solvedB36 - S 36.1 (P 4.13) _ (i) The response to...

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Sheet1 Page 1 S 36.1 (P 4.13) ______________ (i) The response to delta[n-m] is h[n-m]. Therefore the response to x = delta[n+1] - delta[n-1] is y = h[n+1] - h[n-1] The first nonzero value of h occurs at n=0 and the last one at n=4. The first nonzero value of y occurs at n=-1 (produced by h[n+1]), and the last one is at n=5 (produced by h[n-1]). y[-1:5] = [1 -2 0 3 -2 0 0].' - [0 0 1 -2 0 3 -2].' = [1 -2 -1 5 -2 -3 2]' y = 0 for n<-1 and n>5 (ii) The duration of h is 5 time units, that of x is 4 time units, thus the duration of y is 5+4-1=8 time units, from n=0 until n=7. k: 0 ____________________________________________________ h[k]: 1 -2 0 3 -2 y ____________________________________________________ x[-k] -1 3 2 1 1 x[1-k] -1 3 2 1 0 x[2-k] -1 3 2 1 -1 x[3-k] -1 3 2 1 -4 x[4-k] -1 3 2 1 6 x[5-k] -1 3 2 1 5 x[6-k] -1 3 2 1 -9 x[7-k] -1 3 2 1 2 So y[0:7] = [1 0 -1 -4 6 5 -9 2].' (iii) We multiply the polynomials H(z) = 1 - 2*z^(-1) + 3*z^(-3) - 2*z^(-4) and X(z) = 1 + 2*z^(-1) +3*z^(-2) - z^(-3)
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solvedB36 - S 36.1 (P 4.13) _ (i) The response to...

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