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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
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Solucionario Capitulo 16 Sadiku

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Solucionario Capitulo 16 Sadiku

  1. 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 16, Problem 1. Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform. Figure 16.35 For Prob. 16.1. Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 222 )23()21s( 1 1ss 1 s1s1 s1 )s(I ++ = ++ = ++ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = t 2 3 sine 3 2 )t(i 2t- =)t(i A)t866.0(sine155.1 -0.5t 1/s 1 + − 1/s s I(s)
  2. 2. 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 16, Problem 2. Find v x in the circuit shown in Fig. 16.36 given v s .= 4u(t)V. Figure 16.36 For Prob. 16.2.
  3. 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 16, Solution 2. V)t(u)e2e24(v 3 8 j 3 4 s 125.0 3 8 j 3 4 s 125.0 s 25.0 16 )8s8s3(s 2s 16V s 32s16 )8s8s3(V 0VsV)s4s2( s )32s16( )8s4(V 0 s 8 4 0V 2 0V s s 4 V t)9428.0j3333.1(t)9428.0j3333.1( x 2x 2 x x 2 x 2 x xx x −−+− ++−= ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −+ − + ++ − +−= ++ + −= + =++ =+++ + −+ = + − + − + − vx = Vt 3 22 sine 2 6 t 3 22 cose)t(u4 3/t43/t4 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − −− 4 + Vx − s 8/s s 4 + − 2
  4. 4. 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 16, Problem 3. Find i(t) for t > 0 for the circuit in Fig. 16.37. Assume i s = 4u(t) + 2δ (t)mA. (Hint: Can we use superposition to help solve this problem?) Figure 16.37 For Prob. 16.3.
  5. 5. 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 16, Solution 3. In the s-domain, the circuit becomes that shown below. 1 I 4 2 s + 2 0.2s We transform the current source to a voltage source and obtain the circuit shown below. 2 1 I 8 4 s + 0.2s 8 4 20 40 3 0.2 ( 15) 15 s A BsI s s s s s + + = = = + + + + 40 8 15 20 40 52 , 15 3 15 3 x A B − + = = = = − 8/3 52/3 15 I s s = + + 158 52 ( ) ( ) 3 3 t i t e u t−⎡ ⎤ = +⎢ ⎥⎣ ⎦ + _
  6. 6. 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 16, Problem 4. The capacitor in the circuit of Fig. 16.38 is initially uncharged. Find v 0 (t) for t > 0. Figure 16.38 For Prob. 16.4. Chapter 16, Solution 4. The circuit in the s-domain is shown below. I 2 4I + 5 Vo 1/s 1 – 4 5 1/ o o V I I I sV s + = ⎯⎯→ = But 5 2 oV I − = 5 12.5 5 2 5/ 2 o o o V sV V s −⎛ ⎞ = ⎯⎯→ =⎜ ⎟ +⎝ ⎠ 2.5 ( ) 12.5 Vt ov t e− = + _
  7. 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 16, Problem 5. If i s (t) = e t2− u(t) A in the circuit shown in Fig. 16.39, find the value of i 0 (t). Figure 16.39 For Prob. 16.5. Chapter 16, Solution 5. ( ) ( ) A)t(ut3229.1sin7559.0e or A)t(ueee3779.0eee3779.0e)t(i 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 2s 1 )3229.1j5.0s)(3229.1j5.0s)(2s( s 2 Vs I )3229.1j5.0s)(3229.1j5.0s)(2s( s2 2ss s2 2s 1 2 s 2 1 s 1 1 2s 1 V t2 t3229.1j2/t90t3229.1j2/t90t2 o 22 2 o 2 −= ++= −+ ++ +− + ++ −− −− + + = −++++ == −++++ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +++ = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ++ + = − −°−−°−− or io(t) = A)t(ut 2 7 sin 7 2 e t2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −− s s 2 2s 1 + Io 2
  8. 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 16, Problem 6. Find v(t), t > 0 in the circuit of Fig. 16.40. Let v s = 20 V. Figure 16.40 For Prob. 16.6. Chapter 16, Solution 6. For t<0, v(0) = vs = 20 V For t>0, the circuit in the s-domain is as shown below. 10 s 1 10 100 0.1mF F sC s = ⎯⎯→ = 20 2 10 110 sI s s = = ++ 20 10 1 V I s = = + ( ) 20 ( )t v t e u t− = + _ + _ 10 Ωv 20 s I
  9. 9. 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 16, Problem 7. Find v0 (t), for all t > 0, in the circuit of Fig. 16.41. Figure 16.41 For Prob. 16.7.
  10. 10. 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 16, Solution 7. The circuit in the s-domain is shown below. Please note, iL(0) = 0 and vo(0) = o because both sources were equal to zero for all t<0. 1 1 Vo 2/s s 2/s 1/s At node 1 11 1 1 2/ 2 (2 1/ ) 1 1 o o V Vs V V V s V s s −− = + ⎯⎯→ = + − (1) At node O, 1 1 1 (1 / 2) 1/ 1 2/ 2 o o o o V V V s V V s V s s s − + = = ⎯⎯→ = + − (2) Substituting (2) into (1) gives 1 1 2/ (2 1/ )(1 / 2) (2 )o os s s V V s s = + + − + − 2 2 (4 1) ( 1.5 1) 1.5 1 o s A Bs C V s s s s s s + + = = + + + + + 2 2 4 1 ( 1.5 1)s A s s Bs Cs+ = + + + + We equate coefficients. s2 : 0 = A+ B or B = - A s: 4=1.5A + C constant: 1 = A, B=-1, C = 4-1.5A = 2.5 2 22 2 2 3.25 7 47 1 2.5 1 3/ 4 4 1.5 1 7 7 ( 3/ 4) ( 3/ 4) 4 4 o x s s V s s s s s s − + + = + = − + + + ⎛ ⎞ ⎛ ⎞ + + + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3 /4 3 /47 7 ( ) ( ) cos 4.9135 sin 4 4 t t v t u t e t e t− − = − + + _
  11. 11. 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 16, Problem 8. If v0 (0) = -1V,obtain v0 (t) in the circuit of Fig. 16.42. Figure 16.42 For Prob. 16.8.
