Fall 2001 16.3115-6 Consider a siso system with a minimal model a= Ax+ Bu Where a(s)=det(sl-A)and C(sI-A-Ir-_b(s) Then with u(t)=-Klgra(t), closed-loop dynamics are det(sl-A+ bkigr) where the pi=f the left-hand-plane roots of A(s), with A(s=asa(-s)+rb(s)b(s Use this to find the optimal pole locations, and then use those to find the feed back gains required using acker The pole locations can be found using standard root-locus tools △(s)=a(sa(-s)+r-1b(s5b(-s)=0 1+r-G(s)G(-s)=0 The plot is symmetric about the real and imaginary axes Symmetric Root Locus 2n poles are plotted as a function of r The poles we pick are always the n in the LHP I Several leaps made here for now. We will come back to this LQR problem laterFall 2001 16.31 15—6 • Consider a SISO system with a minimal model x˙ = Ax + Bu , y = Cx where a(s) = det(sI − A) and C(sI − A) −1 B ≡ b(s) a(s) • Then1 with u(t) = −Klqrx(t), closed-loop dynamics are: det(sI − A + BKlqr) = Y n i=1 (s − pi) where the pi ={ the left-hand-plane roots of ∆(s)}, with ∆(s) = a(s)a(−s) + r−1 b(s)b(−s) • Use this to find the optimal pole locations, and then use those to find the feedback gains required using acker. • The pole locations can be found using standard root-locus tools. ∆(s) = a(s)a(−s) + r−1 b(s)b(−s)=0 ⇒ 1 + r−1 G(s)G(−s)=0 — The plot is symmetric about the real and imaginary axes. ⇒ Symmetric Root Locus — 2n poles are plotted as a function of r — The poles we pick are always the n in the LHP. 1Several leaps made here for now. We will come back to this LQR problem later