CN105700351A - Active fault tolerance control method for servo system - Google Patents

Abstract

The invention relates to an active fault tolerance control method for a servo system, and the method comprises the steps: 1, enabling a motor servo system to be modeled as a linear model with an unknown input and an executor fault; 2, designing a robust fault estimator, and estimating a system fault; 3, designing a feedback fault tolerance controller according to the estimated system fault. According to the invention, the method obtains the fault information of the system through the fault estimator, enables the motor servo system to be able to adjust the parameter and structure of the controller again according to the fault condition, thereby guaranteeing the stability of the system after a fault, and achieving the active processing of the fault. Moreover, the method is high in flexibility.

Description

The Active Fault-tolerant Control Method of servosystem

Technical field

The present invention relates to technical field of electromechanical control, in particular to the Active Fault-tolerant Control Method of a kind of servosystem。

Background technology

Servomotor, as the main energy sources power-equipment of modern industry, is widely used in the every field in life, such as lathe, printing, security protection with communicate。Along with the constantly progressive of science and technology and development, the complexity of servo-control system becomes more and more higher。Some large-scale servo-control systems, such as cannon positional servosystem, the active tracking system of radar antenna, aviation gun turret servosystem etc., particularity due to working environment and working condition, once break down, huge property loss and casualties will be caused, even result in the paralysis of whole system。Therefore, the safety of servo-control system, reliabilty and availability are had higher requirement by modern industry field and military field。In order to improve the safety and reliability of servo-control system, traditional method focuses primarily upon the robustness how promoting system。But, substantial amounts of practice have shown that nonetheless, the generation of still inevitable fault in system operation。Therefore, it is very important for efficiently reducing the impact that system brought by fault。

Fault Tolerance Control Technology is a kind of extremely important method improving security of system and reliability。Faults-tolerant control thought originates from the Niederlinski integrity control proposed in 1971, refers to if when the executor of system, sensor or other components break down, and system is still stable and has comparatively ideal characteristic。Faults-tolerant control can be divided into passive fault tolerant control and active tolerant control。Passive fault tolerant control refers to when not changing controller architecture and parameter, utilizes robust technique to make system that some fault is had insensitivity。Active tolerant control refer to system after fault occurs according to the failure condition parameter to controller and structural readjustment thus ensureing the stability of post-fault system, the fault that it is capable of occurring carries out active process, has significantly high motility。

Summary of the invention

It is desirable to provide the Active Fault-tolerant Control Method of a kind of servosystem improving robustness。

In order to achieve the above object, the invention provides the Active Fault-tolerant Control Method of servosystem, step includes step 1, and motor servo system is modeled as the linear model with Unknown worm and actuator failures；Step 2, designs robust Fault Estimation device, estimating system fault；Step 3, the system failure according to estimating sets up output feedack fault-tolerant controller。

Further, step 11, set up the kinetic model of motor servo system:

$\begin{array}{c}Rt+L\stackrel{\·}{I}-C_e\stackrel{\·}{{\mathrm{\θ}}_m}=U\\ K_{d}I=J_m\stackrel{\·\·}{{\mathrm{\θ}}_m}+b_m\stackrel{\·}{{\mathrm{\θ}}_m}+k({\mathrm{\θ}}_m-i_m{\mathrm{\θ}}_d)\\ i_{m}k({\mathrm{\θ}}_m-i_m{\mathrm{\θ}}_d)=J_d\stackrel{\·\·}{{\mathrm{\θ}}_d}+b_d\stackrel{\·}{{\mathrm{\θ}}_d}\end{array}---\left(1\right)$

Wherein, U is voltage, and R is stator resistance, and I is stator current, and L is stator inductance, θ_mIt is motor corner,It is motor speed, J_mIt is electric machine rotation inertia, θ_dIt is load corner,It is load speed, J_dRepresent load rotating inertia, C_eIt is back EMF coefficient, K_dIt is electromagnetic torque coefficient, b_mIt is equivalent viscous damping ratio, i_mBeing gear ratio, k is stiffness coefficient；Step 12, makes x₁=i, x₂=θ_m,x₄=θ_d,U=U, is converted into following state-space expression by kinetic model:

$\begin{array}{c}\stackrel{\·}{x}\left(t\right)Ax\left(t\right)+Bu\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(2\right)$

Wherein, x (t) is state vector, and u (t) is input vector, and y (t) is output vector,

