python - 使用Numpy进行傅立叶变换

我正在尝试计算以下高斯的傅立叶变换：

# sample spacing
dx = 1.0 / 1000.0

# Points
x1 = -5
x2 = 5

x = np.arange(x1, x2, dx)

def light_intensity():
    return 10*sp.stats.norm.pdf(x, 0, 1)+0.1*np.random.randn(x.size)

fig, ax = plt.subplots()
ax.plot(x,light_intensity())

我在空间频域中创建了一个新数组（高斯的傅立叶变换是高斯的，因此这些值应该相似）。我绘图并得到这个：

fig, ax = plt.subplots()

xf = np.arange(x1,x2,dx)
yf= np.fft.fftshift(light_intensity())
ax.plot(xf,np.abs(yf))

为什么会分成两个高峰？

最佳答案

这里有一些建议：


使用np.fft.fft
fft从0 Hz开始
归一化/缩放


例：

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

def light_intensity(x, signal_gain, noise_gain):
    signal = norm.pdf(x, 0, 1)
    noise = np.random.randn(x.size)
    return signal_gain * signal + noise_gain * noise

def norm_fft(y, T, max_freq=None):
    N = y.shape[0]
    Nf = N // 2 if max_freq is None else int(max_freq * T)
    xf = np.linspace(0.0, 0.5 * N / T, N // 2)
    yf = 2.0 / N * np.fft.fft(y)
    return xf[:Nf], yf[:Nf]

x1 = 0.0
x2 = 5.0
N = 10000
T = x2 - x1

x = np.linspace(x1, x2, N)
y = light_intensity(x, 10.0, 0.0)
xf, yf = norm_fft(y, T, T / np.pi)

fig, ax = plt.subplots(2)
ax[0].plot(x, y)
ax[1].plot(xf, np.abs(yf))
plt.show()

或者，有噪音：





或者，如果您想享受频域中的对称性：

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

def light_intensity(x, signal_gain, noise_gain):
    signal = norm.pdf(x, 0, 1)
    noise = np.random.randn(x.size)
    return signal_gain * signal + noise_gain * noise

def norm_sym_fft(y, T, max_freq=None):
    N = y.shape[0]
    b = N if max_freq is None else int(max_freq * T + N // 2)
    a = N - b
    xf = np.linspace(-0.5 * N / T, 0.5 * N / T, N)
    yf = 2.0 / N * np.fft.fftshift(np.fft.fft(y))
    return xf[a:b], yf[a:b]

x1 = -10.0
x2 = 10.0
N = 10000
T = x2 - x1

x = np.linspace(x1, x2, N)
y = light_intensity(x, 10.0, 0.0)
xf, yf = norm_sym_fft(y, T, 4 / np.pi)

fig, ax = plt.subplots(2)
ax[0].plot(x, y)
ax[1].plot(xf, np.abs(yf))
plt.show()