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Consider two cosinusoidal signals given by \begin{gather*} z_1(t) = A_1\cos\phi_1(t)\\ z_2(t) = A_2\cos\phi_2(t) \end{gather*}

with

\begin{gather*} \phi_1(t) = (\omega_1 t + \theta_1)\\\phi_2(t) = (\omega_2 t + \theta_2) \end{gather*}

where $\theta$ is the phase shift of the signal compared to some reference signal. My understanding of the concept of phase locking is that for these two signals to be phase locked, the following condition must be met: \begin{gather*} \phi_1(t) = \phi_2(t)\end{gather*}I am currently reading Brain Dynamics by Hermann Haken, and here Haken makes a distinction between phase locking and synchrony. He says that if one were to plot $\phi(t)$ against $t$ for each of these signals, then the curves formed by this plot are considered phase locked if the vertical distance between the curves is the same at all times $t$. He then states that synchrony occurs when the two curves are coincident, i.e. on top of each other.

These definitions do not make sense to me based on my understanding of phase locking. If we have $\phi_1(t) = \phi_2(t)$ which implies phase locking, then the two curves should be coincident on such a plot and therefore synchronized by Haken's definition. Just to be clear, we are speaking only of phase here, and not considering the amplitudes of the signals.

So the question is this: what, if anything, distinguishes phase locking and synchrony?

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  • $\begingroup$ "If we have ϕ1(t)=ϕ2(t)ϕ1(t)=ϕ2(t) which implies phase locking, then the two curves should be coincident on such a plot" : What makes you think so in the presence of A1 and A2? $\endgroup$
    – Peaceful
    Commented May 18, 2017 at 18:24
  • $\begingroup$ @Peaceful As I said near the bottom of my post, we are considering phase separate from amplitude. If we have $A_1 = A_2$ and $\phi_1(t) = \phi_2(t)$, I believe the proper term is coherence. But in my question I am only concerned with the phase of the signal. $\endgroup$
    – nguzman
    Commented May 18, 2017 at 18:30

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By saying $\phi_1(t)$ and $\phi_2(t)$ when plotted against $t$ maintain a constant vertical distance between them, is the same as saying: $$ \phi_1(t)-\phi_2(t) = (\omega_1 t +\theta_1) - (\omega_2 t +\theta_2),$$ then, $$ \tag1 (\omega_1-\omega_2)t + (\theta_1 - \theta_2)=C,$$ where $C$ is a constant, which doesn't change with time. The left-hand side must be the same for any value of $t$, which means the coefficient of $t$ must be zero, which gives: $$\omega_1=\omega_2,$$ which is the condition for 'phase-locking'. This says the two waves have the same frequency. So the phase difference between the two waves is constant in time. (That's just $\phi_1 - \phi_2$, as stated before!)

Here is an example of two "phase-locked" sinusoidal signals. Here $\omega_1=\omega_2 = 1$, and I set $\theta_1 = 0$, $\theta_2 = 0.7$. If I set $\theta_1=\theta_2$, then they would be synchronous. I've also included an example of two waves that are neither synchronous nor phase-locked.

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  • $\begingroup$ Thank you for your answer. I have always considered $\omega_1 = \omega_2$, what you refer to as phase locked , to be what is called frequency locking with the signals having different phase shifts, that is $\theta_1 \neq \theta_2$. It has always been presented to me that phase locking and frequency locking are different concepts. Based on what you have said, it would seem that if I consider frequency locking separate from phase locking, where phase locking implies $\phi_1(t) = \phi_2(t)$ regardless of $\omega$ and $\theta$, then there is no difference between synchrony and phase locking. $\endgroup$
    – nguzman
    Commented May 18, 2017 at 18:41
  • $\begingroup$ The term 'phase-locked' is an unusual one to me - I have taken his definition (which is a constant phase difference), and worked with it to reach this conclusion. I feel this is more question of semantics now - and in his book, you should interpret phase locking as frequency locking. $\endgroup$
    – CDCM
    Commented May 18, 2017 at 18:49
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    $\begingroup$ Thanks, that definitely clears things up for me. It certainly was the semantics tripping me up. I appreciate the effort you put into the answer. $\endgroup$
    – nguzman
    Commented May 18, 2017 at 18:52

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