S 3.1 (P 1.21)
______________

We have x[n] = cos(w*n), where

	w = 2*pi*f/800  (and f is in Hz)

For x[n] to be the same as cos(0.4*pi*n) = cos(-0.4*pi*n) for all n,
it must be that

	f/800 = 0.2 + k   (k integer)
or
	f/800 = -0.2 + k  (k integer)

Thus
	f = 160 + k*800
or
	f = -160 + k*800

The aliases in the range 0 to 3,000 Hz are therefore

	f = 160, 640, 960, 1440, 1760, 2240 and 2560 Hz


S 3.2
_____
	x(t) = cos(2*pi*f*t + 1.8)

where f is between 700 and 800 Hz.  The sampling period is T_s = .01 sec, so

	x[n] = x(n*T_s) = cos(2*pi*f*0.01*n + 1.8) = 

(i) If x[n] also equals cos(2*pi*0.35*n + 1.8), it follows that

	f*0.01 = 0.35 + k

where k is an integer.  Thus

	f = 35 + 100*k

and the only possible value for f in the given interval is f = 735 Hz.

(ii) If x[n] equals cos(2*pi*0.2*n - 1.8), it follows that

	f*0.01 = -0.2 + k

i.e., 
	f = -20 + 100*k

The only possible value for f is 780 Hz.


S 3.3 (P 1.22)
______________

The cyclic frequency of x(t) is 150 Hz.

The sampling rate is f_s = 1/T_s = 500 samples/sec.

The resulting discrete-time sinusoid is 

	x[n] = cos(0.6*pi*n),

i.e, it has w = 0.6*pi.  Note that there is no phase shift.

(i) We want the aliases of 150 Hz with respect to the sampling
rate of 500 samples/sec:

	f = 150 + k*500  Hz
and
	f = -150 + k*500  Hz

In the range 0 to 2000 Hz, we have:
	
	650, 1150, 1650  and 350, 850, 1350, 1850  Hz

(ii) An increase in T_s results in an increase in w.  The next
value of w equivalent to 0.6*pi is

	-0.6*pi + 2*pi = 1.4*pi 

and is obtained for T_s = (1.4*pi)/(300*pi) = 4.67 ms.

Note: This is the same as 6.67 ms (period of x(t)) - 2.0 ms.


S 3.4 (P 2.3)
______________

	[2 ; -1] = [1 ; 0] - [-1 ; 1]

Thus
	B*[2 ; -1] = B*[1 ; 0] - B*[-1 ; 1]

		  = [3 ; -2 ; -1] - [4 ; 7 ; 2]

		  = [-1 ; -9 ; -3]

B is a 3x2 matrix.


S 3.5 (P 2.4)
_____________

	[3 ; 3] = [-1 ; 1] + 2*[2 ; 1]

Thus
	G*[3 ; 3] = G*[-1 ; 1] + 2*G*[2 ; 1]

		  = u + 2*v


S 3.6 (P 2.1)
_____________

A = [1 2 3 4; 5 6 7 8; 9 10 11 12]; 

(i) 
I = [1 3]; 
J = [1 4]; 

A(I,J) 

ans = 

     1     4 
     9    12 

(ii) 
K = [1 4]; 
L = [4 1]; 

B = A; 
B(:,K) = A(:,L) 

B = 
4     2     3     1 
8     6     7     5 
12    10    11     9 

(The choice K = [4 1] , L = [1 4] works also.) 

(iii) 
C = A(:) 

is the concatenation (in order of increasing column index) 
of all columns of A into a single column vector. 

C = 

1 
5 
9 
2 
6 
10 
3 
7 
11 
4 
8 
12 

(iv)  Note that RESHAPE(A,M,N) is the same as taking A(:) and 
parsing it into N columns of size M, to obtain a M by N matrix. 

D = (reshape(A.',6,2)).' 

D = 

1     2     3     4     5     6 
7     8     9    10    11    12 


S 3.7 (P 2.5)
______________

(i) The three unit column vectors are

	e_1 = [1; 0; 0] = [1 0 0].'
	e_2 = [0; 1; 0] = [1 0 0].'
	e_3 = [0; 0; 1] = [1 0 0].'

We have 

	A*e_1 = 1st column of A = [4; -1; 2]
	A*e_2 = 2nd column of A = [-2; 3; -1]
	A*e_3 = 3rd column of A = [1; 0; 3]

Thus
	A = 	 4	-2	 1
		-1	 3	 0
		 2	-1	 3

And 
	A *[2; 5; -1] = 2*[4; -1; 2] + 5*[-2; 3; -1] - [1; 0; 3] 
		= [-3; 13; -4]


(ii) We have

	B*[1; 0; 0] = [2; -4]	(=1st column of B)
	B*[1; 1; 0] = [3; -2]
	B*[1; 1; 1] = [-2; 1]

By linearity of the transformation B,

	B*[0; 1; 0] = B*[1; 1; 0] - B*[1; 0; 0] 
		= [3; -2] - [2; -4]
		= [1; 2]   (=2nd column of B)
and

	B*[0; 0; 1] = B*[1; 1; 1] - B*[1; 1; 0]
		= [-2; 1] - [3; -2]
		= [-5; 3]   (=3rd column of B)

Therefore

	B =	 2	 1	-5
		-4	 2	 3