自动化学报  2017, Vol. 43 Issue (12): 2091-2099   PDF    
离散切换系统观测器存在性讨论及降维观测器设计
朱芳来1, 蔡明1, 郭胜辉1,2     
1. 同济大学电子与信息工程学院 上海 201804;
2. 苏州科技大学电子与信息工程学院 苏州 215009
摘要: 对具有未知输入的离散切换系统讨论了未知输入观测器(Unknown input observer,UIO)设计方法.首先,对一般离散系统的未知输入观测器匹配条件的Lyapunov-type表示,进行了等价性论证;然后,基于不具有未知输入的离散切换系统的稳定性理论,对具有未知输入的离散切换系统提出了一种切换降维观测器设计方法.通过矩阵分块确定出的观测器增益矩阵,使得降维观测器能直接消去未知输入的影响;然后,在此基础上提出了一种未知输入代数重构方法;最后,通过仿真验证了方法的有效性和正确性.
关键词: 离散系统     未知输入估计     观测器匹配条件     切换系统    
Discussions on Existence of Observers and Reduced-order Observer Design for Discrete-time Switched Systems
ZHU Fang-Lai1, CAI Ming1, GUO Sheng-Hui1,2     
1. College of Electronics and Information Engineering, Tongji University, Shanghai 201804;
2. College of Electrical Engineering and Automation, Suzhou University of Science and Technology, Suzhou 215009
Manuscript received : June 16, 2016, accepted: November 23, 2016.
Foundation Item: Supported by National Natural Science Foundation of China (61573256), Shanghai Science and Technology Innovation Fund (16111106502), Zhejiang Province High-end Innovative Carrier Project
Author brief: CAI Ming     Master sutdent at the College of Electronics and Information Engineering, Tongji University. His research interest covers model-based fault detection and isolation;
GUO Sheng-Hui     Ph.D, lecturer at the College of Electronics and Information Engineering, Suzhou University of Science and Technology. His research interest covers model-based fault detection and observer design
Corresponding author. ZHU Fang-Lai     Professor at the College of Electronics and Information Engineering, Tongji University. His research interest covers nonlinear robust control, model-based fault detection and isolation. Corresponding author of this paper
Recommended by Associate Editor SUN Xi-Ming
Abstract: The paper discusses the unknown input observer (UIO) design issue for discrete-time switched systems with unknown inputs. First, a Lyapunov-type equivalent representation of observer matching condition for a general discrete-time system is given and proved. Then, based on the stability theory of discrete-time switched system without unknown inputs, a reduced-order observer is developed for a discrete-time switched system with unknown inputs. The reduced-order observer can eliminate the influences of the unknown inputs directly because of the special selection of the observer gain matrix determined by matrix block computation. Meanwhile, an algebraic unknown input reconstruction method is proposed. Finally, a simulation example is given to verify the correctness and effectiveness of the proposed methods.
Key words: Discrete-time system     unknown input estimation     observer matching condition     switched system    

现代控制理论中, 未知输入观测器(Unknown input observer, UIO)设计具有重要意义, 自上世纪60年代提出以来, 一直受到学者们的广泛的关注和深入研究[1-2].对于连续受控系统, 文献[3]研究了未知输入观测器匹配条件, 并将该条件等价于一个Lyapunov-type矩阵代数方程.之后, 在这一框架下, 未知输入观测器设计得到了广泛的研究[4-10].如文献[7]在匹配条件满足的前提下, 通过系统扩展并结合奇异系统设计手段, 提出了一种滑模观测器设计方法, 以此实现了故障和输出噪声的估计.由于未知输入观测器匹配条件较为苛刻, 很多实际的控制系统并不满足, 因而如何突破该匹配条件的限制, 成为近期UIO设计研究热点之一[11-14], 如文献[12]基于构造辅助输出的方式, 提出了一种解决方案.文献[13]同样通过构造辅助输出, 提出了降维观测器结合高阶滑模观测器的观测器匹配条件突破方法, 并考虑了未知输入重构问题.文献[14]将未知输入分解为代数抑制未知输入和模型未知输入, 以此来达到满足观测器匹配条件的目的.然而, 值得强调的是, 以上这些成果都是基于连续系统讨论的.随着数字系统广泛的应用, 离散系统下的相关研究受到了关注[15-16].基于这样的关注, 本文首先对离散系统未知输入观测器匹配条件进行了讨论, 并将其转换为一个Lyapunov-type代数方程组的等价形式.

切换系统通常含有若干个子系统, 同时具有一个规定子系统之间切换方式的切换律.切换系统可以描述一大类实际的系统, 如脉冲宽度调制(Pulse width modulation, PWM)电路、直流斩波电路、网络控制系统和飞行器姿态等[17-25].早期针对切换系统的研究主要是考虑其稳定性, 如文献[17-18]针对离散切换系统, 讨论了平均驻留时间(Average dwell time, ADT)下的切换系统稳定问题, 而文献[19]对切换系统的稳定性研究成果进行了更为深入的总结.针对切换系统的状态估计的研究相对较少, 如文献[20]针对离散切换系统, 设计了异步滤波器来处理一类具有电力电子现实背景的问题, 并讨论了其稳定性.文献[21]同时针对连续系统和离散系统设计了延迟观测器, 同时达到了异步切换的效果.对于具有未知输入的切换系统, 文献[22]针对广义离散切换系统, 设计了未知输入观测器来估计系统状态, 并基于线性矩阵不等式证明了系统的稳定性.文献[23]则在未知输入匹配条件满足的前提下, 基于系统强可检测性的假设, 针对连续切换系统, 设计了未知输入切换观测器.文献[24]在驻留时间假设下研究了切换系统的指数镇定问题, 并消除了切换超调的不利影响.总之, 目前针对切换系统的观测器设计的成果不多, 降维未知输入切换观测器的成果还鲜有报道.

结合上述背景, 本文首先对离散系统未知输入观测器的存在性进行讨论, 然后将其应用于离散切换系统, 设计了降维未知输入切换观测器, 并提出了未知输入代数重构方法.本文的贡献在于: 1)将连续系统未知输入观测器的秩条件和Lyapunov-type代数方程组两者的等价性结论, 推广到了离散系统; 2)基于等价性条件, 给出了切换系统未知输入观测器存在性的前提条件; 3)提出了一种切换降维观测器设计方法, 该方法可以直接消除未知输入的影响, 并提出了一种未知输入代数重构方法.论文的结构如下:第1节对问题进行了描述, 对一般系统, 讨论了观测器匹配条件与一个Lyapunov-type代数方程的等价性; 第2节提出了降维未知输入切换系统观测器设计方法; 第3节提出了未知输入代数重构; 第4节给出了仿真分析; 第5节对全文进行了总结.

