EE261 Lecture 14

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课程主页：https://see.stanford.edu/Course/EE261

这次回顾第十四讲，这一讲介绍分布的傅里叶变换的性质。

导数定理

回顾函数的傅里叶变换的导数公式

$$f^{\prime }(t)\rightleftharpoons 2\pi isF(s)\text{ and }-2\pi itf(t)\rightleftharpoons F^{\prime }(s)$$

其中

$$f(t)\rightleftharpoons F(s)$$

现在考虑分布的情形：

$$\begin{array}{rl}⟨\mathcal{F}{T}^{\prime },\phi ⟩& =⟨T^{\prime },\mathcal{F}\phi ⟩\\ & =-⟨T,(\mathcal{F}\phi {)}^{\prime }⟩\\ & =-⟨T,\mathcal{F}(-2\pi is\phi )⟩\text{ (from the second formula above) }\\ & =-⟨\mathcal{F}T,-2\pi is\phi ⟩\text{ (moving }\mathcal{F}\text{ back over to }T)\\ & =⟨2\pi is\mathcal{F}T,\phi ⟩\end{array}$$

所以

$$\mathcal{F}{T}^{\prime }=2\pi is\mathcal{F}T$$

类似的，我们有

$$\begin{array}{rl}⟨(\mathcal{F}T{)}^{\prime },\phi ⟩& =-⟨\mathcal{F}T,{\phi }^{\prime }⟩\\ & =-⟨T,\mathcal{F}\left({\phi }^{\prime }\right)⟩\\ & =-⟨T,2\pi is\mathcal{F}\phi ⟩\text{ (from the first formula for functions) }\\ & =⟨-2\pi isT,\mathcal{F}\phi ⟩\\ & =⟨\mathcal{F}(-2\pi isT),\phi ⟩\end{array}$$

所以

$$(\mathcal{F}T{)}^{\prime }=\mathcal{F}(-2\pi isT)$$

例子

考虑${\text{sign}}^{\prime }$

$$\mathcal{F}{\mathrm{sgn}}^{\prime }=2\mathcal{F}\delta =2$$

另一方面，由导数公式可得

$$\mathcal{F}{\mathrm{sgn}}^{\prime }=2\pi is\mathcal{F}\mathrm{sgn}$$

因此

$$2\pi is\mathcal{F}\mathrm{sgn}=2\Rightarrow \mathcal{F}\mathrm{sgn}=\frac{1}{\pi is}$$

实际上我们有更强的结论，原因如下。

注意到如果$sT=0$，那么

$$2\pi is\mathcal{F}\mathrm{sgn}=2+sT$$

由之前结论可得

$$T=c^{\prime }\delta$$

因此

$$\mathcal{F}\mathrm{sgn}=\frac{1}{\pi is}+c\delta$$

但一般情形下，我们主要使用

$$\mathcal{F}\mathrm{sgn}=\frac{1}{\pi is}$$

对上式两边取傅里叶变换可得

$$\mathcal{F}\left(\frac{1}{x}\right)=\pi i\mathcal{F}\mathcal{F}\text{sgn }s=-\pi i\text{ sgn }(s)$$

因为

$$H(t)=\frac{1}{2}(1+\mathrm{sgn}t)$$

所以

$$\mathcal{F}H=\frac{1}{2}(\delta +\frac{1}{\pi is})$$

平移

首先给出如下记号

$$\left({\tau }_b\phi \right)(x)=\phi (x-b)$$

因为

$$⟨{\tau }_{b}f,\phi ⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({\tau }_{b}f\right)(x)\phi (x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)\left({\tau }_{-b}\phi \right)(x)dx=⟨f,{\tau }_{-b}\phi ⟩$$

我们按如下方式定义${\tau }_{b}T$

$$⟨{\tau }_{b}T,\phi ⟩=⟨T,{\tau }_{-b}\phi ⟩$$

来看一个具体例子

$$\begin{array}{rl}⟨{\tau }_a\delta ,\phi ⟩& =⟨\delta ,{\tau }_{-a}\phi ⟩\\ & =\left({\tau }_{-a}\phi \right)(0)\\ & =\phi (a)\\ & =⟨{\delta }_a,\phi ⟩\end{array}$$

