计算科学 - 在 Python 中通过 FFT 和 IFFT 对高斯函数求导 - 吾爱随笔录

我在使用 NumPy 库中的 FFT 和 IFFT 计算高斯函数的导数时遇到问题。我使用的事实是

\frac{d}{d x} f (x) = \frac{1}{\sqrt{2 π}} \int i k e^{i k x} F (f (x)) (k) d k,

$\begin{equation} \frac{d}{dx}f(x) = \frac{1}{\sqrt{2\pi}}\int{ike^{ikx}\mathscr{F}(f(x))(k)dk}, \end{equation}$ 在哪里

F (f (x)) (k) = \frac{1}{\sqrt{2 π}} \int f (x) e^{- i k x} d x

$\mathscr{F}(f(x))(k) = \frac{1}{\sqrt{2\pi}}\int{f(x)e^{-ikx}dx}$ .

$\\$ 我的代码：

import numpy as np
import scipy as sci
import matplotlib.pyplot as plt


def Gauss(x, mean, sigma):
#GAUSS FUNCTION WITH ITS NORMALIZATION FACTOR
    return sci.pi**(-1/4) * sigma**(-1/2) * np.exp(-(x - mean)**2 / (2 * sigma**2))

def Fft(f):
#THAT dx AND np.sqrt(2*sci.pi) ARE SOME NORMALIZATION FACTORS
    return np.fft.fftshift(np.abs(np.fft.fft( f ) * dx / np.sqrt(2*sci.pi)))

def Ifft(fftf):
#THAT dx AND np.sqrt(2*sci.pi) ARE SOME NORMALIZATION FACTORS
    return np.abs(np.fft.fftshift(np.fft.ifft( fftf )).real) / dx * np.sqrt(2 * sci.pi)

def d_Gauss(x, sigma):
#FINAL DIFFERENCE, LATER DENOTED AS EXACT DERIVATIVE
    return (Gauss(x+dx, 0, sigma)-Gauss(x, 0, sigma))/dx

x      = np.linspace(-20, 20, 2**12)
dx     = np.abs(x[1] - x[0])
freq_x = np.fft.fftshift(np.fft.fftfreq(np.shape(x)[0])) / dx * 2 * sci.pi
dk     = np.abs(freq_x[0]-freq_x[1])
sigma  = 3/2
mean   = 0


#COMPUTING EXACT GAUSS DERIVATIVE AND IT'S APPROXIMATION VIA FFT AND IFFT
d_gauss_exact = d_Gauss(x, sigma)
d_gauss       = Gauss(x,mean,sigma)
d_gauss       = Fft(d_gauss)
d_gauss       = 1j*freq_x*d_gauss
d_gauss       = Ifft(d_gauss)


#PLOTTING
plt.figure('Gauss der.')
plt.plot(x, d_gauss,      'b--', label='Gauss der. fft')
plt.plot(x, d_gauss_exact,'r-',  label='Exact Gauss der.')
plt.xlim(-10,10)
plt.xlabel(r'$x$')
plt.ylabel(r'$\frac{d}{dx}g(x)$')
plt.legend()
plt.show()

我得到的是：

该代码适用于高斯函数、调制脉冲和洛伦兹函数的 FFT。在项目的下一部分中，该衍生产品对我至关重要。我将不胜感激任何提示和帮助。