EE261 Lecture 29

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

这次回顾第二十九讲，这一讲介绍了高维$\text{III}$函数。

回顾$\text{III}$函数：

$$\text{III}_{p}(t)=\sum_{k=-\infty}^{\infty} \delta(t-k p)$$

这里对上述内容进行推广。

二维$\text{III}$

二维$\text{III}$函数为

$$\text{III}\left(x_{1}, x_{2}\right)=\sum_{k_{1}, k_{2}=-\infty}^{\infty} \delta\left(x_{1}-k_{1}, x_{2}-k_{2}\right)$$

定义

$$\begin{aligned} \mathbf{e}_{1}&=(1,0)\\ \mathbf{e}_{2}&=(0,1) \\ \mathbf{k}&=k_{1} \mathbf{e}_{1}+k_{2} \mathbf{e}_{2} \end{aligned}$$

那么$\text{III}$函数可以简写为

$$\text{III}_{\mathbf{Z}^{2}}(\mathbf{x})=\sum_{k_{1}, k_{2}=-\infty}^{\infty} \delta\left(\mathbf{x}-k_{1} \mathbf{e}_{1}-k_{2} \mathbf{e}_{2}\right)=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \delta(\mathbf{x}-\mathbf{k})$$

利用$\text{III}$周期化

考虑

$$\Phi(\mathrm{x})=\left(\varphi * \mathrm{III}_{\mathrm{Z}^{2}}\right)(\mathrm{x})=\sum_{\mathrm{k} \in \mathrm{Z}^{2}} \varphi(\mathrm{x}-\mathrm{k})$$

任取$\mathbf{n} \in \mathbf{Z}^{2}$，那么

$$\Phi(\mathrm{x}+\mathrm{n})=\sum_{\mathrm{k} \in \mathrm{Z}^{2}} \varphi(\mathrm{x}+\mathrm{n}-\mathrm{k})=\sum_{\mathrm{k} \in \mathrm{Z}^{2}} \varphi(\mathrm{x}-\mathrm{k})=\Phi(\mathrm{x})$$

$\text{III}$的傅里叶变换

和一维情形类似，我们有

$$\mathcal{F} \text{III}_{\mathbf{Z}^{2}}=\text{III}_{\mathbf{Z}^{2}}$$

证明方法是利用

$$\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \mathcal{F} \psi(\mathbf{k})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \psi(\mathbf{k})$$

证明：考虑如下函数

$$\Phi(\mathrm{x})=\left(\varphi * \Pi \mathrm{II}_{\mathrm{Z}^{2}}\right)(\mathrm{x})=\sum_{\mathbf{n} \in \mathrm{Z}^{2}} \varphi(\mathrm{x}-\mathrm{n})$$

由其周期性，考虑其傅里叶系数展开

$$\Phi(\mathbf{x})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \widehat{\Phi}(\mathbf{k}) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}$$

其中系数为

$$\begin{aligned} \widehat{\Phi}\left(k_{1}, k_{2}\right) &=\int_{0}^{1} \int_{0}^{1} e^{-2 \pi i\left(k_{1} x_{1}+k_{2} x_{2}\right)} \Phi\left(x_{1}, x_{2}\right) d x_{1} d x_{2} \\ &=\int_{0}^{1} \int_{0}^{1} e^{-2 \pi i\left(k_{1} x_{1}+k_{2} x_{2}\right)} \sum_{n_{1}, n_{2}=-\infty}^{\infty} \varphi\left(x_{1}-n_{1}, x_{2}-n_{2}\right) d x_{1} d x_{2} \\ &=\sum_{n_{1}, n_{2}=-\infty}^{\infty} \int_{0}^{1} \int_{0}^{1} e^{-2 \pi i\left(k_{1} x_{1}+k_{2} x_{2}\right)} \varphi\left(x_{1}-n_{1}, x_{2}-n_{2}\right) d x_{1} d x_{2}\\ &=\sum_{n_{1}, n_{2}=-\infty}^{\infty} \int_{-n_{1}}^{1-n_{1}} \int_{-n_{2}}^{1-n_{2}} e^{-2 \pi i\left(k_{1}\left(u+n_{1}\right)+k_{2}\left(v+n_{2}\right))\right.} \varphi(u, v) d u d v \\ &=\sum_{n_{1}, n_{2}=-\infty}^{\infty} \int_{-n_{1}}^{1-n_{1}} \int_{-n_{2}}^{1-n_{2}} e^{-2 \pi i\left(k_{1} n_{1}+k_{2} n_{2}\right)} e^{-2 \pi i\left(k_{1} u+k_{2} v\right)} \varphi(u, v) d u d v \\ &=\sum_{n_{1}, n_{2}=-\infty}^{\infty} \int_{-n_{1}}^{1-n_{1}} \int_{-n_{2}}^{1-n_{2}} e^{-2 \pi i\left(k_{1} u+k_{2} v\right)} \varphi(u, v) d u d v \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2 \pi i\left(k_{1} u+k_{2} v\right)} \varphi(u, v) d u d v \\ &=\mathcal{F} \varphi\left(k_{1}, k_{2}\right) \end{aligned}$$

