How to Determine Phase Difference Between Two Sinusoidal Signals

[STEMucator]

Homework Statement

I have read that the phase difference between two sinusoidal signals is calculated as follows:

$$\theta = \omega \Delta T = \frac{2 \pi}{T} \Delta T$$

Where $\Delta T$ is the time difference. This formula confuses me as it was derived from nowhere.

I am asked to compute the phase difference between the two waveforms shown:

Also how much does wave 1 lead wave 2?

Homework Equations

$$f = \frac{1}{T}$$
$$\omega = 2 \pi f$$

The Attempt at a Solution

I am confused with the question itself. I know I am merely looking for the phase difference $|\theta_1 - \theta_2|$ between the waves.

Wave 1 appears to be a plain old sin wave ($v_1 = 1*sin(\omega t)$). Wave 2 is lagging behind wave 1 ($v_2 = 2*sin(\omega t + \phi)$).

Is that the right approach? Or do I read the periods of each wave off?

 

[olivermsun]

The phase is relative to the wave period $T$, so you need to figure out both $T$ and $\Delta T$ (as you can see from the equation you posted).

 

[STEMucator]

The phase is relative to the wave period $T$, so you need to figure out both $T$ and $\Delta T$ (as you can see from the equation you posted).

$T$ is what confused me as there are two different periods for each wave. Here's what I have so far:

 

[STEMucator]

The phase is relative to the wave period $T$, so you need to figure out both $T$ and $\Delta T$ (as you can see from the equation you posted).

Sorry for the double, but I think I realized something. Is the period of both waves and not just one roughly 6.25? The time difference would be roughly 0.15.

 

[olivermsun]

Look closely — I think both waves have the same period $T$.

 

[gneill]

The formula you wondered about comes about from considering $T$ to be the period and $\Delta T$ to be a portion of $T$. That means $\Delta T / T$ is the fraction of a full cycle of $2 \pi$ radians. So then:
$$\phi = \frac{\Delta T}{T} 2 \pi$$

As near as I can tell both waves have the same period.

One way to do this sort of problem is to identify similar zero-crossing points and use them for the instants of time that you'll be considering. By "similar" I mean where both waves are crossing the zero level in the same direction. Here's your picture with three such crossing points indicated. I've also laid a "ruler" along the zero V axis for convenience.

Note that two time differences are indicated, $T_a$ and $T_b$. One of them is smaller than the other. You generally want to take the smaller one because it will will yield a phase difference less than 180°. But it may make you re-evaluate whether wave 2 is leading or lagging wave 1!

If you are required to consider that wave 1 leads wave 2, then you'll have to live with the larger angular difference in this case.

 

[STEMucator]

Look closely — I think both waves have the same period $T$.

Cool, I now get:

$\theta = \frac{2 \pi}{T} \Delta T = \frac{2 \pi}{6.25} (0.15) = 0.151 rad$

 

[gneill]

I don't see where you're getting the 0.15 second value for $\Delta T$.

 

[STEMucator]

I don't see where you're getting the 0.15 second value for $\Delta T$.

$\Delta T$ was not explained. It was only mentioned as the time difference once.

I think that it's the difference between the peaks? So $\Delta T = 4$.

Then $\theta = 4.02 rad$.

 

[gneill]

$\Delta T$ was not explained. It was only mentioned as the time difference once.

I think that it's the difference between the peaks? So $\Delta T = 4$.

Then $\theta = 4.02 rad$.

It's the difference between any two equivalent points on the waveforms. Peaks are an example. But I find zero crossings can be determined more accurately. That's why I pointed out three choice ones on the figure.

 

[STEMucator]

It's the difference between any two equivalent points on the waveforms. Peaks are an example. But I find zero crossings can be determined more accurately. That's why I pointed out three choice ones on the figure.

Yes I liked your method as well, it actually helped me realize that it was really the difference between two equivalent points.