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）
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
erico vale
2018-6-20
Could you comment the code ?
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?
Anastasios
2014-6-26
0 个投票
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
2014-6-26
Anastasios
2014-6-26
Check the following webpage at Rice University. Hopefully you can find your answer there. 2D Frequency Domain Filtering and the 2D DFT
Fiza
2020-9-8
0 个投票
Hi,
Under what license can i use this code?
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