If I have a two-dimensional discrete dataset (one space, one time), and I create subsets of this dataset by "sampling" (although the original dataset isn't continuous) it at different rates, should I expect that the same filter performed on different subsets would produce the same field (although sampled at different times)?

To elaborate, I have a discrete dataset that contains a four-times daily sampled field. If I choose every fourth data point in time after the first, I can get a discrete sub-dataset of a once-daily sampled field, although the fields in the 4x and 1x datasets are still the same at the same time. The filtering method is then as follows:

  1. Choose the sampling rate of the discrete dataset
  2. Perform an FFT in space and time to get a wavenumber-frequency spectrum
  3. Filter the spectrum to retrieve signals propagating only in a certain direction
  4. Invert the time and space transforms to get the field propagating in that direction

Now, the field I get with the 4x rate and that with the 1x rate are wildly different - not just a matter of magnitude or phase (that I can tell), they are almost entirely incomparable. However, my expectation is that signals propagating in one direction in the 4x dataset should still propagate in that direction in the 1x dataset given that they are actually the same dataset! All relevant signals are well below the Nyquist frequency, and I have accounted for scaling by the dataset length.

I have confirmed that the 1x (sampled from the 4x dataset) and a native 1x-daily sampled dataset produce similar results, as I would expect - and I have a physical reason for believing both 1x results are correct. The output is much different when I try to employ the entire 4x dataset.

My only guess is that there is an extra need to account for the dimensional frequency spectrum being different between the 4x and 1x datasets, but I also thought that the FFT shouldn't care what the time step is. Is there any mathematical source of the difference, and if so, how can I correct it?

  • $\begingroup$ Welcome to DSP.SE! Interesting question. How are you achieving 3. Filter the spectrum to retrieve signals propagating only in a certain direction ? $\endgroup$ – Peter K. Oct 1 '15 at 16:52
  • $\begingroup$ @PeterK. In the wavenumber-frequency domain, I identify signals in the first and third quadrants as those propagating in the positive spatial direction, given that the phase speed $c = \omega/k$ is positive; likewise, signals in the second and fourth quadrants are propagating negatively. As a simple filter, I take these quadrants into their own array and populate the other two quadrants with zeroes. $\endgroup$ – Soular Oct 1 '15 at 18:00
  • $\begingroup$ Clarify me your signal processing chain, please. First you have a data set, say matrix A with time and space dimensions. Next you a) process it directly (4x) or b) resample it by removing 3 of 4 samples (1x). Then you do FFT to achive frequency domain? And next you apply FFT to each frequency-spatial data set to get frequency-wavenumber (or in other words frequency-angular response)? Next you filter signal in this domain and at last perform inverse operations to get frequency/angular filtered signal? Am I right in my assumptions? $\endgroup$ – Serj Oct 1 '15 at 18:20
  • $\begingroup$ @Serj My initial matrix of data is the 4x data. I may proceed using the data as is, or resample as you say to get 1x data. I do the spatial FFT first, then apply a second FFT to the wavenumber-time data to get signal in the wavenumber-frequency domain. Next is the filter as I have described in a previous comment. Following that I apply an inverse FFT in the frequency domain, and lastly the inverse FFT in the wavenumber domain to retrieve the filtered space-time signal. I have done this with both the 1x and 4x data. $\endgroup$ – Soular Oct 1 '15 at 18:56
  • $\begingroup$ So 1) does your input signal satisfy narrowband condition for the array processing? 2) I suppose the inverse spatial FFT is excess because after the first spatial FFT you have the set of spatially filtered signals, beams. 3) when you resample your signal you increase the signal's bandwidth to sampling frequency ratio by 4 because it is decimation. Maybe it could affect your algorithm performance, especially if your signal is wideband. $\endgroup$ – Serj Oct 1 '15 at 19:04

Your Answer

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

Browse other questions tagged or ask your own question.