S 6.1 (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 6.2 (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
in terms of v0, v1 and v5 equals

	s_hat = c0*v0 + c1*v1 + c5*v5,

where, by orthogonality:

	c0 = v0'*s/6 = 5/6

	c1 = v1'*s/6
	   = (3+2*(a-j*b)-(-a-jb)+0-(-a+j*b)+2*(a+jb))/6
	   = (3+6*a)/6 
	   = 1

(v1' is the *conjugate* transpose of v1)

	c5 = v5'*s/6 

Since s is real-valued and v5 is the complex conjugate of v1, it
follows that 

	c5 = (c1)' = 1

Thus
	s_hat = (5/6)*v0 + v1 + v5 

	      = (5/6)*v0 + 2*Real(v)

	      = [17 ; 11 ; -1 ; -7 ; -1 ; 11]/6


S 6.3 (P 3.2)
______________

Since the complex Fourier sinusoids are orthogonal,
the least squares approximation 

	s_hat = c0*v0 + c1*v1 + c7*v7

of s based on v0, v1 and v7 will be the sum of its projections on each
of the vectors

	v0 = [1 1 1 1 1 1 1 1]'

	v1 = [ 	 1 
       		 (1+j)/sqrt(2) 
       		 j 
       		 (-1+j)/sqrt(2) 
      		-1
       		 (-1-j)/sqrt(2)
      		-j
       		 (1-j)/sqrt(2) ]

	v7 = complex conjugate of v1

All vi's have the same norm^2 = 8, so

	c0 = 2

	c1 = (2+sqrt(2))/4

	c2 = c1' (complex conjugate) = c1 since c1 is real.

We know that the error vector s_hat-s is orthogonal to s_hat, 
therefore

	||s||^2 = ||s_hat - s||^2 + ||s_hat||^2

(This identity is true in any least squares approximation problem.)

We have

	||s||^2 = 44

and since v0, v1, v7 are orthogonal with norm^2 = 8:
	
	||s_hat||^2 = (|c_0|^2 + |c_1|^2 + |c_7|^2)*8

		= 38+4*sqrt(2) = 43.657

So 
	||s_hat - s||^2 = 44.0 - 43.657 = 0.343



S 6.4 (P 3.7)
______________

	X = [4  1+j  3-j  z1   z2].'

(i) From the analysis equation (k=0), we have that

	x[1] + x[2] + x[3] + x[4] + x[5] = X[0] = 4

(ii) Since x is real-valued, X has conjugate symmetry about
index k=5/2.  Thus (with ' denoting complex conjugate, as in MATLAB)

	X[3] = X'[2] , or z1 = 3+j
	X[4] = X'[1] , or z2 = 1-j

(iii)  Amplitude spectrum is abs(X):

	[4  sqrt(2) sqrt(10)  sqrt(10)  sqrt(2)].'

	= [4.0000    1.4142    3.1623    3.1623    1.4142].'

Phase spectrum is angle(X):

	[0  atan(1)  atan(-1/3)  atan(1/3)  atan(-1) ].'

	= [0    0.7854   -0.3218    0.3218   -0.7854].'

Therefore:

	x = (1/5)*4

	  + (2/5)*1.4142*cos(2*pi*n/5 + 0.7854)

	  + (2/5)*3.1623*cos(4*pi*n/5 - 0.3218)


S 6.5 (P 3.8)
______________

	x[n] = 3*(-1)^n + 7*cos(pi*n/4 + 1.2) + 2*cos(pi*n/2 - 0.8)


(i) Of the eight Fourier frequencies k*(2*pi/8), (k=0,1,...,7)
only five are present in x[n]: 

	k = 1, 2, 4, 6, 7

(ii) The synthesis equation for a real-valued N-point vector can
be also written as

	x[n] = X[0]/N + (X[N/2]/N)*(-1)^n
		+ sum (2*|X[k]|/N)*cos(2*pi*k*n/N + angle(X[k]))

where the last sum is over 0<k<N/2.  Comparing this to the given
equation for x[n], we have:

	|X[k]| = [0  28  8  0  24  0  8  28].'

	angle(X[k]) = [ 0  1.2  -0.8  0  0  0  0.8  -1.2].'


(The angle of X[4] equals 0 because X[4]>0; it would have been equal
to pi otherwise.)