Digital Signal Processing Reference
In-Depth Information
The function
filter
can also be used to compute the output of a single pole IIR using the following
call syntax:
Output = filter(b,[1,-p],Input)
where
Input
is a signal such as (for example) the unit impulse, unit step, a chirp, etc., and
b
and
p
are as
used in Eq. (4.17).
4.12.4 IMPULSE RESPONSE, UNIT STEP RESPONSE, AND STABILITY
If you are at all familiar with feedback arrangements, you should suspect that if the feedback weight or
gain (or in other words, the pole's magnitude) is too large, the filter will become unstable. To remain
stable, the magnitude of the pole must be less than 1.0. In the following discussion, we use the impulse
and unit step responses corresponding to poles of several magnitudes (
<
1
.
0 and 1
.
0) to explore the issue
of stability in the single-pole IIR.
Impulse Response
Figure 4.30 shows the result from using 0.9 as the value of the pole and an impulse as the input signal.
The resultant output, the impulse response, ultimately decays away, and the filter's response to a bounded
signal (one having only finite values) is stable. The
n
-th value of the impulse response
y
[
n
]
is
p
n
y
[
n
]=
;
n
=
0
:
1
:∞
n
•If
|
p
|
<
1,
|
p
|
→
0 as
n
→∞
n
|
p
| =
|
p
|
=
•If
1,
1 for all
n
n
•If
|
p
|
>
1,
|
p
|
→∞
as
n
→∞
Unit Step Response
Figure 4.31 shows the unit step (i.e., DC or frequency 0) response of an IIR with a pole at 0.9. We can
determine an expression for the steady-state unit step (DC) response of a single-pole filter by observing
the form of the response to the unit step. If the value of the pole is
p
, then the sequence of output values
is 1, 1 +
p
,1+
p
+
p
2
, etc., or at the
N
-th output
N
p
n
y
[
N
]=
(4.19)
n
=
0
If
|
p
|
<
1
.
0, then
y
[
N
]
converges to
1
Y
SS
=
(4.20)
1
−
p
where
Y
SS
,which we'll refer to as the steady state response. In reality, of
course,
N
never reaches infinity, but the difference between the theoretical value of
Y
SS
and
y
[
N
]
is the response when
N
=
∞
can be
made arbitrarily small by increasing
N
. In the case of Fig. 4.31, the steady state value is
1
Y
SS
=
0
.
9
=
10
1
−
