Dynamic properties of the hysteretic Bouc-Wen model

Systems&Control Letters56(2007)197–205/locate/sysconleDynamic properties of the hysteretic Bouc-Wen modelଁFayçal Ikhouane a,b,∗,Víctor Mañosa a,c,JoséRodellar a,ba Departament de Matemàtica Aplicada III,Universitat Politècnica de Catalunya,Spainb Jordi Girona1-3,08034Barcelona,Spainc Colom1,08222Terrassa,SpainReceived6February2006;received in revised form2September2006;accepted2September2006Available online17October2006AbstractThe Bouc-Wen model,widely used in structural and mechanical engineering,gives an analytical description of a smooth hysteretic behavior. In practice,the Bouc-Wen model is mostly used within the following black-box approach:given a set of experimental input–output data,how to adjust the Bouc-Wen model parameters so that the output of the model matches the experimental data.It may happen that a Bouc-Wen model presents a good matching with the experimental real data for a specific input,but does not necessarily keep significant physical properties which are inherent to the real data,independently of the exciting input.This paper presents a characterization of the different classes of Bouc-Wen models in terms of their bounded input-bounded output stability property,and their capability for reproducing physical properties inherent to the true system they are to model.©2006Elsevier B.V.All rights reserved.Keywords:Hysteresis;Bouc-Wen model;Dynamics1.IntroductionThe Bouc-Wen model for smooth hysteresis[13]has received an increasing interest due to its capability to capture in an ana-lytical form a range of shapes of hysteretic cycles which match the behavior of a wide class of hysteretic systems[12].In par-ticular,it has been used experimentally to model piezoelectric elements[9],magnetorheological dampers[14],wood joints [4]and base isolation devices for buildings[10].The obtained models have been used either to predict the behavior of the physical hysteretic element[10]or for control purposes[1,5,6]. In the current literature,the Bouc-Wen model is mostly used within the following black-box approach:given a set of exper-imental input–output data,how to adjust the Bouc-Wen model ଁSupported by CICYT through Grant DPI2005-08668-C03-01.Thefirst author acknowledges the support of the Spanish Ministry of Education and Science through the“Ramón y Cajal”program.The second author acknowl-edges the partial support of the Government of Catalonia’s grant2001SGR-00173.∗Corresponding author.E-mail addresses:faycal.ikhouane@(F.Ikhouane),victor.manosa@(V.Mañosa),jose.rodellar@(J.Rodellar).0167-6911/$-see front matter©2006Elsevier B.V.All rights reserved. doi:10.1016/j.sysconle.2006.09.001parameters so that the output of the model matches the exper-imental data[11]?The use of system identification techniques is one practical way to perform this task.Once an identification method has been applied to tune the Bouc-Wen model param-eters,the resulting model is considered as a“good”appro-ximation of the true hysteresis when the error between the experimental data and the output of the model is small enough. Then this model is used to study the behavior of the true hys-teresis under different excitations.By doing this,it is important to consider the following re-mark.It may happen that a Bouc-Wen model presents a good matching with the experimental real data for a specific input, but does not necessarily keep significant physical properties which are inherent to the real data,independently of the ex-citing input.In this paper we draw the attention to this issue, with the particular focus on the following two properties which are