余弦型振幅光栅（光学信息处理作业）

复振幅透过率：

$$t(x_0,y_0)=[\frac{1}{2}+\frac{m}{2}cos(2\pi f_0x_0)]\cdot rect(\frac{x_0}{l})rect(\frac{y_0}{l})$$

$m\leq 1$称为光栅的调制度
证明：用单位振幅的单色平面光波垂直照明该光栅，可得光栅的频谱为：

$$T(f_x,f_y)=\frac{l^2}2 sinc(lf_y)\cdot\{sinc(lf_x)+\frac{m}2\cdot sinc[l(f_x+f_0)]+\frac{m}2 \cdot sinc[l(f_x-f_0)]\}$$

证明：
令

$$\begin{align} &f(x,y)= \frac{1}{2}+\frac{m}{2}cos(2\pi f_0x_0)\\ \\ &g(x,y)= rect(\frac{x_0}{l})rect(\frac{y_0}{l})\\ \\ &t(x,y)=f(x,y)\cdot g(x,y) \end{align}$$
分别对f(x)​和g(x)​作傅里叶变换得:
$$\begin{align} &F(f_x,f_y)= \frac{1}2 \delta(f_x,f_y)+\frac{m}4 [\delta(f_x+f_0,f_y)+\delta(f_x-f_0,f_y)]\\ \\ &G(f_x,f_y)= l^2\cdot sinc(lf_x)sinc(lf_y)\\ \\ &T(f_x,f_y)=F(f_x,f_y)\ast G(f_x,f_y) \end{align}$$

任意函数$f(x,y)$与$\delta$函数卷积，结果是函数$f(x,y)$本身

$$\begin{align} f(x,y)\ast \delta(x,y)&=\iint_{-\infty}^\infty f(\xi,\eta)\delta(\xi-x,\eta-y)d\xi d\eta \\ \\ &=f(x,y) \end{align}$$
将上式简单推广得到$f(x,y)\ast \delta(x-x_0,y-y_0)=f(x-x_0,y-y_0)​$，这表明$\delta​$函数平移多少距离，原函数就要平移多少距离。

所以

$$\begin{align} T(f_x,f_y)&=F(f_x,f_y)\ast G(f_x,f_y)\\ \\ &=\{\frac{1}2 \delta(f_x,f_y)+\frac{m}4 [\delta(f_x+f_0,f_y)+\delta(f_x-f_0,f_y)]\}\ast [l^2\cdot sinc(lf_x)sinc(lf_y)]\\ \\ &=\frac{ l^2}{2}\cdot sinc(lf_x)sinc(lf_y)+\frac{m}4\cdot l^2sinc[l(f_x+f_0)]sinc(lf_y)+\frac{m}4\cdot l^2sinc[l(f_x-f_0)]sinc(lf_y)\\ \\ &=\frac{ l^2}{2}\cdot sinc(lf_y)\cdot\{sinc(lf_x)+\frac{m}2\cdot sinc[l(f_x+f_0)]+\frac{m}2\cdot sinc[l(f_x-f_0)]\} \end{align}$$

证明完毕

附件：
1、$\delta(t)$ 的傅里叶变换，由$\delta$函数的筛选特性可得：

$$\int_{-\infty}^\infty \delta(t) e^{-j2\pi ft}dt=1$$

2、$cos(2\pi f_0x)$的傅里叶变换：

$$\begin{align} F[cos(2\pi f_0x)]&=\int_{-\infty}^\infty [cos(2\pi f_0x)]\cdot e^{-j2\pi fx}dx\\ \\ &=\int_{-\infty}^\infty\frac{1}2(e^{-j2\pi f_0x}+e^{j2\pi f_0x})\cdot e^{-j2\pi fx}dx\\ \\ &=\frac{1}2\cdot \int_{-\infty}^\infty e^{-j2\pi (f+f_0)x}+e^{-j2\pi (f-f_0)x}dx\\ \\ &=\frac{1}2\cdot [\delta(f+f_0)+\delta(f-f_0)] \end{align}$$

3、$rect(x)$的傅里叶变换：

$$\begin{align} F[rect(x)]&=\int_{-\infty}^\infty rect(x)\cdot e^{-j2\pi fx}dx\\ \\ &=\int_{\frac{-1}2}^{\frac{1}2} e^{-j2\pi fx}dx\\ \\ &=\frac{e^{-j2\pi fx}}{-j2\pi f}|_{\frac{-1}2}^{\frac{1}2}\\ \\ &=\frac{e^{-j2\pi f}-e^{j2\pi f}}{-j2\pi f}\\ \\ &=\frac{sin(\pi f)}{\pi f}\\ \\ &=sinc(f) \end{align}$$