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

$$S(z) \isdef \frac{1-R(z)}{ 1+R(z)}$$ (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,

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

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

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

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.$\Box$