EE261 Lecture 14

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

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

其中

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

现在考虑分布的情形：

$$\begin{aligned} \left\langle\mathcal{F} T^{\prime}, \varphi\right\rangle &=\left\langle T^{\prime}, \mathcal{F} \varphi\right\rangle\\ &=-\left\langle T,(\mathcal{F} \varphi)^{\prime}\right\rangle \\ &=-\langle T, \mathcal{F}(-2 \pi i s \varphi)\rangle \quad \text { (from the second formula above) } \\ &=-\langle\mathcal{F} T,-2 \pi i s \varphi\rangle \quad \text { (moving } \mathcal{F} \text { back over to } T ) \\ &=\langle 2 \pi i s \mathcal{F} T, \varphi\rangle \end{aligned}$$

所以

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

类似的，我们有

$$\begin{aligned} \left\langle(\mathcal{F} T)^{\prime}, \varphi\right\rangle &=-\left\langle\mathcal{F} T, \varphi^{\prime}\right\rangle\\ &=-\left\langle T, \mathcal{F}\left(\varphi^{\prime}\right)\right\rangle \\ &=-\langle T, 2 \pi i s \mathcal{F} \varphi\rangle \quad \text { (from the first formula for functions) } \\ &=\langle- 2 \pi i s T, \mathcal{F} \varphi\rangle \\ &=\langle\mathcal{F}(-2 \pi i s T), \varphi\rangle \end{aligned}$$

所以

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

例子

考虑$\text{sign}’$

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

另一方面，由导数公式可得

$$\mathcal{F} \operatorname{sgn}^{\prime}=2 \pi i s \mathcal{F} \operatorname{sgn}$$

因此

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

实际上我们有更强的结论，原因如下。

注意到如果$sT=0$，那么

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

由之前结论可得

$$T=c' \delta$$

因此

$$\mathcal{F} \operatorname{sgn}=\frac{1}{\pi i s}+c \delta$$

但一般情形下，我们主要使用

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

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

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

所以

$$\mathcal{F} H=\frac{1}{2}\left(\delta+\frac{1}{\pi i s}\right)$$

平移

首先给出如下记号

$$\left(\tau_{b} \varphi\right)(x)=\varphi(x-b)$$

因为

$$\left\langle\tau_{b} f, \varphi\right\rangle=\int_{-\infty}^{\infty}\left(\tau_{b} f\right)(x) \varphi(x) d x=\int_{-\infty}^{\infty} f(x)\left(\tau_{-b} \varphi\right)(x) d x=\left\langle f, \tau_{-b} \varphi\right\rangle$$

我们按如下方式定义$\tau_b T$

$$\left\langle\tau_{b} T, \varphi\right\rangle=\left\langle T, \tau_{-b} \varphi\right\rangle$$

来看一个具体例子

$$\begin{aligned}\left\langle\tau_{a} \delta, \varphi\right\rangle &=\left\langle\delta, \tau_{-a} \varphi\right\rangle \\ &=\left(\tau_{-a} \varphi\right)(0) \\ &=\varphi(a) \\ &=\left\langle\delta_{a}, \varphi\right\rangle \end{aligned}$$

所以

$$\tau_{a} \delta=\delta_{a}$$

接着讨论平移定理

平移定理

$$\mathcal{F}\left(\tau_{b} T\right)=e^{-2 \pi i b x} \mathcal{F} T$$

证明：

首先

$$\left\langle\mathcal{F}\left(\tau_{b} T\right), \varphi\right\rangle=\left\langle\tau_{b} T, \mathcal{F} \varphi\right\rangle=\left\langle T, \tau_{-b} \mathcal{F} \varphi\right\rangle$$

其次

$$\begin{aligned} \tau_{-b}(\mathcal{F} \varphi)(s) &=\mathcal{F} \varphi(s+b) \\ &=\int_{-\infty}^{\infty} e^{-2 \pi i(s+b) x} \varphi(x) d x \\ &=\int_{-\infty}^{\infty} e^{-2 \pi i s x} e^{-2 \pi i b x} \varphi(x) d x\\ &=\mathcal{F}\left(e^{-2 \pi i b x} \varphi\right)(s) \end{aligned}$$

所以

$$\begin{aligned}\left\langle\mathcal{F}\left(\tau_{b} T\right), \varphi\right\rangle &=\left\langle T, \tau_{-b} \mathcal{F} \varphi\right\rangle \\ &=\left\langle T, \mathcal{F}\left(e^{-2 \pi i b x} \varphi\right)\right\rangle \\ &=\left\langle\mathcal{F} T, e^{-2 \pi i b x} \varphi\right\rangle\\ &=\left\langle e^{-2 \pi i b x} \mathcal{F} T, \varphi\right\rangle \end{aligned}$$

所以

$$\mathcal{F}\left(\tau_{b} T\right)=e^{-2 \pi i b x} \mathcal{F} T$$

可以利用该定理推导$\mathcal F \delta_a$：

$$\mathcal F \delta_a=\mathcal{F} \tau_{a} \delta=e^{-2 \pi i a s} \mathcal{F} \delta=e^{-2 \pi i s a}$$

