做网站用的三角形图片,开发公司成本管理,搜索引擎营销的作用,简单企业网站模板免费图像处理作业 第8次
7.11
说明尺度函数ϕ(x)1,0.25≤x0.75\phi(x)1 ,0.25 \le x\lt 0.75ϕ(x)1,0.25≤x0.75并未满足多分辨率分析的第二个要求. ϕ1,0(x)2ϕ(2x)1\phi_{1,0}(x)\sqrt 2 \phi(2x)1ϕ1,0(x)2ϕ(2x)1 当且仅当满足0.125≤x0.3750.125 \le x\lt 0…图像处理作业 第8次
7.11
说明尺度函数ϕ(x)1,0.25≤x0.75\phi(x)1 ,0.25 \le x\lt 0.75ϕ(x)1,0.25≤x0.75并未满足多分辨率分析的第二个要求. ϕ1,0(x)2ϕ(2x)1\phi_{1,0}(x)\sqrt 2 \phi(2x)1ϕ1,0(x)2ϕ(2x)1 当且仅当满足0.125≤x0.3750.125 \le x\lt 0.3750.125≤x0.375 ϕ1,1(x)2ϕ(2x−1)1\phi_{1,1}(x)\sqrt 2 \phi(2x-1) 1ϕ1,1(x)2ϕ(2x−1)1 当且仅当满足0.675≤x0.8750.675 \le x \lt 0.8750.675≤x0.875 可以看出,在0.375≤x0.6750.375 \le x \lt 0.6750.375≤x0.675的位置,显然ϕ(x)\phi(x)ϕ(x)不能由ϕ1,0(x),ϕ1,1(x)\phi_{1,0}(x),\phi_{1,1}(x)ϕ1,0(x),ϕ1,1(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j01j_01j01重新计算函数f(n){1,4,−3,0}f(n)\{1,4,-3,0\}f(n){1,4,−3,0}在区间[0,3][0,3][0,3]中的一维DWT. ϕ(n){1,1,1,1}\phi(n)\{1,1,1,1\}ϕ(n){1,1,1,1} ϕ1,0(n)2ϕ(2n−0)2{1,1,0,0}\phi_{1,0}(n)\sqrt 2 \phi(2n-0)\sqrt2\{1,1,0,0\}ϕ1,0(n)2ϕ(2n−0)2{1,1,0,0} ϕ1,1(n)2ϕ(2n−1)2{0,0,1,1}\phi_{1,1}(n)\sqrt 2 \phi(2n-1)\sqrt2\{0,0,1,1\}ϕ1,1(n)2ϕ(2n−1)2{0,0,1,1}
ψ1,0(n)2ψ(2n−0)2{1,−1,0,0}\psi_{1,0}(n)\sqrt 2 \psi(2n-0)\sqrt2\{1,-1,0,0\}ψ1,0(n)2ψ(2n−0)2{1,−1,0,0} ψ1,1(n)2ψ(2n−1)2{0,0,1,−1}\psi_{1,1}(n)\sqrt 2 \psi(2n-1)\sqrt2\{0,0,1,-1\}ψ1,1(n)2ψ(2n−1)2{0,0,1,−1}
Wϕ(1,0)1/2∑xf(x)ϕ1,0(x)1/2(1∗14∗1−3∗00∗0)52/2W_{\phi}(1,0)1/2 \sum_x f(x)\phi_{1,0}(x)1/2(1*14*1-3*00*0)5\sqrt2/2Wϕ(1,0)1/2∑xf(x)ϕ1,0(x)1/2(1∗14∗1−3∗00∗0)52/2 Wϕ(1,1)1/2∑xf(x)ϕ1,1(x)1/2(1∗04∗0−3∗10∗1)−32/2W_{\phi}(1,1)1/2 \sum_x f(x)\phi_{1,1}(x)1/2(1*04*0-3*10*1)-3\sqrt2/2Wϕ(1,1)1/2∑xf(x)ϕ1,1(x)1/2(1∗04∗0−3∗10∗1)−32/2 Wψ(1,0)1/2∑xf(x)ψ1,0(x)1/2(1∗14∗−1−3∗00∗0)−32/2W_{\psi}(1,0)1/2 \sum_x f(x)\psi_{1,0}(x)1/2(1*14*-1-3*00*0)-3\sqrt2/2Wψ(1,0)1/2∑xf(x)ψ1,0(x)1/2(1∗14∗−1−3∗00∗0)−32/2 Wψ(1,1)1/2∑xf(x)ψ1,1(x)1/2(1∗04∗0−3∗10∗−1)−32/2W_{\psi}(1,1)1/2 \sum_x f(x)\psi_{1,1}(x)1/2(1*04*0-3*10*-1)-3\sqrt2/2Wψ(1,1)1/2∑xf(x)ψ1,1(x)1/2(1∗04∗0−3∗10∗−1)−32/2
B) 使用(A)的结果根据变换值f(1)f(1)f(1)
f(n)1/2(Wϕ(1,0)ϕ1,0(n)Wϕ(1,1)ϕ1,1(n)Wψ(1,0)ψ1,1(n)Wψ(1,0)ψ1,1(n))f(n)1/2( W_{\phi}(1,0)\phi_{1,0}(n)W_{\phi}(1,1)\phi_{1,1}(n)W_{\psi}(1,0)\psi_{1,1}(n)W_{\psi}(1,0)\psi_{1,1}(n))f(n)1/2(Wϕ(1,0)ϕ1,0(n)Wϕ(1,1)ϕ1,1(n)Wψ(1,0)ψ1,1(n)Wψ(1,0)ψ1,1(n))
