Quadratic Stability and Singular SISO Switching Systems

Transcription

1 274 IEEE RANSACIONS ON AUOMAIC CONROL, VOL. 54, NO., NOVEMBER 2009 [25] V. V. Veeravalli,. Başar, and H. V. Poor, Decentralized sequential detection with a fusion center performing the sequential test, IEEE rans. Inform. heory, vol. 39, no. 2, pp , Mar [26] V. V. Veeravalli,. Başar, and H. V. Poor, Decentralized sequential detection with sensors performing sequential tests, Math.. Control, Signals, Syst., vol. 7, no. 4, pp , 994. [27] P. Willett, P. F. Swaszek, and R. S. Blum, he good, ad, and ugly: Distriuted detection of a known signal in dependent gaussian noise, IEEE rans. Signal Process., vol. 48, no. 2, pp , Dec [28] P. Varaiya and J. Walrand, On delayed sharing patterns, IEEE rans. Autom. Control, vol. AC-23, no. 3, pp , Mar [29] A. Gupta and R. Sekar, An approach for detecting self-propagating using anomaly detection, in Recent Advances in Intrusion Detection, ser. Lecture Notes in Computer Science, G. Vigna, E. Jonsson, and C. Kruegel, Eds. Berlin, Germany: Springer-Verlag, 2003, pp Quadratic Staility and Singular SISO Switching Systems Roert Shorten, Martin Corless, Kai Wulff, Steffi Klinge, and Richard Middleton Astract In this note, we consider the prolem of determining necessary and sufficient conditions for the existence of a common quadratic Lyapunov function for a pair of stale linear time-invariant systems whose system matrices are of the form,, and where one of the matrices is singular. A necessary and sufficient condition for the existence of such a function is given in terms of the spectrum of the product ( ). he technical note also contains a spectral characterization of strictly positive real transfer functions which are strictly proper. Examples are presented to illustrate our results. Index erms LI systems. I. INRODUCION Consider a switching system descried y _x = A 0 (t; x)gh x () where the state x(t) and g, h are real vectors, A is a real square matrix, and the scalar switching function satisfies 0 (t; x) : (2) Suppose A is a Hurwitz matrix, that is, all its eigenvalues have negative real parts; then the system corresponding to (t; x) 0, that is, _x = Ax, is gloally asymptotically stale aout the origin of the state space. Suppose also that all the eigenvalues of A 0 gh have negative real parts except for a single eigenvalue at zero. hen the system corresponding to (t; x), that is, _x = (A 0 gh )x, is stale Manuscript received July 6, 2008; revised Decemer 0, First pulished Octoer 6, 2009; current version pulished Novemer 04, his work was supported y Science Foundation Ireland under Grants 07/IW./90 and 07/RPR/I77. Recommended y Associate Editor D. Lierzon. R. Shorten, K. Wulff, S. Klinge, and R. Middleton are with the Hamilton Institute, NUI Maynooth, Maynooth, Ireland ( roert.shorten@nuim.ie; richard.middleton@nuim.ie; kai.wulff@tu-ilmenau.de; mail@steffi-klinge.de). M. Corless is with the School of Aeronautics and Astronautics, Purdue University, West Lafayette, IN USA ( corless@purdue.edu). Digital Oject Identifier 0.09/AC (ut not asymptotically stale) aout the origin and all its solutions are ounded. Consider now the switching system () with any switching function which satisfies the constraint (2). One can show that this system is stale aout the origin and all solutions are ounded if there is a real symmetric positive definite matrix P satisfying the following two Lyapunov matrix inequalities: A P + PA <0 (A 0 gh ) P + P (A 0 gh ) 0: (3) his can e justified as follows. As a candidate Lyapunov function for system () (2) consider the quadratic function V (x) =x Px. hen, along any solution x() of the system we have _V =2x PAx0 2(t; x)x Pgh x: Inequalities (3) imply that V _ 0 when (t; x) equals 0 or. Noting that V _ depends in a linear affine fashion on (t; x), it now follows that _V 0 for any (t; x) in the interval [0; ]. Since V is a positive definite function, standard Lyapunov theory guarantees the claimed staility/ oundedness