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Attitude Control

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					       Attitude Control

                    a        n
       Ricardo S. S´nchez Pe˜a
Sistemas Avanzados de Control - ESAII
                  e
 Universidad Polit´cnica de Catalunya

          August 1-3, 2005
                              Program


1. Preliminaries, reference frames (inertial, body, orbit)


2. Kinematics. Rotation matrices and Euler angles.


3. Dynamics. Angular momentum, Euler equations.


4. Perturbations in satellites.


5. Passive Control: Spin, gravity gradient.


6. Active Control: Sensors, Actuators and Control.


7. Application examples: aircraft, satellites and rockets.
                         Bibliography


1. Wertz, J. ed., Spacecraft Attitude Determination and Control,
   D. Reidel Publishing Company, Dordrecht, Holland, 1978.


2. Griffin, M.D., and French, J.R., Space Vehicle Design, AIAA,
   1991.


3. Brown, C. D. Elements of Spacecraft Design, AIAA, 2002.


4. Piscane, V., Moore, R. Fundamentals of Space Systems, Oxford,
   1994.


    a         n
5. S´nchez Pe˜a R., Alonso R., Control de Veh´ıculos Espaciales,
                                  a              a
   Revista Iberoamericana de Autom´tica e Inform´tica, Vol. 2,
   No. 3, 2005.
Preliminaries
                       Attitude Control


1. What’s attitude control?


2. Why do satellites need it?

   • Communications?

   • Earth observation?

   • Scientific?

   • Military?

   • Bus (Power, antennas, thermal, sensors, etc)?
                  Attitude Control (cont’d)


1. Attitude control?

   • Is the system that determines and controls the orientation
     of the vehicle (S/C or A/C). Oriented w.r.t. what?

   • Passive (no loop).

   • Active (closed loop w/sensors, actuators, algorithm).


2. Why do satellites need it?

   • To orient antennas towards Earth for communications.

   • To orient cameras towards Earth/stars/galaxies/etc for ob-
     servation.

   • Orientation of solar panels to maximize energy reception.
                      Reference Frames


Inertial frame : Non rotating/non accelerated frame w.r.t. ”fixed
    stars”. For near Earth orbits consider Earth centered, Vernal
    pointed frame.


Earth centered Earth fixed frame (ECEF) : North pole, Equa-
   tor/Greenwich intersection direction.


Orbital (aircraft) frame : Roll (velocity vector), Pitch (⊥ orbit),
   Yaw (completes the RH frame: Nadir when orbit is circular).


Body frame : along the symmetry axes of the S/C (principal inertia
  axes).
Kinematics
                           Rotations

Two bodies can rotate with respect to each other (e.g. satellite &
Earth). Represent this by:


 • Attaching a frame to each (rigid) body (frame S to satellite,
   frame E to Earth).


 • Providing both frames share the same origin.


 • Quantify their rotation as an angular velocity vector Wes.
                             Notation


x Physical vector representing position, velocity, acceleration, etc.


Wab Rotation angular velocity of frame B w.r.t frame. A.


  b
 Ca Rotation matrix which transforms (mathematical) vectorcom-
                                                                
                                                   xb          xa
                                                         b     
    ponents in frame A to components in frame B:  yb  = Ca  ya .
                                                   zb          za
                   Rotation parameters


• Remember that translation and rotation can be decoupled if we
  define the CM.

  – Orbital motion – 3 parameters define the position of the CM
    of the S/C.

  – Attitude motion – 3 parameters define the orientation of the
    S/C w.r.t. the CM.


• To quantify rotation between frames (bodies) we may use:

  – Rotation matrices

  – Euler angles
                          Rotation matrices

Question: given a vector v represented in frame A, how do we
represent it in frame B?

                                       b              b
Answer: By means of a rotation matrix Ca, i.e. v b = Cav a,
                            
      1        2 − −     1        
                                  
                            
                                  
     0  =  2 − −  0         
                                  
                                  
                                  
                                  
                                  
      0        2 − −     0        
                                  
                                  
                                  
                                  
                            
                                  
                                                            
         0          − 2 −      0 
                                              1    0      0
                                 b                      
      cos φ  =  − 2 −   1       Ca =  0   cos φ sin φ 
                                  
      − sin φ       − 2 −      0 
                                  
                                              0 − sin φ cos φ
                                  
                                  
                                  
                           
                                  
        0         − − 2      0    
                                  
                           
     sin φ  =  − − 2   0   
                                  
                                  
                                  
