T

T /4

ò

0

T /2

Vdt = -

2 Vdt = 0 T Tò/ 4

T /2

an =

4 f (t ) cos nw tdt T ò0

=

T /4 T /2 ù 4 é ê ò V cos nw tdt - ò V cos nwtdt ú T ê 0 úû T /4 ë

=

4V nwT

é æ nwT ö æ nwT ö æ nwT ö ù êsin çè 4 ÷ø - sin çè 2 ÷ø + sin çè 4 ÷ø ú ë û

=

4V n 2/ p

é np ù / çæ ÷ö ú ê2sin è 2 øû ë

=

4V np 4V sin = ; for n = 1, 5, 9,……… np 2 np

[Q wT = 2p ]

4V ; for n = 3, 7, 11,……… np = 0; for n = 2, 4, 6,………

= -

So,

f(t) =

4V æ 1 1 1 1 cos w t - cos 3w t + cos 5w t - cos 7w t + cos 9w t ¼¼¼÷ö ø 3 5 7 9 p çè

Ans.

8.28

Circuit Theory and Networks

Magnitude Spectra

Phase Spectra

8.2 Find the Fourier series of the function whose periodic waveform is shown in the below figure and plot its frequency spectra.

Solution The function is even \

bn = 0

\

a0 =

2 T ò0

\

an =

4 f (t ) cos nw tdt T 0ò

T /2

T /4

f (t )dt =

2 2V T V ´ = Vdt = T ò0 T 4 2

T /2

T /4

=

4V f (t ) cos nw tdt T ò0

=

4V T

éæ sin nw t ö T / 4 ù êç ÷ ú êëè nw ø 0 úû

[Q wT = 2p ]

8.29

Fourier Series and Fourier Transform

4V nwT

=

4V np 2V ; n = 1, 5, 9 … = .sin 2np 2 np

= \

é æ nwT ö ù êsin çè 4 ÷ø ú ë û

=

f (t ) =

2V ; n = 3, 7, 11, … np

V 2V æ 1 1 1 + cos w t - cos 3w t + cos 5w t - cos 7w t + ...÷ö ø 2 3 5 7 p çè

Ans.

Line Spectra

8.3 Find the Fourier series for the train of pulses shown in the below figure and draw the amplitude and the phase spectra.

Solution Here,

v(t) = V; for 0 < t < T/2 = 0; for

\

a0 =

T

T > fc,

V0 Vi

AF

=

2

= 0.707 AF = –3 dB, f = 45°

@ AF

9.7.2 Filter Design A high-pass active filter can be designed by implementing the following steps: 1. A value of the cut-off frequency, wc (or fc) is chosen. 2. A value of the capacitance C, usually between 0.001 and 0.1 µF, is selected. 3. The value of the resistance R is calculated using the relation, 1 1 = wcC 2p f c C 4. Finally, the values of R1 and Rf are selected depending on the desired pass-band gain, using, Rf ö æ the relation, AF = ç1 + . R1 ÷ø è

R=

Example 9.2 (a) Design a high-pass active filter of cut-off frequency 1 kHz with a pass-band gain of 2. (b) Plot the frequency response of the filter.

9.12

(a) Here, fc = 1kHz, AF = 2 Let, C = 0.01 mF. 1 1 = = 15.9 kW 3 2p f c C 2p ´ 10 ´ 0.01 ´ 10-6

\

R=

\

Rf ö æ Þ Rf = R1 = 10 kW AF = 2 = ç1 + R1 ÷ø è

So, the complete circuit is shown in Fig. 9.12(a).

Figure 9.12(a) Circuit of Example (9.2)

(b) The data for the frequency response plot can be obtained by substituting the input frequency ( f ) values from 100 Hz to 100 kHz in the equation. V0 = Vi

AF ( f / fc ) 1 + ( f / fc )2

2.0

Gain

Solution

Circuit Theory and Networks

1.414

0 1000

Frequency (Hz)

Figure 9.12(b) Filter characteristics of Example (9.2)

9.13

Filter Circuits Frequency (Hz) 100 200 400 700 1,000 3,000 7,000 10,000 30,000 100,000

9.8

Gain

Gain (in dB)

0.20 0.39 0.74 1.15 1.41 1.90 1.98 1.99 2 2

–14.02 –9.13 –2.58 1.19 3.01 5.56 5.93 5.98 6.02 6.02

BAND-PASS ACTIVE FILTER

A band-pass filter has a pass-band between two cut-off frequencies fce (lower cut-off frequency) and fcu (upper cut-off frequency) such that fcu > fcl. Any input frequency outside this pass-band is attenuated.

9.8.1 Bandwidth (BW) The range of frequency between fCL and fCU is called the bandwidth. BW = ( fCU - fCL )

The bandwidth is not exactly centered on the resonant frequency (fr). If fCU and fCL are known, the resonant frequency can be found from, fr =

fCL × fCU

If ‘fr’ and BW are known, cut-off frequencies are found from,

æ f cl = ç è

2

( ) BW 2

ö BW - f r2 ÷ ø 2

( )

f cu = ( f cl + BW)

Figure 9.13 Band pass filter characteristics

9.14

Circuit Theory and Networks

9.8.2 Quality Factor (Q) fr BW Q is a measure of the selectivity. Higher the value of Q, the more selective is the filter, i.e., narrower is the bandwidth. It is defined as the ratio of resonant frequency to bandwidth, i.e., Q =

Example 9.3 A band-pass voice filter has lower and upper cut-off frequencies of 300 and 3000 Hz, respectively. Find (a) Bandwidth, (b) The resonant frequency, (c) The quality factor. Solution

(a) BW = ( fCU – fCL ) = (3000 – 300) = 2700 Hz (b) fr =

fCL fCU =

(c) Q =

300 ´ 3000 = 950 Hz

fr 950 = = 0.35 BW 2700

[Note fr is below the centre frequency Example 9.4

Ans. Ans. Ans.

300 + 3000 = 1650 Hz] 2

A band-pass filter has a resonant frequency of 950 Hz and a bandwidth of 2700 Hz. Find its lower and upper cut-off frequencies. 2 æ ö æ BW ö + f 2 - æ BW ö ç ç ÷ ç ÷ ÷ r ÷ø è 2 ø f CL = çè è 2 ø

Solution

2 æ ö æ 2700 ö + (950)2 - æ 3700 ö = (1650 - 1350) = ç ç ÷ ç ÷ ÷ çè è 2 ø ÷ø è 2 ø

\

= 300 Hz f cu = (300 + 2700) = 3000 Hz

Ans.

9.8.3 Types of Band-Pass Filters 1. Wide Band-Pass Filter wide-band filter has a bandwidth that is two or more times the resonant frequency; i.e., Q £ 0.5. It is made by cascading a low-pass and a high-pass filter circuit. 2. Narrow Band-Pass Filter A narrow band filter has a quality factor, Q > 0.5. It is made by using a single op-amp and multiple feed back circuits.

Wide Band-Pass Active Filter In general, a wide-band filter (Q £ 0.5) is made by cascading a low-and a high-pass filter, provided the cut-off frequency of the low-pass section is greater than that for the high-pass section.

Filter Circuits

9.15

Characteristics (i) The cut-off frequency of low-pass filter should be 10 or more times the cut-off frequency of the high-pass filter. (ii) Each section should have the same pass band gain. (iii) The lower cut-off frequency, fcl, will be determined only by the high-pass filter. (iv) The higher cut-off frequency, fcu, will be determined only by the low-pass filter. (v) Gain will be maximum at the resonant frequency, fr, and equal to the pass-band gain of either filter.

Figure 9.14(a) Wide band-pass active filter circuit

Frequency Response

Figure 9.14(b) Frequency response of wide band-pass active filter circuit

Here, fCL =

1 1 , fCU = 2p R1C1 2p R2 C2

The voltage gain magnitude of the band–pass filter is equal to the product of the voltage gain magnitudes of the high-pass and the low-pass filters.

\

V0 = Vi

AFL AFH ( f / fCL ) [1 + ( f / fCL )2 ] × [1 + ( f / fCU )2 ]

9.16

Circuit Theory and Networks

Where, AFL, AFH = f= f cl = f cu =

Pass-band gain of low-pass and high-pass filter; frequency of input signal (Hz); lower cut-off frequency (Hz); higher cut-off frequency (Hz);

(

At the centre frequency, f r = At f = fCL ,

V0 = Vi V0 = Vi

At f = fCU ,

V0 fCU = K = AFL AFH Vi fCL + fCU

AFL AFH ( fCL / fCL ) [1 + ( fCL / fCL ) 2 ][1 + ( fCL / fCU ) 2 ]

=

AFL AFH (2)[1 + ( fCL / fCU ) 2 ]

AFL AFH fCU 2

2 2 fCL + fCU

V0 A A ( f /f ) = FL FH CU CL = Vi (2)[1 + ( fCL / fCL )2 ]

At f = fCL = fCU , Þ

)

fCL fCU , the Gain is,

Gain,

V0 A A é = FL FH ê Vi 2 ê ë

AFL AFH fCU 2

2 2 fCL + fCU

fCU 2 2 fCL + fCU

ù ú ú û

V0 AFL AFH = Vi 2

Narrow Band-pass Active Filter In general, a narrow band-pass filter is made by using multiple feedback circuit with a single op-amp.

Figure 9.15 Multiple feedback narrow BP active filter

Compared to all other filters, it has some unique features, as given below. (i) It has two feedback paths, hence the name ‘multiple feedback filter’. (ii) The op-amp is used in the inverting mode. (iii) Its centre frequency can be changed without changing the gain or bandwidth.

Filter Circuits

9.17

Performance Equations Writing KCL at (1) (V1 - Vi ) V1 - V0 V1 - 0 V1 + + + =0 R 1/ sC1 1/ sC2 Rr

or

(V1 – Vi) Rr + (V1 – V0) sRrRC1 + V1sRRrC2 + V1R = 0

or

V1 =

Vi Rr + V0 sRRr C1 R + Rr + sRRr (C1 + C2 )

Again, writing KCL at (2), 0 - V0 0 - V1 + =0 Rf 1/ sC2

or

V 0 = –V1 sRf C2

or

é Vi Rr + V0 sRRr C1 ù = -ê ú sR f C2 {by the value of V1 from (9.5)} R R sRR C C ( ) + + + 1 2 û r r ë V0 [R + Rr + sRRr(C1 + C2) + s2RRrRf C1C2] = –Vi sR r Rf C2

\

sRr R f C2 V0 =- 2 Vi s RRr R f C1C2 + sRRr (C1 + C2 ) + R + Rr

So, the gain,

V0 =Vi

( s/RC1 ) æ C + C2 ö R + Rr s2 + s ç 1 + ÷ R C C RR è f 1 2ø r R f C1C2

The general transfer function is of the form, V0 =Vi

æw ö sç r ÷ è Qø s (BW) AF , where, AF = Gain =- 2 æ wr ö s + s (BW) + w r2 2 2 s + s ç ÷ + wr è Qø

æ C + C2 ö 1 (in Hz) {Qw = 2p f } So, here, BW = ç 1 ´ ÷ R C C 2 p è f 1 2ø

With matched capacitor, i.e., C1 = C2 = C

BW = Also,

1 p Rf C

(BW) AF =

Þ

Rf =

Q p fr C

1 1 = RC1 RC | with C1 = C2 = C

(9.5)

9.18

\

Circuit Theory and Networks

R=

Q Q 1 = ÞR= ( BW )CAF w r CAF 2p f r CAF

BW =

1 Hz 2p RCAF

R + Rr | with C1=C2=C RRr R f C 2

Similarly,

w r2 =

or,

4p2fr2 ´ RRr Rf C2 = R + Rr

or

4p 2 f r2 ´

or

2p f r QRr R f C =

or

2p Qf r Rr ´

or

Q Rr éë2Q 2 - AF ùû = 2p f r C

\

Rr =

Also,

Rf 2p frCAF Q = ´ = 2 AF R p fr C Q

Q Q ´ Rr R f C 2 = + Rr 2p f r CAF 2p f r CAF

[Putting the value of Rf ]

Q + Rr AF 2p f r C

Q Q ´C = + Rr AF p fr C 2p f r C

[Putting the value of Rf ]

æ ö AF Q = R ç ÷ 2p f r C (2Q 2 - AF ) è 2Q 2 - AF ø

Rf . So, the gain is a maximum of 1 at fr if Rf = 2R 2R However, the gain must satisfy the condition, AF < 2Q2. So, the narrow-band-pass active filter is designed for specific values of resonant frequency fr and Q (or, fr and BW) by using the relations.

\

AF =

R=

Q , 2p f r CAF

BW =

and

fr =

Rf =

Q , p fr C

Rr =

Q , 2p f r C (2Q 2 - AF )

fr 1 0.1591 = (Hz) = (H ) Q 2p RCAF AF RC Z

1 2 2 RC

1 AF

æ R ö 0.1125 çè1 + R ÷ø = RC r

1 AF

AF =

Rf 2R

(9.6) (9.7)

æ Rö çè1 + R ÷ø r

(9.8)

Note: The resonant frequency can be changed to a frequency fr¢ without changing the gain or BW, 2

æ f ö by, changing Rr to a new value R¢r = so that, Rr¢ = Rr ç r ÷ . è f r¢ ø

9.19

Filter Circuits

Example 9.5

(a) Design a wide band-pass filter with fCL = 200 Hz and fCU = 1kHz, and a passband gain = 4. (b) Draw the frequency response plot of this filter. (c) Calculate the value of Q for the filter.

Solution

(a) To design the low-pass section: fc = 1 kHz Let, C2 = 0.01 mF, R2 =

1 = 15.9 kW 2p f c C2

To design the high-pass section: fc = 200 Hz Let, C1 = 0.05 mF, R1 =

1 = 15.9 kW 2p f c C1

Since the band-pass gain is 4, the gain of both HP and LP sections could be set equal to 2. \

R ¢f ö æ R ¢¢ æ f ö 2 = ç1 + = ç1 + ÷ ÷ ¢ ¢¢ R R è ø è ø

Þ

R ¢f = R ¢¢f = R ¢ = R ¢¢ = 10 kW

Gain

(b) The frequency response will be as shown below.

Frequency (Hz)

Figure 9.16 Frequency response of Example (9.5)

(c) Resonant frequency, f r = So, the quality factor, Q =

200 ´ 1000 = 447.2 Hz

fr 447.2 = = 0.56 BW (1000 - 200)

9.20

Circuit Theory and Networks

Example 9.6

(a) Design a narrow band-pass filter with resonant frequency fr = 1 kHz, Q = 3, and AF = 10. (b) Change the resonant frequency to 1.5 kHz, keeping AF and the bandwidth constant.

Solution

(a) Let, C1 = C2 = 0.01 mF Rf =

Q 3 = = 95.5 kW ; p f r C p ´ 103 ´ 10-8

R=

Q 3 = = 4.77 kW 2p f r CAF 2p ´ 103 ´ 10-8 ´ 10

Rr =

4.77 ´ 10 Q = = 5.97 kW 2p f r C (2Q 2 - AF ) (2 ´ 9 - 10)

(b) To change the resonant frequency, the resistance value will be,

1 R ¢ = 5.97 ´ 103 ´ æç ö÷ = 3.98 kW è 1.5 ø

Gain

The frequency response is shown below.

Frequency

Figure 9.17 Frequency response of Example (9.6)

Example 9.7

A band-pass filter has the component values, R = 21.12 kW, Rf = 42.42 kW, Rr = 3.03 kW and C1 = C2 = 0.015 mF. Find the resonant frequency and the bandwidth.

Solution

Here, since Rf = 2R, so, AF = 1. \

fr =

0.1125 RC

1 æ Rö 0.1125 21.21 @ 1000 Hz 1+ ÷ = 1+ 3.03 AF èç Rr ø 21.21 ´ 103 ´ 0.015 ´ 10- 6

9.21

Filter Circuits

BW =

9.9

0.1591 0.1591 = @ 500 Hz AF RC 1 ´ 21.21 ´ 103 ´ 0.015 ´ 10- 6

BAND-REJECT (NOTCH) ACTIVE FILTER

1. It may be obtained by the parallel connection of a high-pass section with a low-pass section. The cut-off frequency of the high-pass section must be greater than that of the low-pass section. The outputs of HP and LP sections are fed to an adder whose output voltage V0 will have the notch filter characteristics.

Figure 9.18(a) Block diagram of BR filter

Figure 9.18(b)

Frequency response of band reject filter

The circuit of the BR filter is shown in Fig. 9.19. Obviously, the gain of the adder is set at unity; and thus,

æ V ¢ V ¢¢ö V0 = ç 0 + 0 ÷ R4 è R2 R3 ø

Þ R2 = R3 = R4

and

ROM = R2 R3 R4

So,

é j ( f / fCH ) ù é ù 1 V0 = AFH ê + AFL ê ú ú ë1 + j ( f / fCH ) û ë1 + j ( f / fCL ) û

If AFL = AFH = A, then at the center frequency, f r =

fCL fCH , the Gain is K = A ×

2 fCL fCL + fCH

9.22

Circuit Theory and Networks R1

Vy

Rf

–

C

Vx

V 0¢ R2

+

R4

R

–

Vi R3

+

+

RL

R¢ C¢

V 0¢¢

–

R 1¢

V0

R OM

R f¢

Figure 9.19 Band reject active filter circuit using parallel connection of high pass and low pass filters

2. Band-reject filter may also be obtained by using the multiple-feedback band-pass filter circuit with an adder. That is, the notch filter is made by a circuit that subtracts the output of a band pass filter from the original signal.

Figure 9.20 Band reject active filter circuit using multiple feedback band pass filter with an adder

So,

V0¢ = Vi

( s/RC1 ) = T (s) æ C1 + C2 ö R + Rr 2 s + sç + è R f C1C2 ÷ø RRr R f C1C2

Filter Circuits

9.23

Now, writing KCL at (1),

V0¢ V0 V + + i =0 R ¢ R ¢f R ¢¢ æ V ¢ V ¢ö V 0 = - R ¢f ç 1 + 0 ÷ è R ¢¢ R ¢ ø

Þ

T (s) ù é 1 + = - R ¢f Vi ê ¢¢ R R ¢ úû ë At notch frequency, the output is zero (ideally).

