作业信息

1、证明3.44和3.45的正确性

$g(x,y)=\frac{1}{K}{\sum }_{i=1}^K{g}_i(x,y)$

$E[g(x,y)]=f(x,y)+E[n(x,y)]=f(x,y)$

$D[g]=E[(g-E[g]{)}^2]=E[g^2]-E[g{]}^2=E[f^2+\frac{2f(n_1+n_2+...+n_K)+n_1^2+...+n_K^2+{\sum }_{i,j}{n}_i{n}_j}{K^2}]-E[g{]}^2=E[f^2]+\frac{1}{K}E[n^2]-E[f{]}^2=\frac{1}{K}E[n^2]-0=\frac{1}{K}(E[n^2]-E[n{]}^2)=\frac{1}{K}D[n]$

故

${\sigma }_{g(x,y)}^2=\frac{1}{K}{\sigma }_{n(x,y)}^2$

2、 请计算如下两个向量与矩阵的卷积计算结果。

$[1,2,3,4,5,4,3,2,1]\ast [2,0,-2]$

设$a=[1,2,3,4,5,4,3,2,1],b=[2,0,-2]$
设$a$的下标从$0$到$8$，$b$的下标从$0$到$2$，
那么$c=a\ast b$,则$c$的下标从$0$到$11$
根据卷积公式：$c[x]={\sum }_{t=-oo}^{oo}a[t]\ast b[x-t]$
$c[0]=a[0]\ast b[0]=2$
$c[1]=a[0]\ast b[1]+a[1]\ast b[0]=4$
$c[2]=a[0]\ast b[2]+a[1]\ast b[1]+a[2]\ast b[0]=4$
$c[3]=a[1]\ast b[2]+a[2]\ast b[1]+a[3]\ast b[0]=4$
$c[4]=a[2]\ast b[2]+a[3]\ast b[1]+a[4]\ast b[0]=4$
$c[5]=a[3]\ast b[2]+a[4]\ast b[1]+a[5]\ast b[0]=0$
$c[6]=a[4]\ast b[2]+a[5]\ast b[1]+a[6]\ast b[0]=-4$
$c[7]=a[5]\ast b[2]+a[6]\ast b[1]+a[7]\ast b[0]=-4$
$c[8]=a[6]\ast b[2]+a[7]\ast b[1]+a[8]\ast b[0]=-4$
$c[9]=a[7]\ast b[2]+a[8]\ast b[1]=-4$
$c[10]=a[8]\ast b[2]-2$
故$c$数组是$[2,4,4,4,0,-4,-4,-4,-2]$

根据二维卷积公式：

$c[x,y]={\sum }_s{\sum }_{t}a[s,t]\ast b[x-s,y-t]$

可知，卷积结果为

$\left[\begin{array}{ccccccc}-1& 3& -1& 3& -2& 0& 4\\ -3& -6& -4& 4& -4& 2& 11\\ -3& -7& -6& 3& -6& 4& 15\\ -3& -11& -4& 8& -10& 3& 17\\ -7& -11& 2& 5& -10& 6& 15\\ -8& -5& 6& -4& -6& 9& 8\\ -3& -1& 3& -3& -2& 4& 2\end{array}\right]$

3. 证明拉普拉斯变换具有旋转不变形

$\nabla f(x,y)=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$
我们假设新的点为$(x_1,y_1)$
并且假设$(x_1,y_1)$是由$(x,y)$顺时针旋转$\theta$得到的。
那么有公式$x=x_1\ast cos\theta -y_1\ast sin\theta$,$y=x1\ast sin\theta +y_1\ast cos\theta$

从而有$\frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x_1}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x_1}=\frac{\partial f}{\partial x}cos\theta +\frac{\partial f}{\partial y}sin\theta$

$\frac{{\partial }^{2}f}{\partial x_1^2}=\frac{{\partial }^{2}f}{\partial x^2}\frac{\partial x}{\partial x_1}cos\theta +\frac{{\partial }^{2}f}{\partial y^2}\frac{\partial y}{\partial x_1}sin\theta =\frac{{\partial }^{2}f}{\partial x^2}co{s}^2\theta +\frac{{\partial }^{2}f}{\partial y^2}si{n}^2\theta$

$\frac{{\partial }^{2}f}{\partial y_1^2}=-\frac{{\partial }^{2}f}{\partial x^2}\frac{\partial x}{\partial y_1}sin\theta +\frac{{\partial }^{2}f}{\partial y^2}\frac{\partial y}{\partial y_1}cos\theta =\frac{{\partial }^{2}f}{\partial x^2}si{n}^2\theta +\frac{{\partial }^{2}f}{\partial y^2}co{s}^2\theta$

$\frac{{\partial }^{2}f}{\partial x_1^2}+\frac{{\partial }^{2}f}{\partial y_1^2}=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$

证明完成。