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ex1soln

# ex1soln - ECE 202 Linear Circuit Analysis II Exam 1...

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ECE 202 - Linear Circuit Analysis II Exam 1 Solutions September 25, 2008 Solution 1 F ( s ) = ln ( s + 2 s + 1 ) Time shift property gives, L ( f ( t - 2)) = e - 2 s F ( s ) Again, frequency shift property gives, L ( e t f ( t - 2)) = e - 2( s - 1) F ( s - 1) = e - 2 s e - 2 ln ( s + 1 s ) Hence (1) is the correct answer Solution 2 f ( t ) = 1 2 ( t + 1) { u ( t - 1) - u ( t - 3) } = 1 2 [( t - 1) u ( t - 1) - ( t - 3) u ( t - 3) + 2 u ( t - 1) - 4 u ( t - 3)] Application of time shift property gives, F ( s ) = 1 2 [ e - s s 2 - e - 3 s s 2 + 2 { e - s s - 2 e - 3 s s } ] = 1 s ( e - s - 2 e - 3 s ) + 0 . 5 s 2 ( e - s - e - 3 s ) Hence (1) is the correct answer 1

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Solution 3 - + + - I I Vout(s) 1/1us -2/s 1/2us The equivalent circuit when switch is moved to position B is drawn above Writing KVL equations assuming ideal op amp, V out + I 1 μs = 0 - 2 s - I 2 μs = 0 - 2 s = I 2 μs = - V out 2 or V out = 4 s Hence v out ( t ) = 4 u ( t ) = 4 V, t = 0 + Hence (2) is the correct answer Solution 4 - + + - Vin(s) s 1 1/s Vout(s) The voltage at the inverting terminal of the op amp is V in due to virtual ground Applying KCL at the inverting terminal we get, ( V out - V in )( s + 1) = V in s 2
V out = V in ( 1 s 2 + s + 1) or H ( s ) = V out V in = s 2 + s + 1 s 2 + s

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ex1soln - ECE 202 Linear Circuit Analysis II Exam 1...

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