T(s) = - R ¢ R ¢¢ But, at wn (or fn), T(s) = –AF (AF = gain of the BP section) So,

With C1 = C2, Gain for BP section, AF = \ AF =

Rf 2R

Rf R¢ = 2 R R ¢¢

So, the design equations are all those of BP section and this one. Example 9.8

Design a notch filter having a resonant frequency, fr = 400 Hz and Q = 10. Make the resonant frequency gain, AF = 2.

Solution

Here, fr = 400 Hz, Q =10, AF = 2 Let, C = 0.1 mF

Q 10 = = 19.89 kW 2p frCAF 2p ´ 400 ´ 0.1 ´ 10- 6 ´ 2

\

R=

\

Rf =

Q 10 = = 79.58 kW p f rC p ´ 400 ´ 0.1 ´ 10- 6

\

Rr =

RAF 19.89 ´ 2 ´ 103 = = 202 W 200 - 2 2Q 2 - AF

Ans.

Let, R¢ = 1 kW (arbitrary) = R¢f R¢ R² = A = 500 W F

Ans.

9.9.1 Applications of Notch Filters Notch filter is used where unwanted frequencies are to be attenuated while permitting the other signal frequencies to pass through.

9.24

Circuit Theory and Networks

For examples, 50 Hz, 60 Hz, or 400 Hz frequencies from power lines, ripple from a full-wave rectifiers, etc. Example 9.9

Solution

Design an active notch filter to eliminate 120 Hz hum (noise). Take the bandwidth, BW = 12 Hz. 120 = 10 12 The gain of the filter in the pass-band will be maximum of 1, AF = 1. Let , C1 = C2 = 0.1 mF

Hare, fr = 120 Hz, BW = 12Hz, Q =

10 = 132.66 kW 2p ´ 120 ´ 0.1 ´ 10- 6 ´ 1 Rf = 2R = 265.32 kW R=

R Rr = 200 - 1 = 663.3 kW

Now, let R¢ = R¢f = 1 kW (arbitrary) So,

R² =

R¢ AF = 1 kW

Thus the filter will pass all frequencies from (0 – 114) Hz and 126 Hz onwards.

9.10

FILTER APPROXIMATION

In the earlier sections, we saw several examples of amplitude response curves for various filter types. These always included an “ideal” curve with a rectangular shape, indicating that the boundary between the pass-band and the stop-band was abrupt and that the roll-off slope was infinitely steep. This type of response would be ideal because it would allow us to completely separate signals at different frequencies from one another. Unfortunately, such an amplitude response curve is not physically realizable. We will have to settle for the best approximation that will still meet our requirements for a given application. Deciding on the best approximation involves making a compromise between various properties of the filter’s transfer function, such as, filter order, ultimate roll-off rate, attenuation rate near the cut-off frequency, transient response, ripples, etc. If we can define our filter requirements in terms of these parameters, we will be able to design an acceptable filter using standard design methods.

9.10.1 Butterworth Filters The first and probably best-known filter approximation is the Butterworth or maximally-flat response. It exhibits a nearly flat pass-band with no ripple. The roll-off is smooth and monotonic, with a lowpass or high-pass roll-off rate of 20 dB/decade (6 dB/octave) for every pole. Thus, a 5th-order

9.25

Filter Circuits

Butterworth low-pass filter would have an attenuation rate of 100 dB for every factor of ten increase in frequency beyond the cutoff frequency. The general equation for a Butterworth filter’s amplitude response is, H (w ) =

1 æwö 1+ ç ÷ è w0 ø

(9.9)

2n

where n is the order of the filter, and can be any positive whole number (1, 2, 3,…), and w0 is the -3 dB frequency of the filter. Figure 9.21 shows the amplitude response curves for Butterworth low-pass filters of various orders.

Figure 9.21 Amplitude response curves for butterworth low-pass filters of different orders

The coefficients for the denominators of Butterworth filters of various orders are shown in table. Table shows the denominators factored in terms of second-order polynomials. Again, all of the coefficients correspond to a corner frequency of 1 radian/s Table 9.1 Butterworth Polynomials Denominator coefficients for polynomials of the form s n + an -1 s n -1 + an - 2 s n - 2 + ... + a1s + a0 n

a0

a1

a2

a3

a4

a5

a6

a7

a8

1 2 3 4 5 6 7 8 9 10

1 1 1 1 1 1 1 1 1 1

1.414 2.000 2.613 3.236 3.864 4.494 5.126 5.759 6.392

2.000 3.414 5.236 7.464 10.098 13.137 16.582 20.432

2.613 5.236 9.142 14.592 21.846 31.163 42.802

3.236 7.464 14.592 25.688 41.986 64.882

3.864 10.098 21.846 41.986 74.233

4.494 13.137 31.163 64.882

5.126 16.582 42.802

5.759 20.432

a9

6.392

9.26

Circuit Theory and Networks Butterworth Quadratic Factors n

1 2 3 4 5 6 7 8 9 10

(s + 1) (s2 + 1.4142s + 1) (s + 1)(s2 + s + 1) (s2 + 0.7654s + 1)(s2 + 1.8478s + 1) (s + 1)(s2 + 0.6180s + 1)(s2 + 1.6180s + 1) (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9319) (s + 1)(s2 + 0.4450s + 1)(s2 + 1.2470s + 1)(s2 + 1.8019s + 1) (s2 + 0.3902s + 1)(s2 + 1.1111s + 1)(s2 + 1.6629s + 1)(s2 + 1.9616s + 1) (s + 1)(s2 + 0.3479s + 1)(s2 + 1.0000s + 1)(s2 + 1.5321s + 1)(s2 + 1.8794s + 1) (s2 + 0.3129s + 1)(s2 + 0.9080s + 1)(s2 + 1.4142s + 1)(s2 + 1.7820s + 1)(s2 + 1.9754s + 1)

9.10.2 Second Order Low-pass Active Filter The circuit is shown in Figure 9.22.

Figure 9.22 Second order low-pass active filter

Here, Vy =

V0 R and Vx = Vy R1 + R f 1

Writing KCL at node V ¢, V ¢ - Vi V ¢ - V0 V ¢ - Vx + + =0 R R 1/sC or (V ¢ - Vi ) + (V ¢ - V0 ) sRC + (V ¢ - Vx ) = 0 or (-1)Vx + (2 + sRC )V ¢ + (- sRC )V0 = Vi Writing KCL at node x, Vx - V ¢ V + x =0 1/sC R or (1 + sRC )Vx + (-1)V ¢ + (0)V0 = 0

(9.10)

(9.11)

9.27

Filter Circuits

Writing KCL at node y, Vx Vx - V0 + =0 R1 Rf or

( R1 + R f )Vx + (0)V ¢ + (- R1 )V0 = 0

(9.12)

Solving for V0 from equations (9.10), (9.11), and (9.12), we get, (2 + sRC ) Vi -1 (1 + sRC ) 0 -1 ( R1 + R f ) 0 0

( R1 + R f ) 1 ´ 2 2 R1 R C V0 = = Vi æ ( R1 + R f ) ö (2 + sRC ) - sRC -1 1 s 2 + 3sRC - sRC ç ÷ø + R 2 C 2 R1 è (1 + sRC ) 0 -1 ( R1 + R f ) 0 - R1

or,

V0 ( s ) = Vi ( s)

K R C2 2 æ3- Kö æ 1 ö + s2 + s ç ÷ è RC ø è RC ø 2

(9.13)

R1 + R f = DC gain of the amplifier. R1 Substituting s = jw, the transfer function is, V ( jw ) K H ( jw ) = 0 = Vi ( jw ) 1 + j (3 - K ) RCw - R 2 C 2w 2 The magnitude of the transfer function is, K 1 | H ( jw )| = ; where, w c = 2 RC 2 é æ w ö2ù æwö ê1 - ç ÷ ú + [3 - K ]2 ç ÷ è wc ø êë è w c ø úû In the above equation, when w ® 0, | H ( jw )| = K . Thus, the low frequency gain of the filter is K and when w ® ¥, | H ( jw )| = 0, i.e., high frequency gain is zero. From the Table of the Butterworth Filter, the transfer function for second order (n = 2) filter is,

where, K =

T (s) =

K 2

æ s ö æ s ö çè w ÷ø + 1.414 çè w ÷ø + 1 c c

=

K w c2 s 2 + 1.414w c s + w c2

where, wc is the cut-off frequency. Comparing equations (9.13) and (9.14), we get, wc =

1 RC

or,

fc =

1 2p RC

and,

K = (3 - 1.414) = 1.586

(9.14)

9.28

Circuit Theory and Networks

Filter Design 1. Choose a value of the cut-off frequency, wc (or f c). 2. Select a convenient value for the capacitors C, between 100 pF and 0.1 mF. 3. Calculate the value of the resistors R from the relation,

Gain

The frequency response of a second order low-pass active filter is shown in Figure 9.23. It is noted that the filter has very sharp roll-off response.

Frequency

Figure 9.23

Frequency response of the second order low-pass filter

1 2p f c C 4. For minimization of dc offset, the feedback resistor is calculated from the relation, Rf = K (2R) = 3.172R. 5. Calculate the value of the resistor R1 for the value of the gain K = 1.586 from the relation, R1 + R f Rf , i.e., R1 = . K= R1 0.586 R=

Example 9.10 Design a second-order low-pass filter with a gain of 11 and cut-off frequency of 20 kHz. Solution

Let us arbitrarily select C = 200 pF. For a cut-off frequency of 20 kHz, we need R =

1 1 = 2p f c C 2p ´ 20 ´ 103 ´ 200 ´ 10-12

= 39.789 kW If we select a standard resistor of 39 kW for R, then the cut-off frequency is about 20.4 kHz. The dc gain for this filter cannot be anything other than K where K = 1.586. Thus, for a dc gain of 1.586, K = 1 + Rf /R1 = 1.586. This in turn implies that Rf = 0.586 R1. Imposing the dc bias-current balance condition, we obtain 0.586 R1 = 1.586 (2 R) = 123.708 kW. Consequently, R1 = 211.11 kW and Rf = 123.708 kW. Let us select a standard value of 130 kW for Rf. Then R1 should be about 221.8 kW. We need another amplifying stage to obtain the needed gain of 11. The gain of this stage should be 11/K = 6.936. We have chosen to use non-inverting amplifier for this stage. The output amplifier resistors are calculated as, R ö æ 6.936 = ç1 + 2 ÷ and for RA = 100 kW., R2 = 593.6 kW. RA ø è Thus, the final circuit for the second order low-pass active filter becomes as shown in Fig. 9.24.

Filter Circuits

9.29

Figure 9.24 Circuit of Example (9.10)

9.10.3 Second Order High Pass Active Filter The circuit is shown in Figure 9.25.

Figure 9.25 Second order high-pass filter

Here, V y =

V0 R and Vx = Vy R1 + R f 1

Writing KCL at node V ¢,

V ¢ - Vi V ¢ - V0 V ¢ - Vx + + =0 R 1/sC 1/sC

(9.15)

Writing KCL at node x,

Vx - V ¢ Vx + =0 R 1/ sC

(9.16)

9.30

Circuit Theory and Networks

Writing KCL at node y, Vx Vx - V0 + =0 R1 Rf

(9.17)

Solving for V0 from equations (9.15), (9.16), and (9.17), we get, or,

where, K =

V0 ( s ) = Vi ( s )

Ks 2 2 æ3- Kö æ 1 ö s2 + s ç + ÷ è RC ø è RC ø

(9.18)

R1 + R f = DC gain of the amplifier. R1

Note The transfer function of the high-pass filter can also be obtained from the transfer function of

æ s ö the low-pass filter by the transformation ç ÷ è wc ø

æw ö ®ç c÷ è s ø LP

HP

Substituting s = jw, the transfer function is, H ( jw ) =

V0 ( jw ) KR 2 C 2w 2 =Vi ( jw ) 1 + j (3 - K ) RCw - R 2 C 2w 2

The magnitude of the transfer function is,

æwö Kç ÷ è wc ø

| H ( jw )| =

2 ù2

2

2 é æwö æwö ê1 - ç ÷ ú + [3 - K ]2 ç ÷ è wc ø ú è wc ø ëê û

; where, w c =

1 RC

In the above equation, when w ® 0, | H ( jw )| = 0. Thus, the low frequency gain of the filter is zero. When w ® ¥, | H ( jw )| = K, i.e., high frequency gain is K. Here, again, comparing with Butterworth Transfer function, we get, wc =

1 RC

or,

fc =

1 2p RC

K = (3 - 1.414) = 1.586

9.31

Filter Circuits

Example 9.11 A second-order high-pass filter is given in Figure 9.27. Determine its cut-off frequency and high frequency gain. Sketch its gain vs. frequency response.

Gain

The frequency response of a second order low-pass active filter is shown below. It is noted that the filter has very sharp roll-off response. The design procedure for high-pass will be same as low-pass. The frequency response will be a maximally flat one, i.e., having a very sharp roll-off response.

Frequency

Figure 9.26 Gain vs. frequency plot of a second-order high-pass filter

220 kW

220 kW

58.7 kW

100 kW

– – 1 nF

100 kW

1 nF

V0 (s ) +

+

Vin (s )

39 kW

39 kW

Figure 9.27 Circuit of Example (9.11)

Solution

In the second-order filter on the left side of the figure, the gain K 58.7 ö æ = ç1 + = 1.587. è 100 ø÷ Since it is very close to 1.586, we can assume that the filter is maximally flat and its transfer function is as given for Butterworth filters. From the given values of R and C, the cut-off frequency is, 1 1 wc = = = 25, 641 rad/s 3 RC 39 ´ 10 ´ 1 ´ 10- 9 25, 641 The cut-off frequency in Hz, fc = = 4081 Hz 2p 220 ö æ The gain of the non-inverting amplifier, A = ç1 + =2 è 220 ÷ø Hence, the overall gain of the high-pass filter is, AH = 1.587 ´ 2 = 3.174 or approximately 10 dB. The gain vs. frequency will be as shown in Figure 9.26.

9.32

Circuit Theory and Networks

9.10.4 Second Order Band-Pass Active Filter It can be built by the cascade connection of a second order high-pass and a second order low-pass filter.

Figure 9.28 Second-order band-pass active filter circuit

Lower cut-off frequency, w1 =

1 RH CH

Upper cut-off frequency, w 2 =

1 RL CL

é R ¢f ù é R ¢¢f ù Voltage gains, K H = ê1 + and K L = ê1 + ú R¢ û R ¢¢ úû ë ë For maximally flat response (or, Butterworth) filter, KH = KL = 1.586. R ¢f R ¢¢f \ = = 0.586 R ¢ R ¢¢ The overall transfer function is the product of the transfer function of the high-pass and low-pass filters. \

H (s) =

æ s ö KH ç ÷ è w1 ø

´

2

KL 2

æ s ö æ s ö æ s ö æ s ö 1 + ç ÷ + (3 - K H ) ç ÷ 1 + ç ÷ + (3 - K L ) ç ÷ è w1 ø è w1 ø è w2 ø è w2 ø Substituting the values of KH and KL, magnitude of the gain is,

| H ( jw )| =

æwö 2.5154 ç ÷ è w1 ø æwö 1+ç ÷ è w2 ø

4

2

æwö 1+ç ÷ è w1 ø

4

Note In the pass-band, the gain is 2.5154. The frequency response is more flat near the cut-off frequencies.

9.33

Gain

Filter Circuits

w1

w2

Frequency

Figure 9.29 Frequency response of second order band-pass filter

9.10.5 Second Order Band-Reject Active Filter It can be built by the summation of a second order high-pass and a second order low-pass filter. 1 1 and the cut-off frequency of HPF, w 2 = . The cut-off frequency of LPF, w1 = RL C L RH CH The magnitude of the overall transfer function is the sum of the transfer function of the high-pass and low-pass filters,

é 2 ê K æwö H ç è w 2 ø÷ 1 é R ùê + | H ( jw )| = ê1 + 2 ú ê R1 û 2ë 4 ê æwö ê 1 + èç w ø÷ 2 êë

ù ú ú KL ú 4 æwö ú 1+ç ÷ ú è w 1 ø úû

R ¢f ö R ¢¢f ö æ æ where, K H = ç1 + and, K L = ç1 + and for Butterworth filters, KH = KL = 1.586. ÷ R¢ ø R ¢¢ ø÷ è è The roll-off frequency response will be very smooth as shown.

Figure 9.30 Second-order band-reject active filter circuit

9.34

Gain

Circuit Theory and Networks

Frequency

Figure 9.31 Frequency response of second order band-reject filter

9.11

ALL-PASS ACTIVE FILTER

This filter passes all frequency component of the input signal without attenuation and provides some phase shifts between the input and output signals. The circuit of an active all-pass active filter with lagging output is shown in Figure 9.32.

Figure 9.32 Circuit of an all-pass active filter with lagging output

For the circuit, by KCL at node x, Vx - Vi Vx - V0 + =0 Þ R1 R1

Vx =

Vi + V0 2

(9.19)

By KCL at node y, V y - Vi Vy Vi + = 0 Þ Vy = 1/ jw C 1 + jw RC R

Also, from Op-Amp property, Vx = Vy Þ

Vi æ Vi + V0 ö æ ö çè 2 ÷ø = çè 1 + jw RC ÷ø

(9.20)

Filter Circuits

Þ

(Vi + V0 )(1 + jw RC ) = 2Vi

Þ

V0 (1 + jw RC ) = Vi [2 - (1 + jw RC )] = Vi (1 - jw RC )

\

9.35

V0 1 - jw RC = Vi 1 + jw RC

Thus, the amplitude of the gain,

V0 = 1 i.e., |Vout | = |Vin | throughout the entire frequency range Vi Also, the phase shift between the input and the output voltages is, f = –2 tan–1 (wRC ) i.e., phase-shift is a function of frequency

Figure 9.33 Characteristics of all-pass filter

By interchanging the positions of R and C in the circuit, the output can be made leading the input.

