Linear Quadratic Optimal Output Feedback Control For ... - CiteSeerX

Linear Quadratic Optimal Output Feedback Control For Systems With Poles In A Specified Region

(In Print: International Journal of Control, 64, no 6, 1996)

Lisong Yuan * Department of Electrical and Systems Engineering U-157, University of Connecticut Storrs, CT 06269, U.S.A

Luke E. K. Achenie * Department of Chemical Engineering U-222, University of Connecticut Storrs, CT 06269, U.S.A

Weisun Jiang Research Institute of Automatic Control East China University of Science and Technology Shanghai 200237, P.R.China

* Authors to whom correspondence should be addressed.

Abstract: The problem of output feedback LQR synthesis with regional pole constraints is studied. The synthesis algorithm is presented and its convergence is also proved. The effectiveness of the algorithm is illustrated by an example.

Key words: Linear quadratic regulator, output feedback, pole assignment, optimal control.

1

1. Introduction Linear Quadratic Regulator (LQR) theory is no doubt one of the most fundamental control system design methods (Anderson and Moore, 1989). There are many attractive features of the LQR problem. For example, under certain stabilizability and detectability conditions, the resulting closed-loop system is not only asymptotically stable, but it also possesses infinite gain and at least a 60 degree phase margin, which implies that it has good sensitivity and robustness properties (Kalman 1964, Safonov and Athans 1977). Another important problem in control theory and practice is the design of a feedback law that places the closed-loop poles at desired locations. Since many dynamic properties of a system have close relationship with its poles, this problem has been thoroughly studied (Wonham 1979). However, most of the existing approaches deal with the problem of exact pole placement, in which closed-loop poles are required to lie at prescribed locations. It is well known that a feedback controller of a given structure may offer design flexibility beyond pole placement alone. Hence the designer may also specify other closed-loop characteristics, such as the LQ performance (Haddad and Bernstein 1992). From the foregoing, it is very attractive to design control systems by combining the two methods above. Much work has been done on it since the early 1970s. Most of these approaches utilize known relationships between the quadratic performance index and the locations of closed-loop poles. By appropriate selection of weights, these approaches lead to the inverse optimal control problem (see for examples, Saif 1989, Molinari 1973, and the references therin). It should be mentioned that in practice, the performance index must meet many other requirements beyond closed-loop pole placement, which implies that the regulator is to be designed rather than to be selected (Fujii 1987). In (Gutman and Jury 1981, and Gutman and Taub 1989), the authors presented the conditions under which the eigenvalues of a matrix fall in designated regions. These conditions are very similar to the well known Lyaponuv equations. Using these results, Haddad and Bernstein (1992) solved the problem of linear quadratic gaussian (LQG) optimal control with regional pole constraints by introducing an auxiliary minimization problem. As the auxiliary cost is an upper bound on the actual LQG performance index, the minimization of the auxiliary cost is roughly equivalent to the minimization of the actual LQG performance index. But the upper bound estimate may be too large leading to conservatism in the performance index.

2

In this paper, we address the problem of LQR synthesis with regional closed-loop pole constraints. In this problem, the actual LQ performance index function is optimized. A similar problem has been considered by Liu and Yedavalli (1993), but only the degree of stability of the closed-loop system was addressed, and state feedback is used. In the simplest case, the real parts of all closed-loop poles are required to be less than -α, where α>0 is a real number. Generally, the more practically important region for the closed-loop poles is the sector shown in Figure 1.1. Since it directly impacts many dynamic characteristics of the system, such as the degree of stability and damping ratio, this regional pole constraint is more difficult than that in (Liu and Yedavalli 1993) and will be studied in this paper. As it is not always that all states of system can be obtained, we restrict the analysis to the output feedback case. Im

θ

Re

α

Figure 1.1 Sector region s

2. Problem Statement Consider the system described by x& = Ax + Bu y = Cx 

(2-1)

u = Ky

(2-2)

with static output feedback

where x ∈Rns , u ∈Rnc , y ∈Rno ,and A,B,C, K are real matrices. The performance index is linear quadratic function

