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Cooperative Extended State Observer Based Control of Vehicle Platoons With Arbitrarily Small Time Headway

Anquan Liu [email protected]    Tao Li [email protected]    Yu Gu [email protected]    Haohui Dai [email protected] School of Mechatronic Engineering and Automation, Shanghai University, Shanghai, 200072, China. Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice, School of Mathematical Sciences, East China Normal University, Shanghai 200241, China.
Abstract

We study platoon control of vehicles with linear third-order longitudinal dynamics under the constant time headway policy. The controller of each follower vehicle is only based on its own velocity, acceleration, inter-vehicle distance and velocity difference with respect to its immediate predecessor, which are all obtained by on-board sensors. We develop distributed cooperative extended state observers for followers to estimate the acceleration differences between adjacent vehicles. Based on estimates of the acceleration differences, distributed cooperative control laws are designed. By using the stability theory of perturbed linear systems, we show that the control parameters can be properly designed to ensure the closed-loop and 2\mathcal{L}_{2} string stabilities for any given positive time headway. We further show that the proposed control law based on the ideal vehicle model can guarantee the closed-loop and 2\mathcal{L}_{2} string stabilities even if there are small model parameter uncertainties. Also, simulation results demonstrate the robustness of the proposed control law against sensing noises, input delays and parameter uncertainties.

keywords:
Vehicle platoon; Constant time headway; Extended state observer; String stability.
thanks: This paper was partially presented at the 21st IFAC World Congress, Berlin, Germany, July 12-17, 2020. Corresponding author: Tao Li. Tel. +86-21-54342646-318. Fax +86-21-54342609.

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1 Introduction

Vehicle platoon can improve road utilization rate and reduce fuel consumption effectively (Alam, 2011). Therefore, it has attracted worldwide attention (Shladover et al., 1991; Coelingh & Solyom, 2012). From networked control perspective, Li et al. (2015) divided a vehicle platoon system into four basic modules: node dynamics, information flow topology, formation geometry, and distributed controller, among which the formation geometry greatly affects the stability of the vehicle platoon system. Formation geometry is determined by the spacing policy. The constant time headway policy is considered by Klinge & Middleton (2009); Naus et al. (2010); Xiao & Gao (2011); Ploeg et al. (2014); Darbha et al. (2017) and it is well known that a large time headway is conducive to string stabilities of vehicle platoon system (Rajamani & Zhu, 2002; Naus et al., 2010), however, the larger the time headway is, the greater the inter-vehicle distance becomes, which leads to a lower road utilization rate.

It is of interest to guarantee the stability of the vehicle platoon with small time headways. Many researches showed that using the accelerations of the preceding vehicles can reduce the lower bound of the time headway required (Rajamani & Zhu, 2002; Zhou & Peng, 2004; Naus et al., 2010; Darbha et al., 2017; Al-Jhayyish & Schmidt, 2018). All the above works assumed that the accelerations of the preceding vehicles are obtained by the wireless communication network accurately, however, accurate communication doesn’t exist in practical applications. In addition, a control law which relies on communication data runs the risk of failure when the inter-vehicle communication network breaks down under attack. Therefore, a cooperative control law with a small time headway, which can ensure both closed-loop and string stabilities without relying on the inter-vehicle wireless communication network, is of especially significance for practical applications. Ploeg et al. (2015) and Wen & Guo (2019) proposed methods to estimate the acceleration differences between adjacent vehicles, respectively. The control laws in Ploeg et al. (2015) and Wen & Guo (2019) are indeed independent of wireless communication networks. The closed-loop and string stabilities are analyzed by numerical simulations in Ploeg et al. (2015) and Wen & Guo (2019), especially, Ploeg et al. (2015) considered a third-order linear model with input delays, in which the simulation results show that the string stability is not guaranteed if the time headway is small enough.

In this paper, we study platoon control of vehicles with linear third-order longitudinal dynamics under the constant time headway policy. We design a distributed cooperative control law for each follower vehicle only using its own velocity, acceleration, inter-vehicle distance and velocity difference with respect to its immediate predecessor, all of which can be obtained by on-board sensors. Firstly, the models of velocity difference between adjacent vehicles are established, based on which distributed cooperative extended state observers are designed to estimate the acceleration differences between adjacent vehicles. Then based on these estimates, distributed cooperative controllers are designed for follower vehicles. The controller for each follower consists of two parts, where the first part is a feedback term consisting of inter-vehicle distance error and its differential, the other is a feedforward term consisting of the estimate of the acceleration of the preceding vehicle. Thus, the whole cooperative control law only uses the data obtained by on-board sensors without wireless communication networks.

We analyze both closed-loop and 2\mathcal{L}_{2} string stabilities of the vehicle platoon system. The closed-loop system matrix is decomposed into two matrices, one of which is related to feedback parameters of the distributed controllers, and the other is regarded as the perturbation matrix, which is related to feedforward parameters. Then, we give the range of the control parameters to ensure the closed-loop stability by using the stability theory of perturbed linear systems. In the frequency domain, we analyze the transfer functions which describe the inter-vehicle distance error propagation, and give the range of the control parameters to ensure the 2\mathcal{L}_{2} string stability. We show that one can design control parameters properly to ensure both closed-loop and 2\mathcal{L}_{2} string stabilities for any given positive time headway. Our method for estimating the acceleration differences between adjacent vehicles is based on distributed cooperative extended state observers, which is totally different from those in Ploeg et al. (2015) and Wen & Guo (2019). Besides, we give the explicit range of control parameters quantitatively related to the system parameters to ensure both closed-loop and 2\mathcal{L}_{2} string stabilities for any given positive time headway.

We analyze the robustness of the proposed control law by both theoretical study and numerical simulations. (i) The proposed control law guarantees the exponential closed-loop stability, thus it is naturally robust against bounded sensing noises. (ii) We show that the proposed cooperative control law based on the ideal vehicle model can still ensure the closed-loop and 2\mathcal{L}_{2} string stabilities provided the parameter uncertainties in the vehicle model are sufficiently small. (iii) In the numerical simulations, we consider the same vehicle model with input delays and the same time headway as Ploeg et al. (2015), with additional sensing noises and uncertain model parameters. Simulation results show that the closed-loop and 2\mathcal{L}_{2} string stabilities can be guaranteed by the proposed control law based on the ideal vehicle model.

The rest of this paper is organized as follows. The vehicle model and the control objectives are presented in Section II. In Section III, we first establish dynamic models based on velocity differences between every adjacent vehicles, then distributed cooperative extended state observers are designed to estimate the acceleration differences between adjacent vehicles. Finally, distributed cooperative controllers for follower vehicles are designed. In Section IV, we give the range of control parameters for the closed-loop and 2\mathcal{L}_{2} string stabilities. In Section V, we analyze the robustness of the proposed control law against model parameter uncertainties. Numerical simulations are carried out in Section VI. In Section VII, we give some conclusions.

The following notation will be used throughout this paper. For a given matrix AA, its 2-norm and minimum singular value are denoted by A\|A\| and Sn(A)S_{n}(A), respectively; diag(A)diag(A) denotes a block diagonal matrix whose diagonal blocks are all matrix AA; \mathbb{C} denotes the complex domain; \mathbb{R} denotes the real domain; OO and II denote a zero matrix and an identity matrix with an appropriate size, respectively. For a given transfer function G(s)G(s), its \infty-norm and inverse Laplace transform are denoted by G(s)\|G(s)\|_{\infty} and 1[G(s)]\mathscr{L}^{-1}[G(s)], respectively.

2 Problem formulation

We consider the following third-order linear vehicle model which is commonly used in the vehicle platoon control (Rajamani & Zhu, 2002; Ploeg et al., 2015; Zheng et al., 2016; Wen & Guo, 2019).

{p˙i(t)=vi(t),v˙i(t)=ai(t),i=0,1,,N,a˙i(t)=ai(t)/τ+ui(t)/τ,\displaystyle\left\{\begin{aligned} &\dot{p}_{i}(t)=v_{i}(t),\vspace{1ex}\\ &\dot{v}_{i}(t)=a_{i}(t),\vspace{1ex}\hskip 56.9055pti=0,1,...,N,\\ &\dot{a}_{i}(t)=-a_{i}(t)/\tau+u_{i}(t)/\tau,\end{aligned}\right. (1)

where pi(t)p_{i}(t), vi(t)v_{i}(t) and ai(t)a_{i}(t) are the position, velocity and acceleration of the iith vehicle, respectively, u0(t)u_{0}(t) is the control input of the leader vehicle and ui(t)u_{i}(t) is the control input of the iith follower vehicle to be designed, i=1,2,,Ni=1,2,...,N. The constant τ\tau is the inertial delay of vehicle longitudinal dynamics. We consider the constant time headway spacing policy. The expected inter-vehicle distance is denoted by

dr,i(t)=r+hvi(t),i=1,2,,N,\displaystyle d_{r,i}(t)=r+hv_{i}(t),\;i=1,2,...,N, (2)

where the constants rr and hh are the standstill distance and the time headway, respectively. The inter-vehicle distance error is given by,

ei(t)=pi1(t)pi(t)dr,i(t),i=1,2,,N,\displaystyle e_{i}(t)=p_{i-1}(t)-p_{i}(t)-d_{r,i}(t),\;i=1,2,...,N, (3)

which is the difference between the actual inter-vehicle distance and the expected inter-vehicle distance. Denote Gei(s)=i(s)/i1(s)G_{ei}(s)=\mathscr{E}_{i}(s)/\mathscr{E}_{i-1}(s), where i(s)\mathscr{E}_{i}(s) is the Laplace transform of ei(t)e_{i}(t). The control objectives are to design ui(t)u_{i}(t), i=1,2,,Ni=1,2,\ldots,N, for follower vehicles so that the following two objectives are satisfied.

A. closed-loop stability: all the follower vehicles tend to move at the same velocity as the leader vehicle and the inter-vehicle distance errors converge to zero, i.e. limt[vi(t)v0(t)]=0\lim\limits_{t\to\infty}[v_{i}(t)-v_{0}(t)]=0, limtei(t)=0,i=1,2,,N\lim\limits_{t\to\infty}e_{i}(t)=0,i=1,2,\ldots,N.

B. 2\mathcal{L}_{2} string stability: the inter-vehicle distance errors are not amplified during the backward propagation along the platoon, i.e. Gei(s)1,i=2,,N.\|G_{ei}(s)\|_{\infty}\leq 1,\;i=2,...,N.

3 Cooperative control law based on extended state observer

Suppose that there is no wireless communication between vehicles. We consider predecessor following topology, and each follower vehicle in the platoon relies on the on-board sensors to measure its own velocity, acceleration, the inter-vehicle distance and the velocity difference with respect to its immediate predecessor. However, the sensor of each follower vehicle cannot measure the acceleration of its immediate predecessor. Instead, we can design an observer to estimate the acceleration difference between adjacent vehicles. The extended state observer (ESO) was first put forward by Han (1995), whose core idea is to expand the unmodeled dynamics into new state and then according to the new state equation, an extended state observer is designed to estimate all states of the system. However, the extended state observer proposed by Han (1995) is nonlinear, which has difficulty in parameter tuning and stability analysis. Gao (2003) put forward a linear extended state observer, which simplifies parameter tuning and is also beneficial for stability analysis. In this paper, we design a linear extended state observer to estimate the acceleration difference between adjacent vehicles.

According to (1), the models of the velocity difference between adjacent vehicles are given by

{v˙d,i(t)=ad,i(t),a˙d,i(t)=qi(t)+ai(t)/τui(t)/τ,q˙i(t)=wi(t),i=1,2,,N,\displaystyle\left\{\begin{aligned} \dot{v}_{d,i}(t)&=a_{d,i}(t),\vspace{1ex}\\ \dot{a}_{d,i}(t)&=q_{i}(t)+a_{i}(t)/\tau-u_{i}(t)/\tau,\vspace{1ex}\\ \dot{q}_{i}(t)&=w_{i}(t),\ i=1,2,...,N,\end{aligned}\right. (4)

where

vd,i(t)=\displaystyle v_{d,i}(t)= vi1(t)vi(t),\displaystyle v_{i-1}(t)-v_{i}(t),\vspace{1ex} (5)
ad,i(t)=\displaystyle a_{d,i}(t)= ai1(t)ai(t),\displaystyle a_{i-1}(t)-a_{i}(t),\vspace{1ex} (6)
qi(t)=\displaystyle q_{i}(t)= (ai1(t)+ui1(t))/τ,\displaystyle(-a_{i-1}(t)+u_{i-1}(t))/\tau,\vspace{1ex} (7)
wi(t)=\displaystyle w_{i}(t)= (ai1(t)ui1(t)+τu˙i1(t))/τ2.\displaystyle(a_{i-1}(t)-u_{i-1}(t)+\tau\dot{u}_{i-1}(t))/\tau^{2}. (8)

Here, qi(t)q_{i}(t) is the unmodeled dynamics, which contains the control input and the acceleration of i1i-1th vehicle that cannot be measured directly by the iith vehicle. We design a linear extended state observer

{z˙1,i(t)=z2,i(t)+β1(vd,i(t)z1,i(t)),z˙2,i(t)=z3,i(t)+β2(vd,i(t)z1,i(t))+ai(t)/τui(t)/τ,z˙3,i(t)=β3(vd,i(t)z1,i(t)),i=1,2,,N,\displaystyle\left\{\begin{aligned} \dot{z}_{1,i}(t)=&z_{2,i}(t)+\beta_{1}(v_{d,i}(t)-z_{1,i}(t)),\vspace{1ex}\\ \dot{z}_{2,i}(t)=&z_{3,i}(t)+\beta_{2}(v_{d,i}(t)-z_{1,i}(t))+a_{i}(t)/\tau\\ &-u_{i}(t)/\tau,\vspace{1ex}\\ \dot{z}_{3,i}(t)=&\beta_{3}(v_{d,i}(t)-z_{1,i}(t)),\ i=1,2,\ldots,N,\\ \end{aligned}\right. (9)

where z2,i(t)z_{2,i}(t) is the estimate of the acceleration difference ad,i(t)a_{d,i}(t) between the i1i-1th vehicle and the iith vehicle. The constants β1>0\beta_{1}>0, β2>0\beta_{2}>0 and β3>0\beta_{3}>0 are the observer gains to be designed. Then by the acceleration of iith vehicle, the estimate of the acceleration of i1i-1th vehicle is given by z2,i(t)+ai(t)z_{2,i}(t)+a_{i}(t). The controller of iith follower vehicle is designed as

ui(t)=\displaystyle u_{i}(t)= kpei(t)+kv(vd,i(t)hai(t))\displaystyle k_{p}e_{i}(t)+k_{v}(v_{d,i}(t)-ha_{i}(t))
+ka(z2,i(t)+ai(t)),i=1,2,,N,\displaystyle+k_{a}(z_{2,i}(t)+a_{i}(t)),\;i=1,2,...,N, (10)

where kp>0k_{p}>0, kv>0k_{v}>0, ka>0k_{a}>0 are the control parameters to be designed. The controller (3) consists of two parts. The first part kpei(t)+kv(vd,i(t)hai(t))k_{p}e_{i}(t)+k_{v}(v_{d,i}(t)-ha_{i}(t)) is the feedback item, which consists of the inter-vehicle distance error between the adjacent vehicles and its differential; while the second part ka(z2,i(t)+ai(t))k_{a}(z_{2,i}(t)+a_{i}(t)) is the feedforward item, which consists of the estimate of the acceleration of the i1i-1th vehicle. It can be seen that the extended state observer (9) and the controller (3) only use the information obtained by on-board sensors of followers.

4 Stability analysis of vehicle platoon

In reality, the velocity of the leader vehicle will reach at a steady state within a finite time tft_{f}, that is, there exists tft_{f}, such that u0(t)=0u_{0}(t)=0, ttft\geq t_{f}. We make the following assumption.

Assumption 1.

limtu0(t)=0,limtu˙0(t)=0\lim\limits_{t\to\infty}u_{0}(t)=0,\lim\limits_{t\to\infty}\dot{u}_{0}(t)=0.

Denote Ψ(kp,kv)=[Ψ11OΨ21Ψ22]6N×6N\Psi(k_{p},k_{v})={\left[\begin{array}[]{cc}\Psi_{11}&O\\ \Psi_{21}&\Psi_{22}\end{array}\right]}\in\mathbb{R}^{6N\times 6N}, where Ψ11\Psi_{11}, Ψ21\Psi_{21} and Ψ22\Psi_{22} are 3N3N dimensional matrices with Ψ22=diag()\Psi_{22}=diag(\mathcal{H}),

𝒜=[01h001kp/τkv/τ(1kvh)/τ],=[000001000],\mathcal{A}=\begin{bmatrix}0&1&-h\\ 0&0&-1\\ \begin{aligned} k_{p}/\tau\end{aligned}&\begin{aligned} k_{v}/\tau\end{aligned}&\begin{aligned} (-1-k_{v}h)/\tau\end{aligned}\end{bmatrix},\mathcal{B}=\begin{bmatrix}0&0&0\\ 0&0&1\\ 0&0&0\end{bmatrix},
=[00000000kv/τ],=[β110β201β300],\mathcal{E}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&\begin{aligned} -k_{v}/\tau\end{aligned}\end{bmatrix},\mathcal{H}=\begin{bmatrix}-\beta_{1}&1&0\\ -\beta_{2}&0&1\\ -\beta_{3}&0&0\end{bmatrix},
𝒟=[000000kp(1+kvh)/τ2(kv+kv2hkpτ)/τ2(kphτ+kvτ2kvhkv2h21)/τ2].\mathcal{D}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} &k_{p}(1+k_{v}h)\\ &/\tau^{2}\end{aligned}&\begin{aligned} &(k_{v}+k_{v}^{2}h\\ &-k_{p}\tau)/\tau^{2}\end{aligned}&\begin{aligned} &(k_{p}h\tau+k_{v}\tau-2k_{v}h\\ &-k_{v}^{2}h^{2}-1)/\tau^{2}\end{aligned}\end{bmatrix}.
Ψ11=[𝒜𝒜𝒜],Ψ21=[O𝒟O𝒟O𝒟O],\Psi_{11}=\begin{bmatrix}\mathcal{A}&\quad&\quad&\quad\\ \mathcal{B}&\mathcal{A}&\quad&\quad\\ \quad&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{B}&\mathcal{A}\end{bmatrix},\Psi_{21}=\begin{bmatrix}O&\quad&\quad&\quad&\quad\\ \mathcal{D}&O&\quad&\quad&\quad\\ \mathcal{E}&\mathcal{D}&O&\quad&\quad\\ \quad&\ddots&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{E}&\mathcal{D}&O\end{bmatrix},

For the closed-loop and 2\mathcal{L}_{2} string stabilities, we have the following theorems.

Theorem 1.

Suppose that Assumption 1 holds. Consider the system (1) under the control law (9) and (3). For any given h>0h>0, if β1>0\beta_{1}>0, β3>0\beta_{3}>0, β1β2β3>0\beta_{1}\beta_{2}-\beta_{3}>0, kp>0k_{p}>0,

kv>{(((1kph2)2+4kphτ)1/2(1+kph2))/(2h),ifh<τ,0,ifhτ,k_{v}>\left\{\begin{array}[]{ll}\begin{aligned} &(((1-k_{p}h^{2})^{2}+4k_{p}h\tau)^{1/2}-(1+k_{p}h^{2}))\\ &/(2h),\;\hskip 120.92421pt{\rm if}\ h<\tau,\vspace{1ex}\end{aligned}\\ \begin{aligned} 0,\;\hskip 139.4185pt{\rm if}\ h\geq\tau,\end{aligned}\end{array}\right. (11)
0<ka<{rc(Ψ)τ,ifN=1,rc(Ψ)τ2/(kvh+τβ2+4τ+1),ifN=2,(τ(Θ2τ2+4(2N5)rc(Ψ))1/2τ2Θ)/(2(2N5)),ifN3,0<k_{a}<\left\{\begin{array}[]{ll}\begin{aligned} r_{c}(\Psi)\tau\end{aligned},\;\hskip 109.5431pt{\rm if}\ N=1,\vspace{1ex}\\ \begin{aligned} &r_{c}(\Psi)\tau^{2}/(k_{v}h+\tau\beta_{2}+4\tau+1),\end{aligned}\;{\rm if}\ N=2,\vspace{1ex}\\ \begin{aligned} &(\tau(\Theta^{2}\tau^{2}+4(2N-5)r_{c}(\Psi))^{1/2}-\tau^{2}\Theta)\\ &/(2(2N-5)),\;\hskip 82.51299pt{\rm if}\ N\geq 3,\end{aligned}\end{array}\right. (12)

where Θ=(Nτ+(N1)(τβ2+2τ+kvh+1)+(N2)(kp+kv+2kvh+2))/τ2\Theta=(N\tau+(N-1)(\tau\beta_{2}+2\tau+k_{v}h+1)+(N-2)(k_{p}+k_{v}+2k_{v}h+2))/\tau^{2}, rc(Ψ)=minωSn(iωIΨ)r_{c}(\Psi)=\min\limits_{\omega\in\mathbb{R}}S_{n}(i\omega I-\Psi),then vi(t)v0(t)v_{i}(t)-v_{0}(t) and ei(t)e_{i}(t) both converge to zero exponentially, i=1,2,,Ni=1,2,...,N.

The proof of Theorem 1 is given in Appendix B.

Remark 4.1.

Theorem 1 shows that the control law (9) and (3) can be properly designed such that the closed-loop system is exponentially stable, so the proposed control law is naturally robust against sensing noises, i.e. under the control law (9) and (3) with bounded sensing noises in the measurements of vd,i(t)v_{d,i}(t), ei(t)e_{i}(t) and ai(t)a_{i}(t), the inter-vehicle distance errors will converge to a neighborhood of zero whose size is proportional to the noise intensities.

Theorem 2.

