﻿ 基于拓扑结构和个体动态层面的多智能体系统可控性分析
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 智能系统学报  2020, Vol. 15 Issue (2): 264-270  DOI: 10.11992/tis.201901006 0

### 引用本文

CHEN Wanjin, JI Zhijian. Controllability analysis of multi-agent systems based on topological structure and individual dynamic level[J]. CAAI Transactions on Intelligent Systems, 2020, 15(2): 264-270. DOI: 10.11992/tis.201901006.

### 文章历史

Controllability analysis of multi-agent systems based on topological structure and individual dynamic level
CHEN Wanjin , JI Zhijian
School of Automation Engineering, Qingdao University, Qingdao 266071, China
Abstract: The relationship between the controllability of network topology and individual dynamics in the overall controllability of the system is studied, and a new model describing the general linear multi-agent system is proposed. Using the Popov-Belevitch-Hautus (PBH) criterion, the necessary and sufficient conditions for the controllability of a multi-agent system in the network topology and individual dynamic level are obtained and proved, and the effect of repeated eigenvalues in the system matrix on the sufficiency of Theorem 2 is explained with a concrete example. We provide a way to avoid the occurrence of repeated eigenvalues. In particular, the two conditions for judging the uncontrollable system can be determined when the system matrix is a real symmetric matrix under this model; that is, compare the largest eigenvalue algebraic multiplicity in the system matrix with the number of 1 elements in the control matrix. If this condition is satisfied, the system is uncontrollable.
Key words: multi-agent system    controllability    linear time-invariant systems    topology    local interaction    eigenvalue    eigenvector    control theory

1 预备知识

 ${a_{ij}} = \left\{ {\begin{array}{*{20}{l}} {{\rm{ }}{w_{ij}}},&{(j,i) \in E(G)} \\ 0,&{{\text{其他}}} \end{array}} \right.$

 ${l_{ij}} = \left\{ {\begin{array}{*{20}{l}} {\displaystyle\sum\limits_{j = 1,j \ne i}^n {{a_{ij}}} },&{i = j} \\ { - {a_{ij}}},&{i \ne j} \end{array}} \right.$

 ${\rm rank}({{C}}) = \left[ {\begin{array}{*{20}{c}} {{B}}&{{{AB}}}& \cdots &{{{{A}}^{n - 1}}} \end{array}{{B}}} \right] = n$

PBH判据[11]，对于系统 $\dot {{x}} = {{Ax}} + {{Bu}}$ ，当且仅当不存在非零向量v，满足以下等式：

 $\left\{ {\begin{array}{*{20}{c}} {{{{B}}^{\rm{T}}}{{v}} = 0} \\ {{{Av}} = {\textit{λ}} {{v}}} \end{array}} \right.$

2 一般性模型的建立

 ${{{\dot x}}_i} = {{A}}{{{x}}_i} + {{B}}{{{u}}_i}$ (1)

 ${{{\dot x}}_i} = {{A}}{{{x}}_i} + {{B}}{{{u}}_i} + {{C}}{{{z}}_l}$ (2)

 ${{{u}}_i} = - {{{K}}_1}{{{x}}_i} + \sum\limits_{{v_j} \in {N_{(i)}}} {{a_{ij}}} {{{K}}_2}({{{x}}_j} - {{{x}}_i})$ (3)

 ${{\dot x}} = [{{{I}}_n} \otimes ({{A}} - {{B}}{{{K}}_1}) - {{L}} \otimes {{B}}{{{K}}_2}]{{x}} + ({{M}} \otimes {{C}}){{z}}$ (4)

$\otimes$ 表示克罗内克积(Kronecker product)[32]L为图G的拉普拉斯矩阵，M表示领导者与跟随者的身份，定义如下，vl表示领导者：

 ${{{M}}_{il}} = \left\{ {\begin{array}{*{20}{c}} 1&{{\rm{if\; }}i = l} \\ 0&{\text{其他}} \end{array}} \right.$

 $\dot {{x}} = {\overset{\smile} {{L}}} {{x}} + {\overset{\smile} {{M}}} {{z}}$ (5)

