Fa2004 16.3339-6 Thus, by choosing ki and k2, we can put Ai(Ac) anywhere in the complex plane(assuming complex conjugate pairs of poles) To put the poles ats=-5,-6, compare the desired characteristic equation (s+5)(s+6)=s2+118+30=0 with the closed-loop one s2+(k1-3)x+(1-2k+k2)=0 to conclude that k1-3=11 k1=14 1-2k1+k2=30 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 11 0 with the same control approach 1-k11-k2 A=A-Bk so that det(si- ad)=(s-1+k1(s-2)=0 The feedback control can modify the pole at s but it cannot move the pole at s= 2 This system cannot be stabilized with full-state feedback control o What is the reason for this problem?� � � � � � � � � � � � � Fall 2004 16.333 9–6 – Thus, by choosing k1 and k2, we can put λi(Acl) anywhere in the complex plane (assuming complex conjugate pairs of poles). • To put the poles at s = −5, −6, compare the desired characteristic equation (s + 5)(s + 6) = s2 + 11s + 30 = 0 with the closed­loop one 2 s + (k1 − 3)x + (1 − 2k1 + k2) = 0 to conclude that k1 − 3 = 11 k1 = 14 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: 1 1 x˙ = 0 2 1 x + u 0 with the same control approach 1 1 1 � � 1 − k1 1 − k2 Acl = A − BK = k1 k2 = 0 2 − 0 0 2 so that det(sI − Acl) = (s − 1 + k1)(s − 2) = 0 The feedback control can modify the pole at s = 1, but it cannot move the pole at s = 2. • This system cannot be stabilized with full­state feedback control. • What is the reason for this problem?