 as expected for a lowpass filter.  If the
`zi` argument of `lfilter` had not been given, the output would have
shown the transient signal.

>>> from numpy import array, ones
>>> from scipy.signal import lfilter, lfilter_zi, butter
>>> b, a = butter(5, 0.25)
>>> zi = lfilter_zi(b, a)
>>> y, zo = lfilter(b, a, ones(10), zi=zi)
>>> y
array([1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.])

Another example:

>>> x = array([0.5, 0.5, 0.5, 0.0, 0.0, 0.0, 0.0])
>>> y, zf = lfilter(b, a, x, zi=zi*x[0])
>>> y
array([ 0.5       ,  0.5       ,  0.5       ,  0.49836039,  0.48610528,
    0.44399389,  0.35505241])

Note that the `zi` argument to `lfilter` was computed using
`lfilter_zi` and scaled by ``x[0]``.  Then the output `y` has no
transient until the input drops from 0.5 to 0.0.

r