A new method for mixed H
2/H
1control with
regional pole constraints
Jenq-Lang Wu
1,n,yand Tsu-Tian Lee
2,z,}1Department of Electronic Engineering, Hwa Hsia College of Technology and Commerce, 111, Hwa-Shin Street,
Chung-Ho 235, Taipei Country, Taiwan, R.O.C.
2
Department of Electrical and Control Engineering, National Chiao Tung University, 111, Ta-Hseuh Road, Hsinchu 300, Taiwan, R.O.C.
SUMMARY
In this paper, the problem of state feedback mixed H2/H1control with regional pole constraints is studied.
The constraint region is represented by several algebraic inequalities. This constrained optimization problem cannot be solved via the LMI approach. Based on the barrier method, we instead solve an auxiliary minimization problem to get an approximate solution. We shall show that the obtained minimal solution of the auxiliary minimization problem can be arbitrarily close to the infimal solution of the original problem. An example is provided to illustrate the benefits of the approach. Copyright # 2003 John Wiley & Sons, Ltd.
KEY WORDS: mixed H2/H1 control; regional pole placement; barrier method; Lagrange multiplier
method
1. INTRODUCTION
The mixed H2=H1control problem has attracted much attention in recent years (see References [1–7]). The mixed H2=H1 control theory offers a way of combining the design criteria of quadratic performance and disturbance attenuation. But such a controller design method cannot guarantee that the closed-loop systems have good transient responses. The systems’ transient responses are determined mainly by the locations of the systems’ poles. In References [8–12,24], the optimal regional pole placement problem, which involves determining a feedback controller that minimizes a cost functional subject to the requirement that the closed-loop poles
Received 15 October 2001
y
E-mail: [email protected]
n
Correspondence to: Jenq-Lang Wu, Department of Electronic Engineering, Hwa Hsia College of Technology and Commerce, 111, Hwa-Shin Street, Chung-Ho 235, Taipei Country, Taiwan, R.O.C.
z
Fellow, IEEE.
}E-mail: [email protected]
Contract/grant sponsor: National Science Council, Republic of China; Contract/grant number: NSC 89-2213-E-146-008
lie within a specified region, has been studied. In Reference [13], Yedavalli and Liu studied the state feedback H1 control problem with single regional pole constraint via the Lagrange multiplier method. Recently, Wang and Bernstein [14] provided an approach to achieve mixed H2=H1 control with an a-shifted pole constraint via output feedback. Bambang et al. [15] considered the mixed H2=H1control with pole placement in a circular region. In Reference [15], a generalized Riccati equation is employed in order to satisfy H1 -norm and poles’ constraints as well as to obtain an upper bound on the H2cost to be minimized. These approaches have the
disadvantage that they only minimize an ‘upper bound’ of the actual cost but not the actual cost. The obtained solution may be far from the actual solution. Yaz et al. [16] presented an approach to find the controller for a discrete system which simultaneously meets the following three criteria: pole placement in a specified disk, assignment of an upper bound to the H2cost and satisfaction of an H1 disturbance bound. More recently, Bambang et al. [17] provided a unified treatment for pole placement in the mixed H2=H1 optimization problem. The cost function they minimized is also an upper bound of the actual cost. Moreover, the existence of the minimum point of the chosen auxiliary cost cannot be guaranteed. In fact, its infimal solution may lie on the boundary of the admissible solution set and may not be a stationary point. In this case, the infimal solution does not satisfy the obtained necessary conditions in Reference [17]. Chilali and Gahinet [18] used the linear matrix inequality (LMI) approach to solve the mixed H2=H1 problem with regional pole constraint. The constraint region must be convex in LMI approach. Moreover, for tractability in the LMI framework, a single matrix that enforces several constraints must be found. This will lead to the result that the obtained solution may not be the global minimal solution, may not even be the local minimal solution, of the original constraint optimization problem. Furthermore, even in the case that the original constraint optimization problem is solvable, the LMI approach may fail and no solution may be found thus.
