Finding the equivalent filter H(u,v) in the frequency domain of a 3x3 spatial mask

Question

I'm trying to find the equivalent frequency domain filter, $H(u,v)$, of a 3x3 spatial mask that averages all neighbours of a point $(x,y)$ in said 3x3 neighbourhood excluding the point itself. So far, I have been able to calculate:

$$g(x,y) = 1/8 \cdot [f(x+1,y) + f(x+1,y+1) + f(x+1,y-1).... f(x-1,y-1)]$$ -(no $f(x,y)$)- but here is where I get stuck.

I've read that I can use linearity and translation properties to get $H(u,v)$ after performing DFT on $g(x,y)$ but for me that is currently a 'draw the rest of the owl' situation.

How do I get $H(u,v)$ from $g(x,y)$ and what form does it take?
I.e., what will the DFT of $f(x+1,y-1)$ look like and how do I then get $H(u,v)$ from that?

Answer 1

Hint: $g(x,y)$ averages all pixel values around $f(x,y)$, at 8-neighbour connectivity distance 1, except $g(x,y)$. The classical uniform $3\times 3$ average is:

$$ m(x,y) = \frac{1}{9}\sum_{i\in[-1,0,1]}\sum_{j\in[-1,0,1]}f(x-i,y-j)\,.$$

Thus: $$9m(x,y) = 8g(x,y)+f(x,y)\,.$$

Knowing the Fourier transform of the uniform $3\times 3$ mask, and that of the identity, and using linear superposition, you should be on track.