MULTIPLE-CHOICE QUESTIONS 9.1 The two input terminals of an op-amp are labeled as (a) high and low (b) positive and negative (c) inverting and non-inverting (d) differential and non-differential 9.2 Consider the following statements for an ideal op-amp. 1. The differential voltage across the input terminals is zero. 2. The current into the input terminals is zero. 3. The current from the output terminals is zero. 4. The input resistance is zero. 5. The output resistance is zero. Of these statements, those which are not true are (a) 1 and 5 (b) 3 and 4 (c) 2 and 4 (d) 1 and 4 9.3 In a series resonant circuit, to obtain a low-pass characteristic, across which element should the output voltage be taken? (a) Resistor (b) Inductor (c) Capacitor 9.4 In a series resonant circuit, to obtain a high-pass characteristic, across which element should the output voltage be taken? (a) Resistor (b) Inductor (c) Capacitor

9.36

Circuit Theory and Networks

9.5 In a series resonant circuit, to obtain a band-pass characteristic, across which element should the output voltage be taken? (a) Resistor (b) Inductor (c) Capacitor 9.6 A high-pass filter circuit is basically (a) a differentiating circuit with low time constant. (b) a differentiating circuit with large time constant. (c) an integrating circuit with low time constant. (d) an integrating circuit with large time constant. 9.7 The transfer function of an electrical low-pass RC network is RCs RC s 1 (b) (c) (d) (a) 1 + RCs 1 + RCs 1 + RCs 1 + RCs 9.8 For a high-pass RC circuit, when subjected to a unit step input voltage, the voltage across the capacitor will be 9.9

9.10

9.11

9.12

9.13

(a) 1 - e-t / RC (b) e-t / RC (c) et / RC (d) 1 In the magnitude plot of a low-pass filter, at what frequency does the peak of the magnitude characteristic occur? (a) At resonant frequency (b) Below resonant frequency (c) Above resonant frequency (d) At any frequency. In the magnitude plot of a high-pass filter, at what frequency does the peak of the magnitude characteristic occur? (a) At resonant frequency (b) Below resonant frequency (c) Above resonant frequency (d) At any frequency. In the magnitude plot of a band-pass filter, at what frequency does the peak of the magnitude characteristic occur? (a) At resonant frequency (b) Below resonant frequency (c) Above resonant frequency (d) At any frequency. If a filter is de-normalized to a higher frequency, which of the following occurs? (a) Inductors increase in value while capacitors decrease. (b) Inductors decrease in value while capacitors increase. (c) Inductors and capacitors increase in value. (d) Inductors and capacitors decrease in value. V (s) 10 s = is for an active The transfer function 2 V1 ( s ) s 2 + 10s + 100 (a) low pass filter (b) band pass filter (c) high pass filter (d) all pass filter.

s2 belongs to an active s 2 + as + b (b) high pass filter (c) band pass filter

9.14 The transfer function T ( s ) = (a) low pass filter

9.15 The voltage-ratio transfer function of an active filter is given by question is a (a) low pass filter

(b) high pass filter

(d) band reject filter.

V2 ( s ) s2 + d . The circuit in = 2 V1 ( s ) s + a s + d

(c) band pass filter

(d) band reject filter.

9.37

Filter Circuits

9.16 C

R R

– R

+

C

The transfer function of a second order LP filter shown in the figure is (a)

1 R 2 C 2 s 2 + 3 RCs + 1

(b)

RCs R 2 C 2 s 2 + 3 RCs + 1

R2C 2 s 2 + 1 R2C 2 s 2 (d) 2 2 2 2 2 R C s + 3 RCs + 1 R C s + 3 RCs + 1 9.17 An ideal filter should have (a) zero attenuation in the pass band (b) infinite attenuation in the pass band (c) zero attenuation in the attenuation band (d) none of these. 9.18 An RLC series circuit can act as (a) band-pass filter (b) band-stop filter (c) low-pass filter (d) both (a) and (b). 9.19 If R1 = R2 = RA and R3 = R4 = RB , the circuit acts as a/an

(c)

2

R2

C

R1 Vi

– V0 + R3 R4

(a) all pass filter (b) band pass filter (c) high pass filter (d) low pass filter 9.20 The output of the filter in 9.19 is given to the circuit shown in the figure. RA 2

V in

C

V0

9.38

Circuit Theory and Networks

The gain vs frequency characteristic of the output (v0) will be

0

Gain

(b)

Gain

(a)

0

w

0

Gain

(d)

Gain

(c)

w

w

0

w

9.21 In active filter circuits, inductances are avoided mainly because they (a) are always associated with some resistance (b) are bulky and unstable for miniaturisation (c) are non-linear in nature (d) saturate quickly 9.22 The magnitude response of a normalized Butterworth low-pass filter is (a) linear starting with values of unity at zero frequency and 0.707 at the cut-off frequency (b) non-linear all through but with values of unity at zero frequency and 0.707 at the cut-off frequency (c) linear up to the cut-off frequency and non-linear thereafter (d) non-linear up to the cut-off frequency and linear thereafter

EXERCISES 9.1 Design a second order low pass active filter having a cut-off frequency of 5 kHz. [C = 0.03 mF; R = 1 kW; R1 = 10 kW; R2 = 5.86 kW] 9.2 Design a second order band pass active filter that has a centre frequency of 1 kHz and a bandwidth of 100 Hz. Take the centre frequency gain to be 2. [C1 = C2 = 0.02 mF; R1 = 40 kW; R3 = 160 kW; R2 = 400 W] 9.3 Design a second order high pass Butterworth filter with a cut-off frequency of 200 Hz. [C = 0.053 mF; R = 1.5 kW; R1 = 10 kW; R2 = 5.86 kW] 9.4 Design a second order band pass active filter with a centre frequency gain A0 = 50. Given: f0 = 160 Hz and Q = 10. [assuming C1 = C2 = 0.1 mF; R1 = 2 kW; R3 = 200 kW; R2 = 667 W]

9.39

Filter Circuits

SHORT-ANSWER TYPE QUESTIONS 9.1 (a) (b) (c) 9.2 (a)

What is an operational-amplifier? State the characteristics of an op-amp. What is filter? Classify them. Discuss the advantages of an active filter over a passive filter. Briefly discuss the operating principle of an active low-pass filter and derive its gain-frequency characteristics. Explain the design procedure of a low-pass active filter. (b) Briefly discuss the operating principle of an active high-pass filter and derive its gain-frequency characteristics. Explain the design procedure of a high-pass active filter. 9.3 (a) Define the following terms with reference to a band-pass active filter: (i) Bandwidth, (ii) Cut-off frequency, (iii) Quality factor. (b) What are the different types of band-pass filters? Give the salient features and performance equations for the following filters: (i) Wide Band-Pass Active Filter, (ii) Narrow Band-Pass Active Filter. 9.4 Define Notch-frequency. Explain the operational characteristics of an active Notch filter. Where are these filters used?

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 9.1 9.8 9.15 9.22

(c) (a) (c) (b)

9.2 (b) 9.9 (b) 9.16 (a)

9.3 (c) 9.10 (c) 9.17 (a)

9.4 (b) 9.11 (a) 9.18 (a)

9.5 (a) 9.12 (d) 9.19 (c)

9.6 (a) 9.13 (c) 9.20 (d)

9.7 (b) 9.14 (b) 9.21 (b)

CHAPTER

10 Resonance

10.1

INTRODUCTION

Any system having at least a pair of complex conjugate poles has a natural frequency of oscillation. If the frequency of the system driving force coincides with the natural frequency of oscillation, the system resonates and the system response becomes maximum. This phenomenon is known as ‘resonance’ and the frequency at which this phenomenon occurs is known as ‘resonant frequency’. In electrical systems, resonance occurs when the system contains at least one inductor and one capacitor. In this system, the phenomenon of cancellation of reactances when inductor and capacitor are in series or cancellation of susceptances when they are in parallel, is termed as resonance. The circuit under resonance is purely resistive in nature and is termed as ‘resonant circuit’ or ‘tuned circuit’. In this chapter, we consider electrical resonance in details. Electrical resonance is broadly classified into tow categories: (a) Series Resonance, and (b) Parallel Resonance.

10.2

SERIES RESONANCE OR VOLTAGE RESONANCE L

R

VS

C

i

Figure 10.1 Series RLC Resonant Circuit

10.2

Circuit Theory and Networks

The basic series-resonant circuit is shown in Fig. 10.1. We want to observe how the steady state amplitude and the phase angle of the current vary with the frequency of the sinusoidal voltage source. As the frequency of the source changes, the maximum amplitude of the source voltage (Vm) is held constant. Impedance due to capacitor: Z C = -

j ; clearly, as w ® 0, ZC ® ¥ and i ® 0. wC

Impedance due to inductor: ZL = jwL clearly, as w ® ¥, ZL ® ¥ and i ® 0. Therefore, circuits containing inductors and capacitors have responses that are frequency dependent. We analyze in the following steps. i. Current Response

Here, the supply voltage, v s = Vm sin ωt and the current is, i = Im sin (wt + f). The phasor equivalents of vs and i are V and I, respectively. Using phasors, I =

V V = = Z R + jw L - j wC R +

V 1 ö æ j çw L ÷ è wC ø

Vm

Thus, the current magnitude, I =

1 ö æ R2 + çw L ÷ è wC ø

2

(10.1)

(10.2)

From this equation, we have the following observations: · |I| ® 0 as w ® 0; and · |I| ® 0 as w ® ¥. This indicates that there must be a maximum value of the current |I| for some particular value of w. This occurs when the denominator is a minimum. i.e., when wL =

1 Þw = wC

1

(10.3) LC Thus, resonance occurs when the magnitudes of the inductive and capacitive reactances are equal. This frequency is termed as the resonant frequency, w0 of the series RLC circuit.

\ w0 =

1 LC

From equation (10.1), the phase angle of the current is given by, 1 ö æ wL ç wC ÷ f = - tan -1 ç ÷ R ç ÷ è ø

(10.4)

10.3

Resonance

From equation (10.4), it is clear that the phase angle for the current also depends on frequency. We have two observations: æ 1 ö and in this case the current leads the voltage with the phase · As w ® 0, f ® tan -1 ç è w RC ÷ø

relationship being like that of an RC circuit. æw Lö · As w ® ¥, f ® - tan -1 ç and in this case the current lags the voltage with the phase è R ø÷

relationship being like that of an RL circuit. · At w = w0, f = 0, and in this case the current and the voltage are in phase, the circuit behaving like a purely resistive circuit. The current response and phasor diagrams are shown in Fig. 10.2. Im Imax=Vm/R B = R / L = v2

Imax /

v1

1.414

v1 v0 v2

v

Figure 10.2 Frequency Response of a Series – Resonant Circuit Phasor Diagrams The current and voltage phasor diagrams for an RLC series circuit are shown in Fig. 10.3. VL

VL I

VR

0

0 V VC

VC

I V VR RI

VL 0 VC

V I

VR

(a) (b) (c) Figure 10.3 Phasor Diagrams (a) f < f0 (b) f = f0 (c) f > f0 At resonant frequency, the inductive and capacitive reactances are equal so that the current and voltage are in phase. For any frequency lower than the resonant frequency, the inductive reactance is less than the capacitive reactance and hence, the circuit behaves as a capacitive circuit. Similarly, for any frequency higher than the resonant frequency, the inductive reactance is greater than the capacitive reactance and hence, the circuit behaves as an inductive circuit. ii. Bandwidth We define the half-power bandwidth of the RLC circuit as the range of frequencies (or the width of the frequency band) for which the power dissipated in R is greater than or equal to half the maximum power.

10.4

Circuit Theory and Networks

We know that the average power is, P= I

2

R where I =

Im 2

(10.5)

Maximum power will be, Pmax = I 2m R Thus, the half-power points occur when Pmax I or I = m 2 2 At resonance, the circuit is purely resistive, so that P=

Vm R Therefore, at half-power points, Im =

I =

Im

=

(10.6)

Vm

2 2R From equation (10.2) and (10.7), we get,

Vm 2

1 ö æ R2 + çw L ÷ è wC ø

=

(10.7)

Vm 2R

To solve for the frequencies, squaring both sides and equating the denominators, 2

1 ö æ 2 R2 + çw L ÷ = 2R è wC ø 2

Þ

1 ö æ 2 çè w L ÷ =R wC ø

Þ

1 ö æ çè w L ÷ = ±R wC ø 2

Therefore,

w =±

R 1 æ Rö ± ç ÷ + è 2L ø 2L LC

We see that, mathematically, there are 4 possible values of w. Taking the positive roots, the halfpower frequencies are: 2

w1 = -

R 1 æ Rö + ç ÷ + è 2L ø 2L LC

(10.8)

2

R 1 æ Rö w2 = + ç ÷ + è ø 2L 2L LC

(10.9)

Resonance

10.5

By definition, the Bandwidth (BW) is given by, R L Also, from equations (10.8) and (10.9), we get BW = w 2 - w1 =

1 = w 0 2 Þ w 0 = w1w 2 LC Thus, the resonant frequency is the geometric mean of the half-power frequencies. w1w 2 =

(10.10)

(10.11)

iii. Quality Factor or Circuit Magnification Factor (Q) It is defined as, Q = 2p

Maximum Energy Stored Energy Dissipated per cycle

Maximum Energy stored = Electromagnetic energy in inductor or Electrostatic energy in capacitor =

1 1 LI max 2 or CVmax 2 2 2

Therefore,

Q = 2p

1 LI max 2 2 2

æ I max ö R´ 1 è 2ø f

=

2p fL w L = R R

1 1 L wL = = (10.12) R w RC R C Thus, Q is inversely proportional to R. Hence, for series RLC circuit, a high value of quality factor implies low losses and a low value of Q implies high losses. or,

Q=

w0 Resonant Frequency and the = w 2 - w1 Bandwidth selectivity of the circuit is defined as the reciprocal of quality factor, i.e. 1 BW Selectivity = = Q w0 Therefore, a circuit will be highly selective if it has a high value of Q. For series RLC circuit, a high value of quality factor implies a narrow resonant peak and a low value of Q implies broad resonant peak. The variations of magnitude and phase angle of current in RLC series circuit for different values of Quality factor (Q) are shown in Fig. 10.4.

Also, quality factor for a circuit is defined as, Q =

i. Voltage Across Elements Voltage across Resistance

Since V = RI and at resonance VR = Vm, Therefore,

I=

Vm R

(10.13)

10.6

Circuit Theory and Networks

1 0.9 Magnitudes of current

Phase angle of current (degree)

Q =1

0.8

Q =5

0.7

Q = 10

0.6 0.5 0.4 0.3 0.2 0.1

0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Frequency (rad/s)

100 80 60 40 20 0 20 40 60 80 100

Q = 10 Q =5 Q =1

200 400 600 800 1000 1200 1400 1600 1800 2000 Frequency (rad/s)

Figure 10.4 Variation of Magnitude and Phase Angle of Current in RLC Series Circuit for Different Values of Q

Voltage across Inductance

We know, VL = IZL = jwLI, using equation (10.2), we get, VL = w L I =

At resonance, (i.e. w 0 L =

w LVm 1 ö æ R2 + çw L ÷ è wC ø

2

1 w LV ), VL = 0 m = QVm w0C R

(10.14)

To find the frequency at which inductor voltage will be maximum, we have, d é VL ûù = 0 dw ë

Þ

Þ

w LVm d é ê 2 dw 1 ö ê 2 æ ê R + èç w L - w C ø÷ ë wL =

1 LC -

Thus,

ù ú=0 ú ú û

C 2 R2 2

1 æ 1 öæ fL = ç ÷ 2 2 è 2p ø ç ç LC - C R è 2

ö ÷ ÷ ø

(10.15)

Resonance

10.7

Voltage across Capacitance

We know, VC =

I ; using equation (10.2), we get, jw C VC =

I wC

At resonance, (i.e. w 0 L = VC =

Vm

=

1 ö æ R2 + çw L ÷ è wC ø

2

1 wC

1 ), w0C

Vm = QVm w 0 RC

(10.16)

To find the frequency at which capacitor voltage will be maximum, we have, d é VC ûù = 0 dw ë

Þ

Þ

Thus

Vm 1 ù d é ê ú=0 2 dw 1 ö wC ú ê 2 æ ê R + èç w L - w C ø÷ ú ë û

wC =

1 R2 - 2 LC 2 L

R2 ö æ 1 ö æ 1 fC = ç ÷ ç - 2÷ è 2p ø è LC 2 L ø

(10.17)

From equations (10.14) and (10.16) we see that both VL and VC may be very large at resonance and they will add to zero (voltage across L and C are 180° out of phase). At resonant condition, the voltage across the inductor and capacitor are equal in magnitude and opposite in sign, thus canceling each other so that the entire voltage appears across the resistance. For this reason, the resonance in series RLC circuit is known as voltage resonance. Also, at resonant condition, from equations (10.14) and (10.16) we get, VC = VL = QVm Þ Q =

VC Vm

=

VL Vm

Thus, quality factor for series circuit is also defined as the ratio of voltage across the inductor or capacitor at resonance to the supply voltage. For this reason, Q is also known as circuit magnification factor (here, voltage magnification). From equations (10.15) and (10.17), it is observed that fL > fC. The variations of voltage across resistor, capacitor and inductor are shown in Fig. 10.5.

