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# Tuto4_sol - EE3210 Tutorial 4(Solution 1 2 x n h n = 12 n...

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EE3210 Tutorial 4 (Solution) 1. ] 2 [ 2 ] 1 [ 10 ] [ 12 ] [ ] [ + + = n n n n h n x δ δ δ 2. (a) ] 4 [ 2 ] 2 [ 2 ] 1 [ 2 ] [ 4 ] 1 [ 2 ] [ 1 + + + + = n n n n n n y δ δ δ δ δ (b) ] 2 [ ] [ 1 2 + = n y n y (c) ] [ ] [ 2 3 n y n y = 3. The convolution sum is −∞ = −∞ = −∞ = = = = k k n k k n k k k n u k u k n u k u k n h k x n y ] [ ] [ ) / ( ] [ ] [ ] [ ] [ ] [ β α β β α . When n < 0, u [ k ] u [ n - k ] = 0 for all k , implying that y [ n ] = 0. When n 0, u [ k ] u [ n - k ] = 1 for 0 k n. Therefore, α β α β β α β α β β α β = = = + + + = 1 1 1 0 ) / ( 1 ) / ( 1 ) / ( ] [ n n n n n k k n n y . Combining the above two cases, we have ] [ ] [ 1 1 n u n y n n α β α β = + + . 4. We first observe that these two signals have regions of nonzero overlap, regardless of the value of t . When t – 3 0, the product of x ( τ ) and h ( t τ ) is nonzero for – < τ < t – 3, and the convolution integral becomes
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