📄 tfrbud.m
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function [tfr,t,f] = tfrbud(x,t,N,g,h,sigma,trace);
%TFRBUD Butterworth time-frequency distribution.
% [TFR,T,F]=TFRBUD(X,T,N,G,H,SIGMA,TRACE) computes the Butterworth
% distribution of a discrete-time signal X, or the
% cross Butterworth representation between two signals.
%
% X : signal if auto-BUD, or [X1,X2] if cross-BUD.
% T : time instant(s) (default : 1:length(X)).
% N : number of frequency bins (default : length(X)).
% G : time smoothing window, G(0) being forced to 1.
% (default : Hamming(N/4)).
% H : frequency smoothing window, H(0) being forced to 1.
% (default : Hamming(N/4)).
% SIGMA : kernel width (default : 1).
% TRACE : if nonzero, the progression of the algorithm is shown
% (default : 0).
% TFR : time-frequency representation. When called without
% output arguments, TFRBUD runs TFRQVIEW.
% F : vector of normalized frequencies.
%
% Example :
% sig=fmlin(128,0.05,0.3)+fmlin(128,0.15,0.4);
% g=window(9,'Kaiser'); h=window(27,'Kaiser');
% t=1:128; tfrbud(sig,t,128,g,h,3.6,1);
%
% See also all the time-frequency representations listed in
% the file CONTENTS (TFR*)
% F. Auger, May-August 1994, July 1995.
% Copyright (c) 1996 by CNRS (France).
%
% ------------------- CONFIDENTIAL PROGRAM --------------------
% This program can not be used without the authorization of its
% author(s). For any comment or bug report, please send e-mail to
% f.auger@ieee.org
if (nargin == 0),
error('At least 1 parameter required');
end;
[xrow,xcol] = size(x);
if (xcol==0)|(xcol>2),
error('X must have one or two columns');
end
if (nargin <= 2),
N=xrow;
elseif (N<0),
error('N must be greater than zero');
elseif (2^nextpow2(N)~=N),
fprintf('For a faster computation, N should be a power of two\n');
end;
hlength=floor(N/4); hlength=hlength+1-rem(hlength,2);
glength=floor(N/10);glength=glength+1-rem(glength,2);
if (nargin == 1),
t=1:xrow; g = window(glength); h = window(hlength); sigma = 1.0; trace = 0;
elseif (nargin == 2)|(nargin == 3),
g = window(glength); h = window(hlength); sigma = 1.0; trace = 0;
elseif (nargin == 4),
h = window(hlength); sigma = 1.0; trace = 0;
elseif (nargin == 5),
sigma = 1.0; trace = 0;
elseif (nargin == 6),
trace = 0;
end;
[trow,tcol] = size(t);
if (trow~=1),
error('t must only have one row');
end;
[grow,gcol]=size(g); Lg=(grow-1)/2;
if (gcol~=1)|(rem(grow,2)==0),
error('G must be a smoothing window with odd length');
end;
[hrow,hcol]=size(h); Lh=(hrow-1)/2; h=h/h(Lh+1);
if (hcol~=1)|(rem(hrow,2)==0),
error('H must be a smoothing window with odd length');
end;
if (sigma<=0.0),
error('SIGMA must be strictly positive');
end;
taumax = min([round(N/2),Lh]); tau = 1:taumax; points = -Lg:Lg;
BudKer = exp(-kron( abs(points.'), 1.0 ./ (2.0*tau/sqrt(sigma))));
BudKer = diag(g) * BudKer;
tfr= zeros (N,tcol) ;
if trace, disp('Butterworth distribution'); end;
for icol=1:tcol,
ti= t(icol); taumax=min([ti+Lg-1,xrow-ti+Lg,round(N/2)-1,Lh]);
if trace, disprog(icol,tcol,10); end;
tfr(1,icol)= x(ti,1) .* conj(x(ti,xcol));
for tau=1:taumax,
points= -min([Lg,xrow-ti-tau]):min([Lg,ti-tau-1]);
g2 = BudKer(Lg+1+points,tau); g2=g2/sum(g2);
R=sum(g2 .* x(ti+tau-points,1) .* conj(x(ti-tau-points,xcol)));
tfr( 1+tau,icol)=h(Lh+tau+1)*R;
R=sum(g2 .* x(ti-tau-points,1) .* conj(x(ti+tau-points,xcol)));
tfr(N+1-tau,icol)=h(Lh-tau+1)*R;
end;
tau=round(N/2);
if (ti<=xrow-tau)&(ti>=tau+1)&(tau<=Lh),
points= -min([Lg,xrow-ti-tau]):min([Lg,ti-tau-1]);
g2 = BudKer(Lg+1+points,tau); g2=g2/sum(g2);
tfr(tau+1,icol) = 0.5 * ...
(h(Lh+tau+1)*sum(g2 .* x(ti+tau-points,1) .* conj(x(ti-tau-points,xcol)))+...
h(Lh-tau+1)*sum(g2 .* x(ti-tau-points,1) .* conj(x(ti+tau-points,xcol))));
end;
end;
clear BudKer;
if trace, fprintf('\n'); end;
tfr= fft(tfr);
if (xcol==1), tfr=real(tfr); end ;
if (nargout<=1),
tfrqview(tfr,x,t,'tfrbud',g,h,sigma);
elseif (nargout==3),
f=(0.5*(0:N-1)/N)';
end;
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