# Design of a Disturbance Rejection Controller for a Class of Nonholonomic Systems with Uncertainties

Authors: Dianguo Cao; Jiaqian Chen
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

This study investigates the global output feedback stabilization problem for one type of the nonholonomic system with nonvanishing external disturbances. An extended state observer (ESO) is constructed in order to estimate the external disturbance and unmeasurable system states, in which the external disturbance term is seen as a general state. Thus, a new generalized error dynamic system is obtained. Accordingly, a disturbance rejection controller is designed by making use of the backstepping technique. A control law is given to ensure that all the signals in the closed-loop system are globally bounded, while the system states converge to an equilibrium point. The simulation example is proposed to verify that the control algorithm is effective.


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## Body

## 1. Introduction

Within recent decades, the control of nonholonomic systems has always been one of the most popular tasks in control fields since such systems can be frequently found in mechanical systems, for example, car-like vehicles, wheeled mobile robots, knife-edge, and so on. In the theoretical analysis of the nonholonomic system model, some nonlinear feedback controllers for these systems were put forward in the literature to ensure that the systems are asymptotically stable or exponentially regulatable, for example, the studies [1–7] and references therein. By using an input/state scaling technique and switching algorithm, a class of feedback control law was obtained for nonholonomic chained systems with uncertainties to realize exponential stabilization [6, 7], and a switching-based state scaling is designed for prescribed-time stabilization of nonholonomic systems with actuator dead-zones [8]. In practical applications, especially in the research of nonholonomic wheeled mobile robot control, the controller design method to realize the robust stabilization of the system is given [6, 9, 10]. Considering the limitations of the hardware and environment of the actual system, the design method of the controller with saturated input is given in [11, 12]. In order to overcome the external disturbances, the robust tracking control for the wheeled mobile robot is proposed based on the ESO [13, 14].The measurement of full states is usually difficult and sometimes impossible. Moreover, in practical applications, the systems usually contain unknown disturbances, measurement noise, and modeling errors, which are called nonvanishing total disturbances. These disturbances in reality will influence the performance of closed-loop systems. Therefore, it is of great significance to study the output feedback stabilization of nonholonomic systems with nonlinear uncertainties and external disturbances. The output feedback stabilization for nonholonomic systems is more complex and difficult than using the general nonlinear state feedback. The output feedback problem towards asymptotic stability and exponential stability of nonholonomic systems has previously been put forward [15, 16]. In [17–20], the adaptive output feedback global stabilization of a class of nonholonomic systems with parametric uncertainties and strong nonlinear drifts are solved. However, none of the above work considers the existence of disturbance items that do not disappear from the system even though uncertainties or nonlinear drifts exist. This means that the proposed output feedback scheme may be unstable because of the external disturbances. To reject the external disturbances, an output feedback controller has been proposed for nonholonomic systems with nonlinear uncertainties [21] and nonvanishing external disturbances [22–24]. In [23], the external disturbances are considered a generalized system state, and an ESO was constructed. By utilizing the so-called ESO, [25] further investigated the output regulation control problem towards one type of cascade nonlinear systems with the external disturbance, and the output feedback adaptive regulation problem was solved by the time-varying Kalman observer [25]. However, the output regulation controller in [22, 23, 25] requires that the nonlinear uncertainties in the systems are only related to the output of the systems.The ESO in pioneering work [26] is the key creative advancement towards active disturbance rejection control (ADRC). The ESO has the capability for state observation and real-time estimation of generalized disturbances between the controlled object and the model of the controlled system [27, 28]. By using the ESO, this study addresses robust output feedback adaptive control towards one type of nonholonomic chained form systems that have nonvanishing external disturbances in the input channel and uncertain nonlinearity drift. Different from references [22, 23], in the model studied in this study, the upper bound function of nonlinear uncertainties depend not only on the output variables but also on the system state variables, in which such uncertain nonlinearities meet a linearly growing triangular condition.The main contribution of this study is that the extended state observer (ESO) and gain scaling technique [29] are constructed. In order overcome unknown system states and the external disturbance, we reconstruct the system state, and the disturbance is regarded as an extended state. The ESO with dynamic gain is put forward, and the disturbance rejection controller based on an observer is developed by designing a variable observer gain to overcome the uncertainty. The controller design is carried out for one type of the nonholonomic system with nonvanishing external disturbances and uncertain nonlinearities satisfying a linearly growing triangular condition. This approach allows the external disturbances to be a larger class of signals.


## 2. Problem Formulation

In this study, we consider the following nonholonomic system with nonlinear uncertainties and nonvanishing external disturbance:(1)x0.=u0+x0ϕ0dt,x0,xi.=xi+1u0+ϕidt,u0,x0,x,ω0t,xn.=u+ϕndt,u0,x0,x,ω0t+wt,y=x0,x1T,where x0∈R, x=x1,…,xnT∈Rn are the system states, and the initial values are x0t0, xt0, with t0 as the initial moment of the system; u=u0,uT∈R2 is the control input, and y∈R2 is the system output. The functions ϕid⋅,i=1,2,…,n are the uncertainties which represent possible modeling errors and neglected dynamics; w0t, w˙0t, wt∈R, and w˙t are the uncertainties and bounded, where wt∈R is the nonvanishing external disturbance and satisfies that w˙t∈L2. The assumptions and lemmas used in this article are listed as follows.Assumption 1.
For every1≤i≤n, the following inequality holds:(2)ϕ0dt,x0≤α0x0,ϕidt,u0,x0,x,ω0t≤αx0∑j=1ixj,where the nonnegative smooth functions α0x0 and αx0 are known.Lemma 1 (see [30]).
For anyx,y∈R, any scalar k>0, and any positive definite matrix M∈Rn+1×n+1, the following inequality holds:(3)2xTy≤k−1xTMx+kyTM−1y.Lemma 2 (see [31, 32]).
For anyμ>0, there exist positive real numbers d1 and d2, positive definite matrix P, and positive constants ai, such that the following inequality is satisfied:(4)PA+ATP≤−d1In+1,PD+DP≥d2In+1,where Ii is the identity matrix of order i, and A and D are the n+1×n+1 matrices denoted as(5)A=−a1⋮In−an+10⋯0,D=diagμ,1+μ,…,n+μ.


