python实现低通滤波器代码-创新互联

低通滤波器实验代码，这是参考别人网上的代码，所以自己也分享一下，共同进步

# -*- coding: utf-8 -*-

import numpy as np
from scipy.signal import butter, lfilter, freqz
import matplotlib.pyplot as plt


def butter_lowpass(cutoff, fs, order=5):
 nyq = 0.5 * fs
 normal_cutoff = cutoff / nyq
 b, a = butter(order, normal_cutoff, btype='low', analog=False)
 return b, a


def butter_lowpass_filter(data, cutoff, fs, order=5):
 b, a = butter_lowpass(cutoff, fs, order=order)
 y = lfilter(b, a, data)
 return y # Filter requirements.


order = 6
fs = 30.0 # sample rate, Hz
cutoff = 3.667 # desired cutoff frequency of the filter, Hz # Get the filter coefficients so we can check its frequency response.
b, a = butter_lowpass(cutoff, fs, order) # Plot the frequency response.
w, h = freqz(b, a, worN=800)
plt.subplot(2, 1, 1)
plt.plot(0.5*fs*w/np.pi, np.abs(h), 'b')
plt.plot(cutoff, 0.5*np.sqrt(2), 'ko')
plt.axvline(cutoff, color='k')
plt.xlim(0, 0.5*fs)
plt.title("Lowpass Filter Frequency Response")
plt.xlabel('Frequency [Hz]')
plt.grid() # Demonstrate the use of the filter. # First make some data to be filtered.
T = 5.0 # seconds
n = int(T * fs) # total number of samples
t = np.linspace(0, T, n, endpoint=False) # "Noisy" data. We want to recover the 1.2 Hz signal from this.
data = np.sin(1.2*2*np.pi*t) + 1.5*np.cos(9*2*np.pi*t) + 0.5*np.sin(12.0*2*np.pi*t) # Filter the data, and plot both the original and filtered signals.
y = butter_lowpass_filter(data, cutoff, fs, order)
plt.subplot(2, 1, 2)
plt.plot(t, data, 'b-', label='data')
plt.plot(t, y, 'g-', linewidth=2, label='filtered data')
plt.xlabel('Time [sec]')
plt.grid()
plt.legend()
plt.subplots_adjust(hspace=0.35)
plt.show()

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