                                  
      cos φ       − − 2      1
               Rotation matrices & Euler angles

A general rotation between 2 frames can be represented by 3 con-
secutive rotations around 3 different axis:


                                                           
       1    0      0      cos θ 0 − sin θ     cos ψ sin ψ 0
 b                                                       
Cn =  0 cos φ sin φ   0      1   0       − sin ψ cos ψ 0 
       0 − sin φ cos φ    sin θ 0 cos θ          0      0   1
           X−rotation         Y −rotation          Z−rotation


This can also be parameterized in terms of the 3 angles of rotation
φ, θ, ψ. These are called the [Z −Y −X] Euler angles (which is NASA’s
standard).
Dynamics
                         Next steps


• Angular Momentum


• Euler equations: dynamic model of a rigid S/C


• Kinetic energy


• Torque free motion


• Stability [optional]
                        Angular Momentum

Similar to the definition of linear momentum for linear motion (p =
mv), we may define the moment of momentum or angular momen-
tum for a rotating body (around its CM) as:

 H = IbWib                                                             (1)
                                                                    
                  2
                 r2      2
                      + r3   −r1r2    −r1r3              Ixx Ixy Ixz
                                                                    
                                                                    
                                                                    
  Ib =                       2 + r 2 −r r
                             r1                dm =        Iyy   Iyz 
                                         2 3 
          body 
                                  3
                                              
                                                     
                                                     
                                                                       
                                                                       
                                                                    
                                       2    2
                                      r1 + r2                      Izz
where Ib is the moment of inertia (symmetric) matrix of the body (a
rotational version of mass m) and Wib is the angular motion w.r.t.
an inertial frame (a rotational version of linear velocity v).
                      Angular Momentum

Fact: In any rigid body, its body frame can always be selected so
that its inertia matrix is diagonal. This frame defines the principal
axes of the body and in this frame equation (1) is (simpler):


                                                 
                 hx       Ix 0 0     ω1       ω1 Ix
                                             
         H   =  hy  =  0 Iy 0   ω2  =  ω2 Iy            (2)
                 hz       0 0 Iz     ω3       ω3 Iz


                H & Wib need not be colinear!!!
               Newton’s law and BKE [optional]

Newton’s laws are written for a particle and hold in an inertial frame

                             dH
                         T =                                    (3)
                              dt i−f rame

To represent them in the body frame which rotates w.r.t. an inertial
one at an angular rate of Wib, we will use the Basic Kinematic
equation (BKE).
               dx              dx
                            =              + Wib × x             (4)
                dt i−f rame    dt b−f rame
when both origins are coincident or at least fixed w.r.t. each other.

Combining both equations we obtain Newton’s second law repre-
sented in the body frame: Euler equations.
                       dH
                   T =              + Wib × H                   (5)
                        dt b−f rame
                        Euler’s equations

They are the expression of Newton’s 2nd. law for a rotating body:
                          dH
                   T =                 + Wib × H                    (6)
                           dt b−f rame
which represented in the body frame in principal axes [see eqn. (2)]
are:
                                                           
      Tx       ˙
               hx + ω2hz − ω3hy               ˙
                                           Ixω1 + ω2ω3(Iz − Iy )
            ˙                                                
     Ty  =  hy + ω3hx − ω1hz  =       Iy ω2 + ω3ω1(Ix − Iz )  (7)
                                              ˙
      Tz       ˙
               hz + ω1hy − ω2hx               ˙
                                           Iz ω3 + ω1ω2(Iy − Ix)
They can be simulated if we know T . First let’s assume T = 0 and
visualize the torque free motion.
                         Kinetic Energy

The kinetic energy E is a scalar quantity and is defined as the dot
product between the Angular momentum and the angular velocity
of the body as follows:
                                1
                         E =      H · Wib
                                2

When we adopt the principal axis as the body frame, E reduces to:
                         1    2       2       2
                  E =      Ixω1 + Iy ω2 + Iz ω3
                         2
                       Torque free motion

Assumptions:


 • Body frame centered in CM and along the principal axes.


