Basic Engineering Circuit Analysis, Problem Solving Companion

Problem-Solving Companion To accompany

Basic Engineering Circuit Analysis Ninth Edition

J. David Irwin Auburn University

JOHN WILEY & SONS, INC.

Executive Editor Assistant Editor Marketing Manager Senior Production Editor

Bill Zobrist Kelly Boyle Frank Lyman Jaime Perea

Copyright © 2005, John Wiley & Sons, Inc. All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 7504470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212)850-6008, e-mail: [email protected]. ISBN 0-471-74026-8

TABLE OF CONTENTS Preface……………………………………………………………………………………….2 Acknowledgement…………………..……………………………………………………….2 Chapter 1 Problems…………………..……………………………….……………………………………………3 Solutions…………………..…………………………………………………….………………………4

Chapter 2 Problems…………………..…………………………………………………….………………………7 Solutions…………………..…………………………………………………….………………………9

Chapter 3 Problems…………………..…………………………………………………….………………………18 Solutions…………………..…………………………………………………….………………………19

Chapter 4 Problems…………………..…………………………………………………….………………………25 Solutions…………………..…………………………………………………….………………………26

Chapter 5 Problems…………………..…………………………………………………….………………………31 Solutions…………………..…………………………………………………….………………………33

Chapter 6 Problems…………………..…………………………………………………….………………………42 Solutions…………………..…………………………………………………….………………………44

Chapter 7 Problems…………………..…………………………………………………….………………………50 Solutions…………………..…………………………………………………….………………………52

Chapter 8 Problems…………………..…………………………………………………….………………………66 Solutions…………………..…………………………………………………….………………………67

Chapter 9 Problems…………………..…………………………………………………….………………………73 Solutions…………………..…………………………………………………….………………………74

Chapter 10 Problems…………………..…………………………………………………….………………………82 Solutions…………………..…………………………………………………….………………………83

Chapter 11 Problems…………………..…………………………………………………….………………………88 Solutions…………………..…………………………………………………….………………………89

Chapter 12 Problems…………………..…………………………………………………….………………………94 Solutions…………………..…………………………………………………….………………………96

Chapter 13 Problems…………………..…………………………………………………….………………………103 Solutions…………………..…………………………………………………….………………………104

Chapter 14 Problems…………………..…………………………………………………….………………………111 Solutions…………………..…………………………………………………….………………………113

Chapter 15 Problems…………………..…………………………………………………….………………………127 Solutions…………………..…………………………………………………….………………………129

Chapter 16 Problems…………………..…………………………………………………….………………………136 Solutions…………………..…………………………………………………….………………………137 Appendix – Techniques for Solving Linear Independent Simultaneous Equations…………………………...146

2 STUDENT PROBLEM COMPANION To Accompany BASIC ENGINEERING CIRCUIT ANALYSIS, NINTH EDITION By J. David Irwin and R. Mark Nelms PREFACE This Student Problem Companion is designed to be used in conjunction with Basic Engineering Circuit Analysis, 8e, authored by J. David Irwin and R. Mark Nelms and published by John Wiley & Sons, Inc.. The material tracts directly the chapters in the book and is organized in the following manner. For each chapter there is a set of problems that are representative of the end-of-chapter problems in the book. Each of the problem sets could be thought of as a mini-quiz on the particular chapter. The student is encouraged to try to work the problems first without any aid. If they are unable to work the problems for any reason, the solutions to each of the problem sets are also included. An analysis of the solution will hopefully clarify any issues that are not well understood. Thus this companion document is prepared as a helpful adjunct to the book.

3 CHAPTER 1 PROBLEMS 1.1

Determine whether the element in Fig. 1.1 is absorbing or supplying power and how much. -2A

12V + Fig. 1.1 1.2

In Fig. 1.2, element 2 absorbs 24W of power. Is element 1 absorbing or supplying power and how much. 12V + + 6V Fig. 1.2

1.3.

Given the network in Fig.1.3 find the value of the unknown voltage VX. + 4V + 10V 2A 1 2 6A 2A 4A +

12V

+ -

3

8V -

Fig. 1.3

+ -

VX

4 CHAPTER 1 SOLUTIONS 1.1

One of the easiest ways to examine this problem is to compare it with the diagram that illustrates the sign convention for power as shown below in Fig. S1.1(b). -2A i(t)

+

12V

v(t)

+

-

Fig. S1.1(a)

Fig. S1.1(b)

We know that if we simply arrange our variables in the problem to match those in the diagram on the right, then p(t) = i(t) v(t) and the resultant sign will indicate if the element is absorbing (+ sign) or supplying (- sign) power. If we reverse the direction of the current, we must change the sign and if we reverse the direction of the voltage we must change the sign also. Therefore, if we make the diagram in Fig. S1.1(a) to look like that in Fig. S1.1(b), the resulting diagram is shown in Fig. S1.1(c). 2A + (-12V) Fig. S1.1(c) Now the power is calculated as P = (2) (-12) = -24W And the negative sign indicates that the element is supplying power. 1.2

Recall that the diagram for the passive sign convention for power is shown in Fig. S1.2(a) and if p = vi is positive the element is absorbing power and if p is negative, power is being supplied by the element.

5 i + v -

Fig. S1.2(a) If we now isolate the element 2 and examine it, since it is absorbing power, the current must enter the positive terminal of this element. Then P = VI 24 = 6(I) I = 4A The current entering the positive terminal of element 2 is the same as that leaving the positive terminal of element 1. If we now isolate our discussion on element 1, we find that the voltage across the element is 6V and the current of 4A emanates from the positive terminal. If we reverse the current, and change its sign, so that the isolated element looks like the one in Fig. S1.2(a), then P = (6) (-4) = -24W And element 1 is supplying 24W of power. 1.3

By employing the sign convention for power, we can determine whether each element in the diagram is absorbing or supplying power. Then we can apply the principle of the conservation of energy which means that the power supplied must be equal to the power absorbed. If we now isolate each element and compare it to that shown in Fig. S1.3(a) for the sign convention for power, we can determine if the elements are absorbing or supplying power. i +

P = Vi V

-

Fig. S1.3(a) For the 12V source and the current through it to be arranged as shown in Fig. S1.3(a), the current must be reversed and its sign changed. Therefore P12V = (12) (-6) = -72W

6

Treating the remaining elements in a similar manner yields P1 = (4) (6) = 24W P2 = (2) (10) = 20W P3 = (8) (4) = 32W PVX = (VX) (2) = 2VX Applying the principle of the conservation of energy, we obtain And

-72 + 24 + 20 + 32 + 2VX = 0 VX = -2V

7 CHAPTER 2 PROBLEMS 2.1

Determine the voltages V1 and V2 in the network in Fig. 2.1 using voltage division. 2kΩ 12v +-

+

4kΩ

3kΩ

+ V2 -

2kΩ

V1 -

Fig. 2.1 2.2

Find the currents I1 and I0 in the circuit in Fig. 2.2 using current division. 2kΩ I1

6kΩ

3kΩ

12kΩ 9mA

I0

Fig. 2.2 2.3

Find the resistance of the network in Fig. 2.3 at the terminals A-B. 8kΩ 10kΩ 2kΩ A 12kΩ 3kΩ 4kΩ 12kΩ 6kΩ 18kΩ B 6kΩ 3kΩ Fig. 2.3

2.4

Find the resistance of the network shown in Fig. 2.4 at the terminals A-B.

A

B

4kΩ 6kΩ

2kΩ

12kΩ 12kΩ

Fig. 2.4

18kΩ

12kΩ

8 2.5

Find all the currents and voltages in the network in Fig. 2.5. 2kΩ A 10kΩ B I1 48V + -

4kΩ

+ V1 I2

6kΩ I3

I4

+ V2 -

2kΩ 3kΩ I5

I6 + V3 -

4kΩ

Fig. 2.5 2.6

In the network in Fig. 2.6, the current in the 4kΩ resistor is 3mA. Find the input voltage VS. 2kΩ VS

+-

2kΩ

1kΩ 4kΩ 3mA Fig. 2.6

9kΩ 6kΩ

3kΩ

9 CHAPTER 2 SOLUTIONS 2.1

We recall that if the circuit is of the form V1

R1

+ -

+ V0 -

R2

Fig. S2.1(a) Then using voltage division ⎛ R2 V0 = ⎜⎜ ⎝ R1 + R 2

⎞ ⎟⎟ V1 ⎠

That is the voltage V1 divides between the two resistors in direct proportion to their resistances. With this in mind, we can draw the original network in the form 2kΩ 12V

+ -

+ 3kΩ

V1

Fig. S2.1(b)

4kΩ 2kΩ

+ V - 2

The series combination of the 4kΩ and 2kΩ resistors and their parallel combination with the 3kΩ resistor yields the network in Fig. S2.1(c). 12V

2kΩ

+ -

2kΩ

+ V1 -

Fig. S2.1(c) Now voltage division can be sequentially applied. From Fig. S2.1(c). ⎛ 2k ⎞ ⎟⎟12 V1 = ⎜⎜ ⎝ 2k + 2k ⎠ = 6V

Then from the network in Fig. S2.1(b)

10 ⎛ 2k ⎞ ⎟⎟ V1 V2 = ⎜⎜ ⎝ 2k + 4k ⎠ = 2V

2.2

If we combine the 6k and 12k ohm resistors, the network is reduced to that shown in Fig. S2.2(a). I1

2kΩ

3kΩ

4kΩ 9mA

Fig. S2.2(a) The current emanating from the source will split between the two parallel paths, one of which is the 3kΩ resistor and the other is the series combination of the 2k and 4kΩ resistors. Applying current division I1 =

⎞ 9⎛ 3k ⎜⎜ ⎟ k ⎝ 3k + (2k + 4k ) ⎟⎠

= 3mA

Using KCL or current division we can also show that the current in the 3kΩ resistor is 6mA. The original circuit in Fig. S2.2 (b) indicates that I1 will now be split between the two parallel paths defined by the 6k and 12k-Ω resistors. I1 = 3mA 6mA

2kΩ

3kΩ 9mA

6kΩ

12kΩ I0

Fig. S2.2(b) Applying current division again ⎛ 6k ⎞ ⎟⎟ I 0 = I1 ⎜⎜ ⎝ 6k + 12k ⎠ 3 ⎛ 6k ⎞ I0 = ⎜ ⎟ k ⎝ 18k ⎠ = 1mA

Likewise the current in the 6kΩ resistor can be found by KCL or current division to be 2mA. Note that KCL is satisfied at every node.

11 2.3

To provide some reference points, the circuit is labeled as shown in Fig. S2.3(a). 8k 10k 2k A' A" A 12k

4k

3k

18k

6k

12k

B 3k

B'

6k

B"

Fig. S2.3(a) Starting at the opposite end of the network from the terminals A-B, we begin looking for resistors that can be combined, e.g. resistors that are in series or parallel. Note that none of the resistors in the middle of the network can be combined in anyway. However, at the right-hand edge of the network, we see that the 6k and 12k ohm resistors are in parallel and their combination is in series with the 2kΩ resistor. This combination of 6k⎪⎢12k + 2k is in parallel with the 3kΩ resistor reducing the network to that shown in Fig. S2.3(b). 8k 10k A' A" A 12k 4k 2k = 3k (6k 12k + 2k) 18k 6k B 3k B' B" Fig. S2.3(b) Repeating this process, we see that the 2kΩ resistor is in series with the 10kΩ resistor and that combination is in parallel with the12kΩ resistor. This equivalent 6kΩ resistor (2k + 10k)⎪⎢12k is in series with the 3kΩ resistor and that combination is in parallel with the 18kΩ resistor that (6k + 3k)⎪⎢18k = 6kΩ and thus the network is reduced to that shown in Fig. S2.3(c). 8k A' A 4k B

6k

6k B'

Fig. S2.3(c)

12 At this point we see that the two 6kΩ resistors are in series and their combination in parallel with the 4kΩ resistor. This combination (6k + 6k)⎪⎢4k = 3kΩ which is in series with 8kΩ resistors yielding A total resistance RAB = 3k + 8k = 11kΩ. 2.4

An examination of the network indicates that there are no series or parallel combinations of resistors in this network. However, if we redraw the network in the form shown in Fig. S2.4(a), we find that the networks have two deltas back to back. A

B

4k

6k

2k

12k

12k

18k

12k

Fig. S2.4(a) If we apply the ∆→Y transformation to either delta, the network can be reduced to a circuit in which the various resistors are either in series or parallel. Employing the ∆→Y transformation to the upper delta, we find the new elements using the following equations as illustrated in Fig. S2.4(b) 18k 6k R2 R1 R 3 12k Fig. S2.4(b)

R1 =

(6k ) (18k )

= 3kΩ 6k + 12k + 18k (6k ) (12k ) = 2kΩ R2 = 6k + 12k + 18k (12k ) (18k ) = 6kΩ R3 = 6k + 12k + 18k The network is now reduced to that shown in Fig. S2.4(c).

13

A

4k 3k 2k

B

2k

6k 12k

12k

Fig. S2.4(c) Now the total resistance, RAB is equal to the parallel combination of (2k + 12k) and (6k + 12k) in series with the remaining resistors i.e. RAB = 4k + 3k + (14k⎪⎢18k) + 2k = 16.875kΩ If we had applied the ∆→Y transformation to the lower delta, we would obtain the network in Fig. S2.4(d). 4k A 6k 18k 4k 4k B

2k

4k

Fig. S2.4(d) In this case, the total resistance RAB is RAB = 4k + (6k + 4k)⎪⎢(18k + 4k) + 4k +2k = 16.875kΩ which is, of course, the same as our earlier result. 2.5

Our approach to this problem will be to first find the total resistance seen by the source, use it to find I1 and then apply Ohm’s law, KCL, KVL, current division and voltage division to determine the remaining unknown quantities. Starting at the opposite end of the network from the source, the 2k and 4k ohm resistors are in series and that combination is in parallel with the 3kΩ resistor yielding the network in Fig. S2.5(a).

14

10k

A I1 48V +-

I2 + V1 -

2k

6k 4k I3

B I4 + V2 -

2k

Fig. S2.5(a) Proceeding, the 2k and 10k ohm resistors are in series and their combination is in parallel with both the 4k and 6k ohm resistors. The combination (10k + 2k)⎪⎢6k⎪⎢4k = 2kΩ. Therefore, this further reduction of the network is as shown in Fig. S2.5(b). 2k 48 +-

I1

+ V1 -

2k

Fig. S2.5(b) Now I1 and V1 can be easily obtained. I1 =

48 = 12mA 2k + 2k

And by Ohm’s law V1 = 2kI1 = 24V or using voltage division ⎛ 2k ⎞ ⎟⎟ V1 = 48 ⎜⎜ ⎝ 2k + 2k ⎠ = 24V

once V1 is known, I2 and I3 can be obtained using Ohm’s law V1 24 = = 6mA 4k 4k V 24 I3 = 1 = = 4mA 6k 6k I2 =

I4 can be obtained using KCL at node A. As shown on the circuit diagram. I1 = I2 + I3 + I4

15

12 6 4 = + + I4 k k k 2 I 4 = = 2mA k The voltage V2 is then V2 = V1 - 10kI4 ⎛2⎞ = 24 − (10k ) ⎜ ⎟ ⎝k⎠ = 4V or using voltage division ⎛ 2k ⎞ ⎟⎟ V2 = V1 ⎜⎜ ⎝ 10k + 2k ⎠ ⎛1⎞ = 24 ⎜ ⎟ ⎝6⎠ = 4V Knowing V2, I5 can be derived using Ohm’s law

V2 3k 4 = mA 3

I5 =

and also V2 2 k + 4k 2 = mA 3

I6 =

current division can also be used to find I5 and I6. ⎛ 2k + 4 k ⎞ ⎟⎟ I 5 = I 4 ⎜⎜ ⎝ 2k + 4k + 3k ⎠ 4 = mA 3 and

16 ⎛ ⎞ 3k ⎟⎟ I 6 = I 4 ⎜⎜ ⎝ 3k + 2k + 4k ⎠ 2 = mA 3 Finally V3 can be obtained using KVL or voltage division V3 = V2 − 2kI 6 ⎛ 2 ⎞ = 4 − 2k ⎜ ⎟ ⎝ 3k ⎠ 8 = V 3

and ⎛ 4k ⎞ ⎟⎟ V3 = V2 ⎜⎜ ⎝ 4 k + 2k ⎠ 8 = V 3 2.6

The network is labeled with all currents and voltages in Fig. S2.6. + V4 - I5 A + V2 - I3 B 2k VS + -

+ V3 -

I4

1k

2k 3 k

4k 6k

+ V1 I1

9k

I2 3k

Fig. S2.6 Given the 3mA current in the 4kΩ resistor, the voltage ⎛3⎞ V1 = ⎜ ⎟ (4k ) = 12V ⎝k⎠ Now knowing V1, I1 and I2 can be obtained using Ohm’s law as V1 12 = = 2mA 6k 6k V1 12 = = 1mA I2 = 9k + 3k 12k I1 =

Applying KCL at node B

17

3 + I1 + I 2 k = 6mA

I3 =

Then using Ohm’s law V2 = I3 (1k) = 6V KVL can then be used to obtain V3 i.e. V3 = V2 + V1 = 6 + 12 = 18V Then V3 2k = 9mA

I4 =

And I5 = I3 + I 4 6 9 + k k = 15mA =

using Ohm’s law V4 = (2k) I5 = 30V and finally VS = V4 + V3 = 48V

18 CHAPTER 3 PROBLEMS

3.1

Use nodal analysis to find V0 in the circuit in Fig. 3.1. 2mA

1kΩ

+

1kΩ 1kΩ 2kΩ

12V +-

V0 -

Fig. 3.1 3.2

Use loop analysis to solve problem 3.1

3.3

Find V0 in the network in Fig. 3.3 using nodal analysis. 12V -+ 2kΩ 2kΩ 1kIX +1kΩ IX

+ V0 -

Fig. 3.3 3.4

Use loop analysis to find V0 in the network in Fig. 3.4. 2IX 1kΩ

1kΩ +

4mA

1kΩ IX

Fig. 3.4

2kΩ

V0 -

19 CHAPTER 3 SOLUTIONS

3.1

Note that the network has 4 nodes. If we select the node on the bottom to be the reference node and label the 3 remaining non-reference nodes, we obtain the network in Fig. S3.1(a). 2 k V0

V2

V1 12 +-

1k

1k 1k

2k

Fig. S3.1(a) Remember the voltages V1, V2 and V0 are measured with respect to the reference node. Since the 12V source is connected between node V1 and the reference, V1 = 12V regardless of the voltages or currents in the remainder of the circuit. Therefore, one of the 3 linearly independent equations required to solve the network (N – 1, where N is the number of nodes) is V1 = 12 The 2 remaining linearly independent equations are obtained by applying KCL at the nodes labeled V2 and V0 . Summing all the currents leaving node V2 and setting them equal to zero yields V2 − V1 V2 V2 − V0 + + =0 1k 1k 1k

Similarly, for the node labeled V0 , we obtain − 2 V0 − V2 V0 + + =0 k 1k 2k

The 3 linearly independent equations can be quickly reduced to ⎛3⎞ ⎛ 1 ⎞ 12 V2 ⎜ ⎟ − V0 ⎜ ⎟ = ⎝k⎠ ⎝k⎠ k ⎛1⎞ ⎛ 3 ⎞ 2 − V2 ⎜ ⎟ + V0 ⎜ ⎟ = ⎝k⎠ ⎝ 2k ⎠ k

20

or 3V2 – V0 = 12 3 − V2 + V0 = 2 2 40 36 V and V0 = V. 7 7 We can quickly check the accuracy of our calculations. Fig. S3.1(b) illustrates the circuit and the quantities that are currently known. 14 A 7k

Solving these equations using any convenient method yields V2 =

40 V I3 7

84 V 7 + -

1k I1

1k

1k

I2

2k

36 V 7 I4

Fig. S3.1(b) All unknown branch currents can be easily calculated as follows.

I1

I2

I3

I4

84 40 − 7 7 = 44 A = 1k 7k 40 40 = 7 = A 1k 7k 40 36 − 7 7 = 4 A = 1k 7k 36 18 = 7 = A 2k 7 k

KCL is satisfied at every node and thus we are confident that our calculations are correct. 3.2

The network contains 3 “window panes” and therefore 3 linearly independent loop equations will be required to determine the unknown currents and voltages. To begin we arbitrarily assign the loop currents as shown in Fig. S3.2.

