﻿ 基于网络的液压马达伺服位置系统自适应鲁棒积分控制
 上海理工大学学报  2021, Vol. 43 Issue (4): 325-331 PDF

1. 上海理工大学 机械工程学院，上海 200093;
2. 中国船舶集团公司 第七〇三研究所，哈尔滨 150000

Adaptive robust integral control of hydraulic motor servo position system based on network
SHEN Wei1, LIU Shuai1, WU Yi2
1. School of Mechanical Engineering, University of Shanghai for Science and Technology, Shanghai 200093, China;
2. NO. 703 Research Institute of China State Shipbuilding Corporation Limited, Harbin 150000, China
Abstract: For the problems of network delay and uncertainty in modeling the structure of the valve-controlled motor faced by the network-based hydraulic servo control system, a robust integral adaptive controller with error sign synthesized based on Pade's theorem and the backstepping derivation method was proposed. Pade's theorem was used by the controller to approximate the time delay caused by the time-varying network, and reduced the impact of the delay on the tracking performance of the control system. The adaptive rate was applied to approximate the structural uncertainty and delay error values of the system. The error sign control method was used to compensate for the remaining structural uncertainty. By constructing a suitable Lyapunov function, the global stability of the closed-loop system was verified, and the boundedness of all signals of the closed-loop system and the asymptotic convergence of the tracking error were guaranteed. The simulation results show the high-precision tracking performance of the control method.
Key words: electro-hydraulic servo system     adaptive robust integration     backstepping control     time-varying network delay     Lyapunov function

1 问题描述和数学模型

 图 1 基于网络的电液伺服系统原理图 Fig. 1 Schematic diagram of electro-hydraulic servo system based on network

 ${Q_{\rm L}} = {C_{\rm d}}w{x_{\rm v}}\sqrt {\frac{{{P_{\rm{s}}} - {P_{\rm L}} {\rm{sign}} \; {{x_{\rm v}}} }}{\rho }}$ (1)
 ${\rm{sign}}\; x_{\rm v} = \left\{ {\begin{array}{*{20}{c}} 1,&{x \geqslant 0}\\ { - 1,}&{x < 0} \end{array}} \right.$ (2)

 ${Q_{\rm{L}}} - {C_{\rm t}}{P_{\rm{L}}} + Q\left( t \right) = {D_{\rm{m}}}\dot \theta + \frac{V_{\rm t}}{{4{\beta _{\rm e}}}}{{\dot P}_{\rm L}}$ (3)

 $J\ddot \theta {\rm{ = }}{P_{\rm{L}}}{D_{\rm{m}}}{\rm{ - }}B\dot \theta + F\left( t \right)$ (4)

 ${Q_{\rm{L}}}={k_{\rm{t}}}u\sqrt {{P_{\rm{s}}} - {P_{\rm{L}}}{\rm{sign}}\; u }$ (5)
 ${k_{\rm{t}}} = {C_{\rm{d}}}w{k_{\rm{i}}}\sqrt {1/\rho }$ (6)

 ${\boldsymbol X} = {\left( {{x_1}\;\;{x_2}\;\;{x_3}} \right)^{\rm{T}}} = {\left( {\theta \;\;\dot \theta \;\;\frac{{{D_{\rm m}}{P_{\rm L}}}}{J}} \right)^{\rm{T}}}$ (7)

 $\left\{ \begin{array}{l} {{\dot x}_1} {\rm{ = }} {x_2} \\ {{\dot x}_2} = {x_3}{\rm{ + }}b{x_2} + d(t) \\ {{\dot x}_3} = g(u,{x_3})u(t - \tau ) + f({x_2},{x_3}) + {d_2} \end{array} \right.$ (8)
 $\begin{split} & b = \frac{{ - B}}{J}, \;d(t) = \frac{{F(t)}}{J},\\ &g\left( u,{{x}_{3}} \right)=\frac{4{{D}_{\rm m}}{{\beta }_{\rm e}}{{k}_{\rm t}}}{J{{V}_{\rm t}}}\sqrt{{{P}_{\rm s}}-\frac{J}{D_{\rm m}}\left({\rm{sign}}\;u\right){{x}_{3}}},\\ &f({x_2},{x_3}){\rm{ = }} - \frac{{4D_{\rm m}^{\rm{2}}{\beta _{\rm e}}{k_{\rm t}}}}{{J{V_{\rm t}}}}{x_2} - \frac{{4{\beta _{\rm e}}{C_{\rm t}}}}{{{V_{\rm t}}}}{x_3}, \;{d_2} = \frac{{4{D_{\rm m}}{\beta _{\rm e}}Q(t)}}{{JV_{\rm t}}} \end{split}$

a. 不考虑管道的摩擦损失、流体质量。

b. 液压回转马达的每个腔室中的压力在任何地方都是相等的。

c. 液压马达中的每处泄漏都是层流。

d. 干扰 $d(t)$ 在系统模型中是足够平滑的，且满足 $\left| {\dot d(t)} \right| \leqslant {\delta _1}, \left| {\ddot d(t)} \right| \leqslant {\delta _2}$ ${\delta }_{1},{\delta }_{2}$ 均为已知的正数。

 $\begin{split}\phi \left\{ {u(t - \tau )} \right\} = &\exp( - \tau s)\phi \left\{ {u(t)} \right\} =\\ &\frac{{\exp( - \tau s/2)}}{{\exp(\tau s/2)}}\phi \left\{ {u(t)} \right\} = \frac{{1 - \tau s/2}}{{1 + \tau s/2}}\end{split}$ (9)

 $\frac{{1 - \tau s/2}}{{1 + \tau s/2}} = \phi \left\{ {{x_4}(t)} \right\} - \phi \left\{ {u(t)} \right\}$ (10)

 $u - \frac{{\tau \dot u}}{2} = {x_4} + \frac{\tau }{{2{{\dot x}_4}}} - u - \frac{{\tau \dot u}}{2}\quad$ (11)

 ${\dot x_4} = - \frac{2}{\tau }{x_4} + \frac{4}{\tau }u = - \gamma {x_4} + 2\gamma u$ (12)

 $\left\{ \begin{array}{l} {{\dot x}_1} {\rm{ = }} {x_2} \\ {{\dot x}_2} = {x_3}{\rm{ + }}b{x_2} + {d(t)} \\ {{\dot x}_3} = g(u,{x_3})({x_4} - u(t)) + f({x_2},{x_3}) + {d_2} \\ {{\dot x}_4} = - \gamma {x_4}{\rm{ + 2}}\gamma u(t) \end{array} \right.$ (13)
2 自适应鲁棒积分控制器设计

 ${{{e}}_1}{\rm{ = }}{{{x}}_1}{\rm{ - }}{{{x}}_{{\rm{1d}}}}{\rm{ }}$ (14)

 $\begin{split} & {e_2} = {{\dot e}_1} + {k_1}{e_1} = {x_2} - {\alpha _1},\;{\alpha _1} = {{\dot x}_{{\rm{1d}}}} - {k_1}{e_1}, \\ &\quad r = {{\dot e}_2} + {k_2}{e_2},\;{e_3} = {x_3} - {\alpha _2} + \frac{g}{\lambda }{x_4} \end{split}$ (15)

 $\begin{split}r= &{\dot e_2} + {k_2}{e_2}{\rm{ }} = {e_3} + {\alpha _2} - b{x_2} + \\ &{d_1} - {\ddot x_{1d}} + ({k_1} + {k_2}){\dot e_1} + {k_1}{k_2}{e_1}\end{split}$ (16)

 $\begin{split} &{\alpha _2}= {\alpha _{2{\rm a}}} + {\alpha _{2{\rm s}1}} + {\alpha _{2{\rm s}2}},\;{\alpha _{2{\rm a}}} = \int_0^t {(b{{\ddot x}_{1{\rm d}}} + {{\dddot x}_{1{\rm d}}})} {\rm{d}}t,\\ & {\alpha _{2{\rm s}1}}= - {k_{\rm r}}{e_2},\;{\alpha _{2{\rm s}2}} = \int_0^t {( - {k_{\rm r}}{k_2}{e_2} - \hat \beta {\rm{sign}}\;{e_2})} {\rm{d}}t\end{split}$ (17)

 ${\dot{\hat{\beta}}} {\rm{ = - }}\phi r{\rm{sign}}\;{e_2}$ (18)

 $\begin{split}\dot r = {\dot e_3} &+ \mathop {{\alpha _2}}\limits^{.} - b{\dot x_2} + {\dot d_1} - {\dddot x_{1{\rm d}}} +({k_1} + {k_2})r +\\ & k_1^3{e_1} + ( - k_1^2 - k_2^2 - {k_1}{k_2}){e_2}\end{split}$ (19)

 $\begin{split} \dot r = &{{\dot e}_3} + ( - k_1^2b + k_1^3){e_1} - \hat \beta {\rm{sign}}\;{e_2} + {{\dot d}_1} + ({k_1} + {k_2} - b - {k_r})r + \\ &\left( - k_1^2 - k_2^2 - {k_1}{k_2} + {k_1}b + {k_2}b\right){e_2} = g{x_4} - gu +f\left({x_2},{x_3}\right) -\\ & {{\dot \alpha }_2} + \tau \left({x_4},t\right) + \frac{g}{\xi }\Big( -\xi {{x}_{4}}+2\xi u \Big) + ( - k_1^2b + k_1^3){e_1} - \\ &\hat \beta {\rm{sign}}\;{e_2} + {{\dot d}_1} + ({k_1} + {k_2} - b - {k_{\rm r}})r +\\ &\left( - k_1^2 - k_2^2 - {k_1}{k_2} + {k_1}b + {k_2}b\right){e_2}\\[-10pt] \end{split}$ (20)
 $\tau ({x_4},t) = {d_2} + \dfrac{{\dot g}}{\xi }{x_4}$

$\tau ({x_4},t)$ 的自适应更新律

 $\dot {\overset\frown{\tau }} ({x_4},t){\rm{ = - }}\eta {\rm{(}}{{\rm{e}}_3}{\rm{ + r)}}$ (21)

 $\begin{split} &u = {u_{\rm a}} + {u_{\rm s}} ,\quad {u_{\rm a}} = {u_{{\rm a}1}} + {u_{{\rm a}2}},\\ &{u_{{\rm a}1}} = - \frac{{{{\dot \alpha }_2} - f({x_2},{x_3})}}{g} ,\quad {u_{{\rm a}2}} = - \frac{{\hat \tau ({x_4},t)}}{g},\\ &{u_{\rm s}} = - \frac{{{k_3}{e_3}}}{g} \end{split}$ (22)

 $\begin{split} \dot r = &\left({\tau _s} - {{\overset\frown{\tau }}_s}\right) - {k_3}{e_3} + \left( - k_1^2b + k_1^3\right){e_1} -\\ & \hat \beta {\rm{sign}}\;{e_2} +{{\dot d}_1} + \left({k_1} + {k_2} - b - {k_r}\right)r + \\ & \left( - k_1^2 - k_2^2 - {k_1}{k_2} + {k_1}b + {k_2}b\right){e_2} \end{split}$ (23)

 ${\dot e_3} = {\dot x_3} - {\dot \alpha _2} + \frac{{\dot g}}{\xi }{x_4} + \frac{g}{\xi }{\dot x_4} = \left({\tau _{\rm s}} - {{\overset\frown{\tau }} _{\rm s}}\right) - {k_3}{e_3}$ (24)
3 稳定性证明

 $H(t) = r[{\dot d_1} - \beta {\rm{sign}}\; {e_2} ]$ (25)

 $\beta \geqslant {\delta _1} + \frac{{{\delta _2}}}{{{k_2}}}$ (26)

 $P(t) = \beta \left| {{e_2}(0)} \right| - {e_2}(0){\dot d_1}(0) - \int_0^t H (t){\rm d}t$ (27)

Vt）为连续可微的正定函数，定义

 $V = \frac{1}{2}e_1^2 + \frac{1}{2}e_2^2 + \frac{1}{2}e_3^2 + \frac{1}{2}{r^2} + P(t) + \frac{1}{{2\chi }}{\tilde \eta ^2} + \frac{1}{{2\alpha }}{\tilde \beta ^2}$ (28)

 $\begin{split} \dot V = &{e_1}{{\dot e}_1} + {e_2}{{\dot e}_2} + {e_3}{{\dot e}_3} + r\dot r + \dot P + \frac{1}{\chi }\tilde \eta {\dot{\overset\frown{\eta }}} + \frac{1}{\alpha }\tilde \beta {\dot{\hat{\beta}}} =- {k_1}e_1^2 + \\ &{e_1}{e_2} - {k_2}e_2^2 - {k_3}e_3^2 + {e_3}\left({\tau _s} - {{\overset\frown{\tau }}_s}\right) +r\left({\tau _s} - {{\overset\frown{\tau }}_s}\right) + r{{\dot d}_1} - \\ & r\hat \beta {\rm{sign}}\;{e_2} - {k_3}r{e_3} + r\beta {\rm{sign}}\;\;{e_2} - r{{\dot d}_1} + {r^2}( - b - {k_r} + \\ &{k_1} + {k_2}) + {e_1}r\left( - k_1^2b + k_1^3\right)+ {e_2}r(b{k_1} + b{k_2} - k_1^2 -\\ & k_2^2 - {k_1}{k_2} + 1) +\frac{1}{\chi }\tilde \eta {\dot{\overset\frown{\eta }}} + \frac{1}{\alpha }\tilde \beta {\dot{\hat{\beta}}} \\[-15pt] \end{split}$ (29)

 $\dot V \leqslant - {\lambda _{\min}}( {\boldsymbol \varLambda })\left( {e_1^2 + e_2^2 + e_3^2 + {r^2}} \right)$ (30)
 $\begin{split} &{\boldsymbol{\varLambda}} = \left[ {\begin{array}{*{20}{c}} {{k_1}}&{ - 1/2}&{( - 1/2){\sigma _1}}&0 \\ { - 1/2}&{{k_2}}&{( - 1/2){\sigma _2}}&{ - 1/2} \\ {( - 1/2){\sigma _1}}&{( - 1/2){\sigma _2}}&{{k_4}}&{ - {k_3}/2} \\ 0&{ - 1/2}&{ - {k_3}/2}&{{k_3}} \end{array}} \right]\\ &\qquad\quad\;\;{\sigma _2} = b{k_1} + b{k_2} - k_1^2 - k_2^2 - {k_1}{k_2} + 1,\\ &\qquad\quad\;\;{{\sigma }_{1}}=-k_{1}^{2}b+ k_{1}^{3},\qquad {k_4} = - b - {k_r} + {k_1} + {k_2} \end{split}$

4 仿真实验 4.1 系统参数选取

4.2 比较实验结果

 图 2 外界扰动扭矩 Fig. 2 External disturbance torque

a. 自适应鲁棒积分控制器。控制增益给定如下： ${{k}_{1}}=800,{{k}_{2}}{=}800,{{k}_{3}}=100,{{k}_{4}}=100,{{k}_{\rm r}}=900,$ 控制器的自适应增益 $\eta {\rm{ = }}0.5$

b. PID。通过试错法得到控制器增益 ${k}_{\rm{p}}=900,$ ${k}_{\rm{i}}=1\;000,{k}_{\rm{d}}=0$

 图 3 输入信号①时，扰动和网络诱导延时影响下的阀控马达响应 Fig. 3 When the input signal is ①, the response of the valve-controlled motor under the influence of disturbance and network-induced delay

 图 4 输入信号为②时，在扰动和网络诱导延时影响下的阀控马达响应 Fig. 4 When the input signal is ②, the response of the valve-controlled motor under the influence of disturbance and network-induced delay

5 结　论

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