Apply parabolic filter in frequency domain

Question

What would be the effect on initial image of applying this filter:

in the frequency domain?

How can I determine the effect without using computer (on paper)?

This filter (H) is already in frequency domain.

Answer 1

You can try to see what you get for different values.

First thing you can notice is because there is a 1 in the filter function the filter only amplify and don't waken the frequency.

Second if we look at the two other terms: We amplify each spatial frequency as a function of the maximum frequency in each direction. Therefore for small frequencies the amplification will be minimal because (smallFreq/largeFreq) < 1 and the squaring this term will only make this value even smaller.

For the large frequencies the amplification will be bigger. For uMax and vMax the amplification can be by a factor of 3 times the original value.

If you want to know what will be the effect on the original image with out going back to the spatial domain you can say that areas with high frequencies (edges in the image) will have a very high value and areas with low frequencies( areas with a more or less a constant value) will stay the same.

In conclusion it looks like your filter is a sharpening filter.

Answer 2

I assume, u_max and v_max are constant parameters and that by without using a computer you mean not just calculate the effect numerically, but rather prefer an analytical solution.

Filtering/multiplication in Fourier space corresponds to a convolution in real space with the Fourier transformed filter function. In order to know the convolution kernel, i.e. determining the effect, the inverse Fourier transform of the given filter H needs to be contructed if it exists.

Mathematica solves

InverseFourierTransform[1+(u/umax)^2+(v/vmax)^2,{u,v},{x,y}]

to

2 \[Pi] DiracDelta[x] DiracDelta[y]-(2 \[Pi] DiracDelta[y] (DiracDelta^\[Prime]\[Prime])[x])/umax^2-(2 \[Pi] DiracDelta[x] (DiracDelta^\[Prime]\[Prime])[y])/vmax^2

i.e. to a convoluted expression of Dirac delta functions of x and y as well as second derivatives of Dirac delta functions.

It's a bit too difficult for me to imagine the exact shape of this. One problem is that the integral of your filter H is unbounded, so we get a problem with normalization (or existence of the Fourier transform) anyway, but (using a computer) to visualize the results, I found that the convolution kernel is zero for either x or y not zero and is maximal at x=y=0 and falls off sharply along the x and y-axis and is symmetrical with respect to the x- and y-axes.

So, in conclusion, it's a very strange filter that smoothes a tiny bit along the axes. I would have expected a sharpening too like Amitay does in his answer, so it's a bit surprising.