Gyroscopic Effect Decoupling Control of Maglev Rotor Supported by Nonlinear Force
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摘要: 电磁轴承非线性力支承的飞轮转子各自由度之间产生的强耦合,影响轴承转子系统稳定性。为此建立了径向四自由度的非线性电磁力-刚性转子动力学模型。在此基础上,提出了一种自适应径向基神经网络和滑模控制结合的算法(Adaptive RBFNN&SMC)。基于RBFNN对非线性电磁力和陀螺效应进行整体补偿,应用双曲正切函数作为滑模鲁棒项,对滑模控制进行改进,改善了滑模算法的抖振、抑制了质量不平衡扰动和随机扰动。根据Lyapunov稳定性理论证明了系统的渐进稳定性。最后通过仿真将提出的算法与PID算法和
$ \alpha $ 阶逆系统算法对比,结果表明该算法能有效补偿非线性力、解耦系统和改善抖振问题,同时对于外界扰动具有良好的抑制效果。Abstract: The high-speed maglev flywheel supported by nonlinear force shows strong coupling dynamics, which affects the stability of the rotor-bearing system. Therefore, a four-DOF dynamic model of nonlinear magnetic force-rigid rotor was established, and an adaptive radial basis neural network and sliding mode control algorithm (adaptive RBFNN&SMC) was proposed. The nonlinear force and gyro effect was integrally compensated by the RBFNN. The hyperbolic tangent function was used as the sliding mode robust term to improve the sliding mode control, reduce the chattering of the sliding mode algorithm and suppress the mass imbalance disturbance and random disturbance. A comparation of PID, α-order inverse system and the proposed algorithm was simulated. The results shows that the algorithm proposed can effectively compensate the nonlinear force, decouple the system and improve the chattering problem. At the same time, it can suppresses the external disturbance efficiently. -
表 1 系统计算参数
Table 1. System calculation parameters
参数 数值 参数 数值 $\;{{\boldsymbol{\varLambda }}}$ ${\text{diag}}\left( {80,80,80,80} \right)$ ${k_{i} }$ 0 ${{\boldsymbol{\phi}} ^{ - 1} }$ ${\text{diag}}\left( {15,15,15,15} \right)$ $ {k_{\text{d}}} $ 8.34 ${{\boldsymbol{\gamma}} }$ ${\text{diag}}\left( {20,20,20,20} \right)$ ${l_{\text{u}}}/{\text{m}}$ 0.273 ${\boldsymbol{\nu}} $ 20 ${l_{\text{d}}}/{\text{m}}$ 0.282 ${\boldsymbol{\delta}} $ $5 \times {10^{ - 3}}$ $m/{\text{kg}}$ 56.8 ${b_{j} }$ 10 $J/({\text{kg}} \cdot {{\text{m}}^{-2}})$ 1.14 ${ { {\boldsymbol{c} } }_{i} }$ $[ { - 3}\quad { - 2}\quad{ - 1}\quad 0\quad \\ 1\quad 2\quad 3 ] \times {10^{ - 6} }$ ${J_{\textit{z}}}/({\text{kg}} \cdot {{\text{m}}^{-2}})$ 0.93 ${k_{\text{p}}}$ $16\;680 $ $e_0/{\text{m}}$ $1.59 \times {10^{ - 6}}$ -
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