Calculating phase shift between two sines of equal, unknown, amplitude and frequency

Question

I've got a source signal $s_1(t)$ and a system affecting that, yielding a phase-shifted version $s_2(t)$;

\begin{align} \color{blue}{s_1(t)} &= A\sin(t+\phi_1)\\ \color{red}{s_2(t)} &= A\sin(t+\phi_2)\\ \end{align}

The amplitude $A$ is unknown. How many samples (sampling happens periodically) would I need to verify that $\phi_1-\phi_2 = \frac32\pi$?

As a practical example, given these 2 points on each curve: on the purple curve, the original signal:

• point A (231,3.533151)
• point B: (230,3.614015)

and on the the filtered signal: the red curve:

• point C (231,1.398873)
• point D (230,1.174265)

How would you confirm the phase shift of $\frac32\pi$, given an equal -unknown- amplitude and equal -unknown- frequency of the 2 signals? Is a numerical solution possible for online, causal analysis? Or do you need 3 points on each curve? Thank you.

Answer 1

Two samples is sufficient and we can determine if the phase difference is indeed $\frac{3\pi}{2}$ as follows:

Start with the hypothesis that the phase difference $\phi_1-\phi_2$ is $\frac{3\pi}{2}$, which is equivalent to $-\frac{\pi}{2}$

If and only if the phase is in such quadrature, then $s_1(t)$ and $s_2(t)$ will be the real and imaginary components of a single vector on the unit circle with radius A which can be given as $-jAe^{j(t+\phi_1)} = As_1(t) + jAs_2(t)$. This starts at the angle $\phi_1-\pi/2$ when t = 0 and rotates counter-clock-wise with increasing t, which matches the plot shown with the purple line representing the real axis and the red line representing the imaginary.

Starting from a position on the unit circle at time = $t_1$ with position given by the real and imaginary components $s_1(t_1)$ and $s_2(t_1)$, we can move forward in time to $t_2$ over a determined angle which should predict $s_1(t_2)$ and $s_2(t_2)$. If this prediction matches the given result, then this will confirm the hypothesis:

The starting angle using the first sample as given by B and D is given by $tan^{-1}(s_1(t_1)/s_2(t_1))$ $= tan^{-1}(3.614015/1.174265) = 1.256637 $ radians.

The change is angle to the second sample is simply the difference in time since the frequency unit is 1 radian/sec given by $sin(t)$:

$\Delta\phi = t_2-t_1 = 231-230 = 1$ radian (NOTE! This alone clearly does not match the plot visually, so the numbers given cannot be actually derived from the plot! There must be a frequency factor multiplied by t in the actual formulas for this to match the plot).

Therefore if the phase difference was $-\frac{\pi}{2}$ then the second angle would be at $1.256637 + 1$ radian or $2.256637$ radians and the ratio of $(s_1(t_2)/s_2(t_2))$ would be $\tan(2.256637)= -1.22195$

This does not match the ratio for the second sample given by A and C and therefore the phase difference cannot be $\frac{3\pi}{2}$ for the formulas as given.

Answer 2

You've got a system of equations with 2 unknowns ($A$ and the phase offsets). You hence need two equations. That means two samples.