In the heavily cited paper by Harris 1978, there is a section on "Overlap Correlation" where he's using a "fractional overlap r" in his equation. Here is a screen shot from the open source paper linked above:

enter image description here

Can someone please help explain what's going on here? I'm trying to use equation 17 but I don't understand what the indices represent. How can "r" be a fraction? If the summation is taking place over integer values, then the term "rN" is an integer, which in general means that "r" must be an integer. He gives some diagrams, but those don't really help.

  • $\begingroup$ May I recommend Heinzel. It has plots that show "what's going on" ;) $\endgroup$
    – Jdip
    Aug 26, 2022 at 23:04
  • $\begingroup$ "the term $rN$ is an integer, which in general means that $r$ must be an integer" not really: for $r=0.5$ and $N=1000$ you still get $rN=500$. Maybe I didn't get what you want to ask, did I? $\endgroup$
    – AlexTP
    Aug 26, 2022 at 23:14
  • $\begingroup$ @AlexTP that is a trivial special case. It does not hold for all N. If N = 1001 and r = 0.5, then rN is not an integer. That is where my confusion lies. $\endgroup$ Aug 26, 2022 at 23:20
  • $\begingroup$ @Jdip That was where I started :( Now it seems I've come full circle. $\endgroup$ Aug 26, 2022 at 23:29
  • $\begingroup$ Maybe "flooring" the result "rN" is what is intended? I would think they would "put that in writing" if that was the intention, but maybe not. $\endgroup$ Aug 26, 2022 at 23:38

1 Answer 1


OK...I'm going to take a stab at this, but I'm hoping someone here has more experience with this to confirm/refute my answer.

Let the integer $M\in[1,N]$ be the number of discrete overlapping indices. Therefore, $r=M/N$, where $N$ is the total number of indices in the window. In this case the "fractional" aspect remains, but at least we're dealing with integers.

Substituting into the equation (in question) gives

$c_M = \frac{\displaystyle \sum_{n=0}^{M-1} w_n w_{n+N-M}}{\displaystyle \sum_{n=0}^{N-1} w_n^2}$

in which I've indexed $c$ with an integer.

Again, I hope someone has more experience with this, but it makes more sense to me from a programmatic point of view.

  • $\begingroup$ I've compared my results with the work from Heinzel (link in question comments) using a Hann window and the mods above appear to give the same answer. Therefore, I'm going to accept it. $\endgroup$ Aug 27, 2022 at 2:15

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