[Solved] Why is the output of `conv2()` is divided by `sz^2`?
Note that conv2(double(im(:,:,q)),mask,’same’)./sz^2 is the same as conv2(double(im(:,:,q)),mask./sz^2,’same’) This is because the convolution and the multiplication commute. Thus, the convolution operation computes a local mean. Without the division, it would be a local sum. 0 solved Why is the output of `conv2()` is divided by `sz^2`?