CN1605962A - Optimal control method for single frame moment gyro group for spacecraft wide angle maneuver control - Google Patents

Abstract

The invention provides an optimal control method for single frame moment gyro group for spacecraft wide angle maneuver control characterized in that, during the process of attitude control, the gyroscope having the highest frame angular speed is selected, then a minimum frame angular speed is obtained from the gyroscope as the control frame angular speed for rapid large angle power-propelled control, when guaranteeing the gyroscope frame angular speed is lower than the given propulsion generator rotational speed limit, the lost motion frame angular speed and correction coefficient can be designed to give the gyroscope group dynamic formation a predetermined final frame angle quantity, finally the optimum control angular speed and optimum lost motion angular speed are multiplied by the correction coefficient for summing, which can obtain the optimum value of the frame angular speed.

Description

The single frame moment gyro group's of spacecraft large angle maneuver control method in optimal control

Technical field

The present invention relates to spacecraft attitude dynamics and control technology field.

SGCMG：Single?Gimbal?Control?Moment?Gyroscopes。Single frame moment gyro (Fig. 1).Form by the motor that the flywheel of high-speed rotation, the framework of supporting flywheel and driving framework rotate.General, the rotating speed of flywheel is a certain value, and control is what to realize by the rotating speed of Selection Framework drive motor.That is to say that in fact the problem of design SGCMG control law is exactly to select the problem of rational framework rotating speed.

Background technology:

The wide-angle attitude control problem of spacecraft is the key that spacecraft is finished the particular flight task.Its control accuracy and speed directly influence the performance of spacecraft application task.Since the seventies, some high precision large-sized spacecrafts of the U.S., Russia (Soviet Union) (as: the skylab space station of the U.S., the MIR space station of the Soviet Union, Hubble's astronomical telescope, Muscovite resource-DK satellite etc.) have all adopted SGCMG control.Today, the scholar of some research moonlets is also considering to use this technology.As: at the V.J.Lappas of Britain University of Surrey ' Surry Space Centre ', people such as Dr.WH Steyn just are being devoted to design mini-SGCMG, and attempt it is installed in use (V.J.Lappas on the moonlet, Dr.WHSteyn, Dr.C.I.Underwood Attitude control for small satellites using control momentgyros.Acta Astronautica Vol.51, No.1-9, pp.101-111,2002.).Here most important reason just is that SGCMG can guarantee higher attitude control accuracy and speed.

In SGCMG control, core problem is exactly the control law problem of SGCMG the most, promptly chooses the problem of SGCMG frame corners speed.The scholar of China and foreign countries has done number of research projects in this direction.At present at home, up-to-date main pertinent literature is as follows:

1. soup is bright, Xu Shijie.Adopt single frame control-moment gyro group's satellite side-sway automotive Control Study." Aerospace Control ", 2003, the second phase, pp.1-5.

2. Wuzhong.The robust of the uncertain SGCMG of parameter system is handled the rule design.Acta Astronautica, in January, 2004, the 25th volume, the 1st phase, pp.93-97.

3. Wuzhong, ugly Wusheng.Unusual and the avoidance of the motion of single frame control-moment gyro system." Beijing space flight and aviation college journal ", in July, 2003, the 29th volume, the 7th phase, pp.579-582.

4. soup is bright, Jia Yinghong, Xu Shijie.Use the space station attitude control system modeling and simulation of single frame control-moment gyro.Acta Astronautica, in March, 2003, the 24th volume, the 2nd phase, pp.126-131.

5. Zhang Jin river, Li Jisu, Wu Hongxin.Large Spacecraft attitude control system matter emulation research with the single frame control-moment gyro.Acta Astronautica, in July, 2004, the 25th volume, the 4th phase, pp.382-388.

External at present main pertinent literature is as follows:

1.Wie，B.Singularity?analysis?and?visualization?for?single-gimbal?control?moment?gyrosystems.Journal?of?guidance?control?and?dynamics，27(2)：271-282?Mar-Apr，2004.

2.Skelton，CE，Hall，CD.Mixed?control?moment?gyro?and?momentum?wheel?attitude?controlstrategies.ADV?ASTRONAUT?SCI?116：887-899?Part?1-32003.

3.Ahmed，j，Bernstein，DS.Adaptive?control?of?double-gimbal?control-moment?gyro?withunbalanced?rotor.J?guid.Control?dynam.25(1)：105-115?Jan-Feb?2002.

4.Сорокин?А.Б.，БашкееБ?Н.И.，Яременко?В.В.?Гиросиловая?система?ориентациикосмического?аппарата?Ресурс-Дк.IX?Санкт-Петербургская?Междyнароднаяконференцияпo?интегрированным?навигационным?системам.Санкт-Петeрбург：2002.268-274.

Sorokin?A.V.，Bashkeev?N.I.，Yaremenko?V.V.SGCMG?system?for?spacecraft?resource?DK.9 ^th?st.Petersburg-conference?of?internal?navigation?system.St.Petersburg：2002.268-274.(in?Russian)

5.Васильев?В.Н.Управление?кратной?минимальнo?избытоqнойсистемой?гиродинoв.Механика?твердого?тела.1995.3：3-10.

Vasiliev?V.N.Control?of?redundant?SGCMG?system.Mechanical?of?rigid?body.1995，3：3-10.(in?Russian)

Find by concrete analysis domestic and international present Research:

1. lack solution at present SGCMG frame corners speedThe controlling schemes of restricted problem.

SGCMG frame corners speed: the velocity of rotation that provides by the framework drive motor.On spaceborne computer, try to achieve by certain control algolithm calculating.And will offer the framework drive motor of SGCMG as input signal, realize control task.Because rotating speed of motor generally has upper limit requirement, so it should be taken into account in SGCMG CONTROL LAW DESIGN process.

2. though The gyro group configuration is unusualProblem obtained certain concern, but at present for SGCMG ' useful ' Unusual configurationWith ' harmful ' unusual configurationDo not carry out differentiated treatment.Therefore, though singular problem has been had relevant

Research, but most of control law can produce certain negative effect to the speed of spacecraft large angle maneuver control.The gyro group configuration is unusual: SGCMG does not generally use separately as the attitude control assembly of spacecraft, but by certain special layout, the SGCMG more than 3 is combined into specific system (being gyro group) spacecraft is controlled.This special arranged mode is called the static configuration of gyro group.The static configuration of gyro group remains constant in the spacecraft flight process.In control procedure, SGCMG can do the rotation of 360 degree with framework, and this corner is called as the frame corners of SGCMG.The frame corners of SGCMG has been described that sensing of flywheel rotating shaft constantly.In control procedure, the sensing of different moment flywheel rotating shafts is all inequality.That is to say, can have different gyro gimbal angle combinations, the dynamic configuration of promptly different gyro groups at different moment gyro groups.And when being in some dynamic configuration, gyro group can't provide rotating torque for spacecraft on some direction.At this moment we say, gyro group is absorbed in unusual state.

