图像处理作业 第8次
7.11
说明尺度函数
ϕ
(
x
)
=
1
,
0.25
≤
x
<
0.75
\phi(x)=1 ,0.25 \le x\lt 0.75
ϕ(x)=1,0.25≤x<0.75并未满足多分辨率分析的第二个要求.
ϕ
1
,
0
(
x
)
=
2
ϕ
(
2
x
)
=
1
\phi_{1,0}(x)=\sqrt 2 \phi(2x)=1
ϕ1,0(x)=2ϕ(2x)=1 当且仅当满足
0.125
≤
x
<
0.375
0.125 \le x\lt 0.375
0.125≤x<0.375
ϕ
1
,
1
(
x
)
=
2
ϕ
(
2
x
−
1
)
=
1
\phi_{1,1}(x)=\sqrt 2 \phi(2x-1) = 1
ϕ1,1(x)=2ϕ(2x−1)=1 当且仅当满足
0.675
≤
x
<
0.875
0.675 \le x \lt 0.875
0.675≤x<0.875
可以看出,在
0.375
≤
x
<
0.675
0.375 \le x \lt 0.675
0.375≤x<0.675的位置,显然
ϕ
(
x
)
\phi(x)
ϕ(x)不能由
ϕ
1
,
0
(
x
)
,
ϕ
1
,
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
j
0
=
1
j_0=1
j0=1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}
ϕ
1
,
0
(
n
)
=
2
ϕ
(
2
n
−
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}
ϕ
1
,
1
(
n
)
=
2
ϕ
(
2
n
−
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}
ψ
1
,
0
(
n
)
=
2
ψ
(
2
n
−
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}
ψ
1
,
1
(
n
)
=
2
ψ
(
2
n
−
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}
W
ϕ
(
1
,
0
)
=
1
/
2
∑
x
f
(
x
)
ϕ
1
,
0
(
x
)
=
1
/
2
(
1
∗
1
+
4
∗
1
−
3
∗
0
+
0
∗
0
)
=
5
2
/
2
W_{\phi}(1,0)=1/2 \sum_x f(x)\phi_{1,0}(x)=1/2(1*1+4*1-3*0+0*0)=5\sqrt2/2
Wϕ(1,0)=1/2∑xf(x)ϕ1,0(x)=1/2(1∗1+4∗1−3∗0+0∗0)=52/2
W
ϕ
(
1
,
1
)
=
1
/
2
∑
x
f
(
x
)
ϕ
1
,
1
(
x
)
=
1
/
2
(
1
∗
0
+
4
∗
0
−
3
∗
1
+
0
∗
1
)
=
−
3
2
/
2
W_{\phi}(1,1)=1/2 \sum_x f(x)\phi_{1,1}(x)=1/2(1*0+4*0-3*1+0*1)=-3\sqrt2/2
Wϕ(1,1)=1/2∑xf(x)ϕ1,1(x)=1/2(1∗0+4∗0−3∗1+0∗1)=−32/2
W
ψ
(
1
,
0
)
=
1
/
2
∑
x
f
(
x
)
ψ
1
,
0
(
x
)
=
1
/
2
(
1
∗
1
+
4
∗
−
1
−
3
∗
0
+
0
∗
0
)
=
−
3
2
/
2
W_{\psi}(1,0)=1/2 \sum_x f(x)\psi_{1,0}(x)=1/2(1*1+4*-1-3*0+0*0)=-3\sqrt2/2
Wψ(1,0)=1/2∑xf(x)ψ1,0(x)=1/2(1∗1+4∗−1−3∗0+0∗0)=−32/2
W
ψ
(
1
,
1
)
=
1
/
2
∑
x
f
(
x
)
ψ
1
,
1
(
x
)
=
1
/
2
(
1
∗
0
+
4
∗
0
−
3
∗
1
+
0
∗
−
1
)
=
−
3
2
/
2
W_{\psi}(1,1)=1/2 \sum_x f(x)\psi_{1,1}(x)=1/2(1*0+4*0-3*1+0*-1)=-3\sqrt2/2
Wψ(1,1)=1/2∑xf(x)ψ1,1(x)=1/2(1∗0+4∗0−3∗1+0∗−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 ( W ϕ ( 1 , 0 ) ϕ 1 , 0 ( n ) + W ϕ ( 1 , 1 ) ϕ 1 , 1 ( n ) + W ψ ( 1 , 0 ) ψ 1 , 1 ( n ) + W ψ ( 1 , 0 ) ψ 1 , 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 ( 5 2 / 2 ∗ 1 − 3 2 / 2 ∗ 0 − 3 2 / 2 ∗ ( − 1 ) − 3 2 / 2 ∗ 0 ) = 4 f(1)=\sqrt 2/2(5\sqrt2/2*1 -3\sqrt2/2*0 -3\sqrt2/2*(-1) -3\sqrt2/2*0)=4 f(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。
W
ϕ
(
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}
W ψ ( 1 , n ) = { 1 , 3 , 5 , 7 , 4 , 3 , 2 , 1 } ∗ { − 1 / 2 , 1 / 2 } ∣ d o w n 2 = 1 / 2 { − 1 , − 2 , − 2 , − 2 , 3 , 1 , 1 , 1 , 0 } ∣ d o w n 2 = 1 / 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}∣down2=1/2{−1,−2,−2,−2,3,1,1,1,0}∣down2=1/2{−2,−2,1,1}
W ϕ ( 1 , n ) = { 1 , 3 , 5 , 7 , 4 , 3 , 2 , 1 } ∗ { 1 / 2 , 1 / 2 } ∣ d o w n 2 = 1 / 2 { 1 , 4 , 8 , 12 , 11 , 7 , 5 , 3 , 0 } ∣ d o w n 2 = 1 / 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}∣down2=1/2{1,4,8,12,11,7,5,3,0}∣down2=1/2{4,12,7,3}
W ψ ( 0 , n ) = 1 / 2 { 4 , 12 , 7 , 3 } ∗ { − 1 / 2 , 1 / 2 } ∣ d o w n 2 = { − 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}
W ϕ ( 0 , n ) = 1 / 2 { 4 , 12 , 7 , 3 } ∗ { 1 / 2 , 1 / 2 } ∣ d o w n 2 = { 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:
W ϕ ( 1 , n ) = { − 4 , 0 , 2 , 0 } ∗ 1 / 2 { 1 , − 1 } + { 8 , 0 , 5 , 0 } ∗ 1 / 2 { 1 , 1 } = 1 / 2 { − 4 + 8 , 4 + 8 , 2 + 5 , − 2 + 5 } = 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\{-4+8,4+8,2+5,-2+5\}=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{−4+8,4+8,2+5,−2+5}=1/2{4,12,7,3}
f ( n ) = W ϕ ( 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 { − 2 + 4 , 2 + 4 , − 2 + 12 , 2 + 12 , 1 + 7 , − 1 + 7 , 1 + 3 , − 1 + 3 } = { 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\{-2+4,2+4,-2+12,2+12,1+7,-1+7,1+3,-1+3\}=\{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{−2+4,2+4,−2+12,2+12,1+7,−1+7,1+3,−1+3}={1,3,5,7,4,3,2,1}