伸缩

定义

$$\left(\sigma_{a} \varphi\right)(x)=\varphi(a x)$$

注意到

$$\left\langle\sigma_{a} f \varphi\right\rangle=\int_{-\infty}^{\infty}\left(\sigma_{a} f\right)(x) \varphi(x) d x=\int_{-\infty}^{\infty} f(x) \frac{1}{|a|}\left(\sigma_{1 / a} \varphi\right)(x) d x=\left\langle f, \frac{1}{|a|}\left(\sigma_{1 / a} \varphi\right)\right\rangle$$

我们按如下方式定义$\sigma_a T$

$$\left\langle\sigma_{a} T, \varphi\right\rangle=\left\langle T, \frac{1}{|a|} \sigma_{1 / a} \varphi\right\rangle$$

对$\delta$伸缩：

$$\begin{aligned} \left\langle\sigma_{a} \delta, \varphi\right\rangle &=\left\langle\delta, \frac{1}{|a|} \sigma_{1 / a} \varphi\right\rangle\\ &=\frac{1}{|a|}\left(\sigma_{1 / a} \varphi\right)(0) \\ &=\frac{1}{|a|} \varphi(0 / a)\\ &=\frac{1}{|a|} \varphi(0)\\ &=\left\langle\frac{1}{|a|} \delta, \varphi\right\rangle \end{aligned}$$

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

$$\left\langle\mathcal{F}\left(\sigma_{a} T\right), \varphi\right\rangle=\left\langle\sigma_{a} T, \mathcal{F} \varphi\right\rangle=\left\langle T, \frac{1}{|a|} \sigma_{1 / a} \mathcal{F} \varphi\right\rangle$$

其次

$$\frac{1}{|a|}\left(\sigma_{1 / a} \mathcal{F} \varphi\right)(s)=\frac{1}{|a|} \mathcal{F} \varphi\left(\frac{s}{a}\right)=\mathcal{F}\left(\sigma_{a} \varphi\right)(s)$$

所以

$$\left\langle\mathcal{F}\left(\sigma_{a} T\right), \varphi\right\rangle=\left\langle T, \frac{1}{|a|} \sigma_{1 / a} \mathcal{F} \varphi\right\rangle=\left\langle T, \mathcal{F}\left(\sigma_{a} \varphi\right)\right\rangle=\left\langle\mathcal{F} T, \sigma_{a} \varphi\right\rangle=\left\langle\frac{1}{|a|} \sigma_{1 / a}(\mathcal{F} T), \varphi\right\rangle$$

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

$$\begin{aligned}\langle\psi * f, \varphi\rangle &=\int_{-\infty}^{\infty}(\psi * f)(x) \varphi(x) d x \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} \psi(x-y) f(y) d y\right) \varphi(x) d x \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \psi(x-y) \varphi(x) f(y) d y d x \\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} \psi(x-y) \varphi(x) d x\right) f(y) d y\\ &= \left\langle f, \psi^{-} * \varphi\right\rangle \end{aligned}$$

由此给出分布$T$和测试函数$\psi$的卷积的定义

$$\langle\psi * T, \varphi\rangle=\left\langle T, \psi^{-} * \varphi\right\rangle$$

首先讨论$\delta$函数的卷积，结论如下

$$\psi * \delta=\psi$$

证明

$$\begin{aligned}\langle\psi * \delta, \varphi\rangle &=\left\langle\delta, \psi^{-} * \varphi\right\rangle\\ &=\left(\psi^{-} * \varphi\right)(0) \\ &=\int_{-\infty}^{\infty} \psi^{-}(-y) \varphi(y) d y\\ &=\int_{-\infty}^{\infty} \psi(y) \varphi(y) d y\\ &=\langle\psi, \varphi\rangle \end{aligned}$$

接着介绍卷积定理：

卷积定理

$$\begin{aligned} \mathcal{F}(\psi * T)&=(\mathcal{F} \psi)(\mathcal{F} T) \\ \mathcal{F}(\psi T)&=\mathcal{F} \psi * \mathcal{F} T \end{aligned}$$

证明：

结论1：

$$\begin{aligned}\langle\mathcal{F}(\psi * T), \varphi\rangle &=\langle\psi * T, \mathcal{F} \varphi\rangle\\ &=\left\langle T, \psi^{-} * \mathcal{F} \varphi\right\rangle \\ &=\langle T, \mathcal{F} \mathcal{F} \psi * \mathcal{F} \varphi\rangle \\ &=\langle T, \mathcal{F}(\mathcal{F} \psi \cdot \varphi)\rangle \\ &=\langle\mathcal{F} T, \mathcal{F} \psi \cdot \varphi\rangle\\ &=\langle(\mathcal{F} \psi)(\mathcal{F} T), \varphi\rangle \end{aligned}$$