' useful ' unusual configuration: for maximum rotational angular velocity is provided, gyro group will be absorbed in unusual configuration.That is to say that at this moment gyro group can't provide a moment to make the angular velocity of spacecraft surmount this maximal value.In the process of carrying out fast reserve, this configuration is unusual will often to be occurred, and it can improve the motor-driven speed of spacecraft attitude, so we say that this is a kind of useful unusual.

' harmful ' unusual configuration: because the SGCMG CONTROL LAW DESIGN is unreasonable, make spacecraft when slow rotation, the configuration of gyro group has also taken place unusual.This is a kind of situation about must be avoided, because it will influence the quality of control greatly.

3. for the spacecraft of quick, high precision, mobility strong, the configuration the situation initial and finish time of SGCMG is very crucial.How to make SGCMG guarantee that in the finish time configuration is the key that addresses this problem preferably.But, have not yet to see the document of this respect.

Along with the develop rapidly of electronic computer technology, make complicated more control law on spacecraft, to be achieved.This provides the necessary hardware basis for new, more rational controlling schemes.The applicant has designed new SGCMG control law according to the problem of present existence.Its purpose is exactly the particular problem that solves above 3 aspects.

Summary of the invention:

The objective of the invention is to: the method in optimal control that the single frame moment gyro group of the large angle maneuver control that a kind of spacecraft for the quick high accuracy mobility strong uses is provided.

The attitude dynamics modeling problem of spacecraft had detailed discussion (as: [1] Zhang Renwei in a lot of documents.Satellite orbit and attitude dynamics and control." publishing house of BJ University of Aeronautics ﹠ Astronautics ", 1998.)。In [1], also describe the physical model that utilizes SGCMG to realize spacecraft attitude control in detail.Briefly, when utilizing SGCMG to realize spacecraft attitude control, the angular momentum of spacecraft and SGCMG will satisfy equation with angular momentum:

P+H＝C?????????????????????????????(1)

Wherein, the angular momentum of P-spacecraft; H-SGCMG's and angular momentum.In inertial system, when disturbing moment outside not having the external force effect and ignoring, C is a vector that size and Orientation is all constant.Therefore, SGCMG control is exactly in fact spacecraft angular momentum and SGCMG and transfer process angular momentum.By selecting rational H, just can realize control to spacecraft attitude.

Select the problem of rational H Changing Pattern, be called as traditionally and find the solution desirable control moment M ^*Problem.This is the previous work of SGCMG CONTROL LAW DESIGN, and the present invention does not relate to the content of this respect.We suppose M ^*, it is carried out specific implementation and will use SGCMG by trying to achieve someway.

In order to address this problem, briefly learn about earlier the principle of work of SGCMG.The SGCMG system generally (is made up of 4 gyros as the two parallel construction SGCMG system among Fig. 2 by forming more than three moment gyro.), and layout becomes fixing steric configuration (in Fig. 2, when the framework rotating shaft of 4 gyros in one plane and when parallel in twos, to be called two parallel constructions.)。Every flywheel (flywheel rotational angular Ω that gyro is rotated by high speed (as: 20,000 rev/mins) _iDuring for certain value, be called and decide rotating speed SGCMG.), support framework that flywheel rotates, drive the flywheel rotary electric machine and drive the motor that framework rotates to form (as shown in Figure 1).When flywheel with fixed angular speed Ω _iDuring rotation, just produce into an angular momentum h that size is constant _i(h _i=j _iΩ _i, j _iThe inertia matrix of-gyro).The rotation of framework forces the angular momentum h of flywheel _iDirection changes.From the physics angle, the change of flywheel angular momentum will produce gyro moment of reaction M _i, act on (as shown in Figure 1) on the frame base control-torque in forming.Therefore, utilization SGCMG system realizes the attitude control of spacecraft, chooses framework angular velocity exactly (i=1, n, the quantity of gyro in the n-SGCMG system.) problem.

According to moment gyro group's principle of work and the problem that exists at present, following invention scheme is proposed.In order to realize that the consideration of SGCMG frame corners speed limit and the dynamic configuration of gyro group two aspects has been designed following controlling schemes.(1) with SGCMG frame corners speed separated into two parts.(2) first of SGCMG frame corners speed will satisfy the required control moment of large angle maneuver control.Satisfying under the prerequisite of control moment, choosing minimum SGCMG frame corners speed.(3) second portion of SGCMG frame corners speed does not provide any control moment (zero control moment is provided), in order to avoid the interference to controlling.And on this basis, guarantee that the dynamic configuration of gyro group changes to a given perfect condition, thereby realize optimization to the dynamic configuration of gyro group.(4) because the framework acceleration of being asked in (3) is generally bigger, directly use the upper limit that can surpass SGCMG frame corners speed, therefore must be weighted processing.In (4), will find the solution weighting coefficient.

When system design, in order to guarantee that SGCMG still can finish control task under the situation that some gyros lost efficacy, the SGCMG system generally is designed to redundant system (being quantity n 〉=3 of gyro).At this moment its controlling models is as follows:

$L\left(t\right)\stackrel{\·}{\mathrm{\β}}\left(t\right)=M^{*}\left(t\right)---\left(2\right)$

Here, $\stackrel{\·}{\mathrm{\β}}\left(t\right)={[{\stackrel{\·}{\mathrm{\β}}}_1\left(t\right),{\stackrel{\·}{\mathrm{\β}}}_2\left(t\right)...{\stackrel{\·}{\mathrm{\β}}}_n\left(t\right)]}^T,$ The frame corners velocity of gyro; M ^*(t)-the desirable control moment of large angle maneuver control; $L\left(t\right)=\∂H\left(t\right)/\∂\mathrm{\β}\left(t\right)=L\left(\mathrm{\β}\right)=\left[\begin{array}{cccc}\frac{{\∂H}_x}{{\∂\mathrm{\β}}_1}& \frac{{\∂H}_x}{{\∂\mathrm{\β}}_2}& \·\·\·& \frac{{\∂H}_x}{{\∂\mathrm{\β}}_n}\\ \frac{{\∂H}_y}{{\∂\mathrm{\β}}_1}& \frac{{\∂H}_y}{{\∂\mathrm{\β}}_1}& \·\·\·& \frac{{\∂H}_y}{{\∂\mathrm{\β}}_n}\\ \frac{{\∂H}_z}{{\∂\mathrm{\β}}_1}& \frac{{\∂H}_z}{{\∂\mathrm{\β}}_2}& \·\·\·& \frac{{\∂H}_z}{{\∂\mathrm{\β}}_n}\end{array}\right];$ H (t)- Gyro group and angular momentum.Gyro group and angular momentum: in the gyro group system, the vector of all gyro angular momentums and $H\left(t\right)=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{h}_i\left(t\right)=f\left(\mathrm{\β}\left(t\right)\right);$ H _i-gyro group and the projection of angular momentum H (t) on the i axle; I=(x, y, z)-the celestial body coordinate system; β (t)=[β ₁(t), β ₂(t) ... β _n(t)] ^T-SGCMG frame corners vector; The quantity of n-gyro, n 〉=3.

