hw4_solutions

hw4_solutions - Homework 4 Solutions ECE 2B Winter 2011...

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Unformatted text preview: Homework 4 Solutions ECE 2B Winter 2011 H.O. #13 Problem 10.13 Obtain the inverse Laplace transform of each of the following functions: (s + 2)2 (a) F1 (s) = s(s + 1)3 1 (b) F2 (s) = 2 (s + 4s + 5)2 √ 2(s + 1) (c) F3 (s) = 2 s + 6s + 13 −2(s2 + 20) (d) F4 (s) = 2 s(s + 8s + 20) Solution: (a) F1 (s) = (s + 2)2 s(s + 1)3 A B3 B2 B1 =+ + + , s (s + 1)3 (s + 1)2 s + 1 (s + 2)2 (s + 1)3 with A = s F1 (s)|s=0 = = 4, s= 0 B3 = (s + 1)3 F1 (s)|s=−1 = B2 = B1 = d (s + 2)2 ds s = s= − 1 (s + 2)2 s 2(s + 2) (s + 2)2 − s s2 s= − 1 = −1, = −3, d 2(s + 2) (s + 2)2 − ds s s2 s= − 1 2 2(s + 2) 2(s + 2) 2(s + 2)2 = − − + s s2 s2 s3 Hence, F1 (s) = and s= − 1 s= − 1 = −8. 4 1 3 8 − − − , s (s + 1)3 (s + 1)2 s + 1 1 f1 (t ) = 4 − t 2 e−t − 3t e−t − 8e−t u(t ) 2 = 4− (b) F2 (s) = s2 + 4s + 5 = 0 1 (s2 + 4s + 5)2 √ −4 − 16 − 20 s1 = = −2 − j 1 2 s2 = −2 + j1. ©2009 National Technology and Science Press t2 + 3t + 8 e−t u(t ). 2 All rights reserved. Do not reproduce or distribute. By application of the distinct complex-pole formula in Table 10-3, we have f3 (t ) = 2e−3t cos(2t + 45◦ ) u(t ). (d) F4 (s) = s2 + 8s + 20 = 0 −2(s2 + 20) s(s2 + 8s + 20) √ −8 − 64 − 80 s1 = = −4 − j 2 2 s2 = −4 + j2. Hence, F4 (s) = with A = s F4 (s)|s=0 = A B B∗ + = , s s + 4 + j2 s + 4 − j2 −2(s2 + 20) s2 + 8s + 20 s= 0 = −2, = 4e j90 . s=−4− j2 ◦ ◦ B = (s + 4 + j2) F4 (s)|s=−4− j2 = Hence, F4 (s) = and −2(s2 + 20) s(s + 4 − j2) ◦ −2 4e j90 4e− j90 + + , s s + 4 + j2 s + 4 − j2 f4 (t ) = [−2 + 8e−4t cos(2t − 90◦ )] u(t ) = [−2 + 8e−4t sin 2t ] u(t ). Problem 10.8 Determine f (0+ ) and f (!), given that F(s) = s2 + 4 . 2s3 + 4s2 + 10s Solution: f (0+ ) = lim s F(s) = lim s→! s→! s(s2 + 4) 1 = 2s3 + 4s2 + 10s 2 (s2 + 4) 2s2 + 4s + 10 = 0.4. ©2009 National Technology and Science Press All rights reserved. Do (!)reproduce Fr(distribute. f not = lim s o s) = lim s→0 s→0 Problem 10.17 Determine vout (t ) in the circuit in Fig. P10.17, given that vs (t ) = 35u(t ) V, vC1 (0− ) = 20 V, R1 = 1 !, C1 = 1 F, R2 = 0.5 !, and C2 = 2 F. R1 vs(t) _ + vC1 C1 + C2 R2 vout _ Figure P10.17: Circuit for Problem 10.17. Solution: The s-domain circuit is shown in Fig. P10.17(a). R1 + _ 35 s 1 sC _ + V Vs = 20 s + C2 R2 Vout _ Figure P10.17(a) At node V: V− 35 − 20 V s + sC2 V + = 0, 1 R2 R1 + sC1 which leads to V= 15C1 R2 (R2C1 + R2C2 + R1C1 )s 1 R1 R2C1C2 s + + R1 R2C1C2 R1 R2C1C2 7.5 7.5 =2 = . s + 2.5s + 1 (s + 2) s + 1 2 2 Hence, V= A1 A2 + , s+2 s+ 1 2 7.5 s+ 1 2 = 1 s= − 2 A1 = (s + 2)V|s=−2 = A2 = Hence, V= and s+ 1 V 2 7.5 s+2 s= − 2 = −5, = 5. 