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scipy.signal.

# freqz_sos

scipy.signal.freqz_sos(sos, worN=512, whole=False, fs=6.283185307179586)

  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).

      Added in version 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`, `sosfreqz`

  Notes

  This function used to be called `sosfreqz` in older versions (≥ 0.19.0)

  Added in version 1.15.0.

  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()