我知道关于在 python 中使用快速傅立叶变换（FFT）方法存在几个问题，但不幸的是它们都不能帮助我解决我的问题：
我想使用python计算给定二维信号f的快速傅里叶变换，即f(x,y)。 Python 文档帮助很大，解决了 FFT 带来的一些问题，但与我期望它显示的频率相比，我最终得到的频率仍然略有偏移。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：
现在您会发现 x 方向上的频率并不完全位于fq = 3，但稍微向左移动。为什么是这样？
我认为这与以下事实有关：FFT 是一种使用对称参数的算法，并且
half = len(ft) / 2 + 1
用于在适当的位置显示频率。但我不太明白确切的问题是什么以及如何解决它。
编辑：我还尝试使用更高的采样频率（N = 10000.0），这并没有解决问题，而是将频率稍微向右移动了太多。所以我很确定问题不在于采样频率。
注意：我知道泄漏效应会导致这里的非物理振幅，但在这篇文章中我主要对正确的频率感兴趣。
我发现了很多问题
you use 2 * np.pi两次，如果你想要一个很好的整数周期数，你应该选择 linspace 或 arg 到正弦弧度之一
此外np.linspace默认为endpoint=True，给您额外的分数 101 而不是 100
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 端点意味着您有偶数个 fft bin，因此您可以删除+ 1 in the half calc
matshow()似乎有更好的默认值，你的extent = (0, freq_x.max(), 0, freq_y.max()), in imshow似乎 fubar 的 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()
zoomed the plot:
