Homework 3 - -s s s s s s s s s s s s s 2 30 11 2 15 2 2 15...

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Christopher Belcastro I.D. - 904424264 ECE 2704 – Homework #3 27 September 2006 1) i. x(t) = sgn(t) No. This function does not have a Laplace Transform. Reason: The function will equal -1 for t < 0. Laplace Transform must be a positive integer. ii. x(t) = sinc(3t)u s (t) Yes. This function has a Laplace Transform. Reason: Consider that the unit step function is multiplied by the sinc function. For t < 0, the unit step will push x(t) = 0. Therefore the range is x 0 . 2) x(t) = 4tcos(10t)u s (t) 2 2 2 2 2 ) ( ϖ ϖ + - s s { } ( 29 ( 29 4 * ) 10 ( 10 ) ( 2 2 2 2 2 + - = s s t x L 2 2 2 ) 100 ( 400 4 ) ( + - = s s s X The poles of the transform are as followed: ) 100 )( 100 ( 400 4 ) ( 2 2 2 + + - = s s s s X So the poles are + - j10; note that this will occur twice in the denominator. 3) 25 ) 3 ( ) 3 ( 10 ) ( 2 + - - = s s s X 3 , 5 ) ( ) ( 2 2 - = = + + + α ϖ ϖ α α c c s s { } ) ( ) ( ) 5 cos( 10 ) ( 3 1 t x t u t e s X L s t = = - 4) ) ( ) ( , 2 ) 0 ( , 1 ) 0 ( ) ( ) ( 12 ) ( 7 ) ( 2 . . . . t u e t x y y t x t y t y t y s t = = = = + + 2 1 ) ( ) ( ) ( 12 )) 0 ( ) ( ( 7 ) 0 ( ) 0 ( ) ( 2 - = = + - + - - s s X s X s Y y s sY sy y s Y s 2 1 ) ( 12 14 ) ( 7 2 1 ) ( 2 - = + - + - - s s Y s sY s s Y s
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15 2 2 1 ) ( 12 ) ( 7 ) ( 2 + + - = + + s s s Y s sY s Y s ) 15 2 )( 2 ( ) 15 2 )( 2 ( 15 ) 15
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Unformatted text preview: +-+-+ +-+-+ +-+-s s s s s s s s s s s s s 2 30 11 2 ) 15 2 )( 2 ( ) 15 2 )( 2 ( 15 ) 15 2 )( 2 ( 2 ) 15 2 ( 2--+ = +-+-+ +-+ + s s s s s s s s s s s 2 30 11 2 ] 12 7 )[ ( 2 2--+ = + + s s s s s s Y 3 4 2 ) 3 )( 4 )( 2 ( 30 11 2 ) ( 2 + + + +-= + +--+ = s C s B s A s s s s s s Y 2 ) 4 )( 2 ( ) 3 )( 2 ( ) 3 )( 4 ( 30 11 2 2 = → +-+ +-+ + + =-+ s at s s C s s B s s A s s 4 30 30 22 8-= → = ∴ =-+ s at A A 3 7 6 30 44 32-= →-= ∴ =--s at B B 9 ) 5 ( 30 33 18 = ∴-=--C B 3 9 4 7 2 ) ( + + +-+-= s s s s Y t t e e s Y L 3 4 1 9 7 )} ( {---+-= ) ( 9 ) ( 7 ) ( 3 4 t u e t u e t y s t s t--+-= ∴...
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