Timeline for Inverse Fourier Transform Of K-space Image...what is the object space scale?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2013 at 9:56 | answer | added | user1886681 | timeline score: 0 | |
| Jul 11, 2013 at 21:30 | answer | added | Colin McFaul | timeline score: 2 | |
| Jul 10, 2013 at 11:09 | history | tweeted | twitter.com/#!/StackPhysics/status/354920402641883136 | ||
| Jul 10, 2013 at 10:19 | comment | added | user1886681 | ...SHOULD be the size of the group of electrons that emitted the light. The goal is to find the width of the object space image to see how big the electron group/bunch is. This is what I've been struggling with for a while now. Not a hw problem but a personal project I would like to understand completely! | |
| Jul 10, 2013 at 10:17 | comment | added | user1886681 | Thanks for the input! In a sense it can be treated as xray diffraction, but in actuality it is a transition radiation problem.Essentially what is happening is that a group of electrons are passing through a medium, each emitting their own photon/light wave. The emitted light travels some distance (far field) until it reaches a 200x200 pixel CCD. This is the Kspace image above; we know the angle and distance of the brightest band. When we find the IFFT of the image (literally using the Matlab function) assuming we know the phases, we get the object space image. The object space image... | |
| Jul 10, 2013 at 0:52 | comment | added | user10851 | 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. | |
| Jul 10, 2013 at 0:47 | comment | added | user10851 | One interesting point is that the inverse Fourier transform is, for all intents and purposes, the exact same as the forward Fourier transform. That's why few people use the term "inverse" in these contexts. The fact that "inverse" is being specified means the source for this statement probably isn't using the unitary version - i.e. they are using a version where one direction has a $2\pi$ in front and the other doesn't. Without more info, any answer could easily be off by a factor of $2\pi$ or so. | |
| Jul 10, 2013 at 0:08 | history | edited | user1886681 |
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| Jul 9, 2013 at 1:46 | history | edited | user1886681 |
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| Jul 9, 2013 at 1:45 | comment | added | user1886681 | 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. | |
| Jul 9, 2013 at 1:41 | comment | added | tpg2114 | Did you read the link there to see how we define homework? It's not "assigned in a class" type of question per se. | |
| S Jul 9, 2013 at 1:41 | history | suggested | tpg2114 | CC BY-SA 3.0 |
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| Jul 9, 2013 at 1:39 | comment | added | user1886681 | 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. | |
| Jul 9, 2013 at 1:36 | review | First posts | |||
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| Jul 9, 2013 at 1:34 | comment | added | tpg2114 | Is this a homework problem? | |
| Jul 9, 2013 at 1:34 | review | Suggested edits | |||
| S Jul 9, 2013 at 1:41 | |||||
| Jul 9, 2013 at 1:18 | history | asked | user1886681 | CC BY-SA 3.0 |