0
$\begingroup$

My question is: Suppose that we have low pass spatial domain filter that averages 4-connected neighbors of that pixel and it doesn't consider its pixel in averaging. Find its corresponding filter in frequency domain and show that it is a low pass filter.

I know that I must take forward Fourier transform from spatial domain filter and its output is frequency domain filter, but I don't have any idea what is spatial filter is? I think that it is $w(x,y)$ such that $-1 \le x \le 1 $ and $-1 \le y \le 1$ and $w(-1,0) = w(0,-1) = w(0,1) = w(1,0) = 1$ and $w(-1,-1) = w(0,0) = w(1,-1) = w(-1,1) = w(1,1) = 0$ but I don't know how to take Fourier transform from this?

How do I take the Fourier transform of my 3×3 filter kernel?

$\endgroup$
  • 1
    $\begingroup$ So, your question is not actually a filter question, just, "how do I calculate the Fourier transform of a $3\times 3$ matrix?" $\endgroup$ – Marcus Müller Apr 17 '17 at 17:49
  • $\begingroup$ But I am not sure about this whether I am right about this spatial domain filter. $\endgroup$ – user112588 Apr 17 '17 at 18:09
  • 1
    $\begingroup$ now I'm even more confused. Can you answer me exactly what you're asking: a) is my spatial filter correct? OR b) how do I take the Fourier transform of my $3\times 3$ filter kernel? $\endgroup$ – Marcus Müller Apr 17 '17 at 18:13
  • $\begingroup$ Ok;I assume that my spatial domain filter is right ,I doubt because I don't know whether it is low pass spatial domain filter or not? b is my second question? $\endgroup$ – user112588 Apr 17 '17 at 18:29
  • $\begingroup$ you know I am not electrical engineer and I am computer science student and I am learning image processing in order to work on computer vision,and I am not familiar with signals and digital signal processing.I just know signals from image processing. $\endgroup$ – user112588 Apr 17 '17 at 18:33
1
$\begingroup$

So, the $M\times N$-DFT is very well-defined to be (aside from the usual ambiguity about the factor up front)

$$ X[k,l] = \frac{1}{\sqrt{MN}}\sum_{n=0}^{N-1} \left(\sum_{m=0}^{M-1}x[m,n]\cdot e^{-jmk\frac{2\pi}{M}}\right)e^{-jnl\frac{2\pi}{N}} $$

In your $M=N=3$ case, that becomes

$$ X[k,l] = \frac{1}{3}\sum_{n=0}^{2} \left(\sum_{m=0}^{2}x[m,n]\cdot e^{-jmk\frac{2\pi}{3}}\right)e^{-jnl\frac{2\pi}{3}} \tag1 $$

Furthermore,

$$ x = \begin{bmatrix} 0 & 1& 0 \\ 1& 0 & 1\\ 0 & 1&0 \\ \end{bmatrix} $$

so that all there's a lot of zeros.

Simply expand the sum from $(1)$, filling in the values of $x$.

$$\begin{array} \,3X[k,l]&=& &\left(e^{-jk\frac{2\pi}3}\right)e^{j0} & \text{first row}\\ && +& \left(e^{-j0}+e^{-jk\frac{4\pi}3}\right)e^{-jl\frac{2\pi}3}& \text{second row}\\ &&+ &\left(e^{-jk\frac{2\pi}3}\right)e^{-jl\frac{4\pi}3}&\text{third row}\\ &=&&e^{-jk\frac{2\pi}3}\\ &&+&e^{-jl\frac{2\pi}3} + e^{-j\frac{2\pi}3\left(l-2k\right)}\\ &&+&e^{-j\frac{2\pi}3\left(k-2l\right)} \end{array}$$

So,

$$X[k,l]=\frac13\left(e^{-jk\frac{2\pi}3}+e^{-jl\frac{2\pi}3} + e^{-j\frac{2\pi}3\left(l-2k\right)}+e^{-j\frac{2\pi}3\left(k-2l\right)}\right)$$

Set in all values for $k$ and $l$, and get your $3\times3$ DFT output .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.