1、18Two-Port CircuitsAssessment ProblemsAP 18.1 With port 2 short-circuited, we haveI1 = V120 + V15 ; I1V1= y11 = 0.25S; I2 =parenleftbigg2025parenrightbiggI1 = 0.8I1When V2 = 0, we have I1 = y11V1 and I2 = y21V1Therefore I2 = 0.8(y11V1) = 0.8y11V1Thus y21 = 0.8y11 = 0.2SWith port 1 short-circuited, w
2、e haveI2 = V215 + V25 ; I2V2= y22 =parenleftbigg 415parenrightbiggSI1 =parenleftbigg1520parenrightbiggI2 = 0.75I2 = 0.75y22V2181 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 pu
3、blisher 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
4、 Saddle River, NJ 07458.182 CHAPTER 18. Two-Port CircuitsTherefore y12 = (0.75) 415 = 0.2SAP 18.2h11 =parenleftbiggV1I1parenrightbiggV2=0= 20bardbl5 = 4h21 =parenleftbiggI2I1parenrightbiggV2=0= (20/25)I1I1= 0.8h12 =parenleftbiggV1V2parenrightbiggI1=0= (20/25)V2V2= 0.8h22 =parenleftbiggI2V2parenright
5、biggI1=0= 115 + 125 = 875 Sg11 =parenleftbiggI1V1parenrightbiggI2=0= 120 + 120 = 0.1Sg21 =parenleftbiggV2V1parenrightbiggI2=0= (15/20)V1V1= 0.75g12 =parenleftbiggI1I2parenrightbiggV1=0= (15/20)I2I2= 0.75g22 =parenleftbiggV2I2parenrightbiggV1=0= 15bardbl5 = 7520 = 3.75AP 18.3g11 = I1V1vextendsingleve
6、xtendsinglevextendsinglevextendsingleI2=0= 510650103 = 0.1mSg21 = V2V1vextendsinglevextendsinglevextendsinglevextendsingleI2=0= 200 10350 103 = 4g12 = I1I2vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 21060.5106 = 4g22 = V2I2vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 10103
7、0.5106 = 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 m
8、eans, 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.Problems 183AP 18.4 First calculate the b-parameters:b11 = V2V1vextendsinglevextendsinglevex
9、tendsinglevextendsingleI1=0= 1510 = 1.5; b21 = I2V1vextendsinglevextendsinglevextendsinglevextendsingleI1=0= 3010 = 3Sb12 = V2I1vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 105 = 2; b22 = I2I1vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 45 = 0.8Now the z-parameters are calc
10、ulated:z11 = b22b21= 0.83 = 415 ; z12 = 1b21= 13 z21 = bb21= (1.5)(0.8)63 = 1.6; z22 = b11b21= 1.53 = 12 AP 18.5z11 = z22, z12 = z21, 95 = z11(5) + z12(0)Therefore, z11 = z22 = 95/5 = 1911.52 = 19I1 z12(2.72)0 = z12I1 19(2.72)Solving these simultaneous equations for z12 yields the quadratic equation
11、z212 +parenleftbigg7217parenrightbiggz12 613717 = 0For a purely resistive network, it follows that z12 = z21 = 17.AP 18.6 a I2 = Vga11ZL + a12 + a21ZgZL + a22Zg= 50103(5104)(5 103) + 10 + (106)(100)(5 103) + (3102)(100)= 50 10310 = 5mAPL = 12(5103)2(5103) = 62.5mWb ZTh = a12 + a22Zga11 + a21Zg= 10 +
12、 (3102)(100)5104 + (106)(100)= 76104 = 706 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 tran
13、smission 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 07458.184 CHAPTER 18. Two-Port Circuitsc VTh = Vga11 + a21Zg= 5010
14、36104 =5006 VTherefore V2 = 2506 V; Pmax = (1/2)(250/6)2(70/6) 103 = 74.4mWAP 18.7 a For the given bridged-tee circuit, we haveaprime11 = aprime22 = 1.25, aprime21 = 120 S, aprime12 = 11.25The a-parameters of the cascaded networks area11 = (1.25)2 + (11.25)(0.05) = 2.125a12 = (1.25)(11.25) + (11.25)
15、(1.25) = 28.125a21 = (0.05)(1.25) + (1.25)(0.05) = 0.125Sa22 = a11 = 2.125, RTh = (45.125/3.125) = 14.44b Vt = 1003.125 = 32V; therefore V2 = 16Vc P = 16214.44 = 17.73W 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written p
16、ermission 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 De
