Power spectrum of periodic signal

Discussion in 'Homework Help' started by xxxyyyba, Oct 31, 2015.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    This is not really homework question, but this category looks most appropriate to me :)
    We know that periodic function can be writen in terms of complex Fourier coefficients:
    f(t)=F_n_=_0+\sum_{n=-\infty,n\neq 0}^{n=\infty}F_ne^{jnw_0t}, where F_n=\frac{1}{T}\int_{\tau}^{\tau+T}f(t)e^{-jnw_0t}dt and F_n_=_0 is DC component. Power spectrum of signal is defined as S_1_1(nw_0)=\left | F_n \right |^{2}, where \left | F_n \right | is modulus of complex Fourier coefficient F_n.
    We can draw power spectrum as function of both negative and positive discrete frequencies nw0. My question is, if I want to draw it only for positive discrete frequencies, I should multiply \left | F_n \right |^{2} by 2, right? Should \left | F_n_=0 \right |^{2} be also multiplied by 2?
     
    Last edited: Oct 31, 2015
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Yes and no, respectively.

    Having said that, I don't understand your first expression (not that it isn't even an equation).
     
    xxxyyyba likes this.
  3. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    I edited my post, I hope it is clear now :)
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
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    Why exclude n=0 from the sum? Why not just get rid of the first term and let it appear naturally as part of the sum?
     
  5. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Because for n=0 Fn is usually undefined (something/0). For n=0 we calculate integral separately...
     
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