# Convergence Time Calculation for Supertwisting Algorithm and Application for Nonaffine Nonlinear Systems

Authors: Jianhua Zhang; Quanmin Zhu; Yang Li
Journal: Complexity
 (2019)

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## Abstract

In this study, an accurate convergence time of the supertwisting algorithm (STA) is proposed to build up a framework for nonaffine nonlinear systems’ finite-time control. The convergence time of the STA is provided by calculating the solution of a differential equation instead of constructing Lyapunov function. Therefore, precise convergence time is presented instead of estimation of the upper bound of the algorithm’s reaching time. Regardless of affine or nonaffine nonlinear systems, supertwisting control (STC) provides a general solution based on virtual control law skill ensuring the output of the systems converges to the origin point at exact time. Benchmark tests are simulated to demonstrate the effectiveness and efficiency of the algorithm.


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## Body

## 1. Introduction

Sliding mode control (SMC) has become one of the most efficient techniques to control uncertain complex systems and engineering [1–3]. Theoretically, such controllers are able to compensate and match disturbances by confining the systems’ trajectories in a properly chosen hypersurface (the so-called sliding manifold) [4–6] and, under the chosen surface, make the origin of the state space an asymptotically stable equilibrium point for the closed-loop systems [7–9]. Both the convergence to the sliding manifold and convergence to the origin are guaranteed in a finite time interval if the control action is large enough to counteract the effect of the uncertain terms and hypersurface is defined suitably [10–12].STA is a well-known second-order sliding mode (SOSM) algorithm introduced in [13], and it is a possible solution for chattering reduction and widely used for control, observation, and robust exact differentiation. The reaching time estimation and the finite-time control method design are essentially complex problems with the sliding mode control systems’ study [14–16]. Finite-time convergence and robustness of the STA have been proved with some methods, such as geometrical methods [17], homogeneity properties of the algorithm [18], and Lyapunov methods [19]. The Lyapunov stability theorem and Lyapunov function provide a means of determining stability without explicit knowledge on system solutions [20–22]. Traditionally, quadratic Lyapunov functions are constructed to analyze and control design of nonlinear dynamic systems [23–25]. It should be noted that there also exist some other formats of Lyapunov functions, such as integral Lyapunov function, barrier Lyapunov function, and vector Lyapunov function. Such attempts have enhanced Lyapunov function applications in control system design. A strict Lyapunov function is provided to ascertain finite-time convergence, and it would provide an estimate of the convergence time, as well as the robustness of the finite-time or ultimate boundedness for the STA [19, 26]. By a detailed analysis of the Lyapunov function in finite time, robust convergence for the STA is proved, and it is not possible to provide necessary and sufficient conditions to estimate the convergence time from it.However, the form of the estimate of the convergence time contained arbitrary positive matrixes which are related by the algebraic Lyapunov equation (ALE), making it difficult to operate with it for applications or further developments [24, 27, 28]. For nonlinear systems, numerical techniques (open solutions) have played a significant role in the controller design process [29–31]. As general solutions to the complicated nonlinear dynamic problem, the U model uses linear approaches to design nonlinear control [32, 33]. Specifically, it can be concluded that the applicable systems fall into classes of an increasing order of complexity: strict feedback, pure feedback, affine form, and nonaffine systems [34]. Nonaffine systems are difficult to control because of the complexity of the systems.Motivated by the above observations, this paper presents accurate convergence time of the STA without and with perturbation and designs the nonaffine STA finite-time control. The main contributions are listed as follows:(1)
Accurate convergence time is proposed for the STA based on the analytical solution of the differential equation. The main advantages are that the output of the systems converges to the origin point at exact time and the exact time is determined by the designer before the controller is implemented.(2)
To overcome the main obstacle, the trajectory of the STA is analyzed by the analytical solution of the differential equation and LaSalle’s invariance principle instead of Lyapunov stability theory used commonly in the traditional sliding mode control.(3)
Technically, for nonaffine nonlinear systems, sliding mode control based on the STA and backstepping achieves finite-time stabilization. The closed-loop control systems will achieve stability in finite time without violation of the constraint.The rest of this study is organized as follows: In Section2, the problem formulation and preliminaries are presented, which also contains some definitions and lemmas about the STA. They are presented to establish a basis for designing and analyzing the STA. In Section 3, accurate convergence time for finite-time convergence of the STA is developed by the parametric equation in different initial conditions. In Section 4, accurate convergence time is developed by the parametric equation in different initial conditions with perturbation. In Section 5, STA control is designed for nonaffine nonlinear systems based on the backstepping skill. The closed-loop systems’ trajectory is analyzed, and the output can be obtained effectively by the finite-time algorithm. In Section 6, three simulated case studies are conducted to initially demonstrate the efficiency and effectiveness of the procedure. In Section 7, the conclusions are given to summarize the study.


