电路-电路原理-尼尔森Riedel(第九版)(2011)课后习题答案第18章.pdf

18 Two-Port Circuits Assessment Problems AP 18.1 With port 2 short-circuited, we have I1= V1 20 + V1 5 ; I1 V1 = y11= 0.25S;I2= ?20 25 ? I1= 0.8I1 When V2= 0, we haveI1= y11V1andI2= y21V1 ThereforeI2= 0.8(y11V1) = 0.8y11V1 Thusy21= 0.8y11= 0.2S With port 1 short-circuited, we have I2= V2 15 + V2 5 ; I2 V2 = y22= ? 4 15 ? S I1= ?15 20 ? I2= 0.75I2= 0.75y22V2 181 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 182CHAPTER 18. Two-Port Circuits Thereforey12= (0.75) 4 15 = 0.2S AP 18.2 h11= ?V 1 I1 ? V2=0 = 20k5 = 4 h21= ?I 2 I1 ? V2=0 = (20/25)I1 I1 = 0.8 h12= ?V 1 V2 ? I1=0 = (20/25)V2 V2 = 0.8 h22= ? I2 V2 ? I1=0 = 1 15 + 1 25 = 8 75 S g11= ? I1 V1 ? I2=0 = 1 20 + 1 20 = 0.1S g21= ?V 2 V1 ? I2=0 = (15/20)V1 V1 = 0.75 g12= ?I 1 I2 ? V1=0 = (15/20)I2 I2 = 0.75 g22= ?V 2 I2 ? V1=0 = 15k5 = 75 20 = 3.75 AP 18.3 g11= I1 V1 ? ? ? ?I 2=0= 5 106 50 103 = 0.1mS g21= V2 V1 ? ? ? ?I 2=0= 200 103 50 103 = 4 g12= I1 I2 ? ? ? ?V 1=0= 2 106 0.5 106 = 4 g22= V2 I2 ? ? ? ?V 1=0 = 10 103 0.5 106 = 20k 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems183 AP 18.4 First calculate the b-parameters: b11= V2 V1 ? ? ? ?I 1=0 = 15 10 = 1.5;b21= I2 V1 ? ? ? ?I 1=0 = 30 10 = 3S b12= V2 I1 ? ? ? ?V 1=0= 10 5 = 2;b22= I2 I1 ? ? ? ?V 1=0= 4 5 = 0.8 Now the z-parameters are calculated: z11= b22 b21 = 0.8 3 = 4 15 ;z12= 1 b21 = 1 3 z21= b b21 = (1.5)(0.8) 6 3 = 1.6;z22= b11 b21 = 1.5 3 = 1 2 AP 18.5 z11= z22,z12= z21,95 = z11(5) + z12(0) Therefore,z11= z22= 95/5 = 19 11.52 = 19I1 z12(2.72) 0 = z12I1 19(2.72) Solving these simultaneous equations for z12yields the quadratic equation z2 12+ ?72 17 ? z12 6137 17 = 0 For a purely resistive network, it follows that z12= z21= 17. AP 18.6 a I2= Vg a11ZL+ a12+ a21ZgZL+ a22Zg = 50 103 (5 104)(5 103) + 10 + (106)(100)(5 103) + (3 102)(100) = 50 103 10 = 5mA PL= 1 2 (5 103)2(5 103) = 62.5mW b ZTh= a12+ a22Zg a11+ a21Zg = 10 + (3 102)(100) 5 104+ (106)(100) = 7 6 104 = 70 6 k 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 184CHAPTER 18. Two-Port Circuits c VTh= Vg a11+ a21Zg = 50 103 6 104 = 500 6 V ThereforeV2= 250 6 V;Pmax= (1/2)(250/6)2 (70/6) 103 = 74.4mW AP 18.7 a For the given bridged-tee circuit, we have a011= a022= 1.25,a021= 1 20 S,a012= 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 Vt= 100 3.125 = 32V;thereforeV2= 16V c P = 162 14.44 = 17.73W 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems185 Problems