吴利平,郭毓,吴益飞,郭健
(南京理工大学 自动化学院, 南京 210094)
摘要:
为实现太阳帆三轴姿态控制,采用一种新型的帆板-滑块执行机构进行姿态控制.基于滑模控制理论提出一种强鲁棒的姿态控制器,以抑制执行机构工作过程中航天器转动惯量变化对姿态控制的影响.此外,引入自适应律,提出一种自适应抗扰控制律,以抑制光压力矩和引力梯度力矩对姿态的干扰作用.最后,基于执行机构的动力学特性设计了操纵律,解算出帆板转动角度和滑块滑动位移,提供给控制器所需控制力矩.仿真结果表明:采用所提控制律和执行机构操纵律可使太阳帆姿态较快地机动至期望位置,并较好地抑制了转动惯量变化、光压力矩干扰和引力梯度力矩干扰带来的影响,同时使控制力矩、帆板角度和滑块位移均保持在适当的幅值范围内.所提控制策略有效地实现了太阳帆三轴姿态控制.
关键词: 太阳帆 姿态 执行机构 强鲁棒 自适应
DOI:10.11918/j.issn.0367-6234.201707072
分类号:TP273;V488.2
文献标识码:A
基金项目:国家自然科学基金(1,4, 61673219);江苏省重点研发计划(BE2015164, BE2017161);江苏省学术学位研究生创新计划(KYZZ16_0189)
Attitude control for solar-sail using rotating panel-sliding mass actuator
WU Liping,GUO Yu,WU Yifei,GUO Jian
(School of Automation, Nanjing University of Science and Technology, Nanjing 210094, China)
Abstract:
In order to realize the three-axis attitude control of solar sail, this paper employed a new type of actuator, rotating panel-sliding mass, to study its control strategy. A strong robust attitude controller was proposed based on the sliding mode control theory to reduce the influence of the variable inertia moment of the spacecraft on the attitude system. In addition, the adaptive law was introduced and an adaptive disturbance-rejection control law was proposed to suppress the disturbances caused by the solar radiation pressure torque and gravitational gradient torque. Lastly, a steering law was designed for the actuator based on its dynamics to supply the control torque required by the controller, then the rotation angles of rotating panels and sliding displacements of sliding masses were calculated using the steering law. The simulation results showed that the proposed control law and the actuator steering law drove the attitude of solar sail maneuver to desired position quickly, and well resisted the influences of the variation of the inertia moment and the disturbances from solar radiation pressure torque and the gravitational gradient torque. The control torques, the angles of the rotating panels, and the displacements of the sliding masses were kept in proper amplitude range as well. The proposed control strategy effectively realized the three-axis attitude control of solar sail.
Key words: solar sail attitude control actuator strong robust adaptive
吴利平, 郭毓, 吴益飞, 郭健. 采用帆板-滑块执行结构的太阳帆姿态控制[J]. 哈尔滨工业大学学报, 2018, 50(9): 141--1. DOI: 10.11918/j.issn.0367-6234.201707072.
![复制到剪切板](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/copy.gif)
WU Liping, GUO Yu, WU Yifei, GUO Jian. Attitude control for solar-sail using rotating panel-sliding mass actuator[J]. Journal of Harbin Institute of Technology, 2018, 50(9): 141--1. DOI: 10.11918/j.issn.0367-6234.201707072.
![复制到剪切板](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/copy.gif)
基金项目 国家自然科学基金(61773211, 61673214, 61673219);江苏省重点研发计划(BE2015164, BE2017161);江苏省学术学位研究生创新计划(KYZZ16_0189) 作者简介 吴利平(1990—),女,博士研究生;
郭毓(1964—),女,教授,博士生导师 通信作者 郭毓,guoyu@njust.edu.cn 文章历史 收稿日期: 2017-07-13
Contents -->Abstract Full text Figures/Tables PDF
采用帆板-滑块执行结构的太阳帆姿态控制
吴利平, 郭毓
![通讯作者](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/REcor.gif)
![guoyu@njust.edu.cn](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/REemail.gif)
南京理工大学 自动化学院,南京 210094
收稿日期: 2017-07-13
基金项目: 国家自然科学基金(61773211, 61673214, 61673219);江苏省重点研发计划(BE2015164, BE2017161);江苏省学术学位研究生创新计划(KYZZ16_0189)
作者简介: 吴利平(1990—),女,博士研究生;
郭毓(1964—),女,教授,博士生导师
通信作者: 郭毓,guoyu@njust.edu.cn
摘要: 为实现太阳帆三轴姿态控制,采用一种新型的帆板-滑块执行机构进行姿态控制.基于滑模控制理论提出一种强鲁棒的姿态控制器,以抑制执行机构工作过程中航天器转动惯量变化对姿态控制的影响.此外,引入自适应律,提出一种自适应抗扰控制律,以抑制光压力矩和引力梯度力矩对姿态的干扰作用.最后,基于执行机构的动力学特性设计了操纵律,解算出帆板转动角度和滑块滑动位移,提供给控制器所需控制力矩.仿真结果表明:采用所提控制律和执行机构操纵律可使太阳帆姿态较快地机动至期望位置,并较好地抑制了转动惯量变化、光压力矩干扰和引力梯度力矩干扰带来的影响,同时使控制力矩、帆板角度和滑块位移均保持在适当的幅值范围内.所提控制策略有效地实现了太阳帆三轴姿态控制.
关键词: 太阳帆 姿态 执行机构 强鲁棒 自适应
Attitude control for solar-sail using rotating panel-sliding mass actuator
WU Liping, GUO Yu
![Corresponding author](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/REcor.gif)
![guoyu@njust.edu.cn](http://hit.alljournals.cn/html/hitxb_cn/2018/9/images/REemail.gif)
School of Automation, Nanjing University of Science and Technology, Nanjing 210094, China
Abstract: In order to realize the three-axis attitude control of solar sail, this paper employed a new type of actuator, rotating panel-sliding mass, to study its control strategy. A strong robust attitude controller was proposed based on the sliding mode control theory to reduce the influence of the variable inertia moment of the spacecraft on the attitude system. In addition, the adaptive law was introduced and an adaptive disturbance-rejection control law was proposed to suppress the disturbances caused by the solar radiation pressure torque and gravitational gradient torque. Lastly, a steering law was designed for the actuator based on its dynamics to supply the control torque required by the controller, then the rotation angles of rotating panels and sliding displacements of sliding masses were calculated using the steering law. The simulation results showed that the proposed control law and the actuator steering law drove the attitude of solar sail maneuver to desired position quickly, and well resisted the influences of the variation of the inertia moment and the disturbances from solar radiation pressure torque and the gravitational gradient torque. The control torques, the angles of the rotating panels, and the displacements of the sliding masses were kept in proper amplitude range as well. The proposed control strategy effectively realized the three-axis attitude control of solar sail.
Keywords: solar sail attitude control actuator strong robust adaptive
由于无能耗的推进方式,太阳帆航天器在深空飞行中倍受青睐[1-3].太阳帆通过巨大的帆面反射太阳光,从而获得光压力作为轨道动力.调整帆面姿态可改变其所受光压力的幅值与方向,进而改变轨道推进力.太阳帆姿控执行机构巧妙地借助太阳光压力提供姿态控制力矩,实现了无能耗姿态控制.由于结构简单、可行性强,滑动质量块形式的执行机构受到了广泛关注[4-7].然而此方法只能产生俯仰轴和偏航轴控制力矩,需要配合滚转轴执行机构滚使用.目前所采用的滚转轴执行方案有控制小帆和转轴稳定条等,但它们对太阳帆的展开过程影响较大.文献[5]采用小块帆板的转动产生滚转轴控制力矩,其分离式的安装对帆面展开影响小,是一种新型高效的执行机构.
然而,质量块滑动会改变太阳帆质心位置,在提供控制力矩的同时也致使转动惯量发生变化,这就要求太阳帆姿态控制器具有较强的鲁棒性.此外,光压力矩对太阳帆也有着不可忽视的影响,相同质心/压心偏差的情况下,其所受光压力矩幅值是传统航天器的近100倍;在诸如三体问题等复杂飞行环境下,天体对太阳帆的引力产生的梯度力矩对其姿态也构成了显著干扰,且随轨道位置变化.此类外部干扰力矩要求姿态控制器具有较好的自适应抗扰能力.
许多学者在太阳帆姿态控制设计方面进行了大量研究.文献[6]针对以滑块为执行机构的太阳帆计了2自由度的姿态控制律,不记转动惯量变化设计了前馈+反馈的控制策略.文献[7]同样以滑动质量块作为执行机构,忽略其对转动惯量的影响,分别采用LQR和PID算法为太阳帆设计了双闭环姿态控制器.文献[8-9]都采用顶端小帆方式产生姿态控制力矩,且在控制器设计时皆采用PD控制律,通过选取姿态和转速的反馈增益矩阵实现姿态控制.虽然PID等线性反馈控制方法具有结构简单且应用成熟等特点,但是针对上述帆板-滑块结构的太阳帆,在采用此类方法进行姿控设计时,较难快速选取合适参数,且控制器往往不能获得较强的鲁棒性和自适应能力.
本文采用帆板-滑块结构作为太阳帆姿控执行器,针对该结构工作过程中引起的惯量变化问题,同时考虑航天器所受外部干扰,采用非线性设计方法研究具有强鲁棒性和自适应能力的姿态控制律.然后,考虑执行机构力学特性,解算帆板转角和滑块位置,以实现所提控制策略.
1 问题描述 1.1 姿态动力学与运动学本文以方形太阳帆作为研究对象,为描述其姿态,取太阳帆几何中心o为原点建立本体坐标系,如图 1所示. xb轴沿帆面法线n方向,yb轴沿帆面某一对角线方向,zb轴方向符合右手准则.选用3-2-1转序,采用四元数Q描述太阳帆姿态,则系统运动学与动力学方程分别为
$\mathit{\boldsymbol{\dot Q}} = \frac{1}{2}\left[ \begin{array}{l}{q_4}{\mathit{\boldsymbol{I}}_3} + {\mathit{\boldsymbol{q}}^ \times }\\ - {\mathit{\boldsymbol{q}}^{\rm{T}}}\end{array} \right]\mathit{\boldsymbol{\omega }}, $ (1)
$\mathit{\boldsymbol{J\dot \omega }} + \mathit{\boldsymbol{\omega }} \times \mathit{\boldsymbol{J\omega }} = {\mathit{\boldsymbol{\tau }}_{\rm{c}}} + {\mathit{\boldsymbol{\tau }}_{\rm{d}}}.$ (2)
式中:
干扰力矩主要包括光压干扰力矩τs和引力梯度力矩τg,其中τs=ε0×Fs,ε0为生产工艺导致的质心/压心偏差,Fs为帆面所受太阳光压力.采用理想光学模型,有
${\mathit{\boldsymbol{F}}_{\rm{s}}} = {\left[ {\begin{array}{*{20}{c}}{2{P_{\rm{s}}}A{{\cos }^2}\alpha }&0&0\end{array}} \right]^{\rm{T}}}.$
式中:Ps为一个天文单位处的光压辐射常数,A为太阳帆帆面面积,α为帆面法向量与阳光矢量的夹角,也为太阳帆俯仰角,可通过姿态四元数转换而得.此外,引力梯度力矩由附近天体对太阳帆的引力产生,由于姿态机动时间远小于轨道飞行时间,引力梯度力矩在每个轨道位置可视为常值.
1.2 执行机构动力学帆板-滑块形式的执行机构是文献[5]提出的一类分离式安装的执行机构.长方形帆板pi和质量块Mi(i=1, 2, 3, 4)的安装,如图 1所示,4个帆板/质量块规格一致.工作时,帆板沿对应连接杆转动,连接杆长为l,转角为γi(i=1, 2, 3, 4);质量块沿帆面对角线滑动,记其相对o点的位移为di(i=1, 2, 3, 4).
Figure 1
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-1.jpg)
不妨以帆板p3为例,分析移动帆板的工作原理.以帆板质心op为原点建立帆板坐标系oxpypzp,yp轴和zp轴分别平行于帆板两边,xp轴沿帆板法线方向.工作时,p3沿zp轴转动γ3角度,如图 2所示. oxpypzp系与oxbybzb系的变换关系为
$\left[ {\begin{array}{*{20}{c}}{{\mathit{\boldsymbol{x}}_{\rm{b}}}}\\{{\mathit{\boldsymbol{y}}_{\rm{b}}}}\\{{\mathit{\boldsymbol{z}}_{\rm{b}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos {\gamma _3}}&{\sin {\gamma _3}}&0\\{ - \sin {\gamma _3}}&{\cos {\gamma _3}}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{\mathit{\boldsymbol{x}}_{\rm{p}}}}\\{{\mathit{\boldsymbol{y}}_{\rm{p}}}}\\{{\mathit{\boldsymbol{z}}_{\rm{p}}}}\end{array}} \right].$
Figure 2
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-2.jpg)
在oxpypzp系中,p3反射太阳光获得的光压力为
${\mathit{\boldsymbol{F}}_{{\rm{p3}}}} = 2{P_{\rm{s}}}{A_{\rm{p}}}{\cos ^2}\left( {\alpha + {\gamma _3}} \right){\mathit{\boldsymbol{x}}_{\rm{p}}}, $
式中Ap为帆板面积,变换至坐标系oxoyozo中有
${\mathit{\boldsymbol{F}}_{{\rm{p3}}}} = \left[ {\begin{array}{*{20}{c}}{2{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _3}} \right)\cos {\gamma _3}}\\{ - 2{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _3}} \right)\sin {\gamma _3}}\\0\end{array}} \right].$
设op点到o点的距离为l3=lzb,Fp3产生的力矩为
${\mathit{\boldsymbol{\tau }}_{{\rm{p3}}}} = {\mathit{\boldsymbol{l}}_3} \times {\mathit{\boldsymbol{F}}_{{\rm{p3}}}} = \left[ {\begin{array}{*{20}{c}}{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _3}} \right)\sin {\gamma _3}}\\{2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _3}} \right)\cos {\gamma _3}}\\0\end{array}} \right].$
同理,帆板1、2、4产生的力矩分别为
${\mathit{\boldsymbol{\tau }}_{{\rm{p1}}}} = \left[ {\begin{array}{*{20}{c}}{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _1}} \right)\sin {\gamma _1}}\\0\\{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _1}} \right)\cos {\gamma _1}}\end{array}} \right], $
${\mathit{\boldsymbol{\tau }}_{{\rm{p2}}}} = \left[ {\begin{array}{*{20}{c}}{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _2}} \right)\sin {\gamma _2}}\\0\\{2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _2}} \right)\cos {\gamma _2}}\end{array}} \right], $
${\mathit{\boldsymbol{\tau }}_{{\rm{p4}}}} = \left[ {\begin{array}{*{20}{c}}{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _4}} \right)\sin {\gamma _4}}\\{ - 2l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + {\gamma _4}} \right)\cos {\gamma _4}}\\0\end{array}} \right].$
令γ1=γ2,τp1+τp2=-4lPsApcos2(α+γ1)sinγ1xb;令γ3=γ4,τp3+τp4=-4lPsApcos2(α+γ3)sinγ3xb.可见4块帆板力矩之和只产生滚转轴力矩.俯仰轴和偏航轴控制力矩由滑块滑动提供,结合文献[5]可得执行机构输出的控制力矩为
${\mathit{\boldsymbol{\tau }}_{\rm{c}}} = \left[ {\begin{array}{*{20}{c}}{ - 4l{P_{\rm{s}}}{A_{\rm{p}}}\sum\limits_{i = 1, 3} {{{\cos }^2}\left( {\alpha + {\gamma _i}} \right)\sin {\gamma _i}} }\\{ - 2m/{m_{\rm{t}}}\left( {{d_3} + {d_4}} \right){P_{\rm{s}}}A{{\cos }^2}\alpha }\\{2m/{m_{\rm{t}}}\left( {{d_1} + {d_2}} \right){P_{\rm{s}}}A{{\cos }^2}\alpha }\end{array}} \right].$ (3)
式中m为滑块质量,mt为航天器总质量.
执行机构工作时,由于质量块滑动会引起质心变化,转动惯量J也随之变化.设J=diag(Jxb,Jyb,Jzb),则有
$\left\{ \begin{array}{l}{J_{x{\rm{b}}}} = {I_{x{\rm{b}}}} + 2{m_r}d_3^2, \\{J_{y{\rm{b}}}} = {I_{y{\rm{b}}}} + 2{m_r}d_1^2, \\{J_{z{\rm{b}}}} = {I_{z{\rm{b}}}} + 2{m_r}\left( {d_1^2 + d_3^2} \right).\end{array} \right.$ (4)
式中:mr=m(ms+m)/mt,ms=mt-4m,Ixb、Iyb、Izb分别为忽略质心变化时Jxb、Jyb、Jzb的标称值.三者对时间的微分为
$\left\{ \begin{array}{l}{{\dot J}_{x{\rm{b}}}} = 2{m_\text{r}}\left( {{{\dot d}_3}{d_3} + {{\dot d}_4}{d_4}} \right), \\{{\dot J}_{y{\rm{b}}}} = 2{m_\text{r}}\left( {{d_1}{{\dot d}_1} + {d_2}{{\dot d}_2}} \right), \\{{\dot J}_{z{\rm{b}}}} = 2{m_\text{r}}\left( {{{\dot d}_1}{d_1} + {{\dot d}_2}{d_2} + {{\dot d}_3}{d_3} + {{\dot d}_4}{d_4}} \right).\end{array} \right.$ (5)
2 控制器设计本文控制器设计的目标是对一类采用帆板-滑块执行机构的太阳帆航天器,考虑光压干扰力矩和引力梯度力矩等外部干扰、执行机构导致的转动惯量变化,设计具有较强自适应能力和鲁棒性的姿态控制器;由控制器输出u求解执行机构的运动过程,继而输出控制力矩τc;在τc作用下使太阳帆姿态Q可以快速、准确地跟踪给定的期望姿态Qd.姿态控制系统结构如图 3所示.
Figure 3
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-3.jpg)
2.1 姿态控制器设计设期望姿态角为
$\left\{ \begin{array}{l}{\mathit{\boldsymbol{q}}_{\rm{e}}} = {q_{{\rm{d4}}}}\mathit{\boldsymbol{q}} - {\mathit{\boldsymbol{q}}_{\rm{d}}}^ \times \mathit{\boldsymbol{q}} - {q_{{\rm{d4}}}}{\mathit{\boldsymbol{q}}_{\rm{d}}}, \\{q_{{\rm{e4}}}} = \mathit{\boldsymbol{q}}_{\rm{d}}^{\rm{T}}\mathit{\boldsymbol{q}} + {\mathit{\boldsymbol{q}}_4}{q_{{\rm{d4}}}}.\end{array} \right.$
${\mathit{\boldsymbol{\omega }}_{\rm{e}}} = \mathit{\boldsymbol{\omega }} - {\mathit{\boldsymbol{\omega }}_{\rm{d}}}.$
由式(2)得误差模型为
$\mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \omega }}}_{\rm{e}}} = - \mathit{\boldsymbol{\omega }} \times \mathit{\boldsymbol{J\omega }} + \mathit{\boldsymbol{u}} + {\mathit{\boldsymbol{\tau }}_d} - \mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \omega }}}_{\rm{d}}}, $ (6)
${{\mathit{\boldsymbol{\dot Q}}}_{\rm{e}}} = \frac{1}{2}\left[ \begin{array}{l}\mathit{\boldsymbol{T}}\\ - {\mathit{\boldsymbol{q}}_{\rm{e}}}^{\rm{T}}\end{array} \right]{\mathit{\boldsymbol{\omega }}_{\rm{e}}}.$ (7)
式中T=qe4I3+qe×,且‖T‖=1.
为方便控制器设计,首先对太阳帆系统做如下合理假设:
假设1??转动惯量在标称值J0附近变化,其变化量及变化速度有界且已知,即J=J0+ΔJ,‖ΔJ‖≤σJ, σJ≥0,
假设2??外部干扰力矩有界,即‖τd‖≤d,d≥0.
此外给出如下Comparison引理[10]:
引理??函数g,
${{\mathit{\dot V}}_l} \le - a{V_l}\left( t \right) + g\left( t \right), $
那么对任意的a有
${{V}_{l}}\le {{\text{e}}^{-a}}{{V}_{l}}\left( 0 \right)+\int\limits_{0}^{t}{{{\text{e}}^{-a\left( t-\tau \right)}}g\left( \tau \right)\text{d}\tau },\forall t\ge 0.$
为了使太阳帆姿态控制系统对滑块运动引起的惯量变化具有鲁棒性[11],设计太阳帆滑模姿态控制律.取滑模面:
$\mathit{\boldsymbol{s}} = {\mathit{\boldsymbol{\omega }}_{\rm{e}}} + \lambda {\mathit{\boldsymbol{q}}_{\rm{e}}}, \lambda > 0;$ (8)
滑模控制器结构为
$\mathit{\boldsymbol{u}} = {\mathit{\boldsymbol{u}}_{\rm{r}}} + {\mathit{\boldsymbol{u}}_{\rm{e}}}, $ (9)
式中ur为趋近律,其作用是使系统状态到达滑模面;ue为等效控制律,在系统状态到达滑模面后,使其保持在该流形上.设计ur和ue分别为
${\mathit{\boldsymbol{u}}_{\rm{r}}} = - k\mathit{\boldsymbol{s}} - \left( {b + \hat d + \upsilon } \right){\rm{sign}}\left( \mathit{\boldsymbol{s}} \right), $ (10)
${\mathit{\boldsymbol{u}}_{\rm{e}}} = \mathit{\boldsymbol{\omega }} \times {\mathit{\boldsymbol{J}}_0}\mathit{\boldsymbol{\omega + }}{\mathit{\boldsymbol{J}}_0}{{\mathit{\boldsymbol{\dot \omega }}}_{\rm{d}}} - \lambda {\mathit{\boldsymbol{J}}_0}{{\mathit{\boldsymbol{\dot q}}}_{\rm{e}}}.$ (11)
其中:b=σJ(
受投影算法启发,设计自适应律
$\dot {\hat d} = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{\eta }\left\| \mathit{\boldsymbol{s}} \right\|, }&{\left\| \mathit{\boldsymbol{s}} \right\| > {s_{\rm{e}}};}\\{0, }&{\left\| \mathit{\boldsymbol{s}} \right\| \le {s_{\rm{e}}}.}\end{array}} \right.$ (12)
定理??考虑式(6)、(7)描述的系统,若满足假设1和假设2,在控制律(9)~(12)的作用下,
姿态误差qe稳定,且‖qe‖≤
证明??设
$V = \frac{1}{2}\left( {{s^{\rm{T}}}Js + \eta {{\tilde d}^2}} \right), $
对其微分有
$\begin{array}{l}\mathit{\dot V} = {s^{\rm{T}}}J\dot s + \frac{1}{2}{\mathit{\boldsymbol{s}}^{\rm{T}}}\mathit{\boldsymbol{J\dot s + }}\eta \tilde d\dot {\hat d} = {\mathit{\boldsymbol{s}}^{\rm{T}}}\left( {\mathit{\boldsymbol{J}}{{\dot \omega }_{\rm{e}}} + \lambda \mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \beta }}}_{\rm{e}}}} \right) + \\\;\;\;\;\frac{1}{2}{\mathit{\boldsymbol{s}}^{\rm{T}}}\mathit{\boldsymbol{\dot Js + }}\eta \tilde d\dot {\hat d} = {\mathit{\boldsymbol{s}}^{\rm{T}}}\left( { - \mathit{\boldsymbol{\omega }} \times \mathit{\boldsymbol{J\omega }} + \mathit{\boldsymbol{u}} + {\mathit{\boldsymbol{\tau }}_{\rm{d}}} - } \right.\\\;\;\;\;\left. {\mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \omega }}}_{\rm{d}}} + \lambda \mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \beta }}}_{\rm{e}}}} \right) + \frac{1}{2}{\mathit{\boldsymbol{s}}^{\rm{T}}}\mathit{\boldsymbol{\dot Js}} + \eta \tilde d\dot {\hat d}.\end{array}$
当‖s‖>se时, 有
$\begin{array}{l}\mathit{\dot V} = {s^{\rm{T}}}\left( { - \mathit{\boldsymbol{\omega }} \times \Delta \mathit{\boldsymbol{J\omega }} - \Delta \mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \omega }}}_{\rm{d}}} + \lambda \Delta \mathit{\boldsymbol{J}}{{\mathit{\boldsymbol{\dot \beta }}}_{\rm{e}}} - k\mathit{\boldsymbol{s}} - \left( {b + } \right.} \right.\\\;\;\;\;\;\left. {\left. {d + \mathit{\boldsymbol{\upsilon }}} \right){\rm{sign}}\left( s \right) + {\tau _{\rm{d}}}} \right) + \frac{1}{2}{\mathit{\boldsymbol{s}}^{\rm{T}}}\dot Js + \tilde d\left\| \mathit{\boldsymbol{s}} \right\| \le \\\;\;\;\;\; - k{\left\| \mathit{\boldsymbol{s}} \right\|^2} + \frac{1}{2}\left\| {\mathit{\boldsymbol{\dot J}}} \right\|{\left\| \mathit{\boldsymbol{s}} \right\|^2} + \left( {\left\| {\mathit{\boldsymbol{\dot J}}} \right\| - {\delta _J}} \right)\left\| \mathit{\boldsymbol{s}} \right\| - \\\;\;\;\;\;\mathit{\boldsymbol{\upsilon }}\left\| \mathit{\boldsymbol{s}} \right\| - \hat d\left\| \mathit{\boldsymbol{s}} \right\| + d\left\| \mathit{\boldsymbol{s}} \right\| + \tilde d\left\| \mathit{\boldsymbol{s}} \right\| \le - k{\left\| \mathit{\boldsymbol{s}} \right\|^2} + \\\;\;\;\;\;\frac{1}{2}{\delta _j}{\left\| \mathit{\boldsymbol{s}} \right\|^2} - \mathit{\boldsymbol{\upsilon }}\left\| \mathit{\boldsymbol{s}} \right\| + \left( { - \hat d + d} \right)\left\| \mathit{\boldsymbol{s}} \right\| + \tilde d\left\| \mathit{\boldsymbol{s}} \right\| \le \\\;\;\;\;\; - {k_0}{\left\| \mathit{\boldsymbol{s}} \right\|^2} - \mathit{\boldsymbol{\upsilon }}\left\| \mathit{\boldsymbol{s}} \right\| < 0.\end{array}$
可见,在所提控制律作用下,系统状态将沿‖s‖衰减的方向运动,直至‖s‖≤se.
当‖s‖≤se时,取Lyapunov函数
$V' = \frac{1}{2}\mathit{\boldsymbol{q}}_{\rm{e}}^{\rm{T}}{\mathit{\boldsymbol{q}}_{\rm{e}}}, $
沿系统(6)、(7)对其微分,有
$\dot V' = \mathit{\boldsymbol{q}}_{\rm{e}}^{\rm{T}}{{\mathit{\boldsymbol{\dot q}}}_{\rm{e}}} = \frac{1}{2}\mathit{\boldsymbol{q}}_{\rm{e}}^{\rm{T}}\mathit{\boldsymbol{T}}{\mathit{\boldsymbol{\omega }}_{\rm{e}}}, $
滑模面上有s=ωe+λqe,联合T的性质有
$\begin{array}{*{20}{c}}{\dot V' = \frac{1}{2}\mathit{\boldsymbol{q}}_{\rm{e}}^{\rm{T}}\mathit{\boldsymbol{T}}\left( {\mathit{\boldsymbol{s}} - \lambda {\mathit{\boldsymbol{q}}_{\rm{e}}}} \right) \le }\\{ - \frac{\lambda }{2}\mathit{\boldsymbol{q}}_{\rm{e}}^{\rm{T}}{\mathit{\boldsymbol{q}}_{\rm{e}}} + \frac{1}{2}\left\| {{\mathit{\boldsymbol{q}}_{\rm{e}}}} \right\|{s_{\rm{e}}} \le }\\{ - \frac{\lambda }{2}V' + \frac{1}{2}{s_{\rm{e}}}.}\end{array}$
由Comparison引理[10]得
${\left\| {{\mathit{\boldsymbol{q}}_{\rm{e}}}} \right\|^2} \le {{\rm{e}}^{ - \lambda /2}}V'\left( 0 \right) + \frac{1}{2}\int_0^t {{{\rm{e}}^{ - \lambda /2\left( {t - \tau } \right)}}{s_{\rm{e}}}{\rm{d}}\tau } , $
进一步有
证毕.
2.2 执行机构解算利用姿态控制器输出u可求解帆板的转角与滑块的期望位移,通过帆板的转动和滑块的滑动可获得对应的控制力矩τc,继而实现对姿态系统的控制.由式(3)可知,执行机构工作过程中,4块小帆的角度互相影响,4个滑块的位置也存在耦合关系.对此,本文设计了简单的力矩分配规则,即令γ=γ1=γ3=γ2=γ4,d1=d2,d3=d4,则有
${\mathit{\boldsymbol{\tau }}_{\rm{c}}} = \left[ {\begin{array}{*{20}{c}}{ - 8l{P_{\rm{s}}}{A_{\rm{p}}}{{\cos }^2}\left( {\alpha + \gamma } \right)\sin \gamma - 4m/{m_{\rm{t}}}{d_3}{P_{\rm{s}}}A{{\cos }^2}\alpha }\\{4m/{m_{\rm{t}}}{d_1}{P_{\rm{s}}}A{{\cos }^2}\alpha }\end{array}} \right].$ (13)
对帆板,在太阳帆任务设计中,滚转轴所需控制力矩往往较小,帆板转角也在较小范围内变化.据此,对滚转轴力矩表达式进行小角度线性化,即cosγ=0,sinγ=γ,有τc(1)≈8lPsApγcos2α.记γd为帆板期望转角,则有
${\gamma _{\rm{d}}} \approx \mathit{\boldsymbol{u}}\left( 1 \right)/\left( {8l{P_{\rm{s}}}{A_p}\gamma {{\cos }^2}\alpha } \right), $
帆板转角运动过程为
${T_1}\dot \gamma + \gamma = {\gamma _{\rm{d}}}, $
式中T1由帆板角度及角速度限幅决定,即
对滑块,令z=[d1d3]T为滑块位置,zd=[d1d d3d]T为其期望位置,由式(13)可得
${\mathit{\boldsymbol{z}}_{\rm{d}}} = \left[ {\begin{array}{*{20}{c}}{\mathit{\boldsymbol{u}}\left( 3 \right)/\left( {4m/{m_{\rm{t}}}{d_1}{P_{\rm{s}}}A{{\cos }^2}\alpha } \right)}\\{ - \mathit{\boldsymbol{u}}\left( 2 \right)/\left( {4m/{m_{\rm{t}}}{d_3}{P_{\rm{s}}}A{{\cos }^2}\alpha } \right)}\end{array}} \right], $
滑块运动采用模型
${T_2}\mathit{\boldsymbol{\dot z}} + \mathit{\boldsymbol{z}} = {\mathit{\boldsymbol{z}}_{\rm{d}}}.$
式中T2由最大滑动位置zmax和最大滑动速度?max决定,即?max=zmax/T2.
3 数值模拟参考文献[5],本文采用如下太阳帆参数进行数值仿真实验:转动惯量标称值为[6 000??3 000??3 000],Ap=2m2,l=8m,A=1 200 m2,m=2 kg,mt=157 kg.帆板转角和角速度的最大值分别为γmax=80°,
仿真结果如图 4~9所示. 图 4为姿态响应图,由图可知,姿态四元数误差qe趋向于0,太阳帆俯仰角由0°机动至35°时间远小于1 h,机动时间短;角度稳态误差保持在0.01°以内,满足了太阳帆飞行要求. 图 5为4个滑块位移曲线.滑块1、2的最大位移为6.598 m,随着姿态角趋向机动位置,逐渐稳定于3.925 m处以抵抗偏航轴所受干扰.滑块3、4的运动提供了俯仰角机动所需力矩,二者最大位移均为18.442 m;姿态误差收敛后,滑块3、4位置均保持在1.962 m处. 图 6为帆板转动过程,帆板转角最大幅度为31.693°,并稳定在-9.666°以消除滚转轴所受干扰. 图 5、6中的子图还分别显示了滑块和帆板在时间初始段的响应,可见二者运动轨迹平滑无大幅跃变,在实际应用中易于实现.
Figure 4
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-4.jpg)
Figure 5
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-5.jpg)
Figure 6
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-6.jpg)
Figure 7
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-7.jpg)
Figure 8
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-8.jpg)
Figure 9
![点击查看原图](http://hit.alljournals.cn/html/hitxb_cn/2018/9/PIC/hebgydxxb-50-9-141-9.jpg)
图 7给出了干扰力矩上界的估计结果图,图 8为干扰力矩示意图,滚动轴干扰力矩为常值,俯仰轴和偏航轴干扰力矩随俯仰角变化. 图 9为控制器输出u和执行机构输出的控制力矩τc. U为执行机构的期望输入,3轴最大幅值分别为0.327 6、6.486、1.827 mN·m,皆在执行机构可实现范围内.此外,各轴控制量曲线缓和,自适应律的加入在抑制干扰的同时,也有效地抑制了滑模控制器中存在的抖振现象.通过滑块位移和帆板转动可求得实际控制力矩τc,如图中实线所示.可见,通过图 5、6所示的帆板转动和滑块滑动,执行机构较好地实现了所需控制力矩.
综上,所提控制策略在考虑光压力矩干扰、引力梯度干扰和转动惯量变化的情况下,有效地实现了太阳帆三轴姿态控制.
4 结论本文采用了一种新型帆板-滑块结构作为太阳帆航天器的姿控执行机构,在此结构下研究了太阳帆姿态控制律设计与实现问题.针对滑块滑动致使太阳帆转动惯量变化问题,设计了鲁棒滑模姿态控制器.同时,考虑了光压力矩和引力梯度力矩两种干扰,设计了自适应律对干扰上界进行估计.最后,解算执行机构,实现所提控制律.仿真结果表明,所提控制方法可使太阳帆姿态较快机动至期望位置,并使控制力矩、帆板角度和滑块位移均保持在适当的幅值范围内,易于实现.
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