CN101709969A - Method for inhibiting moving-gimbal effects of single gimbal magnetically suspended control moment gyroscope - Google Patents

Abstract

The invention relates to a method for inhibiting moving-gimbal effects of a single gimbal magnetically suspended control moment gyroscope. In the method, a state equation of the single gimbal magnetically suspended control moment gyroscope is set up according to newton second law and a gyro technique equation; an inversion analysis of a system is calculated by utilizing an inverse system method; in a rotor system, moving-gimbal displacement of a magnetically suspended rotor is eliminated through state feedback of rotor displacement, an rotor reflection angular speed and a gimbal angular speed, and in a gimbal system, reactionary torque interference of the rotor to the gimbal system is eliminated through state feedback of the gimbal angular speed and the rotor reflection angular speed; and a robust servocontrol strategy is adopted to improve the robustness of the whole system. In the method, the moving-gimbal displacement of the magnetically suspended rotor is eliminated, the reactionary torque interference of the rotor system to the gimbal system is also eliminated in the gimbal moving process, and the stability and precision of the whole single gimbal magnetically suspended control moment gyroscope are improved. The method belongs to the technical field of aerospace control, and can be applied to high-precision control of magnetically suspended control moment gyroscopes.

Description

A kind of method that suppresses moving-gimbal effects of single gimbal magnetically suspended control moment gyroscope

Technical field

The present invention relates to the method for the moving framework effect of a kind of inhibition magnetic suspension control moment gyro of single framework (Control Moment Gyroscope-CMG), be applicable to the High Accuracy Control of single frame magnetic levitation control CMG.

Technical background

CMG is the crucial topworks of spacecraft attitude controls such as space station, large-scale satellite, quick maneuvering satellite and space maneuver platform.CMG is made up of the rotor and the framework rate servo of constant speed running, when framework rotates, forces rotor angular momentum to change, and outwards exports gyroscopic couple.The CMG output torque is directly proportional with the angular momentum of its high speed rotor.Spacecraft attitude topworks requires to have the advantages that the life-span is long, moment is big, precision is high, volume is little, low in energy consumption, and magnetic levitation CMG just becomes the important development direction of CMG because of having these advantages.

Yet; different with the rigid mechanical supporting is; magnetic suspension bearing is actually the ACTIVE CONTROL springiness supporting that possesses certain rigidity and damping; therefore framework rotates the rotor convected motion that causes and can disturb the motion of rotor with respect to rotor case; simultaneously, the gyroscopic effect of high speed rotor makes the motion of being disturbed rotor complicated more again, makes the rotor axis increasing of beating; thereby have the possibility of crash protection bearing, influenced the stability of magnetic suspension system.The phenomenon that the said frame rotation will make the radial angle displacement of the relative rotor case stator of high-speed magnetic levitation rotor enlarge markedly is referred to as moving framework effect.Thereby the excessive stability that not only can the crash protection bearing threatens magnetic suspension system of rotor relative angular displacement; and influence the directional precision of angular momentum; the rotor relative angular displacement also causes disturbance to frame system simultaneously; reduce framework speed is servo and CMG moment is exported response speed and precision; and then influence the precision and the degree of stability of posture control system, thereby must be suppressed moving framework effect.At the inhibition of moving framework effect, existing method mainly is to adopt the angular speed feedforward control.The angular speed feed forward control method is reduced to the additional moment disturbance of frame system to rotor-support-foundation system to moving framework effect, and the feedforward control by frame corners speed realizes the inhibition to the additonal disturbing force square that brings to rotor-support-foundation system because of the framework rate variation.But, the angular speed feed forward control method has been ignored frame system self dynamic process, and in real system, the dynamic process of frame system is an outwardness, therefore, this method can not compensate the additional moment disturbance of framework rotation to rotor-support-foundation system exactly, thereby can not eliminate the moving framework displacement of rotor fully.Simultaneously, because the angular speed feedforward control has been ignored the moment of reaction disturbance of rotor to frame system, moving framework disturbance unidirectional kinematic constraint and external disturbance have been simplified to, and in the system of reality, rotor will cause the fluctuation of frame corners speed to the moment of reaction of frame system when moving framework, thereby influence the precision of single frame magnetic levitation CMG output torque inevitably.

Summary of the invention

Technology of the present invention is dealt with problems and is: overcome the angular speed feedforward control and can not suppress the deficiency of moving framework effect to the influence of single frame magnetic levitation CMG stability and precision fully, a kind of method that suppresses the moving framework effect of single frame magnetic levitation CMG has been proposed, in rotor-support-foundation system,, in frame system, pass through the feedback of status of frame corners speed and rotor deflection angular speed is eliminated the moment of reaction disturbance of rotor to frame system by the feedback of status of rotor displacement, rotor deflection angular speed and frame corners speed having been eliminated the moving framework displacement of magnetic suspension rotor.

Technical solution of the present invention is: the state equation of setting up single frame magnetic levitation CMG according to Newton second law and gyro technology equation; It is contrary to utilize method of inverse to obtain the parsing of system; In rotor-support-foundation system,, in frame system, pass through the feedback of status of frame corners speed and rotor deflection angular speed is eliminated the moment of reaction disturbance of rotor to frame system by the feedback of status of rotor displacement, rotor deflection angular speed and frame corners speed having been eliminated the moving framework displacement of magnetic suspension rotor; Adopt robust servocontrol strategy to improve the robustness of total system.Specifically may further comprise the steps:

1, the state equation of magnetic levitation CMG when pedestal is motionless is according to Newton second law and the foundation of gyro technology equation:

$\left\{\begin{array}{c}\stackrel{\·}{X}=f(X,U)\\ Y=\mathrm{CX}\end{array}\right.---\left(1\right)$

