Chapter 14, Problem 1. - Howard R. Hughes College of ...

[Pages:114]Chapter 14, Problem 1.

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

Figure 14.68 For Prob. 14.1.

Chapter 14, Solution 1.

H() =

Vo Vi

=

R R + 1 jC

= jRC 1+ jRC

H()

=

j 0 1 + j 0

,

1 where 0 = RC

H = H() = 0 1 + ( 0 )2

=

H ()

=

2

-

tan

-1

0

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

H

1 0.7071

0

0 = 1/RC

90? 45?

0

0 = 1/RC

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 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 =

2+ 1 s/8

= 2+8/s = 1

s+4

Vi 10 + 20 + 1 12 + 8 / s 6 s + 0.6667

s/8

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.

0.2F

1 = 1 = 5 jC s(0.2) s

0.1F 1 = 10 s(0.1) s

The circuit becomes that shown below. 5

2

V1

s

+

Vi

_

10

s

+

5

Vo

_

Let

Z

=

10 s

//(5

+

5 )

s

=

10 (5 + 5) ss 5 + 15

=

10

s 5

5(1+ s s

(3 + s)

)

=

10(s +1) s(s + 3)

s

s

V1

=

Z

Z +

2 Vi

Vo

=

5 5+5/

s V1

=

s

s +

1V1

=

s

s? +1

Z Z+

2 Vi

10(s +1)

H (s)

= Vo Vi

=

s

s? +1

s(s + 3) 2 + 10(s +1)

=

10s 2s(s + 3) +10(s +1)

=

s2

5s + 8s

+5

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( ) = V O /V i of the circuits shown in Fig. 14.71.

Figure 14.71 For Prob. 14.4.

Chapter 14, Solution 4.

1

R

(a) R || jC = 1+ jRC

R

H() =

Vo Vi

=

1+ jRC

jL

+

1

+

R jRC

=

R+

R jL (1+

jRC)

R H() = - 2RLC + R + jL

R + jL

jC (R + jL)

(b) H() = R + jL + 1 jC = 1+ jC (R + jL)

- 2LC + jRC H() = 1 - 2LC + jRC

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 = sRL R + sL

Vo

=

Z

Z + Rs

Vs

sRL

H (s) = Vo = Z = R + sL =

sRL

Vs

Z + Rs

Rs

+

sRL R + sL

RRs + s(R + Rs )L

(b)

Let

Z

=

R

//

1 sC

=

Rx 1 sC

R+ 1

=R 1+ sRC

sC

Vo

=

Z

Z +

sL

Vs

R

H(s) = Vo = Z = 1 + sRC =

R

Vi Z + sL sL + R

s2LRC + sL + R

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 jL = sL = s

Let Z = s //1 = s s +1

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

1

S

+ Is 1 _

+

Vo

Z

_

s

Vo

=

Z

Z +s

+1 (Is x1)

=

s

s +1 +1+

s

Is

=

(s

sI s + 1)2

+

s

=

s2

sI s + 3s +1

s +1

Io

= Vo 1

=

s2

sI s + 3s +1

H (s) = Io = s Is s2 + 3s +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( ) if H dB equals

(a) 0.05dB

(b) -6.2 dB

Chapter 14, Solution 7. (a) 0.05 = 20 log10 H 2.5 ?10-3 = log10 H H = 102.5?10-3 = 1.005773

(b) - 6.2 = 20 log10H - 0.31 = log10H H = 10-0.31 = 0.4898

(c) 104.7 = 20 log10 H 5.235 = log10 H H = 105.235 = 1.718 ? 105

(c) 104.7 dB

Chapter 14, Problem 8.

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

H ( ) equals

(a) 0.05 dB

(b) 125

(c) 10 j 2 + j

Chapter 14, Solution 8. (a) H = 0.05

HdB = 20 log10 0.05 = - 26.02 ,

= 0?

(d) 3 + 6 1 + j 2 + j

(b) H = 125 HdB = 20log10 125 = 41.94 ,

= 0?

(c)

H (1)

=

j10 2+ j

=

4.47263.43?

HdB = 20log10 4.472 = 13.01 ,

= 63.43?

(d) H(1) = 3 + 6 = 3.9 - j2.7 = 4.743 - 34.7? 1+ j 2+ j

HdB = 20 log10 4.743 = 13.521, = ?34.7?

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(

)

=

(1 +

10 j)(10 +

j)

Sketch the Bode plots for the gain.

Chapter 14, Solution 9. 1

H() = (1+ j)(1+ j 10)

HdB = -20 log10 1 + j - 20 log10 1 + j /10

= - tan -1() - tan -1( /10)

The magnitude and phase plots are shown below.

HdB

0.1

1

10 100

1

-20

20 log10 1 + j /10

1

-40

20 log10 1 + j

0.1 -45? -90? -135? -180?

1

10 100

1 arg 1+ j/10

1 arg 1+ j

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.

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