S 7.1 (P 3.10)
_______________

	x = [ 2  1 -1 -2 -3  3 ].'

We have:

	x1 = P*x = [3  2  1 -1 -2 -3].'

	x2 = (P^5)*x = [1 -1 -2 -3  3  2].'

	x3 = P*x + (P^5)*x = [4 1 -1 -4  1  -1].'

	R*x = [2  3 -3 -2 -1  1].'

	x4 = x + R*x = [4  4 -4 -4 -4  4 ].'

	x5 = x - R*x = [0 -2  2  0  -2  2].'

	F^3 = diag([1 -1  1 -1  1 -1]), so

	x6 = (F^3)*x = [2 -1  -1  2  -3  -3].'

	x7 = x - (F^3)*x = [0  2  0 -4  0  6].'


S 7.2 (P 3.11)
_______________


	x = [ a b c d e f ].'

	X = [ A B C D E F ].'

For x_1:

	x_1[n] = (-1)^n * x[n] = v^(3*n) * x[n]

By Property 6 (multiplication by Fourier sinusoid in
time domain), X is circularly shifted by +3 indices:

	X_1 = P^(-3)*X = [ D E F A B C ].'

For x_2:

	x_2 = (x + x_1)/2

By linearity of the DFT (Property 1)

	X_2 = (1/2)*[ A+D B+E C+F D+A E+B F+C ].'

For x_3: 

	x_3 = R*x, i.e., circular reversal of x

By Property 4 (circular reversal in time domain),

	X_3 = R*X = [ A F E D C B ].'

For x_4:

	x_4 = (P^3)*x, i.e., circular delay of x by 3 units

By Property 5 (circular delay in the time domain), X is
multiplied elementwise by w^(3*k) = (-1)^k:

	X_4 = [ A -B C -D E -F ].'

For x_5:

	x_5 = P*x_3, i..e, circular delay of x_3 by 1 unit:

By Property 5, X_3 is multiplied elementwise by w^k, i.e., 
the entries of the vector

	[ 1 r-js -r-js -1 -r+js r+js ].'

where r = 1/2, s = sqrt(3)/2.  The result is

	X_5 = [ A (r-js)*F (-r-js)*E -D (-r+js)*C (r+js)*B ].'

For x_6:

	x_6 = (P + P^5)*x

By Property 5, X is multiplied elementwise by

	w^k + w^(5k) = w^n + w^(-k) = 2*cos(pi*k/3), 

i.e., the entries of the vector

	2*[ 1  r  -r  -1  -r  r ].'

to yield

	X_6 = 2*[ A r*B -r*C -D -r*E r*F ].'

For x_7:

	x_7[n] = sin(pi*n/3) * x[n] 

               = (v^n - v^(-n))*x[n]/(2j)

By Property 6, 

	X_7 = (P*X - P^(-1)*X)/(2j)

	    = (-j/2)*[ F-B A-C B-D C-E D-F E-A ].'

For x_8:

	x_8 = X

so by duality (Property 7),

	X_8 = NR*x = 6*[ a f e d c b ].'


S 7.3 (P 3.13)
_______________

y[n] = x[n]*cos(pi*n/4)

	 = x[n]*(v^n + v^(-n))/2

where v = exp(j*2*pi*n/8) = exp(j*pi*n/4).

In other words,

	y = (F + F^(-1))*x/2

and thus

	Y = (P + P^(-1))*X/2
	  
	  = (1/2)*[0 0 0 0 -1 2 -1 0].' + ...
	    (1/2)*[0 0 -1 2 -1 0 0 0].'

	  = [0 0 -1/2 1 -1 1 -1/2 0].'


S 7.4 (P 3.15)
_______________


(i) y is a circular shift of x by 6 positions:
	
	y = (P^6)*x 

Therefore

	Y = (F^(-6))*x
i.e.,
	Y[k] = exp(-j*2*pi*6*k/12)*X[k]

Since exp(-j*2*pi*6*k/12) = (-1)^k, we have

	Y = [x0 -X1 X2 -X3 X4 -X5 X6 -X7 X8 -X9 X10 -X11].'

(ii) s = -(F^6)*x, i.e., s[n] = -(-1)^n*x[n]

Therefore

	s = -P^6*x

	  = -[X5 X7 X8 X9 X10 X11 X0 X1 X2 X3 X4 X5].'


S 7.5 (P 3.14)
_______________

	x = [4 3 2 1 0 1 2 3].'
	X = [A B C D E F G H].'

(i) x is real and circularly symmetric, thus X
is also real and circularly symmetric.  It follows
that

	B=H, C=G, and D=F

(ii) y = [0 1 2 3 4 3 2 1].' = (P^4)*x
so
	Y[k] = exp(-j*(2*pi/8)*4*k)*X[k]
	     = (-1)^k * X[k]
             = [A -B C -D E -F G -H].'
	     = [A -B C -D E -D C -B].'

S 7.6 (P 3.16)
_______________

(i) X can be also written as

	X = [1 ; ones(10,1); zeros(43,1); ones(10,1)]

and thus

	R*X = X

i.e., X is circularly symmetric.  Since X is also real,
we conclude that the time domain signal has the same
properties, i.e., it is real-valued and circularly
symmetric.  Thus

	x = ifft(X)

has no imaginary part; a small imaginary part resulting
from the IDFT computation is due to round-off.  After 
confirming this fact, x is defined as

	x = real(ifft(X))

ensuring that it is treated as a real-valued quantity from
that point onwards.

(ii)  Since N = 64, it follows that 3*pi/4 = 24*(2*pi/N),
i.e., it is the k=24th Fourier frequency for this vector size.

	cos(3*pi*n/4) = (exp(j*3*pi*n/4) + exp(-j*3*pi*n/4))/2

(iii) cs = cos(3*pi*n/4) is obviously real, but it is
also circularly symmetric since

	cos(3*pi*(64-n)/4) = cos(3*pi*n/4)

Thus y = x.*cs (element-wise product of x and cs) is also
real-valued and circularly symmetric, and its DFT Y has the 
same properties.  As in part (i), any imaginary part in Y
is due to roundoff, and letting 

	Y = real(fft(y))

ensures that it is plotted as a real-valued signal.

(iv) Element-wise multiplication of x by 

	cos(3*pi*n/4) = ( v^(24*n) + v^(-24*n) )/2

in the time domain results in a spectrum equal to 

	Y = ( P^{24}*X + P^{-24}*X )/2

In this case,

	P^{24}*X = [zeros(14,1); ones(21,1); zeros(29,1)]

	P^{-24}*X = [zeros(30,1); ones(21,1); zeros(13,1)]

and so

	Y = [zeros(14,1); 0.5*ones(16,1); ones(5,1); 
                          0.5*ones(16,1); zeros(13,1)]

which is circularly symmetric.