Delta 与 Fourier

• 原文: https://zhuanlan.zhihu.com/p/67187240
• 发布日期: 2019-05-27
• 分类: 数学方法 / 物理数学工具箱

写着写着突然觉得不对劲, 近日忙··· 等哪天有空了再想想办法看能不能让这几个概念自洽

怎么证明 f\left( t \right)=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega t}\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{-{\rm{i}n\omega {t}'}{\rm{d}{t}'} 呢?

Wait, 这个我莫名其妙地突然知道该怎么证明了⋯ 就事箱归一化.

我在群论的文章里还用到了 (11.4 节):

https://zhuanlan.zhihu.com/p/342592239所以我说, 搞不定的就别搞了, 过两天会突然就知道该怎么算的.

Part. I - δ函数[1]的数学表达式

傅里叶变换: g\left( \omega \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{f\left( t \right){\rm{e}^{-{\rm{i}\omega t}{\rm{d}t}.

傅里叶积分: f\left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{g\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }.

将变换式代入积分式:

\begin{align} & f\left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{g\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega } \\ & \ \ \ \ \ \ \ =\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\left[ \frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{f\left( {t}'} \right){\rm{e}^{-{\rm{i}\omega {t}'}{\rm{d}{t}'} \right]{\rm{e}^{\rm{i}\omega t}{\rm{d}\omega } \\ & \ \ \ \ \ \ \ =\int_{-\infty }^{\infty }{f\left( {t}'} \right)\left[ \frac{1}{2\pi }\int_{-\infty }^{\infty }{e}^{\rm{i}\omega \left( t-{t}' \right)}{\rm{d}\omega } \right]{\rm{d}{t}'}. \\ \end{align}

这种选择性实际上就是δ函数的定义, 所以我们可以说:

\delta \left( t-{t}' \right)=\frac{1}{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{\rm{i}\omega \left( t-{t}' \right)}{\rm{d}\omega }.

• 或者如果你总觉得不太接受上面那个就是 δ 函数的定义的话, 还可以这样想:
• 对 δ 函数做傅里叶变换 \tilde{\delta }\left( \omega \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{-{\rm{i}\omega t}\delta \left( t \right){\rm{d}t}=\frac{1}{\sqrt{2\pi } 是个常数.
• 这样再积分回去就有 \delta \left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\tilde{\delta }\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }=\frac{1}{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }.
• Voila! 又回来力!

Part. II - 离散的δ函数: Kronecker delta[2]

克罗内克符号的定义: {\delta }_{t{t}'}=\left\{ \begin{align} & 1,\ \ t={t}'; \\ & 0,\ \ t\ne {t}'. \\ \end{align} \right.

我们将证明它的表达式可以写作 {\delta }_{t{t}'}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}.

离散形式下 $t$ 的取值也是离散的[3]: $t=\frac{2\pi }{N\omega }p其中p\in Z.$

当 $t=t^{\prime }+m\frac{2\pi }{\omega }$其中 $m\in Z$ 时:

{\delta }_{t{t}'}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{0}=1.

当 $t\ne t^{\prime }+n\frac{2\pi }{\omega }$ 其中 $m\in Z$ 时[4]: \begin{align} & {\delta }_{t{t}'}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}=\frac{1}{N+1}\frac{\rm{e}^{-{\rm{i}\frac{N}{2}\omega \left( t-{t}' \right)}\left( 1-{\rm{e}^{\rm{i}N\omega \left( t-{t}' \right)} \right)}{1-{\rm{e}^{\rm{i}\omega \left( t-{t}' \right)} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{N+1}\frac{\rm{e}^{-{\rm{i}\frac{N}{2}\omega \left( t-{t}' \right)}{1-{\rm{e}^{\rm{i}\frac{2\pi \left( p-{p}' \right)}{N}\left( 1-{\rm{e}^{\rm{i}2\pi \left( p-{p}' \right)} \right)=0. \\ \end{align}

综上所述, 在 $t\in \left(-\frac{\pi }{\omega },\frac{\pi }{\omega }\right)$ 或者说 $t\in \left(-\frac{T}{2},\frac{T}{2}\right)$ 时[5]:

{\delta }_{t{t}'}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}=\left\{ \begin{align} & 1,\ \ t={t}'; \\ & 0,\ \ t\ne {t}'. \\ \end{align} \right.

