S 11.1 (P 4.4)
______________

(i) 

	H(z) = 1 - z^(-1} - *z^(-2) + z^(-3)

(ii)  H(exp(j*omega)) = 1 - exp(-j*omega) - exp(-j*2*omega) + exp(-j*3*omega)
	= exp(-j*3*omega/2)...
	 *(exp(j*3*omega/2) - exp(j*omega/2) - exp(-j*omega/2) + exp(-j*3*omega/2))
	= exp(-j*3*omega/2)*(2*cos(3*omega/2)-2*cos(omega/2))

Thus 
	|H(exp(j*omega)| = |2*cos(3*omega/2)-2*cos(omega/2)|
and
	|H(exp(j*omega)|^2 = (2*cos(3*omega/2)-2*cos(omega/2))^2

(Note: An alternative expression for |H(exp(j*omega)|^2 is 
found by expanding

	H(exp(j*omega))*H(exp(-j*omega)) )

To plot |H(exp(j*omega)|:

	h = [1 -1 -1 1];
	f = (0:511)/512;	% 512 points chosen (arbitrarily)
	H = fft(h,512);
	plot(f,abs(H)), title('Amplitude Response'), xlabel('\Omega/2\pi')
	

(iii)
	x[n] = 1 = exp(j*0*n)
	y[n] = H(exp(j*0)*x[n] = 0

	x[n] = (-1)^n = exp(j*pi*n)
	y[n] = H(exp(j*pi))*x[n] = 0 

	x[n] = exp(j*pi*n/4) 
	y[n] = H(exp(j*pi/4))*x[n] 
	     = (1 - exp(-j*pi/4) - exp(-j*2*pi/4) + exp(-j*3*pi/4))*x[n]
	     = (-sqrt(2)+1+j)*exp(j*pi*n/4)

	x[n] = cos(pi*n/4+q) 
	y[n] = |H(exp(j*pi/4))| * cos((pi*n/4) + q + angle(H(exp(j*pi/4)))
	     = (4-2*sqrt(2))*cos((pi*n/4)+ q + 3*pi/8)
	     = |2*cos(3*pi/8)-2*cos(pi/8)|cos((pi*n/4)+ q - 3*pi/8 + pi) 
	     = (sqrt(2+sqrt(2))-sqrt(2-sqrt(2)))cos((pi*n/4)+ q + 5*pi/8)
	     = 1.0824*cos((pi*n/4)+ q + 5*pi/8)


	x[n] = 2^(-n)
	y[n] = H(1/2)*x[n] = (1-2-4+8)*x[n] = 3*(1/2)^n

	x[n] = 2^(-n)cos(pi*n/4) = Real( 2^(-n)*exp(j*pi*n/4) )

	We need H(z) for z = exp(j*pi/4)/2

	H( exp(j*pi/4)/2 ) = -6.0711 - 0.2426*j = 6.0759*exp(-j*3.1016)

	So 
		y[n] = 6.0759*(1/2)^n*cos((pi*n/4) - 3.1016)


S 11.2 (P 4.8)
______________

	H1(z) = 3 + 2*z^(-1) + z^(-2) + 2*z^(-3) + 3*z^(-4)

	H2(z) = 1 - 2*z^(-1) + 2*z^(-2) - z^(-3)

(i) The cascade has system function

	H(z) = H1(z)*H2(z) 
	     = 3 - 4*z^(-1) + 3*z^(-2) + z^(-3) - z^(-4) -3*z^(-5) + 4*z^(-6) - 3*z^(-7)

(ii) The impulse response of the cascade is

	h[0:7] = [3 -4 3 1 -1 -3 4 -3].'
	h[n] = 0 for n<0 and n>7

(iii) We factor out z^(-7/2) = exp(-j*7*w/2), then use the identity

	exp(j*k*w/2) - exp(-j*k*w/2) = 2*j*sin(k*w/2)

to obtain

	H(exp(j*w)) = -2*j*exp(-j*7*w/2)
			*( sin(w/2) + 3*sin(3*w/2) - 4*sin(5*w/2) + 3*sin(7*w/2) )

Taking the modulus of each term on the right-hand side, we have

	|H(exp(j*w)} = 2*| sin(w/2) + 3*sin(3*w/2) - 4*sin(5*w/2) + 3*sin(7*w/2) |


S 11.3 (P 4.9)
______________

(i) H(z) = (1 + z^(-1) + z^(-2) + z^(-3))^2

	= 1 + 2*z^(-1) + 3*z^(-2) + 4*z^(-3) + 3*z^(-4) + 2*z^(-5) + z^(-6)

Coefficient vector of cascade:

	b = [1 2 3 4 3 2 1].'

(ii) The frequency of the sinusoid is w=pi/2.  We have

	H(exp(j*w)) = (1 + exp(-j*w) + exp(-j*2*w) + exp(-j*3*w)).^2

		= (1 - j - 1 + j).^2  =  0

Therefore
		y[n] = 0*cos(n*pi/2 + 0) = 0  for all n

S 11.4 (P 4.3)
______________

Divide by 2*pi to express the three frequencies in cycles per 
sample:

	3/28, 9/35 and 17/48

Since these are rational (i.e., integer fractions), the
signal is periodic.  The period is the smallest integer
L such that each frequency can be expressed as k/L.  
With irreducible fractions (as above), this is given by
the least common multiple (LCM) of the denominators:

	LCM(28,35,48) = 1680


S 11.5 (P 4.7)
______________

The output y is also periodic with period L=7.
If 
	xL = x[0:6]	and 	yL = y[0:6]

then the DFT's of xL and yL are related by

	YL[k] = H(exp(j*2*pi*k/8))*XL[k]

for k=0:7.  The easiest way to obtain the values H(exp(j*2*pi*k/8))
numerically is to zero-pad b to 2*L = 14 points, then take every
*other* coefficient in the DFT.

