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K n k Figure S243 Graphically we see that for n < 0, x(k) h(n k) is zero and consequently y(n) = 0, n < 0 For n > 0 n n y(n) = ak nk n (a,/,)n k=0 k=0 _ln( ax~ n1l nl_ nl 1(a~~3 Consequently for all n, Knlan11 y (n)Show that n xn *hn = xmhn – m m=0 n = (b) Suppose now that you wish to convolve xn with hn and hn is nonzero only for 0,1,2,3, 4 Show that one way to implement the convolution is as shown in the block diagram of Figure 422Y n x n h n x k h n u 1 u nk 1 3 1 * uk1 = 1 for –k1 ≥0 k≤1 0 otherwise Condition 1 – when k≤1 the summation is not 0 & the equation for yn can be writer as ∑ − =−∞ − − − = 1 1 3 1 k k y n u n k unk1 = 1 for n–k1 ≥0 k ≤ n1 0 otherwise
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