Wherein,

$f(X,U)=\left[\begin{array}{c}\stackrel{\·}{x}\\ \stackrel{\·}{\mathrm{\β}}\\ \stackrel{\·}{y}\\ -\stackrel{\·}{\mathrm{\α}}\\ \stackrel{\·}{z}\\ \frac{1}{m}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibx}}{i}_{\mathrm{bx}}+k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\\ \frac{l_m}{J_y}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})+\frac{H}{J_y}(\stackrel{\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\·}{\mathrm{\ω}}}_g\\ \frac{1}{m}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}})\\ \frac{l_m}{J_x}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}}-k_{\mathrm{ibxy}}{i}_{\mathrm{by}}-k_{\mathrm{hby}}{y}_{\mathrm{bm}})+\frac{H}{J_x}(\stackrel{\·}{\mathrm{\β}}-\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\·}{\mathrm{\ω}}}_g\\ \frac{1}{m}(k_{\mathrm{iz}}{i}_z+k_{\mathrm{hz}}z)\\ \frac{n_p{\mathrm{\φ}}_f}{J_{\mathrm{gx}}}{i}_s-\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}}-k_{\mathrm{iay}}{i}_{\mathrm{ay}}-k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\end{array}\right]---\left(2\right)$

$C=\left[\begin{array}{ccccccccccc}1& l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& -l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& -l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]---\left(3\right)$

In the formula, the state variable of system $X={[x,\mathrm{\β},y,-\mathrm{\α},z,\stackrel{\·}{x},\stackrel{\·}{\mathrm{\β}},\stackrel{\·}{y},-\stackrel{\·}{\mathrm{\α}},\stackrel{\·}{z},{\mathrm{\ω}}_g]}^T;$ The input control variable U=[i of system _Ax, i _Bx, i _Ay, i _By, i _z, i _s] ^TThe output variable Y=[x of system _Am, x _Bm, y _Am, y _Bm, z, ω _g] ^TM is a rotor quality; As shown in Figure 3, magnetic suspension rotor is suspended by two radial direction magnetic bearings and two axial magnetic bearings, be called A end radial direction magnetic bearing, B end radial direction magnetic bearing and A end axial magnetic bearing, B and hold axial magnetic bearing, A, B end constitutes radially ax, bx, ay, by passage along the winding of X, Y direction respectively, rotor constitutes axial z passage along the winding of Z direction, and framework is axially perpendicular to rotor axial; J _x, J _yAnd J _zThe X that is respectively rotor to, Y to Z to moment of inertia, and J _x=J _yi _Ax, i _Bx, i _Ay, i _By, i _zBe respectively the Control current of rotor-support-foundation system ax, bx, ay, by, z passage; i _sBe the size of the stator current vector of frame motor, x _Am, x _Bm, y _Am, y _BmBe respectively rotor at transverse bearing A and B place with respect to the displacement of equilibrium position along X-axis and Y direction, α, β are the radially rotational displacement of rotor around X-axis and Y-axis,

Ω is respectively the angle of rotation speed of rotor around X, Y, Z axle; X, y, z are respectively the translation displacement of rotor centroid on X-axis, Y-axis and Z axle;

Be respectively the translation speed of rotor centroid along X-axis, Y-axis and Z axle; l _mThe expression radial direction magnetic bearing is to the distance of rotor center O; k _Iax, k _Ibx, k _Iay, k _IbyAnd k _IzBe respectively the radially current stiffness of four-way and axial passage; k _Hax, k _Hax, k _Hay, k _HbyAnd k _HzBe respectively the radially displacement rigidity of four-way and axial passage; ω _gAngle of rotation speed for framework; J _GxBe the axial rotation inertia of framework; n _pIt is the number of pole-pairs of frame motor; φ _fIt is the magnetic linkage of frame motor.

2, it is contrary to find the solution the parsing of magnetic levitation CMG;

According to resolving contrary computing method, the parsing of magnetic levitation CMG is contrary to be:

$\left\{\begin{array}{c}{i}_{\mathrm{ax}}=-\frac{k_{\mathrm{hax}}}{k_{\mathrm{iax}}}{x}_{\mathrm{am}}-\frac{J_z\mathrm{\Ω}}{2l_m{k}_{\mathrm{iax}}}(\stackrel{\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)-\frac{1}{4k_{\mathrm{iax}}}(m+\frac{J_y}{l_m^2}){\stackrel{\·\·}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iax}}}(m-\frac{J_y}{l_m^2}){\stackrel{\·\·}{x}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iax}}}{\stackrel{\·}{\mathrm{\ω}}}_g\\ i_{\mathrm{bx}}=-\frac{k_{\mathrm{hbx}}}{k_{\mathrm{ibx}}}{x}_{\mathrm{bm}}+\frac{J_z\mathrm{\Ω}}{2l_m{k}_{\mathrm{ibx}}}(\stackrel{\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)-\frac{1}{4k_{\mathrm{ibx}}}(m-\frac{J_y}{l_m^2}){\stackrel{\·\·}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{ibx}}}(m+\frac{J_y}{l_m^2}){\stackrel{\·\·}{x}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{ibx}}}{\stackrel{\·}{\mathrm{\ω}}}_g\\ i_{\mathrm{ay}}=-\frac{k_{\mathrm{hay}}}{k_{\mathrm{iay}}}{y}_{\mathrm{am}}-\frac{J_z\mathrm{\Ω}}{2l_m{k}_{\mathrm{iay}}}(\stackrel{\·}{\mathrm{\β}}-\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)-\frac{1}{4k_{\mathrm{iay}}}(m+\frac{J_y}{l_m^2}){\stackrel{\·\·}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iay}}}(m-\frac{J_y}{l_m^2}){\stackrel{\·\·}{y}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iay}}}{\stackrel{\·}{\mathrm{\ω}}}_g\\ i_{\mathrm{by}}=-\frac{k_{\mathrm{hby}}}{k_{\mathrm{iby}}}{y}_{\mathrm{am}}+\frac{J_z\mathrm{\Ω}}{2l_m{k}_{\mathrm{iby}}}(\stackrel{\·}{\mathrm{\β}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)-\frac{1}{4k_{\mathrm{iby}}}(m-\frac{J_y}{l_m^2}){\stackrel{\·\·}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iby}}}(m+\frac{J_y}{l_m^2}){\stackrel{\·\·}{y}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iby}}}{\stackrel{\·}{\mathrm{\ω}}}_g\\ i_z=\frac{1}{k_{\mathrm{ia}}}(m\stackrel{\·\·}{z}-k_{\mathrm{ha}}z)\\ i_s=\frac{1}{n_p{\mathrm{\φ}}_f}[\frac{\sqrt{2}}{2}(\stackrel{\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)+\frac{\sqrt{2}}{2}(\stackrel{\·}{\mathrm{\β}}-\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)+\frac{\sqrt{2}{J}_y}{4l_m}({\stackrel{\·\·}{x}}_{\mathrm{am}}-{\stackrel{\·\·}{x}}_{\mathrm{bm}}+{\stackrel{\·\·}{y}}_{\mathrm{am}}-{\stackrel{\·\·}{y}}_{\mathrm{bm}})-(J_{\mathrm{gx}}+J_y){\stackrel{\·}{\mathrm{\ω}}}_g]\end{array}\right.---\left(4\right)$

