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
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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
1. A ¼
1
1
2. A ¼
1
1
3. A ¼
1
1
4. A ¼
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
v ðt Þ
v ðt Þ
d
C 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
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
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
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
ð6Þ
ð2Þ
)2
Solving (3) and (4):
2
Solving (5) and (4):
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
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):
a22
1
Det ½A a21
a12
3
d
vC ð t Þ
a22
1
6 dt
7
4
5¼
d
Det ½A a21
i ðt Þ
dt L
a12
a11
i C ðt Þ
vL ð t Þ
¼
vC ð t Þ
ð10Þ
i L ðt Þ
a11
Solving (4), (5), and (10):
2
vC ð t Þ
i L ðt Þ
Equation (11) shows the state equations of a circuit with the state vector of
a22
1
Anew 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
d
i ðt Þ ¼ 0
dt L2
ð1Þ 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
>
>
< ¼
L2
2
) R ¼ 1 Ω, L1 ¼ 1 H, L2 ¼ 2 H, C ¼ 1 F
>
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
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
ð1Þ 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
● Easy
● Small
○ Normal
○ Normal
○ Hard
○ Large
ðsÞ
1
1. VVout
¼ 1þRCs
in ðsÞ
ðsÞ
RCs
2. VVout
¼ 1þRCs
in ðsÞ
ðsÞ
3. VVout
¼ 1þ2RCs
2þRCs
in ðsÞ
ðsÞ
2RCs
4. VVout
¼ 1þRCs
in ðsÞ
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
2
2. dtd 3 iðt Þ þ dtd 2 iðt Þ þ dtd iðt Þ ¼ dtd 2 vðt Þ þ 2 dtd vðt Þ þ vðt Þ
3
2
2
3. 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 þ 3vðt Þ
3
2
2
4. 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 þ vðt Þ
3
2
2
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
3
Problems: Laplace Transform and Network Function
31
þ10s1
1. s 10s
Ω
s2 þ6sþ1
3
2
þ10sþ2
2. s þ10s
Ω
s2 þ6sþ1
3
2
þ1
3. s þ10s
Ω
s2 þ3
3
2
þ8sþ2
4. s 8s
Ω
s2 þ6sþ3
3
2
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.
33
2ðs2 þ1Þ
ðsþ1Þ2
ðs2 þ1Þ
2. ðsþ2Þ2
ðs2 þ1Þ
3. s2 þsþ1
ðs2 þ1Þ
4. 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. dtd 2 va ðt Þ þ 2 dtd va ðt Þ þ 2va ðt Þ ¼ dtd 2 is ðt Þ þ dtd 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
2
2
3. dtd 2 va ðt Þ þ dtd va ðt Þ þ va ðt Þ ¼ dtd 2 is ðt Þ þ dtd is ðt Þ þ is ðt Þ
2
2
4. 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
2
2
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
○ Easy
○ Small
● Normal
● Normal
○ Hard
○ Large
1. dtd 2 iðt Þ þ dtd iðt Þ þ iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ
2
2. dtd 2 iðt Þ þ dtd iðt Þ þ 2iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ
2
3. dtd 2 iðt Þ þ 2 dtd iðt Þ þ 2iðt Þ ¼ dtd vs ðt Þ þ vs ðt Þ
2
4. dtd 2 iðt Þ þ 2 dtd iðt Þ þ 2iðt Þ ¼ 2 dtd vs ðt Þ þ vs ðt Þ
2
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
35
þ3s
1. ðssþ1
Þ2
2
2
Þ
2. ððs1
sþ1Þ2
ðs1Þ
3. ðssþ1
Þ2
sþ3
4. ðsþ1
Þ2
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. ssðsþ2s20
2 þ2sþ2Þ A
2
2. ssðsþ2sþ20
2 þ2sþ2Þ A
2
3. ssðsþ3s20
2 þ2s2Þ A
2
4. ssðs3s20
2 þ2sþ2Þ A
2
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
2. s2 ðs3 þ2s5s2 þ3sþ1Þ V
2
3. ðs3 þs5s2 þsþ1Þ V
4. s2 ðs3 þ2s5s2 þ3sþ1Þ 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. s4 þs3 þs12 þ3sþ1
1
2. s4 þ2s3 þ4s
2 þ3sþ1
1
3. s3 þs2 þ3sþ1
1
4. s4 2s3 3s
2 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. 4.
R1
1
R2 C L
s2 þ
R1
1
L þR2 C
A
1
R2 R1
3.
A
s2
ðs2 þω2 Þ
s2
R
sþLR 1C ðs2 þω2 Þ
V
2
s2
ðs2 þRL1 sþR2 Þðs2 ω2 Þ
1
R2 þR1
V
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:
pffiffiffi
2uðt Þ, and the network
I ðsÞ 4ðs þ 20Þ
H ðsÞ ¼ 2
=
sþ8
I 1 ðsÞ
Difficulty level
Calculation amount
1. 0.1 J
2. 0.15 J
3. 0.2 J
4. 0.05 J
○ Easy
○ Small
○ Normal
● Normal
● 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. (2et 2tet) V
2. (2et + 2tet) V
3. (2et + 2tet + 4e2t) V
4. (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
ð2Þ
1
1
) Z2 F ¼
Ω
Cs
2s
ð3Þ
ZC ¼
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
47
1
V ðsÞ
V ðsÞ
1
1
1
V out ðsÞ ¼ 1 s V in ðsÞ ¼ 2
V in ðsÞ ) out
) H ðsÞ ¼ out
¼ 2
¼ 2
ð
s
Þ
ð
s
Þ
V
V
s
s
s
þ
1
þ
1
þ1
þ
s
in
in
s
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
1
1
) Z1 F ¼ Ω
Cs
s
ð4Þ
ZC ¼
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:
ðs þ 2Þ 4 þ 1s
4s þ 9 þ 2s
1
4s2 þ 9s þ 2
¼sþ
¼sþ
¼sþ 2
Z eq ¼ s þ ðs þ 2Þ 4 þ
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:
1
1
2
) Z 0:5 F ¼
¼ Ω
Cs
0:5s s
ð1Þ
ZR ¼ R ) Z3 Ω ¼ 3 Ω
ð2Þ
Z L ¼ Ls ) Z 1 H ¼ s Ω
ð3Þ
ZC ¼
Applying current division rule:
I L ðsÞ ¼ 2
2
s
sþ3þs
Choice (2) is the answer.
