LMI Methods in Optimal and Robust Control Matthew M. Peet Arizona State University Lecture 20: LMI/SOS Tools for the Study of Hybrid Systems
Stability Concepts There are several classes of problems for which we would like to prove stability: Stability under Arbitrary Switching Sometimes called Differential Inclusions Similar to Time-Varying uncertainty, but discrete ẋ(t) {A 1 x(t), A 2 x(t)} Stability under State-Dependent Switching Alternatively, Does there exist a stabilizing switching law? Often used to stabilize systems such as the inverted pendulum. Stability under Arbitrary Switching with Dwell-Time restrictions Places a lower limit on τi+1 τ i. Often used for hysteresis (e.g. Thermostat problem) There are three tools we will use Quadratic Stability Common non-quadratic Lyapunov functions or Multiple (state-dependent) Lyapunov functions (w/ continuity) Switched Lyapunov functions. M. Peet Lecture 20: 1 / 18
2 Non-intuitive Facts About Switched Stability Fact 1: Stability of each subsystem {A 1, A 2 } does not guarantee stability under arbitrary switching. Figure: 2 Unstable Systems (a,b) can be Stabilized (c) or Destabilized (d) Fact 2: Smart switching can stabilize two unstable subsystems {A 1, A 2 }. Figure: 2 Stable Systems (a,b) can be Stabilized (c) or Destabilized (d) M. Peet Lecture 20: 2 / 18
Stability under Arbitrary Switching Quadratic Stability Theorem 1. The switched system ẋ(t) {A 1 x(t), A 2 x(t)} is stable under arbitrary switching if there exists some P > 0 such that A T 1 P + P A 1 < 0 A T 2 P + P A 2 < 0 and This implies that BOTH A 1 and A 2 are Hurwitz (Necessity). But A 1 and A 2 Hurwitz is not Sufficient. For example, consider A 1 = [ ] 1 1, A 1 1 2 = [ 1 ] 10.1 1 A 1 and A 2 are both Hurwitz. ẋ(t) {A 1 x(t), A 2 x(t)} is stable under arbitrary switching There is no common quadratic Lyapunov function for A 1 and A 2! M. Peet Lecture 20: 3 / 18
Stability under Arbitrary Switching Quadratic Stability and Commuting Matrices Commuting Matrices: If A 1, A 2 are Hurwitz and commute, we have quadratic Stability. Sufficient, not necessary for quadratic stability. Also for larger sets, {A i } if all pairs commute. Easier to check than the LMI Simply Test if A i is Hurwitz and A i A j A j A i = 0 i, j. M. Peet Lecture 20: 4 / 18
Stability under Arbitrary Switching Common Non-Quadratic Lyapunov Functions Theorem 2. The switched system ẋ(t) {f 1 (x(t)), f 2 (x(t))} is stable under arbitrary switching if there exists some V (x) > α x 2 such that V (x) T f 1 (x) < 0 x and V (x) T f 1 (x) < 0 x Converse Result: If ẋ = f p (x(t)) is asymptotically stable and f p is locally Lipschitz, then there exists a common Lyapunov function. If ẋ {A i x(t)} i, then V can be chosen to have the form V (x) = max(c T i x) 2 i However, this is difficult to enforce, and an easier approach is to use SOS: Find V such that V (x) ɛ( x 6 i ) is SOS i V (x) T f i (x) is SOS for all i M. Peet Lecture 20: 5 / 18
Stability under Arbitrary Switching Common Non-Quadratic Lyapunov Functions Example: Consider the System (Boyd) ẋ(t) Co{A 1, A 2 } [ ] 100 0 A 1 =, A 0 1 2 = This system is stable, but not Quadratically Stable. [ 8 ] 9 120 18 Stability can be proven using the Lyapunov function V (x) := max{x T P 1 z, x T P 2 x} [ ] [ ] 14 1 0 0 P 1 =, P 1 1 2 = 0 1 M. Peet Lecture 20: 6 / 18
Stability under State-Dependent Switching Multiple Lyapunov Functions with Continuity Quadratic Stability is mostly useless for Analysis of state-dependent switching. Since it establishes stability under arbitrary switching Consider a simple model: { ẋ(t) = f i (x(t)), x(t) D i. where the D i are disjoint except at the Guard sets G (i,j) for (i, j) E. Recall E := {e i } is the set of possible transitions from domain i to j. Theorem 3. Suppose there exist Lyapunov functions V i (x) such that V i (x) ɛ x 2 for all i and V i (x) T f i (x) 0 for all x D i, i = 1,, k and V i (x) = V j (x) for all x G (i,j) for all i, j : (i, j) E { Basically V (x) = V i (x), x D i is a common, continuous, non-quadratic Lyapunov function. M. Peet Lecture 20: 7 / 18
Multiple Lyapunov Functions with Continuity Consider the System: { A 1 x(t), if x 1 0 ẋ(t) = A 2 x(t), otherwise. [ ] [ ] γ 1 γ 2 A 1 =, A 2 γ 2 = 1 γ where γ < 0 implies each subsystem is stable There is no global common quadratic Lyapunov function. However, if we define P 1 = [ ] 2 0 0 1 P 2 = [ 1 ] 2 0 0 1 Then V 1 (x) = x T P 1 x = x T P 2 x = V 2 (x) when x 1 = 0. Furthermore, A T 1 P 1 + P 1 A 1 < 0 and A T 2 P 2 + P 2 A 2 < 0. M. Peet Lecture 20: 8 / 18
