Fall 2001 6.3113-3 To put the poles at s=-5,-6, compare the desired characteristic equation (s+5)(s+6)=s2+11s+30=0 ith the closed-loop one +(k1-3)x+(1-2k1+k2)=0 to conclude that k1-3=11 1-2k1+k2=30 k2=57 so that K= 14 57, which is called Pole Placement Of course, it is not always this easy, as the issue of controllability must be addressed Example #2: Consider this system C+ 02 0 with the same control approach Ad=A- BK 11 02 [k1k2] 1-k11-k2 that det(s-Aa)=(s-1+k)(s-2)=0 So the feedback control can modify the pole at s= 1, but it cannot move the pole at s=2 The system cannot be stabilized with full-state feed back controlFall 2001 16.31 13–3 • To put the poles at s = −5, −6, compare the desired characteristic equation (s + 5)(s + 6) = s2 + 11s + 30 = 0 with the closed-loop one s2 + (k1 − 3)x + (1 − 2k1 + k2)=0 to conclude that k1 − 3 = 11 1 − 2k1 + k2 = 30  k1 = 14 k2 = 57 so that K = 14 57  , which is called Pole Placement. • Of course, it is not always this easy, as the issue of controllability must be addressed. • Example #2: Consider this system: x˙ =
1 1 0 2 x +
1 0 u with the same control approach Acl = A − BK =
1 1 0 2 −
1 0 k1 k2  =
1 − k1 1 − k2 0 2 so that det(sI − Acl)=(s − 1 + k1)(s − 2) = 0 So the feedback control can modify the pole at s = 1, but it cannot move the pole at s = 2. • The system cannot be stabilized with full-state feed￾back control.