hw7_solutions

hw7_solutions - Homework 7 Solutions ECE 2B Winter2011...

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Unformatted text preview: Homework 7 Solutions ECE 2B Winter2011 H.O. #22 Problem 9.7 For the circuit shown in Fig. P9.7: (a) Obtain an expression for the input impedance Zin (! ). (b) If R1 = R2 = 1 ", C = 1 F, and L = 5 H, at what angular frequency is Zin purely real? (c) Scale the circuit by Km = 20 and write down the new expression for the input impedance. (d) Is the value of ! at which the input impedance of the scaled circuit the same or different from the answer of part (b)? R1 Zin L C R2 Figure P9.7: Circuit for Problem 9.7. Solution: (a) Zin = R1 + j! L = R1 + = R1 + R2 + 1 j! C −! 2 R2 LC + j! L (1 − ! 2 LC) + j! R2C (b) Resonance occurs when imaginary part of Zin is zero, or ! 4 R2 L2C2 + j! L[1 − ! 2 (LC − R2C2 )] 2 . (1 − ! 2 LC)2 + ! 2 R2C2 2 ! L[1 − ! 2 (LC − R2C2 )] = 0, 2 which is satisfied when ! = 0 (trivial resonance) or 1 − ! 2 (LC − R2C2 ) = 0. 2 Hence, 1 LC − R2C2 2 1 = 0.5 rad/s. 5 × 1 − 12 × 12 L = 5 × 20 = 100 H, 1 20 2 !0 = = (c) With Km = 20, R1 = R2 = 20 ", C= C 1 = F. Km 20 100 202 −2 20 20 Zin = 20 + 20 × 1002 × ! 4 + j100! 1 − ! 2 1− 100 2 ! 20 2 + 202 2 ! 202 = 20 + 500! 4 + j100! [1 − 4! 2 ] . (1 − 5! 2 )2 + ! 2 ©2009 National Technology and Science Press All rights reserved. Do not reproduce or distribute. (d) Resonance occurs when or 2 1 − 4!0 = 0, !0 = which is the same as in part (b). 1 = 0.5 rad/s, 4 −20 −36.5 −40 −56.5 −60 −56.5 dB −20 log |1 + jω/500| −20 log |1 + jω/100| Problem 9.13 Generate Bode magnitude and phase plots for the following voltage Magnitude transfer functions: φH(! ) = j100! (a) 10 + j! 90o 0.4(50 + j! )2 (1 + jω/10) (b) H(! ) = ( j! )2 (40 + j80! ) 45o (c) H(! ) = (10 + j50! ) φ(ω) (20 + j5! )(20 + j! ) (d0 H(! ) = ) ω (rad/s) 1 10 j! 100 1000 10000 30(10 + j! ) (e) H(! ) = (200 + j2! )(1000 + j2! ) 1 −45o j100! (1 + jω/500) 1 (f) H(! ) = (100 + j5! )(100 + (1 + 2jω/100) j! ) −90o (200 + j2! ) (g) H(! ) = (50 + j5! )(1000 + j! ) Phase Solution: j100! j100! j100! j10!j10−4 ! f H(! ) = == (a)) H(! ) = = . (10++!5! )10(1 + jj! /10) (1 1 + !! 20)(1 + j! /100)2 (100 + ! )2 + j j / /10 10 0 j j • Constant ferm r 10 4 =⇒ +280 dB t acto 10− =⇒ −0 dB • Zero @ origin • Simple pole with !c = 10 rad/s 2 All rights reserved. Do not reproduce or distribute. • Simple pole with !c = 100 rad/s, of order 2 ©2009 National Technology and Science Press M [dB] = 20 log |H| 20 log ω dB 20 = 20 log 10 + 20 log ! − 20 log |1 + j! /10| 0 1 10 100 1000 −40 log |1 + jω/100| 10000 ω (rad/s) −20 −20 log |1 + jω/20| −40 −54 −60 M [dB] −80 −94 −100 Magnitude φ 90o All rights reserved. Do not reproduce or distribute. −80 dB j ©2009 National Technology and Science Press 45o φ(ω) ω (rad/s) 0 1 1 (1 + jω/100) 10 100 1000 −45o 1 (1 + jω/20) −90o −180o Phase All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press −20 −20 log ω3 −36 −40 Magnitude φ 450o M [dB] −20 log |1 + jω/500| Problem 9.14 Generate Bode magnitude and phase plots for the following voltage transfer functions: 360o 4 × 104 (60 + j6! ) (a) H(! ) = (4 + j2! )(100 + j2! )(4φ(ω) j4! ) 00 + (1 + j0.2! )2 (100 + j2! )2 270o (b) H(! ) = ( j! )3 (500 + j! ) 8 × 10−2 (10 + j10! ) (1 + jω/5)2 (c 180o) H(! ) = j! (16 − ! 