# hw5soln - ECE 202 - Linear Circuit Analysis II Purdue...

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ECE 202 - Linear Circuit Analysis II Purdue University, Spring 2009 Homework Set 2 Solutions Solution 17 3 VC(s) VL(s) 4/s s 2 2 2/s Note that the current sources I in ( s ) and Cv C (0 - ) have been combined into a single current source(2+1=3). From the above simpliﬁed circuit in s-domain, we can write 1 , V C - V L 2 + ( s + 2) V C 4 = 3 V L - V C 2 + 2 s = - V L s Solving using cramer’s rule and ilaplace command in MATLAB, V C = 4(3 s + 4) s 2 + 4 s + 8 v C ( t ) = 12 ± cos(2 t ) - 1 3 sin(2 t ) e - 2 t u ( t ) V L = 8( s - 2) s 2 + 4 s + 8 v L ( t ) = 8[cos(2 t ) - 2sin(2 t )] e - 2 t u ( t ) Solution 18 (a) The equivalent frequency domain circuit taking into account initial conditions is drawn on the next page. (b) We can write the following nodal equations:- V in [ G 1 + C 1 s ] + I d = I in + C 1 v C 1 (0 - ) 1 from now onwards, we will not explicitly write V C ( s ) for a corresponding time function v C ( t ) , upper case would mean s domain 1

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+ - Iin(s)+C1vC1(0-) G1 IC1 C1s C2s vC2(0-)/s Vout(s) G3 Id(s) aIC1 R2 Vin(s) V in - R 2 I d - aI C 1 = V out V out G 3 + V out - v C 2 (0 - ) s C 2 s = I d I C 1 = V in C 1 s - C 1 v C 1 (0 - ) Rearranging and simplifying, we get the following matrix equation, G 1 + C 1 s 0 1 1 - aC 1 s - 1 - R 2 0 G 3 + C 2 s - 1 V in ( s ) V out ( s ) I d ( s ) = I in ( s ) + C 1 v C 1 (0 - ) - aC 1 v C 1 (0 - ) C 2 v C 2 (0 - ) (c) The transfer function is computed using MATLAB as follows:-
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hw5soln - ECE 202 - Linear Circuit Analysis II Purdue...

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