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ex2soln

# ex2soln - ECE 202 Linear Circuit Analysis II Exam 2...

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ECE 202 - Linear Circuit Analysis II Exam 2 Solutions October 30, 2008 Solution 1 V out ( t ) = V in ( t ) * h ( t ) = 30 sin[ π (60 t + 1)] * 2 δ t - 1 40 = 60 sin π 60 t - 1 40 + 1 = 60 sin π 60 t - 1 2 ¶‚ = 60 sin h 60 πt - π 2 i = - 60 cos(60 πt ) Hence (3) is the correct answer Solution 2 A 40 dB/decade fall from 100 implies the term, 1 1 + ( s 100 ) 2 Similarly, the flat portion from 1000 to 100000 implies the term, 1 + s 1000 · 2 A 20 dB/decade rise from 10 5 implies the term, 1 + s 10 5 Finally the flat portion from 10 7 implies the term, 1 1 + s 10 5 To account for the start at 80 dB, we must have a constant term equal to 10 80 / 20 = 10 4 . The gain function can therefore be written as, H ( s ) = 10 4 h 1 + ( s 1000 ) 2 i ( 1 + s 10 5 ) h 1 + ( s 100 ) 2 i ( 1 + s 10 7 ) Hence (8) is the correct answer 1

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Solution 3 V in ( t ) = 10 u ( t ) h ( t ) = 3[ u ( t - 3) - u ( t - 6)] + 8[ u ( t - 6) - u ( t - 9)] = 3 u ( t - 3) + 5 u ( t - 6) - 8 u ( t - 9) I out ( t ) = V in ( t ) * h ( t ) = V (1) in ( t ) * h ( - 1) ( t ) ( convolution algebra ) = 10 δ ( t ) * [3 r ( t - 3) + 5 r ( t - 6) - 8 r ( t - 9)] = 30 r ( t - 3) + 50 r ( t - 6) - 80 r ( t - 9) Hence (5) is the correct answer Solution 4 From the frequency response plot, Q = ω m B ω = ω m ω 2 - ω 1 = 65 k 80 k -
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ex2soln - ECE 202 Linear Circuit Analysis II Exam 2...

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