1 s= − 2 −5 5 + , s+2 s+ 1 2 vout (t ) = v(t ) = 5(e−t /2 − e−2t ) u(t ). ©2009 National Technology and Science Press All rights reserved. Do not reproduce or distribute. Problem 10.18 Determine iL (t ) in the circuit of Fig. P10.18 for t ≥ 0, given that the switch was opened at t = 0, after it had been closed for a long time, vs = 12 mV, R0 = 5 !, R1 = 10 !, R2 = 20 !, L = 0.2 H, and C = 6 mF. R0 t=0 L C R1 iL 12 mV _ + R2 Figure P10.18: Circuit for Problems 10.18 and 10.19. Solution: The s-domain equivalent circuit is shown Fig. P10.18(a). R0 + R1 12 10−3 _ s V iL sL 1 sC + R2 Figure P10.18(a) Prior to t = 0, no currents or voltages existed in the part of the circuit to the right of the branch containing the switch. After t = 0, at node V: V− 12 × 10−3 V s ++ R0 + R1 sL V R2 + 1 sC = 0, which leads to 1.44s + 12 V= 210 1 1000s 7500 2 s+ + 21 21 1.44s + 12 1 = . 210 (s + 38.3)(s + 9.3) V 5V 1.44s + 12 = = . sL s 42s(s + 38.3)(s + 9.3) Hence, IL = Partial fraction expansion gives IL = A1 A2 A3 + + , s s + 38.3 s + 9.3 All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press with A1 = sIL |s=0 = 1.44s + 12 42(s + 38.3)(s + 9.3) = 8.02 × 10−4 , s=−38.3 s=−9.3 s= 0 A2 = (s + 38.3)IL |s=−38.3 = A3 = (s + 9.3)IL |s=−9.3 = Hence, IL = and 1.44s + 12 42s(s + 9.3) = −9.25 × 10−4 , 1.44s + 12 42s(s + 38.3) = 1.23 × 10−4 . 8.02 9.25 1.23 − + × 10−4 , s s + 38.3 s + 9.3 iL (t ) = [8.02 − 9.25e−38.3t + 1.23e−9.3t ] × 10−4 A. Problem 10.42 When excited by a unit step function at t = 0, a system generates the output response y(t ) = [5 − 10t + 20 sin 2t ] u(t ). Determine: (a) the system transfer function, and (b) the impulse response. Solution: (a) Transforming y(t ) into the s-domain gives Y(s) = With 1 , s Y(s) H(s) = X(s) X(s) = = 5− 10 40s 5(s3 + 6s2 + 4s − 8) +2 = . s s +4 s(s2 + 4) 40 5 10 −+ . s s2 s2 + 4 (b) To obtain h(t ), we start with the expression for H(s) given by H(s) = 5 − 10 40s +2 . s s +4 By comparison with entries #1, #2, and #11 in Table 10-2, we have h(t ) = 5! (t ) − 10[1 − 4 cos 2t ] u(t ). All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 10.30 The circuit in Fig. P10.30 is excited by a 10-V, 1-s–long rectangular pulse. Determine i(t ), given that R1 = 1 !, R2 = 2 !, and L = 1/3 H. R1 10 V vs(t) = 0 1s + _ L R2 i(t) Figure P10.30: Circuit for Problems 10.30 and 10.57. Solution: The input excitation is given by vs (t ) = 10[u(t ) − u(t − 1)] In the s-domain, Vs = (V). 10 10 −s − e. s s We will apply the superposition principle to determine I1 due to the first component of Vs and I2 due to the second component. That is, V s1 = 10 For Vs1 = . s current division leads to I1 = and i1 (t ) = 10 For Vs2 = − e−s , s I2 = 10 , s V s2 = − 10e−s . s 10 , 3(s + 2) (A). 10 −2t e u(t ) 3 −10 −s e, 3(s + 2) and by entry #3a in Table 10-2, i2 (t ) = − Hence, i(t ) = 10 −2(t −1) e u(t − 1) 3 (A). 