电路-电路原理-尼尔森Riedel(第九版)(2011)课后习题答案第17章.pdf

17 The Fourier Transform Assessment Problems AP 17.1 a F() = Z 0 /2(Ae jt)dt + Z /2 0 Aejtdt = A j 2 ej/2 ej/2 = 2A j 1 ej/2+ ej/2 2 # = j2A 1 cos(/2) b F() = Z 0 teatejtdt = Z 0 te(a+j)tdt = 1 (a + j)2 AP 17.2 f(t) = 1 2 ?Z 2 34e jt d + Z 2 2e jt d + Z 3 24e jt d ? = 1 j2t 4ej2t 4ej3t+ ej2t ej2t+ 4ej3t 4ej2t = 1 t 3ej2t 3ej2t j2 + 4ej3t 4ej3t j2 # = 1 t(4sin3t 3sin2t) AP 17.3 a F() = F(s) |s=j= Leatsin0ts=j = 0 (s + a)2+ 2 0 ? ? ? ?s=j= 0 (a + j)2+ 2 0 b F() = Lf(t)s=j= 1 (s + a)2 # s=j = 1 (a j)2 171 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. 172CHAPTER 17. The Fourier Transform c f+(t) = teat,f(t) = teat Lf+(t) = 1 (s + a)2 ,Lf(t) = 1 (s + a)2 ThereforeF() = 1 (a + j)2 1 (a j)2 = j4a (a2+ 2)2 AP 17.4 a f0(t) = 2A , 2 t 0;f0(t) = 2A ,0 t 2 . .f0(t) = 2A u(t+ /2) u(t) 2A u(t) u(t /2) = 2A u(t + /2) 4A u(t) + 2A u(t /2) . .f00(t) = 2A ? t + 2 ? 4A + 2A ? t 2 ? b Ff00(t) = ?2A ej/2 4A + 2A ej/2 ? = 4A ej/2+ ej/2 2 1 # = 4A ? cos ? 2 ? 1 ? c Ff00(t) = (j)2F() = 2F();thereforeF() = 1 2 Ff00(t) Thus we haveF() = 1 2 ?4A ? cos ? 2 ? 1 ? AP 17.5 v(t) = Vm ? u ? t + 2 ? u ? t 2 ? F ? u ? t + 2 ? = () + 1 j # ej/2 F ? u ? t 2 ? = () + 1 j # ej/2 ThereforeV () = Vm () + 1 j # h ej/2 ej/2 i = j2Vm()sin ? 2 ? + 2Vm sin ? 2 ? = (Vm)sin(/2) /2 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. Problems173 AP 17.6 a Ig() = F10sgnt = 20 j b H(s) = Vo Ig Using current division and Ohms law, Vo= I2s = ? 4 4 + 1 + s ? (Ig)s = 4s 5 + sIg H(s) = 4s s + 5, H(j) = j4 5 + j c Vo() = H(j) Ig() = j4 5 + j ! 20 j ! = 80 5 + j d vo(t) = 80e5tu(t)V e Using current division, i1(0) = 1 5ig = 1 5(10) = 2A f i1(0+) = ig+ i2(0+) = 10 + i2(0) = 10 + 8 = 18A g Using current division, i2(0) = 4 5(10) = 8A h Since the current in an inductor must be continuous, i2(0+) = i2(0) = 8A i Since the inductor behaves as a short circuit for t 0 f(t) = 2 ? 2 ? = 1 t 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. Problems1713 b Fe|t| = 1 1 + j + 1 1 j = 2 1 + 2 ThereforeFea|t| = (1/a)2 (/a)2+ 1 ThereforeFe0.5|t| = 4 42+ 1, Fe|t| = 2 2+ 1 Fe2|t| = 1/0.252+ 1, yes as “a” increases, the sketches show that f(t) approaches zero faster and F() fl attens out over the frequency spectrum. P 17.15 a Ff(t a) = Z f(t a)ejtdt Let u = t a, then du = dt, t = u + a, and u = when t = . Therefore, Ff(t a) = Z f(u)e j(u+a) du = eja Z f(u)e ju du = ejaF() b Fej0tf(t) = Z f(t)ej(0)tdt = F( 0) c Ff(t)cos0t = F ( f(t) ej0t+ ej0t 2 #) = 1 2F( 0) + 1 2F( + 0) P 17.16 Y () = Z ?Z x()h(t )d ? ejtdt = Z x() ?Z h(t )e jt dt ? d 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. 