3

J = ∫0∞ ( x T R1 X + u T R2 u )dt

(2-3)

where R1 ≥ 0 and R2 > 0 Let

Ac = A + BKC

R = R1 + C T K T R2 KC

Definition 2.1 (Haddad and Bernstein 1992): λ ∈Λ( A ) is a B-controllable eigenvalue if rank [ λI ns − A, B] = ns . λI − A Definition 2.2: λ ∈Λ( A ) is a C-observable eigenvalue if rank  ns  = ns .  C  Lemma 2. 1 (Haddad and Bernstein 1992): The Pair [ A,B] is stabilizable if and only if every eigenvalue of A in the closed right-half plane is B-controllable. Furthermore the pair [A,B] is controllable if and only if every eigenvalue of A is B-controllable. Lemma 2.2: The Pair [C,A] is detectable if and only if every eigenvalue of A in the closed right-half plane is C-observable. Furthermore the pair [C,A] is observable if and only if every eigenvalue of A is Cobservable Let D be an open region on left-half plane, Definition 2.3: The pair [A,B] is D-stabilizable if every eigenvalue of A that is not in D is B-controllable. Definition 2.4: The pair [C,A] is D -detectable if every eigenvalue of A that is not in D is C-detectable. In this paper, the region D is supposed to be the open sector shown in Figure 1.1, subsequently the problem can be stated as: Problem 2.1: Consider the system given by Equation (2-1). Find the output feedback gain K, such that Min

J

s.t.

Λ( Ac ) ⊂ S

Obviously, the above problem is solvable if and only if [Α,Β] is S-stabilizable, and [C, A] is Sdetectable (Anderson and Moore 1989, Wonham 1977, Huang and Li 1989). Therefore, throughout the rest of this paper, these conditions are assumed to be satisfied. In the next section, we consider different realizations of the performance index, J and the constraint set.

4

3. Main Results Let Ks = {K : K ∈Rnc × no ,Re( λ ) < 0, ∀ λ ∈ Λ( A + BKC )}

(3-1)

K = {K : K ∈ Rnc × no , Λ( A + BKC ) ⊂ S}

(3-2)

Following from the well known LQR theory, if K ∈Ks , then lemma 3.1 (Anderson and Moore 1989): Consider the system given by Equation (2-1). If there exits a matrix P > 0, P ∈ Rns × ns , such that AcT P + PAc + R = 0

(3-3)

J=Tr(PX)

(3-4)

then

where X is the variance matrix of initial states, i.e. X = E [ x ( 0) x (0) T ] Let

Aα = A + α I n  A cos θ − Aα sin θ  A$ =  α   Aα sin θ Aα cos θ 

[

C$ 1 = C 0 ny × ns

)  B cos θ  B1 =    B sin θ 

]

− B sin θ  B$ 2 =   Bconθ 

[

C$ 2 = 0 ny × ns C

]

A$ c = A + B$1 KC$ 1 + B$ 2 KC$ 2 Lemma 3.2 (Davison and Ramesh 1970, Anderson et al. 1975): Λ( Ac ) ⊂ S if

and only if A$ c is

asymptotically stable. This also means that for ∀ M>0, M ∈ R ( 2ns ) × ( 2ns ) , there exists a matrix P$ ∈ R ( 2ns ) ×( 2 ns ) and P$ > 0 such that: ) ) $$ AcT P + PA c+M =0 Following the above results, problem 2.1 can been converted into

5

(3-5)

Problem 3.1: Consider the system given by Equation (2-1). Find an output feedback controller gain K such that Min

J=Tr(PX)

s.t.

P>0 and P$ > 0 AcT P + PAc + R = 0 ) ) $$ AcT P + PA c+M =0

Unfortunately, the initial states of the system x(0) are usually unknown, so the optimal LQ performance index can be obtained generally by the minimization of J=Tr(P) instead of J=Tr(PX). Thus problem 3.1 can be restated as: Problem 3.2: Consider the system given by Equation (2-1). Find an output feedback controller gain K such that Min

J=Tr(P)

s.t.

