Fourier Transform

目录

• Fourier Transform
  • 基本的定义
  • 性质
    • 对称性
  • 卷积
  • FFT

Fourier Transform

基本的定义

严格来说, 傅里叶变换是Schwartz space 上的一一映射, 对于$L^1$, 即可积函数我们都可以找到其对应的傅里叶变换.

符号	定义
傅里叶变换: $\hat{f}(u)$	$\int_{-\infty}^{+\infty} f(t) e^{-j2\pi u t} \mathrm{d}t$
逆傅里叶变换: $\check{F}(t)$	$\int_{-\infty}^{+\infty} F(u) e^{j2\pi tu} \mathrm{d}u$​
离散傅里叶变换: $F_m=\hat{f}_m$	$\sum_{n=0}^{N-1} f_n e^{-j2\pi nm/N}$​
离散逆傅里叶变换: $f_n = \check{F}_n$	$\frac{1}{N}\sum_{m=0}^{N-1} F_m e^{j2\pi mn /N}$
卷积: $f \star g (t)$	$\int_{-\infty}^{+\infty} f(\tau) g(t-\tau) \mathrm{d}\tau$
correlation: $f \bullet g(t)$​	$\int_{-\infty}^{+\infty}f^*(\tau)g(\tau + t)\mathrm{d}\tau$
二元傅里叶变换: $\hat{f}(u, v)$	$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(t, z) e^{-j2\pi ut} e^{-j2\pi vz} \mathrm{d}t \mathrm{d}z$
二元傅里叶逆变换: $\check{F}(t, z)$	$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} F(u, v) e^{j2\pi tu} e^{-j2\pi zv} \mathrm{d}t \mathrm{d}z$
二元卷积: $f \star g (t, z)$	$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(\tau, \omega) g(t - \tau, z - \omega) \mathrm{d}\tau \mathrm{d}\omega$
共轭	$f^*(t) = [r(t) + j i(t)]^* = r(t) - ji(t), \quad r(t), i(t) \in \mathbb{R}$

注: 不同版本的傅里叶变换的定义存在系数上的差异.

性质

	$\hat{}$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$\check{}$
1	$f(t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u)$
2	$f \star g (t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) \cdot G (u)$
3	$f \bullet g(t)$​	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F^*(u) \cdot G(u)$
4	$f(t) \cdot g(t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F \star G (u)$
5	$f(t) + g(t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) + G(u)$
6	$f(t + t_0)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$e^{j2\pi t_0 u} F(u)$
7	$f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(-u)$
8	$F(u)=\hat{f}(u)$​	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$f(-t) = \hat{\hat{f}}(t)$
以上	对于	离散	成立
9	$f(at)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$\frac{1}{
10	$f'(t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$j2\pi u F(u)$
11	$-j2\pi tf(t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F'(u)$
12	$f(R \mathbf{t})$, $R^TR=I$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(R\mathbf{u})$

注: 虽然以上性质除11外都是用了标量, 但是实际上对于一般的傅里叶变换也是成立的.

1. pass

2. 卷积:

   $$\begin{array}{ll} (f \star g)^{\hat{}}(u) &= \int f \star g (t) e^{-j2\pi u t} \mathrm{d}t \\ &= \int \int f(t-\tau) g (\tau) \mathrm{d}\tau e^{-j2\pi u t} \mathrm{d}t \\ &= \int \int f(t-\tau) g (\tau) e^{-j2\pi u t} \mathrm{d}t \mathrm{d}\tau \quad \rightarrow \mathrm{Fubini}\\ &= \int \int f(t-\tau) e^{-j2\pi u (t-\tau)} \mathrm{d}t \: g(\tau)e^{-j2\pi u\tau} \mathrm{d}\tau \quad \\ &= F(u) \cdot G(u). \end{array}$$

   $\Leftarrow$是类似的证明.

3. correlation:

   $$\begin{array}{ll} (f \bullet g)^{\hat{}}(u) &= \int f*g(t) e^{-j2\pi ut} \mathrm{d}t \\ &= \int \int f^*(\tau) g(\tau + t) \mathrm{d}\tau e^{-j2\pi ut} \mathrm{d}t \\ &= \int f^*(\tau) \int g(\tau + t) e^{-j2\pi ut} \mathrm{d}t \mathrm{d}\tau \\ &= G(u)\int f^*(\tau) e^{j2\pi u\tau} \mathrm{d}\tau \\ &= G(u)[\int f(\tau) e^{-j2\pi u\tau} \mathrm{d}\tau]^* \\ &= F^*(u)G(u) \end{array}$$

4. 乘积的证明和上面是一样的, 无非是作用于$\check{}$上.

