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Power spectral density question

  1. Feb 14, 2013 #1
    Normally, when we dealing with power signal in frequency domain we usually use power spectral density.

    My question is why is that? Why we define the power P(f) this way.

    http://img21.imageshack.us/img21/6786/powerec.jpg [Broken]

    This makes no sense to me, Why don't we just Fourier transform p(t) therefore we also get P(f)

    http://img198.imageshack.us/img198/8603/power2a.jpg [Broken]

    I just don't get the concept of power spectral density.

    Please help!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 14, 2013 #2

    sophiecentaur

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    You want a practical relevance of P(f)?
    P(f) is just the average power over a long (infinite) time at frequency f. A real signal will vary in time so, to get a useful idea of the amount of power around a particular frequency, you need to measure it and average it over a long time - or you could miss something.
    You could also do an FT on p(t) but that would also involve knowing p(t) over an infinite time interval so how would it necessarily be any easier?

    On a theoretical level, you are just questioning the 'point' of an identity. Well, you could do that for many (all) identities. You could just as easily ask what's the point of the multiple angle formulae.
    An identity is just a useful relationship that sometimes makes calculations easier.
     
  4. Feb 14, 2013 #3
    I don't know what you are studying, but in communications knowing the power of signal means nothing to you.

    Let's see what I mean by this.


    Say you have a 1 W signal. This 1 W of power can be spread over 10 Hz or 10 MHz, you don't know that. Knowing the power spectral density gives you the advantage to know in which part of you spectrum the power is concentrated mostly.

    Then when you integrate the PSD in some frequency range, you get total power in that frequency range.

    For example, take a FT of square pulse.
    Its a sinc function right? Most of its power is concentrated in that first arcade(first lobe). If you would integrate the PSD in the frequency range of the first lobe, you would get that it takes up around 90% of total power of the square pulse.

    Information like this is crucial, when you are designing channels.
     
  5. Feb 14, 2013 #4
    Thanks you for the answers!

    Now I see that knowing power spectral density is useful,
    but why does the Fourier transform of power mean nothing?

    Because the Fourier Transform of p(t) shows power at each frequency, suppose I get sinc function from Fourier Transform p(t), it shows that most of a component that create this power signal signal has frequency of -10hz - 10hz (say first lobe of a since function is -10hz - 10hz)

    This is not useful?
     
  6. Feb 14, 2013 #5

    sophiecentaur

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    I don't understand what this means, actually.
     
  7. Feb 14, 2013 #6

    f95toli

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    I am not sure I understand the question. Most methods for estimating the PSD DOES involve calculating the Fourier transform.

    However, actually estimating the PSD for a real signal is by no means trivial, and there are many different algorithms for doing so (Welch method is perhaps the most widely used one). It is (unfortunately) not as easy as just calculating the FFT.
     
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