I ðsÞ
2
2
I s ðsÞ ¼ 2
I ðsÞ ) H ðsÞ ¼ L ¼ 2
I s ðsÞ s þ 3s þ 2
s 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):
lim sV out ðsÞ ¼ s!1
lim
vout ð0þ Þ ¼ s!1
25
¼0
ðs2 þ 10s þ 125Þ
ð5Þ
Solving (2) and (4):
25
vout ð1Þ ¼ lim sV out ðsÞ ¼ lim
s!0 ðs2 þ 10s þ 125Þ
s!0
¼
25
¼ 0:2
125
ð6Þ
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
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
1
1
) Z2 F ¼
Ω
Cs
2s
ð6Þ
ZC ¼
ZC ¼
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
ðsþ6Þ
4 þ s24þ6sþ4
1
4
s2 þ 6s þ 4
sþ6
¼1þ 3 3
1¼ 2
¼1
sþ5 sþ2
ð s þ 2Þ ð s þ 5Þ
s 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
1
¼ 1 s þ 1 s
sþ1
¼ 1 s þ
!
1
sþ1
2
s þsþ1
¼ 1
sþ1
¼
s2 þsþ1
s2 þ s þ 1
sþ1
¼ 2
s2 þsþ1
s þ 2s þ 2
1 þ sþ1
ð4Þ
Using Ohm’s law in the circuit of Figure 4.9.3:
V a ðsÞ ¼ Z eq I s ðsÞ
ð5Þ
s2 þ s þ 1
V a ðsÞ ¼ 2
I s ðsÞ ) s2 þ 2s þ 2 V a ðsÞ ¼ s2 þ s þ 1 I s ðsÞ
s þ 2s þ 2
ð6Þ
Solving (4) and (5):
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.
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:
1
1
) Z1 F ¼ Ω
Cs
s
ð1Þ
ZR ¼ R ) Z1 Ω ¼ 1 Ω
ð2Þ
ZC ¼
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
1
8s þ 1
2s þ 4
V ðsÞ
V in ðsÞ ¼
1
12s
þ 1 in
2 þ 2s þ 4
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
ð1Þ
2
sþ2
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
1
1
2
) Z 0:5 F ¼
¼ Ω
Cs
0:5s s
ð3Þ
ZC ¼
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
ZR ¼ R ) Z1 Ω ¼ 1 Ω
ð3Þ
ZC ¼
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
5s
) V out ðsÞ ¼ 2 3
V
s ðs 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
s þ 1 þ 1s
1
V A ðsÞ ¼ 2
V A ðsÞ ) V A ðsÞ ¼ s2 þ s þ 1 V 2 ðsÞ
s þsþ1
ð2Þ
4
Solutions of Problems: Laplace Transform and Network Function
67
Solving (1) and (2):
s2 þ s þ 1
s4 þ 2s3 þ 4s2 þ 3s þ 1
V ðsÞ
1
V 1 ðsÞ ¼ s2 þ s þ 1 V 2 ðsÞ ) 2 ¼ 4
3
V 1 ðsÞ s þ 2s þ 4s2 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
As
vs ðt Þ ¼ Acosðωt Þ ¼
¼
¼
¼
) V s ðsÞ ¼ 2
s þ ω2
ð5Þ
Applying voltage division rule for node A:
V A ðsÞ ¼
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 B:
V B ðsÞ ¼
Therefore:
V out ðsÞ = V A ðsÞ 2 V B ðsÞ =
s
s
s þ RL1 s þ R21C
!
s
V s ðsÞ
V s ðsÞ ¼
s2 þ RL1 þ R21C s þ LRR21C
R1
1
R2 C L
ð8Þ
68
4
Solutions of Problems: Laplace Transform and Network Function
Solving (5) and (8):
R1 2
1
A
s
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
R1
1
R2 C L
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
ð5Þ
Eð2s2 þ s þ 1Þ
¼ 2E
s!1 ðs2 þ s þ 2Þ
ð6Þ
s!1
Solving (4) and (5):
ið0þ Þ ¼ lim
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:
1
1
) Z1 F ¼ Ω
Cs
s
ð1Þ
ZR ¼ R ) Z1 Ω ¼ 1 Ω
ð2Þ
ZC ¼
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
pffiffiffi
L
2uð t Þ ¼
¼
¼
¼
) I 1 ðsÞ ¼
pffiffiffi
2
s
I ðsÞ 4ðs þ 20Þ
4ðs þ 20Þ
) I 2 ðsÞ ¼
I ðsÞ
H ðsÞ ¼ 2
=
s
þ
8
sþ8 1
I 1 ðsÞ
ð1Þ
ð3Þ
Solving (1) and (3):
I 2 ðsÞ ¼
pffiffiffi
4ðs þ 20Þ
2
sþ8
s
ð4Þ
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:
2
vs ðt Þ ¼ sin ð2t Þuðt Þ ) V s ðsÞ ¼ 2
s þ4
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):
8
v C ð 0 Þ
>
vC ð0 Þ
vC ð0 Þ
vC ð0 Þ
<
¼2
0 s
s
2
2
2
2
2
þ0¼0) 2
) 2
þ
¼
¼ 2 21 )
1
1
>
s þ4
s þ 4 2s þ Cs
s þ 4 s þ 2C
1
2s þ Cs
:
¼4
2C
) vC ð0 Þ ¼ 4 V, C ¼
1
F
8
Choice (2) is the answer.