shared by most of the hysteretic mechanical and structural systems:Property1.Let us conceptualize a nonlinear hysteretic be-havior as a map x(t)→ s(x)(t),where x represents the time history of an input variable and s(x)describes the time history of the hysteretic output variable.For any bounded input x,the198F .Ikhouane et al./Systems &Control Letters 56(2007)197–205output of the true hysteresis s (x)is bounded.This bounded input-bounded output (BIBO)stability property stems from the fact that we are dealing with mechanical and structural systems that are stable in open loop.Property 2.Let us consider that x is the displacement of a one degree of freedom mechanical system connected to an element or device that supplies to the system a hysteretic restor-ing force s (x).The hysteretic element or device contributes to dissipate the mechanical energy of the system as usually observed in practice.The Bouc-Wen model has to reproduce this energy dissipation property in order to represent adequately the physical behavior of real systems.Fig.1shows an example of a typical hysteretic loop (x, )obtained by the Bouc-Wen model for a specific set of pa-rameters for the signal x(t)=sin (t).However,Example 1shows that there exist other different bounded time histo-ries x for which this Bouc-Wen model delivers unbounded responses (x),which means that this model is not BIBO stable.Example 1.Consider the Bouc-Wen model given by the fol-lowing parameters:D =1,A =1, =0.5, =−1.5and n =2.Take z(0)=0and define the bounded input signal x(t)=( /2)sin (t).The corresponding derivative is ˙x(t)=( /2)cos (t),which isalso bounded.For 0 t2we have ˙x(t) 0.This implies that,during the time interval [0,2],the Bouc-Wen model (5)can take only one of the two forms:˙z =˙x(1+z 2)for z 0,(1)˙z =˙x(1+2z 2)for z 0.(2)In both cases (1)and (2)we have ˙z 0for 0 t 2,which implies that z(t)is a non-decreasing function.Since z(0)=0,this means that z(t) 0,so that ˙z is given by (1).−1−0.500.51−2−112α=0.5; k=1; D=1; A=1; β=0.5; γ= −1.5; n=2InputH y s t e r e t i c o u t p u tFig.1.Example of a Bouc-Wen model that isunstable.−1−0.50.51−0.5−0.4−0.3−0.2−0.100.10.20.30.40.5InputH y s t e r e t i c o u t p u tA = −2; α = 0.5; β = 0.5; γ = 0.1; n = 1.1Fig.2.Example of a Bouc-Wen model that does not dissipate energy.Integrating (1)we obtain d z1+z 2=d x ,(3)which gives arctan (z)=x ,since z(0)=0and x(0)=0.This im-plies that z(t)=tan (x(t)).Observe that lim t → /2z(t)=+∞.Thus,the bounded input signal x(t)has given rise to an un-bounded hysteretic output.A similar construction can be done for any initial condition z(0)=0.In a similar vein,the Bouc-Wen model illustrated in Fig.2is BIBO stable but it does not dissipate the mechanical energy of the system as considered in the above Property 2.These two examples highlight the fact that,while these models may give a good approximation of a true hysteresis loop for a specific input excitation used with parametric identification or tuning purposes,they may not be appropriate to represent the behavior of a true hysteretic system under general input excitations.This paper presents an analytical study with the aim of giving the conditions on the Bouc-Wen model so that it holds the above BIBO stability and dissipation properties.A first study of the stability properties of the Bouc-Wen model has been done in [5].Section 2of this paper completes and enhances the results of [5].Sections 3and 4are new results.2.The BIBO stability Bouc-Wen model properties2.1.The modelConsider a physical system with a hysteretic component thatcan be represented by a map x(t)→ s (x)(t),which is referred to as the “true”hysteresis.The so-called Bouc-Wen model rep-resents the true hysteresis in the formBW (x)(t)= kx(t)+(1− )Dkz(t),(4)˙z =D −1(A ˙x − |˙x ||z |n −1z − ˙x |z |n ),(5)where ˙z denotes the time derivative,n >1,D >0,k >0and 0< <1(the limit cases n =1, =0, =1are treated inF.Ikhouane et al./Systems&Control Letters56(2007)197–205199Section5).Also we consider that + =0,the singular case + =0being treated in Section5.2.2.Problem statementThis study lies in the experimentally based premise that a truephysical hysteretic element is BIBO stable,which means that,for any bounded input signal x(t),the hysteretic response isalso bounded.Thus,the Bouc-Wen model BW should keep theBIBO stability property in order to be considered an adequatecandidate to model real physical systems.Example1gives anexample of a set of parameters A, , ,n such that,for a partic-ular bounded input x(t),the corresponding output BW(x)(t) given by the Bouc-Wen model(4)–(5)is unbounded.Thus,this set of parameters does not correspond to the descriptionof a hysteretic physical element.This motivates the followingproblem:Given the parameters0< <1,k>0,D>0,A, , with + =0and n>1,find the set of initial conditions z(0)for which the Bouc-Wen model(4)–(5)is BIBO stable.Note that when this set is empty,this means that the Bouc-Wen model is not BIBO stable.The solution of this problem will lead to classify different sets of parameters and initial conditions and,additionally,to determine explicit bounds for the hysteretic variable z(t).2.3.Classification of the BIBO stable Bouc-Wen modelsLet us introduce the following set:,k,D,A, , ,n={z(0)∈R such that BW is BIBO stable forall C1input signals x(t)withfixed valuesof the parameters ,k,D,A, , ,n}.(6) If the set ,k,D,A, , ,n is empty,then,for any initial condi-tion z(0),there exists a bounded signal x(t)such that the cor-responding hysteretic output BW(t)is unbounded.That is, the set of parameters{ ,k,D,A, , ,n}does not correspond to a BIBO stable Bouc-Wen model.The emptiness of the set ,k,D,A, , ,n is thus equivalent to the instability of the Bouc-Wen model,and for this reason,the rest of the analysis is devoted to determining explicitly the set ,k,D,A, , ,n as a function of the Bouc-Wen model parameters.Let z(0)be an element of ,k,D,A, , ,n.Then,for any bounded C1input x(t),the output BW(x)(t)is bounded. This implies by Eq.(4)that the output z(t)of the differential equation(5)should be bounded.This means that the set1A, , ,n={z(0)∈R such that z(t)is bounded for any C1 bounded input signal x(t)withfixed values of theparameters A, , ,n}(7) is such that ,k,D,A, , ,n⊂ A, , ,n.The inclusion in the other way is immediate,which shows that ,k,D,A, , ,n= A, , ,n. The importance of this equality stems for the fact that it is1The correct notation would be A,, ,n,D.However,we will see later that this set does not depend on the parameter D.Table1Classification of the BIBO stable Bouc-Wen modelsCase Upper boundon|z(t)|ClassA>0 + >0and − 0R max(|z(0)|,z0)I− <0and 0[−z1,z1]max(|z(0)|,z0)IIA<0 − >0and + 0R max(|z(0)|,z1)III + <0and 0[−z0,z0]max(|z(0)|,z1)IV A=0 + >0and − 0R|z(0)|VAll other cases∅easier to determine the set A, , ,n.Note that an empty set A, , ,n means that,with the chosen parameters A, , ,n,the Bouc-Wen model does not represent adequately the behavior of a real hysteretic system s(see Example1).We also define the following set:A, , ,n={z(0)∈R such that z(t)is bounded for any C1 input signal x(t)withfixed values of theparameters A, , ,n}.(8) Note that A, , ,n⊂ A, , ,n.With the notations introduced above,the main results of this section are given below. Theorem2.Let x(t),t∈[0,∞)be a C1input signal andz0 nA+ and z1nA− .(9) Then,Table1holds.Corollary1. A, , ,n= A, , ,n .Corollary1means that the boundedness of the signal z(t) depends only on the parameters A, , and n,and it is inde-pendent of the boundedness of the input signal x(t).This fact is particularly important for system control theory:when x(t) is a closed loop signal,we cannot assume a priori that it is bounded.The fact that A, , ,n= A, , ,n shows that for every input signal x(t)(under the only assumption that it is C1),the output z(t)is always bounded if the set is non-empty,and if z(0)∈ .Corollary2.In all cases where a non-empty set exists,the parameter is non-negative.Proof.First we check that the differential equation(5)has a unique solution.Eq.(5)may be seen as a non-autonomous lo-cally Lipschitz system where the dependence on time is con-tinuous.The local Lipschitz property is due to the fact that n>1.The time-dependent part of Eq.(5)is the term˙x which is continuous as x(t)has been assumed to be C1.Thus,by [7,Theorem3.1],a unique solution of(5)does exist on some time interval[0,t0].Note that for0 n<1,the differential equation(5)does not verify the local Lipschitz property and thus,the solutions may not be unique.200F .Ikhouane et al./Systems &Control Letters 56(2007)197–205We consider the following three cases:A >0,A <0and A =0,and we focus first on the case A >0.Consider the following three possibilities:P 1: + >0and − 0,P 2: + >0and − <0,P 3: + <0.Let us focus on the case P 1.We consider the Lyapunov function candidate V (t)=z(t)2/2.Its derivative takes different forms depending on the signs of ˙x and z .Indeed,setting Q 1={˙x 0and z 0},and denoting ˙V|Q 1as the expression of the derivative of the function V over the set Q 1,we have ˙V|Q 1=z ˙xD −1(A −( + )z n ).Thus,˙V|Q 1 0for z z 0.Also,if we set Q 2={˙x0and z 0},we have ˙V |Q 2=z ˙xD −1(A +( − )|z |n ).In this case,˙V|Q 2 0for all values of z .The same conclusion is drawn in the case of Q 3={˙x 0and z 0},since ˙V|Q 3=z ˙xD −1(A +( − )z n ).Finally,taking Q 4={˙x 0and z 0},we get ˙V |Q 4=z ˙xD −1(A −( + )|z |n ).Thus,˙V|Q 4 0for |z | z 0.We then conclude that,for all the possibilities of thesigns of ˙x and z ,we have ˙V0for all |z | z 0.By [7,Theorem 4.18]2we conclude that z(t)is bounded for every continuous function ˙x(t)and every initial condition z(0)which meansthatA, , ,n =R .Since A, , ,n ⊂ A, , ,n ,this implies thatA, , ,n = A, , ,n =R .The bounds on z(t)can be derived from Theorem [7,Theorem 4.18]as follows:(i)If the initial condition of z is such that |z(0)| z 0then |z(t)| z 0for all t 0.(ii)If the initial condition of z is such that |z(0)| z 0then|z(t)| |z(0)|for all t 0.We now turn to the case P 2.Again,the derivative of V (t)depends on the signs of ˙x and z .Indeed,˙V0in the following regions:{˙x 0and z 0and z z 0},(10){˙x 0and z 0and |z | z 1},(11){˙x 0and z 0and z z 1},(12){˙x 0and z 0and |z | z 0}.(13)At this point,two subcases have to be distinguished:the subcase P 21: 0and the subcase P 22: <0.We first treat the subcase P 21.The condition 0leadsto z 1 z 0.From (10)–(13)we conclude that ˙V0for every z 0 |z | z 1independently of the sign of ˙x.By [7,Theorem 4.18]we conclude that z(t)is bounded for every continuous function ˙x(t)and any initial state z(0)such that |z(0)| z 1.This means that [−z 1,z 1]⊂ A, , ,n .Now,take z(0)/∈[−z 1,z 1];we claim that there exists a bounded C 1signal x(t)such that the corresponding signal z(t)is unbounded.The construc-tion of such a signal is done in Lemma 1,which means that2The function W 3which appears in [7,Theorem 4.18]is zero in our case.This means that the time T of [7,Equation 4.42]may be