nevanlinna test of stability
TRANSCRIPT
Systems & Control Letters 60 (2011) 892–899
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Systems & Control Letters
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Nevanlinna test of stabilityShodhan Rao ∗
Control Engineering group, Faculty of Electrical Engineering, University of Twente, EL/CE (Room Carre C4430), P.O. Box 217, 7500 AE Enschede, The Netherlands
a r t i c l e i n f o
Article history:Received 1 September 2010Received in revised form28 July 2011Accepted 9 August 2011Available online 28 September 2011
Keywords:Quadratic differential formsSymmetric canonical factorizationBézoutianHermite–Biehler theoremPositive pairInterlacing property
a b s t r a c t
In this paper, a new test of stability for scalar autonomous systems known as Nevanlinna test of stability isintroduced. This test can be derived using the Bézoutian of the characteristic polynomial of a given system.It is shown in this paper that an associated quadratic differential form called the Bézoutian functional ispositive if and only if the given system is asymptotically stable. Nevanlinna test of stability proceeds bychecking if the Bézoutian functional associated with a given system is positive.
© 2011 Elsevier B.V. All rights reserved.
1. Introduction
An autonomous system is a system with no inputs or free vari-ables. For such a system, the future of every trajectory is completelydetermined by its past. A linear time-invariant autonomous systemis asymptotically stable if its characteristic polynomial has all itsroots in the open left half plane, or in other words if it is Hurwitz.The most well-known method for determining stability of an au-tonomous multivariable system involves finding the characteristicpolynomial of the system and then performing a Routh–Hurwitztest on this polynomial in order to determine whether it is Hur-witz. For details on the Routh–Hurwitz test of stability, the readeris referred to [1,2].
The main aim of this paper is to introduce a new test of sta-bility called the Nevanlinna test of stability. This test is linked withthe concept of quadratic differential forms and of the Bézoutian func-tional of a given polynomial which are defined below. Consider theset of quadratic functionals acting on an infinitely differentiablescalar trajectory w of the form
QΦ(w) =
N−h,k=0
dhw
dth
Φh,k
dkw
dtk
(1)
where Φh,k are real numbers and N is a nonnegative integer. Sucha functional is called a quadratic differential form (QDF). Note that
∗ Tel.: +31 534892707; fax: +31 534892223.E-mail addresses: [email protected], [email protected].
0167-6911/$ – see front matter© 2011 Elsevier B.V. All rights reserved.doi:10.1016/j.sysconle.2011.08.001
with the QDF QΦ given by Eq. (1), we can associate a two-variablepolynomial Φ(ζ , η), which is given by
Φ(ζ , η) =
N−h,k=0
Φh,kζhηk. (2)
The Bézoutian (Ψ ) of a given polynomial r with real coefficients, asin [3, p. 1717] is defined by
Ψ (ζ , η) :=r(ζ )r(η) − r(−ζ )r(−η)
2(ζ + η). (3)
The Bézoutian is also referred to as Hermite generating function inthe literature (see for example, [4, p. 314]). Note that the Bézoutianof a given polynomial is always a two-variable polynomial, as thenumerator of the right hand side of Eq. (3) is divisible by its denom-inator for any polynomial r . The QDF associatedwith the Bézoutianof a given polynomial is called the Bézoutian functional of the poly-nomial.
Equipped with the definition of a QDF and of the Bézoutianfunctional of a given polynomial, we now explain the basis forthe Nevanlinna test of stability. It is shown in this paper that theBézoutian functional of a given polynomial r with real coefficientsis positive for all nonzero infinitely differentiable trajectoriesif and only if r is Hurwitz. Nevanlinna test of stability checkswhether a given polynomial is Hurwitz or not, by finding whetherits Bézoutian functional is positive for all nonzero infinitelydifferentiable trajectories or not. It is shown in this paper thatthe algorithm for the Nevanlinna test of stability can also be usedto perform a symmetric canonical factorization of the Bézoutian
S. Rao / Systems & Control Letters 60 (2011) 892–899 893
functional of a given Hurwitz polynomial, which involves writingit as a sum of squares of linearly independent terms.
The Bézoutian of a given polynomial is actually a classical con-cept in stability (see for example, [3, p. 1717] and [5] for refer-ences) and many known stability tests can be derived from it. Anexample is the set of Routh–Hurwitz type of stability tests derivedin [6, pp. 174–175]. In [3, p. 1718], it has been shown that theRouth–Hurwitz test can also be derived from the Bézoutian of agiven polynomial.
It is assumed that the reader is familiar with the calculus ofquadratic differential forms, and interested readers are referredto [3] for a thorough exposition of the concepts and mathematicaltechniques.
The paper is organized as follows. In Section 2, some basic con-cepts of quadratic differential forms that are necessary to under-stand the Nevanlinna test of stability are discussed. In Section 3,theNevanlinna test of stability is described. Section 4 presents con-clusions based on thematerials presented in the previous sections.Notation
The space of n dimensional real vectors is denoted by Rn,and the space of m × n real matrices by Rm×n. The space ofm × m symmetric real matrices is denoted by Rm×m
s . If one of thedimensions is not specified, a bullet • is used; so that for example,R• × n denotes the set of real matrices with n columns and anunspecified number of rows. In order to enhance readability, whendealing with a vector space R• whose elements are denoted withw, the notation Rw (note the typewriter font type!) is used. Thering of polynomials with real coefficients in the indeterminate ξis denoted by R[ξ ]; the ring of polynomials with real coefficientsin the indeterminate ζ and η is denoted by R[ζ , η]. The set ofn × m polynomial matrices in ξ is denoted by Rn× m
[ξ ], andthat consisting of all n × m polynomial matrices in ζ and η byRn × m
[ζ , η]. The set of infinitely differentiable functions from R toRw is denotedbyC∞(R, Rw). deg(p)denotes thedegree of p ∈ R[ξ ].Given twomatrices A and Bwith the same number of columns, wedenote with col(A, B) the matrix obtained by stacking A over B. INstands for identity matrix of size N . 0w×u denotes a matrix of sizew×u consisting of zeros. The set of positive real numbers is denotedwith R+. The set of natural numbers is denoted by N.
