# hw7sol - Purdue University, School of Electrical and...

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #7 Solution (25–28) Problem 25 R 1 = 0 . 5 Ω , R 2 = 1 Ω , L = 0 . 1H, C = 0 . 8F v s 1 ( t )= 3 u ( t ) [V], i s 2 ( t 3 u ( t ) [A] (a) Nodal equations 1 ± : V C ( s ) V s 1 ( s ) R 1 + CsV C ( s ) Cv C ( 0 )+ V C ( s ) V R ( s ) Ls + i L ( 0 ) s = 0 2 ± : i L ( 0 ) s + V R ( s ) V C ( s ) Ls + V R ( s ) R 2 + I s 2 ( s 0 ± G 1 + Cs + 1 / ( Ls ) 1 / ( Ls ) 1 / ( Ls ) 1 / ( Ls G 2 ²± V C ( s ) V R ( s ) ² = ± G 1 V s 1 ( s C ( 0 ) i L ( 0 ) / s I s 2 ( s i L ( 0 ) / s ² 1 s ± 0 . 8 s 2 + 2 s + 10 10 10 s + 10 V C ( s ) V R ( s ) ² = ± 2 V s 1 ( s ) I s 2 ( s ) ² (b) v R , zs ( t ) ZSR 1 s ± 0 . 8 s 2 + 2 s + 10 10 10 s + 10 V C ( s ) V R ( s ) ² = ± 6 / s 3 / s ² V R , zs ( s 3 s 7 . 5 s ( s + 7 . 5 ) v R , zs ( t ( 1 4 e 7 . 5 t ) u ( t ) [V] 1/ 7

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (c) v R , zs ( t ) ZIR 1 s ± 0 . 8 s 2 + 2 s + 10 10 10 s + 10 ²± V C ( s ) V R ( s ) ² = ± 2 . 4 0 ² V R , zi ( s )= 30 ( s + 5 )( s + 7 . 5 ) v R , zi ( t 12 ( e 5 t e 7 . 5 t ) u ( t ) [V] (d) v R ( t ) Complete V R ( s V R , zs ( s )+ V R , zi ( s ) v R ( t ³ 1 | {z ´ steadystate + 12 e 5 t 16 e 7 . 5 t | {z ´ transient µ u ( t ) [V] MATLAB code used: syms s Vnode Vs1 Is2 VC VR VR zs t VR zi vR zs vC0 iL0 R1=0.5; R2=1; L=0.1; C=0.8; G1=1/R1; G2=1/R2; % ZSR Vs1=3/s; Is2=3/s; iL0=0; vC0=0; M=[ G1+C * s+1/(L * s) 1/(L * s) ; * s) 1/(L * s)+G2]; b=[G1 * Vs1+C * vC0
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## This document was uploaded on 04/02/2010.

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hw7sol - Purdue University, School of Electrical and...

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