Linear Algebra and its Applications 320 (2000) 23–36
www.elsevier.com/locate/laa
Generalizations of the Hermite–Biehler theorem:
the complex case聻
Ming-Tzu Ho a , Aniruddha Datta b,∗ , S.P. Bhattacharyya b
a Engineering Science Department, National Cheng Kung University, 1, University Road, Tainan 701,
Taiwan, ROC
b Department of Electrical Engineering, Texas A&M University, College Station, TX 77843-3128, USA
Received 21 December 1999; accepted 30 May 2000
Submitted by H. Schneider
Abstract
The Hermite–Biehler theorem gives necessary and sufficient conditions for the Hurwitz
stability of a polynomial in terms of certain interlacing conditions. In this paper, we extend
our earlier generalization of the Hermite–Biehler theorem for real, not necessarily Hurwitz
polynomials to the domain of polynomials with complex coefficients. This result, which is
of interest in its own right, can also be used to analytically solve an important stabilization
problem in control theory. © 2000 Elsevier Science Inc. All rights reserved.
Keywords: Hermite–Biehler theorem; Generalized interlacing; Complex coefficients; Root distribution
1. Introduction
In [1] we have presented a generalization of the Hermite–Biehler theorem for
polynomials with real coefficients. In this paper, we extend our earlier result to the
domain of polynomials with complex coefficients. Such an extension is not of pure
academic interest and finds application in analytically solving an important problem
in the area of control theory. The subject of this paper is root distribution of poly聻
This work was supported in part by the National Science Foundation under Grant ECS-9903488 and
in part by the Texas Advanced Technology Program under Grant No. 000512-0099-1999.
∗ Corresponding author. Tel.: +1-409-845-5917; fax: +1-409-845-6259.
E-mail addresses: [email protected] (M.-T. Ho), [email protected] (A. Datta),
[email protected] (S.P. Bhattacharyya).
0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved.
PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 1 9 1 - 9
24
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
nomials. The matrix counterpart of such root distribution results is known as inertia
theory and a relevant reference is [2].
The paper is organized as follows. In Section 2, we provide a statement of the
Hermite–Biehler theorem as well as some equivalent characterizations. In Section
3, we state the relationship between the net phase change of the “frequency response” of a complex polynomial as the frequency ω varies from −∞ to ∞ and
the numbers of its roots in the open left-half and open right-half planes. In Section
4, we derive generalizations of the Hermite–Biehler theorem for complex polynomials that are not necessarily Hurwitz. Finally, Section 5 contains some concluding
remarks.
2. The Hermite–Biehler theorem
In this section, we first state the Hermite–Biehler theorem which provides necessary and sufficient conditions for the Hurwitz stability of a given real polynomial.
The proof can be found in [3]; see also [4,5] for an alternative proof using the boundary crossing theorem. We also refer the reader to [6] for several results related to the
Hermite–Biehler theorem.
Theorem 2.1 (Hermite–Biehler theorem). Let δ(s) = δ0 + δ1 s + · · · + δn s n be a given real polynomial of degree n. Write δ(s) = δe (s 2 ) + sδo (s 2 ), where δe (s 2 ), sδo (s 2 )
are the components of δ(s) made up of even and odd powers of s, respectively. Let
ωe1 , ωe2 , . . . denote the distinct non-negative real zeros of δe (−ω2 ) and let ωo1 ,
ωo2 , . . . denote the distinct non-negative real zeros of δo (−ω2 ), both arranged in
ascending order of magnitude. Then δ(s) is Hurwitz stable if and only if all the zeros
of δe (−ω2 ), δo (−ω2 ) are real and distinct, δn and δn−1 are of the same sign, and
the non-negative real zeros satisfy the following interlacing property:
0 < ωe1 < ωo1 < ωe2 < ωo2 < · · ·
(2.1)
When a polynomial is not Hurwitz, it will not satisfy the interlacing property of
the Hermite–Biehler theorem. A logical question that comes to mind is whether it
is possible to relate the violations of interlacing to the degree of “non-Hurwitzness”
of the given polynomial, and thereby generalize the Hermite–Biehler theorem. In
[1], we provided an affirmative answer to this question for the case where the test
polynomial happens to have only real coefficients. In this paper, our objective is to
obtain a similar result applicable to non-Hurwitz polynomials with complex coefficients. Towards this end, we begin by introducing the standard signum function
sgn : R → {−1, 0, 1} defined by

