Homework 4 - Christopher Belcastro I.D 904424264 ECE 2704...

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Christopher Belcastro I.D. - 904424264 ECE 2704 – Homework #4 2 October 2006 1) a. Find the transfer function: ) ( ) ( 3 ) 0 ( 4 ) ( 4 ) 0 ( ) 0 ( ) ( . 2 s X s Y y s sY sy y s Y s = + - + - - ) 0 ( 4 ) 0 ( ) 0 ( ) ( ] 3 4 [ ) ( . 2 y sy y s X s s s Y - - + = + + 0 ) 0 ( , 0 ) 0 ( : . = = y y Assume ) 3 4 ( ) ( ) 0 ( 4 ) 0 ( ) 0 ( 1 ) ( ) ( 2 . + + - - + = s s s X y sy y s X s Y b. Poles at -1 & -3 c. 1 3 ) 3 4 ( ) ( ) 0 ( 4 ) 0 ( ) 0 ( 1 2 . 1 + + + = + + - - + - s B s A s s s X y sy y L { } ) 3 ( ) 1 ( ) ( 1 + + + = - s B s A s H L 2 1 1 2 1 2 1 1 2 3 = = - = - = = - - = B so B s A so A s + - = + + + - - - - ) ( 2 1 ) ( 2 1 1 3 3 2 1 2 1 1 t u e t u e s s L s t s t d. ) ( ) ( 4 t u e t x s t - = 4 1 ) ( + = s s X 4 1 8 2 ) ( 3 ) ( 4 1 ) ( 2 + = - - + + - s s s Y s sY s Y s
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[ ] 9 2 4 1 3 4 ) ( 2 + + + = + + s s s s s Y [ ] ) 9 2 )( 4 ( ) 9 2 )( 4 ( 9 ) 9 2 )( 4 ( ) 9 2 )( 4 ( 2 ) 9 2 ( ) 4 ( ) 9 2 )( 4 ( 3 4 ) ( 2 2 + + + + + + + + + + + + + + = + + s s s s s s s s s s s s s s s s Y [ ] ) 9 2 )( 4 ( ) 9 2 )( 4 ( 9 ) 9 2 )( 4 ( 2 ) 9 2 ( 3 4 ) ( 2 + + + + + + + + + = + + s s s s s s s s s s s Y ) 1 )( 4 )( 3 ( 37 17 2 ) 1 )( 4 )( 3 ( 36 9 8 2 1 ) ( 2 2 + + + + + = + + + + + + + = s s s s s s s s s s s s Y 4 1 3 ) 1 )( 4 )( 3 ( 37 17 2 ) ( 2 + + + + + = + + + + + = s C s B s A s s s s s s Y 3 ) 1 )( 3 ( ) 4 )( 3 ( ) 4 )( 1 ( 37 17 2 2 - = + + + + + + + + = + + s at s s C s s B s s A s s 1 2 2 37 51 18 - = - = - = + - s at A A 4 3 11 6 22 6 37 17 2 - = = = = + - s at B B 3 1 3 37 68 32 = = + - C C 4 1 3 2 ) ( 3 1 3 11 + + + + + - = s s s s Y { } t t t e e e s Y L 4 3 1 3 1 3 11 2 ) ( - - - - + + - = ) ( 3 1 3 11 2 ) ( 4 3 t u e e e t y s t t t + + - = - - - e. Convolution Integral: ( 29 - - - - - - - - - - + - = = + - = λ λ λ λ λ λ d u e t u e e t y t u e t x t u e t u e t h s s t t s t s t s t ) ( 2 1 2 1 ) ( ) ( )
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