2. Quadratic differential forms
In order to make the paper as self-contained as possible, inthis section, we introduce the notion of a quadratic differentialform (QDF) as in [3] and discuss those properties of QDFs that arerelevant for understanding the results of this paper. Consider theset of quadratic functionals acting on an infinitely differentiabletrajectory w of the form
QΦ(w) =
N−h,k=0
dhw
dth
⊤
Φh,k
dkw
dtk
(4)
where Φh,k are w × w-dimensional real matrices, and N is a non-negative integer. Such a functional is called a quadratic differentialform (QDF). With the QDF given by Eq. (4), is associated a two-variable polynomial matrix Φ(ζ , η), which is given by
Φ(ζ , η) =
N−h,k=0
Φh,kζhηk. (5)
A QDF QΦ is called symmetric if its associated two-variable poly-nomial matrix is symmetric, i.e Φ(ζ , η) = Φ(η, ζ )⊤. The ring ofsymmetric w×w two-variable polynomial matrices with real coef-ficients in the indeterminates ζ and η is denoted by Rw × w
s [ζ , η].In this paper, we consider only QDFs acting on scalar trajecto-
ries. Therefore, in the remaining part of this section, we discuss
relevant properties of a QDF, whose associated two-variable poly-nomialmatrix,Φ is such thatΦ ∈ R[ζ , η]. LetΦ be givenbyEq. (5),where Φh,k are real numbers and let N denote the highest powerof ζ or η occurring in Φ . Define mat(Φ) as
mat(Φ) :=
Φ0,0 Φ0,1 . . . Φ0,NΦ1,0 Φ1,1 . . . Φ1,N
......
......
ΦN,0 ΦN,1 . . . ΦN,N
.
Observe that Φ(ζ , η) is given by
Φ(ζ , η) =1 ζ . . . ζ N
mat(Φ)
1η...
ηN
mat(Φ) is called the coefficient matrix of Φ . Note that Φ is sym-metric if and only if mat(Φ) is symmetric. If Φ is symmetric, thenmat(Φ) can be factored asmat(Φ) =mat(M)⊤ΣM mat(M), wheremat(M) is a real matrix and ΣM is a signature matrix of the form
ΣM =
[Iw1 0w1×w2
0w2×w1 −Iw2
]with w1, w2 being nonnegative integers. Thus, the two-variablepolynomial matrix Φ can be written as
Φ(ζ , η) = M(ζ )⊤ΣMM(η) (6)
where M(ξ) = mat(M) col(1, ξ , ξ 2, . . . , ξN). This decompositionof Φ is not unique, but if we impose the condition that the rowsof M are linearly independent over R, then ΣM is unique in Eq. (6)and is denoted by ΣΦ . The resulting factorization
Φ(ζ , η) = M(ζ )⊤ΣΦM(η) (7)
is called a symmetric canonical factorization of Φ . A symmetriccanonical factorization of a given Φ ∈ Rs[ζ , η] is not unique.However, if we assume that (7) is a symmetric canonical factor-ization of Φ , then another symmetric canonical factorization of Φ
can be obtained by replacing M(ξ) in Eq. (7) with UM(ξ), whereU is a real square matrix of the same dimension as ΣΦ , such thatU⊤ΣΦU = ΣΦ .
Defined below are the notions of nonnegativity and positivityof QDFs.
Definition 1. Let Φ ∈ Rs[ζ , η]. QΦ is said to be nonnegative,denoted by QΦ ≥ 0 if QΦ(w) ≥ 0 for all w ∈ C∞(R, R), andpositive denoted by QΦ > 0, if QΦ ≥ 0, and QΦ(w) = 0 impliesw = 0.
The following proposition gives algebraic conditions on thetwo-variable polynomial Φ corresponding to a QDF QΦ underwhich QΦ is nonnegative or positive.
Proposition 2. Let Φ ∈ Rs[ζ , η]. QΦ ≥ 0 iff there exists D ∈ R•[ξ ]
such that Φ(ζ , η) = D(ζ )⊤D(η), and QΦ > 0 iff Φ(ζ , η) =
D(ζ )⊤D(η) and D(λ) has column rank equal to 1 for all λ ∈ C.
Proof. See [3, p. 1712]. �
Note that QΦ ≥ 0 iff mat(Φ) ≥ 0. This implies that QΦ ≥ 0 iffΣΦ is an identity matrix.
3. Nevanlinna test of stability
In this section, we explain the steps of a new test of stabilitywhich are based on the method used for solving the subspaceNevanlinna interpolation problem described in [7]. This test will
894 S. Rao / Systems & Control Letters 60 (2011) 892–899
be referred to as the Nevanlinna test of stability. Although theproblem of subspace Nevanlinna interpolation is not related withthe problem of checking stability of a scalar autonomous system,the idea for the steps used in the Nevanlinna test of stability wasborn out of themethod used for subspaceNevanlinna interpolationproblem in [7] as described in Remark 12.
Before explaining the steps of the Nevanlinna test of stability,we explain the basis for this test. It is shown in the following that acertain QDF associated with a polynomial, known as its Bézoutianfunctional is positive iff the given polynomial is Hurwitz. TheNevanlinna test checks whether a given polynomial is Hurwitz ornot by checkingwhether its Bézoutian functional is positive or not.Below we define the Bézoutian functional of a given polynomial.
Definition 3. Let r ∈ R[ξ ]. Define
Ψ (ζ , η) :=r(ζ )r(η) − r(−ζ )r(−η)
2(ζ + η)(8)
Ψ is called the Bézoutian of r and the QDFQΨ is called the Bézoutianfunctional of r .
The above definition of a Bézoutian of a given polynomial can alsobe found in [3, p. 1717]. It is shown in the following that if QΨ
denotes the Bézoutian functional of a given polynomial r , thenQΨ > 0 iff r is Hurwitz.