P 18.1h11= ?V 1 I1 ? V2=0 = 20k5 = 4 h21= ?I 2 I1 ? V2=0 = (20/25)I1 I1 = 0.8 h12= ?V 1 V2 ? I1=0 = (20/25)V2 V2 = 0.8 h22= ? I2 V2 ? I1=0 = 1 15 + 1 25 = 8 75 S g11= ? I1 V1 ? I2=0 = 1 20 + 1 20 = 0.1S g21= ?V 2 V1 ? I2=0 = (15/20)V1 V1 = 0.75 g12= ?I 1 I2 ? V1=0 = (15/20)I2 I2 = 0.75 g22= ?V 2 I2 ? V1=0 = 15k5 = 75 20 = 3.75 P 18.2y11= I1 V1 ? ? ? ? V2=0 ;y21= I2 V1 ? ? ? ? V2=0 V 1 20 + V 10 + V 4 = 0;soV = 0.125V . .I1= 1 0.125 20 + 1 0 8 = 168.75mA;I2= 0 0.125 4 + 0 1 8 = 156.25mA 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 186CHAPTER 18. Two-Port Circuits y11= I1 V1 ? ? ? ? V2=0 = 168.75mS;y21= I2 V1 ? ? ? ? V2=0 = 156.25mS y12= I1 V2 ? ? ? ? V1=0 ;y22= I2 V2 ? ? ? ? V1=0 V 20 + V 10 + V 1 4 = 0;soV = 0.625V . .I1= 0 0.625 20 + 0 1 8 = 156.25mA;I2= 1 0.625 4 + 1 0 8 = 218.75mA y12= I1 V2 ? ? ? ? V1=0 = 156.25mS;y22= I2 V2 ? ? ? ? V1=0 = 218.75mS Summary: y11= 168.75mSy12= 156.25mSy21= 156.25mSy22= 218.75mS P 18.3 z11= V1 I1 ? ? ? ?I 2=0= 1 + 12 = 13 z21= V2 I1 ? ? ? ?I 2=0 = 12 z22= V2 I2 ? ? ? ?I 1=0= 4 + 12 = 16 z21= V1 I2 ? ? ? ?I 1=0= 12 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems187 P 18.4z = (13)(16) (12)(12) = 64 y11= z22 z = 16 64 = 0.25S y12= z12 z = 12 64 = 0.1875S y21= z21 z = 12 64 = 0.1875S y22= z11 z = 13 64 = 0.203125S P 18.5h11= V1 I1 ? ? ? ? V2=0 ;h21= I2 I1 ? ? ? ? V2=0 V1 I1 = 80k10 + 20k20 = 80k20 = 16 . .h11= 16 I6= 80 80 + 20I1 = 0.8I1 I2= 20 20 + 20I6 = 0.5I6= 0.5(0.8)I1= 0.4I1 . .h21= 0.4 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 188CHAPTER 18. Two-Port Circuits h12= V1 V2 ? ? ? ? I1=0 ;h22= I2 V2 ? ? ? ? I1=0 V2 I2 = 80k20 + 20k90 = 25 . .h22= 1 25 = 40 mS Vx= 20k90 20 + 20k90 V2 V1= 80 80 + 10Vx = 80(20k90) 90(20 + 20k90) V2= 0.4V2 . .h12= 0.4 Summary: h11= 16;h12= 0.4;h21= 0.4;h22= 40 mS P 18.6V2= b11V1 b12I1 I2= b21V1 b22I1 b12= V2 I1 ? ? ? ? V1=0 ;b22= I2 I1 ? ? ? ? V1=0 5k15 = (15/4);10k20 = (20/3) I2= V2 (15/4) + (20/3) = 12V2 125 ;I1= Ib Ia 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems189 Ia= 15 20 I2;Ib= 20 30 I2 I1= ?20 30 15 20 ? I2= 5 60 I2= 1 12 I2 b22= I2 I1 = 12 b12= V2 I1 = V2 I2 ?I 2 I1 ? = 125 12 (12) = 125 b11= V2 V1 ? ? ? ? I1=0 ;b21= I2 V1 ? ? ? ? I1=0 V1= Va Vb;Va= 10 15 V2;Vb= 20 35 V2 V1= 10 15V2 20 35 V2= 2 21 V2 b11= V2 V1 = 21 2 = 10.5 V2= (10 + 5)k(20 + 15)I2= 10.5I2 b21= I2 V1 = ? I2 V2 ? ?V 2 V1 ? = ? 1 10.5 ? (10.5) = 1S P 18.7h11= V1 I1 ? ? ? ?V 2=0 = R1kR2= 4 . . R1R2 R1+ R2 = 4 h21= I2 I1 ? ? ? ?V 2=0= R2 R1+ R2 = 0.8 . .R2= 0.8R1+ 0.8R2soR1= R2 4 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 1810CHAPTER 18. Two-Port Circuits Substituting, (R2/4)R2 (R2/4) + R2 = 4soR2= 20andR1= 5 h22= I2 V2 ? ? ? ?I 1=0= 1 R3k(R1+ R2) = 1 R3k25 = 0.14 . .R3= 10 Summary: R1= 5;R2= 20;R3= 10 P 18.8V1= a11V2 a12I2 I1= a21V2 a22I2 a11= V1 V2 ? ? ? ? I2=0 ;a21= I1 V2 ? ? ? ? I2=0 V1= 103I1+ 104V2= 103(0.5 106)V2+ 104V2 . .a11= 5 104+ 104= 4 104 V2= (50I1)(40 103); . .a21= 1 2 106 = 0.5S 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems1811 a12= V1 I2 ? ? ? ? V2=0 ;a22= I1 I2 ? ? ? ? V2=0 I2= 50I1; . .a22= I1 I2 = 1 50 V1= 1000I1; . .a12= V1 I2 = V1 I1 I1 I2 = (1000)(1/50) = 20 Summary a11= 4 104;a12= 20;a21= 0.5S;a22= 0.02 P 18.9g11= a21 a11 = 0.5 106 4 104 = 1.25mS g12= a a11 = (4 104)(1/50) (0.5 106)(20) 4 104 = 0.005 g21= 1 a11 = 1 4 104 = 2500 g22= a12 a11 = (20) 400 106 = 5 104 P 18.10 For V2= 0: Ia= 50I2 200 = 1 4 I2= I2; . .I2= 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 1812CHAPTER 18. Two-Port Circuits h21= I2 I1 ? ? ? ?V 2=0 = 0 V1= (10 + j20)I1 . .h11= V1 I1 ? ? ? ?V 2=0= 10 + j20 For I1= 0: V1= 50I2;I2= V2 j100 + V2 50I2 200 200I2= j2V2+ V2 50i2 250I2= V2(1 + j2) 50I2= V2 ?1 + j2 5 ? = (0.2 + j0.4)V2 . .V1= (0.2 + j0.4)V2 h12= V1 V2 ? ? ? ?I 1=0= 0.2 + j0.4 h22= I2 V2 ? ? ? ?I 1=0 = 1 + j2 250 = 4 + j8 mS Summary: h11= 10 + j20;h12= 0.2 + j0.4;h21= 0;h22= 4 + j8mS 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems1813 P 18.11 For I2= 0: 50I1 40I2= 1 40I1+ 48I2 3(40)(I1 I2) = 0so 160I1+ 168I2= 0 Solving, I1= 84mA;I2= 80mA V2= 3I2 3(40)(I1 I2) = 0.24V g11= I1 V1 ? ? ? ?I 2=0= 84m 1 = 84mS g21= V2 V1 ? ? ? ?I 2=0= 0.24 1 = 0.24 For V1= 0: 50I1 40I2= 0 40I1+ 48I2+ 3 3(40)(I1 I2) = 0so 160I1+ 168I2= 3 Solving, I1= 60mA;I2= 75mA 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 1814CHAPTER 18. Two-Port Circuits V2= 3(I2+ 1) 3(40)(I1 I2) = 0.975V g12= I1 I2 ? ? ? ?V 1=0 = 60m 1 = 0.06 g22= V2 I2 ? ? ? ?V 1=0 = 0.975 1 = 0.975 P 18.12 For V2= 0: V1= (400 + 1200)I1 h11= V1 I1 ? ? ? ?V 2=0 = 1600 1 = 1600 Vp= 1200(1A) = 1200V = Vn At Vn, 1200 500 + 1200 Vo 1000 = 0soVo= 3600V . .I2= 3600 200 = 18A h21= I2 I1 ? ? ? ?I 1=0 = 18 1 = 18 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems1815 For I1= 0: V1= 0 h12= V1 V2 ? ? ? ?I 1=0 = 0 1 = 0 At Vn, Vn 500 + Vn Vo 100 = 0 But Vn= Vp= 0 so Vo= 0; therefore, I2= 1V 200 = 5mS h22= I2 V2 ? ? ? ?I 1=0 = 5m 1 = 5mS