$\begin{array}{ccc}A=\left[\begin{array}{ccccc}-\frac{R}{L}& 0& -\frac{C_e}{L}& 0& 0\\ 0& 0& 1& 0& 0\\ \frac{K_d}{J_m}& -\frac{k}{J_m}& -\frac{b_m}{J_m}& \frac{{\mathrm{ki}}_m}{J_m}& 0\\ 0& 0& 0& 0& 1\\ 0& \frac{{\mathrm{ki}}_m}{J_d}& 0& -\frac{{\mathrm{ki}}_m^2}{J_d}& -\frac{b_d}{J_d}\end{array}\right]& B=\left[\begin{array}{c}\frac{1}{L}\\ 0\\ 0\\ 0\\ 0\end{array}\right],& C=\left[\begin{array}{ccccc}0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\end{array};$ Step 13, it is assumed that w (t) is Unknown worm vector, and f (t) is actuator failures, obtains failure system model and is:

$\begin{array}{c}\stackrel{\·}{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)+E_{w}w\left(t\right)+E_{f}f\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(3\right)$

Wherein, E_wIt is known suitable dimension matrix, E_fIt is that ffault matrix represents the actuator failures impact on system。

Further, step 21, design the robust Fault Estimation device based on observer, state space equation is as follows:

$\begin{array}{c}\stackrel{\·}{\hat{x}}\left(t\right)=A\hat{x}\left(t\right)+Bu\left(t\right)+E_f\hat{f}\left(t\right)L\left(\hat{y}\right(t)-y(t\left)\right)\\ \hat{y}\left(t\right)=C\hat{x}\left(t\right)\end{array}---\left(4\right)$

Wherein,It is the state vector of estimator,It is the output vector of estimator,Being Fault Estimation vector, L is estimator matrix；Step 22, when $\begin{array}{cc}{e}_x\left(t\right)=\hat{x}\left(t\right)-x\left(t\right),& e_y\left(t\right)=\hat{y}\left(t\right)-y\left(t\right)\end{array},$ $e_f\left(t\right)=\hat{f}\left(t\right)-f\left(t\right),$ Obtain $\stackrel{\·}{e_x}\left(t\right)=(A-LC)e_x\left(t\right)+E_f{e}_f\left(t\right)-E_{w}w\left(t\right)---\left(5\right)$

e_y(t)=Ce_xT () (6) step 23, sets up Fault Estimation strategy:

$\stackrel{\·}{\hat{f}}\left(t\right)=-{\mathrm{Fe}}_y\left(t\right)---\left(7\right)$

Obtain ${\stackrel{\·}{e}}_f\left(t\right)=\stackrel{\·}{\hat{f}}\left(t\right)-\stackrel{\·}{f}\left(t\right)=-{\mathrm{FCe}}_x\left(t\right)-\stackrel{\·}{f}\left(t\right)---\left(8\right)$

According to formula (5) and (8), then error system is:

$\begin{array}{c}\stackrel{\·}{\stackrel{\‾}{e}}\left(t\right)=(\stackrel{\‾}{A}-\stackrel{\‾}{L}\stackrel{\‾}{C})\stackrel{\‾}{e}\left(t\right)-{\stackrel{\‾}{E}}_{w}v\left(t\right)\\ e_f\left(t\right)=\stackrel{\‾}{I}\stackrel{\‾}{e}\left(t\right)\end{array}---\left(9\right)$

Wherein $\begin{array}{cccccc}\stackrel{\‾}{e}\left(t\right)=\left[\begin{array}{c}{e}_x\left(t\right)\\ e_f\left(t\right)\end{array}\right],& v\left(t\right)=\left[\begin{array}{c}w\left(t\right)\\ \stackrel{\·}{f}\left(t\right)\end{array}\right],& \stackrel{\‾}{A}=\left[\begin{array}{cc}A& E_f\\ 0& 0\end{array}\right],& \stackrel{\‾}{L}=\left[\begin{array}{c}L\\ F\end{array}\right],& \stackrel{\‾}{C}=\left[\begin{array}{cc}C& 0\end{array}\right],& {\stackrel{\‾}{E}}_w=\left[\begin{array}{cc}{E}_w& 0\\ 0& I\end{array}\right],\end{array}$ $\stackrel{\‾}{I}=\left[\begin{array}{cc}0& I\end{array}\right].$

Design the fault-tolerant controller of output feedack further:

$\begin{array}{c}\stackrel{\·}{\mathrm{\ζ}}\left(t\right)=A_k\mathrm{\ζ}\left(t\right)+B_{k}y\left(t\right)\\ u\left(t\right)=C_k\mathrm{\ζ}\left(t\right)+D_{k}y\left(t\right)+r\left(t\right)-B^{*}E\hat{f}\left(t\right)\end{array}---\left(15\right)$

