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When getting phase angle difference, do amplitudes have to be positive

  1. Mar 21, 2014 #1
    $$f(x)=32sin(x+60°)\quad and\quad g(x)=-1.5sin(x+45°)$$

    Would the phase angle difference here be 15° or is that incorrect? I think the amplitudes don't have to be the same but I'm not sure if they at least have to be both positive.
     
  2. jcsd
  3. Mar 21, 2014 #2

    mathman

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    -sin(a) = sin(a+π). See what happens.
     
  4. Mar 21, 2014 #3
    n = 180 and a<90, for -sin(a) = sin(a+n)
     
  5. Mar 24, 2014 #4
    $$-1.5sin(x+45°)\\ =1.5sin(x+225°)$$

    Here the phase angle difference is 165°. So which would be more correct? The phase angle difference is simply defined as the difference between the two phase angles in two sinusoids, right? So are they both right?
     
  6. Mar 24, 2014 #5

    mathman

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    When you look at the graphs, the phase difference is the difference between corresponding points (e.g. max) on the the two curves. Therefore they both need to expressed with the same sign. In this case 225° is the correct value for phase difference.
     
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