21 3 A k I3 12V +-

1kΩ

1kΩ 1kΩ

I1

I2

+ 2kΩ

V0 -

Fig. S3.2 The equations for the loop currents are obtained by employing KVL to the identified loops. For the loops labeled I1 and I2, the KVL equations are -12 + 1k(I1 – I3) + 1k (I1 – I2) = 0 and 1k(I2 –I1) + 1k(I2 – I3) + 2kI3 = 0 In the case of the 3rd loop, the current I3 goes directly through the current source and therefore I3 =

2 k

Combining these equations yields 2kI1 – 1kI2 = 14 -1kI1 + 4kI2 = 2 Solving these equations using any convenient method yields I1 =

58 18 A and I2 = A. 7k 7k

Then V0 is simply V0 = 2kI3 36 = V 7 Once again, a quick check indicates that KCL is satisfied at every node. Furthermore, KVL is satisfied around every closed path. For example, consider the path around the two “window panes” in the bottom half of the circuit. KVL for this path is -12 + 1k(I1 – I3) + 1k(I2 – I3) + 2kI3 = 0

22 ⎛ 58 14 ⎞ ⎛ 18 14 ⎞ ⎛ 18 ⎞ − 12 + 1k ⎜ − − ⎟ + 1k ⎜ ⎟ + 2k ⎜ ⎟ = 0 ⎝ 7k 7k ⎠ ⎝ 7k 7k ⎠ ⎝ 7k ⎠ 0=0 3.3

The presence of the two voltage sources indicates that nodal analysis is indeed a viable approach for solving this problem. If we select the bottom node as the reference node, the remaining nodes are labeled as shown in Fig. S3.3(a). V0-12 1kIX V0 -+ 2k 12 + 2k 1k IX Fig. S3.3(a) The node at the upper right of the circuit is clearly V0, the output voltage, and because the 12V source is tied directly between this node and the one in the center of the network, KVL dictates that the center node must be V0 –12 e.g. if V0 = 14V, then the voltage at the center node would be 2V. Finally, the node at the upper left is defined by the dependent source as 1kIX. If we now treat the 12V source and its two connecting nodes as a supernode, the current V − 12 − 1kI X , the current down through the center leaving the supernode to the left is 0 2k V − 12 and the current leaving the supernode on the right edge is leg of the network is 0 2k V0 . Therefore, KCL applied to the supernode yields 1k V0 − 12 − 1kI X V0 − 12 V0 + + =0 2k 2k 1k

Furthermore, the control variable IX is defined as IX =

V0 − 12 2k

combining these two equations yields V0 =

36 V 7

23 The voltages at the remaining non-reference nodes are V0 − 12 =

36 36 84 − 48 − 12 = − = V 7 7 7 7

And ⎛ − 48 ⎞ ⎜ ⎟ − 24 ⎛ V0 − 12 ⎞ 1kI X = 1k ⎜ V ⎟ = 1k ⎜ 7 ⎟ = 7 ⎜ 2k ⎟ ⎝ 2k ⎠ ⎜ ⎟ ⎝ ⎠ The network, labeled with the node voltages, is shown in Fig. S3.3(b) -48 V -24 36 V V 7 7 7 -+ I1 2k I3 I2 12 + 1k 2k

Fig. S3.3(b) Then − 24 ⎛ − 48 ⎞ −⎜ ⎟ 7 ⎝ 7 ⎠ = 12 A I1 = 2k 7k − 48 24 A I2 = 7 = − 2k 7k 36 36 I3 = 7 = 1k 7k Note carefully that KCL is satisfied at every node. 3.4

Because of the presence of the two current sources, loop analysis is a viable solution method. We will select our loop currents (we need 3 since there are 3 “window panes” in the network) so that 2 of them go directly through the current sources as shown in Fig. S3.4(a).

24 2IX I1 4 k

+

1k

1k 1k I2

IX

2k

I3

V0 -

Fig. S3.4(a) Therefore, two of the three linearly independent equations needed are I1 = 2IX = 2(I2 – I3) 4 I2 = k Applying KVL to the third loop yields 1k(I3 –I2) + 1k(I3 – I1) + 2kI3 = 0 combining equations yields I3 = 2mA And then V0 = 4V And I1 = 4mA Using these values, the branch currents are shown in Fig. S3.4(b) 4 k

2 k 2 4 k

2 k

k

Fig. S3.4(b) Although one branch of the network are no current, KCL is satisfied at every node.

25 CHAPTER 4 PROBLEMS

4.1

Derive the gain equation for the nonideal noninverting op-amp configuration and show that it reduces to the ideal gain equation if Ri and A are very large, i.e. greater than 106.

4.2

Determine the voltage gain of the op-amp circuit shown in Fig. 4.2. +

25kΩ 25kΩ vs

+ -

50kΩ

+ vo

50kΩ

3kΩ

6kΩ

-

Fig. 4.2 4.3

Using the ideal op-amp model show that for the circuit shown in Fig. 4.3, the output voltage is directly related to any small change ∆R. R

∆R

R R vs

+ -

+ R

+ vo -

Fig. 4.3 4.4

Given an op-amp and seven standard 12kΩ resistors, design an op-amp circuit that will produce an output of 1 v 0 = - 2v1 - v 2 2

26 CHAPTER 4 SOLUTIONS

4.1

The noninverting op-amp circuit is shown in Fig. S4.1(a). v1N

+

vo

RF RI

Fig. S4.1(a) The nonideal model is RF + v+ v1N

+ ve v-

Ri RI

-

Ro

+ -

Ave

Fig. S4.1(b) or +

v1N

+ -

ve - v1 Ri

RF RI

Ro + -

+ vo

Ave

Fig. S4.1(c) The node equations for this circuit are v1 - v1N v1 v1 - v o + + =0 RF RI Ri

-

27

v o - v1 v o - Av e + =0 RF Ro

v e = v1N - v1 or ⎡1 ⎡ 1 ⎤ v1N 1 ⎤ 1 + ⎢ + ⎥ v1 - ⎢ ⎥ v o = Ri ⎣Ri RI RF ⎦ ⎣RF ⎦ ⎡ 1 ⎡ 1 Av1N 1 ⎤ A⎤ −⎢ − + ⎥ v1 + ⎢ ⎥ vo = Ro ⎣RF Ro ⎦ ⎣RF Ro ⎦

Following the development on page 141 of the text yields ⎡ 1 1 1 ⎤ Av1N A ⎤ v1N ⎡ 1 +⎢ + + ⎥ ⎢ ⎥ RF Ro ⎦ Ri ⎣Ri RI RF ⎦ Ro ⎣ vo = ⎛ 1 1 ⎞ 1 ⎛ 1 A 1 1 ⎞⎛ 1 ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎜ + + + ⎝ Ri RI RF ⎠ ⎝ RF Ro ⎠ RF ⎝ RF Ro

⎞ ⎟⎟ ⎠

assuming Ri → ∞, the equation reduces to ⎛ 1 1 ⎞⎛ A ⎞ ⎟ ⎜⎜ ⎟⎜ + R I R F ⎟⎠ ⎜⎝ R o ⎟⎠ vo ⎝ = v 1N ⎛ 1 1 ⎞ 1 ⎛ 1 A 1 ⎞⎛ 1 ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎜ + + ⎝ RI RF ⎠ ⎝ RF Ro ⎠ RF ⎝ RF Ro

⎞ ⎟⎟ ⎠

Now dividing both numerator and denominator by A and using A → ∞ yields ⎛ 1 1 ⎞ ⎜⎜ ⎟ + R I R F ⎟⎠ vo ⎝ = v 1N ⎛ 1 ⎞⎛ 1 ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ Ro ⎠⎝ RF ⎠ 1 Ro

=1+

which is the ideal gain equation.

RF RI

28 4.2

The network in Fig. 4.2 can be reduced to that shown in Fig. S4.2(a) by combining resistors. +

25kΩ vs

+ -

-

50kΩ

75kΩ

+ vo

2kΩ

-

Fig. S4.2(a) v+ is determined by the voltage divider at the input, i.e. ⎡ 75k ⎤ 3 v+ = vs ⎢ = vs ⎣ 25k + 75k ⎥⎦ 4 The op-amp is in a standard noninverting configuration and the gain is 1 + 50k Therefore

(

v o = (26) 3 v s 4

)

and vo = 19.5 vs

4.3

The node equations for the circuit in Fig. 4.3 are vs - v- vo - v+ =0 R R + ∆R vs - v+ v+ = R R v− = v+

Then

2k

= 26 .

29

v− = v+ =

vs 2

vs - 1 vs vo - 1 vs 2 + 2 =0 R R + ∆R

vs vo vs + =0 2R R + ∆R 2 (R + ∆R ) vo = vs R + ∆R

⎡ 1 1 ⎤ ⎢ 2 (R + ∆R ) - 2R ⎥ ⎣ ⎦

⎡ ⎤ − ∆R = vs ⎢ ⎥ ⎣ (2R )(R + ∆R ) ⎦

⎡ - ∆R ⎤ vo = vs ⎢ ⎣ 2R ⎥⎦ v o - ∆R = vs 2R

4.4

A weighted-summer circuit shown in Fig. S4.4(a) can be used to produce an output of the R R v1 v2 . form v o = R1 R2 R

R1

v1

+ -

v2 + -

R2

+

Fig. S4.4(a) Note that R 1 R = 2 and = R2 2 R1 Therefore if

+ vo -

30

R = 24kΩ (two 12kΩ resistors in series) R1 = 12kΩ R2 = 48kΩ (four 12kΩ resistors in series) then the design conditions are satisfied.

31 CHAPTER 5 PROBLEMS

5.1

Find V0 in the circuit in Fig. 5.1 using the Principle of Superposition. 6kΩ

12V

8kΩ +

+

4kΩ

V0

6mA

-

Fig. 5.1 5.2

Solve problem 5.1 using source transformation.

5.3.

Find V0 in the network in Fig. 5.3 using Thevenin’s Theorem. 4mA 6kΩ +

4kΩ

+ -

3kΩ

2kΩ

V0 -

Fig. 5.3 5.4

Find I 0 in the circuit in Fig. 5.4 using Norton’s Theorem.

12V

+ -

3kΩ 2kΩ

6kΩ

2mA

I0 Fig. 5.4 5.5

For the network in Fig. 5.5, find RL for maximum power transfer and the maximum power that can be transferred to this load.

32

+ VX

-

3kΩ 12V

+ -

6kΩ

2VX

+ -

Fig. 5.5

RL

33 CHAPTER 5 SOLUTIONS

5.1

To apply superposition, we consider the contribution that each source independently makes to the output voltage V0 . In so doing, we consider each source operating alone and we zero the other source(s). Recall, that in order to zero a voltage source, we replace it with a short circuit since the voltage across a short circuit is zero. In addition, in order to zero a current source, we replace the current source with an open circuit since there is no current in an open circuit. Consider now the voltage source acting alone. The network used to obtain this contribution to the output V0 is shown in Fig. S5.1(a). 8kΩ

6kΩ 12V

+

+

V0

4kΩ 6mA

-

Fig. S5.1(a) Then V0 ′ (only a part of V0 ) is the contribution due to the 12V source. Using voltage division ⎞ ⎛ 4k ⎟⎟ V0 = − 12⎜⎜ ⎝ 4k + 6k + 8k ⎠ =

−8 V 3

The current source’s contribution to V0 is obtained from the network in Fig. S5.1(b).

I0 6k

8k

6 k

4k

+ V0′′ -

Fig. S5.1(b) Using current division, we find that I0 =

⎞ 6⎛ 6k ⎟ ⎜⎜ k ⎝ 6k + 8k + 4k ⎟⎠

= 2mA

Then

34

″ V0 = 4kI 0 = 8V Then superposition states that ′ ″ V0 = V0 + V0 = 5.2

−8 16 +8= V 3 3

Recall that when employing source transformation, at a pair of terminals we can exchange a voltage source VS in series with a resistor RS for a current source Ip in parallel with a resistor Rp and vice versa, provided that the following relationships among the parameters exist. VS RS Rp = RS Ip =

Now the original circuit is shown in Fig. S5.2(a). 8kΩ

6kΩ 12V

+

+

V0

4kΩ 6mA

-

Fig. S5.2(a) Note that we have a 12V source in series with a 6kΩ resistor that can be exchanged for a current source in parallel with the resistor. This appears to be a viable exchange since we will then have two current sources in parallel which we can add algebraically. Performing the exchange yields the network in Fig. S5.2(b).

8k 2 k

6k

6 k

+ 4k

V0 -

Fig. S5.2(b) Note that the voltage source was positive at the bottom terminal and therefore the current source points in that direction. Adding the two parallel current sources reduces the network to that shown in Fig. S5.2(c).

35

+

8k 4 k

6k

4k

V0 -

Fig. S5.2(c) At this point we can apply current division to obtain a solution. For example, the current in the 4kΩ resistor is I 4k = =

4 k

⎞ ⎛ 6k ⎟⎟ ⎜⎜ ⎝ 6k + 8k + 4k ⎠

4 mA 3

Then V0 = (I 4 k ) (4k ) =

16 V 3

However, we could also transform the current source and the parallel 6kΩ resistor into a voltage source in series with the 6kΩ resistor before completing the solution. If we make this exchange, then the network becomes that shown in Fig. S5.2(d). 6k 24V

+

8k

+ -

4k

V0 -

Fig. S5.2(d) Then using voltage division

⎞ ⎛ 4k ⎟⎟ V0 = 24 ⎜⎜ ⎝ 4k + 6k + 8k ⎠ 16 = V 3 5.3

Since the network contains no dependent source, we will simply determine the open circuit voltage, V0 c , and with the sources in the network made zero, we will look into the

36 open circuit terminals and compute the resistance at these terminals, RTH. The open circuit voltage is determined from the network in Fig. S5.3(a). 4 k

I1

12V +-

4k

6k 3k

I2

+ V2 -

-

V1

+

+ V0 C

-

Fig. S5.3(a) Note the currents and voltages labeled in the network. First of all, note that V0 C = V1 + V2 Therefore, we need only to determine these voltages. Clearly, the voltage V1 is V1 = I1 (4k) = 16V However, to find V2 we need I2. KVL around the loop I2 yields -12 + 6k (I2 – I1) + 3kI2 = 0 or 4 − 12 + 6k (I 2 − ) + 3kI 2 = 0 k 4 I 2 = = 4mA k Now V0 C = V1 + V2 = 4kI1 + 3kI 2 = 28V

The Thevenin equivalent resistance is found by zeroing all sources and looking into the open circuit terminals to determine the resistance. The network used for this purpose is shown in Fig. S5.3(b).

37 6k

4k

R TH

3k

Fig. S5.3(b) From the network we see that the 6k and 3k Ohm resistors are in parallel and that combination is in series with the 4kΩ resistor. Thus RTH = 4k + 3k⎟ ⎜6k = 6kΩ Therefore, the Thevenin equivalent circuit consists of the 28V source in series with the 6kΩ resistor. If we connect the 2kΩ resistor to this equivalent network we obtain the circuit in Fig. S5.3(c).

28V

+ -

6kΩ

+ 2kΩ

V0 -

Fig. S5.3(c) Then using voltage division ⎛ 2k ⎞ ⎟⎟ V0 = 28 ⎜⎜ ⎝ 2k + 6k ⎠ = 7V

5.4

In this network, the 2kΩ resistor represents the load. In applying Norton’s Theorem we will replace the network without the load by a current source, the value of which is equal to the short-circuit current computed from the network in Fig. S5.4(a), in parallel with the Thevenin equivalent resistance determined from Fig. S5.4(b).

12V

+ -

3k

ISC

Fig. S5.4(a)

6k

2 k

38

3k 6k

R TH Fig. S5.4(b)

with reference to Fig. S5.4(a), all current emanating from the 12V source will go through the short-circuit. Likewise, all the current in the 2mA current source will also go through the short-circuit so that I SC =

12 2 − = 2mA 3k k

If this statement is not obvious to the reader, then consider the circuit shown in Fig. S5.4(c). I ISC

R

Fig. S5.4(c) Knowing that the resistance of the short-circuit is zero, we can apply current division to find ISC ⎛ R ⎞ ⎟⎟ I SC = I ⎜⎜ ⎝R + 0⎠ =I

indicating that all the current in this situation will go through the short-circuit and none of it will go through the resistor. From Fig. S5.4(b) we find that the 3k and 6k Ohm resistors are in parallel and thus RTH = 3k⎟ ⎜6k = 2kΩ Now the Norton equivalent circuit consists of the short-circuit current in parallel with the Thevenin equivalent resistance as shown in Fig. S5.4(d).

2mA

2kΩ Fig. S5.4(d)

39

Remember, at the terminals of the 2kΩ load, this network is equivalent to the original network with the load removed. Therefore, if we now connect the load to the Norton equivalent circuit as shown in Fig. S5.4(e), the load current I 0 can be calculated via current division as I0 =

2 ⎛ 2k ⎞ ⎜ ⎟ k ⎜⎝ 2k + 2k ⎟⎠

= 1 mA

2 k

2k

2k

I0

Fig. S5.4(e) 5.5

The solution of this problem involves finding the Thevenin equivalent circuit at the terminals of the load resistor RL and setting RL equal to the Thevenin equivalent resistance RTH. To determine the Thevenin equivalent circuit, we first find the open circuit voltage as shown in Fig. S5.5(a). + VX′ 12

+ -

3k

+ -

2VX′

6k

+

V0 C -

Fig. S5.5(a) We employ the prime notation on the control variable Vx since the circuit in Fig. S5.5(a) is different than the original network. Applying KVL to the left side of the network yields -12 +Vx′ + 2Vx′ = 0 Vx′ = 4V Then the open circuit voltage is V0 C = 2VX

′

= 8V since there is no current in the 6kΩ resistor and therefore no voltage drop across it.

40

Because of the presence of the dependent source we cannot simply look back into the open circuit terminals, with all independent sources made zero, and determine the Thevenin equivalent resistance. We must determine the short-circuit current, ISC and determine RTH from the expression R TH =

V0 C I SC

ISC is found from the circuit in Fig. S5.5(b). + VX′′ 3k 12

+ -

+ -

6k

2VX′′

ISC

Fig. S5.5(b) Once again, using KVL -12 +Vx″ + 2Vx″ = 0 Vx″ = 4 Then, since the dependent source 2Vx″ = 8V is connected directly across the 6kΩ resistor I SC

″ 2VX 2 = = mA 6k 3

and R TH =

V0 C 8 = = 12kΩ 2 I SC 3k

Hence, for maximum power transfer RL = RTH = 12kΩ And the remainder of the problem involves finding the power absorbed by the 12kΩ load, PL. From the network in Fig. S5.5(c) we find that

41 PL = I 2L R L 2

⎛ 8 ⎞ =⎜ ⎟ (12k ) ⎝ 24k ⎠ = 1.33 mW 12kΩ 8V

IL

+ -

12kΩ = R2

Fig. S5.5(c)

42 CHAPTER 6 PROBLEMS

6.1

If the voltage across a 10µF capacitor is shown in Fig. 6.1, derive the waveform for the capacitor current.

4 2

0

2

4

6

8

t (ms)

Fig. 6.1 6.2

If the voltage across a 100mH inductor is shown in Fig. 6.2, find the waveform for the inductor current. v(t) (mV)

4 0.1 -2

0.2

0

t(s)

Fig. 6.2 6.3

Find the equivalent capacitance of the network in Fig. 6.3 at the terminals A-B. All capacitors are 6µF. A

C eq

B

Fig. 6.3 6.4

Find the equivalent inductance of the network in Fig. 6.4 at the terminals A-B. All inductors are 12mH.

43 A

B

Fig. 6.4

44 CHAPTER 6 SOLUTIONS

6.1

The equations for the waveforms in the 4 two millisecond time intervals are listed below. v (t ) = mt + b 2 = t 2 × 10 −3 =2 2 = −2+ t 2 × 10 −3 4 = + 16 − t 2 × 10 −3 =0

0 ≤ t ≤ 2ms 2 ≤ t ≤ 4ms 4 ≤ t ≤ 6ms 6 ≤ t ≤ 8ms t < 0, t > 8ms

Note that within each interval we have simply written the equation of a straight line using the expression y = mx + b or equivalently v(t) = mt + b where m is the slope of the line and b is the point at which the line intersects the v(t) axis. The equation for the current in a capacitor is i( t ) = C

dv(t ) dt

Using this expression we can compute the current in each interval. For example, in the interval from 0 ≤ t ≤ 2ms i(t ) = (10 × 10 − 6 )

d dt

⎛ 2 ⎜⎜ −3 ⎝ 2 × 10

⎞ t ⎟⎟ ⎠

0 ≤ t ≤ 2ms

= 10mA

i(t ) = (10 × 10 − 6 )

d (2) dt

2 ≤ t ≤ 4ms

=0

i(t ) = (10 × 10 − 6 )

d dt

⎛ 2 ⎜⎜ − 2 + 2 × 10 −3 ⎝

d dt

⎛ 4 ⎜⎜16 − 2 × 10 −3 ⎝

⎞ t ⎟⎟ ⎠

4 ≤ t ≤ 6ms

= 10mA

i(t ) = (10 × 10 − 6 )

⎞ t ⎟⎟ ⎠

6 ≤ t ≤ 8ms

= − 20mA

The waveform for the capacitor current is shown in Fig. S6.1.

45 i(t) (mA) 4 2 -2

0

2

4

6

t(ms)

8

-4

Fig. S6.1 6.2

The general expression for the current in an inductor is i(t ) = i(t 0 ) + ∫t v(x ) dx t

0

In order to evaluate this function we need the equation of the voltage waveform in the two time intervals 0 ≤ t ≤ 0.1s and 0.1 ≤ t ≤ 0.2s. In the first case, the voltage function is a straight line and the function passes through the origin of the graph. The equation of a straight line on this graph is v(t) = mt + b where m is the slope of the line and b is the point at which the line intersects the v(t) axis. 4 × 10 −3 , the equation of the line is Since the slope is 0 .1 4 × 10 −3 t v (t ) = 0 .1

where v(t) is measured in volts and time is measured in seconds i.e., the slope has units of volts/sec. Therefore, i (t ) = i (0 ) +

1 t 4 × 10 −3 × dx ∫0 L 0 .1

since there is no initial current in the inductor i(t) = 0 and

i (t ) = 10∫0 4 × 10 −2 × dx t

or

1 = 10 L

46 x2 t 0 2 = 0.2 t 2 A = 200 t 2 mA

i (t ) = 0.4 ∫0 × dx = 0.4 t

Since the initial current for the second time interval is determined by the value of the current at the end of the first time interval we calculate i (t )

t = 0.1

= 200t 2

t = 0.1

mA

= 2mA Therefore, in the time interval 0.1 ≤ t ≤ 0.2s i (t ) = i (0.1) +

1 t ∫0.1 v (x ) dx L

Note that in this interval v(x) is a constant –2mV or –2 × 10-3V. Hence, i (t ) = 2 × 10 −3 + 10 ∫0.1 (− 2 × 10 −3 ) dx t

= 2 × 10 −3 − 20 × 10 −3 × = (4 − 20 t ) mA

t 0 .1

If we now plot the two functions for the current within their respective time intervals we obtain the plot in Fig. S6.2.