Figure 4.23 The circuit of solution of problem 4.28
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
ð1Þ
L
4s
¼
¼
¼
) V in ðsÞ ¼ 2
vin ðt Þ ¼ 4 cos ð2t Þ ¼
s þ4
ð2Þ
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
Solving (1) and (2):
V out ðsÞ
4s
s2 þ4
=
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
ð1Þ
1
1
) Z2 F ¼
Ω
Cs
2s
ð2Þ
Z L ¼ Ls ) Z 0:5 H ¼ 0:5s Ω
ð3Þ
1
1
) Z3 F ¼
Ω
Cs
3s
ð4Þ
ZC ¼
ZC ¼
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 VVoutin ððssÞÞ if the input and the output are as
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
87
1. 2, j6
2. 2, 3, 0
3. 2, 3, j6
4. 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
ðs2 þ1Þ
2. ðsþ1Þðsþ2Þ Ω
Þðsþ2Þ
Ω
3. ðsþ1
ðs2 þ1Þ
Þðsþ2Þ
4. ðsþ1
2ðs2 þ1Þ 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
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
< 2
2 22 4 2
¼ 1 j ) Poles : 1 j
s þ 2s þ 2 ¼ 0 ) s ¼
2
:
s þ 2 ¼ 0 ) s ¼ 2 ) Zero : 2
Choice (2) is the answer.
Figure 6.4 The circuit of solution of problem 6.4
ð4Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
97
6.5. The circuit of Figure 6.5.2 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 0:1 H ¼ 0:1s Ω
ð2Þ
ZR ¼ R ) Z2 Ω ¼ 2 Ω
ð3Þ
The network function (transfer function) for this problem has been presented in the following form:
I ðsÞ
H ðsÞ ¼ out
I s ðsÞ
ð4Þ
Applying current division rule:
I out ðsÞ ¼
I ðsÞ s þ 10
1 þ 0:1s
s þ 10
I ðsÞ ¼
I ðsÞ ) H ðsÞ ¼ out
¼
1 þ 0:1s þ 2 s
s þ 30 s
s þ 30
I s ðsÞ
Therefore, the zero and pole of the network function (transfer function) are as follows:
s þ 30 ¼ 0 ) s ¼ 30 ) Pole : 30
s þ 10 ¼ 0 ) s ¼ 10 ) Zero : 10
Choice (1) is the answer.
Figure 6.5 The circuit of solution of problem 6.5
6.6. The damping status of a circuit can be identified based on the roots of the characteristic equation of the circuit, where the
characteristic equation is determined by solving the following equation:
detðs½I ½AÞ ¼ 0
ð1Þ
Based on the information given in the problem, we have:
½A ¼
0
1
1 1
ð2Þ
98
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Solving (1) and (2):
1
det s
0
0
1
0
1
1
1
¼ 0 ) det
(
) s2 þ s þ 1 ¼ 0 )
s
1
1
sþ1
¼ 0 ) s ð s þ 1Þ þ 1 ¼ 0
b 6¼ 0
Δ ¼ 12 4 1 ¼ 3 < 0
Therefore, the damping status of the circuit is underdamped. Choice (3) is the answer.
6.7. The circuit of Figure 6.6.2 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
1
1
) Z1 F ¼ Ω
Cs
s
ð3Þ
Z R ¼ R ) Z 4 Ω ¼ 4 Ω
ð4Þ
ZC ¼
As we know, the damping status of a circuit can be determined based on the roots of the characteristic equation of the
circuit. Moreover, the characteristic equation of a circuit can be identified from the determinant of the nodal admittance
matrix ([Ynodal]) of the circuit, as follows:
detð½Y nodal Þ ¼ 0
ð1Þ
The circuit includes two supernodes shown in Figure 6.6.2. The nodal admittance matrix can be determined as follows:
2P
y1j
6 j¼1
½Y nodal ¼ 4
y21
y12
3
2
1 1 1
þ þ
6
7
P 5 ¼ 42 s 2
1
y2j
j¼1
2
3
2
sþ1
7 6 s
5¼4
1
1
1
þsþ
4
2
2
1
2
3
1
2 7
5
4s þ 1
4
Solving (1) and (2):
s þ 1
s
1
2
1 2
2 ¼ 0 ) s þ 1 4s þ 1 1 ¼ 0 ) 4s þ 4s þ 1 ¼ 0 ) ð2s þ 1Þ2 ¼ 0 ) s ¼ 1 , 1
s
4
4
4s
2
2
4s þ 1 4
Since the roots are real and equal, the circuit is in the critically damped status. Choice (2) is the answer.
ð2Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
99
Figure 6.6 The circuit of solution of problem 6.7
6.8. The circuit of Figure 6.7.2 shows the main circuit in Laplace domain. The impedances of the components are as follows:
ZR ¼ R
ð1Þ
Z L ¼ Ls
ð2Þ
Zr ¼ r
ð3Þ
1
Cs
ð4Þ
ZC ¼
As we know, the damping status of a circuit can be determined based on the roots of the characteristic equation of the
circuit. Moreover, the characteristic equation of a circuit can be identified from the determinant of the mesh impedance
matrix ([Zmesh]) of the circuit, as follows:
detð½Z mesh Þ ¼ 0
ð1Þ
The circuit includes two meshes shown in Figure 6.7.2. The mesh impedance matrix can be determined as follows:
2P
z1j
6 j¼1
½Z mesh ¼ 4
z21
3
2
R þ r þ Ls
7
P 5¼4
z2j
ðr þ LsÞ
z12
j¼1
3
ðr þ LsÞ
1 5
r þ Ls þ
Cs
ð2Þ
100
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Solving (1) and (2):
R þ r þ Ls ðr þ LsÞ 1
ðr þ LsÞ r þ Ls þ 1 ¼ 0 ) ðR þ r þ LsÞ r þ Ls þ Cs ðr þ LsÞðr þ LsÞ ¼ 0
Cs ) RLCs2 þ ðL þ RrC Þs þ R þ r ¼ 0
ð3Þ
As we know, to create an undamped status, the roots must be on the imaginary axis. Therefore, the factor of s in (3) must
be zero, as follows:
L þ RrC ¼ 0 ) R ¼ L
rC
Choice (2) is the answer.