−1 if x < 0,
0 if x = 0,
sgn[x] =

1 if x > 0,
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
25
and then recalling the following equivalent characterization of interlacing derived in
[1].
Lemma 2.1. Let δ(s) = δ0 + δ1 s + · · · + δn s n be a given real polynomial of degree
n. Write δ(s) = δe (s 2 ) + sδo (s 2 ), where δe (s 2 ), sδo (s 2 ) are the components of δ(s)
made up of even and odd powers of s, respectively. For every frequency ω ∈ R,
denote δ(jω) = p(ω) + jq(ω), where p(ω) = δe (−ω2 ), and q(ω) = ωδo (−ω2 ). Let
ωe1 , ωe2 , . . . denote the non-negative real zeros of δe (−ω2 ) and let ωo1 , ωo2 , . . .
denote the non-negative real zeros of δo (−ω2 ), both arranged in ascending order of
magnitude. Then the following conditions are equivalent:
(i) δ(s) is Hurwitz stable.
(ii) δn and δn−1 are of the same sign and

sgn[δ0 ] · {sgn[p(0)] − 2 sgn[p(ωo1 )] + 2 sgn[p(ωo2 )] + · · ·




 +(−1)m−1 · 2 sgn[p(ωom−1 )] + (−1)m · sgn[p(∞)]} for n = 2m,
(2.2)
n= sgn[δ0 ] · {sgn[p(0)] − 2 sgn[p(ωo1 )] + 2 sgn[p(ωo2 )] + · · ·


