Advanced Electrical Circuit Analysis: Practice Problems, Methods, and Solutions 3030785394, 9783030785390

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Mehdi Rahmani-Andebili

Advanced Electrical Circuit Analysis Practice Problems, Methods, and Solutions

Advanced Electrical Circuit Analysis

Mehdi Rahmani-Andebili

Advanced Electrical Circuit Analysis Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili Engineering Technology State University of New York Buffalo, NY, USA

ISBN 978-3-030-78539-0 ISBN 978-3-030-78540-6 https://doi.org/10.1007/978-3-030-78540-6

(eBook)

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Electrical circuit analysis is one of the most fundamental subjects of electrical engineering major. This textbook includes the advanced subjects of electrical circuit analysis that have not been covered in the previously published textbooks, that is, DC Electrical Circuit Analysis and AC Electrical Circuit Analysis. The subjects include state equations of electrical circuits, Laplace transform and network function, natural frequencies of electrical circuits, network theorems (Tellegen’s and linear time-invariant network theorems), and two-port networks. Like the previously published textbooks, this textbook includes very detailed and multiple methods of problem solutions. It can be used as a practicing textbook by students and as a supplementary teaching source by instructors. To help students study the textbook in the most efficient way, the exercises have been categorized in nine different levels. In this regard, for each problem of the textbook, a difficulty level (easy, normal, or hard) and a calculation amount (small, normal, or large) have been assigned. Moreover, in each chapter, problems have been ordered from the easiest problem with the smallest calculations to the most difficult problem with the largest calculations. Therefore, students are advised to study the textbook from the easiest problems and continue practicing till they reach the normal and then the hardest ones. On the other hand, this classification can help instructors choose their desirable problems to conduct a quiz or a test. Moreover, the classification of computation amount can help students manage their time during future exams and instructors give the appropriate problems based on the exam duration. Since the problems have very detailed solutions and some of them include multiple methods of solution, the textbook can be useful for the under-prepared students. In addition, the textbook is beneficial for knowledgeable students because it includes advanced exercises. In the preparation of problem solutions, an attempt has been made to use typical methods of electrical circuit analysis to present the textbook as an instructor-recommended one. In other words, the heuristic methods of problem solution have never been used as the first method of problem solution. By considering this key point, the textbook will be in the direction of instructors’ lectures, and the instructors will not see any untaught problem solutions in their students’ answer sheets. The Iranian University Entrance Exam for the master’s and PhD degrees of electrical engineering major is the main reference of the textbook; however, all the problem solutions have been provided by me. The Iranian University Entrance Exam is one of the most competitive university entrance exams in the world that allows only 10% of the applicants to get into prestigious and tuition-free Iranian universities. Buffalo, NY, USA

Mehdi Rahmani-Andebili

v

Contents

1

Problems: State Equations of Electrical Circuits . . . . . . . . . . . . . . . . . . . . . . .

1

2

Solutions of Problems: State Equations of Electrical Circuits . . . . . . . . . . . . .

9

3

Problems: Laplace Transform and Network Function . . . . . . . . . . . . . . . . . .

29

4

Solutions of Problems: Laplace Transform and Network Function . . . . . . . . .

45

5

Problems: Natural Frequencies of Electrical Circuits . . . . . . . . . . . . . . . . . . .

83

6

Solutions of Problems: Natural Frequencies of Electrical Circuits . . . . . . . . .

93

7

Problems: Network Theorems (Tellegen’s and Linear Time-Invariant Network Theorems) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8

Solutions of Problems: Network Theorems (Tellegen’s and Linear Time-Invariant Network Theorems) . . . . . . . . . . . . . . . . . . . . . . . 121

9

Problems: Two-Port Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

10

Solutions of Problems: Two-Port Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

vii

About the Author

Mehdi Rahmani-Andebili is an assistant professor in the Department of Engineering Technology at State University of New York, Buffalo State. He received his first M.Sc. and Ph.D. degrees in electrical engineering (power system) from Tarbiat Modares University and Clemson University in 2011 and 2016, respectively, and his second M.Sc. degree in physics and astronomy from the University of Alabama in Huntsville in 2019. Moreover, he was a postdoctoral fellow at Sharif University of Technology during 2016–2017. As a professor, he has taught many courses such as Essentials of Electrical Engineering Technology, Electrical Circuits Analysis I, Electrical Circuits Analysis II, Electrical Circuits and Devices, Industrial Electronics, and Renewable Distributed Generation and Storage. Dr. Rahmani-Andebili has more than hundred single-author publications including textbooks, books, book chapters, journal papers, and conference papers. His research areas include smart grid, power system operation and planning, integration of renewables and energy storages into power system, energy scheduling and demand-side management, plug-in electric vehicles, distributed generation, and advanced optimization techniques in power system studies.

ix

1

Problems: State Equations of Electrical Circuits

Abstract

In this chapter, state equations are applied to solve the basic and advanced problems of electrical circuit analysis. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 1.1. In the circuit of Figure 1.1, vC(t) and iL(t) are the state variables [1–2]. Write the output voltage (vo(t)) based on the state variables. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. iL ðt Þ þ 16 vC ðt Þ  12 vs ðt Þ 2. 6iL(t)  vC(t)  vs(t) 3. 3iL ðt Þ  12 vC ðt Þ  12 vs ðt Þ 4. 2iL ðt Þ þ 13 vC ðt Þ  vs ðt Þ

Figure 1.1 The circuit of problem 1.1


 iL ðt Þ 1.2. In the circuit of Figure 1.2, if X ¼ is assigned as the state vector, determine matrices A and B in the relation of vC ð t Þ
 i s ðt Þ X_ = AX þ B . vs ð t Þ Difficulty level Calculation amount

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_1

1

2

1 Problems: State Equations of Electrical Circuits




1. 2. 3. 4.

1 A¼ 1
1 A¼ 1
1 A¼ 1
1 A¼ 1


1 1 ,B ¼ 1 1 
1 1 ,B ¼ 1 1 
1 1 ,B ¼ 1 1 
1 0 ,B ¼ 1 1

 0 1  0 1  0 1  1 1

Figure 1.2 The circuit of problem 1.2

1.3. In the circuit of Figure 1.3, the state equations are in the form of X_ = AX. Determine matrix A, for the state vector of
 iL ðt Þ X¼ . vC ð t Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large " # 8 2  1. A ¼ 3 1 0 " # 8 2  2. A ¼ 3 1 0 " # 4 2  3. A ¼ 3 1 0 " # 4 2  4. A ¼ 3 0 1

1

Problems: State Equations of Electrical Circuits

3

Figure 1.3 The circuit of problem 1.3

1.4. Determine matrix A of the state equations (X_ = AX þ Bw) for the circuit of Figure 1.4 if X ¼ Difficulty level ○ Easy Calculation amount ○ Small
 1 1 1. A ¼ 1 3
 1 3 2. A ¼ 1 1
 0:5 0:5 3. A ¼ 0:5 1:5
 0:5 1:5 4. A ¼ 0:5 0:5

● Normal ● Normal

○ Hard ○ Large




vC ð t Þ iL ðt Þ

 .

Figure 1.4 The circuit of problem 1.4

1.5. Determine the state equations of the circuit of Figure 1.5 based on the given voltage of the capacitor and current of the inductor. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2 3 2 3 " # d 3 v ðt Þ 1
vC ð t Þ  1 6 dt C 7 6 2 7 1. 4 þ 1 vs ðt Þ 5¼4 5 1 d  i L ðt Þ  0 iL ðt Þ 2 2 dt

4

1 Problems: State Equations of Electrical Circuits

2

3 2 d 3 vC ð t Þ  6 dt 7 6 2 2. 4 5¼4 1 d  iL ðt Þ 2 dt 3 2 2 d 3 v ðt Þ  6 dt C 7 6 2 3. 4 5¼4 1 d  iL ðt Þ 2 dt 3 2 2 d 3 v ðt Þ  6 dt C 7 6 2 4. 4 5¼4 1 d  iL ðt Þ 2 dt

3 " # 1
vC ð t Þ  1 7 þ 1 vs ðt Þ 5  i L ðt Þ 0 2 3 " # 1
vC ð t Þ  1 7 þ 1 vs ðt Þ 5 i L ðt Þ 0 2 3 " # 1
vC ðt Þ  1 7 þ 1 vs ð t Þ 5 i L ðt Þ 0 2

Figure 1.5 The circuit of problem 1.5

1.6. In the circuit of Figure 1.6, determine matrix A if the state equations are in the form of X_ = AX. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large
 0 2 1. A ¼ 1 1
 1 1 2. A ¼ 2 0 2 3 1 0 1 6 7 3. A ¼ 4 1 0 0 5 1 1 1
 2 2 4. A ¼ 1 0

Figure 1.6 The circuit of problem 1.6

1

Problems: State Equations of Electrical Circuits

5

1.7. If the state equations of the circuit of Figure 1.7 are presented in the form of X_ = AX þ Bis ðt Þ, determine vector B, where
 iL ðt Þ . X¼ vC ð t Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large
 1 1. B ¼ 1
 1 2. B ¼ 1
 0 3. B ¼ 1
 1 4. B ¼ 0

Figure 1.7 The circuit of problem 1.7


1.8. In the circuit of Figure 1.8, by choosing X ¼ X_ = AX þ BW? Difficulty level ○ Easy Calculation amount ○ Small
 1 1 1. A ¼ 1 2 2 3 1 1  6 37 2. A ¼ 4 3 5 1 2 23 3 3 1 1  6 7 3. A ¼ 4 3 3 5 1 2  
3 3 1 1 4. A ¼ 1 2

● Normal ○ Normal

vC ðt Þ



i L ðt Þ ○ Hard ● Large

as the state vector, what is matrix A in the state equations of

6

1 Problems: State Equations of Electrical Circuits

Figure 1.8 The circuit of problem 1.8


1.9. A network includes some resistors, an inductor of 1 H, and a capacitor of 1 F. By choosing the state vector of

 iL ðt Þ , vC ð t Þ

the system matrix of the state equations is as follows:
A¼

a11

a12

a21

a22



Determine
 the updated system matrix of the state equations if the place of the inductor and the capacitor is changed and vC ð t Þ is chosen as the new state vector. iL ðt Þ Difficulty level Calculation amount

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

1. It is impossible to determine Anew.
 a22 a12 2. Anew ¼ Det1½A . a21 a11
 a11 a21 3. Anew ¼ Det1½A . a12 a22 2 1 1 3  a12 7 6 a 4. Anew ¼ Det1½A 4 22 5. 1 1  a21 a11 2

3 iL1 ðt Þ
 is ðt Þ 6 7 1.10. In the circuit of Figure 1.9, the state vector and the input vector are X ¼ 4 iL2 ðt Þ 5 and W ¼ , respectively. If vs ð t Þ vC ð t Þ
 i s ðt Þ _ , determine matrix B, while we have: the state equations of the circuit are written in the form of X = AX þ B v s ðt Þ 2 3 1 1 0 1 17 6 1 A ¼ 4   5 2 2 2 0 1 0 Difficulty level Calculation amount 2 3 1 1 6 1 17 1. B ¼ 4  5 2 2 1 0

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

1

Problems: State Equations of Electrical Circuits

2

1 6 1 2. B ¼ 4  2 0 2 1 6 1 3. B ¼ 4 2 1 2 1 6 1 4. B ¼ 4 2 1

7

3 1 17  5 2 1 3 1 1 7 5 2 0 3 1 17 5 2 0

Figure 1.9 The circuit of problem 1.10




 vC ð t Þ 1.11. In the circuit of Figure 1.10, X ¼ is the state vector. Determine matrix A if the state equations are written in the iL ðt Þ form of X_ = AX þ Bis ðt Þ. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large
 1 0:5 1. A ¼ 0:5 1:5
 1 1 2. A ¼ 1 3
 0:5 0:5 3. A ¼ 0:5 1:5
 1 1 4. A ¼ 1 3

Figure 1.10 The circuit of problem 1.11

8

1 Problems: State Equations of Electrical Circuits

1.12. Determine the state equations of the circuit of Figure 1.11. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 3 2 d 2 32 3 2 3 6 dt i1 ðt Þ 7 4 0 2 i 1 ðt Þ 4 0
 7 6 7 6 76 6d 7 6 7 is ðt Þ 1. 6 i2 ðt Þ 7 ¼ 4 0 4 2 54 i2 ðt Þ 5 þ 4 0 2 5 7 6 dt vs ð t Þ 5 4 1 1 0 vð t Þ 0 0 d vð t Þ 3 2 dt d 2 32 3 2 3 6 dt i1 ðt Þ 7 4 0 2 i 1 ðt Þ 4 0
 7 6 7 6 76 6d 7 6 7 is ðt Þ 2. 6 i2 ðt Þ 7 ¼ 4 0 4 2 54 i2 ðt Þ 5 þ 4 0 2 5 7 6 dt vs ð t Þ 5 4 1 1 0 vð t Þ 0 0 d vð t Þ 3 2 dt d 2 32 3 2 3 i ð t Þ 6 dt 1 7 4 0 2 i 1 ðt Þ 4 0
 7 6 7 6 76 6d 7 6 7 i s ðt Þ 3. 6 i2 ðt Þ 7 ¼ 4 0 4 2 54 i2 ðt Þ 5 þ 4 0 2 5 7 6 dt vs ð t Þ 5 4 1 1 0 vð t Þ 0 0 d vð t Þ 3 2 dt d 2 32 3 2 3 i ð t Þ 6 dt 1 7 0 4 2 i 1 ðt Þ 4 0
 7 6 7 6 76 6d 7 6 7 is ðt Þ 4. 6 i2 ðt Þ 7 ¼ 4 0 4 2 54 i2 ðt Þ 5 þ 4 0 2 5 7 6 dt vs ð t Þ 5 4 1 0 1 vð t Þ 0 0 d vð t Þ dt

Figure 1.11 The circuit of problem 1.12

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

2

Solutions of Problems: State Equations of Electrical Circuits

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods.

2.1. Applying KVL in the right-side mesh [1–2]: 1 2vL ðt Þ  vC ðt Þ þ 3iL ðt Þ þ vL ðt Þ ¼ 0 ) 3vL ðt Þ  vC ðt Þ þ 3iL ðt Þ ¼ 0 ) vL ðt Þ ¼ vC ðt Þ  iL ðt Þ 3

ð1Þ

Applying KVL in the left-side mesh: ð1Þ

vs ðt Þ  vo ðt Þ þ vC ðt Þ  2vL ðt Þ ¼ 0 ) vo ðt Þ ¼ vs ðt Þ þ vC ðt Þ  2




1 v ðt Þ  i L ðt Þ 3 C



1 ) vo ðt Þ ¼ 2iL ðt Þ þ vC ðt Þ  vs ðt Þ 3 Choice (4) is the answer.

Figure 2.1 The circuit of solution of problem 2.1

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_2

9

10

2

Solutions of Problems: State Equations of Electrical Circuits

2.2. Applying KCL in node 1 in the circuit of Figure 2.2.2: is ðt Þ þ iL ðt Þ þ i1 ðt Þ ¼ 0 ) i1 ðt Þ ¼ is ðt Þ  iL ðt Þ

ð1Þ

Applying KCL in node 2 in the circuit of Figure 2.2.2: ð1Þ

i1 ðt Þ  iC ðt Þ þ i2 ðt Þ ¼ 0 ) i2 ðt Þ ¼ i1 ðt Þ þ iC ðt Þ ) i2 ðt Þ ¼ is ðt Þ  iL ðt Þ þ iC ðt Þ

ð2Þ

As we know, the current-voltage relation of inductor and the voltage-current relation of capacitor are as follows: iC ðt Þ ¼ C

C¼1 d d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ dt C dt

ð3Þ

d d L¼1 i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

ð4Þ

v L ðt Þ ¼ L Solving (2) and (3):

i2 ðt Þ ¼ is ðt Þ  iL ðt Þ þ

d v ðt Þ dt C

ð5Þ

Applying KVL in the top mesh: ð1Þ, ð4Þ d vL ðt Þ þ i1 ðt Þ  vC ðt Þ ¼ 0 ) vL ðt Þ ¼ i1 ðt Þ  vC ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iL ðt Þ ¼ iL ðt Þ  vC ðt Þ þ is ðt Þ dt

ð6Þ

Applying KVL in the lower-right mesh: ð2Þ

vs ðt Þ þ vC ðt Þ þ i2 ðt Þ ¼ 0 )  vs ðt Þ þ vC ðt Þ þ is ðt Þ  iL ðt Þ þ iC ðt Þ ¼ 0 ð3Þ

)  vs ðt Þ þ vC ðt Þ þ is ðt Þ  iL ðt Þ þ

d d v ðt Þ ¼ 0 ) v ð t Þ ¼ i L ð t Þ  vC ð t Þ  i s ð t Þ þ vs ð t Þ dt C dt C

The equations of (6) and (7) can be written in the matrices form as follows: 2

3 d  i L ðt Þ 1 6 dt 7 4 5¼ d 1 v ðt Þ dt C

1 1

 X_ = AX þ B  ) A= Choice (3) is the answer.





i L ðt Þ vC ð t Þ

i s ðt Þ vs ð t Þ

1 1 1

1



 þ



1



 ,B=



1 1 

,X ¼

0

i L ðt Þ



vC ð t Þ 1

0

1

1



i s ðt Þ vs ð t Þ



ð7Þ

2

Solutions of Problems: State Equations of Electrical Circuits

11

Figure 2.2 The circuit of solution of problem 2.2

2.3. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: 1 L¼ 2 d 1 d vL ðt Þ ¼ L iL ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i ðt Þ dt 2 dt L

ð1Þ

1 C¼ 3 d 1 d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) i C ðt Þ ¼ iC ðt Þ ¼ C vC ðt Þ ¼ dt 3 dt C

ð2Þ

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

i L ðt Þ vC ð t Þ



Applying KCL in the indicated node of the circuit of Figure 2.3.2: 1 iL ðt Þ þ 2iðt Þ þ iðt Þ ¼ 0 ) iðt Þ ¼ iL ðt Þ 3

ð3Þ

12

2

Solutions of Problems: State Equations of Electrical Circuits

Applying KVL in the indicated loop of the circuit of Figure 2.3.2: 1  i L ðt Þ þ

ð3Þ 1 d 1 d 1 i ðt Þ þ 1  iðt Þ þ vC ðt Þ ¼ 0 ) iL ðt Þ þ i ð t Þ þ i L ð t Þ þ vC ð t Þ ¼ 0 2 dt L 2 dt L 3

)

d 8 i ðt Þ ¼  iL ðt Þ  2vC ðt Þ dt L 3

ð4Þ

Solving (2) and (3) and considering iC(t) ¼ i(t), which is clear in Figure 2.3.2: 1 1 d d i ðt Þ ¼ v ð t Þ ) vC ð t Þ ¼ i L ð t Þ 3 L 3 dt C dt By writing (4) and (5) in the form of matrices, we have: 2

3 " # " # d  8 8 iL ðt Þ i L ðt Þ  2 2  6 dt 7 ) A¼ 3 3 4 5¼ d vC ð t Þ 1 0 1 0 vC ðt Þ dt Choice (1) is the answer.

