Fourier Coefficient Relations

  • Thread starter nickmai123
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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:

Answers and Replies

  • #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.
 
  • #3
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Thanks.
 

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