DeCarlo Ch14b solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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2/23/02 page P14.1 © R. A. DeCarlo, P. M. Lin SOLUTION 14.38. In the s-domain, we break the response up into the part due to the initial condition and the part due to the source with the initial condition set to zero. The transfer function with the initial condition set to zero is H ( s ) = V C ( s ) V in ( s ) = 1 Cs R + 1 Cs = 1 RC s + 1 RC = 0.25 s + 0.25 Using the parallel equivalent circuit for the charged capacitor while setting the source voltage to zero, the capacitor voltage due only the initial condition is: V C , IC ( s ) = 1 1 R + Cs Cv C (0 - ) [ ] = v C (0 - ) s + 0.25 Hence, V C ( s ) = 0.25 s + 0.25 V in ( s ) + v C (0 - ) s + 0.25 and I C ( s ) = V in ( s ) - V C ( s ) 20 = 0.05 1 - 0.25 s + 0.25 V in ( s ) - 0.05 v C (0 - ) s + 0.25 = 0.05 s s + 0.25 V in ( s ) - 0.05 v C (0 - ) s + 0.25 for all inputs and initial conditions. (a) If v in ( t ) = 20 u ( t ) and v C (0 - ) = 10 V, then V in ( s ) = 20 s and V C ( s ) = 5 s ( s + 0.25) + 10 s + 0.25 = 20 s - 10 s + 0.25 v C ( t ) = 20 - 10 e - 0.25 t ( 29 u ( t ) V and I C ( s ) = 1 s + 0.25 - 0.5 s + 0.25 = 0.5 s + 0.25 i C ( t ) = 0.5 e - 0.25 t u ( t ) A (b) If v in ( t ) = 5 e - 0.25 t u ( t ) V , then V in ( s ) = 5 s + 0.25 . Hence, V C ( s ) = 1.25 ( s + 0.25) 2 + 10 s + 0.25 v C ( t ) = 10 + 1.25 t ( 29 e - 0.25 t u ( t ) V and I C ( s ) = 0.25 s ( s + 0.25) 2 - 0.5 s + 0.25 = - 0.25 s + 0.25 - 0.0625 ( s + 0.25) 2 Hence i C ( t ) = - 0.25 + 0.0625 t ( 29 e - 0.25 t u ( t ) A
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