properties. Staility prolems involving systems descried y () (2) arise in a variety of applications; for example, in applications where integrators are switched in and out of feedack loops to achieve certain performance ojectives [] [4]. We refer to a matrix P = P > 0 satisfying (3) as a common Lyapunov matrix for A and A0gh ; the corresponding Lyapunov function V (x) =x Px is referred to as a common quadratic Lyapunov function. Assuming that (A; g) is controllale and (A; h) is oservale, we show that the following simple condition is oth a necessary and sufficient condition for the existence of a common Lyapunov matrix P. he matrix product A(A 0 gh ) has no negative real eigenvalues and exactly one zero eigenvalue. Comment : We have assumed that A 0 gh has an eigenvalue at zero. One can relax this to the requirement that A 0 gh is marginally stale ut not Hurwitz. o see this, one can show that, in order for the Lyapunov inequalities (3) to e satisfied, A 0 gh can have only one eigenvalue at zero and no other eigenvalues on the imaginary axis; see the Appendix (First note). One can also show that if A 0 gh is marginally stale, A is Hurwitz and the conditions on the eigenvalues of A(A 0 gh ) hold, then A 0 gh can have only one eigenvalue at zero and no other eigenvalues on the imaginary axis; see the Appendix (Second Note). In the previous literature, the results most closely related to this technical note are contained in [5], [6]. Both of these papers consider quadratic staility of switched systems where oth constituent matrices A, A 2 are Hurwitz stale. In the first paper, conditions are given for the existence of a CQLF for a pair of second order LI systems in terms of the spectrum of A A 2 and A A 0 2, and in the second paper, conditions are given for a pair of matrices of aritrary order, ut whose rank difference is one. Before proceeding it is useful to note that the CQLF existence prolem for pairs of LI systems, one of which is marginally stale, is sustantially more difficult than the equivalent prolem when oth LI systems are Hurwitz stale. o see that results do not immediately follow from one prolem to another, consider again the prolem treated in the second of the aforementioned papers [6]. Here the authors consider two Hurwitz matrices A ;A 2 in companion form and show that there exists a real matrix P = P > 0 satisfying PA i + A i P < 0 for i =; 2 if and only if the matrix product A A 2 has no negative real eigenvalues. One may e tempted to conclude that one can readily otain the results of the present paper using the result /$ IEEE Authorized licensed use ited to: he Lirary NUI Maynooth. Downloaded on Feruary 2, 200 at :00 from IEEE Xplore. Restrictions apply.

2 IEEE RANSACIONS ON AUOMAIC CONROL, VOL. 54, NO., NOVEMBER in [6] y using the following reasoning. For >0sufficiently small, let A 2 () e a perturation of A 0 gh such that A 2 () is Hurwitz and the product AA 2 () has no negative real eigenvalues. hen, for each > 0 sufficiently small [6] tells us that there is a matrix P () satisfying P ()A + A P () < 0 and P ()A 2 ()+A 2 () P () < 0. Now consider its as goes to zero. he following counterexample illustrates why this proof technique will not work. ) Example: Consider A = A 2 () = : A 0 gh = Here, g = [0 ], h = [0 0] and one can readily verify that (A; g) is controllale and (A; h) is oservale. Clearly A and A 2 () are Hurwitz for all >0 and A 2 (0) = A 0 gh. he characteristic polynomial of the matrix AA 2() is given y d(s) =s 2 + s +.For 0 <<4, this polynomial has complex roots. Hence, for this range of, A A 2 () has no real negative eigenvalues and it follows from [6] that there is a matrix P () =P () > 0 satisfying P ()A + A P () < 0 and P ()A 2 () +A 2 () P () < 0. However the matrix product A(A 0 gh ) = AA 2 (0) has a repeated eigenvalue at zero. Hence, using the result of this technical note, there does not exist a matrix P = P > 0 satisfying (3). In the next section, we present some results on positive real