Formula (2) is described to be a system of linear equations that infinite multiresolution is arranged.It is found the solution, just can be met the gyro gimbal angular velocity of desirable control moment.

How (2) being found the solution is the key of problem.By the problem of three aspects of existence in the SGCMG control of above mentioning,

The present invention proposes following solution.

The first step:

Will Be divided into two parts:

$\stackrel{\·}{\mathrm{\β}}\left(t\right)={\stackrel{\·}{\mathrm{\β}}}_O\left(t\right)+\mathrm{\α}{\stackrel{\·}{\mathrm{\β}}}_z\left(t\right)--\left(3\right)$

-satisfy control moment M ^*(t) SGCMG frame corners speed; -the frame corners speed of zero control moment is provided.α-weighting coefficient, 0≤α≤1.

Second step:

Because of considering the upper limit requirement of gyro gimbal angular velocity

$\left|{\stackrel{\·}{\mathrm{\β}}}_i\right|\≤{\stackrel{\·}{\mathrm{\β}}}_{\max };i=\stackrel{\‾}{1,n}---\left(4\right)$

So in the CONTROL LAW DESIGN process, need reasonably select to SGCMG frame corners speed.In order to satisfy this requirement, propose one New optimization performance index:

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">q</figure-callout>}}_1=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }\underset{i=1,n}{m\underset{\&OverBar;}{a}x}\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}\left(t\right)\right|---\left(5\right)$

The meaning of these performance index is: it can be under the prerequisite that realizes control, and greatly the minimizing of degree (is a gyro gimbal angular velocity to controlling resource

Take.Get maximum value (max) part in the formula (5) and choose of n SGCMG middle frame angular velocity maximum, be designated as Then operate by minimalization (min),

In choose minimum, as optimum solution.Simultaneous (2), (4), (5) obtain formula (6)

(6) formula is found the solution, can be obtained

Be example with the 4-gyrosystem below, and the solution procedure of performance index (4) is described by diagram.

At first, with vector

Separated into two parts: first is one 1 * 3 a vector; Second portion is a scalar.Being about to the frame corners velocity vector is write as ${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)={\left(\underset{1\&times;3}{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o^{\left(1\right)}},{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\right)}^T$ Form; -vector

First three items, be one 1 * 3 vector. -vector

The 4th, be a scalar.Accordingly $L\left(t\right)=\left[\begin{array}{cc}\underset{3\&times;3}{L_{11}}& \underset{3\&times;1}{L_{12}}\\ \underset{1\&times;3}{L_{21}}& \underset{1\&times;1}{L_{22}}\end{array}\right].$ Like this, first formula in the formula (6) can be write as following form:

${{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}^{\left(1\right)}=L_{11}^{-1}(M^{*}\left(t\right)-L_{12}{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4})={\mathrm{<figure-callout\; id="0"\; label="c"\; filenames="A20041000987900262.png,A20041000987900281.png"\; state="\{\{state\}\}">c</figure-callout>}}_0+{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">c</figure-callout>}}_1{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}--\left(7\right)$

Wherein: c ₀Be one 3 * 1 vector, c ₀=L ^-1 ₁₁M ^*(t); c ₁Also be one 3 * 1 vector, c ₁=-L ^-1 ₁₁L ₁₂(7) the described family of straight lines of forming by 3 straight lines of formula.(7) formula is taken absolute value, obtains:

$\left|{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}^{\left(1\right)}\right|=|{\mathrm{<figure-callout\; id="0"\; label="c"\; filenames="A20041000987900262.png,A20041000987900281.png"\; state="\{\{state\}\}">c</figure-callout>}}_0+{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">c</figure-callout>}}_1{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}|--\left(8\right)$

Formula (8) be actually one with Be independent variable, System of linear equations (number of equation is 3 in the system of equations) for variable.If introduce equation again

$\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\right|=\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\right|--\left(9\right)$

And establish $Z={\left(\right|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{<figure-callout\; id="1"\; label="o"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">o</figure-callout>}1}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o2}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o3}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\left|\right)}^T;$ ${\mathrm{<figure-callout\; id="0"\; label="C"\; filenames="A20041000987900262.png,A20041000987900281.png"\; state="\{\{state\}\}">C</figure-callout>}}_0={(c_0^T,0)}^T;$ ${\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">C</figure-callout>}}_1={(c_1^T,1)}^T;$ $x={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}.$ Can obtain system of equations with lower linear:

Z＝|C ₀+C ₁x|?????????????????????????(10)

Z is a system of linear equations of being made up of four equations, in that (x, Z) plane is one group of broken line (as Fig. 3).

Through such processing, system of equations (6) has following form:

$\underset{x}{\mathrm{<figure-callout\; id="1"\; label="min\; x\; q"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">min</figure-callout>}}{q}_{}{}_1=\underset{x}{\min }\underset{i=1,n}{m\underset{\&OverBar;}{a}x}{Z}_i\left(x\right)---\left(11\right)$

Like this, the problem of finding the solution steering order has just changed into the problem that (11) formula is found the solution.

Below we by to the analysis of Fig. 3, the process that specific description is found the solution.

May interval at x

On, the coboundary of asking for n group broken line Zi (x) is asked for the pairing x of minimum value, Z again from the coboundary _i(x) value; Be not difficult to find out, for identical control moment M ^*(t), this control law will farthest reduce taking controlling resource (SGCMG frame corners speed).And the controlling resource of the saving function (the existing strictly according to the facts idle running control that hereinafter will mention) to realize other also.On the other hand, compare with other algorithm, when controlling resource was identical, this algorithm can provide bigger control moment.So be not difficult to find out control law

Also help to improve the speed of attitude maneuver.

