Fal!2001 16.3122-1 Linear Quadratic Regulator(LQR) We have seen the solutions to the lqr problem using the symmetric root locus which defines the location of the closed-loop poles Linear full-state feedback control Would like to demonstrate from first principles that this is the optimal form of the control Deterministic Linear Quadratic Regulator Plant: i(t)=A(tac(t)+ Bu(tu(t), a(to)=ro (t)=C2(t)x(t) Cost LQR 2(t) R2(t)z(t)+u(t)Ru(t)u(t) dt+a(tt)Ptr a(tf) Where P 20, Rz(t)>0 and Ruu(t)>0 Define Rxx=CRnC2≥0 A(t)is a continuous function of time Bu(t), C2(t, Rz(t), Ruu(t) are piecewise continuous functions of time. and all are bounded Problem Statement: Find the input u(t)Vt Eto, te] to mini- mize JLQRFall 2001 16.31 22—1 Linear Quadratic Regulator (LQR) • We have seen the solutions to the LQR problem using the symmetric root locus which defines the location of the closed-loop poles. — Linear full-state feedback control. — Would like to demonstrate from first principles that this is the optimal form of the control. • Deterministic Linear Quadratic Regulator Plant: x˙(t) = A(t)x(t) + Bu(t)u(t), x(t0) = x0 z(t) = Cz(t)x(t) Cost: Z tf £ JLQR = zT (t)Rzz(t)z(t) + uT (t)Ruu(t)u(t) ¤ dt + x(tf )Ptfx(tf ) t0 — Where Ptf ≥ 0, Rzz(t) > 0 and Ruu(t) > 0 — Define Rxx = Cz TRzzCz ≥ 0 — A(t) is a continuous function of time. — Bu(t), Cz(t), Rzz(t), Ruu(t) are piecewise continuous functions of time, and all are bounded. • Problem Statement: Find the input u(t) ∀t ∈ [t0, tf ] to mini￾mize JLQR