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摘要: 针对一类考虑模型非仿射特性和执行机构饱和特性的高超声速飞行器轨迹跟踪控制问题, 提出一种基于backstepping的输出反馈非线性控制方法. 考虑执行机构故障激发的未知非线性动态, 建立了非仿射形式飞行器模型. 为解决实际工程应用中存在的气流角测量值难以使用的问题, 利用高度和速度测量值以及高阶微分器设计了航迹倾角在线估计方法. 基于跟踪微分器设计了模型干扰项的估计方法, 并解决了backstepping方法应用中存在的“微分项爆炸”问题. 引入辅助系统降低控制量饱和带来的不利影响. 基于Lyapunov理论证明了闭环系统的稳定性. 最后, 通过对比仿真实验验证了所提方法的有效性.Abstract: A backstepping-based nonlinear control method is proposed for addressing the trajectory tracking control problem of a non-affine hypersonic vehicle subject to actuator saturation. Considering the unknown nonlinear dynamics invoked by actuator faults, the non-affine model of the flight vehicle is established. For tackling the unavailability of the measurements of the flow angles in practical engineering application, an on-line estimation method for flight path angle is designed utilizing the measurements of velocity and altitude as well as the high-order differentiator. The tracking differentiator is employed to construct the estimation methods for model disturbances and tackle the explosion of differentiation problem existing in the application of backstepping. The auxiliary systems are introduced to reduce the adverse effects caused by the saturation of control inputs. The stability of closed-loop system is proved based on the Lyapunov theory. Finally, the effectiveness of the proposed method is verified via comparative simulation experiments.
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Key words:
- Hypersonic vehicle /
- non-affine /
- output feedback /
- state estimation /
- input constraint /
- tracking differentiator
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图 11 本文方法航迹倾角子系统控制指令曲线
Fig. 11 Curves of flight path angle subsystem control commands under proposed method
图 12 本文方法总干扰项估计曲线
Fig. 12 Estimated curves of the lumped disturbances under proposed method
图 13 本文方法抗饱和补偿信号曲线
Fig. 13 Curves of anti-windup compensated signals under proposed method
表 1 控制量的容许范围
Table 1 Admissible ranges for control inputs
飞行状态/控制量 容许范围 $\phi $ [0.1, 1.2] ${\delta _{\rm{e}}}$ [−15°, 15°] ${\delta _{\rm{c}}}$ [−27°, 27°] 
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表 2 控制器参数设置
Table 2 Parameter settings of controllers
控制方法 控制器参数 本文方法 $\begin{aligned} {k_V} = 5,{k_\gamma } = 1.5,{k_\theta } = {k_Q} = 5,{k_1} = 10,{k_2} = 10, \\ {R_H} = 100,{a_{H1} } = {a_{H2} } = 6,{a_{H3} } = 4,{R_V} = 100,\;\; \\ {R_\theta } = {R_{Qc} } = 10,{R_Q} = 50,{a_{V1} } = {a_{V2} } = 5,\;\;\;\; \qquad\\ {a_{\theta 1} } = {a_{Qc1} } = {a_{Q1} } = {a_{\theta 2} } = {a_{Qc2} } = {a_{Q2} } = 5.\qquad \end{aligned}$ 对比方法[17] $\begin{aligned} { {k'}_1} = 5,{k_1} = 1.5,{k_2} = {k_3} = 5,\varepsilon = \varepsilon ' = 0.25, \;\;\;\;\;\\ \iota = \iota ' = 0.5,{M_0} = 15,{ {M'}_0} = 1.2, \qquad\qquad\qquad\;\;\;\;\\ {\mu _{i1} } = {\mu _{i2} } = 1,i = 2,3,{\theta _{j1} } = {\theta _{j2} } = 1,j = 1,2,3, \\ \theta _1^{'} = \theta _2^{'} = 10,\Delta {t_1} = \Delta {t_2} = \Delta {t_3} = 0.2.\qquad\qquad\;\; \end{aligned}$
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