Static Output Feedback and Estimators Chapter 5 Static Output Feedback and Estimators

Because state feedback uses all the information of the system, it is also called full information feedback. We can often get a perfect result in theory by state feedback . Output feedback uses an incomplete information of the system. Hence, it is much more difficult to stabilize a system or improve the performance by using static output feedback.

In this chapter, we first introduce the knowledge of pole-placement using static output feedback. Then, we will study the design of estimators.

§5-1 Static Output Feedback and Pole Placement 1. Property of static output feedback Consider linear time-invariant system as follows: (5-1) Let u=Ky+v (5-2) In general, Eq. (5-2) is said to be a static output feedback control law, where K is a p×q constant matrix, v is a p-dimensional vector. From Eq. (5-1) and Eq. (5-2), we easily obtain the dynamical equation of the closed-loop system as follows:

Proof Firstly, we have the following equation Closed-loop system Theorem 5-1 Feedback control law (5-2) does not change the observability of the system. Proof Firstly, we have the following equation (5-4)

Because the right hand side of Eq Because the right hand side of Eq. (5-4) is nonsingular, for any s and K, we have Q.E.D Hence, system (A+BKC, C) is observable iff (A, C) is observable, which implies that static output feedback does not change the obervability of the system. Suppose system (A, C) is unobservable. From Eq. (5-5), the s that will reduce rank of the right hand side of Eq.(5-5) also reduces the rank of the left hand side. Thus, we conclude that static output feedback does not change unobservable modes of the system.

Corollary: Feedback control law u=Ky+v does not change the controllability of systems. In chapter 4, it is proved that poles of a controllable system can be arbitrarily assigned. Unfortunately, output feedback as a special case of state feedback does not have such a property in general.

The relation between state feedback and state feedback Poles can be arbitrarily assigned by state feedback. Poles cannot be arbitrarily assigned by output feedback. The relation between state feedback and state feedback

2. Cyclic Matrix 1. Definition of cyclic matrix Definition: n×n square matrix A is called a cyclic matrix if its minimal polynomial is its character polynomial. The following statements are equivalent: 1). There only exists a non-one invariant factor for Smith canonical form of sIA; 2). An eigenvalue only has one Jord block in Jordan-canonical-form of A. Example: Consider matrix It is clear that A is a cyclic matrix.

In particular, we have Suppose the eigenvalues of A are distinct. Then A is a cyclic matrix if there is only one Jordan block for each eigenvalue. A is a cyclic matrix iff there exists a vector b such that are linearly independent, i.e. (A, b) is controllable.

Hence, for single input system (A, b), it is controllable iff A is a cyclic matrix and b is a generated element of A. 2. Cyclic matrix and controllability Corollary 1*: Suppose (A, B) is controllable and A is a cyclic matrix. Then for almost all real number and suitable vector, is controllable. Remark: The above corollary means that if A is a cyclic matrix, the design procedure to obtain can be omitted.

For example, the following system is controllable:

Generally, consider That A is a cyclic matrix means if then . (A, B) is controllable The last row of is nonzero. From the above analysis, we have for almost any , is controllable The last row of is nonzero.

4) Lemma 2. : Suppose (A, B) is controllable 4) Lemma 2*: Suppose (A, B) is controllable. Then for almost all , eigenvalues of A+BK are distinct. Hence, A+BK is a cycle matrix. Refer to (Davison, 1968, IEEE AC, No.6) for the proof.

3. Theorems used in output feedback pole-placement Corollary 5-2 Suppose (A, B) is controllable and A is a cyclic matrix. Then there exists a vector bImB such that (A, b) is controllable. Corollary 5-4 Suppose (A, B, C) is controllable and observable. Then there exists a p×q matrix H such that (A+BHC, B) is controllable and (A+BHC, C) is observable. Moreover, A+BHC is a cyclic matrix, i.e. the order of its minimum polynomial is n.

Example 5-3 Consider system (A, B, C): It is easy to verify that (A+BHC, B) is controllable, (A+BHC, C) is observable and A+BHC is a cyclic matrix. Moreover, the order of its minimum polynomial is 4.

4. Pole placement with static output feedback 1. Conditions that n poles can be assigned for single input systems First of all, let us study a SIMO system to get an idea of the difficulties of pole placement with static output feedback, which do not arise in state feedback systems. （5-11） The dynamical equation of closed-loop system is

Let the characteristic equation of A and A+bKC be △o(s) and △c(s), respectively. If (A, b) is controllable, then it can be transformed into a canonical form with a matrix P

Now the closed-loop system matrix becomes Denote , where is a column matrix and is a scalar. For a set of given closed-loop poles, is determined. What we shall do is to choose a K1q, such that

(5-14) or

If the desired poles satisfy the nq equations, then it is possible to assign the poles with output feedback. Example 5-4 where rankC=q=2, from the Condition 5-15, we have

Substituting (1) and (2) into (3), then we get the restriction as follows: Hence, if the desired poles satisfy the above equation, then they can be assigned. From the above equation we have: It is easy to verify that the set of poles {1, 1, 2} can be assigned by output feedback, but {5/6, 1, 2} can not be assigned by output feedback.

the characteristic equation of the closed-loop system is Substituting i(i=1,2,…,q) into the above equation, we have i.e.

i.e. where Denoting as , it follows that

If is nonsingular, then from we have

Example 5-5 Compute output feedback K: 1）such that the poles are assigned at 1= 1, 2= 2 From the equation: we have

then

2) The desired poles are then

From the equation: Let then