chap 18 solution

8/13/2019 Chap 18 solution

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18Two-Port Circuits

Assessment Problems

AP 18.1 With port 2 short-circuited, we have

I 1 = V 120

+ V 15

; I 1

V 1= y11 = 0.25S; I 2 = 2025 I 1 = 0.8I 1

When V 2 = 0, we have I 1 = y11V 1 and I 2 = y21V 1

Therefore I 2 = 0.8(y11V 1) = 0.8y11V 1Thus y21 = 0.8y11 = 0.2 SWith port 1 short-circuited, we have

I 2 = V 215

+ V 25

; I 2

V 2= y22 =

415

S

I 1 = 1520 I 2 = 0.75I 2 = 0.75y22V 2181

2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


2/43

182 CHAPTER 18. Two-Port Circuits

Therefore y12 = (0.75) 415

= 0.2 S

AP 18.2h11 =

V 1I

1 V 2 =0

= 20 5 = 4

h21 =I 2I 1 V 2 =0

= (20/ 25)I 1

I 1= 0.8

h12 =V 1V 2 I 1 =0

= (20/ 25)V 2

V 2= 0 .8

h22 =I 2V 2 I 1 =0

= 115

+ 125

= 875

S

g11 =I 1V 1 I 2 =0

= 120

+ 120

= 0.1 S

g21 =V 2V 1 I 2 =0

= (15/ 20)V 1

V 1= 0 .75

g12 =I 1I 2 V 1 =0

= (15/ 20)I 2

I 2= 0.75

g22 = V 2

I 2 V 1 =0= 15 5 = 75

20 = 3.75

AP 18.3g11 =

I 1V 1 I 2 =0

= 510650103

= 0.1 mS

g21 = V 2V 1 I 2 =0

= 200103

50 103 = 4

g12 = I 1I 2 V 1 =0= 21060.5 106

= 4

g22 = V 2I 2 V 1 =0

= 101030.5 106

= 20k



3/43

Problems 183

AP 18.4 First calculate the b-parameters:

b11 = V 2V 1 I 1 =0

= 1510

= 1.5 ; b21 = I 2V 1 I 1 =0

= 3010

= 3 S

b12 = V 2I 1 V 1 =0 = 105 = 2; b22 = I 2I 1 V 1 =0 = 45

= 0.8

Now the z -parameters are calculated:

z 11 = b22b21

= 0.8

3 =

415

; z 12 = 1b21

= 13

z 21 = bb21

= (1.5)(0.8) 6

3 = 1.6 ; z 22 =

b11b21

= 1.5

3 =

12

AP 18.5z 11 = z 22, z 12 = z 21 , 95 = z 11(5) + z 12(0)Therefore, z 11 = z 22 = 95/ 5 = 19

11.52 = 19I 1 z 12(2.72)0 = z 12I 1 19(2.72)Solving these simultaneous equations for z 12 yields the quadratic equation

z 212 + 7217z 12 613717 = 0

For a purely resistive network, it follows that z 12 = z 21 = 17 .

AP 18.6 [a] I 2 = V ga11Z L + a12 + a21Z gZ L + a22Z g= 50103

(5 104)(5 103) + 10 + (10 6)(100)(5 103) + ( 3 102)(100)

= 50

103

10 = 5 mAP L =

12

(5 103)2(5 103) = 62 .5 mW

[b] Z Th = a12 + a22Z ga11 + a21Z g

= 10 + (3 102)(100)5 104 + (10 6)(100)

= 7

6 104 =

706

k



4/43


[c] V Th = V g

a11 + a21Z g=

501036 104

= 500

6 V

Therefore V 2 = 250

6 V; P max =

(1/ 2)(250/ 6)2

(70/ 6) 103 = 74.4 mW

AP 18.7 [a] For the given bridged-tee circuit, we have

a11 = a22 = 1.25, a21 = 120

S, a12 = 11.25

The a -parameters of the cascaded networks are

a11 = (1 .25)2 + (11 .25)(0.05) = 2.125

a12 = (1 .25)(11.25) + (11 .25)(1.25) = 28.125

a21 = (0 .05)(1.25) + (1 .25)(0.05) = 0.125S

a22 = a11 = 2.125, RTh = (45 .125/ 3.125) = 14.44

[b] V t = 1003.125

= 32V; therefore V 2 = 16 V

[c] P = 162

14.44 = 17.73 W



5/43

Problems 185

Problems

P 18.1 h11 =V 1I 1 V 2 =0

= 20 5 = 4

h21 =I 2I 1 V 2 =0

= (20/ 25)I 1

I 1= 0.8

h12 =V 1V 2 I 1 =0

= (20/ 25)V 2

V 2= 0 .8

h22 =I 2V 2 I 1 =0

= 115

+ 125

= 875

S

g11 =I 1V 1 I 2 =0

= 120

+ 120

= 0.1 S

g21 =V 2V 1 I 2 =0

= (15/ 20)V 1

V 1= 0 .75

g12 =I 1I 2 V 1 =0

= (15/ 20)I 2

I 2= 0.75

g22 =V 2I 2 V 1 =0 = 15 5 =

7520 = 3.75

P 18.2 y11 = I 1V 1 V 2 =0

; y21 = I 2V 1 V 2 =0

V 120

+ V 10

+ V 4

= 0; so V = 0.125V

. . I 1 = 1 0.12520 + 1 0

8 = 168.75mA; I 2 =

0 0.1254

+ 0 1

8 = 156.25mA



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y11 = I 1V 1 V 2 =0

= 168.75mS; y21 = I 2V 1 V 2 =0

= 156.25mS

y12 = I 1V 2 V 1 =0

; y22 = I 2V 2 V 1 =0

V 20 +

V 10 +

V

1

4 = 0; so V = 0.625V

. . I 1 = 00.62520 + 0 1

8 = 156.25mA; I 2 =

10.6254

+ 1 0

8 = 218.75mA

y12 = I 1V 2 V 1 =0

= 156.25mS; y22 = I 2V 2 V 1 =0

= 218.75mS

Summary:

