Continuous-time Markov jump linear systems by Oswaldo Luiz do Valle Costa

By Oswaldo Luiz do Valle Costa

1.Introduction.- 2.A Few instruments and Notations.- 3.Mean sq. Stability.- 4.Quadratic optimum regulate with entire Observations.- 5.H2 optimum keep an eye on With entire Observations.- 6.Quadratic and H2 optimum keep an eye on with Partial Observations.- 7.Best Linear clear out with Unknown (x(t), theta(t)).- 8.H_$infty$ Control.- 9.Design Techniques.- 10.Some Numerical Examples.- A.Coupled Differential and Algebraic Riccati Equations.- B.The Adjoint Operator and a few Auxiliary Results.- References.- Notation and Conventions.- Index

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Using similar reasoning, we can show that (ii) is equivalent to (iv). Suppose now that the operator eLt is positive for every t ∈ R+ . Clearly, (iii) imn+ plies (iii ) since Hn+ ⊂ Hn+ C . Suppose now that (iii ) holds. For any H ∈ HC , define n+ I(H) ∈ H as follows: I(H) := H1 In , . . , HN In . Clearly, H ≤ I(H). From the fact that eLt is a positive operator we get that eLt (H) ≤ eLt (I(H)), and thus eLt (H) 1 ≤ eLt (I(H)) 1 . From (iii ) we have that eLt (H) 1 ≤ eLt I(H) 1 →0 as t → ∞. Therefore, we have shown that (i), (ii), (iii), and (iii ) are all equivalent.

13 that Re{λ(F )} < 0. 21 it follows that q(t)→0 ˆ as t → ∞, and since q(t) = i∈S qi (t), we have that q(t)→0 as t → ∞. 40) for some a > 0 and b > 0. This shows that (ii) implies (iv) and (v). 13 that (v) implies (ii) and (iv). The next examples, borrowed from [223], illustrate the cases in which (1) each mode is unstable, but the overall system is stable, and (2) each mode is stable, but the overall system is unstable. 1). 16 (Each mode is unstable, but the overall system is stable) Consider an MJLS with A1 = 1 2 0 −1 , −2 A2 = −2 0 −1 1 2 , Π= −β β β , −β so that each mode is unstable.

25), and consider the homogeneous system y(t) ˙ = Ay(t), t ∈ R+ , with initial condition y(0) = ϕ(Q), ˆ n Q ∈ HC . Then, y(t) = eAt y(0) = ϕˆ eLt (Q) . 28) The result also holds replacing A and L by A∗ and T , respectively. Proof We begin by noticing that the solution of the above differential equation is 2 given by y(t) = eAt y(0). Consider any y ∈ CN n and take Y = ϕˆ −1 (y) ∈ HnC . 8 it follows that ϕ(L ˆ k (Y)) = Ak ϕ(Y) ˆ = Ak y. 19) we have that e At y = ∞ k=0 tk k A y= k! r = lim ϕˆ r→∞ k=0 ∞ k=0 tk k ˆ = lim A ϕ(Y) r→∞ k!

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