 +(−1)m−1 · 2 sgn[p(ωom−1 )] + (−1)m · 2 sgn[p(ωom )]}


for n = 2m + 1,
(iii) δn and δn−1 are of the same sign and

sgn[q(ωe1 )] − 2 sgn[q(ωe2 )] + 2 sgn[q(ωe3 )] + · · ·

sgn[δ0 ] · {2


 +(−1)m−2 · 2 sgn[q(ωem−1 )] + (−1)m−1 · 2 sgn[q(ωem )]}
for n = 2m,
n=


sgn[δ

0 ] · {2 sgn[q(ωe1 )] − 2 sgn[q(ωe2 )] + 2 sgn[q(ωe3 )] + · · ·


+(−1)m−1 · 2 sgn[q(ωem )] + (−1)m · sgn[q(∞)]} for n = 2m + 1.
(2.3)
3. Root distribution and net accumulated phase
In this section, we state a fundamental relationship between the net accumulated
phase of the frequency response of a complex polynomial and the difference between
the numbers of roots of the polynomial in the open left-half and open right-half
planes. Let C denote the complex plane, C− the open left-half plane and C+ the
open right-half plane.
First, we focus on polynomials without zeros on the imaginary axis. Consider a
polynomial δ(s) of degree n:
δ(s) =δ0 + δ1 s + δ2 s 2 + · · · + δn s n , δi ∈ C,
such that δ(jω) =
/ 0 ∀ω ∈ (−∞, ∞).
i = 0, 1, . . . , n, δn =
/ 0
Let p(ω) and q(ω) be two functions defined pointwise by p(ω) = Re[δ(jω)],
q(ω) = Im[δ(jω)].
26
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
With this definition, we have
δ(jω) = p(ω) + jq(ω)
∀ω.
1
Furthermore θ (ω) = ∠δ(jω) = arctan q(ω)/p(ω) . Let D∞
−∞ θ denote the net change
in the argument θ (ω) as ω increases from −∞ to ∞ and let l(δ) and r(δ) denote the
numbers of roots of δ(s) in C− and C+ , respectively. Then we can state the following
lemma.
Lemma 3.1. Let δ(s) be a complex polynomial with no imaginary axis roots. Then
D∞
−∞ θ = π(l(δ) − r(δ)).
Proof. As ω goes from −∞ to ∞, each open left-half plane root contributes an
angle of π while each open right-half plane root contributes an angle of −π. 4. Generalizations of the Hermite–Biehler theorem
In this section, we derive generalizations of the Hermite–Biehler theorem by first
developing a procedure for systematically determining the net accumulated phase
change of the “frequency response” of a polynomial. Proceeding as in [1], we first
note that at any given frequency ω,
dθ (ω) q̇(ω)p(ω) − ṗ(ω)q(ω)
=
.
dω
p2 (ω) + q 2 (ω)
(4.1)
Next, we note that for complex polynomials with real leading coefficients, the
frequency response plot can be made to approach either the real or imaginary axis
as ω → ±∞ by scaling the plot of δ(jω) with 1/f (ω) where f (ω) = (1 + ω2 )n/2 .
Accordingly, for complex polynomials with real leading coefficients, we define the
normalized frequency response plot by
δf (jω) = pf (ω) + jqf (ω),
where
pf (ω) :=
p(ω)
,
(1 + ω2 )n/2
qf (ω) :=
q(ω)
.
(1 + ω2 )n/2
The subsequent development in this paper makes use of this normalized frequency
response plot for determining the net accumulated phase change as we move from
ω = −∞ to ω = +∞. Note that the assumption about real leading coefficients is
not restrictive because if the polynomial in question has a complex leading coefficient, one can divide the entire polynomial by its leading coefficient to obtain a new
polynomial whose leading coefficient is unity (and hence real) and whose roots are
the same as those of the original polynomial.
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
27
In view of the preceding discussion, we now consider a complex polynomial δ(s)
of degree n with no zeros on the imaginary axis, and whose leading coefficient is
real:
δ(s) =δ0 + δ1 s + δ2 s 2 + · · · + δn s n , δi ∈ C, i = 0, 1, . . . , n − 1, δn ∈ R,
/ 0, such that δ(jω) =
/ 0 ∀ω ∈ (−∞, ∞).
δn =
Let p(ω), q(ω), pf (ω), qf (ω) be as already defined and let
ω1 < ω2 < · · · < ωm−1
be the real, distinct finite zeros of qf (ω) with odd multiplicities. (Note that the
function qf (ω) does not change sign while passing through a real zero of even
multiplicity; hence such zeros need not be considered while counting the net phase
accumulation.) Also let us define ω0 = −∞ and ωm = +∞.
Then, as in [1], we can make the following simple observations:
1. If ωi , ωi+1 are both zeros of qf (ω), then
ω
Dωi+1
θ = 12 π sgn[pf (ωi )] − sgn[pf (ωi+1 )] · sgn[qf (ωi+ )].
(4.2)
i
2. If ωi is not a zero of qf (ω) while ωi+1 is a zero of qf (ω), a situation possible
only when ωi = −∞ is a zero of pf (ω) and n is odd, then
ω
−
θ =− 12 πsgn[pf (ωi+1 )] · sgn[qf (ωi+1
)]
Dωi+1
i
+
)].
= 12 πsgn[pf (ωi+1 )] · sgn[qf (ωi+1
(4.3)
3. If ωi is a zero of qf (ω) while ωi+1 is not a zero of qf (ω), a situation possible
only when ωi+1 = ∞ is a zero of pf (ω) and n is odd, then
ω
θ = 12 πsgn[pf (ωi )] · sgn[qf (ωi+ )].
Dωi+1
i
(4.4)
+
sgn[qf (ωi+1
)]= −sgn[qf (ωi+ )],
(4.5)
4.
i = 0, 1, . . . , m − 2.
Using (4.5) repeatedly, we obtain
+
)],
sgn[qf (ωi+ )] = (−1)m−i−1 · sgn[qf (ωm−1
i = 0, 1, . . . , m − 1.
(4.6)
Substituting (4.6) into (4.2), we see that if ωi , ωi+1 are both zeros of qf (ω), then
ω
Dωi+1
θ = 12 π sgn[pf (ωi )] − sgn[pf (ωi+1 )] · (−1)m−i−1
i
+
·sgn[qf (ωm−1
)].
(4.7)
The above observations enable us to state and prove the following theorem concerning l(δ) − r(δ). However, to simplify the theorem statement, we first define the
“imaginary signature” σi (δ) of a complex polynomial δ(s) with real leading coefficient.
28
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
Definition 4.1. Let δ(s) be any complex polynomial of degree n with real leading
coefficient. Let ω1 < ω2 < · · · < ωm−1 be the real, distinct finite zeros of qf (ω)
with odd multiplicities. Also define ω0 = −∞ and ωm = +∞.
Then
1
Pm−1
m−1 + 2
m−1−i

i=1 sgn[pf (ωi )] · (−1)
 2 {sgn[pf (ω0 )] · (−1)
−sgn[pf (ωm )]} · sgn[q(∞)] if n is even,
σi (δ):=