即

$$\Phi(\mathbf{x})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \mathcal{F} \varphi(\mathbf{k}) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}$$

令$\mathbf{x} = 0$，我们有

$$\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \mathcal{F} \varphi(\mathbf{k})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \varphi(\mathbf{k})$$

最后利用分布理论证明上述结论，

$$\left\langle\mathcal{F} \text{III}_{\mathbf{Z}^{2}}, \psi\right\rangle=\left\langle \text{III}_{\mathbf{Z}^{2}}, \mathcal{F} \psi\right\rangle=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \mathcal{F} \psi(\mathbf{k})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \psi(\mathbf{k})=\left\langle\text{III}_{\mathbf{Z}^{2}}, \psi\right\rangle$$

格子点上的$\text{III}$

之前讨论的是在整格子点上$\text{III}$函数，现在讨论一般情形，首先定义

$$\mathcal{L}=A\left(\mathbf{Z}^{2}\right)$$

其中$\mathcal{L}$的基底为$u_1=A \mathbf{e}_{1},u_2=A \mathbf{e}_{2}$，$A$为$2$阶可逆矩阵。

考虑

$$\text{III}_{\mathcal{L}}(\mathbf{x})=\sum_{\mathbf{p} \in \mathcal{L}} \delta(\mathbf{x}-\mathbf{p})=\sum_{\mathbf{k} \in \mathbf{Z}^{2}} \delta(\mathbf{x}-A \mathbf{k})$$

由之前的结论，我们有

$$\delta(\mathbf{x}-A \mathbf{k})=\delta\left(A\left(A^{-1} \mathbf{x}-\mathbf{k}\right)\right)=\frac{1}{|\operatorname{det} A|} \delta\left(A^{-1} \mathbf{x}-\mathbf{k}\right)$$

因此

$$\Pi_{\mathcal{L}}(\mathbf{x})=\frac{1}{|\operatorname{det} A|} \text{III}_{\mathbf{Z}^{2}}\left(A^{-1} \mathbf{x}\right)$$

傅里叶变换

利用

$$\mathcal{F} \text{III}_{\mathbf{Z}^{2}}=\text{III}_{\mathbf{Z}^{2}}$$

推导一般情形的傅里叶变换。

$$\begin{aligned} \mathcal{F} \operatorname{III}_{\mathcal{L}}(\boldsymbol{\xi}) &=\frac{1}{|\operatorname{det} A|} \mathcal{F}\left(\operatorname{III}_{\mathrm{Z}^{2}}\left(A^{-1} \mathbf{x}\right)\right) \\ &=\frac{1}{|\operatorname{det} A|} \frac{1}{\left|\operatorname{det} A^{-1}\right|} \mathcal{F} \text{III}_{\mathrm{Z}^{2}}\left(A^{\top} \boldsymbol{\xi}\right) \\ &=\mathcal{F} \operatorname{III}_{\mathbf{Z}^{2}}\left(A^{\top} \boldsymbol{\xi}\right) \\ &=\text{III}_{\mathbf{Z}^{2}}\left(A^{\top} \boldsymbol{\xi}\right) \end{aligned}$$

采样

如上图所示，假设$\mathcal{F} f(\xi)$在$2$维单位方格以外为$0$，类似一维情形，我们有

$$\mathcal{F} f(\boldsymbol{\xi})=\text{III}\left(\xi_{1}\right) \text{III}\left(\xi_{2}\right)\left(\mathcal{F} f * \text{III}_{\mathbf{Z}^{2}}\right)(\boldsymbol{\xi})$$

取逆变换可得

$$\begin{aligned} f(\mathrm{x})=f\left(x_{1}, x_{2}\right) &=\left(\operatorname{sinc} x_{1} \operatorname{sinc} x_{2}\right) *\left(f(\mathrm{x}) \cdot \text{III}_{\mathrm{Z}^{2}}(\mathrm{x})\right) \\ &=\left(\operatorname{sinc} x_{1} \operatorname{sinc} x_{2}\right) *\left(f(\mathrm{x}) \cdot \sum_{k_{1}, k_{2}=-\infty}^{\infty} \delta\left(\mathrm{x}-k_{1} \mathrm{e}_{1}-k_{2} \mathrm{e}_{2}\right)\right) \\ &=\left(\operatorname{sinc} x_{1} \operatorname{sinc} x_{2}\right) * \sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1}, k_{2}\right) \delta\left(x_{1}-k_{1}, x_{2}-k_{2}\right) \\ &=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1}, k_{2}\right) \operatorname{sinc}\left(x_{1}-k_{1}\right) \operatorname{sinc}\left(x_{2}-k_{2}\right) \end{aligned}$$

带有坐标的形式为

$$f(\mathrm{x})=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1} \mathrm{e}_{1}+k_{2} \mathrm{e}_{2}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{e}_{1}-k_{1}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{e}_{2}-k_{2}\right)$$

对于一般的格子，设

$$\begin{aligned} B&=A^{-\top}\\ \mathbf{u}_{1}^{*}&=B \mathbf{e}_{1}\\ \mathbf{u}_{2}^{*}&=B \mathbf{e}_{2} \end{aligned}$$

令

$$g(\mathrm{x})=f(B \mathrm{x})$$

那么

$$\mathcal{F} g(\xi)=\frac{1}{|\operatorname{det} B|} \mathcal{F} f\left(B^{-\top} \xi\right)=|\operatorname{det} A| \mathcal{F} f(A \xi)$$

对$g$使用之前的结论可得

$$g(\mathrm{x})=\sum_{k_{1}, k_{2}=-\infty}^{\infty} g\left(k_{1} \mathrm{e}_{1}+k_{2} \mathrm{e}_{2}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{e}_{1}-k_{1}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{e}_{2}-k_{2}\right)$$

令$\mathrm{y}=B \mathrm{x}$，我们有

$$\begin{aligned} f(\mathbf{y}) &=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(B\left(k_{1} \mathbf{e}_{1}+k_{2} \mathbf{e}_{2}\right)\right) \operatorname{sinc}\left(B^{-1} \mathbf{y} \cdot \mathbf{e}_{1}-k_{1}\right) \operatorname{sinc}\left(B^{-1} \mathbf{y} \cdot \mathbf{e}_{2}-k_{2}\right) \\ &=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1} B \mathbf{e}_{1}+k_{2} B \mathbf{e}_{2}\right) \operatorname{sinc}\left(A^{\top} \mathbf{y} \cdot \mathbf{e}_{1}-k_{1}\right) \operatorname{sinc}\left(A^{\top} \mathbf{y} \cdot \mathbf{e}_{2}-k_{2}\right)\\ &=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1} \mathrm{u}_{1}^{*}+k_{2} \mathrm{u}_{2}^{*}\right) \operatorname{sinc}\left(\mathrm{y} \cdot A \mathrm{e}_{1}-k_{1}\right) \operatorname{sinc}\left(\mathrm{y} \cdot A \mathrm{e}_{2}-k_{2}\right) \\ &=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1} \mathrm{u}_{1}^{*}+k_{2} \mathrm{u}_{2}^{*}\right) \operatorname{sinc}\left(\mathrm{y} \cdot \mathrm{u}_{1}-k_{1}\right) \operatorname{sinc}\left(\mathrm{y} \cdot \mathrm{u}_{2}-k_{2}\right) \end{aligned}$$

更换符号可得

$$f(\mathrm{x})=\sum_{k_{1}, k_{2}=-\infty}^{\infty} f\left(k_{1} \mathrm{u}_{1}^{*}+k_{2} \mathrm{u}_{2}^{*}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{u}_{1}-k_{1}\right) \operatorname{sinc}\left(\mathrm{x} \cdot \mathrm{u}_{2}-k_{2}\right)$$