RMS error of matrices

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darksideofthemoon101
darksideofthemoon101 2011 年 4 月 6 日
コメント済み: Todd Pierce 2022 年 5 月 11 日
Hi,
I have 2 large matrixes (2048x2048), and have taken one away from the other to get a 'difference' matrix. I want to quantify this matrix by using an RMS error, however if I use a standard RMS error formula I get another 2048x2048 matrix.
Is there a way to get a single representative RMS value out?
Thanks,
Richard
  1 件のコメント
Todd Pierce
Todd Pierce 2022 年 5 月 11 日
Hi there,
I am in a simlar position to yourself however, i want to maintain the matrix with all the rms values in, what initial line did you type in to get the rms for every value in the matrix whilst maintaining the shape.

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採用された回答

Walter Roberson
Walter Roberson 2011 年 4 月 6 日
Isn't it usually
sqrt(mean(A(:).^2 - B(:).^2))

その他の回答 (1 件)

darksideofthemoon101
darksideofthemoon101 2011 年 4 月 6 日
I've been using
rms = sqrt((A - B)^2))
How dissimilar are these equations?
  1 件のコメント
Walter Roberson
Walter Roberson 2011 年 4 月 6 日
When A and B are arrays then A-B is an array, and (A-B)^2 is
(A-B)*(A-B) which is a matrix multiplication which will produce an output the same size as A (in this case.) sqrt() of that would then be the same size.
Nothing in your code calculates the _mean_ portion of "RMS". Root MEAN square.
Possibly I should have suggested
sqrt(mean((A(:)-B(:)).^2))
but I am too tired to look up the definition at the moment.

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