Theorem 4. Let r ∈ R[ξ ]. Let QΨ denote the Bézoutian functional ofr. Then QΨ > 0 iff r is Hurwitz.
Proof. In order to prove the theorem, we need the following defi-nition of a positive pair of polynomials from [4, p. 303].
Definition 5 (Positive pair). A pair of real polynomials (u, v) is saidto be a positive pair if the leading coefficients of u and v havethe same sign, the roots ti of u and t ′j of v are all simple, real andnegative and satisfy one of the two interlacing conditions, wherem = deg(u), ℓ = deg(v)
m = ℓ and t ′m < tm < t ′m−1 < · · · < t ′1 < t1 < 0,
m = ℓ + 1 and tm < t ′m−1 < tm−1 < · · · < t ′1 < t1 < 0.
For proving Theorem 4, we also need the following two theorems:
Theorem 6 (Hermite–Biehler). A polynomial r(ξ) = u(ξ 2) + ξv(ξ 2) with real coefficients is Hurwitz if and only if (u, v) is a positivepair.
Proof. See the proof of [4, Theorem 3.4.11, p. 303]. �
Theorem 7. Let u, v ∈ R[ξ ]. Define
Ψ (ζ , η) :=u(ζ 2)v(η2)η + ζv(ζ 2)u(η2)
ζ + η.
Then QΦ > 0 iff (u, v) is a positive pair.
Proof. See the proof of [8, Theorem 2, p. 1527]. �
We now resume the proof of Theorem 4. Decompose r into itseven and odd parts as follows:
r(ξ) = u(ξ 2) + ξv(ξ 2).
Observe that
Ψ (ζ , η) =r(ζ )r(η) − r(−ζ )r(−η)
2(ζ + η)
=u(ζ 2)v(η2)η + ζv(ζ 2)u(η2)
ζ + η.
From Theorems 6 and 7, it now follows that QΨ > 0 iff r isHurwitz. �
We now show how the Nevanlinna test of stability checks forthe positivity of the Bézoutian functional of a given polynomialr ∈ R. Let Ψ ∈ Rs[ζ , η] denote the two-variable polynomialinducing the Bézoutian functional of the given polynomial. Thealgorithm for the Nevanlinna test consists of the following steps:
Step 1: Define T1 := Ψ (λ1, λ1), whereλ1 is an arbitrarily chosennonzero real number. The first step of the Nevanlinna test is tocheck if T1 > 0. Note that for w(t) := eλ1t , QΨ (w)(t) = T1e2λ1t .This implies that if T1 ≤ 0, then QΨ is not positive, which in turnimplies that r is not Hurwitz, and in this case the algorithm for theNevanlinna test stops with the output ‘‘r is not Hurwitz’’.
Step 2: Now assume that T1 > 0. Two cases arise:Case 1: T1 > 0 and deg(r) = 1. Write r as r(ξ) = aξ + b,
where a, b ∈ R, and observe that Ψ (ζ , η) is a constant equalto ab = T1 which is positive. From Proposition 2, it follows thatQΨ > 0. Therefore, the algorithm for the Nevanlinna test stopswith the output ‘‘r is Hurwitz’’.
Case 2: T1 > 0 and deg(r) > 1. In this case, decompose r intoits even and odd parts as follows:r(ξ) = u(ξ 2) + ξv(ξ 2).
Compute the following according to their definitions given below.
p(ξ) :=1
ξ − λ21
[u(ξ) −
u(λ21)v(ξ)
v(λ21)
]q(ξ) :=
1ξ − λ2
1
[ξv(ξ) −
λ21v(λ2
1)u(ξ)
u(λ21)
]r1(ξ) := q(ξ 2) + ξp(ξ 2)
Ψ1(ζ , η) :=q(ζ 2)p(η2)η + ζp(ζ 2)q(η2)
ζ + η.
In Theorem 8 stated further below, it is shown that p, q ∈ R[ξ ];deg(r1) = deg(r) − 1; Ψ1 ∈ R[ζ , η] and (QΨ1 > 0) ⇐⇒ (QΨ >0). Therefore, in order to check if QΨ > 0, the Nevanlinna testchecks if QΨ1 > 0. This is carried out in the next step.
Step 3: Note thatQΨ1 is the Bézoutian functional of r1. Therefore,in order to check if QΨ1 > 0, assign r = r1; Ψ = Ψ1 and λ1 = λ2where λ2 is another arbitrarily chosen nonzero real number and goto Step 1.
Since the degree of r reduces whenever Step 3 is encountered,it is guaranteed that the process comes to a halt at a particular stepwhen either we obtain as output ‘‘r is not Hurwitz’’ at the end ofStep 1, or we obtain as output ‘‘r is Hurwitz’’ under Case 1 of Step 2.
We now state and prove Theorem 8 mentioned earlier underStep 2. In order to keep the proof short, without loss of generality,we assume that r has a positive leading coefficient although thetheorem also holds for a polynomial r with a negative leadingcoefficient.
Theorem 8. Let u, v ∈ R[ξ ] and let λ be an arbitrary nonzero realnumber. Define the following:
r(ξ) := u(ξ 2) + ξv(ξ 2)
Ψ (ζ , η) :=u(ζ 2)v(η2)η + ζv(ζ 2)u(η2)
ζ + η
T1 := Ψ (λ, λ).
Assume that T1 > 0, deg(r) > 1 and the leading coefficient of r haspositive sign. Define
p(ξ) :=1
ξ − λ2
[u(ξ) −
u(λ2)v(ξ)
v(λ2)
](9)
q(ξ) :=1
ξ − λ2
[ξv(ξ) −
λ2v(λ2)u(ξ)
u(λ2)
](10)
r1(ξ) := q(ξ 2) + ξp(ξ 2)
Ψ1(ζ , η) :=q(ζ 2)p(η2)η + ζp(ζ 2)q(η2)
ζ + η.
S. Rao / Systems & Control Letters 60 (2011) 892–899 895
Then(i) p, q ∈ R[ξ ];
(ii) deg(r1) = deg(r) − 1, and the leading coefficient of r1 is equalto that of r;(iii) (QΨ > 0) ⇐⇒ (QΨ1 > 0).
Proof. (i) Observe that T1 = u(λ2)v(λ2) > 0. This implies thatu(λ2) = 0 and v(λ2) = 0. This implies that p and q in the statementof the theorem are properly defined. Define
p1(ξ) := u(ξ) −u(λ2)v(ξ)
v(λ2)(11)
q1(ξ) := ξv(ξ) −λ2v(λ2)u(ξ)
u(λ2)(12)
and observe that p1(λ2) = 0; q1(λ2) = 0. Note that since deg(r) >1, at least one of u and v is not a constant. This implies that p1(ξ)and q1(ξ) are divisible by (ξ − λ2). Since
p(ξ) =p1(ξ)
ξ − λ2and q(ξ) =
q1(ξ)
ξ − λ2,
it follows that p, q ∈ R[ξ ].(ii) Define
x(ξ) :=
[u(ξ 2)
ξv(ξ 2)
]; x1(ξ) :=
[ξp(ξ 2)
q(ξ 2)
].
Observe that
x1(ξ) =1
ξ 2 − λ2
ξ −u(λ2)
v(λ2)
−λ2v(λ2)
u(λ2)ξ
x(ξ).
From the above equation, it can be inferred that the position andcoefficient of the highest degree term in x1 is the same as in x, theonly difference being that the degree of this term is one less thanthat in x. This implies that the leading coefficients of r and r1 areequal. This also implies that the column degree of x1 is one lessthan that of x. It is easy to see that the column degrees of x and x1are respectively equal to the degrees of r and r1. This implies thatdeg(r1) = deg(r) − 1.
(iii) [H⇒] Since QΨ > 0, from Theorem 7, it follows that (u, v)is a positive pair. We prove that (q, p) is also a positive pair, whichimplies that QΨ1 > 0. It will now be assumed that deg(u) >deg(v). The proof for the case, when deg(u) = deg(v) is similarand this will not be given explicitly. We first prove that the leadingcoefficients of p and q have the same sign. From the definition of apositive pair (Definition 5), without loss of generality we can write
u(ξ) = ul(ξ − t1)(ξ − t2) · · · (ξ − tm)
v(ξ) = vl(ξ − t ′1)(ξ − t ′2) · · · (ξ − t ′m−1)
with tm < t ′m−1 < · · · < t ′1 < t1 < 0 and ul, vl nonzero realnumbers having the same sign. Since r is Hurwitz, and its leadingcoefficient is positive, it follows that ul > 0 and vl > 0. Let pland ql denote the leading coefficients of p and q respectively. FromEqs. (9) and (10), it follows that
pl = limξ→∞
p(ξ)
ξm−1= ul > 0
ql = limξ→∞
q(ξ)
ξm−1= vl −
λ2v(λ2)ul
u(λ2)
=λ2v(λ2)ul
u(λ2)
[vlu(λ2)
ulv(λ2)λ2− 1
]=
λ2v(λ2)ul
u(λ2)
[(λ2
− t1)λ2
·(λ2
− t2)(λ2 − t ′1)
×(λ2
− t3)(λ2 − t ′2)
. . .(λ2
− tm)
(λ2 − t ′m−1)− 1
].
Define t ′0 := 0. Observe that (λ2−ti)(λ2−t ′i−1)
> 1 for i = 1, . . . ,m. Also
note that λ2v(λ2)ulu(λ2)
> 0. This implies that ql > 0.We now prove that the roots of q and p are all real and negative,
and they are interlaced in order to prove that (q, p) is a positivepair. Note that u(λ2)
v(λ2)> 0 and λ2v(λ2)
u(λ2)> 0. Using Eq. (11), since p1 is
continuous, it is easy to verify thatp1(t ′1) < 0p1(t2) > 0
⇒ At least 1 real root of p1 exists between t2 and t ′1
p1(t ′2) > 0p1(t3) < 0
⇒ At least 1 real root of p1 exists between t3 and t ′2
...p1(t ′m−1) · · ·
p1(tm) · · ·
⇒ At least 1 real root of p1 exists between tm and t ′m−1.
Observe that deg(r) = 2m. Since deg(r1) = deg(r) − 1 andr1(ξ) = q(ξ 2) + ξp(ξ), it follows that r1 is odd and deg(p) = m−1.The roots of p are the same as those of p1 other than its root at λ2.These arem − 1 in total. Consequently phas exactly one real root ineach of the intervals (tm, t ′m−1), (tm−1, t ′m−2), . . . , (t3, t
′
2), (t2, t′
1).Since q1 is continuous, using Eq. (12), it can be verified that
q1(t1) < 0q1(t ′1) > 0
⇒ At least 1 real root of q1 exists between t ′1 and t1
q1(t2) > 0q1(t ′2) < 0
⇒ At least 1 real root of q1 exists between t ′2 and t2
...q1(tm−1) · · ·
q1(t ′m−1) · · ·
⇒ At least 1 real root of q1exists between t ′m−1 and tm−1.
Notice that the roots of q are the same as those of q1 other than itsroot at λ2. Since deg(r1) = 2m−1, q can have amaximum ofm−1roots in total. Consequently, q has exactly one real root in each ofthe intervals (t ′m−1, tm−1), (t ′m−2, tm−2), . . . , (t ′2, t2), (t
′
1, t1).It is now easy to see that the roots of q and p are all real and
negative, and they are interlaced. Therefore, (q, p) is a positive pair.From Theorem 7, it follows that QΨ1 > 0.