10.8

Circuit Theory and Networks

However, from equations (10.15) and (10.17), it is observed that if R is very small (or Q is very large), both fL and fC approach f0. For circuits with Q ³ 10, the maximum voltages across R, L and C will practically occur at resonant frequency f0. 2.5 Voltage across capacitor

2

Voltage across inductor Voltage across resistor

Voltage

1.5 1 0.5 0

0

2

1

vC vL v0

7 5 6 Frequency, (rad/s)

4

8

Figure 10.5 Variation of Voltage Across Resistor, Capacitor and Inductor with Frequency

Impedances

Variation of Impedance with Frequency The variations of the impedances with frequency are shown in Fig. 10.6. 4 3.5 3 2.5 2 1.5 1 0.5 0 0

L

R

C Z

1

1.5 2

Resonant frequency, (rad/s) 0

2.5 3

Frequency, (rad/s)

Figure 10.6 Variations of Impedances with Frequency

Here,

R—wL—

1 Z =R+ wC

1 ö æ j çw L ÷ è wC ø

i Impedance at Frequencies near Resonant Frequency We introduce a new term, fractional frequency deviation or fractional detuning, d, defined as, w - w0 w d = Þ = (1 + d ) w0 w0 Thus, the impedance of the RLC series circuit at any frequency is given by,

1 ö é æ = R ê1 + z = R + j çw L ÷ è wC ø ë é = R ê1 + êë

1 öù æw L jç è R w RC ø÷ úû

æw L w 1 w0 ö ù jç 0 ú è R w 0 w 0 RC w ÷ø úû

10.9

Resonance

At frequencies near the resonant frequency, d 10), at resonant frequency, the impedance is,

Z 0 » Q 2 RL =

L CRL

(ii) For large values of Q(i.e., Q > 10), for near resonant frequency, the impedance is,

Z0 =

Rd Q 2 RL L [as, Dynamic Resistance, Rd = ] = 1 + j 2Qd 1 + j 2Qd CRL

Z0 1 = Rd 1 + j 2Qd

\

Current in Parallel Tuned Circuit Let, Current delivered by the source = Is, Current in the inductor branch = IL, Current in the capacitor branch = IC, \

Power delivered by the source, Ps = Is2 Rd [Rd = Dynamic Resistance]

Power dissipated in the parallel circuit, P = IL2RL IC

IS

V IL

Figure 10.13 Phasor diagram for Parallel Tune Resonant Circuit with Lossless Capacitor

Now, Ps = P Þ

I s 2 Rd = I L 2 RL

Þ

I L2 = I s2

Þ

IL = Is

Rd L é L ù = Is2 êQ Rd = ú 2 RL CRL û CRL ë

1 RL

L C

Thus, the parallel tuned circuit is a current amplifier. Voltage across the capacitor, VC = IsRd and current through the capacitor, IC = w0CVC \

Is =

VC IC = Rd w 0 CRd

From (10.37), I s 2 Rd = I s 2 RL

(10.37)

10.21

Resonance

Is2

Þ

IL

=

2

RL Rd

IC 2

Þ

2

2

2

w 0 C Rd I L IC 2

Þ

IL

=

2

2

=

RL Rd é L ù êQ Rd = ú CRL û ë

RL ´ w 0 2 C 2 Rd 2 = w 0 2 LC Rd

é æ IC CRL 2 ö ùú CRL 2 ö 1 æ 1 1 = LCw 0 = LC ê = ÷ ç ÷ ê LC çè IL L øú L ø è ë û

Þ

=

1-

é 1 êQ Q = RL ë

1 Q2

Lù ú Cû

IC 1 = 1- 2 IL Q

\

If RL is very large, the currents are not equal. However, for very low value of RL (for very high value of Q, Q ³ 10), both the currents are equal. Higher the value of Q, higher will be IC and IL and lower will be the source current, Is. As Q ® µ, both IC and IL tend to infinity and Is tends to zero. Effect of Source Resistance over Bandwidth and Quality Factor Bandwidth for the parallel tuned circuit is affected by the source resistance as explained below a 1 V 2

RS

a

RL C

L

I⫽V/Rs

RS

a

RL C

I⫽V/Rs

C

L

b

b

C

E

C RL

a Re

E

Le

L b

L

b

a

RS

RL RS

b

Figure 10.14 Simplification of Real Parallel Circuit to get Series Equivalent Circuit

By circuit reduction methods, the parallel tuned circuit is simplified as shown from Fig. 10.14 (i) to Fig. 10.14 (v), Ze =

( RL

+ jw L) Rs

( RL +

jw L ) + Rs

10.22

Circuit Theory and Networks

=

RL Rs + jw LRs (RL + jw L) + Rs

=

( RL Rs + jw LRs ) {( RL + Rs ) - jw L} {( RL + Rs ) + jw L} {( RL + Rs ) - jw L}

=

{R R (R L

s

L

}

+ Rs ) + w 2 L2 Rs + j {w LRs ( RL + Rs ) - w LRs }

( RL + Rs )2 + w 2 L2

= Re + jwLe Therefore, the equivalent resistance is,

RL Rs ( RL + Rs ) + w 2 L2 Rs

Re =

(RL + Rs )2 + w 2 L2

and the equivalent inductance is,

Le =

LRs 2

(RL + Rs )2 + w 2 L2

Since the equivalent circuit is an RLC series circuit, the effective quality factor is, Q=

w Le w LRs 1 1 = = = 2 2 2 2 2 2 Re RL ( RL + Rs ) + w L RL RL + w L 1 æ RL + w 2 L2 ö + +ç ÷ RL wL w LRs Q è RLw LRs ø æ w Lö çèQ Q = R ÷ø L

Q 1 1 = = = 1 æ RL ö Rd 1 1 æ Rd ö 1 + Rd +ç + ÷ Rs Q è w L ø Rs Q Q èç Rs ø÷ æ L L ö 2 2 2 çè as, at resonance RL + w L = C and Rd = CR ÷ø L

wL

Therefore,

Qe =

RL

æ L ö çè1 + CR R ø÷ L s

=

Q æ Rd ö çè1 + R ø÷ s

10.23

Resonance

Therefore, the effective bandwidth will be,

( BW )e =

R ö w0 1 æ = ç1 + d ÷ w 0 Qe Q è Rs ø

æ Rd ö w 0 é L ù çè1 + R ÷ø = Q ê1 + CR R ú s L s û ë Thus, the bandwidth of the circuit depends upon the circuit constants (RL, L and C) and the source resistance (Rs). For a given resonant frequency, the circuit will be less selective and bandwidth will be large if L is large and C is small. For more selectivity, the value of L should be reduced and C increased; however, this may reduce the value of Rd, which is undesirable.

\

( BW )e

=

w0 Q

NB: If the source resistance is infinite (Rs ® ¥), then the bandwidth and quality factor of this parallel tuned circuit is the same as those of the RLC series circuit. To obtain maximum possible value of power delivered from the source to the load, we have, Rs = Rd and under this condition, the bandwidth becomes,

BW =

2 w0 Q

Example 10.4 A coil of inductance 1 H and 10 W resistance is connected in parallel with 100 mF capacitor. If the supply voltage is 200 V, find the resonant frequency and the current at resonance. Sol: Here, RL = 10 W, L = 1 H, C = 100 mF, V = 200 V Resonant frequency is, R 2 1 1 1 102 - L2 = = 15.84 Hz Ans 2p 1 ´ 100 ´ 10-6 LC 12 L At resonance, the impedance of the circuit is, f0 =

1 2p

Z0 =

L 1 = = 1000 W CRL 100 ´ 10 -6 ´ 10

\ Current in the circuit at resonance is, I 0 =

V 200 = = 0.2 A Ans Z 0 1000 a

V

⫹ ⫺

L

RC C

b

Figure 10.15 Real Parallel Resonant Circuit with Lossless Inductor

10.24

Circuit Theory and Networks

Case (2): RL = 0: Here, the resonant frequency is,

ö 1 æ L ç ÷ LC è L - CRC 2 ø

w0 =

Under this condition, the total admittance becomes, from equation (10.35) Y0 =

w 0 2 C 2 RC 1 + w 0 2 C 2 RC 2

From resonant frequency, 1

w 02 =

(

C L - CRC

2

Therefore, the admittance is, Y0 =

)

(

)

Þ 1 + w 0 2 C 2 RC 2 = w 0 2 LC

w 0 2 C 2 RC 2

w 0 LC

=

CRC and the impedance under resonant condition L

becomes, Z0 =

L CRC

This is the Dynamic Resistance of the parallel tuned circuit. Here also, lower the value of resistance RC, higher is the value of dynamic resistance of the parallel circuit. The make-up current i.e., current drawn from the supply at resonance is,

I0 = V

CRC L

and the forced oscillatory current, given by, I =

V w0 L

This circuit is also a rejector circuit as the impedance approaches a maxima and the current a minima. Approaching in the way similar to case (1), we get the equivalent quality factor and bandwidth of this resonating circuit as,

Qe =

and

BW =

Q 1 = æ R ö æ R ö w CRC ç1 + d ÷ ç1 + d ÷ Rs ø è Rs ø è

where, Q =

R ö L ù 1æ 1é 1 + d ÷ w 0 = ê1 + ú w0 ç Qè Rs ø Q ë CRC Rs û

1 w CRC

10.25

Resonance

Case (3): RL = RC = 0: RS C

V

L

Figure 10.16 Tuned Tank Circuit

1 . This circuit with L and C in parallel is termed as tuned LC tank circuit. At resonance, with capacitive and inductive reactances equal to each other, the total impedance increases to infinity, meaning that the tank circuit draws no current from the AC power source. However, for this case, there will be some circulating current, given by, Here, the resonant frequency is, w 0 =

1

I = IC = V w C = V

LC

Rs 10) is given as, Z0 =

Z=

Here, Q =

\

Z0 1 + j 2Qd

w 0 L 2p ´ 106 ´ 250 ´ 10 -6 = = 170 RL 9.24 d =

f - f 0 0.99 - 1 = = -0.01 f0 1

Z=

Z0 267.04 ´ 103 267.04 ´ 103 = = = 75.35Ð73.61° (kW) Ans 1 + j 2Qd 1 + j 2 ´ 170 ´ ( -0.01) 1 - j 3.4

18. Show that the high-Q coil resonant circuit can be approximated as shown in figure. R

R

C

L

C

L

High-Q Coil Resonant Circuit

Approximated Equivalent Circuit

Resonance

10.43

Sol: For the approximated circuit, the resonant frequency is given as,

1

w0 =

(See section 10.3.) LC For the high-Q resonant circuit, the resonant frequency is given as, 1 æ CR 2 ö 1ç LC è L ø÷

w0 =

(i) (See section 10.5)

Therefore, Q-factor of the inductance is, Q= Þ Þ

L CR 2

1 æ CR 2 ö L = -1 1 ç ÷ LC è L ø CR 2

w0 L L = ´ R R

= 1 + Q2

CR 2 1 = L 1 + Q2

Putting this value in (i), we get, w0 =

1 æ 1 ö 1ç LC è 1 + Q 2 ø÷

1 æ CR 2 ö 1 = LC èç L ø÷

æ 1 ö For very high values of Q, the term ç ÷ becomes negligible and we get the resonant è 1 + Q2 ø

frequency as, w 0 =

1 (1 - 0) = LC

1 which is the same as that for the approximated equivalent LC

circuit. 19. A parallel resonant circuit comprising a coil of 150 nH with Q of 20 in parallel with a capacitor. What is the value of capacitor? Find also the resistance of the coil and the circuit impedance at resonance. Take f0 = 1 MHz. Sol: Here, L = 150 nH, Q = 20, f0 = 106 Hz Q

f0 =

Q

Q0 =

1 2p LC

ÞC=

1 2

2

4p f0 L

=

1 2

12

4p ´ 10

´ 150 ´ 10-9

= 0.168 ì F Ans.

wL w L 2p ´ 106 ´ 150 ´ 10-9 Þ R= = = 47.1 mÙ Ans. R Q 20

Circuit Theory and Networks

10.44

20. In a two-branch parallel circuit, calculate the resonant frequency w0 if R1 = 4 W and R2 = 6 W, C = 20 mF and L = 1 mH. If R1 is increased, what is its maximum value for which there is a resonant frequency? Sol: Here, R1 = 4 W, L = 1 ´ 10–3 H, R2 = 6 W and C = 20 ´ 10–6 F

\ w0 =

2 1 æ L - CR2 ö ç ÷ = LC è L - CR12 ø

R1 L

R2

C

æ 10 -3 - 20 ´ 10 -6 ´ 36 ö ç ÷ = 4.537.4 rad/s Ans. 20 ´ 10 -9 C è 10 -3 - 20 ´ 10 -6 ´ 16 ø 1

L , the resonance will occur C at w ® µ. Beyond this value of R1, the quantity within the square root will become imaginary and no real frequency will give resonance. \ Maximum value of R1 is obtained as,

When R1 is increased, resonant frequency will also increase. For R12 =

L 1 ´ 10-3 = = 7.071 W Ans. C 20 ´ 10-6 21. A coil of inductance L and resistance R, in series with a capacitor is supplied at a constant voltage from a variable frequency source. Find the values of that frequency, in terms of R, L and w0 at which the circuit current would be half as much as at resonance. Hence, or otherwise, determine the bandwidth and selectivity of the circuit. Sol: The current at resonance is, R1 =

V R and current at any other frequency is, V I= 2 1 ö æ R2 + çw L ÷ è wC ø For this problem, I0 =

I=

I0 I Þ 0 =2 I 2

1 ö æ R2 + çw L ÷ è wC ø R 2

Þ

2

=2

1 ö æ 2 R2 + çw L ÷ = 4R è wC ø

Resonance

1 ö w 2 L2 æ 1- 2 ÷ 2 ç R è w LC ø

Þ

3=

Þ

2 w 2 L2 æ w 0 ö 3= 1 ç ÷ w2 ø R2 è

2

2

é êQ w 0 = ë

1 ù ú LC û

2

w 2 L2 æ w 2 ö w2 = 0 2 ç1 - 02 ÷ ´ 2 R è w ø w0 æ w w0 ö = Q2 ç w ÷ø è w0 2

Þ

æ w w0 ö 3 = ±ç = ÷ Q è w0 w ø

If the two frequencies are w2 and w1, (w2 > w0 > w1) then, æ w2 w0 ö 3 çè w - w ÷ø = Q 0 2

and Þ

æ w 0 w1 ö 3 çè w - w ÷ø = Q 1 0

ü ï 2Q ï ý Ans. 3w 0 + w 0 3 + 4Q 2 ï & w2 = ï 2Q þ

w1 =

\

- 3w 0 + w 0 3 + 4Q 2

Alternately, 2

1 ö æ 2 çè w L ÷ = 3R wC ø

Þ

1 ö æ çè w L ÷ = ± 3R wC ø

Therefore,

æ 3R ö 3R 1 w =± ± ç + ÷ LC 2L è 2L ø

2

10.45

Circuit Theory and Networks

10.46

Taking the positive roots of w, 2 ü æ 3R ö 3R 1 ï w1 = + ç + ÷ 2L LC ïï è 2L ø ý Ans. 2 æ ö 3R 3R 1 ï ï w2 = + ç + ÷ 2L LC ïþ è 2L ø

3w 0 3R Ans. = Q L

\ Bandwidth BW = (w 2 - w1 ) =

22. A series circuit consisting of a coil and a capacitor is excited by a sinusoidal voltage source of E volt and variable frequency. The resonant frequency of the circuit is f0 and quality factor of the circuit is Q. Calculate the frequency at which the ratio of capacitor voltage to the source voltage is maximum and the maximum value of this ratio. Sol: 1 Here, f 0 = 2p LC w0 L 1 1 L = = R w 0 RC R C E The current, I = 1 ö æ R + j çw L ÷ è wC ø Q=

1 1 é = jw C jw C ê êR + êë

\ Capacitor voltage, V C = I ´

=

E ù = ú 1 ö 1 ö æ æ j çw L ÷ ú jw RC - w C çè w L ÷ è w C ø úû wC ø E

E

(1 - w

2

)

LC + jw RC

1 VC = 2 E 1 - w LC + jw RC

(

)

This ratio will be maximum when the denominator is minimum. \

d éæ 1 - w 2 LC d w êë è

Þ

2 1 - w 2 LC ( -2w LC ) + 2w R 2 C 2 = 0

Þ

-4w LC + 4w 3 L2 C 2 + 2w R 2 C 2 = 0

Þ

(

(

2

)

ù + w 2 R2C 2 ö ú = 0 øû

)

2w 2 L2 C 2 = 2 LC - R 2 C 2

Resonance

1 æ R2 C ö æ Q w0 = 1ç LC è 2 L ø÷ èç

\

w=

Þ

w = w0 1 -

1

1 LC

10.47

and Q =

1 Lö R C ø÷

Ans.

2Q 2

Putting this value, the maximum value of the ratio is, VC E

1

=

(1 - w

max

2

LC

2

)

(i) 2

2

+w R C

2

Now,

(1 - w

2

LC

2

)

2

é æ 1 ù é 1 R2 ö R2 ù + w R C = ê1 - ç - 2 ÷ LC ú + R 2 C 2 ê - 2ú êë è LC 2 L ø úû ë LC 2 L û 2

2

2

æ R2 C 2 ö R2 C R4 C 2 = ç1 - 1 + + L 2 L ø÷ è 2 L2

(1 - w

Þ

=

Thus, from (i),

2

LC

2

)

2

æ 1 ö 1 1 +w R C = ç 2 ÷ + Q 2Q 4 è 2Q ø 2

1 + 4Q3 - 2

VC E

4Q 4

= max

=

2

2

4Q3 - 1 4Q 4

2Q 2 4Q3 - 1

Ans.

23. Impedances Z2 and Z3 in parallel are in series with impedance Z1 across a 100 V, 50 Hz A.C. supply. Z1 = (6.25 + j1.25) W, Z 2 = (5 + j 0) W, Z3 = (5 - jX C ) W

Determine the value of capacitance of XC such that the total current of the circuit will be in phase with the total voltage. What is then the circuit current and power? Sol: Here,

Z1 = (6.25 + j1.25) W, Z 2 = (5 + j 0) W, Z3 = (5 - jX C ) W

Z1

Z2

Z3 100 V, 50Hz

Circuit Theory and Networks

10.48

Total impedance, ZT = Z1 +

5 (5 - jX C ) Z 2 Z3 = (6.25 + j1.25) + Z 2 + Z3 10 - jX C

æ æ 25 X C ö 250 + 5 X C 2 ö = ç 6.25 + - jç - 1.25÷ 2 ÷ 2 100 + X C ø è è 100 + X C ø

The total current will be in phase with the total voltage if the impedance is purely resistive. \

æ 25 X C ö - 1.25÷ = 0 ç 2 è 100 + X C ø

Þ

X C 2 - 20 X C + 100 = 0

Þ

(XC

Þ

X C = 10

Þ

1 = 10 2p fC

Þ

C=

2

- 10) = 0

1 = 318 ì F Ans. 2p ´ 50 ´ 10

Putting XC = 10, impedance, ZT = 6.25 +

250 + 500 = 10 W 200

V 100 = = 10 A Ans. ZT 10

\

Current, I =

\

Power, P = I 2 R = 102 ´ 10 = 1 kW Ans.

