As defined earlier in §5.6 (page ), a
filter is said to be stable if its impulse response
decays to 0 as
goes to infinity.
In terms of poles and zeros, an irreducible filter transfer function
is stable if and only if all its poles are inside the unit circle in
the
plane (as first discussed in §6.8.6). This is because
the transfer function is the z transform of the impulse response, and if
there is an observable (non-canceled) pole outside the unit circle,
then there is an exponentially increasing component of the impulse
response. To see this, consider a causal impulse response of the form
This signal is a damped complex sinusoid when
The signal
has the z transform
where the last step holds for
, which is
true whenever
. Thus, the transfer function consists of a single pole at
, and it exists for
.9.1Now consider what happens when we let
become greater than 1. The
pole of
moves outside the unit circle, and the impulse response
has an exponentially increasing amplitude. (Note
.) Thus, the definition of stability is violated. Since the z transform
exists only for
, we see that
implies that the
z transform no longer exists on the unit circle, so that the frequency
response becomes undefined!
The above one-pole analysis shows that a one-pole filter is stable if and only if its pole is inside the unit circle. In the case of an arbitrary transfer function, inspection of its partial fraction expansion (§6.8) shows that the behavior near any pole approaches that of a one-pole filter consisting of only that pole. Therefore, all poles must be inside the unit circle for stability.
In summary, we can state the following:
Isolated poles on the unit circle may be called marginally stable. The impulse response component corresponding to a single pole on the unit circle never decays, but neither does it grow.9.2 In physical modeling applications, marginally stable poles occur often in lossless systems, such as ideal vibrating string models [86].![]()