[⇐H] Since QΨ1 > 0, from Theorem 7, it follows that (q, p) isa positive pair. We prove that (u, v) is also a positive pair, whichimplies thatQΨ > 0. It will now be assumed that deg(q) > deg(p).The proof for the case, when deg(q) = deg(p) is similar andthis will not be given explicitly. We first prove that the leadingcoefficients of u and v have the same sign. From the definition of apositive pair (Definition 5), without loss of generality we can write
q(ξ) = ql(ξ − s1)(ξ − s2) · · · (ξ − sn) (13)
p(ξ) = pl(ξ − s′1)(ξ − s′2) · · · (ξ − s′n−1) (14)
with sn < s′n−1 < · · · < s′1 < s1 < 0 and ql, pl nonzeroreal numbers having the same sign. Since r1 is Hurwitz, and itsleading coefficient which is the same as the leading coefficient ofr is positive, it follows that ql > 0 and pl > 0. Let ul and vl denotethe leading coefficients of u and v respectively. From Eqs. (9) and(10), it follows that
u(ξ) = ξp(ξ) +u(λ2)q(ξ)
v(λ2)(15)
v(ξ) =λ2v(λ2)p(ξ)
u(λ2)+ q(ξ). (16)
896 S. Rao / Systems & Control Letters 60 (2011) 892–899
Since T1 > 0, it follows that u(λ2)v(λ2) > 0 and u(λ2)v(λ2)
> 0. Observethat
ul = limξ→∞
u(ξ)
ξ n= pl +
u(λ2)qlv(λ2)
> 0
vl = limξ→∞
v(ξ)
ξ n= ql > 0.
We now prove that the roots of u and v are real, negative andinterlaced in order to prove that (u, v) is a positive pair. Since uis continuous, using Eq. (15), it is easy to verify thatu(0) > 0u(s1) < 0
⇒ At least 1 real root of u exists between s1 and 0
u(s′1) < 0u(s2) > 0
⇒ At least 1 real root of u exists between s2 and s′1
...u(s′n−1) · · ·
u(sn) · · ·
⇒ At least 1 real root of u exists between sn and s′n−1.
Since deg(r) = deg(r1) + 1 = 2n + 1, u can have a maximum ofn roots. This implies that exactly one real root of u exists in each ofthe intervals (sn, s′n−1), (sn−1, s′n−2), . . . , (s2, s
′
1), (s1, 0).Since v is continuous, using Eq. (16), it can be verified that
v(s1) > 0v(s′1) < 0
⇒ At least 1 (can be 3, 5, etc.) real root of
v exists between s′1 and s1v(s2) < 0v(s′2) > 0
⇒ At least 1 (can be 3, 5, etc.) real root of v
exists between s′2 and s2...v(sn−1) · · ·
v(s′n−1) · · ·
⇒ At least 1 (can be 3,5, etc.) real root of v
exists between s′n−1 and sn−1.
Since deg(r) = 2n + 1, deg(v) = n. This implies that exactlyone real root of v exists in each of the intervals (s′n−1, sn−1), (s′n−2,sn−2), . . . , (s′2, s2), (s
′
1, s1). If we denote these n − 1 roots byω1, ω2, . . . , ωn−1, then
s′n−1 < ω1 < sn−1 < s′n−2 < ω2 < sn−2
< · · · < s′2 < ω2 < s2 < s′1 < ω1 < s1 < 0.
This leaves one more root of v which we denote by ωn. It iseasy to see that ωn is real as complex roots of a polynomialoccur in conjugate pairs. From Eqs. (13), (14) and (16), it is easyto see that v(y) > 0 for a nonnegative real number y. Alsoobserve that v(sn) and v(s′n−1) both have the same sign as that of(−1)n−1.ωn cannot lie in anyof the intervals (sn, s′n−1), (sn−1, s′n−2),(sn−2, s′n−3), . . . , (s2, s
′
1), (s1, ∞), because v is continuous and hasthe same sign at the end points of the intervals, which implies thatreal roots in each of these intervals must occur in multiples of 2.Therefore, it can be inferred that ωn < sn.
It is now easy to see that the roots of u and v are real and neg-ative, and are interlaced. This implies that (u, v) is a positive pair.Consequently QΨ > 0. �
Algorithm 9 is a formal algorithm for the Nevanlinna test ofstability.
Algorithm 9. Data: A polynomial r ∈ R[ξ ]with degreem > 1 anda positive leading coefficient.
Output: Whether r is Hurwitz or not.
Step 1: Decompose r into its even and odd parts as follows:
r(ξ) = u1(ξ2) + ξv1(ξ
2)
Step 2: Choose real, nonzero numbers λ1, λ2, . . . , λm.Step 3: Assign i = 1.Step 4: Compute Ti = ui(λ
2i )vi(λ
2i ). If Ti ∈ R+, output ‘‘r is not
Hurwitz’’ and stop. Else if i = m, output ‘‘r is Hurwitz’’ and stop.Else go to step 5.Step 5: Compute the following according to their definitions
vi+1(ξ) :=1
ξ − λ2i
[ui(ξ) −
ui(λ2i )vi(ξ)
vi(λ2i )
]ui+1(ξ) :=
1ξ − λ2
i
[ξvi(ξ) −
λ2i vi(λ
2i )ui(ξ)
ui(λ2i )
].
Step 6: Assign i = i + 1 and go to step 4.Proposition 10 proves the correctness of Algorithm 9 bymaking
use of Theorem 8.
Proposition 10. Let u1, v1 ∈ R[ξ ] be such that r defined by r(ξ) :=
u1(ξ2) + ξv1(ξ
2) has a positive leading coefficient and has degreem > 1. For i = 1, . . . ,m − 1, define
vi+1(ξ) :=1
ξ − λ2i
[ui(ξ) −
ui(λ2i )vi(ξ)
vi(λ2i )
]ui+1(ξ) :=
1ξ − λ2
i
[ξvi(ξ) −
λ2i vi(λ
2i )ui(ξ)
ui(λ2i )
]where λi ∈ R is nonzero. Then r is Hurwitz if and only if Ti :=
ui(λ2i )vi(λ
2i ) > 0 for i = 1, . . . ,m.
