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If for a function $f(x)$ only its absolute value $|f(x)|$ and the absolute value $|\tilde f(k)|$ of its Fourier transform $\tilde f(k)=N\int f(x)e^{-ikx} dx$ is known, can $f(x) = |f(x)|e^{i\phi(x)}$ and thus the phase function $\phi(x)$ be extracted? (with e.g. $N=1/(2\pi)$)

As Marek already stated, this is even not uniquely possible for $f(x)=c\in\mathbb C$, since the global phase cannot be re-determined. So please let me extend the question to

Under what circumstances is the phase-retrieval (up to a global phase) uniquely possible, and what ambiguities could arise otherwise?

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    @Tobias Kienzler, @Marek, Thanks. Your comments explain why in different books the conventions are not always the same.2010-11-11

4 Answers 4

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I'm not sure if this applies, but for minimum-phase functions, the phase and magnitude of the Fourier Transform are related. See here for a brief overview. I've never actually used this relationship in practice, so can't really give you much more information.

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    sorry, I somehow missed the notification for your answer, so a very belated than$k$s for your answer. So does this simply state that the phase is the negate Hilbert transform of the absolute value's logarithm (plus some constant)? That almost sounds plausible, since the logarithm is analytical for positive arguments, but what about the essential singularity at 0? At the zeros, a phase is of course meaningless, so that might still be somehow o$k$ay. I wonder if there is some more rigorous work on this, Wikipedia doesn't seem to cite any source for that paragraph...2012-01-12
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The short answer is no. Take constant function $f(x) \equiv C \in \mathbb{C}$. Disregarding normalization, we have $\hat{f} = C \delta$ (in the sense of distributions). Clearly, there is no way to recover the phase of $C$ once we take the absolute value on both sides.

To make this a little more explicit, consider a lot easier version of the problem on the group $G = \mathbb{Z} / N\mathbb{Z}$. Its dual is $\hat{G} = G$. If you'll write out the Fourier transform equations (i.e. $\hat{f}(k) = \sum_{n=0}^{N-1} f(n) \exp(-{i k n \over 2 \pi})$), you'll obtain $2N$ real equations for $2N$ coefficients ($N$ Fourier phases and $N$ original phases). The properties of this system of equations are not clear to me, but the case $N=1$ (this is the same as in the first paragraph, but here we don't need to talk about distributions) already shows that the solutions need not be unique.

I hope someone else can provide more information, I'd be also interested to see what conditions on $f$ one needs to assume to get a unique solution. Even for the case $G = \mathbb{Z} / N\mathbb{Z}$ this looks interesting enough.

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    @crasic: "phase" is just a parlance for the phi(x) part of f(x) = R(x)exp(i phi(x)). For constant function phi doesn't depend on x.2010-11-12
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This is the phase retrieval problem, for which simple iterative, numerical algorithms exist. For an overview, see J. R. Fienup, "Phase retrieval algorithms: a comparison," Appl. Opt. 21, 2758-2769 (1982).

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    @littleO fixed.2016-07-26
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The right way to ask the question is: given a function f\in L²(\mathbb{R}), can $f$ be determined from $|f|$ and $|\widehat{f}|$ up to a multiplicative constant $c$ of modulus $|c|=1$.

This question dates back to Pauli and the answer is no. One can construct counter examples of the form $a\gamma(x-x_0)+b\gamma(x)+c\gamma(x+x_0)$ with $a,b,c$ properly chose ($\gamma(x)=e^{-\pi x²}$ the standard gaussian so that it is ots own Fourier transform). An other construction is as follows:

take $\chi=\mathbf{1}_{[0,1/2]}$ $(a_j)_{j\in\mathbb{Z}}$ a sequence with finite support (to simplify) and $f(x)=\sum_j a_j\chi(x-j)$ so that $\hat f(\xi)=\sum_j a_je^{2i\pi j\xi}\hat\chi(\xi)$.

Now we want to construct a sequence $(b_j)$ such that $|a_j|=|b_j|$ and $\left|\sum_j a_je^{2i\pi j\xi}\right|=\left|\sum_j b_je^{2i\pi j\xi}\right|$. This can be done via a Riesz product: take $\alpha_1,\ldots,\alpha_N$ a finite real sequence, $\varepsilon_1,\ldots,\varepsilon_N$ a finite sequence of $\pm1$ and consider $ \prod_{k=1}^N (1+i\alpha_j\varepsilon_j\sin 2\pi 3^j\xi)=\sum a_j^{(\varepsilon)}e^{2i\pi j\xi}. $

Changing a $\varepsilon_j$ from $+1$ to $-1$ conjugates one of the factors on the left hand side, so it does not change the modulus. Now the same happens for the $a_j^{(\varepsilon)}$: each of them is either $0$ or a product of $i\alpha_j\varepsilon_j$ (up to a constant) -- the point is that it is not the sum of products of $i\alpha_j\varepsilon_j$'s, this is why we took the $3^j$ in the sine!.