Figure 2.3 The circuit of solution of problem 2.3

ð5Þ

2

Solutions of Problems: State Equations of Electrical Circuits

13

2.4. In this problem, only matrix A needs to be determined; therefore, the source can be turned off, as is illustrated in Figure 2.4.2. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: L¼1 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

ð1Þ

C¼1 d d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ dt C dt

ð2Þ

vL ð t Þ ¼ L iC ðt Þ ¼ C

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

vC ð t Þ



i L ðt Þ

Applying KVL in the left-side mesh of the circuit of Figure 2.4.2: 1  iC ðt Þ þ vC ðt Þ þ 1  ðiC ðt Þ  iL ðt ÞÞ ¼ 0 ) iC ðt Þ ¼ 0:5vC ðt Þ þ 0:5iL ðt Þ

ð3Þ

Solving (2) and (3): 2

d d v ðt Þ þ vC ðt Þ  iL ðt Þ ¼ 0 ) v ðt Þ ¼ 0:5vC ðt Þ þ 0:5iL ðt Þ dt C dt C

ð4Þ

Applying KVL in the right-side mesh of the circuit of Figure 2.4.2: 1  ðiL ðt Þ  iC ðt ÞÞ þ

d d i ðt Þ þ 1  iL ðt Þ ¼ 0 ) i ðt Þ ¼ 2iL ðt Þ þ iC ðt Þ dt L dt L

ð5Þ

Solving (3) and (5): d i ðt Þ ¼ 2iL ðt Þ  0:5vC ðt Þ þ 0:5iL ðt Þ ¼ 1:5iL ðt Þ  0:5vC ðt Þ dt L Writing (4) and (6) in the form of matrices: 2

3 d  vC ð t Þ 0:5 6 dt 7 4 5¼ d 0:5 i ðt Þ dt L Choice (3) is the answer.

0:5 1:5



vC ð t Þ i L ðt Þ



 ) A¼

0:5 0:5

0:5 1:5



ð6Þ

14

2

Solutions of Problems: State Equations of Electrical Circuits

Figure 2.4 The circuit of solution of problem 2.4

2.5. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: vL ð t Þ ¼ L

L¼2 d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ 2 i L ðt Þ i ðt Þ ¼ dt L dt

ð1Þ

i C ðt Þ ¼ C

C¼1 d d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ dt C dt

ð2Þ

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

vC ð t Þ



i L ðt Þ

Applying KVL in the indicated loop of the circuit of Figure 2.5.2: vs ðt Þ þ vC ðt Þ þ 2

d d i ðt Þ ¼ 0 ) i ðt Þ ¼ 0:5vs ðt Þ  0:5vC ðt Þ dt L dt L

Applying KCL in the indicated supernode of the circuit of Figure 2.5.2: 2 d i L ðt Þ vC ð t Þ v ðt Þ d  iC ðt Þ þ iL ðt Þ þ dt ¼ 0 )  C  iC ðt Þ þ iL ðt Þ þ 2 iL ðt Þ ¼ 0 2 2 dt 1 ð3Þ v ðt Þ )  C  iC ðt Þ þ iL ðt Þ þ vs ðt Þ  vC ðt Þ ¼ 0 )  1:5vC ðt Þ  iC ðt Þ þ iL ðt Þ þ vs ðt Þ ¼ 0 2 

ð3Þ

2

Solutions of Problems: State Equations of Electrical Circuits ð2Þ

)  1:5vC ðt Þ 

15

d d v ð t Þ þ i L ð t Þ þ vs ð t Þ ¼ 0 ) v ðt Þ ¼ iL ðt Þ þ vs ðt Þ  1:5vC ðt Þ dt C dt C

ð4Þ

Writing (3) and (4) in the form of matrices: 2

3 2 d 3 vC ð t Þ  6 dt 7 6 2 4 5¼4 1 d  i L ðt Þ 2 dt

3 " # 1  vC ð t Þ  1 7 þ 1 vs ð t Þ 5 i L ðt Þ 0 2

Choice (3) is the answer.

Figure 2.5 The circuit of solution of problem 2.5

2.6. First, we should simplify the series connection of the inductors as well as the parallel connection of the capacitors, as can be seen in the circuit of Figure 2.6.2. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: L¼1 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

ð1Þ

C ¼ 0:5 d d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ 0:5 vC ðt Þ dt C dt

ð2Þ

vL ð t Þ ¼ L iC ðt Þ ¼ C

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

vC ð t Þ i L ðt Þ



16

2

Solutions of Problems: State Equations of Electrical Circuits

Applying KCL in the indicated node of the circuit of Figure 2.6.2: iC ðt Þ þ iL ðt Þ ¼ 0 ) iC ðt Þ ¼ iL ðt Þ ð2Þ

) 0:5

ð3Þ

d d v ðt Þ ¼ i L ðt Þ ) v ðt Þ ¼ 2iL ðt Þ dt C dt C

ð4Þ

Applying KVL in the indicated mesh of the circuit of Figure 2.6.2: 

d d i ðt Þ  2iL ðt Þ  vC ðt Þ þ 3  ð2iL ðt Þ  iC ðt ÞÞ ¼ 0 )  iL ðt Þ þ 4iL ðt Þ  vC ðt Þ  3iC ðt Þ ¼ 0 dt L dt ð3Þ

)

d d i ðt Þ þ 4iL ðt Þ  vC ðt Þ  3iL ðt Þ ¼ 0 ) i ðt Þ ¼ i L ðt Þ  v C ðt Þ dt L dt L

Writing (4) and (5) in the form of matrices: 2

3 d  vC ð t Þ 0 6 dt 7 4 5¼ d 1 i ðt Þ dt L

2 1



vC ð t Þ iL ðt Þ



 ) A¼

0 1

Choice (1) is the answer.

Figure 2.6 The circuit of solution of problem 2.6

2 1



ð5Þ

2

Solutions of Problems: State Equations of Electrical Circuits

17

2.7. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: vL ð t Þ ¼ L

L¼2 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ 2 i L ðt Þ dt L dt

i C ðt Þ ¼ C

C¼1 d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ v ðt Þ ¼ dt C dt

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

i L ðt Þ



vC ð t Þ

Applying KCL in the indicated supernode of the circuit of Figure 2.7.2: is ðt Þ þ

d d v ðt Þ þ i1 ðt Þ  αiL ðt Þ ¼ 0 ) i1 ðt Þ ¼ is ðt Þ  vC ðt Þ þ αiL ðt Þ dt C dt

ð1Þ

Applying KVL in the indicated upper mesh of the circuit of Figure 2.7.2: 2

d d i ðt Þ  1  i2 ðt Þ ¼ 0 ) i2 ðt Þ ¼ 2 iL ðt Þ dt L dt

ð2Þ

Applying KCL in the indicated node of the circuit of Figure 2.7.2: i2 ðt Þ þ iL ðt Þ þ i1 ðt Þ  αiL ðt Þ ¼ 0 ) i2 ðt Þ ¼ i1 ðt Þ þ ð1  αÞiL ðt Þ

ð3Þ

Applying KVL in the indicated lower mesh of the circuit of Figure 2.7.2: vC ðt Þ þ 1  i2 ðt Þ þ 0:5  i1 ðt Þ ¼ 0 ) i2 ðt Þ ¼ vC ðt Þ  0:5i1 ðt Þ

ð4Þ

Solving (3) and (4): i1 ðt Þ þ ð1  αÞiL ðt Þ ¼ vC ðt Þ  0:5i1 ðt Þ ) 1:5i1 ðt Þ ¼ vC ðt Þ þ ðα  1ÞiL ðt Þ ¼ 0 2 2 ) i1 ðt Þ ¼ vC ðt Þ þ ðα  1ÞiL ðt Þ 3 3

ð5Þ

1 1 2 1 i2 ðt Þ ¼ vC ðt Þ  vC ðt Þ  ðα  1ÞiL ðt Þ ) i2 ðt Þ ¼ vC ðt Þ  ðα  1ÞiL ðt Þ 3 3 3 3

ð6Þ

Solving (4) and (5):

Solving (1) and (5):
 2 2 d d 2 αþ2 vC ðt Þ þ ðα  1ÞiL ðt Þ ¼ is ðt Þ  vC ðt Þ þ αiL ðt Þ ) v C ð t Þ ¼ i s ð t Þ  vC ð t Þ þ iL ðt Þ 3 3 dt dt 3 3

ð7Þ

Solving (2) and (6): 2

d 2 1 d 1 1 i ðt Þ ¼ vC ðt Þ  ðα  1ÞiL ðt Þ ) i ðt Þ ¼  vC ðt Þ þ ðα  1ÞiL ðt Þ dt L 3 3 dt L 3 6

By writing (7) and (8) in the form of matrices, we have: 2

3 2 d 1 iL ðt Þ ð α  1Þ 6 dt 7 66 4 5¼4 1 d ð α þ 2Þ v ðt Þ 3 dt C Choice (3) is the answer.

3 1       3 7 iL ðt Þ þ 0 i ðt Þ ) B = 0 5 s 2 1 vC ð t Þ 1  3 

ð8Þ

18

2

Solutions of Problems: State Equations of Electrical Circuits

Figure 2.7 The circuit of solution of problem 2.7

2.8. In this problem, only matrix A needs to be determined; therefore, the sources should be turned off, as is illustrated in Figure 2.8.2. As can be seen, the voltage source is replaced by a short circuit branch and the current source is replaced by an open circuit branch. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: L¼1 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

ð1Þ

C¼1 d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ vC ðt Þ ¼ dt dt

ð2Þ

vL ð t Þ ¼ L iC ðt Þ ¼ C

2

Solutions of Problems: State Equations of Electrical Circuits

19

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

vC ð t Þ



i L ðt Þ

Applying KVL in the right-side mesh of the circuit of Figure 2.8.2: 

d d i ðt Þ þ 1  ðiC ðt Þ  iL ðt ÞÞ ¼ 0 )  iL ðt Þ þ iC ðt Þ  iL ðt Þ ¼ 0 dt L dt ð 2Þ

)

d d i ðt Þ þ vC ðt Þ  iL ðt Þ ¼ 0 dt L dt

ð3Þ

Applying KVL in the left-side mesh of the circuit of Figure 2.8.2: 1  i C ð t Þ þ 1  i C ð t Þ þ vC ð t Þ þ ð2Þ

)2

d d i ðt Þ ¼ 0 ) 2iC ðt Þ þ vC ðt Þ þ iL ðt Þ ¼ 0 dt L dt

d d v ðt Þ þ vC ðt Þ þ iL ðt Þ ¼ 0 dt C dt

ð4Þ

Solving (3) and (4): 2

d d d 1 1 v ðt Þ þ vC ðt Þ þ vC ðt Þ  iL ðt Þ ¼ 0 ) v ðt Þ ¼  vC ðt Þ þ iL ðt Þ dt C dt dt C 3 3

ð5Þ

Solving (5) and (4):
 1 1 d d 1 2 i ð t Þ ¼  vC ð t Þ  i L ð t Þ 2  vC ðt Þ þ iL ðt Þ þ vC ðt Þ þ iL ðt Þ ¼ 0 ) 3 3 dt dt L 3 3 Writing (5) and (6) in the form of matrices: 2

3 2 d 1 v C ðt Þ  6 dt 7 6 3 4 5¼4 1 d  iL ðt Þ 3 dt Choice (3) is the answer.

3 2 1  1   3 7 vC ð t Þ ) A ¼ 6 3 5 4 2 1 iL ðt Þ   3 3

3 1 3 7 5 2  3

ð6Þ

20

2

Solutions of Problems: State Equations of Electrical Circuits

Figure 2.8 The circuit of solution of problem 2.8

2.9. Since the network includes some resistors, an inductor of 1 H, and a capacitor of 1 F, it can be modeled like the one shown in Figure 2.9. Based on the information given in the problem, we have:  X¼

i L ðt Þ

 ð1Þ

vC ð t Þ



a11 A¼ a21

a12 a22

 ð2Þ

Therefore, the state equations are as follows: 2

3 d  iL ðt Þ a11 6 dt 7 4 5¼ d a21 v ðt Þ dt C

a12 a22



i L ðt Þ



vC ð t Þ

ð3Þ

As we know, the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: vL ð t Þ ¼ L

L¼1 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

ð4Þ

2

Solutions of Problems: State Equations of Electrical Circuits

iC ðt Þ ¼ C

21

C¼1 d d v ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ dt C dt

ð5Þ

Solving (3)–(5): 

vL ð t Þ iC ðt Þ



 ¼

a11

a12

a21

a22



i L ðt Þ

 ð6Þ

vC ð t Þ

Determining the inverse of the matrix: 

i L ðt Þ vC ð t Þ



 ¼

a11

a12

a21

a22

1 

 a22 1 ¼ Det ½ A  i C ðt Þ a21

vL ðt Þ



a12



a11

vL ðt Þ



iC ðt Þ

ð7Þ

As can be seen in Figure 2.9.2, by changing the place of the inductor and the capacitor, we have: iL ðt Þ ¼ iC ðt Þ

ð8Þ

vL ð t Þ ¼ vC ð t Þ

ð9Þ

Solving (7)–(9): 

i C ðt Þ vL ð t Þ

 ¼

 a22 1 Det ½A a21

a12



vC ð t Þ

 ð10Þ

i L ðt Þ

a11

Solving (4), (5), and (10): 2

3 d  vC ð t Þ a22 1 6 dt 7 4 5¼ d Det ½A a21 i ðt Þ dt L

a12 a11



vC ð t Þ i L ðt Þ 

Equation (11) shows the state equations of a circuit with the state vector of

Anew

 a22 1 ¼ Det ½A a21

a12 a11



vC ðt Þ iL ðt Þ



Choice (2) is the answer.

Figure 2.9 The circuit of solution of problem 2.9

ð11Þ  . Therefore:

22

2

Solutions of Problems: State Equations of Electrical Circuits

2.10. The current-voltage relation of inductor and voltage-current relation of capacitor are as follows: vL ð t Þ ¼ L

d i ðt Þ dt L

i C ðt Þ ¼ C

d v ðt Þ dt C

We need to write the state equations of the circuit based on the state vector and the input vector given in the problem: 2 3 iL1 ðt Þ   i s ðt Þ 6 7 X ¼ 4 iL2 ðt Þ 5, W ¼ vs ð t Þ vC ð t Þ Applying KCL in the supernode of the circuit of Figure 2.10.2: iR ðt Þ þ iL1 ðt Þ  is ðt Þ þ iL2 ðt Þ ¼ 0 ) iR ðt Þ ¼ iL1 ðt Þ  is ðt Þ þ iL2 ðt Þ

ð1Þ

Applying KVL in the indicated left-side mesh of the circuit of Figure 2.10.2: d i ðt Þ ¼ 0 dt L1 ð1Þ d 1 R R R ) iL1 ðt Þ ¼ vs ðt Þ  iL1 ðt Þ þ is ðt Þ  iL2 ðt Þ dt L1 L1 L1 L1 vs ðt Þ þ iR ðt ÞR þ L1

ð2Þ

Applying KVL in the loop of the circuit of Figure 2.10.2: vs ðt Þ þ iR ðt ÞR þ vC ðt Þ þ L2 ð1Þ

)

d i ðt Þ ¼ 0 dt L2

d 1 R R R 1 i ðt Þ ¼ vs ðt Þ  iL1 ðt Þ þ is ðt Þ  iL2 ðt Þ  vC ðt Þ dt L2 L2 L2 L2 L2 L2

ð3Þ

Applying KCL in the node of the circuit of Figure 2.10.2: C

d d 1 1 v ðt Þ  is ðt Þ þ iL2 ðt Þ ¼ 0 ) v ðt Þ ¼  is ðt Þ þ iL2 ðt Þ dt C dt C C C

ð4Þ

Writing (2)–(4) in the form of matrices: 3 2 R d  ð t Þ i 6 dt L1 7 6 L1 7 6 6 7 6 R 6d 6 iL2 ðt Þ 7 ¼ 6  7 6 L2 6 dt 5 4 4 d vC ð t Þ 0 dt 2

R L1 R  L2 1 C 

3 2 R 2 3 7 iL1 ðt Þ 6 L1 7 6 1 76 7 6 R  74 iL2 ðt Þ 5 þ 6 6 L2 L2 7 5 vC ð t Þ 4 1 0  C 0

3 1 L1 7  7 1 7 is ðt Þ 7 L2 7 vs ðt Þ 5 0

ð6Þ

Based on the information given in the problem, we know that: 2

1 1 1 6 1 A ¼ 4  2 2 0 1 By comparing (6) and (7), we can write:

3 0 17  5 2 0

ð7Þ

2

Solutions of Problems: State Equations of Electrical Circuits

23

8 R >  ¼ 1 > > L > 1 > > > > R 1 > >

1 1 > >  ¼  > > L2 2 > > > > > : 1 ¼1 C Solving (6) and (8), we can write: 2

1 6 1 6 6 1 B¼6 6 2 4 1  1

3 1 2 1 17 7 6 1 17 7 ) B¼4 2 27 5 1 0

3 1 17 5 2 0

Choice (4) is the answer.

Figure 2.10 The circuit of solution of problem 2.10

ð8Þ

24

2

Solutions of Problems: State Equations of Electrical Circuits

2.11. The current-voltage relation of inductor and voltage-current relation of capacitor are as follows: v L ðt Þ ¼ L i C ðt Þ ¼ C

L¼1 d d i ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) vL ðt Þ ¼ i L ðt Þ dt L dt

C¼1 d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iC ðt Þ ¼ vC ðt Þ v ðt Þ ¼ dt C dt

We need to write the state equations of the circuit based on the state vector given in the problem:  X¼

vC ð t Þ



i L ðt Þ

Applying KVL in the indicated mesh of the circuit of Figure 2.11.2: 1 1 vC ðt Þ þ 1  iR ðt Þ þ 1  ðiR ðt Þ  iL ðt ÞÞ ¼ 0 )  vC ðt Þ þ 2iR ðt Þ  iL ðt Þ ¼ 0 ) iR ðt Þ ¼ vC ðt Þ þ iL ðt Þ 2 2

ð1Þ

Applying KCL in the indicated node of the circuit of Figure 2.11.2: is ðt Þ þ iR ðt Þ þ

ð1Þ d d 1 1 v ð t Þ ¼ 0 ) vC ð t Þ ¼ i s ð t Þ  vC ð t Þ  i L ð t Þ dt C dt 2 2

ð2Þ

Applying KVL in the indicated loop of the circuit of Figure 2.11.2: vC ðt Þ þ 1  iR ðt Þ þ ð1Þ

)

d d i ðt Þ þ 1  ðiL ðt Þ  is ðt ÞÞ ¼ 0 ) i ðt Þ ¼ vC ðt Þ  iR ðt Þ  iL ðt Þ þ is ðt Þ dt L dt L

d 1 1 1 3 i ðt Þ ¼ vC ðt Þ  vC ðt Þ  iL ðt Þ  iL ðt Þ þ is ðt Þ ¼ vC ðt Þ  iL ðt Þ þ is ðt Þ dt L 2 2 2 2

By writing (2) and (3) in the form of matrices, we have: 2

3 2 d 1 vC ð t Þ  6 dt 7 6 2 4 5¼4 1 d i L ðt Þ 2 dt Choice (3) is the answer.

