0
$\begingroup$

I'm having a set of N real data points that correspond to the autocorrelation sequence at N different lags. Suppose I know that the original set of data from which the autocorrelation was computed itself had only N points, is it possible to retrieve the real signal back?$$r\left(m\right)\ =\sum_{k=0}^{N-m-1}x\left(k\right)x\left(k+m\right), m\ =\ 0,1,2.....(N-1)$$

My attempt till now: I'm sure that the solution is not unique if we don't know that there were exactly N points. And even when we have these N points, say if $\left\{y\left(i\right):\ i=\ 0,1,2...\left(N-1\right)\right\}$ is a solution, so is $\left\{-y\left(i\right):\ i=\ 0,1,2...\left(N-1\right)\right\}$. But that is not a problem as of now. I tried finding alternative solutions for some random autocorrelation sequences. But I always found myself getting only the above mentioned 2 solutions and the others were always complex.

Also, is it possible to say anything about the properties that the signal should possess in order to be able to uniquely retrieve the signal in the above sense? Or is there any other parameter that needs to be present in order to be able to retrieve the sequence uniquely?

$\endgroup$

1 Answer 1

2
$\begingroup$

is it possible to retrieve the real signal back?

I don't believe so.

The spectrum, $R_{xx}(\omega)$, of the autocorrelation is simply the magnitude squared of the signal spectrum $X(\omega)$, i.e.

$$R_{xx}(\omega) = |X(\omega)|^2$$

That means that all signals that have the same magnitude spectrum have the same autocorrelation, regardless of the phase. For example: a unit impulse, a 10 sample delay, the impulse response of an all pass filter, a linear chirp or white noise all have the same autocorrelation although they are totally different signals.

Trivial example: The sequence [1 2 3] has the same autocorrelation as [3 2 1].

If you apply an operation to a sequence that changes it's phase but NOT it's magnitude or length, the autocorrelation will remain the same.

$\endgroup$
5
  • $\begingroup$ Hi Hilmar: That's a lot for me to take in because of statistics background but, to anyone who comes here from statistics, the autocorrelation in your trivial example would not be the same. The autocorrelation at lag zero would be the same. Yet, the autocorrelation at lags 1 and 2 would be the exact opposite of each other in sign. But this is caused by a definition difference between statistics and dsp so I'm not saying anything that you said is incorrect. I'm just giving a heads up to any stat people who may find it puzzling. $\endgroup$
    – mark leeds
    Commented Jun 7, 2021 at 21:52
  • $\begingroup$ Thanks for your comment. It certainly NOT a good idea to use the same word for two related concepts that are have substantial differences. Can you post a link to a crisp definition of what the statistical autocorrelation is. I googled some but most stuff I found was fluffy. $\endgroup$
    – Hilmar
    Commented Jun 8, 2021 at 11:28
  • $\begingroup$ @Hilmar: Thank you very much for the answer. I tried implementing the above idea in MATLAB by taking the fft of a known sequence and then multiplying the entire fft sequence by "i". Then I adjusted some of the constants to ensure the condition for realness after ifft, without changing the magnitude. After I took the ifft, I found the autocorrelation sequence of the newly formed sequence. But it varies from the original autocorrelation by a significant factor, especially at the trailing edge of the sequence. Could you give an idea as to what causes this deviation? $\endgroup$ Commented Jun 8, 2021 at 19:30
  • $\begingroup$ Actually, I became more confused when I tried graphing the case of N=3 using some 3d graphing apps. I found that the only possible solutions, in that case, were the original sequence, its negation, its reverse and the reverse's negation. Then when I tried this code, I found that these deviations existed even though I had the same magnitude responses. $\endgroup$ Commented Jun 8, 2021 at 19:34
  • $\begingroup$ Hi Hilmar: It's definitely confusing for me and at some point I just decided to consider them as two different concepts that happen to have the same The link below isn't bad. The only "negative" ( no pun intended ) is that something went with wrong with the latex so there should be negative signs in the definition of the sample autocorrelation, $\hat{\rho}_{k}$. statlect.com/fundamentals-of-statistics/autocorrelation $\endgroup$
    – mark leeds
    Commented Jun 9, 2021 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.