Solution Manual - Fundamentals of Electric Circuits 3rd Edition Chapter 14 PDF

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Chapter 14, Problem 1. Find the transfer function V o /V i of the RC circuit in Fig. 14.68. Express it using ωo = 1/RC.

Figure 14.68 For Prob. 14.1. Chapter 14, Solution 1. V R jωRC = H(ω) = o = Vi R + 1 jωC 1 + j ωRC jω ω 0 1 , where ω 0 = H (ω) = RC 1 + j ω ω0

H = H(ω) =

ω ω0 1 + ( ω ω0 )

2

φ = ∠H(ω) =

⎛ω⎞ π − tan -1 ⎜ ⎟ 2 ⎝ω 0 ⎠

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that ω 0 = 1 RC . Thus, the sketches of H and φ are shown below. H 1 0.7071

0

ω0 = 1/RC

ω

φ

90° 45° 0

ω0 = 1/RC

ω

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Chapter 14, Problem 2. Obtain the transfer function V o (s)/V i of the circuit in Fig. 14.69.

Figure 14.69 For Prob. 14.2.

Chapter 14, Solution 2.

H (s ) =

Vo Vi

2+ =

1 s /8

10 + 20 +

1 s/8

=

2 +8 / s 1 s + 4 = 12 + 8 / s 6 s + 0.6667

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 3. For the circuit shown in Fig. 14.70, find H(s) = V o /V i (s).

Figure 14.70 For Prob. 14.3. Chapter 14, Solution 3. 1 1 5 = = jωC s (0.2) s 1 10 0.1F ⎯⎯ → = s(0.1) s The circuit becomes that shown below. 0.2F

⎯⎯ →

2

V1

5 s +

Vi

+ _

10 s

5

Vo _

10 5 10 1 + s (5 + ) 5( ) 10 5 s = s s = 10(s + 1) Let Z = //(5 + ) = s 15 5 s s s ( s + 3) 5+ (3 + s ) s s Z V1 = V Z +2 i s s Z 5 Vo = V1 = V1 = Vi • s +1 s +1 Z + 2 5 +5 / s 10(s +1) 10s 5s s s( s + 3) H ( s) = Vo = • = = 2 Vi s + 1 10( s + 1) 2 s( s + 3) + 10( s + 1) s + 8 s + 5 2+ s( s + 3) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 4.

Find the transfer function H( ω ) = V O /V i of the circuits shown in Fig. 14.71.

Figure 14.71 For Prob. 14.4. Chapter 14, Solution 4.

(a)

R ||

1 R = j ωC 1 + jω RC

R R 1 + jωRC V H(ω) = o = = R R + jωL (1 + jωRC) Vi j ωL + 1 + j ωRC

(b)

H (ω) =

R - ω RLC + R + jωL

H (ω) =

R + j ωL jωC ( R + j ωL) = R + jωL + 1 jωC 1 + jωC ( R + jωL)

H (ω) =

- ω2 LC + jωRC 1 − ω 2LC + jωRC

2

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 5. For each of the circuits shown in Fig. 14.72, find H(s) = V o (s)/V s (s).

Figure 14.72 For Prob. 14.5. Chapter 14, Solution 5. (a) Let Z = R // sL =

Vo =

sRL R + sL

Z Vs Z + Rs

sRL Vo Z sRL H (s) = = = R + sL = Vs Z + Rs R + sRL RRs + s (R + Rs )L s R + sL 1 Rx 1 R sC = (b) Let Z = R // = sC R 1 1 + sRC + sC Z Vo = Vs Z + sL

R R V Z + H (s ) = o = = 1 sRC = 2 R Vi Z + sL s LRC + sL + R sL + 1 + sRC PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 6. For the circuit shown in Fig. 14.73, find H(s) = I o (s)/I s (s).

Figure 14.73 For Prob. 14.6. Chapter 14, Solution 6. 1H

⎯⎯ →

Let Z = s //1 =

jω L = sL = s

s s +1

We convert the current source to a voltage source as shown below. 1

Is ⋅ 1

+ _

S

+ Vo

Z

_

s Z sIs sI s+1 = 2 s ( I s x1) = Vo = Is = 2 s (s + 1) + s s + 3s + 1 Z + s +1 s +1 + + s 1 Vo sI s Io= = 2 1 s + 3 s +1 I s H ( s) = o = 2 I s s + 3 s +1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 7. Calculate H(ω ) if H dB equals (a) 0.05dB

(b) -6.2 dB

(c) 104.7 dB

Chapter 14, Solution 7. (a) 0.05 = 20 log 10 H

2.5 × 10-3 = log 10 H H = 102.5×10 = 1.005773 -3

(b)

- 6.2 = 20 log10 H - 0.31 = log 10 H

H = 10-0.31 = 0.4898 (c)

104.7 = 20 log 10 H 5.235 = log10 H H = 105.235 = 1.718 × 10 5

Chapter 14, Problem 8.