Wherein ζ (t) ∈ RⁿBeing the state vector of controller, r (t) is reference input, $\begin{array}{cc}{C}_k=(\stackrel{\‾}{\stackrel{\‾}{C}}-D_{k}CX)M^{-T},& B_k=N^{-1}(\stackrel{\‾}{\stackrel{\‾}{B}}-{\mathrm{YBD}}_k)\end{array},$ $A_k=N^{-1}(\stackrel{\‾}{\stackrel{\‾}{A}}-Y(A+{\mathrm{BD}}_{k}C\left)X\right)M^{-T}-B_k{\mathrm{CXM}}^{-1}-N^{-1}{\mathrm{YBC}}_k,$ And M, N ∈ R^n×nMeet MN^T=I-XY。

The Active Fault-tolerant Control Method of the servosystem according to the present invention, the fault message of system is obtained by fault approximator, make the motor servo system can according to the failure condition parameter to controller and structural readjustment thus ensureing the stability of post-fault system, realize the fault to occurring and carry out active process, there is significantly high motility。

Accompanying drawing explanation

The accompanying drawing constituting the part of the application is used for providing a further understanding of the present invention, and the schematic description and description of the present invention is used for explaining the present invention, is not intended that inappropriate limitation of the present invention。In the accompanying drawings:

Fig. 1 is the structure diagram of motor servo system；

Fig. 2 is the faults-tolerant control structure chart of motor servo system；

Fig. 3 is fault-signal f (t) and Fault Estimation signal

Fig. 4 is output y₁(t) response curve；

Fig. 5 is output y₂(t) response curve。

Detailed description of the invention

Describe the present invention below with reference to the accompanying drawings and in conjunction with the embodiments in detail。

Step 1, the modeling of system

Method according to modelling by mechanism, in conjunction with structure and the physical theorem of motor, sets up the kinetic model of motor servo system:

$\begin{array}{c}RI+L\stackrel{\·}{I}-C_e\stackrel{\·}{{\mathrm{\θ}}_m}=U\\ K_{d}I=J_m\stackrel{\·\·}{{\mathrm{\θ}}_m}+b_m\stackrel{\·}{{\mathrm{\θ}}_m}+k({\mathrm{\θ}}_m-i_m{\mathrm{\θ}}_d)\\ i_{m}k({\mathrm{\θ}}_m-i_m{\mathrm{\θ}}_d)=J_d\stackrel{\·\·}{{\mathrm{\θ}}_d}+b_d\stackrel{\·}{{\mathrm{\θ}}_d}\end{array}---\left(1\right)$

Wherein, U is voltage, and R is stator resistance, and I is stator current, and L is stator inductance, θ_mIt is motor corner,It is motor speed, J_mIt is electric machine rotation inertia, θ_dIt is load corner,It is load speed, J_dRepresent load rotating inertia, C_eIt is back EMF coefficient, K_dIt is electromagnetic torque coefficient, b_mIt is equivalent viscous damping ratio, i_mBeing gear ratio, k is stiffness coefficient。

Make x₁=i, x₂=θ_m,x₄=θ_d,U=U。Then system (1) can be converted into following state-space expression:

$\begin{array}{c}\stackrel{\·}{x}\left(t\right)Ax\left(t\right)+Bu\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(2\right)$

Wherein, x (t) is state vector, and u (t) is input vector, and y (t) is output vector,

$\begin{array}{ccc}A=\left[\begin{array}{ccccc}-\frac{R}{L}& 0& -\frac{C_e}{L}& 0& 0\\ 0& 0& 1& 0& 0\\ \frac{K_d}{J_m}& -\frac{k}{J_m}& -\frac{b_m}{J_m}& \frac{{\mathrm{ki}}_m}{J_m}& 0\\ 0& 0& 0& 0& 1\\ 0& \frac{{\mathrm{ki}}_m}{J_d}& 0& -\frac{{\mathrm{ki}}_m^2}{J_d}& -\frac{b_d}{J_d}\end{array}\right]& B=\left[\begin{array}{c}\frac{1}{L}\\ 0\\ 0\\ 0\\ 0\end{array}\right],& C=\left[\begin{array}{ccccc}0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\end{array}.$

Consider that real system can be subject to external interference and effect of noise, and these impacts are generally modeled as the Unknown worm of system。Further, since actuator failures can cause the mutation of executor, system model considers actuator failures。Make w (t) and f (t) represent Unknown worm vector sum actuator failures respectively, such that it is able to obtain failure system model be:

