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摘要: 针对一类不确定非线性系统, 结合Backstepping方法, 设计重复学习控制方法.采用Lyapunov-like综合, 设计重复学习控制器处理系统中的参数和非参数不确定性, 可实现系统状态在整个作业区间上完全跟踪期望轨迹.分别讨论部分限幅和完全限幅学习机制, 证明闭环系统中各变量的一致有界性以及跟踪误差的一致收敛性.仿真结果验证了所提出控制方法的有效性.Abstract: This paper presents a repetitive learning control approach for a class of uncertain nonlinear systems by means of the backstepping method. Based on Lyapunov-like synthesis, a learning controller is designed for handling parametric and nonparametric uncertainties, such that system state can completely track the reference trajectory over the whole time interval after many iterations. Both the partially-saturated and fully-saturated learning mechanisms are discussed respectively. The uniform boundedness of all variables in the closed-loop system and the uniform convergence of the tracking error are guaranteed. Simulation results demonstrate the effectiveness of the method.
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Key words:
- Repetitive learning control /
- parametrization /
- nonparametrization /
- uncertain systems
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实际场合存在很多在有限时间内重复运作的系统, 学习控制非常适用于这类重复运作的运动对象.它能利用“学习”过程重复的特性, 提高误差的收敛性能, 达到完全跟踪[1].学习控制包括迭代学习控制和重复控制.
在早期的迭代学习控制领域, 主要基于压缩映射的方法, 但这种方法很难将系统的结构、参数以及有效的信息包含进来.这种方法只能处理非常有限的非线性不确定系统.它对系统的要求非常高:初始误差为零、系统满足全局Lipschitz条件、系统的相对阶为零.
为了克服上述方法的局限性, 文献[2]提出了基于Lyapunov-like方法的迭代学习控制技术.这种方法能非常有效地处理非线性不确定系统[3-5].文献[6]针对一类时变参数不确定系统, 在非一致轨迹下, 应用Lyapunov-like方法设计迭代学习控制方法.文献[7]针对带有时变参数不确定性的非线性系统, 提出了一种新的离散时间自适应迭代学习控制方法.文献[8]分别针对定常参数、时变参数以及复合参数的不确定系统, 设计能够实现期望误差轨迹完全跟踪的学习算法.文献[9]针对一类含有时变和时不变参数的高阶非线性系统, 设计自适应迭代学习控制算法, 解决非一致目标跟踪问题.分别针对定常参数、时变参数以及复合参数的不确定系统, 设计能够实现期望误差轨迹完全跟踪的学习算法.
运用迭代学习控制技术, 每次迭代都需要精确的初始定位, 但在实际中很难实施.微小的初始定位误差会导致跟踪误差不能收敛.但是, 重复控制并不要求系统满足初始定位误差为零的条件, 但它要求被学习量满足周期性.文献[10]研究了周期学习方法非常适用于对周期未知量的估计.结合迭代学习控制和重复控制的优点, 文献[11]给出了重复学习控制的提法.重复学习控制像重复控制一样不需要进行初始定位, 只要求每次迭代开始时系统的初始条件是对准的, 即与上次运行终止时的状态相重合; 重复学习控制并不要求被学习量满足周期性, 只要求满足重复性即可.
基于Lyapunov-like综合的迭代学习控制方面已发表的文献大多讨论系统可参数化[12-13], 但是大部分系统都很复杂, 不能完全进行参数化, 因此提出非参数化不确定系统的控制技术非常重要[14].文献[15]针对一类非线性系统, 设计迭代学习控制, 处理系统中的非参数不确定性.文献[16]采用鲁棒自适应迭代学习控制, 讨论含有参数和非参数不确定性的离散非线性系统.文献[17]针对具有参数和非参数不确定系统, 借助BLF (Barrier Lyapunov function)函数, 设计了输出限制系统的迭代学习控制算法.文献[18]讨论了任意初态下的一类含有状态依赖控制增益的非参数不确定系统在预定区间上的完全跟踪控制问题.文献[19]针对非参数不确定系统, 利用时变神经网络, 设计状态受限的自适应迭代学习控制方法.文献[20]研究非参数不确定学习控制系统的初值问题和状态受限问题.因此, 如何通过重复学习控制解决更一般的不确定非线性系统, 特别地, 含有参数和非参数不确定性, 目前相关的研究结果较少.
本文讨论了一类更一般的高阶不确定非线性系统的重复学习控制问题.结合Backstepping方法, 构造Lyapunov-like函数, 设计重复学习控制方法, 处理系统中的参数和非参数不确定性, 可实现系统状态在整个作业区间上完全跟踪期望轨迹.分别讨论了部分限幅和完全限幅学习机制, 证明了限幅学习机制下的系统跟踪误差的一致收敛性.本文研究一类更一般的不确定非线性系统, 同时含有参数和非参数不确定性, 其中控制增益未知并与状态相关.所提出的系统是文献[15-18]所讨论的系统的一种推广, 文献[15]和文献[18]中, 系统只含有非参数不确定性, 文献[16]中, 系统不包含控制增益, 文献[17]中, 系统只含一项不确定项.因此, 本文所给出的控制方法, 适用于更广泛的动态系统.
1. 问题的提出与准备
考虑下述不确定非线性系统
$$ \begin{align} & \dot{x}_{j}\; (t) = x_{j+1}\; (t), j = 1, 2, \cdots, n-1\\& \dot{x}_{n}\; (t) = f_{1}\; (\pmb{x}, t)+g(\pmb{x}, t)(f_{2}\; (\pmb{x}, t)+u(t)) \end{align} $$ (1) 其中, $ \pmb{x} = [x_{1}, x_{2}, \cdots, x_{n}]^{\rm{T}}\in {\bf R}^{n} $系统的状态向量. $ f_{1}\; (\pmb{x}, t)\in{\bf R} $和$ f_{2}\; (\pmb{x}, t)\in{\bf R} $代表系统的不确定性部分, $ g(\pmb{x}, t)\in{\bf R} $是与系统状态和时间有关的未知控制增益, $ u(t)\in{\bf R} $是控制输入.
对系统(1)做如下假设:
假设1. 不确定$ f_{1}\; (\pmb{x}, t) $和$ f_{2}\; (\pmb{x}, t) $可表示为
$$ \begin{align} &f_{1}\; (\pmb{x}, t) = \pmb{\theta}^{\rm{T}}\; (t)\pmb{\varphi}\; (\pmb{x}, t)+\eta_{1}\; (\pmb{x}, t)\\& f_{2}\; (\pmb{x}, t) = \pmb{p}^{\rm{T}}\; (t)\pmb{\psi}\; (\pmb{x}, t)+\eta_{2}\; (\pmb{x}, t) \end{align} $$ (2) 其中, $ \pmb{\theta}\; (t) $和$ \pmb{p}\; (t) $是未知的时变向量, $ \pmb{\varphi}\; (\pmb{x}, t) $和$ \pmb{\psi}\; (\pmb{x}, t) $是与$ \pmb{x} $和$ t $有关的已知向量, 未知非参数化函数$ \eta_{1}\; (\pmb{x}, t) $和$ \eta_{2}\; (\pmb{x}, t) $分别满足
$$ \begin{align} &|\eta_{1}\; (\pmb{x}_{1}, t)-\eta_{1}\; (\pmb{x}_{2}, t)|\leq\beta_{1}\; (\pmb{x}_{1}, \pmb{x}_{2}, t)\|\pmb{x}_{1}-\pmb{x}_{2}\|, \\& \qquad \qquad\forall \pmb{x}_{1}\in {\bf R}^{n}, \forall \pmb{x}_{2}\in {\bf R}^{n}\\& |\eta_{2}\; (\pmb{x}_{1}, t)-\eta_{2}\; (\pmb{x}_{2}, t)|\leq\beta_{2}\; (\pmb{x}_{1}, \pmb{x}_{2}, t)\|\pmb{x}_{1}-\pmb{x}_{2}\|, \\& \qquad \qquad\forall \pmb{x}_{1}\in {\bf R}^{n}, \forall \pmb{x}_{2}\in {\bf R}^{n} \end{align} $$ (3) 其中, $ \beta_{1}\; (\pmb{x}_{1}, \pmb{x}_{2}, t) $和$ \beta_{2}\; (\pmb{x}_{1}, \pmb{x}_{2}, t) $均为已知的连续函数.
假设2. 存在已知的常数$ g_{0} $和已知的连续函数$ g_{m}\; (\pmb{x}, t) $, 满足
$$ \begin{align} & 0<g_{0} \leq g(\pmb{x}, t)\leq g_{m}\; (\pmb{x}, t) \end{align} $$ (4) $ g(\pmb{x}, t) $满足
$$ \begin{align} |g(\pmb{x}_{1}, t)-g(\pmb{x}_{2}, t)|\leq\beta_{3}\; (\pmb{x}_{1}, \pmb{x}_{2}, t)\|\pmb{x}_{1}-\pmb{x}_{2}\| \end{align} $$ (5) 其中, $ \beta_{3}\; (\pmb{x}_{1}, \pmb{x}_{2}, t) $为已知的连续函数.
考虑系统在区间$ [0, T] $上重复作业, $ 0 < T < \infty $, 记$ k $为重复作业次数, $ \pmb{x}_{k}\; (t) $为第$ k $次重复时的系统状态. $ \pmb{r}\; (t) = [r_{1}, r_{2}, \cdots, r_{n}]^{\rm{T}} $为$ [0, T] $上给定的期望轨迹.
本文讨论重复学习控制器的设计, 为此做如下假设:
假设3. 给定的期望轨迹是闭合的, 即满足
$$ \begin{align} \pmb{r}\; (0) = \pmb{r}\; (T) \end{align} $$ (6) 假设4. 实际系统每次运行的终值作为下次运行的初值, 即满足
$$ \begin{align} \pmb{x}_{k}\; (0) = \pmb{x}_{k-1}\; (T) \end{align} $$ (7) 本文的控制目标是对系统(1)设计重复学习控制器$ u_{k} $, 实现系统(1)能跟踪给定的期望轨迹$ \pmb{r}\; (t) $, 期望轨迹$ \pmb{r}\; (t) $是由下式生成
$$ \begin{align} & \dot{r}_{j}\; (t) = r_{j+1}\; (t), \; \; \; \; \; \; \; j = 1, 2, \cdots, n-1\\& \dot{r}_{n} = s(\pmb{r}, t) \end{align} $$ (8) 其中, $ s(\pmb{r}, t) $是关于变量$ t $, $ \pmb{r} $的已知光滑函数.引入未知期望控制输入$ u_{r}\; (t) $,
$$ \begin{align} & u_{r}\; (t) = \frac{1}{g(\pmb{r}, t)}\dot{r}_{n}-\frac{f_{1}\; (\pmb{r}, t)}{g(\pmb{r}, t)}-f_{2}\; (\pmb{r}, t) \end{align} $$ (9) 使得下式成立
$$ \begin{align} \dot{r}_{n} = f_{1}\; (\pmb{r}, t)+g(\pmb{r}, t)(f_{2}\; (\pmb{r}, t)+u_{r}\; (t)) \end{align} $$ (10) 为了下面的分析, 我们假设$ u_{r}(t) $, $ \pmb{r} $, $ f_{1}(\pmb{r}, t) $, $ f_{2}(\pmb{r}, t) $和$ g(\pmb{r}, t) $都是有界的.
为表述简便, 记$ g_{k} = g(\pmb{x}_{k}, t) $, $ f_{1, k} = f_{1}\; (\pmb{x}_{k}, t) $, $ f_{2, k} = f_{2}\; (\pmb{x}_{k}, t) $, $ g_{m, r} = g_{m}\; (\pmb{r}, t) $, $ g_{r} = g(\pmb{r}, t) $, $ \eta_{1, r} = \eta_{1}\; (\pmb{r}, t) $, $ \eta_{2, r} = \eta_{2}\; (\pmb{r}, t) $, $ f_{1, r} = f_{1}\; (\pmb{r}, t) $, $ f_{2, r} = f_{2}\; (\pmb{r}, t) $, $ \eta_{1, k} = \eta_{1}\; (\pmb{x}_{k}, t) $, $ \pmb{\varphi}_{k} = \pmb{\varphi}\; (\pmb{x}_{k}, t) $, $ \pmb{\psi}_{k} = \pmb{\psi}\; (\pmb{x}_{k}, t) $, $ \eta_{2, k} = \eta_{2}\; (\pmb{x}_{k}, t) $, $ \pmb{\varphi}_{r} = \pmb{\varphi}\; (\pmb{r}, t) $, $ \pmb{\psi}_{r} = \pmb{\psi}\; (\pmb{r}, t) $, $ \beta_{1, k} = \beta_{1}\; (\pmb{x}_{k}, \pmb{r}, t) $, $ \beta_{2, k} = \beta_{2}\; (\pmb{x}_{k}, \pmb{r}, t) $, $ \beta_{3, k} = \beta_{3}\; (\pmb{x}_{k}, \pmb{r}, t) $.
2. 控制器设计
定义跟踪误差为$ \pmb{e}_{k} = \pmb{x}_{k}-\pmb{r} $.其中, $ \pmb{e}_{k} = [e_{1, k}, e_{2, k}, \cdots, e_{n, k}]^{\rm{T}} $.下面利用反步设计法来设计系统的控制律.对$ e_{j, k} $做变换, 定义另一变量$ z_{j, k} $,
$$ \begin{align} & z_{1, k} = e_{1, k} \end{align} $$ (11) $$ \begin{align} z_{j, k} = e_{j, k}-\alpha_{j-1, k}\; (z_{1, k}, &z_{2, k}, \cdots, z_{j-1, k}), \\ & j = 2, 3, \cdots, n \end{align} $$ (12) 步骤1.由式(11)可得
$$ \begin{align} & \dot{z}_{1, k} = \dot{e}_{1, k} = z_{2, k}+\alpha_{1, k} \end{align} $$ (13) 设计$ \alpha_{1, k} $为
$$ \begin{align} & \alpha_{1, k}\; (z_{1, k}) = -cz_{1, k} \end{align} $$ (14) 其中, $ c = c_{1}+c_{2} $, $ c_{1}, c_{2} $为正的待设定参数.式(13)可以写成
$$ \begin{align} & \dot{z}_{1, k} = z_{2, k}-cz_{1, k} \end{align} $$ (15) 由式(15), 可得
$$ \begin{align} &z_{1, k}\dot{z}_{1, k} = -cz_{1, k}^{2}+z_{1, k}z_{2, k}\\& \dot{\alpha}_{1, k} = c^{2}z_{1, k}-cz_{2, k} \end{align} $$ (16) 步骤2.由式(12)可得
$$ \begin{align} \dot{z}_{2, k} = \, &\dot{e}_{2, k}-\frac{\partial\alpha_{1, k}}{\partial z_{1, k}}\dot{z}_{1, k} = \\& z_{3, k}+\alpha_{2, k}-\frac{\partial\alpha_{1, k}}{\partial z_{1, k}}\dot{z}_{1, k} \end{align} $$ (17) 设计$ \alpha_{2, k} $为
$$ \begin{align} \alpha_{2, k}\; (z_{1, k}, z_{2, k}) = \, &-cz_{2, k}-z_{1, k}+\frac{\partial\alpha_{1, k}}{\partial z_{1, k}}\dot{z}_{1, k} = \\& -2cz_{2, k}+(c^{2}-1)z_{1, k} \end{align} $$ (18) 式(17)可以写成
$$ \begin{align} & \dot{z}_{2, k} = z_{3, k}-cz_{2, k}-z_{1, k} \end{align} $$ (19) 由式(19), 可得
$$ \begin{align} & z_{2, k}\dot{z}_{2, k} = -cz_{2, k}^{2}-z_{1, k}z_{2, k}+z_{2, k}z_{3, k}\; \; \; \; \; \; \\& \dot{\alpha}_{2, k} = (-c^{3}+3c)z_{1, k}+(3c^{2}-1)z_{2, k}-2cz_{3, k} \end{align} $$ (20) 步骤 $ {\pmb j} \; \; {(\bf3}\leq {\pmb j}\leq {\pmb n}-{\bf1)} $.设计
$$ \begin{align} & \alpha_{j, k}\; (z_{1, k}, \cdots, z_{j, k}) = -cz_{j, k}-z_{j-1, k}+\\& \qquad \sum^{j-1}_{i = 1}\left(\frac{\partial \alpha_{j-1, k}}{\partial z_{i, k}}\dot{z}_{i, k}\right) \end{align} $$ (21) 由式(12)和式(21), 可得
$$ \begin{align} & \dot{z}_{j, k} = -z_{j-1, k}-cz_{j, k}+z_{j+1, k} \end{align} $$ (22) 由式(21)和(22), 可逐步得到$ \dot{\alpha}_{j, k} $, 并且可知$ \dot{\alpha}_{j, k} $是和$ z_{1, k}, z_{2, k}, \cdots, z_{j+1, k} $有关的一次多项式.
步骤 $ \pmb n $.由式(1), (10)和(12), 我们得到
$$ \begin{align} & \dot{z}_{n, k} = \dot{x}_{n, k}-\dot{r}_{n}-\dot{\alpha}_{n-1, k} = f_{1, k}-f_{1, r}+\; \; \; \; \\& \qquad g_{k}f_{2, k}-g_{r}f_{2, r} +g_{k}u_{k}-g_{r}u_{r}-\dot{\alpha}_{n-1, k} \end{align} $$ (23) 由式(11)和(12)定义的变换可表示成
$$ \begin{align} & \pmb{e}_{k} = Z_{c}\pmb{z}_{k} \end{align} $$ (24) 其中, $ \pmb{z}_{k} = [z_{1, k}, z_{2, k}, \cdots, z_{n, k}]^{\rm{T}} $.记矩阵$ Z_{c} $为$ Z_{c} = (a_{pq})_{nn} $, 它的每个元素$ a_{pq} $满足
$$ \begin{align} & a_{pq} = \left\{ \begin{array}{ll} f_{pq}(c), &p>q \\ 1, &p = q \\ 0, &p<q \end{array} \right. \end{align} $$ (25) 其中, $ f_{pq}\; (c) $可表示为$ c $的多项式, 由式(25)可知$ Z_{c} $是主对角线元素为$ 1 $的下三角矩阵.则$ Z_{c} $是可逆矩阵, 利用式(24)可得
$$ \begin{align} \pmb{e}_{k}^{\rm{T}}\pmb{e}_{k} = \pmb{z}_{k}^{\rm{T}}Z_{c}^{\rm{T}}Z_{c}\pmb{z}_{k} \end{align} $$ (26) 其中, $ Z_{c}^{\rm{T}}Z_{c} $是一正定矩阵, 其表达式可写成
$$ \begin{align} Z_{c}^{\rm{T}}Z_{c} = \left[ \begin{array}{ccccc} b_{11} & \ast & \cdots & \ast\\ \ast & b_{22} & \cdots & \ast \\ \vdots & \vdots & \ddots & \vdots \\ \ast & \ast& \cdots& 1 \\ \end{array} \right] \end{align} $$ (27) 其中, $ b_{ii} = 1+\sum\limits^{n}_{p = i+1}a_{pi}^{2}, i = 1, 2, \cdots, n-1 $.
假设$ \lambda_{1}, \lambda_{2}, \cdots, \lambda_{n} $是矩阵$ Z_{c}^{\rm{T}}Z_{c} $的$ n $个特征值, 不妨设
$$ \begin{align} & 0<\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{n} \end{align} $$ (28) 根据矩阵的迹与特征值之间的关系, 可得下面的等式
$$ \begin{align} &\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n} = {\rm tr}\; (Z_{c}^{\rm{T}}Z_{c}) = \; \; \; \; \; \; \; \\& \qquad n+\sum^{n-1}_{q = 1}\sum^{n}_{p = q+1}a_{pq}^{2} \end{align} $$ (29) 可得
$$ \begin{align} \lambda_{n}<n+\sum^{n-1}_{q = 1}\sum^{n}_{p = q+1}a_{pq}^{2} \end{align} $$ (30) 记$ \kappa = n+\sum\nolimits^{n-1}_{q = 1}\sum\nolimits^{n}_{p = q+1}a_{pq}^{2} $.
为设计控制器, 选取正定函数
$$ \begin{align} & V_{k}\; (t) = \sum^{n}_{j = 1}\frac{1}{2}z^{2}_{j, k} \end{align} $$ (31) 对$ V_{k} $求导,
$$ \begin{align} \dot{V}_{k} = \, &-\sum^{n-1}_{j = 1}cz^{2}_{j, k}+z_{n, k}z_{n-1, k}+z_{n, k}\; (f_{1, k}-\; \; \; \; \\& f_{1, r}+ g_{k}f_{2, k}-g_{r}f_{2, r}+g_{k}u_{k}-g_{r}u_{r}-\dot{\alpha}_{n-1, k}) \end{align} $$ (32) 由假设1, 可得
$$ \begin{align} &z_{n, k}\; (f_{1, k}-f_{1, r}) = \; \\& \qquad z_{n, k}\; (\pmb{\theta}^{\rm{T}}\pmb{\varphi}_{k}-\pmb{\theta}^{\rm{T}}\pmb{\varphi}_{r}+\eta_{1, k}-\eta_{1, r})\leq \; \\& \qquad z_{n, k}\pmb{\theta}^{\rm{T}}\; (\pmb{\varphi}_{k}-\pmb{\varphi}_{r})+|z_{n, k}|\beta_{1, k}\|\pmb{e}_{k}\|\leq \\& \qquad z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}+\frac{1}{2}l_{1}z_{n, k}^{2}\beta_{1, k}^{2}+\frac{1}{2l_{1}}\|\pmb{e}_{k}\|^{2} \end{align} $$ (33) 其中, $ l_{1} $是正的待设定参数, $ \delta\pmb{\varphi}_{k} = \pmb{\varphi}_{k}-\pmb{\varphi}_{r} $.由假设2, 可得
$$ \begin{align} & z_{n, k}\; (g_{k}u_{k}-g_{r}u_{r}) = \; \; \; \; \\& \qquad z_{n, k}\; (g_{k}u_{k}-g_{k}\hat{u}_{k}+g_{k}\hat{u}_{k}-g_{r}\hat{u}_{k}+\; \; \; \; \; \; \; \; \\& \qquad g_{r}\hat{u}_{k}-g_{r}u_{r}) = \; \; \; \\& \qquad z_{n, k}\; (g_{k}u_{k}-g_{k}\hat{u}_{k})+z_{n, k}\; (g_{k}-g_{r})\hat{u}_{k}+\; \\& \qquad z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})\leq\; \; \; \; \; \; \; \; \\& \qquad z_{n, k}g_{k}\; (u_{k}-\hat{u}_{k})+z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+\; \; \; \; \; \\& \qquad |z_{n, k}|\beta_{3, k}\|\pmb{e}_{k}\||\hat{u}_{k}|\leq\; \; \; \; \; \; \\& \qquad z_{n, k}g_{k}\; (u_{k}-\hat{u}_{k})+z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+\; \; \; \; \; \\& \qquad \frac{1}{2}l_{2}z_{n, k}^{2}\beta_{3, k}^{2}\hat{u}_{k}^{2}+\frac{1}{2l_{2}}\|\pmb{e}_{k}\|^{2}\; \; \; \; \; \; \; \end{align} $$ (34) 其中, $ l_{2} $是正的待设定参数, $ \hat{u}_{k} $为$ u_{r} $的估计.由假设1, 可得
$$ \begin{align} & z_{n, k}\; (g_{k}f_{2, k}-g_{r}f_{2, r}) = \; \; \; \; \; \; \; \; \\& \qquad z_{n, k}\; (g_{k}\pmb{p}^{\rm{T}}\pmb{\psi}_{k}-g_{r}\pmb{p}^{\rm{T}}\pmb{\psi}_{r}+g_{k}\eta_{2, k}-g_{r}\eta_{2, r}) = \\& \qquad z_{n, k}\; (g_{k}-g_{r})\pmb{p}^{\rm{T}}\pmb{\psi}_{k}+z_{n, k}g_{r}\pmb{p}^{\rm{T}}\; (\pmb{\psi}_{k}-\pmb{\psi}_{r})+\; \; \\& \qquad z_{n, k}g_{k}\; (\eta_{2, k}-\eta_{2, r})+z_{n, k}\; (g_{k}-g_{r})\eta_{2, r}\; \; \end{align} $$ (35) 根据假设2, 可得
$$ \begin{align} & z_{n, k}\; (g_{k}-g_{r})\pmb{p}^{\rm{T}}\pmb{\psi}_{k}\leq|z_{n, k}|\beta_{3, k}\|\pmb{e}_{k}\||\pmb{p}^{\rm{T}}\pmb{\psi}_{k}|\leq\\& \qquad |z_{n, k}|\beta_{3, k}\|\pmb{e}_{k}\|\|\pmb{p}\|\|\pmb{\psi}_{k}\|\leq\; \; \; \; \\& \qquad \frac{1}{2}l_{3}z_{n, k}^{2}\beta_{3, k}^{2}\|\pmb{\psi}_{k}\|^{2}+\frac{1}{2l_{3}}\|\pmb{e}_{k}\|^{2}\|\pmb{p}\|^{2} \end{align} $$ (36) 其中, $ l_{3} $为正的待设定参数.类似于式(13)的推导, 可得
$$ \begin{align} & z_{n, k}g_{k}\; (\eta_{2, k}-\eta_{2, r}) \leq|z_{n, k}|g_{k}\beta_{2, k}\|\pmb{e}_{k}\| \end{align} $$ (37) 和
$$ \begin{align} & z_{n, k}\; (g_{k}-g_{r})\eta_{2, r} \leq |z_{n, k}|\beta_{3, k}\|\pmb{e}_{k}\||\eta_{2, r}|\leq \\& \qquad \frac{1}{2}l_{4}z_{n, k}^{2}\beta_{3, k}^{2}+\frac{1}{2l_{4}}\|\pmb{e}_{k}\|^{2}\eta_{2, r}^{2} \end{align} $$ (38) 其中, $ l_{4} $为正的待设定参数.将式(36)~(38)代入式(35),
$$ \begin{align} & z_{n, k}\; (g_{k}f_{2, k}-g_{r}f_{2, r})\leq \; \; \\& \qquad z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k} +\frac{1}{2}l_{3}z_{n, k}^{2}\beta_{3, k}^{2}\|\pmb{\psi}_{k}\|^{2}+\\& \qquad \frac{1}{2l_{3}}\|\pmb{e}_{k}\|^{2}\|\pmb{p}\|^{2} +|z_{n, k}|g_{k}\beta_{2, k}\|\pmb{e}_{k}\|+\\& \qquad \frac{1}{2}l_{4}z_{n, k}^{2}\beta_{3, k}^{2}+\frac{1}{2l_{4}}\|\pmb{e}_{k}\|^{2}\eta_{2, r}^{2}\; \; \; \; \; \end{align} $$ (39) 其中, $ \delta\pmb{\psi}_{k} = \pmb{\psi}_{k}-\pmb{\psi}_{r} $.将式(33), (34)和(39)代入式(32)
$$ \begin{align} & \dot{V}_{k}\leq-\sum^{n-1}_{j = 1}cz^{2}_{j, k}+\mu_{1}\|\pmb{e}_{k}\|^{2}+z_{n, k}z_{n-1, k}-\\& \qquad z_{n, k}\dot{\alpha}_{n-1, k}+\mu_{2}z_{n, k}^{2}+z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}+\; \; \; \\& \qquad z_{n, k}g_{k}\; (u_{k}-\hat{u}_{k})+z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+\\& \qquad z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k} +|z_{n, k}|g_{k}\beta_{2, k}\|\pmb{e}_{k}\|\; \end{align} $$ (40) 其中, $ \mu_{1} = \frac{1}{2l_{1}}+\frac{1}{2l_{2}}+\frac{1}{2l_{3}}\|\pmb{p}\|^{2}+\frac{1}{2l_{4}}\eta_{2, r}^{2} $和$ \mu_{2} = \frac{1}{2}l_{1}\beta_{1, k}^{2}+\frac{1}{2}l_{2}\beta_{3, k}^{2}\hat{u}_{k}^{2} +\frac{1}{2}l_{3}\beta_{3, k}^{2}\|\pmb{\psi}_{k}\|^{2}+\frac{1}{2}l_{4}\beta_{3, k}^{2} $.
通过上面的推导, 我们采用如下形式的控制器
$$ \begin{align} & u_{k} = \hat{u}_{k} -\frac{z_{n-1, k}{\rm sgn}\; (z_{n-1, k}z_{n, k})}{g_{0}} -\frac{\mu_{2}z_{n, k}}{g_{0}}-\\& \qquad \frac{|\dot{\alpha}_{n-1, k}|{\rm sgn}\; (z_{n, k})}{g_{0}} -\frac{|\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}|{\rm sgn}\; (z_{n, k})}{g_{0}}-\\& \qquad \frac{cz_{n, k}}{g_{0}}-\frac{g_{m, r}}{g_{0}}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}|{\rm sgn}\; (z_{n, k})-\; \; \; \; \\& \qquad \beta_{2, k}\|\pmb{e}_{k}\|{\rm sgn}\; (z_{n, k})\; \end{align} $$ (41) 其中, $ \hat{\pmb{\theta}}_{k} $和$ \hat{\pmb{p}}_{k} $分别为$ \pmb{\theta} $和$ \pmb{p} $的估计.应用控制律(41), 可将式(40)写成
$$ \begin{align} \dot{V}_{k}\leq\, &-\sum^{n}_{j = 1}cz^{2}_{j, k}+\mu_{1}\|\pmb{e}_{k}\|^{2} +z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}+\\& z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k}-\\& |\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}||z_{n, k}| -g_{m, r}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}||z_{n, k}| \end{align} $$ (42) 由于$ c = c_{1}+c_{2} $, 式(42)可以写成
$$ \begin{align} \dot{V}_{k}\leq\, &-\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}-c_{2}\|\pmb{z}_{k}\|^{2}+\mu_{1}\|\pmb{e}_{k}\|^{2}+\\& \qquad z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k} +z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+\\& \qquad z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k}-|\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}||z_{n, k}|-\\& \qquad g_{m, r}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}||z_{n, k}| \end{align} $$ (43) 利用式(28), 可知
$$ \begin{align} &-c_{2}\|\pmb{z}_{k}\|^{2}+\mu_{1}\|\pmb{e}_{k}\|^{2}\leq\; \; \; \; \; \\& \qquad -c_{2}\|\pmb{z}_{k}\|^{2}+\mu_{1}\lambda_{n}\|\pmb{z}_{k}\|^{2} \leq\\& \qquad (-c_{2}+\mu_{1}\kappa)\|\pmb{z}_{k}\|^{2}\; \; \; \; \; \end{align} $$ (44) 设计如下正参数$ c_{2}, l_{1}, l_{2}, l_{3}, l_{4} $, 使得下式成立
$$ \begin{align} -c_{2}+\mu_{1}\kappa\leq0 \end{align} $$ (45) 式(43)可以写成
$$ \begin{align} \dot{V}_{k}\leq\, &-\sum^{n}_{j = 1}c_{1}z^{2}_{j, k} +z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}+\\& z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r})+z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k}-\\& |\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}||z_{n, k}| -g_{m, r}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}||z_{n, k}| \end{align} $$ (46) 3. 性能分析
分别针对部分限幅学习和完全限幅学习, 分析闭环系统的稳定性和收敛性.
3.1 部分限幅学习
在部分限幅情形下, 各参数的估计分别由如下更新律给出
$$ \begin{align} & \hat{\pmb{\theta}}_{k} = {\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})+\gamma_{1}z_{n, k}\delta\pmb{\varphi}_{k} \\& \hat{\pmb{p}}_{k} = {\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k-1})+\gamma_{2}z_{n, k}\delta\pmb{\psi}_{k}\\& \hat{u}_{k} = {\rm sat}_{\bar{u}}\; (\hat{u}_{k-1})-\gamma_{3} z_{n, k} \end{align} $$ (47) 其中, $ \gamma_{1} > 0 $, $ \gamma_{2} > 0 $和$ \gamma_{3} > 0 $为参数更新增益, $ {\rm sat} $为饱和函数, 本文$ {\rm sat}\; (\cdot) $定义为:对于$ a\in {\bf R} $,
$$ \begin{align*} {\rm sat}_{\bar{a}}\; (a) = \left\{ \begin{array}{ll} \bar{a}{\rm sgn}\; (a), &|a|>\bar{a} \\ a, & |a|\leq\bar{a} \\ \end{array} \right. \end{align*} $$ $ \bar{a} $为对应的限幅值.饱和函数仅作用于式(47)右端的部分项, 因此称为部分限幅. $ \bar{\pmb{\theta}} $, $ \bar{\pmb{p}} $和$ \bar{u} $分别为各参数估计的限幅值.这里, 假定各参数真值都处于限幅值之间, 且, $ \hat{\pmb{\theta}}_{0} = \pmb{0} $, $ \hat{\pmb{p}}_{0} = \pmb{0} $, $ \hat{u}_{0} = 0 $.
考虑下述Lyapunov-like函数
$$ \begin{align} & L_{k}\; (t) = V_{k}\; (t)+\frac{1}{2\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}{\rm d}\tau +\; \; \; \; \; \; \; \\& \qquad \frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k}^{\rm{T}}\tilde{\pmb{p}}_{k}{\rm d}\tau+ \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\; (\hat{u}_{k}-u_{r})^{2}{\rm d}\tau \end{align} $$ (48) 其中, $ \tilde{\pmb{\theta}}_{k} = \pmb{\theta}-\hat{\pmb{\theta}}_{k} $, $ \tilde{\pmb{p}}_{k} = \pmb{p}-\hat{\pmb{p}}_{k} $.可将$ V_{k} $表达为
$$ \begin{align} & V_{k} = \int^{t}_{0}\dot{V}_{k}{\rm d}\tau+V_{k}\; (0) \end{align} $$ (49) 由假设3和4, 可得$ \pmb{e}_{k}\; (0) = \pmb{e}_{k-1}\; (T) $.根据式(24), 可得
$$ \begin{align} \pmb{z}_{k}\; (0) = \pmb{z}_{k-1}\; (T) \end{align} $$ (50) 由式(31), 可知
$$ \begin{align} & V_{k}\; (0) = V_{k-1}\; (T) \end{align} $$ (51) $ L_{k}\; (t) $在第$ k $次的迭代差分为
$$ \begin{align} & \Delta L_{k}\; (t) = \int^{t}_{0}\dot{V}_{k}{\rm d}\tau+V_{k}\; (0)-V_{k-1}\; (t)+\; \; \\& \qquad \frac{1}{2\gamma_{1}}\int^{t}_{0}\; (\tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}-\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}){\rm d}\tau +\; \; \; \; \; \; \; \\& \qquad \frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\; (\tilde{\pmb{p}}_{k}^{\rm{T}}\tilde{\pmb{p}}_{k}-\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}){\rm d}\tau+\; \; \; \\& \qquad \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\{(\hat{u}_{k}-u_{r})^{2}-(\hat{u}_{k-1}-u_{r})^{2}\}{\rm d}\tau \end{align} $$ (52) 考察式(52)中右边第4项
$$ \begin{align} & \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}-\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1} = \; \; \; \; \; \; \; \; \\& \qquad \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k} -(\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\; (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))+\; \; \\& \qquad (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\; (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) -\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1} \end{align} $$ (53) 化简式(53)右边的前两项得
$$ \begin{align} & \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k} -(\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\; (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) = \; \; \; \; \\& \qquad -2\tilde{\pmb{\theta}}^{\rm{T}}_{k}\; (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))-\; \; \; \; \; \; \; \; \\& \qquad (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}} (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) \end{align} $$ (54) 化简式(53)右边的后两项得
$$ \begin{align} & (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\; (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) -\; \; \; \; \; \; \; \\& \qquad \tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1} = (\hat{\pmb{\theta}}_{k-1}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\times\; \; \; \; \; \; \; \\& \qquad (2\pmb{\theta}-\hat{\pmb{\theta}}_{k-1}- {\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))\; \; \end{align} $$ (55) 由文献[21]中的引理2.1知
$$ \begin{align} & (\hat{\pmb{\theta}}_{k-1}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}))^{\rm{T}}\times\; \; \; \; \; \; \; \\& \qquad (2\pmb{\theta}-\hat{\pmb{\theta}}_{k-1} -{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) \leq0 \end{align} $$ (56) 由式(53)~(56), 可得
$$ \begin{align} & \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}-\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1} \leq-2\tilde{\pmb{\theta}}^{\rm{T}}_{k}\; (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})) \end{align} $$ (57) 类似地, 可得如下两式
$$ \begin{align} & \tilde{\pmb{p}}_{k}^{\rm{T}}\tilde{\pmb{p}}_{k}-\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1} \leq-2\tilde{\pmb{p}}^{\rm{T}}_{k}\; (\hat{\pmb{p}}_{k}-{\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k-1})) \end{align} $$ (58) $$ \begin{align} & (\hat{u}_{k}-u_{r})^{2}-(\hat{u}_{k-1}-u_{r})^{2}\leq\; \\& \qquad -2(u_{r}-\hat{u}_{k})(\hat{u}_{k}-{\rm sat}_{\bar{u}}\; (\hat{u}_{k-1}))) \end{align} $$ (59) 将式(57) ~ (59)代入式(52)
$$ \begin{align} & \Delta L_{k}\; (t) \leq\int^{t}_{0}\dot{V}_{k}{\rm d}\tau+V_{k}\; (0)-V_{k-1}\; (t)-\; \\& \qquad \frac{1}{\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}^{\rm{T}}_{k}\; (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})){\rm d}\tau-\; \; \\& \qquad \frac{1}{\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k}^{\rm{T}}\; (\hat{\pmb{p}}_{k}-{\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k-1})){\rm d}\tau-\; \; \; \; \; \; \; \\& \qquad \frac{1}{\gamma_{3}}\int^{t}_{0}g_{r}\; (u_{r}-\hat{u}_{k})(\hat{u}_{k}-{\rm sat}_{\bar{u}}\; (\hat{u}_{k-1})){\rm d}\tau \end{align} $$ (60) 将式(46)代入式(60)可得
$$ \begin{align} & \Delta L_{k}\; (t)\leq\; \; \; \; \; \; \; \\& \qquad -\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)-V_{k-1}\; (t)+\\& \qquad \int^{t}_{0}z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}{\rm d}\tau- \int^{t}_{0}|\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}||z_{n, k}|{\rm d}\tau+\\& \qquad \int^{t}_{0}z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k}{\rm d}\tau -\; \; \; \\& \qquad \int^{t}_{0}g_{m, r}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}||z_{n, k}|{\rm d}\tau+\; \; \; \; \; \\& \qquad \int^{t}_{0}z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r}){\rm d}\tau-\; \; \; \; \; \; \; \; \\& \qquad \frac{1}{\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}^{\rm{T}}_{k}\; (\hat{\pmb{\theta}}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1})){\rm d}\tau-\; \; \; \; \\& \qquad \frac{1}{\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k}^{\rm{T}}\; (\hat{\pmb{p}}_{k}-{\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k-1})){\rm d}\tau-\\& \qquad \frac{1}{\gamma_{3}}\int^{t}_{0}g_{r}\; (u_{r}-\hat{u}_{k})(\hat{u}_{k}-{\rm sat}_{\bar{u}}\; (\hat{u}_{k-1}))){\rm d}\tau \end{align} $$ (61) 利用参数更新律(47), 可得
$$ \begin{align} & \Delta L_{k}\; (t) \leq-\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)- V_{k-1}\; (t) \end{align} $$ (62) 由式(62)知
$$ \begin{align} L_{k}\; (t) \leq\, &-\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)-\\& V_{k-1}\; (t)+L_{k-1}\; (t) \end{align} $$ (63) 由于$ L_{k-1}\; (t)-V_{k-1}\; (t) = \frac{1}{2\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}{\rm d}\tau +\frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}{\rm d}\tau+ \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\; (\hat{u}_{k-1}-u_{r})^{2}{\rm d}\tau $, 因此
$$ \begin{align} L_{k}\; (t) \leq\, &-\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)+\\& \frac{1}{2\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\; (\hat{u}_{k-1}-u_{r})^{2}{\rm d}\tau \end{align} $$ (64) 将式(51)代入式(64)得
$$ \begin{align} L_{k}\; (t) \leq\, &-\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k-1}\; (T)+\\& \frac{1}{2\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\; (\hat{u}_{k-1}-u_{r})^{2}{\rm d}\tau \end{align} $$ (65) 由式(65)得
$$ \begin{align} L_{k}\; (t) \leq\, & V_{k-1}\; (T)+\frac{1}{2\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{3}}\int^{t}_{0}g_{r}\; (\hat{u}_{k-1}-u_{r})^{2}{\rm d}\tau\leq\\& V_{k-1}\; (T)+\frac{1}{2\gamma_{1}}\int^{T}_{0}\tilde{\pmb{\theta}}_{k-1}^{\rm{T}}\tilde{\pmb{\theta}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{2}}\int^{T}_{0}g_{r}\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1}{\rm d}\tau+\\& \frac{1}{2\gamma_{3}}\int^{T}_{0}g_{r}\; (\hat{u}_{k-1}-u_{r})^{2}{\rm d}\tau \end{align} $$ (66) 上式右端即为$ L_{k-1}\; (T) $.因此, 对于$ t\in[0, T], $
$$ \begin{align} L_{k}\; (t)\leq L_{k-1}\; (T) \end{align} $$ (67) 在式(65)两端取$ t = T $, 可得
$$ \begin{align} & \Delta L_{k}\; (T)\leq-\sum^{n}_{j = 1}\int^{T}_{0}c_{1}z^{2}_{j, k}{\rm d}\tau \end{align} $$ (68) 至此, 容易给出部分限幅学习下的闭环系统的收敛性结果.
定理1. 对于满足假设1~4的系统(1), 采用控制律(41)以及部分限幅参数更新律(47), 可以保证系统中的所有变量在$ [0, T] $上一致有界; 且当$ k \rightarrow \infty $时, 跟踪误差在$ [0, T] $上一致收敛于零,
$$ \begin{align} \lim\limits_{k\rightarrow\infty}e_{j, k}\; (t) = 0, t\in[0, T], \quad j = 1, 2, \cdots, n \end{align} $$ (69) 证明.首先对$ L_{1} $求导,
$$ \begin{align} & \dot{L}_{1}\; (t) = \sum^{n}_{j = 1}z_{j, 1}\dot{z}_{j, 1} +\frac{1}{2\gamma_{1}}\tilde{\pmb{\theta}}_{1}^{\rm{T}}\tilde{\pmb{\theta}}_{1} +\frac{1}{2\gamma_{2}}g_{r}\tilde{\pmb{p}}_{1}^{\rm{T}}\tilde{\pmb{p}}_{1} +\\& \qquad \frac{1}{2\gamma_{3}}g_{r}\; (\hat{u}_{1}-u_{r})^{2}\leq\; \; \; \; \; \; \\& \qquad -\sum^{n}_{j = 1}c_{1}z^{2}_{j, 1} +z_{n, 1}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{1} +\frac{1}{2\gamma_{1}}\tilde{\pmb{\theta}}_{1}^{\rm{T}}\tilde{\pmb{\theta}}_{1}+\; \; \\& \qquad z_{n, 1}g_{r}\; (\hat{u}_{1}-u_{r})+ z_{n, 1}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{1}-\\& \qquad |\hat{\pmb{\theta}}_{1}^{\rm{T}}\delta\pmb{\varphi}_{1}||z_{n, 1}| -g_{m, r}|\hat{\pmb{p}}_{1}^{\rm{T}}\delta\pmb{\psi}_{1}||z_{n, 1}|+\; \; \; \; \; \; \\& \qquad \frac{1}{2\gamma_{2}}g_{r}\tilde{\pmb{p}}_{1}^{\rm{T}}\tilde{\pmb{p}}_{1}+ \frac{1}{2\gamma_{3}}g_{r}\; (\hat{u}_{1}-u_{r})^{2}\; \; \; \; \; \end{align} $$ (70) 由于, $ \hat{\pmb{\theta}}_{0} = \pmb{0}, \hat{\pmb{p}}_{0} = \pmb{0}, \hat{u}_{0} = 0, $结合式(47),
$$ \begin{align} \dot{L}_{1}\; (t) \leq&-\sum^{n}_{j = 1}c_{1}z^{2}_{j, 1} -\frac{\gamma_{3}}{2}g_{r}z_{n1}^{2}+\frac{1}{2\gamma_{3}}g_{r}u^{2}_{r}-\\& \frac{\gamma_{1}}{2}z_{n1}^{2}\; (\delta\pmb{\varphi}_{1})^{2} -\frac{\gamma_{2}}{2}g_{m, r}z_{n1}^{2}\; (\delta\pmb{\psi}_{1})^{2}+\\& \frac{1}{2\gamma_{1}}\pmb{\theta}^{\rm{T}}\pmb{\theta}+\frac{1}{2\gamma_{2}}\pmb{p}^{\rm{T}}\pmb{p}\leq\\& \frac{1}{2\gamma_{1}}\pmb{\theta}^{\rm{T}}\pmb{\theta}+\frac{1}{2\gamma_{2}}\pmb{p}^{\rm{T}}\pmb{p} +\frac{1}{2\gamma_{3}}g_{r}u^{2}_{r} \end{align} $$ (71) 由$ g_{r} $, $ u_{r} $, $ \pmb{\theta} $和$ \pmb{p} $的有界性可知, $ \dot{L}_{1} $在$ [0, T] $上是有界的, 因此, $ L_{1} $在$ [0, T] $上是有界的, $ L_{1}\; (T) $的有界性得到保证.利用式(67)知, $ L_{k}\; (t) $在$ [0, T] $上一致有界.由$ L_{k}\; (t) $的定义可得$ z_{j, k}\; (j = 1, 2, \cdots, n) $在$ [0, T] $上的一致有界性.因而, $ e_{j, k}\; (j = 1, 2, \cdots, n) $在$ [0, T] $上同样是一致有界的.进一步, 可知$ x_{j, k}\; (j = 1, 2, \cdots, n) $在$ [0, T] $上是一致有界的.
进而可得$ |\dot{\alpha}_{n-1, k}| $是一致有界的.由$ \beta_{1, k} $, $ \beta_{2, k} $和$ \beta_{3, k} $的连续性可知, $ \beta_{1, k} $, $ \beta_{2, k} $和$ \beta_{3, k} $是一致有界的.同理可得$ \pmb{\varphi}_{k} $, $ \delta\pmb{\varphi}_{k} $和$ \delta\pmb{\psi}_{k} $也是一致有界的.由于这里采用了部分限幅的参数学习机制, 故$ \hat{\pmb{\theta}}_{k} $, $ \hat{\pmb{p}}_{k} $, 和$ \hat{u}_{k} $是一致有界的.由控制律(41)可以看出, $ u_{k}\; (t) $在$ [0, T] $上是一致有界的.
为证明$ z_{j, k}\; (t) $的一致收敛性, 对式(68)进行累加,
$$ \begin{align} & \lim\limits_{k\rightarrow\infty}L_{k}\; (T) = L_{1}\; (T)+\lim\limits_{k\rightarrow\infty}\sum^{k}_{i = 2}\Delta L_{i}\; (T)\leq\; \; \; \; \; \; \\& \qquad L_{1}\; (T)-\lim\limits_{k\rightarrow\infty}\sum^{k}_{i = 2}\sum^{n}_{j = 1}\int ^{T}_{0}c_{1}z^{2}_{j, i}{\rm d}\tau \end{align} $$ (72) 由于$ L_{k}\; (T) $是正的并且$ L_{1}\; (T) $是有界的, 可得
$$ \begin{align} & \lim\limits_{k\rightarrow\infty}\int ^{T}_{0}z^{2}_{j, k}{\rm d}\tau = 0, \; \; \; j = 1, 2, \cdots, n \end{align} $$ (73) 由式(15), (19)和式(22), 可得$ \dot{z}_{j, k}\; (j = 1, 2, \cdots, n-1) $是一致有界的.进一步, 由式(23)可知, $ u_{k} $, $ f_{1, k} $, $ f_{1, r} $, $ f_{2, k} $, $ f_{2, r} $, $ g_{k} $, $ \dot{\alpha}_{n-1, k} $, $ u_{r} $, $ g_{r} $的一致有界性可以保证$ \dot{z}_{n, k} $是一致有界的.因此, $ \dot{z}_{j, k}\; (j = 1, 2, \cdots, n) $是一致有界的, 进一步可推知$ \dot{z}_{j, k}\; (j = 1, 2, \cdots, n) $是等度连续.因此, 结合式(73), 可推得, 当$ k\rightarrow\infty $, $ z_{j, k}\; (j = 1, 2, \cdots, n) $在$ [0, T] $上是一致收敛的.
由$ e_{j, k} $与$ z_{j, k} $的关系式$ \pmb{e}_{k} = Z_{c}\pmb{z}_{k} $可知, 当$ k\rightarrow\infty $, $ e_{j, k}\; (j = 1, 2, \cdots, n) $在$ [0, T] $上是一致收敛的.
注1.由上面的分析可知, 当$ k \rightarrow \infty $, $ z_{n-1, k} $, $ z_{n, k} $和$ \dot{\alpha}_{n-1, k} $都将会趋于0.当$ \pmb{e}_{k}\rightarrow \pmb{0} $, 可得$ \delta\pmb{\varphi}_{k}\rightarrow\pmb{0} $和$ \delta\pmb{\psi}_{k}\rightarrow\pmb{0} $.因此, 从控制律(41)中不难看出, 控制量中的震荡现象随着迭代次数的增加便会消除.
注2.定理1给出了部分限幅下闭环系统的稳定性和收敛性结果, 参数更新律中引入饱和函数对参数估值的有界性的证明起到关键作用, 但估值的界不能事先确定.为了能将估值限定在预先确定的界内, 在更新律中需要采取完全限幅学习.
3.2 完全限幅学习
考虑完全限幅学习的情形.控制律形式上与部分限幅情形相同, 各参数估计分别由如下更新律给出
$$ \begin{align} &\hat{\pmb{\theta}}_{k} = {\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k}^{\ast})\\& \hat{\pmb{\theta}}_{k}^{\ast} = {\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}_{k-1}^{\ast})+\gamma_{1}z_{n, k}\delta\pmb{\varphi}_{k} \\& \hat{\pmb{p}}_{k} = {\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k}^{\ast})\\& \hat{\pmb{p}}_{k}^{\ast} = {\rm sat}_{\bar{\pmb{p}}}\; (\hat{\pmb{p}}_{k-1}^{\ast})+\gamma_{2}z_{n, k}\delta\pmb{\psi}_{k}\\& \hat{u}_{k} = {\rm sat}_{\bar{u}}\; (\hat{u}_{k}^{\ast})\\& \hat{u}_{k}^{\ast} = {\rm sat}_{\bar{u}}\; (\hat{u}_{k-1}^{\ast})-\gamma_{3} z_{n, k} \end{align} $$ (74) 其中, $ \gamma_{1} > 0 $, $ \gamma_{2} > 0 $和$ \gamma_{3} > 0 $为可调整参数.饱和函数分别作用于$ \hat{\pmb{\theta}}_{k}^{\ast} $, $ \hat{\pmb{p}}_{k}^{\ast} $和$ \hat{u}_{k}^{\ast} $, 相当于右端的所有项, 因此称为完全限幅.利用等式
$$ \begin{align} & \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}-\tilde{\pmb{\theta}}^{\rm{T}}_{k-1}\tilde{\pmb{\theta}}_{k-1} = -2\tilde{\pmb{\theta}}_{k}^{\rm{T}}\; (\hat{\pmb{\theta}}_{k}-\hat{\pmb{\theta}}_{k-1})-\\& \qquad (\hat{\pmb{\theta}}_{k}-\hat{\pmb{\theta}}_{k-1})^{\rm{T}}\; (\hat{\pmb{\theta}}_{k}-\hat{\pmb{\theta}}_{k-1}) = \; \; \; \\& \qquad -2\tilde{\pmb{\theta}}_{k}^{\rm{T}}\; (\hat{\pmb{\theta}}^{*}_{k}-\hat{\pmb{\theta}}_{k-1})+2\tilde{\pmb{\theta}}_{k}^{\rm{T}}\; (\hat{\pmb{\theta}}^{*}_{k}-\hat{\pmb{\theta}}_{k})-\; \; \; \; \; \; \\& \qquad (\hat{\pmb{\theta}}_{k}-\hat{\pmb{\theta}}_{k-1})^{\rm{T}}\; (\hat{\pmb{\theta}}_{k}-\hat{\pmb{\theta}}_{k-1})\; \; \; \; \; \; \; \; \end{align} $$ (75) 由文献[21]中的引理2.1知
$$ \begin{align} & (\pmb{\theta}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}^{*}_{k}))^{\rm{T}} (\hat{\pmb{\theta}}^{*}_{k}-{\rm sat}_{\bar{\pmb{\theta}}}\; (\hat{\pmb{\theta}}^{*}_{k}))\leq0 \end{align} $$ (76) 由式(75)及式(76), 可得
$$ \begin{align} & \tilde{\pmb{\theta}}_{k}^{\rm{T}}\tilde{\pmb{\theta}}_{k}-\tilde{\pmb{\theta}}^{\rm{T}}_{k-1}\tilde{\pmb{\theta}}_{k-1}\leq -2\tilde{\pmb{\theta}}_{k}^{\rm{T}}\; (\hat{\pmb{\theta}}^{*}_{k}-\hat{\pmb{\theta}}_{k-1}) \end{align} $$ (77) 类似地, 可得如下两式
$$ \begin{align} & \tilde{\pmb{p}}_{k}^{\rm{T}}\tilde{\pmb{p}}_{k}-\tilde{\pmb{p}}_{k-1}^{\rm{T}}\tilde{\pmb{p}}_{k-1} \leq-2\tilde{\pmb{p}}^{\rm{T}}_{k}\; (\hat{\pmb{p}}_{k}^{*}-\hat{\pmb{p}}_{k-1}) \end{align} $$ (78) $$ \begin{align} & (\hat{u}_{k}-u_{r})^{2}-(\hat{u}_{k-1}-u_{r})^{2}\leq\; \; \; \; \; \; \; \\& \qquad -2(u_{r}-\hat{u}_{k})(\hat{u}_{k}^{*}-\hat{u}_{k-1}) \end{align} $$ (79) 将式(77)~(79)代入式(52)
$$ \begin{align} & \Delta L_{k}\; (t)\leq\; \; \; \; \\& \qquad -\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)-V_{k-1}\; (t)+\; \; \; \\& \qquad \int^{t}_{0}z_{n, k}\pmb{\theta}^{\rm{T}}\delta\pmb{\varphi}_{k}{\rm d}\tau- \int^{t}_{0}|\hat{\pmb{\theta}}_{k}^{\rm{T}}\delta\pmb{\varphi}_{k}||z_{n, k}|{\rm d}\tau+\\& \qquad \int^{t}_{0}z_{n, k}g_{r}\pmb{p}^{\rm{T}}\delta\pmb{\psi}_{k}{\rm d}\tau -\; \; \; \; \; \\& \qquad \int^{t}_{0}g_{m, r}|\hat{\pmb{p}}_{k}^{\rm{T}}\delta\pmb{\psi}_{k}||z_{n, k}|{\rm d}\tau+\; \; \; \; \; \; \; \\& \qquad \int^{t}_{0}z_{n, k}g_{r}\; (\hat{u}_{k}-u_{r}){\rm d}\tau-\; \; \\& \qquad \frac{1}{\gamma_{1}}\int^{t}_{0}\tilde{\pmb{\theta}}^{\rm{T}}_{k}\; (\hat{\pmb{\theta}}^{*}_{k}-\hat{\pmb{\theta}}_{k-1}){\rm d}\tau-\; \; \; \; \; \; \; \; \\& \qquad \frac{1}{\gamma_{2}}\int^{t}_{0}g_{r}\tilde{\pmb{p}}_{k}^{\rm{T}}\; (\hat{\pmb{p}}_{k}^{*}-\hat{\pmb{p}}_{k-1}){\rm d}\tau-\; \; \; \; \; \\& \qquad \frac{1}{\gamma_{3}}\int^{t}_{0}g_{r}\; (u_{r}-\hat{u}_{k})(\hat{u}_{k}^{*}-\hat{u}_{k-1}){\rm d}\tau\; \; \; \; \; \; \end{align} $$ (80) 应用参数更新律(74), 可得
$$ \begin{align} & \Delta L_{k}\; (t) \leq-\int^{t}_{0}\sum^{n}_{j = 1}c_{1}z^{2}_{j, k}{\rm d}\tau+V_{k}\; (0)- V_{k-1}\; (t) \end{align} $$ (81) 利用式(51), 并令$ t = T $, 可得
$$ \begin{align} & \Delta L_{k}\; (T)\leq-\sum^{n}_{j = 1}\int^{T}_{0}c_{1}z^{2}_{j, k}{\rm d}\tau \end{align} $$ (82) 依据式(82), 容易证得下述定理.
定理2. 对于满足假设1~4的系统(1), 采用控制律(41)以及完全限幅参数更新律(74), 可以保证系统中的所有变量在$ [0, T] $上一致有界; 且当$ k \rightarrow \infty $时, 跟踪误差在$ [0, T] $上一致收敛于零,
$$ \begin{align} \lim\limits_{k\rightarrow\infty}e_{j, k}\; (t) = 0, t\in[0, T], \quad j = 1, 2, \cdots, n \end{align} $$ (83) 4. 仿真算例
考虑如下二阶非线性系统
$$ \begin{align} \dot{x}_{1} = \, &x_{2}\; \; \; \; \; \; \; \; \\ \dot{x}_{2} = \, &\cos(x_{1})\cos(16\pi t)+\cos(x_{2})\sin(8\pi t)- \\& x_{2}- x_{1}^{3}+ (1+0.1\cos(\sin(t)x_{1})) (\sin(x_{1})\times \\& \cos(2\pi t)+\sin(x_{2})\cos(4\pi t)-x_{1}^{2}+u)\; \; \; \; \; \end{align} $$ (84) 其中, 参数不确定为, $ \pmb{\theta}\; (t) = [\cos(16\pi t)\; \sin(8\pi t)]^{\rm{T}} $, $ \pmb{p}\; (t) = [\cos(2\pi t)\; \cos(4\pi t)]^{\rm{T}} $, 相应的已知向量为, $ \pmb{\varphi}\; (\pmb{x}, t) = [\cos(x_{1})\; \cos(x_{2})]^{\rm{T}} $, $ \pmb{\psi}\; (\pmb{x}, t) = [\sin(x_{1})\; \sin(x_{2})]^{\rm{T}} $.非参数不确定为, $ \eta_{1}\; (\pmb{x}, t) = -x_{2}-x_{1}^{3} $, $ \eta_{2}\; (\pmb{x}, t) = -x_{1}^{2} $.控制增益为, $ g(\pmb{x}, t) = 1+0.1\cos(\sin(t)x_{1}) $.参考信号为$ r_{1} = 1+\sin(2\pi t)+0.25\sin(4\pi t) $, $ r_{2} = 2\pi\cos(2\pi t)+\pi\cos(4\pi t) $.跟踪周期为$ T = 1 $.系统初始状态为$ x_{1}\; (0) = 1, x_{2}\; (0) = 3\pi $, 取$ \beta_{1, k} = 1+|x_{1, k}^{2}+r_{1}^{2}+x_{1, k}r_{1}| $, $ \beta_{2, k} = |x_{1, k}+r_{1}| $, 满足假设1.取$ g_{0} = 0.9 $, $ g_{m, r} = 1.1 $, $ \beta_{3, k} = 0.1 $, 满足假设2.
采用控制律(41)以及参数更新律(47)进行仿真, 各参数取值为, $ c_{1} = 0.5 $, $ c_{2} = 0.5 $, $ l_{1} = 12 $, $ l_{2} = 12 $, $ l_{3} = 24 $, $ l_{4} = 312 $, $ \gamma_{1} = \gamma_{2} = \gamma_{3} = 10 $.系统运行40个周期后, 仿真结果如图 1~7所示.系统在第$ 40 $个学习周期中的状态信息$ x_{1} $和$ x_{2} $分别如图$ 1 $和图$ 2 $所示.可以看出, 经过足够多个周期学习后, 系统状态$ \pmb{x}_{k} $在整个作业区间上完全跟踪上期望轨迹$ \pmb{r} $.在第$ 40 $个学习周期中的中间变量信息$ z_{1} $和$ z_{2} $见图$ 3 $和图$ 4 $.控制量在第$ 40 $个学习周期中的取值情况见图$ 5 $.图$ 6 $表示$ J_{1, k} $随学习周期变化情况, 其中$ J_{1, k}: = {\rm max}_{t \in [0, T]} |e_{1, k}\; (t)| $表示在$ [0, T] $上误差$ e_{1, k} $绝对值的最大值.图$ 7 $表示$ J_{2, k} $的收敛情况, 其中$ J_{2, k}: = {\rm max}_{t \in [0, T]} |e_{2, k}\; (t)| $表示在$ [0, T] $上误差$ e_{2, k} $绝对值的最大值.
5. 结论
本文针对一类高阶不确定非线性系统, 构造Lyapunov-like函数, 设计重复学习控制方法, 处理系统中的参数和非参数不确定性, 能够实现系统状态在整个作业区间完全跟踪期望轨迹.讨论了部分限幅和完全限幅学习机制, 该重复学习控制方法可以保证跟踪误差的一致收敛性.
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