P>0 and P$ > 0 AcT P + PAc + R = 0 ) ) $$ AcT P + PA c+M =0

Using Lagrange multiplier, the above constrained optimization problem can be converted into an unconstrained ones. That is $$ + M ) Q $] Min J = Tr[ P + ( AcT P + PAc + R)Q] + Tr [( A$ cT P$ + PA c

(3-6)

$ ∈ R ( 2ns ) × ( 2ns ) and Q > 0 , Q $ ≥0. with Q ∈ R ns × ns , Q The necessary conditions for the solution of th above problem are given by ∂J = QAcT + Ac Q + I ns = 0 ∂P

(3-7)

∂J $ $T + A $ Q$ = 0 = QA c c ∂P$

(3-8)

∂J $ $ $ + B$ T PQC $ $ $ )=0 = 2[( B T PQC + R2 KCQC T ) + ( B$ 1T PQC 1 2 2 ∂K

(3-9)

Theorem 3.1: If K is the solution of the problem, then K satisfies the following equation B T PQC + R 2 KCQC T = 0

6

(3-10)

Furthermore if rank (C ) = n y , then K = R2−1 B T PQC( CQC ) −1

(3-11)

proof: See appendix. To simplify the problem, in the rest of this paper, we suppose that C is row full rank. If this condition   C' is not satisfied, then C can transformed into the form   , were C ' is row full rank 0  ( n0 − r )× ns  and rank (C ' ) = rank (C ) = r . It can been proved that [ C ' , A] is also S-detectable. Therefore we can use C ’ instead of C to design the system (Huang and Li 1989)., and the same results will be obtained. Next we will present a synthesis algorithm for the optimal K. In order to prove its convergence, let us first present some necessary preliminary results Suppose that P1 , K1 , Q 1 and P2, K 2, Q2 satisfy the following equations ( A + BK1C ) T P1 + P1 ( A + BK1 C ) + R1 + CK1T R2 K1 C = 0

(3-12)

Q1 ( A + BK1C ) T + ( A + BK1C )Q1 + I ns = 0

(3-13)

( A + BK 2 C ) T P2 + P2 ( A + BK2 C ) + R1 + CK 2T R2 K 2 C = 0

(3-14)

Q2 ( A + BK 2 C ) T + ( A + BK 2 C )Q2 + I ns = 0

(3-15)

Let

∆P = P2 − P1

∆K = K 2 − K 1

From Equations (3-12) and (3-14), we obtain T T T T T ( A + BK 2C ) T ∆P + ∆P( A + BK2 C ) + C T ∆K T R2∆KC + ( PB 1 + C K1 R2 ) ∆KC + C ∆K ( B P1 + R2 K1C ) = 0

Combining with Equation (3-15), we obtain ∆J = Tr ( P2 − P1 ) = Tr ( ∆P) = − Tr{∆P[( A + BK 2 C )Q2 + Q2 ( A + BK 2 C ) T ]} = Tr[ 2Q2 ( P1 B + C T K 1T R2 )∆ KC + C T ∆K T R2 ∆ KC]

(3-16)

Since P, Q, and J are all functions of K, the performance index can be rewritten as J ( K ) = Tr[ P( K )] = Tr [( R1 + C T K T R2 KC )Q( K )] Let K 0 be a stabilizing controller of systems (2-1). Define

7

(3-17)

Ω( K 0 ) = {K : K ∈K, J ( K ) ≤ J ( K 0 ) Theorem 3.2: The set Ω( K 0 ) is compact, and contains the minimum of the performance index Tr(P). proof: See appendix. Remark 3.1: The above theorem indicates that if Ks is not empty, then the optimal solution exists. Furthermore if we can find a K 0 ∈ Ks , then the minimum is contained in the level set Ω( K 0 ), The existence of the optimal solution has been guaranteed by Theorem 3.2, next we will present a corresponding algorithm, and establish its convergence. Theorem 3.3: Suppose that K k ∈K , then there always exists an ε k >0 such that for any ε k ∈( 0, ε k ) , we have

K k+1 ∈ K

(3-18)

where

K k +1 = K k + ε k ∆ K k

(3-19)

