3
$\begingroup$

I've sampled a propagating wave so that I know its amplitude at several equidistant points on a line, and at several points in time. That is, I have this discrete function:

f[t,x]

Now, I want to find the frequency of the wave. But there seems to be two frequencies I can calculate: by finding the FFT at different points in time (e.g. FFT{f[1,x]} ) and at different points in space (e.g. FFT{f[t,1]}).

Only the latter ("temporal frequency") gives me a main frequency in Hz. The other one gives one in 1/m, but I suppose that if I knew the wave propagation speed in m/s, I could convert one to the other. Is this true?

All of the above I've just reasoned through myself. Could someone point me to a source where I could read about the two different types of frequency?

Edit: I'll add that the time-space relation is "normal", i.e. the wave propagates like $g(kx-\omega t)$. Hope I'm not being too unclear. I realise that I'm lacking the terminology needed to explain what I mean.

Edit 2: It might be important to know that $f$ is a pulse that is localised in time and space.

Edit 3: I'll put this question here too: Is there a way to take advantage of all the sample points (a 2D matrix) to estimate the central frequency, assuming I know the wave speed?

$\endgroup$
  • $\begingroup$ I added a quick edit to my answer regarding your 3rd point. $\endgroup$ – Sam Maloney May 2 '13 at 16:43
5
$\begingroup$

The two frequencies you are referring to are the spatial frequency and temporal frequency of the wave, and you are correct in your reasoning on converting one to the other. The spatial frequency refers to how many complete periods the signal goes through for a given unit of distance (eg. cylcles/m) while the temporal frequency refers to how many complete periods the signal goes through for a given unit of time (eg. cyles/s or Hz).

The physical meaning of the two frequencies should be fairly easy to understand if you imagine the wave you are sampling is a simple sine wave. Then it is easy to see that for any constant point in time, your signal will be a frozen sin wave which is periodic in space, and thus has a "spatial frequency". Conversely, for any constant point in space, the value of the signal will oscillate over a regular period in time, and thus have a "temporal frequency" component. You can also imagine that the faster the wave was travelling through space, the faster the signal would change at any one point in space, giving rise to a higher temporal frequency and underlining the relationship between the two properties via the wave's speed.

Regarding using all the sample points to estimate the wave frequency, I believe you should be able to use the known speed to scale one of your dimensions to match the other, since $distance=velocity*time$. Then a 2D FFT of the matrix should give you information based on all of the available samples in whichever dimension you choose.

This article gives a brief description of the relationship between the two in the context of radar imaging and this article talks about the use of the 2D FFT for analyzing higher dimensional signals.

$\endgroup$
  • $\begingroup$ Thanks! Is there a way to take advantage of all the data points when estimating the frequency content of the signal? Currently, I've just looked at either the spatial or temporal frequency at chosen points. (I know of the 2D FFT but don't understand how it would be applied here.) $\endgroup$ – Andreas May 2 '13 at 15:57
  • $\begingroup$ @Andreas I added a link at the bottom of the answer that may be helpful. $\endgroup$ – Sam Maloney May 2 '13 at 16:09
  • $\begingroup$ Sam, you imply it but don't come right out and say it: An issue is that the spatial sampling frequency will generally be different from the temporal sampling frequency. So just doing the FFT of F[t,1] and looking for peaks will give a different peak frequency from the FFT of F[1,x]. @Andreas: Do you know what the equidistant distance is? That will help calibrate things. Also, you may need to correct for "approximately equidistant" spatial sampling points. $\endgroup$ – Peter K. May 2 '13 at 16:52
  • 1
    $\begingroup$ @PeterK. The points are really equidistance, and I know the distance. I'll try the scaling that Sam suggested. $\endgroup$ – Andreas May 2 '13 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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