3 1      2 7 vC ðt Þ þ 1 i ðt Þ ) A ¼ 0:5 5 s 3 0:5 i L ðt Þ 1  2 

0:5 1:5



ð3Þ

2

Solutions of Problems: State Equations of Electrical Circuits

25

Figure 2.11 The circuit of solution of problem 2.11

2.12. First, we should use source transformation theorem for the parallel connection of the independent current source and the 2-Ω resistor, as is shown in Figure 2.12.2. Recall that the current-voltage relation of inductor and voltage-current relation of capacitor are as follows: vL ð t Þ ¼ L

d i ðt Þ dt L

i C ðt Þ ¼ C

d v ðt Þ dt C

We need to write the state equations of the circuit based on the state vector given in the problem: 2

i1 ðt Þ

3

6 7 X ¼ 4 i2 ðt Þ 5 vð t Þ

26

2

Solutions of Problems: State Equations of Electrical Circuits

Applying KVL in the left-side mesh of the circuit of Figure 2.12.2: 2is ðt Þ þ 2  i1 ðt Þ þ 0:5 

d d i ð t Þ þ vð t Þ ¼ 0 ) i ðt Þ ¼ 4is ðt Þ  4i1 ðt Þ  2vðt Þ dt 1 dt 1

ð1Þ

Applying KVL in the right-side mesh of the circuit of Figure 2.12.2: vðt Þ þ 0:5 

d d i ðt Þ þ 2i2 ðt Þ þ vs ðt Þ ¼ 0 ) i ðt Þ ¼ 4i2 ðt Þ  2vs ðt Þ þ 2vðt Þ dt 2 dt 2

ð2Þ

Applying KCL in the indicated node of the circuit of Figure 2.12.2: i1 ðt Þ þ i2 ðt Þ þ 1 

d d vð t Þ ¼ 0 ) vð t Þ ¼ i 1 ð t Þ  i 2 ð t Þ dt dt

Writing (1)–(3) in the form of matrices: 3 2 d 2 6 dt i1 ðt Þ 7 4 0 7 6 7 6 6d 6 i2 ðt Þ 7 ¼ 4 0 4 7 6 dt 5 4 1 1 d vðt Þ dt

32

3   76 7 6 7 is ðt Þ 2 54 i2 ðt Þ 5 þ 4 0 2 5 vs ð t Þ 0 vð t Þ 0 0

2

i 1 ðt Þ

3

2

4

0

Choice (3) is the answer.

Figure 2.12 The circuit of solution of problem 2.12

ð3Þ

References

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

27

Problems: Laplace Transform and Network Function

Abstract

In this chapter, Laplace transform and network function (transfer function) are applied to solve the basic and advanced problems of electrical circuit analysis. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 3.1. Determine the network function (transfer function) of the circuit illustrated in Figure 3.1 [1–2]. Difficulty level Calculation amount 1. 2. 3. 4.

V out ðsÞ V in ðsÞ V out ðsÞ V in ðsÞ V out ðsÞ V in ðsÞ V out ðsÞ V in ðsÞ

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

1 ¼ 1þRCs RCs ¼ 1þRCs

¼ 1þ2RCs 2þRCs 2RCs ¼ 1þRCs

Figure 3.1 The circuit of problem 3.1

3.2. In the circuit of Figure 3.2, determine the input impedance in Laplace domain seen from the left side of the circuit. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large sþ2 1. 5sþ1 Ω sþ1 2. sþ2 Ω s Ω 3. 5sþ2 sþ1 4. 5sþ1 Ω

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_3

29

3

30

3 Problems: Laplace Transform and Network Function

Figure 3.2 The circuit of problem 3.2 ðsÞ 3.3. In the circuit of Figure 3.3, determine the network function (transfer function) of H ðsÞ ¼ VVout : in ðsÞ

Difficulty level Calculation amount 1. s2 1þ1 1 2. s1 3. s2 11 1 4. sþ1

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

Figure 3.3 The circuit of problem 3.3

3.4. In an electrical circuit, the relation below is given for the input admittance of the circuit in Laplace domain. Determine the differential equation between the voltage and current. Y in ðsÞ ¼ Difficulty level Calculation amount

● Easy ● Small

○ Normal ○ Normal

I ðsÞ s2 þ 2s þ 3 = 3 V ðsÞ 4s þ 5s2 þ 6s þ 7

○ Hard ○ Large

1. 4 dtd 3 iðt Þ þ dtd 2 iðt Þ þ 6 dtd iðt Þ ¼ dtd 2 vðt Þ þ dtd vðt Þ þ 3vðt Þ 3

2. 3. 4.

2

2

d3 d2 d d2 d dt 3 iðt Þ þ dt 2 iðt Þ þ dt iðt Þ ¼ dt 2 vðt Þ þ 2 dt vðt Þ þ vðt Þ 2 d3 d2 d 4 dt3 iðt Þ þ 5 dt2 iðt Þ þ 6 dt iðt Þ þ 7iðt Þ ¼ dtd 2 vðt Þ þ 2 dtd vðt Þ 3 2 2 4 dtd 3 iðt Þ þ 5 dtd 2 iðt Þ þ 6 dtd iðt Þ þ 7iðt Þ ¼ dtd 2 vðt Þ þ 2 dtd vðt Þ

3.5. In the circuit of Figure 3.4, determine the value of VI ððssÞÞ. Difficulty level Calculation amount

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

þ 3vðt Þ þ vð t Þ

3

Problems: Laplace Transform and Network Function

1. 2. 3. 4.

31

s3 10s2 þ10s1 Ω s2 þ6sþ1 s3 þ10s2 þ10sþ2 Ω s2 þ6sþ1 s3 þ10s2 þ1 Ω s2 þ3 s3 8s2 þ8sþ2 Ω s2 þ6sþ3

Figure 3.4 The circuit of problem 3.5

3.6. In the circuit of Figure 3.5, determine the network function (transfer function) of H ðsÞ ¼ IILððssÞÞ. Difficulty level Calculation amount 2 1. s2 þ2sþ1 2 2. s2 þ3sþ2 1 3. s2 þsþ2 1 4. s2 þ3sþ1

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

Figure 3.5 The circuit of problem 3.6

3.7. In the circuit of Figure 3.6, determine the input impedance in Laplace domain seen by the terminal (Zin(s)). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. 2(s2 + 10s + 5) Ω 1 2. 5ðsþ1 Þ Ω 3. 2(s + 1) Ω 2ðs2 þ10sþ5Þ 4. 5ðsþ1Þ Ω

32

3 Problems: Laplace Transform and Network Function

Figure 3.6 The circuit of problem 3.7

3.8. Which one of the choices is correct for the unit step response of a circuit with the following network function (transfer function)? H ðsÞ ¼ Difficulty level ○ Easy ● Normal Calculation amount ● Small ○ Normal 1. vout(t ¼ 0+) ¼ vout(t ¼ 1) ¼ 0 2. vout(t ¼ 0+) ¼ vout(t ¼ 1) ¼ 0.2 3. vout(t ¼ 0+) ¼ 0, vout(t ¼ 1) ¼ 0.2 4. None of them

V out ðsÞ 25 = 2 V in ðsÞ s þ 10s þ 125

○ Hard ○ Large

3.9. The impulse function of a linear time-invariant (LTI) system is h(t) ¼ (et  e2t)u(t). Determine the output response of the system for the input signal of x(t) ¼ 2e3tu(t). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. y(t) ¼ (et + e2t + e3t)u(t) 2. y(t) ¼ (et  2e2t + e3t)u(t) 3. y(t) ¼ (2et + e2t + e3t)u(t) 4. y(t) ¼ (et + 2e2t + e3t)u(t) 3.10. The impulse function of a linear time-invariant (LTI) system is hðt Þ ¼ 34 ðet þ e3t Þuðt Þ . Determine the output response of the system for the input signal of x(t) ¼ 2δ(t  5). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. y(t) ¼ (e(t  5) + e3(t  5))u(t  5)
 2. yðt Þ ¼ 32 eðt5Þ þ e3ðt5Þ uðt  5Þ
 3. yðt Þ ¼  32 eðt5Þ þ e3ðt5Þ uðt  5Þ 4. y(t) ¼  (e(t  5) + e3(t  5))u(t  5) 3.11. In the circuit of Figure 3.7, calculate the network function (transfer function) of VV 2s ððssÞÞ. Difficulty level Calculation amount

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

3

Problems: Laplace Transform and Network Function

1. 2. 3. 4.

33

2ðs2 þ1Þ ðsþ1Þ2

ðs2 þ1Þ ðsþ2Þ2

ðs2 þ1Þ s2 þsþ1 ðs2 þ1Þ ðsþ1Þ2

Figure 3.7 The circuit of problem 3.11

3.12. In the circuit of Figure 3.8, is(t) is the input and i(t) is the response of the circuit. Determine the impulse response of the circuit. Difficulty level ○ Easy Calculation amount ○ Small
 1. 13 e5t  43 e2t uðt Þ  δðt Þ
 2. 13 e5t þ 43 e2t uðt Þ þ δðt Þ
 3. 13 e5t  43 e2t uðt Þ þ δðt Þ
1 5t 4 2t  4. 3 e þ 3 e uðt Þ  δðt Þ

● Normal ● Normal

○ Hard ○ Large

Figure 3.8 The circuit of problem 3.12

34

3 Problems: Laplace Transform and Network Function

3.13. In the circuit of Figure 3.9, determine the differential equation between is(t) and va(t). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. 2. 3. 4.

d2 d d2 d dt 2 va ðt Þ þ 2 dt va ðt Þ þ 2va ðt Þ ¼ dt 2 is ðt Þ þ dt is ðt Þ þ is ðt Þ d2 d d2 d dt 2 va ðt Þ þ dt va ðt Þ þ 2va ðt Þ ¼ dt 2 is ðt Þ þ dt is ðt Þ þ is ðt Þ d2 d d2 d dt 2 va ðt Þ þ dt va ðt Þ þ va ðt Þ ¼ dt 2 is ðt Þ þ dt is ðt Þ þ is ðt Þ 2 2 2 dtd 2 va ðt Þ þ dtd va ðt Þ þ 2va ðt Þ ¼ dtd 2 is ðt Þ þ dtd is ðt Þ þ is ðt Þ

Figure 3.9 The circuit of problem 3.13

3.14. In the circuit of Figure 3.10, determine the differential equation between vs(t) and i(t). Difficulty level Calculation amount 1. 2. 3. 4.

d2 dt 2 d2 dt 2 d2 dt 2 d2 dt 2

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

iðt Þ þ dtd iðt Þ þ iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ

iðt Þ þ dtd iðt Þ þ 2iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ

iðt Þ þ 2 dtd iðt Þ þ 2iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ

iðt Þ þ 2 dtd iðt Þ þ 2iðt Þ ¼ 2 dtd vs ðt Þ þ vs ðt Þ

Figure 3.10 The circuit of problem 3.14

3.15. In the circuit of Figure 3.11, determine the network function (transfer function) of H ðsÞ ¼ VVoutin ððssÞÞ. Difficulty level Calculation amount

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

3

Problems: Laplace Transform and Network Function

1. 2. 3. 4.

35

s2 þ3s ðsþ1Þ2 ðs1Þ2 ðsþ1Þ2 sðs1Þ ðsþ1Þ2 sþ3 ðsþ1Þ2

Figure 3.11 The circuit of problem 3.15

3.16. In the circuit of Figure 3.12, determine the equivalent impedance in Laplace domain seen by the terminal (Zeq(s)). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. s1 5 Ω sþ11 2. 5 Ω 3. (s  1) Ω 4. 1 Ω

Figure 3.12 The circuit of problem 3.16

3.17. In the circuit of Figure 3.13, determine the network function (transfer function) of H ðsÞ ¼ VVoutin ððssÞÞ. Difficulty level Calculation amount 2sþ1 1. 12sþ1 2sþ1 2. 12sþ4 4sþ1 3. 12sþ4 4sþ1 4. 12sþ1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

36

3 Problems: Laplace Transform and Network Function

Figure 3.13 The circuit of problem 3.17

3.18. The impulse function of a linear time-invariant (LTI) system is h(t) ¼ (et  e2t)u(t). Determine the output response of the system if the input signal is x(t) ¼ 2e2tu(t). Difficulty level ○ Easy ● Normal Calculation amount ○ Small ● Normal 1. y(t) ¼ (2et  e2t + e3t)u(t) 2. y(t) ¼ (2et  4e2t + e3t)u(t) 3. y(t) ¼ (2et  4e2t + 2e3t)u(t) 4. y(t) ¼ (2et  4e2t  2e3t)u(t)

○ Hard ○ Large

3.19. In the circuit of Figure 3.14, calculate the value of IL(s) for vC(0) ¼ 2 V, iL(0) ¼ 1 A. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. 2. 3. 4.

s2 þ2s20 sðs2 þ2sþ2Þ s2 þ2sþ20 sðs2 þ2sþ2Þ s2 þ3s20 sðs2 þ2s2Þ s2 3s20 sðs2 þ2sþ2Þ

A A A A

Figure 3.14 The circuit of problem 3.19

3

Problems: Laplace Transform and Network Function

37

3.20. In the circuit of Figure 3.15, calculate the value of Vout(s). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. s2 ðs3 þs5s 2 þ3sþ1Þ V 2. 3. 4.

5s2 s2 ðs3 þ2s2 þ3sþ1Þ 5s2 ðs3 þs2 þsþ1Þ V 5s s2 ðs3 þ2s2 þ3sþ1Þ

V V

Figure 3.15 The circuit of problem 3.20

3.21. In the circuit of Figure 3.16, calculate the impulse response of vout(t). Difficulty level ○ Easy Calculation amount ○ Small 1. (2et  e0.5t)u(t) V 2. 2tetu(t) V 3. (et  e0.5t)u(t) V 4. 2etu(t) V

● Normal ○ Normal

○ Hard ● Large

Figure 3.16 The circuit of problem 3.21

ðsÞ 3.22. In the circuit of Figure 3.17, determine the network function (transfer function) of VVout . in ðsÞ

Difficulty level Calculation amount

○ Easy ○ Small

● Normal ○ Normal

○ Hard ● Large

38

3 Problems: Laplace Transform and Network Function

1. 2. 3. 4.

1 s4 þs3 þs2 þ3sþ1 1 s4 þ2s3 þ4s2 þ3sþ1 1 s3 þs2 þ3sþ1 1 s4 2s3 3s2 þ3sþ1

Figure 3.17 The circuit of problem 3.22

3.23. In the circuit of Figure 3.18, calculate the value of Vout(s) if vs(t) ¼ Acos(ωt) and the circuit is at zero state. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large   A

1.

ð

1 R2 R1

Þ 

R s2 þ L1 sþR2

2. 

 s2 þ



A

3. 4.

A

s2

ðs2 þω2 Þ

R1 1 R2 C L

R1 1 L þR2 C



s2 R



V

2

s2

ðs2 þRL1 sþR2 Þðs2 ω2 Þ   1 R2 þR1

V

sþLR 1C ðs2 þω2 Þ



1 R2 R1



s2

ðs2 þRL1 sþR2 Þðs2 þω2 Þ

V V

Figure 3.18 The circuit of problem 3.23

3.24. In the circuit of Figure 3.19, the input impedance in Laplace domain is as follows: Z in ðsÞ =

s2 þ s þ 2 2s2 þ s þ 1

While the circuit is in zero state, the switch is closed at t ¼ 0, and then i(0+) ¼ 6 A is measured. Determine the value of E.

3

Problems: Laplace Transform and Network Function

Difficulty level Calculation amount 1. 3 V 2. 6 V 3. 9 V 4. 12 V

○ Easy ● Small

○ Normal ○ Normal

39

● Hard ○ Large

Figure 3.19 The circuit of problem 3.24

3.25. In the circuit of Figure 3.20, vs(t) is a DC voltage source. Determine the time (in second) that the voltage of the capacitor will be twice as the voltage of the source. The primary voltage of the capacitor and the primary current of the inductor are zero. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffiffiffiffi 1. π LC sec . pffiffiffiffiffiffi 2. 2π LC sec . pffiffiffiffiffi 3. πLC sec . 4. No time can be found, as it is impossible.

Figure 3.20 The circuit of problem 3.25

3.26. In the circuit of Figure 3.21, determine the Thevenin equivalent circuit in Laplace domain seen from the terminal. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 3 1. Z Th ðsÞ ¼ sþ1 Ω, V Th ðsÞ ¼ 3 V 2. Z Th ðsÞ ¼ ð3s þ 3Þ Ω, V Th ðsÞ ¼ 3s V 3s 3. Z Th ðsÞ ¼ sþ1 Ω, V Th ðsÞ ¼ 3s V 4. ZTh(s) ¼ (3s + 3) Ω, VTh(s) ¼ 3 V

40

3 Problems: Laplace Transform and Network Function

Figure 3.21 The circuit of problem 3.26

3.27. In the circuit of Figure 3.22, determine the energy stored in the inductor at t ¼ 1 if i1 ðt Þ ¼ function (transfer function) of the circuit is as follows: H ðsÞ ¼ Difficulty level Calculation amount 1. 0.1 J 2. 0.15 J 3. 0.2 J 4. 0.05 J

○ Easy ○ Small

○ Normal ● Normal

pffiffiffi 2uðt Þ, and the network

I 2 ðsÞ 4ðs þ 20Þ = sþ8 I 1 ðsÞ

● Hard ○ Large

Figure 3.22 The circuit of problem 3.27

3.28. In the circuit of Figure 3.23, determine the capacitance (C) and the primary voltage of the capacitor (vC(0)) to have iout(t > 0) ¼ 0. The inductor does not have any primary energy and vs(t) ¼ sin (2t)u(t). Difficulty level ○ Easy Calculation amount ○ Small 1. C ¼ 14 F, vC ð0 Þ ¼ 8 V 2. C ¼ 18 F, vC ð0 Þ ¼ 4 V 3. C ¼ 18 F, vC ð0 Þ ¼ 8 V 4. C ¼ 14 F, vC ð0 Þ ¼ 8 V

○ Normal ● Normal

● Hard ○ Large

3

Problems: Laplace Transform and Network Function

41

Figure 3.23 The circuit of problem 3.28

3.29. In an electrical circuit, the relation below exists. If vin(t) ¼ 4 cos (2t), determine the steady-state output voltage of the circuit: H ðsÞ ¼ Difficulty level ○ Easy ○ Normal Calculation amount ○ Small ● Normal  1. vout(t) ¼ 21.76 cos (2t + 40.6 ) V  2. vout(t) ¼ 20.2 cos (2t + 40.6 ) V  3. vout(t) ¼ 21.76 cos (2t  40.6 ) V  4. vout(t) ¼ 43.52 cos (2t  40.6 ) V

V out ðsÞ 10ðs þ 1Þ = 2 V in ðsÞ s þ 2s þ 3

● Hard ○ Large

3.30. The pulse voltage, shown in the circuit of Figure 3.24.1, is applied on the circuit of Figure 3.24.2. Determine the current if vC(0) ¼ 0. Difficulty level ○ Easy ○ Normal Calculation amount ○ Small ● Normal 1. 20etu(t)  10e(t  2)u(t  1) A 2. 10etu(t)  10e(t  1)u(t  1) A 3. 20etu(t)  20e(t  1)u(t  1) A 4. 20etu(t) + 20e(t  1)u(t  1) A

● Hard ○ Large

Figure 3.24 The circuit of problem 3.30

3.31. In the circuit of Figure 3.25, the primary current of each inductor is 2 A (i1(0) ¼ i2(0) ¼ 2 A), while the primary voltage of each capacitor is zero. Calculate the output voltage (vout(t)) for t  0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

42

3 Problems: Laplace Transform and Network Function

1. 2. 3. 4.

