ME Fall 2001, Fall 2002, Spring I/O Stability. Preliminaries: Vector and function norms

Transcription

1 I/O Stability Preliminaries: Vector and function norms 1. Sup norms are used for vectors for simplicity: x = max i x i. Other norms are also okay 2. Induced matrix norms: let A R n n, (i stands for induced) A i, := sup x 0 Ax x = max j i A ji 3. vector function norms: for x : [0, ) R n, the L norm is used, i.e. x( ) = sup t x(t) = sup t x i (t). M.E. University of Minnesota 84

2 Definition A system (not necessarily linear) F : u( ) y( ) is I/O stable iff there exists k <, s.t. for all bounded u( ), y( ) k u( ) For a linear possibly time varying system, (ignoring initial conditions) the output is given by: y(t) = H(t, τ)u(τ)dτ (8) note that this is a linear map between the input function space and the output function space. Theorem The linear system (8) is I/O stable iff sup t { } H(t, τ) i dτ := k <. (9) Because norms are equivalent on finite dimensional spaces, the norms on vectors (x(t) and y(t)) do not have to be infinity norms. M.E. University of Minnesota 85

3 Specializations For system given by: ẋ(t) = A(t)x(t) + B(t)u(t), y(t) = C(t)x(t) + D(t)u(t) with D(t) bounded, I/O stability is equivalent to: sup t C(t)Φ(t,τ)B(τ)dτ =: k < For LTI systems, y(t) = stability is given by: 0 H(τ) i dτ =: k <. H(t τ)u(τ)dτ, I/O M.E. University of Minnesota 86

4 For LTI system ẋ = Ax + Bu, y = Cx + Du, I/O is equivalent to either 1): 0 Ce At B dt < or 2) the poles of H(s) := C(sI A) 1 B has negative real parts. For LTI systems, since poles form subset of {λ i (A)}, it therefore suffices to check the eigenvalues of A. Why? If poles[h(s)] = {p i } = {λ i (A)}, it is easy to see. For, Ce At B has terms that are of the form t k e λ it, which is absolutely integrable iff p i has negative real parts. If poles[h(s)] is a strict subset of {λ i (A)}, then Ce At B has terms of the form t k e p it. λ k {poles} do not appear (see homework problem). M.E. University of Minnesota 87

5 Proof (Sufficiency) y(t) = H(t,τ)dτ H(t, τ) i u(τ) dτ H(t, τ) i dτ u( ) k u( ). (Necessity - sketch) 1. The function t is equivalent to: H(t,τ)dτ being bounded i,j, t H ij (t,τ) dτ (try to prove it using the property of infinity induced norm). 2. Proof by contraposition. Suppose F : u( ) y( ) is I/O stable but (9) is not satisfied. Then, there M.E. University of Minnesota 88

6 exists i, j such that { lim g ij (t) = t } H ij (t,τ) dτ =. W.L.O.G., let i = j = 1. t 1,t 2,... so that g 11 (t k ) > k. Then, there exists 3. Define a sequence of inputs u l ( ), l = 1,2,3, by u l (t) = (u l 1(t), 0, ) T u l 1(t) = { sign(h 11 (t,τ) t (,t l ] 0 otherwise 4. For each l, u l ( ) = 1, but y l ( ) y l (t l ) = l h 11 (t,τ) dτ > l Thus, as we take l, y l ( ) > l. Therefore, there can be no k < so that y( ) < k u( ). Q.E.D. M.E. University of Minnesota 89

7 Lyapunov (internal) stability ẋ = f(x,t), f(x e,t) = 0 t. (10) Definition: The equilibrium, x = x e is stable in the sense of Lyapunov (i.s.l.) at t 0 iff for any ǫ > 0, there exists δ(ǫ,t 0 ) s.t. if x(t 0 ) x e δ(ǫ,t 0 ), x(t) x e ǫ for all t t 0. It is said to be stable if it is stable for each t 0 0. Remarks: the state can be arbitrarily close to the equilibrium if initial state is sufficient close. Definition: The equilibrium, x = x e is asymptotically stable in the sense of Lyapunov (i.s.l.) at t 0 iff 1. x e is stable at t 0 2. as t, x(t) x e. M.E. University of Minnesota 90

8 For linear systems: x(t) = Φ(t,t 0 )x(t 0 ), 1. x = 0 is stable at arbitrary t 0 iff sup t t0 Φ(t, t 0 ) i = k < 2. x = 0 is asymptotically stable iff Φ(t,t 0 ) 0 (note Φ(t, t 0 ) is finite for finite t. Why?) Remark: x = 0 is stable at t 0 iff it is stable at t 0 = 0, i.e. sup t 0 Φ(t,0) := k 0. x = 0 is asymptotically stable at t 0 iff it is asymptotically stable at t 0 = 0, i.e. Φ(t,0) 0. Hence, if an equilibrium is stable at one initial time, it is stable at any initial time. Proof of remark: x(t) = Φ(t, t 0 )x 0 = Φ(t, 0)Φ(0,t 0 )x 0 Φ(t, 0) i Φ(0, t 0 ) i x 0 Notice that for a given t 0, Φ(0,t 0 ) i is bounded. Therefore, given ǫ > 0, choose δ(ǫ,t 0 ) = ǫ k 0 Φ(0,t 0 ) i. Q.E.D. M.E. University of Minnesota 91

9 Theorem For LTI system, x = 0 is stable if each λ i (A) has non-positive real part, and in the case of λ i (A) = 0, it is not repeated in a Jordan form. asymptotically stable if and only if each λ i (A) has negative real part. Proof This is because Φ(t, 0) = exp(at) = Texp(tΛ)T 1 where Λ is in a Jordan form with diagonal elements being the eigenvalues. Now exp(tλ) consists of terms of the form t k e λit (where k 0 for Jordan form). If Re(λ i ) < 0, then each of these terms 0 as t. If λ i = jω (zero real part) and k = 0, then the exp(tλ i ) is sinusoidal. If k 0, then t k e λit has terms of the form t k sin(ωt+φ) which grows unboundedly. M.E. University of Minnesota 92

10 Example: eigenvalue: The Jordan form with repeated zero A = ( ) has a transition matrix of exp(ta) = unbounded (thus x = 0 is unstable). However, the semi-simple A = ( ) ( ) 1 t 0 1 which is has a repeated zero eigenvalue but its transition matrix is I, which is bounded. Thus, x = 0 is stable. If ( ) 2 0 A = 0 1 ( ) exp( 2t) 0 the transition matrix is exp(ta) =. 0 exp( t) Thus x = 0 is asymptotically stable. M.E. University of Minnesota 93

11 Discrete time systems Equivalent results exist for discrete time systems except that the pole / eigenvalue locations must be considered w.r.t. to the unit circle (instead of the left half plane) in the complex plane. E.g. For x(k + 1) = Ax(k), x = 0 is x = 0 is stable if λ i (A) 1, and with no repeated eigenvalues on the unit circle. x = 0 is asymptotically stable iff it is exponentially stable iff λ i (A) < 1 M.E. University of Minnesota 94