Gonzalez R. C. and Woods R. E. Digital Image Processing (Forth Edition)
基本
酉变换
一维的变换:
\[\mathbf{t} = \mathbf{A} \mathbf{f}, \\
\mathbf{f} = \mathbf{A}^{H} \mathbf{t}, \\
\mathbf{A}^H = {\mathbf{A}^*}^{T}, \mathbf{A}^H\mathbf{A} = \mathbf{I}.
\]
\mathbf{f} = \mathbf{A}^{H} \mathbf{t}, \\
\mathbf{A}^H = {\mathbf{A}^*}^{T}, \mathbf{A}^H\mathbf{A} = \mathbf{I}.
\]
以及二维的变换:
\[\mathbf{T} = \mathbf{A} \mathbf{F} \mathbf{B}^T, \\
\mathbf{F} = \mathbf{A}^H \mathbf{T} \mathbf{B}^*, \\
\mathbf{A}^H\mathbf{A=I}, \mathbf{B}^{T}\mathbf{B}^* =\mathbf{I}.
\]
\mathbf{F} = \mathbf{A}^H \mathbf{T} \mathbf{B}^*, \\
\mathbf{A}^H\mathbf{A=I}, \mathbf{B}^{T}\mathbf{B}^* =\mathbf{I}.
\]
以一维的为例, 实际上就是
\[t_u = \sum_{x = 0}^{N-1} f_x s(x, u) = \mathbf{f}^T \mathbf{s}_u, u=0,1,\cdots, N-1,\\
\mathbf{s}_u = [s(0, u), s(1, u), \cdots, s(N-1, u)]^T.
\]
\mathbf{s}_u = [s(0, u), s(1, u), \cdots, s(N-1, u)]^T.
\]
故
\[\mathbf{A} = [\mathbf{s}_0, \cdots, \mathbf{s}_{N-1}]^{T}.
\]
\]
others
\[\sum_{k=0}^n \sin (kx) = \frac{\cos(\frac{1}{2}x) - \cos (\frac{2n+1}{2}x)}{2 \sin (\frac{x}{2})}, \quad x \in (2K\pi, 2(K+1)\pi)
\]
\]
proof:
\[\begin{array}{ll}
2\sin (\frac{x}{2}) \sum_{k=0}^n \sin (kx)
&=\sum_{k=0}^n [\cos (\frac{2k-1}{2}x) -\cos (\frac{2k+1}{2}x) ]\\
&= \cos(\frac{1}{2}x) - \cos (\frac{2n+1}{2}x).
\end{array}
\]
2\sin (\frac{x}{2}) \sum_{k=0}^n \sin (kx)
&=\sum_{k=0}^n [\cos (\frac{2k-1}{2}x) -\cos (\frac{2k+1}{2}x) ]\\
&= \cos(\frac{1}{2}x) - \cos (\frac{2n+1}{2}x).
\end{array}
\]
类似地
\[\sum_{k=0}^n \cos (kx) = \frac{\sin(\frac{2k+1}{2}x) + \sin (\frac{1}{2}x)}{2 \sin (\frac{1}{2}x)}, \quad x \in (2K\pi, 2(K+1)\pi)
\]
\]
proof:
\[\begin{array}{ll}
2\sin (\frac{x}{2}) \sum_{k=0}^n \cos (kx)
&=\sum_{k=0}^n [\sin (\frac{2k+1}{2}x) -\sin (\frac{2k-1}{2}x) ]\\
&= \sin(\frac{2k+1}{2}x) + \sin (\frac{1}{2}x).
\end{array}
\]
2\sin (\frac{x}{2}) \sum_{k=0}^n \cos (kx)
&=\sum_{k=0}^n [\sin (\frac{2k+1}{2}x) -\sin (\frac{2k-1}{2}x) ]\\
&= \sin(\frac{2k+1}{2}x) + \sin (\frac{1}{2}x).
\end{array}
\]
Fourier-related Transforms
DFT
\[s(x, u) = \frac{1}{\sqrt{N}} e^{\frac{-j2\pi xu}{N}}
\]
\]
\(\mathbf{s}_u^H \mathbf{s}_u = 1\)是显然的, 又注意到
\[\mathbf{s}_u^H \mathbf{s}_{u'} = \frac{1}{N}\sum_{x=0}^{N-1} e^{\frac{-j2\pi x(u-u')}{N}},
\]
\]
又
\[\sum_{n=0}^{N-1} a^n = \frac{1-a^N}{1-a},
\]
\]
由于
\[e^{-j2\pi x (u - u')} = 1, \forall u \not = u'.
\]
\]
DHT
DISCRETE HARTLEY TRANSFORM
\[s(x, u) = \frac{1}{\sqrt{N}}\mathrm{cas}(\frac{2\pi xu}{N}) = \frac{1}{\sqrt{N}}[\cos (\frac{2\pi ux}{N}) + \sin (\frac{2\pi ux}{N})].