Wherein It is respectively the rotor displacement acceleration expectation value of rotor-position servo control unit output;

It is the frame corners acceleration expectation value of framework speed servo control unit output.

3, make up linear subsystem;

Constitute composite controlled object jointly by six current track inverters and single frame magnetic levitation CMG, this composite controlled object is with { i _Ax, i _Bx, i _Ay, i _By, i _z, i _sSix current signals are as input, with each passage displacement of magnetic suspension rotor and frame corners speed as output.Before the inverse system that step 2 is obtained is connected on composite controlled object, just form linear subsystem, this linear subsystem is made of the linear subsystem of five position second order integral forms, the linear subsystem of a speed single order integral form.Thereby eliminated the moving framework disturbance of frame system, eliminated the moment of reaction disturbance of rotor-support-foundation system frame system to rotor-support-foundation system.

4, design linear closed loop controller;

To five location subsystem and a speed subsystem, make up position robust servo controller and speed robust servo controller respectively.Choose suitable robust servo controller structure according to the robust theory of servomechanism.For second order linear subsystem after the decoupling zero and single order linear subsystem, can design respectively as Fig. 1 (is example with the ax passage) and control system shown in Figure 2, wherein the robust compensation device of the robust servo controller of second order linear subsystem is $W_1\left(s\right)=\frac{a_0+a_{1}s}{s},$ Calm compensator is W ₂(s)=k ₀+ k ₁S; The robust compensation device of the speed robust servo controller of single order linear subsystem is $W_3\left(s\right)=\frac{a_{2}s+a_3}{s},$ Calm compensator is W ₄(s)=0.S represents the variable of the mathematical model-transport function of control system in complex field, a in the formula ₀, a ₁Represent constructed robust compensation device W respectively ₁(s) integral coefficient and scale-up factor, k ₀, k ₁Represent calm compensator W respectively ₂(s) scale-up factor and differential coefficient, a ₂, a ₃Represent robust compensation device W respectively ₃(s) scale-up factor and integration number.Utilize the method for undetermined coefficients can solve each parameter value according to system performance index on this basis.

Principle of the present invention is:

The moving framework effect of magnetic levitation CMG is because the dynamics coupling between rotor-support-foundation system and the frame system causes in essence, there is opplied moment in frame system to rotor-support-foundation system, also there is the moment of reaction in rotor-support-foundation system to frame system simultaneously, therefore, to eliminate the influence of gyroscopic effect fully to system stability and precision, just must eliminate the disturbance of frame system, also must eliminate the moment of reaction disturbance of rotor simultaneously framework to rotor.The present invention in rotor-support-foundation system by the feedback of status of rotor displacement, rotor deflection angular speed and frame corners speed having been eliminated the moving framework displacement of magnetic suspension rotor, in frame system,, adopt robust servocontrol strategy to improve the robustness of total system on this basis by the feedback of status of frame corners speed and rotor deflection angular speed is eliminated the moment of reaction disturbance of rotor to frame system.

The structural representation of single frame magnetic levitation CMG as shown in Figure 3, rotor is suspended by two radial direction magnetic bearings and two axial magnetic bearings, be called A end radial direction magnetic bearing, B end radial direction magnetic bearing and A end axial magnetic bearing, B and hold axial magnetic bearing, A, B end constitutes radially ax, bx, ay, by passage along the winding of X, Y direction respectively, rotor constitutes axial z passage along the winding of Z direction, and framework is axially perpendicular to rotor axial.According to Newton second law and gyro technology equation, when pedestal is fixed on ground (being similar to inertial system), when framework with angular speed ω _gThe kinetic model of magnetic suspension rotor system is during rotation:

$\left\{\begin{array}{c}m\stackrel{\·\·}{x}=f_x=f_{\mathrm{ax}}+f_{\mathrm{bx}}\\ J_y(\stackrel{\·\·}{\mathrm{\β}}-\frac{\sqrt{2}}{2}{\stackrel{\·}{\mathrm{\ω}}}_g)-H(\stackrel{\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)=p_y=l_m(f_{\mathrm{ax}}-f_{\mathrm{bx}})\\ m\stackrel{\·\·}{y}=f_y=f_{\mathrm{ay}}+f_{\mathrm{by}}\\ J_x(\stackrel{\·\·}{\mathrm{\α}}+\frac{\sqrt{2}}{2}{\stackrel{\·}{\mathrm{\ω}}}_g)-H(\stackrel{\·}{\mathrm{\β}}-\frac{\sqrt{2}}{2}{\mathrm{\ω}}_g)=p_y=l_m(f_{\mathrm{by}}-f_{\mathrm{ay}})\\ m\stackrel{\·\·}{z}=f_z\end{array}\right.---\left(5\right)$

Wherein, rotor-support-foundation system radially four-way and axial passage magnetic axis load can adopt the linearizing method representation of piecewise approximation to be:

$\left\{\begin{array}{c}{f}_{\mathrm{ax}}=k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}\\ f_{\mathrm{bx}}=k_{\mathrm{ibx}}{i}_{\mathrm{bx}}+k_{\mathrm{hbx}}{x}_{\mathrm{bm}}\\ f_{\mathrm{ay}}=k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}}\\ f_{\mathrm{by}}=k_{\mathrm{iby}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}}\\ m\stackrel{\·\·}{z}=k_{\mathrm{iz}}{i}_{\mathrm{iz}}+k_{\mathrm{hz}}z\end{array}\right.---\left(6\right)$

