在Python中过滤信号频率

到目前为止,这么好

import numpy as np
import matplotlib.pyplot as plt
from scipy.fftpack import fftfreq, irfft, rfft

t = np.linspace(0, 2*np.pi, 1000, endpoint=True)
f = 3.0 # Frequency in Hz
A = 100.0 # Amplitude in Unit
s = A * np.sin(2*np.pi*f*t) # Signal
dt = t[1] - t[0] # Sample Time

W = fftfreq(s.size, d=dt)
f_signal = rfft(s)

cut_f_signal = f_signal.copy()
cut_f_signal[(np.abs(W)>3)] = 0 # cut signal above 3Hz

cs = irfft(cut_f_signal)

fig = plt.figure(figsize=(10,5))
plt.plot(s)
plt.plot(cs)

我的期望

我得到了什么

我真的不知道噪音来自哪里.
我认为这是一些基本的东西,但我不明白.
有人可以向我解释一下吗？

编辑

更多信息

频率

yf = fft(s)
N = s.size
xf = np.linspace(0, fa/2, N/2, endpoint=True)
fig, ax = plt.subplots()
ax.plot(xf,(2.0/N * np.abs(yf[:N//2])))
plt.xlabel('Frequency ($Hz$)')
plt.ylabel('Amplitude ($Unit$)')
plt.show()

fs = 1000
t = np.linspace(0, 1000 / fs, 1000, endpoint=False) # 1000 samples
f = 3.0 # Frequency in Hz
A = 100.0 # Amplitude in Unit
s = A * np.sin(2*np.pi*f*t) # Signal
dt = 1/fs

这里是整个代码：

import numpy as np
import matplotlib.pyplot as plt
from scipy.fftpack import fftfreq, irfft, rfft

fs = 1000
t = np.linspace(0, 1000 / fs, 1000, endpoint=False)
f = 3.0 # Frequency in Hz
A = 100.0 # Amplitude in Unit
s = A * np.sin(2*np.pi*f*t) # Signal
dt = 1/fs

W = fftfreq(s.size, d=dt)
f_signal = rfft(s)

cut_f_signal = f_signal.copy()
cut_f_signal[(np.abs(W)>3)] = 0 # cut signal above 3Hz

cs = irfft(cut_f_signal)

fig = plt.figure(figsize=(10,5))
plt.plot(s)
plt.plot(cs)

并且f = 3.0Hz并且(np.abs(W)> = 3)：

并且f = 1.0 Hz：