Proof. For i = 1, . . . ,m, define the following:
Ψi(ζ , η) :=ui(ζ
2)vi(η2)η + ζvi(ζ
2)ui(η2)
ζ + η
ri(ξ) := ui(ξ2) + ξvi(ξ
2).
(If ): We have Ti > 0 for i = 1, . . . ,m. From Theorem 8,it follows that deg(ri+1) = deg(ri) − 1 for i = 1, . . . ,m − 1.Since deg(r1) = m, it follows that deg(rm) = 1. This implies thatdeg(um) = deg(vm) = 0. This in turn implies that Ψm(ζ , η) =
Tm > 0, or that QΨm > 0. From Theorem 8, it now follows that
(QΨm > 0) H⇒ (QΨm−1 > 0) H⇒ · · · · · · H⇒ (QΨ1 > 0).
Since QΨ1 is the Bézoutian functional of r , from Theorem 4, itfollows that r is Hurwitz.
(Only If ): r is Hurwitz. Since QΨ1 is the Bézoutian functional ofr , from Theorem 4, it follows that QΨ1 > 0. Consider trajectorieswi(t) = eλit . We have QΨi(wi)(t) = Tie2λit . If QΨi > 0, then itfollows that Ti > 0. Since deg(ri) > 1 for i = 1, . . . ,m − 1,from Theorem 8, it follows that (QΨi > 0) H⇒ (QΨi+1 > 0)for i = 1, . . . ,m − 1. We know that QΨ1 > 0. Hence byinduction, it follows that QΨi > 0, and consequently Ti > 0 fori = 1, . . . ,m. �
We now show the application of Algorithm 9 to an example ofa Hurwitz polynomial.
Example 11. Consider the polynomial r(ξ) = ξ 5+ 2ξ 4
+ 6ξ 3+
8ξ 2+ 8ξ + 6. In this case m = 5. The chosen values of λi and
the obtained expressions for ui(ξ), vi(ξ) and values of Ti for i =
1, . . . , 5 are given in Table 1.Observe that Ti > 0 for i = 1, . . . , 5. This implies that r is
Hurwitz.
S. Rao / Systems & Control Letters 60 (2011) 892–899 897
Table 1An example for the Nevanlinna test for stability.
i ui(ξ) vi(ξ) λi Ti
1 2ξ 2+ 8ξ + 6 ξ 2
+ 6ξ + 8 2 3360
2 ξ 2+
158ξ35 +
14435
13ξ24 +
1712 1 15839
840
3 114ξ337 +
282337 ξ +
13447 −1 71676
15839
4 ξ +18166
6181 −2 445
1991
5 6445 1 1 6
445
Remark 12. We now show the similarity between the methodused for the Nevanlinna test of stability and for the problemof subspace Nevanlinna interpolation dealt with in [7]. Withreference to Algorithm 9, define ri(ξ) = ui(ξ
2) + ξvi(ξ) andxi(ξ) := col
ui(ξ
2), ξvi(ξ2)
. Note that xi defines the even and odd
parts of ri and whenever step 4 of Algorithm 9 is performed for theith time, it is checked whether ri is Hurwitz or not. Now it can beverified that
xi+1(ξ) =1
(ξ 2 − λ2i )
ξ −ui(λ
2i )
vi(λ2i )
−λ2vi(λ
2i )
ui(λ2i )
ξ
xi(ξ).
Define
J :=
[0 11 0
]; Si(ξ) :=
ξ −ui(λ
2i )
vi(λ2i )
−λ2vi(λ
2i )
ui(λ2i )
ξ
and observe that
xi+1(ξ) =Si(ξ)xi(ξ)
ξ 2 − λ2i
.
It can be verified that
Ti =xi(λi)
⊤Jxi(λi)
2λi;
and
Si(ξ) = (ξ + λi)I2 − xi(λi)T−1i xi(λi)
⊤J. (17)
The above equation is very similar to Eq. (14), [7, p. 295], whichis written below:
R1(ξ) = (ξ + λ1)Ip+m − V1T−11 V⊤
i J (18)
with
T1 =V⊤
1 JV1
λ1 + λ1
where λ1 ∈ C \ 0, V1 ∈ Cp+m and A stands for the complex conju-gate of Awhenever A is a scalar, vector or a matrix. The idea for theNevanlinna test of stability was actually born out of Eq. (18). Apartfrom the similarity between Eqs. (17) and (18), there is absolutelyno similarity between the methods used for the Nevanlinna test ofstability and for subspace Nevanlinna interpolation. Since there isno relation between the two problems, the problem of subspaceinterpolation method will not be discussed in this paper.
In the remainder of this section, we first show that the algo-rithm used for the Nevanlinna test of stability can be used to ob-tain a symmetric canonical factorization of the Bézoutian of a givenHurwitz polynomial. Theorem 13 gives this method. It will thenbe shown that the entries of the Routh table of a given Hurwitzpolynomial can also be used for obtaining a symmetric canonical
factorization of the Bézoutian of the polynomial, using a result ob-tained in [3], thus showing one similarity between the Nevanlinnatest of stability and the Routh–Hurwitz test. The section ends witha remark on the set of Routh–Hurwitz type of tests derived in [6],which like the Nevanlinna test of stability have also been derivedfrom the Bézoutian of a given polynomial.
Theorem 13. Let u1, v1 ∈ R[ξ ] be such that r defined by r(ξ) :=
u1(ξ2) + ξv1(ξ
2) is Hurwitz, has a positive leading coefficient andhas degree greater than 1. Let m denote the degree of r. For i =
1, . . . ,m − 1, define
vi+1(ξ) :=1
ξ − λ2i
[ui(ξ) −
ui(λ2i )vi(ξ)
vi(λ2i )
]ui+1(ξ) :=
1ξ − λ2
i
[ξvi(ξ) −
λ2i vi(λ
2i )ui(ξ)
ui(λ2i )
]where λi ∈ R is nonzero. Define Ti := ui(λ
2i )vi(λ
2i ); xi(ξ) :=
colui(ξ
2), ξvi(ξ2)
;
J :=
[0 11 0
]; and
X(ξ) := colmi=1
T
−12
i
xi(λi)
⊤Jxi(ξ)
ξ + λi
i−1∏k=1
(ξ − λk)
,
where∏0
k=1(ξ−λk) := 1. Let Ψ ∈ R[ζ , η] denote the Bézoutian of r.Then ΣΨ = Im and Ψ (ζ , η) = X(ζ )⊤X(η) is a symmetric canonicalfactorization of Ψ .