The torque equilibrium equation of (because of coefficient of rolling friction is very little, this being similar to is rational) frame motor can be expressed as under the situation of ignoring frame system friction force:

$J_{\mathrm{gx}}{\stackrel{\·}{\mathrm{\ω}}}_g=T_e-\frac{\sqrt{2}}{2}(p_x-p_y)---\left(7\right)$

In the formula, p _xAnd p _yRepresent the moment that rotor is subjected to respectively on X and Y direction, T _eThe electromagnetic torque of representational framework motor, it can further be expressed as:

N in the formula _pThe number of pole-pairs of representational framework motor; φ _fIt is the magnetic linkage of frame motor; i _sSize for the stator current vector of frame motor.Bring (8) formula into (7) Shi Kede

In conjunction with (5) formula, (6) formula and (9) formula, the kinetic model that can obtain magnetic levitation CMG is

In (10) formula, definition status variable X, input variable U, output variable Y are respectively:

$X={[x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11}]}^T={[x,\mathrm{\β},y,-\mathrm{\α},z,\stackrel{\·}{x},\stackrel{\·}{\mathrm{\β}},\stackrel{\·}{y},-\stackrel{\·}{\mathrm{\α}},\stackrel{\·}{z},{\mathrm{\ω}}_g]}^T---\left(11\right)$

U＝[u ₁，u ₂，u ₃，u ₄，u ₅，u ₆] ^T＝[i _ax，i _bx，i _ay，i _by，i _z，i _s] ^T (12)

Y=[y ₁, y ₂, y ₃, y ₄, y ₅, y ₆] ^T=[x _Am, x _Bm, y _Am, y _Bm, z, ω _g] ^T(13) then system of equations (10) can be write as system state equation shown in (1) formula.System model (1) formula is carried out reversibility Analysis, at first output variable y ₁, y ₂, y ₃, y ₄, y ₅, y ₆Time t differentiate is obviously comprised input variable u in the derivative equation ₁, u ₂, u ₃, u ₄, u ₅, u ₆Till, the Jacobi matrix that can get system is:

$A=\left[\begin{array}{cccccc}\frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_1}{\∂u_6}\\ \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_2}{\∂u_6}\\ \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_3}{\∂u_6}\\ \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_4}{\∂u_6}\\ \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_5}{\∂u_6}\\ \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_1}& \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_2}& \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_3}& \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_4}& \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_5}& \frac{\∂{\stackrel{\·\·}{y}}_6}{\∂u_6}\end{array}\right]$

$=\left[\begin{array}{cccccc}\frac{1}{m}+\frac{l_m^2}{J_y}+\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{1}{m}-\frac{l_m^2}{J_y}-\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{l_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2J_{\mathrm{gx}}}& -\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& 0& \frac{\sqrt{2}{l}_m{n}_p{\mathrm{\&phi;}}_f}{2J_{\mathrm{gx}}}\\ \frac{1}{m}-\frac{l_m^2}{J_y}-\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{1}{m}+\frac{l_m^2}{J_y}+\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& -\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{l_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2J_{\mathrm{gx}}}& 0& -\frac{\sqrt{2}{l}_m{n}_p{\mathrm{\&phi;}}_f}{2J_{\mathrm{gx}}}\\ \frac{l_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2J_{\mathrm{gx}}}& -\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{1}{m}+\frac{l_m^2}{J_y}+\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{1}{m}-\frac{l_m^2}{J_y}-\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& 0& \frac{\sqrt{2}{l}_m{n}_p{\mathrm{\&phi;}}_f}{2J_{\mathrm{gx}}}\\ -\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{l_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2J_{\mathrm{gx}}}& \frac{1}{m}-\frac{l_m^2}{J_y}-\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& \frac{1}{m}+\frac{l_m^2}{J_y}+\frac{{\mathrm{<figure-callout\; id="2"\; label="l\; m"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">l</figure-callout>}}_m^{}}{2J_{\mathrm{gx}}}& 0& -\frac{\sqrt{2}{l}_m{n}_p{\mathrm{\&phi;}}_f}{2J_{\mathrm{gx}}}\\ 0& 0& 0& 0& \frac{k_{\mathrm{iz}}}{m}& 0\\ \frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}& -\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}& \frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}& -\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}& 0& \frac{n_p{\mathrm{\&phi;}}_f}{J_{\mathrm{gx}}}\end{array}\right]---\left(14\right)$

So

$\det \left(A\right)=\frac{16l_m^4{n}_p{\mathrm{\&phi;}}_f{k}_{\mathrm{iax}}{k}_{\mathrm{ibx}}{k}_{\mathrm{iay}}{k}_{\mathrm{iby}}{k}_{\mathrm{iz}}}{m^3{J}_x{J}_y{J}_{\mathrm{gx}}}\&NotEqual;0---\left(15\right)$

The relative rank of system are $\&PartialD;=({\mathrm{\&alpha;}}_1,{\mathrm{\&alpha;}}_2,{\mathrm{\&alpha;}}_3,{\mathrm{\&alpha;}}_4,{\mathrm{\&alpha;}}_5,{\mathrm{\&alpha;}}_6)=\left(\mathrm{2,2,2,2,2,1}\right),$ And satisfy $\underset{i=1}{\overset{6}{\mathrm{\&Sigma;}}}{\mathrm{\&alpha;}}_i=11\&le;n,$ Wherein n is the number of defined state variable X in original system (1) formula, and according to the inverse system theory, original system is reversible.So, utilize and resolve contrary computing method, can obtain inverse system as the formula (4).As shown in Figure 4, form inverter unit by ax channel current following-up type inverter, bx channel current following-up type inverter, ay channel current following-up type inverter, by channel current following-up type inverter, z channel current following-up type inverter and framework circulation passage current track inverter, inverter and single frame magnetic levitation CMG form composite controlled object jointly; This inverse system is connected on before the composite controlled object, can obtains relative rank and be respectively (α ₁, α ₂, α ₃, α ₄, α ₅) linear subsystem, this linear subsystem can be expressed as:

$\left\{\begin{array}{c}{G}_{\mathrm{ax}}=s^{-{\mathrm{\&alpha;}}_1}\\ G_{\mathrm{bx}}=s^{-{\mathrm{\&alpha;}}_2}\\ G_{\mathrm{ay}}=s^{-{\mathrm{\&alpha;}}_3}\\ G_{\mathrm{by}}=s^{-{\mathrm{\&alpha;}}_4}\\ G_z=s^{-{\mathrm{\&alpha;}}_5}\\ G_{{\mathrm{\&omega;}}_g}=s^{-{\mathrm{\&alpha;}}_6}\end{array}\right.---\left(16\right)$

G wherein _Ax, G _Bx, G _Ay, G _ByRepresent the radially transport function of four-way respectively, G _zBe the transport function of axial passage,

Be the transport function of framework passage, s represents the variable of the mathematical model-transport function of control system in complex field, α ₁, α ₂, α ₃, α ₄, α ₅, α ₆The relative rank of representing the linear subsystem after the decoupling zero respectively.A complicated nonlinear systems by formula (10) expression has just become 6 simple integration line style subsystems as the formula (16) like this, thereby eliminated of the moving framework displacement of frame corners rate variation, also eliminated the moment of reaction disturbance of rotor-support-foundation system framework to rotor.In order to make designed system have stronger robustness to model parameter and load disturbance, and the floating characteristic that guarantees the command signal tracking is unaffected, the present invention adopts the robust servo controller that linear subsystem is carried out the linear closed-loop design of Controller, forms the robust servo-control system.

For second order linear subsystem after the decoupling zero and single order linear subsystem, can design respectively as Fig. 1 (is example with the ax passage) and control system shown in Figure 2.Wherein the robust compensation device of the robust servo controller of second order linear subsystem is $W_1\left(s\right)=\frac{a_0+a_{1}s}{s},$ Calm compensator is W ₂(s)=k ₀+ k ₁S.The closed loop transfer function, of system is:

$\mathrm{\&Phi;}\left(s\right)=\frac{a_{1}s+a_0}{s({\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+k_{1}s+k_0)+a_{1}s+a_0}---\left(17\right)$

Order

$\mathrm{\&Phi;}\left(s\right)=\frac{{\mathrm{\&omega;}}_1^2(s+{\mathrm{\&delta;}}_1)}{(s+{\mathrm{\&delta;}}_1)({\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2{\mathrm{\&xi;}}_1{\mathrm{\&omega;}}_{1}s+{\mathrm{\&omega;}}_1^2)}---\left(18\right)$

According to the characteristic of second-order system, be s for the closed loop fundamental function ²+ 2 ξ ω s+ ω ²=0 second-order system, when $\mathrm{\&xi;}=\sqrt{2}/2$ The time, overshoot M _p% ≈ 5%, and transit time t _s≈ 3.5/ ω.Choosing like this ${\mathrm{\&xi;}}_1=\sqrt{2}/2$ Condition under, excessive time requirement can be determined ω according to system ₁, can choose suitable δ to the influence of system stability according to closed loop zero pole distribution ₁It is as follows then can to solve each parameter value according to the method for undetermined coefficients:

$\left\{\begin{array}{c}{a}_0={\mathrm{\&delta;}}_1{\mathrm{\&omega;}}_1^2\\ {\mathrm{<figure-callout\; id="1"\; label="a"\; filenames="H2009102412452E00012.png,H2009102412452E00031.png"\; state="\{\{state\}\}">a</figure-callout>}}_1={\mathrm{\&omega;}}_1^2\\ k_0=2{\mathrm{\&xi;}}_1{\mathrm{\&omega;}}_1{\mathrm{\&delta;}}_1\\ {\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="H2009102412452E00012.png,H2009102412452E00031.png"\; state="\{\{state\}\}">k</figure-callout>}}_1={\mathrm{\&delta;}}_1+2{\mathrm{\&xi;}}_1{\mathrm{\&omega;}}_1\end{array}\right.---\left(19\right)$

For the single order linear subsystem, the robust compensation device of the robust servo controller of single order linear subsystem is $W_3\left(s\right)=\frac{a_{2}s+a_3}{s},$ Calm compensator is W _A(s)=0.The closed loop transfer function, that can get system is

$\mathrm{\&Phi;}\left(s\right)=\frac{a_{2}s+a_3}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+a_{2}s+a_3}---\left(20\right)$

Order

$\mathrm{\&Phi;}\left(s\right)=\frac{2{\mathrm{\&xi;}}_2{\mathrm{\&omega;}}_{2}s+{\mathrm{\&omega;}}_2^2}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="H2009102412452E00021.png,H2009102412452E00031.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2{\mathrm{\&xi;}}_2{\mathrm{\&omega;}}_{2}s+{\mathrm{\&omega;}}_2^2}---\left(21\right)$

In like manner, require to determine ξ according to system performance index ₂And ω ₂, and then can try to achieve according to the method for undetermined coefficients

$\left\{\begin{array}{c}{a}_2=2{\mathrm{\&xi;}}_2{\mathrm{\&omega;}}_2\\ a_3={\mathrm{\&omega;}}_2^2\end{array}\right.---\left(22\right)$

As shown in Figure 6, constitute the linear closed-loop controller jointly by ax passage robust position controller, bx passage robust position controller, ay passage robust position controller, by passage robust position controller, z passage robust position controller and framework passage robust rate controller, realize robust control the linear subsystem after the decoupling zero.

The solution of the present invention is compared with existing scheme, and major advantage is:

(1) existing controlling schemes has been ignored frame system self dynamic process, therefore do not eliminate of the additional moment disturbance of moving framework effect fully to rotor-support-foundation system, the present invention on the basis of considering frame system self dynamic process by the feedback of status of rotor displacement, rotor deflection angular speed and frame corners speed having been eliminated of the additional moment disturbance of moving framework effect to rotor-support-foundation system, thereby eliminated the moving framework displacement of magnetic suspension rotor;

Rotor was to the retroaction of frame system when (2) existing controlling schemes had been ignored moving framework, had a strong impact on the precision of whole single frame magnetic levitation CMG output torque, the present invention passes through the feedback of status of frame corners speed and rotor deflection angular speed is eliminated the moment of reaction disturbance of rotor to frame system in frame system, improve the precision of frame system output angle speed, thereby improved the moment precision of whole single frame magnetic levitation CMG.