1 问题描述

考虑一类具有未知输入的离散线性系统

$\begin{equation} \left\{ \begin{array}{l} x(k+1) = Ax(k) + Bu(k) + D\eta(k) \\ y(k) = Cx(k) \\ \end{array} \right. \end{equation}$ (1)

其中, $x(k) \in {{\bf R}^n}$, $u(k) \in {{\bf R}^m}$是已知控制输入, $y(k) \in {{\bf R}^p}$是可测输出, $\eta(k) \in {{\bf R}^q}$是未知输入.系数矩阵$A \in {{\bf R}^{n \times n}}$, $B \in {{\bf R}^{n \times m}}$, $C \in {{\bf R}^{p \times n}}$$D \in {{\bf R}^{n \times p}}$为已知常数矩阵.不失一般性, 假设$(A, C)$可观测, B, D为列满秩, C为行满秩, 并有${n \geq p \geq q}$.

定义1.定义

$\begin{equation} {{R({z})}}=\left[{\begin{array}{*{20}{c}} {zI-A}&D \\ C&0 \\ \end{array}} \right] \end{equation}$ (2)

为系统(1)的Rosen-brock矩阵.当复数$z_0 $满足秩条件rank$\left( {R\left( {z_0 } \right)} \right)<n+q$时, 称$z_0 $为系统(1)的不变零点.

定义2. 若系统(1)的所有不变零点均在z平面单位圆内, 则称系统(1)为最小相系统.其等价描述为:对任何满足$\left| z \right|\ge 1$$z\in C$, 有rank$\left( {R\left( z \right)} \right)=n+q$.

假设1.系统(1)是最小相系统.

假设2. rank$\left( D \right)={\rm rank}\left( {CD} \right)=q$.

注1.无论是连续系统还是离散系统, 假设1和2都是未知输入观测器设计的基本前提条件.针对连续系统, 文献[3]给出了一个与假设1和2等价的Lyapunov-type描述, 而这样的等价描述, 为未知输入观测器设计带来了极大的便利.接下来我们在引理1中将该等价描述推广到离散系统, 并给出严格的证明和算法.

引理1. 假设1和2成立的充分必要条件是:对于任意对称正定矩阵$Q\in {{\bf R}^{n \times n}}$, 如下Lyapunov矩阵代数方程组

$\begin{equation} \left\{ {\begin{array}{l} \left( {A+LC} \right)^{\rm T}P\left( {A+LC} \right)-P=-Q \\ D^{\rm T}P=GC \\ \end{array}} \right. \end{equation}$ (3)

关于矩阵$L\in {{\bf R}^{n \times p}}$$G\in {{\bf R}^{q \times p}}$和对称正定矩阵$P\in {{\bf R}^{n \times n}}$有解.

证明. 详见附录.

2 切换降维观测器设计

本节将上述结论应用于切换系统观测器设计中, 结合切换系统的稳定性理论, 提出一种切换降维观测器设计方法.为此, 考虑如下具有未知输入的离散切换系统

$\begin{equation} \left\{ \!\!{\begin{array}{l} x\left( {k+1} \right)=A_{\sigma \left( k \right)} x\left( k \right)+B_{\sigma \left( k \right)} u\left( k \right)+D_{\sigma \left( k \right)} \eta \left( k \right) \\ y\left( k \right)=Cx\left( k \right) \\ \end{array}} \right. \end{equation}$ (4)

其中, $x(k) \in {{\bf R}^n}$, $u(k) \in {{\bf R}^m}$是已知控制输入, $y(k) \in {{\bf R}^p}$是可测输出, $\eta(k) \in {{\bf R}^q}$是未知输入.系数矩阵$A_{\sigma \left(k \right)} \in {{\bf R}^{n \times n}}$, $B_{\sigma \left(k \right)} \in {{\bf R}^{n \times m}}$$D_{\sigma \left(k \right)} \in {{\bf R}^{n \times q}}$为常数切换矩阵, $C\in {{\bf R}^{p \times n}}$为已知常数矩阵.不失一般性, 假设$C=\left[ {{\begin{array}{*{20}c} {I_p } \hfill&0 \hfill \\ \end{array} }} \right]$.切换信号$\sigma \left( k \right):N^+\to \ell =\left\{ {1, 2, \cdots, N} \right\}$, 其中N为子系统数.设采样时刻$k_1 , k_2, \cdots, k_v, k_{v+1}, \cdots $为切换系统(4)的切换时刻, 且不妨假设当$k\in \left[{k_v, k_{v+1} } \right)$, 第i个子系统被激活, 即假设当$k\in [k_v, {\kern 1pt}{\kern 1pt}k_{v+1} )$时, $\sigma (k)=i$.

假设3. 系统(4)的任意子系统均为最小相系统, 而且存在常数$0<\varsigma <1$, 对任何满足$\left| z \right|\ge \varsigma $$z\in C$, 有rank$\left( {R_{\sigma (k)} \left( z \right)} \right)=n+q$, 其中

$ R_{\sigma (k)} \left( z \right)=\left[{{\begin{array}{*{20}c} {zI-A_{\sigma (k)} }&{D_{\sigma (k)} } \\ C&0 \\ \end{array} }} \right] $

假设4. rank$\left( {D_{\sigma \left( k \right)} } \right)={\rm rank}\left( {CD_{\sigma \left( k \right)} } \right)=q$.

定义3. 对任意$0\le k_l \le k_s $, 以$N_{\sigma \left( k \right)} \left( {k_l, k_s } \right)$表示区间$\left[{k_l , k_s } \right)$上的切换次数.若存在常数$\tau _a >0$使得

$ N_{\sigma \left( k \right)} \left( {k_l, k_s } \right)\le N_0 +\frac{k_s-k_l }{\tau _a } $

成立, 则称常数$\tau _a $为切换信号$\sigma \left( k \right)$的平均驻留时间(Average dwell time, ADT), 称$N_0 $为抖振界.

引理2[17].考虑离散切换系统

$\begin{equation} x\left( {k+1} \right)=f_{\sigma \left( k \right)} \left( {x\left( k \right)} \right), \quad \sigma \left( k \right)\in \ell \end{equation}$ (5)

若存在常数$0<\alpha <1$, $\beta >1$, 并存在$C^1$函数$V_{\sigma \left( k \right)} :{\bf R}^n\to {\bf R}$, $\sigma ( k )\in \ell $, 和两类$K_\infty $函数$\varepsilon _1 $$\varepsilon _2 $使得$\forall \sigma \left( k \right)=i\in \ell $, 均有:

$\begin{align*} &\varepsilon _1 \left( {\left| {x\left( k \right)} \right|} \right)\le V_i \left( {x\left( k \right)} \right)\le \varepsilon _2 \left( {\left| {x\left( k \right)} \right|} \right) \\& \Delta V_i \left( {x\left( k \right)} \right) := V_i \left( {x\left( {k+1} \right)} \right)-\\&\qquad V_i \left( {x\left( k \right)} \right)\le -\alpha V_i \left( {x\left( k \right)} \right) \end{align*}$

$\forall \left( {\sigma \left( {k_v } \right)=i, \sigma \left( {k_v -1} \right)=j} \right)\in \ell \times \ell $, $i\ne j$,

$ V_i \left( {x\left( {k_v } \right)} \right)\le \beta V_j \left( {x\left( {k_v } \right)} \right) $

则平均驻留时间(ADT)满足

$ \tau _a \ge \tau _a^\ast =-\frac{\ln \beta }{\ln \left( {1-\alpha } \right)} $

的切换系统(5)是全局渐近稳定的.

