Systems, Structure and Control 2012 Part 13 pptx

A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems 233 Since we are concerned with a discrete controller, the discretization of the continuous system (1)-(3) can be described by 1 1 kdkdkdk kdk kkk x Ax Bu Pw wSw eCxRw + + =++ = =− where 0 1 0 0 0 1 0 0 ; ! ; ! ; ! ;; ; (1)! i Ai d i i sA i d i i Si d i i sA ddd i i Ae A i BeBds AB i Se S i CCRRP ePds P i δ δ δ δ δ δ δ δ ∞ = ∞ − = ∞ = + ∞ = == == == === = + ∑ ∑ ∫ ∑ ∑ ∫ where i P can be computed iteratively from 01 ; ; 1,2, i ii PPPAP PSi − ==+ = The classical Robust Regulator Problem with Measurement of the Output for system (1)- (3) consists in finding a dynamic controller () () () () e tFtGet uH t ξξ ξ • =+ = such that the following requirements hold: S) The equilibrium point (, ) (0,0)x ξ = of the closed loop system without disturbances () () () () () () e x tAxtBHt tFtGCxt ξ ξξ • • =+ =+ is exponentially stable. Systems, Structure and Control 234 R) For each initial condition 000 (, ,)xw ξ , the dynamics of the system () () () () () () ( () ()) () () e x tAxtBHtPwt tFtGCxtRwt wt Swt ξ ξξ • • • =+ + =+ − = satisfy that lim ( ) 0. t et →∞ = A solution to this problem can be found in [1]. This solution is stated in terms of the existence of mappings ; ss ss x ww ξ =Π =Σ satisfying the Francis equations 0 e SA BH P SF CR Π=Π+ Σ+ Σ=Σ =Π− (4) for all admissible values of the systems parameters. More precisely, the solution can be stated in terms of the existence of mappings , ss ss x wu w=Π =Γ solving the equations SA B PΠ=Π+Γ+ (5) 0 CR=Π− (6) from which we reckon 1 1 01 1 q q q S S aaS aS − − − Γ ⎛⎞ ⎜⎟ Γ ⎜⎟ ⎜⎟ Σ= ⎜⎟ Γ ⎜⎟ ⎜⎟ −Γ−Γ− − Γ ⎝⎠ # where the polynomial 1 110 0 qq q sas asa − − ++++= is the characteristic polynomial of .S The mapping ss x w=Π represents the steady state zero output subspace and ss uw=Γ is the steady-state input which make invariant that subspace. This steady-state input can be generated, independently of the values of the parameters of the system and thanks to the Cayley-Hamilton Theorem, by the linear dynamical system A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems 235 η η • =Φ (7a) ss uH η = (7b) where 11 { , } ; { , , } mm diag H diag H HΦ= Φ Φ = and 012 1 1 010 001 0 ; 000 1 (1 0 0) . i q iq aaa a H − × ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ Φ= ⎜⎟ ⎜⎟ ⎜⎟ −−− − ⎝⎠ = " " ###%# " " " Defining the transformation 12 ;,zx wz η =−Π = the system can be rewritten as 1 12 zAzBHzBu • =− + (8) 2 2 zz • =Φ (9) [] 1 2 () 0 z et C z ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ (10) Finally, a controller which solves the problem can be constructed as an observer for system (8)-(9), namely 1 0101020 1 2 201 2 2 12 () A GC B H Bu Ge GC Ge uK H ξξξ ξξξ ξξ • • =− − ++ =− +Φ + =+ (11) where 000 ,,ABC are the nominal values of the matrices of the system (1)-(3) and K and 12 ,GG make stable the matrices 00 ()ABK+ and () 1 00 0 2 0. 0 G ABH C G − ⎛⎞ ⎛⎞ − ⎜⎟ ⎜⎟ Φ ⎝⎠ ⎝⎠ (12) When dealing with controllers implemented via digital devices and zero order holders, the sampled data version of the controller could render unstable the closed-loop system. In this Systems, Structure and Control 236 work we will take the approach of designing a hybrid controller consisting in two parts: a discrete sliding mode controller ensuring the stabilization of the closed-loop system, and a continuous part containing the internal model dynamics (internal model) obtained from the continuous model. 