Proof. Define
Si(ξ) :=
ξ −ui(λ
2i )
vi(λ2i )
−λ2vi(λ
2i )
ui(λ2i )
ξ
.
As in Remark 12, it can be verified that
Si(ξ) = (ξ + λi)I2 − xi(λi)T−1i xi(λi)
⊤J
and
xi+1(ξ) =Si(ξ)xi(ξ)
ξ 2 − λ2i
. (19)
By definition,
Ψ (ζ , η) =u1(ζ
2)v1(η2)η + ζv1(ζ
2)u1(η2)
ζ + η=
x1(ζ )⊤Jx1(η)
ζ + η.
Since r is Hurwitz, it follows from Proposition 10 that Ti > 0 fori = 1, . . . ,m. Now observe that
S1(ζ )⊤JS1(η) = [(ζ + λ1)I2 − Jx1(λ1)T−11 x1(λ1)
⊤]J
× [(η + λ1)I2 − x1(λ1)T−11 x1(λ1)
⊤J]
= (ζ + λ1)(η + λ1)J − (ζ + η)Jx1(λ1)T−11 x1(λ1)
⊤J. (20)
Also from Eq. (19), it follows that
x1(ζ )⊤S1(ζ )⊤JS1(η)x1(η)
= (ζ 2− λ2
1)(η2− λ2
1)x2(ζ )⊤Jx2(η). (21)
Hence from Eqs. (20) and (21), it follows that
x1(ζ )⊤Jx1(η)(ζ + λ1)(η + λ1)
= (ζ 2− λ2
1)(η2− λ2
1)x2(ζ )⊤Jx2(η)
+ (ζ + η)x1(ζ )⊤Jx1(λ1)T−11 x1(λ1)
⊤Jx1(η).
898 S. Rao / Systems & Control Letters 60 (2011) 892–899
Dividing the above equation by (ζ +η)(ζ +λ1)(η+λ1), we obtain
x1(ζ )⊤Jx1(η)
ζ + η= (ζ − λ1)(η − λ1)
x2(ζ )⊤Jx2(η)
ζ + η
+x1(ζ )⊤Jx1(λ1)T−1
1 x1(λ1)⊤Jx1(η)
(ζ + λ1)(η + λ1).
Observe that the second term on the right hand side of the aboveequation is a polynomial because x1(λ1)
⊤Jx1(ξ) is divisible by (ξ +
λ1). In a similar way, we can write an expression for x2(ζ )⊤Jx2(η)
ζ+ηin
terms of x3(ζ )⊤Jx3(η)
ζ+η. If this process is continued, since xm(ζ )⊤Jxm(η)
ζ+η=
Tm, we obtain the following expression for Ψ (ζ , η):
Ψ (ζ , η) = X(ζ )⊤X(η). (22)
We now prove that the elements of X are linearly independentover R in order to prove that Eq. (22) is a symmetric canonicalfactorization of Ψ . Let Xi denote the ith element of X . Assume thatthere exist βi ∈ R for i = 1, . . . ,m, such that
m−i=1
βiXi(ξ) = 0. (23)
In order to prove the claimabout linear independence ofXi, wenowprove that βi = 0 for i = 1, . . . ,m. Note that
Xi(ξ) = T−
12
i
xi(λi)
⊤Jxi(ξ)
ξ + λi
i−1∏k=1
(ξ − λk).
Putting ξ = λ1 in Eq. (23), we obtain β1√T 1 = 0, which implies
that β1 = 0, as T1 > 0.Wemake use of induction to prove thatβi = 0 for i = 2, . . . ,m.
Assume that βi = 0 for i = 1, . . . , n, where 1 ≤ n < m. We provethat this implies βn+1 = 0. Note that Eq. (23) now reduces to
m−i=n+1
βiXi(ξ) = 0. (24)
Since Xi(ξ) is divisible by∏n
k=1(ξ − λk) for i = n + 1, . . . ,m, wecan divide both sides of Eq. (24) by
∏nk=1(ξ − λk) to obtain
m−i=n+1
βiYi(ξ) = 0 (25)
where
Yi(ξ) :=Xi(ξ)
n∏k=1
(ξ − λk)
.
Now put ξ = λn+1 in Eq. (25) to obtain βn+1√T n+1 = 0. Since
Tn+1 > 0, it follows that βn+1 = 0. Since β1 = 0, by inductionit follows that βi = 0 for i = 2, . . . ,m. Consequently, fromthe discussion of Section 2, it follows that Eq. (22) is a symmetriccanonical factorization of Ψ and that ΣΨ = Im. �
In order to compute a symmetric canonical factorization of theBézoutian (Ψ ) of a given Hurwitz polynomial r with degree equaltom > 1 and a positive leading coefficient, first compute ui, vi andTi for i = 1, . . . ,m using Algorithm 9 and then use these values tocompute X defined in the statement of Theorem 13. Then Eq. (22)is a symmetric canonical factorization of Ψ .
Remark 14. In [3, p. 1718], it is shown that a symmetric canonicalfactorization of the Bézoutian of a givenHurwitz polynomial r witha positive constant term can also be obtained using the entriesof the Routh table of r . Below, we summarize this result of [3].Decompose r into its even and odd parts as follows:
r(ξ) = E0(ξ 2) + ξE1(ξ 2).
