0208106224 发表于 2012-5-15 09:53

EMD分解问题,急~~~~~

本帖最后由 0208106224 于 2012-5-15 10:45 编辑

下面这段程序我仿真出来是一组数据,现在我想让他呈现出emd图的形式要怎么弄?在后面我试着添加plot那些语句照样不行;各位大侠帮帮忙啊:'( 在emd分解后又如何去实现信号的去噪了???
% EMD:Emprical mode decomposition
%
% imf = emd(x,n)
%
% x   - input signal (must be a column vector)
% n   - number of intrinsic mode functions
%
% imf - Matrix of intrinsic mode functions (each as a row)
%
% See:Huang et al, Royal Society Proceedings on Math, Physical,
%       and Engineering Sciences, vol. 454, no. 1971, pp. 903-995,
%       8 March 1998
%
% Author: Ivan Magrin-Chagnolleau<ivan@ieee.org>
%
function imf = emd(x,n);
load noissin;
x=noissin;
c = x(:)'; % copy of the input signal (as a row vector)
N = length(x);
%-------------------------------------------------------------------------
% loop to decompose the input signal into n successive IMFs
imf = []; % Matrix which will contain the successive IMF, and the residue
n=size(c);
for t=1:n % loop on successive IMFs
   
   %-------------------------------------------------------------------------
   % inner loop to find each imf
   
   h = c; % at the beginning of the sifting process, h is the signal
   SD = 1; % Standard deviation which will be used to stop the sifting process
   
   while SD > 0.3 % while the standard deviation is higher than 0.3 (typical value)
      
      % find local max/min points
      d = diff(h); % approximate derivative
      maxmin = []; % to store the optima (min and max without distinction so far)
      for i=1:N-2
         if d(i)==0                        % we are on a zero
            if sign(d(i-1))~=sign(d(i+1))% it is a maximum
               maxmin = ;
            end
         elseif sign(d(i))~=sign(d(i+1))   % we are straddling a zero so
            maxmin = ;      % define zero as at i+1 (not i)
         end
      end
      
      if size(maxmin,2) < 2 % then it is the residue
         break
      end
      
      % divide maxmin into maxes and mins
      if maxmin(1)>maxmin(2)            % first one is a max not a min
         maxes = maxmin(1:2:length(maxmin));
         mins= maxmin(2:2:length(maxmin));
      else                              % is the other way around
         maxes = maxmin(2:2:length(maxmin));
         mins= maxmin(1:2:length(maxmin));
      end
      
      % make endpoints both maxes and mins
      maxes = ;
      mins= ;
      
      
      %-------------------------------------------------------------------------
      % spline interpolate to get max and min envelopes; form imf
      maxenv = spline(maxes,h(maxes),1:N);
      minenv = spline(mins, h(mins),1:N);
      
      m = (maxenv + minenv)/2; % mean of max and min enveloppes
      prevh = h; % copy of the previous value of h before modifying it
      h = h - m; % substract mean to h
      
      % calculate standard deviation
      eps = 0.0000001; % to avoid zero values
      SD = sum ( ((prevh - h).^2) ./ (prevh.^2 + eps) );
      
   end
   
   imf = ; % store the extracted IMF in the matrix imf
   % if size(maxmin,2)<2, then h is the residue
   
   % stop criterion of the algo. if we reach the end before n
   if size(maxmin,2) < 2
      break
   end
   
   c = c - h; % substract the extracted IMF from the signal
   
end
return

0208106224 发表于 2012-5-15 10:19

{:{19}:}{:{19}:}没人帮忙吗

shnyxpvib 发表于 2012-5-21 18:47

俺也不会,俺也想知道

fred110 发表于 2012-5-27 14:13

一样不会,刚开始接触,工具箱怎么弄,还不明白呢

天涯青驹 发表于 2013-4-7 13:37

{:{19}:}{:{28}:}
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