I need clarification on reshape and conv2 comparison

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mathango
mathango le 6 Mai 2016
Commenté : mathango le 6 Mai 2016
In convmtx2 documentation I found the following description :
T = convmtx2(H,m,n) returns the convolution matrix T for the matrix H. If X is an m-by-n matrix, then T = convmtx2(H,m,n) returns the convolution matrix T for the matrix H. If X is an m-by-n matrix, then reshape(T*X(:),size(H)+[m n]-1) is the same as conv2(X,H).
T = convmtx2(H,[m n]) returns the convolution matrix, where the dimensions m and n are a two-element vector. is the same as conv2(X,H).
T = convmtx2(H,[m n]) returns the convolution matrix, where the dimensions m and n are a two-element vector.
I am trying an alternative for C=conv2(A,B,'same') by using reshape command. Below is my attempt that does not work,
A = rand(n,n);
B = rand(3,3);
C=reshape(T*A(:),size(B)+[n n]-1);
I know that reshape command set up is incorrect. How to fix this?

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Image Analyst
Image Analyst le 6 Mai 2016
That is not correct. Be aware that convolution "flips" the kernel, so unless your kernel is symmetric, which it won't be if you're getting it from rand(), then the answers won't be the same.
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mathango
mathango le 6 Mai 2016
Thanks for the tip. I was not aware of that.

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