 • Satellite–like body (cylinder), i.e. Ix = Iy = I


 • Torque free motion (T = 0)


Under these conditions, the Euler equations (7) are:

                     I ω1 + (Iz − I ) ω2ω3 = 0
                       ˙                                      (8)
                       ˙
                     I ω2 − (Iz − I ) ω1ω3 = 0                (9)
                                        ˙
                                     Iz ω3 = 0               (10)
                 Torque free motion–Remarks


1. Angular momentum H remain inertially fixed (T = 0).


                                                   d    2    2
2. Multiplying (8) by ω1 and (9) by ω2 and adding: dt (ω1 + ω2 ) = 0.
   From equation (10), ω3 = constant =⇒ the magnitude of Wib
   is constant, it only changes direction.


3. Kinetic energy is constant, i.e. H · Wib =const., hence Wib pre-
   serves a constant angle w.r.t. H (remember from (1) that Wib
   is not necessarily colinear with H).


4. Combining both (8) and (9): ω1 + Ω2ω1 = 0, i.e. oscillatory
                                   ¨
   motion (Ω defined in (11)). The same holds for ω2, i.e. Wib
   rotates around the z–axis at a constant speed.
                      Torque free motion

Define:
                             Iz − I
                          Ω=        ω3                      (11)
                                I
as the precession frequency and the nutation θ and wobble γ half–
angles as:
                          h                 ω
                  tan θ =           tan γ =                 (12)
                          hz                ω3
both constants.

Conclusion: H is fixed inertially (defines the space cone) and Wib
rotates around the z–axis at a frequency Ω (defines a body cone).
The angular speed of the body around its main axis is ω3.
            Torque free motion stability [optional]

A body is stable when it remains ”near” it’s previous state of motion .

From the Torque free motion parameters, we may compute the
energy of the body as a function of its nutation angle:
                   1            1
           E(θ) =    H · Wib =     H Wib cos(γ − θ)
                   2            2
                   1
                 =    H Wib (cos γ cos θ + sin γ sin θ)
                   2
                    H 2        Iz − I
                 =        1+          sin2 θ
                    2Iz           I
Hint: apply angle definitions in equation (12).
                  Torque free motion stability

The energy derivative w.r.t. time follows:
                 d           H 2 sin 2θ      Iz − I
                           ˙
                    E(θ) = θ
                 dt            2Iz              I

                               d
Fact: Energy dissipation means dt E < 0.


                     ˙
1. For Iz > I , then θ < 0, the nutation angle decreases and the
   body moves to a stable situation.


                     ˙
2. For Iz < I , then θ > 0, the nutation angle increases, and the
   body motion is unstable.


A spinner is stable when it spins around its major inertia axis
              Further implications for satellites


• Whenever H is not colinear with Wib, the S/C nutates.


• This depends only on the initial conditions, when there is a
  lateral component h = 0.


• This is created by a perturbing impulse T t = h (ring pyros
  asymmetric forces when separating satellite from launcher).


• The nutation angle also depends on the magnitude of the main
  angular momentum hz (see equation (12)).


• The spinner responds to a lateral impulse with a nutation angle
  and to a lateral constant torque with an angular constant velocity
  (like a gyroscope).
Perturbations
Attitude Control
                        Passive control

                                   √
• Spinning satellites (spinners)

                                   √
• Gravity gradient (GG) control

                             √
• Magnetic passive control


• Aerodynamic control


• Solar sail
             Passive control: Spinning satellites


• The spin should be applied over the major axis of inertia for
  stability.


• Take into account for mission design that the spinning axis will
  remain ”almost” inertial.


• Whenever H is not colinear with Wib, the S/C nutates, which
  depends only on the initial conditions.
             Passive control: Spinning satellites


• A lateral component of H can be produced by a disturbing im-
  pulse, e.g. when ejected from launcher T δt = δh . This in turn
  produces a nutation angle.


• The larger the rotation speed, the smaller the nutation angle.


• The spinner responds to lateral disturbance torques with or-
  thogonal angular velocities (gyro equation), but to longitudinal
  torques with colinear angular acceleration.
       Passive control: gravity gradient (GG) [optional]

The idea here is to use Nature to control the orientation of the S/C.
The figure is self explanatory.