infinite and that allwhat we can say about x(t)in [7]is that x(t) max ( x(t 0) , −11( 2( ))which is exactly what we use.z(0)/∈ A, , ,n .This implies that A, , ,n ⊂[−z 1,z 1].SinceA, , ,n ⊂ A, , ,n ,we get A, , ,n = A, , ,n =[−z 1,z 1].Using [7,Theorem 4.18]we can obtain the following bound:|z(t)| max (|z(0)|,z 0).Lemma 1.Take z(0)/∈[−z 1,z 1];then there exists a bounded C 1signal x(t)such that the corresponding signal z(t)is un-bounded .Proof.Assume that − <0, 0and assume that the ini-tial condition z(0)is such that |z(0)|>z 1.Take z(0)>z 1(the construction is similar in the case z(0)<−z 1)and define the signal ˙x =DA +( − )z n.(14)Since z(0)>z 1>0,we have A +( − )z(0)n <0,which means that the solution z(t)of the differential equation (5)is well defined at least during a maximal time interval [0,t 1)in whichwe have z(t)>z 1.For 0 t <t 1we have z >0and ˙x<0.Thus,Eq.(5)reduces to ˙z =D −1˙x(A +( − )z n ).(15)Combining Eqs.(14)and (15),it follows that ˙z =1.(16)Integrating (16)gives for 0 t <t 1:z(t)=t +z(0),(17)˙x(t)=DA +( − )(t +z(0))n.(18)The function z(t)is increasing,so that the conditions of ex-istence of ˙x in (14)and (18)are satisfied for any t 0.This means that t 1=∞,that is ˙x is well defined for all t 0and a solution of the differential equation (5)(given by (17))exists over t ∈R +.From Eq.(18),it follows that ˙x ∈L 1as n >1.This implies by [2,p.233]that x(t)goes to a finite limit as t goes to infinity,which means that x(t)is bounded.Thus,we have constructed a bounded C 1signal x(t)with an unbounded corresponding signal z(t).We have thus proved that,for 0,we have A, , ,n = A, , ,n =[−z 1,z 1]and |z(t)| max (|z(0)|,z 0).Remains the subcase P 22: <0.In this case,define z 1<z 2 nA/( − /2)<z 0and consider that the initial con-dition z(0) 0is such that |z(0)| z 2(the case z(0) 0can be treated in a similar way).Take ˙x(t)=a for some positive constant a ,then we have from equation (5)and from the fact that <0:˙z aD −1(A − |z |n )aD −1(A − z n 2)=aD −1A1−− /2b >0.(19)F.Ikhouane et al./Systems&Control Letters56(2007)197–205201 Eq.(19)shows that z(t)will increase and reach in afinite timethe value z2.At this point,we choose for˙x(t)the expressionin Eq.(14)so that the analysis of Lemma1remains valid.This means that for each initial condition z(0),it is possible toconstruct a bounded C1signal x(t)such that the correspondingsignal z(t)is unbounded.We have thus shown that for <0we obtain A, , ,n=∅.Since A, , ,n⊂ A, , ,n,this impliesthat A, , ,n= A, , ,n=∅.We now move to the case P3.Assume that z(0) 0(a similaranalysis can be done for the case z(0) 0),and definek1=− +D,k2=A+ .Choosing˙x>0we have from Eq.(5):˙z=k1˙x(k2+z n).(20)On the other hand,since n>1the quantity S= ∞z(0)d u/(k2+u n)>0isfinite.Choosing for the signal x(t)any increasing function such that x(0)=0and lim t→∞x(t)=S/k1,it follows from Eq.(20)thatlim t→∞ z(t)z(0)d uk2+u n=S.(21)Eq.(21)shows that lim t→∞z(t)=∞so that we have con-structed a bounded signal x(t)with an unbounded output z(t) for every initial condition z(0).This means that A, , ,n=∅, which implies that A, , ,n= A, , ,n=∅.The cases A<0and A=0can be treated in a similar way, which ends the proof of Theorem2.Note that the case + >0, − <0and 0simplifies to − <0and 0since the former inequality is implied by the latter two.Corollary1follows immediately from the proof of Theorem2,while Corollary2follows from Table1. Table1shows that the classes I–V are BIBO stable.A class is composed of a range for the Bouc-Wen model parameters and a range for the initial condition z(0)of the hysteretic part of the model.For all the classes,it can be seen that the hysteretic variable