\]
\]
\[2\cos (\frac{2\pi ux}{N})
\cos (\frac{2\pi u'x}{N})
=\cos (\frac{2\pi (u-u')x}{N})
+\cos (\frac{2\pi (u+u')x}{N}) \\
2\sin (\frac{2\pi ux}{N})
\sin (\frac{2\pi u'x}{N})
=\cos (\frac{2\pi (u-u')x}{N})
-\cos (\frac{2\pi (u+u')x}{N}) \\
2\sin (\frac{2\pi ux}{N})
\cos (\frac{2\pi u'x}{N})
=\sin (\frac{2\pi (u+u')x}{N})
-\sin (\frac{2\pi (u-u')x}{N}) \\
\]
\cos (\frac{2\pi u'x}{N})
=\cos (\frac{2\pi (u-u')x}{N})
+\cos (\frac{2\pi (u+u')x}{N}) \\
2\sin (\frac{2\pi ux}{N})
\sin (\frac{2\pi u'x}{N})
=\cos (\frac{2\pi (u-u')x}{N})
-\cos (\frac{2\pi (u+u')x}{N}) \\
2\sin (\frac{2\pi ux}{N})
\cos (\frac{2\pi u'x}{N})
=\sin (\frac{2\pi (u+u')x}{N})
-\sin (\frac{2\pi (u-u')x}{N}) \\
\]
故想要证明其为标准正交基, 只需注意到:
\[\sum_{x=0}^{N-1} \sin (\frac{2\pi k x}{N})
=\frac{\cos(\frac{k\pi}{N}) - \cos (\frac{(2N-1)k\pi}{N})}{...},
\]
=\frac{\cos(\frac{k\pi}{N}) - \cos (\frac{(2N-1)k\pi}{N})}{...},
\]
\(k\not=0\)的时候, 有
\[\cos (\frac{(2N-1)k\pi}{N}) = \cos (\frac{k\pi}{N}),
\]
\]
故
\[\sum_{x=0}^{N-1}\sin (\frac{2\pi kx}{N}) =0, k\not=0.
\]
\]
类似可得:
\[\sum_{x=0}^{N-1}\cos (\frac{2\pi kx}{N}) =0, k\not=0.
\]
\]
正交性如此是易证明的, 实际上标准性是显然的.
DCT
DISCRETE COSINE TRANSFORM
\[s(x, u) = \alpha (u) \cos (\frac{(2x + 1)u\pi}{2N}), \\
\alpha (u) =
\left \{
\begin{array}{ll}
\sqrt{\frac{1}{N}}, & u=0, \\
\sqrt{\frac{2}{N}}, & u=1,2,\cdots, N-1. \\
\end{array}
\right .
\]
\alpha (u) =
\left \{
\begin{array}{ll}
\sqrt{\frac{1}{N}}, & u=0, \\
\sqrt{\frac{2}{N}}, & u=1,2,\cdots, N-1. \\
\end{array}
\right .
\]
其标准正交的思路和DHT是如出一辙的.
与DFT的联系
- 定义
\[g(x) =
\left \{
\begin{array}{ll}
f(x), & x = 0, 1, \cdots, N-1, \\
f(2N-x-1), & u=N, N+1, \cdots, 2N-1. \\
\end{array}
\right .
\]
\left \{
\begin{array}{ll}
f(x), & x = 0, 1, \cdots, N-1, \\
f(2N-x-1), & u=N, N+1, \cdots, 2N-1. \\
\end{array}
\right .
\]
此时\(g(x) = g(2N-1-x)\);
- 计算DFT
\[\mathbf{t}_F = \mathbf{A}_F \mathbf{g} =
\left [
\begin{array}{c}
\mathbf{t}_1 \\
\mathbf{t}_2 \\
\end{array}
\right ].
\]
\left [
\begin{array}{c}
\mathbf{t}_1 \\
\mathbf{t}_2 \\
\end{array}
\right ].
\]
- 定义
\[h(u) = e^{-j\pi u / 2N}, u=0,1,\cdots, N-1, \\
\mathbf{s} = [1 / \sqrt{2}, 1, 1, \cdots, 1]^T.
\]
\mathbf{s} = [1 / \sqrt{2}, 1, 1, \cdots, 1]^T.
\]
\[\mathbf{t}_C = \mathrm{Re}\{\mathbf{s\circ h \circ t_1}\}.
\]
\]
其中\(\mathrm{Re}\)表示实部, \(\circ\)表示逐项乘法.
证明是平凡的.
DST
DISCRETE SINE TRANSFORM
\[s(x, u) = \sqrt{\frac{2}{N+1}} \sin (\frac{(x+1)(u+1)\pi}{N+1}).
\]
\]
与DFT的联系
- 定义
\[g(x) =
\left \{
\begin{array}{ll}
0, & x = 0, \\
f(x-1), & x = 1, \cdots, N, \\
0, & x = N + 1, \\
-f(2N-x+1), & u=N+1, \cdots, 2N+1. \\
\end{array}
\right .
\]
\left \{
\begin{array}{ll}
0, & x = 0, \\
f(x-1), & x = 1, \cdots, N, \\
0, & x = N + 1, \\
-f(2N-x+1), & u=N+1, \cdots, 2N+1. \\
\end{array}
\right .
\]
此时\(g(x) = -g(2N + 2 - x)\).
- DFT
\[\mathbf{t}_F = \mathbf{A}_F \mathbf{g} =
\left [
\begin{array}{c}
0 \\
\mathbf{t}_1 \\
0 \\
\mathbf{t}_2 \\
\end{array}
\right ].
\]
\left [
\begin{array}{c}
0 \\
\mathbf{t}_1 \\
0 \\
\mathbf{t}_2 \\
\end{array}
\right ].
\]
\[\mathbf{t}_S = -\mathrm{Imag}\{\mathbf{t}_1\}.
\]
\]
其中\(\mathrm{Imag}\)表虚部.