In image processing, a forward transform of an $M\times N\text{-}$pixel image $f$ from spatial domain coordinates $(x, y)$ to transform domain coordinates $(u, v)$ can be defined as [1, p.12]:

$$T(u,v)=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}f(x,y)\cdot r(x,y,u,v)$$

The transformation kernel $r(x, y, u, v)$ is said to be symmetric if [1, p.13],

$$r(x, y, u, v) = r_1(x, u) \cdot r_1(y, v)$$

The forward kernel in discrete Fourier transform is [1, p.35]:

$$r(x,y,u,v) = e^{-j2\pi\left(\frac{ux}{M}+\frac{vy}{N}\right)}\\(u=0,1,\dots,M-1,\,v=0,1,\dots,N-1)$$

My question is, What is the benefit of using symmetric kernels in Fourier transform?

I know the benefits of separable kernels, already.

[1]: "Fourier Transform" Slide set, George Bebis (UNR). Based on Ch 4 of “Digital Image Processing”, Gonzales and Woods. http://www.cs.umb.edu/~duc/cs447_647/spring13/slides/FourierTransform.pdf (archived copy)

  • $\begingroup$ Hm, maybe you want to define what you're doing here, and even more importantly, why you're applying a kernel. Also, applying a kernel has not really something to do with Fourier transforming. $\endgroup$ – Marcus Müller Jan 29 '17 at 22:06
  • $\begingroup$ Also, I'd call what you describe separable, not symmetric. could you please cite at least one source? compare: dsp.stackexchange.com/questions/36962/… $\endgroup$ – Marcus Müller Jan 29 '17 at 22:07
  • $\begingroup$ @MarcusMüller, cs.umb.edu/~duc/cs447_647/spring13/slides/FourierTransform.pdf see the Page-13. I know the advantage of separable. I am asking for symmetric. $\endgroup$ – user23572 Jan 30 '17 at 2:04
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    $\begingroup$ Ah I seem to have constantly read $r_2$ where there was in fact a second $r_1$! Ok, I'll go ahead and add the info that you know the advantages of separability already (because based on reading the question alone correctly, that would've been what I've pointed you at – separable filters are fast to execute, and symmetric is a special case of separable, here) $\endgroup$ – Marcus Müller Jan 30 '17 at 8:40
  • $\begingroup$ It's still not clear how symmetric kernel and Fourier Transform are related, so I'm afraid until you clarify that, your question remains unclear! $\endgroup$ – Marcus Müller Jan 30 '17 at 16:10

A set of symmetric kernels along Cartesian axes should make the transform consitent with respect to (at least 90 degree) rotations of the input. If the 2D input yields a particular 2D output, rotating that same 2D input by an angle of 90 degrees, should yield the same 2D output rotated by 90 degrees. If the kernels in the two dimensions were not symmetric, the transform would yield different output that could not be rotated back to match the original output.

I'd have to think if the invariance applied to angles that were not multiples of 90 degrees.

You may also want to have a look at the Abel transform which has a circularly symmetric kernel, useful for physical situations with circular symmetry.


Let's see if discrete Fourier transform (DFT) has a symmetric transform kernel. If it does, then:

$$r_1(x,u)=e^{-j2\pi\frac{ux}{N}}$$ $$\begin{align}r(x, y, u, v) &= r_1(x, u) \cdot r_1(y, v)\\ &= e^{-j2\pi\frac{ux}{N}}\cdot e^{-j2\pi\frac{vy}{N}}\\ &= e^{-j2\pi\frac{ux + vy}{N}}.\end{align}$$

This only matches with the question's definition of the DFT's transform kernel:

$$r(x,y,u,v) = e^{-j2\pi\left(\frac{ux}{M}+\frac{vy}{N}\right)},$$

if $M = N.$

To recap, DFT has a symmetric transform kernel if the image and its transform have square dimensions $N\times N,$ which may offer some benefits.


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