transfer functions which are useful in the development of the main result. In particular, heorem 2. provides a simple spectral characterization of strictly positive real transfer functions. We elieve this is a useful result on its own and not just for the purposes of otaining the main result. Section III develops the main result of this technical note. o achieve this we also need the Kalman-Yacuovich-Popov lemma for proper SISO systems that is found in most textooks; see the ook y Boyd [7] or Khalil [8]. hroughout, known results are quoted without proof whereas new results are given with full proofs. II. SPECRAL CHARACERIZAIONS OF SRIC POSIIVE REALNESS Before otaining our main result, we otain some preinary results on strictly positive real (SPR) transfer functions. In everything that follows, A is a real n 2 n matrix and ; c are real n-vectors. Recall that a scalar transfer function H is strictly positive real (SPR) if there exists a scalar >0 such that H is analytic in a region of the complex plane which includes those s for which Re(s) 0 and H(j! 0 ) +H(j! 0 ) 3 0 8! 2 IR: (4) We say H is regular if H(j!)+H(j!) 3 is not identically zero for all! 2 IR. For convenience, we will include regularity as a requirement for SPR. he following standard result provides a more convenient characterization of SPR. It einates. Lemma 2.: [8]: Suppose A is Hurwitz. hen the transfer function H(s) =c (si 0 A) 0 is SPR if and only if H(j!)+H(j!) 3 > 0 8! 2 (5)! 0! 3!2 H(j!)+H(j!) > 0: (6) In checking SPR of a system it is sometimes more convenient to check SPR of a system which is equivalent (from an SPR viewpoint) to the original system. he following lemma provides such an equivalent system and is useful for generating some of the results of this technical note. his system is simply otained y replacing A with A 0. Lemma 2.2: he transfer function H(s) =c (si 0 A) 0 is SPR if and only if H I (s) =c (si 0 A 0 ) 0 is SPR. Proof: Suppose H is SPR. he identity (si 0A 0 ) 0 = s 0 I 0 s 02 (s 0 I 0 A) 0 implies that H I(s) =s 0 c 0 s 02 H(s 0 ) (7) hence we get the equation shown at the ottom of the page. Considering its as! 0! 0 H I (0) + H I (0) 3 = Finally, we note that ~! 0! 3 ~!2 [H(j ~!)+H(j ~!) ] > 0:! 0!!2 [H I (j!)+h I (j!) 3 ]=H(0) + H(0) 3 > 0: he core of our main result is ased on a spectral condition for strict positive realness of strictly proper transfer functions; this is related to corresponding results in [9] for transfer functions which are proper ut not strictly proper. his result makes use of the following lemma. Lemma 2.3: [0] [2]: Let H(s) =d + c (si 0 A) 0 where A is invertile. hen, H(s 0 )= d +c (si 0 A) 0 with A = A 0, = 0A 0, c = c A 0 and d = d 0 c A 0. ) Comment 2: Note that when H is SPR we must have d>0. his follows from the fact that d = H(0) and H(0) + H(0) 3 > 0 since H is SPR. Now we give the aforementioned spectral characterization of strict positive realness. heorem 2.: Suppose A is Hurwitz. hen, the following statements are equivalent. (a) he transfer function H(s) =c (si 0 A) 0 is SPR. () c A 0 < 0 and the matrix product A 0 (A 0 0 A 0 c A 0 =c A 0 ) has no negative real eigenvalues and exactly one zero eigenvalue. (c) c A < 0 and the matrix product A(A 0 Ac A=c A) has no negative real eigenvalues and exactly one zero eigenvalue. Proof: In what follows it is convenient to work with H(s 0 ) as in Lemma 2.3. In particular, the conditions for SPR of H may e restated in terms of the transfer function H(s 0 ). Specifically, conditions (5) and (6) for SPR of H are equivalent to! 0! 0! 0! H(0j!0 )+H(0j! 0 3 ) > 0 (8) H(0j! 0 )+H(0j! 0 ) 3 > 0for! 6= 0 (9)! 2 H(0j!0 )+H(0j! 0 ) 3 > 0: (0) Condition (8) is equivalent to c A 0 <0. Now consider conditions (9) and (0). Since A is invertile, Lemma 2.3 tells us that H(0j! 0 )= d +c (j!i 0 A) 0 () H I(j!)+H I(j!) 3 =! 02 [H(0j! 0 )+H(0j! 0 ) 3 ] > 0 8! 6= 0: Authorized licensed use ited to: he Lirary NUI Maynooth. Downloaded on Feruary 2, 200 at :00 from IEEE Xplore. Restrictions apply.

3 276 IEEE RANSACIONS ON AUOMAIC CONROL, VOL. 54, NO., NOVEMBER 2009 with A,, c, d defined in Lemma 2.3. Using the results in [3], we have c (j!i 0 A) 0 + c (j!i 0 A) 0 3 A matrix P = P > 0 is a strict Lyapunov matrix for A, that is A P + PA < 0 Since d = 0c A 0 >0, we can write hus where H(0j! 0 )+H(0j! 0 ) 3 =2 d det 0 d c (! 2 I + A 2 ) 0 A =2 d det I 0 d (! 2 I + A 2 ) 0 A c =2 d det (! 2 I + A 2 ) 0 2 det! 2 I + A 2 0 d A c : = 02c (! 2 I + A 2 ) 0 A : H(0j! 0 )+H(0j! 0 ) 3 = 2 d det[! 2 I + M ] det! 2 I + A 2 (2) M := A A 0 d c = A 0 A 0 0 A0 c A 0 c A 0 Since A is Hurwitz, all the real eigenvalues of A 2 = A 02 are positive which implies that det[! 2 I + A 2 ] 6= 0 for all!. Noting that det[! 2 I + A 2 ] > 0 for! sufficiently large, it follows from continuity arguments that det[! 2 I + A 2 ] > 0 for all!. Recalling that d>0, it follows from identity (2) that conditions (9) and (0) on H(0j! 0 ) are respectively equivalent to det[! 2 I + M ] > 0 8! 2 IR;! 6= 0! 0! 0! 2 det[!2 I + M ] > 0: Since det[! 2 I + M ] > 0 for large!, the aove conditions are equivalent to det[i 0 M ] 6=0 8 2 IR; < 0 (3) det[i 0 M ] 6=0: (4) 0! 0 Condition (3) is equivalent to the requirement that M has no negative real eigenvalues. Since M =0and 6= 0, the matrix M must have at least one zero eigenvalue; hence det[i 0 M ] = q() and all the other eigenvalues of M are given y the roots of the polynomial q. hus condition (4) is equivalent to q(0) 6= 0, that is, zero is not a root of q. hus (4) is equivalent to the requirement that M has only one eigenvalue at zero. he equivalence etween the first and third statement of the lemma follows from the SPR equivalence of c (si 0 A 0 ) 0 and c (si 0 A) 0 as stated in Lemma ) Comment 3: he literature contains spectral conditions for checking SPR of a strictly proper transfer function [0]; however, these conditions involve the eigenvalues of a 2n 2 2n Hamiltonian matrix. he conditions here involve a matrix of dimension n 2 n. III. MAIN RESUL In everything that follows, A is a real n 2 n matrix and g and h are real n-vectors. hese results make use of the following oservations. : if and only if P is a strict Lyapunov matrix for A 0, that is A 0 P + PA 0 < 0: o see, this post- and pre-multiply the first inequality y A 0 and its transpose. In a similar fashion one can also show that P is a (nonstrict) Lyapunov matrix for A, that is A P + PA 0 if and only if P is a (nonstrict) Lyapunov matrix for A 0, that is, PA 0 + A 0 P 0: he proof of the main result requires the following KYP lemma. Lemma 3.: [7]: Suppose (A; ) is controllale and (A; c) is oservale. hen, the following statements are equivalent. (i) he matrix A is Hurwitz and the transfer function H(s) = c (si 0 A) 0 is SPR. (ii) here exists a matrix P = P > 0 that satisfies the constrained Lyapunov inequality A P + PA <0 (5) P = c: (6) ) Comment 4: A discussion of strictly positive real transfer functions can e found in the ook [4] y Narendra & aylor on frequency domain staility criteria. he assumption that (A; c) is oservale ensures that P is positive definite [5]. heorem 3. (Main heorem): Suppose that A is Hurwitz and all the eigenvalues of A 0 gh have negative real part, except one, which is zero. Suppose also that (A; g) is controllale and (A; h) is oservale. hen, there exists a matrix P = P > 0 such that (5) and (6) hold if and only if the matrix product A(A 0 gh ) has no real negative eigenvalues and exactly one zero eigenvalue. Proof: he proof consists of two parts. First we use an equivalence to show that the conditions on A(A 0 gh ) are sufficient for the existence of a Lyapunov matrix P with the required properties. We then show that these conditions are also necessary. Sufficiency: Let e a right eigenvector of A 0 gh corresponding to the zero eigenvalue; then 6= 0and A = gh = h g. Let c = A 0 h; then c A = h. Since A is Hurwitz, we must have h 6= 0, otherwise A =0. Hence c A 6= 0and, without loss of generality, we assume that is chosen so that c A = 0. In this case g = 0A and h = c A: Controllaility of (A; ) and oservaility of (A; c) follow from controllaility of (A; g) and oservaility of (A; h), respectively. Noting that A 2 := A 0 gh = A 0 Ac A c A it follows from heorem 2. that the conditions on AA 2 imply that the transfer function c (si 0A) 0 is SPR. Consequently, it follows from Lemma 3. that there exists a matrix P = P > 0 such that (5) and (6) hold. Pre- and post- multiplying inequality (5) y A 0 and A 0 shows that this inequality is equivalent to A 0 P + PA 0 < 0: (7) Authorized licensed use ited to: he Lirary NUI Maynooth. Downloaded on Feruary 2, 200 at :00 from IEEE Xplore. Restrictions apply.

