hwsolB20 - 4 2 -2 -4 10 -10 ].' d[0] = (4 + 2-2-4+10-10)/6...

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HW 20 Solutions_______________Let a = sqrt(3) throughout.v0 = [ 1 1 1 1 1 1 ].'v1 = [ 1 (1+j*a)/2 (-1+j*a)/2 -1 (-1-j*a)/2 (1-j*a)/2 ].'v2 = [ 1 (-1+j*a)/2 (- 1-j*a)/2 1 (-1+j*a)/2 (-1-j*a)/2 ].'v3 = [ 1 -1 1 -1 1 -1 ].'v4 = [ 1 (-1- j*a)/2 (-1+j*a)/2 1 (-1-j*a)/2 (-1+j*a)/2 ].' = complex conjugate of v2v5 = [ 1 (1-j*a)/2 (-1-j*a)/2 -1 (-1+j*a)/2 (1+j*a)/2 ].' = complex conjugate of v1(i) x = [ 4 4 1 -2 -2 1].'c[k] = <vk,x>/6 = (vk)'*x/6 (note the complex conjugate in vk) c[0] = (4 + 4 + 1 - 2 - 2 + 1)/6 = 6/6 = 1 c[1] = (4*1 + 4*(1-j*a)/2 + 1*(-1-j*a)/2 + 2 -2(-1+j*a)/2 + 1*(1+j*a)/2)/6 = 3/2 - j*a/2 c[2] = (4*1 + 4*(-1-j*a)/2 + 1*(-1+j*a)/2 - 2 -2(-1-j*a)/2 + 1*(-1+j*a)/2)/6 = 0 c[3] = (4 - 4 + 1 + 2 - 2 - 1)/6 = 0 c[4] = complex conjugate of c[2], since x is real and v4 is the complex conjugate of v2 ; so c[4] = 0 c[5] = complex conjugate of c[1], since x is real and v5 is the complex conjugate of v1 ; so c[5] = 3/2 + j*a/2Since c[2] = c[3] = c[4] = 0, it follows that x is a linear combinationof v0, v1 and v5 only.(ii) y = [
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Unformatted text preview: 4 2 -2 -4 10 -10 ].' d[0] = (4 + 2-2-4+10-10)/6 = 0 d[1] = (4*1 +2(1-j*a)/2 -2(-1-j*a)/2 + 4 + 10*(-1+j*a)/2 - 10*(1+j*a)/2)/6 = d[2] = (4*1 +2(-1-j*a)/2 -2(-1+j*a)/2 - 4 + 10*(-1-j*a)/2 - 10*(-1+j*a)/2)/6 = -j*2*a d[3] = (4 - 2 - 2 + 4 + 10 +10)/6 = 4Since y is real,d[4] = j*2*a, d[5] = 0y is a linear combination of v2, v3 and v4 only.(iii) c = fft(x)/6, d = fft(y)/6(iv) s = x + y = V*e, where e = [ c[0] c[1] d[2] d[3] d[4] c[5] ].' = [ 1 3/2-j*a/2 -j*2*a 4 j*2*a 3/2+j*a/2 ].'Since the columns of V are orthogonal, the least squares approximation ofs in terms of v1, v3 and v5 is given by s_hat = V*[0 3/2-j*a/2 0 4 0 3/2+j*a/2 ].' = c[1]*v1 + d[3]*v3 + c[5]*v5To avoid using complex arithmetic for c[1]*v1 + c[5]*v5 , note that c[1]*v1 + c[5]*v5 = x - c[0]*v0Thus s_hat = x - c[0]*v0 + d[3]*v3 = [ 4 4 1 -2 -2 1].' - [1 1 1 1 1 1].' + [4 -4 4 -4 4 -4].' = [7 -1 4 -7 1 -4].'...
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