17、partment, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 185ProblemsP 18.1 h11 =parenleftbiggV1I1parenrightbiggV2=0= 20bardbl5 = 4h21 =parenleftbiggI2I1parenrightbiggV2=0= (20/25)I1I1= 0.8h12 =parenleftbiggV1V2parenrightbiggI1=0= (20/25)V2V2= 0.8h22 =parenleftbiggI2V2parenrightbiggI1
18、=0= 115 + 125 = 875 Sg11 =parenleftbiggI1V1parenrightbiggI2=0= 120 + 120 = 0.1Sg21 =parenleftbiggV2V1parenrightbiggI2=0= (15/20)V1V1= 0.75g12 =parenleftbiggI1I2parenrightbiggV1=0= (15/20)I2I2= 0.75g22 =parenleftbiggV2I2parenrightbiggV1=0= 15bardbl5 = 7520 = 3.75P 18.2 y11 = I1V1vextendsinglevextends
19、inglevextendsinglevextendsingleV2=0; y21 = I2V1vextendsinglevextendsinglevextendsinglevextendsingleV2=0V 120 +V10 +V4 = 0; so V = 0.125V. . I1 = 1 0.12520 +108 = 168.75mA; I2 =0 0.1254 +0 18 = 156.25mA 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is pro
20、tected 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), wr
21、ite to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.186 CHAPTER 18. Two-Port Circuitsy11 = I1V1vextendsinglevextendsinglevextendsinglevextendsingleV2=0= 168.75mS; y21 = I2V1vextendsinglevextendsinglevextendsinglevextendsingleV2=0= 156.25mSy12 = I1V2vexten
22、dsinglevextendsinglevextendsinglevextendsingleV1=0; y22 = I2V2vextendsinglevextendsinglevextendsinglevextendsingleV1=0V20 +V10 +V 14 = 0; so V = 0.625V. . I1 = 0 0.62520 +018 = 156.25mA; I2 =10.6254 +1 08 = 218.75mAy12 = I1V2vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 156.25mS; y22 = I
23、2V2vextendsinglevextendsinglevextendsinglevextendsingleV1=0= 218.75mSSummary:y11 = 168.75mS y12 = 156.25mS y21 = 156.25mS y22 = 218.75mSP 18.3z11 = V1I1vextendsinglevextendsinglevextendsinglevextendsingleI2=0= 1+ 12 = 13z21 = V2I1vextendsinglevextendsinglevextendsinglevextendsingleI2=0= 12z22 = V2I2
24、vextendsinglevextendsinglevextendsinglevextendsingleI1=0= 4+ 12 = 16z21 = V1I2vextendsinglevextendsinglevextendsinglevextendsingleI1=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 f
25、rom 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, I
26、nc., Upper Saddle River, NJ 07458.Problems 187P 18.4 z = (13)(16) (12)(12) = 64y11 = z22z = 1664 = 0.25Sy12 = z12z = 1264 = 0.1875Sy21 = z21z = 1264 = 0.1875Sy22 = z11z = 1364 = 0.203125SP 18.5 h11 = V1I1vextendsinglevextendsinglevextendsinglevextendsingleV2=0; h21 = I2I1vextendsinglevextendsingleve
27、xtendsinglevextendsingleV2=0V1I1 = 80bardbl10 + 20bardbl20 = 80bardbl20 = 16. . h11 = 16I6 = 8080 + 20I1 = 0.8I1I2 = 2020 + 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 writte
28、n 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
29、 Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.188 CHAPTER 18. Two-Port Circuitsh12 = V1V2vextendsinglevextendsinglevextendsinglevextendsingleI1=0; h22 = I2V2vextendsinglevextendsinglevextendsinglevextendsingleI1=0V2I2 = 80bardbl20 + 20bardbl90 = 25. . h22 = 125 = 40 mSVx = 20bar
30、dbl9020 + 20bardbl90V2V1 = 8080 + 10Vx = 80(20bardbl90)90(20 + 20bardbl90)V2 = 0.4V2. . h12 = 0.4Summary:h11 = 16; h12 = 0.4; h21 = 0.4; h22 = 40 mSP 18.6 V2 = b11V1 b12I1I2 = b21V1 b22I1b12 = V2I1vextendsinglevextendsinglevextendsinglevextendsingleV1=0; b22 = I2I1vextendsinglevextendsinglevextendsi
31、nglevextendsingleV1=05bardbl15 = (15/4); 10bardbl20 = (20/3)I2 = V2(15/4) + (20/3) = 12V2125 ; 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 pr
