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.orgif (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');endif (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==0), tfrqview(tfr,x,t,'tfrridbn',g,h);elseif (nargout==3), f=(0.5*(0:N-1)/N)';end;