Figure 6.7 The circuit of solution of problem 6.8
6.9. Based on the information given in the problem, we have:
L
vin ðt Þ ¼ 5e2t ) V in ðsÞ ¼
L
vout ðt Þ ¼ 3te2t þ 2e3t sin ð6t Þ ) V out ðsÞ ¼ 5
sþ2
d
3
26
3
12
¼
þ
þ
2
2
2
ds s þ 2
ð s þ 3Þ þ 6
ð s þ 2Þ
ðs þ 3Þ2 þ 36
Solving (1) and (2):
V out ðsÞ ðsþ2Þ2 þ ðsþ3Þ2 þ36
3ð5s2 þ 22s þ 61Þ
¼
¼
5
V in ðsÞ
5ðs þ 2Þ ðs þ 3Þ2 þ 36
sþ2
3
12
Poles ) 5ðs þ 2Þ ðs þ 3Þ2 þ 36 ¼ 0 ) s ¼ 2, 3 j6
Choice (4) is the answer.
ð1Þ
ð2Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
101
6.10. The circuit of Figure 6.8.2 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
ð1Þ
ZR ¼ R ) Z3 Ω ¼ 3 Ω
ð2Þ
Z L ¼ Ls ) Z 0:5 H ¼ 0:5s Ω
ð3Þ
ZR ¼ R ) Z2 Ω ¼ 2 Ω
ð4Þ
As we know, the damping status of a circuit can be determined based on the roots of the characteristic equation of the
circuit. In addition, the characteristic equation of a circuit can be identified from the determinant of the mesh impedance
matrix ([Zmesh]) of the circuit, as follows:
detð½Z mesh Þ ¼ 0
ð5Þ
The circuit includes two meshes illustrated in Figure 6.8.2. The mesh impedance matrix cannot be determined by using
the straightforward rule, since there is a dependent source in the circuit.
KVL in the right-side mesh:
0:5sðI 2 ðsÞ I ðsÞÞ þ 2I 2 ðsÞ þ 2αI ðsÞ ¼ 0 ) ð0:5s þ 2αÞI ðsÞ þ ð0:5s þ 2ÞI 2 ðsÞ ¼ 0
ð6Þ
KVL in the left-side mesh:
V s ðsÞ þ 2sI ðsÞ þ 3I ðsÞ þ 0:5sðI ðsÞ I 2 ðsÞÞ ¼ 0 ) ð2:5s þ 3ÞI ðsÞ 0:5sI 2 ðsÞ ¼ V s ðsÞ
ð7Þ
Writing (6) and (7) in the form of [Z][I] ¼ [Es]:
0:5s þ 2α
0:5s þ 2
2:5s þ 3
0:5s
I ðsÞ
I 2 ðsÞ
¼
0
V s ðsÞ
) ½Z mesh ¼
0:5s þ 2α
0:5s þ 2
2:5s þ 3
0:5s
ð8Þ
Solving (5) and (8):
det
0:5s þ 2α
0:5s þ 2
2:5s þ 3
0:5s
¼ 0 ) ð0:5s þ 2αÞð0:5sÞ ð0:5s þ 2Þð2:5s þ 3Þ ¼ 0
) s2 þ ðα þ 6:5Þs þ 6 ¼ 0
ð9Þ
To put the circuit in the critically damped status, the discriminant of the characteristic equation must be zero. Hence:
ðα þ 6:5Þ2 4 1 6 ¼ 0 ) α2 þ 13α þ
)α¼
Choice (1) is the answer.
13 169
73
24 ¼ 0 ) α2 þ 13α þ
¼0
4
4
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffi
132 73 13 4 6
¼
) α 1:6, 11:4
2
2
102
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Figure 6.8 The circuit of solution of problem 6.10
6.11. The circuit of Figure 6.9.2 shows the main circuit in Laplace domain. The impedances of the components are as follows:
ZR ¼ R ) Z1 Ω ¼ 1 Ω
ð1Þ
1
1
) Z1 F ¼ Ω
Cs
s
ð2Þ
ZC ¼
As we know, the damping status of a circuit can be determined based on the roots of the characteristic equation of the
circuit. In addition, the characteristic equation of a circuit can be achieved from the determinant of the mesh impedance
matrix ([Zmesh]) of the circuit, as follows:
detð½Z mesh Þ ¼ 0
ð3Þ
The circuit includes two meshes illustrated in Figure 6.9.2. The mesh impedance matrix cannot be determined by using
the straightforward rule, since there is a dependent source in the circuit.
Defining V(s) based on the mesh currents:
1
V ðsÞ ¼ ðI 1 ðsÞ I 2 ðsÞÞ
s
KVL in the right-side mesh:
1
1
1
2
þ 1 I 2 ðsÞ þ βV ðsÞ ¼ 0
ðI ðsÞ I 1 ðsÞÞ þ 1 I 2 ðsÞ þ I 2 ðsÞ þ βV ðsÞ ¼ 0 ) I 1 ðsÞ þ
s 2
s
s
s
ð4Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Using ð4Þ
1
2
1
þ 1 I 2 ðsÞ þ β ðI 1 ðsÞ I 2 ðsÞÞ ¼ 0 )
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
) I 1 ðsÞ þ
s
s
s
103
β1
2β
I 1 ðsÞ þ
þ 1 I 2 ðsÞ ¼ 0
s
s
ð5Þ
KVL in the left-side mesh:
1
1
1
1 I 1 ðsÞ þ ðI 1 ðsÞ I 2 ðsÞÞ ¼ 0 ) 1 þ I 1 ðsÞ I 2 ðsÞ ¼ 0
s
s
s
ð6Þ
Writing (5) and (6) in the form of [Zmesh][I] ¼ [Es]:
2
β1
6 s
4
1
1þ
s
3
2
2β
β1
þ 1 I 1 ðsÞ 0 7
6 s
s
¼
) ½Z mesh ¼ 4
5
1
1
I 2 ðsÞ
0
1þ
s
s
3
2β
þ1
7
s
5
1
s
ð7Þ
Solving (3) and (7):
02
β1
B6 s
det@4
1
1þ
s
31
2β
þ1
β1
1
2β
1
7C
s
þ1 1þ
¼0
5A ¼ 0 )
s
s
s
s
1
s
) s2 þ ð3 βÞs þ 1 ¼ 0
ð8Þ
To put the circuit in the undamped status, the roots of the characteristic equation must be on the imaginary axis. In other
words, the value of b in the quadratic equation (as2 + bs + c ¼ 0) must be zero. Hence:
3β ¼0)β ¼3
Choice (3) is the answer.