2

0

0.1

0.2

t(s)

Fig. S6.2 6.3

To begin our analysis we first label all the capacitors and nodes in the network as shown in Fig. S6.3(a).

47 C1

A

C2

C

D

C3

C4

C5

C6 B

Fig. S6.3(a) First of all, the reader should note that all the nodes have been labeled, i.e., there are no other nodes. As we examine the topology of the network we find that since C3 and C5 are both connected to node D the network can be redrawn as shown in Fig. S6.3(b). C D A C2 C1 C4 C6

C3

C5

B

Fig. S6.3(b) Clearly, C5 and C6 are in parallel and their combination we will call C56 = C5⎟ ⎜C6. Combining these two capacitors reduces the network to that shown in Fig. S6.3(c). C1 C C 2 A C4

C 56

D

C3

B

Fig. S6.3(c) At this point we find that C2 and C4 are in parallel and their combination, which we call C24 = C2⎟ ⎜C4, reduces the network to that shown in Fig. S6.3(d). C1 C C 24 D A

C56 B

Fig. S6.3(d)

C3

48 If we now use the given capacitor values, the network becomes that shown in Fig. S6.3(e). 12µF 6µF A 6µF

12µF B

Fig. S6.3(e) Starting at the opposite end of the network from the terminals A-B and combining elements we find that 6µF in series with 12µF is 4µF and this equivalent capacitance is in parallel with 12µF yielding 16µF, which in turn is in series with 6µF producing a total capacitance of C eq = 6µF 16µF

= 4.36µF

6.4

To aid our analysis, we will first label all inductors and nodes as shown in Fig. S6.4(a). L1 L2 C A L3

L5

L4 L6 B

Fig. S6.4(a) Note carefully that all the nodes have been labeled. Once readers recognize that there are no other nodes, they are well on their way to reducing the network since this node recognition provides data indicating which elements are in series or parallel. For example, since one end of L4 is connected to node B, the network can be redrawn as shown in Fig. S6.4(b). L1 L2 C A L5 L3

L4 B

Fig. S6.4(b)

L6

49

This diagram clearly indicates that L2 and L5 are in parallel. In addition, L4 and L6 are in parallel. Therefore, if we combine elements so that L25 = L2⎟ ⎜L5 and L46 = L4⎟ ⎜L6, then the circuit can be reduced to that in Fig. S6.4(c). L 25 L1 C A L3

L 46

B

Fig. S6.4(c) However, we note now if we did not see it earlier that L25 is in parallel with L46 so that the network can be reduced to that shown in Fig. S6.4(d). L1 C A L3

L 2456

B

Fig. S6.4(d) Where L2456 = L25⎟ ⎜L46. Since all inductors are 12mH, L2456 = 3mH which is in series with 12mH and that combination is in parallel with 12mH yielding LAB = 12mH⎟ ⎜15mH = 6.66mH

50 CHAPTER 7 PROBLEMS

7.1

Use the differential equation approach to find i 0 (t ) for t > 0 in the circuit in Fig. 7.1 and plot the response including the time interval just prior to opening the switch. 2kΩ -+ t=0 12V 3kΩ 3kΩ 6kΩ 12kΩ 150µ F i 0 (t ) Fig. 7.1

7.2

Use the differential equation approach to find i(t) for t > 0 in the circuit in Fig. 7.2 and plot the response including the time interval just prior to opening the switch. 5kΩ t=0 2kΩ i (t ) 5mA

1kΩ

1kΩ 1mH

Fig. 7.2 7.3

Use the step-by-step technique to find v 0 (t ) for t > 0 in the circuit in Fig. 7.3.

6kΩ

50µF 6kΩ

+ v 6kΩ 0 (t ) -

t=0

6kΩ + 12V Fig. 7.3 7.4

Use the step-by-step method to find v 0 (t ) for t > 0 in the network in Fig. 7.4.

51 1 H 3 t=0

4Ω +

2Ω 12V +-

v 0 (t )

2Ω 12V +

-

Fig. 7.4 7.5

Given the network in Fig. 7.5, find (a) the differential equation that describes the current i(t) (b) the characteristic equation for the network (c) the network’s natural frequencies (d) the type of damping exhibited by the circuit (e) the general expression for i(t) i (t )

vS (t ) +-

2H 14Ω 0.05F

Fig. 7.5 7.6

Find i 0 (t ) for t > 0 in the circuit in Fig. 7.6 and plot the response including the time interval just prior to closing the switch. 1 F 120 t=0

12V

+ -

i 0 (t )

24Ω 24Ω

Fig. 7.6

2.4H

24Ω

52 CHAPTER 7 SOLUTIONS

7.1

We begin our solution by redrawing the network and labeling all the components as shown in Fig. S7.1(a) R 3 i X (t ) -+ t=0 12V R1 R4 R5 C R2 i 0 (t )

Fig. S7.1(a) In order to determine the initial condition of the network prior to switch action, we must determine the initial voltage across the capacitor. A circuit, which can be used for this purpose, is shown in Fig. S7.1(b). iX -+ 12V R1 = 3kΩ

R2 = 12kΩ

+ vC - 150µ F

R6 = 4kΩ

Fig. S7.1(b) Where we have combined the resistors at the right end of the network so that R6 = R3 + R4⎥⎜R5 = 2k + 3k⎥⎜6k = 4kΩ In the steady-state condition before the switch is thrown, the capacitor looks like an opencircuit and therefore vC(0-) is the voltage across the parallel combination of R2 and R6. Using voltage division, the 12V source will produce the voltage ⎛ R2 R6 ⎞ ⎟ v C (0 − ) = 12 ⎜⎜ ⎟ R R R + 2 6 ⎠ ⎝ 1 ⎛ 3k ⎞ ⎟⎟ = 6V = 12 ⎜⎜ ⎝ 3k + 3k ⎠

Now that the initial voltage across the capacitor is known, we can find the initial value of the current i 0 (t ) . From Fig. S7.1(b) we see that i x (0 − ) =

v C (0 − ) 6 = = 1.5mA R6 4k

53 Then, using current division as shown in Fig. S7.1(a), i 0 (0 − ) =

i x (0 − ) (R 5 ) R4 + R5

⎛ 3 ⎞ ⎜ ⎟ (6k ) 2k = 1mA =⎝ ⎠ 3k + 6k

The parameters for t < 0 are now known. For the time interval t > 0, the network is reduced to that shown in Fig. S7.1(c). iX R 2 = 12kΩ

+ vC(t ) -

150µ F

R 6 = 4 kΩ

Fig. S7.1(c) Applying KCL to this network yields C

dv C (t ) v C (t ) v C (t ) + + =0 dt R2 R6

or using the parameter values dv C (t ) 20 + v C (t ) = 0 dt 9

The solution of this differential equations of the form v C (t ) = k 1 + k 2 e

−t τ

Since the differential equation has no constant forcing function, we know that k1 = 0. −t

Therefore, substituting v C (t ) = k 2 e τ into the equation yields −t −t −t 20 k 2e τ + k 2e τ = 0 τ 9

and τ=

9 sec . 20

54 In addition, since vC(0) = 6 = k2e° k2 = 6 Thus v C ( t ) = 6e

−

20 t 9

V

Recall that i 0 ( t ) = i x (t )

R5 R4 + R5

and i x (t ) =

v C (t ) R6

Then ⎛ v (t ) ⎞ ⎛ R 5 ⎞ ⎟⎟ i 0 (t ) = ⎜⎜ C ⎟⎟ ⎜⎜ ⎝ R6 ⎠ ⎝ R4 + R5 ⎠ −

20

t

= 1e 9 mA = 1mA 7.2

t>0 t<0

The network can be redrawn as shown in Fig. S7.2(a). t=0 5 IS = A k

5kΩ

R 1 = 1k

i (t )

1mA

R 2 = 3k

Fig. S7.2(a) In the steady-state time interval prior to switch action, the inductor looks like a shortcircuit. Therefore, in this time period t < 0, the initial inductor current is iL(0-) = IS = 5mA At t = 0 the switch changes positions and hence for t > 0 the network reduces to that shown in Fig. S7.2(b).

55 i (t )

R 1 = 1k

i = 1mH

R 2 = 3k

Fig. S7.2(b) If we let R = R1⎟ ⎜R2 then the differential equation for the inductor current is L

di (t ) + R i (t ) = 0 dt

The solution of this equation is of the form −t

i (t ) = k 1 + k 2 e τ

The differential equation has no constant forcing function and hence k1 = 0. Substituting −t

i(t ) = k 2 e τ into the equation for the current yields

⎛1⎞⎛−t⎞ ⎛ 3k ⎞ ⎜ ⎟ ⎜ ⎟ k 2e τ + ⎜ ⎟ k 2e τ = 0 ⎝k⎠⎝ τ ⎠ ⎝ 4 ⎠ −t

−t

where we have used the circuit parameter values in the equation, i.e., L = R=

3 Ω . This equation produces a τ value of 4k τ=

4 µ sec . 3

Furthermore, since i (0 − ) = 1mA

and i (0 ) = k 2 e −0

we find that k2 = 5mA Therefore,

1 H and k

56

i(t ) = 5mA, = 5e 7.3

t<0

− 7.5 × 10 5 t

mA, t > 0

The circuit is redrawn for convenience in Fig. S7.3(a). C 50µF

R1 6k

R2 6k

+ v 0 (t ) -

R 4 6k

R3 6k

t=0

+ 12V -

Fig. S7.3(a) Before we begin our analysis, we note that resistors R3 and R4 are in parallel and so we first reduce the network to that shown in Fig. S7.3(b). + C R2 + R1 v1 v C 6k v 0 (t ) 50µF 6k + R = 3k t=0

+ -

12

Fig. S7.3(b) Now that the network has been simplified, we begin our analysis +v R 1 = 6 k v1 R 2 = 6k - C + R = 3k

12V

+ -

Fig. S7.3(c)

57 + R 2 = 6k v (0 + ) 0 -

+ 6V

R 1 = 6k

R = 3k + 12V -

Fig. S7.3(d)

6k

6k

R TH 3k Fig. S7.3(e)

Step-1

v 0 (t ) = k 1 + k 2 e

−t τ

Step-2 In steady-state prior to switch action, the capacitor looks like an open-circuit and the 12-V source is directly across the resistor R = 3kΩ. As shown in Fig. S7.3(c) the voltage v1 across R1 is equal and opposite to vC. Since the voltage of the 12-V source is divided between R1 and R2 we can use voltage division to find v1 as ⎛ R1 v 1 = 12 ⎜⎜ ⎝ R1 + R 2

⎞ ⎟⎟ = 6V ⎠

hence v C (0 − ) = − v1 = − 6V = v C (0 + ) Step-3 The new circuit, valid only for t = 0 + is shown in Fig. S7.3(d). Once again, using voltage division, ⎛ R2 ⎞ ⎟⎟ v 0 (0 + ) = − v C (0 + ) ⎜⎜ ⎝ R2 + R ⎠ = − 4V

Step-4 For the period t > 5τ, the capacitor looks like an open-circuit and the source is disconnected. With no source of energy present in the network v 0 (∞ ) = 0

58

Step-5 The Thevenin equivalent resistance obtained by looking into the network from the terminals of the capacitor with all sources made zero is derived from the circuit in Fig. S7.3(e) RTH = (6k)⎟ ⎜(6k + 3k) = 3.6kΩ Then the time constant of the network is τ = RTHC = 0.18 sec. Step-6

Evaluating the constants in the solution, we find that k 1 = v 0 (∞ ) = 0 k 2 = v 0 (0 + ) − v 0 (∞ ) = − 4

Therefore, −t

v 0 (t ) = − 4e 0.18 V

7.4

We begin our analysis of the network with Step-1

The output voltage will be of the form −t

v 0 (t ) = k 1 + k 2 e τ

Step-2 In the steady-state prior to the time the switch is thrown, the inductor acts like a short-circuit and shorts out the 4Ω resistor. The network for this situation is as shown in Fig. S7.4(a). i L (0 − )

12V +-

4Ω 2Ω + 12V

+ 2Ω

v0 -

Fig. S7.4(a) Under these conditions, i L (0 − ) is the current from the 12-V source at the left side of the network, through the short-circuit, with a return path through the 2Ω resistor at the

59 output. What is the contribution of the 12V source in the center of the network? No contribution! Why? If we applied superposition and treated each source independently, we would quickly find that when the left-most source was replaced with a short-circuit, all the current from the other 12-V source would be diverted through this short-circuit. Therefore, i L (0 − ) = Step-3

12 = 6A = i L (0 + ) 2

The new network, valid only for t = 0 + , is shown in Fig. S7.4(b). 6A 4Ω 2Ω 12V +-

2Ω

+

v 0 (0 + ) -

Fig. S7.4(b) If we employ superposition, we find that ⎛ ⎞ ⎛ ⎞ 2 4 ⎟⎟ + 6 ⎜⎜ ⎟⎟ (2 ) v 0 (0 + ) = − 12 ⎜⎜ ⎝2 + 4 + 2⎠ ⎝4 + 2 + 2⎠ = 3V

where in this equation we have used first voltage division in conjunction with current division to obtain the voltage. The two networks employed are shown in Figs. S7.4(c) and (d). 6A 2Ω 12V +-

+

4Ω 2Ω

v′0 (0 + )

4Ω 2Ω

2Ω

-

Fig. S7.4(c)

+ v′0′ (0 + ) -

Fig. S7.4(d)

Step-4 For t > 5τ, the inductor again looks like a short-circuit and the network is of the form shown in Fig. S7.4(e).

60

+

4Ω 2Ω

2Ω

12V +-

v 0 (∞ ) -

Fig. S7.4(e) A simple voltage divider indicates that the output voltage is ⎛ 2 ⎞ ⎟⎟ = − 6V v 0 (∞ ) = − 12 ⎜⎜ ⎝2 + 2⎠

Step-5 The Thevenin equivalent resistance obtained by looking into the circuit from the terminals of the inductor with all sources made zero is derived from the network in Fig. S7.4(f). R TH

4Ω 2Ω 2Ω Fig. S7.4(f) Clearly RTH = 4⎥⎜(2 + 2) = 2Ω Then the time constant is 1 L 3 1 τ = = = sec . R 2 6 Step-6

The solution constants are then k 1 = v 0 (∞ ) = − 6

k 2 = v 0 (0 + ) − v 0 (∞ ) = 3 − (− 6) = 9V

Hence, v 0 (t ) = − 6 + 9e −6 t V

61

7.5

(a) Applying KVL to the closed path yields v S (t ) = Ri (t ) +

di(t ) 1 t ∫t i (x ) dx + L dt C 0

differentiating both sides of the equation we obtain dv S (t ) di (t ) i (t ) d 2 i (t ) =R + +L dt dt C dt 2

By rearranging the terms, the equation can be expressed in the form L

or

d 2 i (t ) di (t ) i (t ) dv S (t ) +R + = 2 dt dt C dt

d 2 i (t ) R di (t ) 1 1 dv S (t ) i (t ) = + + 2 dt L dt RC L dt

Using the circuit component values yields d 2 i (t ) di (t ) 1 dv S (t ) ( ) 7 10 i t + + = dt 2 dt 2 dt

(b) The characteristic equation for the network is s 2 + 7s + 10 = 0

(c) The network’s natural frequencies are the roots of the characteristic equation. The quadratic formula could be used to obtain those roots or we can simply recognize that the equation can be expressed in the form s 2 + 7s + 10 = (s + 2 ) (s + 5) = 0

Therefore, the networks natural frequencies are s=2 s=5 (d) Since the roots of the characteristic equation are real and unequal, the network response is overdamped. (e) Based upon the above analysis, the general expression for the current is

62

i (t ) = k 0 + k 1e −2 t + k 2 e −5 t A where k 0 is the steady-state value and the constants k1 and k2 are determined from initial conditions. 7.6

The network is re-labeled as shown in Fig. S7.6(a). + - v C (t ) i 0 (t )

C

12V

+ -

R2

t=0

R1

i L (t )

+ R3

v 0 (t )

-

Fig. S7.6(a) 1 F . Consider now the conditions of the 120 network at three critical points in time.

where all R = 24Ω, L = 2.4H and C =

At t = 0 − , i.e., the steady-state condition prior to switch action, the capacitor acts like an open-circuit, the inductor acts like a short-circuit and hence v C (0 − ) = 0 , i L (0 − ) = 0 , i 0 (0 − ) = 0 and v 0 (0 − ) = 0 . At t = 0 + , i.e., the instant the switch is thrown, the conditions on the storage elements (L and C) cannot change instantaneously and therefore v C (0 + ) = 0 , i L (0 + ) = 0 , 12 1 i 0 (0 + ) = = A and v 0 (0 + ) = 12V . R3 2 At t = ∞, i.e., the steady-state condition after the switch is thrown, the capacitor acts like an open-circuit, the inductor acts like a short-circuit and hence v C (∞ ) = 12V , 12 1 i L (∞ ) = = A , i 0 (∞ ) = 0 and v 0 (∞ ) = 0 . R2 2 Now applying KCL to the network in the time interval t > 0 , we obtain 12 − v 0 (t ) d (12 − v 0 (t )) 1 t v (t ) +C = ∫0 v 0 (x ) dx + 0 R2 dt L R3

63 v 0 (t ) expressing v 0 (t ) in terms of i 0 (t ) and using the component values R3 reduces the equation to

where i 0 (t ) =

1 di 0 (t ) 1 t − i 0 (t ) − − 10 ∫0 i 0 (x ) dx − i 0 (t ) = 0 5 dt 2

Combining terms and differentiating this expression yields d 2 i 0 (t ) di (t ) + 10 0 + 50 i 0 (t ) = 0 2 dt dt

Therefore, the characteristic equation for the network is s2 + 10s + 50 = 0 Factoring this equation using the quadratic formula or any other convenient means yields s1, s2 = -5 ± j5 = σ ± jω Since the roots of the characteristic equation are complex conjugates, the network is underdamped and the general form of the current i 0 (t ) is i 0 (t ) = k + e − σt (A cos ωt + B sin ωt )

= k + e −5 t (A cos 5t + B sin 5t )

where k is the steady-state term resulting from the presence of the voltage source in the time interval t → ∞. We can now evaluate the constants k, A and B using the known conditions for the network. For example, i 0 (0 + ) =

1 =k+A 2

and i 0 (∞ ) = 0 = k Therefore, k = 0 and A =

1 . 2

64 We need another equation in order to evaluate the constant B. If we return to our original equation and evaluate it at time t = 0 + , we have 12 − 12 1 ⎛ − dv 0 (t ) ⎞ + ⎟ ⎜ R2 120 ⎝ dt ⎠

t = 0+

=0+

12 R3

where v 0 (0 + ) = 12V , the integration interval is zero and the derivative function is our unknown. Therefore, dv 0 (t ) dt

= − 60

t = 0+

or di 0 (t ) dt

t = 0+

= − 2.5

The general form of the solution is

⎛1 ⎞ i 0 (t ) = e −5 t ⎜ cos 5t + B sin 5t ⎟ ⎝2 ⎠ Then di 0 (t ) ⎛1 ⎞ ⎛−5 ⎞ = − 5e −5 t ⎜ cos 5t ⎟ + e −5 t ⎜ sin 5t ⎟ − 5e −5 t B sin 5t + e −5 t 5B cos 5t dt ⎝2 ⎠ ⎝ 2 ⎠ and di 0 (t ) dt

t = 0+

=

−5 + 5B 2

Therefore,

− 2.5 =

−5 + 5B 2

or B=0 The general solution is then

65 i 0 (t ) = 0 =

t<0

1 −5 t e cos 5t 2

t>0

A plot of this function is shown in Fig. S7.6(b).

I0 (t) (A)

0.6 0.4 0.2 0 -0.2

0

0.2

0.4

0.6 t (sec)

Fig. S7.6(b)

0.8

1

66 CHAPTER 8 PROBLEMS

8.1

Find the frequency domain impedance Z, shown in Fig. 8.1. 1Ω 1Ω

-j1Ω

Z

1Ω

j1Ω

Fig. 8.1 8.2

If the impedance of the network in Fig. 8.2 is real at f = 60Hz, what is the value of the inductor L? 1Ω L

2Ω

Z

8.3

10mF

Fig. 8.2 Use nodal analysis to find V0 in the network in Fig. 8.3. -+

12 ∠0° V

1Ω j2Ω

1Ω 2Ω

-j4Ω

+ V0 -

Fig. 8.3 8.4

Find V0 in the network in Fig. 8.4 using (a) loop analysis (b) superposition and (c) Thevenin’s Theorem. 2Ω

-j1Ω

12 ∠0° V +-

4Ω

Fig. 8.4

j2Ω + V0 -

2 ∠0° A

67 CHAPTER 8 SOLUTIONS

8.1.