y11 = 168.75mS y12 =

156.25mS y21 =

156.25mS y22 = 218.75mS

P 18.3

z 11 = V 1I 1 I 2 =0

= 1 + 12 = 13

z 21 = V 2I 1 I 2 =0

= 12

z 22 = V 2I 2 I 1 =0

= 4 + 12 = 16

z 21 = V 1I 2 I 1 =0

= 12



7/43

Problems 187

P 18.4 z = (13)(16) (12)(12) = 64

y11 = z 22 z

= 1664

= 0.25 S

y12 = z 12 z = 1264 = 0.1875S

y21 = z 21 z = 12

64 = 0.1875S

y22 = z 11 z = 1364

= 0 .203125S

P 18.5 h11 = V 1

I 1 V 2 =0; h21 =

I 2

I 1 V 2 =0

V 1I 1

= 80 [10 + 20 20] = 80 20 = 16 . . h11 = 16

I 6 = 8080 + 20I 1 = 0.8I 1

I 2 = 2020 + 20I 6 = 0.5I 6 = 0.5(0.8)I 1 = 0.4I 1 . . h21 = 0.4



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h12 = V 1V 2 I 1 =0

; h22 = I 2V 2 I 1 =0

V 2I 2

= 80 [20 + 20 90] = 25 . . h22 = 125 = 40 mS

V x = 20 9020 + 20 90

V 2

V 1 = 8080 + 10

V x = 80(20 90)

90(20 + 20 90)V 2 = 0.4V 2

. . h12 = 0.4Summary:

h11 = 16; h12 = 0.4; h21 = 0.4; h22 = 40 mS

P 18.6 V 2 = b11V 1 b12I 1I 2 = b21V 1 b22I 1

b12 = V 2I 1 V 1 =0 ; b22 = I 2

I 1 V 1 =0

5 15 = (15/ 4) ; 10 20 = (20/ 3)

I 2 = V 2

(15/ 4) + (20 / 3) =

12V 2125

; I 1 = I b I a 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


9/43

Problems 189

I a = 1520

I 2; I b = 2030

I 2

I 1 =2030

1520

I 2 = 560 I 2 = 1

12 I 2

b22 = I 2I 1 = 12

b12 = V 2I 1 = V 2

I 2I 2I 1

= 125

12 (12) = 125

b11 = V 2V 1 I 1 =0

; b21 = I 2V 1 I 1 =0

V 1 = V a V b ; V a = 1015

V 2; V b = 2035

V 2

V 1 = 1015V 2

2035V 2 =

221V 2

b11 = V 2V 1

= 21

2 = 10.5

V 2 = (10 + 5) (20 + 15) I 2 = 10.5I 2

b21 = I 2V 1

=I 2V 2

V 2V 1

= 110.5

(10.5) = 1 S

P 18.7 h11 = V 1

I 1 V 2 =0= R1 R2 = 4 . . R1R2R1 + R2 = 4

h21 = I 2I 1 V 2 =0

= R2R1 + R2

= 0.8

. . R2 = 0.8R1 + 0 .8R2 so R1 = R24 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


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Substituting,

(R2/ 4)R2(R2/ 4) + R2

= 4 so R2 = 20 and R1 = 5

h22 = I 2

V 2 I 1 =0= 1

R3 (R1 + R2) = 1

R3 25 = 0 .14

. . R3 = 10Summary:

R1 = 5; R2 = 20; R3 = 10

P 18.8 V 1 = a11V 2

a12I 2

I 1 = a21V 2 a22I 2

a11 = V 1V 2 I 2 =0

; a21 = I 1V 2 I 2 =0

V 1 = 103I 1 + 104V 2 = 103(0.5 106)V 2 + 104V 2

. . a11 = 5 104 + 104 = 4 104

V 2 = (50I 1)(40 103); . . a21 = 1

2 106 = 0.5 S



11/43

Problems 1811

a12 = V 1I 2 V 2 =0 ; a22 = I 1

I 2 V 2 =0

I 2 = 50I 1; . . a22 = I 1I 2

= 150

V 1 = 1000I 1; . . a12 = V 1I 2 = V 1

I 1I 1I 2

= (1000)(1/ 50) = 20

Summary

a11 = 4 104; a12 = 20; a21 = 0.5 S; a22 = 0.02

P 18.9 g11 = a21a11

= 0.5 1064 104

= 1.25mS

g12 = aa11 = (4 104

)(1/ 50) (0.5 106

)(20)4 104 = 0.005

g21 = 1a11

= 1

4 104 = 2500

g22 = a12a11

= (20)400106

= 5 104

P 18.10 For V 2 = 0:

I a = 50I 2200

= 14

I 2 = I 2; . . I 2 = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


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h21 = I 2I 1 V 2 =0

= 0

V 1 = (10 + j 20)I 1 . . h11 = V 1I 1 V 2 =0 = 10 + j 20

For I 1 = 0:

V 1 = 50I 2; I 2 = V 2

j 100

+ V 2 50I 2

200

200I 2 = j 2V 2 + V 2 50i2250I 2 = V 2(1 + j 2)

50I 2 = V 21 + j 2

5 = (0 .2 + j 0.4)V 2

. . V 1 = (0 .2 + j 0.4)V 2

h12 = V 1

V 2 I 1 =0= 0 .2 + j 0.4

h22 = I 2V 2 I 1 =0

= 1 + j 2

250 = 4 + j 8 mS

Summary:

h11 = 10 + j 20; h12 = 0.2 + j 0.4; h21 = 0; h22 = 4 + j 8 mS



13/43

Problems 1813

P 18.11 For I 2 = 0:

50I 1 40I 2 = 1

40I 1 + 48I 2 3(40)(I 1 I 2) = 0 so 160I 1 + 168I 2 = 0Solving,

I 1 = 84 mA; I 2 = 80mA

V 2 = 3I 2 3(40)( I 1 I 2) = 0.24 V

g11 = I 1V 1 I 2 =0

= 84 m

1 = 84 mS

g21 = V 2V 1 I 2 =0

= 0.241

= 0.24

For V 1 = 0:

50I 1 40I 2 = 0

40I 1 + 48I 2 + 3 3(40)(I 1 I 2) = 0 so 160I 1 + 168I 2 = 3Solving,

I 1 = 60mA; I 2 = 75mA 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


14/43


V 2 = 3( I 2 + 1) 3(40)( I 1 I 2) = 0 .975V

g12 = I 1I 2 V 1 =0

= 60 m1

= 0.06

g22 = V 2I 2 V 1 =0

= 0.9751

= 0.975

P 18.12 For V 2 = 0:

V 1 = (400 + 1200) I 1

h11 = V 1I 1 V 2 =0

= 1600

1 = 1600

V p = 1200(1 A) = 1200V = V n

At V n ,

1200500

+ 1200V o

1000 = 0 so V o = 3600V

. . I 2 = 3600200

= 18 A

h21 = I 2I 1 I 1 =0

= 181

= 18



15/43

Problems 1815

For I 1 = 0:

V 1 = 0

h12 = V 1V 2 I 1 =0

= 01

= 0

At V n ,

V n500 + V n V o100 = 0

But V n = V p = 0 so V o = 0; therefore,

I 2 = 1 V200

= 5 mS

h22 = I 2V 2 I 1 =0

= 5 m

1 = 5 mS

P 18.13 I 1 = g11V 1 + g12I 2; V 2 = g21V 1 + g22I 2

g11 = I 1V 1 I 2 =0

= 0.25106

20103 = 12.5 106 = 12.5 S

g21 = V 2V 1 I 2 =0

= 520 10

3 = 250

0 = 250(10) + g22(50 106)

g22 = 250050 106 = 50M

200 106 = 12.5 106(10) + g12(50 106)(200 125)106 = g12(50 106)

g12 = 7550

= 1.5



16/43


P 18.14 [a] I 1 = y11V 1 + y12V 2; I 2 = y21V 1 + y22V 2

y21 = I 2V 1 V 2 =0

= 50106

10 = 5 S

0 = y21(20

103) + y22(

5)

. . y22 = 15y21(20 103) = 20nS

200106 = y11(10) so y11 = 20 S0.25 106 = 20 106(20 103) + y12(5)y12 =

0.25 106 0.4 1065

= 30nS

Summary:

y11 = 20 S; y12 = 30nS; y21 = 5 S; y22 = 20nS

[b] y11 = gg22

; y12 = g12g22

; y21 = g21g22 ; y22 = 1g22

g = g11g22 g12g21 = (12.5 106)(50 106) 1.5(250)= 625 + 375 = 1000

y11 = 100050 106

= 20 S; y21 = 2505 106

= 5 S

y12 = 1.550 106 = 30nS; y22 = 15 106 = 20 nS

These values are the same as those in part (a).

P 18.15 I 1 = g11V 1 + g12I 2

V 2 = g21V 1 + g22I 2

V 1 = I 1g11

g12g11

I 2 and I 2 = V 2g22

g21g22

V 1

Substituting,

V 1 = I 1g11

g12g11

V 2g22

g21g22

V 1

V 1 = 1 g12g21g11 g22

= I 1g11

g12g11 g22

V 2



17/43

Problems 1817

V 1 = g22

g11g22 g12g21I 1

g12g11g22 g12g21

V 2

V 1 = h11 I 1 + h12V 2

Therefore,

h11 = g22 g

; h12 = g12 g where g = g11 g22 g12g21

I 2 = V 2g22

g21g22

I 1g11

g12g11

I 2

I 2 = 1 g12g21g11 g22

= V 2g22

g21g11 g22

I 1

I 2 = g11 g

V 2 g21 g

I 1

I 2 = h21I 1 + h22V 2

Therefore,

h21 = g21 g ; h22 = g11 g

P 18.16 V 1 = h11 I 1 + h12V 2; I 2 = h21I 1 + h22V 2Rearranging the rst equation,

V 2 = 1h12

V 1 h11h12

I 1

V 2 = b11V 1 b12I 1Therefore,

b11 = 1h12; b12 = h

11

h12

Solving the second h-parameter equation for I 2:

I 2 = h21I 1 + h22 1h12

V 1 h11h12

I 1



18/43


= I 1 h21 h22h11

h12 +

h22h12

V 1

= hh12

I 1 + h22h12

V 1

I 2 = b21V 1 b22I 1Therefore,

b21 = h22h12

; b22 = hh12

where h = h11 h22 h12h21P 18.17 I 1 = g11V 1 + g12I 2; V 2 = g21V 1 + g22I 2

V 1 = z 11I 1 + z 12I 2; V 2 = z 21I 1 + z 22I 2

I 1 = V 1z 11 z 12z 11 I 2

. . g11 = 1z 11 ; g12 = z 12

z 11

V 2 = z 21 V 1z 11

z 12z 11

I 2 + z 22I 2 = z 21z 11

V 1 +z 11z 22 z 12z 21

z 11I 2

. . g21 = z 21z 11 ; g22 = z z 11

where z = z 11z 22 z 12z 21

P 18.18 For I 2 = 0:

V 2 = V 1

4 + (1 /s )(4) =

4sV 14s + 1

a11 = V 1V 2 I 2 =0= 4s + 1

4s = s + 0.25

s

I 1 = V 1

4 + (1/s ) =

sV 14s + 1

so V 2 = 4I 1 = 4sV 14s + 1

a21 = I 1V 2 I 2 =0

= 14

= 0.25



19/43

Problems 1819

For V 2 = 0:

I 2 = I 1(4)s + 4

a22 = I 1I 2 V 2 =0

= s + 4

4

V 1 = 1

sI 1 +

4s

s + 4I 1 =

1

s +

4s

s + 4I 1 =

1

s +

4s

s + 4(s + 4)

4 I 2

= s + 4

4s + s I 2 =

4s2 + s + 44s

I 2

a12 = V 1I 2 V 2 =0

= s2 + 0 .25s + 1

s

P 18.19 z 11 = V 1I 1 I 2 =0

= (1 + 1/s )(1)

2 + 1/s + s =

s + 12s + 1

+ s

= s + 1 + 2 s2 + 2

2s + 1 =

2s2 + 2 s + 12s + 1

z 22 = z 11 (the circuit is reciprocal and symmetrical)

z 21 = V 2I 1 I 2 =0

V 2 = I 11

2 + 1/s (1) + sI 1; V 2

I 1 = s2s + 1 + s =

s + 2s2 + s2s + 1

z 21 = 2s2 + 2 s

2s + 1 =

2s(s + 1)2s + 1

z 12 = z 21 (the circuit is reciprocal and symmetrical)



20/43


21/43

Problems 1821

For V 2 = 0:

I 1 = V 1

j 50 +

V 1100

+ V 120

= V 1(6 + j 2)

100

I 2 = 0.025V 1 V 120

= 0.025V 1

a12 = V 1I 2 V 2 =0

= 10.025

= 40

a22 = I 1I 2 V 2 =0

= 2V 1(3 + j 1)100(0.025)V 1

= 2 .4 + j 0.8

Summary:

a11 = 3; a12 = 40; a21 = 80 + j 60mS; a22 = 2.4 + j 0.8

P 18.21 h11 = a12a22

= 40

(0.8)(3 + j 1) = 15 j 5

h12 = aa22

a = 3(2.4 + j 0.8) 40(0.08 + j 0.06) = 7.2 + j 2.4 3.2 j 2.4 = 4

h12 = 4

(0.8)(3 + j 1) = 1.5

j 0.50

h21 = 1a22

= 1(0.8)(3 + j 1)

= 0.375 + j 0.125

h22 = a21a22

= 0.08 + j 0.06(0.8)(3 + j 1)

= 0.0375 + j 0.0125 = 37.5 + j 12.5 mS



22/43


P 18.22 [a] h11 = V 1I 1 V 2 =0

; h21 = I 2I 1 V 2 =0

V 1 = ( R + sL)I 1 sMI 20 = sMI 1 + ( R + sL)I 2

=(R + sL) sM

sM (R + sL)

= ( R + sL)2 s2M 2

N 1 =V 1 sM 0 (R + sL)

= ( R + sL)V 1

I 1 = N 1

= (R + sL)V 1

(R + sL)2 s2M 2; h11 =

V 1I 1

= (R + sL)2 s2M 2

R + sL

0 = sMI 1 + ( R + sL)I 2; . . h21 = I 2I 1

= sM R + sL

h12 = V 1V 2 I 1 =0 ; h22 = I 2V 2 I 1 =0

V 1 =

sMI

2; I

2 =

V 2

R + sL

V 1 = sMV 2R + sL ; h12 = V 1V 2

= sM R + sL

h22 = I 2V 2

= 1

R + sL



23/43

Problems 1823

[b] h12 = h21 (reciprocal)h11 h22 h12h21 = 1 (symmetrical, reciprocal)h12 = sM R + sL ; h21 =

sM R + sL

(checks)

h11 h22 h12h21 = (R + sL)2 s2M 2

R + sL 1

R + sL (sM )(sM )

(R + sL)2

= (R + sL)2 s2M 2 + s2M 2

(R + sL)2 = 1 (checks)

P 18.23 First we note that

z 11 = (Z b + Z c)(Z a + Z b )

Z a + 2Z b + Z cand z 22 =

(Z a + Z b )(Z b + Z c)Z a + 2Z b + Z c

Therefore z 11 = z 22 .

z 12 = V 1I 2 I 1 =0

; Use the circuit below:

V 1 = Z bI x Z cI y = Z bI x Z c(I 2 I x ) = ( Z b + Z c)I x Z cI 2

I x = Z b + Z c

Z a + 2Z b + Z cI 2 so V 1 =

(Z b + Z c)2

Z a + 2Z b + Z cI 2 Z cI 2

. . Z 12 = V 1I 2 = (Z b + Z c)2

Z a + 2 Z b + Z c Z c = Z 2b Z a Z cZ a + 2Z b + Z c



24/43


z 21 = V 2I 1 I 2 =0

; Use the circuit below:

V 2 = Z bI x Z cI y = Z bI x Z c(I 1 I x ) = ( Z b + Z c)I x Z cI 1

I x = Z b + Z c

Z a + 2Z b + Z cI 1 so V 2 =

(Z b + Z c)2

Z a + 2Z b + Z cI 1 Z cI 1

. . z 21 = V 2I 1 = (Z b + Z c)2

Z a + 2Z b + Z c Z c = Z 2b Z a Z cZ a + 2Z b + Z c = z 12

Thus the network is symmetrical and reciprocal.

