# Fast Sampling Control of Singularly Perturbed Systems with Actuator Saturation andL2 Disturbance

Authors: Yanzi Miao; Lei Ma; Xiaoping Ma; Linna Zhou
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

We will consider the problem of fast sampling control for singularly perturbed systems subject to actuator saturation andL2 disturbance. A sufficient condition for the existence of a state feedback controller is proposed. Under this controller, the boundedness of the trajectories in the presence of L2 disturbances is guaranteed for any singular perturbation parameter less than or equal to a predefined upper bound. To improve the capacity of disturbance tolerance and disturbance rejection, two convex optimization problems are formulated. Finally, a numerical example is presented to demonstrate the effectiveness of the main results of this paper.


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## Body

## 1. Introduction

Many practical physical systems consist of subsystems operating on different time scales. Applying the normal control methods to these systems usually lead to ill-conditioned numerical problems. To overcome the numerical problem, singular perturbation theory was introduced to the field of control system and widely used in practice [1]. In this framework, a multiple time-scale system is modeled by a singularly perturbed system (SPS) with a small positive parameter such that the degree of separation between fast and slow modes can be determined. For example, in a power system, the flux linkages of the rotor windings are fast variables while the emf behind transient reactance, the generator’s rotor angle in radians, and the actual rotor speed are the slow ones. Thus the power system can be modeled as a SPS, where the torque of the winding represents the perturbation parameter ε [2].Since most of the modern control systems are implemented by computer, sampling control of SPSs has been widely investigated. There are three sampling modes for SPSs: multirate, slow, and fast sampling mode. Under multirate sampling mode, the slow and fast states are measured at different sampling rate. In [3], a multirate sampling model predictive control method is proposed for large-scale nonlinear uncertain systems. In [4], a multirate sampling composite controller is designed. Slow sampling control is usually under the assumption that the fast subsystem is stable [5, 6]. Slow sampling discrete-time SPS is considered in [7, 8]. If the fast subsystem is not stable, fast sampling control is necessary [9, 10]. Fast sampling control of SPSs with disturbance is considered in [10–12]. In [11], H∞ controller design method together with a stability bound optimization method is proposed, and a less conservative method is proposed in [12]. But the above achievements do not take into account the actuator saturation, which is common in practice. It is known that actuator saturation may force the systems to work in the open-loop and thus destroy stability of control systems [13]. Thus many research efforts have been devoted to analysis and design of control systems with actuator saturation [14–19]. Recently, SPSs with actuator saturation are considered. Besides basin of attraction, stability bound is also an important stability index for SPSs with actuator saturation. In [20–24], state feedback controller is designed and the basin of attraction is estimated. The obtained results guarantee the existence of the stability bound but can not present an estimate of the bound. Many results have been proposed for estimating or enlarging the stability bound of the SPSs without actuator saturation [25–32]. In [33], continuous-time SPS with actuator saturation is considered and a state feedback controller is designed to achieve a desired stability bound while the basin of attraction is optimized. To the best knowledge of the authors, the fast sampling control problem of the SPSs with actuator saturation and disturbance has not been considered.This paper focuses on fast sampling control of SPSs subject to actuator saturation andL2 disturbance. First, a state feedback controller design method is proposed such that the trajectories of the closed-loop SPSs starting from a bounded set remain bounded for any allowable singular perturbation parameter and L2 disturbance. Then, a method to enlarge the capacity of disturbance tolerance is proposed in terms of linear matrix inequalities (LMIs). Furthermore, a convex optimization problem is formulated to optimize the disturbance rejection. Finally, an example is given to show the effectiveness of the proposed results.The rest of this paper is organized as follows: Section2 provides the problems under consideration. Section 3 gives the main results of this paper. In Section 4, an example is presented to demonstrate the proposed approaches. Section 5 makes a conclusion of the paper.Notations. The superscript T stands for matrix transpose. For a matrix M, the notation M-T denotes the transpose of the inverse matrix of M and M(i) denotes ith row of M. Let Q∈Rn×n be a positive definite matrix. An ellipsoid Ω(Q,ρ) is defined as Ω(Q,ρ)≜{η∈Rn∣ηTQη≤ρ}.


## 2. Problem Formulation

The fast sampling model of SPSs with actuator saturation andL2 disturbance is described by the following compact form: (1)xk+1=Aεxk+Bεsatuk+Eεwk,zk=Cxk,where (2)xk=x1kx2k∈Rn,Aε=I+εA11εA12A21A22,Bε=εB1B2,Eε=εE1E2,C=C1C2.ε represents the singular perturbation parameter and is assumed to be available for controller design in this paper. x1∈Rn1 and x2∈Rn2 are the state variables, u(k) is the control input, z(k) is the output of the system, A11, A12, A21, A22, B1, B2 and E1, E2 are constant matrices of appropriate dimensions and sat(·) is a componentwise saturation map Rm↦Rm defined by sat(uj)=sign⁡ujmin⁡{1,uj}, j=1,2,…,m, and w(k) is the L2 disturbance which belongs to (3)Wα2≔w:R+⟶Rq:∑k=0+∞wTkwk≤α,where α is a positive number.Under the state feedback controller(4)uk=Kεxk,the closed-loop system can be described by (5)xk+1=Aεxk+BεsatKεxk+Eεwk,zk=Cxk.The problems under consideration are as follows.Problem 1.
Givenε0>0 and α∗>0, design a state feedback controller (4), such that, for any ε∈(0,ε0] and w∈Wα∗2, all the trajectories of the closed-loop system starting from inside in Ω(P-1(ε),1) will remain inside of Ω(P-1(ε),1+ηα∗) with η>0 and P(ε)>0 to be determined.Problem 2.
Givenε0>0, design a state feedback controller (4) to maximize α∗, such that for any ε∈(0,ε0] and w∈Wα∗2 all the trajectories of the closed-loop system starting from the origin are bounded.Problem 3.
Givenε0>0 and α∗>0, design a state feedback controller (4) to minimize γ>0, such that, for any ε∈(0,ε0] and w∈Wα∗2, the restricted L2 gain from output to disturbance of the closed-loop system with zero initial condition is less than γ under the control of (4).Remark 4.
The upper boundε0 characterizes the robustness of the system performance with respect to ε. The disturbance tolerance bound α∗ describes the largest disturbance that can be tolerated. The disturbance rejection γ means the largest ratio between the L2 norms of the output and the disturbance.There are two common tools to handle the saturation nonlinearities, one is the sector bound approach [34], and the other is the convex hull approach [13, 18]. The latter is adopted in this paper since it has been shown to be less conservative than the former [18]. The related preliminaries are recalled in the following.For a given matrixK∈Rm×n, denote the ith row of matrix K as ki and define (6)LK=η∈Rn:kiη≤1,i∈1,m.LetD be the set of m×m diagonal matrices whose diagonal elements are either 1 or 0. There are 2m elements in D. Suppose these elements of D are labeled as Di, i∈[1,2m]. Denote Di-=I-Di. Clearly, Di-∈D if Di∈D.Lemma 5 (see [13]).
LetK,H∈Rm×n. Then, for any x∈L(H), it holds that (7)satKx∈coDiKx+Di-Hx,i∈1,2m,where co stands for the convex hull.


## 3. Main Results

### 3.1. Controller Design

This subsection will present a solution to Problem1.Theorem 6.
Givenε0>0 and α∗>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2, as well as a positive scalar η satisfying (8)P22ϕ1Tϕ2T0⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆P22E2⋆⋆⋆ηI>0,i∈1,2m,(9)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,i∈1,2m,(10)P22Y2rTY2r11+ηα∗>0,r=1,2,…,m,(11)ε0P11ε0P12ε0Y1rTε0P12TP22Y2rTε0Y1rTY2r11+ηα∗>0,r=1,2,…,m,(12)ϕ1=A12P22+B1DiZ2+B1Di-Y2,ϕ2=A22P22+B2DiZ2+B2Di-Y2,ϕ3=A11P11+A12P12T+B1DiZ1+B1Di-Y1,ϕ4=A21P11+B2DiZ1+B2Di-Y1+A22P12T,i∈1,2m,then, for any ε∈(0,ε0] and w∈Wα∗2, all of the trajectories of the closed-loop system (5) starting within Ω(P-1(ε),1) will remain inside of Ω(P-1(ε),1+ηα∗) with P(ε)=εP11εP12εP12TP22>0.
And the state feedback controller gain matrix is given by(13)Kε=Z1Z2P11εP12P12TP22-1.Proof.
From (10) and (11), it follows that, for any ε∈(0,ε0], (14)εP11εP12εY1rTεP12TP22Y2rTεY1rTY2r11+ηα∗>0,r=1,2,…,m.
LetZε=εZ1Z2, Yε=εY1Y2. Then (14) is equivalent to (15)PεYεrTYεr11+ηα∗>0,r=1,2,…,m,which implies that (16)P-1εHεrTHεr11+ηα∗>0,r=1,2,…,m,where Hε=YεP-1(ε).
By Schur complement, it follows from (16) that (17)P-1ε>HεrT1+ηα∗Hεr,r=1,2,…,m,which implies that Ω(P-1(ε),1+ηα∗)⊆L(Hε), ∀ε∈(0,ε0]. Then by Lemma 5, we have (18)Aεxk+BεsatKεxk∈coAεxk+BεDiKε+Di-Hεxk,∀ε∈0,ε0,i∈1,2m.
From (8) and (9), it follows that(19)εP11εP12εϕ3Tεϕ4T0⋆P22εP12TA11T+ϕ1TεP12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Before and after multiplying (19) by (20)I00000I00000εI00000I00000Iand its transpose, respectively, we have (21)εP11εP12ε2ϕ3Tεϕ4T0⋆P22ε2P12TA11T+εϕ1TεP12TA21T+ϕ2T0⋆⋆ε2-ϕ3-ϕ3TεP12-εϕ4TεE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Before and after multiplying (21) by the matrix (22)I00000I000I0I00000I00000Iand its transpose, respectively, we have(23)εP11εP12εP11+ε2ϕ3Tεϕ4T0⋆P22εP12T+ε2P12TA11T+εϕ1TεP12TA21T+ϕ2T0⋆⋆εP11εP12εE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Taking into account the definitions ofAε, Bε, Eε, and P(ε), we can rewrite (23) as (24)PεAεPε+BεDiZε+BεDi-YεT0⋆PεEε⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Define(25)Kε=ZεP-1ε.
Before and after multiplying (24) by diag⁡(P-1(ε),P-1(ε),I), we have (26)P-1εAε+BεDiKε+BεDi-HεTP-1ε0⋆P-1εP-1εEε⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
By Schur complement, inequality (26) is equivalent to (27)Ξ1Ξ2Ξ2TΞ3-ηI<0,where(28)Ξ1=Aε+BεDiKε+BεDi-HεTP-1εAε+BεDiKε+BεDi-Hε-P-1ε,Ξ2=Aε+BεDiKε+BεDi-HεTP-1εEε,Ξ3=EεTP-1εEε,∀ε∈0,ε0,i∈1,2m.
Define anε-dependent Lyapunov function:(29)Vx=xTP-1εx.
Calculating the difference ofV(k) along the trajectories of the closed-loop system (5), and using (18), we have (30)ΔVx=Vxk+1-Vxk=Aεxk+Bεsatu+EεwkTP-1εAεxk+Bεsatu+Eεwk-xTkP-1εxk≤maxi∈1,2mAεxk+BεDiKε+Di-Hεxk+EεwkTP-1εAεxk+BεDiKε+Di-Hεxk+Eεwk-xTkP-1εxk=maxi∈1,2mxTkAε+BεDiKε+Di-1HεTP-1εAε+BεDiKε+Di-1Hε-P-1εxk+xTkAε+BεDiKε+Di-1HεTP-1εEεwk+wTkEεTP-1εAε+BεDiKε+Di-1Hεxk+wTkEεTP-1εEεwk,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2.
It follows that(31)ΔVx-ηwTkwk=ξTkΞ1Ξ2Ξ2TΞ3-ηIξk≤0,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2,with (32)ξk=xkwk.
Then we have(33)ΔVx≤ηwTkwk,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2.
Summing up both sides of (33) from 0 to m, we can get (34)Vxm+1≤Vx0+η∑k=0mwTkwk≤Vx0+αη,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2,which shows that when x(0)∈Ω(P-1(ε),1), we can get V(x(m+1))≤1+ηα∗, that is, x(m+1)∈Ω(P-1(ε),1+ηα∗).
And, according to (25), the controller gain matrix is (35)Kε=ZεP-1ε=εZ1Z2εP11εP12εP12TP22-1=Z1Z2P11εP12P12TP22-1.Remark 7.
According to (33), when the disturbance w(k)=0, it holds that ΔV(x)<0, ∀x≠0, which means that the closed-loop system (5) is locally asymptotically stable. In this case, the ellipsoid Ω(P-1(ε),1) is an estimation of the basin of attraction of the closed-loop system. In addition, when ε is small enough, an ε-independent gain matrix can be computed by (36)K=limε→0+Z1Z2P11εP12P12TP22-1=Z1Z2P110P12TP22-1=Z1-Z2P22-1P12TP11-1Z2P22T.


### 3.2. Disturbance Tolerance

The ability of the closed-loop system to tolerate the disturbance is characterized byα∗. Based on Theorem 6, we have the following corollary which can be used to maximize α∗.Corollary 8.
Givenε0>0 and α>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2 satisfying(37)P22ϕ1Tϕ2T0⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆P22E2⋆⋆⋆I>0,i∈1,2m,(38)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆I>0,i∈1,2m,(39)P22Y2rY2rT1α>0,(40)ε0P11ε0P12ε0Y1rε0P12TP22Y2rε0Y1rTY2rT1α>0,where ϕ1, ϕ2, ϕ3, ϕ4 are defined in Theorem 6, then all the trajectories of the closed-loop system (5) starting from the origin will still remain inside of Ω(P-1(ε),α). And the state feedback controller gain matrix is given by (41)Kε=Z1Z2P11εP12P12TP22-1.Proof.
According to (39) and (40), for any ε∈(0,ε0], we have (42)εP11εP12εY1rεP12TP22Y2rεY1rTY2rT1α>0,which is equivalent to (43)P-1εHεrTHεr1α>0,r=1,2,…,m.
By Schur complement, it follows from (43) that (44)P-1ε>HεrTαHεr,r=1,2,…,m;thus Ω(P-1(ε),α)⊆L(Hε), ∀ε∈(0,ε0].
Similarly to (34) in the proof for Theorem 6, for the trajectories start from the origin, we have (45)Vxm+1≤V0+∑k=0mwTkwk≤V0+α=α.
This complete the proof.From Corollary8, the bigger α means the better disturbance tolerance ability. To get the best disturbance tolerance ability we formulate the following optimization problem:(46)maxP11,P12,P22,Z1,Z2,Y1,Y2αs.t.37,38,39,40,where Ω(S,1) represents the initial condition set.Letμ=1/α. Then inequalities (39) and (40) can be rewritten as (47)P22Y2rTY2rμ>0,r=1,2,…,m,ε0P11ε0P12ε0Y1rTε0P12TP22Y2rTε0Y1rTY2rμ>0,r=1,2,…,m.Then the optimization problem (46) can be converted into (48)minP11,P12,P22,Z1,Z2,Y1,Y2μs.t.37,38,47.Solving the optimization problem (48) yields the minimal value μ∗. Then the largest disturbance that can be tolerated by the closed-loop system at zero initial condition is bounded by α∗=1/μ∗. Therefore, Problem 2 can be solved by the optimization problem (48).


### 3.3. Disturbance Rejection

Problem3 will be considered in this subsection.Theorem 9.
Givenε0>0, α∗>0, and γ>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2 satisfying (49)P22ϕ1Tϕ2T0P22C2T⋆-ϕ3-ϕ3TP12-ϕ4TE1-ϕ5T⋆⋆P22E20⋆⋆⋆γ2I0⋆⋆⋆⋆I>0,(50)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0ε0ϕ5T⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0ε0P12TC1T+P22C2T⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1-ϕ5T⋆⋆⋆P22E20⋆⋆⋆⋆γ2I0⋆⋆⋆⋆⋆I>0,i∈1,2m,(51)P22Y2Y2T1α∗>0,(52)ε0P11ε0P12ε0Y1ε0P12TP22Y2ε0Y1TY2T1α∗>0,where(53)ϕ1=A12P22+B1DiZ2+B1Di-Y2,ϕ2=A22P22+B2DiZ2+B2Di-Y2,ϕ3=A11P11+A12P12T+B1DiZ1+B1Di-Y1,ϕ4=A21P11+B2DiZ1+B2Di-Y1+A22P12T,i∈1,2m,ϕ5=C1P11+C2P12,then the L2 gain from w to z of the closed-loop system (5) with x(0)=0 is less than γ. And the state feedback controller gain matrix is given by (54)Kε=Z1Z2P11εP12P12TP22-1.Proof.
LetPε=εP11εP12εP12TP22>0, Zε=εZ1Z2, Yε=εY1Y2, Hε=YεP-1ε, Kε=ZεP-1(ε).
Similarly to the proof for Theorem6, it follows from (49) and (50) that (55)PεAεPε+BεDiZε+BεDi-YεT0PεCT⋆PεEε0⋆⋆γ2I0⋆⋆⋆I>0,∀ε∈0,ε0,i∈1,2m.
Applying the Schur complement formula to (55), we have (56)Ξ1+CTCΞ2Ξ2TΞ3-γ2I<0,where(57)Ξ1=Aε+BεDiKε+BεDi-HεTP-1εAε+BεDiKε+BεDi-Hε-P-1ε,Ξ2=Aε+BεDiKε+BεDi-HεTP-1εEε,Ξ3=EεTP-1εEε,∀ε∈0,ε0,i∈1,2m.
From (51) and (52), for any ε∈(0,ε0], it follows that (58)εP11εP12εY1εP12TP22Y2εY1TY2T1α∗>0.
From the proof for Theorem6, we can get (59)P-1ε>HεrTαHεr,r=1,2,…,m,which shows that Ω(P-1(ε),α∗)⊆L(Hε),∀ε∈(0,ε0].
Define anε-dependent Lyapunov function (60)Vx=xTP-1εx.
Similarly to proof for Theorem6, it follows from (56) that (61)ΔVxk+zTkzk-γ2wTkwk<0,∀x∈ΩP-1ε,α∗.
Then, summing up left and right of (61), respectively, with x(0)=0, yields that (62)Vxm+1-0+∑k=0mzTkzk-γ2wTkwk<0,which implies (63)∑k=0mzTkzk<γ2∑k=0jwTkwk.Thus the L2 gain from w to z of the closed-loop system with x(0)=0 is less than γ. This completes the proof.By Theorem9, the minimal L2 gain can be obtained by solving the following optimization problem: (64)minP11,P12,P22,Z1,Z2,Y1,Y2γ2s.t.49,50,51,52.Remark 10.
As mentioned in Section1, discretization of a continuous-time SPS can lead to different discrete-time models depending on the sampling rate. Since the structure of fast sampling models is different from that of slow sampling models, it is not easy to generalize the proposed results to slow sampling control of SPSs, as will be considered in our future work.


## 3.1. Controller Design

This subsection will present a solution to Problem1.Theorem 6.
Givenε0>0 and α∗>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2, as well as a positive scalar η satisfying (8)P22ϕ1Tϕ2T0⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆P22E2⋆⋆⋆ηI>0,i∈1,2m,(9)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,i∈1,2m,(10)P22Y2rTY2r11+ηα∗>0,r=1,2,…,m,(11)ε0P11ε0P12ε0Y1rTε0P12TP22Y2rTε0Y1rTY2r11+ηα∗>0,r=1,2,…,m,(12)ϕ1=A12P22+B1DiZ2+B1Di-Y2,ϕ2=A22P22+B2DiZ2+B2Di-Y2,ϕ3=A11P11+A12P12T+B1DiZ1+B1Di-Y1,ϕ4=A21P11+B2DiZ1+B2Di-Y1+A22P12T,i∈1,2m,then, for any ε∈(0,ε0] and w∈Wα∗2, all of the trajectories of the closed-loop system (5) starting within Ω(P-1(ε),1) will remain inside of Ω(P-1(ε),1+ηα∗) with P(ε)=εP11εP12εP12TP22>0.
And the state feedback controller gain matrix is given by(13)Kε=Z1Z2P11εP12P12TP22-1.Proof.
From (10) and (11), it follows that, for any ε∈(0,ε0], (14)εP11εP12εY1rTεP12TP22Y2rTεY1rTY2r11+ηα∗>0,r=1,2,…,m.
LetZε=εZ1Z2, Yε=εY1Y2. Then (14) is equivalent to (15)PεYεrTYεr11+ηα∗>0,r=1,2,…,m,which implies that (16)P-1εHεrTHεr11+ηα∗>0,r=1,2,…,m,where Hε=YεP-1(ε).
By Schur complement, it follows from (16) that (17)P-1ε>HεrT1+ηα∗Hεr,r=1,2,…,m,which implies that Ω(P-1(ε),1+ηα∗)⊆L(Hε), ∀ε∈(0,ε0]. Then by Lemma 5, we have (18)Aεxk+BεsatKεxk∈coAεxk+BεDiKε+Di-Hεxk,∀ε∈0,ε0,i∈1,2m.
From (8) and (9), it follows that(19)εP11εP12εϕ3Tεϕ4T0⋆P22εP12TA11T+ϕ1TεP12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Before and after multiplying (19) by (20)I00000I00000εI00000I00000Iand its transpose, respectively, we have (21)εP11εP12ε2ϕ3Tεϕ4T0⋆P22ε2P12TA11T+εϕ1TεP12TA21T+ϕ2T0⋆⋆ε2-ϕ3-ϕ3TεP12-εϕ4TεE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Before and after multiplying (21) by the matrix (22)I00000I000I0I00000I00000Iand its transpose, respectively, we have(23)εP11εP12εP11+ε2ϕ3Tεϕ4T0⋆P22εP12T+ε2P12TA11T+εϕ1TεP12TA21T+ϕ2T0⋆⋆εP11εP12εE1⋆⋆⋆P22E2⋆⋆⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Taking into account the definitions ofAε, Bε, Eε, and P(ε), we can rewrite (23) as (24)PεAεPε+BεDiZε+BεDi-YεT0⋆PεEε⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
Define(25)Kε=ZεP-1ε.
Before and after multiplying (24) by diag⁡(P-1(ε),P-1(ε),I), we have (26)P-1εAε+BεDiKε+BεDi-HεTP-1ε0⋆P-1εP-1εEε⋆⋆ηI>0,∀ε∈0,ε0,i∈1,2m.
By Schur complement, inequality (26) is equivalent to (27)Ξ1Ξ2Ξ2TΞ3-ηI<0,where(28)Ξ1=Aε+BεDiKε+BεDi-HεTP-1εAε+BεDiKε+BεDi-Hε-P-1ε,Ξ2=Aε+BεDiKε+BεDi-HεTP-1εEε,Ξ3=EεTP-1εEε,∀ε∈0,ε0,i∈1,2m.
Define anε-dependent Lyapunov function:(29)Vx=xTP-1εx.
Calculating the difference ofV(k) along the trajectories of the closed-loop system (5), and using (18), we have (30)ΔVx=Vxk+1-Vxk=Aεxk+Bεsatu+EεwkTP-1εAεxk+Bεsatu+Eεwk-xTkP-1εxk≤maxi∈1,2mAεxk+BεDiKε+Di-Hεxk+EεwkTP-1εAεxk+BεDiKε+Di-Hεxk+Eεwk-xTkP-1εxk=maxi∈1,2mxTkAε+BεDiKε+Di-1HεTP-1εAε+BεDiKε+Di-1Hε-P-1εxk+xTkAε+BεDiKε+Di-1HεTP-1εEεwk+wTkEεTP-1εAε+BεDiKε+Di-1Hεxk+wTkEεTP-1εEεwk,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2.
It follows that(31)ΔVx-ηwTkwk=ξTkΞ1Ξ2Ξ2TΞ3-ηIξk≤0,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2,with (32)ξk=xkwk.
Then we have(33)ΔVx≤ηwTkwk,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2.
Summing up both sides of (33) from 0 to m, we can get (34)Vxm+1≤Vx0+η∑k=0mwTkwk≤Vx0+αη,∀xk∈ΩP-1ε,1+ηα∗,ε∈0,ε0,w∈Wα∗2,which shows that when x(0)∈Ω(P-1(ε),1), we can get V(x(m+1))≤1+ηα∗, that is, x(m+1)∈Ω(P-1(ε),1+ηα∗).
And, according to (25), the controller gain matrix is (35)Kε=ZεP-1ε=εZ1Z2εP11εP12εP12TP22-1=Z1Z2P11εP12P12TP22-1.Remark 7.
According to (33), when the disturbance w(k)=0, it holds that ΔV(x)<0, ∀x≠0, which means that the closed-loop system (5) is locally asymptotically stable. In this case, the ellipsoid Ω(P-1(ε),1) is an estimation of the basin of attraction of the closed-loop system. In addition, when ε is small enough, an ε-independent gain matrix can be computed by (36)K=limε→0+Z1Z2P11εP12P12TP22-1=Z1Z2P110P12TP22-1=Z1-Z2P22-1P12TP11-1Z2P22T.


## 3.2. Disturbance Tolerance

The ability of the closed-loop system to tolerate the disturbance is characterized byα∗. Based on Theorem 6, we have the following corollary which can be used to maximize α∗.Corollary 8.
Givenε0>0 and α>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2 satisfying(37)P22ϕ1Tϕ2T0⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆P22E2⋆⋆⋆I>0,i∈1,2m,(38)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1⋆⋆⋆P22E2⋆⋆⋆⋆I>0,i∈1,2m,(39)P22Y2rY2rT1α>0,(40)ε0P11ε0P12ε0Y1rε0P12TP22Y2rε0Y1rTY2rT1α>0,where ϕ1, ϕ2, ϕ3, ϕ4 are defined in Theorem 6, then all the trajectories of the closed-loop system (5) starting from the origin will still remain inside of Ω(P-1(ε),α). And the state feedback controller gain matrix is given by (41)Kε=Z1Z2P11εP12P12TP22-1.Proof.
According to (39) and (40), for any ε∈(0,ε0], we have (42)εP11εP12εY1rεP12TP22Y2rεY1rTY2rT1α>0,which is equivalent to (43)P-1εHεrTHεr1α>0,r=1,2,…,m.
By Schur complement, it follows from (43) that (44)P-1ε>HεrTαHεr,r=1,2,…,m;thus Ω(P-1(ε),α)⊆L(Hε), ∀ε∈(0,ε0].
Similarly to (34) in the proof for Theorem 6, for the trajectories start from the origin, we have (45)Vxm+1≤V0+∑k=0mwTkwk≤V0+α=α.
This complete the proof.From Corollary8, the bigger α means the better disturbance tolerance ability. To get the best disturbance tolerance ability we formulate the following optimization problem:(46)maxP11,P12,P22,Z1,Z2,Y1,Y2αs.t.37,38,39,40,where Ω(S,1) represents the initial condition set.Letμ=1/α. Then inequalities (39) and (40) can be rewritten as (47)P22Y2rTY2rμ>0,r=1,2,…,m,ε0P11ε0P12ε0Y1rTε0P12TP22Y2rTε0Y1rTY2rμ>0,r=1,2,…,m.Then the optimization problem (46) can be converted into (48)minP11,P12,P22,Z1,Z2,Y1,Y2μs.t.37,38,47.Solving the optimization problem (48) yields the minimal value μ∗. Then the largest disturbance that can be tolerated by the closed-loop system at zero initial condition is bounded by α∗=1/μ∗. Therefore, Problem 2 can be solved by the optimization problem (48).


## 3.3. Disturbance Rejection

Problem3 will be considered in this subsection.Theorem 9.
Givenε0>0, α∗>0, and γ>0, if there exist symmetric matrices P11∈Rn1∗n1, P22∈Rn2∗n2 and matrices P12∈Rn1∗n2, Z1∈Rm∗n1, Z2∈Rm∗n2, Y1∈Rm∗n1, Y2∈Rm∗n2 satisfying (49)P22ϕ1Tϕ2T0P22C2T⋆-ϕ3-ϕ3TP12-ϕ4TE1-ϕ5T⋆⋆P22E20⋆⋆⋆γ2I0⋆⋆⋆⋆I>0,(50)ε0P11ε0P12ε0ϕ3Tε0ϕ4T0ε0ϕ5T⋆P22ε0P12TA11T+ϕ1Tε0P12TA21T+ϕ2T0ε0P12TC1T+P22C2T⋆⋆-ϕ3-ϕ3TP12-ϕ4TE1-ϕ5T⋆⋆⋆P22E20⋆⋆⋆⋆γ2I0⋆⋆⋆⋆⋆I>0,i∈1,2m,(51)P22Y2Y2T1α∗>0,(52)ε0P11ε0P12ε0Y1ε0P12TP22Y2ε0Y1TY2T1α∗>0,where(53)ϕ1=A12P22+B1DiZ2+B1Di-Y2,ϕ2=A22P22+B2DiZ2+B2Di-Y2,ϕ3=A11P11+A12P12T+B1DiZ1+B1Di-Y1,ϕ4=A21P11+B2DiZ1+B2Di-Y1+A22P12T,i∈1,2m,ϕ5=C1P11+C2P12,then the L2 gain from w to z of the closed-loop system (5) with x(0)=0 is less than γ. And the state feedback controller gain matrix is given by (54)Kε=Z1Z2P11εP12P12TP22-1.Proof.
LetPε=εP11εP12εP12TP22>0, Zε=εZ1Z2, Yε=εY1Y2, Hε=YεP-1ε, Kε=ZεP-1(ε).
Similarly to the proof for Theorem6, it follows from (49) and (50) that (55)PεAεPε+BεDiZε+BεDi-YεT0PεCT⋆PεEε0⋆⋆γ2I0⋆⋆⋆I>0,∀ε∈0,ε0,i∈1,2m.
Applying the Schur complement formula to (55), we have (56)Ξ1+CTCΞ2Ξ2TΞ3-γ2I<0,where(57)Ξ1=Aε+BεDiKε+BεDi-HεTP-1εAε+BεDiKε+BεDi-Hε-P-1ε,Ξ2=Aε+BεDiKε+BεDi-HεTP-1εEε,Ξ3=EεTP-1εEε,∀ε∈0,ε0,i∈1,2m.
From (51) and (52), for any ε∈(0,ε0], it follows that (58)εP11εP12εY1εP12TP22Y2εY1TY2T1α∗>0.
From the proof for Theorem6, we can get (59)P-1ε>HεrTαHεr,r=1,2,…,m,which shows that Ω(P-1(ε),α∗)⊆L(Hε),∀ε∈(0,ε0].
Define anε-dependent Lyapunov function (60)Vx=xTP-1εx.
Similarly to proof for Theorem6, it follows from (56) that (61)ΔVxk+zTkzk-γ2wTkwk<0,∀x∈ΩP-1ε,α∗.
Then, summing up left and right of (61), respectively, with x(0)=0, yields that (62)Vxm+1-0+∑k=0mzTkzk-γ2wTkwk<0,which implies (63)∑k=0mzTkzk<γ2∑k=0jwTkwk.Thus the L2 gain from w to z of the closed-loop system with x(0)=0 is less than γ. This completes the proof.By Theorem9, the minimal L2 gain can be obtained by solving the following optimization problem: (64)minP11,P12,P22,Z1,Z2,Y1,Y2γ2s.t.49,50,51,52.Remark 10.
As mentioned in Section1, discretization of a continuous-time SPS can lead to different discrete-time models depending on the sampling rate. Since the structure of fast sampling models is different from that of slow sampling models, it is not easy to generalize the proposed results to slow sampling control of SPSs, as will be considered in our future work.


