【Paper】2019_带有不匹配干扰的多智能体系统有限时间积分滑模控制
LIU Fan, YANG Hong-Yong, YANG Yi-Ze, LI Yu-Ling, LIU Yuan-Shan. Finite-Time Integral Sliding-Mode Control for Multi-Agent Systems With Mismatched Disturbances. ACTA AUTOMATICA SINICA, 2019, 45(4): 749-758. doi: 10.16383/j.aas.c180315
文章目錄
- 1 預備知識
- 引理 1
- 2 二階多智能體系統(tǒng)的有限時間包容控制
- 3 帶有不匹配干擾的多智能體系統(tǒng)的有限時間包容控制
- 3.1 非線性干擾觀測器設計
- 3.2 復合式分布式控制律設計
- 4 數(shù)值仿真
1 預備知識
引理 1
2 二階多智能體系統(tǒng)的有限時間包容控制
3 帶有不匹配干擾的多智能體系統(tǒng)的有限時間包容控制
假設二階受擾多智能體系統(tǒng)的動力學模型為
{x˙i(t)=vi(t)+di1(t)v˙i(t)=ui(t)+di2(t)(9)\left\{\begin{aligned} \dot{x}_i(t) &= v_i(t) + d_{i1}(t) \\ \dot{v}_i(t) &= u_i(t) + d_{i2}(t) \\ \end{aligned}\right. \tag{9}{x˙i?(t)v˙i?(t)?=vi?(t)+di1?(t)=ui?(t)+di2?(t)?(9)
領導者的動力學模型為
{x˙j(t)=vj(t)v˙j(t)=0(10)\left\{\begin{aligned} \dot{x}_j(t) &= v_j(t) \\ \dot{v}_j(t) &= 0 \\ \end{aligned}\right. \tag{10}{x˙j?(t)v˙j?(t)?=vj?(t)=0?(10)
3.1 非線性干擾觀測器設計
根據(jù)引理 6,設計干擾觀測器如下:
{x^˙i(t)=vi+zi1zi1=?λi1sig23(x^i?xi)+d^i1d^˙i1=?λi2sig12(d^i1?zi1)v^˙i(t)=vi+zi2zi2=?λi3sig23(v^i?vi)+d^i2d^˙i2=?λi4sig12(d^i2?zi2)(13)\left\{\begin{aligned} \dot{\hat{x}}_i(t) &= v_i + z_{i1} \\ z_{i1} &= -\lambda_{i1} ~ \text{sig}^{\frac{2}{3}}(\hat{x}_i - x_i) + \hatvt6mr5x_{i1} \\ \dot{\hatvt6mr5x}_{i1} &= -\lambda_{i2} ~ \text{sig}^{\frac{1}{2}}(\hatvt6mr5x_{i1} - z_{i1}) \\\\ \dot{\hat{v}}_i(t) &= v_i + z_{i2} \\ z_{i2} &= -\lambda_{i3} ~ \text{sig}^{\frac{2}{3}}(\hat{v}_i - v_i) + \hatvt6mr5x_{i2} \\ \dot{\hatvt6mr5x}_{i2} &= -\lambda_{i4} ~ \text{sig}^{\frac{1}{2}}(\hatvt6mr5x_{i2} - z_{i2}) \\ \end{aligned}\right. \tag{13}??????????????????????????????x^˙i?(t)zi1?d^˙i1?v^˙i?(t)zi2?d^˙i2??=vi?+zi1?=?λi1??sig32?(x^i??xi?)+d^i1?=?λi2??sig21?(d^i1??zi1?)=vi?+zi2?=?λi3??sig32?(v^i??vi?)+d^i2?=?λi4??sig21?(d^i2??zi2?)?(13)
3.2 復合式分布式控制律設計
ui=?k0sgn(∑j=1naij(si?sj)+∑j=n+1n+maijsi)?k1sigα1(ωix)?k2sigα2(ωiv)?d^i2(16)\begin{aligned} u_i &= -k_0 \text{sgn} (\sum_{j=1}^n a_{ij} (s_i - s_j) + \sum_{j=n+1}^{n+m} a_{ij} s_i) \\ &- k_1 \text{sig}^{\alpha_1} (\omega_i^x) - k_2 \text{sig}^{\alpha_2} (\omega_i^v) - \hatvt6mr5x_{i2} \end{aligned}\tag{16}ui??=?k0?sgn(j=1∑n?aij?(si??sj?)+j=n+1∑n+m?aij?si?)?k1?sigα1?(ωix?)?k2?sigα2?(ωiv?)?d^i2??(16)
4 數(shù)值仿真
以下是論文中的原圖與我復現(xiàn)的結果圖的對比
總結
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