P 18.24 I 1 = y11V 1 + y12V 2; V 1 = V g Z g I 1I 2 = y21V 1 + y22V 2; V 2 = Z L I 2

V 2Z L

= y21V 1 + y22V 2

. . y21V 1 = 1Z L

+ y22 V 2; y21Z L V 1 = (1 + y22Z L )V 2

. . V 2V 1 = y21Z L

1 + y22Z L

P 18.25

V 2 = b11V 1 b12I 1; V 1 = V g I 1Z gI 2 = b21V 1 b22I 1; V 2 = Z L I 2



25/43

Problems 1825

I 2 = V 2Z L

= b11V 1 + b12I 1Z L

b11V 1 + b12I 1Z L

= b21V 1 b22I 1

. . V 1 b11Z L + b21 = b22 + b12Z L

I 1

V 1I 1

= b22Z L + b12b21Z L + b11

= Z in

P 18.26 V 1 = h11 I 1 + h12V 2; V 1 = V g Z gI 1I 2 = h21I 1 + h22V 2; V 2 = Z L I 2. . V g

Z gI 1 = h11 I 1 + h12V 2; V g = ( h11 + Z g )I 1 + h12V 2

. . I 1 = V g h12V 2h11 + Z g

. . V 2Z L

= h21V g h12V 2h11 + Z g

+ h22V 2

V 2(h11 + Z g )Z L

= h21V g h12h21V 2 + h22(h11 + Z g )V 2

V 2(h11 + Z g ) = h21Z L V g h12h21Z L V 2 + h22Z L (h11 + Z g )V 2h21Z L V g = ( h11 + Z g ) [V 2 + h22Z L V 2]h12h21Z L V 2

. . V 2V g = h21Z L

(h11 + Z g )(1 + h22Z L ) h12h21Z LP 18.27 I 1 = g11V 1 + g12I 2; V 1 = V g Z g I 1

V 2 = g21V 1 + g22I 2; V 2 = Z L I 2

Z L I 2 = g21V 1 + g22 I 2; V 1 = I 1 g12I 2

g11

. . Z L I 2 = g21g11

(I 1 g12I 2) + g22I 2

. . Z L I 2 + g12g21

g11I 2 g22I 2 =

g21g11

I 1

. . (Z L g11 + g)I 2 = g21 I 1; . . I 2

I 1= g21

g11Z L + g



26/43


P 18.28 V 1 = z 11I 1 + z 12I 2; V 1 = V g Z gI 1V 2 = z 21I 1 + z 22I 2; V 2 = Z L I 2V Th = V 2

I 2 =0; V 2 = z 21I 1; I 1 =

V 1

z 11=

V g I 1Z gz 11

. . I 1 = V gz 11 + Z g ; . . V 2 = z 21V gz 11 + Z g

= V Th

Z Th = V 2I 2 V g =0

; V 2 = z 21I 1 + z 22I 2

I 1Z g = z 11 I 1 + z 12I 2; I 1 = z 12I 2

z 11 + Z g

. . V 2 = z 21 z 12I 2z 11 + Z g + z 22I 2

. . V 2I 2 = z 22 z 12z 21z 11 + Z g

= Z Th

P 18.29 V 2V g

= bZ L

b12 + b11Z g + b22Z L + b21Z gZ L

b = b11b22

b12b21 = (25)(

40)

(1000)(

1.25) = 250

. . V 2V g = 250(100)

1000 + 25(20) 40(100) 1.25(2000) = 5

V 2 = 5(120/0) = 600/ 180 V(rms)

I 2 = V 2100 = 600/ 180

100 = 6 A(rms)

I 2I 1

= bb11 + b21Z L

= 25025

1.25(100)

= 2.5

. . I 1 = I 22.5 = 62.5

= 2.4A(rms)

. . P g = (120)(2 .4) = 288W; P o = 36(100) = 3600 W

. . P oP g = 3600

288 = 12.5



27/43

Problems 1827

P 18.30 [a] V2V g

= y21Z L

y12y21Z gZ L (1 + y11Z g )(1 + y22Z L )y12y21Z g Z L = (2 106)(100 103)(2500)(70,000) = 351 + y11Z g = 1 + (2

103)(2500) = 6

1 + y22Z L = 1 + ( 50 106)(70 103) = 2.5y21Z L = (100 103)(70 103) = 7000V 2V g

= 7000

35 (6)(2.5) =

7000

20 = 350

V 2 = 350V g = 350(80) 103 = 28V(rms)V 2 = 28 V(rms)

[b] P = |V 2|270,000

= 11 .2

103 = 11 .20mW

[c] I2 = 28/18070,000 = 0.4 103 /180 = 400/0 A

I2I1

= y21

y11 + yZ L

y = (2 103)(50 106) (2 106)(100 103)= 100 109

yZ L = (100)(70)

103

109 = 7

103

y11 + yZ L = 2 103 + 7 103 = 9 103I2I1

= 100103

9 103 =

1009

. . 100I1 = 9I2; I1 = 9(400 106)

100 = 36 A(rms)

P g = (80)103 (36) 106 = 2 .88WP 18.31 [a] Z Th =

1 + y11Z gy22 + yZ g

From the solution to Problem 18.30

1 + y11Z g = 1 + (2 103)(2500) = 6y22 + yZ g = 50 106 + 107(2500) = 200 106

Z Th = 6200 10

6 = 30,000

Z L = Z Th = 30,000



28/43


[b] y21Z L = (100 103)(30,000) = 3000y12y21Z gZ L = (2 106)(100 103)(2500)(30,000) = 151 + y11Z g = 6

1 + y22Z L = 1 + (50 106)(30 103) = 0.5V 2V g

= 3000

15 6(0.5) =

3000

12 = 250

V 2 = 250(80 103) = 20 = 20/ 180 V(rms)P = |V 2|2

30,000 =

40030 10

3 = 13.33mW

[c] I2 = V 230,000

= 20/0

30,000

= 2

3

mA

I2I1

= 100103

2 103 + 107(30,000) =

1001035 103

= 20

I1 = I220

= 2103

3(20) =

130

mA(rms)

