# Instantaneous Frequency Estimation by Hilbert Transform - Theoretical Justification and Proof

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)?

• 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). – Peter K. May 19 '18 at 18:50

• 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. – Jazzmaniac May 19 '18 at 20:01