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Stability Proof

A transfer function $ R_J(s) = N(s)/D(s)$ is stable if there are no poles in the right-half $ s$ plane. That is, for each zero $ s_i$ of $ D(s)$, we must have re$ \left\{s_i\right\}\leq 0$. If this can be shown, along with $ \left\vert R_J(j\omega)\right\vert\leq 1$, then the reflectance $ R_J$ is shown to be passive. We must also study the system zeros (roots of $ N(s)$) in order to determine if there are any pole-zero cancellations (common factors in $ D(s)$ and $ N(s)$).


Since re$ \left\{s{t_{\xi}}\right\}\geq 0$ if and only if re$ \left\{s\right\}\geq 0$, for $ {t_{\xi}}>0$, we may set $ {t_{\xi}}=1$ without loss of generality. Thus, we need only study the roots of
\begin{eqnarray*}
N(s) &=& 1 - e^{-2s} - 2s e^{-2s} \\
D(s) &=& 2s - 1 + e^{-2s}
\end{eqnarray*}
If this system is stable, we have stability also for all $ {t_{\xi}}>0$. Since $ e^{-2s}$ is not a rational function of $ s$, the reflectance $ R_J(s)$ may have infinitely many poles and zeros. Let's first consider the roots of the denominator

$\displaystyle D(s) = 2s - 1 + e^{-2s}.$ (C.159)

At any solution $ s$ of $ D(s)=0$, we must have

$\displaystyle s = \frac{1-e^{-2s}}{2}$ (C.160)

To obtain separate equations for the real and imaginary parts, set $ s=\sigma+j\omega$, where $ \sigma$ and $ \omega $ are real, and take the real and imaginary parts of Eq.$ \,$(C.161) to get
\begin{eqnarray*}
\mbox{re}\left\{D(s)\right\} &=& (2\sigma - 1) + e^{-2\sigma}\...
...}\left\{D(s)\right\} &=& 2\omega - e^{-2\sigma}\sin(2\omega) = 0
\end{eqnarray*}
Both of these equations must hold at any pole of the reflectance. For stability, we further require $ \sigma\leq 0$. Defining $ \tau = 2\sigma$ and $ \nu=2\omega$, we obtain the somewhat simpler conditions
$\displaystyle e^\tau ( 1 - \tau)$ $\displaystyle =$ $\displaystyle \cos(\nu)$ (C.161)
$\displaystyle e^\tau$ $\displaystyle =$ $\displaystyle \frac{\sin(\nu)}{\nu}$ (C.162)

For any poles of $ R_J(s)$ on the $ j\omega $ axis, we have $ \tau=0$, and Eq.$ \,$(C.163) reduces to

$\displaystyle \nu = \sin(\nu)$ (C.163)

It is well known that the ``sinc function'' $ \sin(\nu)/\nu$ is less than $ 1$ in magnitude at all $ \nu$ except $ \nu=0$. Therefore, Eq.$ \,$(C.164) can hold only at $ \omega=\nu=0$. We have so far proved that any poles on the $ j\omega $ axis must be at $ \omega=0$. The same argument can be extended to the entire right-half plane as follows. Going back to the more general case of Eq.$ \,$(C.163), we have

$\displaystyle \frac{\sin(\nu)}{\nu} = e^\tau$ (C.164)

Since $ \left\vert\sin(\nu)/\nu\right\vert\leq 1$ for all real $ \nu$, and since $ \left\vert e^\tau\right\vert>1$ for $ \tau>0$, this equation clearly has no solutions in the right-half plane. Therefore, the only possible poles in the right-half plane, including the $ j\omega $ axis, are at $ s=0$. In the left-half plane, there are many potential poles. Equation (C.162) has infinitely many solutions for each $ \tau<0$ since the elementary inequality $ 1-\tau \leq e^{-\tau}$ implies $ e^\tau ( 1
- \tau) < e^\tau e^{-\tau} = 1$. Also, Eq.$ \,$(C.163) has an increasing number of solutions as $ \tau $ grows more and more negative; in the limit of $ \tau=-\infty$, the number of solutions is infinite and given by the roots of $ \sin(\nu)$ ( $ \nu = k\pi$ for any integer $ k$). However, note that at $ \tau=-\infty$, the solutions of Eq.$ \,$(C.162) converge to the roots of $ \cos(\nu)$ ( $ \nu = (2k+1)(\pi/2)$ for any integer $ k$). The only issue is that the solutions of Eq.$ \,$(C.162) and Eq.$ \,$(C.163) must occur together.
Figure: Locus of solutions to Eq.$ \,$(C.162) (dashed) (and Eq.$ \,$(C.163) (solid). The dotted lines are lines of construction for an analytic proof that there are infinitely many roots in the left-half $ s$-plane.
\includegraphics[width=3.5in]{eps/cylconelhp}
Figure C.49 plots the locus of real-part zeros (solutions to Eq.$ \,$(C.162)) and imaginary-part zeros (Eq.$ \,$(C.163)) in a portion the left-half plane. The roots at $ s=0$ 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 $ \left\vert\nu\right\vert$ 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 $ \nu$, the roots occur near the log-sinc peaks, getting closer to the peaks as $ \left\vert\nu\right\vert$ increases. The log-sinc peaks thus provide a reasonable estimate number and distribution in the left-half $ s$-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 $ \nu$ (as the figure suggests).
  • Prove that for $ \left\vert\nu\right\vert>3$, there are infinitely many extrema of the log-sinc function (imaginary-part root-locus) which have negative curvature and which lie below $ \tau<-1$ (as the figure suggests). The $ \tau=-1$ and $ \nu=\pm 3$ 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 $ \tau>-1$ (again as the figure suggests).
  • Prove that the two root-loci interlace at $ \tau=-\infty$ (already done above).
  • Then topologically, the continuous functions must cross at infinitely many points in order to achieve interlacing at $ \tau=-\infty$.
The peaks of the log-sinc function not only indicate approximately where the left-half-plane roots occur
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Reflectance Magnitude
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Reflectance of the Conical Cap