## Copyright (C) 1999 Paul Kienzle ## ## 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 2 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, write to the Free Software ## Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA ## 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 usage("[n, Wn] = buttord(Wp, Ws, Rp, Rs)"); end if length(Wp) != length(Ws) error("buttord: Wp and Ws must have the same length"); end if length(Wp) != 1 && length(Wp) != 2 error("buttord: Wp,Ws must have length 1 or 2"); end if 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

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