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Power spectral density?

  1. Oct 16, 2009 #1

    KFC

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    Suppose the power spectral density denoted by [tex]P(\omega)[/tex] where $\omega$ is the angular frequency and [tex]\omega = 2\pi \nu[/tex], I wonder how to prove that

    [tex]2\pi P(\omega) = P(\nu)[/tex]

    without know the functional form of [tex]P(\omega)[/tex]. I saw this relation in some book, but I don't know how to prove that.
     
    Last edited: Oct 16, 2009
  2. jcsd
  3. Oct 16, 2009 #2

    mathman

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    It looks like a scale factor needed to insure that the total power (integral) is the same whatever the function argument (integral differential) is.
     
    Last edited: Oct 16, 2009
  4. Oct 16, 2009 #3
    Disproof by Counterexample:

    Suppose [tex]P(\omega) = 1/\omega^{2}[/tex] then [tex]P(\nu) = 1/\nu^{2} = 4\pi^{2}/\omega^{2}[/tex]
    Thus
    [tex]P(\nu) = 4\pi^{2} P(\omega)[/tex]
     
  5. Oct 17, 2009 #4

    KFC

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    But I saw the statement about this somewhere else, like mathman said, when you integrate the power spectral density, no matter on [tex]\omega[/tex] or [tex]\nu[/tex] you should get the same energy, so you will find a relation between [tex]P(\omega)[/tex] and [tex]P(\nu)[/tex]. But as to your example, I don't know why it leads to this ...
     
  6. Oct 17, 2009 #5

    mathman

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    What I was trying to say was in essence:

    [tex]d\omega = 2\pi d\nu[/tex]
     
  7. Oct 17, 2009 #6

    KFC

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    Yes, that's what I mean. But what I am thinking to pseudophonist's example is: [tex]
    2\pi P(\omega) = P(\nu)
    [/tex] is only valid in the sense of integral but not the expression alone, right? Otherwise, how do you explain the counter example?
     
  8. Oct 18, 2009 #7

    cepheid

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    The fallacy in pseudophonist's example is that he assumes that P(ν) and P(ω) are of the same functional form, and that going from one to the other is merely a matter of changing what symbol is used for the argument. However, in order to satisfy the condition that the power in any frequency interval is the same as measured using both functions, it's pretty clear that they can't be of the same functional form (even though we are confusingly using the same symbol, P, for both). So let's not use the same symbol for them. Let's call the power per unit frequency interval Pν(ν) and the power per unit angular frequency interval Pω(ω) (by adding the subscripts, we've used a different symbol for each spectrum, making it clear that these are in fact two different functions).

    First of all, the sloppy physicist's derivation of the result is to talk about the power in any "small" (i.e. infinitesimal) frequency interval being the same regardless of what kind of frequency you're talking about. Hence:

    Pν(ν) dν = Pω(ω)dω ​

    and since

    2πdν = dω ​

    The result follows immediately:

    Pν(ν) = 2πPω(ω) ​

    A more mathematically sensible version of the derivation is to talk about the power over some *finite* frequency interval (which is calculated by integrating the power spectral densities over that interval). It shouldn't matter whether you're using angular frequency or just plain old frequency: you should get the same answer for the total power in that interval:

    [tex] \int_{\omega_1}^{\omega_2} P_{\omega}(\omega)\, d\omega = \int_{\nu_1}^{\nu_2} P_{\nu}(\nu)\, d\nu [/tex] ​

    Again, it's pretty clear that, for this condition to be satisfied, there's no way that Pν(ν) and Pω(ω) can be of the same functional form.
     
  9. Oct 18, 2009 #8

    KFC

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    Thanks. Make sense
     
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