The MATLAB code is

	b = [ 1 2 -2 -1 4 -1 -2 2 1 ].' ;
	B = fft(b,14);
	H = B(1:2:13);
	xL = [ 1 -1 0 3 1 -2 0 ].' ;
	XL = fft(xL);
	YL = H.*XL;
	yL = ifft(YL)
	bar(0:6,yL)

and the resulting yL vector is

	[20 3 -20 10 21 -14 -12]'

Alternatively:  

	b = [ 1 2 -2 -1 4 -1 -2 2 1].' ;
	B = fft(b,14);
	xL = [ 1 -1 0 3 1 -2 0 ].' ;
	xL2 = [xL ; xL]; 	% two periods of xL
	XL2 = fft(xL2);		% odd-indexed entries are zero
	YL2 = B.*XL2;		% no need to sample even-indexed entries
	yL2 = ifft(YL2);	% two periods of yL
	yL = yL2(1:7)



S 11.6 (P 4.13)
______________

(i) The response to delta[n-m] is h[n-m].  Therefore the response to

	x = delta[n+1] - delta[n-1]
is
	y = h[n+1] - h[n-1]

The first nonzero value of h occurs at n=0 and the last one at n=4.

The first nonzero value of y occurs at n=-1 (produced by h[n+1]), 
and the last one is at n=5 (produced by h[n-1]).

	y[-1:5] = [1 -2 0 3 -2 0 0].' - [0 0 1 -2 0 3 -2].'
		= [1 -2 -1 5 -2 -3 2]'

	y = 0 for n<-1 and n>5

(ii) The duration of h is 5 time units, that of x is 4 time units, thus
the duration of y is 5+4-1=8 time units, from n=0 until n=7.

 k:                   0
____________________________________________________
 h[k]:                1  -2   0   3  -2                     y
____________________________________________________
 x[-k]   -1   3   2   1                                     1
 x[1-k]      -1   3   2   1                                 0
 x[2-k]          -1   3   2   1                            -1
 x[3-k]              -1   3   2   1                        -4
 x[4-k]                  -1   3   2   1                     6
 x[5-k]                      -1   3   2   1                 5
 x[6-k]                          -1   3   2   1            -9
 x[7-k]                              -1   3   2   1         2


So 

	y[0:7] = [1  0 -1 -4  6  5 -9 2].'

(iii) We multiply the polynomials

	H(z) = 1 - 2*z^(-1) + 3*z^(-3) - 2*z^(-4)
and
	X(z) = 1 + 2*z^(-1) +3*z^(-2) - z^(-3)

together to obtain

	H(z)X(z) = 1 - z^(-2) - 4*z^(-3) + 6*z^(-4)
                   + 5*z^(-5) - 9*z^(-6) + 2*z^(-7)

This is Y(z), and the coefficients equal y[0:7].


S 11.7 (P 4.14)
______________

(i) x has duration n=0:M, thus conv(x,x) has duration n=0:2*M
(length = 2*M+1).  

For n in 0:2*M:

	y[n] = x[0]*x[n] + x[1]*x[n-1] + ... + x[n-1]*x[1] + x[n]*x[0]

There are a total of n+1 terms (summands) in the above sum.

If n<=M, then all the summands are nonzero; they are in fact equal to 1,
and thus

	y[n] = number of terms in the sum = n+1

If n>M, then 2*(n-M) summands will contain exactly one x[i] with i>M, 
which equals zero.  Thus

	y[n] = (number of terms in the sum) - 2*(n-M) = 2*M-n+1

The result is symmetric about n = M; this is a triangular signal
with peak at n = M.

To illustrate in the case M = 4:

 x[k]:                    1   1   1   1   1                      y
___________________________________________________________    
 x[-k]    1   1   1   1   1                                      1
 x[1-k]       1   1   1   1   1                                  2
 x[2-k]           1   1   1   1   1                              3
 x[3-k]               1   1   1   1   1                          4
 x[4-k]                   1   1   1   1   1                      5
 x[5-k]                       1   1   1   1   1                  4
 x[6-k]                           1   1   1   1   1              3
 x[7-k]                               1   1   1   1   1          2
 x[8-k]                                   1   1   1   1   1      1

(ii) An advance of M time units would make y a symmetric signal.



S 11.8 (P 4.15)
______________

Since both sequences x and h equal zero for n<0, so does y=conv(x,h).
Since h has infinite duration (over the positive time axis), so does
y.  This is expected, since a single impulse in the input generates
a response that never turns off.

For n = 0:3
	
	y[0] = h[0]*x[0] = 1

	y[1] = h[0]*x[1] + h[1]*x[0]  = -1 + a 

	y[2] = h[0]*x[2] + h[1]*x[1] + h[2]*x[0] = 1 -a + a^2

	y[3] = h[0]*x[3] + h[1]*x[2] + h[2]*x[1] + h[3]*x[0] = -1 +a - a^2 + a^3

For n>3, the convolution sum has four nozero terms only:

	y[n] = h[n-3]*x[3] + h[n-2]*x[2] + h[n-1]*x[1] + h[n]*x[0]

		= a^(n-3)*(h[0]*x[3] + h[1]*x[2] + h[2]*x[1] + h[3]*x[0])

		= a^(n-3)*(-1 +a - a^2 + a^3)