余弦型振幅光栅（光学信息处理作业）

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#傅里叶变换 #光学信息处理 #余弦型振幅光栅

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本文详细解析了复振幅透过率的数学表达式，并通过傅里叶变换推导出光栅的频谱公式。利用单位振幅的单色平面光波照明光栅，展示了如何计算光栅的频谱分布。

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复振幅透过率：

t (x 0, y 0) = [1 2 + m 2 c o s (2 π f 0 x 0)] \cdot r e c t (x 0 l) r e c t (y 0 l)

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

$m\leq 1$ 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) = l 2 2 s i n c (l f y) \cdot {s i n c (l f x) + m 2 \cdot s i n c [l (f x + f 0)] + m 2 \cdot s i n c [l (f x - f 0)]}

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

证明：
令

f (x, y) = 1 2 + m 2 c o s (2 π f 0 x 0) g (x, y) = r e c t (x 0 l) r e c t (y 0 l) t (x, y) = f (x, y) \cdot g (x, y) (1) (2) (3) (4) (5)

$\begin{align} &f(x,y)= \frac{1}{2}+\frac{m}{2}cos(2\pi f_0x_0)\\ \\ &g(x,y)= rect(\frac{x_0}{l})rect(\frac{y_0}{l})\\ \\ &t(x,y)=f(x,y)\cdot g(x,y) \end{align}$
分别对f(x)和g(x)作傅里叶变换得:

F (f x, f y) = 1 2 δ (f x, f y) + m 4 [δ (f x + f 0, f y) + δ (f x - f 0, f y)] G (f x, f y) = l 2 \cdot s i n c (l f x) s i n c (l f y) T (f x, f y) = F (f x, f y) * G (f x, f y) (6) (7) (8) (9) (10)

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

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

f (x, y) * δ (x, y) = \iint \infty - \infty f (ξ, η) δ (ξ - x, η - y) d ξ d η = f (x, y) (11) (12) (13)

$\begin{align} f(x,y)\ast \delta(x,y)&=\iint_{-\infty}^\infty f(\xi,\eta)\delta(\xi-x,\eta-y)d\xi d\eta \\ \\ &=f(x,y) \end{align}$
将上式简单推广得到

f(x,y)∗δ(x−x0,y−y0)=f(x−x0,y−y0)f(x,y)∗δ(x−x0,y−y0)=f(x−x0,y−y0) $f(x,y)\ast \delta(x-x_0,y-y_0)=f(x-x_0,y-y_0)$ ，这表明

δδ $\delta$ 函数平移多少距离，原函数就要平移多少距离。

所以

T (f x, f y) = F (f x, f y) * G (f x, f y) = {1 2 δ (f x, f y) + m 4 [δ (f x + f 0, f y) + δ (f x - f 0, f y)]} * [l 2 \cdot s i n c (l f x) s i n c (l f y)] = l 2 2 \cdot s i n c (l f x) s i n c (l f y) + m 4 \cdot l 2 s i n c [l (f x + f 0)] s i n c (l f y) + m 4 \cdot l 2 s i n c [l (f x - f 0)] s i n c (l f y) = l 2 2 \cdot s i n c (l f y) \cdot {s i n c (l f x) + m 2 \cdot s i n c [l (f x + f 0)] + m 2 \cdot s i n c [l (f x - f 0)]} (14) (15) (16) (17) (18) (19) (20)

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

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 δ (t) e - j 2 π f t d t = 1

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

2、 $cos(2\pi f_0x)$ 的傅里叶变换：

F [c o s (2 π f 0 x)] = \int \infty - \infty [c o s (2 π f 0 x)] \cdot e - j 2 π f x d x = \int \infty - \infty 1 2 (e - j 2 π f 0 x + e j 2 π f 0 x) \cdot e - j 2 π f x d x = 1 2 \cdot \int \infty - \infty e - j 2 π (f + f 0) x + e - j 2 π (f - f 0) x d x = 1 2 \cdot [δ (f + f 0) + δ (f - f 0)] (21) (22) (23) (24) (25) (26) (27)

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

3、 $rect(x)$ 的傅里叶变换：

F [r e c t (x)] = \int \infty - \infty r e c t (x) \cdot e - j 2 π f x d x = \int 1 2 - 1 2 e - j 2 π f x d x = e - j 2 π f x - j 2 π f | 1 2 - 1 2 = e - j 2 π f - e j 2 π f - j 2 π f = s i n ( π f ) π f = s i n c (f) (28) (29) (30) (31) (32) (33) (34) (35) (36) (37) (38)

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