1. 4.2 State Feedback for Multivariable Systems
Basic idea：Suppose that a multivariable system (A, B) is controllable, then the basic idea of its pole-placement is to transform the problem into the pole-placement of single variable systems. The steps are as follows.
Choose any
bIm(B) , b0
Then, find a state feedback gain matrix K1 such that the single variable system
(A+BK1, b)
is controllable.

2. Assign the poles of the single variable system (A+BK1, b) to the expected place.
3. Note that
with
Therefore,

3. Theorem 4-4. Suppose that (A, B) is controllable
Theorem 4-4. Suppose that (A, B) is controllable. Then for any nonzero vector b in the value field of B, there exists a state feedback gain matrix K1 Rpn such that
(A+BK1 , b)
is controllable.
The proof includes the following steps:

4. Use the following conclusion.
Lemma. Suppose that (A, B) is controllable. Then we can choose vectors u1, u2, …, un1 such that the n vectors x1, x2, …, xn defined by the following equations are linearly
where b is any nonzero vector in the value field of B , xiRn1 and uiRp1

5. 2). Define K1Rpn:
That is
Note that

6. 3). Prove that (A+BK1, b) is controllable @p5
It is easy to find that

7. Since are linearly independent, we have
is controllable.
Q.E.D

8. Theorem 4-5. Suppose that the system (4-1) is controllable
Theorem 4-5. Suppose that the system (4-1) is controllable. Then there exists a state feedback gain matrix K such that the n eigenvalues of A+BK can be arbitrarily assigned.
Proof . First of all, we choose a nonzero vector L such that
b=BL
From Theorem 4-4, there exists K1 such that
(A+BK1, b)
is controllable. From the theorem about the pole placement of single variable systems, we know that there exists a n dimension vector k such that the poles of
A+BK1+bk

9. can be arbitrarily assigned. Note that
A+BK1+bk
=A+BK1+BLk
= A+B(K1+Lk)
Hence, one only needs to choose
K= K1+Lk
This completes the proof of Theorem 4-5.
Q.E.D

10. Proof of the Lemma. The proof is performed by construction
Proof of the Lemma. The proof is performed by construction. Without loss of generality, suppose that the columns of B=[b1, b2,…., bp] are linearly independent.
Construct n linearly independent columns in the following way:
(Until can be linearly expressed by )
(Until can be linearly expressed by )

11. Continue the above procedure
Continue the above procedure. Since (A, B ) is controllable, there exists
which forms a set of basis of the n dimensional space, where ni = n.

12. 2）Construct n linearly independent vectors xi as follows:

13. It is easy to verify that are linearly independent. In fact,

14. 3). Determine ui. Let
Q.E.D

15. Example 1. Consider the system
Construct K1 such that (A+BK1 , b=BL) is controllable.
Solution. Choose
x1 = BL =b=[ ]T0, L=[1 1]T。
and consider
x1=b, xk+1= A xk+ B uk (k=1, 2, … , n1)，

16. Noticing that
Ax1=[1 1 1]T
is linearly independent of x1, choose
x2= Ax1,
and
u1=[0 0]T
Because Ax2, x1 and x2 are linearly dependent, we can not choose u2 as [0 0]T. Let
u2=[1 1]T，
Then, we have x3= Ax2+Bu2= [ ] T.

17. From K1=[u1 u2 0][x1 x2 x3]1, we have
It can be readily checked that (A+BK1 b) is controllable.

18. Example 2. Consider the following system:
Determine K such that the closed-loop system (A+BK) has eigenvalues 2, 2,  1j.
Solution .
Choose L =[1 0]T, x1=b1=[ ]T;
u1=[-1 0]T, x2=[ ]T;
u2=[0 0]T, x3=[ ]T;
u3=[0 1]T, x4=[ ]T;

19. Then, we have
It is clear that (A+BK1, b1) is controllable. Let k=[k1 k2 k3 k4], we have

20. whose characteristic polynomial is
s4 (1+k3)s3+(k3k2)s2+(k2k1)s+k1 k4,
The expected characteristic polynomial is
s4+6s3+14s2+16s+8,
Comparing the coefficients of the two polynomials, we can obtain
k1 =37, k2=21, k3=7, k4=45
Hence, the feedback gain matrix can be selected as

21. In the above method, once L and ui are selected, then k is determined
In the above method, once L and ui are selected, then k is determined. But L and ui are nonunique, which implies that there exists a family of k to get the same poles. This is one of the main differences between multi-input systems and single-input systems.

22. The other issue in pole-placement of multi-input systems involves “nonlinear equations”. Denoting the elements of K by kij , then the closed-loop polynomial can be rewritten as
where fi (K)’s denote non-linear functions with kij as variables. Let the expected polynomial be

23. Comparing the coefficients, we have
Example. Consider the multivariable system
It is clear that the coefficients of det(sIABK) are nonlinear.

24. det(sIABK) is linear for single-input systems but usually nonlinear for multi-input systems. Theorem 4-4 indicates that when the system is controllable, we can get a set of linear equations of (sIABK) by abandoning some free parameters of K.

25. Example 3. For the system in Example 2, compute K such that fi(K)=i and the poles of the closed-loop system (A+BK) are 2, 2, 1j.
Solution. Since

26. Scheme 1. Let k4=k5=k6=k7=0, 1+k8= 2, From
we have
That is,

27. Scheme 2. Let
k1=k2= 0, k3= 1 , k4=1
From
we have
k5=8, k6=16 , k7=14, k8=7, i.e.

28. Since the feedback gain matrix is nonunique, the transfer function matrix of the closed-loop system is not unique either. Hence, different response may be obtained. We should choose K such that the closed-loop system yields a better response.
The same as the single-input case, the condition that the system is controllable in Theorem 4-4 is sufficient and necessary for arbitrary assignment of the poles. But for a given set of poles, it is only a sufficient condition.

29. 5. Stabilization
1). Stabilization of state feedback systems. Consider the following LTI system
If we can find a state feedback
such that all real parts of the poles of the closed-loop system
are negative, then the system is said to be stabilizable with state feedback.

30. Theorem 4-6. The system (A, B) is stabilizable by state feedback if and only if the real parts of all uncontrollable modes are negative.
In fact, because the state feedback does not change the uncontrollable modes：
the system is stabilizable

31. Uncontrollable systems
2). System classification by stabilizability
The systems can be classified into:
The poles can be assigned arbitrarily
Controllable systems
Stabilizable
Stabilizable: If the real parts of all uncontrollable modes are negative.
Systems
Uncontrollable systems
Unstabilizable: If the real parts of some uncontrollable modes are negative.

32. Conclusion
The condition that pole assignment is possible: The states can be measured.
If the states of the systems can not be completely measured, which is the common feature of most control systems, then we can not use the state feedback method to assign the poles directly. But the result of pole placement is still important in theory and engineering, and is a classical achievement of linear system theory.