HW2_sol

HW2_sol - x(n*h(1 y(2*n-1 = x(n*h(n Therefore y[n =[2-10-11...

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ECE 158 Fall 2010 Gibson Homework #2 Solutions Problem 1: 5.2 b) x[n] = [-1 5 3 0 3] h[n] = [-2 0 5 3 -2] By adding the columns above each Y c component, you should obtain Y c [n] = [1 -1 -2 16 26]. If you have the 4th edition with x[n] = [-1 5 3 0 3] and h[n] = [-2 0 5 3 -2], then Y c [n] = [11 -6 -15 14 -45 24] by using the method shown above. Problem 2: x[n] = [-1 5 3 0 3] h[n] = [-2 0 5 3 -2] Linear Convolution of x[n] and h[n].

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y(1) = x(1)*h(1) y(2) = x(1)*h(2)+x(2)*h(1) y(3) = x(1)*h(3)+x(2)*h(2)+x(3)*h(1) ... y(n) = x(1)*h(n)+x(2)*h(n-1)+ ... +x(n)*h(1)
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Unformatted text preview: .. +x(n)*h(1) ... y(2*n-1) = x(n)*h(n) Therefore y[n] = [2 -10 -11 22 26 -1 9 9 -6] For the 4th edition with x[n] = [-1 5 3 0 3] and h[n] = [-2 0 5 3 -2], then Y c [n] = [6 -3 -4 12 -48 24 5 -3 -11 2 3]. Problem 3: Problem 4: Proving Eq. 5.116 Checking if eq. 5.115 holds we have. .. y c [n] = [6 7 6 5] Y c [k] = [24 -j2 0 j2] = therefore. .. ( 36 + 49 +36 +25) = .25(576 + 4 + 0 + 4) 146 = 146, and Eq. 5.115 holds. Problem 5: G[k] = [4, 1-j, -2, 1+j] (equation 5.29) (equation 5.31) Therefore. .....
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