Solution manual fundamentals of electric circuits 3rd edition chapter14

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

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

ω ω0
1 + (ω ω0 ) 2

φ = ∠H (ω) =

⎛ω⎞
π
− tan -1 ⎜ ⎟

2
⎝ ω0 ⎠

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except
that ω0 = 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.

V
H(s) = o =
Vi

2+

1
s/8

10 + 20 +

1
s/8

=

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

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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.
0.2 F

⎯⎯

→

2

V1

5
s
+

Vi

+
_

10
s

5

Vo
_

10
5
10 1 + s
(5 + )
5(
)
10

5
10( s + 1)
s
s
s
s
=
=
Let Z = //(5 + ) =
15
5
s
s
s ( s + 3)
5+
(3 + s )
s
s
Z
V1 =
Vi
Z +2
5
s
s
Z
Vo =
V1 =
V1 =
•

Vi
5+5/ s
s +1
s +1 Z + 2
10( s + 1)
10s
5s
s
s ( s + 3)
V
•
=
= 2
H (s) = o =
Vi s + 1
10( s + 1) 2s ( s + 3) + 10( s + 1) s + 8s + 5
2+
s ( s + 3)
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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
=
jωC 1 + jωRC

R
Vo
R
1 + jωRC
=
=
H (ω) =
R
Vi
R + jωL (1 + jωRC)
jωL +
1 + jωRC

(b)

H (ω) =

R

- ω RLC + R + jωL

H (ω) =

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

2

- ω 2 LC + jωRC
H (ω) =
1 − ω 2 LC + jωRC

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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
sC = R
(b) Let Z = R //
=
sC R + 1 1 + sRC
sC
Z
Vo =
Vs

Z + sL

V
Z
H(s) = o =
=
Vi Z + sL

R
R
1 + sRC =
R
s 2 LRC + sL + R
sL +
1 + sRC

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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

s +1

sI s
sI
Z
Is =
= 2 s
( I s x1) =
2
s
( s + 1) + s s + 3s + 1
Z + s +1
s +1+
s +1
Vo
sI s
Io = = 2
1 s + 3s + 1
I
s
H (s) = o = 2
I s s + 3s + 1

Vo =

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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 log10 H

2.5 × 10 -3 = log10 H
H = 10 2.5×10 = 1.005773
-3

(b)

- 6.2 = 20 log10 H
- 0.31 = log10 H

H = 10 -0.31 = 0.4898
(c)

104.7 = 20 log10 H
5.235 = log10 H
H = 10 5.235 = 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
+
2 + jω
1 + 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) =

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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 + jω 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 + jω / 10
20 log10

-40

1
1 + jω

φ
0.1
-45°

1

10

ω

100
arg

1
1 + jω / 10

-90°
-135°

arg

1
1 + jω

-180°

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Chapter 14, Problem 10.

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

50
jω (5 + jω )

Chapter 14, Solution 10.

H( jω) =

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°

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Chapter 14, Problem 11.

Sketch the Bode plots for
H( ω ) =

10 + jω
jω ( 2 + jω )

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°

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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 )
,
jω (1 + jω / 10)

20 log 0.1 = −20

The plots are shown below.
|T|

(db)

20

ω

0
0.1

1

10

100

-20
-40

arg T

90o

ω

0
0.1

1

10

100

-90o

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

G dB = -20 + 20 log10 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°

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

50( jω + 1)
jω (−ω 2 + 10 jω + 25)

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

1 + jω
⎛ jω10 ⎛ jω ⎞ 2 ⎞
+ ⎜ ⎟ ⎟⎟
jω⎜⎜1 +
⎝5⎠ ⎠
25
⎝

H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 jω
− 20 log10 1 + jω2 5 + ( jω 5) 2
⎛ ω10 25 ⎞

⎟
φ = -90° + tan -1 ω − tan -1 ⎜
⎝ 1 − ω2 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°
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Chapter 14, Problem 15.
Construct the Bode magnitude and phase plots for

H(s) =

40( s + 1)
,
( s + 2)( s + 10)

s=j ω

Chapter 14, Solution 15.
H (ω) =

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

H dB = 20 log10 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°

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Chapter 14, Problem 16.
Sketch Bode magnitude and phase plots for

H(s) =

10
,
s ( s + s + 16)

s=j ω

2

Chapter 14, Solution 16.
10 /16
0.625
H (ω ) =
=

2
2
⎡
⎡
⎛ jω ⎞ ⎤
⎛ jω ⎞ ⎤
jω ⎢1 + jω + ⎜
⎟ ⎥ jω ⎢1 + jω + ⎜
⎟ ⎥
⎝ 4 ⎠ ⎥⎦
⎝ 4 ⎠ ⎥⎦
⎢⎣
⎢⎣
2

⎛ jω ⎞
H dB = 20 log 0.625 − 20 log | jω | −20 log |1 + jω + ⎜
⎟ |
⎝ 4 ⎠
(20log0,625= –4.082)
The magnitude and phase plots are shown below.
H
20
20 log (jω)
1

10

4

40

100

ω

0.1

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

–40

2

–60

φ

-90
-180

0.4

1

4

10

40

90°

100

-tan-1

ω
ω2

1−

ω

16

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Chapter 14, Problem 17.

Sketch the Bode plots for
G(s) =

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

s=j ω

Chapter 14, Solution 17.
G (ω) =

(1 4) jω
(1 + jω)(1 + jω 2) 2

G dB = -20log10 4 + 20 log10 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 + 14 s + 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

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Chapter 14, Problem 19.

