Relation to Schur Functions

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Relation to Schur Functions

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

Theorem. The function

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

(M.8)

is a Schur function if and only if

is positive real.

Proof.

Suppose is positive real. Then for $\vert z\vert\geq1$ , re $\left\{R(z)\right\}\geq 0 \,\,\Rightarrow\,\,1+$ re $\left\{R(z)\right\}\geq 0 \,\,\Rightarrow\,\,1+R(z)$ is PR. Consequently, is minimum phase which implies all roots of lie in the unit circle. Thus is analytic in $\vert z\vert\leq1$ . Also,

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

By the maximum modulus theorem,

takes on its maximum value in $\vert z\vert\geq1$ on the boundary. Thus

is Schur.

Conversely, suppose is Schur. Solving Eq. $\,$ (M.8) for 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 is not constant, then by the maximum principle, 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 has the same number of zeros as the function in

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Physical Audio Signal Processing
    Passive Impedances
       Positive Real Functions
          Relation to Schur Functions