 1 Pm−1
m−1−i } · sgn[q(∞)] if n is odd,
i=1 sgn[pf (ωi )] · (−1)
2 {2
(4.8)
defines the imaginary signature σi (δ) of δ(s).
Theorem 4.1. Let δ(s) be a given complex polynomial of degree n with real leading
coefficient and no roots on the jω-axis, i.e., the normalized plot δf (jω) does not pass
through the origin. Then
l(δ) − r(δ) = σi (δ).
(4.9)
Proof. First, let us suppose that n is even. Then ω0 = −∞ and ωm = ∞ are zeros
of qf (ω). By repeatedly using (4.7) to determine 1∞
−∞ θ , applying Lemma 3.1, and
+
then using the fact that sgn[qf (ωm−1
)] = sgn[q(∞)], it follows that l(δ) − r(δ) is
equal to the first expression in (4.8). Hence (4.9) holds for n even.
Next let us consider the case that n is odd. Then ω0 = −∞ and ωm = ∞ are not
zeros of qf (ω). Hence,
ω1
D∞
−∞ θ =D−∞ θ +
m−2
X
i=1
ω
Dωi+1
θ + D∞
ωm−1 θ
i
= 12 πsgn[pf (ω1 )] · sgn[qf (ω1+ )]
+
m−2
X
1
2π
sgn[pf (ωi )] − sgn[pf (ωi+1 )]
i=1
+
·(−1)m−1−i sgn[qf (ωm−1
)]
+
+ 12 πsgn[pf (ωm−1 )] · sgn[qf (ωm−1
)]
(using (4.3), (4.7) and (4.4)).
(4.10)
From (4.6) it follows that
+
)].
sgn[qf (ω1+ )] =(−1)m−2 · sgn[qf (ωm−1
(4.11)
Substituting (4.11) into (4.10), applying Lemma 3.1, and then using the fact that
+
sgn[qf (ωm−1
)] = sgn[q(∞)], it follows that l(δ) − r(δ) is equal to the second
expression in (4.8). Hence (4.9) also holds for n odd. M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
29
We now state the result analogous to Theorem 4.1 where l(δ) − r(δ) of a complex
polynomial δ(s) is to be determined using the values of the frequencies where δf (jω)
crosses the imaginary axis. The proof is omitted since it follows along essentially the
same lines as that of Theorem 4.1. Once again, to simplify the theorem statement, we
first define the “real signature” σr (δ) of a complex polynomial δ(s) with real leading
coefficient.
Definition 4.2. Let δ(s) be any given complex polynomial of degree n with real
leading coefficient. Let ω1 < ω2 < · · · < ωm−1 be the real, distinct finite zeros of
pf (ω) with odd multiplicities. Also define ω0 = −∞ and ωm = +∞.
Then
 1 Pm−1
m−1−i } · sgn[p(∞)] if n is even,

− 2 {2 i=1 sgn[qf (ωi )] · (−1)
P
σr (δ):= − 12 {sgn[qf (ω0 )] · (−1)m−1 + 2 m−1
i=1 sgn[qf (ωi )]