10 −2t [e u(t ) − e−2(t −1) u(t − 1)] 3 (A). All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press with B2 = s2 Y2 (s)|s=0 = 36s + 20 s2 + 6s + 5 = 4, s= 0 Section 10-8d Transfer Function3and Imp2slse6)(36son2e ) :2 6 ( u + Resp + s 0 36s + 20 DetermiAe = e s +tput Y2spo|sse y(= , if the input exc=a1i.on is given by: n2 th ( ou 5) re (s) n=−5 t ) it t 6. s2 (s + 1) s=−5 (a) x1 (t ) = u(t ), ( n) , Hebcex2 (t ) = 2t u(t ), (c) x3 (t ) = 2e−4t u(t ),Y2 (s) = 2.4 + 4 − 4 + 1.6 , s s2 s + 1 s + 5 (d) x4 (t ) = [4 cos 4t ] u(t ). and y2 (t ) = [2.4 + 4t − 4e−t + 1.6e−5t ] u(t ). Solution: ) (a) (c B1 = s Y2 (s) s=0 = 2 − = 2.4, 2 2 s c 6s + 5 s iv Problem 10.4d s A system is chara+terized by a t(s n+ ers function g=0en by 1 ra sf 6 + 5) 36s + 20 A1 = (s + 1) Y2 (s)|s=−1H(s) 2 18s + 10 = −4, == s (s s2 + 6s=−1 . + 5) s + 5 1 X1 (s) = 2 , X3 (s) = s , s+4 18s + 10 Y1 (s) = H(s) X1 (s) = 2 36s + 20 Y3 (s) = H(s) X3 (s) = s(s + 6s + 5) . (s + 1)(s + 4)(s + 5) 18s + 10 = . Application of partial fraction expansion gives s(s + 1)(s + 5) x (t ) = u(t ) x3 (t ) =12e−4t u(t ) A1 A A3 Application of partial fraction expansion gives2 Y3 (s) = + + , s+1 s+4 s+5 A1 A2 A3 Y1 (s) = + + , with s s+1 s+5 with Hence, Hence, and and (d) b 36s + 20 4 A1 = (s + 1) Y3 (s)|s=−1 = =− , (s + 4 3 18s + 10 )(s + 5) s=−1 A1 = s Y1 (s)|s=0 = 2 = 2, s + 6s3+s5 s20 6 + =0 124 A2 = (s + 4) Y3 (s)|s=−4 = = , (s + 8s(s +0 ) s=−4 3 11) + 1 5 A2 = (s + 1) Y1 (s)|s=−1 = = 2, 3s(s+ 20 s=−1 6s + 5) A3 = (s + 5) Y3 (s)|s=−5 = = −40. (s + 8s(s +0 ) s=−5 11) + 1 4 A3 = (s + 5) Y1 (s)|s=−5 = = −4. s(s + 1) s=−5 −4 124 40 Y3 (s) = + − , 3(s +21) 2(s + 4)4 s + 5 3 Y1 (s) = + − , s s+1 s+5 4 −t 124 −4t y3 (t ) = − e + −t e −5t40e−5t u(t ). − 3 y1 (t )3 [2 + 2e − 4e ] u(t ). = 4s 2 X4 (X2= ) = 2 , , s) (s 2 s + 16 s 34s(18s0+ 10) 6 +2 . . Y4 (Y2= ) =sHX4 (X2= ) = 2 s) (s H( ) (s) s) (s s )( + 1) )( + 5 j (s + 1(s s + 5(s s + )4)(s − j4) ©2009 National Technology and Science Press x4 (t ) = [4 co(t4t= 2t )u(t ) x2 s ) ] u( Application of partial fraction expansion gives All rights reserved. Do not reproduce or distribute. B1 B2 A1 A2 + 2+ + , Y2 (s) = s s s+1 s+5 All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 10.45 For the circuit shown in Fig. P10.45, determine (a) H(s) = Vo /Vi and (b) h(t ), given that R = 5 !, L = 0.1 mH, and C = 1 µ F. L + Vi + C R Vo _ _ Figure P10.45: Circuit for Problem 10.45. Solution: (a) Application of KVL leads to: H(s) = 1 Vo = Vi 1 + sL/R + s2 LC 1010 s2 + 2 × 105 s + 1010 1010 = . (s + 105 )2 = (b) Use of entry #8 in Table 10-2 leads to h(t ) = t e−at u(t ), with a = 105 . All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press ...
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