3 主要结论

3.1 一般线性多智能体系统可控的必要性与充分性条件

 $\begin{split} {\rm{ }}{({{x}} \otimes {{y}})^{\rm{T}}}& \;{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over L} }} ={({{x}} \otimes {{y}})^{\rm{T}}}({{{I}}_n} \otimes ({{A}} - {{B}}{{{K}}_1} - {{L}} \otimes {{B}}{{{K}}_2})) = \\ & \quad {({{x}} \otimes {{y}})^{\rm{T}}} \otimes ({{{I}}_n} \otimes ({{A}} - {{B}}{{{K}}_1})) - \\ & \qquad {({{x}} \otimes {{y}})^{\rm{T}}} \otimes ({{L}} \otimes {{B}}{{{K}}_2}) = \\ & \;{{{x}}^{\rm{T}}} \otimes {{{y}}^{\rm{T}}}({{A}} - {{B}}{{{K}}_1}) - {{{x}}^{\rm{T}}}{{L}} \otimes {{{y}}^{\rm{T}}}{{B}}{{{K}}_2} = \\ & \quad {{{x}}^{\rm{T}}} \otimes ({{{y}}^{\rm{T}}}({{A}} - {{B}}{{{K}}_1} - {\textit{λ}} {{B}}{{{K}}_2}) = \\ & \qquad {{{x}}^{\rm{T}}} \otimes ({{{y}}^{\rm{T}}}{\rm{\theta }}) = {({{x}} \otimes {{y}})^{\rm{T}}}{\rm{\theta }} \end{split}$

${\rm{\theta}}$ ${\overset{\smile} {{L}}}$ 的一个特征值并且 ${({{x}} \otimes {{y}})^{\rm{T}}}$ 是相应的左特征向量。有 ${({{x}} \otimes {{y}})^{\rm{T}}}{\overset{\smile} {{M}}} = ({{{x}}^{\rm{T}}}{{M}}) \otimes ({{{y}}^{\rm{T}}}{{C}})$ ，根据PBH判据，即 $({\overset{\smile} {{L}}},{\overset{\smile} {{M}}})$ 不可控，这与已知其可控矛盾，因此 $({{L}},{{M}})$ 可控，证毕。

 $\begin{split} {{\tilde L}} = ({{{U}}^{\rm{T}}} \otimes {{{I}}_p})& {\overset{\smile} {{L}}}({{U}} \otimes {{{I}}_p}) = {\rm blockding}({{A}} - {{B}}{{{K}}_1} - {\lambda _s}{{B}}{{{K}}_2}) \\ & {{\tilde M}} = ({{{U}}^{\rm{T}}} \otimes {{{I}}_p}){\overset{\smile} {{M}}} = {{{U}}^T}{{M}} \otimes {{C}} \end{split}$

$s = \left( {1,2, \cdots ,n} \right)$ 。因为 $({\overset{\smile} {{L}}},{\overset{\smile} {{M}}})$ 不可控且 ${{{U}}^{\rm{T}}} \otimes {{{I}}_p}$ 非奇异，所以 $({{\tilde L}},{{\tilde M}})$ 不可控。又因为 ${{\tilde L}}$ 是一个对角块矩阵，并且矩阵块中无重复特征值导致 $({{\tilde L}},{{\tilde M}})$ 不可控的偶然性情况，则存在一个相应的矩阵对 $({{A}} - {{B}}{{{K}}_1} - {\lambda _s}{{B}}{{{K}}_2},{({{{U}}^{\rm{T}}}{{M}})_s} \otimes {{C}})$ 不可控，这里用Ms表示矩阵M的第s行。这表明 $({{L}},{{M}})$ 在的 ${({{{U}}^{\rm{T}}}{{M}})_s} = 0$ 情况下不可控或在 ${({{{U}}^{\rm{T}}}{{M}})_s} \ne 0$ 的情况下 $({{A}} - {{B}}{{{K}}_1} - \lambda {{B}}{{{K}}_2},{{C}})$ 不可控，出现矛盾，所以 $({\overset{\smile} {{L}}},{\overset{\smile} {{M}}})$ 可控，证毕。

1) 智能体状态矩阵 ${{A}} - {{B}}{{{K}}_1}$ 与系统拉普拉斯矩阵L各自不含有重复特征值且两者无任一特征值相同。

2) 内部耦合矩阵 ${{B}}{{{K}}_2}$ 的元素与智能体的状态矩阵 ${{A}} - {{B}}{{{K}}_1}$ 的元素互为相反数。

3) 智能体状态矩阵 ${{A}} - {{B}}{{{K}}_1}$ 的任意两个特征值之间的倍数不等于拉普拉斯矩阵L的任意一个特征值+1。

3.2 重复特征值导致不充分性的原因

 \begin{aligned} & {{A}} = \left[\!\!\! {\begin{array}{*{20}{c}} {0.5}&1\\ 1&{0.5} \end{array}}\!\!\! \right],{{{K}}_1} = \left[\!\!\! {\begin{array}{*{20}{c}} 2&{1.5}\\ {1.5}&2 \end{array}}\!\!\! \right],{{C}} = \left[\!\!\! {\begin{array}{*{20}{c}} {\rm{1}}\\ {\rm{0}} \end{array}}\!\!\! \right]\\ {{B}} = {{{I}}_2},&{{{K}}_2} = \left[\!\!\! {\begin{array}{*{20}{c}} {1.5}&{0.5}\\ {0.5}&{1.5} \end{array}}\!\!\! \right],{{L}} = \left[\!\!\! {\begin{array}{*{20}{r}} 1&{ - 1}&0\\ { - 1}&2&{ - 1}\\ 0&{ - 1}&1 \end{array}}\!\!\! \right], {{M}} = \left[\!\!\! {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}}\!\!\! \right] \end{aligned}

