Integrator Backstepping using Barrier Functions for Systems with Multiple State Constraints

Transcription

1 Integrator Backstepping using Barrier Functions for Systems with Multiple State Constraints Khoi Ngo Dep. Engineering, Australian National University, Australia Robert Mahony Dep. Engineering, Australian National University, Australia Zhong-Ping Jiang Department of Electrical and Computer Engineering Polytechnic University, USA. Multi-state constrained backstepping. 15 December 005 1

2 Motivating problem Consider the idealized model of the longitudinal dynamics of an aircraft: elevator input δ E pitch rate q pitch attitude θ climb rate. h altitude h Control margin provided by elevators is sufficient for all required manoeuvres. Magnitude constraints on: Climb rate due to stall. Pitch attitude and pitch rate due to passenger comfort factor. Control design must guarantee that the constrained states always remain below their limits. Multi-state constrained backstepping. 15 December 005

3 Some flight control design techniques Gain scheduled linear control design: Does not explicitly include constraint limits. Constrained designs tend to be based on low gain arguments. Feedback linearization: Demands accurate an dynamic model. Does not directly incorporate constraints. Model Predictive Control: Requires a good model and is computationally intensive. Forwarding: Incorporates input constraints and can be extended to state constraints. Low gain design. Multi-state constrained backstepping. 15 December 005 3

4 Backstepping Affords control designers great freedom in selecting final control law. Able to accommodate large nonlinearities and uncertainties in system s model, ignored dynamics, input/measurement disturbances. Aircraft longitudinal dynamics can be transformed into lower-triangular feedback form. The lack of gain-limiting feedforward interconnection terms means no gain restriction on intermediate control signals. High gains are required to impose state constraints. Multi-state constrained backstepping. 15 December 005 4

5 Motion systems dynamics ξ 1 = ξ ξ = ξ 3 ξ 3 = ξ 4 ξ 4 = b(ξ) + a(ξ)u Translation position (unconstrained) Translation velocity Attitude Angular velocity Magnitude constraints on system states ξ i (t) Ξ i, i =, 3, 4 The proposed approach can be extended to: Arbitrary lower triangular systems. Systems of arbitrary order. Multi-state constrained backstepping. 15 December 005 5

6 Step I: Growth condition on first state for clf Consider the first scalar system ξ 1 = ξ Since ξ 1 is unbounded the clf design must incorporate a growth bound on its dependence on ξ 1. V 1 (ξ 1 ) = O(ξ 1 ), as ξ 1 A candidate clf is V 1 (ξ 1 ) = k 1 ξ 1 arctan(ξ 1 ), k 1 > 0 The time differential of V 1 is bounded in ξ 1 V 1 = k 1 ξ [arctan(ξ 1 ) + ξ ξ 1 ] Multi-state constrained backstepping. 15 December 005 6

7 Step I: Bounded reference trajectory Define the error variable z 1 z 1 = ξ ξ ref The reference trajectory ξ ref must be bounded. ξ ref = c 1 arctan(ξ 1 ), c 1 > 0 By choosing bounded growth in V 1 and bounded reference trajectory, ξ ref, this ensures bounded propagation of virtual error in the back-stepping procedure. Multi-state constrained backstepping. 15 December 005 7

8 Step I: Error dynamics of ξ 1 -subsystem ξ 1 = c 1 arctan(ξ 1 ) + z 1 V 1 = k 1 c 1 arctan(ξ 1 ) W (ξ 1 ) + k 1 z 1 [ where W 1 (ξ 1 ) is positive-definite in ξ 1, [ arctan(ξ 1 ) + ξ ξ1 [ + k 1 z 1 arctan(ξ 1 ) + ξ 1 arctan(ξ 1 ) + ξ ] ξ1 ] 1 + ξ 1 ] Multi-state constrained backstepping. 15 December 005 8

9 Step II: Barrier function on z 1 Need to choose an augmented clf V that takes V 1 and adds a term for the stability of the z 1 state. We propose to incorporate a barrier condition on the clf V with respect to z 1 z 1 k = V (z 1 ) A candidate clf is V (ξ 1, z 1 ) = V k 3 log ( k k z 1 Local quadratic structure of V ) V k 1 ξ 1 + k 3z 1, for ξ 1, z 1 small Linear growth in ξ 1. Unbounded growth in V as z 1 k. Multi-state constrained backstepping. 15 December 005 9