To begin our analysis, we note that the circuit can be labeled as shown in Fig. S8.1. Z1 Z2

Z

Z3

Fig. S8.1 In this case, Z1 consists of a 1Ω resistor, Z2 is the series combination of a 1Ω resistor and a j1Ω inductor and Z2 consists of a –j1Ω capacitor in series with a 1Ω resistor. Therefore, Z1 = 1Ω Z2 = 1 + j1Ω Z3 = 1 – j1Ω Starting at the opposite end of the network from the terminals at which Z is calculated we note that Z2 and Z3 are in parallel and their combination is in series with Z1. Hence Z = Z1 + Z 2 Z 3 =1+ =1+

(1 + j) (1 − j) 1+ j+1− j 2 2

= 2Ω

8.2

The general expression for the impedance of this network is ⎛ 1 ⎞ ⎟ Z = 1 + 2 ⎜⎜ jωL + jωC ⎟⎠ ⎝

⎛ 1 ⎞ ⎟ must be real, i.e. In order for Z to be purely resistive, the term ⎜⎜ jωL + jωC ⎟⎠ ⎝ ZLC = RLC + j0 However, since ZLC can be written as 1 ⎞ ⎛ Z LC = j ⎜ ωL − ⎟ ωC ⎠ ⎝

68

it is clearly an imaginary term and RLC = 0. Therefore, in order for Z to be resistive

ωL −

1 =0 ωC

or L=

=

1

ω2 C 1

(377 ) (10 ) 2

−2

= 703.6 µH 8.3

The presence of the voltage source indicates that nodal analysis is a viable approach to this problem. The voltage source and its two connecting nodes form a supernode as shown in Fig. S8.3. -+

V1 = V0 − 12 ∠0° j2Ω

12 ∠0° V

V2 1Ω 2Ω

1Ω -j4Ω

+ V0 -

Fig. S8.3 Note that there are three non-reference nodes, i.e., V1, V2 and V0 . Because the voltage source is tied directly between nodes V1 and V0 , V1 = V0 − 12∠0° . This constraint condition is one of our three equations required to solve the network. The two remaining equations are obtained by applying KCL at the supernode and the node labeled V2. For the supernode, KCL yields V V1 V1 − V2 V0 − V2 + + + 0 =0 − j4 j2 1 1 At the node labeled V2, KCL yields V2 − V1 V2 V2 − V0 + + =0 1 2 1

Therefore, the three equations that will provide the node voltages are

69 V1 = V0 − 12 1 1 − j V1 + V1 − V2 + V0 − V2 + j V0 = 0 4 2 1 V2 − V1 + V2 + V2 − V0 = 0 2 Substituting the first equation in for the two remaining equations and combining terms yields 1⎞ ⎛ V0 ⎜ 2 − j ⎟ − 2V2 = 12 − j6 4⎠ ⎝ 5 − 2V0 + V2 = − 12 2 Solving for V2 in this last equation and substituting it into the one above it, we obtain V0 (0.4 − j0.25) = 2.4 − j6 and hence V0 = 13.57 ∠ − 36.2° V 8.4

(a) Since the network has two loops, or in this case two meshes, we will need two equations to determine all the currents. Consider the network as labeled in Fig. S8.4(a). 2Ω 12 ∠0° V +-

-j1Ω

I1

j2Ω

4Ω

2 ∠0° A

+ V0 -

I2

Fig. S8.4(a) Note that since I2 goes directly through the current source, I2 must be 2∠0°A. Hence, one of our two equations is I2 = 2∠0° If we now apply KVL to the loop on the left of the network, we obtain − 12 + I1 (2 − j1) + (I1 − I 2 ) (4 + j2 ) = 0

70 These two equations will yield the currents. Substituting the first equation into the second yields − 12 + I1 (2 − j1 + 4 + j2 ) − 2 (4 + j2 ) = 0

and then I1 =

20 + j4 = 3.35 ∠1.85° A 6 + j1

Finally, V0 = 4 (I1 − I 2 ) ⎞ ⎛ 20 + j4 − 2 ⎟⎟ = 4 ⎜⎜ ⎠ ⎝ 6 + j1 = 5.42 ∠4.57° V

(b) In applying superposition to this problem, we consider each source acting alone. If we zero the current source, i.e., replace it with an open circuit, the circuit we obtain is shown in Fig. S8.4(b). 2Ω

-j1Ω

j2Ω

12 ∠0° V +-

4Ω

+ V0′ -

Fig. S8.4(b) Using voltage division ⎞ ⎛ 4 ⎟⎟ V0′ = 12 ⎜⎜ ⎝ 4 + j2 + 2 − j1 ⎠ 48 V = 6 + j1

Now, if we zero the voltage source, i.e., replace it with a short circuit, we obtain the circuit in Fig. S8.4(c).

71 IX

-j1Ω

j2Ω 2 ∠0° A

2Ω

+ V0′′ -

4Ω

Fig. S8.4(c) Employing current division, the current IX is ⎞ ⎛ 2− j ⎟⎟ I X = − 2 ∠0° ⎜⎜ − + + 2 j 4 j 2 ⎠ ⎝ − 4 + j2 = A 6 + j1

Then, V0′′ = 4I X =

− 16 + j8 6 + j1

And finally, V0 = V0′ + V0′′ − 16 + j8 48 + 6 + j1 6 + j1 32 + j8 = 6 + j1 = 5.42 ∠4.57° V =

(c) In applying Thevenin’s Theorem, we first break the network at the load and determine the open-circuit voltage as shown in Fig. S8.4(d). -j1Ω 2Ω j2Ω

12 ∠0° V +-

+ V0 C -

Fig. S8.4(d)

2 ∠0° A

72 Note that there exists only one closed path and the current in it must be 2∠0° A. Note also that there is no current in the inductor and therefore no voltage across it. Hence VOC is also the voltage across the current source. Hence, V0 C = 12 − 2 (2 − j) = 8 + j2 V The Thevenin equivalent impedance found by zeroing the independent sources and looking into the network at the terminals of the load can be determined from the circuit in Fig. S8.4(e). 2Ω -j1Ω j2Ω Z TH

Fig. S8.4(e) This network indicates that ZTH = 2 – j1 + j2 = 2 + j1Ω If we now form the Thevenin equivalent circuit and re-connect the load, we obtain the network in Fig. S8.4(f). Z TH 2Ω j1Ω V0 C

8+j2V +-

4Ω

+ V0 -

Fig. S8.4(f) Applying voltage division yields ⎛ ⎞ 4 ⎟⎟ V0 = (8 + j2) ⎜⎜ ⎝ 4 + 2 + j1 ⎠ 32 + j8 = 6 + j1 = 5.42 ∠4.57° V

73 CHAPTER 9 PROBLEMS

9.1

Determine the average power supplied by each source in the circuit in Fig. 9.1.

10 ∠0° V +-

-j1Ω

j1Ω 2 ∠30° A

1Ω

Fig. 9.1 9.2

Given the circuit in Fig. 9.2, determine the impedance ZL for maximum average power transfer and the value of the maximum average power transferred to this load. -j1Ω 1Ω

1Ω +

12 ∠0° V +-

ZL 6 ∠0° V

Fig. 9.2 9.3

Calculate the rms value of the waveform shown in Fig. 9.3. v(t) (V)

6

1

2

3

4

5

6

7

t(s)

Fig. 9.3 9.4

Determine the source voltage in the network shown in Fig. 9.4. j0.5Ω 0.1Ω + VS -

+ 40 kW 60 kW 0.85 pf 240 ∠0° V rms 0.78 pf lagging lagging -

Fig. 9.4 9.5

A plant consumes 75 kW at a power factor of 0.70 lagging from a 240-V rms 60 Hz line. Determine the value of the capacitor that when placed in parallel with the load will change the load power factor to 0.9 lagging.

74 CHAPTER 9 SOLUTIONS

9.1

Because the series impedance of the inductor and capacitor are equal in magnitude and opposite in sign, from the standpoint of calculating average power the network can be reduced to that shown in Fig. S9.1. I VS

10 ∠0° V +-

+ V1

I CS 1Ω

2 ∠30° A

-

Fig. S9.1 The general expression for average power is P=

1 VI cos (θ V − θ I ) 2

In the case of the current source V1 = 10V, ICS = 2A, θV = 0° and θI = 30°. Therefore, the average power delivered by the current source is

⎛1⎞ PCS = ⎜ ⎟ (10) (2) cos (− 30°) ⎝2⎠ = 8.66 W In order to calculate the average power delivered by the voltage source, we need the current IVS. Using KCL I VS + 2 ∠30° =

V = 10 ∠0° 1

or IVS = 8.33∠-6.9° A Now 1 (10) (8.33) cos (0° − (− 6.9°)) 2 = 41.34 W

PVS =

Therefore, the total power generated in the network is PT = PCS + PVS = 50 W

75

Let us now calculate the average power absorbed by the resistor. We know that the average power absorbed by the resistor must be Vm2 R ⎛ 10 2 ⎞ ⎟ ⎜ ⎝ 1 ⎠ = 50 W 1 2 1 = 2

PR =

In addition, the average power absorbed by the resistor can also be determined by PR =

1 2 Im R 2

However, we do not know the current in the resistor. Using KCL. IR = IVS + ICS = 8.66∠-6.9° + 2∠30° = 10∠0° A Now 1 (10)2 (1) 2 = 50 W

PR =

Thus, we find that the total average power generated is equal to the average power absorbed. 9.2

We will first determine the Thevenin equivalent circuit for the network without the load attached. The open-circuit voltage, V0C, can be determined from the network in Fig. S9.2(a). -j1Ω 1Ω 12 ∠0° V +6 ∠0° V

1Ω +

I

- V+ + R V0 C -

Fig. S9.2(a) This open-circuit voltage can be calculated in a number of ways. For example, we can compute the current I as

76

I=

12 (0° − (− 6∠0°)) 18 = A 1− j 1− j

Then using KVL, V0 C = 1 I − 6∠0° 12 + 6 j V 1− j or, we could use voltage division to determine the voltage across the 1-Ohm resistor on the right, i.e.,

=

⎛ 1 VR = [12 ∠0° − (− 6 ∠0°)] ⎜⎜ ⎝1 − 18 = V 1− j

⎞ ⎟ j ⎟⎠

Then, once again V0 C = VR − 6∠0° =

12 + 6 j V 1− j

= 9.49∠71.56° V

The Thevenin equivalent impedance is obtained by looking into the open-circuit terminals with all sources made zero. In this case, we replace the voltage sources with short circuits. This network is shown in Fig. S9.2(b). -j1Ω 1Ω

1Ω Z TH

Fig. S9.2(b) Note that the 1-Ohm resistor on the left is shorted and thus the ZTH is Z TH =

(1) (− j) =

1− j 1 1 = −j Ω 2 2

−j Ω 1− j

77

Hence, for maximum average power transfer Z L = Z *TH

or ZL =

1 1 +j Ω 2 2

Therefore, the network is reduced to that shown in Fig. S9.2(c). 1 1 −j 2 2 Ι 1 2

9.49 ∠71.56° V +-

+j

1 2

Fig. S9.2(c) Then 9.49 ∠71.56° 1 1 1 1 −j + +j 2 2 2 2 = 9.49 ∠71.56° A

I=

and the maximum average power transferred to the load is 1 (9.49)2 2 = 90 W

PL =

9.3

⎛1⎞ ⎜ ⎟ ⎝2⎠

In order to calculate the rms value of the waveform, we need the equations for the waveform within each of the distinctive intervals. In the interval 0 ≤ t ≤ 2s, the waveform is a straight line that passes through the origin of the graph. The equation for a straight line in this graph is v(t) = mt + b Where m is the slope of the line and b is the v(t) intercept. Since the line goes through the origin, b = 0. The slope m is

78

m= Therefore, in the interval 0 ≤ t ≤ 2s,

6V =3 2s

v(t) = 3t The waveform has constant values in the intervals 2 ≤ t ≤ 3s and 3 ≤ t ≤ 4s, i.e., 2 ≤ t ≤ 3s 3 ≤ t ≤ 4s

v(t) = 6 v(t) = 0

Since the waveform repeats after 4s, the period of the waveform is T = 4s Now that the data for the waveform is known,

⎡1 4 ⎤ Vrms = ⎢ ∫ v 2 (t ) dt ⎥ ⎣T 0 ⎦

1 2

Therefore, in this case 3 4 ⎡1 2 ⎤ 2 2 2 Vrms = ⎢ ⎡ ∫ (3t ) dt + ∫ (6) dt + ∫ (0) dt ⎤ ⎥ ⎢ ⎥ 0 2 3 ⎣ ⎦ ⎣4 ⎦ 1 ⎡1 ⎤ = ⎢ 3t 3 02 + 36t 32 ⎥ 2 ⎣4 ⎦ 1 ⎡1 ⎤ = ⎢ (24 + 36)⎥ 2 ⎣4 ⎦

[

1 2

]

= (15) 2 = 3.87 V rms 1

9.4

We begin our analysis by labeling the various currents and voltages in the circuit as shown in Fig. S9.4. IS VS +-

I1 0.1Ω j0.5Ω 60 kW 0.85 pf lag

I2 40 kW 0.78 pf lag

Fig. S9.4

+

VL = 240 ∠0° V rms -

79 Our approach to determining VS is straight forward: We will compute the currents I1 and I2; add them using KCL to find IS; determine the voltage across the line impedance and finally use KVL to add the line voltage and load voltage to determine the source voltage. The magnitude of the current I1 is I1 = =

P1 VL (pf 1 ) 60,000 (240) (0.85)

= 294.12 A rms.

And the phase angle is θ I = − cos −1 (0.85) 1

= − 31.79°

The negative sign is a result of the fact that the power factor is lagging. Thus I1 = 294.12∠-31.79° A rms. The magnitude of the current I2 is I2 = =

P2 VL (pf 2 ) 40,000 (240) (0.78)

= 213.68 A rms.

And the phase angle is θ I = − cos −1 (0.78) 2

= − 38.74°

Thus I2 = 213.68∠-38.74° A rms. Using KCL

80 I S = I1 + I 2 = 294.12 ∠ − 31.79° + 213.68 ∠ − 38.74° = 504.1 ∠ − 34.25° A rms.

Then V2 = I S (0.1 + j0.5) + 240∠0°

= (504.1∠ − 34.25°) (0.51 ∠78.7°) + 240 ∠0° = 257.04 ∠44.44° + 240 ∠0° = 460.17 ∠23.02° V rms.

9.5

Since the original power factor is 0.7 lagging the power factor angle is θOLD = cos-1 (0.7) = 45.57° Then QOLD = POLD tan θOLD = 75,000 tan 45.57° = 76.52 kvar Hence SOLD = 75,000 + j76,515 = 107.14∠45.57° kVA The new power factor angle we wish to achieve is θNEW = cos-1 (new power factor) = cos-1 (0.9) = 25.84° Then QNEW = POLD tan θNEW = 75,000 tan 25.84° = 36,324 kvar Now the difference between QNEW and QOLD is achieved by the capacitor, i.e., QCAP = QNEW - QOLD = 36,324 – 76,515 = -40,191 kvar

81

And since QCAP = -ω CV2 Then 40,191 (377 ) (240)2 = 1850.8 µF

C=

82 CHAPTER 10 PROBLEMS

10.1

Find V0 in the network in Fig. 10.1. j1Ω

2Ω

10 ∠0° A

j2Ω

1Ω

j1Ω j2Ω

1Ω

+ V0 -

Fig. 10.1 10.2

Determine the impedance seen by the source in the circuit in Fig. 10.2. j1Ω 1Ω -j1Ω 2Ω

120 ∠0° V + -

j4Ω

j2Ω

j2Ω 1Ω

3Ω

-j2Ω

Fig. 10.2 10.3

Determine I1, I2, V1 and V2 in the circuit in Fig. 10.3. 1Ω 3Ω I1 I2 1:2 + + + 2 ∠0° V 1∠0° V +V1 V2 Ideal

Fig. 10.3 10.4

Given the circuit in Fig. 10.3, determine the two networks obtained by replacing (a) the primary and the ideal transformer with an equivalent circuit and (b) the ideal transformer and the secondary with an equivalent circuit.

83 CHAPTER 10 SOLUTIONS

10.1

Our first step in the solution of this problem is to apply source transformation to the leftend of the network and transform the 10∠0° A source in parallel with the 1Ω resistor into a 10∠0° V source in series with the 1Ω resistor as shown in Fig. S10.1(a). j1Ω j1Ω 2Ω 1Ω 10 ∠0° V +-

j2Ω

1Ω

j2Ω

+ V0 -

Fig. S10.1(a) Let us redraw the network as shown in Fig. S10.1(b). j1Ω 3Ω 10 ∠0° V +-

I1

+ V1

+ V2

-

-

j1Ω

I2

1Ω

+ V0 -

Fig. S10.1(b) The equations for this network are -10 + 3I1 + V1 = 0 -V2 + I2(1 + j1) = 0 We now write the equations for the mutually coupled coils. In order to force the variables in this circuit into our standard form for mutually coupled inductors, we must reverse the signs on V1, I1 and I2. Therefore, the equations that relate V1 and V2 to I1 and I2, in this case, are -V1 = j2(-I1) + j1(-I2) V2 = j2(-I2) + j1(-I1) Combining the equations yields (3 + j2) I1 + j1 I2 = 10 j1 I1 + (1 + j3) I2 = 0 Solving for I1 in the second equation and substituting it into the first equation yields [(3 + j2) (-3 + j1) + j1] I2 = 10 or

84 10 − 11 − j2 = − 0.894∠10.3° A

I2 =

And finally V0 = 1 I 2

= − 0.894 ∠10.3° V 10.2

Let us first determine the total impedance on the right side of the circuit as shown in Fig. S10.2(a). 2Ω j2Ω -j2Ω

ZL 1Ω

Fig. S10.2(a) As the figure indicates Z L = 2 + (1 + j2)

(− j2) (1 + j2) (− j2) =2+ 1 + j2 − j2 = 6 − j2Ω

The original network can now be redrawn in the following form shown in Fig. S10.2(b). j1Ω -j1Ω 1Ω + 120 ∠0° V +-

j4Ω

I1

+

V1 V2 -

6Ω j2Ω

I2

3Ω

Fig. S10.2(b) The two KVL equations for the network in Fig. S10.2(b) are 120 = (4 – j1) I1 + V1 V2 = (6 – j2) I2

-j2Ω

85 In order to force the variables in this circuit into our standard form for mutually coupled inductors, we must reverse the sign on V2. Therefore, the equations that relate V1 and V2 to I1 and I2, in this particular case, are V1 = j4 I1 + j1 I2 -V2 = j2 I2 + j1 I1 Combining all of these equations results in the following two equations. (4 + j3) I1 + j1 I2 = 120 j1 I1 + 6 I2 = 0 Solving the second equation for I2 and substituting this value into the first equation yields

⎛ ⎜ 4 + j3 + ⎝

1⎞ ⎟ I1 = 120 6⎠

Then, the impedance seen by the source is ZS = 10.3

120 = 4.167 + j3Ω I1

The KVL equations for this network are 1∠0° = − I1 (1) + V1 V2 = 3I 2 + 2∠0° If we now force the variables in this circuit into our standard form for the ideal transformer, we must reverse the signs on V1 and I2. Therefore, the equations that relate V1 to V2 and I1 to I2, in this particular case, are

− V1 1 = V2 2

1 I1 + 2 (− I 2 ) = 0

Solving the later equations for V2 and I2 and substituting these values into the first equations yields 1 = − I1 + V1 − 2V1 =

Solving these equations produces

3 I1 + 2 2

86 I1 = 1.142∠180° A V1 = 0.142∠180° V Then, the transformer relationships yield 1 I1 2 V2 = − 2V1 I2 =

Therefore, I2 = 0.571∠180° A V2 = 0.284∠0° V 10.4

As shown in the previous problem, the ideal transformer equations are

− V1 1 = V2 2

1 I1 + 2 (I 2 ) = 0

These two equations and the equation for reflecting impedance from the primary of the transformer to the secondary i.e., 2

⎛N ⎞ Z p = ⎜⎜ 1 ⎟⎟ Z S ⎝ N2 ⎠ 1 = ZS 4 are the necessary equations for developing the equivalent circuits. (a) If we reflect the primary to the secondary, we note that V2 = -2V1 And ZS = 4Zp Therefore, the voltage source in the primary becomes V2 = -2(1∠0°) = 2∠180° V

87 And ZS = 4(1) = 4Ω Therefore, the equivalent circuit in this case is shown in Fig. S10.4(a). I2 2 ∠0° V +

4Ω + 3Ω V2 -

+ 2 ∠0° V -

Fig. S10.4(a) (b) Once again, using the ideal transformer equation to reflect the secondary to the primary we obtain the network in Fig. S10.4(b). 3 Ω 1Ω 4 1 ∠0° V + -

+ V1 -

Fig. S10.4(b)

+ 1 ∠0° V

88 CHAPTER 11 PROBLEMS

11.1

In a three-phase balanced wye-wye system, the source is an abc-sequence set of voltages with Van = 120∠40° V rms. The per phase impedance of the load is 10 + j8Ω. If the line impedance per phase is 0.6 + j0.4Ω, find the line currents and load voltages.

11.2

An abc-sequence set of voltages feeds a balanced three-phase wye-wye system. If Van = 440∠40° V rms, VAN = 410∠39° V rms and the line impedance is 1.5 + j1.0Ω, find the load impedance.

11.3

In a balanced three-phase wye-delta system, the source has an abc-phase sequence and Van = 120∠30° V rms. The line and load impedance are 0.6 + j0.4Ω and 24 + j12Ω, respectively. Find the delta currents in the load.

11.4

A balanced three-phase source serves two loads: Load 1: 32 kVA at 0.85 pf lagging. Load 2: 20 kVA at 0.6 pf lagging. The line voltage at the load is 208 V rms at 60Hz. Determine the line current and the combined power factor at the load.

11.5

In a three-phase balanced system an abc-sequence wye-connected source with Van = 220∠0° V rms supplies power to a wye-connected load that consumes 36 kW of power in each phase at a pf of 0.75 lagging. Three capacitors, each with an impedance of –j2.0Ω, are connected in parallel with the original load in a wye configuration. Determine the power factor of the combined load as seen by the source.

89 CHAPTER 11 SOLUTIONS

11.1

First of all, we note that since this is a balanced system, we need only consider one phase of the system. All currents in the two remaining phases have the same magnitude but are shifted in phase by 120° and 240°. Consider now the circuit for the a-phase shown in Fig. S11.1. a 0.6Ω j0.4Ω A I aA

Line

10Ω Load

Van = 120 ∠40° V rms +-

j8Ω n Fig. S11.1

N

In this circuit, lower case letters represent the source end of the network and capital letters represent the load end of the network. The line current for this a-phase is I aA =

Van Z Line + Z Load

120 ∠40° 10.6 + j8.4 = 8.87∠1.6° A rms.