 ${{M}} = {\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&0&1 \end{array}} \right]^{\rm{T}}}$

3.3 利用特征值重数对系统不可控的判定条件

1） ${\overset{\smile} {{M}}}$ 中只有单个1元素，此时 ${\overset{\smile} {{L}}}$ 出现重复特征值，系统整体不可控。

2) ${\overset{\smile} {{M}}}$ 中有l个1元素，此时 ${\overset{\smile} {{L}}}$ 出现某一特征值代数重数（ $\geqslant$ 3），系统整体不可控。

1）取单领导者，且外部输入控制信号只影响领导者的一种状态。对于一个具有n个个体的多智能体系统，用 $1,2, \cdots n$ 将它们标注。令 ${{A}} - {{B}}{{{K}}_1}$ 为每个智能体的s阶状态矩阵。令第一个智能体为领导者，那么 ${{M}} = {{{e}}_1}$ 。令外部输入控制矩阵 ${{C}} = {{{e}}_i},1 \leqslant i \leqslant s$ ，此时 ${\overset{\smile} {{M}}}$ 共有 $n \times s$ 个元素并且有如下形式：

 ${{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over M} }} = [\begin{array}{*{20}{c}} {\underbrace {0 \cdots 1 \cdots 0}_s}&{\underbrace {0 \cdots 0}_{(n - 1)s}} \end{array}]$

${\overset{\smile} {{L}}}$ 的特征值为 $({{\textit{λ}} _1},{{\textit{λ}} _2}, \cdots ,{{\textit{λ}} _i},\cdots ,{{\textit{λ}} _j},\cdots ,{{\textit{λ}} _{ns}})$ 并且互不相同，其中 ${{\textit{λ}}_j}(1 \leqslant i, j \leqslant ns,i \ne j)$ 的代数重数为2。此时有两个对应于 ${{\textit{λ}}_j}$ 的线性无关的特征向量 ${{{u}}_{j1}}{\text{、}}{{{u}}_{j2}}$ 。对这两个特征向量进行线性组合，得到对应于 ${{\textit{λ}}_j}$ 的一个新的特征向量 ${{{u}}_j}$ ${ u_j} = a{{{u}}_{j1}} + b{{{u}}_{j2}}$ $a,b \in {\rm{R}}$ 并且ab不同时为0。我们总能找到这样一组ab使得 ${{{u}}_j}$ 某一个位置的元素为0，这个位置对应于 ${\overset{\smile} {{M}}}$ 中元素为1的位置。即 ${{{u}}_j}$ 有如下形式：

 ${{{u}}_j} = [\begin{array}{*{20}{c}} {\underbrace { * \cdots \mathop 0\limits_i \cdots * }_s}&{\underbrace { * \cdots * }_{(n - 1)s}} \end{array}]$

2) ${\overset{\smile} {{M}}}$ 中有l个1元素，此时有两种情况。

① 取单领导者系统，外部输入控制信号影响领导者的多种状态。即外部输入控制矩阵有m个元素为1 $(2 \leqslant m \leqslant s)$ ，可以是C中任意m个元素的位置，假设前m个元素为1，则C具有如下形式：

 ${{C}} = [\underbrace {\begin{array}{*{20}{c}} {\underbrace {1 \cdots 1}_m}&{0 \cdots 0} \end{array}}_s]$

 ${\overset{\smile} {{M}}} = [\begin{array}{*{20}{c}} {\underbrace {1 \cdots 1}_m}&{\underbrace {0 \cdots 0}_{s - m}}&{\underbrace {0 \cdots 0}_{(n - 1)s}} \end{array}]$