## 4. Examples

This section will illustrate the proposed results by an example.Consider an inverted pendulum system controlled by DC motor via a gear train. The model, which was first established in [35], is described by (65)x˙1t=x2t,x˙2t=glsin⁡x1t+NKmml2x3t,Lax˙3t=-KbNx2t-Rax3t+ut+wt,where x1(t)=θp(t) denotes the the angle (rad) of the pendulum from the vertical upward, x2(t)=θ˙p(t), x3(t)=Ia(t) denotes the current of the motor, u(t) is the control input voltage, w(t) is the disturbance, Km is the motor torque constant, Kb is the back emf constant, N is the gear ratio, and La is the inductance which is usually a small positive constant. The parameters for the plant are as follows: g=9.8m/s2, N=10, l=1m, m=1kg, Km=0.1 Nm/A, Kb=0.1 Vs/rad, Ra=1 Ω, and La=0.05 H and the input voltage is required to satisfy u≤1. Note that La represents the singular perturbation parameter of the system. Substituting the parameters into (65) and linearizing the equations, we have (66)x˙1t=x2t,x˙2t=9.8x1t+x3t,εx˙3t=-x2t-x3t+ut+wt,where ε=La.The equilibrium point of system (66), that is, xe=000T, corresponds to the upright rest position of the inverted pendulum. We will design a controller to balance the pendulum around its upright rest position.According to [36], we choose the sampling period as Tf=αfε, where αf=0.1, ε=0.05. Then the fast sampling discrete-time model of system (66) is in the form of (1) with (67)A11=00.10.98-0.0048,A12=00.0952,A22=0.9048,B1=00.0048,B2=0.0952,E1=00.0048,E2=0.0952,C1=11,C2=1.Solving the LMIs of Theorem6 with ε0=0.05, α∗=0.2, and η=1, we have (68)P11=23.4260-52.6506-52.6506157.2814,P12=-92.204068.0912,P22=92.9218,Z1=-152.3819187.0950,Z2=-491.8613,Y1=0.1739-0.1772,Y2=-0.1620.Choosingε=0.05∈(0,ε0], then we have (69)Pε=1.1713-2.6325-4.6102-2.63257.86413.4046-4.61023.404692.9218,Zε=-7.61919.3548-491.8613,Yε=0.0087-0.0089-0.1620.Then we can calculate the state feedback controller gain:(70)Kε=-147.2017-43.3206-11.0093.Solving the optimization problem (48) with ε0=0.05, then we get (71)P11=0.0097-0.0277-0.02770.1043,P12=-0.0122-0.0551,P22=0.0710,Z1=-0.07480.0910,Z2=-0.0492,Y1=-0.0442-0.0299,Y2=0.0133.Choosingε=0.05∈(0,ε0], we have (72)Pε=0.0005-0.0014-0.0006-0.00140.0052-0.0028-0.0006-0.00280.0710,Zε=-0.00370.0045-0.0492,Yε=-0.0022-0.00150.0133,Kε=-37.7656-9.9051-1.4045,and μ=0.0641, which means the capacity of disturbance tolerance of the system is α∗=15.5997.To improve the ability of disturbance rejection, solving the optimization problem (64) with ε0=0.05, α∗=15, we have (73)P11=0.0045-0.0133-0.01330.0522,P12=-0.03380.0513,P22=0.0888,Z1=-0.04790.0363,Z2=-0.6921,Y1=-0.0238-0.0405,Y2=-0.0004.Choosingε=0.05∈(0,ε0], we have (74)Pε=0.0002-0.0007-0.0017-0.00070.00260.0026-0.00170.00260.0888,Zε=-0.00240.0018-0.6921,Yε=-0.0012-0.0020-0.0004,Kε=-234.9799-48.4215-10.8701,and γ=1.0369. Thus the L2 gain from w to z of the closed-loop system with x(0)=0 is less than 1.0369.Define a piecewise function as follows:(75)vk=0.3if1≤k≤1000ifk>100.Givenα∗=15<15.5997 and ε0=0.05, it is easy to see that the disturbance w(k)=v(k)∈Wα∗2. As shown in Figure 1, under controller (4), with (76)Kε=-234.9799-48.4215-10.8701the trajectory of the closed-loop system with the disturbance ω(k)=v(k) starting from the origin is bounded and converges to the origin when the disturbance disappears.Figure 1
The ellipsoidΩ(P-1(ε),α) and the trajectory starting from the origin.
(a)
The projection of trajectory tox1x2
(b)
The projection of trajectory tox1x3
(c)
The projection of trajectory tox2x3
(d)
The trajectory starting from the origin


## 5. Conclusion

This paper investigated the problem of fast sampling control for singularly perturbed systems subject to actuator saturation andL2 disturbance. A state feedback controller method was proposed such that all the trajectories of the closed-loop system starting from a bounded set will remain bounded for any singular perturbation parameter less than or equal to a predefined upper bound. Convex optimization problems were formulated to optimize the ability of disturbance tolerance and disturbance rejection, respectively. The presented example has illustrated the significance and validity of the proposed approaches.



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