P g (developed) = (80 103) 130 10

3 = 83

W

P 18.32 [a] For I 2 = 0:

V 2 = j 150I 1 = j 150 V 1

50 + j 50 = j 3V 1

1 + j 1

a11 =

V 1V 2 I 2 =0 =

1 + j 1

j 3 = 1 + j 1

3

a21 = I 1V 2 I 2 =0

= 1

j 150 =

j150

S



29/43

Problems 1829

For V 2 = 0:

V 1 = (50 + j 50)I 1 j 150I 20 = j 150I 1 + (400 + j 800)I 2

=50 + j 50 j 150 j 150 400 + j 800

= 2500(1 + j 24)

N 2 =50 + j 50 V 1

j 150 0= j 150V 1

I 2 = N 2

= j 150V 1

2500(1 + j 24)

a12 = V 1I 2 V 2 =0 = 503

(24 j 1) j 150I 1 = (400 + j 800)I 2

a22 = I 1

I 2 V 2 =0= 83(2 j 1)

[b] V Th = V g

a11 + a21Z g=

260/0(1 + j 1)/ 3 + j 25/ 150

= (260/ 0)6

2 + j 2 + j 1 =

1560/ 0

2 + j 3= 120(2 j 3) = 432.47/ 123.69 V

Z Th = a12 + a22Z ga11 + a21Z g

= [(50/ 3)(24 j 1)] + [(8/ 3)(2 j 1)(25)]

[(1 + j 1)/ 3] + [( j/ 150)(25)]= 100(24 j 1) 16(2 j 1)(25)2 + j 2 + j 1

= 3200 + j 5002 + j 3= 607.69 + j 661.54



30/43


[c] V 2 = 1000

1607.69 + j 661.54(432.67/ 123.69) = 248.88/ 146.06

v2(t) = 248.88 cos(4000t 146.06) V

P 18.33 [a] Z Th = g22 g12g21Z g1 + g11 Z g

g12g21 = 12

+ j12

12 j

12

= j12

1 + g11Z g = 1 + 1 j 1 = 2 j 1

. . Z Th = 1.5 + j 2.5 j 32 j 1

= 2.1 + j 1.3

. . Z L = 2.1 j 1.3 V 2V g

= g21Z L

(1 + g11Z g )(g22 + Z L ) g12g21Z gg21Z L =

12 j

12

(2.1 j 1.3) = 0 .4 j 1.71 + g11Z g = 1 + 1 j 1 = 2 j 1g22 + Z L = 1.5 + j 2.5 + 2.1 j 1.3 = 3.6 + j 1.2g12g21Z g = j 3

V 2V g

= 0.4 j 1.7

(2 j 1)(3.6 + j 1.2) j 3 =

0.4 j 1.78.4 j 4.2

V 2 = 0.4 j 1.78.4

j 4.2

(42/ 0) = 5 j 6 V(rms) = 7 .81/ 50.19 V(rms)The rms value of V 2 is 7.81 V.

[b] I 2 = V 2Z L = 5 + j 6

2.1 j 1.3 = 3 + j 1A(rms)

P = |I2|2(2.1) = 21 W



31/43

Problems 1831

[c] I2I1

= g21g11Z L + g

g =16 j

16

32

+ j52

12 j

12

12

+ j12

= 312

+ j 512 j 312 + 512 j 12 = 23 j 13

g11 Z L =16 j

16

(2.1 j 1.3) = 0.8

6 j3.46

. . g11Z L + g = 0.86 j3.46

+ 46 j

26

= 0.8 j 0.9I2I1

= [(1/ 2) j (1/ 2)]0.8 j 0.9

. . I1 = (0.8 j 0.9)I20.5 + j 0.5 = 1.6 j 1.81 + j 1 I2

= ( 1.7 + j 0.1)(3 + j 1) = 5 j 2A(rms). . P g(developed) = (42)(5) = 210 W

% delivered = 21210

(100) = 10%

P 18.34 V 1 = h11 I 1 + h12V 2

I 2 = h21I 1 + h22V 2

From the rst measurement:

h11 = V 1I 1

= 45 10

3 = 800

h21 = 2005 = 40

. . V 1 = 800I 1 + h12V 2; I 2 = 40I 1 + h22V 2From the second measurement:

h22V 2 = 40I 1

h22 = 40(20 106)

40 = 20 S



32/43


20 103 = 800(20 106) + 40h12

. . h12 = 4103

40 = 104

Summary:

h11 = 800; h12 = 104 ; h21 = 40; h22 = 20 SFrom the circuit,

Z g = 250; V g = 5.25mV

Z Th = h11 + Z gh22Z g + h

h = 800(20 106) + 40 104 = 20 103

Z Th = 800 + 250

20 106(250) + 20 103 = 42k

V Th = h21V gh22Z g + h = 40(5.25 103)

25 103 = 8.4 V

i = 8.484,000

= 0.10mA

P = (0 .10

103)2(42,000) = 420 W

P 18.35 When V 2 = 0

V 1 = 20 V, I 1 = 1 A, I 2 = 1 AWhen I 1 = 0

V 2 = 80 V, V 1 = 400 V, I 2 = 3 A 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


33/43

Problems 1833

h11 = V 1I 1 V 2 =0

= 20

1 = 20

h12 = V 1V 2 I 1 =0

= 400

80 = 5

h21 = I 2I 1 V 2 =0

= 11

= 1

h22 = I 2V 2 I 1 =0

= 380

= 37.5 mS

Z Th = Z g + h11h22Z g + h

= 10

Source-transform the current source and parallel resistance to get V g = 240 V.Then,