2 + j4! ) (1 + jω/50)2 4 × 104 ! 2 (100 − ! 2 + j50! ) (d) H(! ) = (5 + j5! )(200 + j2! )3 90o j5 × 103 ! (20 + j2! ) (e) H(! ) = (2500 − ! 2 + j20! ) 0 ) 512(1 + j! )(4 + j40!10 0.1 1 100 1000 (f) H(! ) = 1 (256 − ! 2 + j32! )2 1 + jω/500 j(10 + j! ) × 108 −90o (g) H(! ) = (20 + j! )2 (500 + j! )(1000 + j! ) Phase j 10,000 ω (rad/s) Solution: (c) (a) H(! ) = = = 8 × 10−2 (10 + j10! ) H(!4 × 104 (60 + j6! ) )= j! (16 − ! 2 + j4! ) (4 + j2! )(100 + j2! )(400 + j4! ) 4 × 100 × 400(1 + j! /2)(1 + j! /50)(1 + j! /100) −2 (1 + j! /2)(1 + j! /50)(1 + j! /100) 5 × 10−2 − j5 × 10 (1 + j! ) = 5(1 + j! /10) 1 ! [1 − (! /4)2 + j! /4] − j8 × 10−2 × 10(1 + j! ) = 4 × 104 × 60(1 + j! /10) 16! [1 − (! /4)2 + j! /4] . • Constant term 15 • Constant factor • Simple pole with !c = 2 rad/s • Pole @ origin 20 • Zero factor with !c = 1 rad/s =⇒ 23.5 dB =⇒ −26 dB • Quadratic pole with !c = 4 rad/s • Simple zero with !c = 10 rad/s •dB imple pole with !c = 50 rad/s S 20 log |1 + jω| − pole ω • Simple 20 logwith !c = 100 rad/s 1 10 100 1000 ω (rad/s) 0 0.1 −6 −20 −26 −40 −26 dB −60 −66 M [dB] −80 Magnitude φ 90o (1 + jω) Quadratic pole 0 0.1 1 10 1 2 + jω/4 1 − (ω/4) 100 1000 ω (rad/s) φ(ω) −90o −j −180o Phase ( d) H(! ) = = 4 × 104 ! 2 (100 − ! 2 + j 50! ) (5 + j 5! )(200 + j 2! )3 0.1! 2 [1 − (! /10)2 + j ! /2] 4 × 104 × 100! 2 [1 − (! /10)2 + j ! /2] = 5 × (200)3 (1 + j ! )(1 + j ! /100)3 (1 + j ! )(1 + j ! /100)3 = ⇒ −20 dB • Constant factor 0.1 1 5 10 50 100 500 1000 −180o Phase Quadratic (f) H(! ) = 512(1 + j! )(4 + j40! ) (256 − ! 2 + j32! )2 512 × 4(1 + j! )(1 + j! /0.1) 2562 [1 − (! /16)2 + j! /8]2 (1 + j! /0.1)(1 + j! ) 32[1 − (! /16)2 + j! /8]2 −30.1 dB = = • Constant term 1/32 =⇒ • Simple zero with !c = 0.1 rad/s • Simple zero with !c = 1 rad/s • Quadratic pole of order 2 with !c = 16 rad/s dB 40 20 16 0.01 0.1 1 10 100 160 1000 0 ω (rad/s) −20 −30.1 dB −40 M [dB] −60 −80 dB/decade −80 Magnitude φ 180o 90o 1.6 0.01 0.1 1 10 100 160 1000 0 ω (rad/s) −90o φ(ω) −180o −270o Quadratic pole, second order −360o Phase (g) H(! ) = j(10 + j! ) × 108 (20 + j! )2 (500 + j! )(1000 + j! ) Problem 9.15 Determine the voltage transfer function H(! ) corresponding to the Bode magnitude plot shown in Fig. P9.15. M [dB] 26 dB 20 dB 10 dB 6 dB 0 0.1 1 100 1000 ω (rad/s) Figure P9.15: Bode magnitude diagram for Problem 9.15. Solution: The transfer function consists of: (1) A constant term K (2) A zero @ origin of order 1 (slope is 20 dB/decade) (3) A simple pole with !c = 1 rad/s (slope reduces to 0 dB at !c = 1 rad/s) (4) A simple pole with !c = 100 rad/s (slope changes to −20 dB/decade at !c = 100 rad/s) Hence, H(! ) = ( j)N K ! . (1 + j! )(1 + j! /100) At ! = 1 rad/s, the only term that contributes to M [dB] is K . Thus, 26 dB = 20 = log K , or K = 1026/20 = 20. The expression for H(! ) is therefore given by H(! ) = = ( j)N 20! (1 + j! )(1 + j! /100) ( j)N 2000! . (1 + j! )(100 + j! ) The value of N depends on the phase of H(! ), which is not available. All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press Problem 9.16 Determine the voltage transfer function H(! ) corresponding to the Bode magnitude plot shown in Fig. P9.16. The phase of H(! ) is 90◦ at ! = 0. M [dB] 60 dB 40 dB 20 dB 0 0.5 5 50 500 ω (rad/s) Figure P9.16: Bode magnitude plot for Problem 9.16. Solution: H(! ) consists of: (1) A constant term K whose dB value is 60 dB, or K = 1060/20 = 1000. (2) A simple pole of order 3 with !c = 5 rad/s (slope = −60 dB/decade) (3) A simple zero of order 6 with !c = 50 rad/s (slope reverses from −60 dB/decade to +60 dB/decade) (4) A simple pole of order 3 with !c = 500 rad/s (slope changes to 0 dB at !c = 500 rad/s). Hence, H(! ) = ( j)N 1000(1 + j! /50)6 j1000(50 + j! )6 = . (1 + j! /5)3 (1 + j! /500)3 (5 + j! )3 (500 + j! )3 Given that the phase of H(! ) is 90◦ at ! = 0, it follows that N = 1. All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press d s (s + 2 − j1) −2 = (s + 2 − j1)3 Hence, F2 (s) = − 1 4 s=−2− j1 = s=−2− j1 j . 4 1 e j90 e− j90 Problem 10.13 Obtain the inverse Lapla+ transform of each of the following ce + 4 s + 2 + j1 s + 2 − j1 functions: (s + 2)2 1 (a) F1 (s) = f2 (t ) = − (t e−2t e− j1t + t e−2t e j1t ) s(s + 1)3 4 1 1 −2t − j(t −90◦ ) ◦ (b) F2 (s) = 2 2 + e−2t e j(t −90 ) ) u(t ) (s + 4s + 5)+ 4 (e e √ 2(s + 1) 1 −2t 1 −2t (c) F3 (s) = 2 ◦ =− s + 6s + 13 t e cos t + e cos(t − 90 ) u(t ) 2 2 −2(s2 + 20) (d) F4 (s) = 2 s(s + 8=+ −01 t cos t + 1 sin t e−2t u(t ). s 2) 2 2 Solution: (c) (a) √ 2(s + 1) (s + 2)2 F1 (s) = F3 (s) = 2 s(s + 1)3 s + 6s + 13 √ −6 − 36 − 52 A B3 B2 B1 s2 + 6s + 13 = 0= + s3 + = = + , −3 − j 2 1 s (s + 1) (s + 1)22 s + 1 s2 = −3 + j2. with Hence, (s + 2)2 A = s F1 (s)|s=0 = = 4, B∗ B ( F3 (s) =s + 1)3 s=0 + , s + 3 + j2 s + 3 − j2 (s + 2)2 B3 = (s + 1)3 F1 (s)|s=−1 = = −1, s s= √ −1 d (s + 2)2 2(s + 2) 2(s+ 2)) (s + 1 2 s= B2B = (s + 3 + j2) F3 (s)|= −3− j2 = −+ 3 − j2 = = −3, s ds s s s2 ss=−3− j2 s= − 1 =−1 √ d 2(s + 2) (s + 2)2 ◦ 2 B1 = − = (1 − j) = e− j45 . 2 ds s s s= − 1 2 = 2 2(s + 2) 2(s + 2) 2(s + 2)◦2 − − e− 245◦ + = −8. j 2 e3j45 s s s. −1 = Fs (s) = +s 3 s + 3 + j2 s + 3 − j2 1 1 + (s + 2 + j1)2 (s + 2 − j1)2 ◦ ◦ with Hence, Hence, 4 1 3 8 All rights reserved. Do not 1eproduce or distribute. − ©2009 National Technology and Science Press F r (s) = − − , By application of the distinct cs mpls x-po3e fo(s + la)2n Table110-3, we have o ( e+ 1)l rmu 1 i s + and f3 (t ) = 2e−3t cos(2t + 45◦ ) u(t ). 1 f1 (t ) = 4 − t 2 e−t − 3t e−t − 8e−t u(t ) (d) 2 t 2 −2 s2 + − = F − ) = + 3t(+ 8 20)t u(t ). 4 (s e 4 2 s(s2 + 8s + 20) √ (b) −8 − 64 − 80 s2 + 8s + 20 = 0 s1 = = −4 − j 2 1 2 F2 (s) = 2 IN 2 SEE THE APPROACH DESCRIBED4−++CLASS ON THE (s 2 + s 4 5)j2. s= √ −4 − 16 − 20 Hence, s2 + 4s + 5 = 0 s1 = = −2 − j 1 2 B∗ Basic Steps: A B F4 (s) = + s = −2 + j1. = , 2 s s + 4 + j2 s + 4 − j2 NEXT PAGE with All rights reserved. square of the second-order denominator (complex Science Press ©2009 National Technology and poles). 1) Complete theDo not reproduce or distribute. −2(function by adjusting the numerator to match the two s2 + 20) 2) Rearrange s F4(stransfer A = the )|s=0 = 2 = −2, s + 8s + 20 s=0 transforms below. 2 B = (s + 4 + j2) F4 (s)|s=−4− j2 = Hence, F4 (s) = and −2(s + 20) s(s + 4 − j2) ◦ = 4e j90 . ◦ 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 ). ...
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