Sign in

Not a member? | Forgot your Password?

Search Online Books

Search tips

Free Online Books


FIR Filter Design Software

See Also

Embedded SystemsFPGA
Chapter Contents:

Search Introduction to Digital Filters


Book Index | Global Index

Would you like to be notified by email when Julius Orion Smith III publishes a new entry into his blog?


Stability Revisited

As defined earlier in §5.6 (page [*]), a filter is said to be stable if its impulse response $ h(n)$ decays to 0 as $ n$ 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 $ z$ 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

$\displaystyle h(n) = R^n e^{j\omega nT}, \qquad n=0,1,2,\ldots\,.

This signal is a damped complex sinusoid when $ 0 < R < 1$. It oscillates with zero-crossing rate $ 2\omega/2\pi=\omega/\pi$ zeros per second, and it has an exponentially decaying amplitude envelope. If $ R>1$, then the amplitude envelope increases exponentially as $ R^n$.

The signal $ h(n)=R^n e^{j\omega n T}$ has the z transform

H(z) &=& \sum_{n=0}^\infty R^n e^{j\omega nT} z^{-n}\\
&=& \... T}z^{-1}\right)^{n}\\
&=& \frac{1}{1-Re^{j\omega T}z^{-1}},

where the last step holds for $ \left\vert R e^{j\omega T}z^{-1}\right\vert<1$, which is true whenever $ R <
\left\vert z\right\vert$. Thus, the transfer function consists of a single pole at $ z
= Re^{j\omega T}$, and it exists for $ \vert z\vert>R$.9.1Now consider what happens when we let $ R$ become greater than 1. The pole of $ H(z)$ moves outside the unit circle, and the impulse response has an exponentially increasing amplitude. (Note $ \left\vert h(n)\right\vert =
R^n$.) Thus, the definition of stability is violated. Since the z transform exists only for $ \left\vert z\right\vert > R$, we see that $ R\geq 1$ 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 expansion6.8) 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:

$\textstyle \parbox{0.8\textwidth}{A necessary and
sufficient condition for the ...
...oles of its irreducible transfer function lie strictly inside the
unit circle.}$
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].

Previous: Graphical Phase Response Calculation
Next: Computing Reflection Coefficients to Check Filter Stability

Order a Hardcopy of Introduction to Digital Filters

About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See for details.


No comments yet for this page

Add a Comment
You need to login before you can post a comment (best way to prevent spam). ( Not a member? )