📄 tfrridbn.m
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function [tfr,t,f] = tfrridbn(x,t,N,g,h,trace);
%TFRRIDBN Reduced Interference Distribution with a binomial kernel.
% [TFR,T,F]=TFRRIDBN(X,T,N,G,H,TRACE) Reduced Interference
% Distribution with a kernel based on the binomial coefficients.
% TFRRIDBN computes either the distribution of a discrete-time
% signal X, or the cross representation between two signals.
%
% X : signal if auto-RIDBN, or [X1,X2] if cross-RIDBN.
% 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/10)).
% H : frequency smoothing window, H(0) being forced to 1.
% (default : Hamming(N/4)).
% TRACE : if nonzero, the progression of the algorithm is shown
% (default : 0).
% TFR : time-frequency representation. When called without
% output arguments, TFRRIDBN runs TFRQVIEW.
% F : vector of normalized frequencies.
%
% Example :
% sig=[fmlin(128,.05,.3)+fmlin(128,.15,.4)]; tfrridbn(sig);
%
% See also all the time-frequency representations listed in
% the file CONTENTS (TFR*)
% F. Auger, June 1996.
% 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 & nargin==6),
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); trace = 0;
elseif (nargin == 2)|(nargin == 3),
g = window(glength); h = window(hlength); trace = 0;
elseif (nargin == 4),
h = window(hlength); trace = 0;
elseif (nargin == 5),
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; g=g/g(Lg+1);
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;
taumax = min([round(N/2)-1,Lh]);
tfr= zeros (N,tcol) ;
if trace, disp('Binomial RID 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)= g(Lg+1) * x(ti,1) .* conj(x(ti,xcol));
Ker=[1];
for tau=1:taumax,
points= -min([tau,Lg,xrow-ti-tau]):min([tau,Lg,ti-tau-1]);
Ker= ([Ker; 0; 0]+2*[0; Ker; 0]+[0; 0; Ker])/4.0;
g2 = g(Lg+1+points) .* Ker(tau+points+1);
if (sum(g2)>eps), g2=g2/sum(g2); end;
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),
Ker=ones(2*tau+1,1);
for p=1:2*tau,
Ker(p+1)=Ker(p)*(2*tau-p+1)/p;
end;
Ker=Ker/sum(Ker);
points= -min([tau,Lg,xrow-ti-tau]):min([tau,Lg,ti-tau-1]);
g2 = g(Lg+1+points) .* Ker(tau+points+1); 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;
if trace, fprintf('\n'); end;
tfr= fft(tfr);
if (xcol==1), tfr=real(tfr); end ;
if (nargout<=1),
tfrqview(tfr,x,t,'tfrridbn',g,h);
elseif (nargout==3),
f=(0.5*(0:N-1)/N)';
end;
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