Why Fourier transform and Stockwell-transform retain the absolute phase information of one signal?

Question

Hello friends am studying the topic of signal processing and the Fourier transform and the s-transform and in most books as for example "Time-Frequency Signal Analysis and Processing. 2nd" of Boashash in section 5.11.2.3 says that these transforms retain the absolute phase information of the signal and I do not understand that it mean with this concept about retain the absolute phase information of the signal and why this is important?.

Answer 1

It simply means that for a signal: $$ x(t) = a(t) e^{j\phi(t)} $$ that the phase information $\phi(t)$ is completely preserved in the transform domain.

Time-frequency representations that use a quadratic form do not necessarily have this property: $$ x(t+\tau)x^*(t-\tau) = a(t+\tau)a(t-\tau)e^{j(\phi(t+\tau) - \phi(t-\tau))} $$ so if $\phi$ is of the form: $$ \phi(t) = f(t) + C $$ where $C$ is a constant, then $C$ is not preserved in $x(t+\tau)x^*(t-\tau)$.

This is important because sometimes you are interested in the absolute phase of a signal, not just its relative phase.