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How Fourier Expansion indicates the Amplitude w.r.t a certain frequency?

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    I wanna know how could I extract the amplitude(of the sinusoid component) of a random continuous wave w.r.t a certain frequency response? The teacher said the Fourier Expansion can do that but I'm really confused by the limits and integrals.


    2. Relevant equations

    [itex]F(\omega)=\int f(t) e^{-j \omega t} dt[/itex]

    3. The attempt at a solution

    I tried the very easy example and wanna extract the amplitude where the frequency matches(say [itex]\omega = \omega_0 [/itex]). [itex]f(t)=A \cdot cos\omega_0t,\hat{f}(t)=A \cdot e^{j \omega_0 t},\hat{F}(\omega)=\int \hat{f}(t) \cdot e^{-j \omega t} dt [/itex], range ([itex]-\infty ,\infty [/itex]), but it turned out to be [itex]Re\{ \hat{F}( \omega ) \}=A \cdot \frac{sin(\omega_0 - \omega ) (t_2-t_1)}{\omega_0 - \omega}[/itex], where [itex]t_2=\infty,t_1=-\infty[/itex] , it's weird if I follow the basic operation of sin function, I got [itex]Re\{ \hat{F}( \omega ) \}=2 \cdot A \cdot \frac{sin(\omega_0 - \omega ) \infty}{\omega_0 - \omega}[/itex], and then although applying that [itex]lim \frac{sinx}{x} -> 1[/itex] while x->0, it's 2A, besides I don't even know if this's right.

    I have no idea what happened...

    Any help will be appreciated !!!
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2
    Terribly sorry that my teacher has corrected my mistakes, the Amplitude should be [itex]|F(\omega)|[/itex] instead of [itex]Re \{ \hat{F}(\omega) \}[/itex],but the calculation becomes even harder, I'm still trying on this question.
     
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