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Homework Help: Fourier Coefficient Relations

  1. Jun 29, 2010 #1
    I have a quick question about the relationship between the complex Fourier coefficient,[tex]\alpha_n[/tex] and the real Fourier coefficients, [tex]a_n[/tex] and [tex]b_n[/tex].

    Given a real-valued function, I could just find the real coefficients and plug them into the relation below, right?


    Fourier Coefficients for periodic functions of period 2a.
    Complex Form:
    [tex]\alpha_n = \frac{1}{2a}\int_{-a}^{a} f\left(t\right)e^{\frac{-jn\pi t}{a}dt[/tex]

    Real Form:
    [tex]a_0 = \frac{1}{a}\int_{-a}^{a} f\left(t\right)dt[/tex]

    [tex]a_n = \frac{1}{a}\int_{-a}^{a} f\left(t\right) cos\left(\frac{n\pi t}{a}\right)dt[/tex]

    [tex]b_n = \frac{1}{a}\int_{-a}^{a} f\left(t\right) sin\left(\frac{n\pi t}{a}\right)dt [/tex]

    Relation
    [tex]\alpha_n = \left\{
    \begin{array}{lr}
    \frac{1}{2}\left(a_n + jb_n\right) & : n < 0\\ \\
    \frac{1}{2}a_0 & : n = 0\\ \\
    \frac{1}{2}\left(a_n - jb_n\right) & : n > 0
    \end{array}
    \right.[/tex]
     
    Last edited: Jun 30, 2010
  2. jcsd
  3. Jun 29, 2010 #2

    vela

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    Yup, except that [itex]\alpha_0 = a_0[/itex]. The factor of 1/2 for that coefficient isn't needed with the formulas you're using.
     
  4. Jun 29, 2010 #3
    Thanks.
     
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