solved11

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/18/2011 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online