## 2. System Description and Preliminaries

The STA can be written as(1)x˙1=−k1x11/2signx1+x2,x˙2=−k2signx1,where x1andx2 are the scalar state variables: x10=aandx20=b, and k1andk2 are gains to be designed.Ifk1>0andk2>0, the states x1andx2 of the system reach the origin in finite time T; therefore, the system undergoes finite-time convergence [19], using x1,x2∣t,k1,k2,a,b to indicate STA (1).Lemma 1 (see [35]).
For differential equation,(2)yyx′−y=Ax+B,A≠0.
The solution is in the parametric form:(3)x=Cexp−∫τdττ2−τ−A−BA,y=Cτexp−∫τdττ2−τ−A.More details about the solution of the parametric equation (3) are presented in Appendix A.


## 3. Finite-Time Convergence of STA

As mentioned, whenk1>0andk2>0, the accurate convergence time is considered. To show the trajectory of the STA, the analytic method is used to describe the system.If the initial conditionx10=a, then x1 will reach the origin point in finite time at x1ta0=0, where ta0 indicates the time of state when x1 reaches the origin, and let Ta0=t∣0<t<ta0; if the initial condition x10=0, then x1 will reach the origin point in finite time at x1t00=0, where t00 indicates the time of state when x1 reaches the origin, and let T00=t∣0<t<t00.The main contribution are the following three facts:Fact 1. For the STS (1), whenk1>0,k2>0, and A≥−1/4 hold, the initial conditions are x10=aandx20=b; if a≠0 holds, the system state x1 arrives at the zero point in finite time.As shown in Figure1, x10=a,x20=b, and x1ta0=0 hold.Fact 2. This is the most important fact; for the STS (1), when k1>0,k2>0, and A≥−1/4 hold, the initial conditions are x10=aandx20=b; if a=0 holds, the system states x1andx2 arrive at the zero point simultaneously.As shown in Figure2, the initial conditions x10=0andx20=b hold, in period T00; then at time t00, x1t00=0andx2t00=0 hold.Fact 3. For the STS (1), when k1>0,k2>0, and A≥−1/4 hold, the initial conditions are x10=aandx20=b, and the system states x1andx2 arrive at the zero point simultaneously.Figure 1
Trajectory of the STS witha≠0.Figure 2
Trajectory of the STS witha=0.As shown in Figure3, the initial conditions x10=aandx20=b hold, and then system states change as in Figure 1 in period Ta0, where state x1 converges to zero at time ta0; after that, as in Figure 2, states x1andx2 converge to zero in period T00 at time ta0+t00, where x1ta0+t00=0andx2ta0+t00=0 hold.Figure 3
Trajectory of STS in the convergence procedure.To prove the facts above, int∈Ta0, let(4)y=−2k1x11/2.Its derivative on the trajectories of (B.2) and (B.6) can be obtained as follows:(5)yy˙−y=At+B,where(6)A=−2k2k12,B=2bk12sign2signa+signb.The details about the parameters of equation (6) are presented in Appendix B.Then, based on the parametric solution in Appendix A and general form of (5), define the solution in different situations by the parameter A and define the trajectory in different parameters and different initial conditions.Case I.