$\begin{array}{c}\stackrel{\·}{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)+E_{w}w\left(t\right)+E_{f}f\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(3\right)$

Wherein, E_wIt is known suitable dimension matrix, E_fIt is that ffault matrix represents the actuator failures impact on system。

Step 2, based on the design of the robust Fault Estimation device of observer

In order to estimate the fault of system, designing a robust Fault Estimation device based on observer, its state space equation is as follows:

$\begin{array}{c}\stackrel{\·}{\hat{x}}\left(t\right)=A\hat{x}\left(t\right)+Bu\left(t\right)+E_f\hat{f}\left(t\right)L\left(\hat{y}\right(t)-y(t\left)\right)\\ \hat{y}\left(t\right)=C\hat{x}\left(t\right)\end{array}---\left(4\right)$

Wherein,It is the state vector of estimator,It is the output vector of estimator,Being Fault Estimation vector, L is estimator matrix。

Order $\begin{array}{ccc}{e}_x\left(t\right)=\hat{x}\left(t\right)-x\left(t\right),& e_y\left(t\right)=\hat{y}\left(t\right)-y\left(t\right),& e_f\left(t\right)=\hat{f}\left(t\right)-f\left(t\right)\end{array},$ Then

$\stackrel{\·}{e_x}\left(t\right)=(A-LC)e_x\left(t\right)+E_f{e}_f\left(t\right)-E_{w}w\left(t\right)---\left(5\right)$

e_y(t)=Ce_x(t)(6)

Design following Fault Estimation strategy:

$\stackrel{\·}{\hat{f}}\left(t\right)=-{\mathrm{Fe}}_y\left(t\right)---\left(7\right)$

Then

${\stackrel{\·}{e}}_f\left(t\right)=\stackrel{\·}{\hat{f}}\left(t\right)-\stackrel{\·}{f}\left(t\right)=-{\mathrm{FCe}}_x\left(T\right)-\stackrel{\·}{f}\left(t\right)---\left(8\right)$

According to formula (5) and (8), then error system is:

$\begin{array}{c}\stackrel{\·}{\stackrel{\‾}{e}}\left(t\right)=(\stackrel{\‾}{A}-\stackrel{\‾}{LC})\stackrel{\‾}{e}\left(t\right)-{\stackrel{\‾}{E}}_{w}v\left(t\right)\\ e_f\left(t\right)=\stackrel{\‾}{I}\stackrel{\‾}{e}\left(t\right)\end{array}---\left(9\right)$

Wherein

$\begin{array}{cccccc}\stackrel{\‾}{e}\left(t\right)=\left[\begin{array}{c}{e}_x\left(t\right)\\ e_f\left(t\right)\end{array}\right],& v\left(t\right)=\left[\begin{array}{c}w\left(t\right)\\ \stackrel{\·}{f}\left(t\right)\end{array}\right],& \stackrel{\‾}{A}=\left[\begin{array}{cc}A& E_f\\ 0& 0\end{array}\right],& \stackrel{\‾}{L}=\left[\begin{array}{c}L\\ F\end{array}\right],& \stackrel{\‾}{C}=\left[\begin{array}{cc}C& 0\end{array}\right],& {\stackrel{\‾}{E}}_w=\left[\begin{array}{cc}{E}_w& 0\\ 0& I\end{array}\right],\end{array}$ $\stackrel{\‾}{I}=\left[\begin{array}{cc}0& I\end{array}\right].$

Bounded Real Lemma according to disc area POLE PLACEMENT USING lemma and continuous system, for given H_∞, if there is a symmetric positive definite matrix in performance indications γ and disc area D (α, τ)Meet with matrix R:

$\left[\begin{array}{cc}-\stackrel{\&OverBar;}{P}& \stackrel{\&OverBar;}{P}\stackrel{\&OverBar;}{A}-R\stackrel{\&OverBar;}{C}-\mathrm{\&alpha;}\stackrel{\&OverBar;}{P}\\ *& -{\mathrm{\&tau;}}^2\stackrel{\&OverBar;}{P}\end{array}\right]<0---\left(13\right)$

$\left[\begin{array}{ccc}\stackrel{\&OverBar;}{P}\stackrel{\&OverBar;}{A}+{\stackrel{\&OverBar;}{A}}^T\stackrel{\&OverBar;}{P}-R{\stackrel{\&OverBar;}{C}}^T{R}^T& R{\stackrel{\&OverBar;}{F}}_w-\stackrel{\&OverBar;}{P}{\stackrel{\&OverBar;}{E}}_w& I\\ *& -\mathrm{\&gamma;}I& 0\\ *& *& -\mathrm{\&gamma;}I\end{array}\right]<0---\left(14\right)$

ThenEigenvalue be positioned in disc area D (α, τ), and error system (9) meets H_∞Performance | | e_f(t)||₂< γ | | v (t) | |₂, gain matrix

Step 3, the design of output feedack fault-tolerant controller

Assume rank (B, E_f)=rank (B), it is equivalent to one matrix B of existence^*∈R^p×nMeet: (I-BB^*)E_f=0。

Design the fault-tolerant controller of following output feedack:

$\begin{array}{c}\stackrel{\&CenterDot;}{\mathrm{\&zeta;}}\left(t\right)=A_k\mathrm{\&zeta;}\left(t\right)+B_{k}y\left(t\right)\\ u\left(t\right)=C_k\mathrm{\&zeta;}\left(t\right)+D_{k}y\left(t\right)+r\left(t\right)-B^{*}E\hat{f}\left(t\right)\end{array}---\left(15\right)$

Wherein ζ (t) ∈ RⁿBeing the state vector of controller, r (t) is reference input, A_k,B_k,C_k,D_kIt it is the unknown matrix needing to solve。

Making r (t)=0, (15) are brought into (1) can be obtained:

$\begin{array}{c}\stackrel{\&CenterDot;}{x}\left(t\right)=Ax\left(t\right)+{\mathrm{BC}}_k\mathrm{\&zeta;}\left(t\right)+{\mathrm{BD}}_{k}Cx\left(t\right)-{\mathrm{BB}}^{*}{E}_f\hat{f}\left(t\right)+E_f\left(t\right)+E_{w}w\left(t\right)\\ =(A+{\mathrm{BD}}_{k}C)x\left(t\right)+{\mathrm{BC}}_k\mathrm{\&zeta;}\left(t\right)-E_f{e}_f\left(t\right)+E_{w}w\left(t\right)\end{array}---\left(16\right)$

Can obtain according to formula (15) and (16):

$\stackrel{\&CenterDot;}{\tilde{x}}\left(t\right)=\tilde{A}\tilde{x}\left(t\right)+\tilde{D}\mathrm{\&upsi;}\left(t\right)---\left(17\right)$

Wherein $\begin{array}{cccc}\tilde{x}\left(t\right)=\left[\begin{array}{c}x\left(t\right)\\ \mathrm{\&zeta;}\left(t\right)\end{array}\right],& \tilde{A}=\left[\begin{array}{cc}A+{\mathrm{BD}}_{k}C& {\mathrm{BC}}_k\\ B_{k}C& A_k\end{array}\right],& \mathrm{\&upsi;}\left(t\right)=\left[\begin{array}{c}w\left(t\right)\\ e_f\left(t\right)\end{array}\right],& \tilde{D}=\left[\begin{array}{cc}{E}_w& -E_f\\ B_k{F}_w& 0\end{array}\right].\end{array}$

A symmetric positive definite matrix is there is in the Bounded Real Lemma according to continuous system it can be seen that and if only ifMake

$\left[\begin{array}{ccc}\tilde{P}\tilde{A}+{\tilde{A}}^T\tilde{P}& \tilde{P}\tilde{D}& I\\ *& -\mathrm{\&gamma;}I& 0\\ *& *& -\mathrm{\&gamma;}I\end{array}\right]<0---\left(22\right)$

Set up, then have $\left|\right|\tilde{x}\left(t\right)|{|}_2<\mathrm{\&gamma;}|\left|\mathrm{\&mu;}\left(t\right)\right|{|}_2.$

Order

$\begin{array}{cc}\tilde{P}=\left[\begin{array}{cc}Y& N\\ B^T& W\end{array}\right],& {\tilde{P}}^{-1}=\left[\begin{array}{cc}X& M\\ M^T& Z\end{array}\right]\end{array}$

Wherein X, Y ∈ R^n×nBeing positive definite symmetric matrices, other matrixes are suitable dimensions。ByCan obtain:

$\tilde{P}\left[\begin{array}{cc}X& I\\ M^T& 0\end{array}\right]=\left[\begin{array}{cc}I& Y\\ 0& N^T\end{array}\right]$

Definition

$\begin{array}{cc}{F}_1=\left[\begin{array}{cc}X& I\\ M^T& 0\end{array}\right],& F_2=\left[\begin{array}{cc}I& Y\\ 0& N^T\end{array}\right]\end{array}$

Then can obtainDue toKnown

$E_1^T\tilde{P}{F}_1=F_2^T{F}_1=\left[\begin{array}{cc}X& I\\ I& Y\end{array}\right]>0---\left(23\right)$

Thus obtain

$\left[\begin{array}{cc}X& I\\ I& Y\end{array}\right]>0---\left(24\right)$

Both sides premultiplication respectively to inequality (22)Diag{F is taken advantage of with the right side₁, I, I}, with season $\begin{array}{cc}\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{A}}=Y(A+{\mathrm{BD}}_{k}C)X+{\mathrm{NB}}_{k}CX+{\mathrm{YBC}}_k{M}^T+{\mathrm{NA}}_k{M}^T,& \stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}={\mathrm{YBD}}_k+{\mathrm{NB}}_k,\end{array}$ $\begin{array}{cc}\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}=D_{k}CX+C_k{M}^T,& \stackrel{\&OverBar;}{\stackrel{\&OverBar;}{D}}=D_k,\end{array}$ Then can obtain following formula:

$\left[\begin{array}{ccccc}{T}_{11}& T_{12}& E_w& -E_f& X\\ *& T_{22}& {\mathrm{YE}}_w& -{\mathrm{YE}}_f& I\\ *& *& -\mathrm{\&gamma;}I& 0& 0\\ *& *& *& -\mathrm{\&gamma;}I& 0\\ *& *& *& *& -\mathrm{\&gamma;}I\end{array}\right]<0---\left(25\right)$

Wherein matrix $\begin{array}{cc}{T}_{11}=AX+{\mathrm{XA}}^T+B\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}}^T{B}^T,& T_{12}={\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{A}}}^T+A+B\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{D}}C\end{array},$ $T_{22}=YA+A^T+\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}C+C^T{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}}^T.$

Known for given γ > 0 further, and if only if exists symmetric positive definite matrix X, Y ∈ R^n×nAnd matrix $\begin{array}{cccc}\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{A}}\&Element;R^{n\&times;n},& \stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}\&Element;R^{n\&times;n},& \stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}\&Element;R^{p\&times;n},& \stackrel{\&OverBar;}{\stackrel{\&OverBar;}{D}}\&Element;R^{p\&times;n}\end{array},$ Meet formula (24) and (25), then system (17) is robust asymptotic stability, and meetsAnd then the parameter matrix that can obtain output feedack fault-tolerant controller is:

$\begin{array}{ccc}{D}_k=\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{D}},& C_k=(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}-D_{k}CX)M^{-T},& B_k=N^{-1}(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}-{\mathrm{YBD}}_k),\end{array}$

$A_k=N^{-1}(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{A}}-Y(A+{\mathrm{BD}}_{k}C\left)X\right)M^{-T}-B_k{\mathrm{CXM}}^{-1}-N^{-1}{\mathrm{YBC}}_k,$

Wherein M, N ∈ R^n×nMeet MN^T=I-XY。

In order to obtain the coefficient matrix of fault-tolerant controller, first have to determine matrix M and N。Due toMN can be obtained^T=I-XY。And then non-singular matrix M and N can be obtained by the singular value decomposition of I-XY or ORTHOGONAL TRIANGULAR DECOMPOSITION。

For apparent, technical scheme is stated, it is considered to the following parameter of electric machine: L=50mH, R=2.6 Ω, C_e=67.2V/KRPM, k=5.6Nm/rad, K_d=1.066N m s/A, i_m=1, J_m=0.003kg m², J_d=0.0026kg m², b_m=0.015Nm s/rad, b_d=0.02Nm s/rad。Assume E_w=0.1* [1；1；1；1；1], E_f=[20；0；0；0；0], r (t)=10, the sampling period is h=0.01。

Assume that actuator failures signal is:

$f\left(t\right)=\left\{\begin{array}{cc}0,& 0s\&le;t<35s\\ 5(1-e^{0.5(t-35)}),& 35s\&le;t<60s\\ 5-10(1-e^{0.5(t-65)}),& 65s\&le;t<100s\end{array}\right.$

Fig. 3 represents fault-signal f (t) and Fault Estimation signalAs can be seen from Figure 3 proposed method of estimation is capable of the accurate estimation to fault。

Bounded Real Lemma according to disc area POLE PLACEMENT USING lemma and continuous system, chooses H_∞Performance indications γ=2.5 and disc area D (-4.6,4.6), the gain matrix that can try to achieve fault approximator is:

$L={10}^3\&times;\left[\begin{array}{cc}1.0170& 1.2514\\ 0.0162& 0.0046\\ -0.1578& -0.1332\\ 0.0194& 0.0010\\ -0.0375& 0.0124\end{array}\right],$ F=[-165.914119.6621]。

