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In a previous thread, it was asked whether we can infer a signal from its autocorrelation, and the answer was a clear no, since the autocorrelation process is lossy.

Still, is there anything we can infer about the "shape" of the original signal from its autocorrelation? I feel like there should be, but I can't think of what that would be.

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Alternatively, it is interesting to consider the analogous continuous problem, that is, find the family of solutions $f(t)$, such that

$$\intop_{-L}^Lf(t)f(\tau+t)dt = g(\tau)$$

Very often, the continuous analog is harder, but not always, and this version would still be interesting for me.

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    $\begingroup$ That family is not particularly hard to identify. Take any member of it (the trivial one would be a good choice) and multiply it with any function of modulus 1 to get another member. Structurally, that is an infinite dimensional unitary Lie group and corresponding Lie algebra. The only thing you can derive from that is a lower bound on the support of $f(t)$. $\endgroup$
    – Jazzmaniac
    Jul 9, 2023 at 9:15
  • $\begingroup$ What you depicted with your integral expression is convolution of $f(t)$ against itself not autocorrelation. You need $f(t)$ to point in the same direction in the integral. $\endgroup$ Jul 9, 2023 at 13:40
  • $\begingroup$ @robertbristow-johnson , you're right. I edited it according to your correction. $\endgroup$ Jul 10, 2023 at 13:09

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I think what we can infer is already given in the answer to your previous question: the power spectral density.

For example, if your autocorrelation is a delta impulse, your signal is white and has a flat spectrum. But that's where it's end: the signal could also be a delta impulse, could be a linear sine sweep, could be white noise, could be many other things.

We can't tell since ALL these signals have the SAME autocorrelation. So whatever different properties these signals have doesn't show up in the autocorrelation.

It's a many-to-one mapping and can't be reversed.

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    $\begingroup$ As a very specific example of what Hilmar is talking about, note that all maximal-length shift-register sequences of period $N = 2^m-1$ have the same periodic autocorrelation function: $$r_x[n] = \begin{cases}N, & \text{if }n\bmod N = 0,\\-1, &\text{if }n\bmod N \neq 0.\end{cases}$$ Thus, it is impossible to determine much about a sequence just from knowledge of its autocorrelation function. $\endgroup$ Jul 9, 2023 at 15:53
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    $\begingroup$ Wait a minute... It is possible to determine the spectral magnitude about a continuous or discrete function from knowledge of its autocorrelation. You don't know diddley about the phases of the frequency components, but you do know about the magnitude and frequency of the content of the signal. That's a lot. Not everything, but a lot. $\endgroup$ Jul 10, 2023 at 15:53
  • $\begingroup$ @DilipSarwate, are there sequences that have the same autocorrelation function? If not, then getting $r_X[n]$ as your autocorrelation does say something (as robert mentioned). Otherwise, maybe there is some other commonality between all such sequences. $\endgroup$ Jul 11, 2023 at 8:57
  • $\begingroup$ @DataPhysicist: I'm puzzled by your comment. My answer gives you three examples of signals that have identically autocorrelation. That commonality is that they have the same magnitude spectrum. They do have different phase spectral. And yes, it's often useful to know whether a signal is, for example, pink or white. This being said, there are infinitely many signals that are all pink (or white), What else are you looking for ? $\endgroup$
    – Hilmar
    Jul 12, 2023 at 15:21
  • $\begingroup$ @DataPhysicist For $N=2^m-1$ with $m\geq 3$, there are at least $2$ distinct maximal-length linear feedback shift register sequences ($6$ for $m=5$ and more for larger $m$). If you want to count shifted versions of each as a different sequence, then multiply the numbers stated above by $2^m-1$. All of them have the same periodic autocorrelation function, and identical spectral magnitudes. $\endgroup$ Jul 12, 2023 at 16:27

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