freqs(b, a, worN=200, plot=None)
Given the M-order numerator b and N-order denominator a of an analog filter, compute its frequency response
b[0]*(jw)**M + b[1]*(jw)**(M-1) + ... + b[M]
H(w) = ----------------------------------------------
a[0]*(jw)**N + a[1]*(jw)**(N-1) + ... + a[N]
Using Matplotlib's "plot" function as the callable for plot produces unexpected results, this plots the real part of the complex transfer function, not the magnitude. Try lambda w, h: plot(w, abs(h)).
Numerator of a linear filter.
Denominator of a linear filter.
If None, then compute at 200 frequencies around the interesting parts of the response curve (determined by pole-zero locations). If a single integer, then compute at that many frequencies. Otherwise, compute the response at the angular frequencies (e.g., rad/s) given in worN.
Compute frequency response of analog filter.
freqz
from scipy.signal import freqs, iirfilter
import numpy as np
b, a = iirfilter(4, [1, 10], 1, 60, analog=True, ftype='cheby1')
w, h = freqs(b, a, worN=np.logspace(-1, 2, 1000))
import matplotlib.pyplot as plt
plt.semilogx(w, 20 * np.log10(abs(h)))
plt.xlabel('Frequency')
plt.ylabel('Amplitude response [dB]')
plt.grid(True)
plt.show()
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