# Inverse Fourier Transform Of K-space Image…what is the object space scale?

Checked around a buch and could not find any help. But I needed help with:

Understanding that if I get the Inverse FT of K-space data, what is the scaling on the X-space (object space) resultant image/data i.e. for every tick on the axis, how do I know the spatial length?

More detailed explanation in the below image.

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Is this a homework problem? – tpg2114 Jul 9 '13 at 1:34
No, not at all. I am geniunely trying to understand this for a week now but cannot. I made the image in powerpoint because, at this point, I am desperate for help. – user1886681 Jul 9 '13 at 1:39
Did you read the link there to see how we define homework? It's not "assigned in a class" type of question per se. – tpg2114 Jul 9 '13 at 1:41
This is not for a class. Its for my overall understanding, but I guess I could tag as such. I'm not necessarily looking for some one to solve it, I dont think they can with the info I gave them...I just need to understand the scales. – user1886681 Jul 9 '13 at 1:45
Even without that ambiguity, "K-space" is not a universally understood physics term. Or rather it is universal - it always means the Fourier transform of something "real." What that real thing is depends on context. What exactly is going on here? Is this a CCD on a camera? Are there optics involved? Is this an X-ray diffraction question? Without context this is unanswerable. – Chris White Jul 10 '13 at 0:52

The units of your X-space are the inverse of the units of your K-space. So if your K-space is in $\mathrm{m}^{-1}$, then your X-space will be in $\mathrm{m}$.

To make the full circuit $f(x) \rightarrow F(k) \rightarrow f(x)$ requires an overall normalization factor of $1/2\pi$ to ensure that you get the function you started with. As Chris White points out in his comment, there are a few different conventions on where exactly to put this normalization factor. Some put it entirely on one of the transformations. Some conventions split it between the two transforms, and put $1/\sqrt{2\pi}$ on each integral; this has the advantage of making the Fourier transform and the inverse Fourier transform perfectly symmetrical with respect to $x$ and $k$.

In addition, some conventions for wavenumber define it as cycles per unit distance (so that $xk = 1$), while some define wavenumber as radians per unit distance (so that $xk = 2\pi$).

Ultimately, you might need to multiply the axes in your X space by $1, \sqrt{2\pi},$ or $2\pi$, depending on the set of conventions your software is using, and the convention you have used to express your $k$ values. You should already know the latter. For the former, you will have to check the documentation for the Fourier transform in your software.

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Ok, so firstly thanks so much for all of your help...

Secondly I have wrote down the solution (ATTACHED PDF) that one of the guys in my group gave me. But to be honest I don't understand the very first relation (in step one).

I specifically don't understand how the width of the peak in pixels fits in? Any guidance?

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what software are you using for the plots and presentations? – WetSavannaAnimal aka Rod Vance Oct 10 '13 at 23:53