Figure 6.9 The circuit of solution of problem 6.11
104
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
6.12. Based on the zero-pole diagram of the input impedance of the one-port network, shown in Figure 6.10, we can
determine the input impedance, as follows:
Z in ðsÞ ¼ k
ðs jÞðs þ jÞ
k ðs2 þ 1Þ
¼
ð s þ 1Þ ð s þ 2Þ ð s þ 1Þ ð s þ 2Þ
ð1Þ
Based on the information given in the problem, we know that if the network is connected to a 1 A current source, 0.5 V
can be measured across that after a while. This means that the input impedance of the network for s ¼ 0 (the input source
is a DC source) is as follows:
Z in ðsÞ ¼
V ðsÞ
0:5
¼ 0:5 Ω
) Z in ðs ¼ 0Þ ¼
1
I ðsÞ
ð2Þ
Solving (1) for s ¼ 0:
Z in ðs ¼ 0Þ ¼
k ð 0 þ 1Þ
k
¼
ð0 þ 1Þð0 þ 2Þ 2
ð3Þ
Solving (2) and (3):
k
¼ 0:5 ) k ¼ 1
2
ð4Þ
Solving (1) and (4):
Z in ðsÞ ¼
ð s 2 þ 1Þ
Ω
ð s þ 1Þ ð s þ 2Þ
Choice (2) is the answer.
Figure 6.10 The circuit of solution of problem 6.12
6.13. The resonance frequency of the parallel LC circuit of the problem can be determined as follows:
1
1
ωr1 ¼ pffiffiffiffiffiffi ) ωr1 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ 1 rad= sec
LC
11
A parallel LC circuit behaves like an open circuit branch in its resonance frequency. Therefore, the parallel LC circuit
will be like an open circuit for the input signal of cos(t). Thus, no voltage or current will be observed at the output
terminal. In other words:
ωr1 ¼ 1 rad= sec ) vout ðt Þ ¼ 0, iout ðt Þ ¼ 0
ð1Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
105
Moreover, the resonance frequency of the series LC circuit of the problem can be calculated as follows:
1
1
ωr2 ¼ pffiffiffiffiffiffi ) ωr2 ¼ qffiffiffiffiffiffiffiffiffiffi ¼ 2 rad= sec
1
LC
1
4
A series LC circuit behaves like a short circuit branch in its resonance frequency. Therefore, the series LC circuit will be
like a short circuit for the input signal of cos(2t). Hence:
ωr2 ¼ 2 rad= sec ) vout ðt Þ ¼ 0
ð2Þ
From (1) and (2), it can be concluded that vout(t) is always zero for the input signal of is(t) ¼ cos (t) + cos (2t). Choice
(2) is the answer.
Figure 6.11 The circuit of solution of problem 6.13
6.14. Based on the information given in the problem, we know that the input admittance includes the zeros at s ¼ 2, 2.5
and the poles at s ¼ 1, 1. Therefore:
Y in ðsÞ ¼ k
ðs þ 2Þðs þ 2:5Þ
ð s þ 1Þ 2
ð1Þ
An inductor behaves like an open circuit at infinite frequency, since its impedance goes to infinity for s ¼ 1, as can be
seen in the following:
Z L ðsÞ ¼ Ls ) Z L ðs ¼ 1Þ ¼ L 1 ! 1
ð2Þ
The circuit of Figure 6.12.2 illustrates the main circuit for s ¼ 1. The input admittance of the circuit is as follows:
Y in ðs ¼ 1Þ ¼
1
¼2
0:5
ð3Þ
On the other hand, the value of input impedance for s ¼ 1 can be determined by using (1), as follows:
Y in ðs ¼ 1Þ ¼ k
ð4Þ
k¼2
ð5Þ
Solving (3) and (4):
106
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
An inductor behaves like a short circuit at zero frequency, since its impedance is zero for s ¼ 0, as can be seen in the
following:
Z L ðsÞ ¼ Ls ) Z L ðs ¼ 0Þ ¼ L 0 ¼ 0
ð6Þ
The circuit of Figure 6.12.3 illustrates the main circuit for s ¼ 0. The input admittance of the circuit is as follows:
Y in ðs ¼ 0Þ ¼ 0:5 R2
1
¼
0:5 þ R2
0:5R2
ð7Þ
On the other hand, the value of input impedance for s ¼ 0 can be determined by using (1), as follows:
Y in ðsÞ ¼ k
ð2Þð2:5Þ
¼ 5k
ð 1Þ 2
Solving (5), (7), and (8):
0:5 þ R2
1
¼ 5 2 ) 0:5 þ R2 ¼ 5R2 ) R2 ¼ Ω
0:5R2
8
Choice (2) is the answer.
Figure 6.12 The circuit of solution of problem 6.14
ð8Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
107
6.15. Based on the information given in the problem, we know that the input impedance includes the poles at s ¼ 1, 3
and the zeros at s ¼ 2, 4. Therefore:
Z in ðsÞ ¼ k
ð s þ 2Þ ð s þ 4Þ
ð s þ 1Þ ð s þ 3Þ
ð1Þ
A capacitor behaves like an open circuit at zero frequency, since its impedance goes to infinity for s ¼ 0, as can be seen
in the following:
Z C ðsÞ ¼
1
1
) Z C ð s ¼ 0Þ ¼
!1
Cs
C0
ð2Þ
The circuit of Figure 6.13.2 illustrates the main circuit for s ¼ 0. The input impedance of the circuit is as follows:
Z in ðs ¼ 0Þ ¼
8
Ω
3
ð3Þ
On the other hand, the value of input impedance for s ¼ 0 can be determined by using (1), as follows:
Z in ðs ¼ 0Þ ¼ k
ð2Þð4Þ
8
¼k
3
ð1Þð3Þ
ð4Þ
Solving (3) and (4):
8 8
k ¼ )k¼1
3 3
ð5Þ
A capacitor behaves like a short circuit at infinite frequency, since its impedance approaches zero for s ¼ 1, as can be
seen in the following:
Z C ðsÞ ¼
1
1
) Z C ð s ¼ 1Þ ¼
!0
Cs
C1
ð6Þ
The circuit of Figure 6.13.3 illustrates the main circuit for s ¼ 1. The input impedance of the circuit is as follows:
Z in ðs ¼ 1Þ ¼
8
R1
8
R1 ¼ 8 3
3
þ
R1
3
ð7Þ
On the other hand, the value of input impedance for s ¼ 1 can be determined by using (1) and (5), as follows:
ð5Þ
Z in ðs ¼ 1Þ ¼ k)Z in ðs ¼ 1Þ ¼ 1
Solving (7) and ((8):
8
8
8
8 5
8
3 R1
¼ 1 ) þ R1 ¼ R1 ) ¼ R1 ) R1 ¼ Ω
8
3
3
3
3
5
þ
R
1
3
Choice (2) is the answer.
