Lyapunov Control for Flat Spin Recovery and Spin Inversion

6348c7765745abd70d165eb3e34eca1b?s=47 Zac Manchester
September 16, 2016

Lyapunov Control for Flat Spin Recovery and Spin Inversion

I developed new control laws for stabilizing a tumbling spacecraft using reaction wheels. This presentation is from the 2016 AIAA Astrodynamics Specialist Conference.

6348c7765745abd70d165eb3e34eca1b?s=128

Zac Manchester

September 16, 2016
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  1. Zac  Manchester Harvard  Agile  Robotics  Lab (Formerly  Cornell  Space  Systems

     Design  Studio) Lyapunov-­‐Based  Control  Laws  for Flat  Spin  Recovery  and  Spin  Inversion
  2. 2 Motivation

  3. 3 Problem: • Achieve  spin  about  minor  axis  of  inertia

     from  arbitrary  initial  conditions • Align  + face  of  the  spacecraft  with  the  angular  momentum  vector Constraints: • Limited  reaction  wheel  torque • Limited  angular  momentum  storage  in  reaction  wheels Motivation
  4. 4 Gyrostat  Dynamics: J = I 1 ˙ h =

    h ⇥ J ·(h ⇢) h = I·! + ⇢ I· ˙ ! + ! ⇥ (I·! + ⇢) + ˙ ⇢ = ⌧ Background
  5. 0 0.5 1 1.5 2 5 Lyapunov Stability: ˙ x

    = f ( x ) V ( x ) 0 ˙ V ( x ) = @V @x d x d t  0 f ( x ⇤) = 0 Background
  6. 6 Some  Candidates: VQ = 1 2 hd ·J ·hd

    1 2 h·J ·h Lyapunov  Functions
  7. 7 Some  Candidates: VL = hd ·J ·hd hd ·J

    ·h Lyapunov  Functions
  8. 8 Periodic  Averaging: V 0 L = hd ·J ·hd

    1 T Z T 0 hd ·J ·h dt Lyapunov  Functions
  9. 9 Periodic  Averaging: V = 3 2 hd ·J ·hd

    1 T Z T 0 hd ·J ·h + 1 2 h·J ·h dt Lyapunov  Functions
  10. 10 Discretization: hk = h(t0 + k t) ⇢k =

    ⇢(t0 + k t) xk = 2 6 4 hk . . . hk+N 1 3 7 5 uk = 2 6 4 ⇢k . . . ⇢k+N 1 3 7 5 xd = 2 6 4 hd . . . hd 3 7 5 Calculating  ̇
  11. 11 Discretization: 1 T Z T 0 hd · J

    · h + 1 2 h · J · h d t ⇡ 1 N x | d ¯ J xk + 1 2 N x | k ¯ J xk ¯ J = 2 6 6 6 4 J 0 · · · 0 0 J · · · 0 . . . . . . ... . . . 0 0 · · · J 3 7 7 7 5 Calculating  ̇
  12. 12 Periodic  System: A = 2 6 6 6 6

    6 6 4 0 13⇥3 0 · · · 0 0 0 13⇥3 . . . 0 . . . . . . ... ... . . . 0 0 · · · 0 13⇥3 13⇥3 0 · · · 0 0 3 7 7 7 7 7 7 5 Bk = t 2 6 6 6 6 4 h⇥ k J 0 · · · 0 0 h⇥ k+1 J 0 . . . . . . ... ... 0 0 · · · 0 h⇥ k+N 1 J 3 7 7 7 7 5 Calculating  ̇ ˙ h = h ⇥ J ·(h ⇢) ! xk+1 = Axk + Bkuk
  13. 13 Continuum  Limit: V = 1 N x | d

    ¯ J xk + 1 2 N x | k ¯ J xk 1 N x | d ¯ J xk+1 1 2 N x | k+1 ¯ J xk+1 = 1 N x | d ¯ J Bk uk 1 N x | k A | ¯ J Bk uk 1 2 N u | k B | k ¯ J Bk uk ˙ V = lim N!1 V t = 1 T Z t0+T t0 (h + hd) · J · h ⇥ J · ⇢ dt Calculating  ̇
  14. 14 Almost-­‐Global  Asymptotic  Stability: ˙ V = 1 T Z

    T 0 (h + hd)·J ·h ⇥ J ·⇢ dt b(h) ⇢ ⇡ ⇢ max sign(b) Control  Laws
  15. 15 Smooth  Control  Law: ⇢ = ⇢ max tanh(↵b) x

    -5 0 5 tanh(x) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Control  Laws
  16. 16 Momentum  Vector  Trajectory: 0.5 0 -0.5 -1 1 0.5

    0 -0.5 0.5 0 -0.5 1 Flat-­‐Spin  Recovery
  17. 0 2 4 6 8 ; 1 -0.01 0 0.01

    0 2 4 6 8 ; 2 -0.01 0 0.01 Time (min) 0 2 4 6 8 ; 3 -0.01 0 0.01 17 Control  Inputs: Flat-­‐Spin  Recovery
  18. 18 Momentum  Vector  Trajectory: 0.5 0 -0.5 -0.5 0 0.5

    -1 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 Spin  Inversion
  19. 0 1 2 3 4 5 6 ; 1 -0.01

    0 0.01 0 1 2 3 4 5 6 ; 2 -0.01 0 0.01 Time (min) 0 1 2 3 4 5 6 ; 3 -0.01 0 0.01 19 Control  Inputs: Spin  Inversion
  20. 20 zmanchester@seas.harvard.edu Questions?