32、ohibited 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、.Problems 189Ia = 1520I2; Ib = 2030I2I1 =parenleftbigg2030 1520parenrightbiggI2 = 560 I2 = 112 I2b22 = I2I1= 12b12 = V2I1= V2I2parenleftbiggI2I1parenrightbigg= 12512 (12) = 125b11 = V2V1vextendsinglevextendsinglevextendsinglevextendsingleI1=0; b21 = I2V1vextendsinglevextendsinglevextendsinglevextend
34、singleI1=0V1 = Va Vb; Va = 1015V2; Vb = 2035V2V1 = 1015V2 2035V2 = 221V2b11 = V2V1= 212 = 10.5V2 = (10 + 5)bardbl(20 + 15)I2 = 10.5I2b21 = I2V1=parenleftbiggI2V2parenrightbiggparenleftbiggV2V1parenrightbigg=parenleftbigg 110.5parenrightbigg(10.5) = 1SP 18.7 h11 = V1I1vextendsinglevextendsinglevexten
35、dsinglevextendsingleV2=0= R1bardblR2 = 4 . . R1R2R1 + R2= 4h21 = I2I1vextendsinglevextendsinglevextendsinglevextendsingleV2=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 wr
36、itten 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 Permiss
37、ions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.1810 CHAPTER 18. Two-Port CircuitsSubstituting,(R2/4)R2(R2/4) + R2 = 4 so R2 = 20 and R1 = 5h22 = I2V2vextendsinglevextendsinglevextendsinglevextendsingleI1=0= 1R3bardbl(R1 + R2)= 1R3bardbl25= 0.14. . R3 = 10Summary:R1 = 5; R2 =
38、20; R3 = 10P 18.8 V1 = a11V2 a12I2I1 = a21V2 a22I2a11 = V1V2vextendsinglevextendsinglevextendsinglevextendsingleI2=0; a21 = I1V2vextendsinglevextendsinglevextendsinglevextendsingleI2=0V1 = 103I1 + 104V2 = 103(0.5106)V2 + 104V2. . a11 = 5104 + 104 = 4104V2 = (50I1)(40 103); . . a21 = 12106 = 0.5S 201
39、0 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
40、, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 1811a12 = V1I2vextendsinglevextendsinglevextendsinglevextendsingleV2=0; a22 = I1I2vextendsinglevext
41、endsinglevextendsinglevextendsingleV2=0I2 = 50I1; . . a22 = I1I2= 150V1 = 1000I1; . . a12 = V1I2= V1I1I1I2 = (1000)(1/50) = 20Summarya11 = 4104; a12 = 20; a21 = 0.5S; a22 = 0.02P 18.9 g11 = a21a11= 0.51064104 = 1.25mSg12 = aa11= (4104)(1/50) (0.5106)(20)4 104 = 0.005g21 = 1a11= 14104 = 2500g22 = a12
42、a11= (20)400 106 = 5 104 P 18.10 For V2 = 0:Ia = 50I2200 = 14I2 = 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, st
43、orage 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 07458.1812 CHAPTER 18. Two-Por
44、t Circuitsh21 = I2I1vextendsinglevextendsinglevextendsinglevextendsingleV2=0= 0V1 = (10 + j20)I1 . . h11 = V1I1vextendsinglevextendsinglevextendsinglevextendsingleV2=0= 10 + j20For I1 = 0:V1 = 50I2; I2 = V2j100 + V2 50I2200200I2 = j2V2 + V2 50i2250I2 = V2(1 + j2)50I2 = V2parenleftbigg1+ j25parenrigh
45、tbigg= (0.2+ j0.4)V2. . V1 = (0.2+ j0.4)V2h12 = V1V2vextendsinglevextendsinglevextendsinglevextendsingleI1=0= 0.2 + j0.4h22 = I2V2vextendsinglevextendsinglevextendsinglevextendsingleI1=0= 1 + j2250 = 4+ j8 mSSummary:h11 = 10 + j20; h12 = 0.2+ j0.4; h21 = 0; h22 = 4 + j8mS 2010 Pearson Education, Inc
46、., 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, photocopyi
47、ng, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 1813P 18.11 For I2 = 0:50I1 40I2 = 140I1 + 48I2 3(40)(I1 I2) = 0 so 160I1 + 168I2 = 0Solving,I1 = 84mA; I2 = 80mAV2 = 3I2 3
48、(40)(I1 I2) = 0.24Vg11 = I1V1vextendsinglevextendsinglevextendsinglevextendsingleI2=0= 84m1 = 84mSg21 = V2V1vextendsinglevextendsinglevextendsinglevextendsingleI2=0= 0.241 = 0.24For V1 = 0:50I1 40I2 = 040I1 + 48I2 + 33(40)(I1 I2) = 0 so 160I1 + 168I2 = 3Solving,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, electro