In this paper, we consider the regional pole constraints mixed H2=H1state feedback control problem. The constraint region is represented by several algebraic inequalities. It may be non-convex and may not be representable in the form of LMI region. In some special cases, it may contain several disjoint subregions. This problem is difficult to solve and its analytic solution has not been presented yet. Based on the barrier method [23], in this paper we instead solve an auxiliary minimization problem to get an approximate solution of the original optimization problem. The cost function of the auxiliary optimization problem is the sum of the cost function of the original problem and a weighted ‘barrier function’. We shall show that if the weighting factor of the barrier function approaches zero, then the optimal solution of the auxiliary minimization problem will approach the infimal solution of the original optimization problem. The necessary conditions for local optimum of the auxiliary problem are derived. Furthermore, a solution algorithm is provided. The advantages of the presented approach are: (1) it is simple, (2) the existence of the minimum point of the auxiliary minimization problem is guaranteed, (3) the auxiliary minimization problem can be solved via some unconstrained search techniques and (4) the obtained solution can be arbitrarily close to the infimal solution of the original problem.
In what follows, Re(l) and Im(l) denote the real part and the imaginary part, respectively, of a complex number l, s(M) is the spectrum of the matrix M, M>0 (M50) means that the matrix M is positive (negative) definite, ||G(s)||1(||G(s)||2) denotes infinity-norm (2-norm) of
the transfer function G(s), denotes Kronecker product, vecðMÞ ½mT
1mT2:::mTn
T; where m
iis
2. PROBLEM FORMULATION AND PRELIMINARIES Consider a linear time-invariant continuous system
’xxðtÞ ¼ AxðtÞ þ B1w1ðtÞ þ B2w2ðtÞ þ B3uðtÞ z1ðtÞ ¼ C1xðtÞ þ D1uðtÞ
z2ðtÞ ¼ C2xðtÞ þ D2uðtÞ
ð1Þ
where w12 Rnw1 and w22 Rnw2 denote the exogenous inputs, x 2 Rn is the state, u 2 Rm is the control, and z12 Rnz1and z22 Rnz2denote the controlled outputs. All matrices are assumed to be
of appropriate dimensions. Define Oi ða þ ibÞ fiða; bÞ ¼
X fi X hi rfihia fibhi50 ( ) i ¼1; 2;. . . ; c
where rfihi is real. Let O
T
i¼1;:::;cOi belong to the open left-half complex plane.
The design objective is to determine the state feedback controller
uðtÞ ¼ FxðtÞ ð2Þ
to achieve the infimum of jjTz2w2ðFÞjj2; i.e.
inf
F JðFÞ jjTz2w2ðFÞjj2 subject to the following constraints:
(1) jjTz1w1ðFÞjj15g
(2) sðA þ B3FÞ O:
This problem is referred as the Q21p problem. &
Let Gs F 2 Rf mnjA þ B3F is stableg; GOi F 2 R mnsðA þ B 3FÞ Oi j f g; GOTi¼1;2;:::;cGOi; G1ðgÞ F 2 RmnjjTz 1w1ðFÞjj15g
; and G GOTG1ðgÞ: Let AC=A+B3F, C1C=C1+D1F,
and C2C=C2+D2F.
Assumption 1
Suppose all the eigenvalues of A, which are outside O, are B3-controllable, i.e.