10 Step II: Choosing stabilising function ξ 3ref The time-derivative of V is V = W 1 (ξ 1 ) + z 1 [k 1 arctan(ξ 1 ) + k 1ξ ξ 1 + k ] 3ż 1 k z 1 The stabilizing function ξ 3ref is chosen as Thus, ξ 3ref = c z 1 c 1ξ 1 + ξ 1 ż 1 = c z 1 (k z 1 ) k 3 (k z 1 ) k 3 [ [ k 1 arctan(ξ 1 ) + k 1ξ ξ 1 k 1 arctan(ξ 1 ) + k 1ξ ξ 1 ] + z ], V = W 1 k 3c z 1 k z 1 + k 3z 1 z k z 1 Multi-state constrained backstepping. 15 December

11 Step II: Boundedness of state ξ Assume for a moment that ξ is directly controlled and one can set z = 0. Then V = W 1 k 3c z 1 k z 1 V (t) < V (0) It follows that for all time t > 0 z 1 < k and consequently that ξ = z 1 + ξ ref < k + π c 1 Choosing k and c 1 such that k + π c 1 Ξ guarantees the first state bound holds for all time. Multi-state constrained backstepping. 15 December

12 Step III: Continue back-stepping process Define the error signal variable z 3 = ξ 4 ξ 4ref A candidate clf is V 3 = V + 1 k 5 log ( k 4 k 4 z The time derivative of V 3 is ) V 3 = W 1 k 3c z 1 k z 1 Choose stabilising function + k 3z 1 z k z 1 + k 5z k4 ż z ξ 4ref = c 3 z + ξ 3ref, c 3 > 0, ż = c 3 z + z 3 Multi-state constrained backstepping. 15 December 005 1

13 Step III: Stability analysis The cross-term k 3z 1 z k z 1 cannot be directly canceled. The clf time derivative is given by V 3 = W 1 k 3c z1 k z 1 z } {{ } need to show negative semi-definite + k 3z 1 z k z 1 k 5c 3 k4 z + k 5z z 3 k 4 z Case I: If (k z 1 ) is large relative to (k 4 z ) then it is possible to choose gains to ensure that the three terms are negative using a completion of squares argument. Case II: If (k z 1 ) is small relative to (k 4 z ) then z 1 >> 0. Since z 4 < k 4, it is possible to choose the gains such that k 3 c z 1 + k 3z 1 z < 0 Multi-state constrained backstepping. 15 December

14 where ξ 1 ż 1 ż = ż3 Step IV: Crank the handle A A 1 c c c 4 A 11 = c 1 arctan(ξ 1 ) A 1 = ξ 1 ( k z 1) k 3 [ ξ 1 z 1 z z 3 k 1 arctan(ξ 1 ) ξ 1 + k ξ 1 ] For suitable choice of c 1 c 4 it is straightforward to show stability of the error system. However, to guarantee the constraints are satisfied at all times then it is necessary to show a non-negative decrease property of constructed clf V 4. Multi-state constrained backstepping. 15 December

15 Step V: State bounds and control tuning Let Z = (k 1,..., k n, c 1,..., c n ) denote the set of gains used. At each back-stepping step there were two bounds derived from the stability analysis k j k i c i [k i α i 1 ] + 1 Φ i (Z) 0 k j+1 c j+1 k j+1 kj [k jα j 1 ] Ψ j (Z) 0 There are also the original bounds on the states that are interpreted as constraints on Z ξ k (t) = z k 1 (t) + ξ kref (t) = X k (k 1,..., k n, c 1,..., c n ) < Ξ k, k =,..., n Optimisation problem: Minimise F (Z) = (Ξ i X i (Z) )/(Ξ i ) over Z subject to {Φ i (Z) 0} and {Ψ j (Z) 0} and { X k (Z) < Ξ k }. Multi-state constrained backstepping. 15 December

16 Simulation for initial position error of 10m 5 ξ ξ ξ ξ Time [s] Multi-state constrained backstepping. 15 December

17 Simulation for initial position error of 100m 50 ξ ξ ξ ξ Time [s] Multi-state constrained backstepping. 15 December

18 Conclusions Use the flexibility of the backstepping procedure to introduce barrier function characteristics in the constructed control Lyapunov function. This introduces parameterized soft bounds on the state error variables in the transformed system. Derive stability conditions on the gain choices at each iteration of back-stepping procedure to guarantee stability. Derive a set of parameter bounds for the control tuning based on the state bounds. Pose a constrained optimisation problem in the control parameters whose solution leads to good closed-loop performance while preserving the state constraints. Multi-state constrained backstepping. 15 December