applying single pole filter to simple signal

Question

How to apply single pole filter to simple signal

fs = 500;
ts = 1/fs;
t = 0:ts:1-ts;
fc = 50;
S = cos(2*pi*fc*t);  

Answer 1

Given the transfer function of a single pole filter (without a zero):
$$H(z)= \frac{Y(z)}{X(z)} = \frac{1}{1-pz^{-1}}$$ Using the inverse z transform leads to the following equation: $$y[n] = x[n] + p y[n-1]$$

Your single pole filter can be implemented using a for-loop.

fs = 500; 
fc = 50; 
K = fs;  % 1 second of data
k = 0:(K-1);
x = cos(2*pi*fc/fs*k);

y = zeros(K,1);

p = 0.5;
y(1) = 0;  % Choose some initialization for y(1)
for n = 2:K
    y(n) = x(n) + p*y(n-1);
end

Another 'simple' test signal that can be used to visualize the behaviour of your filter is a random signal.

x = randn(K,1);

The spectrum of the filtered random signal shows the frequency characteristics of the filter.

Answer 2

If your pole is at $z=p$ (real or complex), this will do:

Y = filter( [1], [1, -p], S );

Free bonus: if you also want a zero at $z = c$:

Y = filter( [1, -c], [1, -p], S);

This last expression corresponds to the filter:

$$H(z) = \frac{1 - c z^{-1}}{1 - p z^{-1}}$$