Description of drawings

Fig. 1 is the robust servocontrol structured flowchart of second order linear subsystem;

Fig. 2 is the robust servocontrol structured flowchart of single order linear subsystem;

Fig. 3 is a single frame magnetic levitation CMG structural representation;

Fig. 4 is for suppressing the schematic diagram of the moving framework effect of single frame magnetic levitation CMG;

Fig. 5 is a theory diagram of the present invention;

Fig. 6 is a process flow diagram of the present invention.

Specific embodiments

Objective for implementation of the present invention as shown in Figure 3, rotor is suspended by two radial direction magnetic bearings and two axial magnetic bearings, be called A end radial direction magnetic bearing, B end radial direction magnetic bearing and A end axial magnetic bearing, B and hold axial magnetic bearing, A, B end constitutes radially ax, bx, ay, by passage along the winding of X, Y direction respectively, rotor constitutes axial z passage along the winding of Z direction, and framework is axially perpendicular to rotor axial.Specific embodiments of the present invention as shown in Figure 5, concrete implementation step is as follows:

1, the state equation of magnetic levitation CMG when pedestal is motionless is according to Newton second law and the foundation of gyro technology equation:

$\left\{\begin{array}{c}\stackrel{\&CenterDot;}{X}=f(X,U)\\ Y=\mathrm{CX}\end{array}\right.---\left(23\right)$

Wherein,

$f(X,U)=\left[\begin{array}{c}\stackrel{\&CenterDot;}{x}\\ \stackrel{\&CenterDot;}{\mathrm{\&beta;}}\\ \stackrel{\&CenterDot;}{y}\\ -\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}\\ \stackrel{\&CenterDot;}{z}\\ \frac{1}{m}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibx}}{i}_{\mathrm{bx}}+k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\\ \frac{l_m}{J_y}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})+\frac{H}{J_y}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ \frac{1}{m}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}})\\ \frac{l_m}{J_x}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}}-k_{\mathrm{ibxy}}{i}_{\mathrm{by}}-k_{\mathrm{hby}}{y}_{\mathrm{bm}})+\frac{H}{J_x}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ \frac{1}{m}(k_{\mathrm{iz}}{i}_z+k_{\mathrm{hz}}z)\\ \frac{n_p{\mathrm{\&phi;}}_f}{J_{\mathrm{gx}}}{i}_s-\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}}-k_{\mathrm{iay}}{i}_{\mathrm{ay}}-k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\end{array}\right]---\left(24\right)$

$C=\left[\begin{array}{ccccccccccc}1& l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& -l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& -l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]---\left(25\right)$

In the formula, the state variable of system $X={[x,\mathrm{\&beta;},y,-\mathrm{\&alpha;},z,\stackrel{\&CenterDot;}{x},\stackrel{\&CenterDot;}{\mathrm{\&beta;}},\stackrel{\&CenterDot;}{y},-\stackrel{\&CenterDot;}{\mathrm{\&alpha;}},\stackrel{\&CenterDot;}{z},{\mathrm{\&omega;}}_g]}^T;$ The input control variable U=[i of system _Ax, i _Bx, i _Ay, i _By, i _z, i _s] ^TThe output variable Y=[x of system _Am, x _Bm, y _Am, y _Bm, z, ω _g] ^TM is a rotor quality; x ₀And z ₀The monolateral magnetic gap of representing radial direction magnetic bearing and axial magnetic bearing respectively; J _x, J _yAnd J _zThe X that is respectively rotor to, Y to Z to moment of inertia, and J _x=J _yi _Ax, i _Bx, i _Ay, i _By, i _zBe respectively each passage of rotor-support-foundation system (ax, bx, ay, by, Control current z); i _sBe the size of the stator current vector of frame motor, x _Am, x _Bm, y _Am, y _BmBe respectively rotor at transverse bearing A and B place with respect to the displacement of equilibrium position along X-axis and Y direction, α, β are the radially rotational displacement of rotor around X-axis and Y-axis,

Ω is respectively the angle of rotation speed of rotor around X, Y, Z axle; X, y, z are respectively the translation displacement of rotor centroid on X-axis, Y-axis and Z axle; Be respectively the translation speed of rotor centroid along X-axis, Y-axis and Z axle; Lm represents that radial direction magnetic bearing arrives the distance of rotor center O; k _Iax, k _Ibx, k _Iay, k _IbyAnd k _IzBe respectively the radially current stiffness of four-way and axial passage; k _Hax, k _Hbx, k _Hay, k _HbyAnd k _HzBe respectively the radially displacement rigidity of four-way and axial passage; ω _gAngle of rotation speed for framework; J _GxBe the axial rotation inertia of framework; n _pIt is the number of pole-pairs of frame motor; φ _fIt is the magnetic linkage of frame motor.

2, it is contrary to find the solution the parsing of magnetic levitation CMG

By being to the contrary inverse system that can obtain magnetic levitation CMG of asking for of the parsing of the original system shown in (20) formula

$\left\{\begin{array}{c}{i}_{\mathrm{ax}}=-\frac{k_{\mathrm{hax}}}{k_{\mathrm{iax}}}{x}_{\mathrm{am}}-\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iax}}}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iax}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iax}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iax}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{bx}}=-\frac{k_{\mathrm{hbx}}}{k_{\mathrm{ibx}}}{x}_{\mathrm{bm}}+\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{ibx}}}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{ibx}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{ibx}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{ibx}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{ay}}=-\frac{k_{\mathrm{hay}}}{k_{\mathrm{iay}}}{y}_{\mathrm{am}}-\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iay}}}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iay}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iay}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iay}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{by}}=-\frac{k_{\mathrm{hby}}}{k_{\mathrm{iby}}}{y}_{\mathrm{am}}+\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iby}}}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iby}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iby}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iby}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_z=\frac{1}{k_{\mathrm{ia}}}(m\stackrel{\&CenterDot;\&CenterDot;}{z}-k_{\mathrm{ha}}z)\\ i_s=\frac{1}{n_p{\mathrm{\&phi;}}_f}[\frac{\sqrt{2}}{2}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}{J}_y}{4l_m}({\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-{\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}+{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}})-(J_{\mathrm{gx}}+J_y){\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g]\end{array}\right.---\left(26\right)$

Wherein

It is respectively the rotor displacement acceleration expectation value of rotor-position servo control unit output; It is the frame corners acceleration expectation value of framework speed servo control unit output.