  12. 12. 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 16, Solution 8. 1 1 2 2 F sC s ⎯⎯→ = We analyze the circuit in the s-domain as shown below. We apply nodal analysis. 0 3 1 4 14 0 21 1 ( 4) o o oV V V ss s s V s s s − − − − − + + = ⎯⎯→ = + 4 o A B V s s = + + 14 18 7 / 2, 9/ 2 4 4 A B= = = = − − 7 / 2 9/ 2 4 oV s s = − + 47 9 ( ) ( ) 2 2 t ov t e u t−⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ 1 Vo 1 + _ + _ + _ 2 s 1 s − 4 s 3 s
  13. 13. 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 16, Problem 9. Find the input impedance Zin (s) of each of the circuits in Fig. 16.43. Figure 16.43 For Prob. 16.9. Chapter 16, Solution 9. (a) The s-domain form of the circuit is shown in Fig. (a). = ++ + =+= s1s2 )s1s(2 )s1s(||2Zin 1s2s )1s(2 2 2 ++ + (b) The s-domain equivalent circuit is shown in Fig. (b). 2s3 )2s(2 s23 )s21(2 )s21(||2 + + = + + =+ 2s3 6s5 )s21(||21 + + =++ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⋅ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = 2s3 6s5 s 2s3 6s5 s 2s3 6s5 ||sZin 6s7s3 )6s5(s 2 ++ + 2 s 1/s (a) (b) 2 s 2/s 1 1
  14. 14. 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 16, Problem 10. Use Thevenin’s theorem to determine v 0 (t), t > 0 in the circuit of Fig. 16.44. Figure 16.44 For Prob. 16.10. Chapter 16, Solution 10. 1 1H s⎯⎯→ and iL(0) = 0 (the sources is zero for all t<0). 1 1 4 4 F sC s ⎯⎯→ = and vC(0) = 0 (again, there are no source contributions for all t<0). To find ZTh , consider the circuit below. 1 s ZTh 2 2 1//( 2) 3 Th s Z s s + = + = +
  15. 15. 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. To find VTh, consider the circuit below. 1 s + 10 2s + VTh 2 - 2 10 10 3 2 3 Th s V s s s + = = + + + The Thevenin equivalent circuit is shown below ZTh + 4/s Vo VTh – 2 3 2 404 4 3 10 40 3 4 4 2 3 6 12 ( 3) ( 3) 3 o Th Th s sV V s s s s sZ s s s = = = = + + + + + ++ + + . 3 ( ) 23.094 sin 3t ov t e t− = + _ + _
  16. 16. 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 16, Problem 11. Solve for the mesh currents in the circuit of Fig. 16.45. You may leave your results in the s-domain. Figure 16.45 For Prob. 16.11.
  17. 17. 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 16, Solution 11. In the s-domain, the circuit is as shown below. 1 2 10 1 (1 ) 4 4 s I sI s = + − (1) 1 2 1 5 (4 ) 0 4 4 sI I s− + + = (2) In matrix form, 1 2 1 10 1 4 4 1 5 0 4 4 4 s s I s I s s ⎡ ⎤ + −⎡ ⎤ ⎢ ⎥ ⎡ ⎤⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥− +⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦ 4s 4 9 s 4 1 2 ++=∆ 1 10 1 40 504 5 4 0 4 4 s s s s − ∆ = = + + 2 10 1 54 1 2 0 4 s s s + ∆ = = − )16s9s(s 160s50 4s25.2s25.0 2 25 s 40 I 22 1 1 ++ + = ++ + = ∆ ∆ = 16s9s 10 4s25.2s25.0 5.2 I 22 2 2 ++ = ++ = ∆ ∆ = 1 Ω 4 + _ I1 I2 1 H 4 1 s 10 s
  18. 18. 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 16, Problem 12. Find v o (t) in the circuit of Fig. 16.46. Figure 16.46 For Prob. 16.12.
  19. 19. 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 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below. o o o sV2 4 V s 3 s V 1s 10 +=+ − + 1s 15s1510 15 1s 10 V)ss25.01( o 2 + ++ =+ + =++ 1s25.0s CBs 1s A )1s25.0s)(1s( 25s15 V 22o ++ + + + = +++ + = 7 40 V)1s(A 1-so =+= = )1s(C)ss(B)1s25.0s(A25s15 22 ++++++=+ Equating coefficients : 2 s : -ABBA0 =⎯→⎯+= 1 s : C-0.75ACBA25.015 +=++= 0 s : CA25 += 740A = , 740-B = , 7135C = 4 3 2 1 s 2 3 3 2 7 155 4 3 2 1 s 2 1 s 7 40 1s 1 7 40 4 3 2 1 s 7 135 s 7 40- 1s 7 40 V 222o +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅+ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + − + = +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= t 2 3 sine )3)(7( )2)(155( t 2 3 cose 7 40 e 7 40 )t(v 2t-2t-t- o =)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t +− s 3/s41/(2s)+ − 10/(s + 1) Vo
  20. 20. 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 16, Problem 13. Determine i 0 (t) in the circuit of Fig. 16.47. Figure 16.47 For Prob. 16.13. Chapter 16, Solution 13. Consider the following circuit. Applying KCL at node o, o oo V 1s2 1s s12 V 1s2 V 2s 1 + + = + + + = + )2s)(1s( 1s2 Vo ++ + = 2s B 1s A )2s)(1s( 1 1s2 V I o o + + + = ++ = + = 1A = , -1B = 2s 1 1s 1 Io + − + = =)t(io ( ) A)t(uee -2t-t − 2s 2 1 1/s 1/(s + 2) Vo Io
  21. 21. 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 16, Problem 14. * Determine i0 (t) in the network shown in Fig. 16.48. Figure 16.48 For Prob. 16.14. * An asterisk indicates a challenging problem. Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a). A5)0(io =− , V0)0(vc =− We now incorporate these conditions in the s-domain circuit as shown in Fig.(b). 4 Ω1 Ω + − 5 V (a) + vc(0) − io 4/s 41 + − 15/s Vo Io (b) 5/s2s
  22. 22. 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. At node o, 0 s44 0V s 5 s2 V 1 s15V ooo = + − +++ − oV )1s(4 s s2 1 1 s 5 s 15 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ++=− o 2 o 22 V )1s(s4 2s6s5 V )1s(s4 s2s2s4s4 s 10 + ++ = + ++++ = 2s6s5 )1s(40 V 2o ++ + = s 5 )4.0s2.1s(s )1s(4 s 5 s2 V I 2 o o + ++ + =+= 4.0s2.1s CBs s A s 5 I 2o ++ + ++= sCsB)4.0s2.1s(A)1s(4 s2 ++++=+ Equating coefficients : 0 s : 10AA4.04 =⎯→⎯= 1 s : -84-1.2ACCA2.14 =+=⎯→⎯+= 2 s : -10-ABBA0 ==⎯→⎯+= 4.0s2.1s 8s10 s 10 s 5 I 2o ++ + −+= 2222o 2.0)6.0s( )2.0(10 2.0)6.0s( )6.0s(10 s 15 I ++ − ++ + −= =)t(io ( )[ ] A)t(u)t2.0sin()t2.0cos(e1015 0.6t- −−
  23. 23. 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 16, Problem 15. Find V x (s) in the circuit shown in Fig. 16.49. Figure 16.49 For Prob. 16.15. Chapter 16, Solution 15. First we need to transform the circuit into the s-domain. 2s 5 VV 2s 5 VV,But 2s s5 V120V)40ss2(0 2s s5 sVVs2V120V40 0 10 2s 5 V s/5 0V 4/s V3V xoox xo 2 oo 2 xo o oxo + +=→ + −= + −−++== + −++− =+ − + − + − We can now solve for Vx. )40s5.0s)(2s( )20s( 5V 2s )20s( 10V)40s5.0s(2 0 2s s5 V120 2s 5 V)40ss2( 2 2 x 2 x 2 xx 2 −++ + −= + + −=−+ = + −−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + +++ 10 3Vx + Vx − s/4 5/s 2s 5 + + −+ Vo
  24. 24. 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 16, Problem 16. * Find i0 (t) for t > 0 in the circuit of Fig. 16.50. Figure 16.50 For Prob. 16.16. * An asterisk indicates a challenging problem.
