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回复 30 # qqzhouqianziyi 的帖子
我的是老师修改过给我的，我也不太清楚好蛋疼
function [im,tt] = toimage(A,f,t,splx,sply,f_max)
% [im,tt] = TOIMAGE(A,f,t,splx,sply) transforms a spectrum made
% of 1D functions (e.g., output of "spectreh") in an 2D image
%
% inputs : - A : amplitudes of modes (1 mode per row of A)
% - f : instantaneous frequencies
% - t : time instants
% - splx : number of columns of the output im (time resolution).
% If different from length(t), works only for uniform
% sampling.
% - sply : number of rows of the output im (frequency resolution).
% - f_max: the highest frequency with respect to sply.
% outputs : - im : 2D image of the spectrum
% - tt : time instants in the image
%
% utilisation : [im,tt] = toimage(A,f);[im,tt] = toimage(A,f,t);[im,tt] = toimage(A,f,sply);
% [im,tt] = toimage(A,f,splx,sply);[im,tt] = toimage(A,f,t,splx,sply);
%DEFSPL = 400;
fs=8000; % revised
%if nargin < 3
% t = 1:size(A,2);
% sply = DEFSPL;
% splx = length(t);
%else
% if length(t) == 1
% tp = t;
% t = 1:size(A,2);
% if nargin < 4
% sply = tp;
% splx = length(t);
% else
% if nargin > 4
% error('too many arguments')
% end
% sply = splx;
% splx = tp;
% end
% else
% lt = length(t);
% if nargin < 5
% sply = splx;
% splx = lt;
% end
%
% if nargin < 4
% sply = DEFSPL;
% splx = lt;
% end
%
% if nargin > 5
% error('too many arguments')
% end
% end
%end
%end
lt=length(t);
im=[];
im(splx,sply) = 0;
f=f*(fs/2/f_max);% revised
for i=1:size(f,1)
for j = 1:lt-2
ff=floor(f(i,j)*2*(sply-1))+1;
if ff <= sply % in case f(i,j) > 0.5
im(floor(j*(splx-1)/lt)+1,ff)=im(floor(j*(splx-1)/lt)+1,ff)+A(i,j);
end
end
end
for i = 1:splx
tt(i) = mean(t(floor((i-1)*lt/(splx))+1:floor(i*lt/(splx))));
end
im=fliplr(im)'; |
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发表于 2011-3-24 12:29
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