若定义$\hbar _i \left( {\sigma \left( k \right)} \right)=\left\{ {{\begin{array}{*{20}c} 1, \hfill&{\sigma \left( k \right)=i} \hfill \\ 0, \hfill&{\sigma \left( k \right)\ne i} \hfill \\ \end{array} }} \right.$, 则系统(5)可表示为

$ \left\{ {\begin{array}{l} x\left( {k+1} \right)=\sum\limits_{i=1}^N {\hbar _i \left( {\sigma \left( k \right)} \right)} \left[ {A_i x\left( k \right)+B_i u\left( k \right)+D_i \eta \left( k \right)} \right] \\ y\left( k \right)=Cx\left( k \right) \\ \end{array}} \right. $

则对应于$k\in \left[{k_v, k_{v+1} } \right)$上的第i个子系统有

$\begin{equation} \left\{ {\begin{array}{l} x\left( {k+1} \right)=A_i^ x\left( k \right)+B_i^ u\left( k \right)+D_i^ \eta \left( k \right) \\ y\left( k \right)=Cx\left( k \right) \\ \end{array}} \right. \end{equation}$ (6)

引理3. 假设3和4的Lyapunov-type等价描述为:存在$0<\gamma _1 <1$, 使得对任意的对称正定矩阵$Q_i \in {{\bf R}^{n \times n}}$, 如下的矩阵方程组

$\begin{equation} \left\{ {\begin{array}{l} \left( {A_i +L_i C} \right)^{\rm T}P_i \left( {A_i +L_i C} \right)-\left( {1-\gamma _1 } \right)P_i =-Q_i \\ D_i^{\rm T} P_i =G_i C \\ \end{array}} \right. \end{equation}$ (7)

关于对称正定矩阵$P_i \in {{\bf R}^{n \times n}}$和矩阵$L_i \in {{\bf R}^{n \times p}}$$G_i \in {{\bf R}^{q \times p}}$有解.

证明.首先, 在假设3下, 由矩阵$A_i^ * = \left( {1/\varsigma } \right){A_i}$, ${C^ * } = \left( {1/\varsigma } \right)C$$D_i^ * = \left( {1{\rm{ }}/\varsigma } \right){D_i}$所确定的未知输入切换系统$v\left( {A_i^ * ,{C^ * },D_i^ * } \right)$的任何第i个系统均是最小相系统.事实上, $\forall \left| z \right|\ge 1$, 由于$\left| z \right|\ge 1$$0<\varsigma <1\Rightarrow \left| {\varsigma z} \right|=\varsigma \left| z \right|\ge \varsigma $, 故在假设3下有:

$\begin{align*} &\mbox{rank}\left[{{\begin{array}{*{20}c} {zI-A_i^\ast }&{D_i^\ast } \\ {C^\ast }&0 \\ \end{array} }} \right]=\\ &\quad {\rm rank}\left[{{\begin{array}{*{20}c} {zI-\frac{1}{\varsigma }A_i }&{\frac{1}{\varsigma }D_i } \\ {\frac{1}{\varsigma }C}&0 \\ \end{array} }} \right] =\\ &\quad \mbox{rank}\left[{{\begin{array}{*{20}c} {\varsigma zI-A_i }&{D_i } \\ C&0 \\ \end{array} }} \right]=n+q \end{align*}$

这表明系统$\left( {A_i^\ast, C_i^\ast, D_i^\ast } \right)$是最小相系统.另一方面, 易知rank$\left( {D_i^\ast } \right)={\rm rank}\left( {C^\ast D_i^\ast } \right)=q$.故由引理1可知, 假设3和4成立的充分必要条件是对任意的对称正定矩阵$Q_i^\ast =\frac{1}{\varsigma ^2}Q_i $.

$\left\{ {\begin{array}{*{20}{l}} {{{\left( {A_i^ * + {L_i}{C^ * }} \right)}^{\rm{T}}}{P_i}\left( {A_i^ * + {L_i}{C^ * }} \right) - {P_i} = - Q_i^ * }\\ {{{\left( {D_i^ * } \right)}^{\rm{T}}}{P_i} = {G_i}{C^ * }} \end{array}} \right.$ (8)

关于对称正定矩阵$P_i \in {{\bf R}^{n \times n}}$和矩阵$L_i \in {{\bf R}^{n \times p}}$$G_i \in {{\bf R}^{q \times p}}$有解.如果取$\gamma _1 =1-\varsigma ^2$, 则有$0<\gamma _1 <1$, 这时式(8)即为式(7).

分解系统向量$x\left( k \right)=\left[{{\begin{array}{*{20}c} {x_1 \left( k \right)} \\ {x_2 \left( k \right)} \\ \end{array} }} \right]$, 其中$x_1 \left( k \right)\in {\bf R}^p$, 同时分解(7)中系数矩阵

$\begin{align*}&A_i =\left[{{\begin{array}{*{20}c} {A_{i, 11} }&{A_{i, 12} } \\ {A_{i, 21} }&{A_{i, 22} } \\ \end{array} }} \right], B_i =\left[{{\begin{array}{*{20}c} {B_{i, 1} } \\ {B_{i, 2} } \\ \end{array} }} \right] \\&D_i =\left[{{\begin{array}{*{20}c} {D_{i, 1} } \\ {D_{i, 2} } \\ \end{array} }} \right], P_i =\left[{{\begin{array}{*{20}c} {P_{i, 11} }&{P_{i, 12} }\\ {P_{i, 12}^{\rm T} }&{P_{i, 22} } \\ \end{array} }} \right]\end{align*}$

其中, $A_{i, 11} \in {{\bf R}^{p \times p}}$, $B_{i, 11} \in {{\bf R}^{p \times m}}$, $D_{i, 11} \in {{\bf R}^{p \times q}}$, $P_{i, 11} \in {{\bf R}^{p \times p}}$.

对系统(4)进行线性变换$\theta \left( k \right)=T_{\sigma \left( k \right)} x\left( k \right)$, 其中$\theta \left( k \right)=\left[ {{\begin{array}{*{20}c} {\theta _1 \left( k \right)} \\ {\theta _2 \left( k \right)} \\ \end{array} }} \right]$, $\theta _1 \left( k \right)\in {\bf R}^p$, $T_{\sigma \left( k \right)} =\left[{{\begin{array}{*{20}c} {I_p }&0 \\ {K_{\sigma \left( k \right)} }&{I_{n-p} } \\ \end{array} }} \right]$, 其中取$K_{\sigma \left( k \right)} =P_{\sigma (k), 22}^{-1} P_{\sigma (k), 12}^{\rm T} \in {{\bf R}^{{(n-p)} \times p}}$, 则有:

$\begin{align*} \theta _1 \left( k \right)=&x_1 \left( k \right)=y\left( k \right) \\ \theta _2 \left( k \right)=&\left[{{\begin{array}{*{20}c} {K_{\sigma \left( k \right)} }&{I_{n-p} } \\ \end{array} }} \right]x\left( k \right)= \\& K_{\sigma \left( k \right)} x_1 \left( k \right)+x_2 \left( k \right)=K_{\sigma \left( k \right)} y\left( k \right)+x_2 \left( k \right) \end{align*}$

