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Tuto5_sol - EE3210 Tutorial 5(Solution 1 Since the input...

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EE3210 Tutorial 5 (Solution) 1. Since the input is δ [ n ] and the system is initially at rest, we can conclude that h [ n ] = 0 for n < 0. Now, ] 1 [ 2 ] [ ] [ = n y n x n y . Therefore, 8 ] 2 [ 2 ] 3 [ ] 3 [ 4 ] 1 [ 2 ] 2 [ ] 2 [ 2 ] 0 [ 2 ] 1 [ ] 1 [ 1 ] 1 [ 2 ] 0 [ ] 0 [ = = = = = = = = h x h h x h h x h h x h and so on. Combining the two cases, we can write the impulse response in closed form as follows: ] [ ) 2 ( ] [ n u n h n = . 2. Since the system is initially at rest, we must have h [ n ] = 0 for n < 0. Next, we consider the following polynomial equation: 0 16 . 0 6 . 0 2 = r r . Factorize it, 0 ) 8 . 0 )( 2 . 0 ( = + r r . The two roots are -0.2 and 0.8. Hence, for n 0, we have ( ) n n C C n h ) 8 . 0 ( 2 . 0 ] [ + = 2 1 . We determine two initial conditions recursively as follows: 3 ] 1 [ 16 . 0 ] 0 [ 6 . 0 ] 1 [ 5 ] 1 [ 5 ] 2 [ 16 . 0 ] 1 [ 6 . 0 ] 0 [ 5 ] 0 [ = + + = = + + = h h h h h h δ δ Now we solve for C 1 and C 2 : 2 1 2 1 8 . 0 2 . 0 3 5 C C C C + = + = Therefore, C 1 = 1 and C 2 = 4. Hence, ( ) ] [ ] ) 8 . 0 ( 4 2 . 0 [ ] [ n u n h n n + = .
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