I 2 = h21V g

(1 + h22Z L)(h11 + Z g ) h12h21Z L= 1.5 A

P = (1.5)2(10) = 22 .5 W

P 18.36 [a] g11 = I 1V 1 I 2 =0

; g21 = V 2V 1 I 2 =0

V 1I 1

= 1sC

sL + 1sC

= [sL + (1 /sC )](1/sC )

sL + (2 /sC )

= sL + (1 /sC )

s2LC + 2 =

s2LC + 1sC (s2LC + 2)

= (1/C )[s2 + (1 /LC )]

s[s2 + (2 /LC )]

. . g11 = Cs[s2 + (2 /LC )])

s2 + (1 /LC )

V 2 = (1/sC )

sL + (1 /sC )V 1 so

V 2V 1

= (1/sC )

sL + (1 /sC ) =

1s2LC + 1

= (1/LC )s2 + (1 /LC )

. . g21 = (1/LC )s2 + (1 /LC ) 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


34/43


g12 = I 1I 2 V 1 =0

; g22 = V 2I 2 V 1 =0

I 1 = (1/sC )sL + (1 /sC ) I 2 so g12 = (1/LC )

s2 + (1 /LC )

g22 = sL (1/sC ) = sL/sC

sL + (1 /sC ) =

sLs2LC + 1

= (1/C )s

s2 + (1 /LC )

Summary:

g11 = Cs[s2 + (2 /LC )])

s2 + (1 /LC ) ; g12 = (1/LC )

s2 + (1 /LC )

g21 = (1/LC )s2 + (1 /LC )

; g22 = (1/C )s

s2 + (1 /LC )

[b] 1LC

= 109

(0.2)(200) = 25 106

g11 = 2107s(s2 + 50 106)

s2 + 25 106

g12 = 25

106

s2 + 25 106g21 =

25106s2 + 25 106

g22 = 5106ss2 + 25 106

V 2V 1

= g21Z Lg22 + Z L

= 2510

6

s 2 +25 106 400

5106 s

(s 2 +25 106 ) + 400

V 2V 1 =

25

106

s2 + 12 ,500s + 25 106 = 25

106

(s + 2500)(s + 10,000)

V 1 = 30

s

V 2 = 750106

s(s + 2500)(s + 10,000) =

30s

40s + 2500

+ 10

s + 10,000

v2 = [30 40e2500 t + 10e10,000 t]u(t) V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


35/43


36/43


V 2 = K 1

s +

K 2s + 1

+ K 3

s + 12 j 72

+ K 3

s + 12 + j 72

K 1 = 25; K 2 = 25; K 3 = 9.45/90

. . v2(t) = [25

25et + 18 .90e0.5t cos(1.32t + 90)]u(t) V

P 18.38 The a parameters of the rst two port are

a11 = hh21 = 5 103

40 = 125 106

a12 = h11h21 = 100040

= 25

a21 = h22h21

= 2540

106 =

625

109 S

a22 = 1h21 = 1

40 = 25 103

The a parameters of the second two port are

a11 = 54

; a12 = 3R

4 ; a21 =

34R

; a22 = 54

or a11

= 1.25; a12

= 54k; a21

= 1

96 mS; a

22 = 1.25

The a parameters of the cascade connection are

a11 = 125106(1.25) + (25)(103 / 96) = 102

24

a12 = 125106(54 103) + ( 25)(1.25) = 38

a21 = 625109(1.25) + (25 103)(103 / 96) = 104

96 S

a22 = 625109(54 103) + ( 25 103)(1.25) = 65 103

V oV g

= Z L

(a11 + a21Z g)Z L + a12 + a22Z g

a21Z g = 104

96 (800) = 102

12 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


37/43

Problems 1837

a11 + a21Z g = 102

24 + 102

12 = 102

8

(a11 + a21Z g)Z L = 102

8 (72,000) = 90

a22Z g = 65 103(800) = 52V oV g

= 72,000

90 38 52 = 400

vo = V o = 400V g = 3.6 VP 18.39 [a] From reciprocity and symmetry

a11 = a22, a = 1; . . 52 24a21 = 1, a21 = 1 SFor network B

a11 = V1V 2 I 2 =0

V 1 = (5 + j 15 j 10)I 1 = (5 + j 5)I 1V 2 = ( j 10 + j 5)I 1 = j 5I1a11 =

5 + j 5

j 5 = 1 + j 1

a21 = I1V 2 I 2 =0

= 1

j 5 = j 0.2 S

a22 = a11 = 1 + j 1 a = 1 = ( 1 + j 1)(1 + j 1) j 0.2a12

. . a

12 =

10 + j 5

Summary:

a11 = 5 a11 = 1 + j 1a12 = 24 a12 = 10 + j 5 a21 = 1 S a21 = j 0.2 S

a22 = 5 a22 = 1 + j 1 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should beobtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,Pearson Education, Inc., Upper Saddle River, NJ 07458.


38/43


[b] a11 = a11a11 + a12a21 = 5 + j 9.8a12 = a11a12 + a12a22 = 74 + j 49 a21 = a21a11 + a22a21 = 1 + j 2 Sa22 = a21a12 + a22a22 = 15 + j 10I2 = V ga11Z L + a12 + a21Z gZ L = 0 .295 + j 0.279AV 2 = 10I 2 = 2.95 j 2.79 V

P 18.40 a11 = z 11z 21

= 35/ 34000/ 3

= 8.75 103

a12 = z z 21

= 25104 / 3

4000/ 3 = 62.5

a21 = 1z 21

= 14000/ 3

= 0.75 103

a22 = z 22z 21

= 10,000/ 3

4000/ 3 = 2.5

a11 = y22y21 = 40 106

800 106 = 0.05 S

a12 = 1y21 = 1

800

106

= 1250S

a21 = yy21 = 4 108

800106 = 50 106 S

a22 = y11y21 = 200 106

800 106 = 0.25 S

a11 = a11a11 + a12a21 = (8 .75 103)(0.05) + (62 .5)(50 106) = 3 .5625103a12 = a11a12 + a12a22 = (8 .75 103)(1250) + (62 .5)(0.25) = 26.5625

a21 = a21a11 + a22a21 = (0 .75 103)(0.05) + (2 .5)(50 10

6) = 162.5 10

6

a22 = a21a12 + a22a22 = (0 .75 103)(1250) + (2 .5)(0.25) = 1.5625V 2 =

Z L V g(a11 + a21Z g )Z L + a12 + a22Z g

= (15,000)(0.03)

[3.5625 103 + (162 .5 106)(10)](15,000) + 26.5625 + (1.5625)(10) = 3.75 V



39/43


40/43


For V Th note that V oc = z 21

Z g + z 11V g since I 2 = 0.