5. pass

6. 平移:

   $$(f(t+t_0))^{\hat{}} = \int f(t+t_0) e^{-j2\pi ut} \mathrm{d}t =e^{j2\pi u t_0}\int f(t+t_0) e^{-j2\pi u(t+t_0)} \mathrm{d}t = e^{j2\pi t_0u} F(u).$$

7. 逆:

   $$(f(-t))^{\hat{}} = \int f(-t) e^{-j2\pi u t} \mathrm{d}t = \int f(t) e^{j2\pi u t} \mathrm{d}t = F(-u).$$

8. $\hat{F}(t) = \int F(u) e^{j2\pi t u} \mathrm{d}u = (F(-u))^{\check{}}(t) = f(-t)$.

9. scaling:

   $$(f(at))^{\hat{}}(u) = \int_{-\infty}^{+\infty} f(at) e^{-j2\pi ut} \mathrm{d}t = \frac{1}{|a|}\int_{-\infty}^{+\infty} f(z) e^{-j2\pi \frac{u}{a}z} dz = \frac{1}{|a|} F(\frac{u}{a}).$$

10. 导函数:

    $$\begin{array}{ll} f'(t) &= \frac{\mathrm{d}}{\mathrm{d}t} \int_{-\infty}^{+\infty} F(u) e^{j2\pi t u} \mathrm{d}u \\ &= \int_{-\infty}^{+\infty} F(u) \frac{\mathrm{d}}{\mathrm{d}t} e^{j2\pi t u} \mathrm{d}u \\ &= \int_{-\infty}^{+\infty} F(u) (j2\pi u) e^{j2\pi t u} \mathrm{d}u \\ &= (j2\pi u F(u))^{\check{}}. \end{array}$$

11. 同上

12. 这个性质说明了, 对原图像加以旋转, 等价地在频域上加以同样的旋转.

    $$\begin{array}{ll} G(\mathbf{u}) &=\int f(R\mathbf{t}) e^{-j2 \pi \mathbf{u}^T \mathbf{t}} \mathrm{d}\mathbf{t} \\ &=\int f(\mathbf{x}) e^{-j2 \pi (R\mathbf{u})^T \mathbf{x}} |R^{-1}|\mathrm{d}\mathbf{x} \\ &=\int f(\mathbf{x}) e^{-j2 \pi (R\mathbf{u})^T \mathbf{x}} \mathrm{d}\mathbf{x} \\ &=F(R\mathbf{u}) \end{array}$$

对称性

主要指的:

1. 奇函数

   $$f(x) = -f(-x).$$

2. 偶函数

   $$f(x) = f(-x).$$

3. 共轭对称

   $$f^*(x) = f(-x).$$

考查$f(t)=r(t) + j i(t)$与$F(u) = R(u) + j I(u)$之间的关系:

	$\hat{ }$	$\Leftrightarrow$	$\check{ }$
1	$f(t) = r(t)$	$\Leftrightarrow$	$F^*(u) = F(-u)$
2	$f(t)=r(t)$	$\Leftrightarrow$	$R(u)=R(-u), I(u)=-I(-u)$
3	$f(t) = ji(t)$	$\Leftrightarrow$	$F^*(-u) = -F(u)$
4	$f(t) = ji(t)$	$\Leftrightarrow$	$R(u)=-R(-u), I(u)=I(-u)$