Equations instead need a little bit more ellaboration. The gravity
gradient torque and the dynamics can be computed as follows:


1. Integrate gravity acceleration ag over S/C:
                              u         u
   Tg = s/c r × ag dm = 3ωo(ˆz × [Is/c]ˆz ).
    u
   (ˆz : yaw axis unit vector, ωo: orbital frequency)
        Passive control: gravity gradient (GG) [optional]


2. Apply the small angle approximation (SAA) and obtain the GG
   torque (φ is roll angle and θ the pitch angle in the orbital frame):
                                             
                                  φ(Iz − Iy )
                              2
                       Tg = 3ωo  θ(Iz − Ix) 
                                              
                                                                  (13)
                                       0


3. Plug Tg into Euler equations and linearize according to the SAA.
   The motion dynamics are oscillations with ωlib frequency in roll
   & pitch around the vertical, represented by equations:
              ¨    2
              φ + ωφlib φ = 0        ,   ¨    2
                                         θ + ωθlib θ = 0          (14)
          Passive control: gravity gradient (GG)

Main facts:


 • Only for Earth observation missions with low precision (±5◦).


 • The torque stabilizes when Iz ≤ min(Ix, Iy ). The design is based
   on the mass distribution of the body.


                                        Iy −Iz
 • Libration frequencies: ωφlib = 2ω0     Ix
                                                                  −I
                                                 , ωθlib = ω0 3 IxIy z ,
   should be damped.


 • Examples: UOSAT (Univ. of Surrey), MSU-1 (Univ. of Bs.As.).
            Sensors & Actuators for active control


• Sensors

  – Star sensors (precise inertial angular detection)

  – Sun sensors (angular detection of solar vector)

  – Earth sensors (angular detection of local vertical)

  – Magnetic sensors (angular detection of local magnetic field)

  – Inertial sensors

    ∗ Gyroscopes: inertial angular velocity

    ∗ Accelerometers: specific force (a − g)
              Sensors & Actuators for active control


• Actuators

  – Thrusters (external torque)

  – Magnetic torquers (external torque)

  – Reaction, momentum wheels (internal torque)

  – Control moment gyroscope (internal torque)
                           Sensors


• Attitude specified by 3 independent parameters (Euler angles),
  or through a rotation matrix (9 parameters - 6 constraints).


• Reference Sensors (local vertical, magnetic field, Sun or star
  direction, GPS)


• Inertial sensors: provide angular rates and precision degrades
  w/time.
                          Sun Sensors


Coarse (CSS): Solar cell which defines Sun vector (S) to within
   1◦ with a cosine–type scale calibration curve: i(θ) = kdAS · n =
                                                                ˆ
   i(0) cos θ. Broad FOV ≈ 180◦. Combinations of 2 CSS provide
   near linear scales. Used to locate Sun initially or in SHM.


Fine (DSS): Digital output in Gray code (one bit change for each
   unit angle). Narrower FOV and higher precision: 64◦/0.125◦
   is typical, but with extra computations can decrease to the arc
   second range (FSS). Can be combined at right angles as a 2-axis
   DSS. Used in Mission Mode.
                        Earth Sensors


• Infrared horizon detector which scans the Earth’s horizon. Mea-
  sures pitch and roll angles (local vertical).


• Nonspining Earth Sensor Assy. (TRW) scans Earth borders
  (around roll and pitch) counting pulses until S/C Z-axis is cen-
  tered w.r.t. to it. Errors of 0.05◦.


• Scan Wheels (Ithaco) places a detector in a rotating wheel which
  provides pulses at times (t1, t2) when crossing Earth’s horizon
  and a S/C reference pulse t . Difference between ∆t = t2 −t1 and
  its nominal value proportional to roll angle. Difference between
  t2 − t and t − t1 proportional to pitch angle. 2 sensors used in
  differential mode to increase accuracy.


• Can be combined with MW’s as a sensor/actuator.
                         Star Sensors


• Star trackers: Fixed to S/C or in a 3 axis stabilized platform.


• Accuracy/FOV: up to 1” and 8◦ × 8◦


• Components:

  – Sunshade: light baffles to protect against Sun or other light
    sources

  – Optical System

  – Detector (CCD): transforms optical to electrical signal

  – Electronics and Computation (basic pattern recognition)
                     Magnetic Sensors


• Senses direction and strength of Earth magnetic field B with
  accuracies of 0.5◦.


• Very robust, low weight and consumption.


• Based on Faraday laws of induction:

  – Search coil: used in spin stabilized S/C

  – Fluxgate: uses 2 saturable cores, detects through 2nd har-
    monic


• Problem: correlation between measurements and B-model.
                       Inertial sensors


• Rotating wheel gyroscopes: T = Wib × H used to measure Wib
  (accuracy of up to 0.01◦/hr)


• Ring Laser gyros (RLG): path length (frequency) of light rays is
  proportional to inertial rotation.