3, make up linear subsystem

As shown in Figure 4, constitute composite controlled object jointly by six current track inverters and single frame magnetic levitation CMG, this composite controlled object is with { i _Ax, i _Bx, i _Ay, i _By, i _z, i _sSix current signals are as input, with each passage displacement of magnetic suspension rotor and frame corners speed as output.Step 2 is obtained inverse system be connected on before the composite controlled object, just constituted a pseudo-linear system, this pseudo-linear system is respectively by 6 transport functions $G_{\mathrm{ax}}=s^{-{\mathrm{\&alpha;}}_1},$ $G_{\mathrm{bx}}=s^{-{\mathrm{\&alpha;}}_2},$ $G_{\mathrm{ay}}=s^{-{\mathrm{\&alpha;}}_3},$ $G_{\mathrm{by}}=s^{-{\mathrm{\&alpha;}}_4},$ $G_z=s^{-{\mathrm{\&alpha;}}_5},$ $G_{{\mathrm{\&omega;}}_g}=s^{-{\mathrm{\&alpha;}}_6}$ Linear subsystem constitute G wherein _Ax, G _Bx, G _Ay, G _By, represent the radially transport function of four-way, G respectively _zBe the transport function of axial passage, Be the transport function of framework passage, s represents the variable of mathematical model one transport function of control system in complex field, α ₁, α ₂, α ₃, α ₄, α ₅, α ₆The relative rank of representing the linear subsystem after the decoupling zero respectively.Such complicated nonlinear systems has just become 6 simple integral linearity systems, thereby has eliminated the moving framework displacement of frame corners rate variation to rotor, has also eliminated the moment of reaction disturbance of rotor-support-foundation system to framework.

4, design linear closed loop controller

To five location subsystem and a speed subsystem, make up position robust servo controller and speed robust servo controller respectively.Choose suitable robust servo controller structure according to the robust theory of servomechanism.For second order linear subsystem after the decoupling zero and single order linear subsystem, can make up respectively as Fig. 1 (is example with the ax passage) and control system shown in Figure 2.Wherein the robust compensation device of the robust servo controller of second order linear subsystem is $W_1\left(s\right)=\frac{a_0+a_{1}s}{s},$ Calm compensator is W ₂(s)=k ₀+ k ₁S; The robust compensation device of the robust servo controller of single order linear subsystem is $W_3\left(s\right)=\frac{a_2(s+{\mathrm{\&delta;}}_2)}{s},$ Calm compensator is W _A(s)=0.According to the second-order system performance index and utilize the method for undetermined coefficients can solve each parameter value.

The content that is not described in detail in the present disclosure belongs to this area professional and technical personnel's known prior art.

Claims (1)

1. method that suppresses moving-gimbal effects of single gimbal magnetically suspended control moment gyroscope is characterized in that: the state equation of setting up magnetic suspension control moment gyro of single framework according to Newton second law and gyro technology equation; It is contrary to utilize method of inverse to obtain the parsing of system; In rotor-support-foundation system,, in frame system, pass through the feedback of status of frame corners speed and rotor deflection angular speed is eliminated the moment of reaction disturbance of rotor to frame system by the feedback of status of rotor displacement, rotor deflection angular speed and frame corners speed being eliminated the moving framework displacement of magnetic suspension rotor; Adopt robust servocontrol strategy to improve the robustness of total system, specifically may further comprise the steps:

(1) state equation of magnetic suspension control torque gyroscope when pedestal is motionless is according to Newton second law and the foundation of gyro technology equation:

$\left\{\begin{array}{c}\stackrel{\&CenterDot;}{X}=f(X,U)\\ Y=\mathrm{CX}\end{array}\right.$ Wherein,

$f(X,U)=\left[\begin{array}{c}\stackrel{\&CenterDot;}{x}\\ \stackrel{\&CenterDot;}{\mathrm{\&beta;}}\\ \stackrel{\&CenterDot;}{y}\\ -\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}\\ \stackrel{\&CenterDot;}{z}\\ \frac{1}{m}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibx}}{i}_{\mathrm{bx}}+k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\\ \frac{l_m}{J_y}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})+\frac{H}{J_y}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ \frac{1}{m}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{1}{m}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}})\\ \frac{l_m}{J_x}(k_{\mathrm{iay}}{i}_{\mathrm{ay}}+k_{\mathrm{hay}}{y}_{\mathrm{am}}-k_{\mathrm{ibxy}}{i}_{\mathrm{by}}-k_{\mathrm{hby}}{y}_{\mathrm{bm}})+\frac{H}{J_x}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ \frac{1}{m}(k_{\mathrm{iz}}{i}_z+k_{\mathrm{hz}}z)\\ \frac{n_p{\mathrm{\&phi;}}_f}{J_{\mathrm{gx}}}{i}_s-\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{ibxy}}{i}_{\mathrm{by}}+k_{\mathrm{hby}}{y}_{\mathrm{bm}}-k_{\mathrm{iay}}{y}_{\mathrm{ay}}-k_{\mathrm{hay}}{y}_{\mathrm{am}})+\frac{\sqrt{2}{l}_m}{2J_{\mathrm{gx}}}(k_{\mathrm{iax}}{i}_{\mathrm{ax}}+k_{\mathrm{hax}}{x}_{\mathrm{am}}-k_{\mathrm{ibx}}{i}_{\mathrm{bx}}-k_{\mathrm{hbx}}{x}_{\mathrm{bm}})\end{array}\right]$

$C=\left[\begin{array}{ccccccccccc}1& l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& -l_m& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& -l_m& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$

