# Phase locking value (PLV) between $x(t)$ and $x(t-T)$

I have been trying to calculate PLV using the hilbert transform (reference). To test if it even works or not, I calculated PLV between $x(t)$ and $x(t-T)$. i find that it drops to zero very quickly as $T$ becomes nonzero as shown in following figure: Is this as expected? I would expect PLV between $x(t)$ and $x(t-T)$ to be 1. In above graph shift is a normalized quantity such that $T=\rm{shift}\times\rm{length}(x)$

EDIT: thinking about this some more. Let us first define what PLV is. If we define PLV as the phase locking between instantaneous phases of two signals $x(t)$ and $y(t)$ where instantaneous phase is defined according to the analytic signal, then I am not able to derive a mathematical relationship asserting that $PLV(x(t),x(t-T)) = 1$. But then what is the PLV good for? What kind of transformation could be applied to $x(t)$ to yield a $y(t)$ s.t. $PLV(x,y)=1$?

• Is your time series here real data, or something artificial like a sine wave? – Josh de Leeuw Oct 28 '14 at 21:21
• my time series is real eeg data over 10sec. You can run the experiment on any general $x(t)$ that has a continuous FT. – morpheus Oct 28 '14 at 21:24
• I asked because if it were artificial then you would expect the PLV to by phasic, but with real data your pattern looks right. – Josh de Leeuw Oct 28 '14 at 21:25
• I want to make sure I'm understanding: shift is essentially the proportion of the 10s window that you shifted the pattern by? So when you get to 1, you should be computing the PLV of the original time series x with itself? – Josh de Leeuw Oct 28 '14 at 21:29
• yes when T=0 I am computing PLV between $x(t)$ and $x(t)$ which gives 1 as expected. But shouldn't the PLV between $x(t)$ and $x(t-T)$ also be 1? If not, could someone explain why not (mathematically) – morpheus Oct 28 '14 at 21:33 