从而

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

结论2：

$$\begin{aligned}\langle\mathcal{F}(\psi T), \varphi\rangle &=\langle\psi T, \mathcal{F} \varphi\rangle\\ &=\left\langle T, \psi \mathcal{F} \varphi\right\rangle \\ &=\langle T, \mathcal{F} \mathcal{F} \psi^{-} \mathcal{F} \varphi\rangle \\ &=\langle T, \mathcal F(\mathcal{F} \psi^{-} * \varphi)\rangle \\ &=\langle \mathcal F T, (\mathcal{F} \psi)^{-} * \varphi)\rangle \\ &=\langle \mathcal{F} \psi * \mathcal FT, \varphi\rangle \end{aligned}$$

所以

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

分布的卷积可以参考课本193页，这里只给出定义

$$\langle S * T, \varphi\rangle=\langle S(y),\langle T(x), \varphi(x+y)\rangle\rangle$$

例1

$$\delta * \delta=\delta$$

证明：

$$\begin{aligned}\langle\delta * \delta, \varphi\rangle &=\langle\delta(y),\langle\delta(x), \varphi(x+y)\rangle\rangle \\ &=\langle\delta(y), \varphi(y)\rangle\\ &=\varphi(0)\\ &=\langle\delta, \varphi\rangle \end{aligned}$$

例2

$$\delta_{a} * \delta_{b}=\delta_{a+b}$$

证明：

$$\begin{aligned}\left\langle\delta_{a} * \delta_{b}, \varphi\right\rangle &=\left\langle\delta_{a}(y),\left\langle\delta_{b}(x), \varphi(x+y)\right\rangle\right\rangle \\ &=\left\langle\delta_{a}(y), \varphi(b+y)\right\rangle\\ &=\varphi(b+a)\\ &=\left\langle\delta_{a+b}, \varphi\right\rangle \end{aligned}$$

卷积的导数

注意到我们有

$$\begin{aligned}(f * g)^{\prime}(x) &=\frac{d}{d x} \int_{-\infty}^{\infty} f(u) g(x-u) d u \\ &=\int_{-\infty}^{\infty} f(u) \frac{d}{d x} g(x-u) d u=\int_{-\infty}^{\infty} f(u) g^{\prime}(x-u) d u\\ &=\left(f * g^{\prime}\right)(x) \end{aligned}$$

上式说明卷积可以让函数变得更光滑。

更一般的，我们有

$$(f * g)^{(n)}(x)=\left(f * g^{(n)}\right)(x)$$

特别的，我们有

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

$\delta $函数

$\delta$函数有如下性质

$$\begin{aligned} \delta * \varphi &=\varphi\\ \varphi \delta &=\varphi(0) \delta \end{aligned}$$

更一般的，我们有

$$\begin{aligned} \delta(x-b)* \varphi(x) &=\varphi(x-b)\\ \delta(x-b)\varphi(x) &=\varphi(b) \delta(x-b) \end{aligned}$$

滤波

这一部分对之前介绍的滤波进行补充。

低滤波

频域

$$\operatorname{Low}(s)=\Pi_{2 \nu_{c}}(s)=\Pi\left(\frac{s}{2 \nu_{c}}\right)=\left\{\begin{array}{ll}{1} & {|s|<\nu_{c}} \\ {0} & {|s| \geq \nu_{c}}\end{array}\right.$$

时域

$$\operatorname{low}(t)=2 \nu_{c} \operatorname{sinc}\left(2 \nu_{c} t\right)$$

高滤波

频域

$$\operatorname{High}(s)=1-\operatorname{Low}(s)=1-\Pi_{2 \nu_{c}}(s)$$

时域

$$\operatorname{high}(t)=\delta(t)-2 \nu_{c} \operatorname{sinc}\left(2 \nu_{c} t\right)$$

假设输入为$v(t)$，那么

$$\begin{aligned} w(t) &=(\operatorname{high} * v)(t) \\ &=\left(\delta(t)-2 \nu_{c} \operatorname{sinc}\left(2 \nu_{c} t\right)\right) * v(t) \\ &=v(t)-2 \nu_{c} \int_{-\infty}^{\infty} \operatorname{sinc}\left(2 \nu_{c}(t-s)\right) v(s) d s \end{aligned}$$

Notch滤波

频域

$$\operatorname{Notch}(s)=1-\left(\operatorname{Low}\left(s-\nu_{0}\right)+\operatorname{Low}\left(s+\nu_{0}\right)\right)$$

时域

$$\begin{aligned} \operatorname{notch}(t) &=\delta(t)-\left(e^{-2 \pi i \nu_{0} t} \operatorname{low}(t)+e^{2 \pi i \nu_{0} t} \operatorname{low}(t)\right) \\ &=\delta(t)-4 \nu_{c} \cos \left(2 \pi \nu_{0} t\right) \operatorname{sinc}\left(2 \nu_{c} t\right) \end{aligned}$$

假设输入为$v(t)$，那么

$$\begin{aligned} w(t) &=\left(\delta(t)-4 \nu_{c} \cos \left(2 \pi \nu_{0} t\right) \operatorname{sinc}\left(2 \nu_{c} t\right)\right) * v(t) \\ &=v(t)-4 \nu_{c} \int_{-\infty}^{\infty} \cos \left(2 \pi \nu_{0}(t-s)\right) \operatorname{sinc}\left(2 \nu_{c}(t-s)\right) v(s) d s \end{aligned}$$