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

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?.

## 1 Answer

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.

• In maybe somewhat simpler terms. I think what Peter is saying is that the Fourier Transform is 1:1 i.e. it is invertible. Your can recover the original signal exactly and there are no ambiguities. If you just have the Power Spectrum, there are many signals that would have that exact Power Spectrum. I'm not quite sure what Peter means by relative phase in this context. – David Apr 1 '16 at 14:00