I need clarification on reshape and conv2 comparison

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mathango 2016-5-6

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评论： mathango 2016-5-6

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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Answer 1

Image Analyst 2016-5-6

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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.