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Stability Proof

First consider the roots of the denominator

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

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

$\displaystyle s = \frac{1-e^{-2s}}{2}
$

To obtain separate equations for the real and imaginary parts, take the real and imaginary parts of $ D(\sigma+j\omega)=0$ to get

\begin{eqnarray*}
\mbox{re}\left\{D(s)\right\} &=& (2\sigma - 1) + e^{-2\sigma}\cos(2\omega) = 0 \\
\mbox{im}\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 simpler conditions

\begin{eqnarray*}
e^\tau ( 1 - \tau) &=& \cos(\nu)
\\
e^\tau &=& \frac{\sin(\nu)}{\nu}
\end{eqnarray*}

For any poles of $ R_J(s)$ on the $ j\omega$ axis, we have $ \tau=0$ , and the second equation reduces to sinc$ (\nu)=1$ . It is well known that the sinc function is less than $ 1$ in magnitude at all $ \nu$ except $ \nu=0$ . Therefore, this relation can hold only at $ \omega=\nu=0$ , and so

$\displaystyle \zbox{\hbox{\emph{Any right-half-plane poles occur at $\omega=0$.}}}
$


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``Stability Proof for a Cylindrical Bore with Conical Cap'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2014-03-24 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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