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Frequency and phase relationship

  1. Sep 10, 2009 #1
    Frequency is the time derivative of phase? But how?
    Can someone explain?
     
  2. jcsd
  3. Sep 10, 2009 #2

    f95toli

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    No, it is not.
    At least not if you use the normal meaning of "phase", in which case it is a parameter (usually a constant) which tells you the relative postion in time of two or more periodic waveforms

    e.g. if you have

    [itex]\sin (2\pi ft+\theta)[/itex])

    then [itex]\theta[/itex] would be the phase. Note that it is only meaningfull to talk about phase when you are comparing waveforms; the "starting point" for a periodic function is arbitrary so there is no such thing as absolute phase.
     
  4. Sep 10, 2009 #3
    Group delay is a derivative of phase with respect to angular frequency:

    [tex] \tau_g = -\frac{d\phi}{d\omega} [/tex]
     
  5. Sep 10, 2009 #4
    Last edited by a moderator: Apr 24, 2017
  6. Sep 10, 2009 #5

    f95toli

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    OK, now I understand where you got that from.
    This is why I was refering to the "normal meaning of phase" above.

    People (meaning EEs) who work with modulations schemes (in this case FM) have a tendency to refer to the argument of the sine function as "phase" ; i.e "the phase" in this case would be [itex]\omega t+\theta[/itex] and if you take the time derivative of this you obviously get [itex]\omega[/itex] (which also happens to be the angular frequency, not the frequency).

    So -unless I am missing something- this is just another case of confusion due to differences between EE and physics terminology.
    The "definition" of phase I wrote above is certainly what you would find in a physics book.
     
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