m−1−i
·(−1)
− sgn[qf (ωm )]} · sgn[p(∞)] if n is odd
(4.12)
defines the real signature σr (δ) of δ(s).
Theorem 4.2. Let δ(s) be a given complex polynomial of degree n with real leading
coefficient and no roots on the jω-axis, i.e., the normalized plot δf (jω) does not pass
through the origin. Then
l(δ) − r(δ) = σr (δ).
(4.13)
Remark 4.1. Theorems 4.1 and 4.2 hold for both real as well as complex polynomials provided the latter have real leading coefficients. If we focus only on real
polynomials δ(s), then the real zeros of qf (ω) and pf (ω) will be symmetrically
distributed about the origin and qf (ω) will have a zero at the origin of odd multiplicity. Thus in this case, it is enough to consider only the non-negative zeros of
qf (ω) as was done in [1]. Furthermore, for a Hurwitz polynomial l(δ) − r(δ) = n.
Using these facts, it is easy to show that (4.9) and (4.13) effectively generalize (2.2)
and (2.3), respectively. Thus, Theorems 4.1 and 4.2 are indeed generalizations of the
Hermite–Biehler theorem.
Theorems 4.1 and 4.2 require that the polynomial δ(s) has no roots on the imaginary axis. We now present the following refinement of Theorem 4.1, which states
that the conclusion of Theorem 4.1 is valid even if δ(s) has jω-axis roots.
Theorem 4.3. Let δ(s) be a given complex polynomial of degree n with real leading
coefficient. Then
l(δ) − r(δ) = σi (δ).
(4.14)
30
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
Proof. Now, δ(s) can be factored as
δ(s) = δ ∗ (s)δ 0 (s),
where δ ∗ (s) contains all the jω axis roots of δ(s), while δ 0 (s) of degree n0 has no jω
axis roots. Also let δ ∗ (s) be of the form
Y
(s − jαim )nim , im = 1, 2, . . . , αim ∈ R, nim > 0.
im
We use an inductive argument to show that multiplying δ 0 (s) by δ ∗ (s) does not affect
(4.14).
Let the induction index u be equal to one and consider1
δ1 (s) = (s − jα1 )n1 δ 0 (s).
Define
δ1 (jω) := p1 (ω) + jq1 (ω)
and δ 0 (jω) := p0 (ω) + jq 0 (ω).
We now consider four different cases, namely n1 = 4l1 , n1 = 4l1 + 1, n1 = 4l1 + 2
and n1 = 4l1 + 3, where l1 is some positive integer. These four cases correspond to
the four different ways in which multiplication by [j(ω − α1 )]n1 affects the real and
imaginary parts of δ 0 (jω).
Case (I): n1 = 4l1 . For n1 = 4l1 , we have
p1 (ω) = (ω − α1 )4l1 p0 (ω),
q1 (ω) = (ω − α1 )
(4.15)
4l1 0
q (ω).
(4.16)
Let ω1 < ω2 < · · · < ωm−1 , be the real, distinct finite zeros of qf0 (ω) with odd multiplicities. Also define ω0 = −∞ and ωm = +∞. First let us assume that δ 0 (s) has
odd degree. Then, from Theorem 4.1, we have
l(δ 0 ) − r(δ 0 )=σi (δ 0 )
)
(m−1
X
0
m−1−i
· sgn[q 0 (∞)].
sgn[pf (ωi )] · (−1)
=
(4.17)
i=1
Now, from (4.16), it follows that ωi , i = 1, 2, . . . , m − 1, are also the real, distinct
finite zeros of q1f (ω) with odd multiplicities. Furthermore, from (4.15) and (4.16),
we have
sgn[pf0 (ωi )] = sgn[p1f (ωi )],
i = 1, 2, . . . , m − 1,
sgn[q 0 (∞)] = sgn[q1(∞)].
1 Note that in this proof, δ (s) is the polynomial being considered in the ith inductive step. This should
i
not be confused with the coefficients δi used earlier to define the polynomial δ(s).
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
