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Relation to Schur Functions



Definition. A Schur function $ S(z)$ is defined as a complex function analytic and of modulus not exceeding unity in $ \vert z\vert\leq 1$ .

Property. The function

$\displaystyle S(z) \isdef \frac{1-R(z)}{ 1+R(z)}$ (C.84)

is a Schur function if and only if $ R(z)$ is positive real.

Proof.

Suppose $ R(z)$ is positive real. Then for $ \vert z\vert\geq 1$ , re$ \left\{R(z)\right\}\geq 0
\,\,\Rightarrow\,\,1+$re$ \left\{R(z)\right\}\geq 0 \,\,\Rightarrow\,\,1+R(z)$ is PR. Consequently, $ 1+R(z)$ is minimum phase which implies all roots of $ S(z)$ lie in the unit circle. Thus $ S(z)$ is analytic in $ \vert z\vert\leq 1$ . Also,

$\displaystyle \left\vert S(e^{j\omega})\right\vert = \frac{1-2\mbox{re}\left\{R(e^{j\omega})\right\} + \left\vert R(e^{j\omega}\right\vert^2
}{ 1+2\mbox{re}\left\{R(e^{j\omega})\right\} + \left\vert R(e^{j\omega})\right\vert^2} \leq 1.
$

By the maximum modulus theorem, $ S(z)$ takes on its maximum value in $ \vert z\vert\geq 1$ on the boundary. Thus $ S(z)$ is Schur.

Conversely, suppose $ S(z)$ is Schur. Solving Eq.$ \,$ (C.84) for $ R(z)$ and taking the real part on the unit circle yields

\begin{eqnarray*}
R(z) &=& \alpha\,\frac{1-S(z)}{ 1+S(z)} \\
\mbox{re}\left\{R(e^{j\omega})\right\} &=&\alpha\,\mbox{re}\left\{\frac{1-S(e^{j\omega})}{ 1+S(e^{j\omega})}{1+S(e^{-j\omega})}{ 1+S(e^{-j\omega})}\right\} \\
&=&\alpha\,\mbox{re}\left\{\frac{1-S(e^{j\omega})+S(e^{-j\omega})-\left\vert S(e^{j\omega})\right\vert^2
}{ \left\vert 1+S(e^{j\omega})\right\vert^2}\right\}\\
&=&\alpha\,\frac{1-\left\vert S(e^{j\omega})\right\vert^2 }{ \left\vert 1+S(e^{j\omega})\right\vert^2}\\
&\geq& 0.
\end{eqnarray*}

If $ S(z)=\alpha$ is constant, then $ R(z)=(1-\vert\alpha\vert^2)/\vert 1+\alpha\vert^2$ is PR. If $ S(z)$ is not constant, then by the maximum principle, $ S(z)<1$ for $ \vert z\vert>1$ . By Rouche's theorem applied on a circle of radius $ 1+\epsilon $ , $ \epsilon>0$ , on which $ \vert S(z)\vert<1$ , the function $ 1+S(z)$ has the same number of zeros as the function $ 1$ in $ \vert z\vert\geq 1+\epsilon $ . Hence, $ 1+S(z)$ is minimum phase which implies $ R(z)$ is analytic for $ z\geq 1$ . Thus $ R(z)$ is PR.$ \Box$


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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA