Example with Repeated Poles

The following Matlab code performs a partial fraction expansion of a filter having three pairs of repeated roots (one real and two complex):^J.2

N = 1000;  % number of time samples to compute
A = [ 1 0 0 1 0 0 0.25];
B = [ 1 0 0 0 0 0 0];

% Compute "trusted" impulse response:
h_tdl = filter(B, A, [1 zeros(1, N-1)]);

% Compute partial fraction expansion (PFE):
[R,P,K] = residuez(B, A);

% PFE impulse response:
n = [0:N-1];
h_pfe = zeros(1,N);
for i = 1:6
  % repeated roots are not detected exactly:
  if i>1 & abs(P(i)-P(i-1))<1E-7
    h_pfe = h_pfe + (R(i) * (n+1) .* P(i).^n);
    disp(sprintf('Pole %d is a repeat of pole %d',i,i-1));
    % if i>2 & abs(P(i)-P(i-2))<1E-7 ...
  else
    h_pfe = h_pfe + (R(i) * P(i).^n);
  end
end

err = abs(max(h_pfe-h_tdl)) % should be about 5E-8

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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.