0
$\begingroup$

The Gabor-Heisenberg uncertainty represents the fundamental limit of time and frequency resolution one can extract from a signal.

$\sigma_t \sigma_f \geq \frac{1}{4\pi}$

My question is: what happen in a noisy environment?

If a bird sings a perfect constant pitch. Depending of the ambient noise, the frequency estimation of this pitch will be affected by the SNR.

Let say you have a time-varying arbitrary tune (like a bird song) played in a arbitrary colored noise environment. Is there a generalized Gabor uncertainty that takes to account the SNR?

I expected something like this:

$\sigma_t \sigma_f \geq function(SNR)$

Given the downvote(s) and the comments, I'll try to rephrase my question:

  1. Frequency estimation is limited by SNR in a noisy environnent.

  2. Frequency estimation is limited by Gabor-uncertainty.

If statements 1 and 2 are correct, how to reconcile the two statements in one general concept? I suppose, there should be a way to compute the 'frequency-estimation precision' as a function of both SNR and time resolution (maybe with some other assumptions).

|improve this question
$\endgroup$
  • 1
    $\begingroup$ Wouldn't such a noise be additive? Let us say that the bird singing is $x$ and a white noise $n$ is present, then the resulting signal would simply be $y=x+n$. That is, the uncertainty $\sigma_t\sigma_f\geq \frac{1}{4\pi}$ Does not change but is applied to $y$, a signal with more frequencies. Does that answer your model? $\endgroup$ – havakok Jul 17 at 12:57
  • 1
    $\begingroup$ +1 for havakok. SNR has nothing to do with achievable time-frequency resolution. $\endgroup$ – Max Jul 17 at 13:01
  • $\begingroup$ in signal processing, a signal isn’t a quantum mechanic quantity. It isn’t a fundamental limit. Model based algorithms often beat the Gabor limit. In noise, Cramer Rao bound, Weis weinstein bound, Ziv Zakai bound, ... $\endgroup$ – Stanley Pawlukiewicz Jul 17 at 13:07
  • $\begingroup$ I think it's is not necessary to invoke QM when concidering the Gabor-Heisenberg uncertainty. This relationship is more general than QM. Regarding the rest of your comment, is this limit beaten because information is added (a priori model)? $\endgroup$ – pierebean Jul 17 at 13:18
  • $\begingroup$ @pierebean there are no probabilities involved. no confidence regions, no p-value, no standard deviations. Nothing is uncertain. It’s a statement about how concentration in one domain effects the concentration is another. In QM, it is a true uncertainty. In a noiseless deterministic signal, nothing is probabilistic. There are no fundamental limits that can’t be exceeded. $\endgroup$ – Stanley Pawlukiewicz Jul 17 at 14:17
0
$\begingroup$

This uncertainty thing is often misapplied. Your statement for #2 is such a misapplication. The bell curve, aka Gaussian, is an eigenfunction of the FT, (only an approximation in the DFT, but that is a different story). The uncertainty principle is saying when a shape is taken back and forth from the FT and the narrowest cluster of information is desired, this is accomplished by the Gaussian. It has nothing to do with parameter estimation, but noise indeed does.

You are conflating different concepts.

My latest blog article gives the actual code for calculating the frequency, amplitude, and phase of a pure tone exactly in the single noiseless pure tone case. It is also robust in the presence of noise.

A Two Bin Solution

Early on, when trying to show "experts" that I had found exact formula frequencies for tones in the DFT, I was often quoted what you said and they would refuse to even look at my math as if I were proposing a perpetual motion machine.

So many misconceptions exist about the DFT, even among supposed learned folks, and that is scary.

Don't believe me?

Here is an actual quote from an expert who shall remain unnamed.

----------------------------------------------------------
An "exact" computation of frequency based on FTs is not possible for
some very simple reasons:  The Fourier Transform is itself an
estimator, and the time-frequency uncertainty principle means
frequency can't be determined exactly without an infinitely long
observation window.  So essentially all frequency computations of time
series or signals are estimates, it's just that some are more accurate
or more efficient than others.

So claiming that you have an exact solution just isn't going to get
the attention of everyone interested in the field.
----------------------------------------------------------

The only truth in there is the last sentence. So, I wrote it up in a blog where the math is indisputable and anyone who wants to can verify it for themselves.

No, I've never gotten an apology.

|improve this answer
$\endgroup$
  • $\begingroup$ If you observe noiy data where your noise-free data follows a parametric model exactly and the mapping from model parameter to observations is unique (read: you made a sufficient number of observations), you can recover the model with a precision that is only limited by the amount of additive noise. Example: A sinusoid of unknown amplitude, phase, and frequency can be exactly recovered from 3 observations (provided they are close enough), 2 for a complex exponential. We teach that to our students in undergrad level. I'm surprised there are researchers in DSP that do not know this. $\endgroup$ – Florian Jul 18 at 14:43
  • $\begingroup$ @Florian Determining the parameters from time domain points is not the same as recovering them from bin values. It is in the latter, that the belief still exists it can't be done. For time domain, what you say I don't think is generally disputed. However your three time domain point example for sinusoids is flawed. If you sample evenly so and the center point is at a zero crossing, the parameters are not recoverable. Check these articles out: dsprelated.com/showarticle/1051.php dsprelated.com/showarticle/1056.php dsprelated.com/showarticle/1074.php $\endgroup$ – Cedron Dawg Jul 18 at 14:54
  • $\begingroup$ It's absolutely the same since the mapping is linear. I could measure with any quite arbitrary functionals, and it doesn't matter if it's Fourier or a randomly drawn one. People in the Compressed Sensing community have been pushing on this for a good decade now (your problem is 1-sparse). As for the sampling, I said, the model needs to be uniquely recoverable. Of course this puts limits on where to place the sampling points. In general, close enough is fine. Hitting the zero crossing with one of them is a probability zero event if the parameters are assumed to live in a continuous space. $\endgroup$ – Florian Jul 18 at 15:02
  • $\begingroup$ @Florian I'm not sure what you are saying. Yes, the DFT is a linear transform, but the relationship between the tone parameters(except amp)and bin values are not. I would say "I have an exact frequency formula". They would say "It's impossible because of the uncertainty principle." I would say, "That's not true, each bin value is a function of three real variables, with three bins (my original formula) I can form a set of three equations, three unknowns and solve for the unknowns. They would reply, "The math around these issues are well known, refer to.." and send me off on a goose chase. $\endgroup$ – Cedron Dawg Jul 18 at 15:29
  • $\begingroup$ @Florian I should have clarified, the issue is really about having a closed form solution vs numerical methods. That's what I meant by "exact equation". $\endgroup$ – Cedron Dawg Jul 18 at 17:19

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.