1714CHAPTER 17. The Fourier Transform Let u = t , du = dt, and u = , when t = . ThereforeY () = Z x() ?Z h(u)e j(u+)du ? d = Z x() ? ej Z h(u)e ju du ? d = Z x()e jH()d = H()X() P 17.17 Ff1(t)f2(t) = Z ? 1 2 Z F1(u)e jtudu ? f2(t)ejtdt = 1 2 Z ?Z F1(u)f2(t)e jtejtudu ? dt = 1 2 Z ? F1(u) Z f2(t)e j(u)tdt ? du = 1 2 Z F1(u)F2( u)du P 17.18 a F() = Z f(t)ejtdt dF d = Z d d h f(t)ejtdt i = j Z tf(t)ejtdt = jFtf(t) Thereforej dF() d = Ftf(t) d2F() d2 = Z (jt)(jt)f(t)e jtdt = (j)2Ft2f(t) Note that(j)n= 1 jn Thus we havejn dnF() dn # = Ftnf(t) b (i)Featu(t) = 1 a + j = F(); dF() d = j (a + j)2 Thereforej dF() d # = 1 (a + j)2 ThereforeFteatu(t) = 1 (a + j)2 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. Problems1715 (ii)F|t|ea|t| = Fteatu(t) Fteatu(t) = 1 (a + j)2 j d d 1 a j ! = 1 (a + j)2 + 1 (a j)2 (iii)Ftea|t| = Fteatu(t) + Fteatu(t) = 1 (a + j)2 + j d d 1 a j ! = 1 (a + j)2 1 (a j)2 P 17.19 a f1(t) = cos0t,F1(u) = (u+ 0) + (u 0) f2(t) = 1,/2 t 0,io= 400e40tA P 17.25 a Vo= IgR(1/sC) R + (1/sC) = IgR RCs + 1 H(s) = Vo Ig = 1/C s + (1/RC) = 2 106 s + 40 H(j) = 2 106 40 + j ;Ig() = 400 106 j Vo() = H(j)Ig() = 400 106 j ! 2 106 40 + j ! = 800 j(40 + j) = 20 j 20 40 + j vo(t) = 10sgn(t) 20e40tu(t)V b Yes, at the time the current source jumps from 200 to +200A the capacitor is charged to 10 V. That is, at t = 0, vo(0) = (50 103)(200 106) = 10 V. At t = the capacitor will be charged to +10 V. That is, vo() = (50 103)(200 106) = 10 V The time constant of the circuit is (50 103)(0.5 106) = 25 ms, so 1/ = 40. The function vo(t) is plotted below: 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. Problems1721 P 17.26 a Vo Vg = H(s) = 4/s 0.5 + 0.01s + 4/s H(s) = 400 s2+ 50s + 400 = 400 (s + 10)(s + 40) H(j) = 400 (j + 10)(j + 40) Vg() = 6 j Vo() = Vg()H(j) = 2400 j(j + 10)(j + 40) Vo() = K1 j + K2 j + 10 + K3 j + 40 K1= 2400 400 = 6;K2= 2400 (10)(30) = 8 K3= 2400 (40)(30) = 2 Vo() = 6 j 8 j + 10 + 2 j + 40 vo(t) = 3sgn(t) 8e10tu(t) + 2e40tu(t)V b vo(0) = 3V c vo(0+) = 3 8 + 2 = 3V d For t 0+: Vo 3/s 0.5 + 0.01s + (Vo+ 3/s)s 4 = 0 Vo ? 100 s + 50 + s 4 ? = 300 s(s + 50) 0.75 Vo= 1200 3s2 150s s(s + 10)(s + 40) = K1 s + K2 s + 10 + K3 s + 40 K1= 1200 400 = 3;K2= 1200 300 + 1500 (10)(30) = 8 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. 1722CHAPTER 17. The Fourier Transform K3= 1200 4800 + 6000 (40)(30) = 2 vo(t) = (3 8e10t+ 2e40t)u(t)V e Yes. P 17.27 a Io= Vg 0.5 + 0.01s + 4/s H(s) = Io Vg = 100s s2+ 50s + 400 = 100s (s + 10)(s + 40) H(j) = 100(j) (j + 10)(j + 40) Vg() = 6 j Io() = H(j)Vg() = 600 (j + 10)(j + 40) = 20 j + 10 20 j + 40 io(t) = (20e10t 20e40t)u(t)A b io(0) = 0 c io(0+) = 0 d Io= 6/s 0.5 + 0.01s + 4/s = 600 s2+ 50s + 400 = 600 (s + 10)(s + 40) = 20 s + 10 20 s + 40 io(t) = (20e10t 20e40t)u(t)A e Yes. 