scipy.signal.sosfreqz¶

scipy.signal.
sosfreqz
(sos, worN=None, whole=False)[source]¶ 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{n1}(z) H(z) =  *  * ... *  A0(z) A1(z) A{n1}(z)
for z = exp(omega*1j), where B{k}(z) and A{k}(z) are numerator and denominator of the transfer function of the kth second order section.
Parameters: sos : array_like
Array of secondorder filter coefficients, must have shape
(n_sections, 6)
. Each row corresponds to a secondorder 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 None (default), then compute at 512 frequencies equally spaced around the unit circle. If a single integer, then compute at that many frequencies. 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 (in radians/sample; must be 1D).whole : bool, optional
Normally, frequencies are computed from 0 to the Nyquist frequency, pi radians/sample (upperhalf of unitcircle). If whole is True, compute frequencies from 0 to 2*pi radians/sample.
Returns: w : ndarray
The normalized frequencies at which h was computed, in radians/sample.
h : ndarray
The frequency response, as complex numbers.
Notes
New in version 0.19.0.
Examples
Design a 15thorder bandpass filter in SOS format.
>>> from scipy import signal >>> 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.sosfreqz(sos, worN=1500)
Plot the response.
>>> import matplotlib.pyplot as plt >>> plt.subplot(2, 1, 1) >>> db = 20*np.log10(np.abs(h)) >>> 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.abs(h)) >>> plt.plot(w/np.pi, db) >>> plt.subplot(2, 1, 2) >>> plt.plot(w/np.pi, np.angle(h)) >>> plt.show()