Event-triggered State Consensus of Heterogeneous Uncertain Second-order Nonlinear Multi-agent Systems
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摘要: 研究异构不确定二阶非线性多智能体系统事件触发状态趋同控制问题. 首先, 为每个智能体设计参数观测器, 用以估计不确定参数, 这些观测器可渐近估计不确定参数. 其次, 为每个跟随智能体设计分布式参数观测器, 渐近估计领导智能体不确定参数, 每个智能体利用邻居智能体触发时刻的采样值估计其邻居智能体的状态. 基于估计的参数和邻居状态, 提出完全不依赖智能体间连续信息传输的事件触发趋同算法. 同时, 证明在所给算法的作用下, 多智能体系统能够达到状态趋同且不存在芝诺现象. 最后, 给出一个多单摆系统, 用以验证事件触发趋同算法. 仿真实验结果表明, 跟随智能体的位置和速度可以渐近跟踪领导智能体的位置和速度, 并且整个多智能体系统平均每秒触发8.825次. 对比仿真实验显示, 所提出的事件触发趋同算法可有效减少事件触发次数.
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关键词:
- 多智能体系统 /
- 事件触发 /
- 不确定二阶非线性系统 /
- 状态趋同 /
- 异构系统
Abstract: This paper studies event-triggered state consensus of heterogeneous uncertain second-order nonlinear multi-agent systems. Firstly, to deal with the uncertain parameter of each agent dynamics, parameter observers are designed for each agent to asymptotically estimate uncertain parameters, these observers can estimate exact values of uncertain parameters. Secondly, distributed parameter observers are designed for each follower agent to asymptotically estimate uncertain parameters of leader agent, and each agent use its neighbors＇sampled information at the trigger instant to estimate its neighbors＇states. Based on estimated parameters and neighbors＇states, an event-triggered consensus algorithm is proposed, which does not rely on any continuous information transmission among agents. Furthermore, it is proved that multi-agent systems can reach state consensus under the proposed algorithm, and there is no Zeno-behavior. Finally, a multi-pendulum system is given to verify event-triggered consensus algorithm. The simulation experiment results show that the position and velocity of follower agents can track that of leader agent asymptotically, and the average trigger number of overall multi-agent systems is 8.825 per second, comparison simulation experiment shows that the proposed event-triggered consensus algorithm can effectively reduce event-triggered number. -
近20年, 多智能体系统的协同控制因其在无人机编队[1]、传感器网络同步[2]、多机器人协作[3]等工程中的广泛应用, 越来越受到控制理论领域学者们的关注. 传统的协同控制算法依赖智能体间的连续信息传输, 即使信息变化很小或没有变化仍然会进行信息传输, 这会造成电能、通信带宽、网络链路的低效利用[4]. 由于事件触发通信机制可以有效地节约能源和通信带宽, 基于事件触发机制的协同控制成为多智能体系统协同控制领域的研究热点[5−6]. 文献[7]给出一些基于事件触发通信机制的多智能体系统协同控制的研究成果.
多智能体系统事件触发协同控制领域的研究成果大多要求系统模型是精确可知的, 然而实际多智能体系统不可避免地存在未知参数、模型不确定、外部噪声等不确定因素. 文献[8]对无向网络的一类不确定非线性多智能体系统的事件触发趋同问题进行了研究. 文献[9]研究无向网络拓扑下一类二阶非线性多智能体系统的自适应事件触发趋同控制问题. 针对未知二阶非线性多智能体系统, 文献[10]利用自适应事件触发控制方法研究完全分布式控制问题. 文献[11]对网络拓扑信息未知的一般线性多智能体系统的完全分布式事件触发趋同问题进行研究. 针对控制方向未知的高阶多智能体系统, 文献[12]利用虚拟控制律设计自适应事件触发跟踪控制器. 文献[13]研究具有时滞和输入饱和的异构多智能体系统, 并给出基于观测器的事件触发趋同算法. 文献[14]利用组合测量事件触发机制, 研究拓扑结构为无向图的未知非线性二阶时滞多智能体系统的自适应趋同控制. 虽然文献[8−10, 14]研究的系统模型与本文相似, 但都采用基于组合测量的事件触发机制, 这种事件触发机制需要连续不断地监测邻居智能体的状态信息用以判断下一次触发时刻, 即算法依赖智能体间连续信息传输. 文献[15−16]利用输出调节理论, 对异构线性多智能体系统的事件触发输出同步问题进行研究. 文献[17]利用分布式内模设计, 研究一类非线性多智能体系统的事件触发全局鲁棒输出调节问题.
上述文献的分布式控制器虽然采用了事件触发机制进行设计, 但是所给的事件触发趋同算法依然依赖智能体间的连续信息传输. 触发函数对邻居智能体状态信息连续监测问题引起了研究人员的注意. 文献[18]利用基于反步法的分布式自适应输出反馈控制策略研究不确定异构线性多智能体系统的事件触发输出同步问题. 针对由一类高阶不确定非线性系统构成的无领导型异构多智能体系统, 文献[19]给出基于事件触发机制的分布式自适应趋同算法. 文献[20]分别对同构和异构线性多智能体系统的事件触发平均跟踪算法进行研究. 针对异构领导−跟随者型多智能体系统, 文献[21]分别给出基于模型和基于数据的事件触发趋同算法. 文献[22]基于动态事件触发机制, 对一般线性多智能体系统的编队包含控制问题进行研究. 针对拓扑为有向网络的不确定下三角非线性多智能体系统, 文献[23]利用神经网络设计分布式自适应异步事件触发趋同算法. 基于输出调节理论, 文献[24]研究异构线性多智能体系统的自适应事件触发输出趋同控制, 文献[25]研究一类异构非线性多智能体系统的分布式事件触发输出趋同控制问题, 文献[26]研究严格反馈非线性多智能体系统的半全局周期事件触发输出调节问题.
受上述文献启发, 本文研究异构不确定二阶非线性多智能体系统的事件触发状态趋同问题, 主要贡献有如下$ 3 $点: 1)本文研究领导−跟随者型异构不确定多智能体系统的状态趋同问题, 不仅跟随智能体的动力学方程存在不确定参数, 领导智能体也存在不确定参数. 文献[10, 15−16, 24−26]中的领导智能体均为完全已知的, 并未考虑领导智能体存在不确定参数的情形. 2)本文基于邻居智能体的观测状态设计事件触发趋同算法, 由于对邻居智能体的状态进行观测, 避免了事件触发函数对邻居智能体的连续监测, 做到控制器与触发函数都不依赖智能体间的连续信息传输. 同样研究异构不确定二阶非线性多智能体系统事件触发控制的文献[9−10], 其事件触发函数需要对邻居智能体的状态进行连续监测. 3)本文不确定参数为矩阵形式而非向量形式, 不同于以往将矩阵转变为向量的处理方法, 本文直接利用矩阵迹的不等式对矩阵自适应参数估计的收敛性进行证明.
1. 问题描述
为方便表示, 本文使用如下向量与矩阵的符号: $ ||\cdot||_{\rm{F }}$和$ ||\cdot|| $分别表示向量或矩阵的Frobenius范数和2范数, $ \otimes$为矩阵的克罗内克积, $ \mathrm{diag}\{a_1,\;\cdots, a_N\} $表示对角元素为$ a_i $的对角矩阵, $ \mathrm{tr}\{A\} $表示方阵$ A $的迹, $ 1_N $表示每个元素都为$ 1 $的$ N $维常向量, I表示单位矩阵, $ \lambda_{1X} $和$ \lambda_{NX} $分别表示$ N $阶对称矩阵$ X $的最小和最大特征根, $ {\cal{A}}(t) $表示渐近收敛到$ \boldsymbol 0 $的函数集合.
