CMSC 426: Image Processing (Computer Vision)

Let’s look at an example of this. Suppose we have an image in which H(x,y) = y. That is, the image will look like:

11111111111111

22222222222222

33333333333333

And suppose there is optical flow of (1,1). The new image will look like:

-----------------------

-1111111111111

-2222222222222

I(3,3) = 2. H(3,3) = 3. So It(3,3) = -1. GRAD I(3,3) = (0,1). So our constraint equation will be: 0 = -1 + <(0,1), (u,v)>, which is 1 = v. We recover the v component of the optical flow, but not the u component. This is the aperture problem.