# Calculating the DFT of the image after using a filter

Studying for my finals in Image Processing course. Trying to solve the following question:

Let $$h$$ be a filter that replaces each pixel value with the average of it's 8 neighbors. Let $$f$$ be a function of the image $$8\times 8$$ and let $$F(u,v)$$ be it's DFT. Let $$G(u,v)$$ be the DFT of the image, after performing the filer on it. Calculate $$G(5,5)$$ if we know that $$F(5,5)=1$$.

The solution:

The filter looks like: $$\frac{1}{8}\begin{bmatrix}1 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix}$$ The DFT of the filter is: $$H(u,v)=\frac{1}{10\cdot 10}\sum_{x=-1}^{1}H(x,v)\exp(-i2\pi u x /10)$$ when $$H(x,v)=\sum_{y=-1}^{1}h(x,y)\exp(-i2\pi v y /10)$$ so we get $$H(-1,v)=H(1,v)=\frac{1}{4}\cos(\pi v/5)+\frac{1}{8}$$ and $$H(0,v)=\frac{1}{4}\cos(\pi v/5)$$ and for all $$x\not\in\{-1,0,1\}$$ we get $$H(x,v)=0$$. So we get: $$H(u,v)=\frac{1}{100}\left[\left(\frac{1}{4}\cos(\pi v/5)+\frac{1}{8}\right)\cos(\pi u/5)+\frac{1}{4}\cos(\pi v/5)\right]$$ Then we get $$G(5,5)=H(5,5)\cdot F(5,5)=\frac{1}{800}$$.

I don't quite understand the solution. I understand why the filter looks like it's mentioned. But I don't understand the DFT part. How do I turn the filter into a DFT function? Also, why they divided by $$10$$? What is $$H(u,v)$$ and what is $$H(x,v)$$? I just need some technical explanation regarding the solution.

• The notation is awkward. They do a 2-D FFT by first doing an FFTs of the columns ($H(x,v)$) and then FFTs of the rows of the result ($H(u,v)$). These should NOT be both called $H()$ since they are different things. Feb 2 at 15:17