P 18.43 [a] V 1 = ( z 11 z 12)I 1 + z 12(I 1 + I 2) = z 11I 1 + z 12I 2V 2 = ( z 21 z 12)I 1 + ( z 22 z 12)I 2 + z 12(I 2 + I 1) = z 21I 1 + z 22I 2

[b] With port 2 terminated in an impedance Z L , the two mesh equations are

V 1 = ( z 11 z 12)I 1 + z 12(I 1 + I 2)0 = Z L I 2 + ( z 21 z 12)I 1 + ( z 22 z 12)I 2 + z 12(I 1 + I 2)Solving for I 1:

I 1 = V 1(z 22 + Z L )

z 11(Z L + z 22) z 12z 21Therefore

Z in = V 1I 1

= z 11 z 12z 21z 22 + Z L

P 18.44 [a] I 1 = y11V 1 + y21V 2 + ( y12

y21)V 2 ; I 2 = y21V 1 + y22V 2

I 1 = y11V 1 + y12V 2; I 2 = y12V 1 + y22V 2 + ( y21 y12)V 1



41/43

Problems 1841

[b] Using the second circuit derived in part [a], we have

where ya = ( y11 + y12) and yb = ( y22 + y12)

At the input port we have

I 1 = yaV 1 y12(V 1 V 2) = y11 V 1 + y12V 2At the output port we haveV 2Z L + ( y21 y12)V 1 + yb V 2 y12(V 2 V 1) = 0Solving for V 1 gives

V 1 =1 + y22Z L

y21Z LV 2

Substituting Eq. (18.2) into (18.1) and at the same time usingV 2 = Z L I 2, we getI 2I 1

= y21

y11 + yZ L

P 18.45 [a] The g -parameter equations are I 1 = g11 V 1 + g12I 2 and V 2 = g21V 1 + g22I 2.These equations are satised by the following circuit:

[b] The g parameters for the rst two port in Fig P 18.38(a) are

g11 = h22 h

= 25106

5 103 = 5 103 S

g12 = h12 h = 5 104

5 103 = 0.10

g21 = h21 h = 40

5 103 = 8000



42/43


g22 = h11 h

= 10005 103

= 200k

From Problem 3.65 Ref = 72 k, hence our circuit reduces to

V 2 = 8000V 1(72)272I 2 = V 272,000 =

8V 1272

vg = 9mV

. . V 1 9 103

800 + V 1(5 103) 0.1

8V 1272

= 0

V 1 9 103 + 4 V 1 80V 1

34 = 0

. . V 1 = 3.4 103

V 2 = 8000(72)272 3.4 103 = 7.2 V

From Problem 3.65V oV 2

= 0 .5; . . V o = 3.6 VThis result matches the solution to Problem 18.38.

P 18.46 [a] To determine b11 and b21 create an open circuit at port 1. Apply a voltageat port 2 and measure the voltage at port 1 and the current at port 2. Todetermine b12 and b22 create a short circuit at port 1. Apply a voltage atport 2 and measure the currents at ports 1 and 2.

[b] The equivalent b-parameters for the black-box amplier can be calculatedas follows:

b11 = 1h12

= 1103

= 1000

b12 = h11h12

= 500103

= 500k

b21 = h22h12

= 0.05103

= 50S



43/43

Problems 1843

b22 = hh12

= 23.5103

= 23,500

Create an open circuit a port 1. Apply 1 V at port 2. Then,

b11 = V 2

V 1 I 1 =0=

1

V 1= 1000 so V 1 = 1 mV measured

b21 = I 2V 1 I 1 =0

= I 2103

= 50S so I 2 = 50 mA measured

Create a short circuit a port 1. Apply 1 V at port 2. Then,

b12 = V 2I 1 V 1 =0

= 1I 1

= 500 k so I 1 = 2 A measured

b22 = I 2I 1 V 1 =0

= I 22 106

= 23,500 so I 2 = 47 mA measured

P 18.47 [a] To determine z 11 and z 21 create an open circuit at port 2. Apply a currentat port 1 and measure the voltages at ports 1 and 2. To determine z 12and z 22 create an open circuit at port 1. Apply a current at port 2 andmeasure the voltages at ports 1 and 2.

[b] The equivalent z -parameters for the black-box amplier can be calculatedas follows:

z 11 = hh22

= 23.50.05

= 470

z 12 = h12h22

= 103

0.05 = 0 .02

z 21 = h21h22

= 15000.05

= 30k

z 22 = 1h22

= 10.05

= 20

Create an open circuit a port 2. Apply 1 mA at port 1. Then,

z 11 = V 1I 1 I 2 =0

= V 10.001

= 470 so V 1 = 470 mV measured

z 21 = V 2

I 1 I 2 =0=

V 2

0.001 =

30,000 so V 2 =

30V measured

Create an open circuit a port 1. Apply 1 A at port 2. Then,

z 12 = V 1I 2 I 1 =0

= V 1

1 = 0.02 so V 1 = 0.02 V measured

V2 V2

chap 18 solution

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