(2et  2tet) V (2et + 2tet) V (2et + 2tet + 4e2t) V (2et  2tet + 4e2t) V

Figure 3.25 The circuit of problem 3.31

3.32. Which one of the choices below is true for the circuit, seen from terminal A–B, shown in Figure 3.26? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1. It is equivalent to a short circuit. 2. It is equivalent to a capacitor with the capacitance of α F. 3. It is equivalent to a resistor with the resistance of α Ω. 4. It is equivalent to an inductor with the inductance of α H.

Figure 3.26 The circuit of problem 3.32

3.33. The circuit shown in Figure 3.27 has been in that situation for a long time. At t ¼ 0, part of the circuit is cut down from the dashed line. Determine vC1(t) for t > 0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1. (4e0.5t + 6) V 2. (4e2t + 6) V 3. (8e2t + 2) V 4. (8e0.5t + 2) V

References

43

Figure 3.27 The circuit of problem 3.33

3.34. In the circuit of Figure 3.28, both switches are simultaneously closed. Calculate the voltage of 2-F capacitor exactly after the switching operation. Difficulty level Calculation amount 1. 3 V 2. 4 V 3. 6 V 4. 9 V

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

Figure 3.28 The circuit of problem 3.34

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

4

Solutions of Problems: Laplace Transform and Network Function

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods.

4.1. The circuit of Figure 4.1.2 shows the main circuit in Laplace domain. The impedances of the components are as follows [1–2]: ZR ¼ R

ð1Þ

1 Cs

ð2Þ

ZC ¼ Applying voltage division rule for the resistor: V out ðsÞ ¼

V ðsÞ R RCs  V in ðsÞ ) out ¼ 1 RCs þ 1 V in ðsÞ R þ Cs

Choice (2) is the answer.

Figure 4.1 The circuit of solution of problem 4.1

4.2. The circuit of Figure 4.2.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_4

ð1Þ

45

46

4

Solutions of Problems: Laplace Transform and Network Function

Z R ¼ R ) Z 0:25 Ω ¼ 0:25 Ω ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

ð2Þ ð3Þ

The right-side impedance (2s1 Ω) does not have any effect on the input impedance seen from the left side of the circuit, since it is located on an open circuit branch. Therefore:
  1   1 þ 0:25 sþ2
1 2s ¼ þ 0:25 ¼ Ω Z in ¼ 1
2s 5s þ 2 1 þ 2s1 þ 0:25 Choice (1) is the answer.

Figure 4.2 The circuit of solution of problem 4.2

4.3. The circuit of Figure 4.3.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð1Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð2Þ

1 1 ) Z1 F ¼ Ω Cs s

ð3Þ

ZC ¼ Using voltage division rule for the output voltage:

4

Solutions of Problems: Laplace Transform and Network Function

V out ðsÞ ¼ 1

1 s

sþs

V in ðsÞ ¼

47

V ðsÞ V ðsÞ 1 1 1 V ðsÞ ) out ) H ðsÞ ¼ out ¼ 2 ¼ 2 V in ðsÞ V in ðsÞ s2 þ 1 in s þ1 s þ1

Choice (1) is the answer.

Figure 4.3 The circuit of solution of problem 4.3

4.4. Based on the information given in the problem, we have: Y in ðsÞ ¼

I ðsÞ s2 þ 2s þ 3 = 3 V ðsÞ 4s þ 5s2 þ 6s þ 7

    ) 4s3 þ 5s2 þ 6s þ 7 I ðsÞ ¼ s2 þ 2s þ 3 V ðsÞ Applying inverse Laplace transform: L1 d3 d2 d d2 d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 4 3 iðt Þ þ 5 2 iðt Þ þ 6 iðt Þ þ 7iðt Þ ¼ 2 vðt Þ þ 2 vðt Þ þ 3vðt Þ dt dt dt dt dt Choice (3) is the answer. 4.5. The circuit of Figure 4.4.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls ) Z 1 H ¼ s Ω

ð1Þ

48

4

Solutions of Problems: Laplace Transform and Network Function

ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð2Þ

ZR ¼ R ) Z4 Ω ¼ 4 Ω

ð3Þ

ZC ¼

1 1 ) Z1 F ¼ Ω Cs s

ð4Þ

In this problem, the value of VI ððssÞÞ is equal to the input impedance seen by the voltage source. Therefore, we only need to calculate the input impedance as follows: Z eq

 
  ðs þ 2Þ 4 þ 1s 4s þ 9 þ 2s 1 4s2 þ 9s þ 2
 ¼sþ ¼sþ ¼ s þ ðs þ 2Þ
4 þ ¼sþ 2 1 1 s s þ 6s þ 1 sþ6þs ð s þ 2Þ þ 4 þ s Z eq ¼

V ðsÞ s3 þ 10s2 þ 10s þ 2 Ω ¼ I ðsÞ s2 þ 6s þ 1

Choice (2) is the answer.

Figure 4.4 The circuit of solution of problem 4.5

4.6. The circuit of Figure 4.5.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZC ¼

1 1 2 ) Z 0:5 F ¼ ¼ Ω Cs 0:5s s

ZR ¼ R ) Z3 Ω ¼ 3 Ω

ð2Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð3Þ

Applying current division rule: I L ðsÞ ¼ 2

2 s

sþ3þs

Choice (2) is the answer.

I s ðsÞ ¼

ð1Þ

I ðsÞ 2 2 I ðsÞ ) H ðsÞ ¼ L ¼ 2 I s ðsÞ s þ 3s þ 2 s2 þ 3s þ 2 s

4

Solutions of Problems: Laplace Transform and Network Function

49

Figure 4.5 The circuit of solution of problem 4.6

4.7. The circuit of Figure 4.6.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð1Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð2Þ

Z L ¼ Ls ) Z 2 H ¼ 2s Ω

ð3Þ

Z L ¼ Ls ) Z 5 H ¼ 5s Ω

ð4Þ

Z R ¼ R ) Z 10 Ω ¼ 10 Ω

ð5Þ

As is shown in Figure 4.6.2, to find the Thevenin impedance, we need to apply a test source (e.g., a test voltage source with the voltage and current of Vt(s) and It(s)) to calculate the value of VI ttððssÞÞ. Applying KVL in the left-side mesh: V t ðsÞ þ 2I t ðsÞ þ sI t ðsÞ  2sI ðsÞ þ sI t ðsÞ þ 2sI ðsÞ ¼ 0 )  V t ðsÞ þ 2ðs þ 1ÞI t ðsÞ ¼ 0 ) ) Z Th ðsÞ ¼ 2ðs þ 1Þ Ω Choice (3) is the answer.

V t ðsÞ ¼ 2ð s þ 1Þ It

50

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.6 The circuit of solution of problem 4.7

4.8. Based on the information given in the problem, we know that: L

vin ðt Þ ¼ uðt Þ)V in ðsÞ = H ðsÞ ¼

1 s

ð1Þ V out ðsÞ 25 25 25 ) V out ðsÞ = 2 V in ðsÞ)V out ðsÞ ¼ 2 = 2 V in ðsÞ s þ 10s þ 125 s þ 10s þ 125 sðs þ 10s þ 125Þ

ð1Þ ð2Þ

From initial value theorem, we know that: lim sF ðsÞ f ð0þ Þ ¼ s!1

ð3Þ

f ð1Þ ¼ lim sF ðsÞ

ð4Þ

From final value theorem, we know that: s!0

4

Solutions of Problems: Laplace Transform and Network Function

51

Solving (2) and (3): 25 ¼0 ðs2 þ 10s þ 125Þ

ð5Þ

25 25 ¼ 0:2 ¼ ðs2 þ 10s þ 125Þ 125

ð6Þ

lim sV out ðsÞ ¼ s!1 lim vout ð0þ Þ ¼ s!1 Solving (2) and (4): vout ð1Þ ¼ lim sV out ðsÞ ¼ lim s!0

s!0

Choice (3) is the answer. 4.9. Based on the information given in the problem, we have:   L ¼ ¼ ¼ )H ðsÞ ¼ hðt Þ ¼ et  e2t uðt Þ¼

1 1  sþ1 sþ2

L ¼ ¼ ¼ ) X ðsÞ ¼ xðt Þ ¼ 2e3t uðt Þ ¼

2 sþ3

ð1Þ ð2Þ

As we know, the network function (transfer function) is defined as follows: H ðsÞ ¼

Y ðsÞ X ðsÞ

ð3Þ

Therefore:    ð1Þ, ð2Þ 1 1 2 2 2  ¼  Y ðsÞ ¼ H ðsÞX ðsÞ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) Y ðsÞ ¼ sþ1 sþ2 sþ3 ðs þ 1Þðs þ 3Þ ðs þ 2Þðs þ 3Þ ) Y ðsÞ ¼

1 2 1  þ sþ1 sþ2 sþ3

  L1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )yðt Þ ¼ et  2e2t þ e3t uðt Þ Choice (2) is the answer. 4.10. Based on the information given in the problem, we have:

hð t Þ ¼

   L 3  t 3 1 1 e þ e3t uðt Þ ¼ þ ¼ ¼ ¼ ) H ðsÞ ¼ 4 4 sþ1 sþ3 L xðt Þ ¼ 2δðt  5Þ ¼ ¼ ¼ ¼ ) X ðsÞ ¼ 2e5s

ð1Þ ð2Þ

As we know, the network function (transfer function) is defined as follows: H ðsÞ ¼ Therefore:

Y ðsÞ X ðsÞ

ð3Þ

52

4

Solutions of Problems: Laplace Transform and Network Function

  3 3 ð1Þ, ð2Þ 3 1 1 þ  2e5s ¼ 2 e5s þ 2 e5s Y ðsÞ ¼ H ðsÞX ðsÞ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) Y ðsÞ ¼ 4 sþ1 sþ3 sþ1 sþ3   L1 3 ðt5Þ e ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )yðt Þ ¼ þ e3ðt5Þ uðt  5Þ 2 Choice (2) is the answer. 4.11. The circuit of Figure 4.7.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls ) Z 2 H ¼ 2s Ω ZC ¼

ð1Þ

1 1 2 ) Z 0:5 F ¼ ¼ Ω Cs 0:5s s

ð2Þ

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð3Þ

ZR ¼ R ) Z4 Ω ¼ 4 Ω

ð4Þ

Z L ¼ Ls ) Z 0:5 H ¼ 0:5s Ω

ð5Þ

ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

ð6Þ

Applying KCL in node 1: V 1 ðsÞ  V s ðsÞ V 1 ðsÞ V ðsÞ  V 2 ðsÞ þ ¼0) þ 1 1 1 4 0:5s þ 2s



 5 2s 1 þ 2 V ðsÞ  V 2 ðsÞ ¼ V s ðsÞ 4 s þ1 1 4

ð7Þ

Applying KCL in node 2: V 2 ðsÞ  V 1 ðsÞ V 2 ðsÞ V 2 ðsÞ  V s ðsÞ
þ þ ¼0
4 1 2s
2 s

  2  1 5 s2 þ 1 s þ1 þ V 2 ðsÞ ¼ V s ðsÞ )  V 1 ðsÞ þ 4 4 2s 2s Solving (7) and (8): V 2 ð s Þ ð s 2 þ 1Þ ¼ V s ðsÞ ðs þ 1Þ2 Choice (4) is the answer.

ð8Þ

4

Solutions of Problems: Laplace Transform and Network Function

53

Figure 4.7 The circuit of solution of problem 4.11

4.12. Based on the information given in the problem, we know that the input (is(t)) is an impulse function. Therefore: i s ðt Þ ¼ δ ðt Þ ) I s ðsÞ ¼ 1

ð1Þ

The circuit of Figure 4.8.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z4 Ω ¼ 4 Ω

ð2Þ

ZR ¼ R ) Z6 Ω ¼ 6 Ω

ð3Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð4Þ

54

4

ZC ¼

Solutions of Problems: Laplace Transform and Network Function

1 1 4 ) Z 0:25 F ¼ ¼ Ω Cs 0:25s s

ð5Þ

4 I ðsÞ 4 þ Z box s

ð6Þ

Applying current division rule: I ðsÞ ¼ where Zbox can be calculated as follows:
4 4ð s þ 6Þ
4 ð6 þ sÞ s Z box ¼ ð6 þ sÞ
¼ ¼ s 6 þ s þ 4s s2 þ 6s þ 4

ð7Þ

Solving (1), (6), and (7): I ðsÞ ¼

4 4þ

4ðsþ6Þ s2 þ6sþ4

1¼

1

4

s2 þ 6s þ 4 sþ6 ¼1þ 3  3 ¼1 sþ5 sþ2 ð s þ 2Þ ð s þ 5Þ s2 þ 7s þ 10

  L1 1 5t 4 2t e  e ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) iðt Þ ¼ uðt Þ þ δðt Þ 3 3 Choice (3) is the answer.

Figure 4.8 The circuit of solution of problem 4.12

4

Solutions of Problems: Laplace Transform and Network Function

55

4.13. The circuit of Figure 4.9.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð1Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð2Þ

1 1 ) Z1 F ¼ Ω Cs s

ð3Þ

ZC ¼

The equivalent impedance seen by the current source can be calculated as follows:

 1

Z eq ¼ 1
s þ
1 s

¼ 1
s þ 1 s



¼ 1
s þ

1 s

!

þ1

1 sþ1




 2
s þsþ1 ¼ 1
sþ1 ¼

1

s2 þsþ1 sþ1 2 þsþ1 þ s sþ1

¼

s2 þ s þ 1 s2 þ 2s þ 2

ð4Þ

Using Ohm’s law in the circuit of Figure 4.9.3: V a ðsÞ ¼ Z eq I s ðsÞ

ð5Þ

Solving (4) and (5): V a ðsÞ ¼

    s2 þ s þ 1 I s ðsÞ ) s2 þ 2s þ 2 V a ðsÞ ¼ s2 þ s þ 1 I s ðsÞ 2 s þ 2s þ 2

Applying inverse Laplace transform on (6): L1 d2 d d2 d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2 va ðt Þ þ 2 va ðt Þ þ 2va ðt Þ ¼ 2 is ðt Þ þ is ðt Þ þ is ðt Þ dt dt dt dt Choice (1) is the answer.

ð6Þ

56

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.9 The circuit of solution of problem 4.13

4.14. The circuit of Figure 4.10.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð1Þ

1 1 ) Z1 F ¼ Ω Cs s

ð2Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð3Þ

ZC ¼

The equivalent impedance, seen by the current source, is calculated as follows: 
 1 1 s2 þ 2s þ 2
1 Z eq ¼ 1 þ 1
þ s ¼ 1 þ s 1 þ s ¼ 1 þ þs¼ s sþ1 sþ1 1þs

ð4Þ

Using Ohm’s law in the circuit of Figure 4.10.3: V s ðsÞ ¼ Z eq I s ðsÞ Solving (4) and (5): V s ðsÞ ¼

  s2 þ 2s þ 2 I s ðsÞ ) ðs þ 1ÞV s ðsÞ ¼ s2 þ 2s þ 2 I s ðsÞ sþ1

L1 d2 d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2 iðt Þ þ 2 iðt Þ þ 2iðt Þ ¼ vs ðt Þ þ vs ðt Þ dt dt dt Choice (3) is the answer.

ð5Þ

4

Solutions of Problems: Laplace Transform and Network Function

57

Figure 4.10 The circuit of solution of problem 4.14

4.15. The circuit of Figure 4.11.2 shows the primary circuit in Laplace domain. The impedances of the components are as follows: ZC ¼

1 1 ) Z1 F ¼ Ω Cs s

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð1Þ ð2Þ

Applying voltage division formula for the 1 Ω resistor in the vertical branch: V ðsÞ ¼

1 s  V in ðsÞ  V in ðsÞ ¼ sþ1 1 þ 1s

ð3Þ

Applying KCL in the output node: V out ðsÞ  2V ðsÞ V out ðsÞ  V in ðsÞ þ ¼ 0 ) V out ðsÞ  2V ðsÞ þ sV out ðsÞ  sV in ðsÞ ¼ 0 1 1 s ) ðs þ 1ÞV out ðsÞ  2V ðsÞ  sV in ðsÞ ¼ 0

ð4Þ

58

4

Solutions of Problems: Laplace Transform and Network Function

Solving (3) and (4): ðs þ 1ÞV out ðsÞ  2

  s 2s  V in ðsÞ  sV in ðsÞ ¼ 0 ) ðs þ 1ÞV out ðsÞ  þ s V in ðsÞ ¼ 0 sþ1 sþ1

) ðs þ 1ÞV out ðsÞ 

2  V ðsÞ s þ3 s2 þ 3 V in ðsÞ ¼ 0 ) out ¼ sþ1 V in ðsÞ ðs þ 1Þ2

Choice (1) is the answer.

Figure 4.11 The circuit of solution of problem 4.15

4

Solutions of Problems: Laplace Transform and Network Function

59

4.16. The circuit of Figure 4.12.2 shows the primary circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls ) Z 1 H ¼ s Ω

ð1Þ

Z L ¼ Ls ) Z 2 H ¼ 2s Ω

ð2Þ

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð3Þ

As is shown in Figure 4.12.3, to find the Thevenin impedance or the equivalent impedance in Laplace domain (Zeq(s)), we need to turn off all the independent sources (herein, the independent voltage source must be replaced by a short circuit branch) and apply a test source (e.g., a test voltage source with the voltage and current of Vt(s) and It(s)) to calculate the value of VI ttððssÞÞ if the circuit includes a dependent source. To simplify the problem, source transformation theorem is applied on the parallel connection of the dependent current source and the resistor to change it to the series connection of the dependent voltage source and the same resistor, as can be seen in Figure 4.12.3. Applying KVL in the left-side mesh of the circuit of Figure 4.12.3: sI ðsÞ þ sðI ðsÞ þ I t ðsÞÞ þ 2sðI ðsÞ þ I t ðsÞÞ þ sI ðsÞ ¼ 0 ) 5sI ðsÞ þ 3sI t ðsÞ ¼ 0 3 ) I ðsÞ ¼  I t ðsÞ 5

ð1Þ

V t ðsÞ þ I t ðsÞ  1 þ 2I ðsÞ þ 2sðI ðsÞ þ I t ðsÞÞ þ sI ðsÞ ¼ 0 )  V t ðsÞ þ ð2s þ 1ÞI t ðsÞ þ ð3s þ 2ÞI ðsÞ ¼ 0 Solving (1) and (2):   3 s1 I ðsÞ ¼ 0 V t ðsÞ þ ð2s þ 1ÞI t ðsÞ þ ð3s þ 2Þ  I t ðsÞ ¼ 0 )  V t ðsÞ þ 5 5 t ) Choice (1) is the answer.