=

Then the load voltage for this phase is VAN = I aA Z Load

= (8.87 ∠1.6°) (10 + j8) = 113.59 ∠40.26° V rms.

The results for the two remaining phases are IbB = 8.87∠-118.4° A rms IcC = 8.87∠-238.4° A rms 11.2

VBN = 113.59∠-79.74° V rms VCN = 113.59∠-199.74° V rms.

The a-phase equivalent circuit for this system is shown in Fig. S10.2.

90 a I aA 1.5Ω

A +

j1Ω

Van +-

Z LOAD

VAN

N

n

Fig. S11.2 We can approach this problem in a couple of ways. For example, note that by employing voltage division, we can write ⎤ ⎡ Z Load VAN = Van ⎢ ⎥ ⎣ Z Load + Z Line ⎦

If we solve this equation for ZLoad, we obtain Z Load =

Z Line Van −1 VAN

where the quantities on the right side of the equation are all given. We can also calculate the line current IaA and use it with VAN to determine ZLoad. In this later case I aA =

Van − VAN 1.5 + j1

= 17.15 ∠19.6° A rms Then Z Load =

VAN I aA

410∠39° 17.15 ∠19.6° = 23.91∠19.4° Ω

=

11.3

To begin we convert the delta load to an equivalent wye. In this balanced case

91 Z∆ 3 = 8 + j4Ω

ZY =

Now the a-phase wye-wye circuit is shown in Fig. S11.3. 0.6Ω

a

j0.4Ω

I aA

A 8Ω

Van = 120 ∠30° V rms +-

j4Ω

n Fig. S11.3

N

Now the line current for this network is 120 ∠30° 8.6 + j4.4 = 12.42 ∠2.9° A rms

I aA =

This is the current in the a-phase of an equivalent wye load. We can now convert this current to the AB phase of the delta. I AB =

I aA 3

∠θ I + 30° aA

Therefore IAB = 7.17∠32.9° A rms The currents in the remaining phases of the delta are IBC = 7.17∠-87.1° A rms and ICA = 7.17∠-207.1° A rms. 11.4

The total complex power at the load is S L 3φ = 32 ∠ cos −1 (0.85) + 20 ∠ cos −1 (0.6) kVA = 32 ∠31.79° + 20 ∠53.13° kVA = 51.15 ∠39.97° kVA Now, we know that

92

S L 3φ = 3 VL I L And hence IL =

51.15k

3 (208) = 141.98 A rms

And the power factor at the load is pfLoad = cos(39.97°) = 0.766 lagging 11.5

The original situation, prior to adding the capacitors is POLD = 36 kW QOLD = POLD tAN θOLD = 36,000 tan 41.41° = 31,749 var where 41.41° = θOLD = cos-1 (0.75). Therefore, SOLD = 36 + j31.749 kVA is the complex power for each phase. If we now add the capacitor, the real power is unaffected by this and thus PNEW = POLD = 36 kW However, QNEW = QOLD + QC Where QC is the reactive power supplied by the capacitor. Q C = − jωCV 2 rms − V 2 rms = ZC

or

93

− (220 ) QC = 2 = − 24.2 k var 2

Therefore, QNEW = 31.79 – 24.2 = 7.59 kvar And hence S NEW = PNEW + jQ NEW = 36 + j7.59 = 36.79 ∠11.9° kVA

And PfNEW = cos θNEW = cos 11.9° = 0.98 lagging

94 CHAPTER 12 PROBLEMS

12.1

Sketch the bode plot for the following network function H ( jω ) =

12.2

Sketch the bode plot for the following network function H ( jω ) =

12.3

36 (0.5 jω + 1) ( jω)2 (0.02 jω + 1)

250 jω ( jω + 10 ) ( jω + 1) ( jω + 50) ( jω + 100)

Given the magnitude characteristic for the network function shown in Fig. 12.3, find the expression for H(jω) H -20dB/decade dB 20dB -20dB/decade -40dB/decade

1

20

100

Log ω (rad/s)

Fig. 12.3 12.4

Given the series circuit shown in Fig. 12.4, determine the following parameters: ω 0 , Q and the BW. If the resistance is changed to 0.1Ω, what is the impact on these parameters. 1Ω 50µF

VS +-

200µH

Fig. 12.4 Sketch the frequency characteristic for the two values of R. What conclusion can be drawn from these two characteristics. 12.5

The network in Fig. 12.5 operates as a band pass filter. (a) Determine the transfer function for the network, (b) find the upper and lower cut off frequencies and the band width and (c) sketch the magnitude characteristic for this transfer function.

95

C = 5µF vS (t ) +-

L = 2H R = 4kΩ Fig. 12.5

+ v 0 (t ) -

96 CHAPTER 12 SOLUTIONS

12.1

First of all, we note that all the poles and zeros are in the standard form, e.g, the simple pole and zero are each in the form (jωτ + 1). At low frequencies the controlling term is the double pole at the origin. This term provides an initial slope for the magnitude characteristic of –40dB/decade. Furthermore, this initial slope will intersect the 0dB line at ω = 36 = 6 rad / s . However, before this initial slope intersects the 0dB line, we 1 1 = 2 rad / s . This term adds a encounter the break frequency of the zero at ω = = τ 0.5 slope of +20dB/decade to the magnitude characteristic and thus the composite characteristic changes from –40dB/decade to –20dB/decade. This characteristic maintains this slope until another break frequency is encountered. The remaining pole 1 1 has a break frequency at ω = = = 50 rad / s . This term adds a slope of – τ 0.02 20dB/decade to the magnitude characteristic, and since there are no more poles or zeros in the network function, the final slope of the magnitude characteristic is –40dB/decade. The composite magnitude characteristic is shown in Fig. S12.1(a). H -40dB/decade dB -20dB/decade 0dB -40dB/decade

2 6

Log ω (rad/s)

50

Fig. S12.1(a) The composite phase characteristic for this network function is shown in Fig. S12.1(b). Phase in degrees

0° − 90° −180° 2

50

Log ω (rad/s)

Fig. S12.1(b) Once again, the initial phase, at low frequencies, is controlled by the double pole at the origin that has a constant phase of -180°. The phase for the zero is an arc tangent curve that provides 45° of phase at the break frequency, ω = 2 rad/s. As the frequency increases beyond the break frequency this term provides 90° of phase so the composite curve approaches -90° of phase. As the frequency increases further, we encounter the

97 simple pole which provides -45° of phase at its break frequency and finally -90° of phase at higher frequencies. Thus the composite phase starts at -180°, moves toward -90° because of the presence of the zero and finally ends up back at -180° because of the last pole. 12.2

We begin the analysis by putting all the terms of the network function in standard form. The function then becomes H ( jω) =

0.5 jω (0.1 jω + 1) ( jω + 1) (0.02 jω + 1) (0.01jω + 1)

At low frequencies the magnitude characteristic is controlled by the zero at the origin. This term provides an initial slope of +20dB/decade and it will intersect the 0dB line at 1 ω= = 2 rad / s . Prior to reaching this frequency we encounter the break frequency of 0.5 1 1 the pole (jω + 1) which occurs at ω = = = 1 rad / s . This term adds a slope of – τ 1 20dB/decade to the magnitude characteristic and therefore the composite characteristic has a net slope of –20 + 20 = 0dB/decade, i.e., the composite characteristic is flat until it encounters another break frequency. The next break frequency is due to the simple zero 1 = 10 rad / s . At this point, the composite curve changes with break frequency at ω = 0.1 slope to +20dB/decade. The remaining two terms in the network function are poles with 1 1 break frequencies at ω = = 50 rad / s and ω = = 100 rad / s . Since each adds a 0.02 0.01 slope of –20dB/decade, the composite characteristic shifts from +20db/decade to 0dB/decade and then to –20dB/decade. The total composite characteristic is shown in Fig. S12.2(a). H

dB 0dB

+20

-20dB/decade

+20dB/ decade 1

2

10

50

100

Log ω (rad/s)

Fig. S12.2(a) The composite phase characteristic for the network function is shown in Fig. S12.2(b).

98

Phase in degrees + 90° 0° − 90° 1

50

10

Log ω (rad/s)

100

Fig. S12.2(b) At low frequencies, the initial phase is +90° due to the zero at the origin. The first break frequency encountered is due to the pole term (jω + 1) with break frequency at ω = 1 rad/s. Thus the phase shifts toward 0° on an arc tangent curve that provides -45° of phase at ω = 1 rad/s. The phase proceeds toward 0° until it encounters the zero with a break 1 frequency of ω = = 10 rad / s . This term shifts the phase toward +90° going through 0.1 +45° at the break frequency. The two remaining poles shift the composite phase back to 0° and finally to -90° as the characteristic indicates. 12.3

Examining the magnitude characteristic we note that at low frequencies the characteristics has an initial slope of –20dB/decade indicating a single pole at the origin. Furthermore, this initial slope passes through the 20dBs at ω = 1 rad/s. Since the slope is –20dB/decade, this initial slope will cross the 0dB line at ω = 10 rad/s. Therefore, the constant term, i.e., gain, in the network function is 10. Since the slope changes at ω = 1 rad/s from –20dB/decade to 0dB/decade, there is a simple zero at this break frequency. At ω = 20 rad/s, the slope changes again. This time the slope shifts from 0dB/decade to – 20dB/decade indicating the presence of a simple pole with break frequency ω = 20 rad/s. Finally, there is another simple pole with break frequency ω = 100 rad/s. Therefore, the composite network function is H ( jω ) =

12.4

10 ( jω + 1) ( jω) ⎛⎜ jω + 1⎞⎟ ⎛⎜ jω + 1⎞⎟ ⎝ 20 ⎠ ⎝ 100 ⎠

For this network, the resonant frequency is ω0 = =

1 LC 1

(200 × 10 ) (50 × 10 ) −6

= 10,000 rad / s

−6

99

The quality factor is ω0 L R (10 4 ) (200 × 10 −6 ) = 1 =2

Q=

And the bandwidth is BW =

ω0 Q

10 4 = 2 = 5000 rad / s If the resistance, R, is now changed from 1Ω to 0.1Ω the resonant frequency is unaffected. However, the Q changes to ω0 L R (10 4 ) (200 × 10 −6 ) = 0.1 = 20

Q=

And the bandwidth is BW =

ω0 Q

10 4 20 = 500 rad / s =

A sketch of the two frequency characteristics is shown in Fig. S12.4.

100

R = 1Ω

R = 0.1Ω

Log ω

10,000 rad/s

Fig. S12.4 Note that the higher value of Q, i.e., lower value of R, produces a more selective circuit with a much smaller bandwidth. 12.5

(a) Using voltage division, we can express the output as ⎤ ⎡ ⎥ ⎢ R ⎥ VS V0 = ⎢ ⎢ R + jω L + 1 ⎥ ⎢⎣ jωC ⎥⎦

or V0 = VS

R 1 ⎞ ⎛ R + j ⎜ ωL − ⎟ ωC ⎠ ⎝

And therefore V0 = VS

[(RCω)

RCω 2

+ (ω 2 LC − 1)

2

]

1 2

(b) The upper and lower cut off frequencies are the roots of the characteristic equation, i.e., the denominator of the transfer function. At the lower cut off frequency ω2LC – 1 = -RCω or ω2 +

R ω − ω 02 = 0 L

101 where, of course, ω 02 =

1 . With the component values, this function becomes LC ω2 + 2000ω – 105 = 0

Solving for ωLO, we obtain ωL 0 =

(2000)

− 2000 +

2

+ 4 × 10 5

2

= 48.8 rad / s At the upper cut off frequency ω2LC – 1 = +RCω or ω2 −

R ω − ω 02 = 0 L

and ωHI is ω HI =

2000 +

(2000)

2

+ 4 × 10 5

2

= 2048.8 rad / s Therefore, the bandwidth is BW = ω HI − ω LO =

R L

= 2048.8 − 48.8 = = 2000 rad / s (c) Since the resonant frequency is ω0 = =

1 LC 1 2 × 5 × 10 − 6

= 316.23 rad / s

4000 2

102

The magnitude characteristic for the function is shown in Fig. S12.5. dB -20dB/decade

+20

48.8

ω0 = 316.33

2048.8

Fig. S12.5

Log ω (rad/s)

103 CHAPTER 13 PROBLEMS

13.1

If f(t) = e-at sin bt, find F(s) using (a) the definition of the Laplace Transform and (b) the fact that L[e-at f(t)] = F(s + a).

13.2

Find f(t) if F(s) is given by the expression F (s ) =

13.3

24s (s + 2) (s + 4) (s + 6)

Find f(t) if F(s) is given by the expression F (s ) =

13.4

Find f(t) if F(s) is given by the expression F (s ) =

13.5

4(s + 4) s (s + 8s + 20) 2

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

Given the function F (s ) =

24 (s + 10) s (s + 2) (s + 4 )

Find the initial and final values of the function by evaluating it in both the s-domain and time domain.

104 CHAPTER 13 SOLUTIONS

13.1

(a) By definition ∞

F (s ) = ∫ f (t ) e −st dt 0

And since f(t) = e-at sin bt ∞

F (s ) = ∫ e − at sin bt e −st dt 0

Using Euler’s identity ∞ ⎡ e jbt − e − jbt ⎤ F (s ) = ∫ e − (s + a ) t ⎢ ⎥ dt 0 2j ⎦ ⎣ s a jb t s a jb ) t ( ) ( − + − − + + ∞e −e =∫ dt 0 2j

Evaluating the integral ⎡ ⎤ 1 1 ⎢ s + a − jb − s + a + jb ⎥ ⎣ ⎦ b = (s + a ) 2 + b 2

=

1 2j

(b) In this case f(t) = sin bt. Then ∞

F (s ) = ∫ e −st sin bt dt 0

Again, using the Euler identity ∞ ⎛ e jbt − e − jbt ⎞ ⎟⎟ dt F (s ) = ∫ e −s t ⎜⎜ 0 2j ⎝ ⎠ ∞ 1 − ( s − jb ) t = − e − (s + jb ) t ) dt ∫ (e 2j 0

Evaluating the integral

105 ⎡ 1 1 ⎤ ⎢ s − jb − s + jb ⎥ ⎣ ⎦ b = 2 s + b2

=

1 2j

Then using the fact that L[e-at f(t)] = F(s + a) where in this case f(t) = sin bt and F (s ) =

b s + b2 2

we find that F (s + a ) = L [e − at f (t )]

= L [e − at sin bt ] b = (s + a ) 2 + b 2

13.2

The expression F (s ) =

24s (s + 2) (s + 4) (s + 6)

can be written in a partial fraction expansion of the form k k k 24s = 1 + 2 + 3 (s + 2) (s + 4) (s + 6) s + 2 s + 4 s + 6

Multiplying the entire equation by the term s + 2 yields k (s + 2 ) k 3 (s + 2 ) 24s + = k1 + 2 (s + 4) (s + 6) s+4 s+6

If we now evaluate each term at s = -2, we find that the last two terms on the right side of the equation vanish and we have 24s (s + 4) (s + 6)

= k1 s = −2

− 6 = k1

Repeating this procedure for the two remaining terms in the denominator, i.e., (s + 4) and (s + 6) yields

106

24s (s + 2) (s + 6)

= k2 s = −4

24 = k 2 And 24s (s + 2) (s + 4)

= k3 s = −6

− 18 = k 3 Now the function F(s) can be written in the form F (s ) =

−6 24 18 + − s+2 s+4 s+6

The reader can check the validity of this expansion by recombining the terms to produce the original expression. Once F(s) is in this latter form, we can use the transform pair

L [e − at ] =

1 s+a

And hence f (t ) = [− 6e −2 t + 24e −4 t − 18e −6 t ] u (t )

13.3

We begin by writing the function in a partial fraction expansion. Therefore, we need to know the roots of the quadratic term. We can either employ the quadratic formula or recognize that s2 + 8s + 20 = s2 + 8s + 16 + 4 = (s + 4)2 + 4 = (s + 4 – j2) (s + 4 + j2) Hence, the function F(s) can be written as F (s ) =

k k1 k 1* 4 (s + 4 ) = 0 + + s (s + 4 − j2 ) (s + 4 + j2 ) s s + 4 − j2 s + 4 + j2

Multiplying the entire equation by s and evaluating it at s = 0 yields

107

4(s + 4) s + 8s + 20

= k0

2

s=0

4 = k0 5 Using the same procedure for k1, we obtain 4(s + 4 ) s (s + 4 + j2 )

= k1 s = − 4 + j2

1 = k1 −2+ j −1 = k1 2− j

− (2 + j) = k1 5 1 5

∠206.56° = k 1

Then, we know that 1 5

∠ − 206.56° = k 1*

Now using the fact that ⎡ k 1 ∠θ k1 ∠ − θ ⎤ − at + L⎢ ⎥ = 2 k 1 e cos (bt + θ) + − s a jb s a jb + + ⎣ ⎦ The function f(t) is ⎡4 ⎤ 2 −4 t f (t ) = ⎢ + e cos (2t + 206.56°)⎥ u (t ) 5 ⎣5 ⎦ 13.4

In order to perform a partial fraction expansion on the function F(s), we need to factor the quadratic term. We can use the quadratic formula or simply note that (s + 1) (s + 1) = s2 + 2s + 1. Therefore, F(s) can be expressed as F (s ) =

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

108

or in the form F (s ) =

k k 12 k 12 (s + 2 ) = 11 + + 2 2 2 (s + 1) (s + 3) s + 1 (s + 1) s + 3

If we now multiply the entire equation by (s + 1)2, we obtain k (s + 1) 12 (s + 2) = k 11 (s + 1) + k 12 + 2 s+3 s+3

2

Now evaluating this equation at s = -1 yields 12 (s + 2 ) s+3

= k 12 s = −1

6 = k 12 In order to evaluate k11 we differentiate each term of the equation with respect to s and evaluate all terms at s = -1. Note that the derivative of k12 with respect to s is zero, the derivative of the last term in the equation with respect to s will still have an (s + 1) term in the numerator that will vanish when evaluated at s = -1, and the derivative of the first term on the right side of the equation with respect to s simply yields k11. Therefore, d ds

⎡12 (s + 2) ⎤ = k 11 ⎢ s+3 ⎥ ⎦ s = −1 ⎣

(s + 3) (12) − 12 (s + 2) (1) (s + 3)

= k 11

2

s = −1

3 = k 11 Finally, 12 (s + 2) (s + 1)2

= k2 s = −3

− 3 = k2 And therefore, F(s) can be expressed in the form F (s ) =

3 6 3 + − 2 s + 1 (s + 1) s+2

109 Using the transform pairs, we find that f (t ) = [3e − t + 6 te − t − 3e −2 t ] u (t )

13.5

First, let us use the Theorems to evaluate the function in the s-domain. The initial value can be derived from the Theorem lim f(t) = lim sF(s) t→0 s→∞ Therefore, lim sF (s ) =

lim ⎡ 24 (s + 10 ) ⎤ s → ∞ ⎢⎣ (s + 2 ) (s + 4 ) ⎥⎦

⎡ 24s + 240 ⎤ s → ∞ ⎢⎣ s 2 + 6s + 8 ⎥⎦ ⎡ 24 240 ⎤ lim ⎢ s + s 2 ⎥ = ⎥ ⎢ 6 8 s→∞ ⎢ 1+ + 2 ⎥ ⎢⎣ s s ⎥⎦ =0 =

lim

The final value is derived from the expression lim f(t) = lim sF(s) t→∞ s→0 Hence, lim sF (s ) lim ⎡ 24 (s + 10 ) ⎤ = s→0 s → 0 ⎢⎣ (s + 2) (s + 4 ) ⎥⎦ 240 8 = 30 =

The time function can be derived from a partial fraction expansion as F (s ) =

where

k k k 24 (s + 10 ) = 0 + 1 + 2 s (s + 2 ) (s + 4 ) s s+2 s+4

110

24 (s + 10 ) = k = 30 (s + 2) (s + 4) s = 0 0 24 (s + 10 ) = k 1 = − 48 s (s + 4) s = − 2 24 (s + 10 ) = k 2 = 18 s (s + 2 ) s = − 4

Hence, F(s ) =

30 48 18 − + s s+2 s+4

and then f (t ) = [30 − 48e −2 t + 18e −4 t ] u (t )

Given this expression, we find that lim f (t ) = [30 − 48 + 18] = 0 t→0 and lim f (t ) = [30 − 0 + 0] = 30 t→∞

111 CHAPTER 14 PROBLEMS

14.1

Find v 0 (t ) , t > 0 in the circuit in Fig. 14.1 using (a) nodal analysis, (b) source transformation and (c) Norton’s Theorem. 2Η

2Ω

2 u(t) V +-

1F

2Ω

2 u(t) A

+ v o (t ) -

Fig. 14.1 14.2

Find i 0 (t ) , t > 0 in the circuit in Fig. 14.2 using (a) loop equations and (b) Thevenin’s Theorem. 2 u(t) A 1F

1H

e − t u (t ) A

1Ω

1Ω

Fig. 14.2 14.3

Find i 0 (t ) , t > 0 in the circuit in Fig. 14.3. t=0

2H 1Ω

12 u(t) V +-

2Ω

i o (t )

t=0

2Ω

+- 12 u(t) V

1F

Fig. 14.3 14.4

V0 (s ) , Vs (s ) (b) the undamped natural frequency, (c) the damping ratio and (d) the general form of the response of the network to a unit step function.

Given the network in Fig. 14.4, determine (a) the voltage transfer function G (s ) =

v s (t )

2Η +-

+ 1F

1Ω v o (t ) -

Fig. 14.4 14.5

Find the steady-state response v 0 (t ) for the network in Fig. 14.5.

