Estimating time in harmonic signal

Question

I hope someone can help me with the following problem:

Assume a periodic signal of the form

$$\begin{align} s(t) &= \sum\limits_{p=1}^P \sin(p\Omega_0t)\\ &= \sum\limits_{p=1}^P \sin(\theta_p(t)), \end{align}$$

with $0 \leq t < \frac{2\pi}{\Omega_0}$ (only one period of the fundamental frequency).

We take measurements $\theta^*_p(t_0)$ of the phases $\theta_p(t_0)$ at a certain (unknown) time $t_0$ for $p \in [2,\ldots,P]$ (i.e. for all harmonics except the fundamental). The measurements are phase wrapped, i.e. projected to the interval $[-\pi,\pi[$: $$ \theta^*_p(t_0) = \theta_p(t_0) + n_p \cdot 2\pi $$ How can we estimate/determine the time $t_0$ at which the measurements have been taken?

Here is a diagram to illustrate the problem:
From the diagram it becomes clear that for $p=2$, the phase value $\theta^*_2(t_0)$ can occur at two points in time (see red marker). For $p=3$, the phase value $\theta^*_3(t_0)$ can occur at three points in time, etc. But only at $t_0$ they all fall together. Can we use this property to estimate $t_0$?

Answer 1

OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 \pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:

$$\left ( \begin{array} \\2 t_0 & 0.33 \pi \\ 3 t_0 & -0.55 \pi \\ 4 t_0 & 0.55 \pi \\ 5 t_0 & -0.39 \pi\end{array} \right )$$

Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 \pi$ to get $1.45 \pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this

$$\left ( \begin{array} \\2 t_0 & 0.33 \pi \\ 3 t_0 & 1.45 \pi \\ 4 t_0 & 2.55 \pi \\ 5 t_0 & 3.61 \pi\end{array} \right )$$

A best fit line takes the form $\theta(p) = -1.844 \pi+1.094 \pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 \pi$ (which makes sense on your graph), so that knowing $\Omega_0$, you may then determine the value of $t_0$. I hope that helps.

Answer 2

If I may have understood your case then

given:

$$2 \Omega_0 t_0 = \theta_2$$ $$3 \Omega_0 t_0 = \theta_3$$ $$\dots$$

and $\theta_2$ and $\theta_3$ known, then

$$\Omega_0=\frac{\theta_2}{2 t_0}=\frac{\theta_3}{3 t_0}=\dots=\frac{\theta_p}{p t_0}$$

So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .

The answer under the given constraints is that we can not specify a specific $t_0$ or ($\Omega_0$).

For the new information you brought in your question we have now

$$\theta^*_p(t_0) - n_p \cdot 2\pi = \theta_p(t_0) $$

hence:

$$\Omega_0=\frac{\theta^*_2(t_0) - n_2 \cdot 2\pi }{2 t_0}=\dots=\frac{\theta^*_p(t_0) - n_p \cdot 2\pi }{p t_0}$$

It does not look better for getting $t_0$.