Consider the system (1) under the control law (9) and (3). Let kp=μpkk_{p}=\mu_{p}k, kv=μvkk_{v}=\mu_{v}k, ka=μakk_{a}=\mu_{a}k, β1=3ωo\beta_{1}=3\omega_{o}, β2=3ωo2\beta_{2}=3\omega_{o}^{2}, β3=ωo3\beta_{3}=\omega_{o}^{3}, where kk, μp\mu_{p}, μv\mu_{v}, μa\mu_{a}, ωo\omega_{o} are positive parameters to be designed. For any given h>0h>0, if μa>0\mu_{a}>0, μp>0\mu_{p}>0,

μv>\displaystyle\mu_{v}> max{3μa/h,2μa/h2},\displaystyle\max\left\{\sqrt{3}\mu_{a}/h,2\mu_{a}/h^{2}\right\}, (13)
ωo>\displaystyle\omega_{o}> 16μvμa/(3h2μv29μa2),\displaystyle 16\mu_{v}\mu_{a}/(3h^{2}\mu_{v}^{2}-9\mu_{a}^{2}), (14)
k\displaystyle k\geq max{θ1,θ2,θ3,θ4,γ5/α5},\displaystyle\max\left\{\theta_{1},\theta_{2},\theta_{3},\theta_{4},\gamma_{5}/\alpha_{5}\right\}, (15)

where the definitions of θi,i=1,2,3,4\theta_{i},i=1,2,3,4, α5\alpha_{5} and γ5\gamma_{5} are given in Appendix A, then Gei(s)1,i=2,,N\|G_{ei}(s)\|_{\infty}\leq 1,\ i=2,...,N.

The proof of Theorem 2 is given in Appendix C.

Remark 4.2.

The control parameters can be properly designed to guarantee closed-loop and 2\mathcal{L}_{2} string stabilities simultaneously. From Theorem 2, we know limμv+μa0max{θ1,θ2,θ3,θ4,γ5/α5}=2/(h2μp)\lim_{\mu_{v}\to+\infty\atop\mu_{a}\to 0}\max\{\theta_{1},\theta_{2},\theta_{3},\theta_{4},\gamma_{5}/\alpha_{5}\}=2/(h^{2}\mu_{p}). We first fix μp>0\mu_{p}>0 and k>2/(h2μp)k>2/(h^{2}\mu_{p}). It is known that there are μv\mu_{v}^{*} and μa\mu_{a}^{*} such that (15) holds for any μv>μv\mu_{v}>\mu_{v}^{*} and μa<μa\mu_{a}<\mu_{a}^{*}. Then choose μv(μv,+)\mu_{v}\in(\mu_{v}^{*},+\infty) and μa(0,μa)\mu_{a}\in(0,\mu_{a}^{*}) such that (11) and (12) hold, respectively.

Remark 4.3.

Theorem 2 shows that the control parameters can be properly designed to ensure the 2\mathcal{L}_{2} string stability for any given positive time headway. There is another definition of string stability, called \mathcal{L}_{\infty} string stability, which means supt0|ei(t)|supt0|ei1(t)|,i=2,,N\sup_{t\geq 0}|e_{i}(t)|\leq\sup_{t\geq 0}|e_{i-1}(t)|,i=2,...,N. According to Eyre et al. (1998), if Gei(s)1\|G_{ei}(s)\|_{\infty}\leq 1 and 1[Gei(s)]0\mathscr{L}^{-1}[G_{ei}(s)]\geq 0, then \mathcal{L}_{\infty} string stability is satisfied.

Remark 4.4.

In reality, the velocity of leader vehicle is positive. According to Eyre et al. (1998), if Gvi(s)1\|G_{vi}(s)\|_{\infty}\leq 1 and 1[Gvi(s)]0\mathscr{L}^{-1}[G_{vi}(s)]\geq 0, where Gvi(s)=𝒱i(s)/𝒱i1(s)G_{vi}(s)=\mathscr{V}_{i}(s)/\mathscr{V}_{i-1}(s), 𝒱i(s)\mathscr{V}_{i}(s) is Laplace transform of vi(t)v_{i}(t), then supt0|vi(t)|supt0|vi1(t)|,i=1,,N\sup_{t\geq 0}|v_{i}(t)|\leq\sup_{t\geq 0}|v_{i-1}(t)|,i=1,...,N, and the vi(t),i=1,,Nv_{i}(t),i=1,...,N have the same sign as v0(t)v_{0}(t). Thus the velocities of follower vehicles are positive. The evolution of |Gvi(jω)||G_{vi}(j\omega)| and 1[Gvi(s)]\mathscr{L}^{-1}[G_{vi}(s)] are investigated by numerical simulation in Section 6.

5 Robustness against parameter uncertainties

The control law (9) and (3) is designed based on the ideal vehicle model (1) with completely known parameters. In practice, the vehicle model may be inaccurate. In this section, we will give the range of control parameters to ensure the closed-loop and 2\mathcal{L}_{2} string stabilities when there are parameter uncertainties in the vehicle model. It is shown that the control law (9) and (3) based on the ideal vehicle model can ensure the closed-loop and 2\mathcal{L}_{2} string stabilities with small model parameter uncertainties.

Suppose that the real vehicles have the following dynamics

{p˙i(t)=vi(t),v˙i(t)=ai(t),i=0,1,,N,a˙i(t)=(1/τ+ϵi)ai(t)+(1/τ+ϵi)ui(t),\displaystyle\left\{\begin{aligned} &\dot{p}_{i}(t)=v_{i}(t),\vspace{1ex}\\ &\dot{v}_{i}(t)=a_{i}(t),\vspace{1ex}\hskip 71.13188pti=0,1,...,N,\\ &\dot{a}_{i}(t)=-\left(1/\tau+\epsilon_{i}\right)a_{i}(t)+\left(1/\tau+\epsilon_{i}\right)u_{i}(t),\end{aligned}\right. (16)

where the definitions of ui(t),i=0,1,,Nu_{i}(t),i=0,1,...,N are the same as in (1). The constant τ\tau is the known nominal inertial delay of vehicle longitudinal dynamics. The constant ϵi\epsilon_{i} is the parameter uncertainty. We assume that |ϵi|ϵ¯|\epsilon_{i}|\leq\overline{\epsilon}, i=0,1,,Ni=0,1,...,N, where ϵ¯[0,1/τ)\overline{\epsilon}\in[0,1/\tau). We have the following theorems.

Theorem 3.

Suppose that Assumption 1 holds. Consider the system (16) under the control law (9) and (3). For any given h>0h>0, if β1>0\beta_{1}>0, β3>0\beta_{3}>0, β1β2β3>0\beta_{1}\beta_{2}-\beta_{3}>0, kp>0k_{p}>0,

kv>{(((1kph2)2+4kphτ)1/2(1+kph2))/(2h),ifh<τ,0,ifhτ,\displaystyle k_{v}>\left\{\begin{array}[]{ll}\begin{aligned} &(((1-k_{p}h^{2})^{2}+4k_{p}h\tau)^{1/2}-(1+k_{p}h^{2}))\\ &/(2h),\;\hskip 120.92421pt{\rm if}\ h<\tau,\vspace{1ex}\end{aligned}\\ \begin{aligned} 0,\;\hskip 139.4185pt{\rm if}\ h\geq\tau,\end{aligned}\end{array}\right. (19)
{Z1(kp,kv)ϵ¯2+Z2(kp,kv)ϵ¯rc(Ψ(kp,kv))<0,ifN=1,Z3(kp,kv)ϵ¯2+Z4(kp,kv)ϵ¯rc(Ψ(kp,kv))<0,ifN=2,Z5(kp,kv)ϵ¯2+Z6(kp,kv)ϵ¯rc(Ψ(kp,kv))<0,ifN3,\displaystyle\left\{\begin{array}[]{lc}\begin{aligned} &Z_{1}(k_{p},k_{v})\overline{\epsilon}^{2}+Z_{2}(k_{p},k_{v})\overline{\epsilon}-r_{c}(\Psi(k_{p},k_{v}))<0,\\ \;&\hskip 184.9429pt{\rm if}\ N=1,\end{aligned}\vspace{1ex}\\ \begin{aligned} &Z_{3}(k_{p},k_{v})\overline{\epsilon}^{2}+Z_{4}(k_{p},k_{v})\overline{\epsilon}-r_{c}(\Psi(k_{p},k_{v}))<0,\\ \;&\hskip 184.9429pt{\rm if}\ N=2,\end{aligned}\vspace{1ex}\\ \begin{aligned} &Z_{5}(k_{p},k_{v})\overline{\epsilon}^{2}+Z_{6}(k_{p},k_{v})\overline{\epsilon}-r_{c}(\Psi(k_{p},k_{v}))<0,\\ \;&\hskip 184.9429pt{\rm if}\ N\geq 3,\end{aligned}\end{array}\right. (23)
0<ka<{(rc(Ψ)Z1ϵ¯2Z2ϵ¯)/(Y1ϵ¯2+Y2ϵ¯+1/τ)ifN=1,[((Θ1+Y3ϵ¯2+Y4ϵ¯)2+4A1(rc(Ψ)Z3ϵ¯2Z4ϵ¯))1/2(Θ1+Y3ϵ¯2+Y4ϵ¯)]/(2A1),ifN=2,[((Θ+Y5ϵ¯2+Y6ϵ¯)2+4A2(rc(Ψ)Z5ϵ¯2Z6ϵ¯))1/2(Θ+Y5ϵ¯2+Y6ϵ¯)]/(2A2),ifN3,\displaystyle 0<k_{a}<\left\{\begin{array}[]{ll}\begin{aligned} &\left(r_{c}(\Psi)-Z_{1}\overline{\epsilon}^{2}-Z_{2}\overline{\epsilon}\right)/(Y_{1}\overline{\epsilon}^{2}+Y_{2}\overline{\epsilon}+1/\tau)\\ &\hskip 145.10905pt{\rm if}\ N=1,\end{aligned}\vspace{1ex}\\ \begin{aligned} &[((\Theta_{1}+Y_{3}\overline{\epsilon}^{2}+Y_{4}\overline{\epsilon})^{2}+4A_{1}(r_{c}(\Psi)-Z_{3}\overline{\epsilon}^{2}\\ &-Z_{4}\overline{\epsilon}))^{1/2}-(\Theta_{1}+Y_{3}\overline{\epsilon}^{2}+Y_{4}\overline{\epsilon})]/(2A_{1}),\\ &\hskip 145.10905pt{\rm if}\ N=2,\end{aligned}\vspace{1ex}\\ \begin{aligned} &[((\Theta+Y_{5}\overline{\epsilon}^{2}+Y_{6}\overline{\epsilon})^{2}+4A_{2}(r_{c}(\Psi)-Z_{5}\overline{\epsilon}^{2}\\ &-Z_{6}\overline{\epsilon}))^{1/2}-(\Theta+Y_{5}\overline{\epsilon}^{2}+Y_{6}\overline{\epsilon})]/(2A_{2}),\\ &\hskip 145.10905pt{\rm if}\ N\geq 3,\end{aligned}\end{array}\right. (27)

where the definitions of Zi,Yi,i=1,2,,6Z_{i},Y_{i},i=1,2,...,6, Θ1\Theta_{1}, A1A_{1} and A2A_{2} are given in Appendix A, then vi(t)v0(t)v_{i}(t)-v_{0}(t) and ei(t)e_{i}(t) both converge to zero exponentially, i=1,2,,Ni=1,2,...,N.

The proof of Theorem 3 is given in Appendix D.

Theorem 4.

Consider the system (16) under the control law (9) and (3). Let kp=μpkk_{p}=\mu_{p}k, kv=μvkk_{v}=\mu_{v}k, ka=μakk_{a}=\mu_{a}k, β1=3ωo\beta_{1}=3\omega_{o}, β2=3ωo2\beta_{2}=3\omega_{o}^{2}, β3=ωo3\beta_{3}=\omega_{o}^{3}, where kk, μp\mu_{p}, μv\mu_{v}, μa\mu_{a}, ωo\omega_{o} are positive parameters to be designed. For any given h>0h>0, if μa>0\mu_{a}>0, μp>4μab¯/(τb¯2h2)\mu_{p}>4\mu_{a}\overline{b}/(\tau\underline{b}^{2}h^{2}),

μv>\displaystyle\mu_{v}> max{4μab¯2/(hb¯2),2μab¯/(τhb¯2)},\displaystyle\max\left\{4\mu_{a}\overline{b}^{2}/(h\underline{b}^{2}),2\mu_{a}\overline{b}/(\tau h\underline{b}^{2})\right\}, (28)
ωo>\displaystyle\omega_{o}> max{(|λ¯224λ¯1λ¯3|1/2λ¯2)/(2λ¯1),(3(2b¯μpμa/τ\displaystyle\max\left\{(|\underline{\lambda}_{2}^{2}-4\underline{\lambda}_{1}\underline{\lambda}_{3}|^{1/2}-\underline{\lambda}_{2})/(2\underline{\lambda}_{1}),(3(2\underline{b}\mu_{p}\mu_{a}/\tau\right.
+2b¯2μpμa+b¯2h2μp2)/(b¯2h2μv2b¯2μa2))1/2},\displaystyle\hskip 22.76219pt\left.+2\underline{b}^{2}\mu_{p}\mu_{a}+\underline{b}^{2}h^{2}\mu_{p}^{2})/(\underline{b}^{2}h^{2}\mu^{2}_{v}-\overline{b}^{2}\mu^{2}_{a}))^{1/2}\right\}, (29)
k\displaystyle k\geq max{θ1,θ2,θ3,θ4,γ¯5/α¯5},\displaystyle\max\left\{\theta_{1},\theta_{2},\theta_{3},\theta_{4},\overline{\gamma}_{5}/\underline{\alpha}_{5}\right\}, (30)

where the definitions of b¯\overline{b}, b¯\underline{b}, α¯5\underline{\alpha}_{5}, γ¯5\overline{\gamma}_{5}, θi,i=1,2,3,4\theta_{i},i=1,2,3,4 and λ¯i,i=1,2,3\underline{\lambda}_{i},i=1,2,3 are given in Appendix A, then Gei(s)1,i=2,,N\|G_{ei}(s)\|_{\infty}\leq 1,\ i=2,...,N.

The proof of Theorem 4 is given in Appendix E.

Remark 5.5.

From Theorem 3, we know that the control law (9) and (3) is robust against model parameter uncertainties. If ϵ¯=0\overline{\epsilon}=0, then (23) holds and (27) degenerates into (12). By the continuity of (23) and (27) with respect to ϵ¯\overline{\epsilon}, we know that if the control law (9) and (3) is designed according to Theorem 1 based on the ideal vehicle model (1), then the closed-loop and 2\mathcal{L}_{2} string stabilities can be ensured provided ϵ¯\overline{\epsilon} is sufficiently small.

Remark 5.6.

A more accurate nonlinear vehicle model is considered in Zheng et al. (2016)

{p˙i(t)=vi(t),v˙i(t)=ai(t),i=0,1,,N,a˙i(t)=ηiTi,des(t)/(miRiτi)(2Civi(t)ai(t)τi+miai(t)+Civi2(t)+migfi)/(miτi),\displaystyle\left\{\begin{aligned} \dot{p}_{i}(t)=&v_{i}(t),\vspace{1ex}\\ \dot{v}_{i}(t)=&a_{i}(t),\hskip 79.66771pti=0,1,\ldots,N,\vspace{1ex}\\ \dot{a}_{i}(t)=&\eta_{i}T_{i,des}(t)/(m_{i}R_{i}\tau_{i})-(2C_{i}v_{i}(t)a_{i}(t)\tau_{i}\vspace{1ex}\\ &+m_{i}a_{i}(t)+C_{i}v_{i}^{2}(t)+m_{i}gf_{i})/(m_{i}\tau_{i}),\end{aligned}\right. (31)

where Ti,des(t)T_{i,des}(t) is the expected driving or braking torque of the iith vehicle at time tt. The constant mim_{i}, fif_{i}, RiR_{i}, CiC_{i}, τi\tau_{i} and ηi\eta_{i} are the mass, the rolling resistance coefficient, the tire radius, the total air resistance coefficient, the inertial delay of longitudinal dynamics and the mechanical efficiency of the drive train of the iith vehicle, respectively. The constant gg is the gravitational acceleration. The feedback linearization law is given by

Ti,des(t)=\displaystyle T_{i,des}(t)= R~i[C~i(vi(t)+ξiv(t))(2τ~i(ai(t)+ξia(t))\displaystyle\widetilde{R}_{i}[\widetilde{C}_{i}(v_{i}(t)+\xi_{i}^{v}(t))(2\widetilde{\tau}_{i}(a_{i}(t)+\xi_{i}^{a}(t))
+(vi(t)+ξiv(t)))+m~igf~i+m~iui(t)]/η~i,\displaystyle+(v_{i}(t)+\xi_{i}^{v}(t)))+\widetilde{m}_{i}g\widetilde{f}_{i}+\widetilde{m}_{i}u_{i}(t)]/\widetilde{\eta}_{i},
i=0,1,,N.\displaystyle\hskip 85.35826pt\;i=0,1,\ldots,N. (32)

where the constants m~i\widetilde{m}_{i}, f~i\widetilde{f}_{i}, R~i\widetilde{R}_{i}, C~i\widetilde{C}_{i}, η~i\widetilde{\eta}_{i} and τ~i\widetilde{\tau}_{i} are the nominal parameters of iith vehicle, vi(t)+ξiv(t)v_{i}(t)+\xi_{i}^{v}(t) and ai(t)+ξia(t)a_{i}(t)+\xi_{i}^{a}(t) are the velocity and acceleration of iith vehicle measured by on-board sensors, ξiv(t)\xi_{i}^{v}(t) and ξia(t)\xi_{i}^{a}(t) are sensing noises . Substituting (5.6) into (31), we obtain

{p˙i(t)=vi(t),v˙i(t)=ai(t),i=0,1,,N,a˙i(t)=ζi(t)(ai(t)+ξia(t))/τ~i+ui(t)/τ~i,\displaystyle\left\{\begin{aligned} \dot{p}_{i}(t)=&v_{i}(t),\vspace{1ex}\\ \dot{v}_{i}(t)=&a_{i}(t),\hskip 62.59605pti=0,1,\ldots,N,\\ \dot{a}_{i}(t)=&\zeta_{i}(t)-(a_{i}(t)+\xi_{i}^{a}(t))/\widetilde{\tau}_{i}+u_{i}(t)/\widetilde{\tau}_{i},\end{aligned}\right. (33)

where ζi(t)=ηiTi,des(t)/(miRiτi)(2Civi(t)ai(t)τi+Civi2(t)+migfi)/(miτi)+(ai(t)+ξia(t))/τ~iui(t)/τ~i\zeta_{i}(t)=\eta_{i}T_{i,des}(t)/(m_{i}R_{i}\tau_{i})-(2C_{i}v_{i}(t)a_{i}(t)\tau_{i}+C_{i}v_{i}^{2}(t)+m_{i}gf_{i})/(m_{i}\tau_{i})+(a_{i}(t)+\xi_{i}^{a}(t))/\widetilde{\tau}_{i}-u_{i}(t)/\widetilde{\tau}_{i}. Compared with (1), the model (33) contains a nonlinear term ζi(t)\zeta_{i}(t). From (33), we know the models of the velocity difference between adjacent vehicles (4) with qi(t)=ζi1(t)ζi(t)+((ai1(t)+ξi1a(t))+ui1(t))/τ~iq_{i}(t)=\zeta_{i-1}(t)-\zeta_{i}(t)+(-(a_{i-1}(t)+\xi_{i-1}^{a}(t))+u_{i-1}(t))/\widetilde{\tau}_{i}. The ESO (9) can still be used to estimate the acceleration differences between adjacent vehicles. The nonlinear term ζi(t)\zeta_{i}(t) can be estimated by designing another ESO

{v^˙i(t)=a^i(t)+β4(vi(t)+ξiv(t)v^i(t)),a^˙i(t)=ζ^i(t)+β5(vi(t)+ξiv(t)v^i(t))(ai(t)+ξia(t))/τ~i+ui(t)/τ~i,ζ^˙i(t)=β6(vi(t)+ξiv(t)v^i(t)),i=1,,N,\displaystyle\left\{\begin{aligned} \dot{\hat{v}}_{i}(t)=&{\hat{a}}_{i}(t)+\beta_{4}(v_{i}(t)+\xi_{i}^{v}(t)-\hat{v}_{i}(t)),\vspace{1ex}\\ \dot{\hat{a}}_{i}(t)=&{\hat{\zeta}}_{i}(t)+\beta_{5}(v_{i}(t)+\xi_{i}^{v}(t)-\hat{v}_{i}(t))\\ &-(a_{i}(t)+\xi_{i}^{a}(t))/\widetilde{\tau}_{i}+u_{i}(t)/\widetilde{\tau}_{i},\vspace{1ex}\\ \dot{\hat{\zeta}}_{i}(t)=&\beta_{6}(v_{i}(t)+\xi_{i}^{v}(t)-\hat{v}_{i}(t)),\ i=1,\ldots,N,\\ \end{aligned}\right. (34)

where v^i(t)\hat{v}_{i}(t), a^i(t)\hat{a}_{i}(t) and ζ^i(t)\hat{\zeta}_{i}(t) are the estimates of vi(t)v_{i}(t), ai(t)a_{i}(t) and ζi(t)\zeta_{i}(t), respectively. The constants β4>0\beta_{4}>0, β5>0\beta_{5}>0 and β6>0\beta_{6}>0 are the observer gains to be designed. The controller of iith follower vehicle is designed as

ui(t)=\displaystyle u_{i}(t)= kp(ei(t)+ξie(t))+kv(vd,i(t)+ξid(t)h(ai(t)\displaystyle k_{p}(e_{i}(t)+\xi_{i}^{e}(t))+k_{v}(v_{d,i}(t)+\xi_{i}^{d}(t)-h(a_{i}(t)
+ξia(t)))+ka(z2,i(t)+ai(t)+ξia(t))\displaystyle+\xi_{i}^{a}(t)))+k_{a}(z_{2,i}(t)+a_{i}(t)+\xi_{i}^{a}(t))
τ~iζ^i(t),i=1,,N,\displaystyle-\widetilde{\tau}_{i}\hat{\zeta}_{i}(t),\;i=1,...,N, (35)

where ei(t)+ξie(t)e_{i}(t)+\xi_{i}^{e}(t) and vd,i(t)+ξid(t)v_{d,i}(t)+\xi_{i}^{d}(t) are the inter-vehicle distance error and velocity difference between adjacent vehicles measured by on-board sensors, ξie(t)\xi_{i}^{e}(t) and ξid(t)\xi_{i}^{d}(t) are sensing noises. The stability analysis of the vehicle platoon system (33) under control law (9), (34) and (5.6) would be an interesting and challenging topic for future investigation.

