Assignment 1, solution

Assignment 1, solution - Assignment1 ELEC442/ELEC6601...

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Assignment 1 ELEC442/ELEC6601 Concordia University 1 1 Consider following two discrete time systems having input and output signals of ] [ n x and ] [ n y , respectively. System 1: ] 2 [ ] [ n x n y = System 2: = odd n even n n x n y 0 2 ] [ a If ] [ n x is periodic, determine whether ] [ n y is periodic or not. Answer this question for both systems. b If ] [ n y is periodic, determine whether ] [ n x is periodic or not. Answer this question for both systems. Solution 1: a System 1: The answer is true. See the proof below: If ] [ n x is periodic with period of “ N ” then we have ] [ ] [ N n x n x + = for all “ n ” and therefore [ ] ] [ ] 2 [ ] 2 [ ] 2 2 [ n y n x N n x N n x N n y = = + = + = + Having [ ] ] [ n y N n y = + proves that ] [ n y is periodic with period of N . However if N is even the smallest period is 2 N since ] [ ] 2 [ ] 2 [ 2 n y n x N n x N n y = = + = + System 2: The answer is true. See the proof below: If ] [ n x is periodic with period of “ N ” then we have ] [ ] [ N n x n x + = for all “ n ” and therefore [ ] ] [ 0 2 0 2 2 n y odd n even n n x odd n even n N n x N n y = = + = + and therefore ] [ n y is periodic with period of N 2 . b System 1: The answer is false . We just have to give an example. Consider ( ) = odd n even n n x n 2 / 1 1 ] [ which is not periodic. Having this input will result in output of 1 ] 2 [ ] [ = = n x n y which has period of 1 and therefore periodic. System 2: The answer is true. See the proof below: If ] [ n y is periodic with period of “ N ” then we have ] [ ] [ N n y n y + = for all “ n ” and therefore + = odd n even n N n x odd n even n n x 0 2 0 2 . This shows that ] [ n y can be only periodic with an even period and so 0 2 N N = . Therefore: + = odd n even n N n x odd n even n n x 0 2 0 2 0 and ] [ ] [ 0 N n x n x + = and ] [ n x is periodic with period of 0 N

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