112 v s (t ) = 6 cos 2t u (t ) V -+ 1Ω 1Ω

1Η

1F 1Ω

Fig. 14.5

+ v o (t ) -

113 CHAPTER 14 SOLUTIONS

14.1

(a) Consider the transformed network in Fig. S14.1(a). V1 (s ) 2s

2 + s -

2

2 s

1 s

2

+ Vo (s ) -

Fig. S14.1(a) A brute force approach to this problem would be to write two nodal equations for the nodes labeled V1 (s ) and V0 (s ) . Using KCL and summing the currents leaving each node yields the two linearly independent equations 2 s − 2 + V1 (s ) − V0 (s ) = 0 1 2s s 2+ s

V1 (s ) −

and V0 (s ) − V1 (s ) V0 + =0 1 2 2+ s Solving these equations for V0 (s ) and then performing the inverse Laplace transform would yield v 0 (t ) . Another approach that might be simpler would be to write a node equation for V1 (s ) , ignoring V0 (s ) , and then use voltage division to derive V0 (s ) once V1 (s ) is known. Applying KCL at V1 (s ) yields 2 s − 2 + V1 (s ) = 0 1 2s s 4+ s

V1 (s ) −

Rearranging terms we obtain ⎡1 s ⎤ 1 2 V1 (s ) ⎢ + ⎥= 2 + s ⎣ 2s 4s + 1⎦ s

114

or ⎡ 2s 2 + 4s + 1⎤ 2s + 1 V1 (s ) ⎢ ⎥= s2 ⎣ 2s (4s + 1) ⎦

Solving for V1 (s ) yields V1 (s ) =

2 (2s + 1) (4s + 1) s (2s 2 + 4s + 1)

Now applying voltage division ⎞ ⎛ ⎜ 2 ⎟ ⎟ V0 (s ) = V1 (s ) ⎜ 1⎟ ⎜ ⎜4+ ⎟ s⎠ ⎝ 4 (2s + 1) = 2 2s + 4s + 1

This function can be written in partial fraction expansion form as 4s + 2 s 2 + 2s +

1 2

=

A B + s + 0.29 s + 1.71

where A=

4s + 2 = 0.59 s + 1.71 s = − 0.29

B=

4s + 2 = 3.41 s + 0.29 s = −1.71

and

Therefore, v 0 (t ) = [0.59 e −0.29 t + 3.41e −1.71t ] u (t ) V

115 (b) Using source transformation we can convert the voltage source in series with the inductor to a current source in parallel with the inductor yielding the network in Fig. S14.1(b). 1 s2

2 s

2s

1 s

2

2

+ Vo (s ) -

Fig. S14.1(b) Adding the current sources that are in parallel produces an equivalent source of I EQ (s ) =

1 2 2s + 1 + = s2 s s2

The network is then reduced to that shown in Fig. S14.1(c). I o (s )

2s + 1 s2

2s

2

1 s

2

+ Vo (s ) -

Fig. S14.1(c) We could, at this point, transform the current source and inductor back to a voltage source in series with the inductor. However, we can simply apply current division at this point with Ohm’s Law and derive the answer immediately. ⎛ ⎞ ⎟ 2s + 1 ⎜⎜ 2s ⎟ I 0 (s ) = 1 s2 ⎜ ⎟ ⎜ 2s + 2 + + 2 ⎟ s ⎝ ⎠ 4s + 2 = 2 2s + 4s + 1

And V0 (s ) = 2 I 0 (s ) =

4s + 2 s 2 + 2s +

which is identical to the expression obtained earlier.

1 2

116 (c) To apply Norton’s Theorem we will break the network to the right of the current source and form a Norton equivalent circuit for the elements to the left of the break as shown in Fig. S14.1(d). 2s

2 + s

I sc (s )

2 s

Fig. S14.1(d) The short-circuit current is 2 2 I sc (s ) = s + 2s s 2s + 1 = s2 And the Thevenin equivalent impedance is derived from the network in Fig. S14.1(e) as ZTH(s) = 2s 2s Z TH (s )

Fig. S14.1(e) Therefore, attaching the Norton equivalent circuit to the remainder of the network yields the circuit in Fig. S14.1(f) which is the same as that in Fig. S14.1(c). 2s + 1 s2

2s

2

1 s

2

Fig. S14.1(f) 14.2

(a) the transformed network is shown in Fig. S14.2(a).

+ Vo (s ) -

117 2 s I1 (s )

1 s

1

s

1

I 2 (s )

I 0 (s )

I 3 (s )

1 s +1

Fig. S14.2(a) Since there are three “window panes” we will need three linearly independent simultaneous equations to calculate the loop currents. Two of the currents go directly through the current sources and therefore two of the three equations are 2 s −1 I 3 (s ) = s +1

I1 (s ) =

The remaining equation is obtained by using KVL around the loop defined by the current I2(s). That equations is 1 I 2 (s ) +

1 [I 2 (s ) − I1 (s )] + 1 [I 2 (s ) − I 3 (s )] = 0 s

Substituting the first two equations into the last equation yields 1 ⎡ 1 ⎤ 2 I 2 (s ) ⎢1 + + 1⎥ = 2 − s +1 ⎣ s ⎦ s

or I 2 (s ) =

Then

− s 2 + 2s + 2 s (s + 1) (2s + 1)

118 I 0 (s ) = I 2 (s ) − I 3 (s )

=

− s 2 + 2s + 2 1 + s (s + 1) (2s + 1) s + 1

s 2 + 3s + 2 s (s + 1) (2s + 1) s+2 = s (2s + 1) 1 (s + 2) 2 = 1⎞ ⎛ s ⎜s + ⎟ 2⎠ ⎝

=

Expressing this function in partial fraction expansion form we obtain 1 (s + 2) A B = + I 0 (s ) = 2 1 1⎞ s ⎛ s+ s⎜s + ⎟ 2 2⎠ ⎝ where 1 (s + 2) 2 =2 A= 1 s+ 2 s=0 1 (s + 2) B= 2 s

=− s = −

3 2

1 2

Therefore, 3 −t ⎤ ⎡ i 0 (t ) = ⎢ 2 − e 2 ⎥ u (t ) A 2 ⎣ ⎦ (b) In order to apply Thevenin’s Theorem, we first break the circuit between the points where the current I 0 (s ) is located as shown in Fig. S14.2(b).

119 2 s I1 (s ) s

1 s

1

1 s +1

+ V0 C (s ) -

I 2 (s )

Fig. S14.2(b) Applying KVL to the closed path in the lower left-hand corner of the network yields 1 I 2 (s ) +

1 [I 2 (s ) − I1 (s )] + V0C (s ) = 0 s

where 2 s −1 I 2 (s ) = s +1 I1 (s ) =

Combining these equations we obtain 1 1 2 + + 2 s + 1 s (s + 1) s s+2 = 2 s

V0 C (s ) =

The Thevenin equivalent impedance obtained by looking into the open circuit terminals with all sources made zero (current sources open-circuited) is derived from the network in Fig. S14.2(c). 1 s s ZTH (s )

Fig. S14.2(c) Clearly,

120 Z TH (s ) =

s +1 1 +1= s s

If the resistor containing the I 0 (s ) is now attached to the Thevenin equivalent circuit we obtain the network in Fig. S14.2(d). s +1 ZTH (s ) = s I 0 (s ) V0 C (s ) =

s+2 + s2

1

Fig. S14.2(d) Then s+2 s2 I 0 (s ) = s +1 +1 s s+2 = s (2s + 1) which is identical to the result obtained earlier. 14.3

To begin, we first determine the initial conditions in the network prior to switch action. In the steady-state period prior to switch action, the capacitor looks like an open-circuit and the inductor acts like a short-circuit. Therefore, in this time interval the circuit appears as that shown in Fig. S14.3(a). i L (0 ) 2Ω + v C (0 ) -

2Ω

+-

12 u(t) V

Fig. S14.3(a) This network indicates that in the steady-state condition for t < 0 i L (0 ) = and

12 = 3A 2+2

121 ⎛ 2 ⎞ ⎟⎟ = 6V v C (0 ) = 12⎜⎜ ⎝2 + 2⎠

These conditions cannot change instantaneously and hence the network for t > 0 is shown in Fig. S14.3(b). 2H 1Ω 12 u(t) V

+ -

2Ω

i L (0 ) = 3A

1F

i 0 (t )

+ v C (0 ) = 6V -

Fig. S14.3(b) The corresponding transformed network is shown in Fig. S14.3(c). 6 2s 1 +1 12 + s -

I1 (s )

2

I 2 (s ) I 0 (s )

s + 1 s

Fig. S14.3(c) Since the current I 0 (s ) is located in the center leg of the circuit, we will employ loop equations and specify them such that one of the loops is the same as I 0 (s ) . The two equations for the loop currents specified in the network are

− 12 + 1(I1 (s ) + I 2 (s )) + 2 I1 (s ) = 0 s 1 − 12 1 + 1(I1 (s ) + I 2 (s )) + 6 + 2s I 2 (s ) + I 2 (s ) + = 0 s s s Solving the second equation for I 2 (s ) yields I 2 (s ) =

11 − 6s − s I1 (s ) 2s 2 + s + 1

Substituting this value into the first equation we obtain 1 (30s 2 6 I1 (s ) = I 0 (s ) = ⎛ s⎜s2 + ⎝

+ s + 12 )

1⎞ 1 s+ ⎟ 2⎠ 3

122 The roots of the quadratic term in the denominator, obtained using the quadratic formula, are s1 , s 2 = −

1 17 ±j 6 6

The expression for the desired current can now be written in partial fraction expansion form as 1 (30 s 2 + s + 12) A B B* 6 = + + ⎛ 1 17 1 17 1 17 ⎞ s ⎟ s− + j s+ + j s ⎜⎜ s + ± j 6 6 6 6 6 6 ⎟⎠ ⎝ where 1 (30 s 2 + s + 12) 6 =A 1 1 2 s + s+ 3 2 s=0 4=A and 1 (30 s 2 + s + 12) 6 ⎛ 1 17 ⎞ ⎟ s ⎜⎜ s + + j ⎟ 6 6 ⎠ ⎝

=B s = −

1 17 + j 6 6

The evaluation of this last term involves a lot of tedious, but straight forward, complex algebra. The result is 1.09 ∠62.74° = B Therefore, knowing the values for A and B we can write the final expression for the current in the time domain as t ⎡ ⎛ 17 ⎞⎤ − 6 i 0 (t ) = ⎢4 + 2 (1.09) e cos ⎜⎜ t + 62.74° ⎟⎟⎥ u (t ) A ⎝ 6 ⎠⎦⎥ ⎣⎢

14.4

(a) The transformed network is shown in Fig. S14.4.

123

Vs (s ) +-

2s

1 s

1

+ V0 (s ) -

Fig. S14.4 Using voltage division, the voltage transfer function can be expressed as

⎛1⎞ 2⎜ ⎟ ⎝s⎠ 1 2+ V0 (s ) s G (s ) = = Vs (s ) ⎛1⎞ 2⎜ ⎟ s 2s + ⎝ ⎠ 1 2+ s 2 = 2 4s + 2s + 2 1 2 = 1 1 s2 + s + 2 2 (b) The denominator, or characteristic equation, is of the form s 2 + 2ζω 0 s + ω 02 Therefore the undamped natural frequency is

ω 02 =

1 2

and 1

ω0 =

2

= 0.707 r / s

(c) The damping ratio is derived from the expression 2ζω 0 = and using the value for ω 0 we obtain

1 2

124

ζ = 0.354 (d) If the input to the network is a unit step function then 1 2 V0 (s ) = 1 1⎞ ⎛ s⎜s2 + s + ⎟ 2 2⎠ ⎝ By employing the quadratic formula, we can write this expression in the form

V0 (s ) =

1 2

⎛ 1 7⎞ ⎟ s ⎜⎜ s + ± j ⎟ 4 4 ⎝ ⎠

and therefore the general form of the response is 1 ⎡ ⎛ 7 ⎞⎤ − t v 0 (t ) = ⎢A + B e 4 cos ⎜⎜ t + θ ⎟⎟⎥ u (t ) v ⎢⎣ ⎝ 4 ⎠⎥⎦

14.5

The transformed circuit is shown in Fig. S14.5. Vs (s ) V (s ) -+

1

s

1

1 s

1

1

+ V0 (s ) -

Fig. S14.5 Although the network contains three non-reference nodes, we will try to simplify the analysis by first using a supernode to find V1(s) and then employing voltage division to determine V0 (s ) . KCL for the supernode containing the voltage source is V1 (s ) − Vs (s ) V1 (s ) − Vs (s ) V1 (s ) V1 (s ) + + + =0 1 1 s 2 s Solving this equation for V1(s) yields

125 ⎞ ⎛ ⎟ ⎜ s 1 + ⎟ Vs (s ) V1 (s ) = ⎜ ⎜ s2 + 3 s + 1⎟ ⎟ ⎜ 2 ⎠ ⎝

And then using voltage division ⎛ 1 ⎞ ⎟⎟ V0 (s ) = V1 (s ) ⎜⎜ ⎝1 + 1⎠

so that ⎤ ⎡ 1 ⎢ 2 (s + 1) ⎥ V0 (s ) = ⎢ ⎥ Vs (s ) ⎢ s 2 + 3 s + 1⎥ ⎥⎦ ⎢⎣ 2

Therefore, 1 (s + 1) H (s ) = 2 3 s2 + s + 1 2 Since vs(t) = 6 cos 2t u(t) V, then VM = 6 and ω 0 = 2 . Hence, 1 ( j2 + 1) 2 H ( j2 ) = ( j2)2 + 3 ( j2) + 1 2 1 − (2.236 ∠63.43°) = 2 4.24 ∠ − 45°

= 0.264 ∠ − 71.57°

and H ( j2 ) = 0.264

φ ( j2 ) = − 71.57°

Therefore,

126 v 0 ss (t ) = VM H ( j2 ) cos (2 t + φ ( j2 ))

= 1.58 cos (2 t − 71.57°) V

127 CHAPTER 15 PROBLEMS

15.1

Find the exponential Fourier series for the waveform in Fig. 15.1. v(t) 2 1 -3

-2

-1

0

1

2

3

4

5

t(s)

Fig. 15.1 15.2

Determine the trigonometric Fourier series for the function shown in Fig. 15.2. v(t)

π

−π

2π

π

0

t

3π

Fig. 15.2 15.3

Find the trigonometric Fourier series for the waveform shown in Fig. 15.3. v(t) A

−π

0

π

t

2π

Fig. 15.3 15.4

Find the steady-state voltage v 0 (t ) in the circuit in Fig. 15.4 if the input voltage is the waveform shown in Fig. 15.3 with A = 1V. 1Ω

v s (t ) +-

1H 2Ω

Fig. 15.4

+ v 0 (t ) -

128 15.5

Given the network in Fig. 15.4 with the input source v s (t ) = 10 e −2 t u (t ) V , use the transform technique to find v 0 (t ) .

129 CHAPTER 15 SOLUTIONS

15.1

An examination of the waveform indicates that the period T = 3 and ω 0 = The Fourier coefficients are determined from the expression cn =

1T − jnω t dt ∫ v (t ) e T0 0

or in this case 2 1 1 c n = ⎡ ∫ 2 e − jnω t dt + ∫ 1 e − jnω t dt ⎤ ⎥⎦ 1 3 ⎢⎣ 0 2 1 −1 2 e − jnω t 0 + e − jnω t 1 = 3 jnω0 0

0

[

0

0

]

=

−1 [2 (e− jnω − 1) + e− j2nω − e− jnω 3 jnω0

=

−1 [e− jnω + e− j2nω − 2] 3 jnω0

0

0

0

0

−1 j2πn

− j 4 πn ⎡ − j23πn ⎤ 3 e e + − 2⎥ ⎢ ⎣ ⎦ − j 2 πn − j 4 πn 1 ⎛ ⎞⎞ ⎛ ⎜⎜ 2 − ⎜ e 3 + e 3 ⎟ ⎟⎟ = j2πn ⎝ ⎠⎠ ⎝ − jnπ ⎡ ⎛ jn3π 3 1 ⎢ e e + ⎜ 2 − 2 e − jπn ⎜ = j2πn ⎢ 2 ⎜ ⎢⎣ ⎝

=

=

1 ⎛ ⎛ nπ ⎞ ⎞ ⎜⎜1 − e − jnπ cos ⎜ ⎟ ⎟⎟ jnπ ⎝ ⎝ 3 ⎠⎠

In addition 1 3 ∫ v (t ) dt T0 2 1 1 = ⎡ ∫ 2dt + ∫ 1dt ⎤ ⎥⎦ 1 3 ⎢⎣ 0 =1

c0 =

Therefore,

0

⎞⎤ ⎟⎥ ⎟⎥ ⎟ ⎠⎥⎦

]

2π 2π = . T 3

130 v (t ) = 1 + ∑ ∞

0

n = −∞ n≠0

15.2

1 ⎛ ⎛ nπ ⎞ ⎞ ⎜⎜1 − e − jnπ cos⎜ ⎟ ⎟⎟ e jnω t jnπ ⎝ ⎝ 3 ⎠⎠

Since the waveform does not exhibit any symmetry, we will have to determine the coefficients a 0 , a n and b n . The a 0 coefficient is a0 =

1 π ∫ v (t ) dt 2π − π

where, of course, v(t) = t in the interval 0 ≤ t ≤ π and zero elsewhere and ω 0 =

2π = 1. T

1 π ∫ t dt 2π 0 π π⎞ 1 ⎛⎜ t 2 = = ⎟ 2π ⎜⎝ 2 0 4 ⎟⎠

a0 =

Recall that a 0 is simply the average value of the waveform and therefore can be calculated by dividing the area under the curve ( Area = interval (2π) which yields

π . 4

1 1 π2 bh = (π) (π ) = ) by the 2 2 2

The an coefficient is an =

2 π ∫ t cos nt dt 2π 0

Using a table of integrals, we find that π

1 1⎡1 ⎤ a n = ⎢ 2 cos nt + t sin nt ⎥ π ⎣n n ⎦0

The second term is zero at t = π and 0 and the first term can be written as an =

[

]

1 (− 1)n − 1 2 πn

since the cosine term will be +1 or –1 depending upon the value of n. Thus

131

(− 1) =

an

−1

n

πn

2

In addition, 2 π ∫ t sin t dt 2π 0

bn =

Once again, using a set of integral tables we find that π

1 1⎡1 ⎤ b n = ⎢ 2 sin nt − t cos nt ⎥ π ⎣n n ⎦0

The first term will be zero at each limit, but the second term is nonzero at the upper limit and thus −π (− 1)n nπ n − (− 1) = n

bn =

Therefore, the Fourier series expansion is n n ( − 1) π ∞ ⎡ (− 1) − 1⎤ v (t ) = + ∑ ⎢ sin nt ⎥ cos nt − 4 n = 1⎣ πn 2 ⎦ n

15.3

To begin our analysis we first note that the waveform is an even function and therefore b n = 0 for all n. Thus, we need to find only the a 0 and an coefficients. For this waveform, we note that T = 2π and ω 0 = a0 =

2π = 1 . a 0 is now T

1 π ∫ v (t ) dt 2π − π

However, recall that a 0 is simply the average value and we can easily compute this number without resorting to solving the above integral. This average value can be obtained by dividing the area by the base, i.e. ⎛1 ⎞ ⎛1 ⎞ Area = 2 ⎜ bh ⎟ = 2 ⎜ (πA )⎟ = πA ⎝2 ⎠ ⎝2 ⎠

132 The base is 2π and therefore a0 =

πA A = 2π 2

Because the function is even, an =

4 πA ∫ t cos nt dt 2π 0 π

where the equation of the straight line function in the interval 0 ≤ t ≤ π is

an =

A t . So, π

2A π ∫ t cos nt dt π2 0

Using a table of integrals, we find that π

1 ⎡1 ⎤ ⎢⎣ n 2 cos nt + n t sin nt ⎥⎦ 0 2A ⎡ 1 ⎤ = 2 ⎢ 2 (cos nπ − 1)⎥ π ⎣n ⎦ 2A (cos nπ − 1) = (πn )2 − 4A for n odd = (πn )2 =0 for n even

2A an = 2 π

Therefore, v (t ) = 15.4

∞ − 4A A + ∑ cos nt 2 = 1 (πn ) 2 nn odd

The input voltage for the circuit in Fig. 14.4 is given by the expression v s (t ) =

∞ 1 −4 + ∑ cos nt 2 = 1 (πn ) 2 nn odd

where ω 0 = 1 . The output voltage for the network can be derived using voltage division as

133 V0 ( jω) =

2 V ( jω ) ( 1) ( jω) s 2+ 1 + jω

⎡ 2 (1 + jω) ⎤ =⎢ ⎥ Vs ( jω) ⎣ 2 + 3 jω ⎦

and since ω 0 = 1 ⎡ 2 (1 + jn ) ⎤ V0 (n ) = ⎢ ⎥ Vs (n ) ⎣ 2 + 3 jn ⎦

Since Vs (dc ) =

1 2 ⎛2⎞ ⎛1⎞ V0 (dc ) = ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ 1 = 2

Furthermore, − 4 ⎡ 2 (1 + j) ⎤ ⎥ = 0.318 ∠168.69° ⎢ π2 ⎣ 2 + 3j ⎦ − 4 ⎡ 2 (1 + j3) ⎤ V0 (3ω 0 ) = 2 ⎢ = 3.09 × 10 − 2 ∠174.09° 9π ⎣ 2 + j9 ⎥⎦

V0 (ω0 ) =

⎡ 2 (1 + j5) ⎤ −2 ⎢ 2 j15 ⎥ = 1.09 × 10 ∠176.28° ⎦ ⎣ + − 4 ⎡ 2 (1 + j7 ) ⎤ −3 V0 (7ω 0 ) = ⎥ = 5.54 × 10 ∠177.31° 49π 2 ⎢⎣ 2 + j21 ⎦

V0 (5ω 0 ) =

−4 25π 2

Hence, v 0 (t ) =

15.5

1 + 0.318 cos (t + 168.69°) + 3.09 × 10 − 2 cos (3t + 174.09°) 2 + 1.09 × 10 − 2 cos (5t + 176.28°) + 5.54 × 10 −3 cos (7 t + 177.31°) + …

The input function to the network can be expressed in the form

134 Vs ( jω) =

10 jω + 2

The transfer function for the network obtained in the previous problem is G ( jω) =

2 (1 + jω) 2 + 3 jω

Then using the time convolution property of the Fourier transform we can express the output of the circuit in the form V0 ( jω) = G ( jω) Vs ( jω) ⎡ 2 (1 + jω) ⎤ ⎡ 10 ⎤ =⎢ ⎥ ⎥⎢ ⎣ 2 + 3 j ω ⎦ ⎣ 2 + jω ⎦ 20 ( jω + 1) 3 = ( jω + 2) ⎛⎜ jω + 2 ⎞⎟ 3⎠ ⎝ which can be written as a partial fraction expansion of the form 20 ( jω + 1) A B 3 = + 2 ( jω + 2) ⎛⎜ jω + 2 ⎞⎟ jω + 2 jω + 3 3⎠ ⎝

Evaluating the constants yields 20 ( jω + 1) 3 =A=5 2⎞ ⎛ ⎜ jω + ⎟ 3 ⎠ jω = − 2 ⎝ 20 ( jω + 1) 3 ( jω + 2 )

=B= jω = −

Therefore,

2 3

5 3

135

V0 ( jω) =

5 + jω + 2

5 3 jω +

2 3

And ⎡ − 2 t 5 − 23 t ⎤ v 0 (t ) = ⎢5 e + e ⎥ u (t ) V 3 ⎣ ⎦

136 CHAPTER 16 PROBLEMS

16.1

Find the Y parameters for the network shown in Fig. 16.1 and then find the output voltage of the two-port when a 4mA current source is connected to the input port and a 4kΩ load is connected to the output port. 4kΩ 4kΩ

4kΩ

Fig. 16.1 16.2

Find the Z parameters for the circuit shown in Fig. 16.2, and then find the current in a – j4Ω capacitor connected to the output port when a 6 ∠0° V source is connected to the input port. j2Ω -j1Ω 2Ω

Fig. 16.2 16.3

Find the hybrid parameters for the circuit shown in Fig. 16.3. What conclusion can be drawn from this result. I1 aΩ I2 + V1 -

bV2 +-

CI1

+ dΩ V2 -

Fig. 16.3 16.4

Find the transmission parameters of the network in Fig. 16.1 by treating the circuit as a cascade interconnection of elements.

