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
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
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
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
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
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
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
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
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
Example: eigenvalue: The Jordan form with repeated zero A = ( ) 0 1 0 0 has a transition matrix of exp(ta) = unbounded (thus x = 0 is unstable). However, the semi-simple A = ( ) 0 0 0 0 ( ) 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
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