ð8Þ
108
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Figure 6.13 The circuit of solution of problem 6.15
6.16. The circuit of Figure 6.14.2 shows the main circuit in time domain. In Laplace domain, we have:
ZR ¼ R ) Z1 Ω ¼ 1 Ω
L
vs ð t Þ ¼ uð t Þ ) V s ð s Þ ¼
ð1Þ
1
s
3
L
3 1
2
iout ðt Þ ¼ 2 e2t uðt Þ ) I out ðsÞ ¼ 2 1
2
s sþ2
ð2Þ
ð3Þ
To determine the stability of the LTI network, we can determine the input admittance and the input impedance of the
network, as follows:
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
109
3
2
I ðsÞ s sþ12
sþ2
Y in ðsÞ ¼ out
¼ 1 ¼
2s
þ1
V s ðsÞ
s
2
) Z in ðsÞ ¼
1
2s þ 1
¼
sþ2
Y in ðsÞ
ð4Þ
ð5Þ
As can be noticed from Figure 6.14:
Z N ðsÞ ¼ Z in ðsÞ 1 ¼
) Y N ðsÞ ¼
2s þ 1
s1
1¼
sþ2
sþ2
1
sþ2
¼
Z N ðsÞ s 1
ð6Þ
ð7Þ
As can be noticed from (6), the input impedance of the network is stable, as its pole is in the left-half of the s-plane.
Therefore, the LTI network is stable in open circuit status.
Moreover, as can be noticed from (7), the input admittance of the network is unstable, since its pole is in the right-half of
the s-plane. Therefore, the LTI network is unstable in short circuit status.
Choice (4) is the answer.
Figure 6.14 The circuit of solution of problem 6.16
6.17. The circuit of Figure 6.15.2 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
ZC ¼
The network function (transfer function) for this problem is defined as follows:
H ðsÞ ¼
V out ðsÞ
I s ðsÞ
ð3Þ
Applying KCL in the left-side supermesh in the circuit of Figure 6.15.2:
I s ðsÞ þ
V ðsÞ V ðsÞ V ðsÞ V out ðsÞ
þ 1 þ
¼ 0 ) ð2s þ 1ÞV ðsÞ sV out ðsÞ ¼ I s ðsÞ
1
1
s
s
ð4Þ
Applying KCL in the right-side supermesh circuit of Figure 6.15.2:
V out ðsÞ V ðsÞ
1
s
2V ðsÞ þ
V out ðsÞ
¼ 0 ) ðs þ 2ÞV ðsÞ þ ðs þ 1ÞV out ðsÞ ¼ 0
1
) V ðsÞ ¼
sþ1
V ðsÞ
s þ 2 out
ð5Þ
110
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Solving (4) and (5):
ð2s þ 1Þ
sþ1
s2 þ s þ 1
V ðsÞ sV out ðsÞ ¼ I s ðsÞ )
V out ðsÞ ¼ I s ðsÞ
s þ 2 out
sþ2
ð6Þ
V out ðsÞ
sþ2
sþ2
) V out ðsÞ ¼ 2
I ðsÞ
¼ 2
I s ðsÞ
s þsþ1
s þsþ1 s
ð7Þ
Solving (3) and (6):
) H ðsÞ ¼
As can be noticed from (7), if the input function (Is(s)) has a pole at s ¼ 2, this pole will be cancelled by the zero of
the network function. Therefore, the natural frequency of the input signal will not be seen in the output (Vout(s)), and
only the natural frequencies of the circuit will be observed. Hence:
I s ðsÞ ¼
1 L1
) is ðt Þ ¼ e2t uðt Þ A
sþ2
Choice (4) is the answer.
Figure 6.15 The circuit of solution of problem 6.17
6.18. Based on the zero-pole diagram of the network function (transfer function) of the circuit, shown in Figure 6.16, the
network function can be determined, as follows:
H ðsÞ ¼
ðs þ 2Þðs 2Þ
kðs þ 2Þðs 2Þ
Output function
¼k
¼
Input function
ðs þ j4Þðs j4Þ
s2 þ 16
) Output function ¼
kðs þ 2Þðs 2Þ
Input function
s2 þ 16
ð1Þ
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
111
As can be noticed from (1), if the input function has a pole at s ¼ 2, this pole will be cancelled by the zero of the
network function. Therefore, the natural frequency of the input function will not be seen in the output. Hence:
Input function ¼
1 L1
) Input function in time domain ¼ e2t uðt Þ
sþ2
ð2Þ
Based on the information given in the problem, we know that:
Input function in time domain ¼ eat uðt Þ
ð3Þ
Solving (2) and (3):
a¼2
Choice (2) is the answer.
Figure 6.16 The circuit of solution of problem 6.18
6.19. To determine the natural frequencies of a circuit, we need to turn off the independent voltage and current sources, as can
be seen in Figure 6.17.2.
By doing this, the parallel connection of the resistor and the inductor (on the left side of the circuit) is short-circuited and
eliminated. Likewise, the parallel connection of the resistor and the capacitor (in the right side of the circuit) is removed,
as is illustrated in Figure 6.17.3.
