# Homework Help: Finding phase difference between two sinusoids

1. Sep 24, 2011

### neshepard

1. The problem statement, all variables and given/known data
I have a graph of two sinusoids which I need to find the frequency of and phase difference between the two. I have the frequency easily enough, I hope, and I found what I believe is the actual formula for each wave. But I have no clue as to finding the difference.

2. Relevant equations
Frequency
V1(t) = 250Hz (period of 4mS)
V2(t)=333.33Hz (period of 3mS)

I have the following formulas for the waves.
V1(t)=4Vcos(2pi250t + pi/3)
V2(t)=-3Vcos(2pi333.33t)

3. The attempt at a solution
From here, the textbook is blank. Is the difference just simply pi/3? And does it matter that the waves have different peak heights? Any and all help is appreciated.

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Last edited: Sep 24, 2011
2. Sep 24, 2011

### SPYazdani

$\theta_{difference}=\theta_{1}-\theta_{2}$

3. Sep 24, 2011

### neshepard

Really, that easy? Do I feel dumb or what? Thanks! Do you know if the wave peak amplitude differences make any difference?

4. Sep 24, 2011

### SPYazdani

What do you think?

5. Sep 25, 2011

### neshepard

I guess I would have to say no. Provided the waves are out of phase by the same θ at all times, then they only differ in the θ.

6. Sep 25, 2011

### SPYazdani

7. Sep 25, 2011

### neshepard

Thanks

8. Sep 25, 2011

### uart

Two problems.

1. The phase of the v2 (as written in the OP) is not 0, it is $\pi$.

2. I don't think it even makes sense to talk of the phase difference when the frequencies differ.

Last edited: Sep 25, 2011
9. Sep 25, 2011

### Staff: Mentor

Indeed, it does not make sense to speak of phase difference unless the frequencies are identical. (Except maybe when talking about PLL's and instantaneous phase errors, but that's another topic for another day.)

10. Sep 25, 2011

### neshepard

Problem is, that's the homework assignment. Mine is not to question why, .......

11. Sep 25, 2011

### Staff: Mentor

The only way you could bring phase into it is, maybe at t=0 you could specify the phase of each relative to a sinewave of the same frequency. This would indicate the initial phase angle of that sinusoid at t=0, and may be as essential as specifying amplitude.

But it is not possible to compare relative phase angles of waves at different frequencies.

12. Sep 25, 2011

### uart

The problem is that phase is a relative measurement. It is the excess radian angle measured with respect to some rotating angular reference. With v1 for example, the phase is $\pi/3$ relative to a cosine angular reference of $500 \pi t$. A phase angle like this only really contains useful information if the reference is known, which implies a particular t=0. This is not possible in all situations.

For the case of two sinusoids of the same frequency however, the notion of phase is more useful, because in taking the difference of the total angles, the two reference phases (2 pi f t) cancel out (by subtraction) leaving a phase difference which is independent of the exact time origin. This is by far the most useful case for the notion of a phase angle.

For the case of sinusoids of different frequency however the situation is much worse. Not only is a naive calculation of phase difference (as per propose early in this thread) relative to a particular t=0 but each of the phase angles are relative to two completely different speed rotating angular references!

It's like if I was traveling on a bus on a Nth-Sth running highway and my N-S position relative to the front of the bus was -2.4m. My friend is also on a bus on the same highway and his N-S position relative to the front of the bus is -3.6m. So I conclude that the "positional difference" between me and my friend is 1.2m. The only problem is that he is on a different bus traveling at a different speed. How much relevance does my "positional difference" calculation hold?

If you must have a phase difference between the two then it would have to be something like $166.67 \pi t + \frac{2 \pi}{3}$, which is neither constant nor independent of the time origin.

Last edited: Sep 25, 2011