By Vlad Ionescu
This complete textual content is exclusive in offering a generalization of the algebraic Riccati concept for the indefinite signal and singular instances from the viewpoint of the Popov functionality strategy. the idea embodies a scientific presentation from the time area, frequency area and state-space standpoints. the entire improvement is orientated in the direction of functions in powerful structures theory.Extension of the distinguished Popov positivity thought to the sport theoretic caseCharacterizations of the ideas to the Riccati equation when it comes to the input-output operator, move functionality matrix and transmission matrix pencil, all linked to the underlying Hamiltonian systemComprehensive therapy of either non-stop and discrete time casesA signature situation established process which unifies the ideas to numerous mathematical difficulties from powerful keep an eye on thought, together with the maxmin, Nehari, H2, H??? regulate and powerful stabilization problemsGraduate scholars up to the mark engineering and people engaged on the math of structures and keep an eye on conception will locate this e-book crucial analyzing. For researchers, this is often an vital paintings, while looking ever extra strong keep an eye on legislation which take uncertainties under consideration.
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Extra info for Generalized Riccati Theory and Robust Control. A Popov Function Approach
Sample text
If qn (x) = x n , it is exactly Nuttall's compact formula. If qn(x)=Pn(x), then we obtain a result which is Chapter 1 36 closely related to the matrix interpretation of Pade approximants. This result will be given later on. 4. t 2k ~ . f(t)-[k -l/k]f(t) = Pk(t) i~ di+kt' where ~ = C(XiPk(X)) = bOci +b1Ci+l + ... +bkCi+k' Proof. Obvious since di = 0 for i = 0, ... 1. 5 is still valid and the next formula too. 3 also holds. • i=O i=O This property will be proved in Chapter 3. New properties are also satisfied by Pade approximants.
11. 1. Proof. Obvious since the Ok'S satisfy the same recurrence relation. 12. 0. Proof. Let us write the recurrence relation for P k + 1 and multiply it by Ok. Then write the recurrence relation for Ok+l and multiply it by Pk • If we subtract these two relations we get Pk(X)Ok+l(X) - OdX)Pk+1(X) = Ck+1[Pk-1(X)Qk (x)- Ok-l(X)Pk(x)]. Thus for k >0 Pk(X)Ok+l(X)- Ok(X)Pk+1(X) = Ck+1Ck··· C 2 [PO(X)01(X)- OO(X)Pl(X)] = Ak+1hk A1h o PO(X)Ol(X). Using the definitions of Al ho, Po and 0 1 we get the result.
12. 4. k Ok(X)P~+1(x)- Ok+1(x)PHx) = Ak+1hk L hi 1Pi (X)Oi(X), i=O O~(X)Pk+l(X)- 0~+1(X)Pk(X) = -A k+1hk k L hi1Pi (X)Oi(X), i=O Proof. 13 and dividing by x - t, we get Ok(t) Pk+1(X) - Pk+1(t) Ok+1(t) Pk(x)- Pk(t) x-t x-t k = Ak+1hk L hi 1Pi (X)Oi(t). i=O Let x - t and the first part of the corollary follows. 12 we get the second result. • Applying c to the first of these relations we get C(OkP~+1) = C(Ok+1 P O· Let k Hdx, t) = L hi1Pi(X)Oi(t), i=O then Let Mk(t) = (Ok«t» Pk t Nk(t) = ( Ok-l«t»), Akt+ Bk -Ck Pk - 1 t 1) 0' for for k = 0,1, ...