本文研究领导−跟随者型异构不确定二阶非线性多智能体系统事件触发趋同控制问题. 第$ i $个跟随智能体的动力学方程为:
$$ \begin{equation} \left\{\begin{aligned} \dot{x}_i(t)& = y_i(t)\\ \dot{y}_i(t)& = \theta_i^{\mathrm{T}}\phi_i(x_i(t),\;y_i(t))+u_i(t) \end{aligned}\right. \end{equation} $$ (1) 式中, $ x_i,\;y_i,\;u_i\in {\bf{R}}^n $分别表示第$ i $个智能体的位置、速度和控制输入; $ \theta_i\in {\bf{R}}^{n_i\times n} $为不确定常矩阵; $ \phi_i: {\bf{R}}^n\times {\bf{R}}^n\rightarrow {\bf{R}}^{n_i} $为已知向量函数.
领导智能体标记为$ 0 $号智能体, 其动力学方程为含有未知输入的二阶积分器型系统:
$$ \begin{equation} \left\{\begin{aligned} \dot{x}_0(t)& = y_0(t)\\ \dot{y}_0(t)& = \theta_0^{\mathrm{T}} \phi(t) \end{aligned}\right. \end{equation} $$ (2) 式中, $ x_0,\;y_0\in {\bf{R}}^n $分别为领导智能体的位置和速度; $ \theta_0\in {\bf{R}}^{n_0\times n} $为不确定常矩阵; $ \phi(t):[0,\;\infty)\rightarrow {\bf{R}}^{n_0} $为已知向量函数.
本文的目标是设计基于事件触发机制的趋同控制算法, 使得$ \lim_{t\rightarrow\infty}x_i(t) = x_0(t),\;y_i(t) = y_0(t). $
领导−跟随者型多智能体系统(1)、(2)的网络拓扑用有向图$ {\cal{G}} = \{{\cal{V}},\;{\cal{E}}\} $描述, 其中$ {\cal{V}} = \{0,\;1,\; \cdots, N\} $为智能体集合, $ {\cal{E}} = {\cal{V}}\times{\cal{V}} $为边集. $ (i,\;j)\in{\cal{E}} $表示一条从智能体$ j $到智能体$ i $的有向边, 相应的邻接权重$ a_{ij}>0 $, 否则$ a_{ij} = 0 $. 有向边序列$ (i_l,\; i_{l-1}), l=1,\; \cdots,\; k\,\; $表示从智能体$ i_0 $到智能体$ i_k $的一条路径. 图$ {\cal{G}} $的拉普拉斯矩阵$ {\cal{L}} $定义为$ l_{ii} = \sum_{j \,\;=\,\; 0}^Na_{ij}, l_{ij} = -a_{ij},\;i\neq j $.
注1. 由于领导智能体不能接收到跟随智能体的信息, 有向图$ {\cal{G}} $的拉普拉斯矩阵$ {\cal{L}} $可表示为:
$$ \begin{equation*} {\cal{L}} = \left[\begin{array}{cc}0&{\bf 0}_{1\times N}\\ *&L \end{array}\right],\; \; L\in {\bf{R}}^{N\times N},\; \; *\in {\bf{R}}^{N} \end{equation*} $$ 由文献[27]的引理3可知, 当假设1成立时, 矩阵$ L $是非奇异的, 并且存在矩阵$ Q = \mathrm{diag}\{1/ q_1, \cdots,\;1/ q_N\} $, $ H = (QL+L^{\mathrm{T}}Q) /{2}$为正定矩阵, 其中$ [q_1,\;\cdots,\; q_N]^{\mathrm{T}} = L^{-1}1_N $.
为证明算法的稳定性, 需要以下假设和引理.
假设1. 对于任意跟随智能体$ i,\;i = 1,\;\cdots,\;N $, 至少存在一条由领导智能体到跟随智能体$ i $的有向路径.
假设2. $ \phi(t) $, $ \phi_i(x_i(t) $, $ y_i(t)) $为不恒等于$ \bf 0 $的有界向量函数.
假设3. 在不确定输入$ \theta_0^{\mathrm{T}}\phi(t) $的作用下, 领导智能体的状态有界.
引理1[28]. 考虑如下系统:
$$ \begin{equation} \dot{x}(t) = f(t,\;x(t),\;u(t)) \end{equation} $$ (3) 式中, $ f:[0,\;\infty)\times {\bf{R}}^n\times {\bf{R}}^m\rightarrow {\bf{R}}^n $对$ t $是分段连续的, 对$ x(t) $和$ u(t) $满足局部Lipschitz条件. 输入$ u(t) $对所有$ t\geq0 $是分段连续且有界的函数. 如果系统(3)是输入状态稳定的且$ u(t)\in{\cal{A}}(t) $, 则亦有状态$ x(t) \in {\cal{A}}(t) $.
2. 事件触发趋同算法的设计
由于领导智能体的参数$ \theta_0 $不确定, 首先为领导智能体设计如下参数观测器:
$$ \begin{equation} \left\{\begin{aligned} \dot{\hat{y}}_0& =( \hat{\theta}_0^0)^{\mathrm{T}}\phi(t)-s_0(\hat{y}_0-y_0)\\ \dot{\hat{\theta}}_0^0& = -\phi(t)(\hat{y}_0-y_0)^{\mathrm{T}} \end{aligned}\right. \end{equation} $$ (4) 式中, $\hat{y}_0 $为领导智能体速度状态的观测值, $s_0>0 $为正数, $ \hat{\theta}_0^0(t) $用以估计参数$ \theta_0 $. 跟随智能体的参数$ \theta_i $同样不确定, 设计如下参数观测器:
$$ \begin{equation} \left\{\begin{aligned} \dot{\hat{y}}_i& = \hat{\theta}_i^{\mathrm{T}}\phi_i(x_i,\;y_i)+u_i-s_i(\hat{y}_i-y_i)\\ \dot{\hat{\theta}}_i& = -\phi_i(x_i,\;y_i)(\hat{y}_i-y_i)^{\mathrm{T}} \end{aligned}\right. \end{equation} $$ (5) 式中, $\hat{y}_i $为第i个智能体速度状态的观测值, $ s_i>0 $为正数, $ \hat{\theta}_i(t) $用以估计参数$ \theta_i $.
由于领导智能体含有不确定控制输入$ \theta_0^{\mathrm{T}}\phi(t) $, 为了使跟随智能体跟踪上领导智能体, 为跟随智能体$ i $设计如下$ \theta_0 $参数的观测器:
$$ \begin{equation} \dot{\hat{\theta}}_0^i(t) = -\mu\sum\limits_{j = 0}^Na_{ij}(\hat{\theta}_0^i(t_k^i)-\hat{\theta}_0^j(t_{k'}^j)) \end{equation} $$ (6) 式中, $ \mu>0 $为常数, $ t_k^i $和$ t_{k'}^j $为智能体$ i $和$ j $的事件触发时刻, 并且有$ t_0^i = t_0^j = 0 $.