## 3. Controller Design and Stability Analysis

### 3.1. Output Feedback Controller Design

Lemma 3 (see [33]).
For the first subsystem of (1), if the first control law u0 is chosen as(6)u0=−λ0x0−x0α0x0,λ0>0,where t0,x0t0 is regarded as the initial condition, x0t0≠0, then as the corresponding solution, x0t,t0,x0t0 exists and |x0t,t0,x0t0|>0, 0≤t0<t. λ0 is designed as a positive constant parameter. Furthermore, u0t>0.Proof.
Substituting (6) into the first formula of system (1), we can obtain(7)x0.=−λ0x0−x0α0x0+x0ϕ0dt,x0.
Integrating this nonlinear equation, the solution isx0t=x00e−∫0tλ0+α0x0τ−ϕ0dτ,x0τdτ. This shows that x0t≠0 at any time if x00≠0, and thus, u0t≠0. Choosing V0x0=1/2x02, it can be obtained from (6) that(8)V0.=−λ0x02−x02α0x0−ϕ0dt,x0≤−λ0x02≤0.
Thus,x0t asymptotically approaches zero. For any bounded x0t because α0,ϕ0d are smooth functions, there is a positive constant M for x0≤1, α0≤M,ϕ0d≤M. This yields that(9)V0.=−λ0x02−x02α0x0−ϕ0dt,x0≥−λ0+M+MV0.
Integrating both sides of this equation, it can be obtained that(10)V0t≥V00e−λ0+2Mt.
This means thatx0t converges to zero, but x0t≠0 at any given moment, so that u0t>0.
We introduce the following input state scaling:(11)ζi=xiu0n−i,i=1,…,n.
Unknown nonvanishing external disturbancewt is treated as a generalized state. To realize symbol consistency, it is defined as(12)ζn+1=wt.
In the new stateζ, system (1) is converted to(13)ζ˙i=ζi+1+ϕ¯it,u0,x0,ζ,ω0t,ζ˙n=ζn+1+ϕ¯nt,u0,x0,ζ,ω0t+u,ζ˙n+1=w˙t≜ht,i=1,…,n−1.Lemma 4.
For any givenu0 in (6), there is a known nonnegative smooth function φ¯0x0, such that u˙0/u0≤φ¯0x0, t≥0.Proof.
The following calculation is completed:(14)u˙0u0=−x0λ0+α0x0−ϕ0dt,x0λ0+α0x0∂α0x0∂x0−λ0−α0x0+ϕ0dt,x0≤λ01+λ0x0λ0+α0x0∂α0x0∂x0≜φ¯0x0.
This completes the proof of the lemma.
We know from Assumption1 that there is a nonnegative smooth function α¯x0:(15)ϕ¯it,u0,x0,ζ,ω0t≤αx0∑j=1ixju0n−i+n−iα¯0x0ζi≤n−iα¯0x0ζi+αx0u0i−1ζ1|+|u0i−2ζ2|+⋯+|ζi≤α¯x0∑j=1iζi.
Considering thatζ1,…,ζn+1 are unmeasurable signals that cannot be used in feedback control, the dynamic observer for (13) is denoted as follows:(16)ζ^˙i=ζ^i+1+aiγiζ1−ζ^1,i=1,…,n−1,ζ^˙n=u+ζ^n+1+anγnζ1−ζ^1,ζ^˙n+1=an+1γn+1ζ1−ζ^1−γcζ^n+1,where γ is the dynamic gain, which will be designed later according to the requirements. The observer error dynamics is defined as(17)ei=ζi−ζ^ii=1,…,n+1.
We can determine from (13), (16), and (17) that(18)e˙i=ei+1+ϕ¯it,u0,x0,ζ,ω0t−aiγiζ1−ζ^1,e˙n=en+1+ϕ¯nt,u0,x0,ζ,ω0t−anγnζ1−ζ^1,e˙n+1=ht−an+1γn+1ζ1−ζ^1+γcζ^n+1.
Introducing dynamic gain scaling,(19)εi=eiγi−1+μ,zi=ζ^iγi−1+μ,i=1,…,n+1,and defining ε=ε1,…,εn+1T,z=z1,…,zn+1T,Φ⋅=ϕ¯1⋅/γμ,…,ϕ¯n⋅/γn−1+μ,0T, we arrive at(20)ε.i=γεi+1+Φit,u0,x0,ζ,ω0t−γaiε1−γ˙γi−1+μεi,ε.n=γεn+1+Φnt,u0,x0,ζ,ω0t−γanε1−γ˙γn−1+μεn,ε.n+1=htγn+μ−γan+1ε1−γ˙γn+μεn+1+γczn+1,(21)z˙i=γzi+1+γaiε1−γ˙γi−1+μzi,z˙n=uγn−1+μ+γzn+1+γanε1−γ˙γn−1+μzn,z˙n+1=γan+1ε1−γ˙γn+μzn+1−γczn+1.
Denotinga=a1,…,an+1T, b=0,…,11×n+1T, F=0,…,γczn+11×n+1T, and h¯t=ht/γn+μ, we have(22)ε.=γAε+Φt,u0,x0,ζ,ω0t−γ˙γDε+bh¯t+F.
Now, using Assumption1 and (15), it follows that(23)Φit,u0,x0,ζ,ω0t≤α¯x0γi−1+μ∑j=1iζj.
ChoosingVε=εTPε, it is then obtained that(24)Vε.=ε.TPε+εTPε.=γAε+Φ−γ˙γDε+bh¯t+FTPε+εTPΦ+γAε−γ˙γDε+bh¯t+F=γεTATP+PAε+2εTPΦ−γ˙γεTDP+PDε+2εTPbh¯t+2εTPF≤2εTPΦ−γd1εTε+2εTPbh¯t+2εTPF−γ˙γd2εTε.
Introducing the following transformation,(25)x^1=z1,x^i=zi−αi−1,αi−1=−gi−1x^i−1,i=2,…,n,x^n+1=zn+1,where gi−1>0 is a constant number that will be given later, because εi=ei/γi−1+μ, and ei=ζi−ζ^i, we have(26)ζi=γi−1+μεi+zi=γi−1+μεi+x^i−gi−1x^i−1.
Now, using Lemma1, it follows that inequality,(27)2εTPΦ≤2εPα¯x0ε+z=2α¯x0Pε2+2α¯x0Pεz≤φ¯1x0ε2+φ¯2x0z2≤φ1x0ε2+φ2x0x^2,holds, where φ1x0 and φ2x0 are the nonnegative smooth functions. On the other hand, by using Young’s inequality, one has(28)2εTPbh¯≤γL1ε2+γh¯2,2εTPF≤γPc1zn+12+γ1c1Pε2.
Correspondingly, we can obtain that(29)Vε.≤−γd1εTε−γ˙γd2εTε+γL1ε2+γPc1zn+12+γ1c1Pε2+φ1x0ε2+φ2x0x^2+γh¯2.Step 1. ChoosingV1=1/2z12=1/2x^12, we have(30)V˙1=x^1γz2+γa1ε1−γ˙γμz1=γx^1z2−α1+α1+γa1x^1ε1−γ˙γμx^1z1=γx^1x^2+γx^1α1+γa1x^1ε1−γ˙γμx^12.By using Young’s inequality, we have(31)γa1x^1ε1≤γM1ε12+γa124M1x^12=γM1ε12+γβ1x^12,WhereM>0,β1=a12/4M1 are the constants. Choosing α1=−g1x^1=−n+β1x^1,δ1=μ, we then obtain(32)V˙1≤−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12.Step 2. ChoosingV2=p1V1+1/2x^22, where p1 is a designed positive constant,(33)V˙2≤p1−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12+x^2z˙2−α˙1=p1−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12+x^2γz3+γa2ε1−γ˙γ1+μz2−x^2∂α1∂z1z˙1=−γp1nx^12+γp1M1ε12+γp1x^1x^2−γ˙γp1δ1x^12+γx^2x^3+γx^2α2+γa2x^2ε1−γ˙γ1+μx^2z2−x^2∂α1∂z1z˙1.
From Lemma1, we can derive the following inequalities:(34)γp1x^1x^2≤γ3p1x^12+3γp14x^22=γ3p1x^12+γβ21x^22γa2x^2ε1≤γθ21ε12+γa224θ21x^22=γθ21ε12+γβ22x^22,−γ˙γ1+μx^2z2=−γ˙γ1+μx^2x^2−g1x^1=−γ˙γ1+μx^22+γ˙γ1+μx^2⋅g1x^1≤−γ˙γ1+μ2x^22+γ˙γ1+μ2g12x^12=−γ˙γ1+μ2x^22+γ˙γN1l2,g1x^12,where β21=3p1/4,l1=μ,l2=1+μ,N1l2,g1=l2/2g12,θ21>0,β22=a22/4θ21 are the constants. Using the relations α1=−g1x^1 and z˙1=γz2+γa1ε1−γ˙/γμz1, we have(35)−x^2∂α1∂z1z˙1=−x^2∂α1∂z1γz2+γa1ε1−γ˙γμz1.
Definingx^0=0,g0=1, it is derived that(36)γz2−γ˙γl1z1=γx^2−g1x^1−γ˙γl1x^1=γx^2−g1x^1−γ˙γl1x^1−g0x^0≤γx^2+g1x^1|+γ˙γl1x^1+g0x^0|≤Ag0,g1γx^2+x^1+γ˙γl1x^1+x^0,where Ag0,g1=max1,g0,g1. Since ∂α1/∂z1≤g1, this implies that(37)−x^2∂α1∂z1γa1ε1≤x^2g1γa1ε1≤γθ22ε12+γg12a124θ22x^22=γθ22ε12+γβ23x^22,−x^2∂α1∂z1γz2−γ˙γl1z1≤x^2g1Ag0,g1γx^2+x^1+γ˙γl1x^1+x^0.
LettingB21=g1Ag0,g1, it can be concluded that(38)γx^2g1Ag0,g1x^2+x^1=γB21x^22+γB21x^2x^1≤γ3p1x^12+γB21+3B2124p1x^22=γ3p1x^12+γβ24x^22,x^2B21γ˙γl1x^1+x^0=x^2B21γ˙γl1x^1≤γ˙γB212l1x^12+γ˙γl14x^22=γ˙γB2x^12+γ˙γl14x^22,where θ22>0,β23=g12a12/4θ22,β24=B21+3B212/4p1,B2=B212l1 are the constants. Furthermore, we have(39)V˙2≤−γp1nx^12−γ˙γp1δ1x^12+γ3p1x^12+γ˙γN1l2,g1x^12+γ3p1x^12+γ˙γB2x^12+γp1M1ε12+γM21ε12+γM22ε12+γβ21x^22+γβ22x^22+γβ23x^22+γβ24x^22−γ˙γ1+μ2x^22+γ˙γl14x^22+γx^2x^3+γx^2α2.
