imgaussfilt asymmetry as a linear operator
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The Gaussian image blurring operation imgaussfilt is not a symmetric linear operator, as the test below shows. This is surprising to me based on my understanding of Gaussian blurring. I am wondering (a) why it is unsymmetric, and (b) given that it is unsymmetric, how I can obtain its adjoint operator. In other words, I would like a function imgaussfilt_adjoint such that sum(sum(X.*imgaussfilt(Y,k)))==sum(sum(imgaussfilt_adjoint(X,k).*Y)). Note that I am not referring to the 'symmetry' boundary option.
>> X = randn(200,200);
>> Y = randn(200,200);
>> Xblur = imgaussfilt(X,100);
>> Yblur = imgaussfilt(Y,100);
>> sum(sum(X.*Yblur))
ans =
3.6114
>> sum(sum(Xblur.*Y))
ans =
4.5137
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Matt J
il 30 Nov 2021
Modificato: Matt J
il 30 Nov 2021
It is because of edge effects. If you add sufficient zero padding you will see symmetric behavior.
3 Commenti
Matt J
il 30 Nov 2021
Modificato: Matt J
il 30 Nov 2021
It also appears that symmetry is present with any of the padding modes except for 'replicate'.
X = randn(200,200);
Y = randn(200,200);
for p=["symmetric","circular","replicate"]
Xblur = imgaussfilt(X,200,'Padding',p);
Yblur = imgaussfilt(Y,200,'Padding',p);
asymmetry=sum(X.*Yblur, 'all')-sum(Xblur.*Y , 'all')
disp ' '
end
Più risposte (2)
Image Analyst
il 30 Nov 2021
It's because they're random numbers. X and Y are not equal, and neither are the blurred versions. So why would you expect two different random matrices multiplied by each other element by element to have the identical integrated gray value? Look, I don't even get the same numbers as you
X = randn(200,200);
Y = randn(200,200);
Xblur = imgaussfilt(X,100);
Yblur = imgaussfilt(Y,100);
sum(sum(X.*Yblur))
sum(sum(Xblur.*Y))
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