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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.

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    $\begingroup$ 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 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 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 at 18:38
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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 $\omega_c$ your mask would select frequencies:

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

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:

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

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

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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.

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