From above analysis as can be known, the problem of finding the solution steering order has changed into the problem of finding the solution extreme value A at last.This is a linear programming problem, can find the solution by simplicial method.But by attempting finding that utilization simplicial method speed is slower.Because when this method of utilization, system of equations must be changed into standard form.For the 4-SGCMG system, this is one and finds the solution the linear programming problem that contains 12 unknown numbers.Under different situations, also there is very big difference in the optimizing time, is not easy to real-time control.

For this reason, the author has designed simple relatively control algolithm and has realized search.This algorithm compares by all intersection points to family of straight lines, seeks optimum solution.Solution procedure sees flow chart (accompanying drawing 6) for details:

This algorithm has shortened search time greatly, and to different situations, search time is identical, is convenient to realize real-time control.

The 3rd step:

Solve the unusual problem of the dynamic configuration of gyro group.When the SGCMG system was in unusual state, SGCMG can't be for spacecraft provides control moment on some direction, so must avoid the dynamically unusual of SGCMG system as far as possible.But not all unusual state all is ' being harmful to '.Sometimes spacecraft is in order to obtain enough big rotational angular velocity, and gyro group must enter unusual configuration.Existing control law is not reasonably distinguished these situations, therefore will limit the unusual configuration of this ' useful ', obtains necessary velocity of rotation thereby hinder spacecraft.

For this reason, inventor New performance index have been designed:

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{q}_2=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}}^2--\left(12\right)$

Wherein, β _T-given final frame corners vector.β _TThe corresponding different application task requirement of value; The SGCMG frame corners of β (t)-current time (t constantly) will be passed through sensor measurement.(8) function of formula is: it can so that the dynamic configuration of gyro group always towards a specified value β _TChange.When spacecraft needed big rotational angular velocity, its effect will be not clearly; And when the rotational angular velocity of spacecraft hour, the dynamic frame framework type that it can the guarantee gyro group feeding definite value β that becomes rapidly _T

In order to find the solution, (12) formula and idle running instruction simultaneous are obtained:

(13) analytic solution of formula are:

${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)=\left\{\right[{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)]/\mathrm{\&Delta;t}-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)\}{\{E}_n-L^T\left(t\right){\left[L\left(t\right)L^T\left(t\right)\right]}^{-1}L\left(t\right)\}--\left(14\right)$

Here E _nThe unit matrix of-n * n.

Find the solution flow process and see accompanying drawing 7 for details.

The 4th step:

Separating of formula (14) Guarantee frame corners,, become β through time Δ t _TBut because

Generally bigger, so need be right

Be weighted correction, make its restriction of satisfying the frame corners speed of SGCMG ((4) formula).

With (3), (4) simultaneous solution obtains:

α ^-≤α≤α ⁺??????????????????????????(15)

Wherein, ${\mathrm{\&alpha;}}^{-}=\underset{i=1,n}{m\underset{\&OverBar;}{a}x}\min \{\frac{-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&le;0,$ ${\mathrm{\&alpha;}}^{+}=\underset{i=1,n}{m\underset{\&OverBar;}{i}x}\max \{\frac{-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&GreaterEqual;0.$ Therefore, α must go up value in interval (15).In order to select optimum α, observe with minor function

${\mathrm{\&gamma;}}_{\mathrm{\&alpha;}}^{*}\left(\mathrm{\&alpha;}\right)={\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+\mathrm{\&alpha;}\&CenterDot;{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}}^2--\left(16\right)$

Wherein Value is separated for (14).When α=1, have ${\mathrm{\&gamma;}}_{\mathrm{\&alpha;}}^{*}\left(1\right)=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{q}_2,$ Function (16) is got minimum value.On interval [0,1], be monotone decreasing because of (16) again, so Promptly work as α ⁺〉=1 o'clock, α=1; Work as α ⁺＜1 o'clock, α=α ⁺By introducing corrected parameter α, just can guarantee q like this ₂Get minimum value (as shown in Figure 4).Solution procedure sees accompanying drawing 8 for details.

Therefore, Under the situation that does not influence control rate, will play auxiliary effect to control.Generally speaking, To as much as possible the frame corners of SGCMG be controlled to given position β _TIf

Need take more resources,

Under the adjusting of weighting coefficient α, reduce occupying to resource.So just SGCMG control rate, frame corners speed limit and unusual problem have been solved.

The invention is characterized in:

It contains following steps successively:

The first step, initialization:

To single frame moment gyro group, i.e. SGCMG group's frame corners velocity vector, promptly

Control computer import the quantity that following physical parameter: n is the single frame moment gyro, n 〉=3, and the steric configuration of gyro group; The following parameter of single frame moment gyro: angular momentum h, units m ²/ s; The upper limit of frame corners speed

Unit degree/s; The upper limit of frame corners acceleration

Unit degree/s ₂The moment of inertia matrix I of spacecraft; The structural parameters of solar array; The boundary condition of attitude maneuver: initial angular velocity omegae ₀, initial attitude hypercomplex number Λ ₀, the end angle speed omega _T, stop attitude quaternion Λ _T, the unusual degree index of gyro group σ _L, 0≤σ _L≤ 1;

This spacecraft need carry out the angle and the angular velocity of quick wide-angle attitude maneuver;

Single frame moment gyro group's control moment vector M ^*(t), in each sampling period, the attitude quaternion that provides according to the attitude observer And attitude angular velocity

Calculate, and offer frame corners speed control computing machine by the control moment control computer;

The gyro gimbal angular velocity vector

Be the vector that each moment gyro frame corners speed is formed, in frame corners speed control computing machine, calculate, and offer the framework drive motor of each gyro by frame corners speed control computing machine;

Frame corners vector β (t) is the vector that the corner of each moment gyro is formed, and records by the frame corners observer, feeds back to described frame corners speed control computing machine then;

The closed-loop system that described control moment control computer, frame corners speed control computing machine, single frame moment gyro group, frame corners observer and attitude observer are formed is installed on the spacecraft that will control;

In addition, also designed following mathematical model with form of software:

Mathematical model with the single frame moment gyro group control of vector form statement:

$L\left(t\right)\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)=M^{*}\left(t\right),$