3. The Continuous Sliding Robust Regulator Analogously to the case of the Robust Regulator Problem, we formulate the Sliding Mode Robust Regulator Problem ([13], [14], [15]) as the problem of finding a sliding surface 1 ( ) 0, ( ( ), , ( )) m col σσ ξ σσ ξ σ ξ === (13) and a dynamic compensator (,) g e ξξ • = (14) with the control action defined as () () 0 { , 1, , () () 0 ii i ii u uim u ξσξ ξσξ + − > == < (15) where the mappings (), i u ξ + () i u ξ − and () i σ ξ are calculated in order to induce an asymptotic convergence to the sliding surface () i σ ξ 0= and such that, for all admissible parameter values in a suitable neighborhood of the nominal parameter vector, the following conditions hold: (SS c ) The equilibrium point (, ) (0,0)x ξ = of the closed-loop system is asymptotically stable. (SM c ) The sliding surface is attractive, namely the state of the closed loop system converges to the manifold () 0. σ ξ = (SR c ) The output tracking error tends asymptotically to zero, namely lim ( ) 0 t et →∞ = Now, to introduce the sliding mode approach into the regulator problem, we will chose the control input ()ut as () s lid eq ut u u=+ instead of 12 ()ut K H ξξ =+ as taken in the controller (11), where we impose that eq u must be equal to 2 H ξ when () 0 σ ξ = . Note that the stabilizing part 1 K ξ will now be substituted by the term s lid u which will be calculated to make attractive the sliding surface. A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems 237 To be more precise, let us consider the switching surface [] 1 0, σ ξξ =Σ =Σ (16) where m σ ∈ℜ , mxn Σ∈ℜ with rank 0 B mΣ= . Differentiating this function, and from the first equation of (11) we reckon 1 0101020 1 0101 02 0 1 [( ) ] () AGC BH BuGe AGC BH Bu Ge σξ ξ ξ ξξ •• =Σ =Σ − − + + =Σ − −Σ +Σ +Σ from which the equivalent control eq u is obtained from the condition 0 σ =  as [ ] 1 00101021 () ( ) eq uBAGCBHGe ξξ − =−Σ Σ − − + Defining the estimation errors as 111 z ε ξ =− and 22 z ε =− 2 ξ , we may substitute eq u into equation (8) at the nominal values of the parameters to get the sliding motion dynamics 11 1 00 0100 010102 [()] ()( ) n zIBB AzBB AGC BH εε • −− =−Σ Σ +Σ Σ− − where the estimation errors satisfy the dynamics 1 010102 2 201 2 () . AGC BH GC εεε εεε • • =− − =− +Φ Note that these dynamics are asymptotically stable thanks to the observability assumption of matrix (12). Lemma 1. [14] Define the operator D as 1 (()) n DIBB − =−ΣΣ . Then the relation ()0DA S PΠ−Π + = (17) is true if and only if there exist matrices Π and Γ such that . A SPBΠ−Π + = Γ (18) Proof. The operator D is a projection operator along the rank of B over the null space of Σ [16], namely 1 11 1 1 1 (())0 ,{ | 0} n n DB I B B B Dz z z z z − =−ΣΣ= =∀∈ℵℵ=∈ℜΣ= Systems, Structure and Control 238 Thus, if condition (18) holds, then it follows that ()0.DA S P DBΠ−Π + = Σ= Conversely, if condition (17) holds, then () A SDΠ−Π + must be in the image of B this is, () A SD BΠ−Π + = Γ for some matrix .Γ ■ A condition for the solution of the Sliding Mode Regulator Problem can be given in the following result. Proposition 2. Assume the following assumptions: H1) The matrix S has all its eigenvalues on the imaginary axis H2) The pair 00 (,)AB is stabilizable H3) The pair [ ] 0 0,C 00 0 ABH− ⎡ ⎤ ⎢ ⎥ Φ ⎣ ⎦ is observable. Then the Sliding Mode Regulator Problem is solvable if there exists a matrix Π solving the equations A SP BΠ−Π + =− Γ (19) 0CRΠ− = (20) for some matriz ,Γ and or all admissible values of the system parameters. Proof . Let us choose the control as () , eq uMsign u σ =− + with (); 0, ii Mdiagmm=> and 1 ( ) [ ( ), , ( )] . T m sign sign sign σσ σ = This control action guarantees a sliding mode motion on the surface 0. σ = Then, assuming that the observer estimation error decays rapidly by appropriate choice of the gains 12 ,GG we have that 1 1 01 0 | z zDAz • Σ= = Since the matrix Σ by assumption H2 can be chosen such that B Σ is invertible, and the ()nm− eigenvalues of 0 DA can be arbitrarily placed in ,C − then 1 () 0zt→ as t →∞ satisfying condition (SS c ). Now, since the tracking error equation is given by 01 () (),et Cz t= then it follows that ()et goes to zero asymptotically, satisfying condition (SR c ). ■ Note that when the state of the system is on the sliding surface, the control signal is exactly eq u which in turn comes to be 2 , eq ss uH u ξ == namely, the steady-state input. This steady -state input guarantees that the output tracking error stays at zero. This property will be used later. A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems 239 4. A Sliding Robust Regulator for Discrete Systems For the discrete case, the problem can be formulated in a similar way to the continuous case. To this end, let us consider the discretization of system (8)-(10), this is 1, 1 1, 2, 1 2, 0 0 kk d d k kk d zz A B u zz + + −Λ ⎡⎤ ⎡⎤ ⎡⎤ ⎡ ⎤ =+ ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ Φ ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣⎦ (21) [] 1, 2, 0 k kd k z eC z ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ (22) where 0 0 0000 0 00 0 ,, ,( ) ( ); ,0 . T AT A dd T d T A d Ae eBHdCC eukT ukT BeBd T θ θ θ θ θθ Φ =Λ= = Φ= + = =≤≤ ∫ ∫ For this system, the Sliding Regulator Problem can be set as the problem of finding a sliding surface k σ and a dynamic controller 1kdkdk FGe ξξ + =+ (23) (,) kdkk ue α ξ = (24) such that, for all admissible parameter values in a suitable neighborhood of the nominal parameter vector, the following conditions hold: (SS d ) The equilibrium point (, ) (0,0)x ξ = of the closed-loop system is asymptotically stable. (SM d ) The sliding surface is attractive, namely the state of the closed loop system converges to the manifold ()0. kk σ ξ = (SR d ) For each initial condition 000 (, ,)xw ξ , the dynamics of the closed-loop system 1 1 1 (,) () kdkddkk k kkdkdk kdk x Ax B e Pw FGCxRw wSw α ξ ξξ + + + =+ + =+ − = where ST d Se= guarantees that lim 0 . kk e →∞ = Assume the following conditions hold: ( d H1 ) All the eigenvalues of matrix d S lie on the unitary circle. Systems, Structure and Control 240 ( d H2 ) The pair { } 00 , dd AB is stabilizable, ( d H3 ) There exists a solution , dd ΠΓ to the regulator equations dd d d d d d SA B PΠ=Π+Γ+ (25) 0 dd d CR=Π− (26) ( d H4 ) The pair [] 0 0 0, 0 d d d A C −Λ ⎡ ⎤ ⎢ ⎥ Φ ⎣ ⎦ is observable. Then, a classic robust regulator can be constructed as 1, 1 0 1 0 1, 2, 0 1 2, 1 2 0 1, 2, 2 1, 2, () kdddk kdkdk kddkdkdk kdk k A GC Bu Ge GC Ge uK H ξξξ ξξξ ξξ + + =− −Λ+ + =− +Φ + =+ (27) where d K and 12 , dd GG make stable the matrices 00 () ddd ABK+ and () 01 0 2 0. 0 dd d dd AG C G −Λ ⎛⎞⎛⎞ − ⎜⎟⎜⎟ Φ ⎝⎠⎝⎠ (28) respectively. For the Discrete Sliding Regulator Problem, we can chose a sliding surface [ ] 1, 0, kd kdk σ ξξ =Σ =Σ (29) and calculate the equivalent control. The following result, which can be proved similarly to the continuous case, gives a solution to the Discrete Sliding Regulator Problem: Proposition 3. Assume that assumptions H 1 d through H 4 d hold. Then the Discrete Sliding Regulator Problem is solvable. Moreover, the controller solving the problem can be chosen as 1 ,00101,2,1 ()( ) . keqk dd dd dd k k dk uu B A GC Ge ξξ − ⎡ ⎤ ==−Σ Σ − −Λ+ ⎣ ⎦ Proof. Calculating 11,1 0101, 2, 0 1 [( ) ] kdk dd dd k k dk dk AGC BuGe σ ξ ξξ ++ =Σ =Σ − −Λ + + we can calculate the equivalent control from the condition 1 0, k σ + = namely: 1 ,00101,2,1 ()( ) . eq k d d d d d d k k d k uBAGC Ge ξξ − ⎡ ⎤ =−Σ Σ − −Λ + ⎣ ⎦ A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems 241 Note that this control makes also the sliding surface attractive, since the same control guarantees that 0 kj σ + = for j 1. Now, substituting eq u in the first equation of (21) we obtain 1 1, 1 0 0 0 1, 1 00 0101, 2, [()] ()( ) kndddddk ddd dd dd k k zIBB Az BB AGC εε − + − =− Σ Σ +Σ Σ − −Λ where 1, 1, 1, ; kk k z ξ =− 2, 2, 2, . kk k z ξ =− As in the continuos case, if the gains 12 , dd GG are appropriately chosen, the estimation errors 1, k and 2, k will converge to zero and then 1, 1 0 1,kdk zDAz + = where 1 00 [()]. nddd d DIB B − =− Σ Σ Since the matrix d Σ by assumption H2 d can be chosen such that 0dd B Σ is invertible, and the (n-m) eigenvalues of 0d DA can be arbitrarily placed inside the unitary circle, then 1, 0 k z → as k →∞ satisfying condition (SS d ). Now, since the tracking error equation is given by 01, , kdk eCz= then it follows that k e goes to zero asymptotically, satisfying condition (SR d ). ■ Note that when the state of the system is on the sliding surface, the control signal is exactly eq u which in turn comes to be 2 , eq ss uH u ξ == namely, the steady-state input. Again note that when the solution of the system is on the sliding surface, the control signal is exactly eq u which in turn, since 0 , d B HΛ= comes to be 1 02,2, () , eq d d d k k uB H ξξ − =Σ ΣΛ = namely, the steady-state input. Clearly, this controller guarantees zero output tracking error only at the sampling instants, but not at the intersampling. To force the output tracking error to converge to zero also in the intersampling time, in the following section we will formulate the a ripple-free sliding regulator problem. 5. A Ripple-Free Sliding Robust Regulator for Sampled Data Linear Systems From the previous discussion it is clear that implementing a Sliding Mode Robust Regulator for the discretization of the continuous linear system, this will guarantee only that the output tracking error will be zeroed only at the sampling instant. In order to eliminate the possible ripple, it is necessary to reproduce the internal model (7) from its discrete time realization. To do this, we note that the solution of (7) can be written as () (0), t te ξξ Φ = and setting tkT θ =+ with [0, )T θ ∈ we have Systems, Structure and Control 242 () () (0) (0) () ()() () kkT ss ke ee ekT ukT HkT He kT δθ θ θ θ ξ δθ ξξ ξ θξθ ξ Φ+ ΦΦ Φ Φ += = = += += which describe exactly the behavior also in the intersampling. The term e θ Φ is known as the exponential holder. We can now formulate the Ripple-Free Sliding Robust Regulator Problem as the problem of finding a sliding surface kk σ ξ =Σ (30) and a dynamic controller 1kkk FGe ξξ + =+ (31) () () ,,; kk ukT e θαξθ += (32) 0.T θ ≤≤ such that, for all admissible parameter values in a suitable neighborhood of the nominal parameter vector, the following conditions hold: (SS r ) The equilibrium point ()() ,0,0 kk x ξ = of the system in closed-loop is asymptotically stable. (SM r ) The sliding surface is attractive, namely the state of the closed loop system converges to the manifold ()0. kk σ ξ = (SR r ) For each initial condition 000 (, ,)xw ξ , the dynamics of the closed-loop system () 1 () () , , () () () () kk kkdkdk x tAxtB e Pwt FGCxRw wt Swt αξ θ ξξ • + =+ + =+ − =  guarantees that () 0.lim t et →∞ = In order to solve the Ripple-Free Sliding Robust Regulator Problem, the following assumptions will be considered: H1) The matrix S has all its eigenvalues on imaginary axis H2) The pair 00 (,)AB is stabilizable H3) The equations (5), (6) have solution ,ΠΓ for all admissible values of the system parameters. [...]... which guarantees that the output 248 Systems, Structure and Control tracking error is zeroed not only at the sampling instants, but also in the intersampling behavior was alsoformulated and a solution was obtained The controller has two components: one of them depending of the discrete dynamics of the system, and the other containing the internal model of the reference and/ or perturbations generator This... B., and Obregon-Pulido, G (2003) Guaranteeing asymptotic zero intersampling tracking error via a discretized regulator and exponential holder for nonlinear systems, J App Reserch & Tech 1, pp 203-214 [10] Yamamoto, A., A function space approach to sampled data control systems and tracking problems, IEEE Trans Aut Control (1994); 350(4), pp 703-712 [11] Utkin, V.I (1981), Sliding modes in control and. .. 