图像处理作业 第8次

图像处理作业 第8次

蔡少斐
2019E8013261007

7.11

说明尺度函数$\varphi (x)=1,0.25\le x<0.75$并未满足多分辨率分析的第二个要求.
${\varphi }_{1,0}(x)=\sqrt{2}\varphi (2x)=1$ 当且仅当满足$0.125\le x<0.375$
${\varphi }_{1,1}(x)=\sqrt{2}\varphi (2x-1)=1$ 当且仅当满足$0.675\le x<0.875$
可以看出,在$0.375\le x<0.675$的位置,显然$\varphi (x)$不能由${\varphi }_{1,0}(x),{\varphi }_{1,1}(x)$二者进行线性组合得到.
因此该尺度函数并未满足多分辨率分析的第二个要求.

7.2

A)
令$j_0=1$重新计算函数$f(n)=\{1,4,-3,0\}$在区间$[0,3]$中的一维DWT.
$\varphi (n)=\{1,1,1,1\}$
${\varphi }_{1,0}(n)=\sqrt{2}\varphi (2n-0)=\sqrt{2}\{1,1,0,0\}$
${\varphi }_{1,1}(n)=\sqrt{2}\varphi (2n-1)=\sqrt{2}\{0,0,1,1\}$

${\psi }_{1,0}(n)=\sqrt{2}\psi (2n-0)=\sqrt{2}\{1,-1,0,0\}$
${\psi }_{1,1}(n)=\sqrt{2}\psi (2n-1)=\sqrt{2}\{0,0,1,-1\}$

$W_{\varphi }(1,0)=1/2{\sum }_{x}f(x){\varphi }_{1,0}(x)=1/2(1\ast 1+4\ast 1-3\ast 0+0\ast 0)=5\sqrt{2}/2$
$W_{\varphi }(1,1)=1/2{\sum }_{x}f(x){\varphi }_{1,1}(x)=1/2(1\ast 0+4\ast 0-3\ast 1+0\ast 1)=-3\sqrt{2}/2$
$W_{\psi }(1,0)=1/2{\sum }_{x}f(x){\psi }_{1,0}(x)=1/2(1\ast 1+4\ast -1-3\ast 0+0\ast 0)=-3\sqrt{2}/2$
$W_{\psi }(1,1)=1/2{\sum }_{x}f(x){\psi }_{1,1}(x)=1/2(1\ast 0+4\ast 0-3\ast 1+0\ast -1)=-3\sqrt{2}/2$

B)
使用(A)的结果根据变换值$f(1)$

$f(n)=1/2(W_{\varphi }(1,0){\varphi }_{1,0}(n)+W_{\varphi }(1,1){\varphi }_{1,1}(n)+W_{\psi }(1,0){\psi }_{1,1}(n)+W_{\psi }(1,0){\psi }_{1,1}(n))$

$f(1)=\sqrt{2}/2(5\sqrt{2}/2\ast 1-3\sqrt{2}/2\ast 0-3\sqrt{2}/2\ast (-1)-3\sqrt{2}/2\ast 0)=4$

7.3

现在假设我们有一个长度为8的信号f=[1 3 5 7 4 3 2 1], 利用哈尔小波进行两层的快速小波变换分解，计算各层的滤波器输出，然后再进行完美重建，请利用与书中例子相同的框图进行计算。

$W_{\varphi }(2,n)=f(n)=\{1,3,5,7,4,3,2,1\}$
$\varphi (n)=\{1/\sqrt{2},1/\sqrt{2}\}$
$\psi (n)=\{1/\sqrt{2},-1/\sqrt{2}\}$

$W_{\psi }(1,n)=\{1,3,5,7,4,3,2,1\}\ast \{-1/\sqrt{2},1/\sqrt{2}\}{|}_{down2}=1/\sqrt{2}\{-1,-2,-2,-2,3,1,1,1,0\}{|}_{down2}=1/\sqrt{2}\{-2,-2,1,1\}$

$W_{\varphi }(1,n)=\{1,3,5,7,4,3,2,1\}\ast \{1/\sqrt{2},1/\sqrt{2}\}{|}_{down2}=1/\sqrt{2}\{1,4,8,12,11,7,5,3,0\}{|}_{down2}=1/\sqrt{2}\{4,12,7,3\}$

$W_{\psi }(0,n)=1/\sqrt{2}\{4,12,7,3\}\ast \{-1/\sqrt{2},1/\sqrt{2}\}{|}_{down2}=\{-4,2\}$

$W_{\varphi }(0,n)=1/\sqrt{2}\{4,12,7,3\}\ast \{1/\sqrt{2},1/\sqrt{2}\}{|}_{down2}=\{8,5\}$

重建:

$W_{\varphi }(1,n)=\{-4,0,2,0\}\ast 1/\sqrt{2}\{1,-1\}+\{8,0,5,0\}\ast 1/\sqrt{2}\{1,1\}=1/\sqrt{2}\{-4+8,4+8,2+5,-2+5\}=1/\sqrt{2}\{4,12,7,3\}$

$f(n)=W_{\varphi }(2,n)=1/\sqrt{2}\{-2,0,-2,0,1,0,1,0\}\ast 1/\sqrt{2}\{1,-1\}+1/\sqrt{2}\{4,0,12,0,7,0,3,0\}\ast 1/\sqrt{2}\{1,1\}=1/2\{-2+4,2+4,-2+12,2+12,1+7,-1+7,1+3,-1+3\}=\{1,3,5,7,4,3,2,1\}$