  25. 25. 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 16, Solution 16. We first need to find the initial conditions. For 0t < , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. Hence, A1- 3 3- i)0(i oL === , V1-vo = V5.2 2 1- )1-)(2(-)0(vc =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= We now incorporate the initial conditions for 0t > as shown in Fig. (b). For mesh 1, 0 2 V s 5.2 I s 1 I s 1 2 2s 5- o 21 =++−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ + (a) 3 V 2 Ω + − io1 H+ Vo/2 1 F + − Vo 1 Ω (b) -1 V 2 − + Io + Vo/2 1/s + − Vo 1 + − 2.5/s s + − 5/(s + 2) I1 I2
  26. 26. 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. But, 2oo IIV == s 5.2 2s 5 I s 1 2 1 I s 1 2 21 − + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + (1) For mesh 2, 0 s 5.2 2 V 1I s 1 I s 1 s1 o 12 =−−+−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ 1 s 5.2 I s 1 s 2 1 I s 1 - 21 −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +++ (2) Put (1) and (2) in matrix form. ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ++ −+ 1 s 5.2 s 5.2 2s 5 I I s 1 s 2 1 s 1- s 1 2 1 s 1 2 2 1 s 3 2s2 ++=∆ , )2s(s 5 s 4 2-2 + ++=∆ 3s2s2 CBs 2s A )3s2s2)(2s( 132s- II 22 2 2 2o ++ + + + = +++ + = ∆ ∆ == )2s(C)s2s(B)3s2s2(A132s- 222 ++++++=+ Equating coefficients : 2 s : BA22- += 1 s : CB2A20 ++= 0 s : C2A313 += Solving these equations leads to 7143.0A = , -3.429B = , 429.5C = 5.1ss 714.2s7145.1 2s 7143.0 3s2s2 429.5s429.3 2s 7143.0 I 22o ++ − − + = ++ − − + = 25.1)5.0s( )25.1)(194.3( 25.1)5.0s( )5.0s(7145.1 2s 7143.0 I 22o ++ + ++ + − + = =)t(io [ ] A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t +−
  27. 27. 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 16, Problem 17. Calculate i0 (t) for t > 0 in the network of Fig. 16.51. Figure 16.51 For Prob. 16.17. Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below. For mesh 3, 0IsI s 1 I s 1 s 1s 2 213 =−−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ + (1) For the supermesh, 0Is s 1 I)s1(I s 1 1 321 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + (2) s 1 1 1/s 2/(s+1) + − I3 I2I1 4/s
  28. 28. 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. Adding (1) and (2) we get, I1 + I2 = –2/(s+1) (3) But –I1 + I2 = 4/s (4) Adding (3) and (4) we get, I2 = (2/s) – 1/(s+1) (5) Substituting (5) into (4) yields, I1 = –(2/s) – (1/(s+1)) (6) Substituting (5) and (6) into (1) we get, 1s 2 I s 1s 1s s 2 )1s(s 1 s 2 3 2 2 + −= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + +− + + js j5.05.1 js j5.05.1 s 2 I3 − + + + − +−= Substituting (3) into (1) and (2) leads to )1s(s )2s2s(2 I s 1 sI s 1 s- 2 2 32 + ++− =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + (4) 232 s )1s(4 I s 1 sI s 1 s2 + −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ (5) We can now solve for Io. Io = I2 – I3 = (4/s) – (1/(s+1)) + ((–1.5+0.5j)/(s+j)) + ((–1.5–0.5)/(s–j)) or io(t) = [4 – e–t + 1.5811e–jt+161.57˚ + 1.5811ejt–161.57˚ ]u(t)A This is a challenging problem. I did check it with using a Thevenin equivalent circuit and got the same exact answer.
  29. 29. 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 16, Problem 18. (a) Find the Laplace transform of the voltage shown in Fig. 16.52(a). (b) Using that value of v s (t) in the circuit shown in Fig. 16.52(b), find the value of v 0 (t). Figure 16.52 For Prob. 16.18. Chapter 16, Solution 18. vs(t) = 3u(t) – 3u(t–1) or Vs = )e1( s 3 s e s 3 s s − − −=− V)]1t(u)e22()t(u)e22[()t(v )e1( 5.1s 2 s 2 )e1( )5.1s(s 3 V VV)5.1s(0 2 V sV 1 VV )1t(5.1t5.1 o ss o so o o so −−−−= −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + −=− + = =+→=++ − −−− −− 1 Ω 1/sVs + − 2 Ω + Vo −
  30. 30. 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 16, Problem 19. In the circuit of Fig. 16.53, let i(0) = 1 A, v 0 (0) and v s = 4e t2− u(t) V. Find v 0 (t) for t > 0. Figure 16.53 For Prob. 16.19.
  31. 31. 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 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below. At the supernode, o 11 sV s 1 s V 2 2 V))2s(4( ++=+ −+ o1 Vs s 1 V s 1 2 1 2 2s 2 ++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=+ + (1) But I2VV 1o += and s 1V I 1 + = 2s 2Vs s)2s( s2V V s )1V(2 VV oo 1 1 1o + − = + − =⎯→⎯ + += (2) Substituting (2) into (1) oo Vs 2s 2 V 2s s s2 2s s 1 2 2s 2 +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + =−+ + oVs 2 1 s 1 s 1 2 2s 2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =+−+ + oV)2/1s( 2s 6s2 )2s( 24s2 += + + = + ++ 2s B 2/1s A )2/1s)(2s( 6s2 Vo + + + = ++ + = 333.3)25.0/()61(A =+−+−= , 3333.1)2/12/()64(B −=+−+−= 2s 3333.1 2/1s 333.3 Vo + − + = Therefore, =)t(vo (3.333e-t/2 – 1.3333e-2t )u(t) V 2 + − s 1/s VoV1 2 I − + 24/(s + 2) 1/s I
  32. 32. 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 16, Problem 20. Find v 0 (t) in the circuit of Fig. 16.54 if v x (0) = 2 V and i(0) = 1A. Figure 16.54 For Prob. 16.20.
  33. 33. 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 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown. At the main non-reference node, KCL gives s 1 s V 1 V s11 Vs2)1s(1 ooo ++= + −−+ s 1s V)s11)(1s(Vs2 1s s oo + +++=−− + oV)s12s2(2 s 1s 1s s ++=− + − + )1s2s2)(1s( 1s4s2- V 2 2 o +++ −− = 5.0ss CBs 1s A )5.0ss)(1s( 5.0s2s- V 22o ++ + + + = +++ −− = 1V)1s(A 1-so =+= = )1s(C)ss(B)5.0ss(A5.0s2s- 222 ++++++=−− Equating coefficients : 2 s : -2BBA1- =⎯→⎯+= 1 s : -1CCBA2- =⎯→⎯++= 0 s : -0.515.0CA5.00.5- =−=+= 222o )5.0()5.0s( )5.0s(2 1s 1 5.0ss 1s2 1s 1 V ++ + − + = ++ + − + = =)t(vo [ ] V)t(u)2tcos(e2e 2-t-t − 1 + − Vo 2/s 1/s 1/s + − 1 s1/(s + 1)
  34. 34. 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 16, Problem 21. Find the voltage v0 (t) in the circuit of Fig. 16.55 by means of the Laplace transform. Figure 16.55 For Prob. 16.21.
