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how to measure he similarity between two 2D complex fields?

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sanjeev a
sanjeev a el 31 de Ag. de 2017
Comentada: Rik el 1 de Sept. de 2017
I have a matrix operation Y=B*A*X; ; where A=exp(1i.*pi*rand(50,50)); B=transpose(A);
X is the input which is random
X=exp(1i.*pi*rand(50,50));
so I have Y calculated.
Now I find another Y say Y1 for another X1=exp(1i.*pi*rand(50,50)); Y and Y1 are complex outputs. How can I measure the similarity between these fileds ?
Can someone please help. I have tried
rsme=sqrt(mean(abs(Y1(:))-abs(Y(:)).^2); But I guess it is a wrong measure ?
  1 comentario
Rik
Rik el 1 de Sept. de 2017
Does that code error? If so, copy the error, if not, you should explain what you mean with similarity. Similarity can mean many things, and guessing what your field's definition is, is wasting time.
Have a read here and here. It will greatly improve your chances of getting an answer.

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Respuestas (1)

David Goodmanson
David Goodmanson el 1 de Sept. de 2017
Editada: David Goodmanson el 1 de Sept. de 2017
Hi sanjeev,
The rms difference is a reasonable measure, but it's not quite what you have. Rms is
sqrt(mean(abs((Y1(:)-Y(:)).^2))) % or
sqrt(mean(abs( Y1(:)-Y(:)).^2))
You are taking absolute value first, so the difference between, say, 1/2 and -1/2 comes out to be zero.
  1 comentario
Rik
Rik el 1 de Sept. de 2017
The RMSe is also used, mostly in the context of comparing an estimation with the ground truth. Then it stands for the root-mean-square of the error.

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