On the other hand, since the nonzero natural frequencies have been requested, we can simplify the circuit. Hence, the
circuit of Figure 6.17.3 is simplified and shown in Figure 6.17.4, since all the components are in parallel.
Now, the resultant circuit is a parallel RLC circuit that its characteristic equation is as follows:
s2 þ 2αs þ ω20 ¼ 0
ð1Þ
where
2α ¼
1
1
2
¼
¼
RC 0:5 3 3
ð2Þ
ω20 ¼
1
1
2
¼
¼
LC 0:5 3 3
ð3Þ
Solving (1)–(3):
pffiffiffi
2
2
1
s2 þ s þ ¼ 0 ) s ¼ 1 j 5
3
3
3
Choice (1) is the answer.
112
6
Solutions of Problems: Natural Frequencies of Electrical Circuits
Figure 6.17 The circuit of solution of problem 6.19
References
Figure 6.17 (continued)
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.
113
Problems: Network Theorems (Tellegen’s and Linear
Time-Invariant Network Theorems)
Abstract
In this chapter, the basic and advanced problems of network theorems, that is, Tellegen’s and linear time-invariant (LTI)
network theorems, 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.
7.1. The three-port network, shown in Figure 7.1, includes linear time-invariant (LTI) resistors and dependent sources
[1–2]. The network is put under two tests as follows.
Test I: vs ¼ 7 V, is ¼ 3 A, iout ¼ 1 A
Test II: vs ¼ 9 V, is ¼ 1 A, iout ¼ 3 A
Determine the value of iout for vs ¼ 15 V, is ¼ 9 A
Difficulty level
○ Easy
● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1. 0.3 A
2. 0.6 A
3. 3 A
4. 6 A
Figure 7.1 The circuit of problem 7.1
# 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_7
115
7
116
7
Problems: Network Theorems (Tellegen’s and Linear Time-Invariant Network Theorems)
7.2. For the resistive linear time-invariant (LTI) two-port network, shown in Figure 7.2, the information below is given.
v1 ðt Þ ¼ 30t V, v2 ðt Þ ¼ 0, i1 ðt Þ ¼ 5t A, i2 ðt Þ ¼ 2t A
Determine the value of i1(t) in Ampere if v1(t) ¼ 30t + 60 V, v2(t) ¼ 60t + 15 V
Difficulty level
○ Easy
● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1. (t + 9) A
2. (9t + 11) A
3. (5t + 10) A
4. (4t 1) A
Figure 7.2 The circuit of problem 7.2
7.3. For the resistive linear time-invariant (LTI) two-port network, shown in Figure 7.3, the information below is given.
Test I: If v1 ¼ 3 V, is ¼ 3 A, iout ¼ 6 A
Test II: v1 ¼ 0 A, is ¼ 2 A, iout ¼ 2 A
Determine the value of iout(t) if v1(t) ¼ 2 V, is(t) ¼ 0 A
Difficulty level
○ Easy
● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1. 6 A
2. 4 A
3. 4 A
4. 6 A
Figure 7.3 The circuit of problem 7.3
7
Problems: Network Theorems (Tellegen’s and Linear Time-Invariant Network Theorems)
117
7.4. Figure 7.4.1 shows a resistive three-port network that a test is done on it. The same network is put under another test
which is illustrated in Figure 7.4.2. Determine the value of ev1 ðt Þ.
Difficulty level
○ Easy
● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1. etu(t) V
2. (t 1)etu(t) V
3. (t 2)etu(t) V
4. (t + 1)etu(t) V
Figure 7.4 The circuit of problem 7.4
7.5. Figure 7.5 shows a linear time-invariant (LTI) two-port network that includes RLC components.
i1 ðt Þ ¼ 2 sin ωt þ 30 A, v1 ðt Þ ¼ 4 sin ωt þ 45 V, v2 ðt Þ ¼ 0, i2 ðt Þ ¼ sin ωt þ 60 A
Determine the value of v1(t) in Ampere if i1(t) ¼ 0.5 sin (ωt + 15 ) A, v2(t) ¼ sin (ωt) V
Difficulty level
○ Easy
● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1. sin(ωt) V
2. sin(ωt 60 ) V
3. 0.5 sin (ωt + 30 ) V
4. 4 sin (ωt 15 ) V
118
7
Problems: Network Theorems (Tellegen’s and Linear Time-Invariant Network Theorems)
Figure 7.5 The circuit of problem 7.5
7.6. The two-port network, illustrated in Figure 7.6, is linear time-invariant (LTI). The tests below have been carried out on
the resistive load. Calculate the impedance of the network.
Test I: If RL ¼ 1, |Vout| ¼ 13 V
Test II: If RL ¼ 3 Ω, |Vout| ¼ 3 V
Test III: If RL ¼ 14 Ω, |Vout| ¼ 9.1 V
Difficulty level
Calculation amount
1. 2 Ω
2. 12 Ω
3. (12 + j2 ) Ω
4. (2 + j12) Ω
○ Easy
○ Small
○ Normal
● Normal
● Hard
○ Large
Figure 7.6 The circuit of problem 7.6
7.7. In the linear time-invariant (LTI) two-port network that only includes resistors, inductors, and capacitors, when an
impulse function is applied on the left-side port, the voltage of v2(t) ¼ 2 V is measured on the right-side port (see
Figure 7.7.1). Now, the same network is evaluated in another test, presented in Figure 7.7.2. Determine the value of ev1 ðt Þ.
Difficulty level
○ Easy
○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1. 2 cos (t) V
2. cos(t) V
3. 2 sin (t) V
4. sin(t) V
References
119
Figure 7.7 The circuit of problem 7.7
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.
8
Solutions of Problems: Network Theorems
(Tellegen’s and Linear Time-Invariant Network
Theorems)
Abstract
In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods.