According to theorem 2, choose H_∞Performance indications γ=6.2, the parameter matrix that can obtain output feedack fault-tolerant controller is:

$A_k=\left[\begin{array}{ccccc}-42.3819& 259.0903& 32.7553& 100.2476& 185.6862\\ 1.5216& -2.2448& 25.8207& 70.1484& 20.8774\\ -0.1529& 4.0154& 6.0280& 19.0938& 20.4006\\ 0.0030& -3.8770& -6.7846& -17.3372& 3.6168\\ -0.0189& -2.0185& -3.6650& -8.7004& -10.1247\end{array}\right],$ $B_k=\left[\begin{array}{cc}1.8626& -1.0862\\ -6.6333& 5.0585\\ 8.0979& 2.6601\\ -1.8540& 6.3596\\ 3.4822& 9.1496\end{array}\right],$ C_k=[-0.0014-0.01830.07210.4150-0.5883], D_k=[-0.2690-0.1460]。

Fig. 4 and Fig. 5 gives the output response curve of system。From simulation result it can be seen that based on online Fault Estimation, the output feedack fault-tolerant controller of design can make system keep good performance when breaking down。

The foregoing is only the preferred embodiments of the present invention, be not limited to the present invention, for a person skilled in the art, the present invention can have various modifications and variations。All within the spirit and principles in the present invention, any amendment of making, equivalent replacement, improvement etc., should be included within protection scope of the present invention。

Claims (4)

1. the Active Fault-tolerant Control Method of a servosystem, it is characterised in that: described control method includes:

Step 1, is modeled as the linear model with Unknown worm and actuator failures by motor servo system；

Step 2, designs robust Fault Estimation device, estimating system fault；

Step 3, designs output feedack fault-tolerant controller according to the system failure of described estimation。

2. control method according to claim 1, it is characterised in that described step 1 includes:

Step 11, sets up the kinetic model of motor servo system:

$\begin{array}{c}RI+L\stackrel{\&CenterDot;}{I}+C_e\stackrel{.}{{\mathrm{\&theta;}}_m}=U\\ K_{d}I=J_m\stackrel{\mathrm{..}}{{\mathrm{\&theta;}}_m}+b_m\stackrel{.}{{\mathrm{\&theta;}}_m}+k({\mathrm{\&theta;}}_m-i_m{\mathrm{\&theta;}}_d)\\ i_{m}k({\mathrm{\&theta;}}_m-i_m{\mathrm{\&theta;}}_d)=J_d\stackrel{\mathrm{..}}{{\mathrm{\&theta;}}_d}+b_d\stackrel{.}{{\mathrm{\&theta;}}_d}\end{array}---\left(1\right)$

Wherein, U is voltage, and R is stator resistance, and I is stator current, and L is stator inductance, θ_mIt is motor corner,It is motor speed, J_mIt is electric machine rotation inertia, θ_dIt is load corner,It is load speed, J_dRepresent load rotating inertia, C_eIt is back EMF coefficient, K_dIt is electromagnetic torque coefficient, b_mIt is equivalent viscous damping ratio, i_mBeing gear ratio, k is stiffness coefficient；

Step 12, makes x₁=i, x₂=θ_m,x₄=θ_d,U=U, is converted into following state-space expression by described kinetic model:

$\begin{array}{c}\stackrel{\&CenterDot;}{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(2\right)$

Wherein, x (t) is state vector, and u (t) is input vector, and y (t) is output vector,

$A=\left[\begin{array}{ccccc}-\frac{R}{L}& 0& -\frac{C_e}{L}& 0& 0\\ 0& 0& 1& 0& 0\\ \frac{K_d}{J_m}& -\frac{k}{J_m}& -\frac{b_m}{J_m}& \frac{{\mathrm{ki}}_m}{J_m}& 0\\ 0& 0& 0& 0& 1\\ 0& \frac{{\mathrm{ki}}_m}{J_d}& 0& -\frac{{\mathrm{ki}}_m^2}{J_d}& -\frac{b_d}{J_d}\end{array}\right],B=\left[\begin{array}{c}\frac{1}{L}\\ 0\\ 0\\ 0\\ 0\end{array}\right],C=\left[\begin{array}{ccccc}0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right];$

Step 13, it is assumed that w (t) is Unknown worm vector, and f (t) is actuator failures, obtains failure system model and is:

$\begin{array}{c}\stackrel{\&CenterDot;}{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)+E_{w}w\left(t\right)+E_{f}f\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}---\left(3\right)$

Wherein, E_wIt is known suitable dimension matrix, E_fIt is that ffault matrix represents the actuator failures impact on system。