Let us denote(40)β2=β21+β22+β23+β24,M2=p1M1+θ21+θ22,thus, it is obtained that(41)V˙2≤−γn−1p1∑j=12x^j2+γM2ε12+γ˙γN1l2,g1+B2x^12+γx^2x^3+γx^2α2−γ˙γp1δ1x^12−γ˙γl24x^22+γβ2+n−1p1x^22.
For simplicity, letp1′=p1δ1−N1−B2>0 and δ2=minl2/4,p1′,α2=−g2x^2=−β2+n−1p1x^2; then,(42)V˙2≤−γn−1p1∑j=12x^j2+γM2ε12+γx^2x^3−γ˙γδ2∑j=12x^j2.
Step i2<i≤n−1. Assume that in Step i−1, we have(43)V˙i−1≤−γn−i+2p1,…,pi−2∑j=1i−1x^j2+γMi−1ε12+γx^i−1x^i−γ˙γδi−1∑j=1i−1x^j2.
Letting theith candidate Lyapunov function be Vi=pi−1Vi−1+1/2x^i2 and defining x^i=zi−αi−1, where p2,…,pi−2 are the designed positive constants,(44)V˙i=pi−1V˙i−1+x^iz˙i−∑j=1i−1∂αi−1∂zjz˙j=pi−1V˙i−1+x^iγzi+1+γaiε1−γ˙γi−1+μzi−x^i∑j=1i−1∂αi−1∂zjz˙j≤−γn−i+2p1,…,pi−1∑j=1i−1x^j2+γpi−1Mi−1ε12−γ˙γpi−1δi−1∑j=1i−1x^j2+γpi−1x^i−1x^i+γx^izi+1+γaix^iε1−x^i∑j=1i−1∂αi−1∂zjz˙j−γ˙γi−1+μx^izi.
By applying Young’s inequality, one has(45)γpi−1x^i−1x^i≤γ3p1,…,pi−1x^i−12+3γpi−14p1,…,pi−2x^22=γ3pi−1x^i−12+γβi1x^i2γaix^iε1≤γMi1ε12+γai24Mi1x^i2=γMi1ε12+γβi2x^i2−γ˙γi−1+μx^izi=−γ˙γi−1+μx^ix^i−gi−1x^i−1=−γ˙γi−1+μx^i2+γ˙γi−1+μx^i⋅gi−1x^i−1≤−γ˙γi−1+μ2x^i2+γ˙γi−1+μ2gi−12x^i−12=−γ˙γli2x^i2+γ˙γNi−1li,gi−1x^i−12,
whereβi1=3pi−1/4p1,…,pi−2,li=i−1+μ,Ni−1li,gi−1=li/2gi−12,Mi1>0,βi2=ai2/4Mi1 are the constants. Using the relations(46)αi−1=−gi−1x^i−1=−gi−1zi−1+gi−2x^i−2=−∑j=1i−1∏s=ji−1gszj=−∑j=1i−1∏s=ji−1gszj,and z˙j=γzj+1+γajε1−γ˙/γj−1+μzj, it follows that(47)−x^i∑j=1i−1∂αi−1∂zjz˙j=−x^i∑j=1i−1∂αi−1∂zjγzj+1+γajε1−γ˙γj−1+μzj,(48)γzj+1−γ˙γljzj=γx^j+1−gjx^j−γ˙γljzj=γx^j+1−gjx^j−γ˙γljx^j−gj−1x^j−1≤γx^j+1+gjx^j+γ˙γljx^j+gj−1x^j−1≤Agj−1,gjγx^j+1+x^j+γ˙γljx^j+x^j.
Since∂αi−1/∂zj≤gj,…,gi−1, this implies that(49)−x^i∑j=1i−1∂αi−1∂zjγajε1≤x^i∑j=1i−1gj,…,gi−1γajε1≤γMi2ε12+γ∑j=1i−1gj,…,gi−1aj24Mi2x^i2=γMi2ε12+γβi3x^i2−x^i∑j=1i−1∂αi−1∂zjγzj+1−γ˙γljzj≤x^i∑j=1i−1gj,…,gi−1Agj−1,gjγx^j+x^j+1+γ˙γljx^j+x^j+1.
DefiningBij=gj,…,gi−1Agj−1,gj,(50)γx^i∑j=1i−1Bijx^j+1+x^j=γx^i∑j=1i−1Bijx^j+1+γx^i∑j=1i−1Bijx^j=γBi,i−1x^i2+γx^i∑j=2i−1Bi,j−1x^j+γx^i∑j=1i−1Bijx^j≤γBi,i−1x^i2+γx^i∑j=1i−1Bi,j−1x^j+γx^i∑j=1i−1Bijx^j=γBi,i−1x^i2+γx^i∑j=1i−1B¯i,jx^j≤γBi,i−1+3∑j=1i−1B¯ij24p1,…,pi−1x^i2+γ3p1,…,pi−1∑j=1i−1x^j2=γβi4x^i2+γ3p1,…,pi−1∑j=1i−1x^j2,x^i∑j=1i−1Bijγ˙γljx^j+x^j−1=γ˙γx^i∑j=1i−1Bijljx^j+γ˙γx^i∑j=1i−1Bijljx^j−1≤γ˙γx^i∑j=1i−1B˜ijx^j≤γ˙γli−1∑j=1i−1B˜ij2x^j2+γ˙γli−14x^i2=γ˙γBi∑j=1i−1x^j2+γ˙γli−14x^i2,
whereB¯ij=Bij+Bi,j−1,B˜ij=Bijlj+Bi,j−1lj−1,Mi2>0,βi3=∑j=1i−1gj,…,gi−1aj2/4Mi2,βi4=Bi,i−1+3∑j=1i−1B¯ij2/4p1,…,pi−1,Bi=li−1∑j=1i−1B¯ij2 are the constants.
Denoting(51)βi=βi1+βi2+βi3+βi4,Mi=pi−1Mi−1+Mi1+Mi2,we have(52)V˙i≤−2γp1,…,pi−1∑j=1ix^j2+γMiε12+γx^ix^i+1+γx^iαi−γ˙γpi−1δi−1∑j=1ix^j2+γ˙γNi−1x^i−12+γ˙γBi∑j=1i−1x^j2−γ˙γli4x^i2+γβi+2p1,…,pi−1x^i2.
Choosingpi−1′=pi−1δi−1−Ni−1−Bi>0,δi=minli/4,pi−1′ and using the formula(53)αi=−gix^i=−βi+n−i+1p1,…,pi−1x^i,we again obtain(54)V˙i≤−γn−i+1p1,…,pi−1∑j=1ix^j2+γMiε12+γx^ix^i+1−γ˙γδi∑j=1ix^j2.
Step n: for the last step, we chooseVn=pn−1Vn−1+1/2x^n2+1/2zn+12, where pn−1 is a designed positive constant, and by using (30), (32), and (43), we have(55)V˙n=pn−1V˙n−1+x^nz˙n−∑j=1n−1∂αn−1∂zjz˙j+zn+1γan+1ε1−γ˙γn+μzn+1−γczn+1≤−2γp1,…,pn−1∑j=1n−1x^j2+γpn−1Mn−1ε12+γx^nzn+1+γpn−1x^n−1x^n−γ˙γpn−1δn−1∑j=1n−1x^j2+γanx^nε1−γ˙γn−1+μx^nzn+x^nuγn−1+μ+zn+1γan+1ε1−γ˙γn+μzn+1−γczn+1−x^n∑j=1n−1∂αn−1∂zjz˙j.
This is similar to stepi, in that(56)γpn−1x^n−1x^n≤γ3p1,…,pn−1x^n−12+3γpn−14p1,…,pn−2x^n2=γ3pn−1x^n−12+γβn1x^n2γanx^nε1≤γMn1ε12+γan24Mn1x^n2=γMn1ε12+γβn2x^n2−γ˙γn−1+μx^nzn=−γ˙γn−1+μx^nx^n−gn−1x^n−1=−γ˙γn−1+μx^n2+γ˙γn−1+μx^n⋅gn−1x^n−1≤−γ˙γn−1+μ2x^n2+γ˙γn−1+μ2gn−12x^n−12=−γ˙γln2x^n2+γ˙γNn−1ln,gn−1x^n−12,where βn1=3pn−1/4p1,…,pn−2,ln=n−1+μ,Nn−1ln,gn−1=ln/2gn−12,Mn1>0,βn2=an2/4Mn1 are the constants. According to the relations(57)αn−1=−gn−1x^n−1=−gn−1zn−1+gn−2x^n−2=−∑j=1n−1∏s=jn−1gszj=−∑j=1i−1∏s=jn−1gszjand z˙j=γzj+1+γajε1−γ˙/γj−1+μzj, we have(58)−x^n∑j=1n−1∂αn−1∂zjz˙j=−x^n∑j=1n−1∂αn−1∂zjγzj+1+γajε1−γ˙γj−1+μzjγzj+1−γ˙γljzj=γx^j+1−gjx^j−γ˙γljzj=γx^j+1−gjx^j−γ˙γljx^j−gj−1x^j−1≤γ|x^j+1|+|gj‖x^j|+γ˙γlj|x^j|+|gj−1‖x^j−1|≤Agj−1,gjγ|x^j+1|+|x^j|+γ˙γlj|x^j|+|x^j−1|.
Because∂αn−1/∂zj≤gj,…,gn−1, it is obtained(59)−x^n∑j=1n−1∂αn−1∂zjγajε1≤x^n∑j=1n−1gj,…gn−1γajε1≤γMn2ε12+γ∑j=1n−1gj,…,gn−1aj24Mn2x^n2=γMn2ε12+γβn3x^n2−x^n∑j=1n−1∂αn−1∂zjγajε1≤x^n∑j=1n−1gj,…,gn−1γajε1≤γMn2ε12+γ∑j=1n−1gj,…,gn−1aj24Mn2x^n2=γMn2ε12+γβn3x^n2−x^n∑j=1n−1∂αn−1∂zjγzj+1−γ˙γljzj≤x^n∑j=1n−1gj,…,gn−1Agj−1,gjγx^j+1+x^j+γ˙γljx^j+x^j−1.
DefiningBnj=gj,…,gn−1Agj−1,gj,(60)γx^n∑j=1n−1Bnjx^j+1+x^j=γx^n∑j=1n−1Bnjx^j+1+γx^n∑j=1n−1Bnjx^j≤γBn,n−1x^n2+γx^n∑j=1n−1Bn,j−1x^j+γx^n∑j=1n−1Bnjx^j=γBn,n−1x^n2+γx^n∑j=1n−1B¯n,jx^j≤γBn,n−1+3∑j=1n−1B¯nj24p1,…,pn−1x^n2+γ3p1,…,pn−1∑j=1n−1x^j2=γβn4x^n2+γ3p1,…,pn−1∑j=1n−1x^j2x^n∑j=1n−1Bnjγ˙γljx^j+x^j−1=γ˙γx^n∑j=1n−1Bnjljx^j+γ˙γx^n∑j=1n−1Bnjljx^j−1≤γ˙γx^n∑j=1n−1B˜njx^j≤γ˙γln−1∑j=1n−1B˜nj2x^j2+γ˙γln−14x^n2=γ˙γBn∑j=1n−1x^j2+γ˙γln−14x^n2,where B¯nj=Bnj+Bn,j−1,B˜nj=Bnjlj+Bn,j−1lj−1,Mn2>0,βn3=∑j=1n−1gj,…,gn−1aj2/4Mn2,βn4=Bn,n−1+3∑j=1n−1B¯nj2/4p1,…,pn−1,Bn=ln−1∑j=1n−1B¯nj2 are the constants.
Denoting(61)βn=βn1+βn2+βn3+βn4,Mn=pn−1Mn−1+Mn1+Mn2,we have(62)V˙n≤−γp1,…,pn−1∑j=1n−1x^j2+γMnε12+γx^nzn+1−γ˙γln4x^n2−γ˙γpn−1δn−1∑j=1n−1x^j2+γ˙γNn−1x^n−12+γ˙γBn∑j=1n−1x^j2+γβnx^n2+x^nuγn−1+μ+γan+1zn+1ε1−γ˙γn+μzn+12−γczn+12.
For simplicity, we definepn−1′=pn−1δn−1−Nn−1−Bn>0,δn=minln/4,pn−1′,n+μ, and then,(63)V˙n≤−γp1,…,pn−1∑j=1nx^j2+γMnε12+γx^nzn+1−γ˙γδn∑j=1nx^j2+γβn+p1,…,pn−1x^n2+x^nuγn−1+μ+γan+1zn+1ε1−γ˙γn+μzn+12−γczn+12γan+1zn+1ε1≤γMn+1ε12+γan+124Mn+1zn+12=γMn+1ε12+γβn+1zn+12,where Mn+1>0,βn+1=an+12/4Mn+1 are the constants. Now, let us choose the final control signal as(64)u=−γn+μβn+p1,…,pn−1x^n−γn+μzn+1=−γn+μβn+p1,…,pn−1x^n−ζn+1.
Finally, it is derived that(65)V˙n≤−γp1,…,pn−1∑j=1nx^j2+γMn+Mn+1ε12−γ˙γδn∑j=1nx^j2+zn+12+γ−c+βn+1zn+12.