8.1. Based on the information given in the problem, we have [1–2]:
Test I : vs ¼ 7 V, is ¼ 3 A, iout ¼ 1 A
ð1Þ
Test II : vs ¼ 9 V, is ¼ 1 A, iout ¼ 3 A
ð2Þ
Test III : vs ¼ 15 V, is ¼ 9 A, iout ¼ Unknown
ð3Þ
Since the network only includes linear time-invariant (LTI) resistors and dependent sources, the following relation for the
output current is held:
iout ¼ αis þ βvs
ð4Þ
Solving (1), (2), and (4):
(
1 ¼ 3α þ 7β
3 ¼ α þ 9β
) α ¼ 0:6, β ¼ 0:4
ð5Þ
Solving (3), (4), and (5):
iout ¼ αis þ βvs ¼ 0:6 9 þ 0:4 15 ¼ 0:6 A
Choice (2) 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_8
121
8 Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .
122
Figure 8.1 The circuit of solution of problem 8.1
8.2. Based on Tellegen’s theorem, we can write the following equation for the network:
v1 ðt Þ ei1 ðt Þ þ v2 ðt Þei2 ðt Þ ¼ ev1 ðt Þði1 ðt ÞÞ þ ev2 ðt Þi2 ðt Þ
ð1Þ
In (1), a negative sign was applied for the polarity of the current of the left-side voltage source, as its current flows toward
the network.
Based on the information given in the problem, we have:
Test I : v1 ðt Þ ¼ 30t V, v2 ðt Þ ¼ 0, i1 ðt Þ ¼ 5t A, i2 ðt Þ ¼ 2t A
ð2Þ
Test II : v1 ðt Þ ¼ 30t þ 60 V, v2 ðt Þ ¼ 60t þ 15 V, i1 ðt Þ ¼ Unknown
ð3Þ
Solving (1), (2), and (3):
30t ei1 ðt Þ þ 0 ei2 ðt Þ ¼ ð30t þ 60Þ ð5t Þ þ ð60t þ 15Þ 2t
) ei1 ðt Þ ¼
150t 2 300t þ 120t 2 þ 30t 30t 2 270t
¼
¼ ð t þ 9Þ A
30t
30t
Choice (1) is the answer.
Figure 8.2 The circuit of solution of problem 8.2
8.3. Based on the information given in the problem, we have:
Test I : If v1 ¼ 3 V, is ¼ 3 A, iout ¼ 6 A
ð1Þ
8
Solutions of Problems: Network Theorems (Tellegen’s and Linear. . .
123
Test II : v1 ¼ 0 A, is ¼ 2 A, iout ¼ 2 A
ð2Þ
Test III : v1 ¼ 2 V, is ¼ 0 A, iout ¼ Unknown
ð3Þ
Since the network is linear time-invariant (LTI), the following relation for the output current is held:
iout ¼ αv1 þ βis
ð4Þ
Solving (1), (2), and (4):
(
6 ¼ 3α þ 3β
2 ¼ 0 α þ ð2Þβ
) α ¼ 3, β ¼ 1
ð5Þ
Solving (3), (4), and (5):
iout ¼ αv1 þ βis ¼ 3 ð2Þ þ ð1Þ 0 ¼ 6 A
Choice (4) is the answer.
Figure 8.3 The circuit of solution of problem 8.3
8.4. Based on Tellegen’s theorem, we can write:
v1 ðt Þei1 ðt Þ þ v2 ðt Þei2 ðt Þ þ v3 ðt Þei3 ðt Þ ¼ ev1 ðt Þi1 ðt Þ þ ev2 ðt Þi2 ðt Þ þ ev3 ðt Þi3 ðt Þ
ð1Þ
From the circuits of Figs. 8.4.1–2, we have:
Fig:1 :
8
v1 ðt Þ ¼ Unknown
>
>
>
>
>
>
>
>
>
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
>
>
>
>
>
>
>
>
<
ev1 ðt Þ ¼ Requested
ev2 ðt Þ ¼ sin ðωt Þ V
>
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
>
>
>
>
>
>
>
<
V ¼0
2
>
I1 ¼ 2e
>
>
>
>
>
>
>
:
I2 ¼ e j60
j30
, Test 2 :
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):
e 1 3 2e j30 þ 1 3 e j60 ) V
e 1 3 2e j30 = e j60 ) V
e 1 = 0:5e j30
4e j45 0:5e j15 þ 0 3 eI2 = V
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
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
>
>
>
>
>
>
>
<
v1 ð t Þ ¼ δ ð t Þ
v2 ð t Þ ¼ 2 V
>
>
i1 ðt Þ ¼ Unknown
>
>
>
>
>
:
i 2 ðt Þ ¼ 0
, Fig:4 :
8
ev1 ðt Þ ¼ 0
>
>
>
>
>
>
>
>
< ev ðt Þ ¼ Unknown
2
>
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
>
>
>
>
>
>
>
<
V1 ¼ 1
V2 ¼ 2
>
>
I1 ¼ Unknown
>
>
>
>
>
:
I2 ¼ 0
, Fig:4 :
8
>
>
>
>
>
>
>
>
<
e1 ¼ 0
V
e 2 ¼ Unknown
V
>
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
2. αþjωR
1þjωR2
2C
3. 1þjωR
1þjω
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 Rj
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
1β
αþRþj
αþRj
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
i 1 ðt Þ
¼
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:
I h21 ¼ 2 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
α
R2 þ jωC
I α þ jωCR2
¼
¼ ) h21 ¼ 2 1
I1 V2 ¼0
1 þ jωCR2
R2 þ jωC
Choice (1) is the answer.
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
y1j
6 j¼1
½Y nodal ¼ 4
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:
v ðt Þ v ðt Þ ðv2 ðt Þ 2v1 ðt ÞÞ
¼ 0 ) i1 ðt Þ ¼ 4v1 ðt Þ v2 ðt Þ
i1 ðt Þ þ 1 þ 1
1
1
ð2Þ
KCL in the right-side node:
v ðt Þ v ðt Þ ðv1 ðt Þ þ 2v1 ðt ÞÞ
¼ 0 ) i2 ðt Þ ¼ 3v1 ðt Þ þ 2v2 ðt Þ
i2 ðt Þ þ 2 þ 2
1
1
ð3Þ
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:
v 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
i1 ðt Þ þ 1
1
1
ð3Þ
10
Solutions of Problems: Two-Port Networks
145
Applying KCL in the right-side node:
v 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
i2 ðt Þ þ 2
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