	$\hat{}$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$\check{}$
5	$f(-t)\in \mathbb{R}$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F^*(u) \in \mathbb{C}$
6	$f(-t) \in \mathbb{C}$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(-u) \in \mathbb{C}$
7	$f^*(t) \in \mathbb{C}$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F^*(-u) \in \mathbb{C}$
8	$f(t) \in \mathbb{R}$, 且$f(t) = f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) \in \mathbb{R}$, 且满足$F(u)=F(-u)$
9	$f(t) \in \mathbb{R}$, 且$f(t) = -f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) = jI(u) \in \mathbb{C}$, 且$F(u)=-F(-u)$
10	$f(t) = ji(t) \in \mathbb{C}$, 且$f(t) = f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) = jI(u) \in \mathbb{C}$, 且$F(u)=F(-u)$
11	$f(t)=ji(t) \in \mathbb{C}$, 且$f(t)=-f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) \in \mathbb{R}$, 且满足$F(u)=-F(-u)$
12	$f(t) \in \mathbb{C}$, 且$f(t) = f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) \in \mathbb{C}$, 且满足$F(u) = F(-u)$
13	$f(t) \in \mathbb{C}$, 且$f(t) = -f(-t)$	$\mathop{\Leftrightarrow} \limits^{\mathcal{F}}$	$F(u) \in \mathbb{C}$, 且满足$F(u) = -F(-u)$

1. real ->($\Leftarrow$采用反证法即可)

   $$\begin{array}{ll} F^*(u) &=[\int f(t) e^{-j2\pi ut} \mathrm{d}t]^* \\ &=\int f^*(t) e^{j2\pi ut} \mathrm{d}t \\ &=\int f(t) e^{-j2\pi (-u)t} \mathrm{d}t \rightarrow f*(t) = r^*(t) = r(t)\\ &=F(-u). \end{array}$$

2. 由1易证.

3. imaginary ->

   $$\begin{array}{ll} F^*(-u) &=[\int f(t) e^{j2\pi ut} \mathrm{d}t]^* \\ &=\int f^*(t) e^{-j2\pi ut} \mathrm{d}t \\ &=\int -f(t) e^{-j2\pi ut} \mathrm{d}t \rightarrow f^*(t) = -ji(t)=-f(t) \\ &=-F(u). \end{array}$$

4. 由3易证.

5. 由性质6以及上述的对称性3可知$(f(-t))^{\hat{}}= F(-u) = F^*(u)$.

6. 由性质6可知$(f(-t))^{\hat{}} = F(-u)$.

7. $$\int f^*(t) e^{-j2\pi ut} \mathrm{d}t = [\int f(t) e^{j2\pi ut} \mathrm{d}t]^* = F^*(-u).$$

8. $$F(u) = (f(t))^{\hat{}} = (f(-t))^{\hat{}} = F(-u) = F^*(u).$$

9. $$F(u) = (f(t))^{\hat{}} = -(f(-t))^{\hat{}} = -F(-u) = -F^*(u).$$

10. $$F(u) = (f(t))^{\hat{}} = (f(-t))^{\hat{}} = F(-u) = -F^*(u).$$

11. $$F(u) = (f(t))^{\hat{}} = -(f(-t))^{\hat{}} = -F(-u) = F^*(u).$$

12. $$F(u) = (f(t))^{\hat{}} = (f(-t))^{\hat{}} = F(-u).$$

13. $$F(u) = (f(t))^{\hat{}} = -(f(-t))^{\hat{}} = -F(-u).$$

卷积

这里特别说明离散卷积的一个重要性质:
注意到

$$h \star f (n) = \sum_{k=0}^{N-1} h(n-k) f(k) = a^T_n f, \\ a(n) = [h(0), h(n-1), \cdots, h(n-N+1)],$$

则

$$A = [a_0, a_1, \cdots, a_{N-1}] = \left [ \begin{array}{cccc} h(0) & h(1) &\cdots & h(N-1) \\ h(-1) & h(0) & \cdots & h(N-2)\\ \vdots & \vdots & \ddots & \vdots \\ h(1-N) & h(2-N) & \cdots & h(0), \end{array} \right ] [h \star f] = A^T f.$$

定义

$$h'(n) = h(-n),$$

则有

$$[h' \star f] = Af.$$

FFT

考虑离散傅里叶变换:

$$F_m = \sum_{n=0}^{N-1} f_n e^{-j2\pi nm/N}, m=0,1,\cdots, N-1.$$

每次计算包含$N$个乘法, 故傅里叶变换的计算量是$N^2$级别的.

