频率为1000Hz的正弦波的频域 [英] Frequency domain of a sine wave with frequency 1000Hz

问题描述

我正在Python上启动DSP，但遇到了一些困难: 我正在尝试定义频率为1000Hz的正弦波

I'm starting DSP on Python and I'm having some difficulties: I'm trying to define a sine wave with frequency 1000Hz

我尝试执行FFT并使用以下代码找到其频率:

I try to do the FFT and find its frequency with the following piece of code:

import numpy as np

import matplotlib.pyplot as plt

sampling_rate = int(10e3)
n = int(10e3)

sine_wave = [100*np.sin(2 * np.pi * 1000 * x/sampling_rate) for x in range(0, n)]

s = np.array(sine_wave)

print(s)

plt.plot(s[:200])
plt.show()

s_fft = np.fft.fft(s)
frequencies = np.abs(s_fft)
plt.plot(frequencies)
plt.show()

因此，第一个情节对我来说很有意义. 第二个图(FFT)显示两个频率: i)1000Hz，这是我在开始时设置的 ii)9000Hz，出乎意料 频率域

So first plot makes sense to me. Second plot (FFT) shows two frequencies: i) 1000Hz, which is the one I set at the beggining ii) 9000Hz, unexpectedly freqeuncy domain

推荐答案

您的数据不尊重

Your data do not respect Shannon criterion. you do not set a correct frequencies axis.

当信号为实信号时，使用 rfft 而不是 fft 也更容易.

It's easier also to use rfft rather than fft when the signal is real.

您的代码可以像这样进行修改:

Your code can be adapted like :

import numpy as np
import matplotlib.pyplot as plt

sampling_rate = 10000
n = 10000
signal_freq = 4000 # must be < sampling_rate/2 
amplitude = 100

t=np.arange(0,n/sampling_rate,1/sampling_rate)
sine_wave = amplitude*np.sin(2 * np.pi *signal_freq*t) 
plt.subplot(211)
plt.plot(t[:30],sine_wave[:30],'ro')

spectrum = 2/n*np.abs(np.fft.rfft(sine_wave))
frequencies = np.fft.rfftfreq(n,1/sampling_rate)
plt.subplot(212)
plt.plot(frequencies,spectrum)
plt.show()

输出:

即使人眼会因时间表示而困扰，也不会丢失任何信息.

There is no information loss, even if a human eye can be troubled by the temporal representation.

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