EE 451 - Homework 1

Solutions

1.
Problem 2.7
(a)
. The smallest k for which N is an integer is 1, so the period is N = 12
(b)
. The smallest k for which N is an integer is 3, so the perios is N = 8.
(c)
This is not periodic. is periodic with period 10. When you divide this by , it is no longer periodic - it gets smaller as n gets bigger, so is not periodic.

     n = -30:30;
wo = pi/5;
x = sin(wo*n)./(pi *n);
x(n==0) = wo/pi;          % L'Hospital's rule
stem(n,x);
grid
title('x[n] = [sin(\pi n/5)]/(\pi n)');
ylabel('x[n]');


(d)
. There is no k which makes N an integer, so this is not periodic.
2.
Problem 2.29
(a)
     n = -20:20;
x = zeros(size(n));
x((n>=-1)&(n<=3)) = 1;
x(n==4) = 1/2;
subplot(321)
stem(n,x)
grid
axis([-10 10 0 1.5]);
title('x[n]');

subplot(322)
na = n+2;
stem(na,x);
grid
axis([-10 10 0 1.5]);
title('x[n-2]');

subplot(323)
nb = 4-n;
stem(nb,x);
grid
axis([-10 10 0 1.5]);
title('x[4-n]');

subplot(324)
nc = -10:10;               % x goes from -20:20, so y will go from -10:10
for k=-10:10               % (e.g., y[10] = x[20])
y(nc==k) = x(n==(2*k));
end
stem(nc,y);
grid
axis([-10 10 0 1.5]);
title('x[2n]');

subplot(325)
y = x.*((2-n)>=0);
stem(n,y);
grid
axis([-10 10 0 1.5]);
title('x[n] u[2-n]');

subplot(326)
ne = n+1;
y = x.*(ne == 3);
stem(ne,y);
grid
axis([-10 10 0 1.5]);
title('x[n-1] \delta[n-3]');

3.
%(a)
subplot(311)
n=0:40;
x = (0.9).^n .* cos(0.1*pi*n+pi/5);
stem(n,x);
grid
ylabel('x[n]');
title('x[n] = 0.9^n cos(0.1 \pi n + \pi/6)');

%(b)
subplot(312)
n=0:40;
x = (0.9).^n .* cos(4.1*pi*n+pi/5);
stem(n,x);
grid
xlabel('n');
ylabel('x[n]');
title('x[n] = 0.9^n cos(4.1 \pi n + \pi/6)');

%(c)
subplot(313)
n=-10:10;
x=(0.75).^n .* (((n+6)>=0) - ((n-5)>=0)) + 5*(n==2);
stem(n,x);
grid
xlabel('n');
ylabel('x[n]');
title('x[n] = 0.75^n (u[n+6] - u[n-5]) + 5 delta[n-2]');

4.
Problem 4
(a)
The system is not time invariant. Putting in a signal of 1 at time 0 will result in an output of 0. Putting in a signal of 1 at time 4 will result in an output of 4, so the same signal at different times gives different outputs.
(b)
       subplot(321)
n = -20:20;
x = (n>=0) - ((n-4)>=0);
stem(n,x);
grid
axis([-10 10 -4 4])
title('x[n] = u[n] - u[n-4]');


(c)
       subplot(322)
y = n.*x;
stem(n,y);
grid
axis([-10 10 -4 4])
title('y[n] = n x[n]');


(d)
       subplot(323)
stem(n+2,y)
grid
axis([-10 10 -4 4])
title('y[n-2]');


(e)
       subplot(324)
n1 = n+2;
x1 = x;
stem(n1,x1);
grid
axis([-10 10 -4 4])
grid
title('x1[n] = x[n-2]');


(f)
       subplot(325)
y1 = n1.*x1;
stem(n1,y1);
grid
axis([-10 10 -4 4])
grid
title('y1[n]');


(g)
.

5.
Problem 5
(a)
The system is time invariant.
(b)
       subplot(321)
n = -30:30;
x = (n>=0) - ((n-4)>=0);
stem(n,x);
grid
axis([-10 10 0 1.5])
title('x[n] = u[n] - u[n-4]');


(c)
       subplot(322)
ny=-15:15;
for k=-15:15
y(ny==k) = (x((n+1)==k) + x(n==k) + x((n-1)==k))/3;
end
stem(ny,y);
grid
axis([-10 10 0 1.5])
title('y[n] = (x[n+1] + x[n] + x[n-1])/3');


(d)
       subplot(323)
stem(ny+2,y)
grid
axis([-10 10 0 1.5])
title('y[n-2]');


(e)
       subplot(324)
n1 = n+2;
x1 = x;
stem(n1,x1);
grid
axis([-10 10 0 1.5])
title('x1[n] = x[n-2]');


(f)
       subplot(325)
ny1 = -15:15;
for k=-15:15
y1(ny1==k) = (x((n1+1)==k) + x(n1==k) + x((n1-1)==k))/3;
end
stem(ny1,y1);
grid
axis([-10 10 0 1.5])
title('y1[n]');


(g)
y1[n] = y[n-2].

Bill Rison
1999-09-08