• Fiber optic gyros (FOG): same principle applied on fiber optics
  (accuracies of 0.1◦ to 10◦/hr).
                          Actuators


• External torquers: changes total momentum (necessary).

  – Thrusters (Gas Jets, electric propulsion)

  – Magnetic torquers (T = M × B)

  – Gravity gradient, solar radiation, aerodynamic (can be con-
    sidered as perturbations or used for control purposes)


• Internal torquers: changes distribution of moments among S/C
  moving parts (optional)

  – Momentum and Reaction wheels (MW, RW)

  – Control moment gyroscope (CMG)
                           Thrusters


• Used for attitude (low T ) and orbit control (high T )


• Independent of altitude, wide variety of magnitudes


• Requires fuel, which limits S/C life


• Exhaust plume contaminants


• Produces disturbance torques (i.e. due to CM tolerance). Can
  be counteracted by spinning S/C or controlling thrust axis.
                      Magnetic torquers


• Used for attitude and Momentum dumping. Acts similar to a
  compass needle.


                                                   u
• Rod-like electromagnets produce T = m × B = N iA(ˆm × B).


• Does not require fuel, nor contaminates. Insensitive to CM vari-
  ations.


• Cannot produce torques along B direction (more problems in
  equatorial than in polar orbits).


• Placed away from other instruments and separated from each
  other when used in triads, to avoid perturbations.


• Used in lower orbits due to B strength.
           Momentum, Reaction wheels & CMG’s


MW : Rotates at a constant nominal speed (Ω). Has a maximum
  speed of 6000 rpm and momentum of 50-200 Nms. It produces
  torques in both directions by increasing/decreasing its speed up
  to ±10% of the nominal.


RW : The nominal speed is zero. Usually has lower maximum
  speeds (3500 rpm) and less momentum (2 Nms).


CMG : The same principle located on gimbals so that a single wheel
  can produce torques around any arbitrary axis of the S/C.
     Momentum, Reaction wheels & CMG’s (cont’d)


• Compromise between wheels weight and power consumption (de-
  pends on Ω2).


• New developments of magnetically levitated wheels produce longer
  lifetime in MW, RW and CMG’s.


• Wheel used as actuators or as Momentum storage can be com-
  bined with Earth sensors.
Feedback systems for active Control
                            Recap


• Compute relation between angular speeds & angles Wib = f1(Θ).


• Compute relation between torques & angular speeds T = f2(Wib).


• Obtain a dynamic model of a rigid S/C: T = g(Wib, Θ).


           T produced by perturbations and control


               Wib, Θ variables to be controlled
              3-axis control w/o Momentum bias

The mission is a space telescope which should have a precise point-
ing to different stars, galaxies, etc, during several hours. Several
maneuvers should be performed to point to different targets, com-
manded from the ground station. Furthermore, the controller should
not destabilize flexible slow modes due to (large) solar panels.
     3-axis control w/o Momentum bias – Assumptions


1. Inertial pointing objective.


2. Small inertial rotations (ωi −→ 0, i = 1, 2, 3) and constant angu-
   lar objectives.


3. Two control modes: stabilization & maneuver.


4. Furthermore consider the S/C has coarse Sun sensors combined
   with a Star sensor to measure angles, 3 orthogonal gyros to
   sense inertial angular speeds and 3 orthogonal reaction wheels
   (no momentum bias) to control each axis separately.


5. The telescope’s orientation is along one of the principal axes of
   the spacecraft.
     3-axis control w/o Momentum bias – First Mission

Achieve the first point of Aries (Vernal direction). Therefore:


 • Select the body frame with it’s x–axis along the telescope’s axis
   and the others along the principal axis, centered at CM.


 • Select the desired frame as the Earth’s inertial equatorial frame.


 • The objective is to drive the body frame to the desired one, i.e.
   take the Euler angles to zero.
         3-axis control w/o Momentum bias – Model

As a consequence, the model is a set of 3 independent ODE’s plus
nonlinear equation which relates Wib with the Euler angles αi, i =
1, 2, 3:
                        ˙
                T1 = I1ω1
                        ˙
                T2 = I2ω2
                        ˙
                T3 = I3ω3
                                                             (15)
                      ˙   ˙
                ω1 = α1 − α3 sin α2
                      ˙                 ˙
                ω2 = α3 cos α2 sin α1 + α2 cos α1
                      ˙
                ω3 = α3 cos α2 cos α1 − α2 sin α1
                                        ˙
Torques Ti, i = 1, 2, 3 correspond to both torques acting on the
body, perturbations and control.
                 3-axis control w/o Momentum bias