3. control method according to claim 2, it is characterised in that described step 2 includes:

Step 21, sets up the robust Fault Estimation device based on observer, and state space equation is as follows:

$\begin{array}{c}\stackrel{\&CenterDot;}{\hat{x}}\left(t\right)=A\hat{x}\left(t\right)+Bu\left(t\right)+E_f\hat{f}\left(t\right)-L\left(\hat{y}\right(t)-y(t\left)\right)\\ \hat{y}\left(t\right)=C\hat{x}\left(t\right)\end{array}---\left(4\right)$

Wherein,It is the state vector of estimator,It is the output vector of estimator,Being Fault Estimation vector, L is estimator matrix；

Step 22, order $e_x\left(t\right)=\hat{x}\left(t\right)-x\left(t\right),e_y\left(t\right)=\hat{y}\left(t\right)-y\left(t\right),e_f\left(t\right)=\hat{f}\left(t\right)-f\left(t\right),$ Obtain $\stackrel{.}{e_x}\left(t\right)=(A-LC)e_x\left(t\right)+E_f{e}_f\left(t\right)-E_{w}w\left(t\right)---\left(5\right)$

e_y(t)=Ce_x(t)(6)

Step 23, sets up Fault Estimation strategy:

$\stackrel{\&CenterDot;}{\hat{f}}\left(t\right)=-{\mathrm{Fe}}_y\left(t\right)---\left(7\right)$

Obtain ${\stackrel{\&CenterDot;}{e}}_f\left(t\right)=\stackrel{\&CenterDot;}{\hat{f}}\left(t\right)-\stackrel{\&CenterDot;}{f}\left(t\right)=-{\mathrm{FCe}}_x\left(t\right)-\stackrel{\&CenterDot;}{f}\left(t\right)---\left(8\right)$

According to formula (5) and (8), then error system is:

$\begin{array}{c}\stackrel{\&CenterDot;}{\stackrel{\&OverBar;}{e}}\left(t\right)=(\stackrel{\&OverBar;}{A}-\stackrel{\&OverBar;}{L}\stackrel{\&OverBar;}{C})\stackrel{\&OverBar;}{e}\left(t\right)-{\stackrel{\&OverBar;}{E}}_{w}v\left(t\right)\\ e_f\left(t\right)=\stackrel{\&OverBar;}{I}\stackrel{\&OverBar;}{e}\left(t\right)\end{array}---\left(9\right)$

Wherein $\stackrel{\&OverBar;}{e}\left(t\right)=\left[\begin{array}{c}{e}_x\left(t\right)\\ e_f\left(t\right)\end{array}\right],v\left(t\right)=\left[\begin{array}{c}w\left(t\right)\\ \stackrel{\&CenterDot;}{f}\left(t\right)\end{array}\right],\stackrel{\&OverBar;}{A}=\left[\begin{array}{cc}A& E_f\\ 0& 0\end{array}\right],\stackrel{\&OverBar;}{L}=\left[\begin{array}{c}L\\ F\end{array}\right],\stackrel{\&OverBar;}{C}=\&lsqb;\begin{array}{cc}C& 0\end{array}\&rsqb;,{\stackrel{\&OverBar;}{E}}_w=\left[\begin{array}{cc}{E}_w& 0\\ 0& I\end{array}\right],$ $\stackrel{\&OverBar;}{I}=\&lsqb;\begin{array}{cc}0& I\end{array}\&rsqb;.$

4. control method according to claim 3, it is characterised in that described step 3 includes:

Set up the fault-tolerant controller of output feedack:

$\begin{array}{c}\stackrel{\&CenterDot;}{\mathrm{\&zeta;}}\left(t\right)=A_k\mathrm{\&zeta;}\left(t\right)+B_{k}y\left(t\right)\\ u\left(t\right)=C_k\mathrm{\&zeta;}\left(t\right)+D_{k}y\left(t\right)+r\left(t\right)-B^{*}E\hat{f}\left(t\right)\end{array}---\left(15\right)$

Wherein ζ (t) ∈ RⁿBeing the state vector of controller, r (t) is reference input,

$C_k=(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{C}}-D_{k}CX)M^{-T},B_k=N^{-1}(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{B}}-{\mathrm{YBD}}_k),$

$A_k=N^{-1}(\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{A}}-Y(A+{\mathrm{BD}}_{k}C\left)X\right)M^{-T}-B_k{\mathrm{CXM}}^{-T}-N^{-1}{\mathrm{YBC}}_k,$ And M, N ∈ R^n×nMeet MN^T=I-XY。