﻿ 水下动平台条件下欠驱动AUV回收方法
 舰船科学技术  2023, Vol. 45 Issue (13): 80-83    DOI: 10.3404/j.issn.1672-7649.2023.13.016 PDF

1. 中国人民解放军91388部队，广东 湛江 524022;
2. 华中科技大学船舶与海洋工程学院，湖北 武汉 430074;
3. 中国舰船研究设计中心，湖北 武汉 430074

Research on recovery of underactuated AUV in the condition of underwater moving carrier
XIE Zhi-qiang1, CHEN Hong-xuan2, BAI Tie-chao3, TANG Guo-yuan2
1. No.91388 Unit of PLA, Zhanjiang 524022, China;
2. School of Naval Architecture and Ocean Engineering, Huazhong University of Science and Technology, Wuhan 430074, China;
3. China Ship Development and Design Center, Wuhan 430074, China
Abstract: In underwater environment, UUV of large scale can carry multiple AUVs as an underwater platform, which expands its capability of performing tasks. In order to solve the recovery problem of underactuated AUV about locomotor recovery station, this paper studies the 3D motion tracking method of AUV. Firstly, this paper establishes the model of AUV, and then proposes a recovery strategy. Furthermore, based on the model, a backstepping controller is designed, and the stability of the system is analyzed by Lyapunov theorem. The simulation results show that the proposed recovery strategy and controller successfully realizes the 3D tracking control of underactuated AUV for recovery, which is a base to solve the recovery problem of underactuated AUV about locomotor recovery station.
Key words: AUV     UUV     load carrying     recovery     backstepping control
0 引　言

1 运动建模

 图 1 地球坐标系和随体坐标系 Fig. 1 Earth fixed frame and body fixed frame

 $\dot {\boldsymbol{\eta}} = {\boldsymbol{J}}({\boldsymbol{\eta}} ){\boldsymbol{\upsilon}}，$ (1)
 ${\boldsymbol{J}}({\boldsymbol{\eta}} ) = \left[ {\begin{array}{*{20}{c}} {{{\boldsymbol{J}}_1}}&{\boldsymbol{O}} \\ {\boldsymbol{O}}&{{{\boldsymbol{J}}_2}} \end{array}} \right],{\dot {\boldsymbol{J}}_1} = {{\boldsymbol{J}}_1}{\boldsymbol{\upsilon}} _2^*,{\boldsymbol{\upsilon}} _2^* = \left[ {\begin{array}{*{20}{c}} 0&{ - r}&q \\ r&0&{ - p} \\ { - q}&p&0 \end{array}} \right] 。$ (2)

 ${\boldsymbol{M}}\dot {\boldsymbol{v}} + {\boldsymbol{C}}({\boldsymbol{v}}){\boldsymbol{v}} + {\boldsymbol{D}}({\boldsymbol{v}}){\boldsymbol{v}} + {\boldsymbol{G}}({\boldsymbol{\eta}} ) + {{\boldsymbol{\tau}} _d} = {\boldsymbol{\tau}} 。$ (3)

2 回收策略

UUV搭载了水声定位装置、惯导、声学多普勒速度仪（DVL），考虑在其背部配备一个回收笼，上面安装LED灯阵列。AUV上搭载了水声定位装置、惯导、水下摄像头、DVL。水声定位装置、视觉定位可测得AUV相对于UUV的位置信息，视觉和惯导加DVL可获得AUV相对于UUV的欧拉角、速度以及角速度。假设UUV始终在水平面内作直线运动，速度恒定，考虑如下回收场景：

 图 2 回收坐标系 Fig. 2 Recovery frame

1)初对齐

2)精确回收

3 控制器设计

 ${{\boldsymbol{\eta }}_{1e}} = {{\boldsymbol{\eta}} _1} - {{\boldsymbol{\eta}} _d} ，$ (4)
 ${{\boldsymbol{e}}_\eta } = {\boldsymbol{J}}_1^{\rm{T}}{{\boldsymbol{\eta}} _{1e}} 。$ (5)

 ${{\boldsymbol{z}}_1} = {{\boldsymbol{e}}_\eta } - {\boldsymbol{\rho}} ，$ (6)

${{\boldsymbol{z}}_1}$ 求导，结合式(2)、式(4)～式(6)可得：

 ${\dot {\boldsymbol{z}}_1} = - {\boldsymbol{\upsilon}} _2^*{{\boldsymbol{e}}_\eta } + {\boldsymbol{J}}_1^{\rm{T}}{\dot {\boldsymbol{\eta}} _{1e}} - \dot {\boldsymbol{\rho}}，$ (7)

 ${\dot {\boldsymbol{z}}_1} = - {\boldsymbol{\upsilon}} _2^*{{\boldsymbol{z}}_1} + {\boldsymbol{P}}{\left[ {\begin{array}{*{20}{c}} {{u}}&q&r \end{array}} \right]^{\rm{T}}} + {\left[ {\begin{array}{*{20}{c}} 0&v&w \end{array}} \right]^{\rm{T}}} - {\boldsymbol{J}}_1^{\rm{T}}{\dot {\boldsymbol{\eta}} _d} - \dot {\boldsymbol{\rho}} 。$ (8)

 ${\boldsymbol{P}} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&0&{ - \rho } \\ 0&\rho &0 \end{array}} \right] 。$ (9)

 ${\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{{\boldsymbol{v}}} _e} = {\left[ {\begin{array}{*{20}{c}} {{u_e}}&{{q_e}}&{{r_e}} \end{array}} \right]^{\rm{T}}} = {\left[ {\begin{array}{*{20}{c}} {u - {u_d}}&{q - {q_d}}&{r - {r_d}} \end{array}} \right]^{\rm{T}}} 。$ (10)

${\left[ {\begin{array}{*{20}{c}} {{u_d}}&{{q_d}}&{{r_d}} \end{array}} \right]^{\rm{T}}}$

 ${\left[ {\begin{array}{*{20}{c}} {{u_d}}&{{q_d}}&{{r_d}} \end{array}} \right]^{\rm{T}}} = {{\boldsymbol{P}}^{ - 1}}( - {\boldsymbol{K}}{{\boldsymbol{z}}_1} - {\left[ {\begin{array}{*{20}{c}} 0&v&w \end{array}} \right]^{\rm{T}}} + {\boldsymbol{J}}_1^{\rm{T}}{\dot {\boldsymbol{\eta}} _d} + \dot {\boldsymbol{\rho}} ) ，$ (11)

 ${\dot {\boldsymbol{z}}_1} = - {\boldsymbol{\upsilon}} _2^*{z_1} + {\boldsymbol{P}}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{{\boldsymbol{v}}} _e} - {\boldsymbol{K}}{{\boldsymbol{z}}_1}。$ (12)

 ${V_1} = \frac{1}{2}{{\boldsymbol{z}}_1}^{\rm{T}}{{\boldsymbol{z}}_1}，$ (13)

 ${\dot {\boldsymbol{V}}_1} = {{\boldsymbol{z}}_1}^{\rm{T}}(P{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{{\boldsymbol{v}}} _e} - {\boldsymbol{K}}{{\boldsymbol{z}}_1}) 。$ (14)

 ${{\boldsymbol{v}}_d} = [\begin{array}{*{20}{c}} {{u_d}}&{{\alpha _2}}&{{\alpha _3}}&{{\alpha _4}}&{{q_d}}&{{r_d}} \end{array}]。$ (15)

 ${{\boldsymbol{z}}_2} = {\boldsymbol{v}} - {{\boldsymbol{v}}_d} ，$ (16)

 ${V_2} = \frac{1}{2}{{\boldsymbol{z}}_2}^{\rm{T}}{\boldsymbol{M}}{{\boldsymbol{z}}_2}，$ (17)

 ${\dot V_2} = {{\boldsymbol{z}}_2}^{\rm{T}}({\boldsymbol{M}}\dot {\boldsymbol{v}} - {\boldsymbol{M}}{\dot {\boldsymbol{v}}_d})。$ (18)

 ${\dot V_2} = {{\boldsymbol{z}}_2}^{\rm{T}}( - {\boldsymbol{C}}({\boldsymbol{v}}){\boldsymbol{v}} - {\boldsymbol{D}}({\boldsymbol{v}}){\boldsymbol{v}} - {\boldsymbol{G}}({\boldsymbol{\eta}} ) - {{\boldsymbol{\tau}} _d} + {\boldsymbol{\tau}} - {\boldsymbol{M}}{\dot {\boldsymbol{v}}_d}) 。$ (19)

 ${\boldsymbol{\omega}} = {\boldsymbol{C}}({\boldsymbol{v}}){\boldsymbol{v}} + {\boldsymbol{D}}({\boldsymbol{v}}){\boldsymbol{v}} + {\boldsymbol{G}}({\boldsymbol{\eta}} ) + {{\boldsymbol{\tau }}_d} 。$ (20)

 ${\dot V_2} = {{\boldsymbol{z}}_2}^{\rm{T}}( - {\boldsymbol{\omega}} + {\boldsymbol{\tau}} - {\boldsymbol{M}}{\dot {\boldsymbol{v}}_d})。$ (21)

 ${\tau _i} = {\omega _i} + {M_{ii}}{\dot {\boldsymbol{v}}_{di}} - {c_i}{z_i},i = 1,5,6 。$ (22)

 ${M_{ii}}\dot \alpha = - {\omega _i} + {c_i}{z_i},i = 2,3,4 ，$ (23)

 ${\dot V_2} = - {z_2}^{\rm{T}}{\boldsymbol{c}}{{\boldsymbol{z}}_2} 。$ (24)

 ${\boldsymbol{\tau}} = 400\frac{{\boldsymbol{\tau}} }{{{{\left\| {\boldsymbol{\tau}} \right\|}_2}}},{\left\| {\boldsymbol{\tau}} \right\|_2} > 400 。$ (25)

4 仿　真

AUV模型参数来源于文献[8]， ${K_1} = 2$ ${K_2} = 10$ ${K_3} = 10$ ${c_1} = 10$ ${c_5} = 4$ ${c_6} = 4$ $\rho = 0.2$ 。UUV的运动速度设置为0.5 m/s, 海流干扰 ${V_{{\rm{current}}}} = 0.5\;{\rm{m}}/{\rm{s}}$ ，方向沿着 $X$ 轴的负方向，期望轨迹为（假设视觉算法在120 s生效）：

 ${{\boldsymbol{\eta}} _d} = \left\{ {\begin{array}{*{20}{c}} {{{[\begin{array}{*{20}{c}} { - 5 + 0.5t}&0&{20}&0&0&0 \end{array}]}^{\rm{T}}},t < 120} ，\\ {{{\boldsymbol{\eta}} _d} = {{[\begin{array}{*{20}{c}} {0.5t}&0&{20}&0&0&0 \end{array}]}^{\rm{T}}},t > 120} 。\end{array}} \right.$

AUV的初始状态为：

 ${\boldsymbol{\eta}} = {[\begin{array}{*{20}{c}} { - 20}&{ - 10}&0&0&0&0 \end{array}]^{\rm{T}}} 。$

 图 3 AUV的三维轨迹运动图 Fig. 3 3D trajectory of AUV

 图 6 控制输入 Fig. 6 Inputs of control

 图 4 期望位置、实际位置的时间曲线 Fig. 4 Desired position and actual position vs time

 图 5 轨迹跟踪误差曲线 Fig. 5 Trajectory tracking error
5 结　语

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