Trick: Redefine the angular objective for i = 1, 2, 3 as θi = αi − γi,
where γi the angular objective. Hence:


1. The objective will always be to achieve θi −→ 0 both for maneu-
   vers and stabilization.


2. If the control succeeds, we can assume θi near zero and apply
   the S.A. approximation.


With this in mind, applied to equations (15) we obtain:

         Tci +        Tpi             ˙      ¨
                                  = Iiωi = Iiθi    i = 1, 2, 3   (16)
       control    perturbations
        3-axis w/o Momentum bias – Controller design

From the dynamic equation (16) we should design a value for Tc
based on the sensor measurement at hand so that it drives Θ to
zero. Let’s try with Physics .

The equation is equivalent to a mass Ii drived by a force Tpi with
its position θi as the output. An electrical interpretation is that of
an electrical charge θi in a coil Ii drived by a voltage Tpi . For both
interpretations, the best way to take the mass (coil) to the origin
(θi = 0) is by adding a spring (capacitor) and a damper (resistance)
as follows (always i = 1, 2, 3):
                               ¨       ˙
                      Tpi = Ii θi + kd θi + kp θi                 (17)
                                         −Tci
        3-axis w/o Momentum bias – Controller design

Now let’s apply Mathematics . Any solution is the addition of the
solution to the homogeneous equation (kernel) plus a particular solu-
tion.
Homogeneous equation. Try θih = est in:
                   ¨       ˙
             0 = Iiθih + kdθih + kpθih
               = est Iis2 + kds + kp = (s + s1)(s + s2)
                                   =0
Being |θih(t)| = |est| =   |ert| · |eωt|   = |ert| we need:

                      lim |θih(t)| −→ 0         ⇐⇒ Re(s) < 0
                     t−→0
               ⇐⇒    s1 + s2 > 0, s1 s2         > 0 ⇐⇒ kp, kd > 0

Solution: θih(t) = aes1t + bes2t with kp, kd > 0 . The constants (a, b)
                                                         ˙
can be obtained by using the initial conditions [θih(0), θih(0)].
        3-axis w/o Momentum bias – Controller design

Particular solution It depends on the form of the perturbation
Tpi . Suppose it is constant (like in many cases with aerodynamic
torques), then we may also select a constant as the solution θip. It’s
value is obtained by replacing in (17):
                                    Tp i
                            θip =
                                    kp
The controller is called PD w.s.f. proportional–derivative, and achieves
the following solution, which does not vanish with time:

                             ˙
                Tci (t) = −kdθi(t) − kpθi(t)
                θi(t) = θip + aes1t + bes2t kp, kd > 0
         3-axis w/o Momentum bias – Compromises

The time response of this closed loop system is:

           θih(t) = aes1t + bes2t
                                                      2
                     (s + s1)(s + s2) = s2 + s2ξωn + ωn
and it depends on the roots (s1, s2). A standard way to parameterize
the design compromises is by computing:


 • Damping factor ξ = kd
                      2
                             1
                            kp Ii



                              kp
 • Natural frequency ωn =     Ii


                          1
 • Step input error ess = kp (scale input perturbation Tpi = 1).
        3-axis w/o Momentum bias – Compromises


• Increasing the damping (kd) gives a slower response but a better
  stability robustness.


• Increasing kp decreases the error but increases the frequencies
  in the response (could excite natural frequencies in the solar
  panels/antennas)
        3-axis w/o Momentum bias – Controller design

Q. How can we still make θi(t) vanish when the perturbations are
constant (like in many practical cases)?

Argument: We already have        all solutions to the previous equation,
hence we should change the       equation. Add a term that remains
constant as t −→ ∞ although      θi(t) −→ 0.

A. Add a term to the controller:

                            ˙                    t
               Tci (t) = −kdθi(t) − kpθi(t) − kI 0 θi(τ )dτ

             ∞
such that kI 0 θi(t)dt = Tpi .

This controller is called PID w.s.f. proportional–integral–derivative.
                 Example [optional]:

SAC-C Earth observation satellite (launched Nov. 2000)
         SAC-C Earth observation satellite [optional]

Objectives: Earth observation, multi–antenna GPS experiment,
precise magnetic field computation.


