# Can the phase of a function be extracted from only its absolute value and its Fourier transform's absolute value?

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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I think you mean $\tilde{f}(k)=N\int f(x)e^{-ikx}dx$. – Américo Tavares Nov 11 '10 at 13:30
@Américo Tavares: correct. although I could swap $x$ and $k$ as well :p – Tobias Kienzler Nov 11 '10 at 13:36
The sign in the argument of e is not important (or multiplying the argument by any real number for that matter -- e.g. 2\pi is a popular choice). It is just a convention. – Marek Nov 11 '10 at 13:50
@Marek: true. But since I use $x$ and $k$, the usual convention is $f(x)=\int \tilde f(k) e^{ikx} dx$ thus Américo's comment is fine with me – Tobias Kienzler Nov 11 '10 at 13:53
@Tobias Kienzler, @Marek, Thanks. Your comments explain why in different books the conventions are not always the same. – Américo Tavares Nov 11 '10 at 15:14

Do the usual numerical phase retrieval algorithms not work?

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+1 Thanks, I didn't know about the Gerchberg–Saxton algorithm. But does that mean phase retrieval is not possible analytically? If yes, is there a proof somewhere? – Tobias Kienzler Nov 18 '10 at 13:03

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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+1 thanks, I should have thought of that example. I extended my question to include yours about conditions on uniqueness. – Tobias Kienzler Nov 11 '10 at 13:59
It is impossible to determine the phase uniquely, since a global phase could always be factored out that would be destroyed by taking absolute values. However the question is now, (when) is it possible to uniquely determine the relative phase, that is, the phase up to a global phase? – Tobias Kienzler Nov 12 '10 at 9:38
Yeah, I also realized that later. Also, solutions even need not exist. Intuitive reason: Fourier transform has to preserve certain absolute values, like |\hat{f}(0)| = |f(0) + f(1) + ... | so that you clearly can't impose arbitrary conditions on both sides (this is more obvious on finite groups). If I can make this a bit more formal later, I'll update my answer. – Marek Nov 12 '10 at 9:53
Just curious. How would one even define the phase of a constant function? – crasic Nov 12 '10 at 9:54
@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. – Marek Nov 12 '10 at 9:57

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!.

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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 thanks 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 okay. I wonder if there is some more rigorous work on this, Wikipedia doesn't seem to cite any source for that paragraph... – Tobias Kienzler Jan 12 '12 at 9:38