傅立叶光学 [傅立叶变换]--二维傅立叶变换简介

最编程 2024-03-01 11:16:13

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三角傅里叶级数

$\begin{array}{c} g\left(x\right) = \frac{a_0}{2}+\sum\limits_{n=1}^\infty \left(a_n\cos2\pi nf_0x+b_n\sin2\pi nf_0x\right)\\[2mm] f_0=\frac{1}{\tau}\\[2mm] a_n = \frac{2}{\tau} \int_0^\tau g\left(x\right)\cos 2\pi nf_0xdx\\[2mm] b_n = \frac{2}{\tau} \int_0^\tau g\left(x\right)\sin 2\pi nf_0xdx \end{array}$

指数傅里叶级数

$\begin{array}{c} g\left(x\right) = \sum\limits_{n=-\infty}^\infty c_n\exp\left(j2\pi nf_0x\right)\\[2mm] f_0=\frac{1}{\tau}\\[2mm] c_n = \frac{1}{\tau}\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} g\left(x\right)\exp\left(-j2\pi nf_0x\right)dx \end{array}$

三角傅里叶系数和指数傅里叶系数的关系：
$c_0 = \frac{a_0}{2} \qquad c_n = \frac{a_n-jb_n}{2} \qquad c_{-n} = \frac{a_n+jb_n}{2}$

傅里叶变换及逆变换

$G\left(f\right) = \int_{-\infty }^\infty g\left(x\right)\exp \left(-j2\pi fx\right)dx\\[2mm] g\left(x\right) = \int_{-\infty }^\infty G\left(f\right)\exp \left(j2\pi fx\right)df$

二维傅里叶变换及逆变换

$\begin{array}{c} F\left (f_x,f_y\right ) = \mathscr{F}\left\{f\left(x,y\right)\right\} = \iint\limits_{-\infty }^\infty f\left(x,y\right) \exp \left[-j2\pi \left(f_xx+f_yy\right)\right]dxdy\\[2mm] f\left(x,y\right) =\mathscr{F}^{-1}\left\{F\left (f_x,f_y\right )\right\} = \iint\limits_{-\infty }^\infty F\left(f_x,f_y\right) \exp \left[j2\pi \left(f_xx+f_yy\right)\right]df_xdf_y \end{array}$

积分变换： $F\left(x\right) =\int_{-\infty}^\infty f\left(\alpha\right)K\left(\alpha,x\right)d\alpha$ 的变换核为 $K$ ，因此，傅里叶变换的核为 $\exp\left(-j2\pi fx\right)$ 。

极坐标系下的二维傅里叶变换

极坐标变换为：
$\begin{array}{l} 空域 & \left|\qquad\begin{array}{l} r = \sqrt{x^2+y^2} &&&\;\; x = r\cos\theta\\ \theta = \tan^{-1}\frac{y}{x} &&&\;\; y = r\sin\theta\\ \end{array}\right.\\[2mm] 频域 & \left|\qquad\begin{array}{l} \rho = \sqrt{f_x^2+f_y^2}\qquad & f_x = \rho\cos\phi\\ \phi = \tan^{-1}\frac{f_y}{f_x} \qquad & f_y = \rho\sin\phi\\ \end{array}\right. \end{array}$

令 $G\left(\rho,\phi\right) = F\left(\rho\cos\phi,\rho\sin\phi\right)$ ， $g\left(r,\theta\right) = f\left(r\cos\theta,r\sin\theta\right)$ ，那么傅里叶变换为：
$G\left(\rho,\phi\right) = \int_0^{2\pi}d\theta\int_0^\infty g\left(r,\theta\right) \exp\left[-j2\pi r\rho\cos\left(\theta-\phi\right)\right]rdr\\[2mm] g\left(r,\theta\right) = \int_0^{2\pi}d\phi\int_0^\infty G\left(\rho,\phi\right) \exp\left[j2\pi r\rho\cos\left(\theta-\phi\right)\right]\rho d\rho$

当 $f\left(x,y\right)$ 具有圆对称性，即 $g\left(r\right)=g\left(r,\theta\right)$ 时，傅里叶变换为傅里叶—贝塞尔变换：
$\begin{array}{c} G\left(\rho\right)=\mathscr{B}\left\{g\left(r\right)\right\} = 2\pi\int_0^\infty g\left(r\right)J_0\left(2\pi r\rho\right)rdr\\[2mm] g\left(r\right)=\mathscr{B}^{-1}\left\{G\left(\rho\right)\right\} = 2\pi\int_0^\infty G\left(\rho\right)J_0\left(2\pi r\rho\right)\rho d\rho \end{array}$