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final09-02-sol - MAE 143A Signals and Systems Prof M Krstic...

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MAE 143A Signals and Systems Prof. M. Krstic FINAL December 11, 2009 Closed book. One sheet (both sides) of handwritten notes allowed. Present your reasoning and calculations clearly. Random or inconsistent etchings will not be graded. Write answers only in the blue book. The problems are not ordered by difficulty. Total points: 65. Problem 9 is optional, it can be done for extra credit of 9 points. Time: 2.5 hours. SOLUTIONS 1
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Problem 1. (8 points) Find the Laplace transforms of the following signals: (a) (4 points) f 1 ( t ) = e - t ( t - 1) 2 1( t - 1) (b) (4 points) f 2 ( t ) = e - t cos(2 t - 2)1( t - 1) Solution: (a) f 1 ( t ) = e - 1 e - ( t - 1) ( t - 1) 2 1( t - 1) Since, L braceleftbig e - t t 2 1( t ) bracerightbig = - d parenleftbigg - d ( 1 s +1 ) ds parenrightbigg ds = 2 ( s + 1) 3 we have that L { f 1 ( t ) } = e - s - 1 2 ( s + 1) 3 (b) f 2 ( t ) = e - 1 e - ( t - 1) cos (2( t - 1)) 1( t - 1) Since, L braceleftbig e - t cos(2 t )1( t ) bracerightbig = s + 1 ( s + 1) 2 + 4 we have that L { f 2 ( t ) } = e - s - 1 s + 1 ( s + 1) 2 + 4 Problem 2. (9 points) Find the inverse Laplace transforms of the following functions: (a) (4 points) F 1 ( s ) = e - 3 s s +6 s 2 +8 s +20 (a) (5 points) F 2 ( s ) = log parenleftBig s s +1 parenrightBig Solution: (a) F 1 ( s ) = e - 3 s s + 4 ( s + 4) 2 + 2 2 + e - 3 s 2 ( s + 4) 2 + 2 2 Thus L - 1 { F 1 ( s ) } = e - 4( t - 3) (cos (2( t - 3)) + sin (2( t - 3))) 1( t - 3) 2
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(b) log parenleftbigg s s + 1 parenrightbigg = log( s ) - log( s + 1) Thus, - dF 2 ( s ) ds = 1 s + 1 - 1 s Since, L - 1 braceleftbigg - dF 2 ( s ) ds bracerightbigg = L - 1 braceleftbigg 1 s + 1 - 1 s bracerightbigg = ( e - t - 1 ) 1( t ) we have that L - 1 { F 2 ( s ) } = e - t - 1 t 1( t ) Problem 3. (8 points) Compute the step response of the following system: H ( s ) = 4 ( s + 2)( s 2 + 2 s + 2) Solution: Y ( s ) = H ( s ) U ( s ) = H ( s ) 1 s = 4 s ( s + 2)( s 2 + 2 s + 2) = C 1 s + C 2 s + 2 + C 3 s + C 4 s 2 + 2 s + 2
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