# How to choose a phase for the deconvolution of an autocorrelation?

Say I have a function, $C=C\left(x\right)$, whose fourier transform is denoted by $c=c\left(k\right)$, i.e. $C\left(x\right)=\sum_{k=-\infty}^{\infty}c\left(k\right)\chi\left(x\right)$, where $\chi\left(x\right)$ is some complex valued basis function.

I know that the autocorrelation is given by $M\left(\Delta x\right)=\sum_{k=-\infty}^{\infty}m\left(k\right)\chi\left(\Delta x\right)$, where $\Delta x = x_2-x_1$ and by the convolution theorem I know that $m\left(k\right)=\left(c\left(k\right)\right)^*c\left(k\right)$. This means that knowing the coefficients of the autocorrelation immediately gives me the squared magnitude of the original coefficients. This implies that any two functions $C'\left(x\right)=C\left(x-x'\right)$, that differ only by a translation, will possess the same autocorrelation, $M\left(\Delta x\right)$, and so if I am trying to obtain the $c\left(k\right)$ from the $m\left(k\right)$, then I can choose an arbitrary phase for the $c\left(k\right)$. However, I don't think I can choose the phase of $c\left(k\right)$ independently for each $k$, it seems like there needs to be some way to consistently choose the phase for all $k$ (which corresponds to specifying a translation for $C\left(x\right)$).

My question is: How do I choose a consistent phase for each of the coefficients for the original function, i.e. for all $k$?

• Your question is basically unanswerable. Barring possible conjugate symmetry constraints that $c(\cdot)$ must satisfy if $C(\cdot)$ is real-valued, all possible choices for $\theta_k$ in $\sqrt{|m(k)|}e^{j\theta_k}$ are equally valid. – Dilip Sarwate Sep 12 '13 at 22:44
• There are in fact such symmetry constraints. $C\left(x\right)$ is real valued, and the reality constraint leads to the following constraint on the coefficients: $$c\left(-k\right)=c\left(k\right)^*$$. Does that help? – okj Sep 13 '13 at 13:26
• No, it does not help since all that happens is that you get to choose the phases $\theta_k$ for $1 \leq k \leq N/2$ and the rest are those that you get via the conjugacy constraints. – Dilip Sarwate Sep 13 '13 at 14:05
• So there is no intrinsic relationship between the $\theta_k$ for different values of $k$? I realize that you can choose any "global" phase, since the autocorrelation is shift-invariant, but my question is, if I specify $\theta_1$ for instance, how do I choose $\theta_2$, $\theta_3$,... etc. in a consistent way? It sounds to me like you're saying I can choose an arbitrary phase independently for all $k$, am I understanding what you're saying correctly? – okj Sep 13 '13 at 14:31
• Many totally different sequences can have the same autocorrelation function. Yes, C and a delayed version of C will have the same autocorrelation, but so will C reversed in time. You can assign phases arbitrarily to the square roots of $|m(k)|$ and get a sequence with autocorrelation $m$. There is no way of getting $C$ back if all you have is $m$; there are infinitely many possible sequences that all have autocorrelation $m$. – Dilip Sarwate Sep 13 '13 at 17:05

For a real-valued signal $C$, the autocorrelation function $M$ is a real-valued even function and the power spectral density $m$ (Fourier transform of $M$) is a real-valued nonnegative even function. Now, $m(k) = c(k)c^*(k)$ where $c$ is the Fourier transform of $C$, as you correctly assert, but given only $m$ and no other information about $c$ (or $C$), it is not possible to determine $c$ at all. Note that $m(k)$ also equals $[c(k)e^{j\theta_k}][c(k)e^{j\theta_k}]^* = c(k)c^*(k) e^{j\theta_k}e^{-j\theta_k}$ for arbitrary choice of $\theta_k$, and the inverse Fourier transform of $c(k)e^{j\theta_k}$ is also a real-valued signal if we are careful to preserve conjugate symmetry so that we choose $\theta_{-k}$ as $-\theta_k$. In other words, there are infinitely many quite different signals that share the same autocorrelation function, and it is not the case that only $C$ and its time-delayed versions have autocorrelation function $M$. As an example, all PN signals, that is, pulse trains generated from a maximal-length linear feedback shift register (LFSR), have an inverted thumbtack autocorrelation, and the distinct PN signals (from different LFSRs) are even two-level signals, a property not shared by all of the infinitely many real signals that have inverted thumbtack autocorrelation.