# Soln5 - EE411 Spring 2008 HW#5 Solution(Q2 7.74 Please note...

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EE411 Spring 2008 HW#5 Solution (Q2) 7.74 Please note that this problem has a more straightforward solution but the way it is presented here is more instructive for analyzing the step response of OPAMP circuits containing one reactive element. Let v = capacitor voltage. For t < 0, 0 ) 0 ( v = For t > 0, A 10 i s μ = . Consider the circuit below. R v dt dv C i s + = ( 1 ) [] τ + = t - e ) ( v ) 0 ( v ) ( v ) t ( v ( 2 ) It is evident from the circuit that 1 . 0 ) 10 50 )( 10 2 ( RC 3 6 = × × = = τ At steady state, the capacitor acts like an open circuit so that s i passes through R. Hence, V 5 . 0 ) 10 50 )( 10 10 ( R i ) ( v 3 6 s = × × = = Then, ( ) V e 1 5 . 0 ) t ( v -10t = ( 3 ) But f s o f o s R i - v R v 0 i = ⎯→ = ( 4 ) Combining (1), (3), and (4), we obtain dt dv C R v R R - v f f o = i s R + R f C + v o i s

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dt dv ) 10 2 )( 10 10 ( v 5 1 - v 6 - 3 o × × = ( ) -10t -2 -10t e 10 ) 5 . 0 )( 10 2 ( e 0.1 -0.1 × + = o v 1 . 0 = o v V (Q5) 8.37
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Soln5 - EE411 Spring 2008 HW#5 Solution(Q2 7.74 Please note...

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