- #1
mnb96
- 711
- 5
Hello,
given an integrable function f(x), and its Fourier transform
\mathcal{F}\{f\}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx,
we consider the phase \mathrm{Ph}_f : \mathbb{R}\rightarrow [-\pi,\pi) which is given by:
\mathrm{Ph}_f (\omega) = \mathrm{arg}(\mathcal{F}\{f\}(\omega))
In general the phase function will have discontinuities (when it wraps from -\pi to \pi, 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
\mathcal{F}\{f\}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx,
we consider the phase \mathrm{Ph}_f : \mathbb{R}\rightarrow [-\pi,\pi) which is given by:
\mathrm{Ph}_f (\omega) = \mathrm{arg}(\mathcal{F}\{f\}(\omega))
In general the phase function will have discontinuities (when it wraps from -\pi to \pi, 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?
