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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2/23/02 page P14.2 © R. A. DeCarlo, P. M. Lin 0 5 10 15 20 0 1 2 3 4 5 6 7 8 9 10 Time in s Input and Capacitor voltages, V TextEnd S OLUTION 14.39 . The figure which accounts for the initial conditions is given below. (a) For the zero-input response, the above circuit reduces to a parallel RLC driven by two current sources. Hence V C (s) equals the total current divided by the total admittance, i.e.,
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2/23/02 page P14.3 © R. A. DeCarlo, P. M. Lin V C ( s ) = Cv C (0 - ) + i L (0 - ) s Cs + 1 R + 1 Ls = sv C (0 - ) + i L (0 - ) C s 2 + 1 RC s + 1 LC = 20 s + 10 s 2 + 250 s + 10 4 = 26.6 s + 200 - 6.6 s + 50 Hence v C (t) = [26.6e –200t – 6.6e –50t ]u(t) V (b) For the zero-state response, the current sources disappear. Executing a source transformation on the remaining voltage source, we obtain a current, I(s) = V in (s)/(Ls), driving a parallel RLC circuit. Hence, the zero input response is V C ( s ) = V in ( s ) Ls Cs + 1 R + 1 Ls = 1 LC V in ( s ) s 2 + 1 RC s + 1 LC = 20000 s 3 + 250 s 2 + 10 4 s = 2 s + 0.6667 s + 200 - 2.6667 s + 50 Hence v C (t) = [2 + 0.6667e –200t – 2.6667e –50t ]u(t) V (c) By superposition, the complete response is the sum of the answers to (a) and (b). Hence v C (t) = [2 + 27.267e –200t – 9.2667e –50t ]u(t) V (d) By linearity and time-invariance, v C (t) = [2 + 0.6667e –200t – 2.6667e –50t ]u(t) + [4 + 1.3334e –200(t–0.01) – 5.3334e –50(t–0.01) ]u(t – 0.01) V
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2/23/02 page P14.4 © R. A. DeCarlo, P. M. Lin SOLUTION 14.40.
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DeCarlo Ch14b solutions - 2/23/02 page P14.1 R. A. DeCarlo,...

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