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. Problems1723 P 17.28 a ig= 3e5|t| . .Ig() = 3 j + 5 + 3 j + 5 = 30 (j + 5)(j + 5) Vo 10 + Vos 10 = Ig . . Vo Ig = H(s) = 10 s + 1 ;H() = 10 j + 1 Vo() = Ig()H() = 300 (j + 1)(j + 5)(j + 5) = K1 j + 1 + K2 j + 5 + K3 j + 5 K1= 300 (4)(6) = 12.5 K2= 300 (4)(10) = 7.5 K3= 300 (6)(10) = 5 Vo() = 12.5 j + 1 7.5 j + 5 + 5 j + 5 vo(t) = 12.5et 7.5e5tu(t) + 5e5tu(t)V b vo(0) = 5V c vo(0+) = 12.5 7.5 = 5V d ig= 3e5tu(t),t 0+ Ig= 3 s + 5 ;H(s) = 10 s + 1 vo(0+) = 5V;C = 0.5 Vo 10 + Vos 10 = Ig+ 0.5 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. 1724CHAPTER 17. The Fourier Transform Vo(s + 1) = 30 s + 5 + 5 Vo= 30 (s + 5)(s + 1) + 5 s + 1 = 7.5 s + 5 + 7.5 s + 1 + 5 s + 1 = 12.5 s + 1 7.5 s + 5 . .vo(t) = (12.5et 7.5e5t)u(t)V e Yes, for t 0+the solution in part (a) is also vo(t) = (12.5et 7.5e5t)u(t)V P 17.29 a Io= Vg 10 + 10/s = Vgs 10s + 10 H(s) = Io Vg = 0.1 s + 1 H(j) = 0.1 j + 1 Vg() = 30 j + 5 + 30 j + 5 Io() = H(j)Vg(j) = 0.1j j + 1 30 j + 5 + 30 j + 5 # = 3j (j + 1)(j + 5) + 3j (j + 1)(j + 5) = K1 j + 1 + K2 j + 5 + K3 j + 1 + K4 j + 5 K1= 3(1) 6 = 0.5;K2= 3(5) 6 = 2.5 K3= 3(1) 4 = 0.75;K4= 3(5) 4 = 3.75 . .Io() = 1.25 j + 1 + 2.5 j + 5 + 3.75 j + 5 io(t) = 2.5e5tu(t) + 1.25et+ 3.75e5tu(t)A b io(0) = 2.5V c io(0+) = 2.5V 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. Problems1725 d Note since io(0+) = 2.5 A, vo(0+) = 30 25 = 5 V. Io= Vg (5/s) 10 + (10/s) = sVg 5 10s + 10 ;Vg= 30 s + 5 . .Io= 25s 25 10(s + 1)(s + 5) = 2.5(s 1) (s + 1)(s + 5) = 1.25 s + 1 + 3.75 s + 5 io(t) = (1.25et+ 3.75e5t)u(t)A e Yes, for t 0+the solution in part (a) is also io(t) = (1.25et+ 3.75e5t)u(t)A P 17.30 Vo Vg 2s + 100Vo s + Vos 100s + 125 104 = 0 . .Vo= s(100s + 125 104)Vg 125(s2+ 12,000s + 25 106) Io= sVo 100s + 125 104 H(s) = Io Vg = s2 125(s2+ 12,000s + 25 106) H(j) = 8 1032 (25 106 2) + j12,000 Vg() = 300( + 5000) + ( 5000) Io() = H(j)Vg() = 2.42( + 5000) + ( 5000) (25 106 2) + j12,000 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. 1726CHAPTER 17. The Fourier Transform io(t) = 2.4 2 Z 2( + 5000) + ( 5000) (25 106 2) + j12,000 ejtd = 1.2 ( 25 106ej5000t j(12,000)(5000) + 25 106ej5000t j(12,000)(5000) ) = 6 12 (ej5000t j + ej5000t j ) = 0.5ej(5000t+90 ) + ej(5000t+90 ) io(t) = 1cos(5000t + 90)A P 17.31 a Vo Vg sL1 + Vo sL2 + Vo R = 0 . .Vo= RVg L1 ? s + R ? 