Fall 2001 163113-7 We then compare this with the desired characteristic equation de- veloped from the desired closed-loop pole locations 重(s)=s3+(a1)s2+(a2)s+(a3)=0 to get that +k1 k1 1 +k To get the specifics of the Ackermann formula, we then Take an arbitrary A, B and transform it to the control canonical form(az=T- r) Solve for the gains K using the formulas above for the state z (=Kz) Then switch back to gains needed for the state so that K- KT-I (u=Kz=Kr) Pole placement is a very powerful tool and we will be using it for most of this courseFall 2001 16.31 13–7 • We then compare this with the desired characteristic equation de￾veloped from the desired closed-loop pole locations: Φd(s) = s3 + (α1)s2 + (α2)s + (α3)=0 to get that a1 + k1 = α1 . . . an + kn = αn    k1 = α1 − a1 . . . kn = αn − an • To get the specifics of the Ackermann formula, we then: – Take an arbitrary A, B and transform it to the control canonical form (x ❀ z = T −1x) – Solve for the gains Kˆ using the formulas above for the state z (u = Kz ˆ ) – Then switch back to gains needed for the state x, so that K = KTˆ −1 (u = Kz ˆ = Kx) • Pole placement is a very powerful tool and we will be using it for most of this course