Why conv2 gives opposite sign

Hi, I have a simple matrix I:

I = [1,2,3,4,5;
    6,7,8,9,10;
    11,12,13,14,15;
    16,17,18,19,20;
    21,22,23,24,25]

My kernel is:

k = [0,0,0;
    1,0,-1;
    0,0,0]

I would like to compute for the non-boundary elements. For each element, I want to compute its left neighbour - its right neighbour,i.e.:

dIx = [~,~,~,~,~;
       ~,6-8,7-9,8-10,~;
       ~,11-13,12-14,13-15,~;
       ~,16-18,17-19,18-20,~;
       ~,~,~,~,~]

I used this line to achieve this:

dIx = conv2(I,k,'valid');

This gives me the result:

dIx = [0,0,0,0;
       0,2,2,2;
       0,2,2,2;
       0,2,2,2]

which is the opposite sign of what I expected. Why is the case?

Thank you very much.

Answer 1

[Matt J]

Remember that a convolution does not slide k over I, but rather a pre-flipped version of k. You can compensate for this by doing,

k = [0,0,0;
    1,0,-1;
    0,0,0]

k=fliplr(flipud(k));

dIx = conv2(I,k,'valid');

although for the kernel you've shown this pre-flip happens to be equivalent to k=-k.

[xiaojuezi]

Thank you very much!

Answer 2

[Image Analyst]

Convolution flips the kernel - that's just how it works, and there are theoretical/mathematical reasons for it. Perhaps you can use imfilter() instead if you don't want to flip the kernel to be flipped.

[xiaojuezi]

Thank you very much! But imfilter() has less control over boundary points.

[Matt J]

filter2 does have that control, however. You could use that to avoid implementing your own kernel flip.

[Image Analyst]

Here are the boundary options for conv2:

Here are the options for imfilter():

It looks to me like imfilter() has more options than conv2(). What control did you want to have over the boundary that conv2() has but imfilter() does not? 'Valid'? You could always just use indexing to crop out that part yet still avoid the flipping of the kernel, which you don't want to do.

outputImage = outputImage(2:end-1, 2:end-1);