Section 10–2 / Pole Placement 731
Multiplying the preceding equations in order by a 3 ,a 2 ,a 1 , and a 0 (wherea 0 =1),
respectively, and adding the results, we obtain
(10–16)
Referring to Equation (10–15), we have
Also, we have
Substituting the last two equations into Equation (10–16), we have
Since we obtain
(10–17)
Since the system is completely state controllable, the inverse of the controllability matrix
exists. Premultiplying both sides of Equation (10–17) by the inverse of the controllability
matrix, we obtain
Premultiplying both sides of this last equation by [0 0 1],we obtain
which can be rewritten as
This last equation gives the required state feedback gain matrix K.
For an arbitrary positive integer n,we have
K=[0 0 p 0 1]CB AB p An-^1 BD-^1 f(A) (10–18)
K=[0 0 1]CB AB A^2 BD-^1 f(A)
[0 0 1]CBABA^2 BD
- 1
f(A)=[0 0 1]C
a 2 K+a 1 KA
+KA
2
a 1 K+KA
K
S =K
CBABA^2 BD-^1 f(A)= C
a 2 K+a 1 KA
+KA
2
a 1 K+KA
K
S
CB AB A^2 BD
=CBABA^2 BDC
a 2 K+a 1 KA
+KA
2
a 1 K+KA
K
S