solved11 - S 11.1 (P 3.4) _ The four columns of V are...

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S 11.1 (P 3.4) ______________ The four columns of V are orthogonal, with squared norm equal to 4. Any s can be expressed as s = c0*v0 + c1*v1 + c2*v2 + c3*v3 where c# = (v#)'*s/4 Thus 4*c0 = [1 1 1 1]*[1 ; 4 ; -2 ; 5] = 8 4*c1 = [1 -j -1 j]*[1 ; 4 ; -2 ; 5] = 3 + j 4*c2 = [1 -1 1 -1]*[1 ; 4 ; -2 ; 5] = -10 4*c3 = [1 j -1 -j]*[1 ; 4 ; -2 ; 5] = 3 - j and c = [2 ; (3+j)/4 ; -5/2 ; (3-j)/4] S 11.2 (P 3.1) ______________ a = 1/2; b = sqrt(3)/2; v = [ 1 a+j*b -a+j*b -1 -a-j*b a-j*b ]; (i) By inspection, if such z exists, it must equal z = a+j*b = exp(j*pi/3) Indeed, z^2 = exp(j*2*pi/3) = -a+j*b z^3 = exp(j*pi) = -1 z^4 = exp(j*4*pi/3) = -a-j*b z^5 = exp(j*5*pi/3) = a-j*b (ii) v = v1 = first Fourier sinusoid of length N=6. The complex conjugate of v equals v5, the fifth Fourier sinusoid of length N=6. Finally, the all-ones vector equals v0, the zeroth Fourier sinusoid of length N=6. The three sinusoids are orthogonal, with squared norm equal to N=6. The least squares approximation of the six-dimensional vector s
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solved11 - S 11.1 (P 3.4) _ The four columns of V are...

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