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Module « scipy.signal »

Fonction freqz_sos - module scipy.signal

Signature de la fonction freqz_sos

def freqz_sos(sos, worN=512, whole=False, fs=6.283185307179586) 

Description

help(scipy.signal.freqz_sos)

Compute the frequency response of a digital filter in SOS format.

Given `sos`, an array with shape (n, 6) of second order sections of
a digital filter, compute the frequency response of the system function::

           B0(z)   B1(z)         B{n-1}(z)
    H(z) = ----- * ----- * ... * ---------
           A0(z)   A1(z)         A{n-1}(z)

for z = exp(omega*1j), where B{k}(z) and A{k}(z) are numerator and
denominator of the transfer function of the k-th second order section.

Parameters
----------
sos : array_like
    Array of second-order filter coefficients, must have shape
    ``(n_sections, 6)``. Each row corresponds to a second-order
    section, with the first three columns providing the numerator
    coefficients and the last three providing the denominator
    coefficients.
worN : {None, int, array_like}, optional
    If a single integer, then compute at that many frequencies (default is
    N=512).  Using a number that is fast for FFT computations can result
    in faster computations (see Notes of `freqz`).

    If an array_like, compute the response at the frequencies given (must
    be 1-D). These are in the same units as `fs`.
whole : bool, optional
    Normally, frequencies are computed from 0 to the Nyquist frequency,
    fs/2 (upper-half of unit-circle). If `whole` is True, compute
    frequencies from 0 to fs.
fs : float, optional
    The sampling frequency of the digital system. Defaults to 2*pi
    radians/sample (so w is from 0 to pi).

    .. versionadded:: 1.2.0

Returns
-------
w : ndarray
    The frequencies at which `h` was computed, in the same units as `fs`.
    By default, `w` is normalized to the range [0, pi) (radians/sample).
h : ndarray
    The frequency response, as complex numbers.

See Also
--------
freqz, sosfilt

Notes
-----
.. versionadded:: 0.19.0

Examples
--------
Design a 15th-order bandpass filter in SOS format.

>>> from scipy import signal
>>> import numpy as np
>>> sos = signal.ellip(15, 0.5, 60, (0.2, 0.4), btype='bandpass',
...                    output='sos')

Compute the frequency response at 1500 points from DC to Nyquist.

>>> w, h = signal.freqz_sos(sos, worN=1500)

Plot the response.

>>> import matplotlib.pyplot as plt
>>> plt.subplot(2, 1, 1)
>>> db = 20*np.log10(np.maximum(np.abs(h), 1e-5))
>>> plt.plot(w/np.pi, db)
>>> plt.ylim(-75, 5)
>>> plt.grid(True)
>>> plt.yticks([0, -20, -40, -60])
>>> plt.ylabel('Gain [dB]')
>>> plt.title('Frequency Response')
>>> plt.subplot(2, 1, 2)
>>> plt.plot(w/np.pi, np.angle(h))
>>> plt.grid(True)
>>> plt.yticks([-np.pi, -0.5*np.pi, 0, 0.5*np.pi, np.pi],
...            [r'$-\pi$', r'$-\pi/2$', '0', r'$\pi/2$', r'$\pi$'])
>>> plt.ylabel('Phase [rad]')
>>> plt.xlabel('Normalized frequency (1.0 = Nyquist)')
>>> plt.show()

If the same filter is implemented as a single transfer function,
numerical error corrupts the frequency response:

>>> b, a = signal.ellip(15, 0.5, 60, (0.2, 0.4), btype='bandpass',
...                    output='ba')
>>> w, h = signal.freqz(b, a, worN=1500)
>>> plt.subplot(2, 1, 1)
>>> db = 20*np.log10(np.maximum(np.abs(h), 1e-5))
>>> plt.plot(w/np.pi, db)
>>> plt.ylim(-75, 5)
>>> plt.grid(True)
>>> plt.yticks([0, -20, -40, -60])
>>> plt.ylabel('Gain [dB]')
>>> plt.title('Frequency Response')
>>> plt.subplot(2, 1, 2)
>>> plt.plot(w/np.pi, np.angle(h))
>>> plt.grid(True)
>>> plt.yticks([-np.pi, -0.5*np.pi, 0, 0.5*np.pi, np.pi],
...            [r'$-\pi$', r'$-\pi/2$', '0', r'$\pi/2$', r'$\pi$'])
>>> plt.ylabel('Phase [rad]')
>>> plt.xlabel('Normalized frequency (1.0 = Nyquist)')
>>> plt.show()

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