在触发时刻$ t_{k'}^j $, 智能体$ j $将其采样信息$ \hat{\theta}_0^j(t_{k'}^j) $, $ x_j(t_{k'}^j) $和$ y_j(t_{k'}^j) $发送给邻居智能体$ i $. 智能体$ i $利用采样信息$ \hat{\theta}_0^j(t_{k'}^j) $, $ x_j(t_{k'}^j) $和$ y_j(t_{k'}^j) $估计智能体$ j $在下一次采样时刻$ t_{(k+1)'}^j $前的位置和速度. 用$ \hat{x}_j^i(t) $和$ \hat{y}_j^i(t) $表示时间段$ [t_{k'}^j,\;t_{(k+1)'}^j) $内智能体$ i $对智能体$ j $的状态信息估计, 状态估计方程为:
$$ \begin{equation} \left\{\begin{aligned} \dot{\hat{x}}_j^i(t)& = \hat{y}_j^i(t)\\ \dot{\hat{y}}_j^i(t)& = (\hat{\theta}_0^{j}(t_{k'}^j))^{\mathrm{T}}\phi(t) \end{aligned}\right. \end{equation} $$ (7) 式中, 初始状态分别为$ \hat{x}_j^i(t_{k'}^j) = x_j(t_{k'}^j) $, $ \hat{y}_j^i(t_{k'}^j) = y_j(t_{k'}^j) $.
同时, 智能体$ j $也将利用其事件触发采样信息估计其自身的状态信息. 如果智能体$ i $和$ l $同时接收到智能体$ j $的事件触发采样信息, 则不难验证智能体$ i $, $l $和$ j $拥有相同状态估计值, 即:
$$ \hat{x}_j^i(t) = \hat{x}_j^l(t) = \hat{x}_j^j(t),\;\hat{y}_j^i(t) = \hat{y}_j^l(t) = \hat{y}_j^j(t) $$ 记$ \hat{\xi}_{ix} = \sum_{j = 0}^N a_{ij}(\hat{x}_i^i - \hat{x}_j^i),\; \hat{\xi}_{iy} = \sum_{j = 0}^N a_{ij}(\hat{y}_i^i - \hat{y}_j^i) $, 为跟随智能体式(1)设计如下事件触发趋同控制器:
$$ \begin{equation} u_i = -\hat{\theta}_i^{\mathrm{T}}\phi_i(x_i,\;y_i)+(\hat{\theta}_0^{i})^{\mathrm{T}}\phi(t)-ck_1\hat{\xi}_{ix}-ck_2\hat{\xi}_{iy} \end{equation} $$ (8) 式中, $ k_1 $, $ k_2>0 $为耦合增益; $ c>0 $为反馈增益. $ k_1 $, $ k_2 $和$ c $可根据下文式(22)选取. 智能体$ i $的第$ k+1 $次事件触发时刻由如下条件给出:
$$ \begin{equation} t_{k+1}^i = \min\{t>t_k^i|T_{i1}(t)\geq0\; \mathrm{or}\; T_{i2}(t)\geq0\} \end{equation} $$ (9) 式中, $T_{i1}(t) = ||\epsilon_i(t)||_{\rm{F}}^2 - f_{i1}(t),\; T_{i2}(t) = ||e_i(t)||^2 \;- f_{i2}(t)$, $ \epsilon_i(t) = \hat{\theta}_0^i(t_k^i)-\hat{\theta}_0^i(t) $, $ e_i(t) = k_1e_{ix}(t)+ k_2e_{iy}(t) $,$ e_{ix}(t) =\hat{x}_i^i(t)-x_i(t) $, $ e_{iy}(t) = \hat{y}_i^i(t)-y_i(t) $, 正函数$ f_{i1}(t),\;f_{i2}(t)\in{\cal{A}}(t) $.
领导智能体$ 0 $的第$ k+1 $次事件触发时刻由如下条件确定:
$$ \begin{equation} t_{k+1}^0 = \min\{t>t_k^0|T_{01}(t)\geq0\; \mathrm{or}\; T_{02}(t)\geq0\} \end{equation} $$ (10) 式中, 各符号定义与式(9)中符号定义类似.
注2. 跟随智能体的控制输入式(8)只依赖其自身状态、邻居智能体的估计状态和估计参数$ \hat{\theta} _0^i(t), \hat{\theta}_i(t) $, 仅需要邻居智能体提供离散的信息 $ \hat{\theta}_0^j(t_{k'}^j) $, $ x_j(t_{k'}^j) $和$ y_j(t_{k'}^j) $, 不依赖邻居智能体的任何连续信息传输. 同样, 事件触发条件(9)、(10)也不依赖邻居智能体的任何连续信息传输. 因此, 本文提出的事件触发趋同算法完全不依赖智能体间的连续信息传输.
3. 事件触发控制器的稳定性分析
命题1. 如果假设2成立, 参数观测器式(4)、式(5) 中的$ \hat{\theta}_0^0(t) $和$ \hat{\theta}_i(t) $可渐近收敛到$ \theta_0 $和$ \theta_i $, 即$ \lim_{t\rightarrow\infty}\hat{\theta}_0^0(t) = \theta_0 $, $ \lim_{t\rightarrow\infty}\hat{\theta}_i(t) = \theta_i. $
证明. 记$ \tilde{y}_i(t) = \hat{y}_i(t)-y_i(t) $, $ \tilde{\theta}_i(t) = \hat{\theta}_i(t)- \theta_i $. 对于观测器式(5), 可得:
$$ \begin{equation} \left\{\begin{aligned} \dot{\tilde{y}}_i(t)& = \tilde{\theta}_i^{\mathrm{T}}(t)\phi_i(x_i(t),\;y_i(t))-s_i\tilde{y}_i(t)\\ \dot{\tilde{\theta}}_i(t)& = -\phi_i(x_i(t),\;y_i(t))\tilde{y}_i^{\mathrm{T}}(t) \end{aligned}\right. \end{equation} $$ (11) 选取如下李雅普诺夫函数:
$$ V_{i1} = \frac{1}{2}\tilde{y}_i^{\mathrm{T}}(t)\tilde{y}_i(t)+\frac{1}{2}\mathrm{tr}\{\tilde{\theta}_i^{\mathrm{T}}(t) \tilde{\theta}_i(t)\} $$ 沿式(11)的轨迹求$ V_{i1} $的导数, 可得:
$$ \dot{V}_{i1} = -s_i\tilde{y}_i^{\mathrm{T}}(t)\tilde{y}_i(t) $$ 这表明$ \lim_{t\rightarrow\infty}\tilde{y}_i(t) = \bf 0 $. 由系统 (11)可知$ \tilde{y}_i(t) {\text{恒等于}}\, \bf 0 $, 可得$ \tilde{\theta}_i^{\mathrm{T}}(t)\phi_i(x_i (t),\;y_i(t))\,{\text{恒等于}}\, \bf 0$. 由假设2可知$ \phi_i (x_i(t),\;y_i(t)) $不恒等于$ \bf 0 $且有界, 从而可得$ \lim_{t\rightarrow\infty} \hat{\theta}_i(t) = \theta_i $. 亦可证明$ \lim_{t\rightarrow\infty} \hat{\theta}_0^0(t) = \theta_0 $.
□ 命题2. 如果假设1和假设2成立, 在事件触发条件(9)、(10) 作用下, 估计参数$ \hat{\theta}_0^i(t) $渐近收敛至$ \theta_0 $.
证明. 记$ \zeta_i(t) = \sum_{j = 0}^Na_{ij}(\tilde{\theta}_0^i(t)-\tilde{\theta}_0^j(t)) $, $ \sigma_i(t) = \sum_{j = 0}^Na_{ij}(\epsilon_i(t)-\epsilon_j(t)) $, $ \tilde{\theta}_0^i(t) = \hat{\theta}_0^i(t)-\theta_0 $. 由式(6)可得:
$$ \begin{equation} \dot{\tilde{\theta}}_0^i(t) = -\mu\zeta_i(t)-\mu\sigma_i(t) \end{equation} $$ (12) 选取如下李雅普诺夫函数:
$$ \begin{equation} V_2 = \sum\limits_{i = 1}^N\frac{1}{2q_i}\mathrm{tr}\{\zeta_i^{\mathrm{T}}(t)\zeta_i(t)\} \end{equation} $$ (13) 由式(12)可得$ V_2 $的导数:
$$ \begin{equation*} \begin{aligned} \dot{V}_2 = \;&-\mu\mathrm{tr}\{\zeta^{\mathrm{T}}((QL)\otimes I_{n_0})\zeta\}\;-\\ &\mu\mathrm{tr}\{\zeta^{\mathrm{T}}((QL)\otimes I_{n_0})\sigma\}\;+\\ &\sum_{i = 1}^N\frac{a_{i0}}{q_i}\mathrm{tr}\{\zeta_i^{\mathrm{T}}\phi(t)\tilde{y}_0^{\mathrm{T}}\} \end{aligned} \end{equation*} $$ 式中, $ \zeta = [\zeta_1^{\mathrm{T}},\;\cdots,\;\zeta_N^{\mathrm{T}}]^{\mathrm{T}} $, $ \sigma = [\sigma_1^{\mathrm{T}},\;\cdots,\;\sigma_N^{\mathrm{T}}]^{\mathrm{T}} $.
对于$ \dot{V}_2 $的第1项, 由附录的引理2可得:
$$ \begin{equation} \begin{split} \mathrm{tr}\{\zeta^{\mathrm{T}}((QL)\otimes I_{n_0})\zeta\} = \;&\mathrm{tr}\{\zeta^{\mathrm{T}}(H\otimes I_{n_0})\zeta\}\;\geq\\ & \lambda_{1H}\sum_{i = 1}^N\mathrm{tr}\{\zeta_i^{\mathrm{T}}\zeta_i\} \end{split} \end{equation} $$ (14) 记$ L_e $为$ L $的增广矩阵, 即$ L_e = [-a_0|L] $, $ a_0 \;= [a_{01},\;\cdots,\; a_{0N}]^{\mathrm{T}} $. 令$ \epsilon(t) = [\epsilon_0^{\mathrm{T}}(t),\;\epsilon_1^{\mathrm{T}}(t),\;\cdots,\; \epsilon_N^{\mathrm{T}}(t)]^{\mathrm{T}} $, $ \Xi = QLL^{\mathrm{T}}Q $, $ \Delta = L_e^{\mathrm{T}}L_e $. 易证$ \sigma(t) = (L_e\otimes I_{n0})\epsilon(t) $. 对于$ \dot{V}_2 $的后2项, 由附录A的引理2和引理3可得:
$$ \begin{split} & -\mathrm{tr}\{\zeta^{\mathrm{T}}((QL)\otimes I_{n_0})\sigma\}\leq\frac{\eta_1}{2}\mathrm{tr}\{\zeta^{\mathrm{T}}(\Xi\otimes I_{n_0})\zeta\}\; +\\ &\frac{1}{2\eta_1}\mathrm{tr}\{\epsilon^{\mathrm{T}}(\Delta\otimes I_{n_0})\epsilon\}\leq \frac{\eta_1\lambda_{N\Xi}}{2}\sum_{i = 1}^N\mathrm{tr}\{\zeta_i^{\mathrm{T}}\zeta_i\} \;+\\ &\frac{\lambda_{N\Delta}}{2\eta_1}\sum_{i = 0}^N\mathrm{tr}\{\epsilon_i^{\mathrm{T}}\epsilon_i\}\\[-1pt] \end{split} $$ (15) $$ \begin{equation} \begin{split} & \sum_{i = 1}^N\frac{a_{i0}}{q_i}\mathrm{tr}\{\zeta_i^{\mathrm{T}}\phi(t)\tilde{y}_0^{\mathrm{T}}\}\leq \frac{\eta_2}{2}\sum_{i = 1}^N\mathrm{tr}\{\zeta_i^{\mathrm{T}}\zeta_i\}\;+\\ &\;\;\;\sum_{i = 1}^N\frac{a_{i0}^2}{2\eta_2q_i^2}\mathrm{tr}\{\tilde{y}_0\phi^{\mathrm{T}}(t)\phi(t)\tilde{y}_0^{\mathrm{T}}\} \end{split} \end{equation} $$ (16) 式中, $ \eta_1\in(0,\;\lambda_{1H}/ \lambda_{N\Xi}) $, $ \eta_2\in(0,\;\mu\lambda_{1H}) $.
将式(14) ~ 式(16)代入$ \dot{V}_2 $, 有:
$$ \begin{equation*} \begin{aligned} \dot{V}_2\leq&-(\mu(\lambda_{1H}-\frac{\eta_1\lambda_{N\Xi}}{2})-\frac{\eta_2}{2}) \sum_{i = 1}^N\mathrm{tr}\{\zeta_i^{\mathrm{T}}\zeta_i\}\;+\\ &\frac{\mu\lambda_{N\Delta}}{2\eta_1}\sum_{i = 0}^N\mathrm{tr}\{\epsilon_i^{\mathrm{T}}\epsilon_i\} +\sum_{i = 1}^N\frac{a_{i0}^2}{2\eta_2q_i^2}\mathrm{tr}\{\tilde{y}_0\phi^{\mathrm{T}}\phi\tilde{y}_0^{\mathrm{T}}\} \end{aligned} \end{equation*} $$ 令$ \kappa = \min\{q_i(\mu(2\lambda_{1H}-\eta_1\lambda_{N\Xi})-\eta_2)\} $. 由事件触发条件(9)、(10)和命题1易知存在一个函数$ b(t)\in{\cal{A}}(t) $, 使得:
$$ \frac{\mu\lambda_{N\Delta}}{2\eta_1}\sum\limits_{i = 0}^N\mathrm{tr}\{\epsilon_i^{\mathrm{T}}\epsilon_i\} +\sum\limits_{i = 1}^N\frac{a_{i0}^2}{2\eta_2q_i^2}\mathrm{tr}\{\tilde{y}_0\phi^{\mathrm{T}}\phi\tilde{y}_0^{\mathrm{T}}\}\leq b(t) $$ 即
$$ \begin{equation*} \dot{V}_2\leq -\kappa V_2+b(t) \end{equation*} $$ 由引理1可得$ V_2(t)\in{\cal{A}}(t) $, 即$ \lim_{t\rightarrow\infty}\zeta(t) = \bf 0 $. 记$ \tilde{\Theta}_0(t) = [(\tilde{\theta}_0^{1})^{\mathrm{T}},\;\cdots,\;(\tilde{\theta}_0^{N})^{\mathrm{T}}]^{\mathrm{T}} $, 易得:
$$ \zeta(t) = (L\otimes I_{n_0})\tilde{\Theta}_0(t)+a_0\otimes \tilde{\theta}_0^0(t) $$ 由命题1可知$ \lim_{t\rightarrow\infty}\tilde{\theta}_0^0(t) = \bf 0 $, 又因$L $为非奇异矩阵, 可得$ \lim_{t\rightarrow\infty}\tilde{\Theta}_0(t) = \bf 0 $, 即$ \hat{\theta}_0^i(t) $渐近收敛至$ \theta_0 $.
□ 注3. 由命题1和命题2可知, 观测器式(4)和式(5)可实现对参数$ \theta_0 $和$ \theta_i $的渐近估计, 分布式观测器式(6)在观测器式(4)基础上, 可渐近收敛到$ \theta_0 $. 只有观测器渐近收敛时, 所设计的事件触发趋同算法才可达到渐近趋同, 否则只能达到一致渐近有界趋同. 此外, 不确定参数$ \theta_0 $和$ \theta_i $均为矩阵而非向量, 命题1和命题2直接采用矩阵迹的不等式进行收敛性证明. 相比转化为扩维向量, 本文算法更简单明了.
定理1. 如果假设1 ~ 3成立, 则事件触发算法式(8)、式(9)可使领导−跟随者型多智能体系统达到状态趋同.
证明. 记$ \xi_{ix} = \sum_{j = 0}^Na_{ij}(x_i - x_j),\; \xi_{iy} = \sum_{j = 0}^N a_{ij} \times\;(y_i-y_j) $为第$ i $个跟随智能体的相对状态信息, 易证:
$$ \begin{equation} \left\{\begin{aligned} \dot{\xi}_{ix} = \;&\xi_{iy}\\ \dot{\xi}_{iy} = \;&\sum_{j = 1}^Na_{ij}(\tilde{\theta}_j^{\mathrm{T}}\phi_j-\tilde{\theta}_i^{\mathrm{T}}\phi_i) -a_{i0}\tilde{\theta}_i^{\mathrm{T}}\phi_i\;+\\ &\sum_{j = 1}^Na_{ij}((\tilde{\theta}_0^{i})^{\mathrm{T}}-(\tilde{\theta}_0^{j})^{\mathrm{T}})\phi(t)+a_{i0}(\tilde{\theta}_0^{i})^{\mathrm{T}}\phi(t)\;-\\ &ck_1\sum_{j = 1}^Na_{ij}(\hat{\xi}_{ix}-\hat{\xi}_{jx})-ck_1a_{i0}\hat{\xi}_{ix}\;-\\ &ck_2\sum_{j = 1}^Na_{ij}(\hat{\xi}_{iy}-\hat{\xi}_{jy})-ck_2a_{i0}\hat{\xi}_{iy} \end{aligned}\right. \end{equation} $$ (17) 记$ \xi_i = k_1\xi_{ix}+k_2\xi_{iy} $. 选取如下李雅普诺夫函数:
$$ \begin{equation} V_3 = \sum\limits_{i = 1}^N\frac{\rho_i}{2}\xi_{ix}^{\mathrm{T}}\xi_{ix}+\sum\limits_{i = 1}^N\frac{1}{2q_i}\xi_i^{\mathrm{T}}\xi_i \end{equation} $$ (18) 式中, $ \rho_i = k_1^2/q_i $.
沿式(17)的轨迹可得$ V_3 $的导数:
$$ \begin{equation*} \begin{aligned} \dot{V}_3 =\; &\sum_{i = 1}^N\left(-\frac{\rho_ik_1}{k_2}\xi_{ix}^{\mathrm{T}}\xi_{ix}+\frac{k_1}{q_ik_2}\xi_i^{\mathrm{T}}\xi_i\right)\;+\\ &k_2\xi^{\mathrm{T}}((QL)\otimes I_n)\tilde{\theta}_{\phi}+k_2\xi^{\mathrm{T}}((QL)\otimes I_n)\tilde{\theta}_{\phi}^{0}\;-\\ &ck_2\xi^{\mathrm{T}}((QL)\otimes I_n)\hat{\xi} \end{aligned} \end{equation*} $$ 式中, $ \xi = [\xi_1^{\mathrm{T}},\;\cdots,\;\xi_N^{\mathrm{T}}]^{\mathrm{T}} $, $ \hat{\xi} = [\hat{\xi}_1^{\mathrm{T}},\;\cdots,\;\hat{\xi}_N^{\mathrm{T}}]^{\mathrm{T}} $, $ \tilde{\theta}_{\phi} = [\phi_1^{\mathrm{T}}\tilde{\theta}_1,\;\cdots,\;\phi_N^{\mathrm{T}}\tilde{\theta}_N]^{\mathrm{T}} $, $ \tilde{\theta}_{\phi}^0 = [\phi^{\mathrm{T}}(t)\tilde{\theta}_0^{1},\;\cdots,\;\phi^{\mathrm{T}}(t)\tilde{\theta}_0^{N}]^{\mathrm{T}} $, $ \hat{\xi}_i = k_1\hat{\xi}_{ix}+k_2\hat{\xi}_{iy} $.
根据Young不等式, 存在$ \gamma_1,\;\gamma_2\in(0,\;1) $, 使得$ \dot{V}_3 $的第2项和第3项满足如下不等式:
$$ \begin{equation} \begin{split} & \xi^{\mathrm{T}}((QL)\otimes I_n)\tilde{\theta}_{\phi} = \frac{\gamma_1}{2}\xi^{\mathrm{T}}(\Xi\otimes I_n)\xi+\frac{1}{2\gamma_1} \tilde{\theta}_{\phi}^{\mathrm{T}}\tilde{\theta}_{\phi}\;\leq\\ &\;\;\;\frac{\gamma_1\lambda_{N\Xi}}{2}\sum_{i = 1}^N\xi_i^{\mathrm{T}}\xi_i+\frac{1}{2\gamma_1}\sum_{i = 1}^N \phi_i^{\mathrm{T}}\tilde{\theta}_i\tilde{\theta}_i^{\mathrm{T}}\phi_i\\[-1pt] \end{split} \end{equation} $$ (19) $$ \begin{equation} \begin{split} \xi^{\mathrm{T}}&((QL)\otimes I_n)\tilde{\theta}_{\phi}^0 = \frac{\gamma_2}{2}\xi^{\mathrm{T}}(\Xi\otimes I_n)\xi+\frac{1}{2\gamma_2} (\tilde{\theta}_{\phi}^{0})^{\mathrm{T}}\tilde{\theta}_{\phi}^0\;\leq\\ &\frac{\gamma_2\lambda_{N\Xi}}{2}\sum_{i = 1}^N\xi_i^{\mathrm{T}}\xi_i+\frac{1}{2\gamma_2}\sum_{i = 1}^N \phi^{\mathrm{T}}\tilde{\theta}_0^{i}(\tilde{\theta}_0^{i})^{\mathrm{T}}\phi \\[-1pt]\end{split} \end{equation} $$ (20) 对于$ \dot{V}_3 $的最后1项, 有如下不等式:
$$ \begin{equation} \begin{split} -\xi^{\mathrm{T}}&((QL)\otimes I_n)\hat{\xi} = -\xi^{\mathrm{T}}((QL)\otimes I_n)\xi\;-\\ &\xi^{\mathrm{T}}((QL^2)\otimes I_n)e\leq-(\lambda_{1H}\;-\\ &\frac{\gamma_3\lambda_{N\Pi}}{2})\sum_{i = 1}^N\xi_i^{\mathrm{T}}\xi_i +\frac{1}{2\gamma_3}\sum_{i = 1}^Ne_i^{\mathrm{T}}e_i \end{split} \end{equation} $$ (21) 式中, $e=[e_1^{\mathrm{T}},\;\cdots,\;e_N^{\mathrm{T}}]^{\mathrm{T}} $, $ \Pi = QL^2(L^{2})^{\mathrm{T}}Q $, $ \gamma_3\in (0, \;2\lambda_{1H}/\lambda_{N\Pi}) $.
将式(19) ~ 式(21)代入$ \dot{V}_3 $, 可得:
$$ \begin{equation*} \begin{aligned} \dot{V}_3\leq&-\sum_{i = 1}^N\frac{\rho_ik_1}{k_2}\xi_{ix}^{\mathrm{T}}\xi_{ix}-\sum_{i = 1}^N\left( \frac{ck_2(2\lambda_{1H}-\gamma_3\lambda_{N\Pi})}{2}\;-\right.\\ &\left.\frac{k_1}{q_ik_2}-\frac{(\gamma_1+\gamma_2)k_2\lambda_{N\Xi}}{2}\right)\xi_i^{\mathrm{T}}\xi_i +\frac{ck_2}{2\gamma_3}\sum_{i = 1}^Ne_i^{\mathrm{T}}e_i\;+\\ &\frac{k_2}{2\gamma_1}\sum_{i = 1}^N\phi_i^{\mathrm{T}}\tilde{\theta}_i\tilde{\theta}_i^{\mathrm{T}}\phi_i +\frac{k_2}{2\gamma_2}\sum_{i = 1}^N\phi^{\mathrm{T}}\tilde{\theta}_0^{i}(\tilde{\theta}_0^{i})^{\mathrm{T}}\phi \end{aligned} \end{equation*} $$ 记$ q_{\min} = \min_{i\in\{1,\;\cdots,\;N\}}q_i $. 选取合适的参数$ k_1,\; k_2,\;\gamma_1,\;\gamma_2>0 $, $ \gamma_3\in(0,\;2\lambda_{1H}/\lambda_{N\Pi}) $, $ c>\bar{c} $, 其中:
$$ \begin{equation} \bar{c} = \frac{(\gamma_1+\gamma_2)k_2^2\lambda_{N\Xi}+\displaystyle\frac{2k_1}{q_{\min}}} {(2\lambda_{1H}-\gamma_3\lambda_{N\Pi})k_2^2} \end{equation} $$ (22) 记$ \alpha = k_2(2\lambda_{1H}-\gamma_3\lambda_{N\Pi})(c-\bar{c})/2 $, 可得:
$$ \begin{equation*} \begin{aligned} \dot{V}_3\leq&-\sum_{i = 1}^N\frac{\rho_ik_1}{k_2}\xi_{ix}^{\mathrm{T}}\xi_{ix}-\sum_{i = 1}^N\alpha\xi_i^{\mathrm{T}}\xi_i +\frac{ck_2}{2\gamma_3}\sum_{i = 1}^Ne_i^{\mathrm{T}}e_i\;+\\ &\frac{k_2}{2\gamma_1}\sum_{i = 1}^N\phi_i^{\mathrm{T}}\tilde{\theta}_i\tilde{\theta}_i^{\mathrm{T}}\phi_i +\frac{k_2}{2\gamma_2}\sum_{i = 1}^N\phi^{\mathrm{T}}\tilde{\theta}_0^{i}(\tilde{\theta}_0^{i})^{\mathrm{T}}\phi \end{aligned} \end{equation*} $$ 由于$ \lim_{t\rightarrow\infty}\tilde{\theta}_i(t) = \bf 0,\;\lim_{t\rightarrow\infty}\tilde{\theta}_0^i (t) = \bf 0 $, $\phi_i(x_i (t), \; y_i (t)) $, $ \phi(t) $有界, 结合触发函数可知存在函数$ \beta(t) \in {\cal{A}}(t) $, 使得:
$$ \begin{aligned} \beta(t)\geq\;&\frac{ck_2}{2\gamma_3}\sum_{i = 1}^Ne_i^{\mathrm{T}}e_i +\frac{k_2}{2\gamma_1}\sum_{i = 1}^N\phi_i^{\mathrm{T}}\tilde{\theta}_i\tilde{\theta}_i^{\mathrm{T}}\phi_i\;+\\ &\frac{k_2}{2\gamma_2}\sum_{i = 1}^N\phi^{\mathrm{T}}\tilde{\theta}_0^{i}(\tilde{\theta}_0^{i})^{\mathrm{T}}\phi \end{aligned} $$ 即
$$ \begin{equation} \dot{V}_3\leq-h V_3+\beta(t) \end{equation} $$ (23) 式中, $ h = \min\{2k_1/k_2,\;2\alpha q_{\min}\} $. 由引理1可知, $ V_3(t) $渐近趋向$ \bf 0 $, 即对任意$ i\in\{1,\;\cdots,\;N\} $, 都有
$$ \lim_{t\rightarrow\infty} \xi_{ix}(t) = \xi_{iy}(t) = \bf 0 $$ 记:
$$ \begin{equation*} \begin{aligned} &\xi_x = [\xi_{1x}^{\mathrm{T}},\;\cdots,\;\xi_{Nx}^{\mathrm{T}}]^{\mathrm{T}},\;\delta_x = [\delta_{1x}^{\mathrm{T}},\;\cdots,\;\delta_{Nx}^{\mathrm{T}}]^{\mathrm{T}}\\ &\xi_y = [\xi_{1y}^{\mathrm{T}},\;\cdots,\;\xi_{Ny}^{\mathrm{T}}]^{\mathrm{T}},\;\delta_y = [\delta_{1y}^{\mathrm{T}},\;\cdots,\;\delta_{Ny}^{\mathrm{T}}]^{\mathrm{T}}\\ &\delta_{ix} = x_i-x_0,\;\delta_{iy} = y_i-y_0,\;i = 1,\;\cdots,\;N \end{aligned} \end{equation*} $$ 由$ \xi_{ix} $ 和$\; \xi_{iy} $的定义易证 $\xi_x = (L\otimes I_n)\delta_x,\;\xi_y = (L\otimes I_n)\delta_y$. 当假设1成立时, 则$ L $非奇异. 由式(23)可得, 对任意$i $有$ \lim_{t\rightarrow\infty} x_i(t) = x_0(t),\; y_i(t) = y_0(t) $.
□ 定理2. 分布式事件触发趋同算法式(8)和式(9)不存在芝诺现象.
证明. 当$ t\in[t_k^i,\;t_{k+1}^i) $时, $ \epsilon_i(t) $的Frobenius范数和$ e_i(t) $的2范数的Dini导数满足如下不等式:
$$ \begin{equation*} {\rm D}^+||\epsilon_i(t)||_{\rm{F}}\leq||\dot{\epsilon}_i(t)||_{\rm{F}},\; {\rm D}^+||e_i(t)||\leq||\dot{e}_i(t)|| \end{equation*} $$ 由式(6)和式(7)可得:
$$ \begin{aligned} \dot{\epsilon}_i(t) =\; &\mu\sum_{j = 0}^Na_{ij}(\hat{\theta}_0^i(t_k^i)-\hat{\theta}_0^j(t_{k'}^j))\\ \dot{e}_i(t) =\; &k_2((\hat{\theta}_0^{i})^{\mathrm{T}}\phi(t)-\theta_i^{\mathrm{T}}\phi_i-u_i)\;+\\ &k_1(\hat{y}_i^i(t)-y_i) \end{aligned} $$ 由假设2、命题1、命题2和定理1可知, 存在有界实数$ \psi_k^i>0,\;\chi_k^i $和$ c>0 $, 使得:
$$ \begin{aligned} &{\mathrm{D}}^+||\epsilon_i(t)||_{\rm{F}}\leq\psi_k^i\\ &{\mathrm{D}}^+||e_i(t)||\leq c||e_i(t)||+\chi_k^i \end{aligned} $$ 在事件触发时刻$ t_k^i $, $ \epsilon_i(t) $和$ e_i(t) $被重置为$ \bf 0 $. 对于$ t\in[t_k^i,\;t_{k+1}^i) $, 由比较原理可得:
$$ \begin{equation} \left\{\begin{aligned} &||\epsilon_i(t)||_{\rm{F}}\leq\psi_k^i(t-t_k^i)\\ &||e_i(t)||\leq\frac{\chi_k^i}{c}(\mathrm{e}^{c(t-t_k^i)}-1) \end{aligned}\right. \end{equation} $$ (24) $ \forall t \in [t_k^i,\;t_{k+1}^i) $, 有$ ||\epsilon_i(t)||_{\rm{F}} < \sqrt{f_{i1}(t)},\;||e_i (t)|| < \sqrt{f_{i2}(t)} $.
当$ {t \rightarrow t_{k+1}^i} $时, 则有$ \lim_{t\rightarrow t_{k+1}^i}||e_i(t)||\geq\sqrt{f_{i2}(t)} $, 或$\lim_{t\rightarrow t_{k+1}^i}||\epsilon_i(t)||_{\rm{F}}\geq\sqrt{f_{i1}(t)}$. 结合式(24), 可得$ t_{k+1}^i - t_k^i \geq \ln\left({c}\sqrt{f_{i2}(t)}/{\chi_k^i} + 1\right)/{c} $ 或 $ t_{k+1}^i - t_k^i \; \geq {\sqrt{f_{i1}(t)}}/ {\psi_k^i} $. 对任意有限时间$ t $, $ f_{i1}(t)>0, \;f_{i2}(t)> 0 $, 即连续2次触发时刻的时间差$ t_{k+1}^i-t_k^i $是严格大于$ 0 $的, 从而证明, 对任意有限时间$ t $, 事件触发趋同算法式(8)和式(9)不存在芝诺现象.
□ 推论1. 事件触发条件(10)所给出的领导智能体的事件触发算法不存在芝诺现象. 证明过程与定理 2 证明类似
4. 数值仿真
本节通过仿真模型验证事件触发控制器式(8)和式(9)的有效性. 考虑包含$ 5 $个智能体的异构不确定二阶非线性多智能体系统, 其中跟随智能体1 ~ 4为无阻尼单摆系统, 其动力学方程为:
$$ \begin{equation} \left\{\begin{aligned} \dot{x}_i& = y_i\\ \dot{y}_i& = -\frac{g}{l_i}\sin(x_i)+u_i \end{aligned}\right. \end{equation} $$ (25) 式中, $ x_i $为单摆的角位移, $ y_i $为角速度, $ g $为重力加速度, $ l_i $为摆长, $ u_i $为控制输入. 由于测量误差原因, 重力加速度$ g $和摆长$\; l_i $的精确值不确定. 领导智能体的动力学方程为:
$$ \begin{equation} \left\{\begin{aligned} \dot{x}_0& = y_0\\ \dot{y}_0& = \theta_0^{\mathrm{T}}\phi(t) \end{aligned}\right. \end{equation} $$ (26) 式中, $ \phi(t) = [\sin(t),\;\cos(2t)]^{\mathrm{T}} $为已知时间向量函数, $ \theta_0\in {\bf{R}}^2 $为未知常向量. 多智能体系统式(25)和式(26)的网络拓扑由如下拉普拉斯矩阵描述:
$$ \begin{equation*} {\cal{L}} =\left[ \begin{array}{*{20}{r}} 0\;\;\;\;\,\,&0\;\;\;\;\,\,&0\;\;\;\;\,\,&0\;\;\;\;\,\,&0\;\;\;\;\,\,\\ 0\;\;\;\;\,\,&0.60&-0.55&0\;\;\;\;\,\,&-0.05\\ -0.50&0\;\;\;\;\,\,&0.55&-0.05&0\;\;\;\;\,\,\\ -0.50&-0.05&0\;\;\;\;\,\,&0.55&0\;\;\;\;\,\,\\ 0\;\;\;\;\,\,&0\;\;\;\;\,\,&0\;\;\;\;\,\,&-0.55&0.55 \end{array}\right] \end{equation*} $$ 根据参数观测器式(4)和式(6), 为每个智能体设计未知向量$ \theta_0 $的观测值$ \hat{\theta}_0^i $; 根据参数观测器式(5), 为跟随智能体设计不确定系数$ -g/l_i $的观测值$ \hat{\theta}_i $, 其中参数$ \mu = 2,\;\rho_i = 1 $; 根据状态估计器式(7), 为每个智能体设计邻居状态估计器. 通过计算, 可求得参数$ q_{\min} = 2.015\ 3,\;\lambda_{1H} = 0.099\ 3,\; \lambda_{N\Xi} = 0.103\ 5,\;\lambda_{N\Pi} = 0.056\ 7 $. 通过选取参数$ \gamma_1 = \gamma_2 = \gamma_3 = 0.5 $, $ k_1 = 0.5,\;k_2 = 2 $, 可求得$ c' = 1.336\ 6 $. 因此, 选取事件触发控制器的参数为$ k_1 = 0.5,\; k_2 = 2, \;c \,= 2 $. 对于触发函数式(9)和式(10), 选取函数$ f_{i1}(t) \,= f_{i2}(t) = 0.1/(1+0.5t) $.
仿真结果如图1 ~ 图3所示. 由图1可知, 跟随智能体的角度和角速度渐近跟踪上领导智能体的状态; 由图2可知, $ \hat{\theta}_0^i $和$ \hat{\theta}_i $分别可以渐近收敛到$ \theta_0 $和$ -g/l_i $; 图3给出了各智能体的事件触发时刻. 表1为在时间段$ [0,\;40] $ s内, 本文算法的事件触发次数. 作为对比, 利用文献[8−10, 14] 所给出的组合测量事件触发算法对系统式(25)和式(26) 进行仿真, 表2为在时间段$ [0,\;40] $ s内, 组合测量事件触发算法的各智能体事件触发次数. 可以看出, 本文基于参数和状态观测器的事件触发控制算法可有效减少事件触发次数.
图 2 $||\tilde{\theta}_0^i||$和$\tilde{\theta}_i$的轨迹Fig. 2 Trajectories of $||\tilde{\theta}_0^i||$ and $\tilde{\theta}_i$表 1 本文算法的事件触发次数Table 1 Event-triggered number of the proposed algorithm智能体 0 1 2 3 4 触发次数 49 84 75 73 72 表 2 组合测量算法的事件触发次数Table 2 Event-triggered number of the combined measurement algorithm智能体 0 1 2 3 4 触发次数 139 258 266 255 249 5. 结束语
本文基于参数估计与事件触发机制, 研究了异构不确定二阶非线性多智能体系统的状态趋同问题, 给出完全不依赖智能体间连续信息传输的事件触发趋同算法. 因为每个智能体均存在不确定参数, 在设计控制器前, 先设计观测器, 估计其不确定参数. 为使跟随智能体跟踪上领导智能体, 设计分布式参数观测器, 使每个跟随智能体可以渐近估计领导智能体不确定参数. 为使算法达到完全不依赖智能体间连续信息传输的目的, 每个智能体利用其邻居智能体发送的事件触发时刻采样信息, 对邻居智能体状态进行重构, 利用重构的状态信息设计控制器和事件触发函数. 进一步证明了所提事件触发趋同算法不存在芝诺现象. 最后, 通过一个多单摆系统验证了所提事件触发趋同算法的有效性, 同时对比组合测量事件触发算法, 本文所提算法可有效减少事件触发次数. 为简化反馈增益参数对拓扑网络全局信息的依赖, 未来可将现有工作推广到完全分布式事件触发状态趋同控制.
附录 A. 矩阵迹的2个引理
引理 2. 对于空间$ {\bf{R}}^{m\times n} $中的矩阵$ X $, 以及空间$ {\bf{R}}^{m\times m} $中的正定矩阵$ A $, 有:
$$ \lambda_{1A}\mathrm{tr}\{X^{\mathrm{T}}X\}\leq\mathrm{tr}\{X^{\mathrm{T}}AX\}\leq\lambda_{mA}\mathrm{tr}\{X^{\mathrm{T}}X\} $$ 证明. 矩阵$ X $可用 $ n $个列向量 $ x_i\in {\bf{R}}^m, \;i = 1,\;\cdots,\; n $表示, 即$ X = [x_1,\;\cdots,\;x_n] $. 因此, 可得:
$$ \begin{equation*} X^{\mathrm{T}}X = \begin{bmatrix} x_1^{\mathrm{T}}x_1 & x_1^{\mathrm{T}}x_2 & \cdots & x_1^{\mathrm{T}}x_n \\ x_2^{\mathrm{T}}x_1 & x_2^{\mathrm{T}}x_2 & \cdots & x_2^{\mathrm{T}}x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_n^{\mathrm{T}}x_1 & x_n^{\mathrm{T}}x_2 & \cdots & x_n^{\mathrm{T}}x_n \\ \end{bmatrix} \end{equation*} $$ 即, $ \mathrm{tr}\{X^{\mathrm{T}}X\} = \sum_{i = 1}^nx_i^{\mathrm{T}}x_i $.
记$ \Lambda = \mathrm{diag}\{\lambda_{1A},\;\cdots,\;\lambda_{mA}\} $. 由于$ A $为正定矩阵, 所以存在单位正交矩阵$ P\in {\bf{R}}^{m\times m} $使$ P^{\mathrm{T}}AP \;= \Lambda $. 矩阵$ P $可用$ m $个列向量$ p_i\in {\bf{R}}^m, \;i = 1,\;\cdots,\;n $表示, 即$ P = [p_1,\;\cdots,\;p_m] $. 对于$ X^{\mathrm{T}}AX $, 有:
$$ X^{\mathrm{T}}AX = X^{\mathrm{T}}PP^{\mathrm{T}}APP^{\mathrm{T}}X = X^{\mathrm{T}}P\Lambda P^{\mathrm{T}}X $$ 通过计算, 可得:
$$ X^{\mathrm{T}}P\Lambda = \begin{bmatrix} \lambda_{1A}x_1^{\mathrm{T}}p_1 & \lambda_{2A}x_1^{\mathrm{T}}p_2 & \cdots & \lambda_{mA}x_1^{\mathrm{T}}p_m \\ \lambda_{1A}x_2^{\mathrm{T}}p_1 & \lambda_{2A}x_2^{\mathrm{T}}p_2 & \cdots & \lambda_{mA}x_2^{\mathrm{T}}p_m \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_{1A}x_n^{\mathrm{T}}p_1 & \lambda_{2A}x_n^{\mathrm{T}}p_2 & \cdots & \lambda_{mA}x_n^{\mathrm{T}}p_m \\ \end{bmatrix} $$ $$ P^{\mathrm{T}}X = \begin{bmatrix} p_1^{\mathrm{T}}x_1 & p_1^{\mathrm{T}}x_2 & \cdots & p_1^{\mathrm{T}}x_n \\ p_2^{\mathrm{T}}x_1 & p_2^{\mathrm{T}}x_2 & \cdots & p_2^{\mathrm{T}}x_n \\ \vdots & \vdots & \ddots & \vdots \\ p_m^{\mathrm{T}}x_1 & p_m^{\mathrm{T}}x_2 & \cdots & p_m^{\mathrm{T}}x_n \\ \end{bmatrix} $$ 通过计算, 可得:
$$ \mathrm{tr}\{X^{\mathrm{T}}AX\} = \sum\limits_{i = 1}^n\sum\limits_{j = 1}^m\lambda_{jA}x_i^{\mathrm{T}}p_jp_j^{\mathrm{T}}x_i $$ 由于向量组 $ \{p_1,\;\cdots,\;p_m\} $ 为空间 $ {\bf{R}}^m $ 中的一组标准正交基, 所以对数量积$ x_i^{\mathrm{T}}p_j $有 $ x_i^{\mathrm{T}}p_j = ||x_i||\cos\theta_{ij} $, 其中$ \theta_{ij} $为向量 $ x_i $与基向量$ p_j $的夹角. 因此有:
$$ \sum\limits_{j = 1}^m\lambda_{jA}x_i^{\mathrm{T}}p_jp_j^{\mathrm{T}}x_i = \sum\limits_{j = 1}^m\lambda_{jA}(\cos^2\theta_{ij})x_i^{\mathrm{T}}x_i $$ 又由于$ \lambda_{1A}\leq\cdots\leq\lambda_{mA} $和$ \sum_{j = 1}^m\cos^2\theta_{ij} = 1 $, 可得$ \lambda_{1A}\mathrm{tr}\{X^{\mathrm{T}}X\}\leq\mathrm{tr}\{X^{\mathrm{T}}AX\}\leq\lambda_{mA}\mathrm{tr}\{X^{\mathrm{T}}X\}. $
□ 引理 3. 对矩阵$ X\in {\bf{R}}^{m\times n} $, $ Y\in {\bf{R}}^{s\times n} $, $ A\in {\bf{R}}^{m\times s} $和正实数$ \eta $, 有:
$$ \mathrm{tr}\{X^{\mathrm{T}}AY\}\leq\frac{\eta}{2}\mathrm{tr}\{X^{\mathrm{T}}AA^{\mathrm{T}}X\}+\frac{1}{2\eta} \mathrm{tr}\{Y^{\mathrm{T}}Y\} $$ 证明. $ X $, $ Y $, $ A $可表示为:
$$ \begin{aligned} X& = [x_1,\;\cdots,\;x_n],\;x_i\in {\bf{R}}^m,\;i\in\{1,\;\cdots,\;n\}\\ Y& = [y_1,\;\cdots,\;y_n],\;y_i\in {\bf{R}}^s,\;i\in\{1,\;\cdots,\;n\}\\ A& = [a_1,\;\cdots,\;a_s],\;a_i\in {\bf{R}}^m,\;i\in\{1,\;\cdots,\;s\}\\ \end{aligned} $$ 记$ y_i = [y_{i1},\;\cdots,\;y_{is}]^{\mathrm{T}} $, 通过计算可得:
$$ \mathrm{tr}\{X^{\mathrm{T}}AY\} = \sum\limits_{i = 1}^n\sum\limits_{j = 1}^sx_ia_jy_{ij} $$ 根据Young不等式, 可知$ x_ia_jy_{ij}\leq {\eta}(x_ia_j)^2/{2}+ y_{ij}^2/ {2\eta} $, 可得:
$$ \mathrm{tr}\{X^{\mathrm{T}}AY\}\leq\frac{\eta}{2}\sum\limits_{i = 1}^n\sum\limits_{j = 1}^s(x_ia_j)^2+\frac{1}{2\eta} \sum\limits_{i = 1}^n\sum\limits_{j = 1}^sy_{ij}^2 $$ 容易验证$ \sum_{i = 1}^n\sum_{j = 1}^s(x_ia_j)^2 \,=\, \mathrm{tr}\{X^{\mathrm{T}}AA^{\mathrm{T}}X\} ,$ $ \sum_{i = 1}^n \sum_{j = 1}^sy_{ij}^2 = \mathrm{tr}\{Y^{\mathrm{T}}Y\} $.
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图 2 $||\tilde{\theta}_0^i||$和$\tilde{\theta}_i$的轨迹
Fig. 2 Trajectories of $||\tilde{\theta}_0^i||$ and $\tilde{\theta}_i$
表 1 本文算法的事件触发次数
Table 1 Event-triggered number of the proposed algorithm
智能体 0 1 2 3 4 触发次数 49 84 75 73 72 
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表 2 组合测量算法的事件触发次数
Table 2 Event-triggered number of the combined measurement algorithm
智能体 0 1 2 3 4 触发次数 139 258 266 255 249
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