z(t)is bounded and that its bound is independent of the bound of the input x(t).This will show to be useful for control purposes.Note that,for the class V,we have z(t)=0for all t 0if z(0)=0.Since the Bouc-Wen model is often used with an initial condition z(0)=0,this means that the class V corresponds to a linear behavior and is thus irrelevant from the point of view of the description of the hysteretic systems.3.Free motion of a hysteretic structural systemSection2has analyzed the stability properties of the Bouc-Wen model.It has shown that,for the Bouc-Wen model to be BIBO stable,it should belong to classes I–V of Table1. The class V has been shown to be irrelevant as,in practice,it corresponds to the description of a linear behavior.For this reason,we consider only the classes I–IV in the present section.3.1.Problem statementWe consider a structural isolation scheme,as illustrated in Fig.3,which is modelled as1degree-of-freedom system with mass m>0and viscous damping c>0plus a restoring force characterizing a hysteretic behavior of the isolator material.This system is described by the second order differential equationm¨x+c˙x+ (x)(t)=f(t),(22) with initial conditions x(0)and˙x(0)and excited by a force f(t),like the one of the form−ma(t)in the case of an earth-quake with ground acceleration a(t).The restoring force is as-sumed to be described by the Bouc-Wen model:(x)(t)= kx(t)+(1− )Dkz(t),(23)˙z=D−1(A˙x− |˙x||z|n−1z− ˙x|z|n),(24) where n>1,D>0,k>0and0< <1.The purpose of this section is the study of the free motion of system(22)–(24),that is with f(t)=0,to analyze its asymptotic trajectories.In real applications,the base-isolation devices are designed to dissipate the energy introduced in the structure by external perturbations.In the absence of disturbances,the structure is in free motion so that when its initial conditions are not zero, the structure dissipates the energy due to the initial conditions and keeps at rest asymptotically.Base-isolation devices have often been modelled by the Bouc-Wen model[10],so that this model has to reproduce some general physical properties of these devices.In particular,for the Bouc-Wen based model (22)–(24)that is supposed to reproduce the behavior of a base-isolation device,it is desirable that the velocity˙x of the mass goes asymptotically to zero,and that its position x goes to a constant,when the system is in free motion.The next section shows that this is the case for the classes I and II.3.2.Asymptotic trajectoriesIn this section,we analyze the asymptotic behavior of the system defined by Eqs.(22)–(24)in the absence of an exter-nal excitation.The main result of this section is given in the following theorem.Theorem3.For every initial conditions x(0)∈R,˙x(0)∈R and z(0)∈ A, , ,n=∅,the following holds:(a)For all the classes I–IV of Table1,the signals x(t),˙x(t)and z(t)are bounded and C1.(b)Assume that the Bouc-Wen model belongs to the classes Ior II.Then,there exist constants x∞and z∞,which depend on the Bouc-Wen model parameters( ,D,k,A, , ,n), the system parameters(m,c)and the initial condi-tions(x(0),˙x(0),z(0)),and there exists a constant¯c202F.Ikhouane et al./Systems&Control Letters56(2007)197–205(a)(b)Fig.3.Base isolation device(a)and its physical model(b). that depends on the parameters m,k,A, , , such thatfor all c ¯c we have:lim t→∞x(t)=x∞,(25)lim t→∞z(t)=z∞,(26)x∞+(1− )Dz∞=0.(27) Furthermore,we have˙x∈L1([0,∞))and limt→∞˙x(t)=0.