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?

回答(3 个)

Jan
Jan 2014-6-26
编辑:Jan 2014-7-7
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 个评论
Eduardo Rey
Eduardo Rey 2020-3-14
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 2014-6-26
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 个评论
John
John 2014-6-26
That's not an option for now, and that decision doesn't come from me. Is there another option you can think of?

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Fiza
Fiza 2020-9-8
Hi,
Under what license can i use this code?

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