f(1)2/2(52/2∗1−32/2∗0−32/2∗(−1)−32/2∗0)4f(1)\sqrt 2/2(5\sqrt2/2*1 -3\sqrt2/2*0 -3\sqrt2/2*(-1) -3\sqrt2/2*0)4f(1)2/2(52/2∗1−32/2∗0−32/2∗(−1)−32/2∗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。
Wϕ(2,n)f(n){1,3,5,7,4,3,2,1}W_{\phi}(2,n) f(n)\{1,3,5,7,4,3,2,1\}Wϕ(2,n)f(n){1,3,5,7,4,3,2,1} ϕ(n){1/2,1/2}\phi(n)\{1/\sqrt2,1/\sqrt2\}ϕ(n){1/2,1/2} ψ(n){1/2,−1/2}\psi(n)\{1/\sqrt2,-1/\sqrt2\}ψ(n){1/2,−1/2}
Wψ(1,n){1,3,5,7,4,3,2,1}∗{−1/2,1/2}∣down21/2{−1,−2,−2,−2,3,1,1,1,0}∣down21/2{−2,−2,1,1}W_{\psi}(1,n)\{1,3,5,7,4,3,2,1\}*\{-1/\sqrt2,1/\sqrt2\}|_{down2}1/\sqrt2\{-1,-2,-2,-2,3,1,1,1,0\}|_{down2}1/\sqrt2\{-2,-2,1,1\}Wψ(1,n){1,3,5,7,4,3,2,1}∗{−1/2,1/2}∣down21/2{−1,−2,−2,−2,3,1,1,1,0}∣down21/2{−2,−2,1,1}
Wϕ(1,n){1,3,5,7,4,3,2,1}∗{1/2,1/2}∣down21/2{1,4,8,12,11,7,5,3,0}∣down21/2{4,12,7,3}W_{\phi}(1,n)\{1,3,5,7,4,3,2,1\}*\{1/\sqrt2,1/\sqrt2\}|_{down2}1/\sqrt2\{1,4,8,12,11,7,5,3,0\}|_{down2}1/\sqrt2\{4,12,7,3\}Wϕ(1,n){1,3,5,7,4,3,2,1}∗{1/2,1/2}∣down21/2{1,4,8,12,11,7,5,3,0}∣down21/2{4,12,7,3}
Wψ(0,n)1/2{4,12,7,3}∗{−1/2,1/2}∣down2{−4,2}W_{\psi}(0,n)1/\sqrt2\{4,12,7,3\}*\{-1/\sqrt2,1/\sqrt2\}|_{down2}\{-4,2\}Wψ(0,n)1/2{4,12,7,3}∗{−1/2,1/2}∣down2{−4,2}
Wϕ(0,n)1/2{4,12,7,3}∗{1/2,1/2}∣down2{8,5}W_{\phi}(0,n)1/\sqrt2\{4,12,7,3\}*\{1/\sqrt2,1/\sqrt2\}|_{down2}\{8,5\}Wϕ(0,n)1/2{4,12,7,3}∗{1/2,1/2}∣down2{8,5}
重建:
Wϕ(1,n){−4,0,2,0}∗1/2{1,−1}{8,0,5,0}∗1/2{1,1}1/2{−48,48,25,−25}1/2{4,12,7,3}W_{\phi}(1,n)\{-4,0,2,0\}*1/\sqrt2\{1,-1\}\{8,0,5,0\}*1/\sqrt2\{1,1\}1/\sqrt 2\{-48,48,25,-25\}1/\sqrt2\{4,12,7,3\}Wϕ(1,n){−4,0,2,0}∗1/2{1,−1}{8,0,5,0}∗1/2{1,1}1/2{−48,48,25,−25}1/2{4,12,7,3}
f(n)Wϕ(2,n)1/2{−2,0,−2,0,1,0,1,0}∗1/2{1,−1}1/2{4,0,12,0,7,0,3,0}∗1/2{1,1}1/2{−24,24,−212,212,17,−17,13,−13}{1,3,5,7,4,3,2,1}f(n)W_{\phi}(2,n)1/\sqrt2\{-2,0,-2,0,1,0,1,0\}*1/\sqrt2\{1,-1\}1/\sqrt2\{4,0,12,0,7,0,3,0\}*1/\sqrt2\{1,1\}1/2\{-24,24,-212,212,17,-17,13,-13\}\{1,3,5,7,4,3,2,1\}f(n)Wϕ(2,n)1/2{−2,0,−2,0,1,0,1,0}∗1/2{1,−1}1/2{4,0,12,0,7,0,3,0}∗1/2{1,1}1/2{−24,24,−212,212,17,−17,13,−13}{1,3,5,7,4,3,2,1}