rank½lI A B3 ¼ n for all l 2 sðAÞ and l =2 O
Assumption 1 guarantees that the set GOis non-empty. From (1) and (2), it is known that the closed-loop system transfer function from w1to z1is given by
Tz1w1ðFÞ ¼ ðC1þ D1FÞðsI A B3FÞ
1B
1 and the closed-loop system transfer function from w2to z2is given by
Tz2w2ðFÞ ¼ ðC2þ D2FÞðsI A B3FÞ
1B
Let Lobe the positive semidefinite solution of the equation
ATcLoþ LoAcþ CT2cC2c¼ 0 ð3Þ
It can be shown that
jjTz2w2ðFÞjj2¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi TrðBT2LoB2Þ q ; if F 2 Gs 1; otherwise 8 < : ð4Þ Lemma 1
Suppose Acis stable. Then the following statements are equivalent:
(1) jjTz1w1ðFÞjj15g:
(2) For any given matrix #QQ1¼ #QQ T
1 > 0; there exists a positive definite solution #PP1¼ #PP T 1 to ðAcþ 1 g2B1B T 1P1ÞT#PP1þ #PP1ðAcþ 1 g2B1B T 1P1Þ þ FTF þ #QQ1 ¼ 0 ð5Þ where P1 ¼ PT1 0 is the stabilizing solution of
ATcP1þ P1Acþ CT1cC1cþ 1 g2P1B1B T 1P1¼ 0 ð6Þ Proof
According to the Bounded Real Lemma (see Reference [19]), it is known that (6) has a stabilizing solution P1 ¼ PT
1 0: Since the matrix FTF þ #QQ1 is positive definite and the matrix Acþ 1=g2B
1BT1P1 is stable, from the Lyapunov stability theorem, the unique solution
#PP1 of (5) is positive definite. &
Moreover, we have the following result. Lemma 2
Trð #PP1Þ approaches infinity as F approaches the boundary of G1ðgÞ: Proof
As F approaches the boundary of G1ðgÞ; there exists at least one eigenvalue l 2 A þ B3F þ ð1=g2Þ B1BT1P1
that will approach the imaginary axis. Suppose v is the normalized eigenvector corresponding to l. Premultiplying and postmultiplying (5) by vn
and v, respectively, and after some manipulations, we obtain
vn#PP 1v ¼ vn ðFTF þ #QQ1Þv 2ReðlÞ Since vn
ðFTF þ #QQ1Þv > 0 and ReðlÞ ! 0; it follows that vn#PP1v ! 1: Note that #PP1 is positive definite. This leads to the conclusion that Trð #PP1Þ ! 1: This completes the proof. &
Suppose l ¼ a þ ib: Then Oi can be represented equivalently by Oi¼ l 2 C gf j iðlÞ50g
where giðlÞ ¼ fiða; bÞja¼ðlþ%llÞ=2;b¼ðl%llÞ=2i¼ P pi P qicpiqil pi%llqi: Note that c piqi ¼%ccqipi (see Reference
[20]). Now define jiða; bÞ ¼Ppi Pqicpiqia
pibqi:
Lemma 3(Bambang et al. [17])
Suppose that for all aj2 Oi (j=1,..., n), the following conditions are satisfied. (1) jiðaj;%aakÞ=0 for all j,k=1,2,...,n,
(2) The Hermitian matrix Ci j1
i ðaj;%aakÞ
n
j;k¼1 is non-negative definite.
Then sðAcÞ Oi if, and only if, for any given positive definite Hermitian matrix #QQ; the unique solution #PP of the following equation
X pi X qi cpiqi A qi c n PApi c þ Qi¼ 0 ð9Þ
is positive definite. &
A class of regions that satisfies conditions (1) and (2) of Lemma 3 include, among the many possibile, hyperbolic, circular, elliptic, parabolic, etc. More examples and details can be found in References [17] and [21], and references therein. The regions discussed in the following are all assumed to belong to this class of regions.
Remark 1
In order to get a real solution F, the region X must be symmetric with respect to the real axis of the complex plane. However, each individual Xiis not required to be symmetric with respect
to the real axis. This class of regions may be non-convex, may even contain several disjoint subregions. It should be noted that the LMI regions discussed in the literature (see e.g. Reference [18]) are restricted to be convex.
Let Ac¼ A þ B3F in (9) and suppose Pi, i=1, ..., c, are the positive definite solutions of (9). It
is known that Piis a function of F. Then we have the following theorem.
