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Can a signal be reconstructed from its cyclic autocorrelation? Specifically, if we know

$$ R^{\alpha}(\tau) = \int{x(t)x^{\ast}(t-\tau)e^{-j2\pi\alpha t}\mathrm{d}t}, $$

can we reconstruct $x(t)\in\mathbb{C}^{N}$? It seems like the answer should be yes, but it's not obvious to me how to do it. I know there has been work on reconstructing a signal from its autocorrelation, and these algorithms of course require some other knowledge about the signal (e.g., Discrete signal reconstruction from its autocorrelation function and one sample). I guess I am hoping knowing the cyclic autocorrelation will at least relax the required assumptions. Moreover, how does the problem change if we also had the conjugate version:

$$ \bar{R}^{\alpha}(\tau) = \int{x(t)x(t-\tau)e^{-j2\pi\alpha t}\mathrm{d}t} $$

Edit: Let me try rewording the question. I understand that autocorrelation is not an isomorphism. Is this also true of cyclic autocorrelations? Specifically, can we recover $x\in\mathbb{C}^N$ from

$$ R(\alpha,m)=\sum_{n=0}^{N-1}x(n)x^{\ast}(n-m)e^{-j2\pi n \alpha}. $$

If we only have knowledge for $\alpha=0$, then we have just the regular autocorrelation and the answer is in general no. However, does knowing $R$ for $\alpha=0,1/N,2/N,\dots$ enable one to recover $x$ or at least reduce the required assumptions necessary for doing so? If not, what if the conjugate version is also available? Can we recover $x$ then?

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    $\begingroup$ since autocorrelation corresponds to power spectrum, and since power spectrum is related to the spectrum via the magnitude-squared operation, which ditches phase, there is no way to recover exactly the spectrum from the power spectrum. and likewise, there is no way to recover the signal from the autocorrelation. this is because you have lost the phase information. $\endgroup$ – robert bristow-johnson Apr 22 '17 at 22:37
  • $\begingroup$ @robertbristow-johnson I understand the issue with autocorrelation, but there should be a tremendous amount of redundancy in the cyclic autocorrelation. $\endgroup$ – AnonSubmitter85 Apr 22 '17 at 23:18
  • $\begingroup$ agreed, and you can get a periodic function with the correct amplitudes of each harmonic from the autocorrelation. but you cannot get the phase of each harmonic, that information has been lost in computing the autocorrelation. this periodic function: $$ x[n] = \sum\limits_{m=1}^{M} a_m \cos(m \omega_0 n + \phi_m) $$ will have an identical autocorrelation $$ R_x[k] \triangleq \lim_{N \to \infty} \frac{1}{2N+1} \sum\limits_{n=-N}^{+N} x[n] \, x[n+k] $$ independent of the values of any of the $\phi_m$. $\endgroup$ – robert bristow-johnson Apr 22 '17 at 23:36
  • $\begingroup$ @robertbristow-johnson Will that be true for all values of $\alpha$? (The autocorrelation corresponds to $\alpha =0$, but the cyclic autocorrelation will have values for $\alpha=0,1/N,2/N,\dots$.) Also, what if the conjugate version is also available? $\endgroup$ – AnonSubmitter85 Apr 22 '17 at 23:52
  • $\begingroup$ @robertbristow-johnson this is a good answer, why don't you include it below? $\endgroup$ – Dan Boschen Apr 23 '17 at 0:15
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No, you cannot reconstruct the original signal from the cyclic autocorrelation. The fundamental reason is that it results from an averaging operation. Like the autocorrelation and PSD, the cyclic autocorrelation is a many-to-one functional (you already noted that I think). For example, all BPSK signals with identical timing parameters (symbol clock phase and carrier phase), identical pulse functions, and IID symbols have the exact same cyclic autocorrelation functions. Yet there are an infinite number of such signals obtained by all the different sequences of transmitted bits. In general, there is no single unique signal that gives rise to a particular cyclic autocorrelation.

I note that none of the equations you wrote are the cyclic autocorrelation. In general, it is the infinite time average of the second-order lag product of the signal multiplied by a complex sine wave. Aside from the missing scale factor $1/N$, your third equation is an estimate of the cyclic autocorrelation for discrete-time signals.

There are special cases, though, such as periodic signals. Such signals are non-random. The time-varying autocorrelation function $R_s(t, \tau)$ for such signals is identically equal to the second-order lag product itself $s(t+\tau/2)s^*(t-\tau/2)$. Since the time-varying autocorrelation is composed of the sum of sine-wave-weighted cyclic autocorrelations, one might recover $s(t)$ from the set of cyclic autocorrelations, to within a scale factor of $\pm 1$.

The cyclic autocorrelation is sensitive to phase in the sense that the cyclic autocorrelation for $s(t)$ differs from that for $s(t-D)$ in general. This is a basis for synchronization algorithms.

I'm curious why it is obvious to you that there "should be a tremendous amount of redundancy in the cyclic autocorrelation." Redundant what?

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  • $\begingroup$ First of all, thank you. The redundancy assumption may be based more on ignorance than anything, but my thinking was along the lines of a 1-D vector being taken to a 2-D array must be recycling the same information over and over. Somewhere in that recycling, I was hoping that the original 1-D signal could be extracted. If the equations are sloppy, then I apologize; scale factors and infinite averages are things that I tend to take for granted. Other than that, if the equations are incorrect, please correct me -- I thought they were consistent across multiple sources. $\endgroup$ – AnonSubmitter85 Apr 23 '17 at 6:16
  • $\begingroup$ As for reconstructing, what I would really like to do is filter the cyclic autocorrelation of a signal vector. Something along the lines of LTI filtering by element-wise multiplication in the frequency domain but with the cyclic autocorrelation function (or spectral correlation density, since they are isomorphic). Is the idea harebrained? Should I abandon it now before wasting time on it? I know of signal reconstruction algorithms when one only knows the magnitude of the spectrum and some other apriori information about the signal. I am hoping that there is something similar for this case. $\endgroup$ – AnonSubmitter85 Apr 23 '17 at 6:28
  • $\begingroup$ @ChadSpooner Welcome to DSP.SE! Thanks for providing this answer! $\endgroup$ – Dan Boschen Apr 23 '17 at 9:50
  • $\begingroup$ No need to apologize! I think the equations that you can find in my post [link] (cyclostationary.blog/2015/09/28/the-cyclic-autocorrelation) are consistent with all literature sources I value. $\endgroup$ – Chad Spooner Apr 23 '17 at 19:33
  • $\begingroup$ Sure you can filter or otherwise operate on the cyclic autocorrelation, but why do you think that will result in the original signal? I tried to look at that link in your original post, but didn't want to pay for it. I'm most curious what the "other prior information" is that enables reconstruction. I suspect it relates to periodicity or some other kind of signal that has little or no randomness. $\endgroup$ – Chad Spooner Apr 23 '17 at 19:39

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