31
Since l(δ1 ) − r(δ1 ) = l(δ 0 ) − r(δ 0 ), it follows that (4.14) is true for δ1 (s) of odd
degree. The fact that (4.14) is also true for δ1 (s) of even degree, can be verified by
proceeding along exactly the same lines.
Case (II): n1 = 4l1 + 1. For n1 = 4l1 + 1, we have
p1 (ω) = −(ω − α1 )4l1 +1 q 0 (ω),
(4.18)
q1 (ω) = (ω − α1 )4l1 +1 p0 (ω).
(4.19)
Once again let us assume that δ 0 (s) has odd degree. Let ω1 < ω2 < · · · < ωm−1
be the real, distinct finite zeros of pf0 (ω) with odd multiplicities. Also define ω0 =
−∞ and ωm = +∞. Then, from Theorem 4.2, we have
l(δ 0 ) − r(δ 0 )=σr (δ 0 )
m−1
X
1
sgn[qf0 (ωi )] · (−1)m−1−i
=− {sgn[qf0 (ω0 )] · (−1)m−1 + 2
2
−sgn[qf0 (ωm )]} · sgn[p0 (∞)].
i=1
(4.20)
Let us assume that α1 ∈ (ωc , ωc+1 ). Then, from (4.18) and (4.19), we have
sgn[qf0 (ωi )] = sgn[p1f (ωi )], i = 0, 1, . . . , c,
sgn[p1f (α1 )] = 0,
sgn[qf0 (ωi )] = −sgn[p1f (ωi )], i = c + 1, c + 2, . . . , m,
(4.21)
sgn[p0 (∞)] = sgn[q1 (∞)].
Since n0 is odd and n1 = 4l1 + 1, it follows that δ1 (s) has even degree. Since l(δ1 ) −
r(δ1 ) = l(δ 0 ) − r(δ 0 ), from (4.20), we have
l(δ1 ) − r(δ1 )=σr (δ 0 )
=− 12 {sgn[qf0 (ω0 )] · (−1)m−1 + 2 sgn[qf0 (ω1 )] · (−1)m−2
..
.
+ 2 sgn[qf0 (ωc )] · (−1)m−(1+c) + 2 sgn[p1f (α1 )] · (−1)m−(c+2)
+ 2 sgn[qf0 (ωc+1 )] · (−1)m−(c+2)
..
.
+ 2 sgn[qf0 (ωm−1 )] − sgn[qf0 (ωm )]} · sgn[p0 (∞)]
(since p1f (α1 ) = 0).
(4.22)
Now, from (4.19), it follows that ω1 , ω2 , . . . , ωc , α1 , ωc+1 , . . . , ωm−1 are the real,
distinct finite zeros of q1f (ω) with odd multiplicities. From (4.22) and using (4.21),
we have
32
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
(
c
X
1
m
l(δ1 ) − r(δ1 ) =
sgn[p1f (ωi )] · (−1)(m−i)
sgn[p1f (ω0 )] · (−1) + 2
2
i=1
+2 sgn[p1f (α1 )] · (−1)m−(c+1) + 2
m−1
X
sgn[p1f (ωi )]
i=c+1
)
·(−1)
m−(1+i)
− sgn[p1f (ωm )] · sgn[q1 (∞)]
=σi (δ1 ),
which shows that (4.14) is true for δ1 (s) of even degree. The fact that (4.14) holds
for δ1 (s) of odd degree, can be verified by proceeding along exactly the same lines.
Case (III): n1 = 4l1 + 2. For n1 = 4l1 + 2, we have
p1 (ω) = −(ω − α1 )4l1 +2 p0 (ω),
q1 (ω) = −(ω − α1 )
(4.23)
4l1 +2 0
q (ω).
(4.24)
Let ωi , i = 0, 1, . . . , m, be as already defined in Case (I). Once again, we assume
that δ 0 (s) has odd degree. Then, from Theorem 4.1, we have
( m−1
)
X
1
sgn[pf0 (ωi )] · (−1)m−1−i · sgn[q 0 (∞)].
2
l(δ 0 ) − r(δ 0 ) = σi (δ 0 ) =
2
i=1
Now, from (4.24), it follows that ωi , i = 1, 2, . . . , m − 1, are also the real, distinct
finite zeros of q1f (ω) with odd multiplicities. Furthermore, from (4.23) and (4.24),
we have
sgn[pf0 (ωi )] = −sgn[p1f (ωi )],
i = 1, 2, . . . , m − 1,
sgn[q 0 (∞)] = −sgn[q1 (∞)].
Since l(δ1 ) − r(δ1 ) = l(δ 0 ) − r(δ 0 ), it follows that (4.14) is true for δ1 (s) of odd
degree. The fact that (4.14) is also true for δ1 (s) of even degree, can be verified by
proceeding along exactly the same lines.
Case (IV): n1 = 4l1 + 3. For n1 = 4l1 + 3, we have
p1 (ω) = (ω − α1 )4l1 +3 q 0 (ω),
(4.25)
q1 (ω) = −(ω − α1 )4l1 +3 p0 (ω).
(4.26)
Once again let us assume that δ 0 (s) has odd degree. Then, from Theorem 4.2, we
have
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
