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

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

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,

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

$\Box$

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

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

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

$$H(z) = z^{-\nu}\frac{b(z)}{a(z)}$$

$$= z^{-\nu}\frac{1 + b_1 z^{-1} + \cdots + b_M z^{-M} }{ 1 + a_1 z^{-1} + \cdots + a_N z^{-N} }$$

$$= z^{-\nu}\frac{ \prod_{i=1}^M (1 - q_i z^{-1}) }{ \prod_{i=1}^N (1 - p_i z^{-1}) }$$

$$= z^{-\nu}\sum_{i=1}^{N_d} \sum_{j=1}^{\mu_i} \frac{z\,K_{i,j} }{ (z-p_i)^j} , \qquad \nu\geq 0,$$ (C.80)

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

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

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

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

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 $\rho$ ,

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

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

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

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

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

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

must satisfy

$$\begin{eqnarray*} H^\prime(q_1) &\neq& 0 \\ \angle{H^\prime(q_1)}&=& 0 . \end{eqnarray*}$$

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,

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

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

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

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.

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

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

$$H(e^{j\phi} + \rho\,e^{j\psi}) \to \frac{C}{ \rho}\, e^{j(\phi - \psi)}, \qquad \phi- \frac{\pi}{2} \leq \psi\leq \phi + \frac{\pi}{2}$$

$$\isdef \;\; \frac{C}{ \rho}\,e^{j\theta}, \qquad \theta \isdefs \phi - \psi, \quad - \frac{\pi}{ 2} \leq \theta \leq \frac{\pi}{ 2}$$

$$\implies \quad f(z) \to e^{- \frac{C}{ \rho}\, e^{j\theta}}$$

$$= e^{- \frac{C}{ \rho}\, \cos\theta} e^{-j \frac{C}{ \rho}\,\sin\theta}$$

$$\to 0, \protect$$ (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],

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

occurs on the unit circle. Consequently,

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

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. $\Box$