第5次图像处理作业

1. 复习理解课本中最佳陷波滤波器进行图像恢复的过程，请推导出w(x,y)最优解的计算过程.

根据公式:

${\sigma }^2=\frac{1}{(2a+1)(2b+1)}\sum \sum [g-w\eta -(\bar{g}-w\bar{\eta }){]}^2$

$\frac{\partial {\sigma }^2}{\partial w}=\frac{2}{(2a+1)(2b+1)}\sum \sum [g-w\eta -(\bar{g}-w\bar{\eta })][-\eta +\bar{\eta }]=0$

$\sum \sum (g-\bar{g})(-\eta +\bar{\eta })+\sum \sum w(-\eta +\bar{\eta })(-\eta +\bar{\eta })=0$

$w=\frac{\sum \sum (g-\bar{g})(-\eta +\bar{\eta })}{\sum \sum (-\eta +\bar{\eta })(-\eta +\bar{\eta })}$

$w=\frac{\frac{1}{(2a+1)(2b+1)}\sum \sum (g-\bar{g})(-\eta +\bar{\eta })}{\frac{1}{(2a+1)(2b+1)}\sum \sum (-\eta +\bar{\eta })(-\eta +\bar{\eta })}$

$w=\frac{\frac{1}{(2a+1)(2b+1)}(\sum \sum (g\bar{\eta }-\bar{g}\bar{\eta })+\sum \sum (-g\eta +\bar{g}\eta ))}{\frac{1}{(2a+1)(2b+1)}(\sum \sum {\eta }^2+\sum \sum {\bar{\eta }}^2-2\sum \sum \eta \bar{\eta })}$

根据

$\frac{1}{(2a+1)(2b+1)}\sum \sum (g\bar{\eta }-\bar{g}\bar{\eta })=0$

$\frac{1}{(2a+1)(2b+1)}(\sum \sum (g\bar{\eta }-\bar{g}\bar{\eta })+\sum \sum (-g\eta +\bar{g}\eta ))=\overline{gn}-\bar{g}\bar{n}$

$\frac{1}{(2a+1)(2b+1)}(\sum \sum {\bar{\eta }}^2-2\sum \sum \eta \bar{\eta })={\bar{\eta }}^2$

那么

$w=\frac{\overline{gn}-\bar{g}\bar{n}}{\overline{{\eta }^2}-{\bar{\eta }}^2}$

2.考虑在x方向均匀加速导致的图像模糊问题。如果图像在t = 0静止，并用均匀加速x0(t) = at2/2加速，对于时间T, 找出模糊函数H(u, v), 可以假设快门开关时间忽略不计。

新图像函数为

$g(x,y)={\int }_0^{T}f[x-x_0(t),y-y_0(t)]dt$

$G(u,v)=\int \int g(x,y)e^{-j2\pi (ux+vy)}dxdy$

$=\int \int ({\int }_0^{T}f[x-x_0(t),y-y_0(t)]dt)e^{-j2\pi (ux+vy)}dxdy$

$={\int }_0^T(\int \int f[x-x_0(t),y-y_0(t)]e^{-j2\pi (ux+vy)}dxdy)dt$

$={\int }_0^{T}F(u,v)e^{-j2\pi (u{x}_0(t)+v{y}_0(t))}dt$

$=F(u,v){\int }_0^T{e}^{-j2\pi (u{x}_0(t)+v{y}_0(t))}dt$

因此模糊函数为

$H(u,v)={\int }_0^T{e}^{-j2\pi (u{x}_0(t)+v{y}_0(t))}dt={\int }_0^T{e}^{-j\pi ua{t}^2}dt$

3.已知一个退化系统的退化函数H(u,v), 以及噪声的均值与方差，请描述如何利用约束最小二乘方算法计算出原图像的估计。

$\hat{F}(u,v)=[\frac{H^{\ast }(u,v)}{|H(u,v){|}^2+\gamma |P(u,v){|}^2}]G(u,v)......\ast$

设定其残差$r=g-H\hat{f}$

应有$||r|{|}^2=||\eta |{|}^2$

(1)给定$\gamma$一个初始值
(2)计算$||r|{|}^2$
(3)若满足$||r|{|}^2-||\eta |{|}^2$处于某一个精度范围之内则结束,否则更新$\gamma$大小,可以采用牛顿法.
(4)使用计算到的$\gamma$值代入$(\ast )$式中计算
(5)通逆傅里叶变换得到图像.

在计算$||r|{|}^2-||\eta |{|}^2$的时候,其中$||\eta |{|}^2$的计算需要依赖于噪声的方差和均值.

$||\eta |{|}^2=MN[{\sigma }^2-m]$