hw6_solutions

hw6_solutions - Homework 6 Solutions ECE 2B Winter 2011...

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Unformatted text preview: Homework 6 Solutions ECE 2B Winter 2011 H.O. #20 Problem 9.2 Determine the resonant frequency of the circuit shown in Fig. P9.2, given that R = 100 !, L = 5 mH, and C = 1 µ F. C Zi L R Figure P9.2: Circuit for Problem 9.2. Solution: Zi = (R 1 j" C j " LR 1 = + R + j" L j" C j " L) + = −" 2 RLC + R + j" L . −" 2 LC + j" RC After a few steps of algebra, the expression simplifies to Zi = " 4 RL2C2 − j" [" 2 (L2C − R2 LC2 ) + R2C] . " 4 L2C2 + " 2 R2C2 At resonance, the imaginary part of Zi is zero. Thus, " [" 2 (L2C − R2 LC2 ) + R2C] = 0, which gives "0 = 0 (trivial resonance) and "0 = = R2 R2 LC − L2 104 104 × 5 × 10−3 × 10−6 − 25 × 10−6 = 2 × 104 rad/s. All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 9.3 For the circuit shown in Fig. P9.3, determine (a) the transfer function H = Vo /Vi , and (b) the frequency !o at which H is purely real. C V1 L1 L2 + Vi + R Vo _ _ Figure P9.3: Circuit for Problem 9.3. Solution: (a) KCL at node V1 gives: V1 − Vi V1 V1 + + = 0, ZC Z L1 R + Z L2 where ZC = 1/ j!C, ZL1 = j! L1 , and ZL2 = j! L2 . Also, voltage division gives Vo = Solving for the transfer function gives H= Vo −! 2 RL1C = . 2 L C ) + j! (L + L − ! 2 L L C ) Vi R(1 − ! 1 1 2 12 V1 R . R + j ! L2 (b) We need to rationalize the expression for H: H= −! 2 RL1C R(1 − ! 2 L1C) + j! (L1 + L2 − ! 2 L1 L2C) × = −! 2 R2 L1C(1 − ! 2 L1C) + j! 3 RL1C(L1 + L2 − ! 2 L1 L2C) . R2 (1 − ! 2 L1C)2 + ! 2 (L1 + L2 − ! 2 L1 L2C)2 R(1 − ! 2 L1C) − j! (L1 + L2 − ! 2 L1 L2C) R(1 − ! 2 L1C) − j! (L1 + L2 − ! 2 L1 L2C) The imaginary part of H is zero if ! = 0 (trivial solution) or if !0 = L1 + L2 . L1 L2C All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 9.21 A series RLC bandpass filter has half-power frequencies at 1 kHz and 10 kHz. If the input impedance at resonance is 6 !, what are the values of R, L, and C? Solution: At resonance, the input impedance of the series RLC circuit is equal to R. Hence, R = 6 !. The bandwidth in Hz is BHz = (10 − 1) kHz = 9 kHz. B = 2" BHz = From Eq. (9.50a): R L =⇒ L= R 6 = = 0.106 mH. 2" BHz 2" × 9 × 103 #c1 = − R + 2L R 2L B 2 2 2 + 1 LC B =− + 2 Converting to Hz, fc1 = − B Hz + 2 2 + #0 . B Hz 2 2 2 + f0 . With fc1 = 1 kHz and BHz = 9 kHz, solution leads to f0 = 3.1623 kHz, and C= 1 1 1 = = = 23.6 µ F. 2 2 #0 L 4" 2 f0 L 4" 2 × (3.1623)2 × 106 × 1.06 × 10−4 Problem 9.24 Design a parallel RLC filter with f0 = 4 kHz, Q = 100, and an input impedance of 25 k! at resonance. Solution: The input impedance of the parallel RLC circuit is: Zin = 1 1 + + j" C R j" L −1 At resonance, "0 L = 1/"0C, and Zin ("0 ) = R. Hence, R = 25 k!. From Table 9-3, All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Q= From R "0 L =⇒ L= R 25 × 103 25 = = "0 Q 2# × 4 × 103 × 100 800# 10 mH. 2 "0 = 1 LC =⇒ C= 1 1 = = 0.16 µ F. 2 "0 L (2# × 4 × 103 )2 × 10−2 F3 (s) = 10 2 − ! = 10# = 10 tan s . Alternatively, using the formal definition for the Laplace transform F3 (s) = Integral table provides: $ sin bt −at −b dt or ta o 1 Problem 10.6 Determine the Laplace transf= m n f the,following functions: te a 0 ◦ ) " (t ) (a) f1 (t ) = 25 cos(4! t + 30 which when applied to F3 (s) l◦ ads to e (b) f2 (t ) = 25 cos(4! t + 30 ) " (t − 0.2) sin 3t 3 (c) f3 (t ) = 10 u(t ) F3 (s) = 10 tan−1 . t s 2 d (d) f4 (t ) = 2 [e−4t u(t )] (d) dt d2 d (t 3= ( e−4t (e) f5 (t ) = [4t e−2t cos(4!ft4+ ) 0◦ )du2t[)] u(t )]. t dt App)icf6t(o) =feproport(4t +n3Ta) lu(t ) -1 leads to (f l a it n o −3t ce s y 7 i 0◦ b e 10 $ 0− 10 sin 3t −st e dt . t (g) f7 (t ) = t 2 [u(t ) − u(t − 4)] 1 2 −4t (h) f8 (t ) = 10 coF(6s)t = s 0◦ )+ 4t− s (2) ) s 4 (! + 3 s " ( − 0. e t =0 − (−4e−4t ) t =0 s2 16 Solution: = −s+4 = . (a) f1 (t ) = 25 cos(4! t + 30◦s)+ 4t ) "( s+4 By Eq. (10.10), (e) # # d −st F1 (s) = f t e−st [ t e− c 2 ( co t + 3 ◦ ) u(◦ f5i((t))= d4t= 2t −os54"s(4! t0+ 30 t )]" (t ) e dt 0− 0 dt = 25 cos(30◦ ) = 21.65. Let: (b) fa (t ) = 4 cos(4" t + 30◦ ) u(t ), f2 (())= 2e−2t sf 4!),+ 30◦ ) " (t − 0.2) f t t = 5 co ( (t t b fc (t = (t . F2 (s))= t fb25)cos(4! t + 30◦ ) " (t − 0.2) e−st dt 0− # a (c) d f5 (t )= 25 cos(4) × 0.2 + 30◦ )e−0.2s = fc t ! dt = 24(s cos 38! − 4"◦sie−002s)= 23.cos(−725.)13 0.2s = −24.86e−0.2s . 5 cos(0. 0◦ + 30 ) n 3 . ◦ 5 46s 1 4◦ e− Fa (s) = =2 s2 + (4" )2 s + 157.9 Fb (s) = Fa (s + 2) = 3.46(s + 2) 1025n13 − si . 3t [Property 5, Table 10-1] 3 (t ) = (sf+ 2) 2 + 157.9 u(t ) t = 1 fc (t ) t All rights reserved. Do not reproduce or distribute.d d 3.46(s + 2) − 25.13 with Fc (s) = − Fb (s) = − d s fc (t ) = 10d sn 3t (s(+,2)2 + 157.9 si u t) By property 9 in Table 10-1, ©2009 National Technology and Science Press 3.46s2 + 36.41s − 633.72 =3 . 3 20 2 Fc (s) = 10 2 [(s + 2) + 157.9] = . s + 9 s2 + 9 Application of property 6 in Table 10-1 leads to Hence, by property 10 in Table 10-1, F5 (s) = s Fc (s) −# c (0− ) f 30 F3 (s) = d. 3 + 36.41s2 − 633s72s 3.46s s . s2 + 9 −0 = [(s + 2)2 + 157.9]2 Integral table provides 1 3.46s3 + 36.411 2 − −1 3.72s s 63 x = x2 + a2 d x = a tan 2 . . [(s + 2)2 + 157.9] a Hence, (f) ! 