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exfinalsoln

# exfinalsoln - ECE 202 Linear Circuit Analysis II Final Exam...

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ECE 202 - Linear Circuit Analysis II Final Exam Solutions December 19, 2008 Solution 1 Breaking F ( s ) into partial fractions, F ( s ) = 4 s 2 - 9 s ( s - 9) = 4 + 1 s + 35 s - 9 f ( t ) = 4 δ ( t ) + [1 + 35 e 9 t ] u ( t ) A = 9 Hence (3) is the correct answer. Solution 2 f ( t ) = 3[ u ( t ) - u ( t - 1)] - 2[ u ( t - 1) - u ( t - 2)] = 3 u ( t ) - 5 u ( t - 1) + 2 u ( t - 2) F ( s ) = 3 s - 5 e - s s + 2 e - 2 s s Hence (1) is the correct answer. Solution 3 Zin Zin1 ia/6 ia 1/3 ohms s/8 2/s V I Consider the impedance Z in 1 looking right at the node with voltage V. We can then write, V = i a 3 - ( I - i a ) s 8 - i a 6 + i a 3 = 0 1

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I = i a 3 3 + 4 s = V 3 + 4 s Z in 1 = V I = s 3 s + 4 Z in = (2 /s ) || Z in 1 = 2 s s 2 + 6 s + 8 Hence (3) is the correct answer. Solution 4 Transfer function is given by the standard expression, H ( s ) = V out ( s ) V in ( s ) = - Z 2 Z 1 Z 2 = (2 /s ) || 1 = 2 s + 2 Z 1 = 1 + (1 /s ) = s + 1 s H ( s ) = - 2 s ( s + 1)( s + 2) Hence (4) is the correct answer. Solution 5 Using voltage division, H ( s ) = V o V i = 1 s 1 s + s + 2 = 1 s 2 + 2 s + 1 = 1 ( s + 1) 2 h ( t ) = te - t u ( t ) Hence (5) is the correct answer. Solution 6 v i ( t ) = 2[ u ( t ) - u ( t - 1)] V i ( s ) = 2(1 - e - s ) 1 s V o ( s ) = H ( s ) V i ( s ) = 2(1 - e - s ) s ( s + 1) 2 = 2(1 - e - s ) 1 s - 1 ( s + 1) 2 - 1 s + 1 v o ( t ) = [2 - 2 te - t - 2 e - t ] u ( t ) - [2 - 2( t - 1) e - ( t - 1) - 2 e - ( t - 1) ] u ( t - 1) 2
Hence (4) is the correct answer. Solution 7 The equivalent impedance of the LC combination at ω =0.5 is zero. Thus voltage at the inverting terminal of the op-amp is v o . By virtual ground principle of ideal op-amp, v o ( t ) = v s ( t ) = 2 sin(0 . 5 t + 30 ) Hence (5) is the correct answer.

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