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Positive Real Functions

Any passive driving-point impedance, such as the impedance of a violin bridge, is positive real. Positive real functions have been studied extensively in the continuous-time case in the context of network synthesis [68,528]. Very little, however, seems to be available in the discrete time case. This section (reprinted from [432]) summarizes the main properties of positive real function in the $ z$ plane (i.e., the discrete-time case).

Definition. A complex valued function of a complex variable $ f(z)$ is said to be positive real (PR) if

  1. $ z\ real\,\implies f(z)\ real$
  2. $ \left\vert z\right\vert\geq 1\implies$   re$ \left\{f(z)\right\}\geq 0$

We now specialize to the subset of functions $ f(z)=p(z)/q(z)$ representable as a ratio of finite-order polynomials in $ z$ , $ p(z)$ and $ q(z)$ . Such ``rational functions'' are meromorphic, meaning that they are analytic for all $ z$ except at a set of isolated points given by the zeros of the denominator polynomial $ q(z)$ . This class of meromorphic functions represents all filter transfer functions for finite-order, time-invariant, linear systems, and we write $ H(z)$ to denote a member of this class. We use the convention that stable, minimum phase systems are analytic and nonzero in the strict outer disk $ \vert z\vert>1$ . Condition (1) implies that for $ H(z)$ to be PR, the polynomial coefficients must be real, and therefore complex poles and zeros must exist in conjugate pairs. We assume from this point forward that $ H(z)\neq 0$ satisfies (1). From (2) we derive the facts below.

Property 1. A real rational function $ H(z)$ is PR iff $ \left\vert z\right\vert\geq 1 \implies \left\vert\angle{H(z)}\right\vert\leq \frac{\pi}{2}$ .

Proof. Expressing $ H(z)$ in polar form gives

&=& \mbox{re}\left\{\left\vert H(z)\right\vert e^{j\angle{H(z)}}\right\}\\
&=& \left\vert H(z)\right\vert\cos(\angle{H(z)}) \\
&\geq& 0 \quad \Leftrightarrow \quad \left\vert H(z)\right\vert=0\;\;\mbox{or}\;\;\left\vert\angle{H(z)}\right\vert\leq \frac{\pi}{2}.

Since the zeros of $ H(z)$ are isolated (because $ H(z)$ is analytic for $ \vert z\vert \geq 1$ ), the phase $ \angle{H(z)}$ is unrestricted only at isolated points (the zeros of $ H(z)$ ). We may define the phase arbitrarily when $ \vert H(z)\vert=0$ , and the most natural definition is its limit as $ z$ approaches the zero; such a definition satisfies the constraint $ \left\vert\angle{H(z)}\right\vert\leq \pi/2$ because every point in any valid limiting sequence satisfies the constraint, and the domain of the constraint is closed (the image of a continuous function of a compact set is compact--if $ \leq$ were replaced by $ <$ , we could have a problem.) Therefore, we may conclude that $ \left\vert\angle{H(z)}\right\vert\leq \pi/2$ for all $ \left\vert z\right\vert \geq 1$ , not just where $ \vert H(z)\vert$ is nonzero. $ \Box$

Property 2. $ H(z)$ is PR iff $ 1/H(z)$ is PR.

Proof. Assuming $ H(z)$ is PR, we have by Property 1,

$\displaystyle \left\vert\angle{H^{-1}(z)}\right\vert = \left\vert-\angle{H(z)}\right\vert = \left\vert\angle{H(z)}\right\vert \leq \frac{\pi}{2},\quad \left\vert z\right\vert\geq 1.

$ \Box$

Property 3. A PR function $ H(z)$ is analytic and nonzero in the strict outer disk.

Proof. (By contradiction)

Without loss of generality, we treat only $ n^{th}$ order polynomials

$\displaystyle \alpha_0z^n + \alpha_1z^{n-1} + \cdots + \alpha_{n-1}z + \alpha_n

which are nondegenerate in the sense that $ \alpha_0,\alpha_n\neq 0$ . Since facts about $ \alpha_0 H(z)$ are readily deduced from facts about $ H(z)$ , we set $ \alpha_0 = 1$ at no great loss.

The general (normalized) causal, finite-order, linear, time-invariant transfer function may be written

$\displaystyle H(z)$ $\displaystyle =$ $\displaystyle z^{-\nu}\frac{b(z)}{a(z)}$  
  $\displaystyle =$ $\displaystyle z^{-\nu}\frac{1 + b_1 z^{-1} + \cdots + b_M z^{-M} }{
1 + a_1 z^{-1} + \cdots + a_N z^{-N} }$  
  $\displaystyle =$ $\displaystyle z^{-\nu}\frac{ \prod_{i=1}^M (1 - q_i z^{-1}) }{
\prod_{i=1}^N (1 - p_i z^{-1}) }$  
  $\displaystyle =$ $\displaystyle z^{-\nu}\sum_{i=1}^{N_d} \sum_{j=1}^{\mu_i} \frac{z\,K_{i,j} }{ (z-p_i)^j}
, \qquad \nu\geq 0,$ (C.80)

where $ N_d$ is the number of distinct poles, each of multiplicity $ \mu_i$ ,and

$\displaystyle \sum_{i=1}^{N_d}\mu_i = \max\{N,M\}.

