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# prc8 - ECSE-210 Winter 2011 Electric Circuits 2 Practice...

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ECSE-210 Winter 2011 Electric Circuits 2 8-1 Practice Set #8 P14.7-10 Steady-state for t<0: From the equation for v o ( t ): ( ) ( ) o 2 6 12 6 V v e ! " " = + = From the circuit: ( ) ( ) o 3 18 3 v R ! = + Therefore: ( ) 3 6 18 6 3 R R = ! = " + Steady-state for t>0: ( ) ( ) 1 18 6 6 2 0 1 2 I s I s C s s s s C ! " # + + # = \$ = % & ' ( + ( ) ( ) o 1 18 1 6 18 12 12 18 12 6 1 1 1 2 2 2 V s I s Cs s Cs s s s s s s s C C C ! " # \$ % % = + = + = + + = + # \$ # \$ + + + # \$ & ' Taking the inverse Laplace transform: ( ) / 2 o 6 12 V for 0 t C v t e t ! = + > Comparing this to the given equation for v o ( t ), we see that 1 2 0.25 F 2 C C = ! = .

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ECSE-210 Winter 2011 Electric Circuits 2 8-2 P14.7-11 We will determine ( ) o V s , the Laplace transform of the output, twice, once from the given equation and once from the circuit. From the given equation for the output, we have ( ) o 10 5 100 V s s s = + + Next, we determine ( ) o V s from the circuit. For 0 t ! , we represent the circuit in the frequency domain using the Laplace transform. To do so we need to determine the initial condition for the capacitor. When 0 t < and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to
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