Butterworth lowpass filtering without signal processing toolbox

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?

Answer 1

[Jan]

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

[erico vale]

Could you comment the code ?

[Eduardo Rey]

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?

Answer 2

[Anastasios]

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

[John]

That's not an option for now, and that decision doesn't come from me. Is there another option you can think of?

[Anastasios]

Check the following webpage at Rice University. Hopefully you can find your answer there. 2D Frequency Domain Filtering and the 2D DFT

Answer 3

[Fiza]

Hi,

Under what license can i use this code?