Zhang X, Liu X. Consensus of linear multi-agent systems with exogenous disturbance generated from heterogeneous exosystems[J]. International Journal of Systems Science, 2017, 48(15): 3147-3159.

文章目录

  • 1. Introduction
  • 2. Preliminaries and problem formulation
    • 2.1. Graph theory
    • 2.2. Input-to-state stability and some useful lemmas
    • 2.3. Problem formulation
  • 3. Consensus analysis with exogenous disturbances generated by linear heterogeneous exosystems
  • 4. Consensus analysis with exogenous disturbances generated by non-linear heterogeneous exosystems
  • 5. Simulation examples

1. Introduction

2. Preliminaries and problem formulation

2.1. Graph theory

2.2. Input-to-state stability and some useful lemmas

2.3. Problem formulation

领航者为
x˙0=Ax0(5)\dot{x}_0 = A x_0 \tag{5}x˙0​=Ax0​(5)

跟随者为
x˙i=Axi+B(ui+di)(6)\begin{aligned} \dot{x}_i = A x_i + B (u_i + d_i) \end{aligned} \tag{6}x˙i​=Axi​+B(ui​+di​)​(6)

3. Consensus analysis with exogenous disturbances generated by linear heterogeneous exosystems

干扰信号 did_idi​ 由以下动力学产生
ξ˙i=A0iξidi=C0iξi(8)\begin{aligned} \dot{\xi}_i = A_{0i} \xi_i \\ d_i = C_{0i} \xi_i \end{aligned} \tag{8}ξ˙​i​=A0i​ξi​di​=C0i​ξi​​(8)

观测器为
z˙i=(A0i−KiBC0i)(zi+Kixi)−Ki(Axi+Bui)ξ^i=zi+Kixid^i=C0iξ^i(9)\begin{aligned} \dot{z}_i &= (A_{0i}-K_iBC_{0i}) (z_i + K_i x_i) - K_i(Ax_i + Bu_i) \\ \hat{\xi}_i &= z_i + K_i x_i \\ \hat{d}_i &= C_{0i} \hat{\xi}_i \end{aligned} \tag{9}z˙i​ξ^​i​d^i​​=(A0i​−Ki​BC0i​)(zi​+Ki​xi​)−Ki​(Axi​+Bui​)=zi​+Ki​xi​=C0i​ξ^​i​​(9)

控制器为
ui=cF[∑j=1Naij(xj−xi)+gi(x0−xi)]−C0iξ^i(15)\begin{aligned} u_i &= cF \left[ \sum_{j=1}^N a_{ij} (x_j - x_i) + g_i(x_0 - x_i) \right] - C_{0i} \hat{\xi}_i \end{aligned} \tag{15}ui​​=cF[j=1∑N​aij​(xj​−xi​)+gi​(x0​−xi​)]−C0i​ξ^​i​​(15)

4. Consensus analysis with exogenous disturbances generated by non-linear heterogeneous exosystems

干扰信号 did_idi​ 由以上动力学产生
ξ˙i=A0iξi+ϕi(ξi)di=C0iξi(35)\begin{aligned} \dot{\xi}_i &= A_{0i} \xi_i \red{+ \phi_i(\xi_i)} \\ d_i &= C_{0i} \xi_i \end{aligned} \tag{35}ξ˙​i​di​​=A0i​ξi​+ϕi​(ξi​)=C0i​ξi​​(35)

观测器为
z˙i=(A0i−KiBC0i)(zi+Kixi)+ϕi(zi+Kixi)−Ki(Axi+Bui)+Qζi−1ζieiζ˙i=eiTeiξ^i=zi+Kixid^i=C0iξ^i(37)\begin{aligned} \dot{z}_i &= (A_{0i}-K_iBC_{0i}) (z_i + K_i x_i) \red{+\phi_i(z_i + K_i x_i)} - K_i(Ax_i + Bu_i) + \red{Q^{-1}_{\zeta_i} \zeta_i e_i} \\ \red{\dot{\zeta}_i} &\red{= e_i^{\text{T}} e_i} \\ \hat{\xi}_i &= z_i + K_i x_i \\ \hat{d}_i &= C_{0i} \hat{\xi}_i \end{aligned} \tag{37}z˙i​ζ˙​i​ξ^​i​d^i​​=(A0i​−Ki​BC0i​)(zi​+Ki​xi​)+ϕi​(zi​+Ki​xi​)−Ki​(Axi​+Bui​)+Qζi​−1​ζi​ei​=eiT​ei​=zi​+Ki​xi​=C0i​ξ^​i​​(37)

论文中的观测器(37),写出来的结果无法运行,因为维度都不匹配。经过我的测试,感觉应该是少了个负号,我给补充上去了,结果就变好了。

ei=ξi−ξ^i(38)\begin{aligned} {e}_i &= \xi_i - \hat{\xi}_i \end{aligned} \tag{38}ei​​=ξi​−ξ^​i​​(38)

5. Simulation examples

Example 5.1

对应程序为 Main_Ex51.m
效果如下:


Example 5.2

对应程序为 Main_Ex52.m
效果如下:

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