### 3.2. Stability Analysis

ChoosingV=Vn+Vε, we can obtain that(66)V˙≤−γp1,…,pn−1∑j=1nx^j2+γMn+Mn+1ε12+γh¯2−γ˙γδn∑j=1nx^j2+zn+12+γ−c+βn+1zn+12−γd1εTε−γ˙γd2εTε+γL1ε2+γPc1zn+12+γ1c1Pε2+φ1x0ε2+φ2x0x^2.Select parameters(67)−c+βn+1+Pc1<−Ca,−d1+Mn+Mn+1+1c1P+L1<−Cb,where Ca>0,Cb>0 denote p1,…,pn−1=D,δn+1=minD,δn,Ca,Cb, such that(68)V˙≤−γD∑j=1nx^j2−γCazn+12−γCbε2−γ˙γδn∑j=1nx^j2+zn+12−γ˙γd2εTε+φ1x0ε2+φ2x0x^2+γh¯2≤−D0γx^2+γε2−γ˙γδn+1x^2−γ˙γδn+1ε2+φ1x0ε2+φ2x0x^2+γh¯2≤−D02γx^2+ε2+γh¯2−D0γ2+γ˙γδn+1D0−φ2x0D0x^2−D0γ2+γ˙γδn+1D0−φ1x0D0x^2.Denotingωx0>1/D0maxφ1x0,φ2x0 and γ˙=max−γ2/2+ωx0δn+1/D0,0, then(69)V˙≤−D02γx^2+ε2+γh¯2.Definingγh¯2≤D=p1,…,pn−1, we can obtain that(70)V˙≤−D02γx^2+ε2+D.x^,ε are bounded. Now, we shall prove by contradiction that γt is bounded. Assume γt is unbounded in t0,tf. We notice that γ˙t≥0, so limt⟶tfγt=+∞, and thus, there exists a finite time T1∈t0,tf, such that ∀t∈T1,tf, and we have(71)γ˙t=−γ22+ωx0δn+1D0≤−D022+ωx0δn+1D0,0≤−D022+ωx0δn+1D0.Integrating both sides of the top equation, whenγ˙=−γ2/2+ωx0δn+1/D0, we can obtain(72)∫T1tf−γ22+ωx0δn+1D0dτ=∫T1tfγ˙τdτ=γtf−γT1=+∞∫T1tf−γ22+ωx0δn+1D0dτ≤∫T1tf−D022+ωx0δn+1D0dτ.Sincex0 is bounded in t0,tf, ωx0 is bounded, ∫T1tf−γ2/2+ωx0δn+1/D0dτ<+∞, which is a contradiction. Thus, γt is bounded in t0,tf. When γ˙=0, one has(73)∫T1tfγ˙τdτ=γtf−γT1=+∞,∫T1tfγ˙τdτ=0.These two equations contradict each other. Thus,γt is bounded in t0,tf. Integrating V˙≤−D0/2γx^2+ε2+γh¯2, we have(74)Vt−V0≤−D02∫0tγx^2+ε2dτ+∫0tγh¯2dτ,namely,(75)Vt≤V0−D02∫0tγx^2+ε2dτ+∫0tγh¯2dτ.Becauseh¯t=w˙t/γn+μ∈L2,∫0tγh¯τ2dτ<+∞,∀t>0. Because ε,x^∈L2, it can be obtained from (19) and (20) that x^˙,ε. is bounded. According to Barbalat’s lemma, limt⟶∞ε=0, and limt⟶∞x^=0.