则当系统处于第i个子系统时, 一方面, 由式(7)的第二式知

$ \left[{{\begin{array}{*{20}c} {P_{i, 11} }&{P_{i, 12} } \\ {P_{i, 12}^{\rm T} }&{P_{i, 22} } \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {D_{i, 1} } \\ {D_{i, 2} } \\ \end{array} }} \right]=\left[{{\begin{array}{*{20}c} {I_p } \\ 0 \\ \end{array} }} \right]G_i $

由此推导出$P_{i, 12}^{\rm T} D_{i, 1} +P_{i, 22} D_{i, 2} =0$, 因而有$K_i D_{i, 1} +D_{i, 2} =0$.另一方面, 由式(6)可得:

$\begin{align*} \theta _2& \left( {k+1} \right)=\left[{{\begin{array}{*{20}c} {K_i }&{I_{n-p} } \\ \end{array} }} \right]x\left( {k+1} \right) =\\&\left[{{\begin{array}{*{20}c} {K_i }&{I_{n-p} } \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {A_{i, 11} }&{A_{i, 12} } \\ {A_{i, 21} }&{A_{i, 22} } \\ \end{array} }} \right]\times\\&\left[{{\begin{array}{*{20}c} {I_p }&0 \\ {-K_i }&{I_{n-p} }\\ \end{array} }} \right] \left[{{\begin{array}{*{20}c} {\theta _1 \left( k \right)} \\ {\theta _2 \left( k \right)} \\ \end{array} }} \right]+\\&\left[{{\begin{array}{*{20}c} {K_i }&{I_{n-p} } \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {B_{i, 1} } \\ {B_{i, 2} } \\ \end{array} }} \right]u\left( k \right) +\\&\left[{{\begin{array}{*{20}c} {K_i }&{I_{n-p} } \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {D_{i, 1} } \\ {D_{i, 2} } \\ \end{array} }} \right]\eta \left( k \right) =\\&\left( {K_i A_{i, 12} +A_{i, 22} } \right)\theta _2 \left( k \right) +\\&\left[{\left( {A_{i, 21}-A_{i, 22} K_i } \right)+K_i \left( {A_{i, 11}-A_{i, 12} K_i } \right)} \right]y\left( k \right) + \\& \left( {K_i B_{i, 1} +B_{i, 2} } \right)u\left( k \right)+\left( {K_i D_{i, 1} +D_{i, 2} } \right)\eta \left( k \right) \end{align*}$

注意到$K_i D_{i, 1} +D_{i, 2} =0$, 因而有:

$\begin{array}{*{20}{l}} {{\theta _2}{\mkern 1mu} }&{\left( {k + 1} \right) = \left( {{K_i}{A_{i,12}} + {A_{i,22}}} \right){\theta _2}\left( k \right) + }\\ {}&{\left[ {\left( {{A_{i,21}} - {A_{i,22}}{K_i}} \right) + {K_i}\left( {{A_{i,11}} - {A_{i,12}}{K_i}} \right)} \right]y\left( k \right) + }\\ {}&{\left( {{K_i}{B_{i,1}} + {B_{i,2}}} \right)u\left( k \right)} \end{array}$ (9)

由于切换系统在切换时刻, 新子系统的状态初值为前一采样时刻的状态值, 即有$x(k_v )=x(k_v -1) $, 所以有:

$ \theta \left( {k_v } \right)=T_{\sigma \left( {k_v } \right)} x\left( {k_v } \right)=T_{\sigma \left( {k_v } \right)} x\left( {k_v-1} \right)= $ $ T_{\sigma \left( {k_v } \right)} T_{\sigma \left( {k_v-1} \right)}^{-1} \theta \left( {k_v-1} \right) $

于是有:

$\begin{align} \theta _2& \left( {k_v } \right)=\nonumber\\&\left( {K_{\sigma \left( {k_v } \right)} -K_{\sigma \left( {k_v -1} \right)} } \right)\theta _1 \left( {k_v -1} \right)+\theta _2 \left( {k_v -1} \right) =\nonumber\\& \left( {K_{\sigma \left( {k_v } \right)} -K_{\sigma \left( {k_v -1} \right)} } \right)y\left( {k_v -1} \right)+\theta _2 \left( {k_v -1} \right) \end{align}$ (10)

在假设3和4下, 由引理3知, 存在常数$0<\gamma _1 <1$, 使得(7)关于对称正定矩阵$P_i \in {{\bf R}^{n \times n}}$和矩阵$L_i \in {{\bf R}^{n \times p}}$$G_i \in {{\bf R}^{q \times p}}$有解.如果对对称正定矩阵$P_i $解作进一步的假设, 则提出如下切换降维观测器设计方法:

定理2.在假设3和4下, 如果还存在常数$\gamma _2>1$使得对于$\forall \left( {\sigma \left( k \right)=i, \sigma \left( {k-1} \right)=j} \right)\in \ell \times \ell $, $i\ne j$, 满足:

$\begin{equation} P_i -\gamma _2 P_j <0 \end{equation}$ (11)

且假设系统(4)切换信号$\sigma \left( k \right)$的平均驻留时间(ADT)满足:

$\begin{equation} \tau _a \ge \tau _a^\ast =-\frac{\ln \gamma _2 }{\ln \left( {1-\gamma _1 } \right)} \end{equation}$ (12)

则系统

$\left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{l}} {{{\hat \theta }_2}\left( {k + 1} \right) = \left( {{K_i}{A_{i,12}} + {A_{i,22}}} \right){{\hat \theta }_2}\left( k \right) + }\\ {\qquad \left[ {\left( {{A_{i,21}} - {A_{i,22}}{K_i}} \right) + {K_i}\left( {{A_{i,11}} - {A_{i,12}}{K_i}} \right)} \right] \times }\\ {\qquad y\left( k \right) + \left( {{K_i}{B_{i,1}} + {B_{i,2}}} \right)u\left( k \right)}\\ {{{\hat \theta }_2}\left( {{k_v}} \right) = \left( {{K_{\sigma \left( {{k_v}} \right)}} - {K_{\sigma \left( {{k_v} - 1} \right)}}} \right)y\left( {{k_v} - 1} \right) + }\\ {\qquad {{\hat \theta }_2}\left( {{k_v} - 1} \right)}\\ {\hat x\left( k \right) = \left[ {\begin{array}{*{20}{c}} {y\left( k \right)}\\ {{{\hat \theta }_2}\left( k \right) - {K_i}y\left( k \right)} \end{array}} \right]} \end{array}} \end{array}} \right.$ (13)

是系统(4)的切换降维未知输入观测器使得$\mathop {\lim }\nolimits_{k\to \infty } (x(k)-\hat {x}(k))=0$.

