使用python进行二维FFT会使频率稍微偏移 [英] Two dimensional FFT using python results in slightly shifted frequency

问题描述

我知道在python中使用快速傅立叶变换(FFT)方法存在一些问题，但不幸的是，这些问题都无法帮助我解决问题:

I know there have been several questions about using the Fast Fourier Transform (FFT) method in python, but unfortunately none of them could help me with my problem:

我想使用python计算给定二维信号f(即f(x，y))的快速傅立叶变换. Python文档提供了很多帮助，解决了FFT带来的一些问题，但是与我希望它显示的频率相比，我最终仍然会稍微改变一下频率.这是我的python代码:

I want to use python to calculate the Fast Fourier Transform of a given two dimensional signal f, i.e. f(x,y). Pythons documentation helps a lot, solving a few issues, which the FFT brings with it, but i still end up with a slightly shifted frequency compared to the frequency i expect it to show. Here is my python code:

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()

运行它时我从中得到的是:

And what i get out of this when running it, is:

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

Now you see that the frequency in x direction is not exactly at fq = 3, but slightly shifted to the left. Why is this? I would assume that is has to do with the fact, that FFT is an algorithm using symmetry arguments and

half = len(ft) / 2 + 1

用于显示正确位置的频率.但是我不太清楚确切的问题是什么以及如何解决.

is used to show the frequencies at the proper place. But I don't quite understand what the exact problem is and how to fix it.

我也尝试过使用更高的采样频率(N = 10000.0)，这不能解决问题，而是将频率稍微向右移了一点.因此，我很确定问题不在于采样频率.

I have also tried using a higher sampling frequency (N = 10000.0), which did not solve the issue, but instead shifted the frequency slightly too far to the right. So i am pretty sure that the problem is not the sampling frequency.

注意:我知道一个事实，泄漏效应会导致此处出现非自然的振幅，但是在这篇文章中，我主要对正确的频率感兴趣.

Note: I'm aware of the fact, that the leakage effect leads to unphysical amplitudes here, but in this post I am primarily interested in the correct frequencies.

推荐答案

我发现了很多问题

您使用2 * np.pi两次，如果您想要一个整数倍的周期，则应该选择linspace或arg中的一个作为弧度正弦值.

you use 2 * np.pi twice, you should choose one of either linspace or the arg to sine as radians if you want a nice integer number of cycles

另外np.linspace默认为endpoint=True，为您提供101而不是100的加分

additionally np.linspace defaults to endpoint=True, giving you an extra point for 101 instead of 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

您可以检查以下问题:

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箱，因此您将+ 1放在half计算中

fixing the linspace endpoint now means you have an even number of fft bins so you drop the + 1 in the half calc

matshow()似乎具有更好的默认值，您在imshow中的extent = (0, freq_x.max(), 0, freq_y.max()),似乎无法使用fft bin编号

matshow() appears to have better defaults, your extent = (0, freq_x.max(), 0, freq_y.max()), in imshow appears to fubar the fft bin numbering

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()

放大了剧情:

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