∆K k = − R2− 1 B T Pk + 1Qk + 1C T ( CQk +1 C T ) − 1 − K k

(3-20)

and Pk +1 > 0 , Qk+1 > 0 are the solutions of following equations ( A + BK k C ) T Pk +1 + Pk +1 ( A + BKk C ) + R1 + C T K kT R2 K k C = 0

(3-21)

( A + BK k C )Qk +1 + Qk + 1 ( A + BK k C ) T + I ns = 0

(3-22)

Proof: See appendix. Theorem 3.4: Suppose that K k ∈K , and ε k >0 is chosen as in Theorem 3.2, then for ∀ε k ∈( 0, ε k ) ,

J ( K k +1 ) ≤ J ( K k ) Proof: See appendix. Remark 3.2: if K k ∈K , then by appropriate choice of ε k , we can always find a K k+1 in the next step such that K k+1 ∈ K as well. Furthermore, the LQ performance J ( K k +1 ) ≤ J ( K k ) . These imply that if the initial value of output feedback gain, K 0 , is chosen such that K 0 ∈K , then in subsequent iteration the value of output feedback gain, K k+1 , can also be guaranteed to be within K. In addition. the LQ performance index is monotonically decreasing, thus the following convergent algorithm results Algorithm ( optimal output feedback gain synthesis):

8

Step 1: Set k=0, choose K 0 ∈K (several methods can be used, for examples see Yuang 1994, and Arzelier et al. 1993) Step 2: Solve for Pk +1 and Qk+1 from Equations (3-21) and (3-22) respectively; Step 3: Calculate ∆K k from Equations (3-20) Step 4: Using any search method find ε k ∈( 0,2) such that the updated feedback gain K k+1 ∈ K Step 5: Calculate J k = Tr ( Pk +1 ) , if k=0 go to step 2, otherwise continue; Step 6 : Let ∆J k = J k −1 − J k , if ∆J k < ζ , then stop, otherwise set k=k+1 and go to step 2, where the real number ζ >0 is a given error bound.

4. Example The example considered in this section was generated with a random number generator. The matrices in Equation (2-1) are .  −168370  −59939 .  . 115173 −  − 140898 .  35444 . A= − 123383 .   − 119758 .  9.4223  .  − 28002  −30798 .

. 95208 . 82689 . −01133 . 132929 . 133665 70855 . 63853 . 85213 . 106254 . 76628 .

. . . . . . 68392 113629 21161 86666 −28971 − 81444 . . . 9.4247 − 6.4245 137898 − 36683 − 29433 −14. 0432 . . 3.4387 0.4451 173487 1.4037 −87862 −1.4956 . . . . . . 40531 96792 69029 40367 −26871 −119361 . . . . . 85240 30201 216319 130137 62821 68353 − − − − − . − 31123 . − 08824 . − 78925 . 142878 . −176991 . −6. 7328 −21776 . 49644 . 28649 . 118870 . −81313 . − 58729 . − 82547 . 13032 . − 158603 . 6.4536 33211 . − 77177 . −2. 5965 111067 . 32124 . − 09093 . 132068 . −111205 . − 95289 . 09261 . − 06048 . 22289 . 115398 . −26919 .

. − 05175  −11933 .  01774 −  .  −1.4048  −18698 . B= 17462 −  . − 01245 . − 02192 .  02416 −  . − 03564 .

. . . 16483 17393 15523   . . 08595 − 1.4729 10105  . . . 10156 09447 −01808  . . 13047 1.4946 17595   . . . −03930 −01014 − 08457  . . 0. 2267 − 09153 − 15796  . . 00378 15358 0.4086 . . 03759 −01873 − 0.4273  . . . 03904 13000 19183   . . . 17877 15109 02042

9

. −177775 . 28717 . −53281 − 20.4462 − 68779 . 126251 . 11506 . − 45366 . −17.4393 −175596 .

. −01490   . − 70662  . −18740   . − 36941  03949 .  −43816 .   −13072 .  31310 .  −11001 .   05719 .

.  03270  00285 .  −03331 . C= .  18840  −03631 .  10467  .