证明.定义误差方程为$\tilde {\theta }_2 \left( k \right)=\theta _2 \left( k \right)-\hat {\theta }_2 \left( k \right)$, 选取Lyapunov函数为

$\begin{equation} V_{\sigma \left( k \right)} \left( k \right)=\tilde {\theta }_2^{\rm T} \left( k \right)\sum\limits_{i=1}^N {\hbar _i \left( {\sigma \left( k \right)} \right)} P_{i, 22} \tilde {\theta }_2 \left( k \right) \end{equation}$ (14)

当系统处于第i个子系统的时刻时, 即当$k\in \left[{k_v, k_{v+1}}\right)$, 由式(14)可得:

$\begin{equation} \lambda _1 \left( {\left\| {\tilde {\theta }_2 \left( k \right)} \right\|^2} \right)\le V_i \left( k \right)\le \lambda _2 \left( {\left\| {\tilde {\theta }_2 \left( k \right)} \right\|^2} \right) \end{equation}$ (15)

$\lambda _1 =\mathop {\min }_{i\in \ell } \left\{ {\lambda _{\min } \left( {P_{i, 22} } \right)} \right\}$, $\lambda _2 =\mathop {\max }_{i\in \ell } \left\{ {\lambda _{\max } \left( {P_{i, 22} } \right)} \right\}$, $\lambda_{\min } \left({P_{i, 22}} \right)$表示$P_{i, 22} $最小的特征值, $\lambda _{\max } \left( {P_{i, 22} } \right)$表示$P_{i, 22}$ $(i=1, 2, \cdots, N)$最大的特征值.

由式(7)第一式, 利用Schur引理, 取$W_i =P_i L_i $, 可得:

$ \left[{{\begin{array}{*{20}c} {-\left( {1-\gamma _1 } \right)P_i }&\ast \\ {P_i A_i +W_i C_i }&{-P_i } \\ \end{array} }} \right]<0 $

分解上式, 有:

$ \left[ \!\!{{\begin{array}{*{20}c} \ast \!\! &\!\! \ast \!&\! \ast \!&\! \ast \\ \ast \!\! &\! \!\!{-\left( {1-\gamma _1 } \right)P_{i,22} } \!\!\! &\! \ast \!&\! {A_{i,22}^{\rm T} P_{i,22} +A_{i,12}^{\rm T} P_{i,12} } \\ \ast \!&\! \ast \!&\! \ast \!&\! \ast \\ \ast \! &\!\! {P_{i,12}^{\rm T} A_{i,12} +P_{i,22} A_{i,22} } \!\!\!&\! \ast \!\!&\!\! {-P_{i,22} } \\ \end{array} }}\!\! \right]\!\!<\!0 $

再经初等行、列变换, 可得:

$ \left[\!\! {{\begin{array}{*{20}c} \ast &\ast &\ast&\ast \\ \ast &\ast &\ast &\ast \\ \ast &\ast &{-\left( {1-\gamma _1 } \right)P_{i,22} } \!\!&\!\!\! {A_{i,22}^{\rm T} P_{i,22} +A_{i,12}^{\rm T} P_{i,12} } \!\!\\ \ast &\ast \!\!\!& \!\!\!{P_{i,12}^{\rm T} A_{i,12} +P_{i,22} A_{i,22} } \!\!&\!\! {-P_{i,22} } \\ \end{array} }} \!\!\right]\!<\!0 $

故有:

$ \left[\!\! {{\begin{array}{*{20}c} {-\left( {1-\gamma _1 } \right)P_{i, 22} } \!\! &\!\! {A_{i, 22}^{\rm T} P_{i, 22} +A_{i, 12}^{\rm T} P_{i, 12} } \\ {P_{i, 12}^{\rm T} A_{i, 12} +P_{i, 22} A_{i, 22} } \!\!&\!\! {-P_{i, 22} } \\ \end{array} }}\!\! \right]\!<\!0 $

由Schur补定理, 可得:

$\begin{align*} \left( {K_i A_{i, 12} +A_{i, 22} } \right)^{\rm T}&P_{i, 22} \left( {K_i A_{i, 12} +A_{i, 22} } \right)-\\&\left( {1-\gamma _1 } \right)P_{i, 22} <0 \end{align*}$

也即

$\begin{align} \left( {K_i A_{i, 12} +A_{i, 22} } \right)^{\rm T}&P_{i, 22} \left( {K_i A_{i, 12} +A_{i, 22} } \right)-\nonumber\\&P_{i, 22} <-\gamma _1 P_{i, 22} \end{align}$ (16)

对于第i个子系统, 式(14)即为

$ V_i \left( k \right)=\tilde {\theta }_2^{\rm T} \left( k \right)P_{i, 22} \tilde {\theta }_2 \left( k \right) $

从式(9)和(13)中的第一式, 可得:

$ \tilde {\theta }_2 \left( {k+1} \right)=\theta _2 \left( {k+1} \right)-\hat {\theta }_2 \left( {k+1} \right) =$ \begin{equation} \left( {K_i A_{i, 12} +A_{i, 22} } \right)\tilde {\theta }_2 \left( k \right) \end{equation} (17)

则有:

$\begin{align} &\Delta V_i \left( k \right)=\tilde {\theta }_2^{\rm T} \left( {k+1} \right)P_{i, 22} \tilde {\theta }_2 \left( {k+1} \right)- \nonumber\\&\quad \tilde {\theta }_2^{\rm T} \left( k \right)P_{i, 22} \tilde {\theta }_2 \left( k \right)= \nonumber\\&\quad \tilde {\theta }_2^{\rm T} \left( k \right)\left( {K_i A_{i, 12} \!+\!A_{i, 22} } \right)^{\rm T}P_{i, 22} \left( {K_i A_{i, 12} \!+\!A_{i, 22} } \right)\times\nonumber\\&\quad\tilde {\theta }_2 \left( k \right) -\tilde {\theta }_2^{\rm T} \left( k \right)P_{i, 22} \tilde {\theta }_2 \left( k \right) \end{align}$ (18)

由式(16)和(18)可得:

$\begin{equation} \Delta V_i \left( k \right)\le -\gamma _1 V_i \left( k \right) \end{equation}$ (19)

用式(10)减去式(13)中第二式, 我们有:

$\begin{align*} \tilde {\theta }_2 \left( {k_v } \right)=&\theta _2 \left( {k_v } \right)-\hat {\theta }_2 \left( {k_v } \right) = \\& \theta _2 \left( {k_v -1} \right)-\hat {\theta }_2 \left( {k_v -1} \right)=\tilde {\theta }_2 \left( {k_v -1} \right) \end{align*}$

对于$\forall \left( {\sigma \left( {k_v } \right)=i, \sigma \left( {k_v -1} \right)=j} \right)\in \ell \times \ell $, $i\ne j$, 从式(11)可得:

$ \left[{{\begin{array}{*{20}c} {P_{i, 11}-\gamma _2 P_{j, 11} }&{P_{i, 12}-\gamma _2 P_{j, 12} } \\ \ast&{P_{i, 22}-\gamma _2 P_{j, 22} } \\ \end{array} }} \right]<0 $

也即

$ P_{i, 22}-\gamma _2 P_{j, 22} <0 $

因此有

$ \tilde {\theta }_2^{\rm T} \left( {k_v } \right)P_{i, 22} \tilde {\theta }_2 \left( {k_v } \right)\le \gamma _2 \tilde {\theta }_2^{\rm T} \left( {k_v-1} \right)P_{j, 22} \tilde {\theta }_2 \left( {k_v-1} \right) $