V t ðsÞ s  1 s1 ) Z eq ðsÞ ¼ Ω ¼ 5 5 I t ðsÞ

ð2Þ

60

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.12 The circuit of solution of problem 4.16

4.17. The circuit of Figure 4.13.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð1Þ

ZR ¼ R ) Z4 Ω ¼ 4 Ω

ð2Þ

4

Solutions of Problems: Laplace Transform and Network Function

ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

61

ð3Þ

Applying voltage division rule for V2(s) in the circuit of Figure 4.13.2: V 2 ðsÞ ¼

þ4 8s þ 1 V ðsÞ  V in ðsÞ ¼ 1 12s þ 1 in 2 þ 2s þ 4 1 2s

ð4Þ

Applying voltage division rule for V1(s) in the circuit of Figure 4.13.2: V 1 ðsÞ ¼

2 4s V ðsÞ  V in ðsÞ ¼ 12s þ 1 in 4 þ 2s1 þ 2

V out ðsÞ ¼ V 2 ðsÞ  V 1 ðsÞ ¼

8s þ 1 4s 4s þ 1 V ðsÞ  V ðsÞ ¼ V ðsÞ 12s þ 1 in 12s þ 1 in 12s þ 1 in )

V out ðsÞ 4s þ 1 ¼ 12s þ1 V in ðsÞ

Choice (4) is the answer.

Figure 4.13 The circuit of solution of problem 4.17

ð5Þ

62

4

Solutions of Problems: Laplace Transform and Network Function

4.18. Based on the information given in the problem, we have:   L hðt Þ ¼ et  e3t uðt Þ ¼ ¼ ¼ ¼ ) H ðsÞ ¼ L ¼ ¼ ¼ ) X ðsÞ ¼ xðt Þ ¼ 2e2t ¼

1 1  sþ1 sþ3

2 sþ2

ð1Þ ð2Þ

As we know, the network function (transfer function) is defined as follows: H ðsÞ ¼

Y ðsÞ X ðsÞ

ð3Þ

Therefore:   ð1Þ, ð2Þ 1 1 2 2 2  ¼  Y ðsÞ ¼ H ðsÞX ðsÞ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) Y ðsÞ ¼ s þ 1 s þ 3 s þ 2 ð s þ 1Þ ð s þ 2Þ ð s þ 2Þ ð s þ 3Þ ) Y ðsÞ ¼

2 4 2  þ sþ1 sþ2 sþ3

  L1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )yðt Þ ¼ 2et  4e2t þ 2e3t uðt Þ Choice (3) is the answer. 4.19. The circuit of Figure 4.14.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð1Þ

Z L ¼ Ls ) Z 1 H ¼ s Ω

ð2Þ

ZC ¼

1 1 2 ) Z 0:5 F ¼ ¼ Ω Cs 0:5s s

ð3Þ

Moreover, the current of the independent current source in Laplace domain is as follows: L 10 10uðt Þ ¼ ¼ ¼ ¼ ) s

ð4Þ

Based on the information given in the problem, we have: vC ð0 Þ ¼ 2 V, iL ð0 Þ ¼ 1 A

ð5Þ

1 Since the capacitor has a nonzero primary voltage, it needs to be modeled by an impedance (Cs ¼ 2s Ω) in series with an  independent voltage source (vC ðs0 Þ ¼ 2s ) in Laplace domain, as is shown in Figure 4.14.2. Likewise, the inductor has a nonzero primary current. Hence, it needs to be modeled by an impedance (Ls ¼ s Ω) in parallel with an independent  current source (iL ð0s Þ ¼ 1s ) in Laplace domain.

To simplify the circuit, source transformation theorem can be applied on the parallel connection of the independent current source and the inductor to change it to the series connection of the independent voltage source and the same inductor, as can be seen in Figure 4.14.3.

4

Solutions of Problems: Laplace Transform and Network Function

Now, by applying KVL in the right-side mesh, we have: 1  sI L ðsÞ  2I L ðsÞ þ



   10 2 2 2 10 2 2  I L ðsÞ þ ¼ 0 ) s  2  I L ðsÞ þ 1 þ  þ ¼ 0 s s s s s s s

    2 20 2 s2 þ 2s þ 20 ) s þ 2 þ I L ðsÞ ¼ 1 þ 2 þ ) I L ðsÞ ¼ 2 A s s s sðs þ 2s þ 2Þ Choice (2) is the answer.

Figure 4.14 The circuit of solution of problem 4.19

63

64

4

Solutions of Problems: Laplace Transform and Network Function

4.20. The circuit of Figure 4.15.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls ) Z 1 H ¼ s Ω

ð1Þ

1 1 ) Z1 F ¼ Ω Cs s

ð2Þ

ZC ¼

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð3Þ

Moreover, the voltage of the independent voltage source in Laplace domain is as follows: L 5 vs ðsÞ ¼ 5t ¼ ¼ ¼ ¼ ) V s ðsÞ ¼ 2 s

ð4Þ

Applying voltage division rule: 

11s 1 1
1s ð 4Þ 1þ1s 5 5 sþ1 
 V s ðsÞ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )V ð s Þ ¼  ¼  2 V out ðsÞ ¼ out 2 1 1
1 11s s s 1 1 s þ þ 1 þ s þ s þ 1 þ 1þ1 s þ s þ 1 þ 1
s s sþ1 s

) V out ðsÞ ¼

5s V s2 ðs3 þ 2s2 þ 3s þ 1Þ

Choice (4) is the answer.

Figure 4.15 The circuit of solution of problem 4.20

4

Solutions of Problems: Laplace Transform and Network Function

65

4.21. The circuit of Figure 4.16.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z2 Ω ¼ 2 Ω

ð1Þ

ZC ¼

1 1 ) Z1 F ¼ Ω Cs s

ð2Þ

ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

ð3Þ

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð4Þ

Based on the information given in the problem, we know that the voltage source (vs(t)) is an impulse function. Therefore: L vs ð t Þ ¼ δ ð t Þ ¼ ¼ ¼ ¼ ) V s ðsÞ ¼ 1 ð5Þ From the circuit of Figure 4.16.2, it is clear that: 1 V out ðsÞ ¼ 2V ðsÞ ) V ðsÞ ¼ V out ðsÞ 2

ð6Þ

Applying voltage division rule for the voltage of 1 Ω resistor: V ðsÞ ¼

      ð6Þ 1 2s 2s þ 1 2s þ 1 V ðsÞ ) V ðsÞ ¼ V A ðsÞ ) V A ðsÞ ¼ V ðsÞ ) V A ðsÞ ¼ V out ðsÞ 1 A 2s þ 1 2s 4s 1 þ 2s

Applying KCL in node A: V A ðsÞ  V s ðsÞ V A ðsÞ V A ðsÞ V A ðsÞ  V out ðsÞ þ 1 þ1 ¼0 þ 2 2 s 2s þ 1  ) 1þsþ

 2s 1 1 V ðsÞ  V s ðsÞ  V out ðsÞ ¼ 0 2s þ 1 A 2 2

 2  2s þ 5s þ 1 1 1 V A ðsÞ  V s ðsÞ  V out ðsÞ ¼ 0 ) 2s þ 1 2 2   ð5Þ, ð7Þ 2s2 þ 5s þ 1 2s þ 1 1 1 V out ðsÞ   1  V out ðsÞ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2s þ 1 4s 2 2 )

 2   2  2s þ 5s þ 1 1 1 2s þ 3s þ 1 1  V out ðsÞ  ¼ 0 ) V out ðsÞ ¼ 4s 2 2 4s 2

) V out ðsÞ ¼

2s 2s 2 2 2 1 ¼  ¼  ¼ 2s2 þ 3s þ 1 ðs þ 1Þð2s þ 1Þ s þ 1 2s þ 1 s þ 1 s þ 12   L1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )vout ðt Þ ¼ 2et  e0:5t uðt Þ V

Choice (1) is the answer.

ð7Þ

66

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.16 The circuit of solution of problem 4.21

4.22. The circuit has been presented in Laplace domain in Figure 4.17. Applying KCL in node A: V A ðsÞ  V 1 ðsÞ V A ðsÞ V A ðsÞ þ 1 þ ¼0) sþ1 s þ 1 þ 1s s )

)



   1 s 1 þsþ 2 V ðsÞ ¼ 0 V A ðsÞ  sþ1 sþ1 1 s þsþ1

ðs2 þ s þ 1Þ þ sðs3 þ s2 þ s þ s2 þ s þ 1Þ þ ðs2 þ sÞ 1 V ðsÞ V A ðsÞ ¼ sþ1 1 ðs þ 1Þðs2 þ s þ 1Þ

s4 þ 2s3 þ 4s2 þ 3s þ 1 1 s2 þ s þ 1 V 1 ðsÞ ) V A ðsÞ ¼ 4 V A ðsÞ ¼ V ðsÞ 2 sþ1 ðs þ 1Þðs þ s þ 1Þ s þ 2s3 þ 4s2 þ 3s þ 1 1

ð1Þ

Applying voltage division rule for V2(s): V 2 ðsÞ ¼

1 s

sþ1þ

1 s

V A ðsÞ ¼

s2

  1 V A ðsÞ ) V A ðsÞ ¼ s2 þ s þ 1 V 2 ðsÞ þsþ1

ð2Þ

4

Solutions of Problems: Laplace Transform and Network Function

67

Solving (1) and (2):

s4

  V ðsÞ s2 þ s þ 1 1 V 1 ðsÞ ¼ s2 þ s þ 1 V 2 ðsÞ ) 2 ¼ 4 3 2 3 V 1 ðsÞ s þ 2s þ 4s2 þ 3s þ 1 þ 2s þ 4s þ 3s þ 1

Choice (2) is the answer.

Figure 4.17 The circuit of solution of problem 4.22

4.23. The circuit of Figure 4.18.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z R1 ¼ R1 Ω

ð1Þ

Z L ¼ Ls Ω

ð2Þ

1 Ω Cs

ð3Þ

Z R2 ¼ R2 Ω

ð4Þ

ZC ¼

Based on the information given in the problem, we know that: L vs ðt Þ ¼ Acosðωt Þ ¼ ¼ ¼ ¼ ) V s ðsÞ ¼

As s 2 þ ω2

ð5Þ

Ls s V ðsÞ ¼ V s ðsÞ Ls þ R1 s s þ RL1

ð6Þ

R2 s V s ðsÞ ¼ V s ðsÞ 1 s þ R21C R2 þ Cs

ð7Þ

Applying voltage division rule for node A: V A ðsÞ ¼ Applying voltage division rule for node B: V B ðsÞ ¼ Therefore:

V out ðsÞ = V A ðsÞ 2 V B ðsÞ =

s s  s þ RL1 s þ R21C

!



  RL1 s   V s ðsÞ V s ðsÞ ¼ s2 þ RL1 þ R21C s þ LRR21C 1 R2 C

ð8Þ

68

4

Solutions of Problems: Laplace Transform and Network Function

Solving (5) and (8): 

   R1 2 1  RL1 s A  R2 C L s As       V V out ðsÞ ¼  2 ¼ 2 s2 þ RL1 þ R21C s þ LRR21C ðs2 þ ω2 Þ s2 þ RL1 þ R21C s þ LRR21C s þ ω 1 R2 C

ð8Þ

Choice (2) is the answer.

Figure 4.18 The circuit of solution of problem 4.23

4.24. Based on the information given in the problem, we have: ið0þ Þ ¼ 6 A Z in ðsÞ =

s2 þ s þ 2 2s2 þ s þ 1

ð1Þ ð2Þ

The circuit of Figure 4.19.2 shows the main circuit in Laplace domain. The voltage of the independent voltage source in Laplace domain is as follows: L E vð t Þ ¼ E ¼ ¼ ¼ ¼ ) V ðsÞ ¼ s

ð3Þ

4

Solutions of Problems: Laplace Transform and Network Function

69

Using Ohm’s law in Figure 4.19.2: I ðsÞ =

E V ðsÞ Eð2s2 þ s þ 1Þ s ¼ ¼ s2 þsþ2 Z in ðsÞ 2s2 þsþ1 s ð s 2 þ s þ 2Þ

ð4Þ

From initial value theorem, we know that: f ð0þ Þ ¼ lim sF ðsÞ s!1

ð5Þ

Solving (4) and (5): Eð2s2 þ s þ 1Þ ¼ 2E s!1 ðs2 þ s þ 2Þ

ið0þ Þ ¼ lim

ð6Þ

Solving (1) and (6): 2E ¼ 6 ) E ¼ 3 V Choice (1) is the answer.

Figure 4.19 The circuit of solution of problem 4.24

4.25. The circuit of Figure 4.20.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls

ð1Þ

1 Cs

ð2Þ

ZC ¼

Moreover, vs(t) is a DC voltage source. Therefore, the voltage of the independent voltage source in Laplace domain is as follows: L E vs ð t Þ ¼ E ¼ ¼ ¼ ¼ ) V s ðsÞ ¼ s Applying voltage division rule:

ð3Þ

70

4 1 Cs

Solutions of Problems: Laplace Transform and Network Function E LC

1 s ¼E V C ðsÞ ¼ 1  2 V s ðsÞ ) V C ðsÞ ¼  2 1 1 s s þ LC s s þ LC Cs þ Ls ð3Þ

!

   L1 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )vC ðt Þ ¼ E 1  cos pffiffiffiffiffiffi t LC

ð4Þ

Based on the information given in the problem, the primary voltage of the capacitor and the primary current of the inductor are zero. In addition, the time (in second) that vC(t) is twice as vs(t) is requested. Thus: vC ðt Þ ¼ 2E

ð5Þ

Solving (4) and (5):        1 1 1 1 E 1  cos pffiffiffiffiffiffi t ¼ 2E ) 1  cos pffiffiffiffiffiffi t ¼ 2 ) cos pffiffiffiffiffiffi t ¼ 1 ) pffiffiffiffiffiffi t ¼ π LC LC LC LC pffiffiffiffiffiffi ) t ¼ π LC sec Choice (1) is the answer.

Figure 4.20 The circuit of solution of problem 4.25

4.26. The circuit of Figure 4.21.2 shows the primary circuit in Laplace domain. The impedances of the components are as follows: ZC ¼

1 1 ) Z1 F ¼ Ω Cs s

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð1Þ ð2Þ

Moreover, the voltage of the independent voltage source and the current of the independent current source in Laplace domain are as follows: V ðsÞ ¼ Lfδðt Þg ¼ 1 V

ð3Þ

I ð s Þ ¼ Lf δ ð t Þ g ¼ 1 A

ð4Þ

To find the Thevenin equivalent circuit, we need to apply a test source (e.g., a test voltage source with the voltage and current of Vt(s) and It(s)) to determine the relation between Vt(s) and It(s) in the form of Vt(s) ¼ αIt(s) + β [1]. Then, ZTh(s) ¼ α and VTh(s) ¼ β. Herein, the independent sources are not shut down.

4

Solutions of Problems: Laplace Transform and Network Function

71

Applying KCL in the supernode: V 1 ðsÞ  1 1 s

þ

V 1 ðsÞ  1  I t ðsÞ ¼ 0 ) sV 1 ðsÞ  s þ V 1 ðsÞ  1  I t ðsÞ ¼ 0 1 ) ðs þ 1ÞV 1 ðsÞ  ðs þ 1Þ  I t ðsÞ ¼ 0

ð5Þ

Defining the voltage of the dependent voltage source based on the node voltages: 1 V t ðsÞ  V 1 ðsÞ ¼ 2V 1 ðsÞ ) V 1 ðsÞ ¼ V t ðsÞ 3 Solving (5) and (6): 1 3 ðs þ 1Þ V t ðsÞ  ðs þ 1Þ  I t ðsÞ ¼ 0 ) V t ðsÞ ¼ I ðsÞ þ 3 3 sþ1 t ) Z Th ðsÞ ¼

3 Ω, V Th ðsÞ ¼ 3 V sþ1

Choice (1) is the answer.

Figure 4.21 The circuit of solution of problem 4.26

ð6Þ

72

4

Solutions of Problems: Laplace Transform and Network Function

4.27. Based on the information given in the problem, we have: i 1 ðt Þ ¼ H ðsÞ ¼

pffiffiffi L 2uð t Þ ¼ ¼ ¼ ¼ ) I 1 ðsÞ ¼

pffiffiffi 2 s

ð1Þ

I 2 ðsÞ 4ðs þ 20Þ 4ðs þ 20Þ ) I 2 ðsÞ ¼ I ðsÞ = s þ 8 sþ8 1 I 1 ðsÞ

ð3Þ

pffiffiffi 4ðs þ 20Þ 2  sþ8 s

ð4Þ

Solving (1) and (3): I 2 ðsÞ ¼ From final value theorem, we know that: f ð1Þ ¼ lim sF ðsÞ

ð5Þ

pffiffiffi pffiffiffi pffiffiffi 4ðs þ 20Þ 4 2ðs þ 20Þ 2  ¼ 10 2 ¼ lim i2 ð1Þ ¼ lim sI 2 ðsÞ ¼ lim s  sþ8 sþ8 s s!0 s!0 s!0

ð6Þ

s!0

Solving (4) and (5):

Moreover, the value of energy, stored in an inductor, can be determined as follows: 1 W L ¼ LðiL Þ2 2

ð7Þ

Solving (6) and (7):  pffiffiffi2 1 1 W L ð1Þ ¼ Lði2 ð1ÞÞ2 ¼  1  103  10 2 ¼ 0:1 J 2 2 Choice (1) is the answer.

Figure 4.22 The circuit of solution of problem 4.27

4.28. The circuit of Figure 4.23.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð1Þ

Z L ¼ Ls ) Z 2 H ¼ 2s Ω

ð2Þ

ZC ¼

1 Cs

ð3Þ

1 ) in series with an independent Since the capacitor has a nonzero primary voltage, it is modeled by an impedance (Cs vC ð0 Þ voltage source ( s ) in Laplace domain, as is shown in Figure 4.23.2.

4

Solutions of Problems: Laplace Transform and Network Function

73

Based on the information given in the problem, we have: vs ðt Þ ¼ sin ð2t Þuðt Þ ) V s ðsÞ ¼

2 s2 þ 4

ð4Þ

iout ðt > 0Þ ¼ 0 ) I out ðsÞ ¼ 0

ð5Þ

Using Ohm’s law for 1 Ω resistor: ð2Þ

V 1 ðsÞ ¼ 1  I out ðsÞ ¼ I out ðsÞ ) V 1 ðsÞ ¼ 0

ð6Þ

Applying KCL in the node 1: 

V 1 ðsÞ  vC ðs0 Þ V s ðsÞ þ þ I out ðsÞ ¼ 0 1 2s þ Cs

ð7Þ

Solving (4)–(7):



2 þ s2 þ 4

 0  vC ðs0 Þ 1 2s þ Cs

þ0¼0)

2 ¼ s2 þ 4

  vC ðs0 Þ 1 2s þ Cs

)

2 ¼ s2 þ 4

) vC ð0 Þ ¼ 4 V, C ¼

  vC ð20 Þ 1 s2 þ 2C

1 F 8

Choice (2) is the answer.