所以

$${\tau }_a\delta ={\delta }_a$$

接着讨论平移定理

平移定理

$$\mathcal{F}\left({\tau }_{b}T\right)=e^{-2\pi ibx}\mathcal{F}T$$

证明：

首先

$$⟨\mathcal{F}\left({\tau }_{b}T\right),\phi ⟩=⟨{\tau }_{b}T,\mathcal{F}\phi ⟩=⟨T,{\tau }_{-b}\mathcal{F}\phi ⟩$$

其次

$$\begin{array}{rl}{\tau }_{-b}(\mathcal{F}\phi )(s)& =\mathcal{F}\phi (s+b)\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i(s+b)x}\phi (x)dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi isx}{e}^{-2\pi ibx}\phi (x)dx\\ & =\mathcal{F}\left(e^{-2\pi ibx}\phi \right)(s)\end{array}$$

所以

$$\begin{array}{rl}⟨\mathcal{F}\left({\tau }_{b}T\right),\phi ⟩& =⟨T,{\tau }_{-b}\mathcal{F}\phi ⟩\\ & =⟨T,\mathcal{F}\left(e^{-2\pi ibx}\phi \right)⟩\\ & =⟨\mathcal{F}T,e^{-2\pi ibx}\phi ⟩\\ & =⟨e^{-2\pi ibx}\mathcal{F}T,\phi ⟩\end{array}$$

所以

$$\mathcal{F}\left({\tau }_{b}T\right)=e^{-2\pi ibx}\mathcal{F}T$$

可以利用该定理推导$\mathcal{F}{\delta }_a$：

$$\mathcal{F}{\delta }_a=\mathcal{F}{\tau }_a\delta =e^{-2\pi ias}\mathcal{F}\delta =e^{-2\pi isa}$$

伸缩

定义

$$\left({\sigma }_a\phi \right)(x)=\phi (ax)$$

注意到

$$⟨{\sigma }_{a}f\phi ⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({\sigma }_{a}f\right)(x)\phi (x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)\frac{1}{|a|}\left({\sigma }_{1/a}\phi \right)(x)dx=⟨f,\frac{1}{|a|}\left({\sigma }_{1/a}\phi \right)⟩$$

我们按如下方式定义${\sigma }_{a}T$

$$⟨{\sigma }_{a}T,\phi ⟩=⟨T,\frac{1}{|a|}{\sigma }_{1/a}\phi ⟩$$

对$\delta$伸缩：

$$\begin{array}{rl}⟨{\sigma }_a\delta ,\phi ⟩& =⟨\delta ,\frac{1}{|a|}{\sigma }_{1/a}\phi ⟩\\ & =\frac{1}{|a|}\left({\sigma }_{1/a}\phi \right)(0)\\ & =\frac{1}{|a|}\phi (0/a)\\ & =\frac{1}{|a|}\phi (0)\\ & =⟨\frac{1}{|a|}\delta ,\phi ⟩\end{array}$$

所以

$${\sigma }_a\delta =\frac{1}{|a|}\delta$$

接着讨论伸缩定理

伸缩定理

$$\mathcal{F}\left({\sigma }_{a}T\right)=\frac{1}{|a|}{\sigma }_{1/a}(\mathcal{F}T)$$

证明：

首先

$$⟨\mathcal{F}\left({\sigma }_{a}T\right),\phi ⟩=⟨{\sigma }_{a}T,\mathcal{F}\phi ⟩=⟨T,\frac{1}{|a|}{\sigma }_{1/a}\mathcal{F}\phi ⟩$$

其次

$$\frac{1}{|a|}\left({\sigma }_{1/a}\mathcal{F}\phi \right)(s)=\frac{1}{|a|}\mathcal{F}\phi \left(\frac{s}{a}\right)=\mathcal{F}\left({\sigma }_a\phi \right)(s)$$