24. In series RLC circuit with variable capacitance, the current is at maximum value with capacitance of 20 mF and the current reduces to 0.707 times maximum value with capacitance of 30 mF. Find the values of R and L. What is the bandwidth of the circuit if supply voltage is 20 sin (6.28 ´ 103) t volt? Sol: 6.28 ´ 103 w = = 1000 Hz 2p 2p We know that the maximum at resonance. At this condition the value of C is, C = 20 mF. Vm = 20, w = 6.28 ´ 103 ; \ f =

Here,

\

f0 =

1 2p LC

Þ L=

1 C (2p f 0 )

2

=

1 20 ´ 10

-6

´ (2p ´ 1000)

2

= 1.2665 mH Ans

Resonance

10.49

Here, with variable capacitance, L and f0 remains constant. At half power frequency, the current becomes 0.707 times the current at resonance with C = 30 mF. At this condition, the resistance of the circuit is equal to the reactance of the circuit. 1 é ù \ R = ( X L - X C ) = ê 2p ´ 1000 ´ 1.2665 ´ 10 -3 = 2.652 W Ans -6 ú 2p ´ 1000 ´ 30 ´ 10 û ë f0 R 2.652 = = = 333.26 Hz Ans Q 2p L 2p ´ 1.2665 ´ 10 -3 25. (a) Show that the sum of energy stored by the inductor and the capacitor connected in series at resonance at any instant is constant and is given by LI 2. (b) Show that the sum of energy stored by the inductor and the capacitor in parallel RLC circuit at resonance at any instant is constant and is given by CV 2. Sol: (a) Let i and v be the instantaneous current through the inductor and the voltage across the capacitor at any instant of time, t. Let, i = Im cos wt

\

Bandwidth, BW =

\

Energy stored in inductor, WL =

\

Energy stored in capacitor, WC =

1 2 1 Li = LI m 2 cos 2 w t 2 2

1 q2 2 C

t ù 1 é = ê ò idt ú 2C êë 0 úû

1 I m2 = 2C I 2 = m 2C

2

ét ù ê ò cos w tdt ú êë 0 úû

é æ sin w t ö t ù êç ú è ø÷ 0 ú ëê w û

2

2

=

I m2 1 ´ 2 sin 2 w t 2C w

=

I m2 L 2 æ sin w t çQ at resonance w 0 = 2 è

1 ö ÷ LC ø

Total energy stored at resonance, W = WL + WC =

(b) Let, v = Vm cos wt

1 1 LI m 2 cos 2 w t + sin 2 w t = LI m 2 = LI 2 2 2

(

)

[Proved]

Circuit Theory and Networks

10.50

\

Energy stored in Capacitor, WC =

\

Energy stored in Inductor, WL =

1 2 1 Cv = CVm 2 cos 2 w t 2 2

1 2 Li 2

t ù 1 é 1 = L ê ò vdt ú 2 êë 0 L úû

1 = Vm 2 2L V 2 = m 2L

2

ét ù ê ò cos w tdt ú êë 0 úû

é æ sin w t ö t ù êç ÷ ú êë è w ø 0 úû

2

2

Vm 2 1 ´ 2 sin 2 w t 2L w æ 1 = CVm 2 sin 2 w t çQ at resonance w 0 = è 2

=

1 ö ÷ LC ø

Total energy stored at resonance, 1 1 CVm 2 cos 2 w t + sin 2 w t = CVm 2 = CV 2 [ Proved ] 2 2 26. Determine the resonant frequency, the source current and the input impedance for the circuit shown in figure for each of the following cases:

(

W = WC + WL =

)

Case I

RL = 150 W

RC = 100 W

Case II

RL = 150 W

RC = 0 W

Case III

RL = 0 W

RC = 0 W

RL 200V

Sol: Case I: Here RL = 150 W, L = 0.24 H, RC = 100 W, and C = 3 mF Resonant frequency,

1 f0 = 2p 1 = 2p

2 1 æ L - CRL ö 1 ç ÷ = 2 LC è L - CRC ø 2p

1 0.24 ´ 3 ´ 10 -6

= 170 Hz Ans.

2 L 1 æ RL - C ö ç ÷ LC çè RC 2 - L ÷ø C

æ 1502 - 0.24 ö 3 ´ 10-6 ÷ ç ç 1002 - 0.24 ÷ è 3 ´ 10-6 ø

RC

0.24H

3mF

Resonance

10.51

Reactances at this frequency, X L = j 2p ´ 170 ´ 0.24 = j 256 W

XC = -

j 2p ´ 170 ´ 3 ´ 10-6

= - j312 W

\

IL =

200 = (0.34 - j 0.582) A 150 + j 256

and

IC =

200 = (0.186 + j 0.582) A 100 - j 312

\ Total source current, I = ( I L + I C ) = (0.34 - j 0.582) + (0.186 + j 0.582) = 0.526 A Ans. 200 = 380 W Ans. 0.526 Case II: Here RL = 150 W, L = 0.24 H, RC = 0, and C = 3 mF Resonant frequency,

\ Input impedance =

f0 =

1 2p

CRL 2 ö 1 æ 1 ç1 ÷ == LC è L ø 2p

1 0.24 ´ 3 ´ 10 -6

æ 3 ´ 10 -61502 ö 1 ç ÷ = 159 Hz Ans. 0.24 è ø

Reactances at this frequency, X L = j 2p ´ 159 ´ 0.24 = j 240 W

XC = -

j 2p ´ 159 ´ 3 ´ 10-6

= - j334 W

\

IL =

200 = (0.374 - j 0.598) A 150 + j 240

and

IC =

200 = j 0.598 A - j 334

\ Total source current, I = ( I L + I C ) = (0.374 - j 0.598) + j 0.598 = 0.374A Ans. 200 = 535 W Ans. 0.374 Case III: Here RL = 0, L = 0.24 H, RC = 0, and C = 3 mF Resonant frequency,

\ Input impedance =

f0 =

1 2p

1 1 == LC 2p

1 0.24 ´ 3 ´ 10 -6

= 188 Hz Ans.

Circuit Theory and Networks

10.52

Reactances at this frequency, X L = j 2p ´ 188 ´ 0.24 = j 283 W

XC = -

j 2p ´ 188 ´ 3 ´ 10-6

\

IL =

200 = - j 0.706 A j 283

and

IC =

200 = j 0.706 A - j 283

= - j 283 W

\ Total source current, I = ( I L + I C ) = 0 A Ans. 200 = ¥ Ans. 0 27. A voltage of v = 2000 sin wt + 400 sin 3 wt + 100 sin 5 wt is applied to a series circuit having R = 10 W and C = 30 mF and a variable inductance. (i) Find the value of inductance so as to give resonance at 3rd harmonic frequency. (ii) What are the r.m.s. values of voltage and current with this inductance in circuit? Take w = 300 rad/s. Sol: Here, w = 300 rad/s, R = 10 W, C = 30 ´ 10–6 F \ Resonant frequency w0 = 3 ´ w = 900 rad/s

\ Input impedance =

(i) Q

1 LC

= 900 Þ L =

1 900 ´ 30 ´ 10 -6 2

= 41.152 mH Ans.

(ii) For 1st Harmonic (w = 300 rad/s):

R = 10W, X L1 = 300 ´ 41.152 ´ 10-3 = 12.3456 W X C1 =

-j 300 ´ 30 ´ 10-6

= - j111.11 W

\

Z1 = 10 + j (12.3456 - 11.11) = (10 - j98.77) W = 99.27Ð95.78° (W)

\

I1 =

2000 2000 = = 20.15Ð - 95.78° ( A ) Z1 99.27Ð95.78°

For 3rd Harmonic (w = 900 rad/s): \ R = 10W, X L3 = 900 ´ 41.152 ´ 10-3 = 37.037 W, X C3 = \

Z3 = 10 + j (37.037 - 37.037) = 10 W

\

I3 =

400 400 = = 40Ð0° ( A ) Z 3 10Ð0°

-j 900 ´ 30 ´ 10-6

= - j 37.037 W = X L

Resonance

10.53

For 5th Harmonic (w = 1500 rad/s):

R = 10 W, X L5 = 1500 ´ 41.152 ´ 10-3 = 61.728 W, X C5 =

-j 1500 ´ 30 ´ 10-6

= - j 22.22 W = X L

\

Z5 = 10 + j (61.728 - 22.22) = (10 + j39.506) W = 40.752Ð75.795° (W)

\

I5 =

100 100 = = 2.454Ð - 75.795° ( A) Z5 40.752Ð75.795°

\ RMS value of the current,

I12 + I32 + I52 = 2 \ RMS value of the voltage, I rms =

Vrms

V 2 + V32 + V52 = 1 = 2

(20.15)2 + 402 + (2.454)2 2

(2000)2 + (400)2 + (100)2 2

= 31.72 ( A) Ans.

= 1041.63 (V) Ans.

SUMMARY 1. In an electrical system, the phenomenon of cancellation of reactances when inductor and capacitor are in series or cancellation of susceptances when they are in parallel, is termed as resonance. 2. In series resonance, the current at resonance is the maximum, whereas is case of parallel resonance, the current at resonance is the minimum. 3. Quality factor for series resonant circuit is

w0 L 1 L 1 or or , whereas quality factor R C R w 0 RC

R C or w0RC or R . w0 L L 4. Series resonant circuit acts as a voltage amplifier whereas parallel resonant circuit acts as a current amplifier. 5. Under resonant condition, all circuits act as resistive so that the power factor of the circuit is unity. for parallel circuit is

6. Both for series and parallel resonant circuit, resonant frequency is w 0 = is BW =

1 LC

and bandwidth

f0 . Q

7. The concept resonant is useful in oscillator circuits, voltage amplifier, current amplifier, in RF amplifier and other filter circuits.

Circuit Theory and Networks

10.54

EXERCISES 1. A generator supplies a variable frequency voltage of constant amplitude 100 V(rms) to a series RLC circuit where R = 5 W, L = 4 mH and C = 0.01 mF. The frequency is to be varied until maximum current flows. Predict the maximum current, the frequency at which it occurs, and the resulting voltage across the inductance and the capacitance. [20 A(rms); 7,958 Hz; 4 kV] 2. If the bandwidth of a resonant circuit is 10 kHz and the lower half power frequency is 120 kHz, find out the value of the upper half power frequency and the quality factor of the circuit. [130 kHz, 12.5] 3. A series RLC circuit consists of a 100 W resistor, an inductor of 0.318 H and a capacitor of unknown value. When this circuit is energized by 230Ð0° V, 50Hz sinusoidal ac supply, the current was found to be 2.3Ð0° A. Find (i) the value of capacitor in micro-farad (ii) the voltage across the inductor (iii) the total power consumed. [31.86 mF; 230Ð90° V (leading); 529 W] 4. For a series RLC circuit the inductor is variable. Source voltage is 200 2 sin 100 p t . Maximum current obtainable by varying the inductance is 0.314 A and the voltage across the capacitor then is 300V. Find the circuit element values. [900 W; 3.332 mF; 3.04 H] 5. It is desired to design a series resonant circuit with the following specifications. C = 250 ´ 10-12 F, f 0 = 600 kHz, BW = 20 kHz

Calculate Q0, R and L of the circuit. Also, calculate the current at 500 kHz as a fraction of the current at resonance. [30; 35.37 W; 0.28 mH; 0.09 Ir] 6. A 10 mH coil is connected in series with a loss free capacitor to a variable frequency source of 20 V. The current in the circuit has the maximum value of 0.2 A at a frequency of 100 kHz. Calculate: (i) the value of capacitance (ii) the Q-factor of the coil (iii) the half power frequencies. [253.3 pF, 62.83, 99.204 Kz, 100.796 Hz] 7. A series RLC circuit is excited from a constant voltage variable frequency source. The current in the circuit becomes maximum at a frequency at

600 Hz and falls to half the maximum value 2p

400 Hz. If the resistance in the circuit is 3 W, find L and C. 2p

[10.4 mH; 267.3 mF] 8. An RLC series circuit has R = 100 ohm, L = 500 mH and C = 40 mF. Calculate: (i) The resonant frequency; (ii) Lower half power frequency;

Resonance

10.55

(iii) Upper half power frequency; (iv) Bandwidth; (v) Q factor. Also derive the expression for the above. 9.

10.

11.

12.

[223.6 rad/s; 22.5 rad/s; 222.5 rad/s; 200; 1.118] Calculate the value of C which results in resonance for the circuit shown in figure when frequency is 1000 Hz and find Q-factor for each branch. 4⍀ 5⍀ [31.83 mF, 2 (for R-L), 1 (for R-C)] A capacitor is connected in parallel with a coil having L = 5.52mH and R j 8⍀ C = 10 W, to a 100 V, 50 Hz supply. Calculate the value of the capacitance for which the current taken from the supply is in phase with the voltage. [53.6 mF] A parallel circuit has fixed C and variable L. Quality factor of the inductor R ⫹ is Q = 4. Find the values of L and C for the circuit impedance of (100 + j0) C ⫺ ohm at f = 2.4 MHz. What is the bandwidth at matched condition? L [1.557 mH, 2.648 nF, 1.2 MHz] Find the resonant frequency for the circuit shown in figure. [263 Hz] 5V

2V 40 mF

10 mH

13. For a practical tank circuit shown in figure the resonance occurs at 1 MHz. Assuming a highQ coil, find the quality factor of the high-Q coil at resonance frequency. 10⍀ 50pF L

[2000] 14. Calculate the impedance of the parallel tuned circuit, as shown in figure at a frequency of 500 kHz and for the bandwidth of operation equal to 20 kHz. The resistance of the coil is 5 W. [3.13 kW] 15. A coil of resistance R and inductance L is shunted by a capacitor C. Show that: (a) the resonant frequency is, w 0 =

R2 1 - 2 . (b) the LC L

C effective resistance is L CR . (c) the circulating current is V . L

V

⫹ ⫺

R C L

Circuit Theory and Networks

10.56

16. For the parallel resonant circuit, prove that: L

RL

RC

XC V

(a) The resonant frequency is 1 LC

RL 2 - L C . RC 2 - L C

(b) The impedance is independent of frequency if RL = RC = L . C 17. What is the condition for resonance in the circuit shown in figure? For what value of L the circuit will resonate at all frequencies? L

⫹

V1

⫺

1⍀

1⍀

1H

1F

[w = 0; L = 0] 18. Find the resonant frequency for the circuits. Also, find the quality factor, Q. What will be the impedances at resonance? C

L

C

(i)

(ii)

R

1 1 R , w 0 RC , 2 2 2 LC R 2 C 2 1+w R C R

(

L CR 2 - L

)

,

R L , w 0 L CR

L

R

Resonance

10.57

QUESTIONS 1. (a) What is resonance in an ac circuit? (b) Discuss the effects of resonance in electrical systems. 2. Discuss briefly the phenomenon of electrical resonance in simple RLC circuits. Derive an expression for the condition for resonance in RLC circuits. Also, draw the phasor diagrams. Or, State and explain the condition of resonance in a series RLC circuit of an ac circuit. Draw the phasor diagram. 3. Show that w r = w l w h for a series RLC circuit, where w r = the resonant frequency w l = lower half power frequency w h = upper half power frequency of the circuit. Or, Show that the resonant frequency w0 of a series RLC circuit is the geometric mean of w1 and w2, the lower and upper half power frequencies respectively. 4. (a) Define the terms Q factor and bandwidth. (b) Derive expression for Q factor for RL and RC series circuit. (c) Define the Q factor for the series resonant circuit and express it in terms of the circuit parameters. (d) What is the relationship between bandwidth and quality factor for a RLC circuit? Or, Show that:

f 2 - f1 1 Or, = f0 Q0

Quality factor =

Resonant frequency Bandwidth

Where, f2 and f1 are half power frequencies, f0 is the resonant frequency and Q0 is the Q factor at resonant frequency.

5. 6. 7. 8.

(e) Show that: (i) f1 f2 = f02 and (ii) f12 + f22 ³ 2f02 where f0 is the resonance frequency and f1 , f2 are the half power frequencies of a series resonant circuit (a) Define Selectivity and half power frequency. (b) Show that a circuit must have a large value of Q0 (Q factor at resonance frequency) to be highly selective. Prove that in a series resonant circuit, the voltage across capacitor and inductor is Q factor times the supply voltage. In a series RLC circuit, the voltage across L and C at resonance may exceed even the supply voltage. Why? Explain the effect of increase in L/C ratio on the following factors: (a) resonant frequency, (b) Q, (c) bandwidth of an RLC series circuit.

Circuit Theory and Networks

10.58

9. In an RLC series circuit, the source frequency is varied from zero to infinity. How do the values of voltage across L and C change? Draw curve showing these variations. Derive expression for maximum values of these voltages and the frequencies at which the maximum values occur. 10. (a) Describe the phenomenon of resonance in parallel circuits and explain its Q factor. (b) Prove that the Q factor of parallel resonant circuit is reciprocal of that for a series resonant circuit. (c) Find an expression for impedance at the antiresonance of a parallel tuned circuit and also sketch the variation of the impedance of the same circuit with frequency. 11. (a) Compare the properties of series and parallel resonance. (b) The series resonant circuit is often regarded as the acceptor circuit and the parallel resonant as the rejector circuit. Explain. (c) The series resonance is called voltage resonance and the parallel resonance the current resonance- why? 12. At resonance, the current is maximum in series circuit and minimum in parallel circuit. Why? 13. For the circuit shown, draw the phasor diagram. Derive the condition for the two branch currents, IL and IC to be in quadrature. RL V

⫹ ⫺

IL

RC IC

L

C

14. The shape of resonance curve depends on Q of the coil. Why? 15. Derive an expression for the resonant frequency of a parallel circuit consisting of inductance L and resistance R in one branch and a capacitance C in the other. 16. A coil of resistance R and inductance L is shunted by a capacitor. Show that for rejector (parallel) resonance, the effective resistance is

V

L . Show also that the circulating current is CR

C so long as the resistance is small. L

Or, A resistive inductive coil is connected in parallel with a condenser and the combination is connected with a sinusoidal emf. Derive the expression for current flowing through the two branches and hence find the condition for obtaining the minimum input current and the value of the maximum impedance of the circuit. 17. A series combination of a capacitance C and resistance RC is shunted by an inductive coil having resistance RL and inductance L and the combination is connected to an ac source. Derive the expression for the resonant frequency of the circuit. Also, show that the impedance of the circuit will be independent of frequency (or, the circuit will be under resonant condition at any frequency) if RC = RL =

L . C

Resonance

10.59

18. (a) Show that the sum of energy stored by the inductor and the capacitor connected in series at resonance at any instant is constant and is given by LI2. (b) Show that the sum of the energy stored by the inductor and the capacitor in parallel RLC circuit at any instant is constant at resonance frequency and is equal to CV2.