33
l(δ 0 ) − r(δ 0 )=σr (δ 0 )
m−1
X
1
=− {sgn[qf0 (ω0 )] · (−1)m−1 + 2
sgn[qf0 (ωi )] · (−1)m−1−i
2
i=1
−sgn[qf0 (ωm )]} · sgn[p0 (∞)],
(4.27)
where ωi , i = 0, 1, . . . , m, are as already defined in Case (II).
Let us assume that α1 ∈ (ωc , ωc+1 ). Then, from (4.25) and (4.26), we have
sgn[qf0 (ωi )] = −sgn[p1f (ωi )], i = 0, 1, . . . , c,
sgn[p1f (α1 )] = 0,
sgn[qf0 (ωi )] = sgn[p1f (ωi )], i = c + 1, c + 2, . . . , m,
sgn[p0 (∞)] = −sgn[q1 (∞)].
(4.28)
Since n0 is odd and n1 = 4l1 + 3, it follows that δ1 (s) has even degree. Since l(δ1 ) −
r(δ1 ) = l(δ 0 ) − r(δ 0 ), from (4.27), we have
l(δ1 ) − r(δ1 )=σr (δ 0 )
=− 12 {sgn[qf0 (ω0 )] · (−1)m−1 + 2 sgn[qf0 (ω1 )] · (−1)m−2
..
.
+ 2 sgn[qf0 (ωc )] · (−1)m−(1+c) + 2 sgn[p1f (α1 )] · (−1)m−(c+2)
+ 2 sgn[qf0 (ωc+1 )] · (−1)m−(c+2)
..
.
+ 2 sgn[qf0 (ωm−1 )] − sgn[qf0 (ωm )]} · sgn[p0 (∞)]
(since p1f (α1 ) = 0).
(4.29)
Now, from (4.26), it follows that ω1 , ω2 , . . . , ωc , α1 , ωc+1 , . . . , ωm−1 are the real,
distinct finite zeros of q1f (ω) with odd multiplicities.
From (4.29) and using (4.28), we have
c
X
1
sgn[p1f (ωi )] · (−1)(m−i)
l(δ1 ) − r(δ1 ) = {sgn[p1f (ω0 )] · (−1)m + 2
2
i=1
+ 2 sgn[p1f (α1 )] · (−1)m−(c+1) + 2
m−1
X
sgn[p1f (ωi )]
i=c+1
·(−1)m−(i+1) − sgn[p1f (ωm )]} · sgn[q1 (∞)]
=σi (δ1 ),
which shows that (4.14) is true for δ1 (s) of even degree. The fact that (4.14) holds
for δ1 (s) of odd degree, can be verified by proceeding along exactly the same lines.
34
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
This completes the first step of the induction argument. Note that by using similar
arguments, it is possible to show that δ1 (s) also satisfies (4.13).
Now let u = k and consider
δk (s)=
k
Y
(s − jαim )nim δ 0 (s).
im =1
Assume that (4.14) and (4.13) are true for δk (s) (inductive assumption). Now
δk+1 (s) =
k+1
Y
(s − jαim )nim δ 0 (s)
(4.30)
im =1
=(s − jαk+1 )nk+1 δk (s).
(4.31)
Define
δk (jω) = pk (ω) + jqk (ω),
δk+1 (jω) = pk+1 (ω) + jqk+1 (ω).
Once again the proof can be completed by considering four different cases, namely
nk+1 = 4lk+1 , nk+1 = 4lk+1 + 1, nk+1 = 4lk+1 + 2 and nk+1 = 4lk+1 + 3, where
lk+1 is some positive integer. Since each of these cases can be handled by proceeding
along similar lines, we do not treat all of the cases here. Instead, we focus on a
representative case, say nk+1 = 4lk+1 + 1, and provide a detailed treatment for it.
Now, for nk+1 = 4lk+1 + 1, we have
pk+1 (ω) = −(ω − αk+1 )4lk+1 +1 qk (ω),
qk+1 (ω) = (ω − αk+1 )
4lk+1 +1
(4.32)
pk (ω).
(4.33)
First let us assume that δk (s) has odd degree. Let ω1 < ω2 < · · · < ωm−1 be the
real, distinct finite zeros of pkf (ω) with odd multiplicities. Also define ω0 = −∞
and ωm = +∞. Then, since (4.13) holds for δk (s), we have
l(δk ) − r(δk )=σr (δk )
m−1
X
1
sgn[qkf (ωi )]
=− {sgn[qkf (ω0 )] · (−1)m−1 + 2
2
i=1
·(−1)m−1−i − sgn[qkf (ωm )]} · sgn[pk (∞)].
(4.34)
Let us now assume that αk+1 ∈ (ωc , ωc+1 ). Then, from (4.32) and (4.33), we
have
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
sgn[qkf (ωi )] = sgn[pk+1f (ωi )],
35
i = 0, 1, . . . , c,
sgn[pk+1f (αk+1 )] = 0,
sgn[qkf (ωi )] = −sgn[pk+1f (ωi )],
i = c + 1, c + 2, . . . , m,
