ece205: add complex fouriers

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eggy 2023-11-21 12:37:35 -05:00
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@ -453,6 +453,42 @@ Then, for any $x\in[-L,L]$:
$$\int^x_{-L}f(t)dt=\int^x_{-L}\frac{a_0}{2}dt+\sum^\infty_{n=1}\int^x_{-L}(a_n\cos(\frac{n\pi t}{L})+b_n\sin(\frac{n\pi t}{L}))dt$$
### Complex Fourier series
By employing Euler's theorem, sine and cosine can be transformed into exponential forms.
$$
\cos(\frac{n\pi x}{L})=\frac{e^{i\frac{n\pi x}{L}} + e^{-i\frac{n\pi x}{L}}}{2} \\
\sin(\frac{n\pi x}{L})=\frac{-ie^{i\frac{n\pi x}{L}} + ie^{-i\frac{n\pi x}{L}}}{2}
$$
Thus the **complex Fourier series** is given by:
$$
f(x)=\sum^\infty_{n=-\infty}c_ne^{i\frac{n\pi x}{L}} \\
c_n=\frac{1}{2L}\int^L_{-L}f(x)e^{-i\frac{n\pi x}{L}}dx = \frac 1 2(a_n-ib_n)
$$
To convert it to a real Fouier series:
- $a_0=2c_0$
- $a_n=c_n+\overline{c_n}$
- $b_n=i(c_n-\overline{c_n})$
!!! example
The complex Fourier series for the sawtooth wave function: $f(x)=x,-1<x<1,f(x+2)=f(x)$. Thus we have a period of 2 and $L=1$.
\begin{align*}
c_0&=\frac 1 2\int^1_{-1}\underbrace{xe^{0}}_\text{odd}dx \\
&=0 \\
\\
c_n&=\frac 1 2\int^1_{-1}xe^{-in\pi x}dx \\
\tag{IBP}&=\frac 1 2\left[\frac{xe^{-in\pi x}}{-in\pi}-\int\frac{1}{-in\pi}e^{-in\pi x}dx\right]^1_{-1} \\
&=\frac 1 2\left[\frac{xe^{-n\pi x}}{-in\pi}+\frac{1}{n^2\pi^2}e^{-in\pi x}\right]^1_{-1} \\
&=\frac{(-1)^ni}{n\pi} \\
\\
\therefore f(x)&=\sum^\infty_{\substack{n=-\infty \\ n\neq0}}\frac{(-1)^ni}{n\pi}e^{in\pi x}
\end{align*}
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