Lemma 4
Tr(Pi) approaches infinity as F approaches the boundary of GOi;
Proof
As F approaches the boundary of GOi; then
P
pi
P
qicpiqil
pi%llqi ! 0 for some l 2 sðA þ BFÞ:
Let v be the normalized eigenvector corresponding to l. Premultiplying and postmultiplying (9) by vn
and v, respectively, and after some manipulations, we obtain vn Piv ¼ vn Qiv giðlÞ Since vn
Qiv > 0 and giðlÞ ! 0; it follows that v
n
Piv ! 1. Note that Piis positive definite, this
implies Tr(Pi) approaches infinity as F approaches the boundary of GOi: &
3. THE MAIN RESULTS
The Q21p problem is a constrained optimization problem. The infimal solution of the Q21p problem may lie on the boundary of G and may not be a stationary point. Thus far, no analytic
solution to this problem has been derived. In this section, we instead solve an auxiliary problem, denoted by the problem Q21pauxto obtain an approximate solution of the problem Q21p: The cost function of the problem Q21pauxis defined as
Jauxðg; FÞ ¼ JðFÞ þ w Jbðg; FÞ; if F 2 G
1; otherwise
(
ð10Þ
where w>0 is the weighting factor and Jb(F) is defined by
Jbðg; FÞ ¼ l1Trð #PP1Þ þ Xc
i¼1
wiTrðPiÞ
where l1> 0 and wi> 0; i ¼ 1; . . . ; c; are the weighting factors, and #PP1 and Pi; i ¼ 1; . . . ; c; are the solutions of matrix equations (5) and (9).
The problem Q21paux is to find the feedback matrix F to minimize the cost function (10). Lemma 5
Jb(F) approaches infinity as F approaches the boundary of G:
Proof
This lemma can be proved directly from Lemmas 2 and 4. &
Now we shall show that the cost function Jaux(F) has a minimum point in the interior of the
set G if D2has full column rank.
Lemma 6
Suppose D2has full column rank. If the admissible set G is non-empty, then the cost function
Jaux(F) has a minimum point in the interior of the set G:
Proof
Define a level set
GlðF0Þ F 2 G Jf j auxðFÞ JauxðF0Þ; for F02 Gg
The set GlðF0Þ is bounded since if D2 has full column rank, then JðFÞ ! 1 as jjFjj ! 1:
Moreover, since Jaux(F) is continuous in the set G and unbounded on the boundary of G; GlðF0Þ is closed. As a result, the level set GlðF0Þ is compact. From the Weiestrass theorem [10], there is a Fopt2 GlðF0Þ such that
JauxðFoptÞ JauxðFÞ; for all F 2 GlðF0Þ This implies
JauxðFoptÞ JauxðFÞ; for all F 2 GO
and completes the proof. &
Since the minimum point of the auxiliary cost function Jaux(F) lies in the interior of the
employed to derive the necessary conditions for local optimum of the Q21paux problem to satisfy.
Theorem 1
The optimal solution F of the Q21paux problem must satisfy Fgradðg; FÞ 2DT2D2FL2þ D1TD1FðL1þ LT1Þ þ 2F #LL1þ 2 Xc i¼1 FSi þ 2BT 3LoL2þ 2DT2C2L2þ BT3P1LT1þ BT3P1L1þ DT1C1LT1þ DT1C1L1 þ 2BT3#PP1#LL1þX c i1 X pi X qi cpiqi X pi1 k¼0 BT3ðAn cÞ kP iAqciSiðA n cÞ pi1k " " þX qi1 k¼0 BT3ðAn cÞ kP iApciSiðA n cÞ qi1k ## ¼ 0 ð11Þ
where Lo, L2, P1; L1; #PP1#LL1; Pi; i ¼ 1; . . . ; c; and Si; i ¼ 1; . . . ; c; satisfy the following matrix equations: ATcLoþ LoAcþ CT2cC2c¼ 0 ð12Þ AcL2þ L2ATc þ12 TrðB T 2LoB2Þ 1=2 B2BT2 ¼ 0 ð13Þ AcP1þ P1ATc þ C T 1cC1cþ 2 g2P1B1B T 1P1¼ 0 ð14Þ L1ðAcþ 1 g2B1B T 1P1Þ þ ðAcþ 1 g2B1B T 1P1ÞTL1þ 1 g2#LL1#PP1B1B T 1 ¼ 0 ð15Þ ðAcþ 1 g2B1B T 1P1ÞT#PP1þ #PP1ðAcþ 1 g2B1B T 1P1Þ þ FTF þ #QQ1 ¼ 0 ð16Þ ðAcþ 1 g2B1B T 1P1Þ #LL1þ #LL1ðAcþ 1 g2B1B T 1P1ÞTþ w l1 I ¼ 0 ð17Þ X pi X qi cpiqiðA n cÞ qiP iApciþ F TF þ Q i¼ 0 i ¼1; 2; :::; c ð18Þ X pi X qi cpiqiA pi cSiðA n cÞqiþ w wi I ¼ 0 i ¼1; 2; :::; c ð19Þ
Proof
Define the Hamiltonian Hamby
Ham¼ Tr BT 2LoB2 1=2 þw l1Tr #PP1 þ Tr X c i¼1 wiPi !! þ Tr L2 ATcLoþ LoAcþ CT2cC2c þ Tr L1 AcP1þ P1ATc þ CT1cC1cþ 1 g2P1B1B T 1P1 þ Tr #LL1 ðAcþ 1 g2B1B T 1P1ÞT#PP1þ #PP1ðAcþ 1 g2B1B T 1P1Þ þ FTF þ #QQ1 þ Tr X c i¼1 Si Xnpi pi¼0 Xnqi qi¼0 cpiqiðA n cÞ qiP iApciþ FTF þ Qi !!
where L2; L1; #LL1; Siði ¼ 1; 2; :::; cÞ are the Lagrange multipliers. Then the necessary conditions for local optimum are @Ham=@L2¼ 0; @Ham=@Lo¼ 0; @Ham=@L1¼ 0; @Ham=@P1¼ 0; @Ham=@ #LL1 ¼ 0; @Ham=@ #PP1 ¼ 0; @Ham=@Si¼ 0; @Ham=@Pi¼ 0; and @Ham=@F ¼ 0: After some algebraic
manipulations, results of (11)–(19) can be derived. &
For a fixed weighting factor w, suppose the optimal solution of problem Q21paux is Fopt(w).
Suppose the infimal solution of the Q21pproblem is Fn. Then we have the following result.
Proposition 1
limw!0þJðFoptðwÞÞ ! JðFnÞ:
Proof
For any e>0, define the set Ge by
Ge F 2 G JðFÞ JðFnÞ51
2e
This set is non-empty since J(F) is continuous in G: Set we¼ e=2JbðFaÞ; for some Fa2 Ge: Note that J(F) is continuous over the set G; and J(F)>0 is bounded for any F 2 G; so we> 0: If choose wsuch that 05w5we; then w JbðFaÞ51
2e: Moreover,
JauxðFaÞ JðFnÞ ¼ JðFaÞ þ w JbðFaÞ JðFnÞ5e
That is, for any e > 0; we can find w satisfying 05w5we such that minF2GJauxðFÞ JðFnÞ JauxðFaÞ JðFnÞ5e
It should be noted that JauxðFoptðwÞÞ is decreasing as w is decreasing. So limw!0þJauxðFoptðwÞÞ ! JðFnÞ
Since JauxðFoptðwÞÞ > JðFoptðwÞÞ JðFnÞ; for all w>0, it can be concluded that
limw!0þJðFoptðwÞÞ ! JðFnÞ
This completes the proof. &
This proposition shows that the minimum solution of the problem Q21paux will converge to the infimal solution of the Q21p problem if w ! 0þ:
The coupling equations (13)–(21) are not easy to solve. Based on steepest descent method, solution algorithms are provided in Appendix A. It should be noted that the problem Q21paux may have several local optimal solutions, and the solution obtained via the proposed algorithms may not be the global optimal solution. However, in the following example we will see that the proposed approach is useful.
4. AN ILLUSTRATIVE EXAMPLE Consider the following system:
’xxðtÞ ¼ 1 0 1 0 2 0 0 1 3 2 6 6 4 3 7 7 5xðtÞ þ 1 1=2 1 0 1 1 2 6 6 4 3 7 7 5w1ðtÞ þ 1 0 1 2 6 6 4 3 7 7 5w2ðtÞ þ 0 1 1 0 1 3=5 2 6 6 4 3 7 7 5uðtÞ z1ðtÞ ¼ 2=3 1=3 1=3 0 1=3 1=3 " # xðtÞ þ 1 0 0 1 " # uðtÞ z2ðtÞ ¼ 0 1 1 1 1 1 " # xðtÞ þ 1 0 0 1 " # uðtÞ
Suppose the constraint region O (shown in Figure 1) is defined by O ðx þ iyÞ x > 50; x 2þ y2> 9; 10x þ y250
The objective is to find u(t)=Fx(t) to achieve the infimun of jjTz2w2ðFÞjj2under the constraints
(1) jjTz1w1ðFÞjj151; and (2) sðA þ B3FÞ O:
It should be noted that the region O is not convex. It is not a LMI region.
For comparison, we consider seven cases. In case 1, we consider the H2 optimal control
problem without constraints (1) and (2). In case 2, we consider the H2optimal control problem
under constraint (1) and without constraint (2). In case 3, we consider the H2optimal control
problem under constraint (2) and without constraint (1). In cases 4–7, we consider the H2
optimal control problem under both constraints (1) and (2) with different weighting factors. Let the matrix Qi(i=1, 2, and 3) in Equation (10) and the matrix #QQ1 in Equation (5) be identity matrices. Without loss of generality, let the weighting factor w=1 in cases 2–7. By applying the algorithms shown in Appendix A, the final results are shown in Table I, and the locations of the resultant closed-loop poles are shown in Figure 2.
For case 1, we can see that the resultant jjTz2w2ðFoptÞjj2=3.9572 is the smallest value among all
the seven cases. However, some of the resultant closed-loop poles are outside the region O and the resultant jjTz1w1ðFoptÞjj1is larger than 1 since we did not put these constraints into the design
procedure. In case 2, we only put the H1-norm constraint into the controller design procedure. Letting l1¼ 1: It should be noted that the constraint of jjTz1w1ðFoptÞjj151 is satisfied. However,
some of the closed-loop poles are outside the region O: In case 3, we only put the poles’ constraint into the design procedure. It can be verified that all the closed-loop poles lie in the desired region O: However, the resultant jjTz1w1ðFoptÞjj1 is larger than 1. In cases 4–7, both the
pole constraint and the H1 -norm constraint are considered. From Table I, we can see that for these four cases, all the pole constraint and H1-norm constraint are satisfied. In general, if both the weighting factors l1and wiare decreasing, the resultant value of jjTz2w2ðFoptÞjj2will decrease. Note that the weighting factors in the Case 7 are nearly zero, we can expect that the infimum of jjTz2w2ðFoptÞjj2; under the constraints that jjTz1w1ðFoptÞjj151 and all the closed-loop poles lying in
the region O; is about 6.1988.
5. CONCLUSIONS
A new method for the regional pole constraint mixed H2=H1state feedback control problem is provided. The considered constrained region, which is represented by several inequalities, may be non-convex. In some special case, it may contain several disjoint subregions. The solution of the considered problem is approximately obtained via solving an auxiliarily optimization problem. We have proved that the obtained solution can be arbitrarily close to the infimal solution of the original constrained optimization problem. Moreover, solution algorithms are provided. Furthermore, an illustrative example is included to demonstrate the presented
-50 -40 -30 -20 -10 0 -25 -20 -15 -10 -5 0 5 10 15 20 25 Real Part Im ag in ar y P ar t
The constraint region Ω
Table I. The results (w =1) Fopt s (A c ) jjT z2 w2 jj2 jjT z1 w1 jj1 Case 1 1 :7705 0 :2080 6 :1090 0 :4566 2 :1091 3 :4309 0 :7 þ i 0 :3317 ; 0 :7 i 0 :3317 ; 1 3.9572 5.4430 Case 2 l1 =1 5 :8505 6 :1651 17 :2029 1 :0439 7 :5974 12 :8610 0 :5931 ; 2 :3139 ; 8 :8035 6.5900 0.7030 Case 3 wi =1 (i =1,2,3) 8 :7555 11 :5623 19 :1110 1 :1783 14 :2953 15 :4660 2 :4409 þ i 3 :5040 ; 2 :4409 i 3 :5040 ; 7 :1248 10.3340 1.3307 Case 4 l1 =1000, wi =1 (i =1,2,3) 5 :4808 11 :7021 19 :9334 5 :3472 18 :5253 19 :1625 3 :0456 þ i 0 :7725 ; 3 :0456 i 0 :7725 ; 12 :9847 7.9628 0.8508 Case 5 l1 =1, wi =1000 7 :7090 22 :3760 30 :6759 3 :2914 23 :3656 24 :7824 3 :2894 þ i 2 :4050 ; 3 :2894 i 2 :4050 ; 13 :8820 10.4491 0.9987 Case 6 l1 =10 3, wi =10 3 1 :6795 15 :2044 17 :0812 12 :3278 21 :2104 7 :8760 2 :8048 þ i 2 :0075 ; 2 :8048 i 2 :0075 ; 11 :4169 7.5411 0.9450 Case 7 l1 =10 5 , wi =10 5 1 :4664 5 :1435 9 :4430 5 :7785 10 :6113 7 :8760 2 :7411 þ i 1 :5956 ; 2 :7411 i 1 :5956 ; 3 :3215 6.1988 0.9959
approach. Although we only consider the state feedback case, in fact, the output feedback case can be solved without any difficult via the same procedure.
