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# Assign5_sol - EE 3210 Assignment 5 Solution Question 1 a...

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EE 3210 Assignment 5: Solution Question 1 a) The frequency response of this system can be found as follows: ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω j j j j j j j j X j Y j H j X j X j j Y j Y j j Y j + + + = + + + = + + = = + = + + 2 2 / 3 4 2 / 3 ) 2 )( 4 ( 3 9 6 8 3 9 ) ( ) ( ) ( ) ( 9 ) ( ) ( 3 ) ( 8 ) ( ) ( 6 ) ( ) ( 2 2 b) Taking its inverse Fourier transform, we obtain [ ] ) ( 2 3 ) ( 2 4 t u e e t h t t + = . c) Taking Fourier transform of x ( t ), we obtain ) 3 )( 1 ( ) 2 ( 2 3 1 1 1 ) ( ω ω ω ω ω ω j j j j j j X + + + = + + + = . The Fourier transform of the output signal is ω ω ω ω ω ω ω j j j j j X j H j Y + + = + + = = 4 2 1 2 ) 4 )( 1 ( 6 ) ( ) ( ) ( . Taking its inverse Fourier transform, we obtain [ ] ) ( 2 ) ( 4 t u e e t y t t + = . Question 2 a) Consider x [ n ] = e j ω n and y [ n ] = H ( e j ω ) e j ω n . Then we have ω ω ω ω ω ω ω ω j j n j j n j j n j j e e H e e H e e e e H = = 4 4 ) ( ) ( 4 1 ) ( b) The fundamental period of x [ n ] is 10. The fundamental frequency of x [ n ] is π /5. [ ] 5 / 6 / 5 / 6 / ) 6 / 5 / ( ) 6 / 5 / ( 2 2 2 2 1 2 ] [ n j j n j j n j n j e e e e e e n x π π π π π π π π + + = + + = Therefore, the non-zero Fourier series coefficients are 2 , 2 , 2 6 / 1 6 / 1 0 π π j j e a e a a = = = .

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c) The output signal is: 5 / 5 / 6 / 5 / 5 / 6 / 5 / 5 / 6 / 5 / 5 / 6 / 4 2 4 2 3 8 4 4 2 4 4 2 ) 3 4 ( 2 ] [ n j j j n j j j n j j j n j j j e e e e e e e e e e e e n y π π π π π π π π π π π π
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