  35. 35. 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 16, Solution 21. The s-domain version of the circuit is shown below. 1 s V1 Vo + 2/s 2 1/s - At node 1, oo o V s VsV s s VV V s )1 2 ()1(10 21 10 2 1 1 1 −++=⎯→⎯+ − = − (1) At node 2, )1 2 ( 2 2 1 1 ++=⎯→⎯+= − s s VVsV V s VV oo oo (2) Substituting (2) into (1) gives ooo VsssV s Vsss )5.12()1 2 ()12/)(1(10 2 2 2 ++=−++++= 5.12)5.12( 10 22 ++ + += ++ = ss CBs s A sss Vo CsBsssA ++++= 22 )5.12(10 BAs +=0:2 CAs += 20: -40/3C-20/3,B,3/205.110:constant ===⎯→⎯= AA ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++ − ++ + −=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++ + −= 22222 7071.0)1( 7071.0 414.1 7071.0)1( 11 3 20 5.12 21 3 20 ss s sss s s Vo Taking the inverse Laplace tranform finally yields [ ] V)t(ut7071.0sine414.1t7071.0cose1 3 20 )t(v tt o −− −−= 10/s
  36. 36. 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 16, Problem 22. Find the node voltages v1 and v 2 in the circuit of Fig. 16.56 using the Laplace transform technique. Assume that i s = 12e t− u(t) A and that all initial conditions are zero. Figure 16.56 For Prob. 16.22. Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2 1s 12 + 1 2 3/s At node 1, s4 V s4 1 1V 1s 12 s4 VV 1 V 1s 12 2 1 211 −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += + ⎯→⎯ − += + (1) At node 2, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=⎯→⎯+= − 1s2s 3 4 VVV 3 s 2 V s4 VV 2 212 221 (2) Substituting (2) into (1), 2 22 2 V 2 3 s 3 7 s 3 4 s4 1 s4 1 11s2s 3 4 V 1s 12 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++= + ) 8 9 s 4 7 s( CBs )1s( A ) 8 9 s 4 7 s)(1s( 9 V 22 2 ++ + + + = +++ = )1s(C)ss(B) 8 9 s 4 7 s(A9 22 ++++++=
  37. 37. 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. Equating coefficients: BA0:s2 += A 4 3 CCA 4 3 CBA 4 7 0:s −=⎯→⎯+=++= -18C-24,B,24AA 8 3 CA 8 9 9:constant ===⎯→⎯=+= 64 23 ) 8 7 s( 3 64 23 ) 8 7 s( )8/7s(24 )1s( 24 ) 8 9 s 4 7 s( 18s24 )1s( 24 V 222 2 ++ + ++ + − + = ++ + − + = Taking the inverse of this produces: [ ] )t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t 2 −−− +−= Similarly, ) 8 9 s 4 7 s( FEs )1s( D ) 8 9 s 4 7 s)(1s( 1s2s 3 4 9 V 22 2 1 ++ + + + = +++ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ = )1s(F)ss(E) 8 9 s 4 7 s(D1s2s 3 4 9 222 ++++++=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ Equating coefficients: ED12:s2 += D 4 3 6FFD 4 3 6orFED 4 7 18:s −=⎯→⎯+=++= 0F4,E,8DD 8 3 3orFD 8 9 9:constant ===⎯→⎯=+= 64 23 ) 8 7 s( 2/7 64 23 ) 8 7 s( )8/7s(4 )1s( 8 ) 8 9 s 4 7 s( s4 )1s( 8 V 222 1 ++ − ++ + + + = ++ + + = Thus, [ ] )t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t 1 −−− −+=
  38. 38. 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 16, Problem 23. Consider the parallel RLC circuit of Fig. 16.57. Find v(t) and i(t) given that v(0) = 5 and i(0) = -2 A. Figure 16.57 For Prob. 16.23.
  39. 39. 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 16, Solution 23. The s-domain form of the circuit with the initial conditions is shown below. At the non-reference node, sCV sL V R V C5 s 2 s 4 ++=++ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++= + LC 1 RC s s s CV s sC56 2 LC1RCss C6s5 V 2 ++ + = But 8 8010 1 RC 1 == , 20 804 1 LC 1 == 22222 2)4s( )2)(230( 2)4s( )4s(5 20s8s 480s5 V ++ + ++ + = ++ + = =)t(v V)t(u))t2sin(e230)t2cos(e5( -4t-4t + )20s8s(s4 480s5 sL V I 2 ++ + == 20s8s CBs s A )20s8s(s 120s25.1 I 22 ++ + += ++ + = 6A = , -6B = , -46.75C = 22222 2)4s( )2)(375.11( 2)4s( )4s(6 s 6 20s8s 75.46s6 s 6 I ++ − ++ + −= ++ + −= =)t(i 0t),t(u))t2sin(e375.11)t2cos(e66( -4t-4t >−− -2/sR sL 1/sC4/s V I 5C
  40. 40. 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 16, Problem 24. The switch in Fig. 16.58 moves from position 1 to position 2 at t =0. Find v(t), for all t > 0. Figure 16.58 For Prob. 16.24. Chapter 16, Solution 24. When the switch is position 1, v(0)=12, and iL(0) = 0. When the switch is in position 2, we have the circuit as shown below. s/4 + 100/s v 12/s – 1 100 10 0.01mF F sC s = ⎯⎯→ = 2 12/ 48 / 4 100/ 400 s I s s s = = + + , 2 12 4 400 s s V sLI I s = = = + ( ) 12cos20 , 0v t t t= > + _
  41. 41. 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 16, Problem 25. For the RLC circuit shown in Fig. 16.59, find the complete response if v(0) = 2 V when the switch is closed. Figure 16.59 For Prob. 16.25. Chapter 16, Solution 25. For 0t > , the circuit in the s-domain is shown below. s 2/s 9/s 6 + − (2s)/(s2 + 16) + − I + V −
  42. 42. 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. Applying KVL, 0 s 2 I s 9 s6 16s s2 2 =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +++ + − )16s)(9s6s( 32 I 22 +++ − = )16s()3s(s 288 s 2 s 2 I s 9 V 22 ++ − +=+= 16s EDs )3s( C 3s B s A s 2 22 + + + + + + ++= )s48s16s3s(B)144s96s25s6s(A288- 234234 ++++++++= )s9s6s(E)s9s6s(D)s16s(C 232343 ++++++++ Equating coefficients : 0 s : A144288 =− (1) 1 s : E9C16B48A960 +++= (2) 2 s : E6D9B16A250 +++= (3) 3 s : ED6CB3A60 ++++= (4) 4 s : DBA0 ++= (5) Solving equations (1), (2), (3), (4) and (5) gives 2A −= , 202.2B = , 84.3C = , 202.0-D = , 766.2E = 16s )4)(6915.0( 16s s202.0 )3s( 84.3 3s 202.2 )s(V 222 + + + − + + + = =)t(v {2.202e-3t + 3.84te-3t – 0.202cos(4t) + 0.6915sin(4t)}u(t) V
  43. 43. 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 16, Problem 26. For the op amp circuit in Fig. 16.60, find v 0 (t) for t > 0. Take v s = 3e t5− u(t) V. Figure 16.60 For Prob. 16.26.
  44. 44. 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 16, Solution 26. Consider the op-amp circuit below. At node 0, sC)V0( R V0 R 0V o 2 o 1 s −+ − = − ( )o 2 1s V-sC R 1 RV ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += 211s o RRCsR 1- V V + = But 2 10 20 R R 2 1 == , 1)1050)(1020(CR 6-3 1 =××= So, 2s 1- V V s o + = )5s(3Ve3V s -5t s +=⎯→⎯= 5)2)(s(s 3- Vo ++ = 5s B 2s A 5)2)(s(s 3 V- o + + + = ++ = 1A = , -1B = 2s 1 5s 1 Vo + − + = =)t(vo ( ) )t(uee -2t-5t − − + 1/sC R1 R2 + − Vs 0 + Vo −
  45. 45. 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 16, Problem 27. Find I1 (s) and I 2 (s) in the circuit of Fig. 16.61. Figure 16.61 For Prob. 16.27.
