当前位置：网站首页>Derivation of Fourier transform

Derivation of Fourier transform

2022-07-03 09:20:00 【fishfuck】

function $f_L(t)$ With $L$ $(L > 0)$ For cycles , In the interval $[-\frac L2,\frac L2]$ Continuous on （ Or meet the Dirichlet condition ）, be $f_L(t)$ stay $[-\frac L2,\frac L2]$ Can be expanded into Fourier series

$f_{L}(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega t+b_{n} \sin n \omega t\right) \text {. }$

among

$\begin{aligned} &\omega=\frac{2\pi}{L} \\ &a_{n}=\frac{2}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f_{L}(t) \cos n \omega t d t . \quad n=0,1,2 \ldots \\ &b_{n}=\frac{2}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f_{L}(t) \sin { n \omega t } d t . \quad n=1,2, \cdots \end{aligned}$

So there are

$\cos t=\frac{e^{i t}+e^{-i t}}{2}$
$\sin t=-i \frac{e^{i t}-e^{-i t}}{2}$

be
$\begin{aligned} f_{L}(t) &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega t+b_{n} \sin n \omega t\right) \\ &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \frac{e^{i n \omega t}+e^{-i n \omega t}}{2}-i b_{n} \frac{e^{i n \omega t}-e^{-i n \omega t}}{2}\right) \\ &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(\frac{a_{n}-i b_{n}}{2} e^{i n \omega t}+\frac{a_{n}+i b_{n}}{2 i} e^{-i n \omega t}\right) \end{aligned}$

We make
$c_{0}=\frac{a_{0}}{2}=\frac{1}{L} \int_{-\infty}^{\infty} f_{L}(t) d t$

$c_{n}=\frac{a_{n}-i b_{n}}{2}$

${c_{-n}}=\frac{a_{n}+i b_{n}}{2}$

So there are

$C_{n}=\frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f_{L}(t) e^{-i n \omega t t} d t \quad n=0, \pm 1,\pm 2, \ldots$

$\begin{aligned} f_{L}(t) &=c_{0}+\sum_{n=1}^{\infty}\left(c_{n} e^{i n \omega t}+c_{i n} e^{-i n \omega t}\right) \\ &=\sum_{n=-\infty}^{\infty} C_{n} e^{i n u t t} \\ &=\sum_{n=-\infty}^{\infty}\left(\frac{1}{2} \int_{-\frac{L}{2}}^{\frac{L}{2}} f_{L}(\tau) e^{-i n \omega t} d \tau \right) e^{i n \omega t} \end{aligned}$

We consider nonperiodic functions in the interval $[-\frac L2,\frac L2]$ On the situation . Make

$f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{ {in\omega t }}\left(-\frac{L}{2}<t<\frac{L}{2}\right). \quad*$

among

$c_{n}=\frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f(t) e^{-i n \omega \tau} d \tau .$

We noticed that $(*)$ The right end of the equation defines a period of $L$ Function of $f_{L_{1}}(t)=\left\{\begin{array}{l}f(t) \quad-\frac{L}{2}<t<\frac{L}{2} \\ \frac{\left.f\left(-\frac{L }{2}\right)+f(\frac{L}{2}\right)}{2} \quad t=\pm \frac{L}{2}\end{array}\right.$

Make $\rightarrow+\infty$ Then periodic function $f_{L_{1}}(t)$ The limit is $f (t)$ , That is to say

$\lim _{L \rightarrow+\infty} f_{L_{1}}(t)=f(t)$

here , To any $-\infty<t<+\infty$ , Yes

$f(t)=\lim _{L \rightarrow+\infty} \sum_{n=-\infty}^{+\infty} c_{n} e^{ {in\omega t. }}$

$(*)$ The right side of the formula is $\sum_{n=-\infty}^{+\infty} c_{n} e^{ {in\omega t. }}$ , among $\omega=\frac{2 \pi}{L}$ .

Make

$g_{n}=c_{n} L=\int_{-\frac{L}{2}}^{\frac{L}{2}} f(t) e^{\frac{-in 2 \pi t} L} d t$

$\begin{aligned} f_{L}(t) &=\frac{1}{2\pi} \sum_{n=-\infty}^{\infty} g_{n} e^{-\frac{i n 2 \pi t}{L}} \frac{[(n+1)-n] 2\pi}{L} \\ &=\sum_{n=-\infty}^{\infty}\left(\frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f_{L}(\tau) e^{-i n \omega \tau} d \tau\right) e^{ {in\omega t }} \end{aligned}$

remember $\omega_{n}=\frac{n 2 \pi}{L}$ , We make $\mathscr{F}\left[f(t) \right]_{L}(u)=\int_{-\frac{L}{2}}^{\frac{L}{2}} f(t) e^{-i \omega t} d t$ , be ：

$f_{L}(t)=\frac{1}{2 \pi} \sum_{n=-\infty}^{\infty} \mathscr{F}[f(t)]_{L}\left(\omega_{n}\right) e^{ {i\omega n } t}\left(\omega_{n+1}-\omega_{n}\right) \text {. } \quad**$

When $\rightarrow \infty$ when , $\mathscr{F}[f(t)]_{L}\left(\omega\right)$ Naturally, it tends to a frequency domain function $\mathscr{F}[f(t)](\omega)=\int_{-\infty}^{\infty} f(t) e^{-i \omega t} d t$ . So we finished the time domain function $f (t)$ To frequency domain function $\mathscr{F}[f(t)]\left(\omega\right)$ Transformation of . We call $\mathscr{F}[f(t)]\left(\omega\right)$ It's Fourier transform .

With $\rightarrow \infty$ , $\Delta \omega_{n}=w_{n+1}-w_{n} \rightarrow 0$ , be $(* *)$ The formula can be regarded as $\frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathscr{F}[f(t)](\omega) e^{i \omega t} d\omega$ Riemann sum of , So export