EE261 Lecture 29

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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 $III$ 函数。

回顾 $III$ 函数：

{III}_{p} (t) = \sum_{k = - \infty}^{\infty} δ (t - k p)

这里对上述内容进行推广。

二维 $III$

二维 $III$ 函数为

III (x_{1}, x_{2}) = \sum_{k_{1}, k_{2} = - \infty}^{\infty} δ (x_{1} - k_{1}, x_{2} - k_{2})

定义

\begin{aligned} e_{1} & = (1, 0) \\ e_{2} & = (0, 1) \\ k & = k_{1} e_{1} + k_{2} e_{2} \end{aligned}

那么 $III$ 函数可以简写为

{III}_{Z^{2}} (x) = \sum_{k_{1}, k_{2} = - \infty}^{\infty} δ (x - k_{1} e_{1} - k_{2} e_{2}) = \sum_{k \in Z^{2}} δ (x - k)

利用 $III$ 周期化

考虑

Φ (x) = (φ * {III}_{Z^{2}}) (x) = \sum_{k \in Z^{2}} φ (x - k)

任取 $n \in Z^{2}$ ，那么

Φ (x + n) = \sum_{k \in Z^{2}} φ (x + n - k) = \sum_{k \in Z^{2}} φ (x - k) = Φ (x)

$III$ 的傅里叶变换

和一维情形类似，我们有

F {III}_{Z^{2}} = {III}_{Z^{2}}

证明方法是利用

\sum_{k \in Z^{2}} F ψ (k) = \sum_{k \in Z^{2}} ψ (k)

证明：考虑如下函数

Φ (x) = (φ * Π {II}_{Z^{2}}) (x) = \sum_{n \in Z^{2}} φ (x - n)

由其周期性，考虑其傅里叶系数展开

Φ (x) = \sum_{k \in Z^{2}} \hat{Φ} (k) e^{2 π i k \cdot x}

其中系数为

\begin{aligned} \hat{Φ} (k_{1}, k_{2}) & = \int_{0}^{1} \int_{0}^{1} e^{- 2 π i (k_{1} x_{1} + k_{2} x_{2})} Φ (x_{1}, x_{2}) d x_{1} d x_{2} \\ = \int_{0}^{1} \int_{0}^{1} e^{- 2 π i (k_{1} x_{1} + k_{2} x_{2})} \sum_{n_{1}, n_{2} = - \infty}^{\infty} φ (x_{1} - n_{1}, x_{2} - n_{2}) d x_{1} d x_{2} \\ = \sum_{n_{1}, n_{2} = - \infty}^{\infty} \int_{0}^{1} \int_{0}^{1} e^{- 2 π i (k_{1} x_{1} + k_{2} x_{2})} φ (x_{1} - n_{1}, x_{2} - n_{2}) 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 π i (k_{1} (u + n_{1}) + k_{2} (v + n_{2}))} φ (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 π i (k_{1} n_{1} + k_{2} n_{2})} e^{- 2 π i (k_{1} u + k_{2} v)} φ (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 π i (k_{1} u + k_{2} v)} φ (u, v) d u d v \\ = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- 2 π i (k_{1} u + k_{2} v)} φ (u, v) d u d v \\ = F φ (k_{1}, k_{2}) \end{aligned}

即

Φ (x) = \sum_{k \in Z^{2}} F φ (k) e^{2 π i k \cdot x}

令 $x = 0$ ，我们有

\sum_{k \in Z^{2}} F φ (k) = \sum_{k \in Z^{2}} φ (k)

最后利用分布理论证明上述结论，

⟨ F {III}_{Z^{2}}, ψ ⟩ = ⟨ {III}_{Z^{2}}, F ψ ⟩ = \sum_{k \in Z^{2}} F ψ (k) = \sum_{k \in Z^{2}} ψ (k) = ⟨ {III}_{Z^{2}}, ψ ⟩

格子点上的 $III$

之前讨论的是在整格子点上 $III$ 函数，现在讨论一般情形，首先定义

L = A (Z^{2})

