7
$\begingroup$

Assume you have a signal, and within it, some pulses are present. A pulse is a simple tone. You know the pulses' duration and shape. (Let us assume that a pulse is made of a couple of cycles, and then to which all those cycles are multiplied by a hamming window. So the final pulse may look like the blue plot below:

something like this

What we do not know are its frequency. (You know its frequency to within $\pm 100\textrm{ Hz}$).

The question is:

Does performing a match-filtering of a signals' absolute magnitude spectrogram with a 2-D version of your pulse in the time-frequency domain, confer upon you any advantages, versus performing a match-filtering of the signals' (shown in red as an example), against the known envelope of the pulse, in the time-domain?

[envelope]2*

For the TF-domain method, assume:

  • STFT analysis.
  • I am using an analysis window equal to the expected pulse length.
  • Percent Overlap: Whatever you want, I do not think it matters for this case.

I am really on the fence on this one because on the one hand, you cannot create information out of nothing, so taking your problem to the time-frequency space seems redundant, while on the other hand, going into the time-frequency space allows you to, perhaps, create 2-D filters that better match your pulse, and/or, ignore noise from other bands which are (perhaps?) not ignored in the time-domain match-filtering case?

My biggest point of confusion is that, inherent to going into the TF domain, we now have both time and frequency localization ambiguity, (based on our choice of the analysis window we use). In contrast, in the time domain, we are $100\%$ sure of our time localization. How - or why - would trading in $100\%$ time-locatlization unambiguity for some joint time-frequency ambiguity help? I am not seeing it.

EDIT:

Another way to look at the problem is with this rephrase: When would one want to do match filtering in only the time-domain ($0\%$ time ambiguity, $100\%$ frequency ambiguity), vs doing it in the joint TF-domain, (x% time ambiguity, (1-x)% frequency ambiguity).

I had a broader question but broke it down into this one first.

| improve this question | |
$\endgroup$
  • 3
    $\begingroup$ A 2-D approach should be able to perform better; there is more information present in the time-frequency plane than just along the time axis only. However, this sort of processing would typically be done on raw data (not from a spectrogram) via the cross-ambiguity function. $\endgroup$ – Jason R May 17 '12 at 22:02
  • $\begingroup$ @JasonR Thanks, I have clarified the question based on your comment. I am not clear on why this is true from a 'you cant create information that wasnt already there' perspective. $\endgroup$ – Spacey May 17 '12 at 22:25
  • $\begingroup$ @JasonR I agree with Mohammad. There is no new information in the 2-D time/frequency plane. The information that was already in the time-domain signal may be in a more useful form, though, when you convert it to a time/frequency plane. $\endgroup$ – Jim Clay May 18 '12 at 13:54
  • $\begingroup$ I disagree. As a counterexample, consider a constant-envelope signal (such as a phase- or frequency-modulated tone). Its envelope is constant; since it has zero bandwidth, so you cannot localize it in time at all via cross-correlation (i.e. matched filtering). However, its spectrogram would not necessarily be constant along the time axis, so you would have a hope for detecting it there. As I noted before, a more appropriate technique involves full calculation of the ambiguity function, which can be thought of as cross-correlation across a time-frequency plane. $\endgroup$ – Jason R May 18 '12 at 14:01
  • 2
    $\begingroup$ @JasonR a phase-modulated tone (which generalizes frequency modulated tones) does not have zero bandwidth. Further, it absolutely can be "localized in time" using matched filtering techniques. $\endgroup$ – Bryan May 18 '12 at 15:10
3
$\begingroup$

Think of the time - frequency ambiguity of your matched filter like so:

  • Frequency ambiguity means it will respond to a range of frequencies
  • Time ambiguity means the response will be 'smeared' around the spatial location.

If you you have 0% frequency ambiguity, the matched filter must look like a sine wave and go on for ever, which in the frequency spectrum looks like a dirac delta.

0% time ambiguity is a dirac delta in the time domain.

So if you have a matched filter that is more than 1 sample wide in the time domain, then it already is ambiguous in both time and frequency domains.

If you are doing matched filtering of the envelope, then you are just looking at the modulating signal, and there is no need to look at the 2D time-frequency spectrogram.

If you want to match the envelop (modulating signal) and the base frequency, then you need a quadrature filter with a bandwidth around the range of frequencies you expect. A quadrature filter is required because it will make the response invariant to the phase of the base signal.

If you don't know the base frequency, a 2D time-frequency spectrogram will be useful as it will show you what frequency is being modulated. Essentially, the spectrogram is response of the signal - time axis - to a bunch of different (centre) frequency quadrature filters - frequency axis.

TLDR:

The premise that the evelope-matched filter in the time domain is 100% localised is incorrect.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your post, I can agree with what you are saying. $\endgroup$ – Spacey Oct 1 '12 at 13:44

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.