−01383 . 08534 . 06373 . 06373 . −03333 . −1.4767 − 12371 . −12371 . −11224 . 11351 . 11351 . −01169 . 02894 . 09395 . 09395 . −13816 . −01145 . −12660 . − 12660 . −09703 . −0.4962 − 08594 . −08594 . −0. 0386

− 05077 . −12058 . 19313 . − 09453 . − 00910 . −11362 .

03743 . −19435 . 15817 . −08795 .   − 07786 . 01348 . − 02985 . − 10738 .  − 13406 . 0. 7779 − 02940 . −00414 .  − 01885 . 18005 . − 05072 . 08761 .   0.5743 0. 5629 − 12835 . 07928 .  − 02088 . 12436 . 08989 . 09731 . 

the open-loop poles are Λ( A) = {− 183477 . , −8.6142 ± 213195 . i,− 6.4108 ± 21102 . i, −5.9960, −3. 5712,− 2.9440,9.8861 ± 9.4888i) Obviously, the open-loop system is unstable. Let the weights in the linear quadratic performance function be R 1 = 5I10 and R2 = 05 . I 4 . Suppose that the closed-loop poles are to be placed on various sectors, using the algorithm presented in section 3, the results are shown in Table 4.1 to Table 4.3. From these tables, it is evident as α or θ increases, the optimal performance index become worse (i.e. increases). This indicates that the algorithm presented is valid and efficient.

Table 4.1: Results with α=3

J opt

K

Λ( Ac )

π 3

25.5360

. −3.5986 −0.0054 −3.2702 −4.6458 − 2. 0555 − 16249  −11741 . − 0.0385 −4.2452 −0.1973 −3.0250 0.4255  2. 5156 0.4472 4.7642 0.7976 − 18821 . −2.8589  . 0.3451 0.2297 5.8205 18803 . − 0. 0930 18613 

i , −20. 5234 ± 88444 i . −14.4931 ± 66355 i . . −76930 ± 11482 . ,− 61101 . −410052 . , −30379 . −36500

π 4

17.8287

. 0.4769 − 21636 . − 21358 .  −1.4026 − 2.6801 − 39459  − 0. 7611 11225 . − 33931 . −0.1293 −1. 8092 − 0. 5180  1. 0672 − 0.2426 2.9430 0. 7185 − 0. 0263 − 2. 0607   . 1. 9392  2. 2640 − 0.3044 2. 0499 −0.8002 61392

i . . −233676 ± 203674 . −198479 ± 10. 3168i i . . −70722 ± 28258 . ,− 61484 . −80303 . ,− 30747 . −38549

π 6

16.3329

. −2.3466 −2. 7085 0.0145 −2.4020 −1. 2112  −15230 − 14350 , −0. 2623 −2.2840 − 0.5571 −2. 7697 − 0.8939  3. 0158 12689 . 3. 2428 2.0268 −0.5631 −0.4392   . − 0. 3233 2. 7345 −0.5268 83344 . 13348 .   33886

i . . −223255 ± 108528 i . . −175968 ± 58186 −10. 2074 ± 12.4835i i . . −61956 ± 06288 . ,− 30248 . −39878

θ

10

Table 4.2: Results with α=2

θ

J opt

π 3

20.6121

π 4

17.6803

π 6

15.5829

Λ( Ac )

K . − 15634 − 0.4563  27855  . .  19529

− 17417 . −38956 . − 01076 . −43602 . 01573 . 4.4588 17388 . 1.4352

.  −10234 − 0. 0220  15690  .  3. 9379

− 2.0824 − 0.0069 − 11230 . −0.2211

. −21792 . − 19134 − 01683 . 00704 .  29093 11155 .  . 0 . 7532 − 06098 . 

−01722 . − 1.4371 −02298 . −37668 . 1.4015 −06794 . −07735 . 75673 .

− 3.4287 − 3.4357 3.4990 2.4980

0. 0539 − 0. 0085 16207 . −0.3520

−37990 .   00481 . −0. 7865 02469 . 

−18423 . −16975 . −0. 7402 7.1645

−41004 . −02407 . −20333 . − 27895 . −0.4715 −1.4324 23089 . 11505 . −13204 . 30898 . 00687 . 5.4491

− 1.4462  − 0.2443 − 0.7465 2.3871 

−22768 .   − 11080 .  −02999 .  17218 . 

i . −23.4612 ± 122678 i . −8.4106 ± 37011 −29. 7251,− 12.4751 . ,− 59956 . −88019 . ,− 30406 . −38300 . −172000 ± 5.4667i i . −14. 5258 ± 125254 i . . −108028 ± 35454 . −63452 ± 0.4792i . , −30481 . −39759 i . . −238140 ± 92512 i . . −128990 ± 88776 i . . −81487 ± 37537 . −7.4993,−62504 . , −30696 . −37120

Table 4.3: Results with α=1

θ

J opt

π 3

19.8378

π 4

π 6

17.4626

15.3045

Λ( Ac )

K . − 23007  00631 .  13431  . .  31850

− 21949 . 0.4834 1.4206 −0. 5667

− 32147 . −37194 . 36408 . 27155 .

0. 2649 −03588 . 0. 7178 − 02981 .

−2.4441 −10230 . −06817 . 70481 .

−17362 .   − 11419 . −1.4300  23305 . 

. − 23784 . −31142 . 03733 . − 18392 . − 12113 .  −19395  − 0.4489 00405 . −38345 . −03292 . −2. 2438 −0. 0547   . 09896 . 30315 . 06947 . 0. 2594 − 26308 .  −02531   38962  . −1.4179 218821 . −0.4485 78930 . 20553 . . − 15230 − 1.4350  30158  . .  33886

−2.3466 −0.2623 1. 2689 −0.3233

−2.7085 − 2.2840 3.2428 2.7345

11

0.0145 −0.5571 2.0268 −0.5268

− 2.4020 −2.7697 −0.5631 8.3344

− 12112 .  −0.8939  −0.4392  13348 . 

i . . −238000 ± 131636 i . . −191317 ± 26353 . −93128 ± 4. 2383i . , −61242 . −79036 . , − 30834 . −38968 −25.4970 ± 24.4967i −9.0174 ± 6.3953i −4.7156 ± 0.4423i −24.5230, −8. 3203 ,. −5.889131036 i . . −223255 ± 108528 i . . −175968 ± 58186 −10. 2074 ± 12.4835i i . . −61956 ± 06288 . ,− 30240 . −39878

5. Conclusions The problem of LQR synthesis with regional poles assignment is studied . The novelty in our approach lies in the fact that, unlike the LQ inverse optimal problem, the actual LQ performance is optimized. This makes it possible to meet the requirments of poles and other characteristics of closed-loop system simultaneously. The output feedback case is considered. The synthesis algorithm for the optimal feedback gain is developed, and its convergence is also proved. The region considered here is more practical than that in (Liu and Yedavalli 1993), and the formulation is more general. The effective performance of the algorithm is illustrated by an example. Since the proposed algorithm is based on the gradient search technique, the convergence may not be global. Thus a further studies is to find a global convergent algorithm for this problem.

Notation E (•) : denotes the expectation; R: denotes denote the real number space; Rn : denotes real vector space with dimension n;

Rn× m : denotes real matrix space with dimension n × m ;

I n : denotes matrix identity with dimension n × n ; 0 n×m : denotes the n × m zeros matrix whose entries are all zero; (•) T : denotes the transpose of a matrix; Tr(• ) : denotes the trace of a square matrix; Λ(• ) : denotes the set of all its eigenvalues of a square matrix;

λ(• ) ∈ Λ( •) : is a eigenvalue of a square matrix; λ max ( •) , λ min (• ) : are the maximum and minimum eigenvalue of a square matrix respectively; P>0, ≥ 0,0 that also implies that CQC is invertible. By rearrangement of Equation (3-10), we can get Equation (3-11). This completes the proof . Proof of theorem 3.2: Since the eigenvalues of A c and J(K) are continuous functions of K, let

Ω1 ( K 0 ) = {K : K ∈ Ks , J ( K ) ≤ J ( K 0 )} because K and Ω 1 ( K 0 ) is two closed sets, Ω( K 0 ) = KI Ω 1 ( K 0 ) is also closed. Since for ∀K ∈ Ω( K 0 ) , A+BKC is asymptotically stable, and 0