4 IEEE RANSACIONS ON AUOMAIC CONROL, VOL. 54, NO., NOVEMBER Inequality (7) and (6) imply that that is A 0 0c 0 Since c A = 0 6= 0 A 0 P + PA 0 P0 c P 0 c 0 P 0 P (8) A 0 0c 0 0: A 0 0 0c = A 0 Ac A 0 A c A c A 0 c A c A c A = A 0 gh 0g 0h 0 : Post- and pre-multiplying inequality (8) y the aove inverse and its transpose results in that is, A 0 0 0c 0 P 0 P A 0 0 0c 0: 0 (A 0 gh ) P + P (A 0 gh ) 0Pg 0 h 0g P 0 h 02 0: (9) It immediately follows that for the aove inequality to hold, we must have (A 0 gh ) P + P (A 0 gh ) 0: (20) Necessity: We first show that if there exists a matrix P = P > 0 satisfying conditions (5) (6), then AA 2 cannot have a negative real eigenvalue. Note that the conditions on P are equivalent to Hence, for any > 0 A 0 P + PA 0 < 0 (2) A 2 P + PA 2 0: (22) (A 2 + A 0 ) P + P (A 2 + A 0 ) < 0: Since P = P > 0, this Lyapunov inequality implies that A 2 + A 0 must e Hurwitz and hence, nonsingular. hus, AA 2 + I is nonsingular for all >0. his means that AA 2 cannot have a negative real eigenvalue. We now show that AA 2 cannot have a zero eigenvalue whose multiplicity is greater than one. o this end, introduce the matrix ~A(k) =A 2 + kgh : hen A = ~ A () and inequalities (5) (6) hold if and only if ~A(k) P + P ~ A(k) < 0 (23) A 2 P + PA 2 0 (24) hold for all k sufficiently close to one. As we have seen aove, this implies that ~ A(k)A2 cannot have negative real eigenvalues for all k sufficiently close to one. We shall show that AA 2 having an eigenvalue at the origin whose multiplicity is greater than one contradicts this statement. By assumption, A 2 has a single eigenvalue at zero; a corresponding eigenvector is the vector. Clearly, is also an eigenvector corresponding to a zero eigenvalue of A(k)A ~ 2. Now choose any nonsingular matrix whose first column is. hen, 0 A(k)A2 ~ = 0 3 (25) 0 S + krs and the eigenvalues of ~ A(k)A 2 consist of zero and the eigenvalues of S + krs. Note that the matrix S must e invertile since 0 A 2 2 = 0 ~ A(0)A2 = S and A 2 2 has only a single eigenvalue at zero. Now suppose that AA 2 = ~A()A 2 has an eigenvalue at the origin whose multiplicity is greater than one. hen S+rs must have a eigenvalue at zero; hence, det S+ rs =0. Since S is invertile det S + krs = det[s] det I + ks 0 rs = det[s](+ks S 0 r) and we must have +s S 0 r =0which implies that s S 0 r = 0. Hence Suppose det[s] > 0. hen det S + krs = det[s]( 0 k): det S + krs < 0 for k>. Since det S + krs is the product of all the eigenvalues of S + krs and complex eigenvalues occur in complex conjugate pairs, S + krs must have at least one real negative eigenvalue when k >. his yields the contradiction that A(k)A2 ~ has a negative real eigenvalue when k>. he conclusion is the same for det[s] < 0. 2) Comment 5: Let G(s) =h (si 0 A) 0 g +. hen one can readily show that satisfaction of the Lyapunov inequalities (3) is also equivalent to the following frequency domain conditions. G(j!)+G(j!) 3 > 0 for! 6= 0! 0! [G(j!)+G(j!)3 0! 2 ] > 0: (26) One can view this as a KYP result for the systems under consideration. o demonstrate this result, recall Lemma 2.3 and the definitions of and c in the proof of heorem 3. to otain that G(s 0 )=c (si 0 A 0 ) 0 : he proof of heorem 3. and Lemma 2.2 tell us that satisfaction of the Lyapunov inequalities (3) is equivalent to the transfer function c (si 0 A 0 ) 0 eing SPR. Using the fact that c A = 0, the desired result now follows from Lemma 2.. IV. EXAMPLES In this section we present two examples to illustrate the main features of our result. ) Example (No Common Quadratic Lyapunov Function): Consider A = 0 0 ; A 2 = with A 2 = A 0 gh and g = [0 0 ] and h = [0 0 0]. Note that A is Hurwitz; whereas A 2 is singular with all its eigenvalues in the open left half of the complex plane, except one at the origin. Note also that (A; g) and (A; h) are Authorized licensed use ited to: he Lirary NUI Maynooth. Downloaded on Feruary 2, 200 at :00 from IEEE Xplore. Restrictions apply.

