scipy.signal.freqs(b, a, worN=200, plot=None)[source]#

Compute frequency response of analog filter.

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]

Numerator of a linear filter.


Denominator of a linear filter.

worN{None, int, array_like}, optional

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.

plotcallable, optional

A callable that takes two arguments. If given, the return parameters w and h are passed to plot. Useful for plotting the frequency response inside freqs.


The angular frequencies at which h was computed.


The frequency response.

See also


Compute the frequency response of a digital filter.


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)).


>>> 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)