  46. 46. 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 16, Solution 27. Consider the following circuit. For mesh 1, 221 IsII)s21( 3s 10 −−+= + 21 I)s1(I)s21( 3s 10 +−+= + (1) For mesh 2, 112 IsII)s22(0 −−+= 21 I)1s(2I)s1(-0 +++= (2) (1) and (2) in matrix form, ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++ ++ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + 2 1 I I )1s(2)1s(- )1s(-1s2 0 )3s(10 1s4s3 2 ++=∆ 3s )1s(20 1 + + =∆ 3s )1s(10 2 + + =∆ Thus = ∆ ∆ = 1 1I )1s4s3)(3s( )1s(20 2 +++ + = ∆ ∆ = 2 2I )1s4s3)(3s( )1s(10 2 +++ + 2 I1 = 2s 110/(s + 3) + − I1 2s 1I2 s
  47. 47. 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 16, Problem 28. For the circuit in Fig. 16.62, find v 0 (t) for t > 0. Figure 16.62 For Prob. 16.28.
  48. 48. 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 16, Solution 28. Consider the circuit shown below. For mesh 1, 21 IsI)s21( s 6 ++= (1) For mesh 2, 21 I)s2(Is0 ++= 21 I s 2 1-I ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += (2) Substituting (2) into (1) gives 2 2 22 I s 2)5s(s- IsI s 2 12s)-(1 s 6 ++ =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++= or 2s5s 6- I 22 ++ = 4.561)0.438)(s(s 12- 2s5s 12- I2V 22o ++ = ++ == Since the roots of 02s5s2 =++ are -0.438 and -4.561, 561.4s B 438.0s A Vo + + + = -2.91 4.123 12- A == , 91.2 4.123- 12- B == 561.4s 91.2 0.438s 2.91- )s(Vo + + + = =)t(vo [ ] V)t(uee91.2 t438.0-4.561t − s 2I2s2s 1 + − 6/s I1 + Vo −
  49. 49. 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 16, Problem 29. For the ideal transformer circuit in Fig. 16.63, determine i0 (t). Figure 16.63 For Prob. 16.29. Chapter 16, Solution 29. Consider the following circuit. Let 1s2 8 s48 )s4)(8( s 4 ||8ZL + = + == When this is reflected to the primary side, 2n, n Z 1Z 2 L in =+= 1s2 3s2 1s2 2 1Zin + + = + += 3s2 1s2 1s 10 Z 1 1s 10 I in o + + ⋅ + =⋅ + = 5.1s B 1s A )5.1s)(1s( 5s10 Io + + + = ++ + = -10A = , 20B = 5.1s 20 1s 10- )s(Io + + + = =)t(io [ ] A)t(uee210 t-1.5t − − 84/s 1 10/(s + 1) + − Io 1 : 2
  50. 50. 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 16, Problem 30. The transfer function of a system is H(s) = 13 2 +s s Find the output when the system has an input of 4e 3/t− u(t). Chapter 16, Solution 30. )s(X)s(H)s(Y = , 1s3 12 31s 4 )s(X + = + = 22 2 )1s3( 34s8 3 4 )1s3( s12 )s(Y + + −= + = 22 )31s( 1 27 4 )31s( s 9 8 3 4 )s(Y + ⋅− + ⋅−= Let 2 )31s( s 9 8- )s(G + ⋅= Using the time differentiation property, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=⋅= 3t-3t-3t- eet 3 1- 9 8- )et( dt d 9 8- )t(g 3t-3t- e 9 8 et 27 8 )t(g −= Hence, 3t-3t-3t- et 27 4 e 9 8 et 27 8 )t(u 3 4 )t(y −−+= =)t(y 3t-3t- et 27 4 e 9 8 )t(u 3 4 +−
  51. 51. 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 16, Problem 31. When the input to a system is a unit step function, the response is 10 cos 2tu(t). Obtain the transfer function of the system. Chapter 16, Solution 31. s 1 )s(X)t(u)t(x =⎯→⎯= 4s s10 )s(Y)t2cos(10)t(y 2 + =⎯→⎯= == )s(X )s(Y )s(H 4s s10 2 2 +
  52. 52. 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 16, Problem 32. A circuit is known to have its transfer function as H(s) = 54 3 2 ++ + ss s Find its output when: (a) the input is a unit step function (b) the input is 6te t2− u(t). Chapter 16, Solution 32. (a) )s(X)s(H)s(Y = s 1 5s4s 3s 2 ⋅ ++ + = 5s4s CBs s A )5s4s(s 3s 22 ++ + += ++ + = CsBs)5s4s(A3s 22 ++++=+ Equating coefficients : 0 s : 53AA53 =⎯→⎯= 1 s : 57-A41CCA41 =−=⎯→⎯+= 2 s : 53--ABBA0 ==⎯→⎯+= 5s4s 7s3 5 1 s 53 )s(Y 2 ++ + ⋅−= 1)2s( 1)2s(3 5 1 s 6.0 )s(Y 2 ++ ++ ⋅−= =)t(y [ ] )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t −−
  53. 53. 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. (b) 2 2t- )2s( 6 )s(Xet6)t(x + =⎯→⎯= 22 )2s( 6 5s4s 3s )s(X)s(H)s(Y + ⋅ ++ + == 5s4s DCs )2s( B 2s A )5s4s()2s( )3s(6 )s(Y 2222 ++ + + + + + = +++ + = Equating coefficients : 3 s : -ACCA0 =⎯→⎯+= (1) 2 s : DBA2DC4BA60 ++=+++= (2) 1 s : D4B4A9D4C4B4A136 ++=+++= (3) 0 s : BA2D4B5A1018 +=++= (4) Solving (1), (2), (3), and (4) gives 6A = , 6B = , 6-C = , -18D = 1)2s( 18s6 )2s( 6 2s 6 )s(Y 22 ++ + − + + + = 1)2s( 6 1)2s( )2s(6 )2s( 6 2s 6 )s(Y 222 ++ − ++ + − + + + = =)t(y [ ] )t(u)tsin(e6)tcos(e6et6e6 -2t-2t-2t-2t −−+
  54. 54. 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 16, Problem 33. When a unit step is applied to a system at t = 0 its response is y(t) = ⎥⎦ ⎤ ⎢⎣ ⎡ +−+ −− )4sin34cos2( 2 1 4 23 ttee tt u(t) What is the transfer function of the system? Chapter 16, Solution 33. )s(X )s(Y )s(H = , s 1 )s(X = 16)2s( )4)(3( 16)2s( s2 )3s(2 1 s 4 )s(Y 22 ++ − ++ − + += == )s(Ys)s(H 20s4s s12 20s4s )2s(s2 )3s(2 s 4 22 ++ − ++ + − + +
  55. 55. 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 16, Problem 34. For the circuit in Fig. 16.64, find H(s) = V0 (s)/V s (s). Assume zero initial conditions. Figure 16.64 For Prob. 16.34. Chapter 16, Solution 34. Consider the following circuit. Using nodal analysis, s10 V 4 V 2s VV ooos += + − o 2 os V)s2s( 10 1 )2s( 4 1 1V 10 s 4 1 2s 1 )2s(V ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++++=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ + += ( ) o 2 s V30s9s2 20 1 V ++= = s o V V 30s9s2 20 2 ++ s 4 2 + − Vs 10/s Vo + Vo(s) −
  56. 56. 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 16, Problem 35. Obtain the transfer function H(s) = V0 /V s for the circuit of Fig. 16.65. Figure 16.65 For Prob. 16.35. Chapter 16, Solution 35. Consider the following circuit. At node 1, 3s V II2 1 + =+ , where s2 VV I 1s − = 3s V s2 VV 3 11s + = − ⋅ 1s 1 V 2 s3 V 2 s3 3s V −= + s1 V 2 s3 V 2 s3 3s 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + s21 V 2s9s3 )3s(s3 V ++ + = s21o V 2s9s3 s9 V 3s 3 V ++ = + = == s o V V )s(H 2s9s3 s9 2 ++ s 32I 2/s + − Vs V1 I + Vo −
  57. 57. 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 16, Problem 36. The transfer function of a certain circuit is H(s) = 1 5 +s - 2 3 +s + 4 6 +s Find the impulse response of the circuit. Chapter 16, Solution 36. Taking the inverse Laplace transform of each term gives ( )2 4 ( ) 5 3 6 ( )t t t h t e e e u t− − − = − +
  58. 58. 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 16, Problem 37. For the circuit in Fig. 16.66, find: (a) I1 /V s (b) I 2 /V x Figure 16.66 For Prob. 16.37. Chapter 16, Solution 37. (a) Consider the circuit shown below. For loop 1, 21s I s 2 I s 2 3V −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += (1) For loop 2, 0I s 2 I s 2 s2V4 12x =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ But, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= s 2 )II(V 21x So, 0I s 2 I s 2 s2)II( s 8 1221 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++− 21 Is2 s 6 I s 6- 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+= (2) + − Vs I2I1 2s3 + 4Vx2/s + Vx −
