Homework 3

Solution [PDF]

Problem 1: The plot below verifies the solution of the LCCDE by comparing the result of evaluating the solution for n=0..10 with the result of filtering the input directly. As you can see, they are quite close; the error is probably due to numerical precision error. The matlab diary that produced the plot is also show below.

% verifying solution to LCCDE
n=0:10;
xn = 2.^n;
b = [1 .5];
a = [1 -.7 .1];
yn = filter(b,a,xn);
% now with my solution
ym = -1.0567*(.5.^n) + .1417*(.2.^n) + 1.8152*(2.^n);
subplot(211)
plot(n,yn)
hold on
plot(n,ym,'r')
legend('filter output','LCCDE solution')
subplot 212
plot(n,yn-ym)
title('Error (Filter output - LCCDE solution)')

M2.12: The figure below shows the plots of the autocorrelation of signals x[n], y[n], and w[n] from Problem 2-1. And the figure below shows the cross-correlation between x[n] and y[n], and between x[n] and w[n].

M2.13. The figure below shows the autocorrelation function of the same signal x[n] as above, with various levels of random distortion added. The largest peak is of course at zero lag.

The matlab function used to produce the plots is as follows. Note that the book asks for uniformly distributed random noise, but says to use the matlab function randn, which generates normally distributed random noise. Here I use rand.

function r = plotac_dist(x,a)
% plotac_dist(x,a)
% Computation of autocorrelation sequence of distorted signal
% x - the signal
% a - scalar amplitude of noise (uniformly distributed)

% generate the noise sequence
n = length(x);
d = a.*rand(1,n);  % use uniformly distributed randomness
y = x+d;
n = n-1;

% Compute the correlation sequence
r = conv(y, fliplr(y));
k = (-n):n';
stem(k,r);
xlabel('Lag index'); ylabel('Amplitude');
v = axis;
axis([-n n v(3:end)]);

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