Power of periodic signal

Discussion in 'Homework Help' started by xxxyyyba, Nov 1, 2015.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
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    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.
    In book, they gave us some periodic signal to write it in terms of complex Fourier coefficients and calculate power of first three harmonics. What is power of first three harmonics? Is it \left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2}?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What does your text have to say about that?
     
  3. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    There is no such a information in book... But I think it is like I wrote
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    4,805
    So what about F_(n-1), F_(n-2), and F_(n-3)?

    If you sum up all the power in all the harmonics, does it equal the power in the signal?
     
  5. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    It must equal power in signal.
    Here is formula I found in book:
    P=\frac{1}{T}\int_{\tau}^{\tau+T}f^{2}(t)dt=F^{2}_n_=_0+\sum_{n=-\infty,n\neq 0}^{n=\infty}\left | Fn \right |^{2}
    It equals F^{2}_n_=_0+\sum_{n=1}^{n=\infty}2\left | Fn \right |^{2}
    So power of first three harmonics should be (hopefully):
    2(\left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2})
     
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