进一步地

$\begin{align*} \sum\limits_{i=1}^N \sum\limits_{j=1}^N &\hbar _i ( {\sigma ( {k_v } )} )\hbar _j ( {\sigma ( {k_v -1} )} )\tilde {\theta }_2^{\rm T} ( {k_v } )\times \\&P_{i, 22} \tilde {\theta }_2 ( {k_v } ) \le \gamma _2 \sum\limits_{i=1}^N \sum\limits_{j=1}^N \hbar _i ( {\sigma ( {k_v } )})\times \\&\hbar _j ( {\sigma t( {k_v -1} )} )\tilde {\theta }_2^{\rm T} ( {k_v -1} )\times \\&P_{j, 22} \tilde {\theta }_2 ( {k_v -1} ) \end{align*}$

$\sum_{i=1}^N {\hbar _i ( {\sigma ( {k_v } )})} =\sum\nolimits_{j=1}^N {\hbar _j ( {\sigma ( {k_v -1} )} )} =1$, 故

$\begin{align*}& \tilde {\theta }_2^{\rm T} \left( {k_v } \right)\sum\limits_{i=1}^N {\hbar _i \left( {\sigma \left( {k_v } \right)} \right)} P_{i, 22} \tilde {\theta }_2 \left( {k_v } \right)\le \\&\quad \gamma _2 \tilde {\theta }_2^{\rm T} \left( {k_v -1} \right)\sum\limits_{j=1}^N {\hbar _j \left( {\sigma \left( {k_v -1} \right)} \right)} P_{j, 22} \tilde {\theta }_2 \left( {k_v -1} \right) \end{align*}$

也即

$\begin{equation} V_i \left( {\theta _2 \left( {k_v } \right)} \right)\le \gamma _2 V_j \left( {\theta _2 \left( {k_v } \right)} \right) \end{equation}$ (20)

注意到式(15)、(18)、(20)和式(12), 根据引理2推知观测器误差方程(17)是渐近稳定的, 这表明式(13)是式(4)的切换降维观测器.

3 未知输入代数重构

本节对未知输入提出一种代数延迟重构方法, 式(4)两边同时左乘矩阵C, 有:

$ Cx(k+1) =CA_{\sigma (k)} x(k)+CB_{\sigma (k)} u(k)+CD_{\sigma (k)} \eta (k) $

$ CD_{\sigma \left( k \right)} \eta \left( k \right)=y\left( {k+1} \right)-CA_{\sigma \left( k \right)} x\left( k \right)-CB_{\sigma \left( k \right)} u\left( k \right) $

另由假设4可知, rank$\left( {CD_{\sigma \left( k \right)} } \right)=q$, 故, $\left( {CD_{\sigma \left( k \right)} } \right)^{\rm T}\left( {CD_{\sigma \left( k \right)} } \right)$是非奇异矩阵, 因此有:

$\begin{array}{l} \eta \left( k \right) = {\left( {C{D_{\sigma \left( k \right)}}} \right)^ + }\left( {y\left( {k + 1} \right) - C{A_{\sigma \left( k \right)}}x\left( k \right)} \right) - \\ \quad \quad \quad \quad {\left( {C{D_{\sigma \left( k \right)}}} \right)^ + }C{B_{\sigma \left( k \right)}}u\left( k \right) \end{array}$ (21)

其中

$ \left( {CD_{\sigma \left( k \right)} } \right)^+=\left( {\left( {CD_{\sigma \left( k \right)} } \right)^{\rm T}CD_{\sigma \left( k \right)} } \right)^{-1}\left( {CD_{\sigma \left( k \right)} } \right)^{\rm T} $

定理3. 在假设3和4的前提下,

$\begin{array}{l} \hat \eta \left( k \right) = {\left( {C{D_{\sigma \left( k \right)}}} \right)^ + }\left( {y\left( {k + 1} \right) - C{A_{\sigma \left( k \right)}}\hat x\left( k \right)} \right) - \\ \quad \quad \quad \quad {\left( {C{D_{\sigma \left( k \right)}}} \right)^ + }C{B_{\sigma \left( k \right)}}u\left( k \right) \end{array}$ (22)

是未知输入$\eta \left( k \right)$的渐近收敛延迟重构.

证明.由式(21)和(22)可得, 未知输入重构误差

$ \tilde {\eta }\left( k \right)=\eta \left( k \right)-\hat {\eta }\left( k \right)=-\left( {CD_{\sigma \left( k \right)} } \right)^+CA_{\sigma \left( k \right)} \tilde {x}\left( k \right) $

由于$\mathop {\lim }_{k\to \infty } \tilde {x}\left( k \right)=0$, 故有$\mathop {\lim }_{k\to \infty } \tilde {\eta }\left( k \right)=0$.

4 仿真

考虑离散切换系统(4)的系数矩阵为

$ A_1 =\left[{{\begin{array}{*{20}c} {0.39}&{0.08}&{0.07} \\ {-0.14}&{0.66}&{-0.20} \\ {-0.16}&{-0.40}&{0.66} \\ \end{array} }} \right], B_1 =\left[{{\begin{array}{*{20}c} {0.2} \\ {0.6} \\ {0.4} \\ \end{array} }} \right], $ $ D_1 \!=\!\left[{{\begin{array}{*{20}c} {-0.1} \\ {0.2} \\ {0.2} \\ \end{array} }} \right], A_2\! =\!\left[{{\begin{array}{*{20}c} {0.49}&{0.10}&{0.06} \\ {-0.32}&{0.95}&{-0.23} \\ {-0.25}&{-0.06}&{0.63} \\ \end{array} }} \right], $ $ B_1 \!=\!\left[{{\begin{array}{*{20}c} {0.4} \\ {0.4} \\ {0.5} \\ \end{array} }} \right], ~ D_1 \!=\!\left[{{\begin{array}{*{20}c} {-0.1} \\ {0.2} \\ {-0.1} \\ \end{array} }} \right], ~C\!=\!\left[{{\begin{array}{*{20}c} 1&0&0 \\ 0&1&0 \\ \end{array} }} \right] $

假设输入为$u\left( k \right)=5\sin \left( {2k} \right)$, 未知输入为$\eta \left( k \right)=\sin \left( {50k} \right)$, $\gamma _1 =0.6$, $\gamma _2 =3.8$.可以算出:

$ P_1 =10^5\times \left[{{\begin{array}{*{20}c} {1.612} \hfill&{0.560} \hfill&{0.104} \hfill \\ {0.560} \hfill&{0.236} \hfill&{0.041} \hfill \\ {0.104} \hfill&{0.041} \hfill&{0.011} \hfill \\ \end{array} }} \right] $ $ P_2 =10^4\times \left[{{\begin{array}{*{20}c} {4.757} \hfill&{1.516} \hfill&{0.278} \hfill \\ {1.516} \hfill&{1.030} \hfill&{0.209} \hfill \\ {0.278} \hfill&{0.209} \hfill&{0.140} \hfill \\ \end{array} }} \right] $

以及$\tau _a^\ast =1.457$.切换序列如图 1所示, 图 2给出状态估计和未知输入重构, 由图 2可以看出, 所提方法是正确有效的.