6 NUMERICAL SIMULATIONS

Suppose there are 1 leader vehicle and 5 follower vehicles with the following dynamics with parameter uncertainties

{p˙i(t)=vi(t),v˙i(t)=ai(t),i=0,1,5,a˙i(t)=(1/τ+ϵi)ai(t)+(1/τ+ϵi)ui(t),\displaystyle\left\{\begin{aligned} &\dot{p}_{i}(t)=v_{i}(t),\vspace{1ex}\\ &\dot{v}_{i}(t)=a_{i}(t),\vspace{1ex}\hskip 99.58464pti=0,1,...5,\\ &\dot{a}_{i}(t)=-\left(1/\tau+\epsilon_{i}\right)a_{i}(t)+\left(1/\tau+\epsilon_{i}\right)u_{i}(t),\end{aligned}\right.

where τ=0.1\tau=0.1. The parameter uncertainties are given by ϵ0=0.8\epsilon_{0}=-0.8, ϵ1=0.1\epsilon_{1}=0.1, ϵ2=0.5\epsilon_{2}=0.5, ϵ3=0.2\epsilon_{3}=-0.2, ϵ4=0.65\epsilon_{4}=0.65, ϵ5=0.3\epsilon_{5}=-0.3.

To make the behavior of the leader vehicle close to a real vehicle, we construct a virtual vehicle in CarSim and record its acceleration. Then we use the recorded acceleration as the expected acceleration of the leader vehicle in the following simulations. The initial velocities are given by vi(0)=10m/sv_{i}(0)=10\,m/s, i=0,1,,5i=0,1,...,5. The initial accelerations are given by ai(0)=0m/s2a_{i}(0)=0\,m/s^{2}, i=0,1,,5i=0,1,...,5. The initial positions are taken as p0(0)=30mp_{0}(0)=30\,m, p1(0)=24mp_{1}(0)=24\,m, p2(0)=18mp_{2}(0)=18\,m, p3(0)=12mp_{3}(0)=12\,m, p4(0)=6mp_{4}(0)=6\,m, p5(0)=0mp_{5}(0)=0\,m. The standstill distance is given by r=3mr=3\,m. Let h=0.3sh=0.3s. We choose kp=8k_{p}=8, kv=40k_{v}=40, ka=1.2k_{a}=1.2 and ωo=15\omega_{o}=15, β1=45\beta_{1}=45, β2=675\beta_{2}=675 and β3=3375\beta_{3}=3375. The evolution of |Gvi(jω)||G_{vi}(j\omega)| and 1[Gvi(s)]\mathscr{L}^{-1}[G_{vi}(s)] are shown in Fig.1.

Refer to caption
Figure 1: (a)
Refer to caption
(b)

The actual and the estimated acceleration differences between the 3rd and 4th follower vehicles are shown in Fig.2(a). The evolution of vehicles’ accelerations and velocities are shown in Fig.2(b) and Fig.2(c), respectively. The evolution of inter-vehicle distance errors are shown in Fig.2(d).

Refer to caption
Figure 2: (a)
Refer to caption
(b)
Refer to caption
(c)
Refer to caption
(d)
Vehicle model with parameter uncertainties. (a) The actual and the estimated acceleration differences between the 3rd and 4th follower vehicles. (b) Accelerations of the vehicles. (c)Velocities of the vehicles. (d) Inter-vehicle distance errors.

Fig.1 shows that Gvi(s)1\|G_{vi}(s)\|_{\infty}\leq 1 and 1[Gvi(s)]0,i=1,2,,5\mathscr{L}^{-1}[G_{vi}(s)]\geq 0,i=1,2,...,5. Then from Remark 4.4, we know that supt0|vi(t)|supt0|vi1(t)|\sup_{t\geq 0}|v_{i}(t)|\leq\sup_{t\geq 0}|v_{i-1}(t)| and the vi(t),i=1,,5v_{i}(t),i=1,...,5 have the same sign as v0(t)v_{0}(t), which are also shown in Fig.2.(c). From Fig. 2.(a), it is shown that although the ESO (9) are designed based on the nominal τ\tau, the acceleration differences between adjacent vehicles ad,i(t)a_{d,i}(t) can be estimated well by the ESO (9). Fig.2.(b) and Fig.2.(c) show that the accelerations of follower vehicles and velocity differences between adjacent vehicles converge to zero as the acceleration of leader vehicle goes to zero, respectively. From Fig.2.(d), it is observed that the inter-vehicle distance errors converge to zero and they are not amplified in the backward propagation along the platoon.

Next, we consider the vehicle model with input delays, i.e. the ui(t)u_{i}(t) is replaced by ui(tϕ)u_{i}(t-\phi), where ϕ=0.2\phi=0.2 is the input delay. In practical applications, the velocity differences between adjacent vehicles vd,i(t)v_{d,i}(t), i=1,,Ni=1,...,N, measured by on-board sensors are usually corrupted by random noises. In the numerical simulations, we implement the sampled-data version of the control law (9) and (3) with vd,i(kσ)v_{d,i}(k\sigma) replaced by vd,i(kσ)+ξid(kσ)v_{d,i}(k\sigma)+\xi_{i}^{d}(k\sigma), where σ=0.002s\sigma=0.002s is the sampling period and {ξid(kσ),k=0,1,}\{\xi_{i}^{d}(k\sigma),k=0,1,...\} is a sequence of random variables with the uniform distribution U(0.005,0.005)U(-0.005,0.005). We choose kp=0.05k_{p}=0.05, kv=0.6k_{v}=0.6, ka=0.8k_{a}=0.8 and ωo=10\omega_{o}=10, β1=30\beta_{1}=30, β2=300\beta_{2}=300 and β3=1000\beta_{3}=1000. The actual and the estimated acceleration differences between the 3rd and 4th follower vehicles are shown in Fig.3.(a). The evolution of vehicles’ accelerations, velocities and inter-vehicle distance errors are shown in Fig.3.(b), Fig.3.(c) and Fig.3.(d), respectively.

Refer to caption
Figure 3: (a)
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(b)
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(c)
Refer to caption
(d)
Vehicle model with parameter uncertainties, sensing noises and input delays. (a) The actual and the estimated acceleration differences between the 3rd and 4th follower vehicles. (b) Accelerations of the vehicles. (c)Velocities of the vehicles. (d) Inter-vehicle distance errors.

From Fig.3.(a), it is shown that although the velocity differences vd,i(t)v_{d,i}(t) are corrupted by sensing noises, the sensing noises are not significantly amplified and the output of the ESO z2i(t)z_{2i}(t) can still track the acceleration differences between adjacent vehicles ad,i(t)a_{d,i}(t). Fig.3.(b), Fig.3.(c) and Fig.3.(d) show that the accelerations of follower vehicles, velocity differences between adjacent vehicles and inter-vehicle distance errors converge to a small neighborhood of zero. Fig.3 shows the robustness of the proposed control law against parameter uncertainties, sensing noises and input delays.

Then let the time headway hh change to 0.01s0.01s. By Fig.4.(a), it is observed that the control law (9) and (3) with the previous parameters, i.e. kp=8k_{p}=8, kv=40k_{v}=40, ka=1.2k_{a}=1.2, β1=45\beta_{1}=45, β2=675\beta_{2}=675 and β3=3375\beta_{3}=3375, cannot ensure the 2\mathcal{L}_{2} string stability any longer, but the closed-loop stability is still guaranteed.

We reselect kp=0.01k_{p}=0.01, kv=0.2k_{v}=0.2, ka=0.8k_{a}=0.8, β1=45\beta_{1}=45, β2=675\beta_{2}=675 and β3=3375\beta_{3}=3375. It can be seen from Fig.4.(b) that the closed-loop and 2\mathcal{L}_{2} string stabilities are both ensured by reselecting the control parameters. It is shown that for a smaller time headway hh, a smaller kpk_{p} can be chosen to ensure string stability at the cost of slower convergence.

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Figure 4: (a)
Refer to caption
(b)
Vehicle platoon under the control law (9) and (3) with h=0.01sh=0.01s. (a) kp=8k_{p}=8, kv=40k_{v}=40, ka=1.2k_{a}=1.2, β1=45\beta_{1}=45, β2=675\beta_{2}=675 and β3=3375\beta_{3}=3375. (b) kp=0.01k_{p}=0.01, kv=0.2k_{v}=0.2, ka=0.8k_{a}=0.8, β1=45\beta_{1}=45, β2=675\beta_{2}=675 and β3=3375\beta_{3}=3375.

The transient performance of the closed-loop system can be investigated from the structure of the compound controller (3) containing proportional differential feedback and feedforward terms. The convergence rate of the closed-loop system is mainly determined by the proportional differential term. The feedforward term reduces the influence of disturbances.

Refer to caption
Figure 5: The evolution of inter-vehicle distance errors between the leader vehicle and the first follower vehicle with different control parameters.

Let any two of the control parameters kpk_{p}, kvk_{v} and kak_{a} be fixed and the other one be changed. The evolution of inter-vehicle distance errors between the leader and the first follower under different control parameters are shown in Fig.5. Fig.5 shows that a larger kpk_{p} leads to a faster convergence and a larger kvk_{v} leads to a slower convergence with smaller inter-vehicle distance error. The parameter kak_{a} has little effect on the convergence rate, and a larger kak_{a} brings a smaller inter-vehicle distance error.

7 Conclusion

We have considered the platoon control for homogeneous vehicles with third-order linear dynamics model. The constant time headway spacing policy is adopted. Firstly, the distributed cooperative extended state observers are designed to estimate the acceleration differences between adjacent vehicles. Then the controller of each follower vehicle is designed by its own velocity, acceleration, velocity difference and estimated acceleration difference with respect to its immediate predecessor. The information required by the control law can be obtained by on-board sensors. The closed-loop stability of the vehicle platoon system is analyzed by the stability theory of perturbed linear systems, and the sufficient conditions to ensure the closed-loop stability are given. Also the 2\mathcal{L}_{2} string stability is analyzed in the frequency domain and the range of the control parameters that guarantee the 2\mathcal{L}_{2} string stability is presented. It is shown that for any given positive time headway, control parameters can be properly designed to guarantee both closed-loop and 2\mathcal{L}_{2} string stabilities of the vehicle platoon system. In addition, it has been shown that the closed-loop and 2\mathcal{L}_{2} string stabilities of the vehicle platoon system can be guaranteed with small model parameter uncertainties.

Acknowledgements

This work was supported by the National Natural Science Foundation of China under Grant 61977024 and the Basic Research Project of Shanghai Science and Technology Commission under Grant 20JC1414000.

Appendix A

Parameter definitions in Theorem 2

θi={(γi24αiρiγi)/(2αi),ifγi24αiρi0,0,ifγi24αiρi<0,i=1,2,3,4,\theta_{i}=\left\{\begin{array}[]{ll}\begin{aligned} (\sqrt{\gamma_{i}^{2}-4\alpha_{i}\rho_{i}}-\gamma_{i})/(2\alpha_{i}),\,{\rm if}\ &\gamma_{i}^{2}-4\alpha_{i}\rho_{i}\geq 0,\vspace{1ex}\end{aligned}\\ \begin{aligned} 0,\hskip 109.5431pt{\rm if}\ &\gamma_{i}^{2}-4\alpha_{i}\rho_{i}<0,\end{aligned}\\ \hskip 156.49014pti=1,2,3,4,\end{array}\right.
ρ1=\displaystyle\rho_{1}= 3τ2ω02+1,ρ2=3τ2ω04+3ωo2,\displaystyle 3\tau^{2}\omega_{0}^{2}+1,\rho_{2}=3\tau^{2}\omega_{0}^{4}+3\omega_{o}^{2},
ρ3=\displaystyle\rho_{3}= τ2ωo6+3ωo4,ρ4=ωo6,\displaystyle\tau^{2}\omega_{o}^{6}+3\omega_{o}^{4},\rho_{4}=\omega_{o}^{6},
α1=\displaystyle\alpha_{1}= h2μv2,α2=3h2μv2ωo2+h2μp2,\displaystyle h^{2}\mu_{v}^{2},\alpha_{2}=3h^{2}\mu_{v}^{2}\omega_{o}^{2}+h^{2}\mu_{p}^{2},
α3=\displaystyle\alpha_{3}= (3h2μv29μa2)ωo416μaμvωo3+(3h2μp26μpμa)ωo2,\displaystyle(3h^{2}\mu_{v}^{2}-9\mu_{a}^{2})\omega_{o}^{4}-16\mu_{a}\mu_{v}\omega_{o}^{3}+(3h^{2}\mu_{p}^{2}-6\mu_{p}\mu_{a})\omega_{o}^{2},
α4=\displaystyle\alpha_{4}= (h2μv2μa2)ωo6+(3h2μp2+12μaμp)ωo4,\displaystyle(h^{2}\mu_{v}^{2}-\mu_{a}^{2})\omega_{o}^{6}+(3h^{2}\mu_{p}^{2}+12\mu_{a}\mu_{p})\omega_{o}^{4},
α5=\displaystyle\alpha_{5}= (h2μp2+2μaμp)ωo6,\displaystyle(h^{2}\mu_{p}^{2}+2\mu_{a}\mu_{p})\omega_{o}^{6},
γ1=\displaystyle\gamma_{1}= 2(hτ)μv2hτμp,\displaystyle 2(h-\tau)\mu_{v}-2h\tau\mu_{p},
γ2=\displaystyle\gamma_{2}= [6(hτ)μv6hτμp]ωo22μp,\displaystyle[6(h-\tau)\mu_{v}-6h\tau\mu_{p}]\omega_{o}^{2}-2\mu_{p},
γ3=\displaystyle\gamma_{3}= 6(hμvτμvhτμp)ωo46μpωo2,\displaystyle 6(h\mu_{v}-\tau\mu_{v}-h\tau\mu_{p})\omega_{o}^{4}-6\mu_{p}\omega_{o}^{2},
γ4=\displaystyle\gamma_{4}= 2(hμvτμvhτμp)ωo66μpωo4,γ5=2μpωo6.\displaystyle 2(h\mu_{v}-\tau\mu_{v}-h\tau\mu_{p})\omega_{o}^{6}-6\mu_{p}\omega_{o}^{4},\gamma_{5}=2\mu_{p}\omega_{o}^{6}.

Parameter definitions in Theorem 3

Θ1=\displaystyle\Theta_{1}= (kvh+τβ2+4τ+1)/τ2,A1=ϵ¯2+ϵ¯/τ,\displaystyle(k_{v}h+\tau\beta_{2}+4\tau+1)/\tau^{2},A_{1}=\overline{\epsilon}^{2}+\overline{\epsilon}/\tau,
A2=\displaystyle A_{2}= (2N5)/τ2+(4N8)ϵ¯2+(5N11)ϵ¯/τ,\displaystyle(2N-5)/\tau^{2}+(4N-8)\overline{\epsilon}^{2}+(5N-11)\overline{\epsilon}/\tau,
Y1=\displaystyle Y_{1}= kvh+1,Y2=kvh/τ+β2+2,\displaystyle k_{v}h+1,Y_{2}=k_{v}h/\tau+\beta_{2}+2,
Y3=\displaystyle Y_{3}= kp+kv+5kvh+5,\displaystyle k_{p}+k_{v}+5k_{v}h+5,
Y4=\displaystyle Y_{4}= (kp+kv+6kvh+3)/τ+3β2+6,\displaystyle(k_{p}+k_{v}+6k_{v}h+3)/\tau+3\beta_{2}+6,
Y5=\displaystyle Y_{5}= (2N3)(kp+kv+2kvh+2)+(2N1)(kvh+1),\displaystyle(2N-3)(k_{p}+k_{v}+2k_{v}h+2)+(2N-1)(k_{v}h+1),
Y6=\displaystyle Y_{6}= N(kvh/τ+β2+2)+(N1)((kp+kv+4kvh+3)\displaystyle N(k_{v}h/\tau+\beta_{2}+2)+(N-1)((k_{p}+k_{v}+4k_{v}h+3)
/τ+β2+2)+(N2)(4+2kp+2kv+4kvh)/τ,\displaystyle/\tau+\beta_{2}+2)+(N-2)(4+2k_{p}+2k_{v}+4k_{v}h)/\tau,
Z1(kp,kv)=\displaystyle Z_{1}(k_{p},k_{v})= kv2h2+(2kv+kpkv+kv2)h+kp+kv+1,\displaystyle k_{v}^{2}h^{2}+(2k_{v}+k_{p}k_{v}+k_{v}^{2})h+k_{p}+k_{v}+1,
Z2(kp,kv)=\displaystyle Z_{2}(k_{p},k_{v})= kv2h2/τ+(kv2+kpkv+kv)h/τ+kph+kvh\displaystyle k_{v}^{2}h^{2}/\tau+(k_{v}^{2}+k_{p}k_{v}+k_{v})h/\tau+k_{p}h+k_{v}h
+2kp+2kv+1,\displaystyle+2k_{p}+2k_{v}+1,
Z3(kp,kv)=\displaystyle Z_{3}(k_{p},k_{v})= 3kv2h2+(3kv2+3kpkv+6kv)h+3kp\displaystyle 3k_{v}^{2}h^{2}+(3k_{v}^{2}+3k_{p}k_{v}+6k_{v})h+3k_{p}
+3kv+3,\displaystyle+3k_{v}+3,
Z4(kp,kv)=\displaystyle Z_{4}(k_{p},k_{v})= 4kv2h2/τ+((4kv2+6kv+4kpkv)h+2kp\displaystyle 4k_{v}^{2}h^{2}/\tau+((4k_{v}^{2}+6k_{v}+4k_{p}k_{v})h+2k_{p}
+2kv)/τ+(3kp+2kv)h+5kp+6kv+4,\displaystyle+2k_{v})/\tau+(3k_{p}+2k_{v})h+5k_{p}+6k_{v}+4,
Z5(kp,kv)=\displaystyle Z_{5}(k_{p},k_{v})= (2N1)(kpkvh+kp+kv+kv2h+1\displaystyle(2N-1)(k_{p}k_{v}h+k_{p}+k_{v}+k_{v}^{2}h+1
+2kvh+kv2h2),\displaystyle+2k_{v}h+k_{v}^{2}h^{2}),
Z6(kp,kv)=\displaystyle Z_{6}(k_{p},k_{v})= N((kpkvh+kv2h+kvh+kv2h2)/τ+1+kvh\displaystyle N((k_{p}k_{v}h+k_{v}^{2}h+k_{v}h+k_{v}^{2}h^{2})/\tau+1+k_{v}h
+2kp+2kv+kph)+(N1)((2kv+2kp\displaystyle+2k_{p}+2k_{v}+k_{p}h)+(N-1)((2k_{v}+2k_{p}
+2kv2h+2kpkvh+2kv2h2+4kvh)/τ+kp\displaystyle+2k_{v}^{2}h+2k_{p}k_{v}h+2k_{v}^{2}h^{2}+4k_{v}h)/\tau+k_{p}
+kph+2kv+2)+(N2)kv/τ.\displaystyle+k_{p}h+2k_{v}+2)+(N-2)k_{v}/\tau.