1 L1 + 1 L2 ? Io= Vo sL2 . . Io Vg = H(s) = R/L1L2 s(s + R(1/L1) + (1/L2) R L1L2 = 12 105 R ? 1 L1 + 1 L2 ? = 3 104 . .H(s) = 12 105 s(s + 3 104) H(j) = 12 105 j(j + 3 104) Vg() = 125( + 4 104) + ( 4 104) Io() = H(j)Vg() = 1500 105( + 4 104) + ( 4 104) j(j + 3 104) 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. Problems1727 io(t) = 1500 105 2 Z ( + 4 104) + ( 4 104)ejt j(j + 3 104) d io(t) = 750 105 ( ej40,000t j40,000(30,000 j40,000) + ej40,000t j40,000(30,000 + j40,000) ) = 75 106 4 108 ( ej40,000t j(3 + j4) + ej40,000t j(3 + j4) ) = 75 400 ( ej40,000t 5/ 143.13 + ej40,000t 5/143.13 ) = 0.075cos(40,000t 143.13)A io(t) = 75cos(40,000t 143.13)mA b In the phasor domain: Vo 125 j200 + Vo j800 + Vo 120 = 0 12Vo 1500 + 3Vo+ j20Vo= 0 Vo= 1500 15 + j20 = 60/ 53.13V Io= Vo j800 = 75 103/ 143.13A io(t) = 75cos(40,000t 143.13)mA P 17.32 a (Vo Vg)s 106 + Vo 4s + Vo 800 = 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. 1728CHAPTER 17. The Fourier Transform . .Vo= s2Vg s2+ 1250s + 25 104 Vo Vg = H(s) = s2 (s + 250)(s + 1000) H(j) = (j)2 (j + 250)(j + 1000) vg= 45e500|t|;Vg() = 45,000 (j + 500)(j + 500) . .Vo() = H(j)Vg() = 45,000(j)2 (j + 250)(j + 500)(j + 1000)(j + 500) = K1 j + 250 + K2 j + 500 + K3 j + 1000 + K4 j + 500 K1= 45,000(250)2 (250)(750)(750) = 20 K2= 45,000(500)2 (250)(500)(1000) = 90 K3= 45,000(1000)2 (750)(500)(1500) = 80 K4= 45,000(500)2 (750)(1000)(1500) = 10 . .vo(t) = 20e250t 90e500t+ 80e1000tu(t) + 10e500tu(t)V b vo(0) = 10V;Vo(0+) = 20 90 + 80 = 10V vo() = 0V c IL= Vo 4s = 0.25sVg (s + 250)(s + 1000) H(s) = IL Vo = 0.25s (s + 250)(s + 1000) H(j) = 0.25(j) (j + 250)(j + 1000) IL() = 0.25(j)(45,000) (j + 250)(j + 500)(j + 1000)(j + 500) = K1 j + 250 + K2 j + 500 + K3 j + 1000 + K4 j + 500 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. Problems1729 K4= (0.25)(500)(45,000) (750)(1000)(1500) = 5mA iL(t) = 5e500tu(t); . .iL(0) = 5mA K1= (0.25)(250)(45,000) (250)(750)(750) = 20mA K2= (0.25)(500)(45,000) (250)(500)(1000) = 45mA K3= (0.25)(1000)(45,000) (750)(500)(1500) = 20mA . .iL(0+) = K1+ K2+ K3= 20 + 45 20 = 5mA Checks, i.e.,iL(0+) = iL(0) = 5mA At t = 0: vC(0) = 45 10 = 35V At t = 0+: vC(0+) = 45 10 = 35V d We can check the correctness of out solution for t 0+by using the Laplace transform. Our circuit becomes Vo 800 + Vo 4s + (Vo Vg)s 106 + 35 106+ 5 103 s = 0 . .(s2+ 1250s + 24 104)Vo= s2Vg (35s + 5000) vg(t) = 45e500tu(t)V;Vg= 45 s + 500 . .(s + 250)(s + 1000)Vo= 45s2 (35s + 5000)(s + 500) (s + 500) . .Vo= 10s2 22,500s 250 104 (s + 250)(s + 500)(s + 1000) = 20 s + 250 90 s + 500 + 80 s + 1000 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected