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摘要: 面向空间攻防等任务的航天器通常安装微波、激光等大功率对抗载荷, 未来航天器需要装备大型挠性太阳能帆板. 针对挠性航天器姿态机动过程中存在外部干扰、执行机构饱和及挠性附件振动且挠性模态不易直接测量等问题, 提出带挠性附件航天器的全驱姿态控制方法. 首先, 建立挠性航天器全驱姿态控制模型. 其次, 基于扩展非线性观测器(Extended nonlinearity observer, ENO)与努斯鲍姆增益调节设计一种抗饱和的姿态控制鲁棒算法. 将外部扰动、挠性振动和输入饱和函数饱和估计误差作为复合干扰, 采用非线性干扰观测器对其进行有效补偿. 在直接参数设计线性控制参数基础上, 扩展非线性观测器负责对挠性航天器产生的挠性振动等非线性进行实时估计和补偿, 努斯鲍姆函数辅助控制器输出力矩避免饱和, 并利用李雅普诺夫方法严格证明闭环系统的稳定性. 最后通过数学仿真验证该方法不仅能够实现执行机构饱和约束条件下的姿态控制, 还能有效抑制挠性结构的振动, 为探索未来带有大型挠性附件航天器姿态控制新的方法提供参考.Abstract: For space missions such as space attack and defense, spacecrafts are usually equipped with high-power counter loads such as microwaves and lasers, thus future spacecrafts will need to be equipped with large-scale flexible solar panels. To solve the problems of external disturbances, actuator saturation, and vibrations of flexible attachments difficult to measure directly during attitude maneuver of flexible spacecraft, a fully actuated-based attitude control method for flexible spacecraft is proposed. First, a fully actuated attitude model for flexible spacecraft is established. Second, a robust attitude control algorithm against saturation is designed based on the extended nonlinear observer (ENO) and Nussbaum gain adjustment. The external disturbances, flexible vibrations, and the approximation error of the input saturation function are considered composite disturbances, which are efficiently compensated by the nonlinear disturbance observer. Based on the direct parametric design of linear control parameters, the ENO is responsible for the real-time estimation and compensation of nonlinearities such as flexible vibrations generated by the flexible spacecraft. And the Nussbaum function is used for the adjustment of the controller output torque magnitude to avoid actuator saturation. The stability of the closed-loop system is rigorously demonstrated using the Lyapunov method. Finally, it is verified by mathematical simulation that this method can not only realize the attitude control under the actuator saturation constraint but also effectively suppress the flexible vibrations, which contributes to exploring a novel method for the attitude control of spacecraft with large-scale flexural attachments in the future.
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表 1 仿真参数
Table 1 Simulation parameters
物理参数 值 转动惯量矩阵$ (\text{kg}\cdot\text{m}^{2}) $ ${\boldsymbol{J} }=\text{diag}\{40,\;150,\;160\}$ 耦合矩阵 $ {\boldsymbol{\delta}}= \left[\begin{array}{*{20}{r}} 1.352\ 3 & 1.278\ 4 & 2.155\ 3\\-1.151\ 9 & 1.017\ 6 & -1.272\ 4\\2.216\ 7 & 1.589\ 1 & -0.832\ 4\\1.236\ 4 & -1.653\ 7 & 1.225\ 1\end{array}\right] $ 挠性模态数 $ N=4 $ 固有频率(rad/s) $ {\boldsymbol{\Lambda}}=[1.20, \; 2.48 ,\; 3.37, \; 7.47] $ 阻尼比 $ \xi_1 = \xi_2 = \xi_3 = \xi_4 = 0.01 $ 外部干扰矩阵(N·m) $ d=10^{-4}\begin{bmatrix}3\cos(0.1t)+4\\1.5\sin(0.1t)+3\cos(0.1t)\\3\sin(0.1t)+1\end{bmatrix} $ 表 2 各控制器参数
Table 2 Parameters of controllers
控制方法 控制律 控制参数 PD + 前馈补偿 $ {\boldsymbol{u}} = {{\boldsymbol{B}}^{ - 1}}({{\boldsymbol{A}}_p}{\boldsymbol{\sigma}} {\rm{ + }}{{\boldsymbol{A}}_d}\dot{{\boldsymbol{\sigma}}} ) - {{\boldsymbol{B}}^{ - 1}}({{\boldsymbol{f}}_{\rm{2}}} + {{\boldsymbol{f}}_{\rm{3}}}) $ $\begin{array}{l} { {\boldsymbol{A} }_p} = {\rm{diag}}\{ - 0 .12 ,\; - 0.12 ,\; - 0.12 \}\\ { {\boldsymbol{A} }_d} = {\rm{diag\{ - 0.7 ,\; - 0.7,\; - 0.7} }\} \end{array}$ TSMC $ {\boldsymbol{u}} = - { {{\beta q} \over p}}{{\boldsymbol{B}}^{ - 1}}{\dot{{\boldsymbol{\sigma}}} ^{{\rm{(2}} - { {p \over q}}{\rm{)}}}} - {{\boldsymbol{B}}^{ - 1}}({{\boldsymbol{f}}_{\rm{2}}} + {{\boldsymbol{f}}_{\rm{3}}}) + \varepsilon \text{sgn} (s) $ $ \beta = 0.8,\;\varepsilon = 0.002,\; p = 5,\;q = 3 $ FAESO $ {\boldsymbol{u}} {\rm{ = }}{{\boldsymbol{B}}^{ - 1}}{\boldsymbol{Z}}{{\boldsymbol{F}}^2}{{\boldsymbol{V}}^{ - 1}}{[\sigma, {\rm{ }}\dot \sigma ]^{\rm{T}}} - {{\boldsymbol{B}}^{ - 1}}{\rm{(}}{\hat {{\boldsymbol{z}}}_3}{\rm{ + }}{{\boldsymbol{f}}_1}{\rm{)}} $ $\begin{array}{c} {\boldsymbol{F} } = \left[ \begin{array}{l} {\rm{diag} }\{ - 0.09,\; - 0.1,\; - 0.08\}\quad\ { {\boldsymbol{O} }_{3 \times 3} } \\ { {\boldsymbol{O} }_{3 \times 3{\rm{ } } } }\quad {\rm{diag} }\{ - 0.35,\; - 0.4,\; - 0.35\} \end{array} \right] \\ {\boldsymbol{Z} } = [{ {\boldsymbol{I} }_{3 \times 3} },\;\ { {\boldsymbol{I} }_{3 \times 3} }] \\ \left[\beta_1,\; \beta_2,\; \beta_3\right]=\left[50,\; 150,\; 500\right] \end{array}$ 本文方法 式(15)$ \sim $式(17) $\begin{array}{c} {\boldsymbol{F} } = \left[ \begin{array}{l} {\rm{diag} }\{ - 0.09,\; - 0.1,\; - 0.08\}\; { {\boldsymbol{O} }_{3 \times 3} }\\{ {\boldsymbol{O} }_{3 \times 3{\rm{ } } } }\; {\rm{diag} }\{ - 0.35,\; - 0.4,\; - 0.35\} \end{array} \right]\\{\boldsymbol{P} } = \left[ {\begin{array}{*{20}{c} } { {\rm{diag} }\{35,\;35,\;35\}\; {\rm{diag} }\{80,\;80,\;70\}}\\{ {\rm{diag} }\{80,\;80,\;80\}\;{\rm{diag} }\{150,\;150,\;150\}} \end{array} } \right]\\kv = 0.8,\;k = 0.6,\;\varsigma = 0.6,\; {\boldsymbol{Z} } = [{ {\boldsymbol{I} }_{3 \times 3} }{\rm{ } },\; { {\boldsymbol{I} }_{3 \times 3} }],\; {\boldsymbol{\chi} } (0){\rm{ = } }{\bf{0} }\\ \left[\beta_1,\; \beta_2,\; \beta_3\right]=\left[50,\; 150,\; 500\right],\;{ {\boldsymbol{z} }_{\rm{1} } } = {\bf{0} },\; { {\boldsymbol{z} }_{\rm{2} } } = {\bf{0} },\; { {\boldsymbol{z} }_{\rm{3} } } = {\bf{0} } \end{array}$ -
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