Suppose there is a pole of multiplicity $ m$ outside the unit circle. Without loss of generality, we may set $ \mu_1=m$ , and $ p_1 = R\,e^{j\phi}$ with $ R>1$ . Then for $ z$ near $ p_1$ , we have

z^\nu H(z) &=&\frac{ z\,K_{1,m} }{ (z-R\,e^{j\phi})^m} +
\frac{ z\,K_{1,m-1} }{ (z-R\,e^{j\phi})^{m-1}} + \cdots\\
& \approx \frac{z\, K_{1,m} }{ (z-R\,e^{j\phi})^m}.

Consider the circular neighborhood of radius $ \rho$ described by $ z=R\,e^{j\phi}+ \rho\,e^{j\psi},\;-\pi\leq \psi<\pi$ . Since $ R>1$ we may choose $ \rho < R-1$ so that all points $ z$ in this neighborhood lie outside the unit circle. If we write the residue of the factor $ (z-R\,e^{j\phi})^m$ in polar form as $ K_{1,m}=C\,e^{j\xi}$ , then we have, for sufficiently small $ \rho$ ,

$\displaystyle z^\nu H(z) \approx \frac{K_{1,m}R\,e^{j\phi}}{ (z-R\,e^{j\phi})^m} = \frac{K_{1,m}R\,e^{j\phi}}{ \rho^m\,e^{jm\psi}} = \frac{C\,R}{ \rho^m}\, e^{j(\phi + \xi - m\psi)}.$ (C.81)

Therefore, approaching the pole $ R\,e^{j\phi}$ at an angle $ \psi$ gives

$\displaystyle \lim_{\rho\to 0}\left\vert\angle{H(R\,e^{j\phi}+\rho\,e^{j\psi})}\right\vert = \left\vert\phi(1-\nu) + \xi - m\psi \right\vert,
\qquad -\pi \leq \psi <\pi

which cannot be confined to satisfy Property 1 regardless of the value of the residue angle $ \xi$ , or the pole angle $ \phi$ ($ m$ cannot be zero by hypothesis). We thus conclude that a PR function $ H(z)$ can have no poles in the outer disk. By Property 2, we conclude that positive real functions must be minimum phase. $ \Box$

Corollary. In equation Eq.(C.80), $ \nu=0$ .

Proof. If $ \nu>0$ , then there are $ \nu$ poles at infinity. As $ \left\vert z\right\vert\to \infty$ , $ H(z)\to z^{-\nu} \implies \left\vert\angle{H(z)}\right\vert \to
\left\vert\nu\angle{z}\right\vert$ , we must have $ \nu=0$ . $ \Box$

Corollary. The log-magnitude of a PR function has zero mean on the unit circle.

This is a general property of stable, minimum-phase transfer functions which follows immediately from the argument principle [299,329].

Corollary. A rational PR function has an equal number of poles and zeros all of which are in the unit disk.

This really a convention for numbering poles and zeros. In Eq.(C.80), we have $ \nu=0$ , and all poles and zeros inside the unit disk. Now, if $ M>N$ then we have $ M-N$ extra poles at $ z=0$ induced by the numerator. If $ M<N$ , then $ N-M$ zeros at the origin appear from the denominator.

Corollary. Every pole on the unit circle of a positive real function must be simple with a real and positive residue.

Proof. We repeat the previous argument using a semicircular neighborhood of radius $ \rho$ about the point $ p_1 = e^{j\phi}$ to obtain

$\displaystyle \lim_{\rho\to 0}\left\vert\angle{H(e^{j\phi} + \rho e^{j\psi})}\right\vert = \left\vert\phi + \xi - m\psi \right\vert,\qquad \phi-\frac{\pi}{2} \leq \psi\leq \phi + \frac{\pi}{2}.$ (C.82)

In order to have $ \left\vert\angle{H(z)}\right\vert\leq \pi/2$ near this pole, it is necessary that $ m=1$ and $ \xi=0$ . $ \Box$

Corollary. If $ H(z)$ is PR with a zero at $ z=q_1=e^{j\phi}$ , then

$\displaystyle H^\prime(z) \isdef \frac{H(z)}{ (1-q_1 z^{-1})}

must satisfy

H^\prime(q_1) &\neq& 0 \\
\angle{H^\prime(q_1)}&=& 0 .

Proof. We may repeat the above for $ 1/H(z)$ .

Property. Every PR function $ H(z)$ has a causal inverse z transform $ h(n)$ .

Proof. This follows immediately from analyticity in the outer disk [345, pp. 30-36] However, we may give a more concrete proof as follows. Suppose $ h(n)$ is non-causal. Then there exists $ k>0$ such that $ h(-k)\neq 0$ . We have,

H(z)\;&\isdef & \sum_{n=-\infty }^\infty h(n)\,z^{-n}\\
&=& h(-k)z^k + \sum_{n \neq -k} h(n)\,z^{-n}.