  59. 59. 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. In matrix form, (1) and (2) become ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1s I I s2s6s6- s2-s23 0 V ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=∆ s 2 s 6 s2 s 6 s 2 3 4s6 s 18 −−=∆ s1 Vs2 s 6 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=∆ , s2 V s 6 =∆ s 1 1 V s64s18 )s2s6( I −− − = ∆ ∆ = = −− − = 32s9 ss3 V I s 1 9s2s3 3s 2 2 −+ − (b) ∆ ∆ = 2 2I ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆ ∆−∆ =−= 21 21x s 2 )II( s 2 V ∆ = ∆ −− = ss x V4-)s6s2s6(Vs2 V == s s x 2 4V- Vs6 V I 2s 3-
  60. 60. 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 16, Problem 38. Refer to the network in Fig. 16.67. Find the following transfer functions: (a) H1 (s) = V0 (s)/V s (s) (b) H 2 (s) = V0 (s)/I s (s) (c) H3 (s) = I0 (s)/I s (s) (d) H 4 (s) = I0 (s)/V s (s) Figure 16.67 For Prob. 16.38.
  61. 61. 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 16, Solution 38. (a) Consider the following circuit. At node 1, s VV Vs 1 VV o1 1 1s − += − o1s V s 1 V s 1 s1V −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++= (1) At node o, ooo o1 V)1s(VVs s VV +=+= − o 2 1 V)1ss(V ++= (2) Substituting (2) into (1) oo 2 s Vs1V)1ss)(s11s(V −++++= o 23 s V)2s3s2s(V +++= == s o 1 V V )s(H 2s3s2s 1 23 +++ (b) o 2 o 23 1ss V)1ss(V)2s3s2s(VVI ++−+++=−= o 23 s V)1s2ss(I +++= == s o 2 I V )s(H 1s2ss 1 23 +++ (c) 1 V I o o = ==== )s(H I V I I )s(H 2 s o s o 3 1s2ss 1 23 +++ (d) ==== )s(H V V V I )s(H 1 s o s o 4 2s3s2s 1 23 +++ s 1 1 + − Vs 1/s V1 Vo Io + Vo − 1/s Is
  62. 62. 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 16, Problem 39. Calculate the gain H(s) = V0 /V s in the op amp circuit of Fig. 16.68. Figure 16.68 For Prob. 16.39. Chapter 16, Solution 39. Consider the circuit below. Since no current enters the op amp, oI flows through both R and C. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += sC 1 RI-V oo sC I- VVV o sba === = + == sC1 sC1R V V )s(H s o 1sRC + − + Va Vb R 1/sC + − Vs Io + Vo −
  63. 63. 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 16, Problem 40. Refer to the RL circuit in Fig. 16.69. Find: (a) the impulse response h(t) of the circuit. (b) the unit step response of the circuit. Figure 16.69 For Prob. 16.40. Chapter 16, Solution 40. (a) LRs LR sLR R V V )s(H s o + = + == =)t(h )t(ue L R LRt- (b) s1)s(V)t(u)t(v ss =⎯→⎯= LRs B s A )LRs(s LR V LRs LR V so + += + = + = 1A = , -1B = LRs 1 s 1 Vo + −= =−= )t(ue)t(u)t(v L-Rt o )t(u)e1( L-Rt −
  64. 64. 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 16, Problem 41. A parallel RL circuit has R = 4Ω and L = 1 H. The input to the circuit is i s (t) = 2e t− u(t)A. Find the inductor current i L (t) for all t > 0 and assume that i L (0) = -2 A. Chapter 16, Solution 41. Consider the circuit as shown below. 2 4 o o s V V I s s = + − But 2 1 SI s = + 2 1 1 2 4 2 2 4 2 ( ) 1 4 4 1 ( 1) o o s s V V s s s s s s s s + +⎛ ⎞ = + − ⎯⎯→ = + =⎜ ⎟ + + +⎝ ⎠ 8(2 1) ( 1)( 4) o s V s s + = + + 8(2 1) ( 1)( 4) 1 4 o L V s A B C I s s s s s s s + = = = + + + + + + 8(1) 8( 2 1) 8( 8 1) 2, 8/3, 14/3 (1)(4) ( 1)(2) ( 4)( 3) A B C − + − + = = = = = = − − − − 2 8/3 14/3 1 4 o L V I s s s s − = = + + + + 48 14 ( ) 2 ( ) 3 3 t t Li t e e u t− −⎛ ⎞ = + −⎜ ⎟ ⎝ ⎠ = A)t(ue 3 14 e 3 8 2 t4t ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+ −− 4 Is s I2 Vo 2 s −
  65. 65. 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 16, Problem 42. A circuit has a transfer function H(s) = 2 )2)(1( 4 ++ + ss s Find the impulse response. Chapter 16, Solution 42. 2 2 4 ( ) ( 1)( 2) 1 2 ( 2) s A B C H s s s s s s + = = + + + + + + + 2 2 2 4 ( 2) ( 1)( 2) ( 1) ( 2 4) ( 3 2) ( 1)s A s B s s C s A s s B s s C s+ = + + + + + + = + + + + + + + We equate coefficients. s2 : 0=A+B or B=-A s: 1=4A+3B+C=B+C constant: 4=4A+2B+C =2A+C Solving these gives A=3, B=-3, C=-2 2 3 3 2 ( ) 1 2 ( 2) H s s s s = − − + + + 2 2 ( ) (3 3 2 ) ( )t t t h t e e te u t− − − = − −
  66. 66. 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 16, Problem 43. Develop the state equations for Prob. 16.1. Chapter 16, Solution 43. First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: ' CC vi;0'ivi)t(u ==+++− Thus, )t(uivi iv C ' ' C +−−= = Finally we get, [ ] [ ] )t(u0 i v 10)t(i;)t(u 1 0 i v 11 10 i v CCC +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −− = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ′ ′ 1F 1Ω + − u(t) 1H i(t)
  67. 67. 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 16, Problem 44. Develop the state equations for Prob. 16.2. Chapter 16, Solution 44. First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get: LCxxLC ' C C ' C Cx x ' L xL ' C ' Cx L i3333.1v3333.0vor;v2i4v 2 v v 8 v 4vv v)t(u4i v4i8vor;0 8 v 2 v i +=−+=+=+= −= −==++− LC ' L LCLCL ' C i3333.1v3333.0)t(u4i i666.2v3333.1i333.5v3333.1i8v −−= +−=−−= Now we can write the state equations. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −− − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ L C x L C ' L ' C i v 3333.1 3333.0 v;)t(u 4 0 i v 3333.13333.0 666.23333.1 i v 4Ω + vx − 1H 1/8 F )t(u4 + − 2Ω
  68. 68. 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 16, Problem 45. Develop the state equations for the circuit shown in Fig. 16.70. Figure 16.70 For Prob. 16.45. Chapter 16, Solution 45. First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get: 2o ' L oL ' CL o ' C vvi v2i4vor0i 2 v 4 v −= +==++− 1Co vvv +−= 21C ' L 1CL ' C vvvi v2v2i4v −+−= +−= [ ] [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ′ ′ )t(v )t(v 01 v i 10)t(v; )t(v )t(v 02 11 v i 24 10 v i 2 1 C L o 2 1 C L C L
  69. 69. 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 16, Problem 46. Develop the state equations for the circuit shown in Fig. 16.71. Figure 16.71 For Prob. 16.46. Chapter 16, Solution 46. First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get: sC ' L sLC ' Cs C' CL vvi iiv25.0vor0i 4 v vi +−= ++−==−++− Thus, ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ s s L C o s s ' L ' C ' L ' C i v 00 00 i v 0 1 )t(v; i v 01 10 i v 01 125.0 i v
  70. 70. 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 16, Problem 47. Develop the state equations for the circuit shown in Fig. 16.72. Figure 16.72 For Prob. 16.47.
  71. 71. 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 16, Solution 47. First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables. Letting i1 = 4 v' C and i2 = iL and applying KVL we get: Loop 1: 1CL ' CL ' C C1 v2v2i4vor0i 4 v 2vv +−== ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −++− Loop 2: 21C2 1CL L ' L 2 ' L ' C L vvvv 2 v2v2i4 i2i or0vi 4 v i2 −+−=− +− +−= =++ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − 1CL 1CL 1 v5.0v5.0i 4 v2v2i4 i +−= +− = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ′ ′ )t(v )t(v 00 05.0 v i 01 5.01 )t(i )t(i ; )t(v )t(v 02 11 v i 24 10 v i 2 1 C L 2 1 2 1 C L C L
  72. 72. 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 16, Problem 48. Develop the state equations for the following differential equation. 2 2 )( dt tyd + dt tdy )(4 + 3y(t) = z(t) Chapter 16, Solution 48. Let x1 = y(t). Thus, )t(zx4x3yxandxyx 21 ' 22 '' 1 +−−=′′=== This gives our state equations. [ ] [ ] )t(z0 x x 01)t(y;)t(z 1 0 x x 43 10 x x 2 1 2 1 ' 2 ' 1 +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −− = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ Chapter 16, Problem 49. * Develop the state equations for the following differential equation. 2 2 )( dt tyd + dt tdy )(5 + 6y(t) = dt tdz )( z(t) * An asterisk indicates a challenging problem. Chapter 16, Solution 49. zxyorzyzxxand)t(yxLet 2 ''' 121 +=−=−== Thus, z3x5x6zz2z)zx(5x6zyx 21 '' 21 '' 2 −−−=−+++−−=−′′= This now leads to our state equations, [ ] [ ] )t(z0 x x 01)t(y;)t(z 3 1 x x 56 10 x x 2 1 2 1 ' 2 ' 1 +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − +⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −− = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡
  73. 73. 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 16, Problem 50. * Develop the state equations for the following differential equation. 3 3 )( dt tyd + 2 2 )(6 dt tyd + dt tdy )(11 + 6y(t) = z(t) * An asterisk indicates a challenging problem. Chapter 16, Solution 50. Let x1 = y(t), x2 = .xxand,x ' 23 ' 1 = Thus, )t(zx6x11x6x 321 " 3 +−−−= We can now write our state equations. [ ] [ ] )t(z0 x x x 001)t(y;)t(z 1 0 0 x x x 6116 100 010 x x x 3 2 1 3 2 1 ' 3 ' 2 ' 1 + ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡
  74. 74. 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 16, Problem 51. * Given the following state equation, solve for y(t): . x = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − 02 44 x+ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 0 u(t) y(t) = [ ]01 x * An asterisk indicates a challenging problem. Chapter 16, Solution 51. We transform the state equations into the s-domain and solve using Laplace transforms. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=− s 1 B)s(AX)0(x)s(sX Assume the initial conditions are zero. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +++ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −+ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =− − )s/2( 0 4s2 4s 8s4s 1 s 1 2 0 s2 44s )s(X s 1 B)s(X)AsI( 2 1 222222 221 2)2s( 2 2)2s( )2s( s 1 2)2s( 4s s 1 8s4s 4s s 1 )8s4s(s 8 )s(X)s(Y ++ − + ++ +− += ++ −− += ++ −− += ++ == y(t) = ( )( ) )t(ut2sint2cose1 t2 +− −
  75. 75. 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 16, Problem 52. * Given the following state equation, solve for y1 (t). . x = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − −− 42 12 x + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 04 11 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ )(2 )( tu tu y = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − −− 01 22 x + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −10 02 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ )(2 )( tu tu * An asterisk indicates a challenging problem. Chapter 16, Solution 52. Assume that the initial conditions are zero. Using Laplace transforms we get, ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + −+ ++ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +− + = − s/4 s/3 2s2 14s 10s6s 1 s/2 s/1 04 11 4s2 12s )s(X 2 1 22221 1)3s( 8.1s8.0 s 8.0 )1)3s((s 8s3 X ++ −− += ++ + = 2222 1)3s( 1 6. 1)3s( 3s 8.0 s 8.0 ++ + ++ + −= )t(u)tsine6.0tcose8.08.0()t(x t3t3 1 −− +−= 22222 1)3s( 4.4s4.1 s 4.1 1)3s((s 14s4 X ++ −− += ++ + = 2222 1)3s( 1 2.0 1)3s( 3s 4.1 s 4.1 ++ − ++ + −= )t(u)tsine2.0tcose4.14.1()t(x t3t3 2 −− −−= )t(u)tsine8.0tcose4.44.2( )t(u2)t(x2)t(x2)t(y t3t3 211 −− −+−= +−−= )t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t3 12 −− +−−=−=
  76. 76. 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 16, Problem 53. Show that the parallel RLC circuit shown in Fig. 16.73 is stable. Figure 16.73 For Prob. 16.53. Chapter 16, Solution 53. If oV is the voltage across R, applying KCL at the non-reference node gives o o o o s V sL 1 sC R 1 sL V VsC R V I ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=++= RLCsRsL IsRL sL 1 sC R 1 I V 2 ss o ++ = ++ = RsLRLCs IsL R V I 2 so o ++ == LC1RCss RCs RsLRLCs sL I I )s(H 22 s o ++ = ++ == The roots LC 1 )RC2( 1 RC2 1- s 22,1 −±= both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.