## 3.1. Output Feedback Controller Design

Lemma 3 (see [33]).
For the first subsystem of (1), if the first control law u0 is chosen as(6)u0=−λ0x0−x0α0x0,λ0>0,where t0,x0t0 is regarded as the initial condition, x0t0≠0, then as the corresponding solution, x0t,t0,x0t0 exists and |x0t,t0,x0t0|>0, 0≤t0<t. λ0 is designed as a positive constant parameter. Furthermore, u0t>0.Proof.
Substituting (6) into the first formula of system (1), we can obtain(7)x0.=−λ0x0−x0α0x0+x0ϕ0dt,x0.
Integrating this nonlinear equation, the solution isx0t=x00e−∫0tλ0+α0x0τ−ϕ0dτ,x0τdτ. This shows that x0t≠0 at any time if x00≠0, and thus, u0t≠0. Choosing V0x0=1/2x02, it can be obtained from (6) that(8)V0.=−λ0x02−x02α0x0−ϕ0dt,x0≤−λ0x02≤0.
Thus,x0t asymptotically approaches zero. For any bounded x0t because α0,ϕ0d are smooth functions, there is a positive constant M for x0≤1, α0≤M,ϕ0d≤M. This yields that(9)V0.=−λ0x02−x02α0x0−ϕ0dt,x0≥−λ0+M+MV0.
Integrating both sides of this equation, it can be obtained that(10)V0t≥V00e−λ0+2Mt.
This means thatx0t converges to zero, but x0t≠0 at any given moment, so that u0t>0.
We introduce the following input state scaling:(11)ζi=xiu0n−i,i=1,…,n.
Unknown nonvanishing external disturbancewt is treated as a generalized state. To realize symbol consistency, it is defined as(12)ζn+1=wt.
In the new stateζ, system (1) is converted to(13)ζ˙i=ζi+1+ϕ¯it,u0,x0,ζ,ω0t,ζ˙n=ζn+1+ϕ¯nt,u0,x0,ζ,ω0t+u,ζ˙n+1=w˙t≜ht,i=1,…,n−1.Lemma 4.
For any givenu0 in (6), there is a known nonnegative smooth function φ¯0x0, such that u˙0/u0≤φ¯0x0, t≥0.Proof.
The following calculation is completed:(14)u˙0u0=−x0λ0+α0x0−ϕ0dt,x0λ0+α0x0∂α0x0∂x0−λ0−α0x0+ϕ0dt,x0≤λ01+λ0x0λ0+α0x0∂α0x0∂x0≜φ¯0x0.
This completes the proof of the lemma.
We know from Assumption1 that there is a nonnegative smooth function α¯x0:(15)ϕ¯it,u0,x0,ζ,ω0t≤αx0∑j=1ixju0n−i+n−iα¯0x0ζi≤n−iα¯0x0ζi+αx0u0i−1ζ1|+|u0i−2ζ2|+⋯+|ζi≤α¯x0∑j=1iζi.
Considering thatζ1,…,ζn+1 are unmeasurable signals that cannot be used in feedback control, the dynamic observer for (13) is denoted as follows:(16)ζ^˙i=ζ^i+1+aiγiζ1−ζ^1,i=1,…,n−1,ζ^˙n=u+ζ^n+1+anγnζ1−ζ^1,ζ^˙n+1=an+1γn+1ζ1−ζ^1−γcζ^n+1,where γ is the dynamic gain, which will be designed later according to the requirements. The observer error dynamics is defined as(17)ei=ζi−ζ^ii=1,…,n+1.
We can determine from (13), (16), and (17) that(18)e˙i=ei+1+ϕ¯it,u0,x0,ζ,ω0t−aiγiζ1−ζ^1,e˙n=en+1+ϕ¯nt,u0,x0,ζ,ω0t−anγnζ1−ζ^1,e˙n+1=ht−an+1γn+1ζ1−ζ^1+γcζ^n+1.
Introducing dynamic gain scaling,(19)εi=eiγi−1+μ,zi=ζ^iγi−1+μ,i=1,…,n+1,and defining ε=ε1,…,εn+1T,z=z1,…,zn+1T,Φ⋅=ϕ¯1⋅/γμ,…,ϕ¯n⋅/γn−1+μ,0T, we arrive at(20)ε.i=γεi+1+Φit,u0,x0,ζ,ω0t−γaiε1−γ˙γi−1+μεi,ε.n=γεn+1+Φnt,u0,x0,ζ,ω0t−γanε1−γ˙γn−1+μεn,ε.n+1=htγn+μ−γan+1ε1−γ˙γn+μεn+1+γczn+1,(21)z˙i=γzi+1+γaiε1−γ˙γi−1+μzi,z˙n=uγn−1+μ+γzn+1+γanε1−γ˙γn−1+μzn,z˙n+1=γan+1ε1−γ˙γn+μzn+1−γczn+1.
Denotinga=a1,…,an+1T, b=0,…,11×n+1T, F=0,…,γczn+11×n+1T, and h¯t=ht/γn+μ, we have(22)ε.=γAε+Φt,u0,x0,ζ,ω0t−γ˙γDε+bh¯t+F.
Now, using Assumption1 and (15), it follows that(23)Φit,u0,x0,ζ,ω0t≤α¯x0γi−1+μ∑j=1iζj.
ChoosingVε=εTPε, it is then obtained that(24)Vε.=ε.TPε+εTPε.=γAε+Φ−γ˙γDε+bh¯t+FTPε+εTPΦ+γAε−γ˙γDε+bh¯t+F=γεTATP+PAε+2εTPΦ−γ˙γεTDP+PDε+2εTPbh¯t+2εTPF≤2εTPΦ−γd1εTε+2εTPbh¯t+2εTPF−γ˙γd2εTε.
Introducing the following transformation,(25)x^1=z1,x^i=zi−αi−1,αi−1=−gi−1x^i−1,i=2,…,n,x^n+1=zn+1,where gi−1>0 is a constant number that will be given later, because εi=ei/γi−1+μ, and ei=ζi−ζ^i, we have(26)ζi=γi−1+μεi+zi=γi−1+μεi+x^i−gi−1x^i−1.
Now, using Lemma1, it follows that inequality,(27)2εTPΦ≤2εPα¯x0ε+z=2α¯x0Pε2+2α¯x0Pεz≤φ¯1x0ε2+φ¯2x0z2≤φ1x0ε2+φ2x0x^2,holds, where φ1x0 and φ2x0 are the nonnegative smooth functions. On the other hand, by using Young’s inequality, one has(28)2εTPbh¯≤γL1ε2+γh¯2,2εTPF≤γPc1zn+12+γ1c1Pε2.