For the supertwisting system, whenA=−1/4anda≠0, the trajectory is(7)ts=Csexps−BA,ys=C2s+1exps,where the parameter C is determined by initial conditions s=s0, t=0, and y=a1=−2/k1a1/2 as follows:(8)0=Cs0exps0−BA,a1=C2s0+1exps0.
Therefore, the initial parameter can be obtained as(9)s0=B2a1A−B,and then the parameter C is obtained as(10)C=BAs0exps0.
Then, the whole trajectory can be obtained for the parametric equation, whens1=−1, s2=−∞, and y=0:(11)ts1=−Cexp−1−BA,ts2=−BA.
The statex1 will arrive at the origin point at ta0 where the following conditions are met:(a1)
Whens2<s1<s0, ta0=ts1.(a2)
Whens2<s0<s1, if ∂t/∂s>0, ta0=ts1.(a3)
Whens2<s0<s1, if ∂t/∂s<0, ta0=ts2.
Therefore, for the STS at timeta0,(12)x1ta0=0,x2ta0=−k2signata0+b.Remark 1.
Ifta0=b/k2signa holds, x2ta0=0 in (12), and the system converges to the origin point at time ta0; if not, the system changes into another STA with initial conditions a=0 and b≠0 because of the time invariance of the STA.Case II.
For the supertwisting system, whenA=−1/4anda=0,(13)ts=Csexps−BA,ys=C2s+1exps.
For the initial conditions, (13) becomes(14)0=Cs0exps0−BA,0=C2s0+1exps0.
Therefore, the initial parameter can be obtained as(15)s0=−1,and then the parameter C can be obtained as(16)C=−BAexp−1.
Then, for the parametric equation, whens1=−∞ and y=0,(17)t00=−BA.
According to system (7) and convergence time (17), the states of the system reach the origin point at the same time:(18)x1t00=0,x2t00=0.
Therefore, the system reaches finite-time stability.Remark 2.
If the initial conditiona=0, based on the convergence time t00, both the states reach the origin point at the same time.
The supertwisting system is the time invariance system, and states reach finite-time stability whenk1,k2>0; therefore, if the initial condition a≠0, based on the convergence time ta0, by calculating both the states at ta0, a new supertwisting system with the initial condition a=0 can be obtained. Combining process initial conditions a=0 and a≠0, the supertwisting system convergence time can be obtained.Theorem 1.
The supertwisting system (1) reaches finite-time stability, if gains k1>0andk2>0 satisfy k12=8k2 and system’s initial condition x10=aandx20=b; the finite time T satisfies T=ta0+t00, where ta0 is the time when state x1 reaches the origin point in first time x2=b1≠0, and then time t00=−B1/A is elapsed, when both states reach the origin point at the same time; therefore, the STS reaches the origin point at time T where the following conditions are met:(b1)
Whens2<s1<s0, ta0=−Cexp−1−B/A.(b2)
Whens2<s0<s1, if C<0, ta0=−Cexp−1−B/A.(b3)
Whens2<s0<s1, if C>0, ta0=−B/A.
Here,(19)A=−2k2k12,B=2bk12sign2signa+signb,B1=2b1k12signb1,b1=−k2signata0+b,s0=−B4/k1a1/2A+B,s1=−1,s2=−∞,C=BAs0exps0.Proof of Theorem 1.
Based on the supertwisting system and gainsk1andk2, if A=−1/4, the parametric solution can be obtained in the form of (7); based on the initial condition, parameters s0andC can be obtained by (9) and (10), and then ta0 can be obtained in following different situations: (b1)–(b3).