(28) Proof.Part(a):A state-space system realization of(22)–(24)is ˙x1=x2,(29)˙x2=m−1(−cx2− kx1−(1− )kDz),(30)˙z=D−1(Ax2− |x2||z|n−1z− x2|z|n),(31) where x1=x.Since(29)–(31)is locally Lipschitz,then a C1so-lution(x1(t),x2(t),z(t))exists over some time interval[0,t0). We have seen(Corollary1of Section2.3)that z(t)given by (24)is bounded for every C1signal x(bounded or not)once z(0)belongs to A, , ,n=∅.Thus,Eq.(22)can be written in the formm¨x+c˙x+ kx=−(1− )Dkz,(32) which may be seen as an exponentially stable second order linear system excited by a bounded external input signal−(1− )Dkz.This implies that the signals x1(t),x2(t)and z(t)are bounded[2,p.59]and thus t0=∞,which proves part(a)of Theorem3.Let us move to part(b)of Theorem3.We proceed in two steps.First,we show that˙x belongs to L1.Second,we use this property to complete the proof of the theorem.First part of the proof of part(b):We have the two cases P1:|z(t)|>z0for all t 0.P2:There exists some t0<∞such that|z(t0)| z0.Wefirst treat the case ing the results of Section2it follows that the time function z(t)2is non-increasing.Since it is bounded,it goes to a limit z2∞ z20.Consider the case where z(0)>0(the analysis is similar in the case z(0)<0).Then,by continuity of z,we have z(t) z∞ z0>0for all t 0.Take >0,then there exists some t1<∞such thatz n∞ z(t)n z n∞+ for all t t1.(33)Multiplying by x2and integrating both parts of(33),the fol-lowing is obtained for any T 0:z n∞t1+Tt1x2(t)d t−t1+Tt1|x2(t)|d tt1+Tt1x2(t)z(t)n d t z n∞t1+Tt1x2(t)d t+t1+Tt1|x2(t)|d t.(34)On the other hand,using the fact that 0(see Corollary2of Section2.3),the following is obtained from Eqs.(31)and(33): |x2|z n∞ |x2|z n=−D˙z+Ax2− x2z n.(35) Here we need to discuss two subcases:P11: >0,P12: =0.Let us focusfirst on the subcase P11.Integrating both parts of inequality(35)and using Eq.(34)it follows thatt1+Tt1|x2(t)|d t −Dn∞(z(t1+T)−z(t1))+Az n∞−(x1(t1+T)−x1(t1))+| |n∞t1+Tt1|x2(t)|d t.(36)If =0,then using part(a)of Theorem3,it follows from Eq.(36)that x2∈L1as T is arbitrary.If =0,choosing = z n∞/2| |in Eq.(36)shows that x2∈L1.We now consider the subcase P12.Note that the parameter can be zero only for the class II.In this case,Table1shows that >0.By Assumption P1,we have z(t)>z0for all t 0, thus it follows that z n>z n0=A/ or equivalently A− z n<0. Using Eq.(31)with =0leads to x2=D˙z/(A− z n).Since z is non-increasing,˙z 0,which implies that x2(t) 0for all t 0.We then gett|x2( )|d =tx2( )d =x1(t)−x1(0) which,using part(a)of Theorem3,shows that x2∈L1.We now turn to the case P2.Taking the derivative of Eq.(30) we get2¨x2+ 1˙x2+ 0x2=−˙z,(37)F .Ikhouane et al./Systems &Control Letters 56(2007)197–205203where 2=m(1− )Dk,1=c(1− )Dk,0=(1− )D.(38)By Assumption P 2,we have |z(t 0)| z ing the bound on |z(t)|of Table 1for A >0,it follows that |z(t)| z 0for all t t 0.From Eq.(31)we get for all t t 0:0 D ˙z x 2=(A −( + )|z |n )x 22 Ax 22for zx 2 0,Ax 22 D ˙zx 2=(A +( − )|z |n )x 22 2A + x 22for zx 2 0(Class I Bouc-Wen model ),2A x 22 D ˙z x 2=(A +( − )|z |n )x 22 Ax 22for zx 2 0(Class II Bouc-Wen model )so that we have in all cases:0 ˙z x 2 x 22,(39)where =A D max1,2 +.Note that the system (29)–(31)can be viewed as a feedback connection as in [7,Fig.7.1]where the reference r =0,the signal u =−˙z and y =x 2.By Eq.(39),the nonlinearity (·)in [7,Fig.7.1]which corresponds to the Bouc-Wen model,be-longs to the sector [0, ].The idea is to use this fact to prove that the feedback connection is such that the state (x 2,˙x 2)goes exponentially to zero using [7,Theorem 7.1.1].It is to be noted that the Bouc-Wen nonlinearity (·)is not memory-less in our case;however,since we know that z is bounded,it can be checked easily that the stability proof of the feed-back connection is a small variation of that in [7,Theorem 7.1.1].In the next,we use the simplified version of [7,Theo-rem 7.1.1]given in [7,Example 7.1].We then need to check the conditions under which we have Re [1+ G(j )]>0for all ∈(−∞,∞)where the transfer function G(s)is given by G(s)=1/( 2s 2+ 1s + 0).This is equivalent to checking the conditions under which we havef ( )= 22 2+ (−2 0 2+ 21− 2)+ 20+ 0>0for all = 2(40)Eq.