DFT, DHT, DCT, DST

目录

• 基本
  • 酉变换
  • others
• Fourier-related Transforms
  • DFT
  • DHT
  • DCT
    • 与DFT的联系
  • DST
    • 与DFT的联系

Gonzalez R. C. and Woods R. E. Digital Image Processing (Forth Edition)

基本

酉变换

一维的变换:

$$\mathbf{t} = \mathbf{A} \mathbf{f}, \\ \mathbf{f} = \mathbf{A}^{H} \mathbf{t}, \\ \mathbf{A}^H = {\mathbf{A}^*}^{T}, \mathbf{A}^H\mathbf{A} = \mathbf{I}.$$

以及二维的变换:

$$\mathbf{T} = \mathbf{A} \mathbf{F} \mathbf{B}^T, \\ \mathbf{F} = \mathbf{A}^H \mathbf{T} \mathbf{B}^*, \\ \mathbf{A}^H\mathbf{A=I}, \mathbf{B}^{T}\mathbf{B}^* =\mathbf{I}.$$

以一维的为例, 实际上就是

$$t_u = \sum_{x = 0}^{N-1} f_x s(x, u) = \mathbf{f}^T \mathbf{s}_u, u=0,1,\cdots, N-1,\\ \mathbf{s}_u = [s(0, u), s(1, u), \cdots, s(N-1, u)]^T.$$

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:

$$\begin{array}{ll} 2\sin (\frac{x}{2}) \sum_{k=0}^n \sin (kx) &=\sum_{k=0}^n [\cos (\frac{2k-1}{2}x) -\cos (\frac{2k+1}{2}x) ]\\ &= \cos(\frac{1}{2}x) - \cos (\frac{2n+1}{2}x). \end{array}$$

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:

$$\begin{array}{ll} 2\sin (\frac{x}{2}) \sum_{k=0}^n \cos (kx) &=\sum_{k=0}^n [\sin (\frac{2k+1}{2}x) -\sin (\frac{2k-1}{2}x) ]\\ &= \sin(\frac{2k+1}{2}x) + \sin (\frac{1}{2}x). \end{array}$$

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

$$s(x, u) = \frac{1}{\sqrt{N}}\mathrm{cas}(\frac{2\pi xu}{N}) = \frac{1}{\sqrt{N}}[\cos (\frac{2\pi ux}{N}) + \sin (\frac{2\pi ux}{N})].$$

$$2\cos (\frac{2\pi ux}{N}) \cos (\frac{2\pi u'x}{N}) =\cos (\frac{2\pi (u-u')x}{N}) +\cos (\frac{2\pi (u+u')x}{N}) \\ 2\sin (\frac{2\pi ux}{N}) \sin (\frac{2\pi u'x}{N}) =\cos (\frac{2\pi (u-u')x}{N}) -\cos (\frac{2\pi (u+u')x}{N}) \\ 2\sin (\frac{2\pi ux}{N}) \cos (\frac{2\pi u'x}{N}) =\sin (\frac{2\pi (u+u')x}{N}) -\sin (\frac{2\pi (u-u')x}{N}) \\ $$

故想要证明其为标准正交基, 只需注意到:

$$\sum_{x=0}^{N-1} \sin (\frac{2\pi k x}{N}) =\frac{\cos(\frac{k\pi}{N}) - \cos (\frac{(2N-1)k\pi}{N})}{...},$$

$k\not=0$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

$$s(x, u) = \alpha (u) \cos (\frac{(2x + 1)u\pi}{2N}), \\ \alpha (u) = \left \{ \begin{array}{ll} \sqrt{\frac{1}{N}}, & u=0, \\ \sqrt{\frac{2}{N}}, & u=1,2,\cdots, N-1. \\ \end{array} \right .$$

其标准正交的思路和DHT是如出一辙的.

与DFT的联系

1. 定义

$$g(x) = \left \{ \begin{array}{ll} f(x), & x = 0, 1, \cdots, N-1, \\ f(2N-x-1), & u=N, N+1, \cdots, 2N-1. \\ \end{array} \right .$$

此时$g(x) = g(2N-1-x)$;

2. 计算DFT

$$\mathbf{t}_F = \mathbf{A}_F \mathbf{g} = \left [ \begin{array}{c} \mathbf{t}_1 \\ \mathbf{t}_2 \\ \end{array} \right ].$$

3. 定义

$$h(u) = e^{-j\pi u / 2N}, u=0,1,\cdots, N-1, \\ \mathbf{s} = [1 / \sqrt{2}, 1, 1, \cdots, 1]^T.$$

$$\mathbf{t}_C = \mathrm{Re}\{\mathbf{s\circ h \circ t_1}\}.$$

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

1. 定义

$$g(x) = \left \{ \begin{array}{ll} 0, & x = 0, \\ f(x-1), & x = 1, \cdots, N, \\ 0, & x = N + 1, \\ -f(2N-x+1), & u=N+1, \cdots, 2N+1. \\ \end{array} \right .$$

此时$g(x) = -g(2N + 2 - x)$.

2. DFT

$$\mathbf{t}_F = \mathbf{A}_F \mathbf{g} = \left [ \begin{array}{c} 0 \\ \mathbf{t}_1 \\ 0 \\ \mathbf{t}_2 \\ \end{array} \right ].$$

$$\mathbf{t}_S = -\mathrm{Imag}\{\mathbf{t}_1\}.$$

其中$\mathrm{Imag}$表虚部.