Whent=ta0, the states of the system change as in (12), and then the system will elapse the period t00, given by (17), which the parameter B satisfies:(20)B=2b1k12signb1,where(21)b1=−k2signata0+b.
Then, att00=−B/A, the states of the supertwisting system reach the origin point:(22)x1t00=0,x2t00=0.
This completes the proof of Theorem 1.Case III.
For the supertwisting system, whenA>−1/4anda≠0, the trajectory is(23)ts=Cs1+psq−BA,ys=C1+12s+p2s1+psq,where p=4A+1 (0<p<1) and q=1/2p−1/2 (q>0).(24)0=Cs01+ps0q−BA,a1=C1+12s0+p2s01+ps0q.
Therefore,(25)s0=1a1A/B−1/2−p/2,(26)C=BAs01+ps0q.
Then, the whole trajectory can be obtained for the parametric equation, whens1=−2/1+p,s2=−1/p, y=0, and s2<s1:(27)ts1=Cp/2−1/2q−p/2−1/2q+1−BA,(28)ts2=−BA.
The statex1 will reach the origin point at ta0, where the following conditions are met:(c1)
Whens2<s1<s0, ta0=t1.(c2)
Whens0<s2<s1, ta0=t1.(c3)
Whens2<s0<s1, if ∂t/∂ss=s0>0, ta0=ts1.(c4)
Whens2<s0<s1, if ∂t/∂ss=s0>0, ta0=ts2.
Then, the states of the supertwisting system at timeta0 become(29)x1ta0=0,x2ta0=−k2signata0+b.Case IV.
For the supertwisting system, whenA>−1/4anda=0, the trajectory is(30)ts=Cs1+psq−BA,ys=C1+12s+p2s1+psq.
For the initial conditionst=0andy=0,(31)0=Cs01+ps0q−BA,0=C1+12s0+p2s01+ps0q.
Therefore,(32)s0=−21+p,(33)C=BAs01+ps0q,for the parametric equation(34)s2=−1p.
Therefore, the convergence time is(35)t00=−BA,(36)x1t00=0,x2t00=0.
Therefore, the system reaches finite-time stability.Theorem 2.
The supertwisting system (1), if gains k1>0andk2>0 satisfy k12>8k2 and system initial conditions x10=aandx20=b, reaches finite-time stability; the finite time T satisfies T=ta0+t00, where ta0 is the time when state x1 reaches the origin point in first time x2=b1≠0, and then time t00=−B1/A is elapsed, when both states reach the origin point at the same time.(d1)
Whens2<s1<s0, ta0=ts1.(d2)
Whens0<s2<s1, ta0=ts1.(d3)
Whens2<s0<s1, if ∂t/∂ss=s0>0, ta0=ts1.(d4)
Whens2<s0<s1, if ∂t/∂ss=s0<0, ta0=ts2.
Here,(37)A=−2k2k12,B=2bk12sign2signa+signb,B1=2b1k12signb1,b1=−k2signata0+b.Proof of Theorem 2.
Based on the supertwisting system and gainsk1andk2, if A>−1/4, the parametric solution can be obtained in the form of (23); based on the initial condition, parameters s0andC can be obtained by (25) and (26), and then ta0 can be obtained in following different situations: (d1)–(d4).
Whent=ta0, the states of the system change as in (29), and then the system will elapse period t00, given by (35), which the parameter B satisfies:(38)B=2b1k12signb1,where(39)b1=−k2signata0+b.
Then,t00=−B/A, and the states of the supertwisting system reach the origin point at time T:(40)x1T=0,x2T=0.
This completes the proof of Theorem 2.Remark 3.
From Theorem 1 and Theorem 2, the convergence time isT=ta0+t00 and then k1andk2 exist in both ta0 and t00 in the form of B/A; therefore, the gain parameter k1 is not influenced by the convergence time.