下面的就别看了, 我都不知道写的什么几把, 而且已经解决了, 本文一开始就说了.

Part. III - Kronecker delta 与傅里叶展开[6]

$f\left(t+T\right)=f\left(t\right),T=\frac{2\pi }{\omega }.$

周期函数的傅里叶变换: {g}_{n}=\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( t \right){\rm{e}^{-{\rm{i}n\omega t}{\rm{d}t}. ------[1]

周期函数的傅里叶展开: f\left( t \right)=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega t}{g}_{n}. -------------[2]

仿照 Part. I

\begin{align} & f\left( t \right)=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega t}\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{-{\rm{i}n\omega {t}'}{\rm{d}{t}'} \\ & \ \ \ \ \ \ \ =\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right)\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \\ & \ \ \ \ \ \ \ =f\left( t \right)\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \\ & \ \ \ \ \ \ \ =f\left( t \right)\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}. \\ \end{align}

莫非... \frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}=1 真的成立?

分 $t=t^{\prime }$ 与 $t\ne t^{\prime }$ 两种情况, 前者不难看出发散了, 是无穷个 $1$ 相加.

至于 $t\ne t^{\prime }$这种情况:

\begin{align} & \ \ \ \ \ \int_{-\frac{T}{2}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}=T \\ & \Rightarrow \int_{0}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}+\int_{-\frac{T}{2}^{0}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}=T. \\ & \Rightarrow \int_{0}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}-\int_{0}^{-\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}=T. \\ \end{align}

两边对 $T$ 求导试试:

$\frac{1}{2}\sum _{n=-\infty }^{\infty }{e}^{in\omega \left(t-\frac{T}{2}\right)}+\frac{1}{2}\sum _{n=-\infty }^{\infty }{e}^{in\omega \left(t+\frac{T}{2}\right)}=1$

\begin{align} & \ \ \ \ \ \frac{1}{2}\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-\frac{T}{2} \right)}+\frac{1}{2}\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t+\frac{T}{2} \right)} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{2}\left( {\rm{e}^{\rm{i}n\omega \left( t-\frac{T}{2} \right)}+{\rm{e}^{\rm{i}n\omega \left( t+\frac{T}{2} \right)} \right)} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{2}\left( {\rm{e}^{\rm{i}n\omega \left( t-\frac{\pi }{\omega } \right)}+{\rm{e}^{\rm{i}n\omega \left( t+\frac{\pi }{\omega } \right)} \right)} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{2}\left( {\rm{e}^{-{\rm{i}n\pi }+{\rm{e}^{\rm{i}n\pi } \right){\rm{e}^{\rm{i}n\omega t}=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega t}\cos n\pi }. \\ \end{align}

无解, 评论区有提到两次运用分部积分, 试了一下:

\begin{align} & f\left( t \right)=\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right)\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \\ & \ \ \ \ \ \ \ =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \\ & \ \ \ \ \ \ \ =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{-{\rm{i}n\omega }\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{d}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)} \\ & \ \ \ \ \ \ \ =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{-{\rm{i}n\omega }\left[ \left. f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)} \right|_{-\frac{T}{2}^{\frac{T}{2}-\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}f\left( {t}'} \right)}{\rm{d}t}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \right]} \\ & \ \ \ \ \ \ \ =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{\rm{i}n\omega }\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}f\left( {t}'} \right)}{\rm{d}t}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}. \\ \end{align}

可以计算 \sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{\rm{i}n\omega }\int_{-\frac{T}{2}^{\frac{T}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}f\left( {t}'} \right)} 或 \sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{n}^{2}{\omega }^{2}\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}f\left( {t}'} \right)}{\rm{d}t}{\rm{d}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}.

