Computing the impulse response of a system with complex poles (Example 1)

This example shows three different ways to compute the impulse response.

Contents

Define system coefficients

Difference equation: y(n) = x(n) - 2.5 x(n-1) + y(n-1) - 0.7 y(n-2)

b = [1 -2.5];
a = [1 -1 0.7];

Compute impulse response directly from different equation

Use the MATLAB function 'filter' to compute the impulse response

L = 30;
x = [1 zeros(1, L)];
h1 = filter(b, a, x);

figure(1)
clf
n = 0:L;
stem(n, h1, '.');
title('Impulse response computed using difference equation')
ax = [-1 L -2.5 2.5];
axis(ax)
box off

Compute impulse response using partial fraction expansion

Use MATLAB function 'residue' to find the poles and residues.

[r,p,k] = residue(b,a)

% r =
%    0.5000 + 1.4907i
%    0.5000 - 1.4907i
%
% p =
%    0.5000 + 0.6708i
%    0.5000 - 0.6708i
%
% k =
%      []

h2 = r(1)*p(1).^n + r(2)*p(2).^n;

figure(1)
clf
stem(n, h2, '.')
title('Impulse response computed using partial fraction expanstion')
axis(ax)
box off

r =

   0.5000 + 1.4907i
   0.5000 - 1.4907i


p =

   0.5000 + 0.6708i
   0.5000 - 0.6708i


k =

     []

Verify that h2 is the same as h1

max(abs(h1 - h2))
% This is zero to machine precision

ans =

   8.8818e-16

Compute impulse response using cosine formula

h3 = 2*abs(r(1))*(abs(p(1)).^n).*cos(angle(p(1))*n+angle(r(1)));

figure(1)
clf
stem(n, h3, '.')
title('Impulse response computing using cosine formula')
axis(ax)
box off

Verify that h3 is the same as h1

max(abs(h1 - h3))
% This is zero to machine precision

ans =

   9.8532e-16

Plot all

figure(2)
clf
subplot(3, 1, 1)
stem(n, h1, 'k.')
title('Impulse response computed using difference equation')
axis(ax)
box off

subplot(3, 1, 2)
stem(n, h2, 'k.')
title('Impulse response computed using partial fraction expanstion')
axis(ax)
box off

subplot(3, 1, 3)
stem(n, h3, 'k.')
title('Impulse response computing using cosine formula')
axis(ax)
box off

orient tall
print -dpdf ComplexPoles_1

Published with MATLAB® 7.12