图 1 切换序列 Figure 1 Switched sequence
图 2 系统状态估计和未知输入重构 Figure 2 stimated states and unknown input reconstruction
5 总结

本文对具有未知输入的离散切换系统, 提出了一种能直接消去未知输入影响的切换降维观测器设计方法.为了达到设计目的, 事先将文献[3]中有关连续系统观测器匹配条件的等价性条件的相关结论, 推广到一般的离散系统, 针对离散系统给出了类似于连续系统中观测器匹配条件的Lyapunov-type等价表示形式, 并给出了严格的推导过程.然后, 将所得结论应用于离散切换系统, 提出了降维切换观测器设计方法, 并以此讨论了切换观测器存在的前提条件.在同样的前提下, 提出了一种未知输入代数重构方法.对同时具有未知输入和测量噪声的切换系统, 设计切换观测器并讨论存在性, 是值得进一步研讨的议题.

附录

为了证明引理1的结论, 首先引入下面引理.

引理A1[3].假设2成立的充分必要条件是:存在可逆矩阵$T\in {\bf R}^{n\times n}$, $S\in {\bf R}^{p\times p}$, 使得:

$\begin{align*}& \bar {A}:=T^{-1}AT=\left[{{\begin{array}{*{20}c} {\bar {A}_{11} }&{\bar {A}_{12} } \\ {\bar {A}_{21} }&{\bar {A}_{22} } \\ \end{array} }} \right]\\& \bar {B}:=T^{-1}B=\left[{{\begin{array}{*{20}c} {\bar {B}_1 } \\ {\bar {B}_2 } \\ \end{array} }} \right] \\& \bar {D}:=T^{-1}D=\left[{{\begin{array}{*{20}c} {\bar {D}_1 } \\ 0 \\ \end{array} }} \right] \\& \bar {C}:=S^{-1}CT=\left[{{\begin{array}{*{20}c} {\bar {C}_{11} }&0 \\ 0&{\bar {C}_{22} } \\ \end{array} }} \right] \end{align*}$

其中, $\bar {A}_{11} \in {{\bf R}^{q \times q}}$, $\bar {B}_1 \in {{\bf R}^{q \times m}}$, $\bar {D}_1 \in {{\bf R}^{q \times q}}$, $\bar {C}_{11} \in {{\bf R}^{q \times q}}$, $\bar {C}_{22} \in {{\bf R}^{(p-q) \times (n-q)}}$, 且$\bar {D}_1 , \bar {C}_{11} $是可逆的.进一步地, 假设1成立当且仅当$ ( {\bar {A}_{22}, \bar {C}_{22} } )$可检测.

引理A2. 对于任意对称正定矩阵$Q\in {{\bf R}^{n \times n}}$, 矩阵代数方程组(3)关于$L\in {{\bf R}^{n \times p}}$, $G\in {{\bf R}^{q \times p}}$$P\in {{\bf R}^{n \times n}}$有解的充分必要是: $\bar {P}=T^{\rm T}PT$, $\bar {L}=T^{-1}LS$, $\bar {G}=GS$$\bar {Q}=T^{\rm T}QT$是如下的代数矩阵方程

$\left\{ {\begin{array}{*{20}{l}} {{{\left( {\bar A + \bar L\bar C} \right)}^{\rm{T}}}\bar P\left( {\bar A + \bar L\bar C} \right) - \bar P = - \bar Q < 0}\\ {{{\bar D}^{\rm{T}}}\bar P = \bar G\bar C} \end{array}} \right.$ (A1)

的解.

证明. 直接验证即可得知. {引理1的证明}: (必要性:假设1和2成立$\Rightarrow $式(3)有解).定义

$ \bar {Q}_3 =\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)+\varsigma I_{n-p} $

可知

$ \bar {Q}_3 \in {{\bf R}^{(n-q) \times (n-q)}} $

其中, $\varsigma >0$是任意的正常数, $\bar {L}_{12} \in {{\bf R}^{q \times (p-q)}}$为任意矩阵, 显然, $\bar {Q}_3 $是对称正定矩阵.在假设1和2的前提下, 由引理4可知, $\left( {\bar {A}_{22}, \bar {C}_{22} } \right)$可检测, 因此, 存在矩阵$\bar {L}_{22} \in {{\bf R}^{(n-q) \times (p-q)}}$, 使得如下Lyapunov代数方程

$ \left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} =-\bar {Q}_3 $

有对称正定解$\bar {P}_{22} $, 即有:

$\begin{array}{l} \left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)+ \\ \quad \left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} =-\varsigma I_{n-p} \\ \end{array}$

$\bar {P}=\left[{{\begin{array}{*{20}c} {I_q }&0 \\ 0&{\bar {P}_{22} } \\ \end{array} }} \right]$$\bar {L}=\left[{{\begin{array}{*{20}c} {-\bar {A}_{11} \bar {C}_{11}^{-1} }&{\bar {L}_{12} } \\ {-\bar {A}_{21} \bar {C}_{11}^{-1} }&{\bar {L}_{22} } \\ \end{array} }} \right]$

则有式(A2) (见页底).

因此, 对于对称正定矩阵$\bar {Q}=\left[{{\begin{array}{*{20}c} {I_q }&0 \\ 0&{\varsigma I_{n-p} } \\ \end{array} }} \right]$, 存在对称正定矩阵$\bar {P}=\left[ {{\begin{array}{*{20}c}{I_q }&0 \\ 0&{\bar {P}_{22} } \\ \end{array} }} \right]$$\bar {L}=\left[{{\begin{array}{*{20}c} {-\bar {A}_{11} \bar {C}_{11}^{-1} }&{\bar {L}_{12} } \\ {-\bar {A}_{21} \bar {C}_{11}^{-1} }&{\bar {L}_{22} } \\ \end{array} }} \right]$, 使得$\left( {\bar {A}+\bar {L}\bar {C}} \right)^{\rm T}\bar {P}\left( {\bar {A}+\bar {L}\bar {C}} \right)-\bar {P}=-\bar {Q}<0, $$\bar {D}^{\rm T}\bar {P}=\left[{{\begin{array}{*{20}c} {\bar {D}_1^{\rm T} }&0 \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {I_q }&0 \\ 0&{\bar {P}_{22} } \\ \end{array} }} \right]=\left[{{\begin{array}{*{20}c} {\bar {D}_1^{\rm T} }&0 \\ \end{array} }} \right]$, 若取$\bar {G}=\left[{{\begin{array}{*{20}c} {\bar {D}_1^{\rm T} \bar {C}_{11}^{-1} }&0 \\ \end{array} }} \right]$, 则有$\bar {G}\bar {C}=\left[{{\begin{array}{*{20}c} {\bar {D}_1^{\rm T} \bar {C}_{11}^{-1} }&0 \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} {\bar {C}_{11} }&0 \\ 0&{\bar {C}_{22} } \\ \end{array} }} \right]=\left[{{\begin{array}{*{20}c} {\bar {D}_1^{\rm T} }&0 \\ \end{array} }} \right], $因此式(23)有解, 再由引理5易知, 式(3)有解.

