Step response of digital filter
[h,t] = stepz(b,a)
[h,t] = stepz(sos)
[h,t] = stepz(d)
[h,t] = stepz(...,n)
[h,t] = stepz(...,n,fs)
stepz(...)
[h,t] = stepz(b,a)
returns
the step response of the filter with numerator coefficients, b
,
and denominator coefficients, a
. stepz
chooses
the number of samples and returns the response in the column vector h
and
sample times in the column vector t
(where t = [0:n-1]'
, and n
= length(t)
is computed
automatically).
[h,t] = stepz(sos)
returns the step response
of the second order sections matrix, sos
. sos
is
a K-by-6 matrix, where the number of sections, K,
must be greater than or equal to 2. If the number of sections is less
than 2, stepz
considers the input to be the numerator
vector, b
. Each row of sos
corresponds
to the coefficients of a second-order (biquad) filter. The ith
row of the sos
matrix corresponds to [bi(1)
bi(2) bi(3) ai(1) ai(2) ai(3)]
.
[h,t] = stepz(d)
returns the step response
of the digital filter, d
. Use designfilt
to generate d
based
on frequency-response specifications.
[h,t] = stepz(...,n)
computes
the first n samples of the step response when n is an integer (t
= [0:n-1]'
). If n
is
a vector of integers, the step response is computed only at those
integer values with 0 denoting the time origin.
[h,t] = stepz(...,n,fs)
computes
n samples and produces a vector t
of length n
so
that the samples are spaced 1/fs
units apart. fs
is
assumed to be in Hz.
stepz(...)
with no output
arguments plots the step response of the filter. If you input the
filter coefficients or second order sections matrix, the current figure
window is used. If you input a digitalFilter
, the step response is displayed in fvtool
.
Note:
If the input to |
stepz
filters a length n
step
sequence using
filter(b,a,ones(1,n))
and plots the results using stem
.
To compute n
in the auto-length case, stepz
either
uses n = length(b)
for
the FIR case or first finds the poles using p = roots(a)
, if length(a)
is
greater than 1.
If the filter is unstable, n
is chosen to
be the point at which the term from the largest pole reaches 10^{6} times
its original value.
If the filter is stable, n
is chosen to be
the point at which the term due to the largest amplitude pole is 5 × 10^{–5} of
its original amplitude.
If the filter is oscillatory (poles on the unit circle only), stepz
computes
five periods of the slowest oscillation.
If the filter has both oscillatory and damped terms, n
is
chosen to equal five periods of the slowest oscillation or the point
at which the term due to the pole of largest nonunit amplitude is 5 × 10^{–5} times
its original amplitude, whichever is greater.
stepz
also allows for delays in the numerator
polynomial. The number of delays is incorporated into the computation
for the number of samples.
designfilt
| digitalFilter
| freqz
| grpdelay
| impz
| phasez
| zplane