Stability Proof

A transfer function is stable if there are no poles in the right-half plane. That is, for each zero of , we must have re . If this can be shown, along with , then the reflectance is shown to be passive. We must also study the system zeros (roots of ) in order to determine if there are any pole-zero cancellations (common factors in and ).

Since re if and only if re , for , we may set without loss of generality. Thus, we need only study the roots of

If this system is stable, we have stability also for all . Since is not a rational function of , the reflectance may have infinitely many poles and zeros.

Let's first consider the roots of the denominator

(C.156) |

At any solution of , we must have

To obtain separate equations for the real and imaginary parts, set , where and are real, and take the real and imaginary parts of Eq.(C.158) to get

Both of these equations must hold at any pole of the reflectance. For
stability, we further require
. Defining
and
, we obtain the somewhat simpler conditions

For any poles of on the axis, we have , and Eq.(C.160) reduces to

It is well known that the ``sinc function'' is less than in magnitude at all except . Therefore, Eq.(C.161) can hold only at .

We have so far proved that any poles on the axis must be at .

The same argument can be extended to the entire right-half plane as follows. Going back to the more general case of Eq.(C.160), we have

(C.161) |

Since for all real , and since for , this equation clearly has no solutions in the right-half plane. Therefore, the only possible poles in the right-half plane, including the axis, are at .

In the left-half plane, there are many potential poles. Equation (C.159) has infinitely many solutions for each since the elementary inequality implies . Also, Eq.(C.160) has an increasing number of solutions as grows more and more negative; in the limit of , the number of solutions is infinite and given by the roots of ( for any integer ). However, note that at , the solutions of Eq.(C.159) converge to the roots of ( for any integer ). The only issue is that the solutions of Eq.(C.159) and Eq.(C.160) must occur together.

Figure C.49 plots the locus of real-part zeros (solutions to Eq.(C.159)) and imaginary-part zeros (Eq.(C.160)) in a portion the left-half plane. The roots at can be seen at the middle-right. Also, the asymptotic interlacing of these loci can be seen along the left edge of the plot. It is clear that the two loci must intersect at infinitely many points in the left-half plane near the intersections indicated in the graph. As becomes large, the intersections evidently converge to the peaks of the imaginary-part root locus (a log-sinc function rotated 90 degrees). At all frequencies , the roots occur near the log-sinc peaks, getting closer to the peaks as increases. The log-sinc peaks thus provide a reasonable estimate of the number and distribution of the roots in the left-half -plane. An outline of an analytic proof is as follows:

- Rotate the loci in Fig.C.49 counterclockwise by 90 degrees.
- Prove that the two root loci are continuous, single-valued functions of
(as the figure suggests).
- Prove that for
, there are infinitely many extrema
of the log-sinc function (imaginary-part root-locus) which have
negative curvature and which lie below
(as the figure
suggests). The
and
lines are shown in the
figure as dotted lines.
- Prove that the other root locus (for the real part) has
infinitely many similar extrema which occur for
(again as
the figure suggests).
- Prove that the two root-loci interlace at
(already done above).
- Then topologically, the continuous functions must cross at
infinitely many points in order to achieve interlacing at
.

[How to cite this work] [Order a printed hardcopy] [Comment on this page via email]

Copyright ©

Center for Computer Research in Music and Acoustics (CCRMA), Stanford University