P 18.13 I1= g11V1+ g12I2;V2= g21V1+ g22I2 g11= I1 V1 ? ? ? ?I 2=0 = 0.25 106 20 103 = 12.5 106= 12.5S g21= V2 V1 ? ? ? ?I 2=0 = 5 20 103= 250 0 = 250(10) + g22(50 106) g22= 2500 50 106 = 50M 200 106= 12.5 106(10) + g12(50 106) (200 125)106= g12(50 106) g12= 75 50 = 1.5 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 1816CHAPTER 18. Two-Port Circuits P 18.14 a I1= y11V1+ y12V2;I2= y21V1+ y22V2 y21= I2 V1 ? ? ? ?V 2=0= 50 106 10 = 5S 0 = y21(20 103) + y22(5) . .y22= 1 5 y21(20 103) = 20nS 200 106= y11(10)soy11= 20S 0.25 106= 20 106(20 103) + y12(5) y12= 0.25 106 0.4 106 5 = 30nS Summary: y11= 20S;y12= 30nS;y21= 5S;y22= 20nS b y11= g g22 ;y12= g12 g22 ;y21= g21 g22 ;y22= 1 g22 g = g11g22 g12g21= (12.5 106)(50 106) 1.5(250) = 625 + 375 = 1000 y11= 1000 50 106 = 20S;y21= 250 5 106 = 5S y12= 1.5 50 106 = 30nS;y22= 1 5 106 = 20nS These values are the same as those in part (a). P 18.15 I1= g11V1+ g12I2 V2= g21V1+ g22I2 V1= I1 g11 g12 g11 I2andI2= V2 g22 g21 g22 V1 Substituting, V1= I1 g11 g12 g11 V2 g22 g21 g22 V1 # V1= 1 g12g21 g11g22 ! = I1 g11 g12 g11g22 V2 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. Problems1817 V1= g22 g11g22 g12g21 I1 g12 g11g22 g12g21 V2 V1= h11I1+ h12V2 Therefore, h11= g22 g ;h12= g12 g whereg = g11g22 g12g21 I2= V2 g22 g21 g22 I1 g11 g12 g11 I2 # I2= 1 g12g21 g11g22 ! = V2 g22 g21 g11g22 I1 I2= g11 g V2 g21 g I1 I2= h21I1+ h22V2 Therefore, h21= g21 g ;h22= g11 g P 18.16 V1= h11I1+ h12V2;I2= h21I1+ h22V2 Rearranging the fi rst equation, V2= 1 h12 V1 h11 h12 I1 V2= b11V1 b12I1 Therefore, b11= 1 h12 ;b12= h11 h12 Solving the second h-parameter equation for I2: I2= h21I1+ h22 1 h12 V1 h11 h12 I1 ! 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained 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. 1818CHAPTER 18. Two-Port Circuits = I1 h21 h22h11 h12 ! + h22 h12 V1 = h h12 I1+ h22 h12 V1 I2= b21V1 b22I1 Therefore, b21= h22 h12 ;b22= h h12 whereh = h11h22 h12h21 P 18.17 I1= g11V1+ g12I2;V2= g21V1+ g22I2 V1= z11I1+ z12I2;V2= z21I1+ z22I2 I1= V1 z11 z12 z11 I2 . .g11= 1 z11 ;g12= z12 z11 V2= z21 ? V1 z11 z12 z11 I2 ? + z22I2= z21 z11 V1+ ?z 11z22 z12z21 z11 ? I2 . .g21= z21 z11 ;g22= z z11 wherez = z11z22 z12z21 P 18.18 For I2= 0: V2= V1 4 + (1/s) (4) = 4sV1 4s + 1 a11= V1 V2 ? ? ? ?I 2=0 = 4s + 1 4s = s + 0.25 s I1= V1 4 + (1/s) = sV1 4s + 1 soV2= 4I1= 4sV1 4s + 1 a21= I1 V2 ? ? ? ?I 2=0 = 1 4 = 0.25 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is