所以

$$⟨\mathcal{F}\left({\sigma }_{a}T\right),\phi ⟩=⟨T,\frac{1}{|a|}{\sigma }_{1/a}\mathcal{F}\phi ⟩=⟨T,\mathcal{F}\left({\sigma }_a\phi \right)⟩=⟨\mathcal{F}T,{\sigma }_a\phi ⟩=⟨\frac{1}{|a|}{\sigma }_{1/a}(\mathcal{F}T),\phi ⟩$$

即

$$\mathcal{F}\left({\sigma }_{a}T\right)=\frac{1}{|a|}{\sigma }_{1/a}(\mathcal{F}T)$$

卷积

首先考虑如下事实

$$\begin{array}{rl}⟨\psi \ast f,\phi ⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(\psi \ast f)(x)\phi (x)dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\psi (x-y)f(y)dy)\phi (x)dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\psi (x-y)\phi (x)f(y)dydx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\psi (x-y)\phi (x)dx)f(y)dy\\ & =⟨f,{\psi }^{-}\ast \phi ⟩\end{array}$$

由此给出分布$T$和测试函数$\psi$的卷积的定义

$$⟨\psi \ast T,\phi ⟩=⟨T,{\psi }^{-}\ast \phi ⟩$$

首先讨论$\delta$函数的卷积，结论如下

$$\psi \ast \delta =\psi$$

证明

$$\begin{array}{rl}⟨\psi \ast \delta ,\phi ⟩& =⟨\delta ,{\psi }^{-}\ast \phi ⟩\\ & =({\psi }^{-}\ast \phi )(0)\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }^{-}(-y)\phi (y)dy\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\psi (y)\phi (y)dy\\ & =⟨\psi ,\phi ⟩\end{array}$$

接着介绍卷积定理：

卷积定理

$$\begin{array}{rl}\mathcal{F}(\psi \ast T)& =(\mathcal{F}\psi )(\mathcal{F}T)\\ \mathcal{F}(\psi T)& =\mathcal{F}\psi \ast \mathcal{F}T\end{array}$$

证明：

结论1：

$$\begin{array}{rl}⟨\mathcal{F}(\psi \ast T),\phi ⟩& =⟨\psi \ast T,\mathcal{F}\phi ⟩\\ & =⟨T,{\psi }^{-}\ast \mathcal{F}\phi ⟩\\ & =⟨T,\mathcal{F}\mathcal{F}\psi \ast \mathcal{F}\phi ⟩\\ & =⟨T,\mathcal{F}(\mathcal{F}\psi \cdot \phi )⟩\\ & =⟨\mathcal{F}T,\mathcal{F}\psi \cdot \phi ⟩\\ & =⟨(\mathcal{F}\psi )(\mathcal{F}T),\phi ⟩\end{array}$$

从而

$$\mathcal{F}(\psi \ast T)=(\mathcal{F}\psi )(\mathcal{F}T)$$

结论2：

$$\begin{array}{rl}⟨\mathcal{F}(\psi T),\phi ⟩& =⟨\psi T,\mathcal{F}\phi ⟩\\ & =⟨T,\psi \mathcal{F}\phi ⟩\\ & =⟨T,\mathcal{F}\mathcal{F}{\psi }^{-}\mathcal{F}\phi ⟩\\ & =⟨T,\mathcal{F}(\mathcal{F}{\psi }^{-}\ast \phi )⟩\\ & =⟨\mathcal{F}T,(\mathcal{F}\psi {)}^{-}\ast \phi )⟩\\ & =⟨\mathcal{F}\psi \ast \mathcal{F}T,\phi ⟩\end{array}$$

所以

$$\mathcal{F}(\psi T)=\mathcal{F}\psi \ast \mathcal{F}T$$

分布的卷积可以参考课本193页，这里只给出定义

$$⟨S\ast T,\phi ⟩=⟨S(y),⟨T(x),\phi (x+y)⟩⟩$$

例1

$$\delta \ast \delta =\delta$$

证明：

$$\begin{array}{rl}⟨\delta \ast \delta ,\phi ⟩& =⟨\delta (y),⟨\delta (x),\phi (x+y)⟩⟩\\ & =⟨\delta (y),\phi (y)⟩\\ & =\phi (0)\\ & =⟨\delta ,\phi ⟩\end{array}$$

