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I would like to better understand why the instantaneous frequency estimation by Hilbert transformation works (and especially why it doesn't work / lead to precise results in many cases).

The motivation is to estimate signal $x(t)$ by decomposing it into an amplitude envelope $m(t)$ and phase of cosine $\omega_c (t)$ (or carrier waveform):

$$x(t) = m(t) \cos\left(\omega_c (t)\right)$$

Now, assume that $x(t)$ indeed is a result from such a process.

Questions:

1) There are two "parameters" to be estimated for any $t$, as such some constraints are needed. What are the constraints regarding $m(t)$ and $\omega_c (t)$ that are selected when applying the Hilbert transform decomposition?

2) Is there a proof available somewhere that given the constraints, the estimation indeed finds the correct amplitude envelope and carrier (for continuous and also discrete case)?

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  • $\begingroup$ I believe you’d have to assume that $m(t)$ is very slowly varying with respect to the “center” frequency of $\omega_c(t)$ — which is a little mis-named. As you’ve written it, it’s a phase, not a frequency (which is usually what $\omega$ is used for). $\endgroup$
    – Peter K. ♦
    May 19, 2018 at 18:50

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See my earlier comments here: Meaning of Hilbert transform

Common fractal noise isn't an analytic signal (infinitely differentiable). And a Hilbert transform re-creates the imaginary component of an analytic signal if you have the real component of the complex analytic signal (which one rarely has from real-world data).

However, sufficiently band-pass filtered data might be similar to a finite length segment of something that looks like an infinite length analytic signal (e.g. is from a source whose behavior can be modeled or estimated by a 2nd order linear differential equation).

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    $\begingroup$ You are giving the definition for an analytic function, which is not identical to that of analytic signal. There is a non-obvious relation between the two, but they are not identical. For example $t\mapsto\exp(-t^2) \sin(t)$ is an analytic function, but it's not an analytic signal. $\endgroup$
    – Jazzmaniac
    May 19, 2018 at 20:01
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    $\begingroup$ I think the OP asked about the actual process of estimating the instantaneous frequency using Hilbert Transform. I think you should point into that in your answer. $\endgroup$
    – Royi
    Jun 18, 2018 at 20:40

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