Python 高通滤波器

Question

我使用这段代码在 python 中实现了一个高通滤波器:

from scipy.signal import butter, filtfilt
import numpy as np

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

def butter_highpass_filter(data, cutoff, fs, order=5):
    b, a = butter_highpass(cutoff, fs, order=order)
    y = filtfilt(b, a, data)
    return y

rawdata = np.loadtxt('sampleSignal.txt', skiprows=0)
signal = rawdata
fs = 100000.0

cutoff = 100
order = 6
conditioned_signal = butter_highpass_filter(signal, cutoff, fs, order)

我将此滤波器应用于 100 kHz 电压信号，它适用于 >= 60 Hz 的截止频率。但它在下面不起作用。我想切断所有低于 10 Hz 的频率。任何提示我的错误在哪里？我观察到的是滤波器的阶数越低，截止频率就越低。

The sample Signal can be found here.

Answer 1

希望对您有所帮助

import numpy as np
import pandas as pd
from scipy import signal
import matplotlib.pyplot as plt
def sine_generator(fs, sinefreq, duration):
    T = duration
    nsamples = fs * T
    w = 2. * np.pi * sinefreq
    t_sine = np.linspace(0, T, nsamples, endpoint=False)
    y_sine = np.sin(w * t_sine)
    result = pd.DataFrame({ 
        'data' : y_sine} ,index=t_sine)
    return result

def butter_highpass(cutoff, fs, order=5):
    nyq = 0.5 * fs
    normal_cutoff = cutoff / nyq
    b, a = signal.butter(order, normal_cutoff, btype='high', analog=False)
    return b, a

def butter_highpass_filter(data, cutoff, fs, order=5):
    b, a = butter_highpass(cutoff, fs, order=order)
    y = signal.filtfilt(b, a, data)
    return y

fps = 30
sine_fq = 10 #Hz
duration = 10 #seconds
sine_5Hz = sine_generator(fps,sine_fq,duration)
sine_fq = 1 #Hz
duration = 10 #seconds
sine_1Hz = sine_generator(fps,sine_fq,duration)

sine = sine_5Hz + sine_1Hz

filtered_sine = butter_highpass_filter(sine.data,10,fps)

plt.figure(figsize=(20,10))
plt.subplot(211)
plt.plot(range(len(sine)),sine)
plt.title('generated signal')
plt.subplot(212)
plt.plot(range(len(filtered_sine)),filtered_sine)
plt.title('filtered signal')
plt.show()