## 4. Finite-Time Convergence of STA with Perturbation

The STA with perturbation can be written as(41)z˙1=−h1z11/2signz1+z2+ρ1t,z˙2=−h2signz1+ρ2t,where z1andz2 are the scalar state variables, with initial conditions z10=aandz20=b; h1,h2>0 are gains to be designed; and ρ1andρ2 are the perturbation terms. It is well known that the STA is robustly stable to perturbations globally bounded, and the equilibrium point is 0−ρ1; therefore, suppose the perturbation terms of the system.Assume thatρ1≤d1,ρ2≤d2,ρ˙1≤d3,andρ˙2≤d4.Let(42)x1=z1,x2=z2+ρ1,then system (41) can be changed as(43)x˙1=−h1x11/2signx1+x2,x˙2=−h2signx1+ρ2+ρ˙1.Next, let(44)k1=h1,k2=h2−ρ2+ρ˙1signx1,then the system can be written as(45)x˙1=−k1x11/2signx1+x2,x˙2=−k2signx1.The initial conditions arex10=aandb¯<x20<b¯ and parameters are k1=h1andk¯2<k2<k¯2, where b¯=bz−d1,b¯=bz+d1, k¯2=h2−d2−d3,andk¯2=h2+d2+d3.Under the conditions on STA without perturbation in Theorem 2, the robust finite time is presented for STA with perturbation. AssumeA=−2k2/k12>−1/4, then h12>8h2−d2−d3>0. The finite time is presented by TP=tPa0+tP00, where tPa0 and tP00 can be obtained by Theorem 2 in different initial conditions and gains.In Theorem 2,tPa0 is dependent on k1,k2,a,andb and is then calculated in different situations to get the maximum. To compare the influence of perturbation, three STAs are built:(46)STA1:x1,x2t,k1,k2,a,b,STA2:x¯1,x¯2t,k1,k¯2,a,b¯,STA3:x¯1,x¯2t,k1,k¯2,a,b¯.Without loss of generality, assume thata>0, in time phasing t∈0,tPa0.For the perturbation of the STS at different gainsk2 and different initial conditions b, assume states x2, x¯2, and x¯2 as x2=b−k2t, x¯2=b¯−k¯2t, and x¯2=b¯−k¯2t; therefore, x¯2≤x2≤x¯2 and x¯1≤x1≤x¯1 in t∈0,tPa0.The convergence time satisfiesta0k1,k¯2,a,b¯≤tPa0≤ta0k1,k¯2,a,b¯.Assume thata=0andb>0; to compare the influence of perturbation, three STAs are built:(47)STA1:x1,x2t,k1,k2,a,b,STA2:x¯1,x¯2t,k1,k¯2,a,b,STA3:x¯1,x¯2t,k1,k¯2,a,b.Choose a sliding mode surface ass=c1x1+c2x2, s¯=c1x¯1+c2x¯2, and s¯=c1x¯1+c2x¯2, where c1>0andc2>0; therefore,(48)s˙=c1x˙1+c2x˙2=−c1k1x11/2signx1+c1x2−c2k2signx1=−c1k1x11/2+c1b−c1k2t−c2k2,because of(49)−c1k1x11/2+c1b−c1k¯2t−c2k¯2≤s˙≤−c1k1x11/2+c1b−c1k¯2t−c2k¯2.Therefore,(50)s¯˙≤s˙≤s¯˙.Because trajectoriesx¯1,x¯2,x¯1,x¯2,s¯,ands¯ converge to origin in finite time, the trajectory s reaches origin in finite time. Parameters c1andc2 are arbitrary positive constants; therefore, s=0 could be equivalent to x1=0andx2=0. The convergence time satisfies(51)ta0k1,k¯2,a,b≤tP00≤ta0k1,k¯2,a,b.Theorem 3.
For the supertwisting algorithm with perturbation (41), if h1,h2>0 and h12>8h2−d2−d3>0 hold, the system converges to the origin point in finite time, and the maximum convergence time is TP=tPa0+tP00, where tPa0=ta0k1,k¯2,a,b¯ and tP00=maxb¯,b¯−k¯2tPa0/k¯2.Proof of Theorem 3.
For STA with perturbation, when the initial conditiona≠0, in the first stage, the state x1 would arrive at the origin point in finite time tPa0=ta0k1,k¯2,a,b¯, and then when the system is in the second stage, both states x1andx2 arrive at the origin point in finite time:(52)tP00=maxb¯,b¯−k¯2tPa0k¯2.
Then, the maximum finite time is presented asTP=tPa0+tP00.


## 5. STC for Nonaffine Systems

To illustrate the STA, nonlinear system STC is proposed for the nonaffine system:(53)y˙=fy,u.Using the coordinate transform,(54)x1=y−yd.The virtual control law is(55)α=−k1x11/2signz1+u−fy,u+y˙d.Then,(56)x˙1=−k1x11/2signx1+x2,where(57)x2=u−α.Choose the ideal control law(58)v=−k2signx1+α˙,then(59)x˙2=−k2signx1.Therefore, the system satisfies Theorem 1 and Theorem 2 by choosing the gain parametersk1andk2.To overcome the nonaffinity, the virtual control and backstepping skill are used. They are extremely important to ensure the successful design of the proposed STC.