Let m denote the degree of r . Iteratively compute Ek for k =
2, . . . ,m using the equation given below:
Ek(ξ) =Ek−1(0)Ek−2(ξ) − Ek−2(0)Ek−1(ξ)
ξ.
Note that for k = 0, 1, . . . ,m, the coefficients of Ek forms theentries of the kth row of the Routh table of r , with Ek(0) being thefirst entry. For k = 1, . . . ,m, define
αk :=Ek−1(0)k∏
i=1Ei(0)
.
Note that since r is Hurwitz, the entries of the first column of theRouth table of r are all positive, i.e Ei(0) > 0 for i = 0, 1, . . . ,m.This implies that αk > 0 for k = 1, . . . ,m. Define
D(ξ) := colmk=1√
αkξk−1Ek(ξ 2).
It is shown in [3] that if Ψ denotes the Bézoutian of r , then
Ψ (ζ , η) = D(ζ )⊤D(η). (26)
We now show that the elements of D are linearly independentover R, thereby proving that Eq. (26) is a symmetric canonicalfactorization of Ψ . Let Di denote the ith element of D. Assume thatthere exist βi ∈ R for i = 1, . . . ,m, such that
m−i=1
βiDi(ξ) = 0. (27)
In order to prove the claim about linear independence of Di, wenow prove that βi = 0 for i = 1, . . . ,m. Putting ξ = 0 in Eq. (27),we get β1
√α1E1(0) = 0. Since α1 > 0, and E1(0) > 0, we obtain
β1 = 0.Wemake use of induction to prove thatβi = 0 for i = 2, . . . ,m.
Assume that βi = 0 for i = 1, . . . , n, where 1 ≤ n < m. We provethat this implies βn+1 = 0. Note that Eq. (27) now reduces to
m−i=n+1
βiDi(ξ) = 0. (28)
Since Di(ξ) is divisible by ξ n for i = n + 1, . . . ,m, we can divideboth sides of Eq. (28) by ξ n to obtain
m−i=n+1
βi√
αiξi−n−1Ei(ξ 2) = 0. (29)
Now put ξ = 0 in Eq. (29) to obtain βn+1√
αn+1En+1(0) = 0. Sinceαn+1 > 0 and En+1(0) > 0, it follows that βn+1 = 0. Since β1 = 0,by induction it follows that βi = 0 for i = 2, . . . ,m. Consequently,it follows that Eq. (26) is a symmetric canonical factorization of Ψ .
In [3], it is further shown that the Routh–Hurwitz test can bederived from the Bézoutian of a given polynomial.
Remark 15. In [6], a set of Routh–Hurwitz type of stability testshave been derived from the Bézoutian of a given polynomial.Below, we summarize the salient points of the principle behindthe stability tests in [6]. Let Ψ denote the Bézoutian of a givenpolynomial r with degree m. Let Ψ ∈ Rm × m be such that
Ψ (ζ , η) = (1 ζ · · · ζm−1)Ψ
1η...
ηm−1
Ψ is called theHermitematrix associatedwith r . The principle usedin the stability tests of [6] is an extension of Hermite’s Theorem
S. Rao / Systems & Control Letters 60 (2011) 892–899 899
in [9] (see [4, Theorem 3.4.29, p. 317] for an English version),according to which if π, η and ν respectively denote the numberof positive, zero and negative eigenvalues of Ψ , then r is Hurwitzif and only if η = 0 and ν = 0. Note that the set (π, η, ν) iscalled the inertia of Ψ . In [6], the authors have presented amethodfor efficient computation of the inertia of Ψ via its triangularfactorization, thereby checking whether r is Hurwitz or not. Recallthat the triangular factorization of a given symmetric matrix M ∈
Rm × ms of the form
M = LDLT (30)
is such that D ∈ Rm×m is a diagonal matrix with the same inertiaas that of M and L ∈ Rm×m is lower triangular with unity diagonalelements. Thus one can obtain the inertia of a given matrix M byfirst obtaining a triangular factorization of M of the form (30) andthen checking the number of positive, zero and negative diagonalentries of D. The stability tests presented in [6] are based ontriangular factorization of the Hermite matrix associated with agiven polynomial in order to determine its inertia. These tests havethe advantage that the triangular factorization involved in them iscarried out without explicitly evaluating the Bézoutian.
Remark 16. The Nevanlinna test of stability cannot be generalizedin order to determine the inertia of the Hermite matrix associatedwith a given polynomial r if r is not Hurwitz. The reason for this isthat Step 5 of Algorithm 9 works only if the polynomials ui and viare both not equal to zero. The iterations of Algorithm 9 cannotproceed any further if at any step ui or vi is equal to zero. Forexample, if r(ξ) := ξ 4
+4, Algorithm9gives T1 = 0, v1(ξ) = 0, andconsequently the Algorithm cannot be generalized to perform therecursions required to determine the inertia of the Hermite matrixassociated with r .
4. Conclusion
In this paper, a new test of stability for scalar autonomous linearsystemshas beenpresented. This test is derived from theBézoutianof the characteristic polynomial of the given system. It is based
on a result proved in this paper which states that the Bézoutianfunctional of a polynomial is positive if and only if the givenpolynomial is Hurwitz. Note that there are other stability tests thatproceed in a recursive manner and involve factor cancellation atevery recursion as in the case of the Nevanlinna test of stability.One example of such a test is the Schur–Cohn test of stability in thediscrete-time case. The algebra of univariate polynomial matricesand bivariate polynomials has been used throughout as a tool forproving many of the results of this paper.
The Nevanlinna test of stability has the following disadvantageas compared to thewell knownRouth–Hurwitz stability test. Usingthe Routh–Hurwitz stability test, one can find the number of rootsof a given polynomial in the right half plane, left half plane, and onthe imaginary axis, whereas the Nevanlinna test of stability helpsin only finding whether the given polynomial is Hurwitz or not.However, the main purpose of presenting the Nevanlinna test is todemonstrate the role of QDFs in deriving a new test of stability.
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