使用python的二维FFT导致频率稍微偏移

Question

我知道在python中使用快速傅里叶变换（FFT）方法有几个问题，但遗憾的是，它们都不能帮我解决问题：

我想用python来计算给定二维信号f的快速傅里叶变换，即f（x，y）。 Pythons文档有很多帮助，解决了FFT带来的一些问题，但与我期望它显示的频率相比，我的最终频率仍然略有偏移。这是我的python代码：

from scipy.fftpack import fft, fftfreq, fftshift
import matplotlib.pyplot as plt
import numpy as np
import math

fq = 3.0 # frequency of signal to be sampled
N = 100.0 # Number of sample points within interval, on which signal is considered
x = np.linspace(0, 2.0 * np.pi, N) # creating equally spaced vector from 0 to 2pi, with spacing 2pi/N
y = x
xx, yy = np.meshgrid(x, y) # create 2D meshgrid
fnc = np.sin(2 * np.pi * fq * xx) # create a signal, which is simply a sine function with frequency fq = 3.0, modulating the x(!) direction
ft = np.fft.fft2(fnc) # calculating the fft coefficients

dx = x[1] - x[0] # spacing in x (and also y) direction (real space)
sampleFrequency = 2.0 * np.pi / dx
nyquisitFrequency = sampleFrequency / 2.0

freq_x = np.fft.fftfreq(ft.shape[0], d = dx) # return the DFT sample frequencies 
freq_y = np.fft.fftfreq(ft.shape[1], d = dx)

freq_x = np.fft.fftshift(freq_x) # order sample frequencies, such that 0-th frequency is at center of spectrum 
freq_y = np.fft.fftshift(freq_y)

half = len(ft) / 2 + 1 # calculate half of spectrum length, in order to only show positive frequencies

plt.imshow(
    2 * abs(ft[:half,:half]) / half,
    aspect = 'auto',
    extent = (0, freq_x.max(), 0, freq_y.max()),
    origin = 'lower',
    interpolation = 'nearest',
)
plt.grid()
plt.colorbar()
plt.show()

运行它时我得到的是：

现在您看到x方向的频率不完全在fq = 3，但略微向左移动。为什么是这样？ 我认为这与事实有关，即FFT是一种使用对称参数和

的算法

half = len(ft) / 2 + 1

用于显示适当位置的频率。但我不太明白究竟是什么问题以及如何解决它。

编辑：我也尝试使用更高的采样频率（N = 10000.0），这并没有解决问题，而是将频率稍微偏移到右边。所以我很确定问题不是采样频率。

注意：我意识到泄漏效应会导致非物理振幅，但在这篇文章中，我主要对正确的频率感兴趣。

Answer 1

我发现了一些问题

你使用2 * np.pi两次，如果你想要一个很好的整数个周期，你应该选择linspace或arg中的一个作为弧度正弦

另外np.linspace默认为endpoint=True，为您提供101而非100的额外积分

fq = 3.0 # frequency of signal to be sampled
N = 100 # Number of sample points within interval, on which signal is considered
x = np.linspace(0, 1, N, endpoint=False) # creating equally spaced vector from 0 to 2pi, with spacing 2pi/N
y = x
xx, yy = np.meshgrid(x, y) # create 2D meshgrid
fnc = np.sin(2 * np.pi * fq * xx) # create a signal, which is simply a sine function with frequency fq = 3.0, modulating the x(!) direction

你可以检查这些问题：

len(x)
Out[228]: 100

plt.plot(fnc[0])

现在修复linspace端点意味着你有一个偶数个fft箱，所以你将+ 1放在half calc

中

matshow()似乎有更好的默认设置，extent = (0, freq_x.max(), 0, freq_y.max()),中的imshow似乎会影响fft bin编号

from scipy.fftpack import fft, fftfreq, fftshift
import matplotlib.pyplot as plt
import numpy as np
import math

fq = 3.0  # frequency of signal to be sampled
N = 100  # Number of sample points within interval, on which signal is considered
x = np.linspace(0, 1, N, endpoint=False)  # creating equally spaced vector from 0 to 2pi, with spacing 2pi/N
y = x
xx, yy = np.meshgrid(x, y)  # create 2D meshgrid
fnc = np.sin(2 * np.pi * fq * xx)  # create a signal, which is simply a sine function with frequency fq = 3.0, modulating the x(!) direction

plt.plot(fnc[0])

ft = np.fft.fft2(fnc)  # calculating the fft coefficients

#dx = x[1] - x[0]  # spacing in x (and also y) direction (real space)
#sampleFrequency = 2.0 * np.pi / dx
#nyquisitFrequency = sampleFrequency / 2.0
#
#freq_x = np.fft.fftfreq(ft.shape[0], d=dx)  # return the DFT sample frequencies 
#freq_y = np.fft.fftfreq(ft.shape[1], d=dx)
#
#freq_x = np.fft.fftshift(freq_x)  # order sample frequencies, such that 0-th frequency is at center of spectrum 
#freq_y = np.fft.fftshift(freq_y)

half = len(ft) // 2  # calculate half of spectrum length, in order to only show positive frequencies

plt.matshow(
            2 * abs(ft[:half, :half]) / half,
            aspect='auto',
            origin='lower'
            )
plt.grid()
plt.colorbar()
plt.show()

放大了情节：