Python高通Fi

Question

我使用以下代码在python中实现了高通过滤器：

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，它在截止频率>；=60 Hz时工作良好。但在下面不行。我想切断所有低于10赫兹的频率。有什么线索表明我错在哪里吗？我观察到滤波器阶数越低，截止频率就越低。

The sample Signal can be found here.

Answer 1

我添加了验证上述Konstantin Purtov代码的函数，这样您就可以看到顺序和截止频率之间的关系。代码主要来自Warren Weckesser。

import scipy
import sys  
from scipy import signal
from scipy import pi
from scipy.io.wavfile import write
import matplotlib.pyplot as plt
import numpy as np    
from scipy.signal import butter, lfilter, freqz   


# ----- ----- ----- -----    
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


# ----- -----    
# (1)
def foo(sel):
    if (sel == 1):
        # Filter requirements.
        order = 6
        fs = 300.0  # sample rate, Hz
        cutoff = 10  # desired cutoff frequency of the filter, Hz

        # Get the filter coefficients so we can check its frequency response.
        b, a = butter_highpass(cutoff, fs, order)

        # Plot the frequency response.
        w, h = freqz(b, a, worN=8000)
        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("High Filter Frequency Response")
        plt.xlabel('Frequency [Hz]')
        plt.grid()

        # Demonstrate the use of the filter.
        # First make some data to be filtered.
        T = 0.5  # seconds
        n = int(T * fs)  # total number of samples
        t = np.linspace(0, T, n, endpoint=False)
        # "Noisy" data.  We want to recover the 20 Hz signal from this.
        data = np.sin(1.2 * 2 * np.pi * t) + 1.5 * np.cos(5 * 2 * np.pi * t) + 0.5 * np.sin(20.0 * 2 * np.pi * t)

        # Filter the data, and plot both the original and filtered signals.
        y = butter_highpass_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()
    else:
        print ('Please, choose among choices, thanks.')


# ----- -----
def main():
    sel = int (sys.argv[1])
    foo(sel)


# ----- ----- ----- ----- ----- -----
if __name__ == '__main__':
    main()

Answer 2

我希望这能帮助你：

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()

Answer 3

由于我的声誉很低，我无法对你的问题发表评论：“截止和过滤顺序之间的关系是什么？”这不是你原来问题的答案。

对于FIR滤波器，对于给定的截止频率，对于高阶滤波器，脉冲响应图的斜率（| H（f）| vs f）更陡。因此，为了在不需要的频率范围内获得更高的衰减，可以增加滤波器阶数。但是，当滤波器阶数如此之高以至于脉冲响应是一个理想的盒函数时，会发生什么呢？你会看到一种类似符号间干扰（数字通信中的ISI）的效果。当截止频率与采样频率之比变小时（考虑频域中盒函数宽度与sinc函数主瓣宽度之间的关系），这种影响的强度增大。

我第一次观察到这一点是在TI DSP微控制器上实现一个非常窄带的低通IIR滤波器时。TI库将滤波器实现为级联的双四元结构，以处理众所周知的截断效应。这仍然不能解决这个问题，因为这个问题并不仅仅是由于截断造成的。我解决这个问题的方法是使用抗混叠滤波器，然后对输入信号进行下采样，然后使用我想要的低通IIR滤波器。

我知道你正在实现一个HPF，这是一个在频域中翻译的LPF。希望这能回答你的一些问题。让我知道下采样对你是否有效。