16.5

Check the validity of the answers obtained in problems 16.1 and 16.4 by using the parameter conversion formulas to convert the Y parameters in problem 16.1 to the transmission parameters in problem 16.4.

137 CHAPTER 16 SOLUTIONS

16.1

The equations for a two-port in terms of the Y parameters are I1 = y11 V1 + y12 V2 I2 = y21 V1 + y22 V2 Since y 11 =

I1 with V2 = 0 , the network in Fig. S16.1(a) is used to find y11. V1 4kΩ I1 + V1 -

4kΩ

4kΩ

V2 = 0

Fig. S16.1(a) Since V2 is made zero with the short-circuit, the 4kΩ resistor on the right is shorted and V1 = I1 (4k 4k )

or I1 V1

= y 11 = V2 = 0

1 S 2k

The parameter y12 is found from the expression y12 =

I1 V2

V1 = 0

The circuit in Fig. S16.1(b) is used to determine this parameter 4kΩ I1 I2 V1 = 0

4kΩ

+ V 4kΩ 2 -

Fig. S16.1(b) Note in this case, the 4kΩ resistor on the left is shorted and -I1 (4k) = V2 or

138

I1 V2

= y 12 = V1 = 0

−1 S 4k

We could continue this procedure and determine y21 and y22 in the exact same manner, −1 S however, since the network looks the same from either port, we know that y 21 = 4k 1 S . Therefore, the two-port equations for this network in terms of the Y and y 22 = 2k parameters are 1 1 V2 V1 − 4k 2k 1 −1 I2 = V1 + V2 2k 4k I1 =

If we now connect a 4mA current to the input and 4kΩ load to the output, the terminal conditions are 4 A k V2 = − 4k I 2 I1 =

The two-port equations now become 4 1 1 = V1 − V2 k 2k 4k − V2 1 1 V2 V1 + =− 2k 4k 4k or 1 4 1 V2 V1 − = 4k k 2k 1 3 0=− V1 + V2 4k 4k Simplifying 16 = 8V1 − V2 0 = − V1 + 3V2

139 or 16 = 8 (3V2 ) − V2 V2 =

16 V 23

The network with the terminal conditions attached is shown in Fig. S16.1(c). V1 V2 4kΩ 4kΩ

4mA

4kΩ

+ 4kΩ V2 -

Fig. S16.1(c) The nodal equations for this network are 1 ⎞ ⎛ 1 ⎞ 4 ⎛ 1 V1 ⎜ + ⎟ − V2 ⎜ ⎟ = ⎝ 4k ⎠ k ⎝ 4k 4k ⎠ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + + − V1 ⎜ ⎟ + V2 ⎜ ⎟=0 ⎝ 4k 4k 4k ⎠ ⎝ 4k ⎠ Note that these equations are identical to those obtained earlier. 16.2

The equations for a two-port in terms of the Z parameters are V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 Since z 11 =

V1 with I 2 = 0 , the network in Fig. S16.2(a) is used to derive z11. I1 I1 -j1Ω

+ V1 -

j2Ω 2Ω

I2 = 0

Fig. S16.2(a) Note that with the output terminals open-circuited, I 2 = 0 . Then z 11 =

V1 I1

= 2 − j1Ω I2 = 0

140 Likewise, the parameter z12 is found from the expression z 12 =

V1 I2

I1 = 0

The circuit used to derive this parameter is shown in Fig. S16.2(b). j2Ω -j1Ω + V1 -

I2

2Ω

+ V2 -

Fig. S16.2(b) With the input terminals open-circuited, I1 = 0 . Since I1 = 0 , there is no current in the capacitor and therefore no voltage across it. Then V1 is the voltage across the 2Ω resistor and V1 = 2I2 and hence z 12 =

V1 I2

= 2Ω I1 = 0

In a similar manner, we find that z21 = 2Ω z22 = 2 + j2Ω Therefore, the two-port equations in terms of the Z parameters are V1 = (2 − j1) I1 + 2 I 2

V2 = 2 I1 + (2 + j2) I 2 If we now apply the terminal conditions, the network is shown in Fig. S16.2(c). j2Ω -j1Ω I 2

6 ∠0° V +-

I1

2Ω

Fig. S16.2(c)

-j4Ω

+ V2 -

141 The terminal conditions are V1 = 6 ∠0° V

V2 = − (− j4) I 2

And the two-port equations are 6 ∠0° = (2 − j1) I1 + 2 I 2 0 = 2 I1 + (2 − j2 ) I 2

Solving the second equation for I1 and substituting this value into the first equation yields 6 = (-1 + j) (2 – j) + 2I2 or 7 − 3j 2 = 3.81 ∠ − 23.2° A

I2 =

16.3

The network is redrawn as shown in Fig. S16.3. I1 aΩ + V1 -

bV2 +-

CI1

I2

+ V2 -

dΩ

Fig. S16.3 The two-port equations in terms of the hybrid parameters are V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 and thus h 11 = h 21 =

V1 I1 I2 I1

h 12 = V2 = 0

h 22 = V2 = 0

V1 V2

I1 = 0

I2 V2

I1 = 0

Let us now apply these definitions to the network. Note that

142

h 11 =

V1 I1

= aΩ V2 = 0

and V1 V2

h 12 =

=b I1 = 0

In a similar manner h 21 =

I2 I1

=C V2 = 0

and h 22 =

I2 V2

= I1 = 0

1 S d

Note carefully the match between the network in Fig. S16.3 and the hybrid parameters. This network is actually the hybrid model for the basic transistor and given the hybrid parameters for a transistor, the model can be constructed immediately. 16.4

The network in Fig. 16.1 can be redrawn in the following manner as shown in Fig. S16.4(a). 4kΩ 4kΩ

4kΩ

Fig. S16.4(a) In this form we see that the original network can be drawn as a cascade connection of three networks. The general form of the transmission parameters is V1 = A V2 – B I2 I1 = C V2 – D I2 Consider the network in Fig. S16.4(b).

143 I2

I1 + V1 -

+ 4kΩ V2 -

Fig. S16.4(b) For this network A= B= C= D=

V1 V2

=1 I2 = 0

V1 − I2 I1 V2

=0 V2 = 0

= I2 = 0

I1 − I2

1 4k =1

V2 = 0

Next consider the network in Fig. S16.4(c). I1 + V1 -

I2

+ V2 -

4k

Fig. S16.4(c) In this case A= B= C= D=

V1 V2

=1 I2 = 0

V1 − I2 I1 V2

= 4k V2 = 0

=0 I2 = 0

I1 − I2

=1 V2 = 0

144 Since the transmission parameters for the resistor on the right are the same as those for the resistor on the left, we have all the parameters for the individual networks. Now the transmission parameters for the entire network are ⎡A B ⎤ ⎡ 1 ⎢ C D⎥ = ⎢ 1 ⎣ ⎦ ⎢⎣ 4k

0⎤ ⎥ 1⎥ ⎦

⎡1 =⎢ 1 ⎢⎣ 4k

0⎤ ⎥ 1⎥ ⎦

⎡1 =⎢ 1 ⎢⎣ 4k

0⎤ ⎥ 1⎥ ⎦

⎡1 4 k ⎤ ⎡ 1 ⎢0 1 ⎥ ⎢ 1 ⎣ ⎦ ⎢⎣ 4k

0⎤ ⎥ 1⎥ ⎦

⎡ ⎛ 1 ⎞ ⎢(1)(1) + (4k ) ⎜ 4k ⎟ ⎝ ⎠ ⎢ ⎛ ⎢ (0 )(1) + (1) ⎜ 1 ⎞⎟ ⎢⎣ ⎝ 4k ⎠ ⎡ 2 4k ⎤ ⎢1 ⎥ ⎢⎣ 4k 1 ⎥⎦

(1)(0) + (4k )(1)⎤⎥ ⎥ (0)(0) + (1)(1) ⎥⎥ ⎦

⎡ ⎛ 1 ⎞ (1)(4k ) + (0)(1) ⎤⎥ ⎢ (1)(2 ) + (0 ) ⎜ 4k ⎟ ⎝ ⎠ ⎥ =⎢ 1 ⎢⎛⎜ ⎞⎟ (2 ) + (1) ⎛⎜ 1 ⎞⎟ ⎛⎜ 1 ⎞⎟ (4k ) + (1)(1)⎥ ⎥⎦ ⎢⎣⎝ 4k ⎠ ⎝ 4k ⎠ ⎝ 4k ⎠ ⎡2 =⎢ 3 ⎢⎣ 4k

16.5

4k ⎤ ⎥ 2⎥ ⎦

The conversion formulas necessary to convert the Y parameters to the transmission parameters are as follows. ⎡ − y 22 ⎡A B ⎤ ⎢ y 21 ⎢ C D ⎥ = ⎢ − ∆y ⎣ ⎦ ⎢ ⎢⎣ y 21

−1 ⎤ y 21 ⎥ ⎥ − y 11 ⎥ y 21 ⎥⎦

where ∆y = y 11 y 22 − y 12 y 21 . From the results of problem 16.1 ⎡ y 11 ⎢y ⎣ 21

and ∆y =

⎡ 1 y 12 ⎤ ⎢ 2k =⎢ y 22 ⎥⎦ ⎢− 1 ⎣ 4k

1 1 3 − = . Hence, 2 2 4k 16k 16k 2

1⎤ 4k ⎥ 1 ⎥ ⎥ 2k ⎦

−

145 1 − y 22 = 2k = 2 A= −1 y 21 4k −1 −1 = = 4k B= y 21 − 1 4k −3 − ∆y 16k 2 3 = = C= −1 4k y 21 4k −1 − y 11 2k D= = =2 −1 y 21 4k −

These results check with those obtained in problem 16.4.

146 APPENDIX: Techniques for Solving Linearly Independent Simultaneous Equations

In the solution of various circuit problems we encounter a system of simultaneous equations of the form a 11 x 1 + a 12 x 2 +

+ a 1n x n = b 1

a 21 x 1 + a 22 x 2 +

+ a 2n x n = b 2

a n1 x 1 + a n 2 x 2 +

+ a nn x n = b n

(A.1)

where the x’s and b’s are typically voltages and currents or currents and voltages, respectively. As the title implies, we assume that the equations are linearly independent. As a brief reminder of the meaning of linear independence, consider the following KCL equations written for each node of a three-node network: 3 1 V1 − V2 − 4 = 0 (A.2) 2 2 1 5 − V1 + V2 + 5 = 0 (A.3) 2 6 1 − V1 − V2 − 1 = 0 (A.4) 3 where V1 and V2 are two node voltages that are measured with respect to the third (reference) node. Linear independence implies that we cannot find constants a1, a2, and a3 such that 1 5 1 ⎛3 ⎞ ⎛ 1 ⎛ ⎞ ⎞ a 1 ⎜ V1 − V2 − 4 ⎟ + a 2 ⎜ − V1 + V2 + 5 ⎟ + a 3 ⎜ − V1 − V2 − 1⎟ = 0 2 6 3 ⎝2 ⎠ ⎝ 2 ⎠ ⎝ ⎠

(A.5)

However, in this case if we select a1 = a2 = a3 = 1, we obtain +

1 5 1 1 3 V1 − V2 − 4 − V1 + V2 + 5 − V1 − V2 − 1 = 0 3 6 2 2 2 0=0

Said another way, this means, for example, that Eqs. (A.2) and (A.3) can be used to obtain Eq. (A.4), and therefore, Eq. (A.4) is linearly dependent on Eqs. (A.2) and (A.3). Furthermore, any two of the equations could be used to obtain the third equation. Therefore, only two of the three equations are linearly independent.

147 We will now describe three techniques for solving linearly independent simultaneous equations-Guassian elimination, determinants and matrices. Our presentation will be very brief and deal only with the elements of these techniques that are needed in this student problem companion. A.1

Gaussian Elimination

The following example will serve to demonstrate the steps involved in applying this technique. Example A.1

Let us find the solution to the following set of equations: 7 X 1 − 4X 2 − X 3 = 4 − 4X 1 + 7 X 2 − 2X 3 = 0 − X 1 − 2X 2 + 3X 3 = − 1

(A.6) (A.7) (A.8)

Solution The algorithm (i.e., step-by-step procedure) for applying the Gaussian elimination method proceeds in the following systematic way. First, we solve Eq. (A.6) for the variable X1 in terms of the other variables in X2 and X3.

X1 =

4 4 1 + X2 + X3 7 7 7

(A.9)

We then substitute this result into Eqs. (A.7) and (A.8) to obtain 33 18 16 X2 − X3 = 7 7 7 18 20 3 − X2 + X3 = − 7 7 7

(A.10) (A.11)

Continuing the reduction we now solve Eq. (A.10) for X2 in terms of X3: 16 18 + X3 33 33 Substituting this expression for X2 into Eq. (A.11) yields X2 =

(A.12)

336 189 X3 = 231 231 or X3 = 0.563

(A.13)

148 Now backtracking through the equations, we can determine X2 from Eq. (A.12) as X2 = 0.792 and X1 from Eq. (A.9) as X1 = 1.104 In this simple example we have not addressed such issues as zero coefficients or the impact of round-off errors. We have, however, illustrated the basic procedure. A.2

Determinants

A determinant of order n is a square array of elements aij arranged as follows:

∆=

a 11

a 12

a 1n

a 21

a 22

a 2n

a n1

a n2

a nn

(A.14)

The cofactor cij of the element aij is given by the expression cij = (-1)i+jAij

(A.15)

where Aij is the determinant that remains after the ith row and jth column are deleted. Example A.2

Given the determinant a 11

a 12

a 13

∆ = a 21 a 31

a 22

a 23

a 32

a 33

find the cofactor of the element a21. Solution The cofactor of c21 for the element a21 is c 21 = (− 1)

2 +1

a 12

a 13

a 32

a 33

The numerical value of the determinant is equal to the sum of products of the elements in any row or column and their cofactors.

149

Example A.3

Let us determine the value of the determinant in Example A.2 using the first row. Solution

∆ = a 11c11 + a 12 c12 + a 13 c13

= a 11 (− 1) A 11 + a 12 (− 1) 1+1

= a 11

1+ 2

a 22

a 23

a 32

a 33

− a 12

A 12 + a 13 (− 1)

a 21

a 23

a 31

a 33

1+ 3

+ a 13

A 13

a 21

a 22

a 31

a 32

Although the 2-by-2 determinants can be evaluated in the same manner, as illustrated above, the result is simply a b = ad − cb c d

(A.16)

Therefore, ∆ is ∆ = a 11 (a 22 a 33 − a 32 a 23 ) − a 12 (a 21a 33 − a 31a 23 ) + a 13 (a 21a 32 − a 31a 22 ) We could evaluate the determinant using any row or column. The method of solving the set of simultaneous equations of the form shown in Eq. (A.1) using determinants is known as Cramer’s rule. Cramer’s rule states that if ∆ ≠ 0 (that is, the equations are linearly independent), the value of the variable x1 in Eq. (A.1) is given by the expression b1 a 12 b 2 a 22 x1 =

b ∆1 = n ∆

a 2n ∆

a 1n a 2n

(A.17)

a nn

Where ∆1 is the determinant ∆ in which the first column is replaced with the column of coefficients. In the general case, xi is given by an expression similar to Eq. (A.17) with the ith column replaced by the column of coefficients. Example A.4

Let us solve the following equations using determinants.

150 2I1 – 4I2 = 8 -4I1 + 6I2 = -4 Solution In this case, ∆ defined by Eq. (A.16) is

∆=

2 −4 = (2 ) (6 ) − (− 4 ) (− 4 ) = − 4 −4 6

Then using Eq. (A.17)

I1 =

8 −4 −4 6 −4

=

(8) (6) − (− 4) (− 4) = − 8

=

(2) (− 4) − (− 4) (8) = − 6

−4

and

I2 =

2 8 −4 −4 −4

−4

A.3

Matrices

A matrix is defined to be a rectangular array of numbers arranged in rows and columns and written in the form ⎡ a 11 ⎢a ⎢ 21 ⎢ ⎢ ⎣ a m1

a 12 a 22 a m2

a 1n ⎤ a 2 n ⎥⎥ ⎥ ⎥ a mn ⎦

This array is called an m by n (m × n) matrix because is has m rows and n columns. A matrix is a convenient way of representing arrays of numbers; however, one must remember that the matrix itself has no numerical value. In the preceding array the numbers or functions aij are called the elements of the matrix. Any matrix that has the same number of rows as columns is called a square matrix. Example A.5

Are the following matrices?

151 ⎡a ⎤ ⎢b⎥ 1 3 4 3 2 1 ⎤ ⎡ ⎤ ⎢ ⎥, ⎡ ,⎢ , [1 2 3] ⎢ ⎥ ⎢ c ⎥ ⎣2 4⎦ ⎣5 6 7 8⎥⎦ ⎢ ⎥ ⎣d ⎦ Solution Yes.

The identity matrix is a diagonal matrix in which all diagonal elements are equal to one. Example A.6

Are the following identity matrices?

⎡1 0 0 ⎤ ⎡1 0 ⎤ ⎢ ⎥ ⎢0 1 ⎥ , ⎢0 1 0 ⎥ , ⎣ ⎦ ⎢0 0 1 ⎥ ⎣ ⎦

⎡1 ⎢0 ⎢ , ⎢0 ⎢ ⎢ ⎢⎣0

0⎤ 0⎥⎥ 0⎥ ⎥ ⎥ 1⎥⎦

0 . . 1 0 . 0 1 0 0

.

.

Solution Yes.

Consider now the multiplication of two matrices. If we are given an m × n matrix A and an n × r matrix B, the product AB is defined to be an m × r matrix C whose elements are given by the expression n

c ij = ∑ a ik b kj , i = 1, … , m, j = 1, … , r

(A.18)

k=1

Note that the product AB is defined only when the number of columns of A is equal to the number of rows of B. Multiplication is a “row-by-column” operation. In other words, each element in a row of the first matrix is multiplied by the corresponding element in a column of the second matrix and then the products are summed. This operation is diagrammed as follows: ⎡ c11 ⎢ ⎢ ⎢ c m1 ⎣

c ij

c1p ⎤ ⎡ a 11 ⎥ ⎢ ⎥ = ⎢ a i1 c mp ⎥⎦ ⎢⎣a m1

a 1n ⎤ a in ⎥⎥ a mn ⎥⎦

⎡ b11 ⎢ ⎢ ⎢ b n1 ⎣

b ij b nj

b 1p ⎤ ⎥ ⎥ b np ⎥⎦

The following examples will illustrate the computational technique.

(A.19)

152 Example A.7

If ⎡1 3 ⎤ ⎡ 2 1⎤ and B = ⎢ A=⎢ ⎥ ⎥ ⎣ 2 4⎦ ⎣3 5⎦ ⎡1 2⎤ ⎡1 ⎤ and D = ⎢ ⎥ C=⎢ ⎥ ⎣3 4⎦ ⎣ 2⎦ Find AB and CD. Solution

⎡ (1)(2) + (3)(3) AB = ⎢ ⎣(2 )(2) + (4 )(3)

(1)(1) + (3)(5) ⎤ ⎡11 = (2)(1) + (4)(5)⎥⎦ ⎢⎣16

16 ⎤ 22⎥⎦

⎡ (1)(1) + (2)(2)⎤ ⎡ 5 ⎤ CD = ⎢ ⎥=⎢ ⎥ ⎣(3)(1) + (4)(2)⎦ ⎣11⎦ The matrix of order n × m obtained by interchanging the rows and columns of an m × n matrix A is called the transpose of A and is denoted by AT. Example A.8

If ⎡1 4⎤ B = ⎢⎢2 5⎥⎥ ⎢⎣3 6⎥⎦

Find BT. Solution

⎡1 2 3 ⎤ BT = ⎢ ⎥ ⎣ 4 5 6⎦ As defined for determinants, the cofactor Aij of the element aij of any square matrix A is equal to the product (-1)i+j and the determinant of the submatrix obtained from A by deleting row i and column j.