Multiple Lyapunov Functions with Continuity The S-Procedure... Again Our goal is to find V 1 (x) = x T P 1 x and V 2 (x) = x T P 2 x with P 1, P 2 > 0 such that x T P 1 x x T P 2 x = 0 x {x : x 1 = 0 x T (A T 1 P 1 + P 1 A 1 )x 0 x {x : x T Sx 0} x T (A T 2 P 2 + P 2 A 2 )x 0 x {x : x T Sx 0} Of course, we could use the P-Satz... But lets put away the big guns for now... Lets consider the constraints separately. Instead, recall the S-procedure: S-Procedure: x T P x 0 for all x such that x T Sx 0 if there exists a τ > 0 such that P τs 0. So... x T (A T 1 P 1 + P 1 A 1 )x 0 x {x : x T Sx 0} if there exists τ > 0 such that A T 1 P 1 + P 1 A 1 + τs < 0 M. Peet Lecture 20: 9 / 18
Multiple Lyapunov Functions with Continuity The S-Procedure... Nullstellensatz Form! Now, lets examine the constraint x T P 1 x x T P 2 x = 0 x {x : c T x = 0} If P 1 P 2 + tc T + ct T = 0 Then x T (P 1 P 2 )x = x T (P 1 P 2 + tc T + ct T )x = 0 when c T x = 0 Therefore, if we can divide the guard set into lines, then we can enforce continuity Consider: x : x 1 x 2 = 0 This can be represented as the union of x 1 = 0 and x 2 = 0. M. Peet Lecture 20: 10 / 18
Stabilization by State-Dependent Switching Theorem 4. Suppose there exists some α [0, 1] such that αa 1 + (1 α)a 2 is Hurwitz then there exists a state-dependent switching law which quadratically stabilizes the systems. Note: This is also Necessary for Quadratic Stabilization. Switching Law: To test the condition, we find some P > 0 such that α(a T 1 P + P A 1 ) + (1 α)(a T 2 P + P A 2 ) < 0 This is bilinear in α and P but can be done by gridding α. Since α > and (1 α) > 0 this implies that for any x, either x T (A T 1 P + P A 1 )x < 0 or x T (A T 1 P + P A 1 )x < 0 Then choose the switching law { A 1 x(t) x T (A T 1 P + P A 1 )x < 0 ẋ(t) = A 2 x(t) otherwise M. Peet Lecture 20: 11 / 18
Stabilization by State-Dependent Switching Multiple Subsystems Can be extended to multiple subsystems: Find λ i such that i λ i = 1, λ i 0 and α i A i is Hurwitz i No subsystem need be stable. M. Peet Lecture 20: 12 / 18
Arbitrary Switching with Dwell-Time Restrictions Multiple Lyapunov Functions without Continuity We can extend the concept of multiple Lyapunov functions to relax continuity. Associate one Lyapunov function to each mode. Require that each function is decreasing at sequential points of activation. Theorem 5. Let each mode ẋ = f q (x) be globally asymptotically stable with Lyapunov functions V q. The switched system is stable if for every execution (I, T, p, C) V q (x(τ j )) V q (x(τ k )) < 0 for every i, j I such that p i = p j = q and p k q for any i < k < j. M. Peet Lecture 20: 13 / 18
Arbitrary Switching with Dwell-Time Restrictions This only works if we can: Bound the decrease during interval T i = [τ i, τ i+1 ] Bound the increase during intervals T i+1,, T j 1 M. Peet Lecture 20: 14 / 18
Controlling Uncontrollable Systems by Switching The Inverted Pendulum The inverted pendulum cannot be stabilized by continuous feedback. The Domain (circle) is not a contractible set. Requires a continuous function with H(0, θ) = θ and H(1, θ) = 0. A Smooth path from any point to the origin. Model: ẍ = u J θ = mgl sin θ ml cos θu Instead, we define 2 control laws 1. Energy maximization when θ is large (bottom) 2. Linearized Control when θ is small (top) M. Peet Lecture 20: 15 / 18
The Inverted Pendulum Energy Maximization To get to the top, pendulum needs 2mgl of Energy. Ignoring the cart, the Energy of the System is E = 1 2 J θ 2 + mgl (1 + cos θ) Taking the derivative Ė = ml θ cos θu The input which maximizes this energy gain is u(t) = sat(u)sign( θ cos θ) where sat(u) is the maximum acceleration. Discontinuous, due to sign function. Ė = 0 if cos θ = 0 May require multiple swings. M. Peet Lecture 20: 16 / 18
The Inverted Pendulum Multiple Swings and Multiple Pendula The Controller is: u(t) = sat(u)sign( θ cos θ) If Energy 2mgl is not achieved prior to θ = π 2, we need multiple swings. Controller reverses when θ = 0. Don t Forget Linear Control at the top! Multi-Swing Geometry: y Link: Double Inverted Pendulum Angle n=2.1 1 Link: Triple Inverted Pendulum 3 0 2 4 6 0 Normalized Energy C ϕ+θ 0 θ 0 ϕ x 1 2 0 2 4 6 Control Signal 2 A 1 B 0 0 2 4 6 M. Peet Lecture 20: 17 / 18
Conclusion! Stuff we didn t get to: Systems with Delay. Control of PDE Systems. The rest of Nonlinear Control, Hybrid Systems, etc. For more details on these and other topics, consult the recommended texts and references therein. I hope you have enjoyed this class. Please support our ability to teach this and similar courses by completing your student evaluations online. These evaluations are taken seriously by the University and Administration. Thank You For Your Support! M. Peet Lecture 20: 18 / 18