42, No 6, pp 864-868 [6] Kabamba, P T (1987), Control of Linear Systems using generalized sample-data hold funtions, IEEE Trans Aut Control, Vol AC-32, No 9, pp 772-782 [7] Loukianov, Alexander G., Castillo-Toledo, B and García, R (1999) , Output Regulation in Sliding Mode , Proc.of the American Control Conference, pp 1037-1041 [8] Castillo-Toledo, B., and Di Gennaro, S (2002), On the nonlinear ripple... ξ 2, k + Bd 0uk + Gd 1ek ξ 2, k +1 = −Gd 2Cd 0ξ1, k + Φ d ξ 2, k + Gd 2ek (37) 244 Systems, Structure and Control Defining a switching function as σ k = [Σd 0] ξ k = Σ d ξ1, k and proceeding as in the discrete case, we may show that by a proper choice of the gains Gd 1,Gd 2 , the estimation errors converge to zero and the matrix −1 D = [ I n − Bd 0 ( Σ d Bd 0 ) Σ d ] ek → 0 DAd zk where has all the... Castillo-Toledo, B and García, R (1999) , On the sliding mode regulator problem, Proc of the 14th IFAC World Congress, pp 61-66 [13] Utkin V., Castillo-Toledo B., Loukianov A., Espinoza-Guerra O.(2002), On robust VSS nonlinear servomechanism problem, in Variable Structure Systems: Towards the 21st Century, Springer Verlag, Lecture Notes in Control and Information Scie ncies, vol 274, Berlín, , X Yu and J-X Xu... Castillo-Toledo B., , and J Rivera (2004), Sliding mode regulador design, in Variable Structure Systems: from Principles to implementation, The Institution of Electrical Engineers, IEE Control Engineering Series, vol 66, Sabanovi A., Fridman L and Spurgeon S Eds., , pp 19-44, ISBN 0 86341 350 1 [15] El-Chesawi, O.M.E., Zinober, A.S.I., Billings, S.A (1983), Analysis and design of variable structure systems... Regulator, the output tracking error is zero at the sampling instant, but different from zero in the intersampling times Constructing now the controller (33) with an exponential holder we obtain Gd = [1.802 0.156 1.199 −0.992 −35.054] T 246 Systems, Structure and Control ⎡ −0.531 1.0201 ⎢ 0.0634 −0.1218 ⎢ Fd = ⎢ −1.1993 0 ⎢ 0 ⎢ 0.9922 ⎢ 35.0536 0 ⎣ 0 0 0 ⎤ 0 0 0 ⎥ ⎥ 1 0.1994 0.0371⎥ ⎥ 0 0.0707 0.1994... inertia L is the armature inductance and kt is the torque constant x2 = ia , and assuming that τ 1 is a known constant we have: of the motor, rotor and load, Defining x1 = wm and λo ⎡ ⎡• ⎤ ⎢ 0 x1 ⎥ ⎢ =⎢ ⎢ • ⎥ ⎢ λ0 ⎣ x2 ⎦ ⎢− ⎣ L kt ⎤ ⎡ τ⎤ ⎡0⎤ − J ⎥ ⎡ x1 ⎤ ⎢ ⎥ ⎥⎢ ⎥+ 1 u+⎢ J ⎥ ⎢ ⎥ ⎢ ⎥ R x − ⎥⎣ 2⎦ ⎣L⎦ ⎣ 0 ⎦ L⎥ ⎦ • w = Sw A Sampled-data Regulator using Sliding Modes and Exponential Holder for Linear Systems... of the controller on a digital device An illustrative example shows the performance of the presented scheme 8 Bibliography Isidori, A., (1995), Nonlinear Control System Third Edition Ed Springer-Verlag [1] Francis, B A and Wonham, W M., (1976), The internal model principle of control theory Automatica Vol 12 pp 457-465 [2] Francis, B.A (1977), The linear multivariable regulator problem SIAM J Control. .. Servomechanism Controller Using Exponential Hold IEEE Transactions on Automatic Control, Vol 39, No 6, pp 1287-1291 [4] Franklin, G F & Emami-Naeini, A (1986), Design of Ripple Free Multivariable Robust Servomechanism, IEEE Trans Aut Control, Vol AC-31, No 7, pp 661-664 [5] Castillo-Toledo, B., Di Gennaro, S., Monaco, S & Normand-Cyrot (1997), On regulation under sampling, EEE Trans Aut Control, Vol . the controller could render unstable the closed-loop system. In this Systems, Structure and Control 236 work we will take the approach of designing a hybrid controller consisting in two parts:. (1)-(3) and K and 12 ,GG make stable the matrices 00 ()ABK+ and () 1 00 0 2 0. 0 G ABH C G − ⎛⎞ ⎛⎞ − ⎜⎟ ⎜⎟ Φ ⎝⎠ ⎝⎠ (12) When dealing with controllers implemented via digital devices and zero. that the solution of (7) can be written as () (0), t te ξξ Φ = and setting tkT θ =+ with [0, )T θ ∈ we have Systems, Structure and Control 242 () () (0) (0) () ()() () kkT ss ke ee ekT ukT