Parameter definitions in Theorem 4

b¯=\displaystyle\overline{b}= 1τ+ϵ¯,b¯=1τϵ¯,ρ¯1=3ωo2+b¯2,\displaystyle\frac{1}{\tau}+\overline{\epsilon},\underline{b}=\frac{1}{\tau}-\overline{\epsilon},\overline{\rho}_{1}=3\omega_{o}^{2}+\overline{b}^{2},
ρ¯2=\displaystyle\overline{\rho}_{2}= 3ωo4+3b¯2ωo2,ρ¯3=ωo6+3b¯2ωo4,ρ¯4=b¯2ωo6,\displaystyle 3\omega_{o}^{4}+3\overline{b}^{2}\omega_{o}^{2},\overline{\rho}_{3}=\omega_{o}^{6}+3\overline{b}^{2}\omega_{o}^{4},\overline{\rho}_{4}=\overline{b}^{2}\omega_{o}^{6},
λ¯1=\displaystyle\overline{\lambda}_{1}= 3μvh(b¯2hμv2b¯μa/τ)+3h2μa(2b¯2μvh3b¯2μa),\displaystyle 3\mu_{v}h(\overline{b}^{2}h\mu_{v}-2\underline{b}\mu_{a}/\tau)+3h^{2}\mu_{a}(2\overline{b}^{2}\mu_{v}h-3\underline{b}^{2}\mu_{a}),
λ¯1=\displaystyle\underline{\lambda}_{1}= 3μvh(b¯2hμv2b¯μa/τ)+3h2μa(2b¯2μvh3b¯2μa),\displaystyle 3\mu_{v}h(\underline{b}^{2}h\mu_{v}-2\overline{b}\mu_{a}/\tau)+3h^{2}\mu_{a}(2\underline{b}^{2}\mu_{v}h-3\overline{b}^{2}\mu_{a}),
λ¯2=\displaystyle\overline{\lambda}_{2}= 16b¯2hμaμp16(b¯μaμv+b¯hμaμp)/τ,\displaystyle 16\overline{b}^{2}h\mu_{a}\mu_{p}-16(\underline{b}\mu_{a}\mu_{v}+\underline{b}h\mu_{a}\mu_{p})/\tau,
λ¯2=\displaystyle\underline{\lambda}_{2}= 16b¯2hμaμp16(b¯μaμv+b¯hμaμp)/τ,\displaystyle 16\underline{b}^{2}h\mu_{a}\mu_{p}-16(\overline{b}\mu_{a}\mu_{v}+\overline{b}h\mu_{a}\mu_{p})/\tau,
λ¯3=\displaystyle\overline{\lambda}_{3}= 3b¯2h2μp2+6b¯2μaμp12b¯μpμa/τ,\displaystyle 3\overline{b}^{2}h^{2}\mu_{p}^{2}+6\overline{b}^{2}\mu_{a}\mu_{p}-12\underline{b}\mu_{p}\mu_{a}/\tau,
λ¯3=\displaystyle\underline{\lambda}_{3}= 3b¯2h2μp2+6b¯2μaμp12b¯μpμa/τ,\displaystyle 3\underline{b}^{2}h^{2}\mu_{p}^{2}+6\underline{b}^{2}\mu_{a}\mu_{p}-12\overline{b}\mu_{p}\mu_{a}/\tau,
α¯1=\displaystyle\overline{\alpha}_{1}= (μa/τb¯μa+b¯hμv)2,α¯1=(μa/τb¯μa+b¯hμv)2,\displaystyle(\mu_{a}/\tau-\underline{b}\mu_{a}+\overline{b}h\mu_{v})^{2},\underline{\alpha}_{1}=(\mu_{a}/\tau-\overline{b}\mu_{a}+\underline{b}h\mu_{v})^{2},
α¯2=\displaystyle\overline{\alpha}_{2}= ((12b¯hμaμv18b¯μa2)/τ12b¯2hμaμv+3b¯2h2μv2\displaystyle((12\overline{b}h\mu_{a}\mu_{v}-18\underline{b}\mu_{a}^{2})/\tau-12\underline{b}^{2}h\mu_{a}\mu_{v}+3\overline{b}^{2}h^{2}\mu_{v}^{2}
+9μa2/τ2+9b¯2μa2)ωo2+b¯2h2μp2+2b¯2hμaμp/τ\displaystyle+9\mu_{a}^{2}/\tau^{2}+9\overline{b}^{2}\mu_{a}^{2})\omega_{o}^{2}+\overline{b}^{2}h^{2}\mu_{p}^{2}+2\overline{b}^{2}h\mu_{a}\mu_{p}/\tau
2b¯μaμp/τ,\displaystyle-2\underline{b}\mu_{a}\mu_{p}/\tau,
α¯2=\displaystyle\underline{\alpha}_{2}= ((12b¯hμaμv18b¯μa2)/τ12b¯2hμaμv+3b¯2h2μv2\displaystyle((12\underline{b}h\mu_{a}\mu_{v}-18\overline{b}\mu_{a}^{2})/\tau-12\overline{b}^{2}h\mu_{a}\mu_{v}+3\underline{b}^{2}h^{2}\mu_{v}^{2}
+9μa2/τ2+9b¯2μa2)ωo2+b¯2h2μp2+2b¯2μaμp/τ\displaystyle+9\mu_{a}^{2}/\tau^{2}+9\underline{b}^{2}\mu_{a}^{2})\omega_{o}^{2}+\underline{b}^{2}h^{2}\mu_{p}^{2}+2\underline{b}^{2}\mu_{a}\mu_{p}/\tau
2b¯μaμp/τ,\displaystyle-2\overline{b}\mu_{a}\mu_{p}/\tau,
α¯3=\displaystyle\overline{\alpha}_{3}= λ¯1ωo4+λ¯2ωo3+λ¯3ωo2,α¯3=λ¯1ωo4+λ¯2ωo3+λ¯3ωo2,\displaystyle\overline{\lambda}_{1}\omega_{o}^{4}+\overline{\lambda}_{2}\omega_{o}^{3}+\overline{\lambda}_{3}\omega_{o}^{2},\underline{\alpha}_{3}=\underline{\lambda}_{1}\omega_{o}^{4}+\underline{\lambda}_{2}\omega_{o}^{3}+\underline{\lambda}_{3}\omega_{o}^{2},
α¯4=\displaystyle\overline{\alpha}_{4}= (b¯2h2μv2b¯2μa2)ωo6+(3b¯2h2μp2+6b¯2μpμa\displaystyle(\overline{b}^{2}h^{2}\mu^{2}_{v}-\underline{b}^{2}\mu^{2}_{a})\omega_{o}^{6}+(3\overline{b}^{2}h^{2}\mu_{p}^{2}+6\overline{b}^{2}\mu_{p}\mu_{a}
+6b¯μpμa/τ)ωo4,\displaystyle+6\overline{b}\mu_{p}\mu_{a}/\tau)\omega_{o}^{4},
α¯4=\displaystyle\underline{\alpha}_{4}= (b¯2h2μv2b¯2μa2)ωo6+(3b¯2h2μp2+6b¯2μpμa\displaystyle(\underline{b}^{2}h^{2}\mu^{2}_{v}-\overline{b}^{2}\mu^{2}_{a})\omega_{o}^{6}+(3\underline{b}^{2}h^{2}\mu_{p}^{2}+6\underline{b}^{2}\mu_{p}\mu_{a}
+6b¯μpμa/τ)ωo4\displaystyle+6\underline{b}\mu_{p}\mu_{a}/\tau)\omega_{o}^{4}
α¯5=\displaystyle\overline{\alpha}_{5}= (b¯2h2μp2+2b¯2μaμp)ωo6,γ¯5=2b¯2μpωo6,\displaystyle(\overline{b}^{2}h^{2}\mu_{p}^{2}+2\overline{b}^{2}\mu_{a}\mu_{p})\omega_{o}^{6},\underline{\gamma}_{5}=2\underline{b}^{2}\mu_{p}\omega_{o}^{6},
γ¯1=\displaystyle\overline{\gamma}_{1}= 2b¯μa/τ+2b¯2hμv2b¯2μa2b¯hμp2b¯μv,\displaystyle 2\overline{b}\mu_{a}/\tau+2\overline{b}^{2}h\mu_{v}-2\underline{b}^{2}\mu_{a}-2\underline{b}h\mu_{p}-2\underline{b}\mu_{v},
γ¯1=\displaystyle\underline{\gamma}_{1}= 2b¯μa/τ+2b¯2hμv2b¯2μa2b¯hμp2b¯μv,\displaystyle 2\underline{b}\mu_{a}/\tau+2\underline{b}^{2}h\mu_{v}-2\overline{b}^{2}\mu_{a}-2\overline{b}h\mu_{p}-2\overline{b}\mu_{v},
γ¯2=\displaystyle\overline{\gamma}_{2}= 16(μa/τb¯μa)ωo3+(12b¯μa/τ+6b¯2hμv12b¯2μa\displaystyle 16(\mu_{a}/\tau-\underline{b}\mu_{a})\omega_{o}^{3}+(12\overline{b}\mu_{a}/\tau+6\overline{b}^{2}h\mu_{v}-12\underline{b}^{2}\mu_{a}
6b¯μv6b¯hμp)ωo22b¯2μp,\displaystyle-6\underline{b}\mu_{v}-6\underline{b}h\mu_{p})\omega_{o}^{2}-2\underline{b}^{2}\mu_{p},
γ¯2=\displaystyle\underline{\gamma}_{2}= 16(μa/τb¯μa)ωo3+(12b¯μa/τ+6b¯2hμv12b¯2μa\displaystyle 16(\mu_{a}/\tau-\overline{b}\mu_{a})\omega_{o}^{3}+(12\underline{b}\mu_{a}/\tau+6\underline{b}^{2}h\mu_{v}-12\overline{b}^{2}\mu_{a}
6b¯μv6b¯hμp)ωo22b¯2μp,\displaystyle-6\overline{b}\mu_{v}-6\overline{b}h\mu_{p})\omega_{o}^{2}-2\overline{b}^{2}\mu_{p},
γ¯3=\displaystyle\overline{\gamma}_{3}= (6b¯2μa6b¯μv6b¯hμp6b¯μa/τ+6b¯2hμv)ωo4\displaystyle(6\overline{b}^{2}\mu_{a}-6\underline{b}\mu_{v}-6\underline{b}h\mu_{p}-6\underline{b}\mu_{a}/\tau+6\overline{b}^{2}h\mu_{v})\omega_{o}^{4}
+16b¯2μaωo3/τ6b¯2μpωo2,\displaystyle+16\overline{b}^{2}\mu_{a}\omega_{o}^{3}/\tau-6\underline{b}^{2}\mu_{p}\omega_{o}^{2},
γ¯3=\displaystyle\underline{\gamma}_{3}= (6b¯2μa6b¯μv6b¯hμp6b¯μa/τ+6b¯2hμv)ωo4\displaystyle(6\underline{b}^{2}\mu_{a}-6\overline{b}\mu_{v}-6\overline{b}h\mu_{p}-6\overline{b}\mu_{a}/\tau+6\underline{b}^{2}h\mu_{v})\omega_{o}^{4}
+16b¯2μaωo3/τ6b¯2μpωo2,\displaystyle+16\underline{b}^{2}\mu_{a}\omega_{o}^{3}/\tau-6\overline{b}^{2}\mu_{p}\omega_{o}^{2},
γ¯4=\displaystyle\overline{\gamma}_{4}= (2b¯2hμv2b¯hμp2b¯μv)ωo66b¯2μpωo4,\displaystyle(2\overline{b}^{2}h\mu_{v}-2\underline{b}h\mu_{p}-2\underline{b}\mu_{v})\omega_{o}^{6}-6\underline{b}^{2}\mu_{p}\omega_{o}^{4},
γ¯4=\displaystyle\underline{\gamma}_{4}= (2b¯2hμv2b¯hμp2b¯μv)ωo66b¯2μpωo4,\displaystyle(2\underline{b}^{2}h\mu_{v}-2\overline{b}h\mu_{p}-2\overline{b}\mu_{v})\omega_{o}^{6}-6\overline{b}^{2}\mu_{p}\omega_{o}^{4},
θi=\displaystyle\theta_{i}= ((max{|γ¯i|,|γ¯i|})2+4α¯iρ¯i+max{|γ¯i|,\displaystyle(\sqrt{(\max\{|\overline{\gamma}_{i}|,|\underline{\gamma}_{i}|\})^{2}+4\overline{\alpha}_{i}\overline{\rho}_{i}}+\max\{|\overline{\gamma}_{i}|,
|γ¯i|})/(2α¯i),i=1,2,3,4.\displaystyle|\underline{\gamma}_{i}|\})/(2\underline{\alpha}_{i}),\;i=1,2,3,4.

Appendix B Proof of Theorem 1

The proof of Theorem 1 needs the following lemmas.

Lemma B.7.

(Hinrichsen & Pritchard, 1986; Guo, 1993) Suppose x˙(t)=Ax(t)\dot{x}(t)=Ax(t) is exponentially stable, where An×nA\in\mathbb{C}^{n\times n}. Denote rc(A)=minωSn(iωIA)r_{c}(A)=\min\limits_{\omega\in\mathbb{R}}S_{n}(i\omega I-A). If B<rc(A)\|B\|<r_{c}(A), then x˙(t)=(A+B)x(t)\dot{x}(t)=(A+B)x(t) is exponentially stable. Further, there exists Bn×nB\in\mathbb{C}^{n\times n} with B=rc(A)\|B\|=r_{c}(A), such that x˙(t)=(A+B)x(t)\dot{x}(t)=(A+B)x(t) is not asymptotically stable.

Lemma B.8.

For any AA \in n×n\mathbb{R}^{n\times n}, Ai=1nj=1n|aij|\|A\|\leq\sum^{n}_{i=1}\sum^{n}_{j=1}|a_{ij}|, where aija_{ij} is the element of the iith row and jjth column of AA.

Proof. Denote

bpqij={aij,p=i,q=j,0,otherwise.b_{pq}^{ij}=\left\{\begin{array}[]{ll}a_{ij},&p=i,q=j,\vspace{1ex}\\ 0,&\rm{otherwise}.\end{array}\right. (36)

Define Aij=[bpqij]n×nA_{ij}=\left[b_{pq}^{ij}\right]_{n\times n}, where bpqijb_{pq}^{ij} is the element of the ppth row and qqth column of AijA_{ij}.

By (36) and the definition of AijA_{ij}, we have

A=i=1nj=1nAij.\displaystyle A=\sum^{n}_{i=1}\sum^{n}_{j=1}A_{ij}.

This together with the triangle inequality of matrix norm leads to

Ai=1nj=1nAij.\displaystyle\|A\|\leq\sum^{n}_{i=1}\sum^{n}_{j=1}\|A_{ij}\|. (37)

By the definition of the 2-norm of matrix, we have

Aij=λmax(AijTAij),\displaystyle\|A_{ij}\|=\sqrt{\lambda_{max}(A_{ij}^{T}A_{ij})}, (38)

where λmax(AijTAij)\lambda_{max}(A_{ij}^{T}A_{ij}) is the maximum eigenvalue of AijTAijA_{ij}^{T}A_{ij}.

By (36) and the definition of AijA_{ij}, we get

(AijTAij)pq\displaystyle(A_{ij}^{T}A_{ij})_{pq} =k=1nbkpijbkqij={aij2,p=j,q=j,0,otherwise.\displaystyle=\sum^{n}_{k=1}b_{kp}^{ij}b_{kq}^{ij}=\left\{\begin{array}[]{ll}a_{ij}^{2},&p=j,q=j,\vspace{1ex}\\ 0,&\rm{otherwise}.\end{array}\right.

This together with (38) gives Aij=|aij|\|A_{ij}\|=|a_{ij}|. Then by (37), we get Ai=1nj=1n|aij|\|A\|\leq\sum^{n}_{i=1}\sum^{n}_{j=1}|a_{ij}|. ∎

Proof of Theorem 1. Denote

Xi(t)=\displaystyle X_{i}(t)= [pi(t),vi(t),ai(t)]T,i=0,1,2,,N,\displaystyle[p_{i}(t),v_{i}(t),a_{i}(t)]^{T},\;i=0,1,2,...,N,\vspace{1ex} (39)
Ei(t)=\displaystyle E_{i}(t)= [e1,i(t),e2,i(t),e3,i(t)]T,i=1,2,,N,\displaystyle[e_{1,i}(t),e_{2,i}(t),e_{3,i}(t)]^{T},\;i=1,2,...,N,\vspace{1ex} (40)
Fi(t)=\displaystyle F_{i}(t)= [ei(t),vd,i(t),ai(t)]T,i=1,2,N,\displaystyle[e_{i}(t),v_{d,i}(t),a_{i}(t)]^{T},\;i=1,2,...N,\vspace{1ex} (41)
W(t)=\displaystyle W(t)= [F1T(t),F2T(t),,FNT(t),E1T(t),E2T(t),\displaystyle[F_{1}^{T}(t),F_{2}^{T}(t),\cdots,F_{N}^{T}(t),E_{1}^{T}(t),E_{2}^{T}(t),
,ENT(t)]T,\displaystyle\cdots,E_{N}^{T}(t)]^{T},\vspace{1ex} (42)
Δ(t)=\displaystyle\Delta(t)= [δ1T(t),0,,0,ζ1T(t),ζ2T(t),ζ3T(t),0,\displaystyle[\delta_{1}^{T}(t),0,\cdots,0,\zeta_{1}^{T}(t),\zeta_{2}^{T}(t),\zeta_{3}^{T}(t),0,
,0]T,\displaystyle\cdots,0]^{T}, (43)

where

e1,i(t)=\displaystyle e_{1,i}(t)= z1,i(t)vd,i(t),\displaystyle z_{1,i}(t)-v_{d,i}(t), (44)
e2,i(t)=\displaystyle e_{2,i}(t)= z2,i(t)ad,i(t),\displaystyle z_{2,i}(t)-a_{d,i}(t), (45)
e3,i(t)=\displaystyle e_{3,i}(t)= z3,i(t)qi(t),\displaystyle z_{3,i}(t)-q_{i}(t), (46)
δ1(t)=\displaystyle\delta_{1}(t)= [0,a0(t),kaa0(t)/τ]T,\displaystyle\left[0,a_{0}(t),-k_{a}a_{0}(t)/\tau\right]^{T}, (47)
ζ1(t)=\displaystyle\zeta_{1}(t)= [0,0,(u0(t)a0(t)τu˙0(t))/τ2]T,\displaystyle\left[0,0,(u_{0}(t)-a_{0}(t)-\tau\dot{u}_{0}(t))/\tau^{2}\right]^{T}, (48)
ζ2(t)=\displaystyle\zeta_{2}(t)= [0,0,(kaa0(t)τkva0(t)kakvha0(t)\displaystyle\left[0,0,(k_{a}a_{0}(t)-\tau k_{v}a_{0}(t)-k_{a}k_{v}ha_{0}(t)\right.
τkaa˙0(t))/τ2]T,\displaystyle\left.-\tau k_{a}\dot{a}_{0}(t))/\tau^{2}\right]^{T}, (49)
ζ3(t)=\displaystyle\zeta_{3}(t)= [0,0,ka2a0(t)/τ]T.\displaystyle\left[0,0,k_{a}^{2}a_{0}(t)/\tau\right]^{T}. (50)

From (6), (3) and (45), we know

ui(t)=\displaystyle u_{i}(t)= kpei(t)+kv(vd,i(t)hai(t))\displaystyle\;k_{p}e_{i}(t)+k_{v}(v_{d,i}(t)-ha_{i}(t))
+ka(ai1(t)+e2i(t)),i=1,2,,N.\displaystyle+k_{a}(a_{i-1}(t)+e_{2i}(t)),\;i=1,2,\ldots,N. (51)

This together with (1), (3), (5), (39) and (40) leads to

Xi˙(t)={A0Xi(t)+B0ui(t),i=0,AXi(t)+BXi1(t)+CEi(t)+Lr,i=1,,N,\displaystyle\dot{X_{i}}(t)=\left\{\begin{array}[]{ll}A_{0}X_{i}(t)+B_{0}u_{i}(t),&i=0,\vspace{1ex}\\ AX_{i}(t)+BX_{i-1}(t)+CE_{i}(t)&+Lr,\\ &i=1,\ldots,N,\end{array}\right. (55)

where

A0=[010001001τ],B0=[001τ],L=[00kpτ],A_{0}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&\begin{aligned} -\frac{1}{\tau}\end{aligned}\end{bmatrix},B_{0}=\begin{bmatrix}0\\ 0\\ \frac{1}{\tau}\end{bmatrix},L=\begin{bmatrix}0\\ 0\\ \begin{aligned} -\frac{k_{p}}{\tau}\end{aligned}\end{bmatrix},
A=[010001kpτkv+kphτ1+kvhτ],A=\begin{bmatrix}0&1&0\\ 0&0&1\\ \begin{aligned} -\frac{k_{p}}{\tau}\end{aligned}&\begin{aligned} -\frac{k_{v}+k_{p}h}{\tau}\end{aligned}&\begin{aligned} -\frac{1+k_{v}h}{\tau}\end{aligned}\end{bmatrix},
B=[000000kpτkvτkaτ],C=[0000000kaτ0].B=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} \frac{k_{p}}{\tau}\end{aligned}&\begin{aligned} \frac{k_{v}}{\tau}\end{aligned}&\begin{aligned} \frac{k_{a}}{\tau}\end{aligned}\end{bmatrix},C=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} \frac{k_{a}}{\tau}\end{aligned}&0\end{bmatrix}.

From (3), (39) and (41), we get

Fi(t)=PXi1(t)QXi(t)L1r,i=1,2,,N,\displaystyle F_{i}(t)=PX_{i-1}(t)-QX_{i}(t)-L_{1}r,\;i=1,2,\ldots,N, (56)

where

P=[100010000],Q=[1h0010001],L1=[100].P={\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&0\end{array}\right]},Q={\left[\begin{array}[]{ccc}1&h&0\\ 0&1&0\\ 0&0&-1\end{array}\right]},L_{1}={\left[\begin{array}[]{ccc}1\\ 0\\ 0\end{array}\right]}.

From (47), (55) and (56), we obtain

Fi˙(t)={𝒜Fi(t)+𝒢Ei(t)+δi(t),i=1,𝒜Fi(t)+1Fi1(t)+𝒢Ei(t),i=2,,N,\displaystyle\dot{F_{i}}(t)=\left\{\begin{array}[]{ll}\mathcal{A}F_{i}(t)+\mathcal{G}E_{i}(t)+\delta_{i}(t),&i=1,\vspace{1ex}\\ \mathcal{A}F_{i}(t)+\mathcal{B}_{1}F_{i-1}(t)+\mathcal{G}E_{i}(t),&i=2,\ldots,N,\end{array}\right. (59)

where

1=[00000100kaτ],𝒢=[0000000kaτ0].\mathcal{B}_{1}=\begin{bmatrix}0&0&0\\ 0&0&1\\ 0&0&\begin{aligned} \frac{k_{a}}{\tau}\end{aligned}\end{bmatrix},\mathcal{G}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} \frac{k_{a}}{\tau}\end{aligned}&0\end{bmatrix}.

By (3)-(9), (40), (41), (44)-(B) and (59), we have

Ei˙(t)={Ei(t)+ζi(t),i=1,𝒟Fi1(t)+Ei1(t)+ζi(t),i=2,𝒟Fi1(t)+1Fi2(t)+Ei1(t)+𝒥Ei2(t)+ζi(t),i=3,𝒟Fi1(t)+1Fi2(t)+Fi3(t)+Ei1(t)+𝒥Ei2(t),i=4,,N,\displaystyle\dot{E_{i}}(t)=\left\{\begin{array}[]{ll}\mathcal{H}E_{i}(t)+\zeta_{i}(t),&i=1,\vspace{1ex}\\ \mathcal{D}F_{i-1}(t)+\mathcal{I}E_{i-1}(t)+\zeta_{i}(t),&i=2,\vspace{1ex}\\ \mathcal{D}F_{i-1}(t)+\mathcal{E}_{1}F_{i-2}(t)+\mathcal{I}E_{i-1}(t)\\ +\mathcal{J}E_{i-2}(t)+\zeta_{i}(t),&i=3,\vspace{1ex}\\ \mathcal{D}F_{i-1}(t)+\mathcal{E}_{1}F_{i-2}(t)+\mathcal{F}F_{i-3}(t)\\ +\mathcal{I}E_{i-1}(t)+\mathcal{J}E_{i-2}(t),&i=4,\ldots,N,\end{array}\right. (66)

where

𝒥=[0000000ka2τ20],=[00000000ka2τ2],\mathcal{J}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} -\frac{k_{a}^{2}}{\tau^{2}}\end{aligned}&0\end{bmatrix},\mathcal{F}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&\begin{aligned} -\frac{k_{a}^{2}}{\tau^{2}}\end{aligned}\end{bmatrix},
=[000000β2kaτ(1+kvh)kaτ2kaτ].\mathcal{I}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} \frac{\beta_{2}k_{a}}{\tau}\end{aligned}&\begin{aligned} \frac{(1+k_{v}h)k_{a}}{\tau^{2}}\end{aligned}&\begin{aligned} -\frac{k_{a}}{\tau}\end{aligned}\end{bmatrix}.
1=[000000kpkaτ2kvkaτ2kvτ+2(1+kvh)kaτ2],\mathcal{E}_{1}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} -\frac{k_{p}k_{a}}{\tau^{2}}\end{aligned}&\begin{aligned} -\frac{k_{v}k_{a}}{\tau^{2}}\end{aligned}&\begin{aligned} -\frac{k_{v}}{\tau}+\frac{2(1+k_{v}h)k_{a}}{\tau^{2}}\end{aligned}\end{bmatrix},

From (42), (43), (59) and (66), we know

W˙(t)=(Ψ+Ψ^)W(t)+Δ(t),\displaystyle\dot{W}(t)=(\Psi+\hat{\Psi})W(t)+\Delta(t), (67)

where Ψ^=[Ψ^11Ψ^12Ψ^21Ψ^22]\hat{\Psi}=\begin{bmatrix}\hat{\Psi}_{11}&\hat{\Psi}_{12}\\ \hat{\Psi}_{21}&\hat{\Psi}_{22}\end{bmatrix} and

Ψ^11=[O2O2O],Ψ^22=[OO𝒥O𝒥O],\hat{\Psi}_{11}=\begin{bmatrix}O&\quad&\quad&\quad\\ \mathcal{B}_{2}&O&\quad&\quad\\ \quad&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{B}_{2}&O\\ \end{bmatrix},\hat{\Psi}_{22}=\begin{bmatrix}O&\quad&\quad&\quad&\quad\\ \mathcal{I}&O&\quad&\quad&\quad\\ \mathcal{J}&\mathcal{I}&O&\quad&\quad\\ \quad&\ddots&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{J}&\mathcal{I}&O\end{bmatrix},
Ψ^21=[OOO2OO2OO2OO],Ψ^12=[𝒢𝒢],\hat{\Psi}_{21}=\begin{bmatrix}O&\quad&\quad&\quad&\quad&\quad\\ O&O&\quad&\quad&\quad&\quad\\ \mathcal{E}_{2}&O&O&\quad&\quad&\quad\\ \mathcal{F}&\mathcal{E}_{2}&O&O&\quad&\quad\\ \quad&\ddots&\ddots&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{F}&\mathcal{E}_{2}&O&O\end{bmatrix},\hat{\Psi}_{12}=\begin{bmatrix}\mathcal{G}&\quad&\quad\\ \quad&\ddots&\quad\\ \quad&\quad&\mathcal{G}\\ \end{bmatrix},
2=1,2=1.\displaystyle\mathcal{B}_{2}=\mathcal{B}_{1}-\mathcal{B},\mathcal{E}_{2}=\mathcal{E}_{1}-\mathcal{E}.

Firstly, we analyze the stability of Ψ\Psi. The eigenvalues of Ψ\Psi are only related to 𝒜\mathcal{A} and \mathcal{H}. Calculating the characteristic polynomial of 𝒜\mathcal{A}, we obtain

|sI𝒜|=s3+(1+kvhτ)s2+(kv+kphτ)s+kpτ.\displaystyle\left|sI-\mathcal{A}\right|=s^{3}+\left(\frac{1+k_{v}h}{\tau}\right)s^{2}+\left(\frac{k_{v}+k_{p}h}{\tau}\right)s+\frac{k_{p}}{\tau}. (68)

The Rouse table corresponding to (68) is given by

s3s^{3} 1 kv+kphτ\begin{aligned} \frac{k_{v}+k_{p}h}{\tau}\end{aligned}
s2s^{2} 1+kvhτ\begin{aligned} \frac{1+k_{v}h}{\tau}\end{aligned} kpτ\begin{aligned} \frac{k_{p}}{\tau}\end{aligned}
s1s^{1} hkv2+(1+h2kp)kv+(hτ)kpτ+τkvh\begin{aligned} \frac{hk_{v}^{2}+(1+h^{2}k_{p})k_{v}+(h-\tau)k_{p}}{\tau+\tau k_{v}h}\end{aligned} 0
s0s^{0} kpτ\begin{aligned} \frac{k_{p}}{\tau}\end{aligned}

By kp>0k_{p}>0 and (11), we know that the elements of the first column of the Rouse table corresponding to (68) are all greater than zero. From Rouse criterion, 𝒜\mathcal{A} is stable. Calculating the characteristic polynomial of \mathcal{H}, we get

|sI|=s3+β1s2+β2s+β3.\displaystyle\left|sI-\mathcal{H}\right|=s^{3}+\beta_{1}s^{2}+\beta_{2}s+\beta_{3}. (69)

The Rouse table corresponding to (69) is given by

s3s^{3} 1 β2\beta_{2}
s2s^{2} β1\beta_{1} β3\beta_{3}
s1s^{1} β1β2β3β1\begin{aligned} \frac{\beta_{1}\beta_{2}-\beta_{3}}{\beta_{1}}\end{aligned} 0
s0s^{0} β3\beta_{3}

By β1>0\beta_{1}>0, β3>0\beta_{3}>0, β1β2β3>0\beta_{1}\beta_{2}-\beta_{3}>0, we know that the elements of the first column of the Rouse table corresponding to (69) are all greater than zero. From Rouse criterion, \mathcal{H} is stable. Then Ψ\Psi is stable. From the definition of Ψ^\|\hat{\Psi}\| and Lemma 2, we know

Ψ^{ka/τ,ifN=1,ka(kvh+τβ2+4τ+1)/τ2,ifN=2,(2N5)ka2/τ2+Θka,ifN3.\displaystyle\|\hat{\Psi}\|\leq\left\{\begin{array}[]{lc}\begin{aligned} k_{a}/\tau\end{aligned},&$\rm{if}$\ N=1,\vspace{1ex}\\ \begin{aligned} k_{a}(k_{v}h+\tau\beta_{2}+4\tau+1)/\tau^{2}\end{aligned},&$\rm{if}$\ N=2,\vspace{1ex}\\ \begin{aligned} (2N-5)k_{a}^{2}/\tau^{2}+\Theta k_{a}\end{aligned},&$\rm{if}$\ N\geq 3.\end{array}\right. (73)

From (12) and (73), we get Ψ^<rc(Ψ)\|\hat{\Psi}\|<r_{c}(\Psi). It is known from the definition of Δ(t)\Delta(t) and Assumption 1 that limtΔ(t)=0\lim\limits_{t\to\infty}\Delta(t)=0. By Lemma 1 and (67), we know W(t)W(t) converges to zero exponentially. Then vi1(t)vi(t)v_{i-1}(t)-v_{i}(t) and ei(t)e_{i}(t) both converge to zero exponentially, i=1,2,,Ni=1,2,...,N, which implies vi(t)v0(t)v_{i}(t)-v_{0}(t) converges to zero exponentially, i=1,2,,Ni=1,2,...,N. ∎

Appendix C Proof of Theorem 2

Proof of Theorem 2. By (3) and (5), we get

ai(t)=vd,i(t)e˙i(t)h.\displaystyle a_{i}(t)=\frac{v_{d,i}(t)-\dot{e}_{i}(t)}{h}. (74)

Taking the Laplace transform of (74), we have

𝒜i(s)=𝒱d,i(s)si(s)h,\displaystyle\mathscr{A}_{i}(s)=\frac{\mathscr{V}_{d,i}(s)-s\mathscr{E}_{i}(s)}{h}, (75)

where 𝒜i(s)\mathscr{A}_{i}(s) and 𝒱d,i(s)\mathscr{V}_{d,i}(s) are the Laplace transform of ai(t)a_{i}(t), vd,i(t)v_{d,i}(t), respectively. From (1), we know

ui(t)=τa˙i(t)+ai(t).\displaystyle u_{i}(t)=\tau\dot{a}_{i}(t)+a_{i}(t). (76)

This together with (74) leads to

ui(t)=τv˙d,i(t)e¨i(t)h+vd,i(t)e˙i(t)h.\displaystyle u_{i}(t)=\tau\frac{\dot{v}_{d,i}(t)-\ddot{e}_{i}(t)}{h}+\frac{v_{d,i}(t)-\dot{e}_{i}(t)}{h}. (77)

Taking the Laplace transform of (77), we get

𝒰i(s)=τs𝒱d,i(s)s2i(s)h+𝒱d,i(s)si(s)h,\displaystyle\mathscr{U}_{i}(s)=\tau\frac{s\mathscr{V}_{d,i}(s)-s^{2}\mathscr{E}_{i}(s)}{h}+\frac{\mathscr{V}_{d,i}(s)-s\mathscr{E}_{i}(s)}{h}, (78)

where 𝒰i(s)\mathscr{U}_{i}(s) is the Laplace transform of ui(t)u_{i}(t). Taking the Laplace transform of (9), we have

{s𝒵1,i(s)=𝒵2,i(s)+β1(𝒱d,i(s)𝒵1,i(s)),s𝒵2,i(s)=𝒵3,i(s)+β2(𝒱d,i(s)𝒵1,i(s))+𝒜i(s)/τ𝒰i(s)/τ,s𝒵3,i(s)=β3(𝒱d,i(s)𝒵1,i(s)),\displaystyle\left\{\begin{array}[]{lll}s\mathscr{Z}_{1,i}(s)=&\mathscr{Z}_{2,i}(s)+\beta_{1}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s)),\vspace{1ex}\\ s\mathscr{Z}_{2,i}(s)=&\mathscr{Z}_{3,i}(s)+\beta_{2}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s))+\mathscr{A}_{i}(s)/\tau\\ &-\mathscr{U}_{i}(s)/\tau,\vspace{1ex}\\ s\mathscr{Z}_{3,i}(s)=&\beta_{3}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s)),\vspace{1ex}\end{array}\right. (83)

where 𝒵1,i(s)\mathscr{Z}_{1,i}(s), 𝒵2,i(s)\mathscr{Z}_{2,i}(s) and 𝒵3,i(s)\mathscr{Z}_{3,i}(s) are the Laplace transform of z1,i(t)z_{1,i}(t), z2,i(t)z_{2,i}(t) and z3,i(t)z_{3,i}(t), respectively. Substituting (78) into (83), we obtain

𝒵2,i(s)=\displaystyle\mathscr{Z}_{2,i}(s)= ((s3+(hβ2β1)s2+hβ3s)𝒱d,i(s)+(s4\displaystyle\;((-s^{3}+(h\beta_{2}-\beta_{1})s^{2}+h\beta_{3}s)\mathscr{V}_{d,i}(s)+(s^{4}
+β1s3)i(s)/(h(s3+β1s2+β2s+β3))\displaystyle+\beta_{1}s^{3})\mathscr{E}_{i}(s)/(h(s^{3}+\beta_{1}s^{2}+\beta_{2}s+\beta_{3})) (84)

By (3), (74) and (76), we get

τa˙i(t)+ai(t)=\displaystyle\tau\dot{a}_{i}(t)+a_{i}(t)= kpei(t)+kve˙i(t)\displaystyle\;k_{p}e_{i}(t)+k_{v}\dot{e}_{i}(t)
+ka(z2,i(t)+ai(t)).\displaystyle+k_{a}(z_{2,i}(t)+a_{i}(t)). (85)

Taking the Laplace transform of (C), we have

τs𝒜i(s)+𝒜i(s)=\displaystyle\tau s\mathscr{A}_{i}(s)+\mathscr{A}_{i}(s)= kpi(s)+kvsi(s)+ka𝒜i(s)\displaystyle\;k_{p}\mathscr{E}_{i}(s)+k_{v}s\mathscr{E}_{i}(s)+k_{a}\mathscr{A}_{i}(s)
+ka𝒵2,i(s).\displaystyle+k_{a}\mathscr{Z}_{2,i}(s). (86)

Denote H(s)=𝒱d,i(s)i(s)H(s)=\frac{\mathscr{V}_{d,i}(s)}{\mathscr{E}_{i}(s)}. By (75), (E) and (C), we get

H(s)=τs5+n4s4+n3s3+n2s2+n1s+n0τs4+d3s3+d2s2+d1s+d0,\displaystyle H(s)=\frac{\tau s^{5}+n_{4}s^{4}+n_{3}s^{3}+n_{2}s^{2}+n_{1}s+n_{0}}{\tau s^{4}+d_{3}s^{3}+d_{2}s^{2}+d_{1}s+d_{0}}, (87)

where

n0=kphβ3,\displaystyle n_{0}=k_{p}h\beta_{3},
n1=kphβ2+(1ka+kvh)β3,\displaystyle n_{1}=k_{p}h\beta_{2}+(1-k_{a}+k_{v}h)\beta_{3},
n2=kphβ1+(1ka+kvh)β2+τβ3,\displaystyle n_{2}=k_{p}h\beta_{1}+(1-k_{a}+k_{v}h)\beta_{2}+\tau\beta_{3},
n3=kph+(1+kvh)β1+τβ2,\displaystyle n_{3}=k_{p}h+(1+k_{v}h)\beta_{1}+\tau\beta_{2},
n4=1+kvh+τβ1,\displaystyle n_{4}=1+k_{v}h+\tau\beta_{1},
d0=(1ka)β3,\displaystyle d_{0}=(1-k_{a})\beta_{3},
d1=(1ka)β2+(τkah)β3,\displaystyle d_{1}=(1-k_{a})\beta_{2}+(\tau-k_{a}h)\beta_{3},
d2=β1+(τkah)β2,\displaystyle d_{2}=\beta_{1}+(\tau-k_{a}h)\beta_{2},
d3=τβ1+1.\displaystyle d_{3}=\tau\beta_{1}+1.

By (3) and (5), we get

e˙i1(t)e˙i(t)=vd,i1(t)vd,i(t)hv˙d,i(t).\displaystyle\dot{e}_{i-1}(t)-\dot{e}_{i}(t)=v_{d,i-1}(t)-v_{d,i}(t)-h\dot{v}_{d,i}(t). (88)

Taking the Laplace transform of (88), we have

si1(t)si(t)=𝒱d,i1(t)𝒱d,i(t)hs𝒱d,i(t).\displaystyle s\mathscr{E}_{i-1}(t)-s\mathscr{E}_{i}(t)=\mathscr{V}_{d,i-1}(t)-\mathscr{V}_{d,i}(t)-hs\mathscr{V}_{d,i}(t).

This together with 𝒱d,i(s)=H(s)i(s)\mathscr{V}_{d,i}(s)=H(s)\mathscr{E}_{i}(s) leads to

Gei(s)=sH(s)s(hs+1)H(s).\displaystyle G_{ei}(s)=\frac{s-H(s)}{s-(hs+1)H(s)}.

This together with (87) leads to

Gei(s)=\displaystyle G_{ei}(s)= (kvs4+n¯3s3+n¯2s2+n¯1s+kpβ3)/(τs6\displaystyle\;(k_{v}s^{4}+\overline{n}_{3}s^{3}+\overline{n}_{2}s^{2}+\overline{n}_{1}s+k_{p}\beta_{3})/(\tau s^{6}
+d¯5s5+d¯4s4+d¯3s3+d¯2s2+d¯1s+kpβ3),\displaystyle+\overline{d}_{5}s^{5}+\overline{d}_{4}s^{4}+\overline{d}_{3}s^{3}+\overline{d}_{2}s^{2}+\overline{d}_{1}s+k_{p}\beta_{3}), (89)

where

n¯1=\displaystyle\overline{n}_{1}= kpβ2+kvβ3,\displaystyle k_{p}\beta_{2}+k_{v}\beta_{3},
n¯2=\displaystyle\overline{n}_{2}= kpβ1+kvβ2+kaβ3,\displaystyle k_{p}\beta_{1}+k_{v}\beta_{2}+k_{a}\beta_{3},
n¯3=\displaystyle\overline{n}_{3}= kvβ1+kaβ2+kp,\displaystyle k_{v}\beta_{1}+k_{a}\beta_{2}+k_{p},
d¯1=\displaystyle\overline{d}_{1}= kpβ2+(kph+kv)β3,\displaystyle k_{p}\beta_{2}+(k_{p}h+k_{v})\beta_{3},
d¯2=\displaystyle\overline{d}_{2}= kpβ1+(kph+kv)β2+(1+kvh)β3,\displaystyle k_{p}\beta_{1}+(k_{p}h+k_{v})\beta_{2}+(1+k_{v}h)\beta_{3},
d¯3=\displaystyle\overline{d}_{3}= (kph+kv)β1+(1+kvh)β2+τβ3+kp,\displaystyle(k_{p}h+k_{v})\beta_{1}+(1+k_{v}h)\beta_{2}+\tau\beta_{3}+k_{p},
d¯4=\displaystyle\overline{d}_{4}= (1+kvh)β1+τβ2+kph+kv,\displaystyle(1+k_{v}h)\beta_{1}+\tau\beta_{2}+k_{p}h+k_{v},
d¯5=\displaystyle\overline{d}_{5}= τβ1+kvh+1.\displaystyle\tau\beta_{1}+k_{v}h+1.

Substituting s=jωs=j\omega into (C), we get

Gei(jω)=xn(ω)+yn(ω)jxd(ω)+yd(ω)j,\displaystyle G_{ei}(j\omega)=\frac{x_{n}(\omega)+y_{n}(\omega)j}{x_{d}(\omega)+y_{d}(\omega)j}, (90)

where xn(ω)=kpβ3n¯2ω2+kvω4x_{n}(\omega)=k_{p}\beta_{3}-\overline{n}_{2}\omega^{2}+k_{v}\omega^{4}, yn(ω)=n¯1ωn¯3ω3y_{n}(\omega)=\overline{n}_{1}\omega-\overline{n}_{3}\omega^{3}, xd(ω)=kpβ3d¯2ω2+d¯4ω4τω6x_{d}(\omega)=k_{p}\beta_{3}-\overline{d}_{2}\omega^{2}+\overline{d}_{4}\omega^{4}-\tau\omega^{6}, yd(ω)=d¯1ωd¯3ω3+d¯5ω5y_{d}(\omega)=\overline{d}_{1}\omega-\overline{d}_{3}\omega^{3}+\overline{d}_{5}\omega^{5}.

By μp>0\mu_{p}>0 and μa>0\mu_{a}>0, we know α5>0\alpha_{5}>0 and γ5>0\gamma_{5}>0. From (15), we obtain kγ5/α5k\geq\gamma_{5}/\alpha_{5}. This together with α5>0\alpha_{5}>0 and γ5>0\gamma_{5}>0 leads to

α5k2γ5k0.\displaystyle\alpha_{5}k^{2}-\gamma_{5}k\geq 0. (91)

From (13) and μa>0\mu_{a}>0, we know μv>μa/h\mu_{v}>\mu_{a}/h. This together with μp>0\mu_{p}>0 leads to α4>0\alpha_{4}>0. From (15), we know kθ4k\geq\theta_{4}. This together with α4>0\alpha_{4}>0 and ρ4>0\rho_{4}>0 leads to

α4k2+γ4k+ρ40.\displaystyle\alpha_{4}k^{2}+\gamma_{4}k+\rho_{4}\geq 0. (92)

By (13), we know μv>3μa/h\mu_{v}>\sqrt{3}\mu_{a}/h. This together with μa>0\mu_{a}>0 leads to 3h2μv29μa2>03h^{2}\mu_{v}^{2}-9\mu_{a}^{2}>0. By (13), we get μv>2μa/h2\mu_{v}>2\mu_{a}/h^{2}. This leads to 3h2μp26μpμa>03h^{2}\mu_{p}^{2}-6\mu_{p}\mu_{a}>0. By (14), we know ωo>16μvμa/(3h2μv29μa2)\omega_{o}>16\mu_{v}\mu_{a}/(3h^{2}\mu_{v}^{2}-9\mu_{a}^{2}). This together with 3h2μv29μa2>03h^{2}\mu_{v}^{2}-9\mu_{a}^{2}>0 leads to α3>0\alpha_{3}>0. From (15), we know kθ3k\geq\theta_{3}. This together with α3>0\alpha_{3}>0 and ρ3>0\rho_{3}>0 leads to

α3k2+γ3k+ρ30.\displaystyle\alpha_{3}k^{2}+\gamma_{3}k+\rho_{3}\geq 0. (93)

From (15), we obtain kθ2k\geq\theta_{2}. This together with α2>0\alpha_{2}>0 and ρ2>0\rho_{2}>0 leads to

α2k2+γ2k+ρ20.\displaystyle\alpha_{2}k^{2}+\gamma_{2}k+\rho_{2}\geq 0. (94)

From (15), we obtain kθ1k\geq\theta_{1}. This together with α1>0\alpha_{1}>0 and ρ1>0\rho_{1}>0 leads to

α1k2+γ1k+ρ10.\displaystyle\alpha_{1}k^{2}+\gamma_{1}k+\rho_{1}\geq 0. (95)

By (91)-(95), we know

(α5k2γ5k)ω2+(α4k2+γ4k+ρ4)ω4\displaystyle(\alpha_{5}k^{2}-\gamma_{5}k)\omega^{2}+(\alpha_{4}k^{2}+\gamma_{4}k+\rho_{4})\omega^{4}
+(α3k2+γ3k+ρ3)ω6+(α2k2+γ2k+ρ2)ω8\displaystyle+(\alpha_{3}k^{2}+\gamma_{3}k+\rho_{3})\omega^{6}+(\alpha_{2}k^{2}+\gamma_{2}k+\rho_{2})\omega^{8}
+(α1k2+γ1k+ρ1)ω10+τ2ω120,ω.\displaystyle+(\alpha_{1}k^{2}+\gamma_{1}k+\rho_{1})\omega^{10}+\tau^{2}\omega^{12}\geq 0,\;\forall\;\omega\in\mathbb{R}. (96)

Through calculation, we get

{α5k2γ5k=2kpβ3n¯2+d¯122kpβ3d¯2n¯12,α4k2+γ4k+ρ4=2n¯1n¯3+2kpβ3d¯4+d¯22n¯222kpkvβ32d¯1d¯3,α3k2+γ3k+ρ3=2n¯2kv+2d¯1d¯5+d¯32n¯322kpβ3τ2d¯2d¯4,α2k2+γ2k+ρ2=d¯42+2d¯2τkv22d¯3d¯5,α1k2+γ1k+ρ1=d¯522d¯4τ.\displaystyle\left\{\begin{array}[]{rll}\alpha_{5}k^{2}-\gamma_{5}k=&2k_{p}\beta_{3}\overline{n}_{2}+{\overline{d}^{2}_{1}}-2k_{p}\beta_{3}\overline{d}_{2}-\overline{n}^{2}_{1},\\ \alpha_{4}k^{2}+\gamma_{4}k+\rho_{4}=&2\overline{n}_{1}\overline{n}_{3}+2k_{p}\beta_{3}\overline{d}_{4}+\overline{d}^{2}_{2}-\overline{n}^{2}_{2}\\ &-2k_{p}k_{v}\beta_{3}-2\overline{d}_{1}\overline{d}_{3},\\ \alpha_{3}k^{2}+\gamma_{3}k+\rho_{3}=&2\overline{n}_{2}k_{v}+2\overline{d}_{1}\overline{d}_{5}+\overline{d}^{2}_{3}-\overline{n}^{2}_{3}\\ &-2k_{p}\beta_{3}\tau-2\overline{d}_{2}\overline{d}_{4},\\ \alpha_{2}k^{2}+\gamma_{2}k+\rho_{2}=&\overline{d}^{2}_{4}+2\overline{d}_{2}\tau-k_{v}^{2}-2\overline{d}_{3}\overline{d}_{5},\\ \alpha_{1}k^{2}+\gamma_{1}k+\rho_{1}=&\overline{d}^{2}_{5}-2\overline{d}_{4}\tau.\end{array}\right.

This together with (C) leads to

(2kpβ3n¯2+d¯122kpβ3d¯2n¯12)ω2\displaystyle(2k_{p}\beta_{3}\overline{n}_{2}+{\overline{d}^{2}_{1}}-2k_{p}\beta_{3}\overline{d}_{2}-\overline{n}^{2}_{1})\omega^{2}
+(2n¯1n¯3+2kpβ3d¯4+d¯22n¯222kpkvβ32d¯1d¯3)ω4\displaystyle+(2\overline{n}_{1}\overline{n}_{3}+2k_{p}\beta_{3}\overline{d}_{4}+\overline{d}^{2}_{2}-\overline{n}^{2}_{2}-2k_{p}k_{v}\beta_{3}-2\overline{d}_{1}\overline{d}_{3})\omega^{4}
+(2n¯2kv+2d¯1d¯5+d¯32n¯322kpβ3τ2d¯2d¯4)ω6\displaystyle+(2\overline{n}_{2}k_{v}+2\overline{d}_{1}\overline{d}_{5}+\overline{d}^{2}_{3}-\overline{n}^{2}_{3}-2k_{p}\beta_{3}\tau-2\overline{d}_{2}\overline{d}_{4})\omega^{6}
+(d¯42+2d¯2τkv22d¯3d¯5)ω8\displaystyle+(\overline{d}^{2}_{4}+2\overline{d}_{2}\tau-k_{v}^{2}-2\overline{d}_{3}\overline{d}_{5})\omega^{8}
+(d¯522d¯4τ)ω10+τ2ω120,ω.\displaystyle+(\overline{d}^{2}_{5}-2\overline{d}_{4}\tau)\omega^{10}+\tau^{2}\omega^{12}\geq 0,\;\forall\;\omega\in\mathbb{R}. (97)

Through calculation, we know

xd2(ω)+yd2(ω)xn2(ω)yn2(ω)\displaystyle x_{d}^{2}(\omega)+y_{d}^{2}(\omega)-x_{n}^{2}(\omega)-y_{n}^{2}(\omega)
=\displaystyle= (2kpβ3n¯2+d¯122kpβ3d¯2n¯12)ω2\displaystyle(2k_{p}\beta_{3}\overline{n}_{2}+{\overline{d}^{2}_{1}}-2k_{p}\beta_{3}\overline{d}_{2}-\overline{n}^{2}_{1})\omega^{2}
+(2n¯1n¯3+2kpβ3d¯4+d¯22n¯222kpkvβ32d¯1d¯3)ω4\displaystyle+(2\overline{n}_{1}\overline{n}_{3}+2k_{p}\beta_{3}\overline{d}_{4}+\overline{d}^{2}_{2}-\overline{n}^{2}_{2}-2k_{p}k_{v}\beta_{3}-2\overline{d}_{1}\overline{d}_{3})\omega^{4}
+(2n¯2kv+2d¯1d¯5+d¯32n¯322kpβ3τ2d¯2d¯4)ω6\displaystyle+(2\overline{n}_{2}k_{v}+2\overline{d}_{1}\overline{d}_{5}+\overline{d}^{2}_{3}-\overline{n}^{2}_{3}-2k_{p}\beta_{3}\tau-2\overline{d}_{2}\overline{d}_{4})\omega^{6}
+(d¯42+2d¯2τkv22d¯3d¯5)ω8\displaystyle+(\overline{d}^{2}_{4}+2\overline{d}_{2}\tau-k_{v}^{2}-2\overline{d}_{3}\overline{d}_{5})\omega^{8}
+(d¯522d¯4τ)ω10+τ2ω12\displaystyle+(\overline{d}^{2}_{5}-2\overline{d}_{4}\tau)\omega^{10}+\tau^{2}\omega^{12}

This together with (C) leads to

xd2(ω)+yd2(ω)xn2(ω)yn2(ω)0,ω.\displaystyle x_{d}^{2}(\omega)+y_{d}^{2}(\omega)-x_{n}^{2}(\omega)-y_{n}^{2}(\omega)\geq 0,\;\forall\;\omega\in\mathbb{R}. (98)

By (98), we know

xn2(ω)+yn2(ω)xd2(ω)+yd2(ω)1,ω.\displaystyle\frac{x_{n}^{2}(\omega)+y_{n}^{2}(\omega)}{x_{d}^{2}(\omega)+y_{d}^{2}(\omega)}\leq 1,\;\forall\;\omega\in\mathbb{R}. (99)

Through calculation, we obtain

|xn(ω)+yn(ω)jxd(ω)+yd(ω)j|=xn2(ω)+yn2(ω)xd2(ω)+yd2(ω).\displaystyle\left|\frac{x_{n}(\omega)+y_{n}(\omega)j}{x_{d}(\omega)+y_{d}(\omega)j}\right|=\frac{\sqrt{x_{n}^{2}(\omega)+y_{n}^{2}(\omega)}}{\sqrt{x_{d}^{2}(\omega)+y_{d}^{2}(\omega)}}.

This together with (99) leads to

|xn(ω)+yn(ω)jxd(ω)+yd(ω)j|1,ω.\displaystyle\left|\frac{x_{n}(\omega)+y_{n}(\omega)j}{x_{d}(\omega)+y_{d}(\omega)j}\right|\leq 1,\;\forall\;\omega\in\mathbb{R}. (100)

From (90) and (100), we know |Gei(jω)|1|G_{ei}(j\omega)|\leq 1 for any ω\omega\in\mathbb{R}. That is Gei(s)1\|G_{ei}(s)\|_{\infty}\leq 1.∎

Appendix D Proof of Theorem 3

Proof of Theorem 3. For simplicity of presentation, we denote bi=1τ+ϵib_{i}=\frac{1}{\tau}+\epsilon_{i}. According to (16), the models of the velocity difference between adjacent vehicles are given by

{v˙d,i(t)=ad,i(t),a˙d,i(t)=q1i(t)+ai(t)/τui(t)/τ,i=1,2,,N,q˙1i(t)=w1i(t),\displaystyle\left\{\begin{aligned} \dot{v}_{d,i}(t)&=a_{d,i}(t),\vspace{1ex}\\ \dot{a}_{d,i}(t)&=q_{1i}(t)+a_{i}(t)/\tau-u_{i}(t)/\tau,\quad\quad i=1,2,...,N,\vspace{1ex}\\ \dot{q}_{1i}(t)&=w_{1i}(t),\end{aligned}\right. (101)

where the definitions of ad,i(t)a_{d,i}(t) and vd,i(t)v_{d,i}(t) are the same as in (4),

q1i(t)=\displaystyle q_{1i}(t)= bi1ai1(t)+bi1ui1(t)+ϵiai(t)\displaystyle-b_{i-1}a_{i-1}(t)+b_{i-1}u_{i-1}(t)+\epsilon_{i}a_{i}(t)
ϵiui(t),\displaystyle-\epsilon_{i}u_{i}(t),\vspace{1ex} (102)
w1i(t)=\displaystyle w_{1i}(t)= bi12ai1(t)bi12ui1(t)+bi1u˙i1(t)\displaystyle b_{i-1}^{2}a_{i-1}(t)-b_{i-1}^{2}u_{i-1}(t)+b_{i-1}\dot{u}_{i-1}(t)
ϵi2ai(t)+ϵi2ui(t)ϵiu˙i(t).\displaystyle-\epsilon_{i}^{2}a_{i}(t)+\epsilon_{i}^{2}u_{i}(t)-\epsilon_{i}\dot{u}_{i}(t). (103)

Denote

Xi(t)=\displaystyle X_{i}(t)= [pi(t),vi(t),ai(t)]T,i=0,1,2,,N,\displaystyle[p_{i}(t),v_{i}(t),a_{i}(t)]^{T},\;i=0,1,2,...,N,\vspace{1ex} (104)
Ei(t)=\displaystyle E_{i}(t)= [e1,i(t),e2,i(t),e3,i(t)]T,i=1,2,,N,\displaystyle[e_{1,i}(t),e_{2,i}(t),e_{3,i}(t)]^{T},\;i=1,2,...,N,\vspace{1ex} (105)
Fi(t)=\displaystyle F_{i}(t)= [ei(t),vd,i(t),ai(t)]T,i=1,2,N,\displaystyle[e_{i}(t),v_{d,i}(t),a_{i}(t)]^{T},\;i=1,2,...N,\vspace{1ex} (106)
W(t)=\displaystyle W(t)= [F1T(t),F2T(t),,FNT(t),E1T(t),E2T(t),\displaystyle[F_{1}^{T}(t),F_{2}^{T}(t),\cdots,F_{N}^{T}(t),E_{1}^{T}(t),E_{2}^{T}(t),
,ENT(t)]T,\displaystyle\cdots,E_{N}^{T}(t)]^{T},\vspace{1ex} (107)
Δ(t)=\displaystyle\Delta(t)= [δ1T(t),0,,0,ζ1T(t),ζ2T(t),ζ3T(t),0,\displaystyle[\delta_{1}^{T}(t),0,\cdots,0,\zeta_{1}^{T}(t),\zeta_{2}^{T}(t),\zeta_{3}^{T}(t),0,
,0]T,\displaystyle\cdots,0]^{T}, (108)

where

e1,i(t)=\displaystyle e_{1,i}(t)= z1,i(t)vd,i(t),\displaystyle z_{1,i}(t)-v_{d,i}(t), (109)
e2,i(t)=\displaystyle e_{2,i}(t)= z2,i(t)ad,i(t),\displaystyle z_{2,i}(t)-a_{d,i}(t), (110)
e3,i(t)=\displaystyle e_{3,i}(t)= z3,i(t)qiϵ(t),\displaystyle z_{3,i}(t)-q_{i\epsilon}(t), (111)
δ1(t)=\displaystyle\delta_{1}(t)= [0,a0(t),b1kaa0(t)]T,\displaystyle\left[0,a_{0}(t),b_{1}k_{a}a_{0}(t)\right]^{T}, (112)
ζ1(t)=\displaystyle\zeta_{1}(t)= [0,0,b02a0(t)+b02u0(t)b0u˙0(t)+ϵ12kaa0(t)\displaystyle\left[0,0,-b_{0}^{2}a_{0}(t)+b_{0}^{2}u_{0}(t)-b_{0}\dot{u}_{0}(t)+\epsilon_{1}^{2}k_{a}a_{0}(t)\right.
kvϵ1a0(t)+kvkaϵ1b1ha0(t)+ϵ1kaa˙0(t)]T,\displaystyle\left.-k_{v}\epsilon_{1}a_{0}(t)+k_{v}k_{a}\epsilon_{1}b_{1}ha_{0}(t)+\epsilon_{1}k_{a}\dot{a}_{0}(t)\right]^{T}, (113)
ζ2(t)=\displaystyle\zeta_{2}(t)= [0,0,b12kaa0(t)+ka2ϵ2b1a0(t)kvb1a0(t)\displaystyle\left[0,0,b_{1}^{2}k_{a}a_{0}(t)+k_{a}^{2}\epsilon_{2}b_{1}a_{0}(t)-k_{v}b_{1}a_{0}(t)\right.
+kvkahb12a0(t)b1kaa˙0(t)]T,\displaystyle\left.+k_{v}k_{a}hb_{1}^{2}a_{0}(t)-b_{1}k_{a}\dot{a}_{0}(t)\right]^{T}, (114)
ζ3(t)=\displaystyle\zeta_{3}(t)= [0,0,ka2b1b2a0(t)]T.\displaystyle\left[0,0,k_{a}^{2}b_{1}b_{2}a_{0}(t)\right]^{T}. (115)

From (6), (3) and (110), we know

ui(t)=\displaystyle u_{i}(t)= kpei(t)+kv(vd,i(t)hai(t))\displaystyle\;k_{p}e_{i}(t)+k_{v}(v_{d,i}(t)-ha_{i}(t))
+ka(ai1(t)+e2i(t)),i=1,2,,N.\displaystyle+k_{a}(a_{i-1}(t)+e_{2i}(t)),\;i=1,2,\ldots,N. (116)

This together with (16), (3), (5), (104) and (105) leads to

Xi˙(t)={A0Xi(t)+B0ui(t),i=0,AiXi(t)+BiXi1(t)+CiEi(t)+L1ir,i=1,,N,\displaystyle\dot{X_{i}}(t)=\left\{\begin{array}[]{ll}A_{0}X_{i}(t)+B_{0}u_{i}(t),&i=0,\vspace{1ex}\\ A_{i}X_{i}(t)+B_{i}X_{i-1}(t)+C_{i}E_{i}(t)+L_{1i}r,\\ &i=1,\ldots,N,\end{array}\right. (120)

where

A0=[01000100b0],B0=[00b0],L1i=[00kpbi],A_{0}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&\begin{aligned} -b_{0}\end{aligned}\end{bmatrix},B_{0}=\begin{bmatrix}0\\ 0\\ b_{0}\end{bmatrix},L_{1i}=\begin{bmatrix}0\\ 0\\ \begin{aligned} -k_{p}b_{i}\end{aligned}\end{bmatrix},
Ai=[010001kpbikphbikvbibikvhbi],A_{i}=\begin{bmatrix}0&1&0\\ 0&0&1\\ \begin{aligned} -k_{p}b_{i}\end{aligned}&\begin{aligned} -k_{p}hb_{i}-k_{v}b_{i}\end{aligned}&\begin{aligned} -b_{i}-k_{v}hb_{i}\end{aligned}\end{bmatrix},
Bi=[000000kpbikvbikabi],Ci=[0000000kabi0].B_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} k_{p}b_{i}\end{aligned}&\begin{aligned} k_{v}b_{i}\end{aligned}&\begin{aligned} k_{a}b_{i}\end{aligned}\end{bmatrix},C_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} k_{a}b_{i}\end{aligned}&0\end{bmatrix}.

From (3), (104) and (106), we get

Fi(t)=PXi1(t)QXi(t)L2ir,i=1,2,,N,\displaystyle F_{i}(t)=PX_{i-1}(t)-QX_{i}(t)-L_{2i}r,\;i=1,2,\ldots,N, (121)

where

P=[100010000],Q=[1h0010001],L2i=[100].P={\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&0\end{array}\right]},Q={\left[\begin{array}[]{ccc}1&h&0\\ 0&1&0\\ 0&0&-1\end{array}\right]},L_{2i}={\left[\begin{array}[]{ccc}1\\ 0\\ 0\end{array}\right]}.

From (112), (120) and (121), we obtain

Fi˙(t)={𝒜1iFi(t)+𝒢iEi(t)+δi(t),i=1,𝒜1iFi(t)+1iFi1(t)+𝒢iEi(t),i=2,,N,\displaystyle\dot{F_{i}}(t)=\left\{\begin{array}[]{ll}\mathcal{A}_{1i}F_{i}(t)+\mathcal{G}_{i}E_{i}(t)+\delta_{i}(t),&i=1,\vspace{1ex}\\ \mathcal{A}_{1i}F_{i}(t)+\mathcal{B}_{1i}F_{i-1}(t)+\mathcal{G}_{i}E_{i}(t),&i=2,\ldots,N,\end{array}\right. (124)

where

𝒜1i=[01h001kpbikvbi(1kvh)bi],\mathcal{A}_{1i}=\begin{bmatrix}0&1&-h\\ 0&0&-1\\ \begin{aligned} k_{p}b_{i}\end{aligned}&\begin{aligned} k_{v}b_{i}\end{aligned}&\begin{aligned} (-1-k_{v}h)b_{i}\end{aligned}\end{bmatrix},
1i=[00000100bika],𝒢i=[0000000bika0].\mathcal{B}_{1i}=\begin{bmatrix}0&0&0\\ 0&0&1\\ 0&0&\begin{aligned} b_{i}k_{a}\end{aligned}\end{bmatrix},\mathcal{G}_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} b_{i}k_{a}\end{aligned}&0\end{bmatrix}.

By (3), (5), (6), (9), (16), (102)-(106), (109)-(D) and (124), we have

Ei˙(t)={𝒞iFi(t)+1iEi(t)+ζi(t),i=1,𝒞iFi(t)+𝒟1iFi1(t)+1iEi(t)+iEi1(t)+ζi(t),i=2,𝒞iFi(t)+𝒟1iFi1(t)+1iFi2(t)+1iEi(t)+iEi1(t)+𝒥iEi2(t)+ζi(t),i=3,𝒞iFi(t)+𝒟1iFi1(t)+1iFi2(t)+iFi3(t)+1iEi(t)+iEi1(t)+𝒥iEi2(t),i=4,,N,\displaystyle\dot{E_{i}}(t)=\left\{\begin{array}[]{ll}\mathcal{C}_{i}F_{i}(t)+\mathcal{H}_{1i}E_{i}(t)+\zeta_{i}(t),&i=1,\vspace{1ex}\\ \mathcal{C}_{i}F_{i}(t)+\mathcal{D}_{1i}F_{i-1}(t)+\mathcal{H}_{1i}E_{i}(t)\\ +\mathcal{I}_{i}E_{i-1}(t)+\zeta_{i}(t),&i=2,\vspace{1ex}\\ \mathcal{C}_{i}F_{i}(t)+\mathcal{D}_{1i}F_{i-1}(t)+\mathcal{E}_{1i}F_{i-2}(t)\\ +\mathcal{H}_{1i}E_{i}(t)+\mathcal{I}_{i}E_{i-1}(t)+\mathcal{J}_{i}E_{i-2}(t)\\ +\zeta_{i}(t),&i=3,\vspace{1ex}\\ \mathcal{C}_{i}F_{i}(t)+\mathcal{D}_{1i}F_{i-1}(t)+\mathcal{E}_{1i}F_{i-2}(t)\\ +\mathcal{F}_{i}F_{i-3}(t)+\mathcal{H}_{1i}E_{i}(t)+\mathcal{I}_{i}E_{i-1}(t)\\ +\mathcal{J}_{i}E_{i-2}(t),&i=4,\ldots,N,\end{array}\right. (134)

where

1i=[β110β201β3kaβ2ϵikaϵi2kvkahbiϵikaϵi],\mathcal{H}_{1i}=\begin{bmatrix}-\beta_{1}&1&0\\ -\beta_{2}&0&1\\ -\beta_{3}-k_{a}\beta_{2}\epsilon_{i}\ &\begin{aligned} -k_{a}\epsilon_{i}^{2}-k_{v}k_{a}hb_{i}\epsilon_{i}\end{aligned}&k_{a}\epsilon_{i}\end{bmatrix},
i=[00000000ka2bi1bi2],𝒥i=[0000000ka2bi1bi20],\mathcal{F}_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&\begin{aligned} -k_{a}^{2}b_{i-1}b_{i-2}\end{aligned}\end{bmatrix},\\ \mathcal{J}_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&\begin{aligned} -k_{a}^{2}b_{i-1}b_{i-2}\end{aligned}&0\end{bmatrix},
𝒞i=[000000kpkvhbiϵikpϵi2kpϵikvϵi2kv2hbiϵiϵi2+kvhϵi2kvϵikphϵi+kvhbiϵi+kv2h2biϵi],\mathcal{C}_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} &-k_{p}k_{v}hb_{i}\epsilon_{i}\\ &-k_{p}\epsilon_{i}^{2}\end{aligned}&\begin{aligned} &k_{p}\epsilon_{i}-k_{v}\epsilon_{i}^{2}\\ &-k_{v}^{2}hb_{i}\epsilon_{i}\end{aligned}&\begin{aligned} &\epsilon_{i}^{2}+k_{v}h\epsilon_{i}^{2}-k_{v}\epsilon_{i}\\ &-k_{p}h\epsilon_{i}+k_{v}hb_{i}\epsilon_{i}\\ &+k_{v}^{2}h^{2}b_{i}\epsilon_{i}\end{aligned}\end{bmatrix},\\
𝒟1i=[000000kpkvhbi12+kpkaϵibi1+kpbi12kvkaϵibi1+kv2hbi12kpbi1+kvbi12kphbi1+kvbi1bi12kaϵibi1kv2h2bi12kaϵi2+kvϵikvkahϵibikvkahϵibi12kvhbi12],\mathcal{D}_{1i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} &k_{p}k_{v}hb_{i-1}^{2}\\ &+k_{p}k_{a}\epsilon_{i}b_{i-1}\\ &+k_{p}b_{i-1}^{2}\end{aligned}&\begin{aligned} &k_{v}k_{a}\epsilon_{i}b_{i-1}\\ &+k_{v}^{2}hb_{i-1}^{2}\\ &-k_{p}b_{i-1}\\ &+k_{v}b_{i-1}^{2}\end{aligned}&\begin{aligned} &k_{p}hb_{i-1}+k_{v}b_{i-1}\\ &-b_{i-1}^{2}-k_{a}\epsilon_{i}b_{i-1}\\ &-k_{v}^{2}h^{2}b_{i-1}^{2}-k_{a}\epsilon_{i}^{2}\\ &+k_{v}\epsilon_{i}-k_{v}k_{a}h\epsilon_{i}b_{i}\\ &-k_{v}k_{a}h\epsilon_{i}b_{i-1}\\ &-2k_{v}hb_{i-1}^{2}\end{aligned}\end{bmatrix},
1i=[000000kpkabi1bi2kvkabi1bi2kvkahbi1bi2+kvkahbi12+kabi1bi2+ka2ϵibi1+kabi12kvbi1],\mathcal{E}_{1i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} -k_{p}k_{a}b_{i-1}b_{i-2}\end{aligned}&\begin{aligned} -k_{v}k_{a}b_{i-1}b_{i-2}\end{aligned}&\begin{aligned} &k_{v}k_{a}hb_{i-1}b_{i-2}\\ &+k_{v}k_{a}hb_{i-1}^{2}\\ &+k_{a}b_{i-1}b_{i-2}\\ &+k_{a}^{2}\epsilon_{i}b_{i-1}\\ &+k_{a}b_{i-1}^{2}\\ &-k_{v}b_{i-1}\end{aligned}\end{bmatrix},
i=[000000kaβ2bi1kabi12+kvkahbi12+ka2ϵibi1kabi1].\mathcal{I}_{i}=\begin{bmatrix}0&0&0\\ 0&0&0\\ \begin{aligned} k_{a}\beta_{2}b_{i-1}\end{aligned}&\begin{aligned} &k_{a}b_{i-1}^{2}+k_{v}k_{a}hb_{i-1}^{2}\\ &+k_{a}^{2}\epsilon_{i}b_{i-1}\end{aligned}&\begin{aligned} -k_{a}b_{i-1}\end{aligned}\end{bmatrix}.

From (107), (108), (124) and (134), we know

W˙(t)=(Ψ+Ψ^)W(t)+Δ(t),\displaystyle\dot{W}(t)=(\Psi+\hat{\Psi})W(t)+\Delta(t), (135)

where Ψ^=[Ψ^11Ψ^12Ψ^21Ψ^22]\hat{\Psi}=\begin{bmatrix}\hat{\Psi}_{11}&\hat{\Psi}_{12}\\ \hat{\Psi}_{21}&\hat{\Psi}_{22}\end{bmatrix} and

Ψ^11=[𝒜2122𝒜222N𝒜2N],Ψ^12=[𝒢1𝒢N],\hat{\Psi}_{11}=\begin{bmatrix}\mathcal{A}_{21}&\quad&\quad&\quad\\ \mathcal{B}_{22}&\mathcal{A}_{22}&\quad&\quad\\ \quad&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{B}_{2N}&\mathcal{A}_{2N}\\ \end{bmatrix},\hat{\Psi}_{12}=\begin{bmatrix}\mathcal{G}_{1}&\quad&\quad\\ \quad&\ddots&\quad\\ \quad&\quad&\mathcal{G}_{N}\\ \end{bmatrix},
Ψ^21=[𝒞1𝒟22𝒞223𝒟23𝒞3424𝒟24𝒞4N2N𝒟2N𝒞N],\hat{\Psi}_{21}=\begin{bmatrix}\mathcal{C}_{1}&\quad&\quad&\quad&\quad&\quad\\ \mathcal{D}_{22}&\mathcal{C}_{2}&\quad&\quad&\quad&\quad\\ \mathcal{E}_{23}&\mathcal{D}_{23}&\mathcal{C}_{3}&\quad&\quad&\quad\\ \mathcal{F}_{4}&\mathcal{E}_{24}&\mathcal{D}_{24}&\mathcal{C}_{4}&\quad&\quad\\ \quad&\ddots&\ddots&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{F}_{N}&\mathcal{E}_{2N}&\mathcal{D}_{2N}&\mathcal{C}_{N}\end{bmatrix},
Ψ^22=[21222𝒥3323𝒥NN2N],\hat{\Psi}_{22}=\begin{bmatrix}\mathcal{H}_{21}&\quad&\quad&\quad&\quad\\ \mathcal{I}_{2}&\mathcal{H}_{22}&\quad&\quad&\quad\\ \mathcal{J}_{3}&\mathcal{I}_{3}&\mathcal{H}_{23}&\quad&\quad\\ \quad&\ddots&\ddots&\ddots&\quad\\ \quad&\quad&\mathcal{J}_{N}&\mathcal{I}_{N}&\mathcal{H}_{2N}\end{bmatrix},
𝒜2i=𝒜1i𝒜,2i=1i,𝒟2i=𝒟1i𝒟,\displaystyle\mathcal{A}_{2i}=\mathcal{A}_{1i}-\mathcal{A},\mathcal{B}_{2i}=\mathcal{B}_{1i}-\mathcal{B},\mathcal{D}_{2i}=\mathcal{D}_{1i}-\mathcal{D},
2i=1i,2i=1i.\displaystyle\mathcal{E}_{2i}=\mathcal{E}_{1i}-\mathcal{E},\mathcal{H}_{2i}=\mathcal{H}_{1i}-\mathcal{H}.

Firstly, we analyze the stability of Ψ\Psi. The eigenvalues of Ψ\Psi are only related to 𝒜\mathcal{A} and \mathcal{H}. Calculating the characteristic polynomial of 𝒜\mathcal{A}, we obtain

|sI𝒜|=s3+(1+kvhτ)s2+(kv+kphτ)s+kpτ.\displaystyle\left|sI-\mathcal{A}\right|=s^{3}+\left(\frac{1+k_{v}h}{\tau}\right)s^{2}+\left(\frac{k_{v}+k_{p}h}{\tau}\right)s+\frac{k_{p}}{\tau}. (136)

The Rouse table corresponding to (136) is given by

s3s^{3} 1 kv+kphτ\begin{aligned} \frac{k_{v}+k_{p}h}{\tau}\end{aligned}
s2s^{2} 1+kvhτ\begin{aligned} \frac{1+k_{v}h}{\tau}\end{aligned} kpτ\begin{aligned} \frac{k_{p}}{\tau}\end{aligned}
s1s^{1} hkv2+(1+h2kp)kv+(hτ)kpτ+τkvh\begin{aligned} \frac{hk_{v}^{2}+(1+h^{2}k_{p})k_{v}+(h-\tau)k_{p}}{\tau+\tau k_{v}h}\end{aligned} 0
s0s^{0} kpτ\begin{aligned} \frac{k_{p}}{\tau}\end{aligned}

By kp>0k_{p}>0 and (19), we know that the elements of the first column of the Rouse table corresponding to (136) are all greater than zero. From Rouse criterion, 𝒜\mathcal{A} is stable. Calculating the characteristic polynomial of \mathcal{H}, we get

|sI|=s3+β1s2+β2s+β3.\displaystyle\left|sI-\mathcal{H}\right|=s^{3}+\beta_{1}s^{2}+\beta_{2}s+\beta_{3}. (137)

The Rouse table corresponding to (137) is given by

s3s^{3} 1 β2\beta_{2}
s2s^{2} β1\beta_{1} β3\beta_{3}
s1s^{1} β1β2β3β1\begin{aligned} \frac{\beta_{1}\beta_{2}-\beta_{3}}{\beta_{1}}\end{aligned} 0
s0s^{0} β3\beta_{3}

By β1>0\beta_{1}>0, β3>0\beta_{3}>0, β1β2β3>0\beta_{1}\beta_{2}-\beta_{3}>0, we know that the elements of the first column of the Rouse table corresponding to (137) are all greater than zero. From Rouse criterion, \mathcal{H} is stable. Then Ψ\Psi is stable.

From the definition of Ψ^\|\hat{\Psi}\| and Lemma 2, we know

Ψ^{(1/τ+Y1ϵ¯2+Y2ϵ¯)ka+Z1ϵ¯2+Z2ϵ¯,ifN=1,A1ka2+(Θ1+Y3ϵ¯2+Y4ϵ¯)ka+Z3ϵ¯2+Z4ϵ¯,ifN=2,A2ka2+(Θ+Y5ϵ¯2+Y6ϵ¯)ka+Z5ϵ¯2+Z6ϵ¯,ifN3.\displaystyle\|\hat{\Psi}\|\leq\left\{\begin{array}[]{lc}\begin{aligned} &\left(1/\tau+Y_{1}\overline{\epsilon}^{2}+Y_{2}\overline{\epsilon}\right)k_{a}+Z_{1}\overline{\epsilon}^{2}+Z_{2}\overline{\epsilon},\\ &\hskip 153.64487pt{\rm if}\ N=1,\end{aligned}\vspace{1ex}\\ \begin{aligned} &A_{1}k_{a}^{2}+\left(\Theta_{1}+Y_{3}\overline{\epsilon}^{2}+Y_{4}\overline{\epsilon}\right)k_{a}+Z_{3}\overline{\epsilon}^{2}+Z_{4}\overline{\epsilon},\\ &\hskip 153.64487pt{\rm if}\ N=2,\end{aligned}\vspace{1ex}\\ \begin{aligned} &A_{2}k_{a}^{2}+(\Theta+Y_{5}\overline{\epsilon}^{2}+Y_{6}\overline{\epsilon})k_{a}+Z_{5}\overline{\epsilon}^{2}+Z_{6}\overline{\epsilon},\\ &\hskip 153.64487pt{\rm if}\ N\geq 3.\end{aligned}\end{array}\right. (141)

From (23), (27) and (141), we get Ψ^<rc(Ψ)\|\hat{\Psi}\|<r_{c}(\Psi). It is known from the definition of Δ(t)\Delta(t) and Assumption 1 that limtΔ(t)=0\lim\limits_{t\to\infty}\Delta(t)=0. By Lemma 1 and (135), we know W(t)W(t) converges to zero exponentially. Then vi(t)v0(t)v_{i}(t)-v_{0}(t) and ei(t)e_{i}(t) both converge to zero exponentially, i=1,2,,Ni=1,2,...,N. ∎

Appendix E Proof of Theorem 4

Proof of Theorem 4. For simplicity of presentation, we denote bi=1τ+ϵib_{i}=\frac{1}{\tau}+\epsilon_{i}. By (3) and (5), we get

ai(t)=vd,i(t)e˙i(t)h.\displaystyle a_{i}(t)=\frac{v_{d,i}(t)-\dot{e}_{i}(t)}{h}. (142)

Taking the Laplace transform of (142), we have

𝒜i(s)=𝒱d,i(s)si(s)h,\displaystyle\mathscr{A}_{i}(s)=\frac{\mathscr{V}_{d,i}(s)-s\mathscr{E}_{i}(s)}{h}, (143)

where 𝒜i(s)\mathscr{A}_{i}(s) and 𝒱d,i(s)\mathscr{V}_{d,i}(s) are the Laplace transform of ai(t)a_{i}(t), vd,i(t)v_{d,i}(t), respectively. From (16), we know

ui(t)=a˙i(t)bi+ai(t).\displaystyle u_{i}(t)=\frac{\dot{a}_{i}(t)}{b_{i}}+a_{i}(t). (144)

This together with (142) leads to

ui(t)=v˙d,i(t)e¨i(t)bih+vd,i(t)e˙i(t)h.\displaystyle u_{i}(t)=\frac{\dot{v}_{d,i}(t)-\ddot{e}_{i}(t)}{b_{i}h}+\frac{v_{d,i}(t)-\dot{e}_{i}(t)}{h}. (145)

Taking the Laplace transform of (145), we get

𝒰i(s)=s𝒱d,i(s)s2i(s)bih+𝒱d,i(s)si(s)h,\displaystyle\mathscr{U}_{i}(s)=\frac{s\mathscr{V}_{d,i}(s)-s^{2}\mathscr{E}_{i}(s)}{b_{i}h}+\frac{\mathscr{V}_{d,i}(s)-s\mathscr{E}_{i}(s)}{h}, (146)

where 𝒰i(s)\mathscr{U}_{i}(s) is the Laplace transform of ui(t)u_{i}(t). Taking the Laplace transform of (9), we have

{s𝒵1,i(s)=𝒵2,i(s)+β1(𝒱d,i(s)𝒵1,i(s)),s𝒵2,i(s)=𝒵3,i(s)+β2(𝒱d,i(s)𝒵1,i(s))+𝒜i(s)/τ𝒰i(s)/τ,s𝒵3,i(s)=β3(𝒱d,i(s)𝒵1,i(s)),\displaystyle\left\{\begin{array}[]{lll}s\mathscr{Z}_{1,i}(s)=&\mathscr{Z}_{2,i}(s)+\beta_{1}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s)),\vspace{1ex}\\ s\mathscr{Z}_{2,i}(s)=&\mathscr{Z}_{3,i}(s)+\beta_{2}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s))+\mathscr{A}_{i}(s)/\tau\\ &-\mathscr{U}_{i}(s)/\tau,\vspace{1ex}\\ s\mathscr{Z}_{3,i}(s)=&\beta_{3}(\mathscr{V}_{d,i}(s)-\mathscr{Z}_{1,i}(s)),\vspace{1ex}\end{array}\right. (151)

where 𝒵1,i(s)\mathscr{Z}_{1,i}(s), 𝒵2,i(s)\mathscr{Z}_{2,i}(s) and 𝒵3,i(s)\mathscr{Z}_{3,i}(s) are the Laplace transform of z1,i(t)z_{1,i}(t), z2,i(t)z_{2,i}(t) and z3,i(t)z_{3,i}(t), respectively. Substituting (146) into (151), we obtain

𝒵2,i(s)=\displaystyle\mathscr{Z}_{2,i}(s)= ((s3+(τbihβ2β1)s2+τbihβ3s)𝒱d,i(s)\displaystyle\;((-s^{3}+(\tau b_{i}h\beta_{2}-\beta_{1})s^{2}+\tau b_{i}h\beta_{3}s)\mathscr{V}_{d,i}(s)
+(s4+β1s3)i(s))/(τbih(s3+β1s2\displaystyle+(s^{4}+\beta_{1}s^{3})\mathscr{E}_{i}(s))/(\tau b_{i}h(s^{3}+\beta_{1}s^{2}
+β2s+β3))\displaystyle+\beta_{2}s+\beta_{3})) (152)

By (3), (142) and (144), we get

a˙i(t)bi+ai(t)=\displaystyle\frac{\dot{a}_{i}(t)}{b_{i}}+a_{i}(t)= kpei(t)+kve˙i(t)\displaystyle\;k_{p}e_{i}(t)+k_{v}\dot{e}_{i}(t)
+ka(z2,i(t)+ai(t)).\displaystyle+k_{a}(z_{2,i}(t)+a_{i}(t)). (153)

Taking the Laplace transform of (E), we have

s𝒜i(s)bi+𝒜i(s)=\displaystyle\frac{s\mathscr{A}_{i}(s)}{b_{i}}+\mathscr{A}_{i}(s)= kpi(s)+kvsi(s)+ka𝒜i(s)\displaystyle\;k_{p}\mathscr{E}_{i}(s)+k_{v}s\mathscr{E}_{i}(s)+k_{a}\mathscr{A}_{i}(s)
+ka𝒵2,i(s).\displaystyle+k_{a}\mathscr{Z}_{2,i}(s). (154)

Denote Hi(s)=𝒱d,i(s)i(s)H_{i}(s)=\frac{\mathscr{V}_{d,i}(s)}{\mathscr{E}_{i}(s)}. By (143), (E) and (E), we get

Hi(s)=s5+n4is4+n3is3+n2is2+n1is+n0is4+d3is3+d2is2+d1is+d0i,\displaystyle H_{i}(s)=\frac{s^{5}+n_{4i}s^{4}+n_{3i}s^{3}+n_{2i}s^{2}+n_{1i}s+n_{0i}}{s^{4}+d_{3i}s^{3}+d_{2i}s^{2}+d_{1i}s+d_{0i}}, (155)

where

n0i=\displaystyle n_{0i}= bikphβ3,\displaystyle b_{i}k_{p}h\beta_{3},
n1i=\displaystyle n_{1i}= bikphβ2+(bibika+bikvh)β3,\displaystyle b_{i}k_{p}h\beta_{2}+(b_{i}-b_{i}k_{a}+b_{i}k_{v}h)\beta_{3},
n2i=\displaystyle n_{2i}= bikphβ1+(bibika+bikvh)β2+β3,\displaystyle b_{i}k_{p}h\beta_{1}+(b_{i}-b_{i}k_{a}+b_{i}k_{v}h)\beta_{2}+\beta_{3},
n3i=\displaystyle n_{3i}= bikph+(ka/τ+bibika+bikvh)β1+β2,\displaystyle\ b_{i}k_{p}h+(k_{a}/\tau+b_{i}-b_{i}k_{a}+b_{i}k_{v}h)\beta_{1}+\beta_{2},
n4i=\displaystyle n_{4i}= ka/τ+bibika+bikvh+β1,\displaystyle k_{a}/\tau+b_{i}-b_{i}k_{a}+b_{i}k_{v}h+\beta_{1},
d0i=\displaystyle d_{0i}= (bibika)β3,\displaystyle(b_{i}-b_{i}k_{a})\beta_{3},
d1i=\displaystyle d_{1i}= (bibika)β2+(1bikah)β3,\displaystyle(b_{i}-b_{i}k_{a})\beta_{2}+(1-b_{i}k_{a}h)\beta_{3},
d2i=\displaystyle d_{2i}= (ka/τbika+bi)β1+(1bikah)β2,\displaystyle(k_{a}/\tau-b_{i}k_{a}+b_{i})\beta_{1}+(1-b_{i}k_{a}h)\beta_{2},
d3i=\displaystyle d_{3i}= ka/τbika+bi+β1.\displaystyle k_{a}/\tau-b_{i}k_{a}+b_{i}+\beta_{1}.

By (3) and (5), we get

e˙i1(t)e˙i(t)=vd,i1(t)vd,i(t)hv˙d,i(t).\displaystyle\dot{e}_{i-1}(t)-\dot{e}_{i}(t)=v_{d,i-1}(t)-v_{d,i}(t)-h\dot{v}_{d,i}(t). (156)

Denote Gei(s)=i(s)i1(s)G_{ei}(s)=\frac{\mathscr{E}_{i}(s)}{\mathscr{E}_{i-1}(s)}. Taking the Laplace transform of (156), we have

si1(t)si(t)=𝒱d,i1(t)𝒱d,i(t)hs𝒱d,i(t).\displaystyle s\mathscr{E}_{i-1}(t)-s\mathscr{E}_{i}(t)=\mathscr{V}_{d,i-1}(t)-\mathscr{V}_{d,i}(t)-hs\mathscr{V}_{d,i}(t).

This together with 𝒱d,i(s)=Hi(s)i(s)\mathscr{V}_{d,i}(s)=H_{i}(s)\mathscr{E}_{i}(s) leads to

Gei(s)=sHi(s)s(hs+1)Hi(s).\displaystyle G_{ei}(s)=\frac{s-H_{i}(s)}{s-(hs+1)H_{i}(s)}.

This together with (155) leads to

Gei(s)=\displaystyle G_{ei}(s)= (bikvs4+n¯3is3+n¯2is2+n¯1is+bikpβ3)\displaystyle\;(b_{i}k_{v}s^{4}+\overline{n}_{3i}s^{3}+\overline{n}_{2i}s^{2}+\overline{n}_{1i}s+b_{i}k_{p}\beta_{3})
/(s6+d¯5is5+d¯4is4+d¯3is3+d¯2is2+d¯1is\displaystyle/(s^{6}+\overline{d}_{5i}s^{5}+\overline{d}_{4i}s^{4}+\overline{d}_{3i}s^{3}+\overline{d}_{2i}s^{2}+\overline{d}_{1i}s
+bikpβ3),\displaystyle+b_{i}k_{p}\beta_{3}), (157)

where

n¯1i=\displaystyle\overline{n}_{1i}= bikpβ2+bikvβ3,\displaystyle b_{i}k_{p}\beta_{2}+b_{i}k_{v}\beta_{3},
n¯2i=\displaystyle\overline{n}_{2i}= bikpβ1+bikvβ2+bikaβ3,\displaystyle b_{i}k_{p}\beta_{1}+b_{i}k_{v}\beta_{2}+b_{i}k_{a}\beta_{3},
n¯3i=\displaystyle\overline{n}_{3i}= bikvβ1+bikaβ2+bikp,\displaystyle b_{i}k_{v}\beta_{1}+b_{i}k_{a}\beta_{2}+b_{i}k_{p},
d¯1i=\displaystyle\overline{d}_{1i}= bikpβ2+(bikph+bikv)β3,\displaystyle b_{i}k_{p}\beta_{2}+(b_{i}k_{p}h+b_{i}k_{v})\beta_{3},
d¯2i=\displaystyle\overline{d}_{2i}= bikpβ1+(bikph+bikv)β2+(bi+bikvh)β3,\displaystyle b_{i}k_{p}\beta_{1}+(b_{i}k_{p}h+b_{i}k_{v})\beta_{2}+(b_{i}+b_{i}k_{v}h)\beta_{3},
d¯3i=\displaystyle\overline{d}_{3i}= (bikph+bikv)β1+(bi+bikvh)β2\displaystyle(b_{i}k_{p}h+b_{i}k_{v})\beta_{1}+(b_{i}+b_{i}k_{v}h)\beta_{2}
+β3+bikp,\displaystyle+\beta_{3}+b_{i}k_{p},
d¯4i=\displaystyle\overline{d}_{4i}= (ka/τbika+bi+bikvh)β1+β2\displaystyle(k_{a}/\tau-b_{i}k_{a}+b_{i}+b_{i}k_{v}h)\beta_{1}+\beta_{2}
+bikph+bikv\displaystyle+b_{i}k_{p}h+b_{i}k_{v}
d¯5i=\displaystyle\overline{d}_{5i}= β1+ka/τbika+bi+bikvh.\displaystyle\beta_{1}+k_{a}/\tau-b_{i}k_{a}+b_{i}+b_{i}k_{v}h.

Substituting s=jωs=j\omega into (E), we get

Gei(jω)=xni(ω)+yni(ω)jxdi(ω)+ydi(ω)j,\displaystyle G_{ei}(j\omega)=\frac{x_{ni}(\omega)+y_{ni}(\omega)j}{x_{di}(\omega)+y_{di}(\omega)j}, (158)

where xni(ω)=bikpβ3n¯2iω2+bikvω4x_{ni}(\omega)=b_{i}k_{p}\beta_{3}-\overline{n}_{2i}\omega^{2}+b_{i}k_{v}\omega^{4}, yni(ω)=n¯1iωn¯3iω3y_{ni}(\omega)=\overline{n}_{1i}\omega-\overline{n}_{3i}\omega^{3}, xdi(ω)=bikpβ3d¯2iω2+d¯4iω4ω6x_{di}(\omega)=b_{i}k_{p}\beta_{3}-\overline{d}_{2i}\omega^{2}+\overline{d}_{4i}\omega^{4}-\omega^{6}, ydi(ω)=d¯1iωd¯3iω3+d¯5iω5y_{di}(\omega)=\overline{d}_{1i}\omega-\overline{d}_{3i}\omega^{3}+\overline{d}_{5i}\omega^{5}.

Through calculation, we get

{α5ik2γ5ik=2bikpβ3n¯2i+d¯1i22bikpβ3d¯2in¯1i2,α4ik2+γ4ik+ρ4i=2n¯1in¯3i+2bikpβ3d¯4i+d¯2i2n¯2i22bi2kpkvβ32d¯1id¯3i,α3ik2+γ3ik+ρ3i=2n¯2ibikv+2d¯1id¯5i+d¯3i2n¯3i22bikpβ32d¯2id¯4i,α2ik2+γ2ik+ρ2i=d¯4i2+2d¯2ibi2kv22d¯3id¯5i,α1ik2+γ1ik+ρ1i=d¯5i22d¯4i.\displaystyle\left\{\begin{array}[]{rll}\alpha_{5i}k^{2}-\gamma_{5i}k=&2b_{i}k_{p}\beta_{3}\overline{n}_{2i}+{\overline{d}^{2}_{1i}}-2b_{i}k_{p}\beta_{3}\overline{d}_{2i}\\ &-\overline{n}^{2}_{1i},\\ \alpha_{4i}k^{2}+\gamma_{4i}k+\rho_{4i}=&2\overline{n}_{1i}\overline{n}_{3i}+2b_{i}k_{p}\beta_{3}\overline{d}_{4i}+\overline{d}^{2}_{2i}\\ &-\overline{n}^{2}_{2i}-2b_{i}^{2}k_{p}k_{v}\beta_{3}-2\overline{d}_{1i}\overline{d}_{3i},\\ \alpha_{3i}k^{2}+\gamma_{3i}k+\rho_{3i}=&2\overline{n}_{2i}b_{i}k_{v}+2\overline{d}_{1i}\overline{d}_{5i}+\overline{d}^{2}_{3i}-\overline{n}^{2}_{3i}\\ &-2b_{i}k_{p}\beta_{3}-2\overline{d}_{2i}\overline{d}_{4i},\\ \alpha_{2i}k^{2}+\gamma_{2i}k+\rho_{2i}=&\overline{d}^{2}_{4i}+2\overline{d}_{2i}-b_{i}^{2}k_{v}^{2}-2\overline{d}_{3i}\overline{d}_{5i},\\ \alpha_{1i}k^{2}+\gamma_{1i}k+\rho_{1i}=&\overline{d}^{2}_{5i}-2\overline{d}_{4i}.\end{array}\right. (167)

where ρ1i=3ωo2+bi2\rho_{1i}=3\omega_{o}^{2}+b_{i}^{2}, ρ2i=3ωo4+3bi2ωo2\rho_{2i}=3\omega_{o}^{4}+3b_{i}^{2}\omega_{o}^{2}, ρ3i=ωo6+3bi2ωo4\rho_{3i}=\omega_{o}^{6}+3b_{i}^{2}\omega_{o}^{4}, ρ4i=bi2ωo6\rho_{4i}=b_{i}^{2}\omega_{o}^{6} and

λ1i=\displaystyle\lambda_{1i}= 3biμvh(bihμv2μa/τ)+3bi2μa(2μvh3μa),\displaystyle 3b_{i}\mu_{v}h(b_{i}h\mu_{v}-2\mu_{a}/\tau)+3b_{i}^{2}\mu_{a}(2\mu_{v}h-3\mu_{a}),
λ2i=\displaystyle\lambda_{2i}= 16bi2hμaμp(16biμaμv+16bihμaμp)/τ,\displaystyle 16b^{2}_{i}h\mu_{a}\mu_{p}-(16b_{i}\mu_{a}\mu_{v}+16b_{i}h\mu_{a}\mu_{p})/\tau,
λ3i=\displaystyle\lambda_{3i}= 3bi2h2μp2+6bi2μaμp12biμpμa/τ\displaystyle 3b^{2}_{i}h^{2}\mu_{p}^{2}+6b^{2}_{i}\mu_{a}\mu_{p}-12b_{i}\mu_{p}\mu_{a}/\tau
α1i=\displaystyle\alpha_{1i}= (μa/τbiμa+bihμv)2,\displaystyle(\mu_{a}/\tau-b_{i}\mu_{a}+b_{i}h\mu_{v})^{2},
γ1i=\displaystyle\gamma_{1i}= 2biμa/τ+2bi2hμv2bi2μa2bihμp2biμv,\displaystyle 2b_{i}\mu_{a}/\tau+2b_{i}^{2}h\mu_{v}-2b_{i}^{2}\mu_{a}-2b_{i}h\mu_{p}-2b_{i}\mu_{v},
α2i=\displaystyle\alpha_{2i}= ((12bihμaμv18biμa2)/τ+9μa2/τ2+9bi2μa2\displaystyle((12b_{i}h\mu_{a}\mu_{v}-18b_{i}\mu_{a}^{2})/\tau+9\mu_{a}^{2}/\tau^{2}+9b^{2}_{i}\mu_{a}^{2}
+3bi2h2μv212bi2hμaμv)ωo2+b2ih2μp2\displaystyle+3b^{2}_{i}h^{2}\mu_{v}^{2}-12b^{2}_{i}h\mu_{a}\mu_{v})\omega_{o}^{2}+b^{2}_{i}h^{2}\mu_{p}^{2}
+2bi2μaμp2biμaμp/τ\displaystyle+2b^{2}_{i}\mu_{a}\mu_{p}-2b_{i}\mu_{a}\mu_{p}/\tau
γ2i=\displaystyle\gamma_{2i}= (16μa/τ16biμa)ωo3+(12biμa/τ+6bi2hμv\displaystyle(16\mu_{a}/\tau-16b_{i}\mu_{a})\omega_{o}^{3}+(12b_{i}\mu_{a}/\tau+6b_{i}^{2}h\mu_{v}
12bi2μa6bihμp6biμv)ωo22bi2μp\displaystyle-12b_{i}^{2}\mu_{a}-6b_{i}h\mu_{p}-6b_{i}\mu_{v})\omega_{o}^{2}-2b_{i}^{2}\mu_{p}
α3i=\displaystyle\alpha_{3i}= λ1ωo4+λ2ωo3+λ3ωo2,\displaystyle\lambda_{1}\omega_{o}^{4}+\lambda_{2}\omega_{o}^{3}+\lambda_{3}\omega_{o}^{2},
γ3i=\displaystyle\gamma_{3i}= (6bi2μa6biμv6bihμp6biμa/τ+6bi2hμv)ωo4\displaystyle(6b_{i}^{2}\mu_{a}-6b_{i}\mu_{v}-6b_{i}h\mu_{p}-6b_{i}\mu_{a}/\tau+6b_{i}^{2}h\mu_{v})\omega_{o}^{4}
6bi2μpωo2,\displaystyle-6b_{i}^{2}\mu_{p}\omega_{o}^{2},
α4i=\displaystyle\alpha_{4i}= bi2(h2μv2μa2)ωo6+(3bi2h2μp2+6bi2μpμa+\displaystyle b^{2}_{i}(h^{2}\mu^{2}_{v}-\mu^{2}_{a})\omega_{o}^{6}+(3b_{i}^{2}h^{2}\mu_{p}^{2}+6b_{i}^{2}\mu_{p}\mu_{a}+
6biμpμa/τ)ωo4,\displaystyle 6b_{i}\mu_{p}\mu_{a}/\tau)\omega_{o}^{4},
γ4i=\displaystyle\gamma_{4i}= (2bi2hμv2bihμp2biμv)ωo66bi2μpωo4,\displaystyle(2b_{i}^{2}h\mu_{v}-2b_{i}h\mu_{p}-2b_{i}\mu_{v})\omega_{o}^{6}-6b_{i}^{2}\mu_{p}\omega_{o}^{4},
α5i=\displaystyle\alpha_{5i}= (bi2h2μp2+2bi2μaμp)ωo6,γ5i=2bi2μpωo6.\displaystyle(b_{i}^{2}h^{2}\mu_{p}^{2}+2b^{2}_{i}\mu_{a}\mu_{p})\omega_{o}^{6},\gamma_{5i}=2b^{2}_{i}\mu_{p}\omega_{o}^{6}.

By μp>0\mu_{p}>0 and μa>0\mu_{a}>0, we know α¯5>0\overline{\alpha}_{5}>0 and γ¯5>0\underline{\gamma}_{5}>0. From (30), α¯5>α5i\overline{\alpha}_{5}>\alpha_{5i} and γ5i>γ¯5\gamma_{5i}>\underline{\gamma}_{5}, we obtain kγ5i/α5ik\geq\gamma_{5i}/\alpha_{5i}. This together with α5i>0\alpha_{5i}>0 and γ5i>0\gamma_{5i}>0 leads to

α5ik2γ5ik0.\displaystyle\alpha_{5i}k^{2}-\gamma_{5i}k\geq 0. (168)

From (28) and μa>0\mu_{a}>0, we know μv>b¯μa/(b¯h)\mu_{v}>\overline{b}\mu_{a}/(\underline{b}h). By (29), we know ωo>((3b¯2h2μp2+6b¯2μpμa+6b¯μpμa/τ)/(b¯2h2μv2b¯2μa2))1/2\omega_{o}>((3\underline{b}^{2}h^{2}\mu_{p}^{2}+6\underline{b}^{2}\mu_{p}\mu_{a}+6\underline{b}\mu_{p}\mu_{a}/\tau)/(\underline{b}^{2}h^{2}\mu^{2}_{v}-\overline{b}^{2}\mu^{2}_{a}))^{1/2}. This together with μv>b¯μa/(b¯h)\mu_{v}>\overline{b}\mu_{a}/(\underline{b}h) leads to α¯4>0\underline{\alpha}_{4}>0. This together with α¯4<α4i<α¯4\underline{\alpha}_{4}<\alpha_{4i}<\overline{\alpha}_{4}, we know α4i>0\alpha_{4i}>0. From (30) and ρ¯4>ρ4i>0\overline{\rho}_{4}>\rho_{4i}>0, we know kmax{0,(γ4i+|γ4i24α4iρ4i|)/(2α4i)}k\geq\max\{0,(-\gamma_{4i}+\sqrt{|\gamma_{4i}^{2}-4\alpha_{4i}\rho_{4i}|})/(2\alpha_{4i})\}. This together with α4i>0\alpha_{4i}>0 and ρ4i>0\rho_{4i}>0 leads to

α4ik2+γ4ik+ρ4i0.\displaystyle\alpha_{4i}k^{2}+\gamma_{4i}k+\rho_{4i}\geq 0. (169)

By (28), we know μv>max{3μab¯2/(2hb¯2),2μab¯/(τhb¯2)}\mu_{v}>\max\{3\mu_{a}\overline{b}^{2}/(2h\underline{b}^{2}),2\mu_{a}\overline{b}/(\tau h\underline{b}^{2})\}. This together with μa>0\mu_{a}>0 leads to λ¯1>0\underline{\lambda}_{1}>0. By μp>4μab¯/(τb¯2h2)\mu_{p}>4\mu_{a}\overline{b}/(\tau\underline{b}^{2}h^{2}), we know 3μp(b¯2h2μp4μab¯/τ)>03\mu_{p}(\underline{b}^{2}h^{2}\mu_{p}-4\mu_{a}\overline{b}/\tau)>0. This together with μa>0\mu_{a}>0 leads to λ¯3>0\underline{\lambda}_{3}>0. By (29), we know ωo>max{0,(λ¯2+|λ¯224λ¯1λ¯3|)/(2λ¯1)}\omega_{o}>\max\{0,(-\underline{\lambda}_{2}+\sqrt{|\underline{\lambda}_{2}^{2}-4\underline{\lambda}_{1}\underline{\lambda}_{3}|})/(2\underline{\lambda}_{1})\}. This together with λ¯1>0\underline{\lambda}_{1}>0 and λ¯3>0\underline{\lambda}_{3}>0 leads to α¯3>0\underline{\alpha}_{3}>0. This together with α¯3<α3i<α¯3\underline{\alpha}_{3}<\alpha_{3i}<\overline{\alpha}_{3}, we know α3i>0\alpha_{3i}>0. From (30) and ρ¯3>ρ3i>0\overline{\rho}_{3}>\rho_{3i}>0, we know kmax{0,(γ3i+|γ3i24α3iρ3i|)/(2α3i)}k\geq\max\{0,(-\gamma_{3i}+\sqrt{|\gamma_{3i}^{2}-4\alpha_{3i}\rho_{3i}|})/(2\alpha_{3i})\}. This together with α3i>0\alpha_{3i}>0 and ρ3i>0\rho_{3i}>0 leads to

α3ik2+γ3ik+ρ3i0.\displaystyle\alpha_{3i}k^{2}+\gamma_{3i}k+\rho_{3i}\geq 0. (170)

By (28), we know μv>4μab¯2/(hb¯2)\mu_{v}>4\mu_{a}\overline{b}^{2}/(h\underline{b}^{2}). This leads to 3b¯2h2μv212b¯2hμaμv>03\underline{b}^{2}h^{2}\mu_{v}^{2}-12\overline{b}^{2}h\mu_{a}\mu_{v}>0. By (28), we know μv>3μab¯/(2hb¯)\mu_{v}>3\mu_{a}\overline{b}/(2h\underline{b}). This leads to (12b¯hμaμv18b¯μa2)/τ>0(12\underline{b}h\mu_{a}\mu_{v}-18\overline{b}\mu_{a}^{2})/\tau>0. By μp>2μab¯/(b¯2h2)\mu_{p}>2\mu_{a}\overline{b}/(\underline{b}^{2}h^{2}), we know μp(b¯2h2μp2b¯μa/τ)>0\mu_{p}(\underline{b}^{2}h^{2}\mu_{p}-2\overline{b}\mu_{a}/\tau)>0. The above together with μp>0\mu_{p}>0,μa>0\mu_{a}>0 lead to α¯2>0\underline{\alpha}_{2}>0. This together with α¯2<α2i<α¯2\underline{\alpha}_{2}<\alpha_{2i}<\overline{\alpha}_{2}, we know α2i>0\alpha_{2i}>0. From (30) and ρ¯2>ρ2i>0\overline{\rho}_{2}>\rho_{2i}>0, we obtain kmax{0,(γ2i+|γ2i24α2iρ2i|)/(2α2i)}k\geq\max\{0,(-\gamma_{2i}+\sqrt{|\gamma_{2i}^{2}-4\alpha_{2i}\rho_{2i}|})/(2\alpha_{2i})\}. This together with α2i>0\alpha_{2i}>0 and ρ2i>0\rho_{2i}>0 leads to

α2ik2+γ2ik+ρ2i0.\displaystyle\alpha_{2i}k^{2}+\gamma_{2i}k+\rho_{2i}\geq 0. (171)

By (28), we know μv>μab¯/(hb¯)\mu_{v}>\mu_{a}\overline{b}/(h\underline{b}). This together with μa>0\mu_{a}>0, we know μa/τb¯μa+b¯hμv>0\mu_{a}/\tau-\overline{b}\mu_{a}+\underline{b}h\mu_{v}>0 and α¯1<α1i<α¯1\underline{\alpha}_{1}<\alpha_{1i}<\overline{\alpha}_{1}. From (30) and ρ¯1>ρ1i>0\overline{\rho}_{1}>\rho_{1i}>0, we obtain kmax{0,(γ1i+|γ1i24α1iρ1i|)/(2α1i)}k\geq\max\{0,(-\gamma_{1i}+\sqrt{|\gamma_{1i}^{2}-4\alpha_{1i}\rho_{1i}|})/(2\alpha_{1i})\}. This together with α1i>0\alpha_{1i}>0 and ρ1i>0\rho_{1i}>0 leads to

α1ik2+γ1ik+ρ1i0.\displaystyle\alpha_{1i}k^{2}+\gamma_{1i}k+\rho_{1i}\geq 0. (172)

By (168)-(172), we know

(α5ik2γ5ik)ω2+(α4ik2+γ4ik+ρ4i)ω4\displaystyle(\alpha_{5i}k^{2}-\gamma_{5i}k)\omega^{2}+(\alpha_{4i}k^{2}+\gamma_{4i}k+\rho_{4i})\omega^{4}
+(α3ik2+γ3ik+ρ3i)ω6+(α2ik2+γ2ik+ρ2i)ω8\displaystyle+(\alpha_{3i}k^{2}+\gamma_{3i}k+\rho_{3i})\omega^{6}+(\alpha_{2i}k^{2}+\gamma_{2i}k+\rho_{2i})\omega^{8}
+(α1ik2+γ1ik+ρ1i)ω10+ω120,ω.\displaystyle+(\alpha_{1i}k^{2}+\gamma_{1i}k+\rho_{1i})\omega^{10}+\omega^{12}\geq 0,\;\forall\;\omega\in\mathbb{R}. (173)

By (167) and (E), we get

(2bikpβ3n¯2i+d¯1i22bikpβ3d¯2in¯1i2)ω2\displaystyle\hskip 11.38109pt(2b_{i}k_{p}\beta_{3}\overline{n}_{2i}+{\overline{d}^{2}_{1i}}-2b_{i}k_{p}\beta_{3}\overline{d}_{2i}-\overline{n}^{2}_{1i})\omega^{2}
+(2n¯1in¯3i+2bikpβ3d¯4i+d¯2i2n¯2i22bi2kpkvβ3\displaystyle+(2\overline{n}_{1i}\overline{n}_{3i}+2b_{i}k_{p}\beta_{3}\overline{d}_{4i}+\overline{d}^{2}_{2i}-\overline{n}^{2}_{2i}-2b_{i}^{2}k_{p}k_{v}\beta_{3}
2d¯1id¯3i)ω4+(2n¯2ibikv+2d¯1id¯5i+d¯3i2n¯3i2\displaystyle-2\overline{d}_{1i}\overline{d}_{3i})\omega^{4}+(2\overline{n}_{2i}b_{i}k_{v}+2\overline{d}_{1i}\overline{d}_{5i}+\overline{d}^{2}_{3i}-\overline{n}^{2}_{3i}
2bikpβ32d¯2id¯4i)ω6+(d¯4i2+2d¯2ibi2kv2\displaystyle-2b_{i}k_{p}\beta_{3}-2\overline{d}_{2i}\overline{d}_{4i})\omega^{6}+(\overline{d}^{2}_{4i}+2\overline{d}_{2i}-b_{i}^{2}k_{v}^{2}
2d¯3id¯5i)ω8+(d¯5i22d¯4i)ω10+ω120,ω.\displaystyle-2\overline{d}_{3i}\overline{d}_{5i})\omega^{8}+(\overline{d}^{2}_{5i}-2\overline{d}_{4i})\omega^{10}+\omega^{12}\geq 0,\;\forall\;\omega\in\mathbb{R}. (174)

Through calculation, we know

xdi2(ω)+ydi2(ω)xni2(ω)yni2(ω)\displaystyle x_{di}^{2}(\omega)+y_{di}^{2}(\omega)-x_{ni}^{2}(\omega)-y_{ni}^{2}(\omega)
=\displaystyle= (2bikpβ3n¯2i+d¯1i22bikpβ3d¯2in¯1i2)ω2+(2n¯1in¯3i\displaystyle(2b_{i}k_{p}\beta_{3}\overline{n}_{2i}+{\overline{d}^{2}_{1i}}-2b_{i}k_{p}\beta_{3}\overline{d}_{2i}-\overline{n}^{2}_{1i})\omega^{2}+(2\overline{n}_{1i}\overline{n}_{3i}
+2bikpβ3d¯4i+d¯2i2n¯2i22bi2kpkvβ32d¯1id¯3i)ω4\displaystyle+2b_{i}k_{p}\beta_{3}\overline{d}_{4i}+\overline{d}^{2}_{2i}-\overline{n}^{2}_{2i}-2b_{i}^{2}k_{p}k_{v}\beta_{3}-2\overline{d}_{1i}\overline{d}_{3i})\omega^{4}
+(2n¯2ibikv+2d¯1id¯5i+d¯3i2n¯3i22bikpβ3\displaystyle+(2\overline{n}_{2i}b_{i}k_{v}+2\overline{d}_{1i}\overline{d}_{5i}+\overline{d}^{2}_{3i}-\overline{n}^{2}_{3i}-2b_{i}k_{p}\beta_{3}
2d¯2id¯4i)ω6+(d¯4i2+2d¯2ibi2kv22d¯3id¯5i)ω8\displaystyle-2\overline{d}_{2i}\overline{d}_{4i})\omega^{6}+(\overline{d}^{2}_{4i}+2\overline{d}_{2i}-b_{i}^{2}k_{v}^{2}-2\overline{d}_{3i}\overline{d}_{5i})\omega^{8}
+(d¯5i22d¯4i)ω10+ω12\displaystyle+(\overline{d}^{2}_{5i}-2\overline{d}_{4i})\omega^{10}+\omega^{12}

This together with (E) leads to

xdi2(ω)+ydi2(ω)xni2(ω)yni2(ω)0,ω.\displaystyle x_{di}^{2}(\omega)+y_{di}^{2}(\omega)-x_{ni}^{2}(\omega)-y_{ni}^{2}(\omega)\geq 0,\;\forall\;\omega\in\mathbb{R}. (175)

By (175), we know

xni2(ω)+yni2(ω)xdi2(ω)+ydi2(ω)1,ω.\displaystyle\frac{x_{ni}^{2}(\omega)+y_{ni}^{2}(\omega)}{x_{di}^{2}(\omega)+y_{di}^{2}(\omega)}\leq 1,\;\forall\;\omega\in\mathbb{R}. (176)

Through calculation, we obtain

|xni(ω)+yni(ω)jxdi(ω)+ydi(ω)j|=xni2(ω)+yni2(ω)xdi2(ω)+ydi2(ω).\displaystyle\left|\frac{x_{ni}(\omega)+y_{ni}(\omega)j}{x_{di}(\omega)+y_{di}(\omega)j}\right|=\frac{\sqrt{x_{ni}^{2}(\omega)+y_{ni}^{2}(\omega)}}{\sqrt{x_{di}^{2}(\omega)+y_{di}^{2}(\omega)}}.

This together with (176) leads to

|xni(ω)+yni(ω)jxdi(ω)+ydi(ω)j|1,ω.\displaystyle\left|\frac{x_{ni}(\omega)+y_{ni}(\omega)j}{x_{di}(\omega)+y_{di}(\omega)j}\right|\leq 1,\;\forall\;\omega\in\mathbb{R}. (177)

From (158) and (177), we know |Gei(jω)|1|G_{ei}(j\omega)|\leq 1 for any ω\omega\in\mathbb{R}. That is Gei(s)1\|G_{ei}(s)\|_{\infty}\leq 1.∎

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