Wherein, $\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)={[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_1\left(t\right),{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_2\left(t\right)...{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_n\left(t\right)]}^T,n\&GreaterEqual;3,L\left(t\right)=\&PartialD;H\left(t\right)/\&PartialD;\mathrm{\&beta;}\left(t\right)=L\left(\mathrm{\&beta;}\right)=\left[\begin{array}{cccc}\frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_2}& \&CenterDot;\&CenterDot;\&CenterDot;& \frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_n}\\ \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_1}& \&CenterDot;\&CenterDot;\&CenterDot;& \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_n}\\ \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_2}& \&CenterDot;\&CenterDot;\&CenterDot;& \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_n}\end{array}\right],$ I=(x, y z) are the celestial body coordinate system, $H\left(t\right)=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{h}_i\left(t\right)=f\left(\mathrm{\&beta;}\left(t\right)\right),$ H (t) be the single frame moment gyro group's and angular momentum, promptly described each moment gyro angular momentum and, and h _i=j _iΩ _i, j _iInertia matrix for gyro i; β (t)=[β ₁(t), β ₂(t) ... β _n(t)] ^T, be described single frame moment gyro group's frame corners vector;

The mathematical model of frame corners speed:

$\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)+\mathrm{\&alpha;}{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right),$

Wherein, Be the frame corners speed of steering order decision, i.e. pilot angle speed; Be the frame corners speed of idle running instruction decision, angular velocity promptly dallies; α is idle running angular velocity Weighting coefficient, 0≤α≤1;

The mathematical model of the optimum single frame moment gyro control of the quick large angle maneuver control of spacecraft:

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{<figure-callout\; id="1"\; label="o\; q"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">o</figure-callout>}}}{\min }{q}_{}{}_1=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }\underset{i=1,n}{m\underset{\&OverBar;}{a}x}\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}\left(t\right)\right|,$

That is, from n single frame moment gyro, choose earlier a moment gyro of the absolute value maximum of framework angular velocity, choose minimum frame corners speed more therein;

Guarantee the dynamic configuration of the gyro group fixed final frame corners vector β of feeding that becomes _TMathematic model for optimal control:

In the process of spacecraft wide-angle attitude control, by minimizing β _TWith (t+ Δ t) frame corners constantly $\{\mathrm{\&beta;}\left(t\right)+[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}$ Difference find the solution the frame corners speed that idle running instruction is determined

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{q}_2=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}}^2,$

Wherein, β (t) is measured by the frame corners observer, Δ t sampling time interval; Second step, set n=4, according to following simultaneous equations:

In the stage of the quick large angle maneuver control of spacecraft, from n single frame moment gyro, choose earlier a moment gyro of the absolute value maximum of framework angular velocity, choose minimum frame corners speed more therein as optimum solution, it contains following steps successively:

The 2.1st step, n the frame corners velocity vector that steering order is definite

Be expressed as:

${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)={(\underset{1\&times;3}{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o^{\left(1\right)}},{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4})}^T,$

Wherein,

Be vector First three items, be one 1 * 3 vector; Be vector

The 4th, be a scalar; Accordingly $L\left(t\right)=\left[\begin{array}{cc}\underset{3\&times;3}{L_{11}}& \underset{3\&times;1}{L_{12}}\\ \underset{1\&times;3}{L_{21}}& \underset{1\&times;1}{L_{22}}\end{array}\right];$

The 2.2nd step, order $x={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4},$ $Z={\left(\right|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{<figure-callout\; id="1"\; label="o"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">o</figure-callout>}1}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o2}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o3}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\left|\right)}^T,$ Be the absolute value of the pairing frame corners velocity vector of steering order, satisfy following equation:

Z＝|C ₀+C ₁x|，

Wherein, ${\mathrm{<figure-callout\; id="0"\; label="C"\; filenames="A20041000987900262.png,A20041000987900281.png"\; state="\{\{state\}\}">C</figure-callout>}}_0={(c_0^T,0)}^T;$ ${\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">C</figure-callout>}}_1={(c_1^T,1)}^T;$ c ₀=L ^-1 ₁₁M ^*(t) c ₀, be one 3 * 1 vector; c ₁=-L ^-1 ₁₁L ₁₂, be one 3 * 1 vector; (x, Z) on the plane, the ABS function Z of n frame corners speed is a broken line family; Obtain thus

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{<figure-callout\; id="1"\; label="o\; q"\; filenames="A20041000987900252.png,A20041000987900262.png"\; state="\{\{state\}\}">o</figure-callout>}}}{\min }{q}_{}{}_1=\underset{x}{\min }\underset{i=1,n}{m\underset{\&OverBar;}{a}x}{Z}_i\left(x\right),$

That is, will find the solution

Transform into: (x, Z) on the plane, the possible interval of x On, ask for n group broken line Z _i(x) coboundary is asked for the pairing x of minimum value, Z again from the coboundary _i(x) value;

This steps in sequence contains following each step:

The 2.2.1 step in frame corners speed control computing machine, calculates all intersection points of the Z of broken line family, data is deposited among the array Q again;

The 2.2.2 step, find the solution the maximal value of the function Z of all intersection point positions, obtain coboundary s, deposit among the array D;

In the 2.2.3 step, compare maximal value array D, and select wherein minimum value conduct

Separate;

The 3rd step is according to following simultaneous equations

Choose idle running frame corners speed

Make gyro gimbal angle amount $\{\mathrm{\&beta;}\left(t\right)+[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}$ With final frame corners vector β _TMinimum in (t+ Δ t) difference constantly; And with the frame corners speed of this single frame moment gyro

As the optimum solution of the frame corners speed under the idle running instruction, it has following analytical form:

${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)=[({\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right))/\mathrm{\&Delta;t}-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)][E_n-L^T\left(t\right){\left(L\left(t\right)L^T\left(t\right)\right)}^{-1}L\left(t\right)]$

Wherein, E _nBe the unit matrix of n * n, sampling time interval Δ t is default, and β (t) is measured by the frame corners observer;

In the 4th step, ask the pairing frame corners speed of idle running instruction by following formula

Weighting coefficient α, single frame moment gyro group's frame corners speed is satisfied:

$\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_i\left(t\right)\right|\&le;{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max };i=\stackrel{\&OverBar;}{1,n},$

Be that α should satisfy

α ^-≤α≤α ⁺，

Wherein, ${\mathrm{\&alpha;}}^{-}=\underset{i=1,n}{m\underset{\&OverBar;}{a}x}\min \{\frac{-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&le;0,$ ${\mathrm{\&alpha;}}^{+}=\underset{i=1,n}{m\underset{\&OverBar;}{i}x}\max \{\frac{-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&GreaterEqual;0;$

In the 5th step, obtain revised And α,

$\underset{{\mathrm{\&beta;}}_z}{\min }{q}_2=\underset{{\mathrm{\&beta;}}_z}{\min }{\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_0\left(t\right)+\mathrm{\&alpha;}\&CenterDot;{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}}^2,$

Optimizing α is:

The α that calculates by following formula has guaranteed q ₂Get minimum value, promptly α is the optimum solution of the pairing frame corners speed of idle running instruction

Optimal weighting coefficients;

In the 6th step, obtain by step 3～6:

Single frame moment gyro group's frame corners velocity vector

$\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)+{\mathrm{\&alpha;}\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right);$

In the 7th step, frame corners speed control computing machine is with the frame corners speed of being asked Each frame corners drive motor that offers single frame moment gyro group is controlled; And then measure and feedback framework angle amount, and enter the next sampling period.

In sum, this invention is to design at the high precision spacecraft of using SGCMG control.By considering problems such as gyro gimbal angular velocity, gyro group are unusual, new design proposal has been proposed.Investigate the engineering practice requirement, proposed new performance index, and designed new SGCMG optimal control law on this basis.Find that by theoretical analysis and computer simulation experiment new control law will help to improve the precision, speed of the large angle maneuver control of spacecraft and maneuverability continuously.

Description of drawings:

Fig. 1 .SGCMG principle of work sketch.

Fig. 2. two parallel construction SGCMG-4 synoptic diagram.

Fig. 3. steering order angular velocity is found the solution synoptic diagram.

Wherein four thin broken lines are represented the Z of broken line family (x); Thick broken line s represents the coboundary of Z (x).

Fig. 4. choose the synoptic diagram of weighting coefficient α.

Fig. 5. control program process flow diagram of the present invention.

Fig. 6. steering order program flow diagram of the present invention.

Fig. 7. idle running instruction solver process flow diagram of the present invention.

Fig. 8. find the solution the program flow diagram of weighting coefficient α.

Fig. 9. spacecraft attitude control principle block diagram.

Figure 10. the change curve of spacecraft angular velocity (control law A).

Wherein mark (

▲) represent spacecraft angular velocity in body coordinate system (x, y, z) projection components under respectively.

Figure 11. the change curve of spacecraft angular velocity (control law B).

Wherein mark (

▲) represent spacecraft angular velocity in body coordinate system (x, y, z) projection components under respectively.

Figure 12. the change curve of attitude quaternion (control law A).

Wherein mark (●,

▲) represent four component (λ of spacecraft attitude hypercomplex number respectively ₀, λ ₁, λ ₂, λ ₃).

Figure 13. the change curve of attitude quaternion (control law B).

Wherein mark (●, ▲) represent four component (λ of spacecraft attitude hypercomplex number respectively ₀, λ ₁, λ ₂, λ ₃).

Figure 14. the change curve at gyro gimbal angle (control law A).

Wherein mark (●, ▲) represent gyro gimbal angle (β respectively ₁, β ₂, β ₃, β ₄).

Figure 15. the change curve at gyro gimbal angle (control law B).

Wherein mark (●, ▲) represent gyro gimbal angle (β respectively ₁, β ₂, β ₃, β ₄).

Figure 16. the change curve of the unusual degree of gyro group (control law A).

Figure 17. the change curve of the unusual degree of gyro group (control law B).

Figure 18. the change curve of gyro gimbal angular velocity (control law A).

Wherein mark (●, ▲) represent gyro gimbal angular velocity respectively

Figure 19. the change curve of gyro gimbal angular velocity (control law B).

Wherein mark (●,

▲) represent gyro gimbal angular velocity respectively

Figure 20. the Changing Pattern of angular velocity mould in the particular flight task.

Figure 21. the Changing Pattern of the unusual degree of gyro group in the particular flight task.

Embodiment:

The general flow of spacecraft attitude control: at first determine the attitude quaternion that spacecraft is current by attitude and heading reference system Attitude angular velocity In calculation control moment module, by to the current state amount With required quantity of state (Λ ^*(t), ω ^*(t)) comparison calculates required control moment vector M ^*(t).With signal M ^*(t) pass to SGCMG control law module.Find the solution by designed new control law

Then will

Submit to hardware, realize the rotation control of spacecraft by SGCMG.Last attitude and heading reference system will be determined new attitude parameter, and repeat this process (see Fig. 9, SGCMG control law module is a content of the present invention among the figure).

The purpose of simulation calculation is to check the rationality and the feasibility of SGCMG control law.Though test does not relate to particular hardware, but by and the contrast of existing control law and simulate specific control task, realized evaluation and test to the control law performance.

The simulation calculation program realizes calculation control moment, the SGCMG control law module among Fig. 9.Spacecraft attitude is determined and hardware realizes that module is considered to desirable link.Comparing element in the emulation, relatively the performance situation of different control laws under the same terms.And particular task simulation link, in the time of will providing spacecraft to carry out 7 large angle maneuver control continuously, the performance situation of new control law.

(1) with the contrast of existing SGCMG control law

Select widely used a kind of SGCMG control law at present, this control law is with the guaranteed performance index $\underset{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}{\min }\left(0.5{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}^T\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\right).$ At this moment can obtain analytic solution:

$\stackrel{\&CenterDot;}{\mathrm{\&beta;}}=L^T\left(t\right){\left(L\left(t\right)L^T\left(t\right)\right)}^{-1}{M}^{*}\left(t\right)---\left(17\right)$

Below by l-G simulation test to this two kinds of control laws (control law of invention-control law A; Control law (17)-control law B) compares.

Suppose that spacecraft need carry out 180 ° fast (the wide-angle attitude maneuver of 1 ° of average rotational angular velocity/s).The boundary condition of attitude maneuver is: initial angular velocity omegae ₀=(0,0,0) °/s; The initial attitude hypercomplex number ${\mathrm{\&Lambda;}}_0={\mathrm{\&lambda;}}_0+{\mathrm{\&lambda;}}_0=(\mathrm{0,1}/\sqrt{3},1/\sqrt{3},1/\sqrt{3});$ The end angle speed omega _T=(0,0,0) °/s; Stop attitude quaternion Λ _T=(1,0,0,0).The spacecraft moment of inertia matrix $I=\left[\begin{array}{ccc}1500& 0& 0\\ 0& 3000& 0\\ 0& 0& 4000\end{array}\right]$ The structural parameters of solar array (elastic rod) are: ratio of damping ε ₁=0.4 1/s, ε ₂=0.35 1/s; Vibration frequency w ₁=7.75 1/s, w ₂=8.37 1/s; Coupling coefficient $v_{}$ ${}_1=4.6\sqrt{\mathrm{kg}}\&CenterDot;m,v_2=0.51\sqrt{\mathrm{kg}}\&CenterDot;m.$ The parameter of SGCMG is: h=25kg*m ²/ s, ${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }=10o/s,$ ${\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&beta;}}}_{\max }=30o/s^2;$ Gyro group: 4-SGCMG, two parallel constructions.The unusual degree index of gyro group ${\mathrm{\&sigma;}}_L=\frac{\det \left(L\left(t\right)L^T\left(t\right)\right)}{P},$ Wherein $P=\underset{\mathrm{\&beta;}}{\max }\det \left(L\left(\mathrm{\&beta;}\right)L^T\left(\mathrm{\&beta;}\right)\right).$ Therefore, 0≤σ _L≤ 1.The unusual degree of gyro group σ _LBig more, the dynamic configuration of gyro group is good more, otherwise then poor more.In accompanying drawing 10-19, provided in the spacecraft attitude mobile process, use different control laws (A, relevant parameter change curve B):

