Is it possible to combine two convolution kernels (convolution in terms of image processing, so it's actually a correlation) into one, so that covnolving the image with the new kernel gives the same output as convolving it with the first, and then the second kernel?
For example, if I want to convolve an image with 1st 3×3 kernel and then 2nd 3×3 kernel, apparently I should be able to combine those two kernels and convolve my image with this new kernel. What is the size of my new kernel? Is it 3×3? If so, there's something worrying me about that. Let me explain.
Let's say I'm convolving Input image with Kernel 1, then convolving the result with Kernel 2.
Pixel marked X depends on value of pixel P2 (convolving the image with kernel 2). Pixel P2 lies on the new image that was produced by convolving input image with Kernel 1. So the value of P2 was calculated with taking P1 into account.
Now, X relies on values of P1 and P2.
That's what stops me from believing I could achieve the same result with convolving my original image with a 3×3 kernel combining kernel 1 and kernel 2. Pixel X cannot "know" anything about pixel at location P1.
Best Answer
Let's assume the original image is $x[n_1,n_2]$ and two kernels are $h_1[n_1,n_2]$ and $h_2[n_1,n_2]$. Convolution associative law says that:
$$(x[n_1,n_2]*h_1[n_1,n_2])*h_2[n_1,n_2] = x[n_1,n_2]*(h_1[n_1,n_2]*h_2[n_1,n_2])$$
So covnolving the image with the new kernel will always produce the same output as convolving it with the first, and then the second kernel. To get the new kernel just use the 2-dimensional convolution definition:
$$h_3[n_1,n_2]=h_1[n_1,n_2]*h_2[n_1,n_2]=\sum_{i=-\infty}^{+\infty} \sum_{j=-\infty}^{+\infty} h_1[i,j] h_2[n_1-i,n_2-j] $$