## 6. Simulation

Several simulated examples were selected to conduct bench tests of the STA and STC. Example 1–Example 6 were testing simple STA, and the purpose of testing these examples was to investigate whether the convergence time is solved from the parametric equation. Examples 7–10 were designed for the test of STC for nonaffine nonlinear systems and demonstrated the superiority of STC and convergence time of the STA.Example 1.
Consider the STS given by (1), with k1=1, k2=1/8, a=2, and b=−1 in Case I; based on (9) and (10), we can get s0=−0.5858 and C=−24.5332, respectively, and then we can get ta0=1.0253; therefore, the state x1 reaches the zero point at time 1.0253.Example 2.
Consider the STS given by (1), with k1=1, k2=1/8, a=0, and b=−1.1282 in Case II; based on (16) and (17), we can get s0=−1 and C=−B/Aexp−1, respectively, and then we can get t00=9.0256; therefore, the states x1andx2 reach the zero point at time 9.0256.Example 3.
Consider the STS given by (1), with k1=3, k2=1, a=3, and b=−3 in Case III; based on (25) and (26), we can get s0=−0.9510 and C=−4.6188, respectively, and then we can get ta0=0.4641; therefore, the state x1 reaches the zero point at time 0.4641.Example 4.
Consider the STS given by (1), with k1=3, k2=1, a=0, and b=−3.4641 in Case IV; based on (32) and (33), we can get s0=−2/1+p and C=B/As01+ps0q, respectively, and then we can get t00=3.4641; therefore, the states x1andx2 reach the origin point at time 3.4641.Example 5.
Consider the supertwisting system given by(60)x˙1=−x11/2signx1+x2,x˙2=−18signx1,with k1=1,k2=1/8,andA=−1/4 and initial conditions x10=2andx20=−1.
Based on Theorem 1, the system (60) reaches finite-time stability and calculates the time from parametric equation (7) step by step. Firstly, the initial parameter s0=−0.5858 is calculated from (9) and C=−24.5332 from (10) because s2<s1<s0 holds; therefore, ta0=ts1 and ts1 can be obtained from (11) as ta0=1.0253 and x2ta0=−1.1282 in Figure 4. Secondly, t00=9.0256 is calculated from (17) in Figure 5. Finally, the reaching time t=ta0+t00=10.0509 in Figure 6.Figure 4
Finite-time convergence to the origin point with the nonzero initial condition of the state (A = (−1/4)).Figure 5
Finite-time convergence with the zero initial condition of the state (A = (−1/4)).Figure 6
Finite-time convergence (A=−1/4).Example 6.
Consider the supertwisting system given by(61)x˙1=−3x11/2signx1+x2,x˙2=−signx1,with k1=3,k2=1,andA=−2/9 and initial conditions x10=3andx20=−3.
Firstly, the initial parameters0=−0.9510 is calculated from (25) and C=−4.6188 from (26) because s1=−1.5,s2=−3, and s2<s1<s0 hold; therefore, ta0=ts1 and ts1 can be obtained from (27) as ta0=0.4641 and x2ta0=−3.4641 in Figure 7. Secondly, t00=3.4641 is calculated from (35) in Figure 8. Finally, the reaching time t=ta0+t00=3.9282 in Figure 9.Figure 7
Finite-time convergence to the origin point with the nonzero initial condition of the state (A = (−2/9)).Figure 8
Finite-time convergence with the zero initial condition of the state (A = (−2/9)).Figure 9
Finite-time convergence (A=−2/9).Example 7.
Consider the STA with perturbation as given by (45), as k1=4,0.5≤k2≤1.5,a=3,and−5≤b≤−1:(62)STA1:x˙1=−4x11/2signx1+x2,x˙2=−signx1+0.3cost+0.2cos0.1t,x10=3,x20=−3,STA2:x¯˙1=−4x¯11/2signx¯1+x¯2,x¯˙2=−0.5signx¯1,x¯10=3,x¯20=−1,STA3:x˙¯1=−4x11/2signx¯1+x¯2,x˙¯2=−1.5signx¯1,x¯10=3,x¯20=−5.
In Figure10, states of three STAs are presented to indicate stability in finite time, and convergence time is ta0=0.4099, t¯a0=0.5816, and t¯a0=0.3135, respectively. The trajectory indicates states x1, x¯1, and x¯1 with the same initial condition reach zero at different time because of different x2, x¯2, and x¯2. The trajectories x1andx2 are in between x¯1andx¯1and  betweenx¯2andx¯2, respectively. State x¯1 is the last state that reaches zero; therefore, the maximum convergence time t¯a0=0.5816. In addition, the maximum of the state x2 is x¯2t¯a0=5.8723 and absolutely necessary to estimate finite time t00 in the next step.Figure 10
Finite-time convergence of the output with perturbation.Example 8.
