Impulse response of digital filter
[h,t]
=
impz(b,a)
[h,t] = impz(sos)
[h,t] = impz(d)
[h,t] =
impz(...,n)
[h,t] =
impz(...,n,fs)
impz(...)
[h,t]
returns
the impulse response of the filter with numerator coefficients, =
impz(b,a)b
,
and denominator coefficients, a
. impz
chooses
the number of samples and returns the response in the column vector, h
,
and the sample times in the column vector, t
. t = [0:n-1]'
and n
= length(t)
is computed
automatically.
[h,t] = impz(sos)
returns the impulse response
for 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, impz
considers the input to be a numerator
vector. 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] = impz(d)
returns the impulse response
of a digital filter, d
. Use designfilt
to generate d
based
on frequency-response specifications.
[h,t]
computes =
impz(...,n)n
samples
of the impulse response when n
is an integer (t
= [0:n-1]'
). If n
is
a vector of integers, impz
computes the impulse
response at those integer locations, starting the response computation
from 0 (and t
= n
or t
= [0 n]
). If, instead
of n
, you include the empty vector, []
,
for the second argument, the number of samples is computed automatically.
[h,t]
computes =
impz(...,n,fs)n
samples
and produces a vector t
of length n
so
that the samples are spaced 1/fs
units apart.
impz(...)
with no output arguments
plots the impulse response of the filter.
impz
works for both real and complex input
systems.
Note:
If the input to |
impz
filters a length n
impulse
sequence using
filter(b,a,[1 zeros(1,n-1)])
and plots the results using stem
.
To compute n
in the auto-length case, impz
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), impz
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} of
its original amplitude, whichever is greater.
impz
also allows for delays in the numerator
polynomial. The number of delays is incorporated into the computation
for the number of samples.
designfilt
| digitalFilter
| impulse
| stem