📂Fourier Analysis

Fourier Series in Complex Notation

Formulas

The complex Fourier series of a function $f$ defined over the interval $[-L,\ L)$ is given by:

$$f(t) = \sum \limits_{n=-\infty}^{\infty} c_{n} e^{i\frac{n\pi t}{L}}$$

Here, the complex Fourier coefficients are as follows:

$$c_{n} = \dfrac{1}{2L}\int_{-L}^{L}f(t)e^{-i\frac{n \pi t}{L} }dt$$

The Fourier coefficients satisfy the following equation:

$$\begin{align*} a_{0} & = 2 c_{0} \\ a_{n} &= c_{n}+c_{-n} \\ b_{n} &= i(c_{n}-c_{-n}) \\ c_{n} &= \frac{1}{2} (a_{n}-ib_{n}) \\ c_{-n} &= \frac{1}{2} (a_{n}+ib_{n}) \end{align*}$$

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It’s a form that is used more frequently than the trigonometric form because it is simpler.

Proof

Fourier series

$$\begin{equation} f(t) = \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n}\sin\dfrac{n\pi t }{L} \right) \end{equation}$$

$$\begin{align*} \text{where}\quad a_{0} =&\ \dfrac{1}{L}\int_{-L}^{L}f(t)dt \\ a_{n} =&\ \dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{n\pi t}{L} dt \\ b_{n} =&\ \dfrac{1}{L}\int_{-L}^{L}f(t)\sin\dfrac{n\pi t}{L}dt \\ \end{align*}$$

Using the Euler’s formula, the cosine and sine functions can be expressed as complex exponential functions as follows:

$$\begin{align*} \cos \dfrac{n\pi t}{L} &= \dfrac{e^{i\frac{n\pi t}{L}} + e^{-i\frac{n\pi t}{L}} }{2} \\ \sin \dfrac{n \pi t}{L} &= \dfrac{e^{i\frac{n\pi t}{L}} - e^{-i\frac{n\pi t}{L}} }{2i} \end{align*}$$

Substituting this into $(1)$ yields:

$$f(t)=\dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( a_{n} \dfrac{e^{i\frac{n\pi t}{L}} + e^{-i\frac{n\pi t}{L}} }{2} + b_{n}\dfrac{e^{i\frac{n\pi t}{L}} - e^{-i\frac{n\pi t}{L}} } {2i} \right)$$

Grouping terms based on the exponential function gives:

$$f(t) = \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( \dfrac{1}{2}\left(a_{n}-ib_{n} \right)e^{i\frac{n\pi t}{L}} +\dfrac{1}{2}\left(a_{n} + ib_{n} \right) e^{-i\frac{n\pi t}{L} } \right)$$

If we denote $c_{0}=\dfrac{a_{0}}{2}$, $c_{n}=\dfrac{1}{2}\left(a_{n}-ib_{n} \right)$, and $c_{-n}=\dfrac{1}{2}\left(a_{n} + ib_{n} \right)$ as follows:

$$f(t) =c_{0}+\sum \limits_{n=1}^{\infty} \left( c_{n}e^{i\frac{n\pi t}{L}} +c_{-n} e^{-i\frac{n\pi t}{L} } \right)$$

And organize the indices into a single one:

$$f(t) = \sum \limits_{n=-\infty}^{\infty} c_{n} e^{i\frac{n \pi t}{L} }$$

Also, if we compute $c_{0}$, $c_{n}$, and $c_{-n}$:

$$\begin{align*} c_{0} &=\dfrac{a_{0}}{2}=\dfrac{1}{2L}\int_{-L}^{L}f(t)dt \\ c_{n}&=\dfrac{1}{2}\left(a_{n}-ib_{n} \right)=\dfrac{1}{2L}\int_{-L}^{L}f(t)e^{-i\frac{n \pi t}{L} }dt \quad (n=1,\ 2,\ \cdots ) \\ c_{-n}&=\dfrac{1}{2}\left(a_{n} + ib_{n} \right)=\dfrac{1}{2L}\int_{-L}^{L}f(t)e^{i\frac{n \pi t}{L} }dt \quad (n=-1,\ -2,\ \cdots) \end{align*}$$

Then:

$$c_{n} = \dfrac{1}{2L}\int_{-L}^{L}f(t)e^{-i\frac{n \pi t}{L} }dt \quad (n=0,\ \pm 1,\ \pm 2,\ \cdots)$$

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