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Phase difference question

  1. Oct 1, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    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:

    Screen Shot 2014-10-01 at 4.18.47 PM.png

    Also how much does wave 1 lead wave 2?

    2. Relevant equations

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

    3. 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?
     
  2. jcsd
  3. Oct 1, 2014 #2

    olivermsun

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    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).
     
  4. Oct 1, 2014 #3

    Zondrina

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

    IMG_0387.jpg
     
  5. Oct 1, 2014 #4

    Zondrina

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    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.
     
  6. Oct 1, 2014 #5

    olivermsun

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    Look closely — I think both waves have the same period ##T##.
     
  7. Oct 1, 2014 #6

    gneill

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    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.

    waves.png

    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.
     
  8. Oct 1, 2014 #7

    Zondrina

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    Cool, I now get:

    ##\theta = \frac{2 \pi}{T} \Delta T = \frac{2 \pi}{6.25} (0.15) = 0.151 rad##
     
  9. Oct 1, 2014 #8

    gneill

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    I don't see where you're getting the 0.15 second value for ##\Delta T##.
     
  10. Oct 1, 2014 #9

    Zondrina

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    ##\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##.
     
  11. Oct 1, 2014 #10

    gneill

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    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.
     
  12. Oct 1, 2014 #11

    Zondrina

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