Relation to Schur Functions

Definition. A Schur function $S(z)$ is defined as a complex function analytic and of modulus not exceeding unity in $\vert z\vert\leq1$.

Theorem. The function

$$S(z) \isdef \frac{1-R(z)}{ 1+R(z)}$$ (K.8)

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

Proof.

Suppose $R(z)$ is positive real. Then for $\vert z\vert\geq1$, re$\left\{R(z)\right\}\geq 0 \,\,\implies\,\,1+$re$\left\{R(z)\right\}\geq 0 \,\,\implies\,\,1+R(z)$ 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 $\vert z\vert\leq1$. Also,

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

By the maximum modulus theorem, $S(z)$ takes on its maximum value in $\vert z\vert\geq1$ on the boundary. Thus $S(z)$ is Schur.

Conversely, suppose $S(z)$ is Schur. Solving Eq. (K.8) for $R(z)$ and taking the real part on the unit circle yields

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

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.$\Box$