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a2_solution

# a2_solution - Instructors Solutions to Assignment 2 Problem...

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Unformatted text preview: Instructors Solutions to Assignment 2 Problem 3.2 (i) 4 ( ) 4 ( ) 8 ( ) ( ) ( ) with ( ) ( ), (0) 0, and (0) 0. t y t y t y t x t x t x t e u t y y ! + + = + = = = !! ! ! ! (a) Particular solution: The particular solution for input x ( t ) = exp( − 4 t ) u ( t ) is of the form ) ( 4 t u Ke (t) y t p ! = . Substituting the particular solution in the differential equation for system (i) and solving the resulting equation gives K = − 3/8. (b) Homogeneous solution: The characteristic equation of the LTIC system (i) is 8 4 2 = + + s s , which has roots at s = − 2 ± j 2. The zero-input response is given by ) 2 sin( ) 2 cos( ) ( 2 2 t Be t Ae t y t t h ! ! + = for t ≥ 0, with A and B being constants. (c) Overall response of the system: The overall response of the system is obtained by summing up the above two responses, and use initial conditions to derive A and B, and it is given by ( ) ) ( ) 2 sin( ) 2 cos( ) ( 4 2 2 8 3 t u e t e t e t y t t t ! ! ! ! ! = . (iii) [ ] ( ) 2 ( ) ( ) ( ) with ( ) cos( ) sin(2 ) ( ), (0) 3, and (0) 1. y t y t y t x t x t t t u t y y + + = = + = = !! ! !! ! (a) Particular solution: The particular solution for input x ( t ) = [cos( t ) + sin( t )] u ( t ) is of the form ) 2 sin( ) 2 cos( ) sin( ) cos( 4 3 2 1 t K t K t K t K (t) y p + + + = . Substituting the particular solution in the differential equation for system (iii) and solving the resulting equation gives ( ) ( ) ( ) ) 2 sin( 4 ) cos( ) 2 sin( ) 2 cos( ) sin( ) cos( 1 ) 2 cos( 2 ) 2 sin( 2 ) cos( ) sin( 2 ) 2 sin( 4 ) 2 cos( 4 ) sin( ) cos( 4 3 2 1 4 3 2 1 4 3 2 1 t t t K t K t K t K t K t K t K t K t K t K t K t K ! ! = + + + + + ! + ! + ! ! ! ! Collecting the coefficients of the cosine and sine terms, we get ( ) ( ) ( ) ( ) ) 2 sin( 4 4 4 ) 2 cos( 4 4 ) sin( 2 ) cos( 1 2 4 3 4 3 4 3 2 1 2 1 2 1 = + + ! ! + + + ! + + ! ! + + + + ! t K K K t K K K t K K K t K K K which gives K 1 = 0, K 2 = − 0.5, K 3 = 0.64, and K 4 = 0.48. (b) Homogeneous solution: The characteristic equation of the LTIC system (iii) is 1 2 2 = + + s s , Assignment 2 CSE3451: Signals and Systems 2 which has roots at s = − 1, − 1 . The zero-input response is given by t t zi Bte Ae t y ! ! + = ) ( for t ≥ 0, with A and B being constants. (c) Overall response of the system: The overall response of the system is obtained by summing up the above two responses, and use initial conditions to determine A and B, it is given by ( ) ( ) ) ( ) 2 sin( 48 . ) 2 cos( 64 . ) sin( 5 . 1 . 1 64 . ) ( 4 3 ) ( t u t t t te e t u te e t y t t t t + + ! ! ! + + = ! ! ! ! Problem 3.5 (ii) The output y ( t ) is given by ! ! " # " " # \$ # \$ = \$ # \$ \$ # = # % # = ) ( ) ( ) ( ) ( ) ( ) ( d t u d t u u t u t u t y ....
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a2_solution - Instructors Solutions to Assignment 2 Problem...

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