S 2.1
_____

The sinusoid x[n] = cos(w*n) will satisfy 

	x[n] = x[n+16]

provided

	w = (k/16)*2*pi

for some integer k.  Depending on the value of k,
the fundamental period may equal 16 or a submultiple
of it - namely 1,2,4 or 8.  It will equal 16
if and only if k and 16 have no factors in common.

Thus the fundamental period will equal 16 for

	k = 1, 3, 5 and 7

i.e.,  when
	w = pi/8, 3*pi/8, 5*pi/8 and 7*pi/8.
	    

S 2.2 (P 1.15)
______________

(i) A discrete-time sinusoid of frequency w rad/sample is
periodic if and only if

	w = (k/N)*(2*pi)

where k and N are integers.  The fundamental period is the
smallest integer value of N for which the above holds.  If
that value equals N=4, then the nontrivial values of k are
k=1 and k=3, corresponding to 

	w = pi/2  and   w = 3*pi/2

Thus w = pi/2 is the only such frequency in [0,pi].

(ii) We have:

	v[0] = A*cos(q)
	v[1] = A*cos(q+pi/2) = -A*sin(q)
	v[2] = A*cos(q+pi) = -A*cos(q)
	v[3] = A*cos(q+3*pi/2) = A*sin(q)

Thus
	cos(q) = 1/A,  sin(q) = -1/A

where A>0.  This means that

	q = arctan(-1) = -pi/4

(where the other possibility, namely q = -pi/4 + pi, is
ruled out by the fact that cos(q)>0 and sin(q)<0).

Therefore

	A = 1/cos(-pi/4) = sqrt(2)


S 2.3 (P 1.17)
______________

(i) Since w = 7*pi/9 = (7/18)*2*pi, it follows that sinusoid is
periodic with period N = 18.

MATLAB:

	n = 0:71;	%(four periods)
	w = 7*pi/9;
	x = cos(w*n + pi/6);
	bar(n,x)	%(bar graph)

(ii) We have

	cos(pi*n) = (-1)^n,	sin(pi*n) = 0

Therefore 

	cos(a + pi*n) = cos(a)*cos(pi*n) - sin(a)*sin(pi*n) 
		= cos(a)*cos(pi*n)

Taking a = 7*pi*n/9, we have

	cos(16*pi*n/9 + pi/6) = x1[n]*cos(pi*n) = x2[n]

To express x2[n] using a frequency w in [0,pi]:

	x2[n] = cos(16*pi*n/9 + pi/6)
		= cos(-2*pi*n/9 + pi/6)  (reduce w by 2*n)
		= cos(2*pi*n/9 - pi/6)   (cos(t) = cos(-t))

Thus w = 2*pi/9.  We also have

	cos(2*pi*n/9 - pi/6) = cos(2*pi*n/9 + 11*pi/6)

and now the phase is also in [0,2*pi).


S 2.4 (P 1.16)
______________

(i) We have

	cos(w*(n+1)+q) = cos(w*n+q)*cos(w) - sin(w*n+q)*sin(w)

and

	cos(w*(n-1)+q) =  cos(w*n+q)*cos(-w) - sin(w*n+q)*sin(-w)
		= cos(w*n+q)*cos(w) + sin(w*n+q)*sin(w)

Adding the two equations together, we have

	cos(w*(n+1)+q) + cos(w*(n-1)+q) = 2*cos(w*n+q)*cos(w)

(ii) We have

	x[1] = cos(W*2 + q),  x[2] = cos(W*3 + q),  x[3] = cos(W*4 + q)

The formula derived in part (i) gives (n = 2):

	x[1] + x[3] = 2*cos(w)*x[2]

and thus 

	cos(w) = (x[1] + x[3])/(2*x[2]) = 0.3624

and 

	w = arccos(0.3624) = 1.200 (between 0 and pi).

Therefore

	x[1] = A*cos(1.2 + q) = 1.7740

	x[2] = A*cos(2.4 + q) = 3.1251

	x[3] = A*cos(3.6 + q) = 0.4908

We have

	x[2]/x[1] = cos(2.4+q)/cos(1.2+q) = 3.1251/1.7740 = 1.7616

and using 

	cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)

we obtain

	cos(2.4)*cos(q) - sin(2.4)*sin(q) = 
		= 1.7616*(cos(1.2)*cos(q) - sin(1.2)*sin(q))

Dividing both sides by cos(q) we obtain the following equation
in tan(q):

	cos(2.4) - sin(2.4)*tan(q) 
		= 1.7616*(cos(1.2) - sin(1.2)*tan(q))
or:

	0.9664*tan(q) = 1.3757

leading to
	
	tan(q) = 1.4235

Since arctan(q) = 0.9584, there are two possible values for q:

	q = 0.9584  and   q = 0.9584 + pi = 4.1270

A phase shift of pi reverses the sign of cos(.), so
both values of q are acceptable solutions (same x[2]/x[1] ratio).
We take the one consistent with a positive value of A:

	cos(1.200 + 0.9584) = -0.5544

	cos(1.200 + 4.1270) = 0.5544

and thus 

	q = 4.127, 	A = 1.7740/0.5544 = 3.200


S 2.5 (P 1.19)
______________

(i) We know that w = W*T_s, i.e.,

	w = 150*pi*(3/1000) = 0.45*pi

(ii) w is a rational multiple of pi, therefore x[n]
is periodic.  The period N is determined by expressing
w as 
	(r/N)*2*pi

where r and N have no common factors.  In this case,

	w = (9/40)*2*pi

so N=40.

(iii) x[n] is constant for all n if 

	w = 2*k*pi

i.e., 
	T_s = (2*k*pi)/(150*pi) = k*(13.3333...) ms
or
	f_s = 1/T_s = 75/k samples/sec

for some integer k.

x[n] alternates in value between cos(q) and -cos(q) if

	w = pi + 2*k*pi

i.e.,
	T_s = (0.5 + k)*(13.3333...) ms
or
	f_s = 1/T_s = 75/(k+0.5) samples/sec

for some integer k.

S 2.6
_____

(i) By inspection, the period T is given by 0.5 seconds. Thus 
W = 2*pi/T = 4*pi. Again by inspection, A=3.0, and since 
x(0) = 3*sqrt(2)/2, it follows that

	3*cos(0*t + q) = 3*sqrt(2)/2 = sqrt(2)/2

i.e., cos(q) = -pi/4 or pi/4.  Since the waveform is 
increasing at t=0, it follows that q = -pi/4.

(ii) x[n] = 3*cos(W*T_s*n + q) 
          = 3*cos(4*pi*0.05*n + q) 
          = 3*cos(pi*n/5 + q)

thus w = pi/5, which is a rational multiple of pi.  Since 

	w = pi/5 = (2*pi*1)/10

it follows that the period is given by N = 10.

(iii) x[0:4] represents the first half of a period of the 
discrete-time sinusoid, and x[5:9] represents the second half. We have

	x[n+5] = 3*cos(pi*(n+5)/5 + q) = 3*cos(pi*n/5 + pi + q)

We know that cos(theta + pi) = -cos(theta).  Thus

	x[n+5] = -x[n]

and thus also x[5:9] = -x[0:4].  This can be also seen by 
plotting the ten samples on the continuous-time graph given 
in the problem.

S 2.7
_____

(i) The period of x(t) equals 2*pi/W = 1/600 sec 

x(0) = cos(1200*pi*0) = 1, so the initial phase of x(t) equals
zero.  The zeros of x(t) will therefore occur at times

	t = 1/(2400) + k/(1200)

where k is an integer.  These zeros will be preserved in x[.] 
if and only if t = 1/2400 is a sampling instant, i.e., if

	T_s = 1/(2400*M)

for some integer M greater than or equal to 1.  The resulting
signal will be

	x[n] = cos(pi*n/(2*M))

and will be periodic for all choices of M.

(ii) Note that the choice M = 1 gives a sequence of samples
x[n] = cos(pi*n/2), where the abs. difference between 
consecutive samples equals 1.  

As M increases (i.e., T_s decreases), the absolute difference 
between consecutive samples is reduced, though its value
will depend on the position of the two samples in the cycle.  
The maximum difference will be observed at zero crossings, 
where the cosine function attains the maximum absolute value
in its derivative. (Recall that (d/dt)(cos(theta)) = sin(theta) 
which is maximized in abs. value at theta = (pi/2)+k*pi.) 
See also the attached figure020601.pdf (where M=10).

The maximum difference will therefore be

	d = cos(pi/2-pi/(2*M)) - cos(pi/2) = cos(pi/2-pi/(2*M))

	  = sin(pi/(2*M))

To find the minimum value of M (corresponding to the maximum
value of T_s) such that d <= 0.01, we compute

	arcsin(0.01) = 0.010

(not surprising, since sin(theta) ~ theta for small theta).  
Solving for the threshold value of M, we have

	pi/(2*M) = 0.010

or M = 157.1

Since M is constrained to be an integer, the least value of M
is 158, and the resulting maximum value of T_s is 

	1/(2400*158) sec = 2.64 microseconds