Hence, $ H(z)$ has at least one pole at infinity and cannot be PR by Property 3. Note that this pole at infinity cannot be cancelled since otherwise

&& h(-k)z^k = \sum_{l\neq -k}\alpha(l)\,z^{-l}\\
&&\implies h(-k)\delta(n+k) = \sum_{m\neq -k}\alpha(m)\,\delta(m-n)\\
&&\implies h(-k) = 0

which contradicts the hypothesis that $ h(n)$ is non-causal. $ \Box$

Property. $ H(z)$ is PR iff it is analytic for $ \left\vert z\right\vert>1$ , poles on the unit circle are simple with real and positive residues, and re$ \left\{\\ right\}{H(e^{j\theta})\}\geq 0$ for $ 0\leq \theta\leq \pi$ .

Proof. If $ H(z)$ is positive real, the conditions stated hold by virtue of Property 3 and the definition of positive real.

To prove the converse, we first show nonnegativity on the upper semicircle implies nonnegativity over the entire circle.

\mbox{re}\left\{H(e^{j\theta})\right\} &\geq& 0, \qquad 0 \leq \theta \leq \pi \\
\implies\quad \mbox{re}\left\{H(e^{-j\theta})\right\} &\isdef & \mbox{re}\left\{\sum_{n=-\infty }^\infty h(n)e^{jn\theta}\right\}\\
&=&\sum_{n=-\infty }^\infty h(n)\cos(n\theta)\\
&\geq& 0, \qquad 0 \leq \theta \leq \pi \\
\implies\quad \mbox{re}\left\{H(e^{j\theta})\right\}&\geq& 0, \qquad -\pi<\theta\leq \pi.

Alternatively, we might simply state that $ h(n)$ real implies re$ \left\{H(e^{j\theta})\right\}$ is even in $ \theta$ .

Next, since the function $ e^z$ is analytic everywhere except at $ z=\infty $ , it follows that $ f(z)=e^{-H(z)}$ is analytic wherever $ H(z)$ is finite. There are no poles of $ H(z)$ outside the unit circle due to the analyticity assumption, and poles on the unit circle have real and positive residues. Referring again to the limiting form Eq.(C.81) of $ H(z)$ near a pole on the unit circle at $ e^{j\phi}$ , we see that, as $ {\rho \to 0}$ , we have

$\displaystyle H(e^{j\phi} + \rho\,e^{j\psi})$ $\displaystyle \to$ $\displaystyle \frac{C}{ \rho}\, e^{j(\phi - \psi)},
\qquad \phi- \frac{\pi}{2} \leq \psi\leq \phi + \frac{\pi}{2}$  
  $\displaystyle \isdef$ $\displaystyle \;\; \frac{C}{ \rho}\,e^{j\theta},
\qquad \theta \isdefs \phi - \psi,
\quad - \frac{\pi}{ 2} \leq \theta \leq \frac{\pi}{ 2}$  
$\displaystyle \implies \quad f(z)$ $\displaystyle \to$ $\displaystyle e^{- \frac{C}{ \rho}\, e^{j\theta}}$  
  $\displaystyle =$ $\displaystyle e^{- \frac{C}{ \rho}\, \cos\theta} e^{-j \frac{C}{ \rho}\,\sin\theta}$  
  $\displaystyle \to$ $\displaystyle 0,
\protect$ (C.83)

since the residue $ C$ is positive, and the net angle $ \theta$ does not exceed $ \pm\pi/2$ . From Eq.(C.83) we can state that for points $ z,z^\prime $ with modulus $ \geq 1$ , we have that for all $ \epsilon>0$ , there exists $ \delta>0$ such that $ \left\vert z-z^\prime \right\vert<\delta \;\implies\;
\left\vert f(z)-f(z^\prime )\right\vert<\epsilon$ . Thus $ f(z)$ is analytic in the strict outer disk, and continuous up to the unit circle which forms its boundary. By the maximum modulus theorem [83],

$\displaystyle \sup_{\left\vert z\right\vert\geq 1} \left\vert f(z)\right\vert \isdef \sup_{\left\vert z\right\vert\geq 1} \left\vert e^{-H(z)}\right\vert =
\sup_{\left\vert z\right\vert\geq 1} e^{-\mbox{re}\left\{H(z)\right\}} = \inf_{\left\vert z\right\vert\geq 1} \mbox{re}\left\{H(z)\right\}

occurs on the unit circle. Consequently,

$\displaystyle \inf_{-\pi<\theta\leq \pi}$re$\displaystyle \left\{H(e^{j\theta})\right\}\geq 0
\quad\implies\quad\inf_{\left\vert z\right\vert\geq 1}$re$\displaystyle \left\{H(z)\right\}\geq 0
\quad\implies\quad H(z)\;PR\,.

For example, if a transfer function is known to be asymptotically stable, then a frequency response with nonnegative real part implies that the transfer function is positive real.

Note that consideration of $ 1/H(z)$ leads to analogous necessary and sufficient conditions for $ H(z)$ to be positive real in terms of its zeros instead of poles. $ \Box$

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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
Copyright © 2023-08-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University