Hw18sol - Purdue University School of Electrical and Computer Engineering Prof DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II(S P ’10 Homework#18

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Unformatted text preview: Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #18 Solution (69–72) Problem 69 1/Cs V in R S + V out _ 2 : 1 M Z 1 R s = 10 Ω , C = . 25 F, v in ( t ) = 20cos ( 2 t ) [V] (a) Impulse response Using impedance transformation property of an ideal transformer, Z 1 in the above diagram is Z 1 = 2 1 2 1 Cs = 16 s Using Voltage division, V 1 ( s ) = 16 / s 16 / s + 10 = 1 . 6 s + 1 . 6 V in ( s ) V out ( s ) = V 1 2 ⇒ H ( s ) = V out ( s ) V in ( s ) = . 8 s + 1 . 6 (69.1) h ( t ) = . 8 e- 1 . 6 t u ( t ) [V] (b) Step Response v out , step = Z t h ( τ ) d τ = . 5 e- 1 . 6 τ t = . 5 1- e- 1 . 6 t u ( t ) [V] 1/ 12 Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (c) Zero-input response (ZIR) R S + V out _ 2 : 1 Cv C (0-) 1/Cs Z out V out , zir ( s ) = Cv C (- ) ( 1 / Cs ) || Z out Z out = 1 2 2 ( 10 ) = 2 . 5 Ω = ( . 25 )( 16 ) ( 4 / s ) || 2 . 5 = 4 4 s + 1 . 6 v out , zir ( t ) = 16 e- 1 . 6 t u ( t ) [V] (d) Zero-state response (ZSR) Using eq.(69.1), V out , zsr ( s ) = . 8 s + 1 . 6 V in ( s ) = . 8 s + 1 . 6 20 s s 2 + 4 v out , zsr ( t ) =- 3 . 90 e- 1 . 6 t u ( t ) | {z } transient +- 3 . 90cos ( 2 t )+ 4 . 88sin ( 2 t ) u ( t ) | {z } steady state [V] 2/ 12 Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo Problem 70 L 1 s L 2 s V in M R S + V out (s) _ I in Z 1 C R L R s = 20 Ω , L 1 = 1 . 5 mH , M = 3 mH , L 2 = 6 mH , R L = 320 Ω (a) Coupling coefficient, k k = M √ L 1 L 2 = 3 p ( 1 . 5 )( 6 ) = 1 (b) Y 1 ( s ) and Y 1 ( j ω ) as a function of C V in R S + V out (s) _ Z p 1/Cs R L L 1 s L 2 1 : L 1 Z 1 Z p = 1 p L 2 / L 1 !...
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Hw18sol - Purdue University School of Electrical and Computer Engineering Prof DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II(S P ’10 Homework#18

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