例2

$${\delta }_a\ast {\delta }_b={\delta }_{a+b}$$

证明：

$$\begin{array}{rl}⟨{\delta }_a\ast {\delta }_b,\phi ⟩& =⟨{\delta }_a(y),⟨{\delta }_b(x),\phi (x+y)⟩⟩\\ & =⟨{\delta }_a(y),\phi (b+y)⟩\\ & =\phi (b+a)\\ & =⟨{\delta }_{a+b},\phi ⟩\end{array}$$

卷积的导数

注意到我们有

$$\begin{array}{rl}(f\ast g{)}^{\prime }(x)& =\frac{d}{dx}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(u)g(x-u)du\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(u)\frac{d}{dx}g(x-u)du={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(u)g^{\prime }(x-u)du\\ & =(f\ast g^{\prime })(x)\end{array}$$

上式说明卷积可以让函数变得更光滑。

更一般的，我们有

$$(f\ast g{)}^{(n)}(x)=(f\ast g^{(n)})(x)$$

特别的，我们有

$${\delta }^{(n)}\ast f=f^{(n)}$$

$\delta$函数

$\delta$函数有如下性质

$$\begin{array}{rl}\delta \ast \phi & =\phi \\ \phi \delta & =\phi (0)\delta \end{array}$$

更一般的，我们有

$$\begin{array}{rl}\delta (x-b)\ast \phi (x)& =\phi (x-b)\\ \delta (x-b)\phi (x)& =\phi (b)\delta (x-b)\end{array}$$

滤波

这一部分对之前介绍的滤波进行补充。

低滤波

频域

$$\mathrm{Low}(s)={\mathrm{\Pi }}_{2{\nu }_c}(s)=\mathrm{\Pi }\left(\frac{s}{2{\nu }_c}\right)=\{\begin{array}{ll}1& |s|<{\nu }_c\\ 0& |s|\ge {\nu }_c\end{array}$$

时域

$$\mathrm{low}(t)=2{\nu }_c\mathrm{sinc}\left(2{\nu }_{c}t\right)$$

高滤波

频域

$$\mathrm{High}(s)=1-\mathrm{Low}(s)=1-{\mathrm{\Pi }}_{2{\nu }_c}(s)$$

时域

$$\mathrm{high}(t)=\delta (t)-2{\nu }_c\mathrm{sinc}\left(2{\nu }_{c}t\right)$$

假设输入为$v(t)$，那么

$$\begin{array}{rl}w(t)& =(\mathrm{high}\ast v)(t)\\ & =(\delta (t)-2{\nu }_c\mathrm{sinc}\left(2{\nu }_{c}t\right))\ast v(t)\\ & =v(t)-2{\nu }_c{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{sinc}(2{\nu }_c(t-s))v(s)ds\end{array}$$

Notch滤波

频域

$$\mathrm{Notch}(s)=1-(\mathrm{Low}(s-{\nu }_0)+\mathrm{Low}(s+{\nu }_0))$$

时域

$$\begin{array}{rl}\mathrm{notch}(t)& =\delta (t)-(e^{-2\pi i{\nu }_{0}t}\mathrm{low}(t)+e^{2\pi i{\nu }_{0}t}\mathrm{low}(t))\\ & =\delta (t)-4{\nu }_c\cos \left(2\pi {\nu }_{0}t\right)\mathrm{sinc}\left(2{\nu }_{c}t\right)\end{array}$$

假设输入为$v(t)$，那么

$$\begin{array}{rl}w(t)& =(\delta (t)-4{\nu }_c\cos \left(2\pi {\nu }_{0}t\right)\mathrm{sinc}\left(2{\nu }_{c}t\right))\ast v(t)\\ & =v(t)-4{\nu }_c{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cos (2\pi {\nu }_0(t-s))\mathrm{sinc}(2{\nu }_c(t-s))v(s)ds\end{array}$$