Circuit Theory & Networks (EE301) Solution of 2011 WBUT Paper GROUP–A (Multiple-Choice Questions) 1. Choose the correct alternatives for any ten of the following: (i) The internal impedance of an ideal current source is (a) zero (b) infinite (c) both (a) and (b)

10 × 1 = 10 (d) none of these

(ii) In the figure given below, the value of the resistance R in ohms is

(a) 10

(b) 20

(c) 30

(d) 40

(iii) Time constant of the network shown below is

(a) CR (b) 2CR (c) CR/4 (d) CR/2 (iv) For a series RC circuit, when subjected to a unit step input voltage, the voltage across the capacitor will be (a) 1 - e

-t

RC

(b) e

-t

RC

(c) e

t

RC

(d) 1

SQP.2

Circuit Theory and Networks

(v) In the figure given below, the value of the load Z which maximizes the power delivered to it is

(a) 60 + j 40

(b) 60 – j40

(c) 60

(d) none of these

–at

(vi) If the unit step response of a network is (1 – e ), the unit impulse response will be 1 1 -a t (c) e -a t ae a (vii) The resistances R1, R2 and R3 are respectively (a) ae–at

(b)

(a) 1, 3/2 and 3

(b) 3, 3/2 and 6

(c) 9, 3 and 1

(d) (1 – a)e– at

(d) 2, 1 and 9

È 0.9 0.2 ˘ (viii) The z-matrix of a 2-port network is given by Í ˙ . The element Y22 of the correspondÎ0.2 0.6 ˚ ing Y-matrix of the same network is given by (a) 1.2 (b) 0.4 (c) −0.4 (d) 1.8 (ix) The transfer function of an electric low-pass RC network is 1 RCs RC (a) (b) (c) 1 + RCs 1 + RCs 1 + RCs (x) How many branches can be connected to a node? (a) 1 (b) 2 (c) 3

(d)

s 1 + RCs

(d) any number

(xi) When a number of 2-port networks are connected in cascade, the individual (b) YSC matrices are added (a) ZOC matrices are added (c) chain matrices are multiplied (d) H matrices are multiplied

SQP.3

Solution of 2011 WBUT Paper

(xii) The tie-set matrix gives the relation between (a) branch currents and link currents (b) branch voltages and link currents (c) branch currents and link voltages (d) none of these Solutions (i) (b) infinite

(ii) (b) 20

(iii) (a) CR

(iv) (a) 1 - e

-t

RC

–at

(v) (b) 60 – j40 (vi) (a) ae (vii) (a) 1, 3/2 and 3 (viii) (d) 1.8 1 (ix) (b) (x) (d) any number (xi) (a) ZOC matrices are added 1 + RCs (xii) (a) branch currents and link currents

GROUP–B (Short-Answer-Type Questions) Answer any three of the following questions.

3 × 5 = 15

2. Convert the current sources into voltage sources (equivalent) and find the voltage V0.

Solution: Converting the current sources into voltage sources, we get the following circuit.

\ \

i= -

20 10 =- A 6 3

Ê 10 ˆ 10 V0 = 2i + 10 = 2 ¥ Á - + 10˜ = = 3.33 Volt Ë 3 ¯ 3

Ans.

SQP.4

Circuit Theory and Networks

3. For the network given below, determine the X-parameters.

Solution: (NOTE: Since “X-parameters” do not exist, we find Z-parameters.) We consider two cases: When I2 = 0: The modified circuit is shown in Fig. (a).

Fig. (a) By KVL for the middle mesh, we get, I + 3V1 + I – 2 × (I1 – I) = 0 3 ˆ Ê1 I = Á I1 - V1 ˜ Ë2 4 ¯ By KVL for the left mesh, we get, fi

3 ˆ Ê1 V1 = 2 ¥ ( I1 - I ) = 2 I1 - 2 Á I1 - V1 ˜ {by Eq. (i)} Ë2 4 ¯ or,

V1 = –2I1

\

z11 =

V1 I1

= -2 W I2 = 0

Also, by KVL for the right mesh, we get, 3 ˆ 1 3 Ê1 V1 = I = Á I1 - V1 ˜ = I1 - ( -2 I1 ) = 2 I1 Ë2 4 ¯ 2 4 \

z21 =

V2 I1

=2W I2 = 0

(i)

Solution of 2011 WBUT Paper

SQP.5

When I1 = 0: The modified circuit is shown in Fig. (b).

Fig. (b) By KVL for the middle mesh, we get, I – 3V1 + 2I – 1 × (I2 – I) = 0 fi

3 ˆ Ê1 I = Á I 2 + V1 ˜ Ë4 4 ¯

(ii)

By KVL for the left mesh, we get,

fi

3 ˆ 1 3 Ê1 V1 = 2 I = 2 ¥ Á I 2 + V1 ˜ = I 2 + V1 Ë4 4 ¯ 2 2 V1 = –I2

\

z12 =

V1 I2

= -1 W I1 = 0

Also, by KVL for the right mesh, we get, 3 ˆ Ê1 V2 = 1 ¥ ( I 2 - I ) = I 2 - Á I 2 + V1 ˜ Ë4 4 ¯ = \

z22 =

V2 I2

{by Eq. (ii)}

3 3 I 2 - ¥ ( - I 2 ) = 1.5 I 2 4 4

= 1.5 W I1 = 0

Therefore, the z-parameters of the network are, È -2 -1 ˘ [z]= Í ˙ (W) Î 2 1.5˚

Ans.

4. In the circuit given below, the switch is initially in the position 1 until the steady state is reached. At t = 0, the switch is moved to the position 2. Find i(t), the loop current.

SQP.6

Circuit Theory and Networks

Solution:

When the switch is in the position 1, steady state exists and the initial voltage across the capacitor is, v(0–) = 10 V After the switch is moved to the position 2, the KVL gives, in Laplace transform, 1 10 =0 I (s) + 20 ¥ I (s) + -6 s 10 ¥ 10 s 0.5 I ( s) = or, s + 5000 Taking inverse Laplace transform, i(t) = 0.5e–5000t (A); t > 0 5. (a) Define incidence matrix. (b) For the graph shown below, find the complete incidence matrix.

Ans. 2 3

SQP.7

Solution of 2011 WBUT Paper

Solution: (a) Incidence Matrix: The incidence matrix symbolically describes a network. It also facilitates the testing and identification of the independent variables. Incidence matrix is a matrix which represents a graph uniquely. For a given graph with ‘n’ nodes and ‘b’ branches, the complete incidence matrix Aa is a rectangular matrix of order n × b, whose elements have the following values: Number of columns in [A] = Number of branches = b Number of rows in [A] = Number of nodes = n Aij = 1, if branch j is associated with node i and oriented away from node j. = –1, if branch j is associated with node i and oriented towards node j. = 0, if branch j is not associated with node i. Example:

Fig. (a) Network

Fig. (b) Graph of Network

Incidence matrix A Branches

Nodes Reference node

1

2

3

4

5

6

a

1

0

0

−1

0

0

b

0

1

0

1

−1

0

c

0

0

1

0

1

−1

d

−1

−1

−1

0

0

1

¸ ˝ ˛

Reduced incidence matrix AI

(b) The given graph does not have any orientation. We assume the orientation as shown in the figure below.

SQP.8

Circuit Theory and Networks

For the graph shown in the figure, the complete incidence matrix can be written as, a

b

c

d

e

f

1 È -1 1 0 0 0 1 ˘ 2 Í 0 -1 -1 -1 0 0 ˙˙ Aa = Í 3 Í 0 0 0 1 -1 -1˙ Í ˙ 4Î 1 0 1 0 1 0 ˚ 6. Find the Fourier transform for the following gate function:

Solution: The pulse is,

f(t) = 1, -

T T 0

For t > 0, the circuit becomes as shown. By KVL, E R1I(s) + sLI(s) – Li(0–) = s E EL + fi [R1 + sL]I(s) = s R1 + R2 fi

È ˘ ˙ E Í 1 E I(s) = Í ˙+ R R1 Í Ê R1 + R2 s s+ 1 ˆ˙ L ¯ ˙˚ ÍÎ Ë

Ê ˆ Á ˜ 1 Á ˜ Á Ê s + R1 ˆ ˜ L¯ ¯ ËË

SQP.4

Circuit Theory and Networks

Taking inverse Laplace transform, i(t) =

R R - ÊÁ 1 ˆ˜ t ˆ - ÊÁ 1 ˆ˜ t - (1 )t ˆ E Ê E 10 Ê 10 -(11 )t L Ë L¯ e Ë ¯ = Á1 - e 1 ˜ + e Á1 - e ˜+ ¯ 1+ 2 R1 Ë 1 Ë ¯ R1 + R2

= 10 (1 - e - t ) +

10 - t 20 - t e = 10 e 3 3

= 10 - 6.67e - t

( A), t > 0

Ans.

3. Find the Laplace transform of the triangular waveform shown in the figure.

Solution: The equation of the given waveform can be written as, f(t) =

2 4 Ê Tˆ 2 r (t ) - r Á t - ˜ + r (t - T ) 2¯ T T T Ë T - s

2 1 4e 2 2e -Ts + F(s) = T s2 Ts2 Ts2

=

2 Ts2

2 2 È -Ts -Ts ˘ È 2 + e - Ts ˘ = 2 e e 1 2 1 ÍÎ ˙˚ Ts2 ÍÎ ˙˚

\ F ( s) =

2 Ts2

-Ts ˘ È 2 ÍÎ1 - e ˙˚

2

4. Find the y-parameters for the following networks shown in the figure.

Solution: Since this is a T-network, the z-parameters are easily obtained as, z11 = (10 + 40) = 50 W z12 = z21 = 40 W

SQP.5

Solution of 2012 WBUT Paper

z13 = (5 + 40) = 45 W Hence, the y-parameters are obtained as, y11 =

z22 45 9 = = 2 Dz 50 ¥ 45 - 40 130

y12 = y21 = y22 =

z12 40 4 ==Dz 650 65

z11 50 1 = = 2 Dz 50 ¥ 45 - 40 13

5. Define incidence matrix of a graph matrix. È 0 -1 1 Í [A] = Í 0 0 -1 ÍÎ -1 0 0

and draw the orientation graph from the reduced incidence 1 0˘ -1 -1˙˙ 0 1 ˙˚

Solution: �

Incidence Matrix: The incidence matrix symbolically describes a network. It also facilitates the testing and identification of the independent variables. Incidence matrix is a matrix which represents a graph uniquely. For a given graph with n nodes and b branches, the complete incidence matrix Aa is a rectangular matrix of order n ¥ b, whose elements have the following values: Number of columns in [A] = Number of branches = b Number of rows in [A] = Number of nodes = n Aij = 1, if branch j is associated with node i and oriented away from node i. = –1, if branch j is associated with node i and oriented towards node i. = 0, if branch j is not associated with node i. Example

(a) Network

(b) Graph of Network

SQP.6

Circuit Theory and Networks Branches

Nodes Reference node

1

2

3

4

5

6

a

1

0

0

−1

0

0

b

0

1

0

1

−1

0

c

0

0

1

0

1

−1

d

−1

−1

−1

0

0

1

\

Reduced incidence matrix AI

Reduced incidence matrix [A]: The matrix obtained from Aa by eliminating one of the rows is called reduced incidence matrix. In other words, suppression of the datum node (reference node) from the incidence matrix results in the reduced incidence matrix. For the graph shown above, the reduced incidence matrix is given as, È1 0 0 -1 0 0 ˘ Í ˙ A = Í0 1 0 1 -1 0 ˙ ÍÎ0 0 1 0 1 -1˙˚ �

Solution to Numerical Problem: From the property that for complete incidence matrix, the summation of all entries in any column must be zero, the complete incidence matrix is obtained as, 1

2

3

4

5

A È 0 -1 1 1 0 ˘ B Í 0 0 -1 -1 -1˙˙ Aa = Í 1˙ C Í -1 0 0 0 Í ˙ D Î 1 1 0 0 0˚ The graph is shown in the figure. 6. For the circuit shown in the figure, find the value of the current i.

Solution: By KVL for the meshes, we get,

Solution of 2012 WBUT Paper

1 × I1 + 1 × I1 = 5 fi I1 = 2.5 A 3 × i + 1 × i = 4Vab fi Vab Now, from the middle loop, we have, I 1 × I1 – 1 × i – Vab = 0 fi i = 1 = 1.25 A 2 \ i = 1.25 A 7. Explain under what condition, an RC series circuit behaves as (i) Low-pass filter (ii) Integrator

SQP.7

Ans.

Solution: (i) RC Series Circuit as Low-Pass Filter: If the RC series circuit is supplied with a frequencyvarying source then it will act as a low-pass filter if the output is taken as the voltage across the capacitor. The voltage across the capacitor is IXC = 1/wC. The voltage across the series combination Ê 1 ˆ is: IZ = I R 2 + Á Ë w C ˜¯

2

So the gain is V IX g ∫ out = C = Vin IZ

g= \

1 wC Ê 1 ˆ R2 + Á Ë w C ˜¯

2

1 1 + (w RC )2

Here, at low frequencies, capacitive reÊ ˆ actance XC = 1 is very high and, ÁË j 2p fC ˜¯ therefore, the circuit can be considered an open circuit. Under these conditions, the input signal is equal to the output signal. At very Ê 1 ˆ high frequencies, the capacitive reactance Á XC = is very low and, therefore, the j 2p fC ˜¯ Ë output signal is very small as compared with the input signal. Thus, the circuit acts as a low-pass filter with the frequency characteristics as shown in the figure. (ii) RC Series Circuit as Integrator: We have an ac source with voltage vin(t), input to an RC series circuit. The output is the voltage across the capacitor. We consider only high frequencies w >> 1/RC, so that the capacitor has insufficient time to charge up, its voltage is small. So the input voltage approximately equals the voltage across the resistor.

SQP.8

Circuit Theory and Networks

Ê 1 ˆ Vin = IZ = I R + Á Ë w C ˜¯

2

2

1 , so Vin @ IR R 1 For frequencies, w >> , Vin @ VR RC But

\

w C >>

Vout = VC = Vout @

1 1 V idt @ Ú in dt Ú C C R

1 Vin dt RC Ú

\ Thus, the voltage vC is the integration of the input voltage and, hence, the RC series circuit acts as an integrator. GROUP–C (Long-Answer-Type Questions) Answer any three questions.

3 × 15 = 45

8. (a) Find the Z-parameter and ABCD parameter of the circuit given in the figure.

(b) Express h-parameter in terms of Y-parameter of a two-port network. (c) What is the cascade connection between two 2-port networks? Explain with a diagram. 7+4+4 Solution: (a) We consider two cases: Case (1): When I2 = 0 By KVL, we get, 5I1 – j5I1 = V1 fi z11 =

V1 I1

= (5 - j 5) W I2 = 0

\ V2 = I1 ¥ ( - j 5) fi z21 ==

V2 I1

= - j5 W I2 = 0

Solution of 2012 WBUT Paper

SQP.9

Case (2): When I1 = 0 By KVL, we get, I2 (1 + j2 – j5) + 3I2 = V2 fi z22 =

\

V2 I2

= (4 - j 3) W I1 = 0

V1 = I 2 ¥ ( - j 5) + 3I 2 fi z12 ==

V1 I2

= (3 - j 5) W I1 = 0

By interrelationship, the ABCD parameters are obtained as, A=

z11 5 - j 5 = = (1 + j1) z21 - j5

B=

Dz (5 - j 5) ¥ (4 - j 3) - (3 - j 5) ¥ ( - j 5) = = (4 + j 6) W z21 - j5

C=

1 1 = = j 0.2 z21 - j 5

D=

z22 (4 - j 3) = = (0.6 + j 0.8) z21 - j5

Hence, the required parameters are: È(5 - j 5) (3 - j 5) ˘ È(1 + j1) (4 + j 6) W ˘ Z= Í ˙ (W) and T = Í ˙ j 5 (4 j 3) Î ˚ (0.6 + j 0.8)˚ Î j 0.2

Ans.

(b) h-parameter in terms of y-parameters The h-parameter equations are V1 = h11 I1 + h12V2 I 2 = h21 I1 + h22V2

(1)

The y-parameter equations are: I1 = y11V1 + y12V2 I 2 = y21V1 + y 22V2

(2)

We have to express Eq. (2) in the form of Eq. (1). From the first equation of Eq. (2), we get, Ê 1 ˆ Ê y ˆ V1 = Á ˜ I1 + Á - 12 ˜ V2 Ë y11 ¯ Ë y11 ¯

(3)

SQP.10

Circuit Theory and Networks

Replacing this in the second equation of Eq. (2), we have, ÈÊ 1 ˆ Êy ˆ ˘ Êy ˆ Ê Dy ˆ I2 = y21V1 + y 22V2 = y21 ÍÁ ˜ I1 - Á 12 ˜ V2 ˙ + y 22V2 = Á 21 ˜ I1 + Á ˜ V2 (4) Ë y11 ¯ ˚˙ Ë y11 ¯ Ë y11 ¯ ÎÍË y11 ¯ (where, Dy, = y11 y22 – y12 y21) Comparing Eq. (1), (3) and (4), we get, h11 =

1 y11

h12 = -

y12 y11

h21 =

y21 y11

h22 =

Dy y11

(c) Cascade Connection between Two 2-Port Networks: A cascade connection is defined when the output of one network becomes the input to the next network.

It can be easily seen that Ir2 = Is1 and Vr2 = Vs1 Therefore, it can easily be seen that the ABCD parameters are the most suitable to be used for this connection. È Ar ÈVr1 ˘ Í I ˙ = ÍC Î r Î r1 ˚

Br ˘ ÈVr 2 ˘ ÈVs1 ˘ È As = , Dr ˙˚ ÍÎ I r 2 ˙˚ ÍÎ I s1 ˙˚ ÍÎCs

ÈVr1 ˘ È Ar ÈV1 ˘ Í I ˙ = Í I ˙ = ÍC Î r1 ˚ Î r Î 1˚ ÈA = Í r ÎCr

Br ˘ È As Dr ˙˚ ÍÎCs

Br ˘ ÈVr 2 ˘ È Ar = Dr ˙˚ ÍÎ I r 2 ˙˚ ÍÎCr

Bs ˘ ÈVs 2 ˘ Ds ˙˚ ÍÎ I s 2 ˙˚ Br ˘ ÈVs1 ˘ È Ar = Dr ˙˚ ÍÎ I s1 ˙˚ ÍÎCr

Br ˘ È As Dr ˙˚ ÍÎCs

Bs ˘ ÈVs 2 ˘ Ds ˙˚ ÍÎ I s 2 ˙˚

Bs ˘ ÈV2 ˘ Ds ˙˚ ÍÎ I 2 ˙˚

Thus, it is seen that the overall ABCD matrix is the product of the two individual ABCD matrices. This is a very useful property in practice, especially when analyzing transmission lines. ÈA B˘ È Ar Br ˘ È As Bs ˘ ÍC D ˙ = Í ˙ ˙Í Î ˚ ÎCr Dr ˚ ÎCs Ds ˚ 9. (a) Draw the circuit diagram of a first-order high-pass filter and find out the expression for the cutoff frequency. (b) Draw and explain the characteristics of an ideal band-pass and an ideal band-stop filter. (c) The circuit shown in the figure is a second-order low-pass filter. Analyze the circuit and find the cut-off frequency.

