pythonfilterbutterworth

Convert 2nd order Butterworth to 1st order - Part II -


I'm trying to convert my working 2nd-order Butterworth low pass filter to 1st-order in python, but it gives me very big numbers, like flt_y_1st[299]: 26198491071387576370322954146679741443295686950912.0. Here's my 2nd-order and 1st-order Butterworth:

import math
import numpy as np
import matplotlib.pyplot as plt

from scipy.signal import lfilter
from scipy.signal import butter

def butter_lowpass(cutoff, fs, order=1):
    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=1):
    b, a = butter_lowpass(cutoff, fs, order=order)
    y = lfilter(b, a, data)
    return y

def bw_2nd(y, fc, fs):
    filtered_y = np.zeros(len(y))

    omega_c = math.tan(np.pi*fc/fs)
    k1 = np.sqrt(2)*omega_c
    k2 = omega_c**2
    a0 = k2/(1+k1+k2)
    a1 = 2*a0
    a2 = a0
    k3 = 2*a0/k2
    b1 = -(-2*a0+k3)
    b2 = -(1-2*a0-k3)

    filtered_y[0] = y[0]
    filtered_y[1] = y[1]

    for i in range(2, len(y)):
        filtered_y[i] = a0*y[i]+a1*y[i-1]+a2*y[i-2]-(b1*filtered_y[i-1]+b2*filtered_y[i-2])

    return filtered_y


def bw_1st(y, fc, fs):
    filtered_y = np.zeros(len(y))

    omega_c = math.tan(np.pi*fc/fs)
    k1 = np.sqrt(2)*omega_c
    k2 = omega_c**2
    a0 = k2/(1+k1+k2)
    a1 = 2*a0
    k3 = 2*a0/k2
    b1 = -(-2*a0+k3)
#    b1 = -(-2*a0)  # <= Removing k3 makes better, but still not perfect

    filtered_y[0] = y[0]

    for i in range(1, len(y)):
       filtered_y[i] = a0*y[i]+a1*y[i-1]-(b1*filtered_y[i-1])

    return filtered_y

f = 100
fs = 2000
x = np.arange(300)
y = np.sin(2 * np.pi * f * x / fs)

flt_y_2nd = bw_2nd(y, 120, 2000)
flt_y_scipy = butter_lowpass_filter(y, 120, 2000, 1)
flt_y_1st = bw_1st(y, 120, 2000)

for i in x:
    print('y[%d]: %6.3f flt_y_2nd[%d]: %6.3f flt_y_scipy[%d]: %6.3f flt_y_1st[%d]: %8.5f' % (i, y[i], i, flt_y_2nd[i], i, flt_y_scipy[i], i, flt_y_1st[i]))

plt.subplot(1, 1, 1)
plt.xlabel('Time [ms]')
plt.ylabel('Acceleration [g]')
lines = plt.plot(x, y, x, flt_y_2nd, x, flt_y_scipy, x, flt_y_1st)
l1, l2, l3, l4 = lines
plt.setp(l1, linewidth=1, color='g', linestyle='-')
plt.setp(l2, linewidth=1, color='b', linestyle='-')
plt.setp(l3, linewidth=1, color='y', linestyle='-')
plt.setp(l4, linewidth=1, color='r', linestyle='-')
plt.legend(["y", "flt_y_2nd", "flt_y_scipy", "flt_y_1st"])
plt.grid(True)
plt.xlim(0, 150)
plt.ylim(-1.5, 1.5)
plt.title('flt_y_2nd vs. flt_y_scipy vs. flt_y_1st')
plt.show()

... I removed all [i-2]s, which were a feed-forward and a feed-back.

b1 = -(-2*a0+k3)

However, it seems that's not enough. I think I need to change some equations in a0, b1, etc. For example, when I remove '+k3' from b1, I get a plot like this (looks better, doesn't it?):

b1 = -(-2*a0)

I'm not specialized in filters, but at least I know that this 1st order differs from that of scipy.butter. So, please help me find correct coefficients. Thank you in advance.

Here's my reference: filtering_considerations.pdf


Solution

  • Let me answer to myself.

    The final coefficients are:

    omega_c = math.tan(np.pi*fc/fs)
    k1 = np.sqrt(2)*omega_c
    a0 = k1/(math.sqrt(2)+k1)
    a1 = a0
    b1 = -(1-2*a0)
    

    Here's how. I reverse-engineered them from scipy.butter, as @sizzzzlerz suggested (thanks). scipy.butter spits out these coefficients:

    b:  [ 0.16020035  0.16020035]
    a:  [ 1.        -0.6795993]
    

    Note that b and a are reversed from my reference. They're gonna be:

    a0 = 0.16020035
    a1 = 0.16020035
    b0 = 1
    b1 = -0.6795993
    

    Then, applied these coefficients to my incomplete formula:

    a1 = a0 = 0.16020035
    b1 = -(1-2*a0) = -{1-2*(0.16020035)} = -(0.6795993)
    

    So far, so good. Incidentally:

    k1 = 0.2698
    k2 = 0.0364
    

    So:

    a0 = k2/(1+k1+k2) = 0.0364/(1+0.2698+0.0364) = 0.0279
    

    ... This is far from 0.16020035. At this point, I eliminated k2 and put like this:

    a0 = k1/(1+k1+x)
    

    When x = 0.4142, I got 0.16020164. Close enough.

    a0 = k1/(1+k1+0.4142) = k1/(1.4142+k1)
    

    ... 1.4142 ...!? I've ever seen this number before ...:

    = k1/(math.sqrt(2)+k1)
    

    Now the plot looks like this (flt_y_scipy is completely covered by flt_y_1st):

    k1/(math.sqrt(2)+k1)

    You can search for the keywords "first order" butterworth "low pass filter" "sqrt(2)", etc.

    This is the end of my Sunday DIY. ;-)