[Numpy-discussion] invalid value treatment, in filter_design

[R Schumacher]

In an attempt to computationally invert the effect of an analog RC 
filter on a data set and reconstruct the "true" signal, a co-worker 
suggested: "Mathematically, you just reverse the a and b parameters. 
Then the zeros become the poles, but if the new poles are not inside 
the unit circle, the filter is not stable."

So, to "stabilize" the poles' issue seen, I test for the DIV/0 error 
and set it to 2./N+0.j in scipy/signal/filter_design.py ~ line 244
     d = polyval(a[::-1], zm1)
     if d[0]==0.0+0.j:
         d[0] = 2./N+0.j
     h = polyval(b[::-1], zm1) / d

- Question is, is this a mathematically valid treatment?
- Is there a better way to invert a Butterworth filter, or work with 
the DIV/0 that occurs without modifying the signal library?
- Should I post to *-users instead?

I noted d[0] > 2./N+0.j makes the zero bin result spike low; 2/N 
gives a reasonable "extension" of the response curve. This whole 
tweak causes a zero offset however, which I remove.

An example attached...


Ray Schumacher
Programmer/Consultant
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import numpy as np
from scipy.signal import butter, lfilter, freqz
import matplotlib.pyplot as plt


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

def butter_inv_highpass(cutoff, fs, order=5):
    nyq = 0.5 * fs
    normal_cutoff = cutoff / nyq
    b, a = butter(order, Wn=normal_cutoff, btype='highpass', analog=False)
    ## swap the components
    return a, b

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

def butter_inv_highpass_filter(data, cutoff, fs, order=5):
    b, a = butter_inv_highpass(cutoff, fs, order=1)
    offset = data.mean()
    y = lfilter(b, a, data)
    ## remove new offset
    y -= (y.mean() - offset)
    return y


# Filter requirements.
order = 1
fs = 1024.0       # sample rate, Hz
cutoff = 11.6  # desired cutoff frequency of the filter, Hz
nyquist = fs/2.

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


# Plot the frequency response.
plt.subplot(2, 1, 1)
w, h = freqz(b, a, worN=8000)
plt.plot(0.5*fs*w/np.pi, np.abs(h), 'g', label='high pass resp')
wi, hi = freqz(bi, ai, worN=8000)
plt.plot(0.5*fs*wi/np.pi, np.abs(hi), 'r', label='inv. high pass resp')
plt.plot(cutoff, 0.5*np.sqrt(2), 'ko')
plt.axvline(cutoff, color='k')
plt.xlim(0, 0.05*fs)
plt.ylim(0, 5)
plt.title("Lowpass Filter Frequency Response")
plt.xlabel('Frequency [Hz]')
# add the legend in the middle of the plot
leg = plt.legend(fancybox=True)
# set the alpha value of the legend: it will be translucent
leg.get_frame().set_alpha(0.5)
plt.subplots_adjust(hspace=0.35)
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_highpass_filter(data, cutoff, fs, order)
yi = butter_inv_highpass_filter(y, cutoff, fs, order)

plt.subplot(2, 1, 2)
plt.plot(t, data, 'b-', label='1.2Hz "real" data')
plt.plot(t, y, 'g-', linewidth=2, label='blue box data')
plt.plot(t, yi, 'r--', linewidth=2, label='round-trip data')
plt.xlabel('Time [sec]')
plt.grid()
#plt.legend()
# add the legend in the middle of the plot
leg = plt.legend(fancybox=True)
# set the alpha value of the legend: it will be translucent
leg.get_frame().set_alpha(0.5)
plt.subplots_adjust(hspace=0.35)
plt.show()