Figure 4.23 The circuit of solution of problem 4.28

8  > <  vC ð 0 Þ ¼ 2 2 ) > 1 : ¼4 2C

74

4

Solutions of Problems: Laplace Transform and Network Function

4.29. Based on the information given in the problem, we have: H ðsÞ ¼

V out ðsÞ 10ðs þ 1Þ = 2 V in ðsÞ s þ 2s þ 3

L ¼ ¼ ¼ ) V in ðsÞ ¼ vin ðt Þ ¼ 4 cos ð2t Þ ¼

s2

4s þ4

ð1Þ ð2Þ

Solving (1) and (2): V out ðsÞ 4s

s2 þ4

=

10ðs þ 1Þ 10ðs þ 1Þ 4s ) V out ðsÞ ¼ 2  s2 þ 2s þ 3 s þ 2s þ 3 s2 þ 4

ð4Þ

The steady-state response has been requested; therefore, we can use s ¼ jω ¼ j2, as ω ¼ 2 rad/sec is the angular frequency of the power source. Hence: V out ð j2Þ ¼

10ð j2 þ 1Þ 4ð j2Þ  ¼ 16:47  j14:12 ¼ 21:76ej40:6 2 2 ð j2Þ þ 2ð j2Þ þ 3 ð j2Þ þ 4   ) vout ðt Þ ¼ 21:76 cos 2t  40:6 V

Choice (3) is the answer. 4.30. The circuit of Figure 4.24.3 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z R ¼ R ) Z 0:5 Ω ¼ 0:5 Ω

ð1Þ

1 1 ) Z2 F ¼ Ω Cs 2s

ð2Þ

ZC ¼

The pulse voltage, shown in the circuit of Figure 4.24.1, can be mathematically formulated as follows:   1 1 L vs ðt Þ ¼ 10ðuðt Þ  uðt  1ÞÞ ¼ ¼ ¼ ¼ ) V s ðsÞ ¼ 10  es s s

ð3Þ

Applying Ohm’s law in the circuit of Figure 4.24.3:   V s ðsÞ 10 1s  1s es 20ð1  es Þ 20 20es ¼  ¼ ¼ I ðsÞ = sþ1 sþ1 sþ1 Z in ðsÞ 0:5 þ 2s1 Applying inverse Laplace transform on (4): L1

) iðt Þ ¼ 20et uðt Þ  20eðt1Þ uðt  1Þ A

Choice (3) is the answer.

ð4Þ

4

Solutions of Problems: Laplace Transform and Network Function

75

Figure 4.24 The circuit of solution of problem 4.30

4.31. The circuit of Figure 4.25.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

Z L ¼ Ls ) Z 0:5 H ¼ 0:5s Ω ZC ¼

1 1 ) Z3 F ¼ Ω Cs 3s

ð1Þ ð2Þ ð3Þ ð4Þ

Based on the information given in the problem, we have: i 1 ð 0 Þ ¼ i 2 ð 0 Þ ¼ 2 A

ð5Þ

Since each inductor has a nonzero primary current, each of them is modeled by an impedance (Ls ¼ 0.5s Ω) in parallel  with an independent current source (iL ð0s Þ ¼ 2s A) in Laplace domain, as is shown in Figure 4.25.2. To simplify the circuit, source transformation theorem can be applied on the parallel connection of the independent current source and the inductor to change it to the series connection of the independent voltage source and the same inductor, as can be seen in Figure 4.25.3. By looking at the circuit of Figure 4.25.3, it is noticed that the circuit is symmetric. Therefore, the nodes A and B have equal voltages. Hence, no current flows through the middle capacitor. Thus, the circuit can be simplified, as is illustrated in Figure 4.25.4. Applying voltage division rule for right-side 1 Ω resistor: V out ðsÞ ¼

1 2s 2 2  1 ) V out ðsÞ ¼ 2  ¼ s þ 2s þ 1 s þ 1 ðs þ 1Þ2 1 þ 0:5s þ 2s1

76

4

Solutions of Problems: Laplace Transform and Network Function

Applying inverse Laplace transform: L1

) vout ðt Þ ¼ ð2et  2tet Þ V

Choice (1) is the answer.

Figure 4.25 The circuit of solution of problem 4.31

4

Solutions of Problems: Laplace Transform and Network Function

77

Figure 4.25 (continued)

4.32. The circuit of Figure 4.26.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Zα Ω ¼ α Ω ZC ¼

1 1 α ) Z α1 F ¼ 1 ¼ Ω Cs s αs

ð1Þ ð2Þ

Applying KCL in the supernode: I s ðsÞ þ

V s ðsÞ  V 1 ðsÞ V ðsÞ  V 1 ðsÞ þ 2I s ðsÞ ¼ 0 ) s þ I s ðsÞ ¼ 0 ) V 1 ðsÞ ¼ V s ðsÞ þ αI s ðsÞ α α

ð3Þ

Applying KCL in node A: 2I s ðsÞ þ

V 1 ðsÞ  V s ðsÞ V 1 ðsÞ V 1 ðsÞ  2V 1 ðsÞ þ þ ¼0 α α α sþα 

 1 1 s 1 )  2I s ðsÞ þ þ  V ðsÞ  V s ðsÞ ¼ 0 α α αðs þ 1Þ 1 α )  2I s ðsÞ þ ð3Þ

)  2I s ðsÞ þ

sþ2 1 V ðsÞ  V s ðsÞ ¼ 0 α αðs þ 1Þ 1

sþ2 1 ðV ðsÞ þ αI s ðsÞÞ  V s ðsÞ ¼ 0 α αðs þ 1Þ s

    sþ2 sþ2 1 ) 2 þ I ðsÞ þ  V ðsÞ ¼ 0 sþ1 s αðs þ 1Þ α s ) 

s 1 s 1 I ðsÞ þ V ðsÞ ¼ 0 ) I ðsÞ ¼ V ðsÞ sþ1 s sþ1 s αðs þ 1Þ s αðs þ 1Þ s )

s

V s ðsÞ ¼ αs ¼ sþ1 1 I s ðsÞ αðsþ1 Þ

ð4Þ

78

4

Solutions of Problems: Laplace Transform and Network Function

As we know, the value of VI ssððssÞÞ presents the equivalent impedance seen by the source. Therefore: ) Z AB ðsÞ ¼ αs

ð5Þ

Moreover, the impedance of an inductor can be determined as follows: Z L ¼ Ls

ð6Þ

By comparing (5) and (6), it can be concluded that the circuit, seen from terminal A–B, is equivalent to an inductor with the inductance of α H. Choice (4) is the answer.

Figure 4.26 The circuit of solution of problem 4.32

4

Solutions of Problems: Laplace Transform and Network Function

79

4.33. Before t ¼ 0, the circuit has reached its steady-state condition. Therefore, each capacitor is like an open circuit branch. Figure 4.27.2 shows the status of the primary circuit for t ¼ 0. Applying KVL in the indicated loop: 20  30 þ 10ið0 Þ þ 10ið0 Þ þ 10 ¼ 0 ) ið0 Þ ¼ 2 A

ð1Þ

Applying KVL in mesh 1: ð1Þ

30 þ 10ið0 Þ þ vC1 ð0 Þ þ 15  0 ¼ 0 ) vC1 ð0 Þ ¼ 30  10ið0 Þ ) vC1 ð0 Þ ¼ 10 V

ð2Þ

Applying KVL in mesh 2: 20 þ 15  0 þ vC2 ð0 Þ ¼ 0 ) vC2 ð0 Þ ¼ 20

ð3Þ

Figure 4.27.3 shows the main circuit for t ¼ 0+ in Laplace domain. Each capacitor is modeled by an impedance in series  with an independent voltage source (vC ðs0 Þ) in Laplace domain, since they have a nonzero primary voltage. Moreover, the impedances of the components are as follows: Z R ¼ R ) Z 10 Ω ¼ 10 Ω

ð4Þ

Z R ¼ R ) Z 15 Ω ¼ 15 Ω

ð5Þ

1 1 ) Z1 F ¼ Ω Cs s

ð6Þ

1 1 4 ) Z 0:25 F ¼ ¼ Ω Cs 0:25s s

ð7Þ

ZC ¼ ZC ¼

Applying KVL in the indicated mesh in Figure 4.27.3: 20

10I ðsÞ þ

10 20 4 10 1 4  þ I ðsÞ  þ I ðsÞ ¼ 0 ) I ðsÞ ¼ s 5 ¼ s s s s s 10 þ s 2s þ 1

ð8Þ

From the circuit, it is clear that:   1 10 ð8Þ 1 4 10 4 4 10 4 6 V C1 ðsÞ ¼  I ðsÞ þ )V C1 ðsÞ ¼  þ ¼ þ ¼ þ þ s s s 2s þ 1 s s s þ 12 s s þ 12 s Applying inverse Laplace transform on (9):   ) vC1 ð0 Þ ¼ 6 þ 4e0:5t V

L1

Choice (1) is the answer.

ð9Þ

80

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.27 The circuit of solution of problem 4.33

4

Solutions of Problems: Laplace Transform and Network Function

81

4.34. The circuit of Figure 4.28.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZC ¼

1 1 ) Z4 F ¼ Ω Cs 4s

ð1Þ

ZC ¼

1 1 ) Z3 F ¼ Ω Cs 3s

ð2Þ

ZC ¼

1 1 ) Z2 F ¼ Ω Cs 2s

ð3Þ

Moreover, the voltage of each independent voltage source in Laplace domain is as follows: L 5 5V ¼ ¼ ¼ ¼ ) s

ð4Þ

L 10 10 V ¼ ¼ ¼ ¼ ) s

ð5Þ

1 ) in series with an Since each capacitor has a nonzero primary voltage, they need to be modeled by an impedance (Cs  vC ð0 Þ independent voltage source ( s ) in Laplace domain, as is shown in Figure 4.28.2. Therefore:

L 1 1V ¼ ¼ ¼ ¼ ) V s

ð6Þ

L 5 5V ¼ ¼ ¼ ¼ ) V s

ð7Þ

L 6 6V ¼ ¼ ¼ ¼ ) V s

ð8Þ

Applying KCL in the indicated node, which is for the voltage of 2 F capacitor: V ðsÞ  1s  5s 1 4s

þ

V ðsÞ  6s 1 2s

þ

V ðsÞ  5s  10s 1 3s

¼0



     1 5 6 5 10 ) 4s V ðsÞ   þ 2s V ðsÞ  þ 3s V ðsÞ   ¼0 s s s s s )4sV(s)  4  20 + 2sV(s)  12 + 3sV(s)  15  30 ¼ 0 ) 9sV(s) ¼ 81 ) V(s) ¼ 9 Applying inverse Laplace transform on (9): L1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )vðt Þ ¼ 9uðt Þ ) vð0þ Þ ¼ 9 V Choice (4) is the answer.

(9)

82

4

Solutions of Problems: Laplace Transform and Network Function

Figure 4.28 The circuit of solution of problem 4.34

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

5

Problems: Natural Frequencies of Electrical Circuits

Abstract

In this chapter, the basic and advanced problems concerned with the determination of natural frequencies of the electrical circuits are presented. The problems of this chapter include calculating the zero and nonzero natural frequencies of a circuit, determining the zeros and poles of the network function (transfer function) of a circuit, and identifying the damping status of a circuit. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 5.1. Determine the order and the number of nonzero natural frequencies of the circuit shown in Figure 5.1 [1–2]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. 11 and 7 2. 11 and 10 3. 8 and 6 4. 8 and 8

Figure 5.1 The circuit of problem 5.1

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_5

83

84

5 Problems: Natural Frequencies of Electrical Circuits

5.2. In the circuit of Figure 5.2, determine the number of state variables. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. 3 2. 4 3. 5 4. 6

Figure 5.2 The circuit of problem 5.2

5.3. Which one of the following choices is true about the circuit of Figure 5.3? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1. It has 5 nonzero natural frequencies and 3 zero natural frequencies. 2. It has 4 nonzero natural frequencies and 3 zero natural frequencies. 3. It has 5 nonzero natural frequencies and 2 zero natural frequencies. 4. It has 4 nonzero natural frequencies and 2 zero natural frequencies.

Figure 5.3 The circuit of problem 5.3

5.4. Determine the zero and poles of the network function (transfer function) of VVouts ððssÞÞ in the circuit of Figure 5.4. Difficulty level Calculation amount
Poles : 1  j 1. Zero : 2
Poles : 1  j 2. Zero : 2
Poles : 1 þ j 3. Zero : 2 4. None of them

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

85

Figure 5.4 The circuit of problem 5.4

5.5. In the circuit of Figure 5.5, determine the zero and poles of the network function (transfer function), defined as IIouts ððssÞÞ. Difficulty level Calculation amount
Pole : 30 1. Zero : 10
Pole : 30 2. Zero : 10
Pole : 30 3. Zero : 10 4. None of them

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

Figure 5.5 The circuit of problem 5.5

5.6. The state equations of a linear time-invariant (LTI) circuit is given in matrix form in the following. Determine the damping status of the circuit. 2

3 d  v C ðt Þ 0 6 dt 7 4 5¼ d 1 i ðt Þ dt L Difficulty level ○ Easy Calculation amount ● Small 1. Overdamped 2. Critically damped 3. Underdamped 4. Undamped

● Normal ○ Normal

1 1

○ Hard ○ Large



vC ð t Þ i L ðt Þ

"

 þ

1 1

# vs ð t Þ

86

5 Problems: Natural Frequencies of Electrical Circuits

5.7. In the circuit of Figure 5.6, the capacitor and the inductor have nonzero primary voltage and current, respectively. Which one of the following choices is true about its status? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. Overdamped 2. Critically damped 3. Underdamped 4. Undamped

Figure 5.6 The circuit of problem 5.7

5.8. Which one of the following conditions can create an undamped status for the circuit shown in Figure 5.7? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. R ¼  p1ffiffiffiffiffi LC

L 2. R ¼  rC 3. R ¼  rC L 4. R ¼  r

Figure 5.7 The circuit of problem 5.8

5.9. In a circuit, determine the poles of the network function (transfer function) of follows: vin ðt Þ ¼ 5e2t vout ðt Þ ¼ 3te2t þ 2e3t sin ð6t Þ Difficulty level Calculation amount

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

V out ðsÞ V in ðsÞ

if the input and the output are as

87

1. 2. 3. 4.

2,  j6 2, 3, 0 2, 3,  j6 2,  3  j6

5.10. For what value of α is the circuit of Figure 5.8 in the critically damped status? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1. 1.6 2. 0 3. 1 4. None of them

Figure 5.8 The circuit of problem 5.10

5.11. For what value of β is the circuit of Figure 5.9 in the undamped status? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1. 1 2. 2 3. 3 4. 4

Figure 5.9 The circuit of problem 5.11

88

5 Problems: Natural Frequencies of Electrical Circuits

5.12. The zero-pole diagram concerned with the input impedance of a linear time-invariant (LTI) one-port network is shown in Figure 5.10. If the network is connected to a 1 A current source, 0.5 V can be measured across that after a while. Determine the input impedance of the network. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 2ðs2 þ1Þ 1. ðsþ1Þðsþ2Þ Ω 2. 3. 4.

ðs2 þ1Þ ðsþ1Þðsþ2Þ ðsþ1Þðsþ2Þ ðs2 þ1Þ ðsþ1Þðsþ2Þ 2ðs2 þ1Þ

Ω Ω Ω

Figure 5.10 The circuit of problem 5.12

5.13. Figure 5.11 illustrates two networks that only include the linear time-invariant (LTI) resistors. If is(t) ¼ cos (t) + cos (2t), which one of the choices is correct? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1. io(t) will be zero. 2. vo(t) will be zero. 3. io(t) and vo(t) will be zero. 4. None of them.

Figure 5.11 The circuit of problem 5.13

5.14. In the circuit of Figure 5.12, the input admittance has the zeros at s ¼  2,  2.5 and the poles at s ¼  1,  1. Determine the resistance of R3. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1. 14 Ω 2. 18 Ω 3. 1 Ω 4. 2 Ω

89

Figure 5.12 The circuit of problem 5.14

5.15. In the circuit of Figure 5.13, the input impedance has the poles at s ¼  1,  3 and the zeros at s ¼  2,  4. Determine the resistance of R1. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1. 64 55 Ω 2. 85 Ω 3. 1 Ω 4. 83 Ω

Figure 5.13 The circuit of problem 5.15

5.16. Figure 5.14 illustrates the result of a test carried out on the linear time-invariant (LTI) network, where:   3 1 iout ðt Þ ¼ 2  e2t uðt Þ 2 v s ð t Þ ¼ uð t Þ Which one of the following choices is true about the stability of the LTI network? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1. The LTI network is stable in both short circuit and open circuit statuses. 2. The LTI network is unstable in both short circuit and open circuit statuses. 3. The LTI network is stable in short circuit status but unstable in open circuit status. 4. The LTI network is stable in open circuit status but unstable in short circuit status.

90

5 Problems: Natural Frequencies of Electrical Circuits

Figure 5.14 The circuit of problem 5.16

5.17. Which one of the inputs below will only show the natural frequencies in the output of the circuit of Figure 5.15? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1. is(t) ¼ e1.5tu(t) A 2. is(t) ¼ e0.5tu(t) A 3. is(t) ¼ etu(t) A 4. is(t) ¼ e2tu(t) A

Figure 5.15 The circuit of problem 5.17

5.18. The location of zeros and poles of a network are shown in Figure 5.16. For what value of a > 0 will applying an input with the form of eatu(t) not create an output in the same form? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1. 1 2. 2 3. 4 4. For no value of a

Figure 5.16 The circuit of problem 5.18

References

91

5.19. In the circuit of Figure 5.17, determine the nonzero natural frequencies. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large p ffiffi ffi  1. 13 1  j 5 pffiffiffi  2. 1  j 5 ,  j pffiffiffi  3. 1  j 5 ,  1 pffiffiffi  4. 13 1  j 5 ,  j,  1

Figure 5.17 The circuit of problem 5.19

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

6

Solutions of Problems: Natural Frequencies of Electrical Circuits

Abstract

In this chapter, the problems of the fifth chapter are fully solved, in detail, step-by-step, and with different methods.

6.1. The number of energy-saving components (capacitors and inductors) of the circuit of Figure 6.1 is 11 [1–2]. However, there are two inductor cut-sets and one capacitor loop. Therefore, the number of the natural frequencies or the order of the circuit is 11
2
1 ¼ 8. On the other hand, there are two capacitor cut-sets. Hence, two out of the eight natural frequencies are zero natural frequencies. Consequently, the number of nonzero natural frequencies are 8
2 ¼ 6. Choice (3) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_6

93

94

6

Solutions of Problems: Natural Frequencies of Electrical Circuits

Figure 6.1 The circuit of solution of problem 6.1

6.2. The number of state variables of a circuit is equal to the number of energy-saving components (capacitors and inductors). However, as can be seen in Figure 6.2, there is one inductor cut-set and one capacitor loop in the circuit. Moreover, the voltage of C3 depends on the current of L2. Therefore, the number of state variables of the circuit is 6
1
1
1 ¼ 3. Choice (1) is the answer.

6

Solutions of Problems: Natural Frequencies of Electrical Circuits

95

Figure 6.2 The circuit of solution of problem 6.2

6.3. The number of energy-saving components (capacitors and inductors) of the circuit of Figure 6.3 is nine. However, there is one inductor cut-set and one capacitor loop. Therefore, the number of the natural frequencies or the order of the circuit is 9
1
1 ¼ 7. On the other hand, since there is one capacitor cut-set and two inductor loops, three out of the seven natural frequencies are zero natural frequencies. Consequently, the number of nonzero natural frequencies is four (7
3 ¼ 4). Choice (2) is the answer.

Figure 6.3 The circuit of solution of problem 6.3

96

6

Solutions of Problems: Natural Frequencies of Electrical Circuits

6.4. The circuit of Figure 6.4.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: Z L ¼ Ls ) Z 1 H ¼ s Ω

ð1Þ

ZR ¼ R ) Z1 Ω ¼ 1 Ω

ð2Þ

1 1 2 ) Z 0:5 F ¼ ¼ Ω Cs 0:5s s

ð3Þ

ZC ¼

The network function (transfer function) for this problem has been defined as follows: H ðsÞ ¼

V out ðsÞ V s ðsÞ

Applying voltage division rule in the circuit of Figure 6.4.2: V out ðsÞ ¼

1 þ 2s V ðsÞ sþ2 sþ2 V s ðsÞ ¼ 2 V s ðsÞ ) H ðsÞ ¼ out ¼ 2 2 V s ðsÞ s þ 2s þ 2 s þ 2s þ 2 1þsþsþ1

Therefore, the zero and poles of the network function (transfer function) are as follows: 8
> > > > >   > > > v2 ðt Þ ¼ et  e2t uðt Þ V > > > > > > > < v3 ðt Þ ¼ 0 > > > > > > > > > > > > > > > > :

i1 ðt Þ ¼ e2t uðt Þ A i2 ðt Þ ¼ 0   i3 ðt Þ ¼ 2e2t  et uðt Þ A

, Fig:2 :

8 ev1 ðt Þ ¼ 0 > > > > > > > > ev2 ðt Þ ¼ Unknown > > > > > > > > < ev3 ðt Þ ¼ et uðt Þ V ei1 ðt Þ ¼ 0 > > > > > > > > > ei2 ðt Þ ¼ et uðt Þ A > > > > > > > : e i3 ðt Þ ¼ Unknown

ð2Þ

In (2), based on Tellegen’s theorem, a negative sign was applied for the polarity of the currents flowing toward the network. Now, for this problem, we should transfer to Laplace domain. Therefore, the quantities are as follows:

8 Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .

124

8 > V 1 ðsÞ ¼ Unknown > > > > > > 1 1 > > V 2 ðsÞ ¼  > > s þ 1 s þ 2 > > > > > > < V 3 ðsÞ ¼ 0 Fig:1 : 1 > I 1 ðsÞ ¼  > > s þ 2 > > > > > > I 2 ðsÞ ¼ 0 > > > > > > > > : I 3 ðsÞ ¼ 2  1 sþ2 sþ1

8 e 1 ðsÞ ¼ 0 > V > > > > > > e 2 ðsÞ ¼ Unknown > V > > > > > > > > e 3 ðsÞ ¼ 1 < V sþ1 , Fig:2 : eI 1 ðsÞ ¼ 0 > > > > > > > > > eI 2 ðsÞ ¼  1 > > > sþ1 > > > > : e I 3 ðsÞ ¼ Unknown

ð3Þ

For this problem, we should use Tellegen’s theorem in Laplace domain, as follows: e 1 ðsÞI 1 ðsÞ þ V e 2 ðsÞI 2 ðsÞ þ V e 3 ðsÞI 3 ðsÞ V 1 ðsÞeI 1 ðsÞ þ V 2 ðsÞeI 2 ðsÞ þ V 3 ðsÞeI 3 ðsÞ ¼ V Solving (3) and (4): V 1 ðsÞ  0 þ



1 1  sþ1 sþ2 e 1 ðsÞ )V







      1 1 2 1 e 2 ðsÞ  0 þ e 1 ðsÞ  1 þ 0  eI 3 ðsÞ ¼ V þV  sþ1 sþ2 sþ1 sþ2 sþ1

    L1 1 1 1 e 1 ðsÞ ¼ 1 ) v1 ðt Þ ¼ et uðt Þ V ¼ )V sþ2 sþ1 sþ2 sþ1

Choice (1) is the answer.

Figure 8.4 The circuit of solution of problem 8.4

ð4Þ

8

Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .

125

8.5. Based on Tellegen’s theorem, we can write the following equation for the network: v1 ðt Þei1 ðt Þ þ v2 ðt Þei2 ðt Þ ¼ ev1 ðt Þi1 ðt Þ þ ev2 ðt Þi2 ðt Þ

ð1Þ

Based on the information given in the problem, we have:   8 v1 ðt Þ ¼ 4 sin ωt þ 45 V > > > > > > > < v2 ð t Þ ¼ 0 Test 1 :   > i1 ðt Þ ¼ 2 sin ωt þ 30 A > > > > > >  :  i2 ðt Þ ¼ sin ωt þ 60 A

, Test 2 :

8 > > > > > > > >
ei1 ðt Þ ¼ 0:5 sin ωt þ 15 A > > > > > > > : ei2 ðt Þ ¼ Unknown

ð2Þ

We need to solve the problem in phasor domain. Therefore, the quantities will be as follows:

Test 1 :

8  > V1 ¼ 4e j45 > > > > > > > < V2 ¼ 0 > I1 ¼ 2e > > > > > > > : I2 ¼ e

j30

, Test 2 :

j60

8 e 1 ¼ Requested V > > > > > > > > e ¼1 < V 2

ð3Þ

> eI1 ¼ 0:5e j15 > > > > > > > : e I2 ¼ Unknown

Tellegen’s theorem in phasor domain for this problem is as follows: e 1 I1 þ V e 2 I2 V1eI1 þ V2eI2 ¼ V

ð4Þ

Solving (3) and (4): 4e

j45

 0:5e

j15

e 1 3 2e þ 0 3 eI2 = V

j30

þ 13e

j60

e 1 3 2e )V

j30

By transferring to time domain, we have:   ev1 ðt Þ ¼ 0:5 sin ωt þ 30 V Choice (3) is the answer.

Figure 8.5 The circuit of solution of problem 8.5

=e

j60

e 1 = 0:5e )V

j30

8 Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .

126

8.6. Based on the information given in the problem, we have: Test I : If RL ¼ 1, jVout j ¼ 13 V

ð1Þ

Test II : If RL ¼ 3 Ω, jVout j ¼ 3 V

ð2Þ

Test III : If RL ¼ 14 Ω, jVout j ¼ 9:1 V

ð3Þ

Since the network is linear time-invariant (LTI), it can be replaced by its Thevenin equivalent circuit, as can be seen in Figure 8.6.2. By using Figure 8.6.2 and (1), we can conclude: jVTh j ¼ jVout j ¼ 13 V

ð4Þ

Using Figure 8.6.2 and applying voltage division and (2) and (4):             RL 3 13     ) j3 þ a þ jbj = 13   3 13 ) 1 ¼  3 VTh  ) 3 ¼  jVout j =  3 þ a þ jb 3 þ a þ jb RL þ a þ jb ) ð3 þ aÞ2 þ b2 ¼ 169

ð5Þ

Using Figure 8.6.2 and applying voltage division and (3) and (4):             RL 14 182     ) j14 þ a þ jbj = 20   3 13 ) 9:1 ¼  3 VTh  ) 9:1 ¼  jVout j =  14 þ a þ jb 14 þ a þ jb RL þ a þ jb ) ð14 þ aÞ2 þ b2 ¼ 400 Solving (5) and (6): a ¼ 2, b ¼ 12 ) Z Th ¼ ð2 þ j12Þ Ω Choice (4) is the answer.

Figure 8.6 The circuit of solution of problem 8.6

ð6Þ

8

Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .

127

8.7. First, we should define a new network with the boundary shown in Figs. 8.7.3–4. Then, based on Tellegen’s theorem, we can write: v1 ðt Þei1 ðt Þ þ v2 ðt Þei2 ðt Þ ¼ ev1 ðt Þi1 ðt Þ þ ev2 ðt Þi2 ðt Þ

ð1Þ

From the circuits of Figs. 8.7.3–4 as well as based on the information given in the problem, we have:

Fig:3 :

8 > > > > > > >
> i1 ðt Þ ¼ Unknown > > > > > : i 2 ðt Þ ¼ 0

, Fig:4 :

8 ev1 ðt Þ ¼ 0 > > > > > > > > < ev2 ðt Þ ¼ Unknown > ei1 ðt Þ ¼ Unknown > > > > > > > :e i2 ðt Þ ¼  cos ðt Þ A

ð2Þ

In (2), based on Tellegen’s theorem, a negative sign was applied for the polarity of the current flowing toward the network. Now, we need to solve the problem in phasor domain. Therefore, the quantities will be as follows:

Fig:3 :

8 > > > > > > >
> I1 ¼ Unknown > > > > > : I2 ¼ 0

, Fig:4 :

8 > > > > > > > >
eI1 ¼ Unknown > > > > > > > :e I2 ¼ 1e j0 ¼ 1

ð3Þ

Tellegen’s theorem in phasor domain for this problem is as follows: e 1 I1 þ V e 2 I2 V1eI1 þ V2eI2 ¼ V

ð4Þ

e 2  0 ) eI1 ¼ 2 1  eI1 þ 2  ð1Þ ¼ 0  I1 þ V

ð5Þ

Solving (3) and (4):

By transferring to time domain, we have: ei1 ðt Þ ¼ 2 cos ðt Þ A Using (6) and Ohm’s law in Figure 8.7.4: ev1 ðt Þ ¼ 1  ei1 ðt Þ ¼ 2 cos ðt Þ V Choice (1) is the answer.

ð6Þ

128

8 Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .

Figure 8.7 The circuit of solution of problem 8.7

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

9

Problems: Two-Port Networks

Abstract

In this chapter, the basic and advanced problems concerned with the determination of different matrices of a two-port network, that is, impedance matrix, admittance matrix, hybrid matrix, and transmission matrix as well as the series and parallel connection of the two-port networks, are presented. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 9.1. For the two-port network, shown in Figure 9.1, determine the impedance matrix ([Z]) [1–2]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large
 1 1 1. 1 1

1 1 2. 1
1
 1
1 3.
1 1

1
1 4.
1
1

Figure 9.1 The circuit of problem 9.1

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_9

129

130

9

9.2. For the circuit, shown in Figure 9.2, determine the transmission matrix ([T]). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

1 0 1.
1
1

1 0 2. 1
1
 0 1 3. 1 0
 1 0 4. 1 1

Figure 9.2 The circuit of problem 9.2

9.3. In the circuit of Figure 9.3, determine the hybrid parameter of h21. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 2C 1.
αþjωR 1þjωR2 C 2.

αþjωR2 1þjωR2 1þjωR2 C 1þjω

3. 4. None of them

Figure 9.3 The circuit of problem 9.3

Problems: Two-Port Networks

9

Problems: Two-Port Networks

131

9.4. For the two-port network, shown in Figure 9.4, determine the admittance matrix ([Y]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large
 2ð s þ 1Þ
ð s þ 1Þ 1.
ð s þ 1Þ 2ð s þ 1Þ 2  3 2 1 þ2
þ1 6 7 s 2. 4 s 5  1 2
þ1 þ2 s s 2 3 sþ1 2
ð s þ 1Þ 6 7 s 3. 4 5 sþ1
ð s þ 1Þ 2 s 4. None of them

Figure 9.4 The circuit of problem 9.4

9.5. For the two-port network, shown in Figure 9.5, determine the impedance matrix ([Z]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2 3 β αþR R
j 6 Cω 7 1. 4 5 β αþR Rþj Cω 3 2 β αþR j 6 Cω 7 2. 4 5 β α Rþj Cω " # αþR αþR 3. β 1
β αþRþj αþR
j Cω Cω 4. None of them

132

9

Problems: Two-Port Networks

Figure 9.5 The circuit of problem 9.5

9.6. Which one of the following choices is true about the impedance matrix ([Z]), the admittance matrix ([Y]), and the hybrid matrix ([H]) of the two-port network shown in Figure 9.6? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. All of them are available. 2. Only [Z] is not available. 3. Only [Y] is not available. 4. Only [H] is not available.

Figure 9.6 The circuit of problem 9.6

9.7. For the circuit, shown in Figure 9.7, determine the admittance matrix ([Y]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large
 4
1 1.
3 2
 2
1 2.
3 4
 4
3 3.
1 2 4. None of them

9

Problems: Two-Port Networks

133

Figure 9.7 The circuit of problem 9.7

9.8. For what value of α the two-port network, shown in Figure 9.8, does not have the admittance matrix ([Y])? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1. 11 6 2.
11 6 6 3. 11 6 4.
11

Figure 9.8 The circuit of problem 9.8

9.9. For what value of α and β the two-port network, shown in Figure 9.9, has the admittance matrix in the following form?
½Y  ¼ Difficulty level ○ Easy Calculation amount ○ Small 1. α ¼ 1, β ¼
1 2. α ¼ 1, β ¼ 1 3. α ¼
1, β ¼
1 4. α ¼
1, β ¼ 1

● Normal ● Normal

2 0


2
2



○ Hard ○ Large

Figure 9.9 The circuit of problem 9.9

134

9

Problems: Two-Port Networks

9.10. Determine the admittance matrix of the configuration shown in Figure 9.10. The admittance matrix of each network is as follows:
½Y 1  ¼ Difficulty level Calculation
 amount 2 4 1. 4 9
 0 0 2. 0 3
 1 1 3. 1 1 4. None of them

○ Easy ● Small

○ Normal ○ Normal


1 2 1 , ½Y 2  ¼ 2 3 2

2 6



● Hard ○ Large

Figure 9.10 The circuit of problem 9.10

9.11. Which one of the choices presents the correct relation between the input and output voltages and currents for the configuration of Figure 9.11 if the impedance matrices of the networks are as follows?



 1 2 5 6 ½Z 1  ¼ , ½Z 2  ¼ 2 3 6 7 Difficulty level ○ Easy Calculation amount ●


Small v1 10 8 i1 1. ¼ 8 5 i2 v
2

 v1 4 8 i1 2. ¼ v 8 6 i2
2

 v1 8 6 i1 3. ¼ 6 10 i2 v2


 v1 6 8 i1 4. ¼ 8 10 i2 v2

○ Normal ○ Normal

● Hard ○ Large

9

Problems: Two-Port Networks

135

Figure 9.11 The circuit of problem 9.11

9.12. For the two-port network, shown in Figure 9.12, determine the hybrid matrix ([H]). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 2 3 1 þ β1 0 6 1 þ β2 7 7 1. 6 4 1 þ α1 5
0 1 þ α2 2 3 1 þ β2 0 6 1 þ β1 7 7 2. 6 4 1 þ α2 5
0 1 þ α1 2 3 1 þ α1 0 6 1 þ α2 7 7 3. 6 4 1 þ β1 5
0 1 þ β2 2 3 1 þ α2 0 6 1 þ α1 7 7 4. 6 4 1 þ β2 5
0 1 þ β1

Figure 9.12 The circuit of problem 9.12

136

9

Problems: Two-Port Networks

9.13. For the circuit, shown in Figure 9.13, determine the transmission matrix ([T]). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large
 26 15 1. 7 6
 26 15 2. 40 20
 26 15 3. 21 15
 26 15 4. 45 26

Figure 9.13 The circuit of problem 9.13

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

Solutions of Problems: Two-Port Networks

10

Abstract

In this chapter, the problems of the ninth chapter are fully solved, in detail, step-by-step, and with different methods.

10.1. The impedance matrix ([Z]) of a network in time domain is as follows [1–2]:


v1 ðt Þ






v2 ðt Þ

¼

z11

z12

z21

z22




i 1 ðt Þ i 2 ðt Þ




 ) ½Z  ¼

z11

z12

z21

z22

 ð1Þ

Applying KVL in the left-side mesh: v1 ðt Þ þ 1  ði1 ðt Þ þ i2 ðt ÞÞ ¼ 0 ) v1 ðt Þ ¼ i1 ðt Þ þ i2 ðt Þ

ð2Þ

Applying KVL in the right-side mesh: v2 ðt Þ þ 1  ði1 ðt Þ þ i2 ðt ÞÞ ¼ 0 ) v2 ðt Þ ¼ i1 ðt Þ þ i2 ðt Þ

ð3Þ

Solving (1), (2), and (3):



v1 ð t Þ 1 ¼ v2 ð t Þ 1

1 1





i 1 ðt Þ 1 ) ½Z  ¼ i 2 ðt Þ 1

1 1



Choice (1) is the answer.

Figure 10.1 The circuit of solution of problem 10.1

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6_10

137

138

10

Solutions of Problems: Two-Port Networks

10.2. The transmission matrix ([T]) of a network is defined as follows:


v1 ðt Þ




 ¼

i 1 ðt Þ

t 11

t 12

t 21

t 22




v2 ð t Þ




 ) ½T  ¼

i 2 ðt Þ

t 11

t 12

t 21

t 22

 ð1Þ

Applying KVL in the loop: v1 ðt Þ þ v2 ðt Þ ¼ 0 ) v1 ðt Þ ¼ v2 ðt Þ

ð2Þ

Applying KVL in the right-side mesh: v2 ðt Þ þ 1  ði1 ðt Þ  i2 ðt ÞÞ ¼ 0 ) i1 ðt Þ ¼ v2 ðt Þ þ i2 ðt Þ

ð3Þ

Solving (1), (2), and (3):


v1 ð t Þ




¼

i 1 ðt Þ

1

0

1

1




v2 ð t Þ




) ½T  ¼

i 2 ðt Þ

1

0

1

1



Choice (4) is the answer.

Figure 10.2 The circuit of solution of problem 10.2

10.3. Hybrid matrix ([H]) in phasor domain is defined as follows:


V1 I2




¼

h11

h12

h21

h22




I1



V2


) ½H  ¼

h11

h12

h21

h22

 ð1Þ

The hybrid parameter of h21 can be determined as follows: h21 ¼

 I2  I1 V2 ¼0

ð2Þ

Figure 10.3.2 illustrates the main circuit, while the second port of the network has been short-circuited (V2 ¼ 0). By applying KVL in the right-side mesh, we can write:     1 α 1 ðI þ αI1 Þ þ R2 ðI1 þ I2 Þ ¼ 0 ) R2 þ I þ R2 þ I ¼0 jωC 2 jωC 1 jωC 2 ) h21 ¼ Choice (1) is the answer.

 α R2 þ jωC I2  α þ jωCR2 ¼  ¼  1 I1 V2 ¼0 1 þ jωCR2 R2 þ jωC

10

Solutions of Problems: Two-Port Networks

139

Figure 10.3 The circuit of solution of problem 10.3

10.4. The circuit of Figure 10.4.2 shows the main circuit in Laplace domain. The impedances of the components are as follows: ZR ¼ R ) Z1 Ω ¼ 1 Ω ZC ¼

1 1 ) Z1 F ¼ Ω Cs s

For this two-port network, the admittance matrix ([Y]) is the same as the nodal admittance matrix ([Ynodal]). The circuit includes two supernodes shown in Figure 10.4.2. The nodal admittance matrix can be determined as follows: 2P 6 ½Y nodal  ¼ 4

y1j

j¼1

y21

y12 P

3



sþ1þsþ1  ð s þ 1Þ 2ð s þ 1Þ 7 ¼ ¼ 5 y2j  ð s þ 1Þ sþ1þsþ1 ðs þ 1Þ

j¼1


½Y  ¼ ½Y nodal  ¼ Choice (1) is the answer.