前者没什么意义, 又变回去了:

\begin{align} & \ \ \ \ \sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{\rm{i}n\omega }\int_{-\frac{T}{2}^{\frac{T}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}f\left( {t}'} \right)} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{\rm{i}n\omega }\left[ \left. f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)} \right|_{-\frac{T}{2}^{\frac{T}{2}+{\rm{i}n\omega \int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \right]} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} .\\ \end{align}

至于后者:

\begin{align} & \ \ \ \sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{n}^{2}{\omega }^{2}\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}f\left( {t}'} \right)}{\rm{d}t}{\rm{d}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)} \\ & {\rm{=}\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{n}^{2}{\omega }^{2}\left[ \left. \frac{\rm{d}f\left( {t}'} \right)}{\rm{d}t}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)} \right|_{-\frac{T}{\rm{2}^{\frac{T}{2}-\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}^{2}f\left( {t}'} \right)}{\rm{d}{t}^{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'} \right]} \\ & =\sum\limits_{n=-\infty }^{\infty }{\frac{1}{T}\frac{1}{n}^{2}{\omega }^{2}\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}^{2}f\left( {t}'} \right)}{\rm{d}{t}^{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}. \\ \end{align}

这说明 \sum\limits_{n=-\infty }^{\infty }{\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( {t}'} \right){\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}=\sum\limits_{n=-\infty }^{\infty }{\frac{1}{n}^{2n}{\omega }^{2n}\int_{-\frac{T}{2}^{\frac{T}{2}{\frac{\rm{d}^{2n}f\left( {t}'} \right)}{\rm{d}{t}^{2n}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}{\rm{d}{t}'}.

··· I don't really··· eh?

其实从 **[2]**出发通过正交性会很容易推导出 [1]:

\begin{align} & \ \ \ \ f\left( t \right)=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}n\omega t}{g}_{n} \\ & \Rightarrow f\left( t \right){\rm{e}^{-{\rm{i}{n}'\omega t}=\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}\left( n-{n}' \right)\omega t}{g}_{n} \\ & \Rightarrow \int_{-\frac{T}{2}^{\frac{T}{2}{f\left( t \right){\rm{e}^{-{\rm{i}{n}'\omega t}{\rm{d}t}=\int_{-\frac{T}{2}^{\frac{T}{2}{\sum\limits_{n=-\infty }^{\infty }{\rm{e}^{\rm{i}\left( n-{n}' \right)\omega t}{g}_{n}{\rm{d}t}. \\ \end{align}

由于 \left\{ \left. {\rm{e}^{\rm{i}n\omega t} \right|n\in Z \right\} 是一组完备正交函数所以有: \begin{align} & \ \ \ \frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( t \right){\rm{e}^{-{\rm{i}{n}'\omega t}dt}=\sum\limits_{n=-\infty }^{\infty }{g}_{n}\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{\rm{e}^{\rm{i}\left( n-{n}' \right)\omega t}{\rm{d}t} \\ & \Rightarrow {g}_{n}=\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{f\left( t \right){\rm{e}^{-{\rm{i}{n}'\omega t}{\rm{d}t}. \\ \end{align}

也就是得到了 [1], 但这里只说明有 {\delta }_{n}'n}=\frac{1}{T}\int_{-\frac{T}{2}^{\frac{T}{2}{\rm{e}^{\rm{i}\left( n-{n}' \right)\omega t}{\rm{d}t} 是一个平凡的结论.

能不能得到 {\delta }_{t{t}'}=\frac{1}{N+1}\sum\limits_{n=-\frac{N}{2}^{\frac{N}{2}{\rm{e}^{\rm{i}n\omega \left( t-{t}' \right)}=\left\{ \begin{align} & 1,\ \ t={t}'; \\ & 0,\ \ t\ne {t}' \\ \end{align} \right. 呢?

参考

• ^一般指 Dirac delta function.
• ^也称克罗内克符号.
• ^但通过增大 N 可以提高 t 的精细程度以达到准连续.
• ^也就是差了周期整数倍时.
• ^即在一个周期之内.
• ^傅里叶级数展开就是离散的傅里叶变换.