Fall 2001 16.3116-9 Another approach Note that the poles of(A-LC) and(A-LC)are identical Also we have that(A-LC)=A So designing l for this transposed system looks like a standard regulator problem(A- BK)where B→Cr F→L So we can use Ke= acker(A',C,P), L=K e Note that the estimator equivalent of Ackermann's formula is that L=重(s)M 0:01Fall 2001 16.31 16—9 • Another approach: — Note that the poles of (A − LC) and (A − LC) T are identical. — Also we have that (A − LC) T = AT − CTLT — So designing LT for this transposed system looks like a standard regulator problem (A − BK) where A ⇒ AT B ⇒ CT K ⇒ LT So we can use Ke = acker(AT , CT , P) , L ≡ KT e • Note that the estimator equivalent of Ackermann’s formula is that L = Φe(s)M−1 o      0 . . . 0 1     