Fall 2001 6.31142 Example #1: Consider 10|x Already designed K=[14 57 Then the closed-loop system is =(A- BK)x+Br y=C. Which gives the transfer function Rls C(SI-(A-Bk)B =[101/+13561-1 1s-2 0s2+11s+30 Assume that r(t) is a step, then by the FVT R(S)|=030×1! So our step response is quite poorFall 2001 16.31 14–2 • Example #1: Consider: x˙ = 1 1 1 2 x + 1 0 u y =  1 0  x – Already designed K =  14 57  – Then the closed-loop system is x˙ = (A − BK)x + Br y = Cx – Which gives the transfer function Y (s) R(s) = C (sI − (A − BK))−1 B =  1 0  s + 13 56 −1 s − 2 −1 1 0 = s − 2 s2 + 11s + 30 • Assume that r(t) is a step, then by the FVT Y (s) R(s)



s=0 = −2 30 = 1 !! – So our step response is quite poor.