Determine the magnitude (in dB) and the phase (in degrees) of H( ω ) = at ω = 1 if H (ω ) equals (a) 0.05 dB

(b) 125

(c)

Chapter 14, Solution 8. (a) H = 0.05 H dB = 20 log10 0.05 = - 26.02 ,

(b)

(c)

H = 125 H dB = 20 log10 125 = 41.94 ,

(d)

3 6 + 1 + jω 2 + jω

φ = 0°

φ = 0°

H(1) = H dB

(d)

j10 = 4.472∠63.43° 2+j = 20 log10 4.472 = 13.01 ,

10 jω 2 + jω

φ = 63.43°

3 6 + = 3.9 − j2.7 = 4.743∠ - 34.7° 1+ j 2 + j H dB = 20 log10 4.743 = 13.521, φ = –34.7˚

H(1) =

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 9.

A ladder network has a voltage gain of H( ω ) =

10 (1 + jω )(10 + jω )

Sketch the Bode plots for the gain.

Chapter 14, Solution 9. H(ω) =

1 (1 + jω)(1 + j ω 10)

H dB = -20 log10 1 + jω − 20 log10 1 + jω / 10 φ = - tan -1 (ω) − tan -1 (ω / 10) The magnitude and phase plots are shown below.

HdB 0.1

1

10

ω

100 20 log 10

-20

1 1 + j ω / 10 20 log 10

-40

1 1 + jω

φ 0.1 -45°

1

10

ω

100 arg

1 1+ jω / 10

-90° arg

-135°

1 1 + jω

-180°

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Chapter 14, Problem 10.

Sketch the Bode magnitude and phase plots of: H(j ω ) =

50 j ω (5 + j ω )

Chapter 14, Solution 10.

H( j ω) =

50 = jω(5 + jω)

10 jω ⎞ ⎛ 1 jω⎜ 1 + ⎟ 5⎠ ⎝

HdB 40 20 log1

20 10 0.1 -20

1

100

⎛ ⎜ 1 20 log ⎜⎜ jω ⎜ 1+ 5 ⎝

⎛ 1 ⎞ ⎟ 20 log⎜⎜ ⎟ ⎝ jω ⎠

-40

φ 0.1 -45°

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

ω

1 10

ω

100 arg

1 1 + jω / 5

-90° arg

-135°

1 jω

-180°

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 11.

Sketch the Bode plots for 10 + j ω H( ω ) = j ω( 2 + j ω ) Chapter 14, Solution 11. 5 (1 + jω 10) H (ω) = jω (1+ jω 2)

H dB = 20 log10 5 + 20 log10 1 + jω 10 − 20 log10 jω − 20 log10 1+ jω 2 φ = -90° + tan -1 ω 10 − tan -1 ω 2

The magnitude and phase plots are shown below. HdB 40 34 20 14 0.1 -20

1

10

100

ω

1

10

100

ω

-40

φ 90°

45° 0.1 -45° -90°

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 12. A transfer function is given by T (s ) =

s +1 s( s + 10)

Sketch the magnitude and phase Bode plots.

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Chapter 14, Solution 12. T ( w) =

0.1(1 + jω ) , j ω (1 + j ω / 10)

20 log 0.1 = −20

The plots are shown below. |T|

(db)

20

ω

0 0.1

1

10

100

1

10

100

-20 -40

arg T

90o

ω

0 0.1 -90o

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 13. Construct the Bode plots for

G (s ) =

s +1 , s (s + 10)

s=j ω

2

Chapter 14, Solution 13. G(ω ) =

1 + jω (1 10)(1 + j ω) = 2 ( j ω) (10 + j ω) ( j ω) 2 (1 + j ω 10)

G dB = -20 + 20 log 10 1 + jω − 40 log10 jω − 20 log10 1 + jω 10 φ = -180° + tan -1ω − tan -1 ω 10

The magnitude and phase plots are shown below. GdB 40 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40

φ

90° 0.1 -90° -180°

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 14. Draw the Bode plots for H( ω ) =

50( jω +1) j ω( −ω 2 + 10 j ω + 25)

