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# scipy.signal.sosfreqz

`scipy.signal.``sosfreqz`(sos, worN=None, whole=False)

  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 None (default), then compute at 512 frequencies equally spaced around the unit circle. If a single integer, then compute at that many frequencies. If an array_like, compute the response at the frequencies given (in radians/sample). whole : bool, optional Normally, frequencies are computed from 0 to the Nyquist frequency, pi radians/sample (upper-half of unit-circle). 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.

  See also

  `freqz`, `sosfilt`

  Notes

  New in version 0.19.0.

  Examples

  Design a 15th-order 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()

  (Source code)

  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()