其中 $L$ 的基底为 $u_{1} = A e_{1}, u_{2} = A e_{2}$ ， $A$ 为 $2$ 阶可逆矩阵。

考虑

{III}_{L} (x) = \sum_{p \in L} δ (x - p) = \sum_{k \in Z^{2}} δ (x - A k)

由之前的结论，我们有

δ (x - A k) = δ (A (A^{- 1} x - k)) = \frac{1}{| \det A |} δ (A^{- 1} x - k)

因此

Π_{L} (x) = \frac{1}{| \det A |} {III}_{Z^{2}} (A^{- 1} x)

傅里叶变换

利用

F {III}_{Z^{2}} = {III}_{Z^{2}}

推导一般情形的傅里叶变换。

\begin{aligned} F {III}_{L} (ξ) & = \frac{1}{| \det A |} F ({III}_{Z^{2}} (A^{- 1} x)) \\ = \frac{1}{| \det A |} \frac{1}{| \det A^{- 1} |} F {III}_{Z^{2}} (A^{⊤} ξ) \\ = F {III}_{Z^{2}} (A^{⊤} ξ) \\ = {III}_{Z^{2}} (A^{⊤} ξ) \end{aligned}

采样

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

F f (ξ) = III (ξ_{1}) III (ξ_{2}) (F f * {III}_{Z^{2}}) (ξ)

取逆变换可得

\begin{aligned} f (x) = f (x_{1}, x_{2}) & = (sinc x_{1} sinc x_{2}) * (f (x) \cdot {III}_{Z^{2}} (x)) \\ = (sinc x_{1} sinc x_{2}) * (f (x) \cdot \sum_{k_{1}, k_{2} = - \infty}^{\infty} δ (x - k_{1} e_{1} - k_{2} e_{2})) \\ = (sinc x_{1} sinc x_{2}) * \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1}, k_{2}) δ (x_{1} - k_{1}, x_{2} - k_{2}) \\ = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1}, k_{2}) sinc (x_{1} - k_{1}) sinc (x_{2} - k_{2}) \end{aligned}

带有坐标的形式为

f (x) = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1} e_{1} + k_{2} e_{2}) sinc (x \cdot e_{1} - k_{1}) sinc (x \cdot e_{2} - k_{2})

对于一般的格子，设

\begin{aligned} B & = A^{- ⊤} \\ u_{1}^{*} & = B e_{1} \\ u_{2}^{*} & = B e_{2} \end{aligned}

令

g (x) = f (B x)

那么

F g (ξ) = \frac{1}{| \det B |} F f (B^{- ⊤} ξ) = | \det A | F f (A ξ)

对 $g$ 使用之前的结论可得

g (x) = \sum_{k_{1}, k_{2} = - \infty}^{\infty} g (k_{1} e_{1} + k_{2} e_{2}) sinc (x \cdot e_{1} - k_{1}) sinc (x \cdot e_{2} - k_{2})

令 $y = B x$ ，我们有

\begin{aligned} f (y) & = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (B (k_{1} e_{1} + k_{2} e_{2})) sinc (B^{- 1} y \cdot e_{1} - k_{1}) sinc (B^{- 1} y \cdot e_{2} - k_{2}) \\ = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1} B e_{1} + k_{2} B e_{2}) sinc (A^{⊤} y \cdot e_{1} - k_{1}) sinc (A^{⊤} y \cdot e_{2} - k_{2}) \\ = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1} u_{1}^{*} + k_{2} u_{2}^{*}) sinc (y \cdot A e_{1} - k_{1}) sinc (y \cdot A e_{2} - k_{2}) \\ = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1} u_{1}^{*} + k_{2} u_{2}^{*}) sinc (y \cdot u_{1} - k_{1}) sinc (y \cdot u_{2} - k_{2}) \end{aligned}

更换符号可得

f (x) = \sum_{k_{1}, k_{2} = - \infty}^{\infty} f (k_{1} u_{1}^{*} + k_{2} u_{2}^{*}) sinc (x \cdot u_{1} - k_{1}) sinc (x \cdot u_{2} - k_{2})

二维III

利用III周期化

III的傅里叶变换

格子点上的III

傅里叶变换

采样

微信扫一扫：分享

二维 $III$

利用 $III$ 周期化

$III$ 的傅里叶变换

格子点上的 $III$