0
$\begingroup$

I'm trying to understand 2d Fourier Transforms as applied to images.

Given an image, is a 2d transform equivalent to taking an infinite amount of slices of the 2d image to form a one dimensional function and then applying a 1d Fourier Transform on each slice/one variable function?

Also, if you can, could you also explain how my description would be represented algebraically, and if there are any relations to the 2d Fourier Transform (if my explanation is inaccurate)?

$\endgroup$

closed as off-topic by user21820, B. Mehta, JonMark Perry, Claude Leibovici, user223391 May 6 '18 at 16:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, B. Mehta, JonMark Perry, Claude Leibovici, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't see why this question should be closed. I think the idea in it is wrong, but that's all the more reason it should be answered. $\endgroup$ – David K May 4 '18 at 12:32
1
$\begingroup$

No; it's a composition of $2$ different $1$-dimensional Fourier transforms, viz. $$\int dx dy f(x,\,y)\exp i(kx+qy)=\int dy \bigg[\int dx f(x,\,y) \exp ikx\bigg]\exp iqy.$$

$\endgroup$
  • $\begingroup$ So what does this composition mean intuitively? $\endgroup$ – Goldname Apr 26 '18 at 22:42
  • $\begingroup$ @Goldname If you define the Fourier transform of $f(x)$ with $d$-dimensional $x$ as $\int (2\pi)^{-d/2}f(y)\exp -ix\cdot y$, which is also a function of $x$, on the space of such functions the transform is unitary so it's like a rotation. If a composition of two Fourier transforms is still Fourier, that's because we've found two rotations of a certain type whose composition is a third such rotation. $\endgroup$ – J.G. Apr 27 '18 at 6:06

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