5 278 IEEE RANSACIONS ON AUOMAIC CONROL, VOL. 54, NO., NOVEMBER 2009 controllale and oservale, respectively. he eigenvalues of the matrix product A A 2 are (0; 0; 3). Hence, from the results of our main theorem, there cannot exist a P = P > 0 such that A P +PA < 0 and A 2 P + PA ) Example 2 (Quadratic Staility): Consider A = :9 0:9 02:9 ; A 2 = with A 2 = A 0 gh and g = [0 0 ] and h = [0:9 0:9 0:9]. Note that A is Hurwitz; whereas A 2 is singular with all its eigenvalues in the open left half of the complex plane, except one at the origin. Note also that (A; g) and (A; h) are controllale and oservale, respectively. he eigenvalues of the matrix product A A 2 are (0; 0:0349; 2:865). Hence, from the main theorem, there exist a P = P > 0 such that A P + PA < 0 and A 2 P + PA 2 0. V. CONCLUSION In this note, we have derived necessary and sufficient conditions for the existence of a common quadratic Lyapunov function for a pair of stale linear time-invariant systems whose system matrices are of the form A, A 0 gh, and where one of the matrices is singular. As a preinary result, we otained a spectral characterization of strictly positive real transfer functions which are strictly proper. Future work will involve extending our results to nonquadratic Lyapunov functions such as those which arise in the application of Popov s criterion. APPENDIX First Note: Here we show that satisfaction of the Lyapunov inequalities (3) imply that the matrix A0gh can have at most one zero eigenvalue and cannot have a nonzero imaginary eigenvalue. Consider first nonzero imaginary eigenvalues. Suppose on the contrary that, j! is a nonzero imaginary eigenvalue of A 0 gh with eigenvector v and let Q 2 =(A 0 gh ) P + P (A 0 gh ). hen v 0 Q 2 v =[(A 0 gh )v] 0 Pv + v 0 P [(A 0 gh )v] =[j!v] 0 Pv + v 0 P [j!v] =0: Since A 0 gh is real, 0j! is an eigenvalue of A 0 gh with eigenvector v; hence v 0 Q 2v =0. he second inequality in (3) tells us that Q 2 0; hence, the set of vectors for x which x 0 Q 2 x =0is the same as the null space of Q 2. Since the linearly independent vectors v and v are in this null space, this space has dimension greater than one. Let Q = A P + PA. hen Q = Q 2 + Pgh + hg P and when x is in the null space of h, it should e clear that x 0 Q x = x 0 Q 2 x. Since the dimension of the null space of h is n0 and the dimension of the null space of Q 2 is greater than one, these two spaces have a nonzero common element x. For this x, wehave x 0 Q x = x 0 Q 2x =0: and we otain the contradiction that the first inequality in (3) does not hold. Hence A 0 gh cannot have nonzero imaginary eigenvalues. Now suppose that A 0 gh has multiple eigenvalues at zero. It follows from the second inequality in (3) that the system _x =(A0gh )x is stale; hence the algeraic multiplicity and geometric multiplicity of the zero eigenvalue are the same. his implies that there are at least two linearly independent eigenvectors for the zero eigenvalue. Proceeding now as in the case of the nonzero imaginary eigenvalues, one can otain a contradiction. Hence, A 0 gh cannot have a repeated eigenvalue at zero. Second Note: Here we show that if A0gh is marginally stale, A is Hurwitz and the conditions on A(A 0 gh ) hold then, the matrix A0gh can only have one zero eigenvalue and cannot have a nonzero imaginary eigenvalue. Consider first nonzero imaginary eigenvalues. Suppose on the contrary that, j! is a nonzero imaginary eigenvalue of A 0 gh with eigenvector v. hen 0j! is a nonzero imaginary eigenvalue of A 0 gh with eigenvector v. From this, it follows that the linear independent vectors v and v are eigenvectors of (A0gh ) 2 with eigenvalue 0! 