Solution of 2012 WBUT Paper

SQP.11

5+5+5 Solution: (a) First-Order High-Pass Active Filter: The circuit is shown in the figure.

The filtering is done by the CR network and the op-amp is connected as a unity-gain follower. The feedback resistor, Rf is included to minimize dc offset. Here,

Vy = V0

R1 R1 + R f

Voltage across the resistor R, Vx =

R Vi = R + Xc

R 1 R+ jw C

Vi =

jw RC Vi 1 + jw RC

Since op-amp gain is infinite, Vx = Vy fi fi

V0 R1 jw RC = Vi R f + R1 1 + jw RC V0 Ê R f + R1 ˆ Ê jw RC ˆ j 2p fRC =Á = AF ¥ ˜ Á ˜ 1 + j 2p fRC Vi Ë R1 ¯ Ë 1 + jw RC ¯

(2)

SQP.12

Circuit Theory and Networks

where,

AF = (1 + Rf /R1) = pass-band gain of the filter, f = frequency of the input signal (Hz), fc =

1 cut-off frequency of the filter (Hz). 2p RC

The gain magnitude, V0 AF (2p fRC ) w RC = = AF . Vi 1 + (2p fRC )2 1 + w 2 R 2C 2 For this magnitude to be AF 2

or,

=

fc =

AF 2

at f = fc, we have,

AF (2p c fRC ) 1 + (2p c fRC )2 1 2p RC

This is the cut-off frequency of the high-pass filter. (b) Band–Pass Filter: It is a circuit that passes a band of frequencies and attenuates all frequencies outside the band.

The bandwidth of a band-pass filter is the difference between the upper and lower cut-off frequencies. Depending on the value of bandwidth, band-pass filters are of two types: 1. Wide Band-Pass Filter: This is characterized by high bandwidth or low Q-factor, i.e., Q £ 0.5. 2. Narrow Band-Pass Filters: This is characterized by small bandwidth or high Q-factor, i.e., Q > 0.5. An ideal band-pass filter should have the following characteristics: 1. It should have a completely flat pass-band (e.g., with no gain attenuation throughout). 2. It should completely attenuate all frequencies outside the pass-band. 3. The transition out of the pass-band should be instantaneous in frequency.

Solution of 2012 WBUT Paper

SQP.13

In practice, no band-pass or band-stop filter is ideal. The filter does not attenuate all frequencies outside the desired frequency range completely; in particular, there is a region just outside the intended pass-band where frequencies are attenuated, but not rejected. This is known as the filter roll-off, and it is usually expressed in dB of attenuation per octave or decade of frequency. Band Stop Filter: It rejects a specified band of frequencies while passing all other frequencies outside the band. If a band-stop filter has a narrow stop-band (i.e., high Q-factor) then the filter is known as a notch filter.

An ideal band-stop filter should have the following characteristics. 1. It should have a completely flat pass-band (e.g., with no gain attenuation throughout). 2. It should completely attenuate all frequencies within the stop-band. 3. The transition from pass-band to stop band should be instantaneous in frequency. (c) [WBUT 2011 Q.11 (a)] 10. (a) Find the inverse Laplace transform of F(s). F(s) =

s +1 s(s + 4s + 4) 2

(b) The circuit in the figure was in steady state with the switch in position 1. Find current i(t) for t > 0 if the switch is moved from position 1 to 2 at t = 0.

SQP.14

Circuit Theory and Networks

(c) Determine the Laplace transform of the periodic square pulse train of amplitude as shown in the figure.

Solution: (a)

Let,

F(s) =

s +1 s +1 = s(s + 4s + 4) s(s + 2)2

F(s) =

s +1 A B C = + + 2 2 s (s + 2) (s + 2) s(s + 2)

2

By residue method, A=

C=

\

F(s) =

s +1 (s + 2)2

= s=0

1 s +1 1 ;B= = ; 4 s s = -2 2

d È s + 1˘ 1 =- 2 Í ˙ ds Î s ˚ s = -2 s

=s = -2

1 4

s +1 1 1 1 = + 2 2 4s 2(s + 2) 4(s + 2) s(s + 2)

Taking inverse Laplace transform, 1 1 È1 ˘ f(t) = Í u(t ) - e -2 t + te -2 t ˙ 4 2 Î4 ˚

Ans.

(b) When the switch is in position 1, steady state exists and the initial current through the inductor is, 20 i(0 –) = =2A 10 After the switch is moved to position 2, the KVL gives, in Laplace transform, 10I(s) + 0.4sI(s) – 0.4 × 2 =

60 s

Solution of 2012 WBUT Paper

SQP.15

150 2 1 ˘ 2 È1 + = 6Í ˙ + s + 25 s(s + 25) s + 25 s s + 25 Î ˚ Taking inverse Laplace transform, or,

I(s) =

i(t) = 6 – 4e–25t (A); t > 0; (c) The given waveform can be written as,

Ans.

Ê Tˆ Ê 3T ˆ f (t ) = au(t ) - Au Á t - ˜ + Au(t - T ) - Au Á t + Au(t - 2T ) - ... Ë ¯ Ë 2 2 ˜¯ Taking Laplace transform, A A -Ts A A -3Ts A F (s) = - e 2 + e -Ts - e 2 + e -2Ts - ... s s s s s Ts 3 Ts AÈ ˘ = Í1 - e 2 + e -Ts - e 2 + e -2Ts - ...˙ sÎ ˚ ˘ 1 1 ¸ AÈ Ï = Í ˙ Ans Ì since summation of an infinite GP series is = ˝ Ts s ÎÍ 1 - e - 2 ˚˙ 1 - CR ˛ Ó \ F ( s) =

˘ AÈ 1 Í ˙ Ts s ÍÎ 1 - e - 2 ˙˚

11. (a) Find the Fourier expansion of the following waveform shown in the figure.

(b) Determine the Fourier transform and sketch the amplitude and phase spectrums of the function f (t ) = Ve - t / a for t ≥ 0 8+7 =0 for t £ 0 Solution: (a) The waveform has both the odd and half-wave symmetry. \ a0 = 0 an = 0 Also, the waveform will contain only the odd harmonics. \

bn =

4 T

T /2

Ú 0

f (t )sin nw tdt; n is odd only

SQP.16

Circuit Theory and Networks T

4 = T

2

Ú 0

T /2

4V È -t cos nw t cos nw t ˘ V +Ú dt ˙ t sin nw tdt = Í pT Î nw nw p ˚0

4V È T Ê nw T ˆ sin nw t Í= cos Á + 2 2 Ë 2 ˜¯ p T Í 2nw nw Î =

4V p 2p

= =

sin np ˘ È 2p Í - 2n cos np + n2 ˙ Î ˚

2V cos np np

2V np

T /2 ˘

˙ ˙˚

0

(

T = 2p , \ w = 1)

(

sin np = 0, for all n)

(

cos np = -1 for all n)

Hence, the Fourier series of the given waveform is, V(t) =

2V p

1 1 1 Ê ˆ ÁË sin w t + 3 sin 3w t + 5 sin 5w t + 7 sin 7w t + ...˜¯

Ans.

(b) Fourier transform of the function is, •

F(jw) =

Ú

-•

•

f (t )e - jw t dt = Ú Ve

a e - jw t dt

e Ê1 ˆ - Á + jw ˜ Ëa ¯

•

=V

Ú

e

Ê1 ˆ - Á + jw ˜ t Ëa ¯

dt

-•

0

Ê1 ˆ - Á + jw ˜ t Ëa ¯

= V

-t

•

=

Va 1 + jw a

Ans.

0

The amplitude and phase are |F(jw)| =

Va 1+ w a

2 2

and f ( jw ) = - tan -1 (w a)

Solution of 2012 WBUT Paper

SQP.17

12. (a) What is the oriented graph of a network? Explain with a suitable example. (b) Develop at least three trees for your considered network. Mark the twigs and links. (c) For the network in the figure, draw the oriented graph, develop the incidence matrix, choose a tree and considering the tree, develop the tie-set matrix.

Solution: (a) Oriented Graph: A graph whose branches are oriented, i.e. the branch current directions are shown by arrowheads, is called a directed or oriented graph. For example, for the circuit shown in the figure, the oriented graph is shown in the figure.

(b) The three trees of the considered network are shown in the figure below. The solid lines represent the twigs and the dashed lines represent the links.

SQP.18

Circuit Theory and Networks

(c) The oriented graph of the network is shown below.

The incidence matrix is obtained as, 1 2 3 4

5

6

A È -1 1 0 0 1 0 ˘ Aa = B ÍÍ 0 -1 1 1 0 0 ˙˙ C Í 0 0 0 -1 -1 -1˙ Í ˙ D Î 1 0 -1 0 0 1 ˚ A tree has been chosen as shown in the figure. This tree creates three loops as shown in the figure. The tie-set matrix is obtained as, 1 2 3 4 5 6 Ba =

L1 È1 1 1 0 0 0 ˘ L2 ÍÍ0 0 1 -1 0 1 ˙˙ L3 ÍÎ0 -1 0 -1 1 0 ˙˚

Circuit Theory & Networks (EE301) Solution of 2013 WBUT Paper GROUP–A (Multiple-Choice Questions) 1. Choose the correct alternatives for any ten of the following: (i) Unit step function is the first derivative of (a) ramp function (c) gate function

10 × 1 = 10

(b) impulse function (d) parabolic function

(ii) A practical current source is usually represented by (a) a resistance in series with an ideal current source (b) a resistance in parallel with an ideal current source (c) a resistance in parallel with an ideal voltage source (d) none of these (iii) A two-port network is defined by the relations I1 = 2V1 + V2 and I2 = 2V1 + 3V2 , then Z12 is 1 1 (d) - ohm (a) −2 ohm (b) −1 ohm (c) - ohm 2 4 È 0.9 0.2 ˘ (iv) The Z-matrix of a 2-port network is given by Í ˙ . The element Y22 of the correspondÎ0.2 0.6 ˚ ing Y-matrix of the same network is given by (a) 1.2 (b) 0.4 (c) −0.4 (d) 1.8 (v) The Fourier series of the function f(x) = sin2x is (a) sin x + sin 2x (b) 1 – cos 2x (c) sin 2x + cos 2x (vi) A rectangular pulse of duration t and magnitude I Ê Iˆ I (a) (b) Á ˜ e - sT (c) Ë s¯ s

(d) 0.5 – 0.5 cos 2x

has the Laplace transform Ê Iˆ - sT Ê I ˆ sT (d) Á ˜ (1 - e ) ÁË s ˜¯ e Ë s¯

(vii) The Laplace transform of a delayed unit impulse function d(t – 2) is (d) s (a) 1 (b) 0 (c) e–2s

SQP.2

Circuit Theory and Networks

(viii) The convolution of f(t) and g(t) is t

t

(a)

Ú f (t )g (t - t )dt

Ú

(b)

t

f (t )g(t - t )dt (c)

0

0

Ú

t

f (t - t )g(t )dt

(d)

0

Ú f (t )g(t - t )dt 0

(ix) When applying the superposition theorem to any circuit, (a) the voltage source is shorted, the current source is opened (b) the voltage source is opened, the current source is shorted (c) both are opened (d) both are shorted (x) A high-pass filter circuit is basically (a) a differentiating circuit with low time constant (b) a differentiating circuit with large time constant (c) an integrating circuit with low time constant (d) an integrating circuit with large time constant (xi) The Thevenin’s equivalent with respect to the terminals A and B would be only a resistance Rth equal to 4W

A

i 2i

+ –

8W B

(a) 2.66 Ω

(b) 3.2 Ω

(c) 8 Ω

(d) 12 Ω

Solution: (i) (a) ramp function (ii) (b) a resistance in parallel with an ideal current source 1 (iii) (c) - ohm (iv) (d) 1.8 (v) (d) 0.5–0.5cos2x 2 Ê Iˆ - sT (vi) (d) Á ˜ (1 - e ) (vii) (c) e–2s Ë s¯

t

(viii) (b)

Ú f (t )g(t - t )dt 0

(ix) (a) the voltage source is shorted, the current source is opened (x) (a) a differentiating circuit with low time constant

(xi) (b) 3.2 Ω

GROUP–B (Short-Answer-Type Questions) Answer any three of the following questions. 3 × 5 = 15 2. State and prove maximum power transfer theorem for ac network. Solution: Maximum Power Transfer Theorem (for ac network): Statement: Maximum active power will be delivered from a source to a load when the load impedance is the complex conjugate of the source impedance. Proof: Let V be the voltage source, (RS + jXS) be the internal impedance of the source and (RL + jXL) be the load impedance.

SQP.3

Solution of 2013 WBUT Paper

∴

current, I =

V V = Z S + Z L ( RS + RL ) + j ( X S + X L )

(1)

Power delivered to the load is, V 2 RL

2

P = I RL =

(2)

( RS + RL )2 + ( X S + X L )2

where, ZS = RS + jXS, ZL = RL + jXL

(Rs + jXs)

∂P For maximum power, must be zero. ∂X L Now,

V

∂P -2 (V ) RL ( X L + X S ) = =0 2 ∂X L È( RL + RS )2 + ( X L + X S )2 ˘ Î ˚

+ –

(RL + jXL)

2

From which,

XL + XS = 0 or X L = - X S

Load impedance with variable resistance and variable reactance

i.e. the reactance of the load impedance is of opposite sign to the reactance of the source impedance. Putting XL = –XS in Eq. (2) For maximum power, or,

P=

V 2 RL ( RL + RS )2

V 2 ( RL + RS )2 - 2V 2 RL ( RL + RS ) ∂P = =0 ∂RL ( RL + RS )4

V2(RL + RS) – 2V2RL = 0 or RL = RS \ ZL = ZS *

3. What is time constant of an R-L series circuit and what does it signify? Explain it graphically. 2+3 Solution: L Time Constant of an R-L Series Circuit: The quantity t = in an R-L series circuit is known R as the time-constant of the circuit and it is defined in three ways as follows. Definitions of time-constant (τ) 1. It is the time taken for the current to reach 63% of its final value. Thus, it is a measure of the rapidity with which the steady state is reached. Also, at t = 5t , i = 0.993is; the transient is, therefore, said to be practically disappeared in five time constants. R - tˆ VÊ V 2. The tangent to the equation i = Á 1 - e L ˜ at t = 0, intersects the straight line, i = at RË R ¯ L t = t = . Thus, time-constant is the time in which steady state would be reached if the R current increases at the initial rate.

SQP.4

Circuit Theory and Networks

Physically, time-constant represents the speed of the response of a circuit. A low value of time-constant represents a fast response and a high value of time-constant represents a sluggish response. Thus, unit of time constant is the unit of time, i.e. second. Graphical Representation of Time Constant The current in an R-L series circuit with a dc voltage of V is given as, -Á ˜ t ˆ VÊ i(t) = i(t ) = Á 1 - e Ë L ¯ ˜ R ÁË ˜¯ Ê Rˆ

From the current equation at t = t =

L V V , i = (1 - e -1 ) = 0.63 = 0.63is R R R

This is the time constant of the circuit as shown in the figure below.

Variation of current with time in R-L series circuit with step input 4. Find the equivalent p-network for the T-network as shown in the figure. 2.5 W

2W 1

2 5W

1¢

2¢

Solution: Let the equivalent p-network have YC as the series admittance and YA and YB as the shunt admittances at port-1 and port-2, respectively. Now, the z-parameters are given as z11 = (ZA + ZC) = 7 W, z12 = z21 = ZC = 5 W, z22 = (ZB + ZC) = 7.5 W \

Dz = (7 × 7.5 – 5 × 5) = 27.5 W2

\

y11 =

z22 7.5 = Dz 27.5

SQP.5

Solution of 2013 WBUT Paper

y12 = y21 = y22 =

I1

zC 5 =27.5 Dz

+

z11 7 = Dz 27.5

V1

\

YA = ( y11 + y12 ) =

2.5 1 = 27.5 11

\

YB = ( y22 + y12 ) =

2 27.5

and

YC = - y21 =

I2

YC

YA

+ V2

YB

–

–

5 2 = 27.5 11

Thus, the impedances of the equivalent p-networks are ¸ 1 ZA = = 11 W, Ô YA Ô ÔÔ 1 = 13.75 W ˝ ZB = Ans. YB Ô Ô 1 = 5.5 W ZC = Ô YC Ô˛

5.5 W

I1

I2 +

+ 11 W

V1

11.75 W

V2 –

–

Equivalent p-network

NB: This problem can also be solved by using the result of star-delta conversion technique. 2¥5 ZA = 2 + 5 + = 11 W 2.5 2.5 ¥ 5 = 13.75 W 2 2 ¥ 2.5 ZC = 2 + 2.5 + = 5.5 W 5 ZB = 2.5 + 5 +

1 5. Prove that the Laplace transform of a periodic function with period T0 is equal to times 1 - e -T0 s the Laplace transform of the first cycle. Solution: Let, f(t) − be the periodic function, T0 − the time period, f1(t), f2(t), … , fn(t) − the functions representing the first, second, …, nth cycle, respectively \

f(t) = f1(t) + f2(t) + ...+ fn(t) + ... = f1(t) + f1(t – T0) + f1(t – 2T0) + ...

