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Determining phase of resultant from partial interference of waves

  1. Mar 20, 2012 #1
    When you have waves that are out of phase by some fraction of a cycle, e.g not exactly in phase and not exactly 180 degrees anti-phase, how do you determine the phase (relative to the original component waveforms) of the resultant?

    Specifically, is there an equation that solves for phase given the amplitudes and phase difference of the original 2 waves, and is there some rule of thumb that is a useful approximation, like for example the phase of the resultant (relative to the first 2 waves) will be the phase-difference between the original 2 waves?
  2. jcsd
  3. Mar 20, 2012 #2
    Hi Franco1991,

    Take the Fourier Transform of both waves using the same spatial and time base. The phase for each wave can then be calculated from the relative weights of the real and imaginary components.

    In general, if you're modeling a wave analytically (not measuring it from an experiment) then you need to set the initial conditions (and possibly boundary conditions) for the wave equation you're using. The initial conditions fix the phase for a particular solution.
    Last edited: Mar 20, 2012
  4. Mar 20, 2012 #3
    Not in general. The phase of the resultant will depend on the phases and amplitudes of the components.
    I think the easiest way to treat this kind of problem is phasor representation.
  5. Mar 21, 2012 #4


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    If you are considering waves of the same frequency (if you are talking of simple phase relationship then you must be) then the answer is just the sum of the two waveforms. This can be worked out using some pages of Trig (rusty there bit it's doable) or, as nasu says, use vectors / phasors. Mathematica etc. could also do it for you.
  6. Mar 26, 2012 #5
    I am in fact using waves of equal frequency. I'll try the trig first. What do you mean by the sum of two waves? The sum of their phases relative to a reference phase?

    Could you possibly provide an example, solving for the phase of the resultant given the two amplitudes and the phase difference between for 2 waves of the same frequency, or alternately provide a link to something similar?
  7. Mar 26, 2012 #6


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    I mean the algebraic sum of their displacements. You just add these together to get the result. Why would you expect to need an FT to find the result? The bits that are bobbing up and down just follow what each wave tells them to do as it goes past.
    To see what I mean, use a spreadsheet. Calculate the displacement for each wave over time at a particular place and add them together. You will get a new wave with different amplitude an phase from either of the contributions.
    Complex trig is not needed - it just makes things easier (as long as you are happy with the more sophisticated approach).

    [edit: in fact, plot all three waveforms together and get a real idea of what happens - and if you can't use a spreadsheet, this is an excellent reason why you should learn. It's v. easy]
    Last edited: Mar 26, 2012
  8. Mar 26, 2012 #7
    Your two waves [itex]\psi_1[/itex] and [itex]\psi_2[/itex] have the components

    [itex]\psi_1(f, t) = Acos(2 \pi ft) + Bsin(2 \pi ft) \ \ \ \ [/itex] [itex]\psi_2(f, t) = Ccos(2 \pi ft) + Dsin(2 \pi ft)[/itex]

    The phase of each wave will be

    [itex]\phi_1 = arctan(\frac{B}{A}) \ \ \ \ [/itex] [itex]\phi_2= arctan(\frac{D}{C})[/itex]
    Last edited: Mar 26, 2012
  9. Mar 27, 2012 #8


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    As with vectors, you 'can resolve' but it us only a ruse for convenience. There is an infinite choice of sin and cos waves to use. Incredibly useful, of course, as a tool, because it gives a great way of getting a grasp on an otherwise tricky thing to visualise.
  10. Apr 2, 2012 #9
    The spreadsheet idea is a great intuitive way of understanding the process (thank you sophiecentaur!), but I'd also like to be able to do it comparatively quickly using some sort of math. Ive used the following trig to find the resultant phase (using X to denote phase difference between the interfering waves):

    θ= Tan –1 [b(sin)X/a+b(cos)X)]

    Ive used the following equation to find the resultant amplitude:

    Using the equation R =√(0.25^2+0.2g^2+2(0.25)(0.25)cosX

    Using 90° as the phase difference between the waves, I get 0.35 as the amplitude and 45° as the resultant phase.

    I have two questions:

    1.) Is this correct (am I using the equations correctly)

    2.) What is this phase in reference to? I'd like to compare it to the phases of the first two interfering waves (or a non-interfering reference wave that can be compared to the original waves and the resultant) so that I can say the resultant wave has a phase difference of _ relative to wave A, and a phase difference of _ relative to wave B (if they still "existed" in their original form, as they are now technically the resultant).
    If the phase difference between the two original waves was 90°, and the resultant phase is 45°, does this mean that (if the original waves still "existed") the resultant wave would have a phase shift of 45° relative to the original waves?