2. his implies that the eigenspace of (A 0 gh ) 2 associated with 0! 2 is greater than one. Since the dimension of the null space of h is n 0 and the dimension of the aove eigenspace is greater than one, these two spaces have a nonzero common element x. For this x, wehave A(A 0 gh )Ax = A(A 0 gh ) 2 x = 0! 2 Ax: Since A is Hurwitx, Ax is nonzero and the aove yields the contradiction that A(A 0 gh ) has a negative real eigenvalue. Hence A 0 gh cannot have nonzero imaginary eigenvalues. Now suppose that A 0 gh has multiple eigenvalues at zero. Since the system _x = (A 0 gh )x is marginally stale; the algeraic multiplicity and geometric multiplicity of the zero eigenvalue must e the same. his implies that there are at least two linearly independent eigenvectors for the zero eigenvalue of A 0 gh. hese are also eigenvectors for the zero eigenvalue of A(A 0 gh ). his contradicts the fact that A(A 0 gh ) has a single eigenvalue at zero. Hence, A 0 gh cannot have a repeated eigenvalue at zero. REFERENCES [] M. V. Kothare and M. Morari, Multiplier theory for staility analysis of anti-windup control systems, Automatica, vol. 35, pp , 999. [2] B. G. Romanchuk and M. C. Smith, Incremental gain analysis of piecewise linear systems and application to the antiwindup prolem, Automatica, vol. 35, pp , Jul [3] M. J. Gomes da Silva and S. arouriech, Antiwindup design with guaranteed regions of staility: An LMI-ased approach, IEEE rans. Autom. Control, vol. 50, no., pp. 06, Jan [4]. Loquen, S. arouriech, and C. Prieur, Staility analysis for reset systems with input saturation, in Proc. 46th IEEE Conf. Decision and Control, Dec. 2007, pp [5] R. N. Shorten, K. S. Narendra, and O. Mason, A result on common quadratic Lyapunov functions, IEEE rans. Autom. Control, vol. 48, no., pp. 0 3, Jan [6] R. N. Shorten and K. S. Narendra, On common quadratic Lyapunov functions for pairs of stale LI systems whose system matrices are in companion form, IEEE rans. Autom. Control, vol. 48, no. 4, pp , Apr [7] S. Boyd, L. El Ghaoui, E. Feron, and V. Balakrishnan, Linear Matrix Inequalities in System and Control heory. Philadelphia, PA: SIAM, 994. [8] H. Khalil, Nonlinear Systems, 3rd ed. Englewood Cliffs, NJ: Prentice-Hall, [9] R. Shorten and C. King, Spectral conditions for positive realness of SISO systems, IEEE rans. Autom. Control, vol. 49, no. 0, pp , Oct [0] R. Shorten, P. Curran, K. Wulff, and E. Zehe, A note on spectral conditions for positive realness of transfer function matrices, IEEE rans. Autom. Control, vol. 53, no. 6, pp , Jun [] Z. Bai and W. Freund, Eigenvalue ased characterization and test for positive realness of scalar transfer functions, IEEE rans. Autom. Control, vol. 45, no. 2, pp , Dec [2] G. Muscato, G. Nunnari, and L. Fortuna, Singular perturation approximation of ounded real and positive real transfer matrices, in Proc. ACC, Baltimore, MD, 994, pp [3] R. E. Kalman, Lyapunov functions for the prolem of Lur e in automatic control, Proc. National Academy of Sciences, vol. 49, no. 2, pp , 963. [4] K. Narendra and J. aylor, Frequency Domain Criteria for Asolute Staility. New York: Academic, 973. [5] A. Acikmese and M. Corless, Staility analysis with quadratic Lyapunov functions: Some necessary and sufficient multiplier conditions, Syst. Control Lett., vol. 57, pp , Authorized licensed use ited to: he Lirary NUI Maynooth. Downloaded on Feruary 2, 200 at :00 from IEEE Xplore. Restrictions apply.