 • Rigid boom along the yaw (z) axis


 • 2 MW at CM rotated −θ1 and θ2 w.r.t. roll (x) axis


 • Rotation of wheel j around its principal axe yj


 • Inertial frame I


 • Frames S for the satellite, R1 and R2 for the wheels (fixed to
   the wheel plane but non rotating)
                           Total Momentum

The momentum of the satellite (included the rigid body model of
the boom) is:
                                                     
                                                  ω1
                                         s          
                   Hs = [Is]Wis     ;   Wis   =  ω2                  (18)
                                                  ω3
and for frame Rj of wheel j = 1, 2:
                                                            
                               0                              0
        r            r                 r    s        
       Hj j   = Irj Wirj +  Ωj  = Irj Cs j Wis +  Ωj           (19)
                         j
                               0                              0


 • Wis = Wirj for j = 1, 2 (both frames fixed to S/C and centered
                     r      r
   in its CM) and Wirj = Cs j Wis,
                               s
                       j



 • Ωj : j-wheel rotation velocity w.r.t. yj axis of frame Rj ,
 • Is: satellite+boom inertia matrix,


 • Irj : j-wheel inertia matrix.


therefore
                                     
                         1   0  0
             r                            r
 s
Hj =     s
        Crj Hj j             − sin θj  · Hj j
                   =  0 cos θj       
                              cos θj
                       0 sin θj
                                                              
       Ixj     0            0                        ω1
                                                              
   =  0 Iyj cos θj −Izj sin θj   ω2 cos θj + ω3 sin θj + Ωj 
        0 Iyj sin θj Izj cos θj            −ω2 sin θj + ω3 cos θj
                                                             
       Ixj ω1                                           0
                 2                                           
   =  Iyj ω2 cos θj + ω3 sin θj cos θj  +  Iyj Ωj cos θj 
                                           
       Iyj ω2 sin θj cos θj + ω3 sin2 θj          Iyj Ωj sin θj
                                            
           0
                                             
          Izj ω2 sin2 θj − ω3 sin θj cos θj 
        +                                                      (20)
                                              
                      2 θ − ω sin θ cos θ
           Izj ω3 cos j       2      j      j
The total angular momentum H = Hs + H1 + H2 is:
                                                                       
                                                 0
                                                                    
                              ω1                                      
                H s = I +I  ω +                  2                 
                                                         Iyj Ωj cos θj       (21)
                      [ s w]  2                   j=1               
                              ω3                                      
                                                                      
                                                     2
                                                     j=1 Izj Ωj sin θj
with Iw defined as
                                                                                    
    Ix1 + Ix2                   0                                   0
                                                                                    
                                                                                    
                                                                                    

      0          2
                  j=1 Iyj   cos2 θj + Izj sin2 θj       2
                                                        j=1  Iyj − Izj cos θj sin θj 
                                                                                     
                                                                                    
                                                                                    
                                                                                    
       0          2     Iyj − Izj cos θj sin θj         2          2 θ + I cos2 θ
                  j=1                                   j=1 Iyj sin j     zj       j


Assumptions: Identical wheels, symmetric w.r.t. yj axis, i.e. Ix1 =
Ix2 = Iz1 = Iz2 = Imin and Iy1 = Iy2 = Imax, −θ1 = θ2 = θ ≥ 0,
hence
                                  
                      ω1          0
                                   
   H s = (Is + Iw )  ω2  +  hy (Ω) 
                      ω3       hz (Ω)
   Iw = 2 diag Imin    Imax c2θ + Imin s2θ   Imax s2θ + Imin c2θ

   hy (Ω) = Imax (Ω1 + Ω2) cos θ < 0
   hz (Ω) = Imin (Ω2 − Ω1) sin θ


The total momentum of both wheels is opposite to the pitch axis:
θ1 and θ2 are in the 3rd. and 2nd. quadrants, i.e. wheel’s and
orbital momentum are colinear. NOTE: actually wheel-1 is physically
located at (180-θ1) and runs at −Ω1 < 0 because it’s also an Earth–
scanner.
                            Dynamics

We apply Newton’s second law in terms of the rotation with respect
                              ˙    ˙
to the satellite’s frame T = iH = sH + Wis × H with T the vector of
external torques on the system. The dynamics in this same frame
is:
        ˙
   T = sH + Wis × H =                                 
                                
                                     0                  
                                                         
        = [Iw + Is]   ˙
                    sW s + W × ·  h (Ω)  + I + I W s − T s(22)
                                            [ w
                       is   is  y              s ] is  w
                                 h (Ω)                  
                                    z
where
                                               
                                0       −ω3 ω2
                    ×                          
                   Wis =  ω3            0  −ω1              (23)
                           −ω2          ω1   0
                                         
                                    0
                     s      ˙      
                    Tw = −  hy (Ω)                          (24)
                             ˙
                             hz (Ω)
The torques are:
 • Tw : Wheel controls (internal), depends on the variation of the
   angular momentum of the wheels produced by the velocity change.