Correspondingly, we can obtain that(29)Vε.≤−γd1εTε−γ˙γd2εTε+γL1ε2+γPc1zn+12+γ1c1Pε2+φ1x0ε2+φ2x0x^2+γh¯2.Step 1. ChoosingV1=1/2z12=1/2x^12, we have(30)V˙1=x^1γz2+γa1ε1−γ˙γμz1=γx^1z2−α1+α1+γa1x^1ε1−γ˙γμx^1z1=γx^1x^2+γx^1α1+γa1x^1ε1−γ˙γμx^12.By using Young’s inequality, we have(31)γa1x^1ε1≤γM1ε12+γa124M1x^12=γM1ε12+γβ1x^12,WhereM>0,β1=a12/4M1 are the constants. Choosing α1=−g1x^1=−n+β1x^1,δ1=μ, we then obtain(32)V˙1≤−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12.Step 2. ChoosingV2=p1V1+1/2x^22, where p1 is a designed positive constant,(33)V˙2≤p1−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12+x^2z˙2−α˙1=p1−γnx^12+γM1ε12+γx^1x^2−γ˙γδ1x^12+x^2γz3+γa2ε1−γ˙γ1+μz2−x^2∂α1∂z1z˙1=−γp1nx^12+γp1M1ε12+γp1x^1x^2−γ˙γp1δ1x^12+γx^2x^3+γx^2α2+γa2x^2ε1−γ˙γ1+μx^2z2−x^2∂α1∂z1z˙1.
From Lemma1, we can derive the following inequalities:(34)γp1x^1x^2≤γ3p1x^12+3γp14x^22=γ3p1x^12+γβ21x^22γa2x^2ε1≤γθ21ε12+γa224θ21x^22=γθ21ε12+γβ22x^22,−γ˙γ1+μx^2z2=−γ˙γ1+μx^2x^2−g1x^1=−γ˙γ1+μx^22+γ˙γ1+μx^2⋅g1x^1≤−γ˙γ1+μ2x^22+γ˙γ1+μ2g12x^12=−γ˙γ1+μ2x^22+γ˙γN1l2,g1x^12,where β21=3p1/4,l1=μ,l2=1+μ,N1l2,g1=l2/2g12,θ21>0,β22=a22/4θ21 are the constants. Using the relations α1=−g1x^1 and z˙1=γz2+γa1ε1−γ˙/γμz1, we have(35)−x^2∂α1∂z1z˙1=−x^2∂α1∂z1γz2+γa1ε1−γ˙γμz1.
Definingx^0=0,g0=1, it is derived that(36)γz2−γ˙γl1z1=γx^2−g1x^1−γ˙γl1x^1=γx^2−g1x^1−γ˙γl1x^1−g0x^0≤γx^2+g1x^1|+γ˙γl1x^1+g0x^0|≤Ag0,g1γx^2+x^1+γ˙γl1x^1+x^0,where Ag0,g1=max1,g0,g1. Since ∂α1/∂z1≤g1, this implies that(37)−x^2∂α1∂z1γa1ε1≤x^2g1γa1ε1≤γθ22ε12+γg12a124θ22x^22=γθ22ε12+γβ23x^22,−x^2∂α1∂z1γz2−γ˙γl1z1≤x^2g1Ag0,g1γx^2+x^1+γ˙γl1x^1+x^0.
LettingB21=g1Ag0,g1, it can be concluded that(38)γx^2g1Ag0,g1x^2+x^1=γB21x^22+γB21x^2x^1≤γ3p1x^12+γB21+3B2124p1x^22=γ3p1x^12+γβ24x^22,x^2B21γ˙γl1x^1+x^0=x^2B21γ˙γl1x^1≤γ˙γB212l1x^12+γ˙γl14x^22=γ˙γB2x^12+γ˙γl14x^22,where θ22>0,β23=g12a12/4θ22,β24=B21+3B212/4p1,B2=B212l1 are the constants. Furthermore, we have(39)V˙2≤−γp1nx^12−γ˙γp1δ1x^12+γ3p1x^12+γ˙γN1l2,g1x^12+γ3p1x^12+γ˙γB2x^12+γp1M1ε12+γM21ε12+γM22ε12+γβ21x^22+γβ22x^22+γβ23x^22+γβ24x^22−γ˙γ1+μ2x^22+γ˙γl14x^22+γx^2x^3+γx^2α2.
Let us denote(40)β2=β21+β22+β23+β24,M2=p1M1+θ21+θ22,thus, it is obtained that(41)V˙2≤−γn−1p1∑j=12x^j2+γM2ε12+γ˙γN1l2,g1+B2x^12+γx^2x^3+γx^2α2−γ˙γp1δ1x^12−γ˙γl24x^22+γβ2+n−1p1x^22.
For simplicity, letp1′=p1δ1−N1−B2>0 and δ2=minl2/4,p1′,α2=−g2x^2=−β2+n−1p1x^2; then,(42)V˙2≤−γn−1p1∑j=12x^j2+γM2ε12+γx^2x^3−γ˙γδ2∑j=12x^j2.
Step i2<i≤n−1. Assume that in Step i−1, we have(43)V˙i−1≤−γn−i+2p1,…,pi−2∑j=1i−1x^j2+γMi−1ε12+γx^i−1x^i−γ˙γδi−1∑j=1i−1x^j2.
Letting theith candidate Lyapunov function be Vi=pi−1Vi−1+1/2x^i2 and defining x^i=zi−αi−1, where p2,…,pi−2 are the designed positive constants,(44)V˙i=pi−1V˙i−1+x^iz˙i−∑j=1i−1∂αi−1∂zjz˙j=pi−1V˙i−1+x^iγzi+1+γaiε1−γ˙γi−1+μzi−x^i∑j=1i−1∂αi−1∂zjz˙j≤−γn−i+2p1,…,pi−1∑j=1i−1x^j2+γpi−1Mi−1ε12−γ˙γpi−1δi−1∑j=1i−1x^j2+γpi−1x^i−1x^i+γx^izi+1+γaix^iε1−x^i∑j=1i−1∂αi−1∂zjz˙j−γ˙γi−1+μx^izi.
By applying Young’s inequality, one has(45)γpi−1x^i−1x^i≤γ3p1,…,pi−1x^i−12+3γpi−14p1,…,pi−2x^22=γ3pi−1x^i−12+γβi1x^i2γaix^iε1≤γMi1ε12+γai24Mi1x^i2=γMi1ε12+γβi2x^i2−γ˙γi−1+μx^izi=−γ˙γi−1+μx^ix^i−gi−1x^i−1=−γ˙γi−1+μx^i2+γ˙γi−1+μx^i⋅gi−1x^i−1≤−γ˙γi−1+μ2x^i2+γ˙γi−1+μ2gi−12x^i−12=−γ˙γli2x^i2+γ˙γNi−1li,gi−1x^i−12,
whereβi1=3pi−1/4p1,…,pi−2,li=i−1+μ,Ni−1li,gi−1=li/2gi−12,Mi1>0,βi2=ai2/4Mi1 are the constants. Using the relations(46)αi−1=−gi−1x^i−1=−gi−1zi−1+gi−2x^i−2=−∑j=1i−1∏s=ji−1gszj=−∑j=1i−1∏s=ji−1gszj,and z˙j=γzj+1+γajε1−γ˙/γj−1+μzj, it follows that(47)−x^i∑j=1i−1∂αi−1∂zjz˙j=−x^i∑j=1i−1∂αi−1∂zjγzj+1+γajε1−γ˙γj−1+μzj,(48)γzj+1−γ˙γljzj=γx^j+1−gjx^j−γ˙γljzj=γx^j+1−gjx^j−γ˙γljx^j−gj−1x^j−1≤γx^j+1+gjx^j+γ˙γljx^j+gj−1x^j−1≤Agj−1,gjγx^j+1+x^j+γ˙γljx^j+x^j.
Since∂αi−1/∂zj≤gj,…,gi−1, this implies that(49)−x^i∑j=1i−1∂αi−1∂zjγajε1≤x^i∑j=1i−1gj,…,gi−1γajε1≤γMi2ε12+γ∑j=1i−1gj,…,gi−1aj24Mi2x^i2=γMi2ε12+γβi3x^i2−x^i∑j=1i−1∂αi−1∂zjγzj+1−γ˙γljzj≤x^i∑j=1i−1gj,…,gi−1Agj−1,gjγx^j+x^j+1+γ˙γljx^j+x^j+1.
