how can I find coherence of a measurement matrix?
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coherence =A
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  Stephen23
      
      
 on 8 Feb 2016
				@Anthony Uwaechia: You might like to accept John's Answer, if it solved your problem.
Accepted Answer
  John BG
      
 on 8 Feb 2016
        
      Edited: John BG
      
 on 8 Feb 2016
  
      Hi Anthony
have a look:
P=16 
N =512 
L=50 
M=16 
d=floor(N/L)
for k=1:1:d
      P=ones(1,M)+d*[0:1:M-1]  % you may want this line outside the main for loop
      P2=Pk
      for m=1:1:M
%         [k k+d k+2d .. =<N]
%             k=1 [1 1+d 1+2d .. =<512]  =  [1 11 21 31 .. 501 511]
%             k=2 [2 2+d 2+2d .. =<512]  =  [2 12 22 32 .. 502 512]
%             k=3 [3 3+d 3+2d .. =<512]  =  [3 13 23 33 .. 503]
%             ..
%                    k*ones(1,floor((N-k)/d)+1) +  [0 d 2d .. =<(N-k)]
           n_range=k*ones(1,floor((N-k)/d)+1) + d*[0:1:floor((N-k)/d)]
           for n=n_range
               if (sum(ismember(n,Pk))==0)
                   % change Pk upper limit from (M-1)*d-1 to (n-1)*d-1 and redefine P2
                     P3=ones(1,n)+d*[0:1:n-1]  % you may want this line outside the main for loop
                     if cov(P3)<cov(P2)  
                         % or C is Carrier2Noise ratio and Ci(P2)<Ci(Pk) means all P2 carriers below Pk carriers
                         % then perhaps if max(P3)<min(Pk) ..
                         Pk=P3
                     end
               end
           end
% repeatition % for m=[M:-1:1] % % end
      end
end
2 things :
1.- when Pk is generated again with n instead of M as upper limit, either the comment expression circled

should be [1,2, ... Pm] or it means replace the initial elements, leaving the upper ones, whatever left as initially defined
2.- Without further information, C() may mean covariance cov() or it's Carrier2Noise ratio and Ci(P2)<Ci(Pk) means all P2 carriers are below Pk carriers, only then replace
Are you using this loop to generate OFDM (ADSL, DVB-T, WAN, ..) carrier indices?
Just for the time taken, if the lines above are of any help, would you click on the thumbs-up vote above? thanks in advance
John
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