Sketch the asymptotic Bode plots of the magnitude and phase for
H(s) =

100s
,
( s + 10)( s + 20)( s + 40)

s=j ω

Chapter 14, Solution 19.

H (ω ) =

100 jω
jω / 80
=
j
jω
jω
ω
( jω + 10)( jω + 20)( jω + 40) (1 +
)(1 +
)(1 +
)
10
20
40

H dB = 20 log(1/ 80) + 20 log | jω /1| −20 log |1 +

jω
jω
jω
| −20 log |1 +
| −20 log |1 +
|
10
20
40

(20log(1/80) = -38.06)
The magnitude and phase plots are shown below.
H

20 log jω

ω

0.1

1

20

20 log

10

20

40

1
80

100

20 log 1 +

40
60
φ

90
0.1

ω
1

2

4

10

20

40

100 200 400

-90

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jω
10


Chapter 14, Problem 20.

Sketch the magnitude Bode plot for the transfer function
H( ω ) =

10 jω
( jω + 1)( jω + 5) 2 ( jω + 40)

Chapter 14, Solution 20.

H (ω ) =

10 jω
jω /100

=
2
(25)(40)(1 + jω )(1 + jω / 5) (1 + jω / 40) (1 + jω )(1 + jω / 5)2 (1 + jω / 40)

20log(1/100) = -40
The magnitude plot is shown below.

20 log jω

40

1

20 log
1+

20

0.1

jω
10

ω
1

5

10

50

100
20 log

-20
20 log

1
100

1
1 + jω

-40

-60

20 log

1
jω ⎞
⎛
⎜1 +
⎟
5 ⎠
⎝

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2


Chapter 14, Problem 21.

Sketch the magnitude Bode plot for
H(s) =

s ( s + 20)
,
( s + 1)( s 2 + 60s ) = (400)

Chapter 14, Solution 21.
jω ( jω + 20)
=
H (ω ) =
( jω + 1)(−ω 2 + 60 jω + 400)

H(ω) =

0.05 jω(1 + jω / 20)
⎛ 6 jω ⎛ jω ⎞ 2 ⎞
(1 + jω)⎜1 +
+⎜ ⎟ ⎟
⎜
40 ⎝ 20 ⎠ ⎟

⎠
⎝

s=j ω

20 jω (1 + jω / 20)
2

⎛ jω ⎞
400( jω + 1)(1 + 60 jω / 400 + ⎜
⎟ )
⎝ 20 ⎠

H dB = 20 log(0.05) + 20 log jω + 20 log 1 +

jω
j6ω ⎛ jω ⎞
− 20 log 1 + jω − 20 log 1 +
+⎜ ⎟
20
40 ⎝ 20 ⎠

2

The magnitude plot is as sketched below.
H&B
40

20log|jω|
20 log |1+jω/20|

20

0.1

–20

–40

–60

ω
1

10

20

100

20 log 0.05
–20 log 1 + jω

j 6ω ⎛ jω ⎞
+⎜
–20 log 1 +
⎟
40 ⎝ 20 ⎠

–80

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
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2


Chapter 14, Problem 22.

Find the transfer function H( ω ) with the Bode magnitude plot shown in Fig. 14.74.

Figure 14.74
For Prob. 14.22.
Chapter 14, Solution 22.
20 = 20 log10 k

⎯
⎯→ k = 10

A zero of slope + 20 dB / dec at ω = 2 ⎯
⎯→ 1 + jω 2
1
A pole of slope - 20 dB / dec at ω = 20 ⎯
⎯→
1 + jω 20
1
A pole of slope - 20 dB / dec at ω = 100 ⎯
⎯→

1 + jω 100

Hence,
H (ω) =

10 (1 + jω 2)
(1 + jω 20)(1 + jω 100)

10 4 ( 2 + jω)
H (ω) =
( 20 + jω)(100 + jω)

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,
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Chapter 14, Problem 23.

The Bode magnitude plot of H( ω ) is shown in Fig. 14.75. Find H( ω ).

Figure 14.75
For Prob. 14.23.

Chapter 14, Solution 23.

A zero of slope + 20 dB / dec at the origin

⎯
⎯→

jω

1
1 + jω 1
1
A pole of slope - 40 dB / dec at ω = 10 ⎯
⎯→
(1 + jω 10) 2

A pole of slope - 20 dB / dec at ω = 1 ⎯
⎯→

Hence,
H (ω) =

jω
(1 + jω)(1 + jω 10) 2

H (ω) =

100 jω
(1 + jω)(10 + jω) 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 24.

The magnitude plot in Fig. 14.76 represents the transfer function of a preamplifier. Find
H(s).

Figure 14.76
For Prob. 14.24.

Chapter 14, Solution 24.

40 = 20 log10 K
⎯⎯
→ K = 100
There is a pole at ω=50 giving 1/(1+jω/50)
There is a zero at ω=500 giving (1 + jω/500).
There is another pole at ω=2122 giving 1/(1 + jω/2122).
Thus,
1
40 x
( s + 500)
40(1 + jω / 500)
500
H (ω ) =
=
(1 + jω / 50)(1 + jω / 2122) 1 x 1 ( s + 50)( s + 2122)
50 2122
or

H (s) =

8488( s + 500)
( s + 50)( s + 2122)

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.