Taking Laplace transform, L[f(t)] = F(s) = L[f1(t)] + L[f1(t – T0)] + L[f1(t – 2T0)] + ... = F1(s) + e–T0s F1(s) + e–2T0s F1(s) + ... = F1 (s) + Í1 + e -T0 s + e -2T0 s + e -3T0 s + Î

˙ ˚

SQP.6

Circuit Theory and Networks

È 1 ˘ F (s) = F1 (s) Í -T s ˙ Î1 - e 0 ˚ Therefore, it is proved that the Laplace transform of a periodic function with period T0 is equal 1 to times the Laplace transform of the first cycle. 1 - e -T0 s 6. Draw the oriented graph of a network with fundamental cut-set matrix given below:

Twigs

Q=

Ê1 Á1 Á Á0 Á0 Á ÁË 0

2 0 1 0 0

3 0 0 1 0

Links 5 -1 1 0 0

4 0 0 0 1

6 0 0 1 1

7ˆ 0˜ ˜ 1˜ 1˜ ˜ 0˜¯

Solution: The graph has a total of seven branches out which four are tree branches (twigs) and the other three are links. It is obvious that the graph must have 5 nodes. Hence, the graph is shown below: 1

4

C1

5

6

3

2

C4

7 C2

C3

GROUP–C (Long-Answer-Type Questions) Answer any three questions. 3 × 15 = 45 7. (a) What are ABCD parameters? Prove that DT = (AD – BC) = 1. 7 (b) Find the z-parameter for the network shown in the figure below. Hence, find the h-parameter for the same network. 8

SQP.7

Solution of 2013 WBUT Paper

Solution: (a) ABCD Parameters: The ABCD parameters represent the relation between the input quantities and the output quantities in a two-port network. They are thus voltage-current pairs.

The figure shows two-port current and voltage variables for calculation of transmission line parameters. However, as the quantities are defined as an input-output relation, the output current is marked as going out rather than as coming into the port. The transmission parameter matrix may be written as V1 = AV2 - BI 2 ÈV1 ˘ È A B ˘ È V2 ˘ or, ÍI ˙ = Í ˙ ˙Í I1 = CV2 - DI 2 Î 1˚ ÎC D ˚ Î - I 2 ˚ The parameters A, B, C, D can be defined in a similar manner with either Port 2 on short circuit or Port 2 on open circuit. V = Open circuit reverse voltage gain A= 1 V2 I = 0 2

B= -

C=

V1 I2

I1 V2

D= -

= Short circuit transfer impedance V2 = 0

= Open circuit transfer admittance I2 = 0

I1 I2

= Short circuit reverse current gain V2 = 0

• To Prove AD – BC = 1 I1

1

I2

+ –

Vs

2

I¢2

1¢

1

I1

I2 N

I¢1 2¢

1¢

(a) Reciprocal network

2 + –

Vs 2¢

(b) Reciprocal network

From Fig. (a), writing the ABCD-parameter equations, Vs = A.0 - B( - I 2¢ ) = BI 2¢ I1 = C.0 - D( - I 2¢ ) = DI 2¢

fi

I 2¢ 1 = Vs B

(1)

SQP.8

Circuit Theory and Networks

From Fig. (b), writing the ABCD-parameter equations, 0 = AVs - BI 2 - I1¢ = CVs - DI 2

I1¢ AD - BC = Vs B

fi

(2)

From the principle of reciprocity from Eq. (1) and (2), we get, 1 AD - BC = B B

fi

AD - BC = 1

\ AD - BC = 1 (b) We consider two cases: Case (1): When I2 = 0 The circuit is modified as shown in Fig. (a). I1

4W

V1

10 W

3W

2W

2W

V1

V2

V2

0.1 I2

Fig. (a)

Fig. (b)

By KVL, we have, V1 = I1 (4 + 2 + 10 ) = 16 I1 fi z11 = \

V2 = 10 ¥ I1

fi z21 =

V1 I1 V2 I1

= 16 W I2 = 0

= 10 W I2 = 0

Case (2): When I1 = 0 The circuit is modified as shown in Fig. (b). By KVL, we have, V2 = 3I 2 + 10 ¥ ( I 2 + 0.1I 2 ) = 14 I 2 fi z22 = \

V1 = 2 ¥ 0.1I 2 + 10 ¥ ( I 2 + 0.1 I 2 ) fi z12 =

Hence, the z-parameters are given as, È16 11.2 ˘ [z] = Í ˙ (W) Î10 14 ˚

V2 I2

V1 I2

I2

= 14 W I1 = 0

= 11.2 W I2 = 0

SQP.9

Solution of 2013 WBUT Paper

By interrelationships, the h-parameters are obtained as, h11 =

Dz 16 ¥ 14 - 11.2 ¥ 10 = =8W z22 14

h12 =

z12 11.2 = = 0.8 z22 14

h13 = h22 =

z21 10 = - = 0.714 z22 14

1 1 = = 0.0714 mho z22 14

8. (a) State and explain Millman’s theorem. Calculate the load current I in the circuit in the figure by Millman’s theorem. 2+6 2W

2W

5W

I 15 W

+ –

2V

3V

+ –

+ –

5V

(b) What is the power loss in the 10-ohm resistor? Use Thevenin’s theorem in the figure below: 7 R3 = 10 ohms

I1

R1 = 2 ohms

2A

R2 = 4 ohms

R5 = 2 ohms

R4 = 1 ohms

I2

5A

Solution: (a) ∑ Millman’s Theorem (I) This theorem states that if several ideal voltage sources (V1, V2, …) in series with impedances (Z1, Z2,…) are connected in parallel then the circuit may be replaced by a single ideal voltage source (V) in series with an impedance (Z), where, n

V=

 ViYi i =1 n

 Yi i =1

and,

Z=

1 n

 Yi i =1

SQP.10

Circuit Theory and Networks

(II) If several ideal current sources (I1, I2,…) in parallel with impedances (Z1, Z2, …) are connected in series, then the circuit may be replaced by a single ideal current source (I) in parallel with an impedance (Z), where, n

I=

 i =1 n

 i =1

Ii

Yi

and,

1 Yi

Y=

1 n

 i =1

1 Yi

n

or,

Z = Â Zi i =1

∑ Solution to Numerical Problem By Millman’s theorem, 2 3 5 + + 2 2 5 = 35 = 2.91667 V = 1 1 1 12 + + 2 2 5 1 1 10 Z= = = = 0.833 W 1 1 1 12 Y Â + + 2 2 5

 EY V= ÂY

\

I=

2.91667 V = = 0.184 A Z + 15 0.833 + 15

Ans.

(b) We find the Thevenin’s equivalent circuit with respect to terminals a and b where the 10 Ω resistance is connected. Thevenin equivalent resistance is obtained as, 2¥4 7 RTh = +1= W 2+4 3

SQP.11

Solution of 2013 WBUT Paper

Thevenin voltage (open circuit voltage) is obtained as follows. Changing the current source into voltage source by source transformation, we simplify the circuit. No, by KVL, we get, -VOC + 1 ¥ 5 - 8 = 0 3 7 fi VOC = V 3

b 4/3 8/3 V

a

– VOC +

+ –

5A

1 ohm

So, the current through the 10 Ω resistance is given as, 7 VOC 3 = 7 A = I= RTh + RL 7 + 10 37 3 So, the power loss in the 10 Ω resistance is, 2

Ê 7ˆ P = I 2 RL = Á ˜ ¥ 10 = 0.358 W Ë 37 ¯ 9. (a) What is a tree? Discuss with a suitable example. 2 (b) A graph is shown in the figure below. Find the tie-set and cut-set matrices and obtain the KCL and KVL equation [bold lines indicate twigs and dotted lines the links]. 6 1

1

2

2

3

6 7 4

3

5

6

7 5

4

(c) Explain odd symmetry and even symmetry of periodic waveforms. (d) Find the Fourier transform of f(t) = e–a|t|

4 3

SQP.12

Circuit Theory and Networks

Solution: (a) Tree: For a given connected graph of a network, a connected subgraph is known as a tree of the graph if the subgraph has all the nodes of the graph without containing any loop. The branches of a tree are called twigs or tree branches. The number of branches or twigs, in any selected tree is always one less than the number of nodes, i.e. Twigs = (n – 1), where n is the number of nodes of the graph. For this case, twigs = (4 – 1) = 3 twigs. These are shown by solid lines in Fig. (b). R1 R2 1

R4

2

3

+ v1

is

R3

R5

– 4

(a) Circuit

(b) Trees and links of circuit of Fig. (a)

If a graph for a network is known and a particular tree is specified, the remaining branches are referred to as the links. The collection of links is called a co-tree. So, co-tree is the complement of a tree. These are shown by dotted lines in Fig. ((b). The branches of a co-tree may or may not be connected, whereas the branches of a tree are always connected. (b) The graph has three loops.

The tie-set matrix is given as, 1 2 3 4

5

6

7

L1 È1 0 1 1 0 0 0 ˘ Í ˙ Ba = L2 Í0 0 0 1 -1 0 1 ˙ L3 ÎÍ0 1 0 0 -1 -1 0 ˙˚

SQP.13

Solution of 2013 WBUT Paper

The graph has 4 fundamental cut-sets. The cut-set matrix is obtained as, 1 C1 È -1 Q = C2 ÍÍ -1 C3 Í 0 Í C4 Î 0

KCL is

È -1 Í -1 fiÍ Í0 Í Î0

2 3 4 5 6

7

0 0˘ 0 -1˙˙ 0 1˙ ˙ 1 0˚ written in terms of cut-set matrix as, QIb = 0 È i1 ˘ Íi ˙ 2 -i1 + i3 = 0 ¸ 0 1 0 0 0 0 ˘Í ˙ Í ˙ i Ô 3 ˙ -i1 + i4 - i7 = 0Ô 0 0 1 0 0 -1˙ Í ˙ i = fi 0 ˝ 4 1 0 0 1 0 1 ˙ ÍÍ ˙˙ i2 + i5 + i7 = 0 Ô i ˙ 5 Ô 1 0 0 0 1 0 ˚Í ˙ i2 + i6 = 0 ˛ Íi6 ˙ Íi ˙ Î 7˚ 0 0 1 1

1 0 0 0

0 1 0 0

0 0 1 0

Ans

KVL is written in terms of tie-set matrix as, BaVb = 0 È Vb1 ˘ ÍV ˙ Í b2 ˙ 1 0 1 1 0 0 0 È ˘ ÍVb3 ˙ Í ˙ Í fi Í0 0 0 1 -1 0 1 ˙˙ ÍVb 4 ˙ = 0 fi ÍÎ0 1 0 0 -1 -1 0 ˙˚ ÍVb 5 ˙ Í ˙ ÍVb6 ˙ ÍV ˙ Î b7 ˚

Vb1 + Vb3 + Vb 4 = 0 ¸ Ô Vb 4 - Vb 5 + Vb 7 = 0˝ Vb 2 - Vb 5 - Vb6 = 0Ô˛

Ans

(c) Odd Symmetry: A function f(x) is said to be odd if, f(x) = – f(–x) We know that the Fourier coefficients are given as, T È 0 ˘ T 2 1 1Í ˙ + a0 = Ú f (t )dt = f t dt f t dt ( ) ( ) Ú ˙ T0 T Í -TÚ 0 Î 2 ˚ T

2 an = T

T 2

Ú 0

2 [f ( x) + f (- x)]cos nw xdx and bn = T T

Hence, for odd functions a0 = an = 0 and bn =

1 T

2

Ú [f ( x) - f (- x)]sin nw xdx 0

2

Ú 0

f ( x )sin nw x .

SQP.14

Circuit Theory and Networks

Thus, the Fourier series expansion of an odd function contains only the sine terms, the constant and the cosine terms being zero. Even Symmetry A function f(x) is said to be even, if f(x) = f(–x) T

\

2 a0 = T

Ú

f ( x )dx

0

T

4 an = T

2

2

Ú

f ( x )cos nw xdx

0

Even function and bn = 0 Thus, the Fourier series expansion of an even periodic function contains only the cosine terms plus a constant, all sine terms being zero. (d) Fourier Transform of f (t) = e– a | t |, for all values of t F(jw) = F Èe Î

˘= ˚

•

Úe

-a t

e - jw t dt

-•

•

0

=

-a t

Ú

-•

e( a - jw )t dt + Ú e - ( a + jw )t dt 0

=

1 1 + a - jw a + jw

=

2a a + w2

Ans.

2

10. (a) Define Fourier transform. How does Fourier transform differ from Laplace transform? (b) What is impulse function? Find its Laplace transform. (c) For the square wave shown in the figure, find the exponential Fourier series.

5 3 7

v(t) V t 0

T/2

T

3T/2

2T

Solution: (a) Definition of Fourier Transform: The Fourier Transform or the Fourier integral of a function f(t) is denoted by F(jw) and is defined by, •

F(jω) = F [f(t)] =

Ú

-•

f (t )e - jw t dt

(i)

SQP.15

Solution of 2013 WBUT Paper

and the inverse Fourier transform is defined by, • • 1 jw t j 2p f F j w e d w = f(t) = F –1 [F(jω)] = ( ) Ú Ú F ( j 2p f )e df 2p -• -•

(ii)

Equations (i) and (ii) form the Fourier transform pair. Difference between Laplace Transform and Fourier Transform: The defining equations are, •

•

0

-•

F (s) = Ú f (t )e - st dt and F ( jw ) =

Ú

f (t )e - jw t dt

The following are some differences and similarities: 1. Laplace transform is one-sided in the interval 0 < t < • and Fourier transform is double-sided in the interval −• < t < •. Thus, Laplace transform is applicable for positive time functions, f(t), t > 0; while Fourier Transform is applicable for functions defined for all times. 2. Laplace transform includes the initial conditions and is applicable for transient analysis; while Fourier transform is only applicable for steady-state analysis. •

3. For functions f(t) = 0 for t < 0 and

Ú

f (t ) dt < • , the two transforms are related as,

0

F ( jw ) = F (s) s = jw . Thus, Laplace transform is associated with the entire s-plane, while, Fourier transform is restricted to the imaginary (jw) axis. 4. Laplace transform is applicable to a wider range of functions than the Fourier transform. On the other hand, Fourier transforms exist for signals that are not physically realizable and have no Laplace transform. (b) Impulse Function: It is a function of a real variable t, such that the function is zero everywhere except at the instant t = 0. Physically, it is a very sharp pulse of infinitesimally small width and very large magnitude, the area under the curve being unity. We consider a gate function as shown, f(t) 3/a 2/a 1/a

0

a/3

a/2

a

t

The function is compressed along the time-axis and stretched along the y-axis, keeping area 1 under the pulse unity. As a → 0, the value of Æ • and the resulting function is known as a impulse. It is defined as, δ(t) = 0 for t π 0

SQP.16

Circuit Theory and Networks •

and

Ú d (t )dt = 1

-•

Also,

1 δ(t) = lim [u(t ) - u(t - a)] aÆ0 a

Laplace Transform of Impulse Function: The Laplace transform of the impulse function is obtained as, 1 È 1 e - as ˘ 1 - e - as Ï1 ¸ lim = L[d (t)] = lim L Ì [u(t ) - u(t - a)]˝ = lim Í ˙ aÆ0 Ó a s ˚ aÆ0 as ˛ aÆ0 a Î s se - as [by L'Hospital's rule] aÆ0 s

= lim

=1 (c) [WBUT 2012 Q.10 (c)] 11. (a) What are the advantages of active filter over passive filter? 4 (b) Design a high-pass active filter of cut-off frequency 1 kHz with a pass-band gain of 2. 5 (c) Draw the circuit diagram of a first order low-pass filter and find out the expression of the cut-off frequency. 6 Solution: (a) Advantages of Active Filter over Passive Filter 1. Less Cost: Active filters are inexpensive as compared to passive filters, due to the variety of cheaper op-amp and the absence of costly inductors. 2. Gain and Frequency Adjustment Flexibility: Since an op-amp is capable of providing a gain (which may also be variable), the input signal is not attenuated as it is in a passive filter. In addition, the active filter is easier to tune or adjust. 3. No Loading Problem: Active filters provide an excellent isolation between the individual stages due to the high input impedance (ranging from a few kW to a several thousand MW) and low output impedance (ranging from less than 1 W to a few hundred W). So, the active filter does not cause loading of the source or load. 4. Size and Weight: Active filters are small in size and less bulky (due to the absence of bulky ‘L’) and are rugged. 5. Non-floating Input and Output: Active filters generally have single-ended inputs and outputs which do not ‘float’ with respect to the system power supply or common. This property is different from that of the passive filters. (b) Here, fc = 1 kHz, AF = 2 Let, C = 0.01 µF 1 1 \ R= = = 15.9 kW 3 2p fcC 2p ¥ 10 ¥ 0.01 ¥ 10 -6 ∵

Ê R ˆ AF = 2 = Á 1 + f ˜ fi R f = R1 = 10 kW R Ë 1¯

SQP.17

Solution of 2013 WBUT Paper

So, the complete circuit is shown in the figure. 10 kW

10 kW

– 0.01 mF

V0 +

Vi 15.9 kW

(c) First-Order Low-Pass Active Filter: The circuit of the figure is a commonly used low-pass active filter.

The filtering is done by the RC network, and the op-amp is used as a unity-gain amplifier. The resistor Rf (= R) is included for dc offset. Here, all the voltages Vi, Vx, Vy, Vo are measured with respect to ground. Since the input impedance of the op-amp is infinite, no current will flow into the input terminals. V0 ¥ R1 (1) Vy = R1 + R f According to the voltage-divider rule, the voltage across the capacitor, Vx = =

=

Xc Vi ; R + Xc

Xc =

1 1 = jw C j 2p fC

1/j 2p fC Vi 1 R+ j 2p fC Vi 1 + j 2p fRC

(2)

SQP.18

Circuit Theory and Networks

Since the op-amp gain is infinite, \ Vx = Vy V0 R1 Vi = or R1 + R f 1 = j 2p fRC

fi where,

Rf ˆ Ê ÁË 1 + R ˜¯ V0 1 = Vi 1 + j 2p fRC Rf ˆ Ê AF = Á 1 + = pass-band gain of the filter R1 ˜¯ Ë f = frequency of the input signal AcL = closed-loop gain of the filter as a function of frequency

The gain magnitude, AcL =

V0 AF AF = = Vi 1 + w c 2 R 2C 2 1 + 4p 2 fc 2 R 2C 2

For this magnitude to be

AF 2

AF 2 or

=

fc =

at f = fc, we have, AF

1 + (2p fc RC )2 1 2p RC

This is the cut-off frequency of the low-pass filter.