Draw to draw a conclusion by analysis simulation result:

1. find by analyzing accompanying drawing 10-13, use control law A or B all can realize desired high precision, large angle maneuver control fast.Average 1 °/s of rotational angular velocity, departure: attitude error ε _Λ＜0.1 °; Angular velocity error ε _Λ＜0.01 °.

2. find by the attribute (accompanying drawing 14-19) of analyzing the gyro group running, use different SGCMG control laws that sizable difference will be arranged.Control law A finishes under the prerequisite of control, has also guaranteed the good dynamically configuration of gyro group when control finishes, σ _L=0.81 (accompanying drawing 16).And control law B makes the dynamic configuration of gyro group become no good when control finishes, σ _L=0.01 (accompanying drawing 17).

3. in the zone of 40～100s, reach bigger rotational angular velocity in order to make spacecraft, no matter use control law A or B, gyro group all will enter useful unusual state, and this is a kind of useful, inevitable trend.Different is that control law A leads the dynamic configuration of gyro group after 100s towards a good direction, and has finally realized controlling the good dynamically configuration when finishing.And control law B can not realize this effect.

4. owing to use control law A can guarantee to control the good dynamically configuration of gyro group when finishing, therefore, the attitude maneuver next time of spacecraft will start from a kind of comparatively desirable state.This has improved the continuous attitude maneuver ability of spacecraft greatly.

(2) particular flight task analog simulation

This l-G simulation test is intended to show the stepless control ability of new control law.Starting condition sees Table 1, and the task performance sees Table 2, and accompanying drawing 20,21 has provided the Changing Pattern of angular velocity mould and the Changing Pattern of the unusual degree of gyro group respectively.Dotted line among the figure is the sign of each attitude maneuver control initial time.By analyzing the performance of continuous 7 times attitude maneuver control, find that new SGCMG control law has quite high continuous maneuverability.

Task	????ω 0, degree/s	????ω T, degree/s	Corner	Note
					????1	????(0，0，0)	????(0，0，0)	????30°	??ω 0＝ω T＝0
????2	????(0，0，0)	????(0.07，0.1，0)	????45°	??ω 0＝0，ω T＝0
					????3	????(0.07，0.1，0)	????(-0.035，-0.05，0)	????120°	??(ω 0‖ω T)⊥λ 0
????4	????(-0.035，-0.05，0)	????(-0.1，0.07，0)	????60°	??ω 0⊥ω T⊥λ 0
					????5	????(-0.1，0.07，0)	????(0，0，0.1)	????0°	??ω 0⊥ω T，λ 0＝(0，0，0)
????6	????(0，0，0.1)	????(0，0，-0.05)	????180°	??ω 0‖ω T‖λ 0
					????7	????(0，0，-0.05)	????(0，0，0)	????90°	??ω 0⊥λ 0，ω T＝0

The starting condition of table 1 attitude maneuver control

The performance of table 2 attitude maneuver control

Claims (1)

1. the single frame moment gyro group's of spacecraft large angle maneuver control method in optimal control is characterized in that it contains following steps successively:

The first step, initialization:

To single frame moment gyro group, i.e. SGCMG group's frame corners velocity vector, promptly

Control computer import following physical parameter:

N is the quantity of single frame moment gyro, n 〉=3, and the steric configuration of gyro group;

The following parameter of single frame moment gyro:

Angular momentum h, units m ²/ s; The upper limit of frame corners speed Unit degree/s; The upper limit of frame corners acceleration, Unit degree/s ²The moment of inertia matrix I of spacecraft; The structural parameters of solar array;

The boundary condition of attitude maneuver:

Initial angular velocity omegae ₀, initial attitude hypercomplex number Λ ₀, the end angle speed omega _T, stop attitude quaternion Λ _T, the unusual degree index of gyro group σ _L, 0≤σ _L≤ 1;

This spacecraft need carry out the angle and the angular velocity of quick wide-angle attitude maneuver:

Single frame moment gyro group's control moment vector M ^*(t), in each sampling period, the attitude quaternion that provides according to the attitude observer And attitude angular velocity

Calculate, and offer frame corners speed control computing machine by the control moment control computer;

The gyro gimbal angular velocity vector

Be the vector that each moment gyro frame corners speed is formed, in frame corners speed control computing machine, calculate, and offer the framework drive motor of each gyro by frame corners speed control computing machine;

Frame corners vector β (t) is the vector that the corner of each moment gyro is formed, and records by the frame corners observer, feeds back to described frame corners speed control computing machine then;

The closed-loop system that described control moment control computer, frame corners speed control computing machine, single frame moment gyro group, frame corners observer and attitude observer are formed is installed on the spacecraft that will control;

In addition, also designed following mathematical model with form of software:

Mathematical model with the single frame moment gyro group control of vector form statement:

$L\left(t\right)\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)=M^{*}\left(t\right),$

Wherein, $\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)=[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_1\left(t\right),{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_2\left(t\right)...{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_n\left(t\right){]}^T,n\&GreaterEqual;3,L\left(t\right)=\&PartialD;H\left(t\right)/\&PartialD;\mathrm{\&beta;}\left(t\right)=L\left(\mathrm{\&beta;}\right)=\left[\begin{array}{cccc}\frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_2}& ...& \frac{{\&PartialD;H}_x}{{\&PartialD;\mathrm{\&beta;}}_n}\\ \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_2}& ...& \frac{{\&PartialD;H}_y}{{\&PartialD;\mathrm{\&beta;}}_n}\\ \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_1}& \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_2}& ...& \frac{{\&PartialD;H}_z}{{\&PartialD;\mathrm{\&beta;}}_n}\end{array}\right],$