(40)is a second-order algebraic equation in the variable .If its discriminant is negative,then the function f ( )is always positive.If 0,then f ( )=0has real roots 1and 2that have the same sign.If 1<0and 2<0,then f ( )>0for all values of = 2 0.The only case where we can have f ( )<0with 0occurs when the sum of roots is positive.This condition can be written asc < mk A(1− )max 1,2+ +2¯c .(41)Thus,we have proved that c ¯c implies that we have Re [1+ G(j )]>0for all ∈(−∞,∞).Then,using[7,Theorem 7.2],it follows that the state x 2goes exponentiallyto zero.This implies that x 2∈L 1.Second part of the proof of part (b):The fact that x 2∈L 1along with Eq.(29)shows that x 1goes to a finite limit [2,p.233]which establishes Eq.(25).Since x 2is bounded and ˙x 2is bounded (by Eq.30)and x 2∈L 1,then by Barbalat’s lemma [8,p.491,Corollary A.7]it follows that lim t →∞x 2(t)=0.Now,taking the derivative of Eq.(30),we get m ¨x 2+c ˙x 2+ kx 2=−(1− )kD ˙z ,(42)which may be seen as a stable second order system excited bythe input −(1− )kD ˙z .Since x 2∈L 1,by part (a)of Theorem 3and Eq.(31),it follows that ˙z ∈L 1L ∞.From Eq.(42)we then conclude that ˙x 2∈L 1[2,p.59].We now have that˙x 2is bounded (by Eq.(30)),¨x2is bounded (by Eq.(42))and ˙x 2∈L 1.Thus,by application of Barbalat’s lemma,it follows that lim t →∞˙x 2(t)=ing this property in Eq.(30)along with Eq.(28)we obtainlim t →∞( kx 1(t)+(1− )kDz(t))=0.(43)Since x 1goes to a finite limit,this establishes Eqs.(27)and (26),which completes the proof of Theorem 3.Theorem 3shows that,for the classes I and II,the states x and z go to constant values asymptotically and that the velocity ˙x goes to zero.This means that both classes are good candidates for the description of the real physical behavior of a base-isolation device.Crucial to the proof of the theorem is the fact that ˙x ∈L 1.It is to be noted that the condition c ¯c is only a sufficient condition for the validity of Theorem 3.It is not necessary as can be checked using numerical simulations.Theorem 3demonstrates that the Bouc-Wen based model (22)–(24)behaves in accordance with observed experiments for real base-isolation devices when the model belongs to the classes I and II of Table 1.This is not the case for the classes III and IV .Indeed,consider the class III Bouc-Wen based model (22)–(24)given by the following parameters: =0.5,k =2,D =1,A =−2, =1, =0,n =1.1,m =1,c =1.Fig.4gives the solution of the differential equation (22)–(24)with initial conditions x(0)=0,˙x(0)=0.1,z(0)=0.We observe that,after a transient,a limit cycle occurs (a mathematical proof of this property is very difficult).In fact,it can be checked using numerical simulations that,for a large number of values of A <0(that is for the classes III and IV),and for arbitrary small initial conditions,limit cycles are observed.Such a behavior has not been observed for real structures like base-isolation devices.For this reason,we consider that the negative values of the parameter A do not correspond to a physical behavior of the Bouc-Wen model.The conclusion that can be drawn from this section is that,from the classes I–IV of Table 1,only the two classes I and II are relevant from the point of view of the description of the physical phenomena.The rest of the classes of Table 1,that is classes III–V are irrelevant in practice.。