Consider the STA with perturbation as given by (45), as k1=4,0.5≤k2≤1.5,a=0,andb=−5.8723:(63)STA1:x˙1=−4x11/2signx1+x2,x˙2=−signx1+0.3cost+0.2cos0.1t,x10=0,x20=−5.8723,STA2:x¯˙1=−4x¯11/2signx¯1+x¯2,x¯˙2=−0.5signx¯1,x¯10=0,x¯20=−5.8723,STA3:x˙¯1=−4x11/2signx¯1+x¯2,x˙¯2=−1.5signx¯1,x¯10=0,x¯20=−5.8723.
In Figure11, states of three STAs are presented to indicate stability in finite time, and convergence time is t00=5.6589, t¯00=11.7447, and t¯00=3.9149, respectively. The trajectories x1,x2,ands are in between x¯1andx¯1,x¯2andx¯2,ands¯ands¯, respectively. State x¯1 is the last state that reaches zero; therefore, the maximum convergence time t¯00=11.7447. In addition, states of the sliding mode surface indicate the state s is in between s¯ands¯, providing evidence that t¯00 is the maximum convergence time.Figure 11
Finite-time convergence of the output and sliding mode surface.
(a)
(b)
(c)
(d)Example 9.
Consider the supertwisting system with perturbation given by(64)z˙1=−h1z11/2signz1+z2+ρ1t,z˙2=−h2signz1+ρ2t,where z1andz2 are the scalar state variables, with initial conditions z10=3andz20=−3; h1=4andh2=1 are gains; and ρ1t=2sin0.1t and ρ2t=0.3cost are perturbation terms. Therefore, the system (64) changes as(65)x˙1=−h1x11/2signx1+x2,x˙2=−h2signx1+ρ2+ρ˙1,where x1=z1andx2=z2+ρ1; therefore, parameters are k1=4 and 0.5≤k2≤1.5, and initial conditions are a=3and−5≤b≤−1; based on Theorem 3, the convergence time can be obtained.
The whole convergence process can be split into three stages. Firstly, the statex1 from initial condition x10 reaches zero in finite time tpa0, and at this time, x2tpa0≠0. Secondly, both states x1andx2 reach zero at the same time and elapse time tp00 in this stage. Finally, after last two stages, the states x1andx2 would stay at zero.
From Figure12, it can be seen that, in the first stage, state x1 from the initial condition reaches zero in finite time as in Figure 10 for Example 7, and the maximum convergence time ta0=0.5816 is calculated, and in this stage, the maximum of state x2 is x2ta0=5.8723. Therefore, in the next stage, the initial condition is x20=−5.8723, such as in Figure 11 for Example 8, and then the maximum convergence time t00=11.7447 is calculated. Finally, the total finite convergence time t=12.3263; therefore, the system converges to zero after 12.3263.Figure 12
Finite-time convergence of the output and maximum stage.
(a)
(b)Example 10.
Consider a nonaffine nonlinear system(66)y˙=y+y3+2u+sinu,where yandu are the output and input, respectively, and the reference output is taken as yd. The control objective is to design the STC for the system such that the output follows the given reference signal yd in finite time.
Choose statesx1=y−yd and x2=u−α with virtual control where(67)α=−3x11/2signx1+yd−y−y3−u−sinu,v=−signx1+α˙.
The states satisfy the STA conditionA=−2/9 and initial conditions x10=3andx20=−3. Therefore, the states x1andx2 satisfy the condition in Figure 9 and output and reference output in Figure 10.
It can be observed that, for the nonaffine system, the closed-loop system has been controlled in finite time effectively in Figures13–15, and time can be calculated and designed by control gain and initial condition.Figure 13
Finite-time convergence of the output and reference signal in control.Figure 14
State of virtual control.Figure 15
State of the controller.


## 7. Conclusion

A method to calculate the accurate convergence time of the STA has been provided, and an STC method for nonaffine nonlinear systems based on the STA is given. The trajectory of the STA is solved in the form of the parametric equation, and reaching time is calculated accurately. The result indicated the gain parameterk1 is not influenced by the convergence time. For the application of the STA, STC is designed to solve the nonaffine systems’ control problem by backstepping skill.Furthermore, after a certain prefixed gain and initial conditions, the reaching time of the STA is also useful to design STC. In the future, there still exist some problems to face with, such asA<−1/4 in the STA and higher order twisting algorithm.



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*Source: 6235190-2019-10-20.xml*