圖像處理作業二

作業信息

學號：2019E8013261007
姓名：蔡少斐
班級：705
培養單位：計算技術研究所

1、證明3.44和3.45的正確性

$g(x,y)=\frac{1}{K}\sum_{i=1}^K g_i(x,y)$

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

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

故

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

2、 請計算如下兩個向量與矩陣的卷積計算結果。

$[1,2,3,4,5,4,3,2,1] * [2,0,-2]$

設$a = [1,2 ,3, 4, 5, 4, 3, 2, 1],b =[2,0,-2]$
設$a$的下標從$0$到$8$，$b$的下標從$0$到$2$，
那麼$c = a*b$,則$c$的下標從$0$到$11$
根據卷積公式：$c[x] = \sum_{t = -oo}^{oo} a[t]*b[x-t]$
$c[0] = a[0]*b[0] = 2$
$c[1] = a[0]*b[1]+a[1]*b[0] = 4$
$c[2] = a[0]*b[2]+a[1]*b[1]+a[2]*b[0] = 4$
$c[3] = a[1]*b[2]+a[2]*b[1]+a[3]*b[0] = 4$
$c[4] = a[2]*b[2]+a[3]*b[1]+a[4]*b[0] = 4$
$c[5] = a[3]*b[2]+a[4]*b[1]+a[5]*b[0] = 0$
$c[6] = a[4]*b[2]+a[5]*b[1]+a[6]*b[0] = -4$
$c[7] = a[5]*b[2]+a[6]*b[1]+a[7]*b[0] = -4$
$c[8] = a[6]*b[2]+a[7]*b[1]+a[8]*b[0] = -4$
$c[9] = a[7]*b[2]+a[8]*b[1] = -4$
$c[10] = a[8]*b[2] -2$
故$c$數組是$[2,4,4,4,0,-4,-4,-4,-2]$

根據二維卷積公式：

$c[x,y] = \sum_s\sum_ta[s,t]*b[x-s,y-t]$

可知，卷積結果爲

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

3. 證明拉普拉斯變換具有旋轉不變形

$\nabla f(x,y)=\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$
我們假設新的點爲$(x_1,y_1)$
並且假設$(x_1,y_1)$是由$(x,y)$順時針旋轉$\theta$得到的。
那麼有公式$x = x_1*cos\theta-y_1*sin\theta$,$y=x1*sin\theta+y_1*cos\theta$

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

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

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

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

證明完成。