153

Example A.9

Given the matrix ⎡ a 11 A = ⎢⎢a 21 ⎢⎣a 31

a 12 a 22 a 32

a 13 ⎤ a 23 ⎥⎥ a 33 ⎥⎦

Find the cofactors A11, A12, and A22. Solution The cofactors A11, A12, and A22 are

A 11 = (− 1)

2

A 12 = (− 1)

3

A 22 = (− 1)

4

a 22 a 32

a 23 = a 22 a 33 − a 32 a 23 a 33

a 21 a 31

a 23 = − (a 21a 33 − a 31a 23 ) a 33

a 11 a 31

a 13 = a 11a 33 − a 31a 13 a 33

The adjoint of the matrix A (adj A) is the transpose of the matrix obtained from A by replacing each element aij by its cofactors Aij. In others words, if ⎡ a 11 ⎢a A = ⎢ 21 ⎢ ⎢ ⎣a n 1

a 12 a 22

a 1n ⎤ . ⎥⎥ ⎥ ⎥ a nn ⎦

then ⎡ A 11 ⎢A adj A = ⎢ 12 ⎢ ⎢ ⎣ A 1n

A 21 A 22

A n1 ⎤ . ⎥⎥ ⎥ ⎥ A nn ⎦

If A is a square matrix and if there exists a square matrix A-1 such that A-1A = AA-1 = I

(A.20)

154 Then A-1 is called the inverse of A. It can be shown that the inverse of the matrix A is equal to the adjoint divided by the determinant (written here as |A|); that is A −1 =

adj A A

(A.21)

Example A.10

Given ⎡2 3 1 ⎤ B = ⎢⎢1 2 3⎥⎥ ⎢⎣3 1 2⎥⎦

Find B-1. Solution

2 3 3 1 ⎡3 1⎤ −1 +3⎢ ⎥ 1 2 1 2 ⎣2 3⎦ = 2 − 5 + 21 = 18

B =2

and ⎡ 1 −5 7 ⎤ adj B = ⎢⎢ 7 1 − 5⎥⎥ ⎢⎣− 5 7 1 ⎥⎦

Therefore, ⎡ 1 −5 7 ⎤ 1 ⎢ 7 1 − 5⎥⎥ B = ⎢ 18 ⎢⎣− 5 7 1 ⎥⎦ −1

We now have the tools necessary to solve Eqs. (A.1) using matrices. The following example illustrates the approach. Example A.11

The node equations for a network are 2V1 + 3V2 + V3 = 9

155 V1 + 2V2 + 3V3 = 6 3V1 + V2 +2V3 = 8 Let us solve this set of equations using matrix analysis. Solution Note that this set of simultaneous equations can be written as a single matrix equation in the form ⎡2 3 1 ⎤ ⎡ V1 ⎤ ⎡9⎤ ⎢1 2 3 ⎥ ⎢ V ⎥ = ⎢6 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣3 1 2⎥⎦ ⎢⎣ V3 ⎥⎦ ⎢⎣8⎥⎦

or AV = I

Multiplying both sides of the preceding equation through A-1 yields A-1AV = A-1I

or V = A-1I A-1 was calculated in Example A.10. Employing that inverse here, we obtain ⎡ 1 − 5 7 ⎤ ⎡9 ⎤ 1 ⎢ 7 1 − 5⎥⎥ ⎢⎢6⎥⎥ V= ⎢ 18 ⎢⎣− 5 7 1 ⎥⎦ ⎢⎣8⎥⎦

or ⎡ V1 ⎤ ⎢V ⎥ = 1 ⎢ 2 ⎥ 18 ⎢⎣ V3 ⎥⎦

⎡35⎤ ⎢29⎥ ⎢ ⎥ ⎢⎣ 5 ⎥⎦

and hence, V1 =

35 29 5 , V2 = , and V3 = 18 18 18

E–1

EQUATIONS CHAPTER 1 Electric current-charge relationship i(t) =

dq(t) dt

q(t) =

or

Power t

3-q

dw = p = vi dt

i(x) dx

Voltage-energy relationship v =

Energy

dw dq

¢w =

t2

3t1

p dt =

t2

3t1

vi dt

CHAPTER 2 Ohm’s law

Two series resistors & voltage divider

v(t)=R*i(t) or i(t)=Gv(t) where G =

1 R

i(t) =

Power p(t)=v(t)i(t) = Ri2(t) =

v2(t) R

v(t) R1 + R2

vR1 =

R1 v(t) R1 + R2

vR2 =

R2 v(t) R1 + R2

Kirchhoff’s Current Law (KCL)

i(t)

N

a ij(t) = 0

+

j=1

R1

vR 1 –

v(t)

Kirchhoff’s Voltage Law (KVL)

+

N

a vj(t) = 0

R2

j=1

vR 2 –

Multiple series resistors & voltage divider RS = R1 + R2 + p + RN i(t) = vRi =

v(t) RS Ri v(t) RS

+

vR1

–

R1

+

vR2 R2

i(t)

–

+

vR3

–

R3

+ R4

vR4 i(t)

– v(t) + R5 RN –v

RN

vR5 –

+

v(t)

RS = R1 + R2 + R3 + … + RN

E–2

Two parallel resistors & current divider Rp =

R1 R2 R1 + R2 +

R1 R2 v(t) = Rp i(t) = i(t) R1 + R2 i(t)

R2 i1(t) = i(t) R1 + R2

R2 v(t)

R1 i1(t)

i2(t) –

R1 i2(t) = i(t) R1 + R2 Multiple parallel resistors & current divider N 1 1 = a Rp R i=1 i

ij(t) =

Rp Rj

+ v(t)

io(t)

io(t)

i1(t)

i2(t)

iN(t)

R1

R2

RN

+ v(t)

io(t)

–

Rp

–

Delta-to-wye resistance conversion Ra =

R1 R2 R1 + R2 + R3

Rb =

R2 R3 R1 + R2 + R3

Rc =

R1 R3 R1 + R2 + R3

Delta-to-wye resistance conversion (Special case: Identical resistors) R Y = 13 R ¢

Wye-to-delta resistance conversion R1 =

Ra Rb + Rb Rc + Ra Rc Rb

R2 =

Ra Rb + Rb Rc + Ra Rc Rc

R3 =

Ra Rb + Rb Rc + Ra Rc Ra

a a Ra

R1

R2 Rc

c

b

c

R3

Wye-to-delta resistance conversion (Special case: Identical resistors) R ¢ = 3R Y

Rb b

E–3 CHAPTER 4

CHAPTER 3 Ohm’s law expressed in node voltages i =

Ideal op amp

vm - v N R

i+ = i- = 0 v+ = v-

R

Node m i

+ vm

Node N +

+(vm–vN)-

v+

i+

v–

i–

vN

–

–

Ohm’s law expressed in loop currents v3 = Ai1 - i2 B R3 A

+

v1

vS2 B

–

R1 i1

vS1

+ v3

R3

–

R2 F

C

– v + 2

E

+ i2

v4

R4

R5

–

– v + 5

D

CHAPTER 5 Equivalent circuit forms

R1 R1 + R2

R2

R1

R1 R2 ———— R1 + R2

R2

V1 V1 – V2 V2

∞

E–4

Equivalent circuit forms (continued)

I1

I2

I1 – I2

+ VS

R

Io = IS

R

Vo = VS

IS

–

Thévenin & Norton equivalent circuits A

i RTh voc

i

+

A +

Circuit B

vo

voc = RTh isc

isc

RTh

vo

–

–

B

B

i = isc -

vo = voc - RTh i

Circuit B

vo RTh

Maximum power transfer theorem (Thévenin v and R fixed, load RL variable) RL = R Pload =

R

i

v2 4R

v

RL

v & R fixed, RL variable

CHAPTER 6 Parallel-plate capacitor-Capacitance C =

Charge stored on a capacitor

eo A d

q = Cv Current-voltage relationship of a capacitor

dq i = —— dt d A

+ + q(t) –

v(t) Dielectric (a)

–

i = C C

v(t) = (b)

dv dt

t 1 i(x) dx C 3-q

Energy stored in a capacitor

N 1 1 = a CS C i=1 i

1 q2(t) = 21 Cv2(t) J 2 C

wC(t) =

Current-voltage relationship of an inductor

Capacitors connected in parallel N

di(t) v(t) = L dt i(t) =

E–5

Capacitors connected in series

Cp = a Ci i=1

t 1 v(x) dx L 3-q

Inductors connected in series N

LS = a Li

i(t)

i=1

+ v(t)

L

–

Inductors connected in parallel N 1 1 = a Lp L i=1 i

Energy stored in an inductor wL(t) = 21 Li2(t) J

CHAPTER 7 First-order circuits

Second-order circuits

The unit step function

Characteristic equation of a second-order circuit

u(t) = b

0 1

t 6 0 t 7 0

s2 + 2
0 s + 20 = 0 Roots of the characteristic equation

General form of the step response of a first-order circuit excited at t = t0 x(t) = x(q) + Cx At0 B - x(q)De-At - t0B兾, t = t0 where x At0 B is the initial value and x (q) is the final value.

s1 = -
0 + 0 2
2 - 1 s2 = -
0 - 0 2
2 - 1 Overdamped response (i.e.,
> 1) x(t) = K1 e-A
0 - 0 2


2

- 1Bt

+ K2 e-A
0 + 0 2


Time constant of a first-order capacitive circuit  = R Th C

Critically damped response (i.e.,
= 1) x(t) = B1 e-
0 t + B2 te-
0 t

Time constant of a first-order inductive circuit  =

L RTh

Underdamped response (i.e.,
< 1) x(t) = e-t AA 1 cos d t + A 2 sin d tB, where  =
0 , and d = 0 21 -
2

2

- 1Bt

E–6 CHAPTER 8 General form of a sinusoidal waveform x(t) = XM sin (t + ) where  =

Admittance

2 = 2f T

Y =

Conversion between sine and cosine functions

Admittances of R, L, and C

 cos t = sin a t + b 2 sin t = cos a t -

1 I = Z V

 b 2

YR =

1 = G R

YL =

1 1 = / 90° jL L

Impedance YC = jC = C / 90°

VM / v VM V = Z = = / v - i = Z / z I IM / i IM

Admittances connected in series

Impedances connected in series

N 1 1 = a YS Y i=1 i

N

ZS = a Zi i=1

Admittances connected in parallel

Impedances connected in parallel

N

Yp = a Yi i=1

N 1 1 = a Zp i = 1 Zi

Impedances of R, L, and C Passive element

Impedance

R

Z = R

L

Z = jL = jXL = L / 90° , XL = L Z =

C

1 1 1 = jXC = / 90° , XC = jC C C

CHAPTER 9 Average (real) power absorbed by an impedance (watts) P =

1 2 VM IM cos Av

- i B

P = I2rms R =

Maximum average power transfer theorem (When VOC and ZTh fixed, load ZL variable) ZL = RL + jXL = RTh - jXTh = Z*Th Z Th Voc

Power factor (pf) pf = cos Av - i B = cos ZL

S = P + jQ = Vrms / v Irms / -i

+

–

V2rms R

Complex power (volt-amperes)

IL

VL

ac circuit

Average power absorbed by a resistor

ZL

= Vrms Irms / v - i = I 2rms Z

Voc & Z Th fixed, ZL variable Maximum average power transfer theorem (Special case: XL = 0) RL =

2R2Th

+

P = Re(S) = Vrms Irms cos Av - i B = I 2rms Re(Z) Reactive power (vars) Q = Im(S) = Vrms Irms sin Av - i B = I 2rms Im(Z)

X2Th

RMS value of a sinusoidal waveform Irms =

E–7

Average (real) power (watts)

Power triangle relationship tan Av - i B = tan Z =

IM 12

Q P

Im ▲ ▲

▲

Average power absorbed in terms of rms values

S

P = Vrms Irms cos Av - i B

+Q

▲

Re –Q

▲

S

▲

θv– θi P θv– θi

▲

▲

CHAPTER 10 Magnetic flux, voltage and current relationships

Voltage-current relationships for mutually coupled coils

=N=Li webers v =

d di = L dt dt i

v1(t) = L1

di1(t) di2(t) + M dt dt

v2(t) = M

di1(t) di2(t) + L2 dt dt

+

i1(t) v

N



i2(t)

+ v1(t)

–

+

M L1

v2(t)

L2

–

–

Phasor voltage-current relationships for mutually coupled coils

Ideal transformer equations v1 i1 N1 = - = v2 i2 N2

V1 = jL1 I1 + jMI2 V2 = jL2 I2 + jMI1 Energy stored in magnetically coupled inductors

The coefficient of coupling

i1(t) v1(t)+ –

φ

▼

2

▼

2

•

A N2

N1 φ

▼

k =

M 2L1 L2

where 0 k 1

Ideal transformer equations in phasor form V1 I2 N1 = = V2 I1 N2

The turns ratio of a transformer n =

N2 N1

•

i2(t) ▼

w(t) = 21 L1 Ci1(t)D + 21 L2 Ci2(t)D ; Mi1(t)i2(t)

+ v (t) –2

E–8 CHAPTER 11 Three-phase terminology Quantity

Wye

Delta

Line current AIL B Ia , Ib , Ic

Phase current AIp B

Line-to-neutral voltage AVp B

Van , Vbn , Vcn

Phase voltage AVp B

Line-to-line, phase-to-phase, line voltage AVL B Phase voltage AVp B

Vab , Vbc , Vca

Phase current AIp B

Iab , Ibc , Ica

Voltage, current, and impedance relationhips of Y and ⌬ configurations

⌬

Y

Line voltage AVab or VAB B

13 Vp /  + 30°

VL /  + 30°

= VL /  + 30°

Line current IaA

IL / 

IL / 

Phase voltage

Vp /  AVan or VAN B

13 Vp /  + 30°

Phase current

IL / 

Load impedance

ZY /  - 

IL 13

/  + 30°

3 ZY /  - 

E–8 CHAPTER 12 Resonant frequency of a series or parallel RLC circuit 0 =

1 1LC

BW = HI - LO =

Quality factor of a series RLC circuit

Q =

Bandwidth of a parallel RLC circuit

where

0 L 1 1 L = = R 0 CR R BC

Bandwidth of a series RLC circuit BW = HI - LO =

0 Q

where

 LO = -

1 1 1 + + 2 2RC B (2RC) LC

and  HI =

1 1 1 + + LC 2RC B (2RC)2

Half-power (break) frequency of a first-order RC filter

 LO = 0 c  HI = 0 c

1 RC

1 1 2 + a b + 1d 2Q B 2Q

1 1 2 + a b + 1d 2Q B 2Q

 =

1 1 =  RC

Bandwidth of a series RLC bandpass filter BW = HI - LO =

and

R L

20 = LO HI L

Quality factor of a parallel RLC circuit 0 C = R Q = BW AL

V1

+

C R

Vo –

E–10 CHAPTER 13 Laplace transforms of some special functions

Laplace transform of a function f(t) LCf(t)D = F(s) =

q

30

f(t)

f(t)e-st dt

The unit impulse function At - t0 B = 0

t Z t0

and

F(s)

(t)

1

u(t)

1 s

e-at

1 s + a 1 s2

t t0 +

3t0 -

At - t0 B dt = 1

7 0

tn n!

Sampling property of the unit impulse function f At B f(t)At - t0 B dt = b 0 0 3t1 t2

t1 6 t0 6 t2 t0 6 t1 , t0 7 t2

1 sn + 1 1 (s + a)2

te-at tne-at n!

1 (s + a)n + 1

sin bt

b s 2 + b2

cos bt

s s 2 + b2

The initial-value theorem lim f(t) = lim sF(s)

t S0

SSq

e-at sin bt

b (s + a)2 + b2

e-at cos bt

s + a (s + a)2 + b2

The final-value theorem lim f(t) = lim sF(s)

tSq

S S0

Some properties of Laplace transform Property Number

f(t)

F(s)

1.

Af(t)

AF(s)

2.

f1(t) ; f2(t)

F1(s) ; F2(s)

3.

f(at)

1 s Fa b, a 7 0 a a

4.

fAt - t0 BuAt - t0 B, t0 0

e-t0 sF(s)

5.

f(t)uAt - t0 B

e-t0 sLCfAt + t0 B D

6.

e-atf(t)

F(s + a)

n

7. 8. 9.

d f(t) dt n tf(t) f(t) t t

10. 11.

30 30

f() d t

f1()f2(t - ) d

snF(s) - sn - 1f(0) - sn - 2f1(0) p s0fn - 1(0) -

dF(s) ds q

F() d 3s 1 F(s) s F1(s)F2(s)

E–11 CHAPTER 14

The voltage-current relationship of a resistor in the s-domain

The voltage-current relationship of a inductor in the s-domain V(s) = sLI(s) - Li(0)

V(s) = RI(s)

I(s) =

The voltage-current relationship of a capacitor in the s-domain V(s) =

I(s) v(0) + s sC

V(s) i(0) + s sL

Transfer or network function Yo(s) = H(s) Xi(s)

I(s) = sCV(s) - Cv(0)

CHAPTER 15

Trigonometric Fourier series of a periodic function f(t)

Fourier transform pair F() = FCf(t)D =

q

f(t) = a0 + a Dn cos An0 t + n B

q

3-q

f(t)e-jt dt

n=1

f(t) = F-1 CF()D =

q

= a0 + a an cos n0 t + bn sin n0 t n=1

q 1 F()ejt d 2 3-q

where t1 + T0 2 an = f(t) cos n0 t dt T0 3t1

bn =

2 T0 3t1

Fourier transform of some special functions f(t)

F()

t1 + T0

f(t) sin n0 t dt

and Dn / n = an - jbn Exponential Fourier series of a periodic function f(t) q

f(t) = a cn ejn0 t n = -q

(t - a)

e-ja

A

2A()

ej0 t

2A - 0 B

cos 0 t

A - 0 B + A + 0 B

sin 0 t

jA + 0 B - jA - 0 B

e-at u(t), a 7 0

1 a + j

e- ∑t∑, a 7 0

2a a 2 + 2

where t1 + T0 1 cn = f(t)e-jn0 t dt T0 3t1

Relationships between various Fourier series coefficients Dn / n = 2cn = an - jbn

e-at cos 0 tu(t), a 7 0 e-at sin 0 tu(t), a 7 0

j + a (j + a)2 + 20 0 (j + a)2 + 20

E–12

Some properties of Fourier transform f(t)

F()

Property

Af(t)

AF()

f1(t) ; f2(t)

F1() ; F2()

f(at)

1  Fa b, a 7 0 a a

Time-scaling

fAt - t0 B

e-jt0 F()

Time-shifting

ejt0 f(t)

FA - 0 B

Modulation

dn f(t)

(j)n F()

dtn

(j)n

tnf(t) q

3-q

Linearity

f1(x)f2(t - x) dx

dn F()

Differentiation

dn

F1()F2() Convolution q 1 F (x)F2( - x) dx 2 3-q 1

f1(t)f2(t)

Convolution property of the Fourier transform Vo() = H()Vi()

CHAPTER 16 Two-port network admittance equations

Two-port network impedance equations

I y y V B 1 R = B 11 12 R B 1 R I2 y21 y22 V2 I1

V1

B

Impedance parameters

I2 Linear network

V1 z z I R = B 11 12 R B 1 R V2 z21 z22 I2

V2

z11 =

V1 2 I1 I2 = 0

z12 =

V1 2 I2 I1 = 0

z21 =

V2 2 I1 I2 = 0

z22 =

V2 2 I2 I1 = 0

Admittance parameters I1 2 y11 = V1 V2 = 0 y21 =

I2 2 V1 V2 = 0

y12

I1 2 = V2 V1 = 0

y22 =

I2 2 V2 V1 = 0

Two-port network hybrid equations B

V1 h h12 I R = B 11 RB 1R I2 h21 h22 V2

Hybrid parameters h11 =

V1 2 I1 V2 = 0

h12 =

V1 2 V2 I1 = 0

A =

V1 2 V2 I2 = 0

h21 =

I2 2 I1 V2 = 0

h22 =

I2 2 V2 I1 = 0

B =

V1 2 -I2 V2 = 0

C =

I1 2 V2 I2 = 0

D =

I1 2 -I2 V2 = 0

Two-port network transmission equations B

E–13

Transmission parameters

V1 A B V R = B R B 2R C D I1 -I2

Two-port network parameter conversions

z z B 11 12 R z21 z22

y22 ¢ D Y -y21 ¢Y

z22 ¢ D Z -z21 ¢Z

-z12 ¢Z T z11 ¢Z

B

z11 z21 D 1 z21

¢Z z21 T z22 z21

-y22 y D 21 - ¢Y y21

¢Z z D 22 -z21 z22

z12 z22 T 1 z22

1 y11 D y21 y11

-y12 ¢Y T y11 ¢Y

y11 y12 R y21 y22

-1 y21 T -y11 y21 -y12 y11 T ¢Y y11

A C D 1 C D B D 1 B

A c C

B D D 1 D

¢T C T D C

¢H h22 D -h21 h22

- ¢T B T A B

1 h11 D h21 h11

B d D

¢T D T C D

- ¢H h D 21 -h22 h21

B

h12 h22 T 1 h22 -h12 h11 T ¢H h11 -h11 h21 T -1 h21

h11 h12 R h21 h22