  77. 77. 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 16, Problem 54. A system is formed by cascading two systems as shown in Fig. 16.74. Given that the impulse response of the systems are h1 (t)= 3e t− u(t), h 2 (t) = e t4− u(t) (a) Obtain the impulse response of the overall system. (b) Check if the overall system is stable. Figure 16.74 For Prob. 16.54. Chapter 16, Solution 54. (a) 1s 3 )s(H1 + = , 4s 1 )s(H2 + = )4s)(1s( 3 )s(H)s(H)s(H 21 ++ == [ ] ⎥⎦ ⎤ ⎢⎣ ⎡ + + + == 4s B 1s A )s(H)t(h 1-1- LL 1A = , 1-B = =)t(h )t(u)ee( -4t-t − (b) Since the poles of H(s) all lie in the left half s-plane, the system is stable.
  78. 78. 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 16, Problem 55. Determine whether the op amp circuit in Fig. 16.75 is stable. Figure 16.75 For Prob. 16.55. Chapter 16, Solution 55. Let 1oV be the voltage at the output of the first op amp. sRC 1 R sC1 V V s 1o − = − = , sRC 1 V V 1o o − = 222 s o CRs 1 V V )s(H == 22 CR t )t(h = ∞= ∞→ )t(hlim t , i.e. the output is unbounded. Hence, the circuit is unstable.
  79. 79. 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 16, Problem 56. It is desired to realize the transfer function )( )( 1 2 sV sV = 62 2 2 ++ ss s using the circuit in Fig. 16.76. Choose R = 1 kΩ and find L and C. Figure 16.76 For Prob. 16.56. Chapter 16, Solution 56. LCs1 sL sC 1 sL sC 1 sL sC 1 ||sL 2 + = + ⋅ = RsLRLCs sL LCs1 sL R LCs1 sL V V 2 2 2 1 2 ++ = + + += LC 1 RC 1 ss RC 1 s V V 21 2 +⋅+ ⋅ = Comparing this with the given transfer function, RC 1 2 = , LC 1 6 = If Ω= k1R , == R2 1 C F500 µ == C6 1 L H3.333
  80. 80. 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 16, Problem 57. Design an op amp circuit, using Fig. 16.77, that will realize the following transfer function: )( )(0 sV sV i = - )000,4(2 000,1 + + s s Choose C1 =10 ;Fµ determine R1 , R 2 , and C 2 Figure 16.77 For Prob. 16.57.
  81. 81. 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 16, Solution 57. The circuit is transformed in the s-domain as shown below. 1/sC2 1/sC1 R2 Vi – Vo R1 + Let 1 11 1 1 1 1 1 1 1 1 1 // 1 1 R RsC Z R sC sR CR sC = = = ++ 2 22 2 2 2 2 2 2 2 1 1 // 1 1 R RsC Z R sC sR CR sC = = = ++ This is an inverting amplifier. 2 2 2 1 1 12 2 1 1 1 1 11 1 2 2 2 1 1 2 2 2 2 1 1 1 ( ) 1 1 1 o i R s s Z R R C CsR C R C R CV H s V RZ R R C Cs s sR C R C R C ⎡ ⎤ ⎡ ⎤ − + +⎢ ⎥ ⎢ ⎥−+ ⎢ ⎥ ⎢ ⎥= = − = = − = ⎢ ⎥ ⎢ ⎥+ + ⎢ ⎥ ⎢ ⎥+ ⎣ ⎦ ⎣ ⎦ Comparing this with ( 1000) ( ) 2( 4000) s H s s + = − + we obtain: 1 2 1 2 1/ 2 2 20 C C C F C µ= ⎯⎯→ = = 1 3 6 1 1 1 1 1 1 1000 100 1000 10 10 10 R R C C x x − = ⎯⎯→ = = = Ω 2 3 6 2 2 2 1 1 1 4000 12.8 4000 4 10 20 10 R R C C x x x − = ⎯⎯→ = = = Ω12.5Ω
  82. 82. 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 16, Problem 58. Realize the transfer function )( )(0 sV sV s = 10+s s using the circuit in Fig. 16.78. Let Y1 = sC1 , Y 2 = 1/R1 , Y3 = sC 2 . Choose R1 = 1kΩ and determine C1 and C 2 . Figure 16.78 For Prob. 16.58. Chapter 16, Solution 58. We apply KCL at the noninverting terminal at the op amp. )YY)(V0(Y)0V( 21o3s −−=− o21s3 V)YY(-VY += 21 3 s o YY Y- V V + = Let 11 sCY = , 12 R1Y = , 23 sCY = 11 12 11 2 s o CR1s CsC- R1sC sC- V V + = + = Comparing this with the given transfer function, 1 C C 1 2 = , 10 CR 1 11 = If Ω= k1R1 , === 421 10 1 CC F100 µ
  83. 83. 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 16, Problem 59. Synthesize the transfer function )( )(0 sV sV in = 62 6 10100 10 ++ ss using the topology of Fig. 16.79. Let Y1 = 1/ 1R , Y 2 = 1/R 2 , Y3 = sC1 , Y 4 sC 2 . Choose R1 = 1kΩ and determine C1 , C 2 , and R 2 . Figure 16.79 For Prob. 16.59. Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV = and o32 VVV == . − + Vo + − Vin V2 V1 Y1 Y2 Y4 Y3
  84. 84. 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. At node 1, 4o12o111in Y)VV(Y)VV(Y)VV( −+−=− )YY(V)YYY(VYV 42o42111in +−++= (1) At node 2, 3o2o1 Y)0V(Y)VV( −=− o3221 V)YY(YV += o 2 32 1 V Y YY V + = (2) Substituting (2) into (1), )YY(VV)YYY( Y YY YV 42oo421 2 32 1in +−++⋅ + = )YYYYYYYYYYYYYY(VYYV 42 2 243323142 2 221o21in −−+++++= 43323121 21 in o YYYYYYYY YY V V +++ = 1Y and 2Y must be resistive, while 3Y and 4Y must be capacitive. Let 1 1 R 1 Y = , 2 2 R 1 Y = , 13 sCY = , 24 sCY = 21 2 2 1 1 1 21 21 in o CCs R sC R sC RR 1 RR 1 V V +++ = 2121221 212 2121 in o CCRR 1 CRR RR ss CCRR 1 V V +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⋅+ = Choose Ω= k1R1 , then 6 2121 10 CCRR 1 = and 100 CRR RR 221 21 = + We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is =2R Ωk1 , =1C nF50 , =2C F20 µ
  85. 85. 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 16, Problem 60. Obtain the transfer function of the op amp circuit in Fig. 16.80 in the form of )( )(0 sV sV i = cbss as ++2 where a, b, and c are constants. Determine the constants. Figure 16.80 For Prob. 16.67.
  86. 86. 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 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);
  87. 87. 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 16, Problem 61. A certain network has an input admittance Y(s). The admittance has a pole at s = -3, a zero at s = -1, and Y(∞) = 0.25 S. (a) Find Y(s). (b) An 8-V battery is connected to the network via a switch. If the switch is closed at t = 0, find the current i(t) through Y(s) using the Laplace transform. Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);
  88. 88. 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 16, Problem 62. A gyrator is a device for simulating an inductor in a network. A basic gyrator circuit is shown in Fig. 16.81. By finding Vi (s)/I0 (s), show that the inductance produced by the gyrator is L = CR 2 . Figure 16.81 For Prob. 16.69. Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);

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