DefiningBij=gj,…,gi−1Agj−1,gj,(50)γx^i∑j=1i−1Bijx^j+1+x^j=γx^i∑j=1i−1Bijx^j+1+γx^i∑j=1i−1Bijx^j=γBi,i−1x^i2+γx^i∑j=2i−1Bi,j−1x^j+γx^i∑j=1i−1Bijx^j≤γBi,i−1x^i2+γx^i∑j=1i−1Bi,j−1x^j+γx^i∑j=1i−1Bijx^j=γBi,i−1x^i2+γx^i∑j=1i−1B¯i,jx^j≤γBi,i−1+3∑j=1i−1B¯ij24p1,…,pi−1x^i2+γ3p1,…,pi−1∑j=1i−1x^j2=γβi4x^i2+γ3p1,…,pi−1∑j=1i−1x^j2,x^i∑j=1i−1Bijγ˙γljx^j+x^j−1=γ˙γx^i∑j=1i−1Bijljx^j+γ˙γx^i∑j=1i−1Bijljx^j−1≤γ˙γx^i∑j=1i−1B˜ijx^j≤γ˙γli−1∑j=1i−1B˜ij2x^j2+γ˙γli−14x^i2=γ˙γBi∑j=1i−1x^j2+γ˙γli−14x^i2,
whereB¯ij=Bij+Bi,j−1,B˜ij=Bijlj+Bi,j−1lj−1,Mi2>0,βi3=∑j=1i−1gj,…,gi−1aj2/4Mi2,βi4=Bi,i−1+3∑j=1i−1B¯ij2/4p1,…,pi−1,Bi=li−1∑j=1i−1B¯ij2 are the constants.
Denoting(51)βi=βi1+βi2+βi3+βi4,Mi=pi−1Mi−1+Mi1+Mi2,we have(52)V˙i≤−2γp1,…,pi−1∑j=1ix^j2+γMiε12+γx^ix^i+1+γx^iαi−γ˙γpi−1δi−1∑j=1ix^j2+γ˙γNi−1x^i−12+γ˙γBi∑j=1i−1x^j2−γ˙γli4x^i2+γβi+2p1,…,pi−1x^i2.
Choosingpi−1′=pi−1δi−1−Ni−1−Bi>0,δi=minli/4,pi−1′ and using the formula(53)αi=−gix^i=−βi+n−i+1p1,…,pi−1x^i,we again obtain(54)V˙i≤−γn−i+1p1,…,pi−1∑j=1ix^j2+γMiε12+γx^ix^i+1−γ˙γδi∑j=1ix^j2.
Step n: for the last step, we chooseVn=pn−1Vn−1+1/2x^n2+1/2zn+12, where pn−1 is a designed positive constant, and by using (30), (32), and (43), we have(55)V˙n=pn−1V˙n−1+x^nz˙n−∑j=1n−1∂αn−1∂zjz˙j+zn+1γan+1ε1−γ˙γn+μzn+1−γczn+1≤−2γp1,…,pn−1∑j=1n−1x^j2+γpn−1Mn−1ε12+γx^nzn+1+γpn−1x^n−1x^n−γ˙γpn−1δn−1∑j=1n−1x^j2+γanx^nε1−γ˙γn−1+μx^nzn+x^nuγn−1+μ+zn+1γan+1ε1−γ˙γn+μzn+1−γczn+1−x^n∑j=1n−1∂αn−1∂zjz˙j.
This is similar to stepi, in that(56)γpn−1x^n−1x^n≤γ3p1,…,pn−1x^n−12+3γpn−14p1,…,pn−2x^n2=γ3pn−1x^n−12+γβn1x^n2γanx^nε1≤γMn1ε12+γan24Mn1x^n2=γMn1ε12+γβn2x^n2−γ˙γn−1+μx^nzn=−γ˙γn−1+μx^nx^n−gn−1x^n−1=−γ˙γn−1+μx^n2+γ˙γn−1+μx^n⋅gn−1x^n−1≤−γ˙γn−1+μ2x^n2+γ˙γn−1+μ2gn−12x^n−12=−γ˙γln2x^n2+γ˙γNn−1ln,gn−1x^n−12,where βn1=3pn−1/4p1,…,pn−2,ln=n−1+μ,Nn−1ln,gn−1=ln/2gn−12,Mn1>0,βn2=an2/4Mn1 are the constants. According to the relations(57)αn−1=−gn−1x^n−1=−gn−1zn−1+gn−2x^n−2=−∑j=1n−1∏s=jn−1gszj=−∑j=1i−1∏s=jn−1gszjand z˙j=γzj+1+γajε1−γ˙/γj−1+μzj, we have(58)−x^n∑j=1n−1∂αn−1∂zjz˙j=−x^n∑j=1n−1∂αn−1∂zjγzj+1+γajε1−γ˙γj−1+μzjγzj+1−γ˙γljzj=γx^j+1−gjx^j−γ˙γljzj=γx^j+1−gjx^j−γ˙γljx^j−gj−1x^j−1≤γ|x^j+1|+|gj‖x^j|+γ˙γlj|x^j|+|gj−1‖x^j−1|≤Agj−1,gjγ|x^j+1|+|x^j|+γ˙γlj|x^j|+|x^j−1|.
Because∂αn−1/∂zj≤gj,…,gn−1, it is obtained(59)−x^n∑j=1n−1∂αn−1∂zjγajε1≤x^n∑j=1n−1gj,…gn−1γajε1≤γMn2ε12+γ∑j=1n−1gj,…,gn−1aj24Mn2x^n2=γMn2ε12+γβn3x^n2−x^n∑j=1n−1∂αn−1∂zjγajε1≤x^n∑j=1n−1gj,…,gn−1γajε1≤γMn2ε12+γ∑j=1n−1gj,…,gn−1aj24Mn2x^n2=γMn2ε12+γβn3x^n2−x^n∑j=1n−1∂αn−1∂zjγzj+1−γ˙γljzj≤x^n∑j=1n−1gj,…,gn−1Agj−1,gjγx^j+1+x^j+γ˙γljx^j+x^j−1.
DefiningBnj=gj,…,gn−1Agj−1,gj,(60)γx^n∑j=1n−1Bnjx^j+1+x^j=γx^n∑j=1n−1Bnjx^j+1+γx^n∑j=1n−1Bnjx^j≤γBn,n−1x^n2+γx^n∑j=1n−1Bn,j−1x^j+γx^n∑j=1n−1Bnjx^j=γBn,n−1x^n2+γx^n∑j=1n−1B¯n,jx^j≤γBn,n−1+3∑j=1n−1B¯nj24p1,…,pn−1x^n2+γ3p1,…,pn−1∑j=1n−1x^j2=γβn4x^n2+γ3p1,…,pn−1∑j=1n−1x^j2x^n∑j=1n−1Bnjγ˙γljx^j+x^j−1=γ˙γx^n∑j=1n−1Bnjljx^j+γ˙γx^n∑j=1n−1Bnjljx^j−1≤γ˙γx^n∑j=1n−1B˜njx^j≤γ˙γln−1∑j=1n−1B˜nj2x^j2+γ˙γln−14x^n2=γ˙γBn∑j=1n−1x^j2+γ˙γln−14x^n2,where B¯nj=Bnj+Bn,j−1,B˜nj=Bnjlj+Bn,j−1lj−1,Mn2>0,βn3=∑j=1n−1gj,…,gn−1aj2/4Mn2,βn4=Bn,n−1+3∑j=1n−1B¯nj2/4p1,…,pn−1,Bn=ln−1∑j=1n−1B¯nj2 are the constants.
Denoting(61)βn=βn1+βn2+βn3+βn4,Mn=pn−1Mn−1+Mn1+Mn2,we have(62)V˙n≤−γp1,…,pn−1∑j=1n−1x^j2+γMnε12+γx^nzn+1−γ˙γln4x^n2−γ˙γpn−1δn−1∑j=1n−1x^j2+γ˙γNn−1x^n−12+γ˙γBn∑j=1n−1x^j2+γβnx^n2+x^nuγn−1+μ+γan+1zn+1ε1−γ˙γn+μzn+12−γczn+12.
For simplicity, we definepn−1′=pn−1δn−1−Nn−1−Bn>0,δn=minln/4,pn−1′,n+μ, and then,(63)V˙n≤−γp1,…,pn−1∑j=1nx^j2+γMnε12+γx^nzn+1−γ˙γδn∑j=1nx^j2+γβn+p1,…,pn−1x^n2+x^nuγn−1+μ+γan+1zn+1ε1−γ˙γn+μzn+12−γczn+12γan+1zn+1ε1≤γMn+1ε12+γan+124Mn+1zn+12=γMn+1ε12+γβn+1zn+12,where Mn+1>0,βn+1=an+12/4Mn+1 are the constants. Now, let us choose the final control signal as(64)u=−γn+μβn+p1,…,pn−1x^n−γn+μzn+1=−γn+μβn+p1,…,pn−1x^n−ζn+1.
Finally, it is derived that(65)V˙n≤−γp1,…,pn−1∑j=1nx^j2+γMn+Mn+1ε12−γ˙γδn∑j=1nx^j2+zn+12+γ−c+βn+1zn+12.


