- #1
mnb96
- 715
- 5
Hello,
given an integrable function f(x), and its Fourier transform
[tex]\mathcal{F}\{f\}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx[/tex],
we consider the phase [tex]\mathrm{Ph}_f : \mathbb{R}\rightarrow [-\pi,\pi)[/tex] which is given by:
[tex]\mathrm{Ph}_f (\omega) = \mathrm{arg}(\mathcal{F}\{f\}(\omega))[/tex]
In general the phase function will have discontinuities (when it wraps from [itex]-\pi[/itex] to [itex]\pi[/itex], and there are algorithms that attempts to recover a continuous phase function.
My question is: why should the phase be a continuous function? What is the condition/theorem that guarantees that the phase is always continuous?
given an integrable function f(x), and its Fourier transform
[tex]\mathcal{F}\{f\}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx[/tex],
we consider the phase [tex]\mathrm{Ph}_f : \mathbb{R}\rightarrow [-\pi,\pi)[/tex] which is given by:
[tex]\mathrm{Ph}_f (\omega) = \mathrm{arg}(\mathcal{F}\{f\}(\omega))[/tex]
In general the phase function will have discontinuities (when it wraps from [itex]-\pi[/itex] to [itex]\pi[/itex], and there are algorithms that attempts to recover a continuous phase function.
My question is: why should the phase be a continuous function? What is the condition/theorem that guarantees that the phase is always continuous?