- #1
nickmai123
- 78
- 0
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]
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: