Suppose I apply 2D DFT to an image with dimensions $H{\times}W$ where $H \neq W$, then shift the DC component to the center. Why does a circular mask capture the lowest frequency components, i.e. why is it not an ellipse given that the image is rectangular? My concern is that for rectangular images, the K lowest frequencies might be arranged in a non-circular pattern.

  • 1
    $\begingroup$ Why shouldn't it? I honestly don't understand where your question is coming from, so maybe explaining the reason why you're asking this would help us point you in the right direction. $\endgroup$ – Marcus Müller Oct 11 '19 at 16:05
  • $\begingroup$ my concern was that for rectangular images, the K lowest frequencies might be arranged in a non-circular pattern, but this is mostly based on my poor understanding of what the shifting actually does. I'm looking into it now while awaiting responses here $\endgroup$ – tmakino Oct 11 '19 at 18:12
  • $\begingroup$ ah, that would actually make a very good edit to your question (so that people don't have to read our comments to make out what your concern is)! Question from my side: If you have a pattern of frequency $F_\text{diag}$ in the image that lies diagonal, what is its frequency in H- and W-direction? $\endgroup$ – Marcus Müller Oct 11 '19 at 18:38
  • $\begingroup$ hi! I prepared an answer for your question on units, but you deleted it before I could post the answer. So if you wish you can repost it and I could provide the answer. $\endgroup$ – Fat32 Nov 23 '19 at 0:13
  • $\begingroup$ Hi @Fat32, I apologize for that - I deleted it because I think I've figured it out. I'll post what I think is the answer (in the form of a question), and it would be great if you could post your answer there. $\endgroup$ – tmakino Nov 24 '19 at 18:46

For simplicity, let's not do any shifting and only consider non-negative frequencies.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $f_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < f_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2f_c^2}u^2 + \frac{f_s^2}{H^2f_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{f_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{f_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

| improve this answer | |

In one dimension, there is one natural way to arrange frequencies. A filter can be devised, to reduce frequencies $\omega \le \omega_c $. In 2D, there are many ways. Indeed, it depends a lot on the information content in each pixel. For instance: are the width and the height of each pixel comparable? What relative weights ($w_h$, $w_v$) should we give to horizontal or vertical frequencies, to create a global "2D" frequency? One shall remind that, for the human visual system, vertical frequencies are better detected than horizontal frequencies. In image processing, it is common to combine frequencies under $\ell_p$ quasinorms or norms: $$(w_h\omega_h^p+w_v\omega_v^p )^{1/p}\le \omega_c $$

For $p=2$, you get an ellipse (or circular way). With $p=1$, axes are weighted along diagonals. When $p$ is large, only the maximum matters.

| improve this answer | |

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.