<p>I have been trying to calculate PLV using the hilbert transform <a href="http://www.nbtwiki.net/doku.php?id=tutorial:phase_locking_value#.VFAAPvnF99o" rel="nofollow noreferrer">(reference)</a>. 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:
<img src="https://i.sstatic.net/4RViz.png" alt="enter image description here"></p>

<p>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 $T=\rm{shift}\times\rm{length}(x)$</p>

<p><strong>EDIT:</strong> 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 <a href="http://en.wikipedia.org/wiki/Analytic_signal" rel="nofollow noreferrer">analytic signal</a>, then I am not able to derive a mathematical relationship asserting that $PLV(x(t),x(t-T)) = 1$. But then <strong>what is the PLV good for? What kind of transformation could be applied to $x(t)$ to yield a $y(t)$ s.t. $PLV(x,y)=1$?</strong></p>

CC BY-SA 3.0
2014-10-28T21:21:02.087Z
Is your time series here real data, or something artificial like a sine wave?
Josh
CC BY-SA 3.0
2014-10-28T21:24:46.210Z
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
CC BY-SA 3.0
2014-10-28T21:25:32.817Z
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
CC BY-SA 3.0
2014-10-28T21:29:56.233Z
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
CC BY-SA 3.0
2014-10-28T21:33:44.220Z
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
CC BY-SA 3.0
2014-10-28T21:40:49.403Z
When T=1, then $x(t) = x(t-T)$, right?
Josh
CC BY-SA 3.0
2014-10-28T21:41:46.547Z
No, when $T=0$, $x(t)=x(t-T)$
morpheus
CC BY-SA 3.0
2014-10-28T21:48:14.877Z
But, also when $T=1$ because $T$ is equal to the length of $x$, and shifting $x$ by the length of $x$ yields $x$?
Josh
CC BY-SA 3.0
2014-10-28T21:57:12.930Z
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/18234/discussion-between-josh-and-morpheus).
Josh
<eeg>
Phase locking value (PLV) between $x(t)$ and $x(t-T)$
morpheus
# Phase locking value (PLV) between $x(t)$ and $x(t-T)$

(Asked by: username_0 on 2014-10-28)

I have been trying to calculate PLV using the hilbert transform [(reference)](http://www.nbtwiki.net/doku.php?id=tutorial:phase_locking_value#.VFAAPvnF99o). 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:
![enter image description here](https://i.sstatic.net/4RViz.png)

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](http://en.wikipedia.org/wiki/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. $PLV(x,y)=1$?**

**Tags:** eeg

> **Comments:**
> **username_1**: Is your time series here real data, or something artificial like a sine wave?  
> **username_0**: my time series is real eeg data over 10sec. You can run the experiment on any general $x(t)$ that has a continuous FT.  
> **username_1**: I asked because if it were artificial then you would expect the PLV to by phasic, but with real data your pattern looks right.  
> **username_1**: 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?  
> **username_0**: 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)  
> **username_1**: When T=1, then $x(t) = x(t-T)$, right?  
> **username_0**: No, when $T=0$, $x(t)=x(t-T)$  
> **username_1**: But, also when $T=1$ because $T$ is equal to the length of $x$, and shifting $x$ by the length of $x$ yields $x$?  
> **username_1**: Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/18234/discussion-between-josh-and-username_0).  

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## 0 answers