${\overset{\smile} {{L}}}$ 的特征值为 $({{\textit{λ}} _1}, \cdots ,{{\textit{λ}} _{ns - {\rm{2}}}})$ 并且互不相同，其中 ${{\textit{λ}}_j}(1 \leqslant j \leqslant ns - 2)$ 的代数重数为3，对应于 ${{\textit{λ}}_j}$ 有3 个线性无关的特征向量 ${{{u}}_{j1}}{\text{、}}{{{u}}_{j2}}{\text{、}}{{{u}}_{j3}}$ 。对 ${{{u}}_{j1}}{\text{、}}{{{u}}_{j2}}$ 进行线性组合，得到新的特征向量 ${{{u}}_{j12}} = a{{{u}}_{j1}} + b{{{u}}_{j2}}$ ，选取合适的一组不全为0的数a,b， ${\rm{a}}, {\rm{b}} \in {\rm{R}}$ ，使得 ${{{u}}_{j12}}$ 的第一个元素为0。同样地，分别对 ${{{u}}_{j1}}{\text{、}}{{{u}}_{j{\rm{3}}}}$ ${{{u}}_{j{\rm{2}}}}{\text{、}}{{{u}}_{j{\rm{3}}}}$ 进行线性组合，得到 ${{{u}}_{j{\rm{13}}}}{\text{、}}{{{u}}_{j2{\rm{3}}}}$ ，它们的第一个元素均为0。至此，得到了3个新的对应于 ${\lambda _j}$ 的特征向量，并且它们的第1个元素均为0。现在考虑特征向量的第2个元素。与找第一个元素的方法相同，对这3个特征向量中的任意两个进行线性组合，然后得到另外3个第1和第2个元素都为0的特征向量。重复以上步骤，直到得到一个前m个元素都为0的特征向量，将它记为 ${{{u}}_{\rm need}}$ ，它有如下形式：

 ${{u}}_{\rm need}^{\rm{T}} = [\begin{array}{*{20}{c}} {\underbrace {0 \cdots 0}_m}&{\underbrace { * \cdots * }_{ns - m}} \end{array}]$

② 多领导者系统。对于具有l个领导者的多智能体系统，假设前l个智能体为领导者，分别标注为 ${n_1},{n_2}, \cdots ,{n_l}$ 。那么 ${{M}}$ 具有如下形式：

 ${{M}} = [\begin{array}{*{20}{c}} {{{{e}}_1}}&{{{e}}_2}& \cdots &{{{{e}}_l}} \end{array}]$

 $\begin{array}{l} {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over M} }} = [\begin{array}{*{20}{c}} {{{{m}}_1}}&{{{{m}}_2}}& \cdots &{{{{m}}_l}} \end{array}]\\ {{m}}_{\rm{1}}^{\rm{T}} = [\begin{array}{*{20}{c}} {\underbrace {1 \cdots 1}_m}&{\underbrace {0 \cdots 0}_{s - m}}&{\underbrace {0 \cdots 0}_{(n - 1)s}} \end{array}]\\ {{m}}_2^{\rm{T}} = [\begin{array}{*{20}{c}} {\underbrace {0 \cdots 0}_s}&{\underbrace {1 \cdots 1}_m}&{\underbrace {0 \cdots 0}_{s - m}}&{\underbrace {0 \cdots 0}_{(n - 2)s}} \end{array}]\\ \vdots \\ {{m}}_l^{\rm{T}} = [\begin{array}{*{20}{c}} {\underbrace {0 \cdots 0}_{(n - 1)s}}&{\underbrace {1 \cdots 1}_m}&{\underbrace {0 \cdots 0}_{s - m}} \end{array}] \end{array}$

 $\begin{gathered} {{A}} = \left[\!\!\! {\begin{array}{*{20}{r}} 1&2&0 \\ { - 1}&1&2 \\ 0&3&1 \end{array}} \!\!\!\right],{{{K}}_1} = \left[\!\!\! {\begin{array}{*{20}{r}} { - 1}&3&0 \\ 1&{ - 2}&3 \\ 0&4&{ - 1} \end{array}} \!\!\!\right],{{C}} = \left[\!\!\! {\begin{array}{*{20}{c}} 1 \\ 1 \\ 0 \end{array}}\!\!\! \right],{{B}} = {{{I}}_3} \\ {{{K}}_2} = \left[\!\!\! {\begin{array}{*{20}{r}} { - 2}&1&0 \\ 1&{ - 3}&1 \\ 0&1&{ - 2} \end{array}} \right],{{L}} = \left[ {\begin{array}{*{20}{r}} 1&{ - 1}&0 \\ { - 1}&2&{ - 1} \\ 0&{ - 1}&1 \end{array}} \!\!\!\right],{{M}} = \left[\!\!\! {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}}\!\!\! \right] \\ \end{gathered}$

 ${{M}} = {\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&0&1 \end{array}} \right]^{\rm{T}}}$

${\overset{\smile} {{M}}}$ 中有4个元素为1。此时系统拓扑结构和单个智能体动态均是能控的，但是系统整体并不可控，可控性判别矩阵为8<9。

4 结束语

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