HW4%20solutions_EG

# HW4%20solutions_EG - MAE 3780 MECHATRONICS FALL 2009 H O M...

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MAE 3780 MECHATRONICS FALL 2009 1 HOMEWORK 4 SOULUTIONS 1. From the op amp rules we can say: + = V V 0 = = + i i KCL at A yields f S i i = Applying Ohm’s Law: 3 3 1 2 2 3 R V V i V V R i R V V i V V R i A out f A out f A S A S = = = = Substituting: 3 1 2 R V V R V V A out A = 0 = = = + A V V V 3 1 2 R V R V out = 2 1 3 V R R V out = Plugging in the given values: Sketch Vout: 2 2 2 1 2 V V M M V out = Ω Ω = i s i f A t Vout

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MAE 3780 MECHATRONICS FALL 2009 2 KCL at A yields f i i i = + 2 1 Substituting using Ohm’s Law: 3 1 2 2 1 R V V R V V R V V A out A A = + 0 = = = + V V V A 3 1 2 2 1 R V R V R V out = + 2 1 3 1 2 3 V R R V R R V out = Plugging in the given values this becomes 2 1 2 2 V V V out = To sketch the graph we will need to evaluate Vout at each segment: for 1 0 t 0 2 2 1 1 2 1 = + = = = out V V V for 2 1 t 2 ) 2 ( 2 ) 1 ( 2 2 1 2 1 = + = = + = t t V t V t V out i f i 2 i 1 A
FALL 2009 3 for 3 2 t 10 4 ) 2 ( 2 ) 3 ( 2 2 3 2 1 + = = = = t t t V t V t V out for 3 > t 0 0 0 2 1 = = = out V V V Sketch: 1 1 R V i S = dt dV C i out f

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## HW4%20solutions_EG - MAE 3780 MECHATRONICS FALL 2009 H O M...

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