30 −1 s # s F3 (s) = tan = 10 − tan−1 . f6 (3) = e−3t cos(4t + 30◦ ) u(t ) 3 t 3s 2 and its transform is with = e−3t fa (t ), All rights reserved. Do not reproduce or distribute. fa (t ) = cos(4t + 30 ) u(t ). ◦ ©2009 National Technology and Science Press The Laplace transform of fa (t ) is s cos 30◦ − 4 sin 30◦ [Entry #12 in Table 10-2]. s2 + 16 Application of property 5 in Table 10-1 leads to Fa (s) = F f (s) = Fa (s + 3) = (g) f7 (t ) = t 2 [u(t ) − u(t − 4)] = t 2 u(t ) − t 2 u(t − 4). (s + 3) cos 30◦ − 2 0.866(s + 3) − 2 = . (s + 3)2 + 16 (s + 3)2 + 16 To convert the second term into a form f (t − T ) u(t − T ), we add and subtract from it (8t − 16): f7 (t ) = t 2 u(t ) − (t − 4)2 u(t − 4) − 8(t − 4) u(t − 4) − 16u(t − 4). F7 (s) = (h) f8 (t ) = 10 cos(6! t + 30◦ ) " (t − 0.2) # Hence, 2 2e−4s 8e−4s 16e−4s − 3− 2− . s3 s s s Problem 10.14 functions: (a) F1 (s) = 2 + (b) F2 (s) = (c) F3 (s) = (d) F4 (s) = (ec , Hen) eF5 (s) = Obtain the inverse Laplace transform of each of the following (f) F6 (s) = and −2t −10t ) u(t ) + [e−2(t −4) − 25e−10(t −4) ] u(t − 4). Solution:f4 (t ) = (−e + 25e a (e) s(s − 8)e−6s Fs( ) 2 4(s − 4) F1 (5 )s== +s + 2)(s2 + 16) ( s2 + 16 with Fe (s) = = e−6s F4(s), 4) e− = 2+ (s + j4)(s − j4) 4(s − 4) s2 + 16 4 4s + s s2 + 9 (s + 5)e−2s (s + 1)(s + 3) (1 − e−4s )(24s + 40) (s + 2)(s + 10) s(s − 8)e−6s (s + 2)(s2 + 16) 1 25 e−4s 25e−4s F4 ( 4s = + + − 4s(2 − e−s) ) − s + 2 s + 10 s + 2 s + 10 s2 + 9 with A B B∗ =) 4(s − 4s + 2 + s + j4 + s − j4 , Fa (s) = (s + j4)(s − j4) where B B∗ s = + , (s − 8) A = (s + 2) +ej(4 )|s=−2 = 2 F s s − j4 = 1, s s + 16 s=−2 4(s − 4) ◦ B = (s + j4) Fa (s)| =(s − 8) = 2(−− () =42.88) − j45 . 1 j4 j − j − 3e s B = (s + j4) Fe (s)|s=− j4 s=− j4 (s − j4) s=− j4 = = = − j. (s + 2)(s − j4) s=− j4 (− j4 + 2)(− j8) Hence, Hence, ◦ − 45◦ 2.83e− jj90◦ 2.j83◦e j45 F1 (−) s= 2 + s6 + e 90 , 1 e s F5 (s) = e ++ j 4 + s − j 4 s + 2 s + j4 s − j4 and t) ◦◦ f5 (t )f= t[) −2(2−6(t+ 2 5os64(ts− 6) + 90) )](u). − 6) 1 ( e = ! ) + c.6 ( co (4t + 45 u t (t = [e−2(t −6) − 2 sin(4(t − 6))] u(t − 6). =2 s(s − 8) + Fa (s) s(s − 8) = (s + 2)(s2 + 16) (s + 2)(s + j4)(s − j4) (b) (f) with with 4 4s F2 (s) = + −4s ss 4s(2 − e 2 +)9 F6 (s) = 42 + 9 = s + Fb (s), s = (2 − e−4s ) F f (s), F fb(s) = F where 4s 4s = 2 s2 + 9 (s + j3)(s − j3) B B∗ = + , All rights reserved. Do not reproduce or distribute. s + j3 s − j3 B = (s + j3) F f (s) s= − j 3 ©2009 National Technology and Science Press = Hence, F6 (s) = (2 − e−4s ) and 4s s − j3 = 2. s= − j 3 2 2 + , s + j3 s − j3 f6 (t ) = 8 cos 3t u(t ) − 4 cos 3(t − 4) u(t − 4). All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 10.34 Given the current-source waveform displayed in Fig. P10.34, determine iL (t ) in the circuit of Fig. P10.33, given that R1 = 10 !, R2 = 5 !, L = 0.6196 H, and LC = 1/15 s. is(t) 6A 6e−2t 0 0 t Figure P10.34: Waveform for Problem 10.34. Solution: The current source is given by is (t ) = 6A 6e−2t A for t ≤ 0 for t ≥ 0. The fact that is (t ) was not zero prior to t = 0 implies that the initial conditions for C and L may not be zero. Setting C as an open circuit and L as a short circuit, the circuit state at t = 0− is as shown in Fig. P10.34(a). + 6A R1 vC(0−_ ) R2 iL(0−) L Figure P10.34(a) We deduce that vC (0− ) = 6R1 = 6 × 10 = 60 V, iL (0− ) = 0. Figure P10.34(b) shows the s-domain circuit. 1 sC 6 s+2 60 s + I1 IL R2 I 2 sL R1 Figure P10.34(b) I1 R1 + 1 6R1 60 + R2 − I2 R2 = − sC s+2 s ©2009 National Technology and Science Press All rights reserved. Do not reproduce or distribute. _ −I1 R2 + I2 (R2 + sL) = 0. Simultaneous solution leads to IL = I2 = [6R1 − 60)s − 120]R2C (s + 2)[(R1 + R2 )LCs2 + (L + R1 R2C)s + R2 ] −64.56 = (s + 2)(s2 + 6s + 5) −64.56 = . (s + 2)(s + 1)(s + 5) Partial fraction expansion leads to IL = with A1 = (s + 2) IL |s=−2 = A2 = (s + 1) IL |s=−1 = A3 = (s + 5) IL |s=−5 = Hence, IL = and −64.56 (s + 1)(s + 5) −64.56 (s + 2)(s + 5) −64.56 (s + 1)(s + 2) = 21.52, s= − 2 s= − 1 s= − 5 A1 A2 A3 + + , (s + 2) (s + 1) (s + 5) = −16.14, = −5.38. 21.52 16.14 5.38 − − , (s + 2) (s + 1) (s + 5) (A). iL (t ) = (21.52e−2t − 16.14e−t − 5.38e−5t ) u(t ) Problem 10.47 For the circuit shown in Fig. P10.47, determine (a) H(s) = Vo /Vs and (b) h(t ), given that R1 = R2 = 100 ! and C1 = C2 = 1 µ F. Op-amp circuit for Problem 10.47. C1 _ R2 C2 R1 Vs _ + + + Vo _ Figure P10.47: Op-amp circuit for Problem 10.47. Solution: (a) Application of KVL and KCL to the circuit under ideal op-amp conditions leads to All rights reserved. Do not reproduce or distribute. Vo H(s) = Vs = = 1 1 + (R1 + R2 )C2 s + R1 R2C1C2 s2 108 s2 + 2 × 104 s + 108 h(t ) = t e−at u(t ), = 108 . (s + 104 )2 ©2009 National Technology and Science Press (b) Per entry #8 in Table 10-2, with a = 104 s−1 . ...
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This note was uploaded on 04/19/2011 for the course ECE 2B taught by Professor York during the Spring '07 term at UCSB.

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