Butterworth lowpass filtering without signal processing toolbox

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Hi,
I'm trying to accomplish butterworth lowpass filtering but do not have the signal processing toolbox. Is it possible to do this type of filtering without this toolbox?

Answers (3)

Jan
Jan on 26 Jun 2014
Edited: Jan on 7 Jul 2014
This is a fair method to determine the coefficients for a Butterworth filter:
function [Z, P, G] = myButter(n, W, pass)
% Digital Butterworth filter, either 2 or 3 outputs
% Jan Simon, 2014, BSD licence
% See docs of BUTTER for input and output
% Fast hack with limited accuracy: Handle with care!
% Until n=15 the relative difference to Matlab's BUTTER is < 100*eps
V = tan(W * 1.5707963267948966);
Q = exp((1.5707963267948966i / n) * ((2 + n - 1):2:(3 * n - 1)));
nQ = length(Q);
switch lower(pass)
case 'stop'
Sg = 1 / prod(-Q);
c = -V(1) * V(2);
b = (V(2) - V(1)) * 0.5 ./ Q;
d = sqrt(b .* b + c);
Sp = [b + d, b - d];
Sz = sqrt(c) * (-1) .^ (0:2 * nQ - 1);
case 'bandpass'
Sg = (V(2) - V(1)) ^ nQ;
b = (V(2) - V(1)) * 0.5 * Q;
d = sqrt(b .* b - V(1) * V(2));
Sp = [b + d, b - d];
Sz = zeros(1, nQ);
case 'high'
Sg = 1 ./ prod(-Q);
Sp = V ./ Q;
Sz = zeros(1, nQ);
case 'low'
Sg = V ^ nQ;
Sp = V * Q;
Sz = [];
otherwise
error('user:myButter:badFilter', 'Unknown filter type: %s', pass)
end
% Bilinear transform:
P = (1 + Sp) ./ (1 - Sp);
Z = repmat(-1, size(P));
if isempty(Sz)
G = real(Sg / prod(1 - Sp));
else
G = real(Sg * prod(1 - Sz) / prod(1 - Sp));
Z(1:length(Sz)) = (1 + Sz) ./ (1 - Sz);
end
% From Zeros, Poles and Gain to B (numerator) and A (denominator):
if nargout == 2
Z = G * real(poly(Z'));
P = real(poly(P));
end
  2 Comments
Eduardo Rey
Eduardo Rey on 14 Mar 2020
Jan, I tried using this code to get coefficents for a low-pass response using n=1, w = 0.04 since fs=2K and fc = 40Hz but it gave me -1. Am I doing something wrong?

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Anastasios
Anastasios on 26 Jun 2014
Hi John,
You can download a 30-day free trial if you want to do something for now
https://www.mathworks.com/programs/trials/trial_request.html?prodcode=SG&eventid=616177282&s_iid=main_trial_SG_cta2
Tasos
  2 Comments
John
John on 26 Jun 2014
That's not an option for now, and that decision doesn't come from me. Is there another option you can think of?
Anastasios
Anastasios on 26 Jun 2014
Check the following webpage at Rice University. Hopefully you can find your answer there. 2D Frequency Domain Filtering and the 2D DFT

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Fiza
Fiza on 8 Sep 2020
Hi,
Under what license can i use this code?

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