S 8.1 (P 3.20)
_______________

	X = [1 0 1 -1].'

	Y = [3 5 8 -4].'

(i) Neither X nor Y has circular conjugate symmetry,
therefore neither x nor y is real-valued.

(ii) Both X and Y are real-valued, therefore both x and y
have conjugate circular symmetry.

(iii) S1 = (1/4)*circular convolution of X and Y.

	X = [1 0 1 -1].'

	R*Y = [3 -4 8 5].'   
	4*S1[0] = X.'*R*Y = 6

	P*R*Y = [5 3 -4 8].'
	4*S1[1] = X.'*P*R*Y = -7

	(P^2)*R*Y = [8 5 3 -4].'
	4*S1[2] = X.'*(P^2)*R*Y = 15

	(P^3)*R*Y = [-4 8 5 3].'
	4*S1[3] = X.'*(P^3)*R*Y = -2

(iv) S2 = element-wise product of the spectra X and Y.

	S2 = [3 0 8 4].'


S 8.2 (P 3.21)
_______________

We have 

	[Y2 Y3 Y0 Y1].' = (P^2)*Y

therefore the inverse DFT (time-domain signal) of [Y2 Y3 Y0 Y1] is

	(F^2)*y = [1 1 2 4].'

Since S is the element-wise product of X and (P^2)*Y, it follows that
the time domain signal s is the circular convolution of x 
and [1 1 2 4].' :

	[ 1  4  2  1   * [2 0 1 3].'  = [7 12 17 12].'    
	  1  1  4  2  
	  2  1  1  4  
	  4  2  1  1 ]


S 8.3 (P 3.26)
_______________

(i) x is obtained from y by zero-padding to double its length.  Thus

	Y = [ X0 X2 X4 X6 X8 X10 ].'

(ii) x is also obtained from [a b c].' by zero-padding to four times
its length.  Thus

	[ a b c ].' <--> [ X0 X4 X8 ].'

S is a periodic extension of [a b c].' to four times its length.  Thus

	S = 4*[ X0 0 0 0 X4 0 0 0 X8 0 0 0 ].'



S 8,4 (P 3.27)
_______________

For x1=s :

	X1 = [X[0] X[4] X[8] X[12]].'

(since the zero-padded vector has length =  4*length
of s).

For x2 = [s ; s] :

	X2 = [2*X[0] 0 2*X[4] 0 2*X[8] 0 2*X[12] 0].'

(periodic extension of s to twice its length)

For x3 = [s ; s ; s ; s ; s] :

	X3 = [5*X[0]; z4 ; 5*X[4] ; z4 ; 5*X[8] ; z4 ; 5*X[12] ; z4]

For x4 = [s; z4] :

	X4 = [X[0] X[2] X[4] X[6] X[8] X[10] X[12] X[14]].'

(since x has exactly twice as many entries as [s;z4])

For x5 = [z4 ; s] :

	X4 = [X[0] -X[2] X[4] -X[6] X[8] -X[10] X[12] -X[14]].'

(circular shift of x4 by 4=8/2 units, so multiplication by (-1)^k
in the frequency domain)

For x6 = [s ; z4 ; s ; z4] :

	X6 = [2*X[0] 0 2*X[2] 0 2*X[4] 0 2*X[6] 0 ...
	            2*X[8] 0 2*X[10] 0 2*X[12] 0 2*X[14] 0 ].'

(periodic extension of [s;z4] to twice its length)

For x7 = [s ; s ; z4 ; z4] :

	x7 = x + (P^4)*x

and thus 

	X7 = X + F^(-4)*X

Since 4*(2*pi/16) = pi/2, we have

	X7[k] = (1+(-j)^k)*X[k]

for every k (k=0:15) .


S 8.5 (P 3.29)
_______________

(i) The discrete-time (sampled) signal consists of two Fourier
sinusoids at frequencies

	7*(2*pi/32)  and   11*(2*pi/32)

The frequencies of the continuous-time components are found by
multiplying the above by fs=500 samples/sec:

	7*(2*pi/32)*500  and   11*(2*pi/32)*500  rad/sec
or
	(7/32)*500 = 109.4  and   (11/32)*500 = 171.9  Hz

(ii) The phase spectrum is not given in this problem, so we cannot
write an equation for the time-domain signal.  If the phase spectrum
were given, and the phases at the two frequencies (k=7 and k = 11)
were equal to q1 and q2, respectively, then the continuous time 
signal would be given by the equation

	x(t) = (252/32)*cos(7000*pi*t/32 + q1) + ...
	       (168/32)*cos(11000*pi*t/32 + q2)

S 8.6 (P 3.30)
_______________

The two peaks at k = 13 (roughly) and k = 20
correspond to angular frequencies

	(13/80)*(2*pi) = 0.16*(2*pi)
and
	(20/80)*(2*pi) = 0.25*(2*pi)

for the sampled signal.  Since the sampling
rate equals 800 samples/sec, the corresponding
frequencies in Hz (for the continuous-time signal)
are approximately

	(13/80)*800 = 130 Hz

and

	(20/80)*800 = 200 Hz

(Note: Sampling at a rate >2*f1 and >2*f2
guarantees that both discrete sinusoids
have frequencies in [0,pi).)

S 8.7 (P 3.31)
_______________

(i) f1 = 164/640 and f2 = 182/640 cycles/sample.

(ii) For N = 200, the Fourier frequencies will be the multiples 
of 2*pi/200, or 2*pi*0.005.

f1 = 164/640 = 0.25625, so nearest Fourier frequency is 2*pi*(0.255),
corresponding to k = 51

f2 = 182/640 = 0.284375, so nearest Fourier frequency is 2*pi*(0.285),
corresponding to k = 57

(iii) We need 164/640 and 182/640 to both be of the form k/N, where
N = M+200

	164/640 = 41/160  and 182/640 = 91/320

so the smallest value of N for which both frequencies are expressible
as k/N is 320.  This means that M = 120.  Then 

f1 = 82/320 (i.e., k=82)   and f2 = 91/320 (i.e., k=91).