# 2D DFT of a filter kernel

for understanding i want to compute the DFT out of the 3x3 filter kernel using the shifting property:

$$\begin{bmatrix}0& 1 & 0\\1 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix} \cdot \frac{1}{4}$$

My computation is $$y(x,y) = \frac{1}{4}\cdot (f(x+1,y)+f(x-1,y)+f(x,y-1)+f(x,y+1))$$ yields

$$\frac{1}{4}\cdot F(u,v)\cdot (\exp(i2\pi u/3)+\exp(-i2\pi u/3)+\exp(i2\pi v/3)+ \exp(-i2\pi v/3))$$

So $$H(u,v) =\frac{1}{4}\cdot (\exp(i2\pi u/3)+\exp(-i2\pi u/3)+\exp(i2\pi v/3)+ \exp(-i2\pi v/3))$$?

Or is it wrong?

• It depends on the length of your DFT. You've chosen a DFT length of 3, which is the same size as your kernel. Does that make sense? – Peter K. Feb 25 at 15:30