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本章主要讲解频域域滤波的技术，主要技术用到是大家熟悉的傅里叶变换与傅里叶反变换。这里有比较多的篇幅讲解的傅里叶的推导进程，用到Numpy傅里叶变换。本章理论基础比较多，需要更多的耐心来阅读，有发现有错误，可以与我联系。谢谢！

import sysimport numpy as npimport cv2import matplotlib import matplotlib.pyplot as pltimport PILfrom PIL import Imageprint(f"Python version: {sys.version}")print(f"Numpy version: {np.__version__}")print(f"Opencv version: {cv2.__version__}")print(f"Matplotlib version: {matplotlib.__version__}")print(f"Pillow version: {PIL.__version__}")

Python version: 3.6.12 |Anaconda, Inc.| (default, Sep  9 2020, 00:29:25) [MSC v.1916 64 bit (AMD64)]Numpy version: 1.16.6Opencv version: 3.4.1Matplotlib version: 3.3.2Pillow version: 8.0.1

def normalize(mask):    return (mask - mask.min()) / (mask.max() - mask.min() + 1e-8)

背景

傅里叶级数和变换简史

内容比较多，请自行看书，我就实现一维的傅里叶变换先。

卷积用大小为$m\times n$元素的核对大小为$M\times N$的图像进行滤波时，需要运算次数为$MNmn$。如果核是可分享的，那么运算次数为$MN(m+N)$，而在频率域执行等交的滤波所需要的运算次数为$2MN{\text{log}}_{2}MN$，2表示计算一次正FFT和一次反FFT。

$$C_n(m)=\frac{M^2{m}^2}{2M^2{\text{log}}_2}{M}^2=\frac{m^2}{4{\text{log}}_{2}M}\text{\hspace{1em}(4.1)}$$

如果是可分离核，则变为

$$C_s(m)=\frac{M^2{m}^2}{2M^2{\text{log}}_2{M}^2}=\frac{m}{2{\text{log}}_{2}M}\text{\hspace{1em}(4.2)}$$

当$C(m)>1$时，FFT的方法计算优势更大；而$C(m)\le 1$时，空间滤波的优势更大

# FFT 计算的优势M = 2048m = np.arange(0, 1024, 1)c_n = m**2 / (4 * np.log2(M))c_s = m / (2 * np.log2(M))fig = plt.figure(figsize=(10, 5))ax_1 = fig.add_subplot(1, 2, 1)ax_1.plot(c_n)ax_1.set_xlim([0, 1024])ax_1.set_xticks([3, 255, 511, 767, 1023])ax_1.set_ylim([0, 25*1e3])ax_1.set_yticks([0, 5*1e3, 10*1e3, 15*1e3, 20*1e3, 25*1e3])ax_2 = fig.add_subplot(1, 2, 2)ax_2.plot(c_s)ax_2.set_xlim([0, 1024])ax_2.set_xticks([3, 255, 511, 767, 1023])ax_2.set_ylim([0, 5])ax_2.set_yticks([0, 10, 20, 30, 40, 50])plt.show()

def set_spines_invisible(ax):    ax.spines['left'].set_color('none')    ax.spines['right'].set_color('none')    ax.spines['top'].set_color('none')    ax.spines['bottom'].set_color('none')

# 不同频率的叠加x = np.linspace(0, 1, 500)t = 50A = 1y_1 = A * np.sin(t * 2 * np.pi * x)t = 20A = 2.5y_2 = A * np.sin(t * 2 * np.pi * x)t = 5A = 3y_3 = A * np.sin(t * 2 * np.pi * x)t = 2A = 20y_4 = A * np.sin(t * 2 * np.pi * x)y_5 = y_1 + y_2 + y_3 + y_4fig = plt.figure(figsize=(8, 8))ax_1 = fig.add_subplot(5, 1, 1)plt.plot(x, y_1), plt.xticks([]), plt.yticks([])set_spines_invisible(ax_1)ax_2 = fig.add_subplot(5, 1, 2)plt.plot(x, y_2), plt.xticks([]), plt.yticks([])set_spines_invisible(ax_2)ax_3 = fig.add_subplot(5, 1, 3)plt.plot(x, y_3), plt.xticks([]), plt.yticks([])set_spines_invisible(ax_3)ax_4 = fig.add_subplot(5, 1, 4)plt.plot(x, y_4), plt.xticks([]), plt.yticks([])set_spines_invisible(ax_4)ax_5 = fig.add_subplot(5, 1, 5)plt.plot(x, y_5), plt.xticks([]), plt.yticks([])set_spines_invisible(ax_5)plt.tight_layout()plt.show()

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