There are two papers I'm working through that explain applications of autocorrelation.

The first - by Monti - is quite clear to me:

autocorrelation formula

It runs a sum over the time axis.

The second one by Brown makes no sense to me:

narrowed autocorrelation

How is time $t$ collapsed as a dimension? Is there an implicit sum? As it stands, the only way I can interpret this is as a multi-dimensional surface with two independent variables, $t$ and $\tau$.


The autocorrelation function of a random process $x(t)$ is defined as $$R(t,\tau) = {\mathbb E}\{x(t) x(t+\tau)\}.$$

For a stationary random process, this function does not depend on $t$, i.e., we have $R(t,\tau) = R(\tau) \; \forall t$. The correlation only depends on the time difference between two samples, not the "absolute" time $t$ when the correlation is measured.

For a stationary and ergodic process, ensemble averaging can be replaced by time averaging and we have $$R(\tau) = \lim_{T\rightarrow \infty} \int_T x(t) x(t+\tau) {\rm d}t,$$ which in the discrete case collapses to the sum you were showing.

The expression mentioned by Brown is a function of $t$ and $\tau$, no doubt about that. However, Brown argues (I haven't followed through the arguments, but they should be in the second paper he was citing) that the expression can be used to compute a "narrowed autocorrelation function" (what ever that is) and that if viewed as a function of $\tau$ it has peaks when $\tau$ is equal to multiples of the period $T$ of the periodic signal $f(t)$. This sounds like the particular choice of $t$ is irrelevant, which makes sense if we consider $f(t)$ to be stationary and ergodic. Then again, the notation if not really rigorous. It is worth reading some more though because a few equations later he introduces the bracket notation $\langle \cdot \rangle$ for an average over a period $T$. That would make a lot of sense for his definition of $S(\tau)$ too. In fact, if we do open Brown et al 1989 in equation (1b) he does include the $\langle \cdot \rangle$ notation. Maybe it was forgotten in the other paper?

  • $\begingroup$ the discrete case collapses to the sum you were showing - I see. That helps; the part I'm missing is how the second (generalized) form of $N$ summed terms turns out to actually multiply them. I think it's because of the squaring; the language in the paper says: When $N$ is equal to 2 in the above equation, the conventional or standard autocorrelation function is given by the cross term with the other two terms contributing a constant shift - so in the end I'll have an outer sum over $t$ of an inner sum representing the expansion of a squared N-term series. $\endgroup$
    – Reinderien
    Oct 8 '20 at 17:34
  • $\begingroup$ I recommend reading the Brown 1989 paper on this - I just updated the link to one without a paywall. It explains more nicely what is the intuition between the expression. Note that it is not the same as the autocorrelation function. It is a "narrowed" version that really looks differently (on purpose). $\endgroup$
    – Florian
    Oct 8 '20 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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