 • Tg : Gravitational torque (external) is due to the effect of the
   rigid body model of the boom.


 • Tm: Magnetic control (external) produced by the Magnetic torque
   coils (MTC).


 • Tp: Perturbation torques (external) due to the aerodynamic drag
   (almost constant in pitch and yaw along the orbit), etc.


compared with the former. The dynamic equations of satellite and
wheels are:
                                                          
         control                             0              
                                                            
   s     s
  Tp + (Tw + Tm) s            ˙
                           s W s + W × ·  h (Ω)  + [I ]W s − T s
                     = [It] is      is  y           t  is   g
                                         h (Ω)              
                                            z
with the total inertia defined as It = Iw + Is.

Spacecraft in an Earth observation attitude, i.e. desired frame D has
x along velocity vector (roll), y opposite to orbital rotation (pitch)
and z along local vertical (yaw).
                           s      s    s d
                     W = Wis = Wds + Cd Wid
                                      
                                    0
                               s      
                       = wd + Cd  −ω0                           (25)
                                    0
             µ
Here ω0 = R3 is the orbital frequency and w represents the angular
rates around the desired frame. The gravity torque in these new
variables can be computed as follows:
                               2
                        Tg = 3ω0 (v3 × Itv3)                      (26)
                                   
                                  0
                              s    
                        v3 = Cd  0                              (27)
                                  1
where v3 is the nadir (vertical) direction in the satellite’s frame.
                         Linear Model

General nonlinear dynamic equation of the system:
                               ˙
         Tw + Tm +Tg + Tp = ItsW + W × ItW + W × h          (28)
          control                               
                                    0            0
                         s                    ˙ 
                        Tw =      Twy    = −  hy        (29)
                                   Twz           ˙
                                                 hz
                                        
                                    0
                        s               
                       Tm =       Tmy                     (30)
                                   Tmz


Control torques applied only in the pitch and yaw, because of the
coupling between roll and yaw (they exchange their roles every 41

orbit).

Linearize around equilibrium condition (desired frame). Assume
small angles θR = φ, θP = θ, θY = ψ. With It = diag Ix Iy Iz :
                                     
                        1     θY −θP
             s                          
            Cd =    −θY       1     θR                         (31)
                       θP −θR         1
                                  
                     θ˙R − θY ω0
                    ˙             
            W =     θP − ω0                                    (32)
                     ˙
                     θY + θR ω0
                                                              
                     −hy (Ω)(θ                        ˙
                                 ˙Y + θR ω0) + hz (Ω)(θP − ω0)
                                                 ˙            
     W × h(Ω) =                  hz (Ω)(θY ω0 − θR )           (33)
                                           ˙
                                  hy (Ω)(θR − θY ω0)
                                        
                           θR (Iz − Iy )
            Tg =      2
                   3ω0  θP (Iz − Ix) 
                                         
                                                                 (34)
                                 0
The angular rotation around the desired frame is wT = θR θP θY ,
                                                      ˙  ˙  ˙
and the momentum of the wheels have been linearized around the
nominal: hyo in the −y axis, hzo = 0.
                       hy (Ω) = hyo + δhy (Ω)                   (35)
                       hz (Ω) = δhz (Ω)                         (36)
The final linearized equations are as follows:
                     ˙       ¨      2
       Tm2 + Tp2 − +δhy = Iy θP − 3ω0 (Iz − Ix)θP

                             ¨      2
      Tm1 + Tp1 + ω0δhz = IxθR − 4ω0 (Iz − Iy ) + hyoω0 θR
                                                      ˙
                          + [ω0(Iy − Ix − Iz ) − hyo] θY
                     ˙       ¨     2
        Tm3 + Tp3 − δhz = Iz θY + ω0 (Iy − Ix) − hyoω0 θY
                                                          ˙
                              − [ω0(Iy − Ix − Iz ) − hyo] θR
and Tp is much larger in the pitch and yaw axes, due to the higher
aerodynamic torques.

Note: Roll–Yaw dynamics are coupled but independent from Pitch.

				
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