1. Pole Placement and Decoupling by State Feedback
Chapter 4
Pole Placement and Decoupling by State Feedback

2. §4-1 State feedback and pole placement
Consider an open-loop system described by
and a set of expected poles of close-loop system which is real numbers or conjugated complex numbers. These expected poles have close relation with the performance of control systems.
The pole placement by state feedback studies the problem that, for a given open-loop system, how to construct a state feedback such that the closed-loop system has expected poles.

3. 1. The effect of state feedback on system controllability and observability
1). State feedback does not affect the controllability
Consider the following dynamic equations
The linear state feedback control law is
where v is the reference input and K is a p×n matrix called the state feedback gain. Expressing Equation (4-1) and (4-2) with the block diagram in figure 4-1, we obtain a closed-loop system.

4. x y B C A K v u
Figure 4-1. closed-loop system
By introducing the state feedback, the state space equation becomes:
where A+BK is the system matrix of the closed-loop system.

5. Before:
After:
Theorem 4-1. Equation (4-3) is controllable Equation (4-1) is controllable.
Proof. For any K, we have
Hence,
Therefore, the state feedback does not affect the controllability.
The state feedback does not change uncontrollable modes, i.e. the uncontrollable modes of the open-loop system keep unchanged in the closed-loop system.

6. 2). State feedback does not change the uncontrollability subspace
The controllability matrix before introducing the state feedback is defined as
The controllability matrix after introducing the state feedback is
We have the following conclusion.
Theorem 4-2. The state feedback does not change the controllable subspace, i.e.,

7. Proof. For any , i.e. ,
Similarly, it can be proved that
Hence, we have
That is,
Q.E.D

8. 3). The state feedback may change the observability.
Example. Consider the following system:
Open-loop:
Closed-loop:

9. C.L.S — closed-loop system.
The following table gives the relationship among the observability, the matrix C and state gain vector K.
Arbitrary
[1 1]
[1 2]
[0 1]
Observable 1
C.L.S
O.S. c2 c1 K
Unobservable
O. S — original system;
C.L.S — closed-loop system.

10. 2. Pole placement of single-variable systems
Open-loop:
Introduce the state feedback law
Closed-loop:
Theorem 4-3. If (4-5) is controllable, then the eigenvalues of the system matrix A+bk of the closed-loop system (4-7) can be arbitrarily assigned, provided that complex conjugate eigenvalues are coupled.

11. Proof. Sufficiency.
Since the system (4-5) is controllable, there exists a nonsingular matrix P which transforms it into a canonical form, given by
@p14
where

12. The state feedback can be written as
Note that
Hence, the eigenvalues of are the same as those of
2）Suppose that the expected polynomial is
where are expected poles.

13. Choose
and consider the
Appendix 31
The characteristic polynomial of (4-11) is
Therefore, (4-11) has the expected poles.

14. Recalling that and have the same eigenvalues, we can assign A+bk any given n eigenvalues by (4-10) and (4-11).
Necessity. Suppose that the eigenvalues of (4-5) can be arbitrarily assigned, we now prove that (4-5) is controllable. The proof is performed by seeking a contradiction. If (A,b) is uncontrollable, its controllability decomposition is given by

15. From the above equation we can see that the eigenvalues of A4 do not rely on , i.e., some eigenvalues of A+bk are independent of k, which contradicts the hypothesis. 　　　　　 　 Q.E.D
k can be computed by the sufficiency proof:
1) Compute the characteristic polynomial of A
From the n given eigenvalues , compute the expected polynomial

16. 3) Compute
4) Compute the transformation matrix P;
5) Compute the feedback gain matrix ,
where

17. The above steps include the canonical transformation
The above steps include the canonical transformation. But we can directly compute k without this step:
Express k as [k1,k2,….,kn];
Compute det(sIAbk), with n coefficients to be fixed:
Compare this polynomial with the expected polynomial, and let the coefficients of s with the same order be equivalent:

18. Now, we get a linear equation with n variables
Now, we get a linear equation with n variables. Under the condition that the system is controllable, k is determined uniquely by the equation. Sometimes, this direct method is more convenient.
From the problem 4-1, we have
where U is the controllability matrix, and the constant matrix G is given by (1-46).

19. Let
Then we have or

20. Single input systems
If (A, b) is controllable, then there exist n linear equations with ki as the variables and the solution is unique. For “arbitrary pole-placement” the above equation has a solution if and only if the system is controllable.
That the system is controllable is a sufficient and necessary condition of that the poles can be arbitrarily placed. But for a given set of eigenvalues, it is only a sufficient condition.
Corollary: A given set of poles can be assigned by state feedback if and only if it includes all the uncontrollable modes.

21. Suppose that the system is uncontrollable with the controllable canonical decomposition
Introducing the state feedback u=kx+v, we have
Because the state feedback does not change the uncontrollable modes, it is clear that a set of poles can be assigned only if the uncontrollable modes are included.

22. Example. Consider the following dynamic equation:
Check that whether the two set of poles {2, 3, 4} and {1, 2, 3} can be assigned by state feedback.
The eigenvalues of A are –1, –0.5j 0.53. We can verify that –1 is uncontrollable and the other two are controllable. Because the set {–1, –2, –3} includes the uncontrollable mode –1, we can assign it by state feedback, but we can not assign {–2,–3,–4}.

23. Now, we compute k by the direct method. Let k=[k1 k2 k3 ]. Then,
The expected polynomial is
(s+1) (s+2)(s+3)=s3+6s2+11s+6
Comparing the coefficients of the above equation with (*), we have

24. which has a unique solution.

25. We can compute k1, k2, k3 from the :
k1+ k2=4 , k3+ k2=1
Equation (*) can be rewritten as
(s+1)[s2+ (k1+ k2+1)s+ k1+ k3+ 2k2+1]
Comparing it with (s+2)(s+3), we can get the same results. The above equation also indicates that the uncontrollable modes can not be changed by state feedback.

26. 3. Effect of sate feedback on zeros
Open-loop:
We assume that is the controllable canonical form. Then,
Applying the some transformation to the system with state feedback, we have

27. Conclusion: Generally speaking, state feedback only changes the places of poles but does not change the zeros. But after introducing state feedback, the system may involve pole-zero cancellation. In this case, the state feedback not only affects the zeros but also results in that the canceled poles become unobservable

28. Appendix:

30. Then, we have
which is In particular,
which is