    Likewise, if the phase difference between the original waves were say 120°, and I get a resultant phase of 20°, what would the resultant wave have a phase difference (relative to the original wave) of 100° (120-20), 140° (120+20), or would it be 160° (180-20)?

    Or is it in reference to the same "base-point" the original waves were referenced to, so that if wave A had a phase of say 50°, and wave b has a phase of 100° (so a 50° phase difference between the two), and the resultant is say 20°, would the following be correct: Resultant has a phase difference of 30° relative to wave A and an 80° phase difference relative to wave B?

    Thank you in advance for anyone that can clear this up.
  11. Apr 2, 2012 #10
    Not quite. I can't follow your formula for the solution but there is no need to insert either sin() or cos() into the arctan() value for the phase of either wave or some combination of A + B. Just use the relative weight or multiplication factor of each component [itex] \frac{B}{A}[/itex] or [itex] \frac{D}{C}[/itex]

    From the results of your example, a 45° phase value means that A = B.
    ie. to fulfill arctan(B/A) = 45° then arctan(1) = 45° because tan(45°) = 1

    But the amplitude of a wave expressed in the form I gave is [itex] \sqrt{(A^2 + B^2)}[/itex]

    Since A = B then [itex] \sqrt{(A^2 + B^2)} = \sqrt{(2(A^2))} = .35[/itex] according to your results so that A = B = .24748

    The phase is in reference to the real component of the wave (the cos part) If you set A = 1 and B = 0 then no amount of the imaginary component (the sin part) contributes to the phase while the real component contributes 100%. This works for every value that you insert into cos() and sin() simultaneously because the phase difference between them will always be 90°.
    Last edited: Apr 2, 2012
  12. Apr 14, 2012 #11
    I'm still not quite getting what I'm looking for. I want to be able to say that the resultant has a phase difference of ___° relative to either the original waves (that combine to create the resultant) or in reference to a non-interfering wave that can be compared to both (i.e the reference wave has a phase difference of x° relative to the original waves before interference, and a phase difference of y° relative to the resultant wave.)

    I want to be able to say how much of a phase shift the resultant has in reference to either the original interfering waves, or to a non-interfering wave that can be compared to both. Because I can figure out the phase of the resultant, but if I can't compare it to the phase of the original waves, it doesn't really mean anything, it's just a number.

    For example, if the two interfering waves has phases of x and y, say 20° and 40°, I want to say what the resultant's phase is relative to the same "base-point" the original waves' phases were in reference to (and in this way determine the phase difference between the original waves and the resultant, even though the original waves are now technically the resultant.)
  13. Apr 14, 2012 #12


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  14. Apr 14, 2012 #13


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    It doesn't matter what your reference is - you can choose it at your own convenience. If you are interested in an interference pattern and need to know its phase as well as its magnitude, then you could choose some point on that pattern and refer all phases to that. Or to the source. Or to anywhere - as long as you're consistent.
  15. Apr 16, 2012 #14
    Isn't just as simple as this?

    The phase of your first wave [itex]\psi_1[/itex] is [itex]arctan(\frac{B}{A})[/itex]
    The phase of your second wave [itex]\psi_2[/itex] is [itex]arctan(\frac{D}{C})[/itex]

    The phase of the second wave relative to the first is [itex]\psi_2 - \psi_1[/itex]

    This assumes that your measuring device is configured so that is doesn't move in time or space with respect to the initial pulse of the wave generator and that the frequency of both waves is exactly the same (as you originally specified).

    Or maybe you want the phase of both waves combined? I believe that would be [itex]arctan(\frac{B + D}{A + C})[/itex]

    P.S. The last value is the same as equation 3.10 of jtbell's reference for a wave with components having a single frequency.
    Last edited: Apr 16, 2012
  16. Apr 16, 2012 #15


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    That looks right. If you move across the interference pattern, this phase will oscillate from +π to -π (relative to one of the waves) as the resultant vector goes through max's and min's. If you draw the vector diagram, you can see one vector rotating around the end of the other (frozen) vector (phasor). I can't 'see' in my head, how the "arctan" function directly translates to this because it's easier, in may ways, to think in Amplitude and phase terms although the resolution method is easier for computation. It's all the same in the end.
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