I=(x, y z) are the celestial body coordinate system, $H\left(t\right)=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{h}_i\left(t\right)=f\left(\mathrm{\&beta;}\left(t\right)\right),$ H (t) be the single frame moment gyro group's and angular momentum, promptly described each moment gyro angular momentum and, and h _i=j _iΩ _i, j _iInertia matrix for gyro i; β (t)=[β ₁(t), β ₂(t) ... β _n(t)] ^T, be described single frame moment gyro group's frame corners vector;

The mathematical model of frame corners speed:

$\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)+\mathrm{\&alpha;}{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right),$

Wherein,

Be the frame corners speed of steering order decision, i.e. pilot angle speed; Be the frame corners speed of idle running instruction decision, angular velocity promptly dallies; α is idle running angular velocity Weighting coefficient, 0≤α≤1;

The mathematical model of the optimum single frame moment gyro control of the quick large angle maneuver control of spacecraft:

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }{q}_1=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }\underset{i=\stackrel{\&OverBar;}{1,n}}{\max }\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}\left(t\right)\right|,$

That is, from n single frame moment gyro, choose earlier a moment gyro of the absolute value maximum of framework angular velocity, choose minimum frame corners speed more therein;

Guarantee the dynamic configuration of the gyro group fixed final frame corners vector β of feeding that becomes _TMathematic model for optimal control:

In the process of spacecraft wide-angle attitude control, by minimizing β _TWith (t+ Δ t) frame corners constantly $\{\mathrm{\&beta;}\left(t\right)+[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}$ Difference find the solution the frame corners speed that idle running instruction is determined

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }{q}_2=\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z}{\min }\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}{\}}^2,$

Wherein, β (t) is measured by the frame corners observer, Δ t sampling time interval;

Second step, set n=4, according to following simultaneous equations:

In the stage of the quick large angle maneuver control of spacecraft, from n single frame moment gyro, choose earlier a moment gyro of the absolute value maximum of framework angular velocity, choose minimum frame corners speed more therein as optimum solution, it contains following steps successively:

The 2.1st step, n the frame corners velocity vector that steering order is definite Be expressed as:

${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)={(\underset{1\&times;3}{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o^{\left(1\right)}},{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4})}^T,$

Wherein, Be vector First three items, be one 1 * 3 vector; Be vector

The 4th, be a scalar; Accordingly $L\left(t\right)=\left[\begin{array}{cc}\underset{3\&times;3}{L_{11}}& \underset{3\&times;1}{L_{12}}\\ \underset{1\&times;3}{L_{21}}& \underset{1\&times;1}{L_{22}}\end{array}\right],$

The 2.2nd step, order $x={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4},$ $Z={\left(\right|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o1}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o2}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o3}|,|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{o4}\left|\right)}^T,$ Be the absolute value of the pairing frame corners velocity vector of steering order, satisfy following equation:

Z＝|C ₀+C ₁x|，

Wherein, $C_0={(c_0^T,0)}^T;C_1={(c_1^T,1)}^T;$ c ₀=L ^-1 ₁₁M ^*(t) c ₀, be one 3 * 1 vector; c ₁=-L ^-1 ₁₁L ₁₂, be one 3 * 1 vector;

(x, Z) on the plane, the ABS function Z of n frame corners speed is a broken line family;

Obtain thus

$\underset{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o}{\min }{q}_1=\underset{x}{\min }\underset{i=\stackrel{\&OverBar;}{1,n}}{\max }{Z}_i\left(x\right),$

That is, will find the solution

Transform into: (x, Z) on the plane, the possible interval of x

On, ask for n group broken line Z _i(x) coboundary is asked for the pairing x of minimum value, Z again from the coboundary _i(x) value;

This steps in sequence contains following each step:

The 2.2.1 step in frame corners speed control computing machine, calculates all intersection points of the Z of broken line family, data is deposited among the array Q again;

The 2.2.2 step, find the solution the maximal value of the function Z of all intersection point positions, obtain coboundary s, deposit among the array D;

In the 2.2.3 step, compare maximal value array D, and select wherein minimum value conduct Separate;

The 3rd step is according to following simultaneous equations

Choose idle running frame corners speed Make gyro gimbal angle amount $\{\mathrm{\&beta;}\left(t\right)+[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}\}$ With final frame corners vector β _TMinimum in (t+ Δ t) difference constantly; And with the frame corners speed of this single frame moment gyro As the optimum solution of the frame corners speed under the idle running instruction, it has following analytical form:

${\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)=[({\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right))/\mathrm{\&Delta;t}-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)][E_n-L^T\left(t\right){\left(L\left(t\right)L^T\left(t\right)\right)}^{-1}L\left(t\right)]$

Wherein, E _nBe the unit matrix of n * n, sampling time interval Δ t is default, and β (t) is measured by the frame corners observer;

In the 4th step, ask the pairing frame corners speed of idle running instruction by following formula Weighting coefficient α, single frame moment gyro group's frame corners speed is satisfied:

$\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_i\left(t\right)\right|\&le;{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max };i=\stackrel{\&OverBar;}{1,n},$

Be that α should satisfy

α ^-≤α≤α ⁺，

Wherein, ${\mathrm{\&alpha;}}^{-}=\underset{i=\stackrel{\&OverBar;}{1,n}}{\max }\min \{\frac{{-\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&le;0,$ ${\mathrm{\&alpha;}}^{+}=\underset{i=\stackrel{\&OverBar;}{1,n}}{\min }\max \{\frac{-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|};\frac{{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\max }-{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_{\mathrm{oi}}}{\left|{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\right|}\}\&GreaterEqual;0;$

In the 5th step, obtain revised

And α,

$\underset{{\mathrm{\&beta;}}_Z}{\min }{q}_2=\underset{{\mathrm{\&beta;}}_Z}{\min }\{{\mathrm{\&beta;}}_T-\mathrm{\&beta;}\left(t\right)-[{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_o\left(t\right)+\mathrm{\&alpha;}\&CenterDot;{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right)]\&CenterDot;\mathrm{\&Delta;t}{\}}^2,$

Optimizing α is:

The α that calculates by following formula has guaranteed q ₂Get minimum value, promptly α is the optimum solution of the pairing frame corners speed of idle running instruction Optimal weighting coefficients;

In the 6th step, obtain by step 3～6:

Single frame moment gyro group's frame corners velocity vector

$\stackrel{\&CenterDot;}{\mathrm{\&beta;}}\left(t\right)={\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_O\left(t\right)+\mathrm{\&alpha;}{\stackrel{\&CenterDot;}{\mathrm{\&beta;}}}_z\left(t\right);$

In the 7th step, frame corners speed control computing machine is with the frame corners speed of being asked

Each frame corners drive motor that offers single frame moment gyro group is controlled; And then measure and feedback framework angle amount, and enter the next sampling period.

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