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given an integrable functionf(x), and its Fourier transform

[tex]\mathcal{F}\{f\}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx[/tex],

we consider thephase[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 acontinuous phasefunction.

My question is: why should the phase be a continuous function? What is the condition/theorem that guarantees that the phase is always continuous?

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# Fourier phase (unwrapping problem)

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