Fal!2001 16.3122-10 Simple example from before: A scalar system with a=a+ bu with cost(Rxx>0 and Ru>0) (Rxxi (t)+ Ruuu(t)) e Then the steady-state P solves 2aP+ Rxx-P2b2/Ru=0 which gives that P=a+va2+b Rxx/Ruu Then u(t)=-Kx(t)where K=RubP a+√a2+b2Rx/Ru The closed-loop dynamics are (a-bk)a=(a-i(a+va2+b2 Rxx/Ru)ac √a2+b2Bx/Rux=Ax(t) · Note that as Rxx/Ru→∞,A≈-| bIVRx/Ru And as Rxy/Bu→0,K≈(a+)/b Ifa<0(open- loop stable,K≈0 and Acl=a-b≈a Ifa>0( OL unstable,K≈2a/ b and acl=a-bK≈-aFall 2001 16.31 22—10 • Simple example from before: A scalar system with x˙ = ax + bu with cost (Rxx > 0 and Ruu > 0) J = Z ∞ 0 (Rxxx2 (t) + Ruuu2 (t)) dt • Then the steady-state P solves 2aP + Rxx − P2 b2 /Ruu = 0 which gives that P = a+ √a2+b2Rxx/Ruu > 0 R−1 uu b2 • Then u(t) = −Kx(t) where uu bP = a + pa2 + b2Rxx/Ruu K = R−1 b • The closed-loop dynamics are x˙ = (a − bK)x = µ a − b b (a + pa2 + b2Rxx/Ruu) ¶ x = − pa2 + b2Rxx/Ruu x = Aclx(t) • Note that as Rxx/Ruu → ∞, Acl ≈ −|b| pRxx/Ruu • And as Rxx/Ruu → 0, K ≈ (a + |a|)/b — If a < 0 (open-loop stable), K ≈ 0 and Acl = a − bK ≈ a — If a > 0 (OL unstable), K ≈ 2a/b and Acl = a − bK ≈ −a