`scipy.signal._signaltools:lfilter_zi`

source: build-install/usr/lib/python3.14/site-packages/scipy/signal/_signaltools.py:4502

Signature

def lfilter_zi(b, a)

Summary

Construct initial conditions for lfilter for step response steady-state.

Extended Summary

Compute an initial state zi for the lfilter function that corresponds to the steady state of the step response.

A typical use of this function is to set the initial state so that the output of the filter starts at the same value as the first element of the signal to be filtered.

Parameters

b : array_like: The numerator coefficient vector as a 1-D sequence.
a : array_like: The denominator coefficient vector as a 1-D sequence. If a[0] is not 1, then both a and b are normalized by a[0]. Hence, a[0] != 0 must hold.

Returns

zi : 1-D ndarray: The initial state for the filter.

Raises

: ValueError: If a[0] == 0 (invalid denominator polynomial) or sum(a) == 0 (unstable filter).

Notes

The parameters b and a represent a transfer function $H (z) = Y (z) / X (z)$ which is defined in the tutorial_signal_TransferFunctionRepresentation section of the user_guide. As discussed in ^[1], the final value of filtering a step response $X (z) = z / (z - 1)$ , i.e., steady state, is given by

y_{\infty} := k \to \infty lim y [k] = z \to 1 lim (z - 1) Y (z) = \frac{\sum _{i = 0}^{M} b _{i}}{\sum _{j = 0}^{N} a _{j}} .

If the denominator is zero, $H (z)$ has a pole at $z_{\infty} = 1$ , which makes the filter unstable. For the transposed Direct Form II, which is implemented in lfilter, the initialization values $z_{k}$ for $H (z)$ can be determined by the recurrence equation

z_{k} = z_{k + 1} + x [0] (b_{k} - y_{\infty} a_{k}),

with $x [0]$ being the height of the input step function. Note that $a_{0} = 1$ is assumed here, which is incorporated into this function by performing a normalization step.

Examples

The following code creates a lowpass Butterworth filter to filter a signal made up of ones. As expected of a lowpass filter, the output is also all ones. If the `zi` argument of `lfilter` had not been given, a transient signal would have been produced. The second signal illustrates that using the parameter `zi` suppresses transients at the beginning of the output signal:

import numpy as np
from scipy.signal import lfilter, lfilter_zi, butter
b, a = butter(5, 0.25)
zi = lfilter_zi(b, a)
y0, zi0 = lfilter(b, a, np.ones(10), zi=zi)

✓

y0

✗

array([1., 1., 1., 1., 1., 1., 1., 1., 1., 1.])

x = np.array([0.5, 0.5, 0.5, 0.0, 0.0, 0.0, 0.0])
y1, zi1 = lfilter(b, a, x, zi=zi*x[0])

✓

y1

✗

array([0.5       , 0.5       , 0.5       , 0.49836039, 0.48610528,
       0.44399389, 0.35505241])

Note that the `zi` argument to `lfilter` is computed using `lfilter_zi` and scaled by ``x[0]``. As a result, the output `y1` has no transient until the input drops from 0.5 to 0.

Aliases

scipy.signal.lfilter_zi