scipy.signal.

# tf2zpk#

scipy.signal.tf2zpk(b, a)[source]#

Return zero, pole, gain (z, p, k) representation from a numerator, denominator representation of a linear filter.

Parameters:
barray_like

Numerator polynomial coefficients.

aarray_like

Denominator polynomial coefficients.

Returns:
zndarray

Zeros of the transfer function.

pndarray

Poles of the transfer function.

kfloat

System gain.

Notes

If some values of b are too close to 0, they are removed. In that case, a BadCoefficients warning is emitted.

The b and a arrays are interpreted as coefficients for positive, descending powers of the transfer function variable. So the inputs $$b = [b_0, b_1, ..., b_M]$$ and $$a =[a_0, a_1, ..., a_N]$$ can represent an analog filter of the form:

$H(s) = \frac {b_0 s^M + b_1 s^{(M-1)} + \cdots + b_M} {a_0 s^N + a_1 s^{(N-1)} + \cdots + a_N}$

or a discrete-time filter of the form:

$H(z) = \frac {b_0 z^M + b_1 z^{(M-1)} + \cdots + b_M} {a_0 z^N + a_1 z^{(N-1)} + \cdots + a_N}$

This “positive powers” form is found more commonly in controls engineering. If M and N are equal (which is true for all filters generated by the bilinear transform), then this happens to be equivalent to the “negative powers” discrete-time form preferred in DSP:

$H(z) = \frac {b_0 + b_1 z^{-1} + \cdots + b_M z^{-M}} {a_0 + a_1 z^{-1} + \cdots + a_N z^{-N}}$

Although this is true for common filters, remember that this is not true in the general case. If M and N are not equal, the discrete-time transfer function coefficients must first be converted to the “positive powers” form before finding the poles and zeros.

Examples

Find the zeroes, poles and gain of a filter with the transfer function

$H(s) = \frac{3s^2}{s^2 + 5s + 13}$
>>> from scipy.signal import tf2zpk
>>> tf2zpk([3, 0, 0], [1, 5, 13])
(   array([ 0.               ,  0.              ]),
array([ -2.5+2.59807621j ,  -2.5-2.59807621j]),
3.0)