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Discrete Fourier transform in k and 1/k

by gnulinger
Tags: 1 or k, discrete, fourier, transform
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gnulinger
#1
Nov11-12, 03:04 AM
P: 30
Say you have some function that is periodic in a parameter k. The discrete Fourier transform from a sampling may be found in the usual way, giving the frequency spectrum in k. But what if I want to find the frequency spectrum in 1/k ?

I'm not really sure what this is called, and so I've had a hard time Google searching for it. Any links or help would be appreciated. Thanks.
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chiro
#2
Nov11-12, 05:34 AM
P: 4,573
Hey gnulinger.

When you say frequency spectrum are you talking about integer frequencies?

I do know that there are ways to get fractional frequencies that are based on fractional derivatives and subseqent integrals:

http://mathworld.wolfram.com/FractionalDerivative.html

But if you are talking about just having a transfer function to get something in F(1/k) instead of F(k), then I think this is going to be a bit more involved and you should probably outline the reason why you want the function in terms of 1/k as opposed to the linear transform space k.
gnulinger
#3
Nov11-12, 12:11 PM
P: 30
Quote Quote by chiro View Post
But if you are talking about just having a transfer function to get something in F(1/k) instead of F(k), then I think this is going to be a bit more involved and you should probably outline the reason why you want the function in terms of 1/k as opposed to the linear transform space k.
I am talking about the latter, and yes, I think it will be fairly involved. I have a function that is periodic in 1/k, and I am wondering if there is some way of mapping the DFT in k to that in 1/k.

rbj
#4
Nov11-12, 12:16 PM
P: 2,251
Discrete Fourier transform in k and 1/k

i know a lot about the DFT, it's definition, the theorems, how it is related to the continuous fourier transform. but i cannot decode at all what you're talking about. what do you mean that it is "periodic in 1/k" ? try tossing up equations to be clear.

BTW, even though i get in fights about this on comp.dsp, i maintain that the DFT is nothing other than the Discrete Fourier Series. the DFT maps one discrete and periodic sequence of length N to another discrete and periodic sequence of the same length. and the inverse DFT maps it back. dunno if that answers your question.
gnulinger
#5
Nov11-12, 02:35 PM
P: 30
Part of the problem is that I too am unclear on this subject, so it is hard for me to ask the right questions. I was hoping that someone may have heard of something related to what I was asking about, and could have pointed me in the right direction.

In the De Haas-van Alphen effect, wikipedia link, the magnetic moment of a crystal oscillates with period related to 1/B, where B is the magnetic field. The DFT would ostensibly give you a frequency spectrum in 1/B.

This is similar to what I want to do.
rbj
#6
Nov11-12, 09:15 PM
P: 2,251
so, are you sampling the magnetic moment function of time somehow? where do the numbers that go into the DFT get set to some value?
marcusl
#7
Nov11-12, 10:38 PM
Sci Advisor
PF Gold
P: 2,081
The simplest way is to plot the results against 1/k on a nonlinear scale. You often see optical spectra plotted this way--the calculation is done for frequency but the plot is done against lambda.
gnulinger
#8
Nov12-12, 03:22 PM
P: 30
Quote Quote by marcusl View Post
The simplest way is to plot the results against 1/k on a nonlinear scale. You often see optical spectra plotted this way--the calculation is done for frequency but the plot is done against lambda.
Do plot your data or the DFT against 1/k?
mdo
#9
Nov19-12, 03:40 AM
P: 10
You have a function f(t) that has a Fourier transform, F(ω), that is null or almost null for |ω| > Ω. f(t) is sampled at every multiple of a given interval h to obtain a discrete signal f(n) = f(nh), where h should be < π/Ω. When you compute the DFT, F(μ), you are working on a normalized domain 0≤μ<π, but you can express it in "real" ω by multiplying by Ω or by 2π/h.
It's indifferent.


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