下面假设$N = P * Q, P, Q \in \mathbb{N}^+$, 则

$$m = uQ + v, \quad u=0,1,\cdots, P-1, v=0,1,\cdots, Q-1, \\ n = sP + t, \quad s=0,1,\cdots, Q-1, t=0,1,\cdots, P-1. \\ $$

注意到:

$$nm = (sP + t)(uQ + v) = su N + svP + tm.$$

此时, $m, n$与唯一的$(u,v), (s, t)$匹配, 则傅里叶变换可以表示为:

$$\begin{align} F_m &= \sum_{n=0}^{N-1}f_n e^{-j2\pi nm / N} \\ &= \sum_{n=0}^{N-1}f_n e^{-j2\pi (sP+t)(uP+v)/ N} \\ &= \sum_{n=0}^{N-1}f_n e^{-j2\pi (suN + svP + tm)/ N} \\ &= \sum_{n=0}^{N-1}f_n e^{-j2\pi (svP + tm)/ N} \\ &= \sum_{t=0}^{P-1}e^{-j2\pi tm / N}\sum_{s=0}^{Q-1} f_{sP+t} e^{-j2\pi sv/ Q} \\ \end{align}$$

令

$$F_v^t := \sum_{s=0}^{Q-1} f_{sP+t} e^{-j2\pi sv /Q}, \quad v = 0,1, \cdots, Q-1, t=0,1,\cdots, P-1.$$

显然总共需要$QN$次乘法, 而

$$F_m = \sum_{t=0}^{P-1} e^{-j2\pi tm / N} F_v^t, \quad m=0,1,\cdots, N-1, \\ F_{uQ+v} = \sum_{t=0}^{P-1} e^{-j2\pi tv /N}F_v^t e^{-j2\pi tu / P} .$$

总共有$PN$次乘法, 故总共有$(P+Q) N$次乘法运算.
由于每个$F^t$都是一个独立的傅里叶变换过程, 故倘若$Q$能够进一步分解, 计算量可以进一步降低.
甚至, 若假设$N = P^J$, 则可以证明最后的计算量可以分解为

$$PJN = P N\log_P N,$$

特别的, 常常$N=2^J$, 则计算量是$N\log_2 N$量级的.

注: 卷积运算$[f\star g]_n, n=0,\cdots, N-1$, 计算量也是$N^2$级别的, 但是通过FFT, 应当是$(PN\log_P + 1)N$​乘法次数, 也是可以降低计算量的.

逆傅里叶变换实际上就是

$$\hat{F^*}^* = [\sum_{m=0}^{N-1} F_m^* e^{-j2\pi mn/N}]^*,$$

虽然下面的代码我并没有采取这种策略.

from typing import Iterable, Union
import numpy as np



def factorization(N: int):
    for P in range(2, N + 1):
        if N % P is 0:
            return P

def dft(arr: Iterable, m: int):
    N = len(arr)
    basis = np.arange(N) * m * 2 * np.pi * 1j * -1
    basis = np.exp(basis / N)
    return arr @ basis

def idft(arr: Iterable, m: int):
    N = len(arr)
    basis = np.arange(N) * m * 2 * np.pi * 1j
    basis = np.exp(basis / N) / N
    return arr @ basis

def basis_dft(N: int):
    basis = np.outer(np.arange(N), np.arange(N))
    basis = basis * 2 * np.pi * 1j * -1
    return np.exp(basis / N)

def basis_idft(N: int):
    basis = np.outer(np.arange(N), np.arange(N))
    basis = basis * 2 * np.pi * 1j
    return np.exp(basis / N) / N

def fft(arr: Iterable) -> np.ndarray:
    N = len(arr)
    P = factorization(N)
    if P is N:
        return arr @ basis_dft(N)
    Q = N // P
    Fvt = np.array([fft(arr[t::P]) for t in range(P)]).T # Q x P
    dummy = np.outer(np.arange(Q), np.arange(P))
    dummy = np.exp((dummy * 2 * np.pi * 1j * -1) / N)
    Fvt_ext = Fvt * dummy
    return (Fvt_ext @ basis_dft(P)).T.flatten()

def ifft(arr: Iterable) -> np.ndarray:
    N = len(arr)
    P = factorization(N)
    if P is N:
        return arr @ basis_idft(N)
    Q = N // P
    Fvt = np.array([ifft(arr[t::P]) for t in range(P)]).T # Q x P
    dummy = np.outer(np.arange(Q), np.arange(P))
    dummy = np.exp((dummy * 2 * np.pi * 1j) / N)
    Fvt_ext = Fvt * dummy
    return (Fvt_ext @ basis_idft(P)).T.flatten()