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066_Frame_C26 Page 19 Wednesday, January 9, 2002 1:59 PM

By combining (26.24) and (26.25), it can be proven that the feedback system has no roots in the closed
right half plane when h < π/2. Furthermore, the system is unstable if h ≥ π/2. In particular, for h = p/2
there are two roots on the imaginary axis, at ± j1. It is also easy to show that, for any h > 0 as k → ∞,
the roots converge to

r k → 1  2kp 
--------- j2kp±
--- −ln
h  h 
As h → 0, the magnitude of the roots converge to ∞.
As illustrated by the above example, property (iii) implies that for any given real number s there are
only finitely many r k ’s in the region of the complex plane

=
s : { s ∈ : Re s() ≥ σ}

In particular, with σ = 0, this means that the quasi-polynomial χ(s) can have only finitely many roots in
the right half plane. Since the effect of the closed-loop system poles that have very large negative real
parts is negligible (as far as closed-loop systems’ input–output behavior is concerned), only finitely many
“dominant” roots r k , for k = 1,…,m, should be computed for all practical purposes.

Dominant Roots of a Quasi-Polynomial

Now we discuss the following problem: given N(s), D(s), and h ≥ 0 , find the dominant roots of the quasi-
polynomial

χ s() = Ds() + e Ns()
−hs

,
For each fixed h > 0, it can be shown that there exists σ max such that χ(s) has no roots in the region s
max
see [11] for a simple algorithm to estimate σ max , based on Nyquist criterion. Given h > 0 and a region
of the complex plane defined by σ min ≤ Re(s) ≤ s max , the problem is to find the roots of χ(s) in this region.
Clearly, a point r = σ + jω in is a root of χ(s) if and only if

D s + jw) = e – −hs −jhω N s + jw)
(
(
e
Taking the magnitude square of both sides of the above equation, χ(r) = 0 implies

A s x() : D s +(= x)D s – x) e −2hs N s + x)N s – x) = 0
(
(
(
–
where x = jω. The term D(s + x) stands for the function D(s) evaluated at σ + x. The other terms of
A σ (x) are calculated similarly. For each fixed σ, the function A σ (x) is a polynomial in the variable x. By
symmetry, if x is a zero of A σ , then (−x) is also a zero.
If A σ (x) has a root x whose real part is zero, set r = s + x . Next, evaluate the magnitude of χ(r ); if
it is zero, then n is a root of χ(s). Conversely, if A σ (x) has no root on the imaginary axis, then χ(s) cannot
have a root whose real part is the fixed value of σ from which A σ is constructed.

Algorithm
Given N(s), D(s), h, σ min , and σ max :
Step 1. Pick σ values σ 1 ,…,σ M between σ min and s max such that σ min = σ 1 , σ i < s i+1 , and σ M = σ max . For
each σ i perform the following.

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