充分性:式(3)有解$\Rightarrow $假设1和2成立).由于$P\in {{\bf R}^{n \times n}}$为对称正定矩阵, 故矩阵D$D^{\rm T}PD$有相同的零空间, 即rank$\left( {D^{\rm T}PD} \right)={\rm rank}\left( D \right)$, 由$D^{\rm T}P=GC$, 得$D^{\rm T}PD=GCD$, 即

$\begin{align*} &{\rm rank}\left( {D^{\rm T}PD} \right)={\rm rank}\left( {GCD} \right)\le {\rm rank}\left( {CD} \right)\le \\ &\quad {\rm rank}\left( D \right) \end{align*}$

因此有${\rm rank}\left( D \right)\le {\rm rank}\left( {CD} \right)\le {\rm rank}\left( D \right)$, 即有${\rm rank}\left( {CD} \right)={\rm rank}\left( D \right)$, 也即假设2成立.

在假设2成立的前提下, 由引理A1可知, 存在可逆矩阵$T\in {{\bf R}^{n \times n}}$, $S\in {{\bf R}^{p \times p}}$, 使得(3)成立.对于引理A2中定义的矩阵$\bar {P}$, $\bar {L}$$\bar {G}$进行如下的分块

$\begin{align*}&\bar {P}=\left[{{\begin{array}{*{20}c} {\bar {P}_{11} }&{\bar {P}_{12} } \\ {\bar {P}_{21} }&{\bar {P}_{22} } \\ \end{array} }} \right], \bar {L}=\left[{{\begin{array}{*{20}c} {\bar {L}_{11} }&{\bar {L}_{12} } \\ {\bar {L}_{21} }&{\bar {L}_{22} } \\ \end{array} }} \right], \\&\bar {G}=\left[{\begin{array}{*{20}c} {\bar {G}_1 }&{\bar {G}_2 } \\ \end{array} } \right], \bar {Q}=\left[{\begin{array}{*{20}c} {\bar {Q}_{11} }&{\bar {Q}_{12} } \\ {\bar {Q}_{12}^{\rm T} }&{\bar {Q}_{22} } \end{array} } \right]\end{align*}$

其中, 由引理A2可知, 式(3)有解意味着式(A1)有解, 对式(A1)按分块展开

$\begin{align*}\left( {\bar {A}+\bar {L}\bar {C}} \right)^{\rm T}& \bar {P}\left( {\bar {A}+\bar {L}\bar {C}} \right)-\bar {P}=\\&\left[ {{\begin{array}{*{20}c} \ast&\ast \\ \ast&{\Omega _1 } \\ \end{array} }} \right]-\left[{{\begin{array}{*{20}c} {\bar {P}_{11} }&{\bar {P}_{12} } \\ {\bar {P}_{21} }&{\bar {P}_{22} } \\ \end{array} }} \right]=\\&\left[{{\begin{array}{*{20}c} \ast&\ast \\ \ast&{\Omega _2 } \\ \end{array} }} \right] \end{align*}$
$ \begin{array}{l} \left( {\bar {A}+\bar {L}\bar {C}} \right)^{\rm T}\bar {P} \left( {\bar {A}+\bar {L}\bar {C}} \right)-\bar {P}= \\ \quad \left[{{\begin{array}{*{20}c} 0&0 \\ {\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}}&{\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} } \\ \end{array} }} \right]\left[{{\begin{array}{*{20}c} 0& {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \\ 0&{\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \\ \end{array} }} \right]-\left[{{\begin{array}{*{20}c} {I_q }&0 \\ 0&{\bar {P}_{22} } \\ \end{array} }} \right]= \\ \quad \left[{{\begin{array}{*{20}c} {-I_q }&0 \\ 0&{\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)+\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} } \\ \end{array} }} \right]= \\ \quad \left[{{\begin{array}{*{20}c} {-I_q }&0 \\ 0&{-\varsigma I_{n-p} } \\ \end{array} }} \right]=-\bar {Q} \end{array} $ (A2)

其中

$\begin{align*} \Omega _1 =&\Big[\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{11} +\\&\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{21} \Big]\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)+ \\& \Big[\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{12} +\\&\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \Big]\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right) \\ \Omega _2 =&\Big[\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{11} +\\ &\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{21} \Big]\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right) +\\&\Big[\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{12} +\\ &\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \Big]\times\\ &\left( {\bar {A}_{22} +\bar {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} \end{align*}$

也即有

${\Omega _2} = - {{\bar Q}_{22}}$ (A2)

由式(A2)中第二式可得$\bar {D}_1^{\rm T} \bar {P}_{11} =\bar {G}_1 \bar {C}_{11} $$\bar {D}_1^{\rm T} \bar {P}_{12} =\bar {G}_2 \bar {C}_{22} $.因$\bar {D}_1 $是可逆的, 故有$\bar {P}_{11} =\left( {\bar {D}_1^{\rm T} } \right)^{-1}\bar {G}_1 \bar {C}_{11} $, $\bar {P}_{12} =\left( {\bar {D}_1^{\rm T} } \right)^{-1}\bar {G}_2 \bar {C}_{22} $, 另$\bar {P}$是对称矩阵, 即有$\bar {P}_{21} =\left( {\left( {\bar {D}_1^{\rm T} } \right)^{-1}\bar {G}_2 \bar {C}_{22} } \right)^{\rm T}$, 代入式(1)中并化简

$\begin{align*} &\left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} +\\ &\qquad\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {P}_{11} -\bar {P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21} } \right)\times\\&\qquad\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)=-\bar {Q}_{22} \end{align*}$

其中, $\tilde {L}_{22} =\bar {L}_{22} +\bar {P}_{22}^{-1} ( {( {\bar {D}_1^{\rm T} })^{-1}\bar {G}_2 \bar {C}_{22} } )^{\rm T}( \bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} )\bar {C}_{22}^{-1} $.进一步地, 可得:

$\begin{align*} &\left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} =\\ &\quad-\bar {Q}_{22} -\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {P}_{11} -\bar {P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21} } \right)\times\\&\quad\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right) \end{align*}$

由于$\bar {P}_{11} -\bar {P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21}$$\bar {P}$的Schur补, 且$\bar {P}$为对称正定矩阵, 因此$\bar {P}_{11} -\bar {P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21} $为正定矩阵, 故$\left( {\bar{A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {P}_{11} -\bar{P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21} }\right)\left( {\bar {A}_{12}+\bar {L}_{12} \bar {C}_{22} } \right)$为半正定矩阵, 所以

$\begin{align*} -\bar {Q}_{22}& -\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)^{\rm T}\left( {\bar {P}_{11} -\bar {P}_{21}^{\rm T} \bar {P}_{22}^{-1} \bar {P}_{21} } \right)\times\\ &\left( {\bar {A}_{12} +\bar {L}_{12} \bar {C}_{22} } \right)<0 \end{align*}$

$\begin{align*} \left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)^{\rm T}\bar {P}_{22} \left( {\bar {A}_{22} +\tilde {L}_{22} \bar {C}_{22} } \right)-\bar {P}_{22} <0 \end{align*}$

$\left( {\bar {A}_{22} , \bar {C}_{22} } \right)$是可检测的, 由引理A1可知假设1成立.

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