Chapter 14, Solution 14. 50 H (ω) = 25

1 + jω ⎛ j ω10 ⎛ jω ⎞2 ⎞ jω⎜⎜ 1+ + ⎜ ⎟ ⎟⎟ 25 ⎝ 5 ⎠ ⎠ ⎝

H dB = 20 log 10 2 + 20 log 10 1 + jω − 20 log10 j ω − 20 log 10 1 + jω2 5 + ( jω 5) 2 ⎛ ω10 25 ⎞ ⎟ φ = -90° + tan -1 ω − tan -1 ⎜ ⎝1 − ω 2 5 ⎠ The magnitude and phase plots are shown below.

HdB 40 26 20 6 0.1 -20

1

10

100

ω

1

10

100

ω

-40 φ 90° 0.1 -90° -180° PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 15. Construct the Bode magnitude and phase plots for 40( s +1) , s=j ω H (s ) = ( s + 2)(s + 10) Chapter 14, Solution 15. H (ω) =

40 (1 + jω) 2 (1+ jω ) = (2 + jω)(10 + jω) (1 + j ω 2)(1 + j ω 10)

H dB = 20 log 10 2 + 20 log10 1 + jω − 20 log10 1 + jω 2 − 20 log10 1 + jω 10 φ = tan -1 ω − tan -1 ω 2 − tan -1 ω 10 The magnitude and phase plots are shown below. HdB 40 20 6 0.1 -20

1

10

100

ω

1

10

100

ω

-40 φ 90° 45° 0.1 -45° -90°

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 16. Sketch Bode magnitude and phase plots for 10 , s=j ω H (s ) = 2 s (s + s + 16) Chapter 14, Solution 16. 10 /16 0.625 H (ω ) = = 2 2 ⎡ ⎡ ⎛ jω ⎞ ⎤ ⎛ jω ⎞ ⎤ jω ⎢1 + jω + ⎜ ⎟ ⎥ jω ⎢ 1 + j ω + ⎜ 4 ⎟ ⎥ ⎝ 4 ⎠ ⎥⎦ ⎝ ⎠ ⎥⎦ ⎢⎣ ⎣⎢

⎛ jω ⎞ HdB = 20log 0.625 − 20log | jω | −20log |1 + jω + ⎜ ⎟ | ⎝ 4 ⎠ (20log0,625= –4.082) 2

The magnitude and phase plots are shown below. H 20 20 log (j ω) 1

4

10

100

40

ω

0.1

–4.082 –20 ⎛ jω ⎞ 20 log 1 + jω + ⎜ ⎟ ⎝ 4 ⎠

–40 –60

φ

-90 -180

ω 0.4

1

4

10

40

90°

100

-tan-1

ω ω2

1−

16

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2

Chapter 14, Problem 17.

Sketch the Bode plots for G(s) =

s , (s + 2) + (s + 1) 2

s=jω

Chapter 14, Solution 17. G (ω ) =

(1 4 ) jω (1 + jω)(1 + jω 2)2

G dB = -20log10 4 + 20 log 10 jω − 20 log10 1 + jω − 40 log10 1 + jω 2 φ = -90° - tan -1ω − 2 tan -1 ω 2 The magnitude and phase plots are shown below.

GdB 20 0.1

1

10

100 ω

-12 -20 -40

φ 90° 0.1

1

10

100

ω

-90° -180°

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Chapter 14, Problem 18.

A linear network has this transfer function H(s) =

7s 2 + s + 4 , (s 3 + 8s 2 + 14s + 5)

s=j ω

Use MATLAB or equivalent to plot the magnitude and phase (in degrees) of the transfer function. Take 0.1 < ω < 10 rads/s. Chapter 14, Solution 18. The MATLAB code is shown below.

>> w=logspace(-1,1,200); >> s=i*w; >> h=(7*s.^2+s+4)./(s.^3+8*s.^2+14*s+5); >> Phase=unwrap(angle(h))*57.23; >> semilogx(w,Phase) >> grid on 60

40

H (jw ) P h a s e

20

0

-2 0

-4 0

-6 0 -1 10

10 w

0

10

1

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Now for the magnitude, we need to add the following to the above, >> H=abs(h); >> HdB=20*log10(H); >> semilogx(w,HdB); >> grid on

0

-5

HdB

-1 0

-1 5

-2 0

-2 5 -1 10

10 w

0

10

1

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 14, Problem 19. Sketch the asymptotic Bode plots of the magnitude and phase for

H(s) ...