--- /dev/null
+## Copyright (C) 1999 Paul Kienzle <pkienzle@users.sf.net>
+##
+## This program is free software; you can redistribute it and/or modify it under
+## the terms of the GNU General Public License as published by the Free Software
+## Foundation; either version 3 of the License, or (at your option) any later
+## version.
+##
+## This program is distributed in the hope that it will be useful, but WITHOUT
+## ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
+## FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more
+## details.
+##
+## You should have received a copy of the GNU General Public License along with
+## this program; if not, see <http://www.gnu.org/licenses/>.
+
+## Compute butterworth filter order and cutoff for the desired response
+## characteristics. Rp is the allowable decibels of ripple in the pass
+## band. Rs is the minimum attenuation in the stop band.
+##
+## [n, Wc] = buttord(Wp, Ws, Rp, Rs)
+## Low pass (Wp<Ws) or high pass (Wp>Ws) filter design. Wp is the
+## pass band edge and Ws is the stop band edge. Frequencies are
+## normalized to [0,1], corresponding to the range [0,Fs/2].
+##
+## [n, Wc] = buttord([Wp1, Wp2], [Ws1, Ws2], Rp, Rs)
+## Band pass (Ws1<Wp1<Wp2<Ws2) or band reject (Wp1<Ws1<Ws2<Wp2)
+## filter design. Wp gives the edges of the pass band, and Ws gives
+## the edges of the stop band.
+##
+## Theory: |H(W)|^2 = 1/[1+(W/Wc)^(2N)] = 10^(-R/10)
+## With some algebra, you can solve simultaneously for Wc and N given
+## Ws,Rs and Wp,Rp. For high pass filters, subtracting the band edges
+## from Fs/2, performing the test, and swapping the resulting Wc back
+## works beautifully. For bandpass and bandstop filters this process
+## significantly overdesigns. Artificially dividing N by 2 in this case
+## helps a lot, but it still overdesigns.
+##
+## See also: butter
+
+function [n, Wc] = buttord(Wp, Ws, Rp, Rs)
+ if nargin != 4
+ print_usage;
+ elseif length(Wp) != length(Ws)
+ error("buttord: Wp and Ws must have the same length");
+ elseif length(Wp) != 1 && length(Wp) != 2
+ error("buttord: Wp,Ws must have length 1 or 2");
+ elseif length(Wp) == 2 && (all(Wp>Ws) || all(Ws>Wp) || diff(Wp)<=0 || diff(Ws)<=0)
+ error("buttord: Wp(1)<Ws(1)<Ws(2)<Wp(2) or Ws(1)<Wp(1)<Wp(2)<Ws(2)");
+ end
+
+ if length(Wp) == 2
+ warning("buttord: seems to overdesign bandpass and bandreject filters");
+ end
+
+ T = 2;
+
+ ## if high pass, reverse the sense of the test
+ stop = find(Wp > Ws);
+ Wp(stop) = 1-Wp(stop); # stop will be at most length 1, so no need to
+ Ws(stop) = 1-Ws(stop); # subtract from ones(1,length(stop))
+
+ ## warp the target frequencies according to the bilinear transform
+ Ws = (2/T)*tan(pi*Ws./T);
+ Wp = (2/T)*tan(pi*Wp./T);
+
+ ## compute minimum n which satisfies all band edge conditions
+ ## the factor 1/length(Wp) is an artificial correction for the
+ ## band pass/stop case, which otherwise significantly overdesigns.
+ qs = log(10^(Rs/10) - 1);
+ qp = log(10^(Rp/10) - 1);
+ n = ceil(max(0.5*(qs - qp)./log(Ws./Wp))/length(Wp));
+
+ ## compute -3dB cutoff given Wp, Rp and n
+ Wc = exp(log(Wp) - qp/2/n);
+
+ ## unwarp the returned frequency
+ Wc = atan(T/2*Wc)*T/pi;
+
+ ## if high pass, reverse the sense of the test
+ Wc(stop) = 1-Wc(stop);
+
+endfunction