## 3.2. Stability Analysis

ChoosingV=Vn+Vε, we can obtain that(66)V˙≤−γp1,…,pn−1∑j=1nx^j2+γMn+Mn+1ε12+γh¯2−γ˙γδn∑j=1nx^j2+zn+12+γ−c+βn+1zn+12−γd1εTε−γ˙γd2εTε+γL1ε2+γPc1zn+12+γ1c1Pε2+φ1x0ε2+φ2x0x^2.Select parameters(67)−c+βn+1+Pc1<−Ca,−d1+Mn+Mn+1+1c1P+L1<−Cb,where Ca>0,Cb>0 denote p1,…,pn−1=D,δn+1=minD,δn,Ca,Cb, such that(68)V˙≤−γD∑j=1nx^j2−γCazn+12−γCbε2−γ˙γδn∑j=1nx^j2+zn+12−γ˙γd2εTε+φ1x0ε2+φ2x0x^2+γh¯2≤−D0γx^2+γε2−γ˙γδn+1x^2−γ˙γδn+1ε2+φ1x0ε2+φ2x0x^2+γh¯2≤−D02γx^2+ε2+γh¯2−D0γ2+γ˙γδn+1D0−φ2x0D0x^2−D0γ2+γ˙γδn+1D0−φ1x0D0x^2.Denotingωx0>1/D0maxφ1x0,φ2x0 and γ˙=max−γ2/2+ωx0δn+1/D0,0, then(69)V˙≤−D02γx^2+ε2+γh¯2.Definingγh¯2≤D=p1,…,pn−1, we can obtain that(70)V˙≤−D02γx^2+ε2+D.x^,ε are bounded. Now, we shall prove by contradiction that γt is bounded. Assume γt is unbounded in t0,tf. We notice that γ˙t≥0, so limt⟶tfγt=+∞, and thus, there exists a finite time T1∈t0,tf, such that ∀t∈T1,tf, and we have(71)γ˙t=−γ22+ωx0δn+1D0≤−D022+ωx0δn+1D0,0≤−D022+ωx0δn+1D0.Integrating both sides of the top equation, whenγ˙=−γ2/2+ωx0δn+1/D0, we can obtain(72)∫T1tf−γ22+ωx0δn+1D0dτ=∫T1tfγ˙τdτ=γtf−γT1=+∞∫T1tf−γ22+ωx0δn+1D0dτ≤∫T1tf−D022+ωx0δn+1D0dτ.Sincex0 is bounded in t0,tf, ωx0 is bounded, ∫T1tf−γ2/2+ωx0δn+1/D0dτ<+∞, which is a contradiction. Thus, γt is bounded in t0,tf. When γ˙=0, one has(73)∫T1tfγ˙τdτ=γtf−γT1=+∞,∫T1tfγ˙τdτ=0.These two equations contradict each other. Thus,γt is bounded in t0,tf. Integrating V˙≤−D0/2γx^2+ε2+γh¯2, we have(74)Vt−V0≤−D02∫0tγx^2+ε2dτ+∫0tγh¯2dτ,namely,(75)Vt≤V0−D02∫0tγx^2+ε2dτ+∫0tγh¯2dτ.Becauseh¯t=w˙t/γn+μ∈L2,∫0tγh¯τ2dτ<+∞,∀t>0. Because ε,x^∈L2, it can be obtained from (19) and (20) that x^˙,ε. is bounded. According to Barbalat’s lemma, limt⟶∞ε=0, and limt⟶∞x^=0.


## 4. Simulation Results

In this section, we consider a simulation example to prove that the controller design in this study is effective. Consider the following three-dimensional uncertain nonholonomic system with nonvanishing external disturbance:(76)x0.=u0+x02x0,x1.=x2u0+x02x1cos2u0,x2.=u+0.01x02x1u0+|x2|+sintt+1+1.This example shows that the nonlinear uncertainty in third equation is related to the unknown system statex2, so that the assumption condition in [23] is not satisfied, and thus, the given method in that of article is not available to deal with this model. However, this example satisfies the given Assumption 1. By designing the controller, execution simulation algorithm, and choosing parameters u0=−λ0x0−x0x02+1,u1=−2γ3x^2−γ3z3,λ0=μ=0.01,a1=a2=a3=2,g1=10,c=100,k=0.01, we have(77)ζ˙1=ζ2+x02cos2u0ζ1−91+x023+x02ζ1,ζ˙2=u1+0.01x02ζ1+ζ2+ζ3,ζ˙3=costt+1−sintt+12,ζ^˙1=ζ^2+2γζ1−ζ^,ζ^˙2=u1+2γ2ζ1−ζ^+ζ^3,ζ^˙3=2γ3ζ1−ζ^−100γζ^3,z˙1=γz2+2γ0.99ζ1−ζ^−0.01γ˙γz1,z˙2=γz3+2γ0.99ζ1−ζ^−1.01γ˙γz2+u1γ1.01,z˙3=2γ0.99ζ1−ζ^−2.01γ˙γz3−100γz3,x^˙1=z˙1,x^˙2=z˙2+10x^˙1,x^˙3=z^˙3,where the initial states are x0=1,x1=2,x2=7,ζ1=−1,ζ2=7,ζ3=0,ζ^1=1,ζ^2=1,ζ^3=5,z1=1,z2=10,z3=1,x^1=1,x^2=10,x^3=1, and γ=15, and dynamic gain is selected as γ˙=−γ2/2+20x02ζ1+ζ2. In Figure 1, the simulation results are shown. This study presents an output feedback control scheme that realizes stability control, and the control inputs u0 and u are bounded, as shown in Figure 2.Figure 1
State response curve of the system.Figure 2
Input response curve of the system.


## 5. Conclusion

This study solves the problems of output feedback control for one type of the nonholonomic system with nonvanishing external disturbances and nonlinear uncertainties for which the strong uncertainties are restricted by a generalized lower triangular linearly growing condition. The system is reconstructed by introducing a new extended state observer. The external disturbance is viewed as a general state. An adjustable varying gain scaling transformation and the extended state observer are used to carry out output feedback control and overcome the uncertainties and disturbances. The output of the system and states of the system go to zero, and all signals of the closed-loop system are guaranteed to be bounded. Simulation examples show that the control algorithm is effective. How to reduce the uncertainty and external disturbance assumptions of the model (1) and make the types of the models more extensive will be further considered.



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