(4.35)
sgn[pk (∞)] = sgn[qk+1(∞)].
Since δk (s) has odd degree and nk+1 = 4lk+1 + 1, it follows that δk+1 (s) has even
degree. Since l(δk+1 ) − r(δk+1 ) = l(δk ) − r(δk ), from (4.34), we have
l(δk+1 ) − r(δk+1 ) =σr (δk )
=− 12 {sgn[qkf (ω0 )] · (−1)m−1 + 2 sgn[qkf (ω1 )] · (−1)m−2
..
.
+ 2 sgn[qkf (ωc )] · (−1)m−(c+1) + 2 sgn[pk+1f (αk+1 )]
·(−1)m−(c+2) + 2 sgn[qkf (ωc+1 )] · (−1)m−(c+2)
..
.
+ 2 sgn[qkf (ωm−1 )] − sgn[qkf (ωm )]}
·sgn[pk (∞)] (since pk+1f (αk+1 ) = 0).
(4.36)
Now, from (4.33), it follows that ω1 , ω2 , . . . , ωc , αk+1 , ωc+1 , . . . , ωm−1 are the
real, distinct finite zeros of qk+1f (ω) with odd multiplicities.
From (4.36) and using (4.35), we have
c
X
1
sgn[pk+1f (ωi )]
l(δk+1 ) − r(δk+1 ) = {sgn[pk+1f (ω0 )] · (−1)m + 2
2
i=1
·(−1)
+2
m−i
m−1
X
+ 2sgn[pk+1f (αk+1 )] · (−1)m−(c+1)
sgn[pk+1f (ωi )] · (−1)m−(i+1)
i=c+1
−sgn[pk+1f (ωm )]} · sgn[qk+1 (∞)]
=σi (δk+1 ),
which shows that (4.14) is true for δk+1 (s) of even degree. The fact that (4.14) holds
for δk+1 (s) of odd degree, can be verified by proceeding along exactly the same lines.
This completes the induction argument and hence the proof. Remark 4.2. By using similar arguments as in the proof of Theorem 4.3, it can
be shown that the conclusion of Theorem 4.2 is valid even when δ(s) has jω axis
roots.
36
M.-T. Ho et al. / Linear Algebra and its Applications 320 (2000) 23–36
5. Concluding remarks
In this paper, we have presented generalizations of the Hermite–Biehler theorem
applicable to complex and not necessarily Hurwitz polynomials. These results are not
of mere academic interest and can be used for solving important stabilization problems in control theory. Indeed, a special case of these results has been successfully
used in [7] to obtain analytical results on constant gain stabilization with guaranteed
damping. It should, however, be pointed out that a graphical solution to the same
problem can be obtained using standard root locus techniques [8].
References
[1] M.T. Ho, A. Datta, S.P. Bhattacharyya, Generalizations of the Hermite–Biehler theorem, Linear
Algebra Appl. (Special Issue to honour Hans Schneider) 302&303 (1999) 135–153.
[2] B.N. Datta, Stability and inertia, Linear Algebra Appl. (Special Issue to honour Hans Schneider)
302&303 (1999).
[3] F.R. Gantmacher, The Theory of Matrices, Chelsea, New York, 1959.
[4] H. Chapellat, M. Mansour, S.P. Bhattacharyya, Elementary proofs of some classical stability criteria,
IEEE Transactions on Education 33 (3) (1990).
[5] S.P. Bhattacharyya, H. Chapellat, L.H. Keel, Robust Control: The Parametric Approach, Prentice Hall,
Englewood Cliffs, NJ, 1995.
[6] M. Mansour, Robust stability in systems described by rational functions, in: C.T. Leondes (Ed.),
Control and Dynamic Systems, vol. 51, Academic Press, New York, 1992, pp. 79–128.
[7] M.T. Ho, A. Datta, S.P. Bhattacharyya, Constant gain stabilization with a specified damping ratio and
damped natural frequency, in: Proceedings of the IFAC World Congress, Beijing, People’s Republic
of China, July 1999.
[8] G.F. Franklin, J.D. Powell, A. Emami-Naeini, Feedback Control of Dynamic Systems, AddisonWesley, Reading, MA, 1994.