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How to get phase difference between two signals?

  1. Feb 25, 2010 #1
    Hello all. I have two functions in time[tex] f_1(t) [/tex] and [tex] f_2(t) [/tex] composed of the same set of frequencies such that, say, [tex] f_1(t) = \sum a_n \cos(\omega t + d_1(t)) [/tex] and [tex] f_2(t) = \sum b_n cos(\omega t) [/tex] and I would like to find out the value of the phase difference (I've set the phase in [tex]f_2[/tex] equal to zero) at each point in time.

    Can I get this by comparing the complex parts of the two signal's Fourier transforms? What is the "meaning" of the complex part anyway? Whenever I've used the FT in the past, it's been the absolute value that represents the amplitude of the field at a given frequency, so what are the "meanings" of the real and imaginary parts separately? What is the "meaning" of the phase constructed from the arctan of the ratios of the real and imaginary parts?

    Thanks for any help you can give me.
  2. jcsd
  3. Feb 25, 2010 #2


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    Staff: Mentor

    I could be wrong, but it sure seems like your a_n will have to equal your b_n for all n, in order for "phase difference" to have a stationary meaning.
  4. Feb 25, 2010 #3
    Really? I was thinking that if I calculated the phase for [tex]f_1(t)[/tex] by writing the Fourier transform as [tex] \sqrt(Re(F_1(\omega))^2+Im(F_1\ometa)^2)e^{i\theta}[/tex] where [tex] \theta_1=arctan(Im/Re)[/tex] and then the phase difference at time t would just be
    [tex] d(t)=\theta_1(t)-\theta_2(t)[/tex]. Wouldn't the problem of the differences in amplitudes be removed by either the fact that we're looking at a ratio for each signal or the fact that the arctan is bounded?
  5. Feb 25, 2010 #4
    Last edited by a moderator: Apr 24, 2017
  6. Feb 25, 2010 #5
    No, they're a lot more complicated than that. They're actually two independently shaped UV laser fields that form an elliptically polarized pulse (with ellipiticity and phase changing with time) when you form the resultant field. I can plot them both together vs. time and see how the polarization changes, but I want to be able to get the numerical value of the phase at each point in time just from the real amplitudes.
    Last edited: Feb 25, 2010
  7. Feb 26, 2010 #6
    Are we talking about a phase difference for two signals at one particular frequency? You could do that easily by using integration with cos(f2pi*t) to single out specific frequency components independently in each signal. Then multiply those frequency components and use trig identities to calculate the phase difference.

    The FT of a signal has an entire spectrum of frequency components and the phase of a single component can be different at any point on the spectrum. That means the phase difference between the two signals must be a function of frequency. Is that what you want?
  8. Feb 26, 2010 #7
    Although there are many frequency harmonics, they are all exactly cosine-like and have the same phase delay, as stated in the OP. Doesn't this imply that if the phase delay is determined for the fundamental harmonic, it is known for all?

    Do you really mean

    f1(t) = ∑ancos[(nω0t) + θn] or ∑ancos[(nω0{t-tn})]

    with a different phase shift for every harmonic? Please rewrite your equations for f1(t) and f2(t).

    Bob S
    Last edited: Feb 26, 2010
  9. Feb 27, 2010 #8
    Okay, I think what I was trying isn't going to work. Maybe it would help if I was more general.

    What I have is two sets of data that are the real amplitudes of two shaped laser fields at each time point. What I want to do is find the phase difference between the two, which pretty much means (I think) that I have to find the phase of each of them. I only have the real part of [tex]E(t)=E^{re}+iE^{im}[/tex] and so I can't construct the phase of either field from the Fourier Transform. What I HOPE will now work is that I can get it from the Kramers-Kronig relation. That is, that because the real part of the field (which I know) isn't independent from the imaginary part (which I don't) that I can use the Kramers-Kronig relation to calculate it.

    Is this the correct use of the Kramers-Kronig relations or am I barking up the wrong tree here?
  10. Mar 1, 2010 #9
    I have used the Kramers Kroning (Dispersion relations) relations and real part sufficiency to calculate the imaginary part in the frequency domain, but not time domain. It may work. There is a complete page of Kramers-Kronig relations for EE circuits in Bode's book Network Analysis and Feedback Amplifier Design (1945).

    Bob S
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