In the formula, the state variable of system $X={[x,\mathrm{\&beta;},y,-\mathrm{\&alpha;},z,\stackrel{\&CenterDot;}{x},\stackrel{\&CenterDot;}{\mathrm{\&beta;}},\stackrel{\&CenterDot;}{y},-\stackrel{\&CenterDot;}{\mathrm{\&alpha;}},\stackrel{\&CenterDot;}{z},{\mathrm{\&omega;}}_g]}^T;$ The input control variable U=[i of system _Ax, i _Bx, i _Ay, i _By, i _z, i _s] ^TThe output variable Y=[x of system _Am, x _Bm, y _Am, y _Bm, z, ω _g] ^TM is a rotor quality; J _x, J _yAnd J _zBe respectively rotor X to, Y to Z to moment of inertia, and J _x=J _yi _Ax, i _Bx, i _Ay, i _Yb, i _zBe respectively the radially Control current of ax, bx, ay, by and axial z passage of rotor-support-foundation system; i _sBe the size of the stator current vector of frame motor, x _Am, x _Bm, y _Am, y _BmBe respectively rotor at transverse bearing A and B place with respect to the displacement of equilibrium position along X-axis and Y direction, α, β are the radially rotational displacement of rotor around X-axis and Y-axis,

Ω is respectively the angle of rotation speed of rotor around X, Y, Z axle; X, y, z are respectively the translation displacement of rotor centroid on X-axis, Y-axis and Z axle; Be respectively the translation speed of rotor centroid along X-axis, Y-axis and Z axle; l _mThe expression radial direction magnetic bearing is to the distance of rotor center O; k _Iax, k _Ibx, k _Iay, k _IbyAnd k _IzBe respectively the radially current stiffness of four-way and axial passage; k _Hax, k _Hbx, k _Hay, k _HbyAnd k _HzBe respectively the radially displacement rigidity of four-way and axial passage; ω _gAngle of rotation speed for framework; J _GxBe the axial rotation inertia of frame; n _pIt is the number of pole-pairs of frame motor; φ _fIt is the magnetic linkage of frame motor;

It is the frame corners acceleration expectation value of framework speed servo control unit output;

(2) it is contrary to find the solution the parsing of magnetic suspension control torque gyroscope

According to resolving contrary computing method, the parsing that can get magnetic suspension control torque gyroscope is against being:

$\left\{\begin{array}{c}{i}_{\mathrm{ax}}=-\frac{k_{\mathrm{hax}}}{k_{\mathrm{iax}}}{x}_{\mathrm{am}}-\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iax}}}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iax}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iax}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iax}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{bx}}=-\frac{k_{\mathrm{hbx}}}{k_{\mathrm{ibx}}}{x}_{\mathrm{bm}}+\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{ibx}}}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{ibx}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{ibx}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{ibx}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{ay}}=-\frac{k_{\mathrm{hay}}}{k_{\mathrm{iay}}}{y}_{\mathrm{am}}-\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iay}}}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iay}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iay}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}}+\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iay}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_{\mathrm{by}}=-\frac{k_{\mathrm{hby}}}{k_{\mathrm{iby}}}{y}_{\mathrm{am}}+\frac{J_z\mathrm{\&Omega;}}{2l_m{k}_{\mathrm{iby}}}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)-\frac{1}{4k_{\mathrm{iby}}}(m-\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-\frac{1}{4k_{\mathrm{iby}}}(m+\frac{J_y}{l_m^2}){\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}}-\frac{\sqrt{2}{J}_y}{4l_m{k}_{\mathrm{iby}}}{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g\\ i_z=\frac{1}{k_{\mathrm{ia}}}(m\stackrel{\&CenterDot;\&CenterDot;}{z}-k_{\mathrm{ha}}z)\\ i_s=\frac{1}{n_p{\mathrm{\&phi;}}_f}[\frac{\sqrt{2}}{2}(\stackrel{\&CenterDot;}{\mathrm{\&alpha;}}+\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}}{2}(\stackrel{\&CenterDot;}{\mathrm{\&beta;}}-\frac{\sqrt{2}}{2}{\mathrm{\&omega;}}_g)+\frac{\sqrt{2}{J}_y}{4l_m}({\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}}-{\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}}+{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}}-{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}})-(J_{\mathrm{gx}}+J_y){\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_g]\end{array}\right.$

Wherein ${\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{am}},{\stackrel{\&CenterDot;\&CenterDot;}{x}}_{\mathrm{bm}},{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{am}},{\stackrel{\&CenterDot;\&CenterDot;}{y}}_{\mathrm{bm}},\stackrel{\&CenterDot;\&CenterDot;}{z}$ It is respectively the rotor displacement acceleration expectation value of rotor-position servo control unit output; It is the frame corners acceleration expectation value of framework speed servo control unit output;

(3) make up linear subsystem

Constitute composite controlled object jointly by six current track inverters and single frame magnetic levitation CMG, this composite controlled object is with { i _Ax, i _Bx, i _Ay, i _By, i _z, i _sSix current signals are as input, with each passage displacement of magnetic suspension rotor and frame corners speed as output, before the inverse system that step (2) is obtained is connected on composite controlled object, just form linear subsystem, this linear subsystem is made of the linear subsystem of five position second order integral forms, the linear subsystem of a speed single order integral form;

(4) make up the linear closed-loop controller

To the linear subsystem of five position second order integral forms and the linear subsystem of a speed single order integral form, make up position robust servo controller and speed robust servo controller respectively, wherein the robust compensation device of position robust servo controller is $W_1\left(s\right)=\frac{a_0+a_{1}s}{s},$ Calm compensator is W ₂(s)=k ₀+ k ₁S; The robust compensation device of speed robust servo controller is $W_3\left(s\right)=\frac{a_{2}s+a_3}{s},$ Calm compensator is W ₄(s)=0; S represents the variable of mathematical model one transport function of control system in complex field, a ₀, a ₁Represent constructed robust compensation device W respectively ₁(s) integral coefficient and scale-up factor, k ₀, k ₁Represent calm compensator W respectively ₂(s) scale-up factor and differential coefficient, a ₂, a ₃Represent robust compensation device W respectively ₃(s) scale-up factor and integration number.

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