# [SciPy-User] how to use signal.lfiltic?

Skipper Seabold jsseabold@gmail....
Wed Nov 24 09:39:14 CST 2010

```This is mainly for my own understanding of going back and forth
between signal processing language and time series econometrics.

I don't see how to use lfiltic.

Say I have a (known) output vector of errors and an input y vector.  y
follows a mean zero ARMA(p,q) process given by b and a below where p =
q = 2.  If I want to use lfilter to recreate the errors, forcing the
first p outputs (errors) to be zero, then I need to solve the
difference equations for the zi that do this which is given by zi
below.  But if I try to recreate this using lfiltic, it doesn't work.
Am I missing the intention of lfiltic?  Matlab's documentation, which
a lot of the signal stuff seems to be taken from, suggests that y and
x in lfiltic need to be reversed, but this also doesn't give the zi I
want.

from scipy import signal
import numpy as np

errors = np.array([ 0.,  0.,  0.00903417,  0.89064639,  1.51665674])
y = np. array([-0.60177354, -1.60410646, -1.16619292, 0.44003132, 2.36214611])

b = np.array([ 1.        , -0.8622494 ,  0.34549996])
a = np.array([ 1.        ,  0.07918344, -0.81594865])

# zi I want to produce errors = 0,0,...

zi = np.zeros(2)
zi[0] = -b[0] * y[0]
zi[1] = -b[1] * y[0] - b[0] * y[1]

zi
# array([ 0.60177354,  1.08522758])

e = signal.lfilter(b, a, y, zi=zi)
e[0]
# array([ 0.        ,  0.        ,  0.00903417,  0.89064639,  1.51665674])

zi2 = signal.lfiltic(b,a, errors[:2], y[:2])

zi2
# array([-0.03533984, -0.20791273])

e2 = signal.lfilter(b, a, y, zi=zi2)

e2[0]
# array([-0.63711338, -1.24269149, -0.41241704, -0.0899541 ,  1.25042151])

Skipper
```