2ðs þ 1Þ ðs þ 1Þ

ðs þ 1Þ 2ð s þ 1Þ



ðs þ 1Þ 2ð s þ 1Þ



140

10

Solutions of Problems: Two-Port Networks

Figure 10.4 The circuit of solution of problem 10.4

10.5. Impedance matrix ([Z]) in phasor domain is defined as follows:


V1



V2


¼

z11

z12

z21

z22




I1 I2




) ½Z  ¼

z11

z12

z21

z22

 ð1Þ

Figure 10.5.2 shows the main circuit in phasor domain. By applying KCL in the cut-set, we can write: I ¼ I1 þ I2

ð2Þ

Using ð2Þ V1 ¼ αI þ RI ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) V1 ¼ ðα þ RÞI1 þ ðα þ RÞI2

ð3Þ

Applying KVL in the left-side mesh:

Applying KVL in the indicated loop:     Using ð2Þ 1 β 1β V2 ¼ I þ αþRþ I ðI  βIÞ þ αI þ RI ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) V2 = α þ R  jωC 2 jωC 1 jωC 2

ð4Þ

10

Solutions of Problems: Two-Port Networks

141

Combining (3) and (4) in the matrix form:


V1 V2

"

 ¼

αþR αþRþj

β ωC

#
 " αþR αþR I1 ) ½Z  ¼ 1β β αþRj αþRþj I2 ωC ωC

αþR 1β αþRj ωC

#

Choice (3) is the answer.

Figure 10.5 The circuit of solution of problem 10.5

10.6. The impedance matrix ([Z]), the admittance matrix ([Y]), and the hybrid matrix ([H]) in time domain are defined in the following forms:


v1 ðt Þ v2 ðt Þ




 ¼

z11

z12

z21

z22




i 1 ðt Þ i 2 ðt Þ




 ) ½Z  ¼

z11

z12

z21

z22

 ð1Þ

142

10


i 1 ðt Þ y11 ¼ i 2 ðt Þ y21

y12 y22


h11 v1 ð t Þ ¼ i 2 ðt Þ h21

h12 h22










Solutions of Problems: Two-Port Networks


 v1 ð t Þ y11 ) ½Y  ¼ v2 ð t Þ y21

y12 y22


 i1 ðt Þ h11 ) ½H  ¼ v2 ð t Þ h21

h12 h22




 ð2Þ  ð3Þ

From the left-side mesh of the circuit of Figure 10.6, it is clear that: v1 ðt Þ ¼ αi2 ðt Þ

ð4Þ

1 ) i2 ðt Þ ¼ v1 ðt Þ α

ð5Þ

Likewise, from the right-side mesh of the circuit of Figure 10.6, it is seen that: v2 ðt Þ ¼ αi1 ðt Þ

ð6Þ

1 ) i1 ðt Þ ¼  v2 ðt Þ α

ð7Þ

Solving (1), (4), and (6):


v1 ð t Þ v2 ð t Þ




¼

0

α

α

0




i 1 ðt Þ i 2 ðt Þ




) ½Z  ¼

0

α

α

0

 ð8Þ

Solving (2), (5), and (7):




2 0

i1 ðt Þ 6 ¼4 1 i2 ðt Þ α

3 2 1
 0 α 7 v1 ð t Þ ) ½ Y  ¼ 6 5 4 1 v2 ð t Þ 0 α



3 1 α7 5 0



ð9Þ

However, as can be seen from (3), (4), and (5), it is impossible to form the hybrid matrix for the given network. Therefore, the hybrid matrix of the network is not available. Consequently, [Z] and [Y] are available but [H] is not available. Choice (4) is the answer.

Figure 10.6 The circuit of solution of problem 10.6

10

Solutions of Problems: Two-Port Networks

143

10.7. The admittance matrix ([Y]) in time domain is defined as follows:


i 1 ðt Þ




 ¼

i 2 ðt Þ

y11

y12

y21

y22




v1 ð t Þ




 ) ½Y  ¼

v2 ð t Þ

y11

y12

y21

y22

 ð1Þ

KCL in the left-side node: v1 ðt Þ v1 ðt Þ  ðv2 ðt Þ  2v1 ðt ÞÞ þ ¼ 0 ) i1 ðt Þ ¼ 4v1 ðt Þ  v2 ðt Þ 1 1

ð2Þ

v2 ðt Þ v2 ðt Þ  ðv1 ðt Þ þ 2v1 ðt ÞÞ þ ¼ 0 ) i2 ðt Þ ¼ 3v1 ðt Þ þ 2v2 ðt Þ 1 1

ð3Þ

i1 ðt Þ þ KCL in the right-side node: i2 ðt Þ þ Solving (1), (2), and (3):




i1 ðt Þ



i2 ðt Þ


¼

4

1

3

2




v1 ð t Þ




) ½Y  ¼

v2 ð t Þ

4

1

3

2



Choice (1) is the answer.

Figure 10.7 The circuit of solution of problem 10.7

10.8. The admittance matrix ([Y]) of a circuit is not available if the determinant of the inverse matrix of the admittance matrix ([Y]1) or impedance matrix (([Z]) is zero. In other words: detð½Z Þ ¼ 0

ð1Þ

The impedance matrix ([Z]) in time domain is as follows:


v1 ðt Þ v2 ðt Þ




 ¼

z11

z12

z21

z22




i 1 ðt Þ i 2 ðt Þ




 ) ½Z  ¼

z11

z12

z21

z22

 ð2Þ

Applying KVL in the left-side mesh of the circuit of Figure 10.8.2: v1 ðt Þ þ 1  i1 ðt Þ þ 3ðð1 þ aÞi1 ðt Þ þ i2 ðt ÞÞ ¼ 0 ) v1 ðt Þ ¼ ð4 þ 3αÞi1 ðt Þ þ 3i2 ðt Þ

ð3Þ

Applying KVL in the right-side loop of the circuit of Figure 10.8.2: v2 ðt Þ þ 2  i2 ðt Þ þ 3ðð1 þ aÞi1 ðt Þ þ i2 ðt ÞÞ ¼ 0 ) v2 ðt Þ ¼ ð3 þ 3αÞi1 ðt Þ þ 5i2 ðt Þ

ð4Þ

144

10

Solutions of Problems: Two-Port Networks

Solving (2), (3), and (4):


v1 ð t Þ v2 ð t Þ




¼

4 þ 3α

3

3 þ 3α

5




i 1 ðt Þ i 2 ðt Þ




) ½Z  ¼

4 þ 3α

3

3 þ 3α

5

 ð5Þ

Solving (1) and (5): 
det

4 þ 3α

3

3 þ 3α

5

 ¼ 0 ) ð4 þ 3αÞð5Þ  ð3Þð3 þ 3αÞ ¼ 0 ) 6α þ 11 ¼ 0 ) α ¼ 

11 6

Choice (2) is the answer.

Figure 10.8 The circuit of solution of problem 10.8

10.9. Based on the information given in the problem, the admittance matrix of the circuit is as follows:
½Y  ¼

2

2

0

2

 ð1Þ

The admittance matrix ([Y]) of a circuit is defined as follows:



i 1 ðt Þ y11 ¼ i 2 ðt Þ y21

y12 y22





 v1 ð t Þ y11 ) ½Y  ¼ v2 ð t Þ y21

y12 y22

 ð2Þ

Applying KCL in the left-side node: i1 ðt Þ þ

v1 ðt Þ  βv2 ðt Þ v1 ðt Þ  v2 ðt Þ þ ¼ 0 ) i1 ðt Þ ¼ 2v1 ðt Þ  ð1 þ βÞv2 ðt Þ 1 1

ð3Þ

10

Solutions of Problems: Two-Port Networks

145

Applying KCL in the right-side node: i2 ðt Þ þ

v2 ðt Þ  ðαv1 ðt ÞÞ v2 ðt Þ  v1 ðt Þ þ ¼ 0 ) i2 ðt Þ ¼ ðα  1Þv1 ðt Þ þ 2v2 ðt Þ 1 1

ð4Þ

Solving (2), (3), and (4):



i 1 ðt Þ 2 ¼ i 2 ðt Þ α1

 ð1 þ β Þ 2





v1 ð t Þ 2 ) ½Y  ¼ v2 ð t Þ α1

 ð1 þ β Þ 2

Comparing (1) with (5): 

ð1 þ βÞ ¼ 2 ) β ¼ 1 α1¼0)α¼1

Choice (2) is the answer.

Figure 10.9 The circuit of solution of problem 10.9

10.10. The two networks have been connected in parallel. Therefore:
½Y  ¼ ½Y 1  þ ½Y 2  ¼

1

2

2

3




þ

1

2

2

6




¼

2 4 4 9

Choice (1) is the answer.

Figure 10.10 The circuit of solution of problem 10.10



 ð5Þ

146

10

Solutions of Problems: Two-Port Networks

10.11. The two networks have been connected in series. Therefore:
½Z  ¼ ½Z 1  þ ½Z 2  ¼

1

2

2

3




þ

5

6

6

7




¼

6

8

8

10



As we know, the impedance matrix ([Z]) of a network is defined as follows:


v1 ð t Þ




 ¼

v2 ð t Þ

z11

z12

z21

z22




i 1 ðt Þ



i 2 ðt Þ

Therefore:


v1 ð t Þ v2 ð t Þ




¼

6

8

8

10




i 1 ðt Þ



i 2 ðt Þ

Choice (4) is the answer.

Figure 10.11 The circuit of solution of problem 10.11

10.12. The hybrid matrix ([H]) in time domain is as follows:



v1 ð t Þ h11 ¼ i 2 ðt Þ h21

h12 h22





 i1 ðt Þ h11 ) ½H  ¼ v2 ð t Þ h21

h12 h22

 ð1Þ

Applying KVL in the indicated loop: v1 ðt Þ þ β2 v2 ðt Þ þ v2 ðt Þ  β1 v1 ðt Þ ¼ 0 ) ð1 þ β2 Þv2 ðt Þ ¼ ð1 þ β1 Þv1 ðt Þ ¼ 0 ) v1 ðt Þ ¼

1 þ β2 v ðt Þ 1 þ β1 2

ð2Þ

Applying KCL in the indicated supernode: i1 ðt Þ þ α1ei1 ðt Þ þ α2ei2 ðt Þ  i2 ðt Þ ¼ 0 ) i1 ðt Þ þ i2 ðt Þ ¼ α1ei1 ðt Þ þ α2ei2 ðt Þ

ð3Þ

Applying KCL in node A: i1 ðt Þ þ α1ei1 ðt Þ þ ei1 ðt Þ ¼ 0 ) ei1 ðt Þ ¼

1 i ðt Þ 1 þ α1 1

ð4Þ

10

Solutions of Problems: Two-Port Networks

147

Applying KCL in node B: ei2 ðt Þ  α2ei2 ðt Þ þ i2 ðt Þ ¼ 0 ) ei2 ðt Þ ¼

1 i ðt Þ 1 þ α2 2

ð5Þ

Solving (3), (4), and (5): i1 ðt Þ þ i2 ðt Þ ¼

α1 α2 1 1 1 þ α2 i ðt Þ þ i ðt Þ ) i ðt Þ þ i ðt Þ ¼ 0 ) i2 ðt Þ ¼  i ðt Þ 1 þ α1 1 1 þ α2 2 1 þ α1 1 1 þ α2 2 1 þ α1 1

ð6Þ

Solving (1), (2), and (6):


2



6 v1 ð t Þ ¼6 4 i 2 ðt Þ

0 

1 þ α2 1 þ α1

3 2 1 þ β2
 0 6 i ðt Þ 1 þ β1 7 7 1 6 ) ½ H  ¼ 5 v ðt Þ 4 1 þ α2 2 0  1 þ α1

3 1 þ β2 1 þ β1 7 7 5 0

Choice (2) is the answer.

Figure 10.12 The circuit of solution of problem 10.12

10.13. The transmission matrix ([T]) of a circuit is defined as follows:



v1 ðt Þ t 11 ¼ i 1 ðt Þ t 21

t 12 t 22





 v2 ð t Þ t 11 ) ½T  ¼ i 2 ðt Þ t 21

t 12 t 22

 ð1Þ

The transmission matrix of the circuit shown in Figure 10.13.2 can be determined as follows: Applying KVL in the loop: v1 ðt Þ þ v2 ðt Þ ¼ 0 ) v1 ðt Þ ¼ v2 ðt Þ

ð2Þ

Applying KVL in the right-side mesh: v2 ðt Þ þ Rði1 ðt Þ  i2 ðt ÞÞ ¼ 0 ) i1 ðt Þ ¼

1 v ðt Þ þ i2 ðt Þ R 2

ð3Þ

Solving (1), (2), and (3):


 "1 v1 ð t Þ ¼ 1 i 1 ðt Þ R

# " 1 0
v2 ð t Þ   ) Tp ¼ 1 1 i 2 ðt Þ R

0 1

# ð4Þ

148

10

Solutions of Problems: Two-Port Networks

The transmission matrix of the circuit shown in Figure 10.13.3 can be determined as follows: Applying KVL in the loop: v1 ðt Þ þ Ri2 ðt Þ þ v2 ðt Þ ¼ 0 ) v1 ðt Þ ¼ Ri2 ðt Þ þ v2 ðt Þ

ð5Þ

From the circuit, it is clear that: i1 ðt Þ ¼ i2 ðt Þ

ð6Þ

Solving (1), (5), and (6):



v1 ð t Þ 1 ¼ i1 ðt Þ 0

R 1





 1 v 2 ðt Þ ) ½T s  ¼ 0 i 2 ðt Þ

R 1

 ð7Þ

The circuit of Figure 10.13.1 can be assumed like the series connection of seven small circuits. The total transmission matrix of such a circuit can be determined as follows:     ½T  ¼ T p1  ½T s1   T p2  ½T s1   T p2  ½T s1   T p1 " ¼

1 1 1


¼

0

#

 " 1 0#
1 1  1  1 1 0 0:5




1  1 0 1 0



1 1




1

1

0

1






1

0

2

1






1

1

0

1




 "1 1 1   1 1 0 1 1

 " 1 1  1 1 0:5

26 ) ½T  ¼ 45




 15 26

0

1

0

2

1

#









1 1 0 1



Choice (4) is the answer.

Figure 10.13 The circuit of solution of problem 10.13






0 1

1

0

1

1



#

References

References 1. Rahmani-Andebili, M. (2020). DC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature. 2. Rahmani-Andebili, M. (2020). AC Electrical circuit analysis: Practice problems, methods, and solutions, Springer Nature.

149

Index

A Admittance matrix, 139, 141, 143, 144 Angular frequency, 74 C Capacitor, 39–43, 70 D DC voltage source, 39, 69 Dependent source, 59 Dependent voltage source, 59, 71 Differential equation, 30, 34 E Electrical circuits, solutions of problems capacitor, current-voltage relation, 10, 11, 13 capacitor, voltage-current relation, 24 inductor, current-voltage relation, 10, 13, 15, 18 input vector, 22 KCL, 10, 17 KVL, 9, 10, 13, 24, 26 matrices, 15, 22 network, 20 source transformation theorem, 25 state vector, 11, 14, 15, 17, 22 Equivalent impedance, 55, 56, 59, 78 H Hybrid matrix, 141 I Impedance in Laplace domain, 35 Impedances, 45–47 Impulse function, 32, 36, 65 Impulse response, 33, 37 Independent current source, 62 Independent sources, 59 Independent voltage source, 59, 62, 64, 68–70, 72, 75, 79, 81 Inductor, 39–42, 62, 70, 72 Input admittance, 30 Input impedance, 29, 31, 38, 46, 48 Input signal, 32, 36 Inverse Laplace transform, 55, 74

L Laplace domain, 29–31, 35, 38, 39, 45–49, 52, 53, 55–57, 59, 60, 62, 64–70, 72, 74, 75, 77, 79, 81, 123, 124, 139 Laplace transform and network function, 29–32, 34, 35, 37, 40, 47, 51, 62 Linear time-invariant (LTI), 85, 88, 89, 116–118, 123 resistors, 115, 121 system, 32, 36 Loop, 79 M Mesh, 49, 59, 63, 79 N Natural frequencies capacitor/inductor, 86, 95 characteristic equation, 101, 103 damped status, 87, 101 damping status, 98, 99, 102 energy-saving components, 93 impedances, 96, 97 infinite frequency, 105, 107 input admittance, 88, 105, 109 input impedance, 89 Laplace domain, 96–99, 102 LC circuit, 104 loops, 95 LTI network, stability, 108 mesh impedance matrix, 99, 101, 102 nodal admittance matrix, 98 nonzero, 83, 91, 93, 111 nonzero primary voltage and current, 86 one-port network, 104 parallel RLC circuit, 111 resonance frequency, 105 RLC circuit, 112 series LC circuit, 105 state variables, 84, 94 supermesh, 109 time domain, 108 transfer function, 84, 86, 96, 97, 109 undamped status, 87, 100 voltage and current sources, 111 zero and pole, 97 zero frequency, 107 zero/poles, 84, 90, 104

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Advanced Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-78540-6

151

152 Network function, 29–32, 34, 35, 37, 40, 51, 62 Node, 52, 57, 65–67, 71, 73, 77, 81 Nonzero primary current, 62 Nonzero primary voltage, 62, 72, 79, 81 O Ohm’s law, 55, 56, 69, 74, 127 Open circuit, 46 Output response of system, 32, 36 Output voltage, 46 P Phasor domain, 125, 127 Primary circuit, 70 Primary current of inductor, 39, 41 Primary energy, 40 Primary voltage, 39, 40, 70 R Resistive load, 118 Resistor, 42 Response of circuit, 32, 33 S Source transformation theorem, 25, 59, 62, 75 State equations capacitor/inductor, 3 circuit, 8 input vector, 7 matrix, 4 resistors, 6 state vector, 1

Index system matrix, 6 variables, 1 vector, 2, 5 Steady-state output, 41 Steady-state response, 74 Supernode, 71, 77 Switching operation, 43 Symmetric circuit, 75 T Tellegen’s theorem, 122–125, 127 Thevenin equivalent circuit, 70, 126 Thevenin impedance, 49, 59 Three-port network, 115, 117 Transfer function, 29–32, 34, 35, 37, 40, 51, 62, 96 Two-port network, 116–118, 133 admittance matrix, 131–134, 139, 143, 144 hybrid matrix, 132, 135, 138, 142, 146 hybrid parameter, 130, 138 impedance matrix, 129, 131, 132, 137, 141, 143, 146 parallel, 145 phasor domain, 140 series connection, 148 transmission matrix, 130, 136, 138, 147, 148 V Voltage division formula, 57 Voltage division rule, 45, 46, 61, 64–67, 69, 75 Voltage source, 48, 65 Z Zero state, 38