APPENDIX A
Here we provide a solution algorithm for the Q21pauxproblem. Note that the Fgrad(g,F) defined
in (13) is the gradient of Jaux(F) with respect to F under the constraints of (4)–(6) and (9) (see
chapter 1 of [22]). If Fopt is a minimal solution, then Fgrad(g,Fopt)=0. Based on the steepest
descent method, a solution algorithm is proposed below. Suppose e>0 is a specified small number. The algorithm will stop if jjFgradðFÞjj5e:
Algorithm 1
Find the optimal solution Foptof the auxiliary minimization problem.
(1) Choose a F(0) 2 G: Set k=0.
(2) Solving (12)–(19), by substituting F by F(k), yield Lo(k), L2(k), P1ðkÞ; L1ðkÞ; #PP1ðkÞ; #LL1ðkÞ; PiðkÞ; SiðkÞ:
(3) Find Fgradðg; FðkÞÞ:
(4) If jjFgradðg; FðkÞÞjj5e; then Fopt=F(k), end.
else find d(k)>0 via line search techniques such that Fðk þ 1Þ ¼ FðkÞ dðkÞ Fgradðg; FðkÞÞ shall minimize J(F(k+1)). Let k=k+1, go to (2).
The step (1) in the Algorithm 1 is not a trivial task. In the following, we will provide an algorithm to find a F 2 G: -16 -14 -12 -10 -8 -6 -4 -2 0 2 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Part Im ag in ar y P ar
t The constraint region Ω
Figure 2. The locations of the resultant closed-loop poles of the example (+: case 1, : case 2, *: case 3, *: case 4, ,: case 5, &: case 6,n: case 7).
Algorithm 2
Find F such that F 2 G: (1) Choose any Fð0Þ 2 GO:
If jjðC1þ D1Fð0ÞÞðsI A B3Fð0ÞÞ1B1jj15g; then let F=F(0), end.
else choose #ggð0Þ > jjðC1þ D1Fð0ÞÞðsI A B3Fð0ÞÞ1B1jj1 and set k=1, go to (2). (2) Solving (12)–(19), by substituting F by F(k) and g by #ggðkÞ; yield Lo(k), L2(k), P1ðkÞ;
L1ðkÞ; #PP1ðkÞ; #LL1ðkÞ; PiðkÞ; SiðkÞ: (3) Find Fgradð#ggðkÞ; FðkÞÞ:
(4) Find d(k)>0, via line search technique, such that Fðk þ 1Þ ¼ FðkÞ dðkÞ Fgradðg; FðkÞÞ shall minimize Jauxð#ggðkÞ; Fðk þ 1ÞÞ:
(5) Let *gg ¼ ðkÞjjðC1þ D1Fðk þ 1ÞÞðsI A B3Fðk þ 1ÞÞ1B1jj1: If *ggðkÞ5g; then F ¼ Fðk þ 1Þ 2 G; end
else set #ggðk þ 1Þ ¼ #ggðkÞ Zð#ggðkÞ *ggðkÞÞ; for 05Z51; and let k=k+1, go to (2).
ACKNOWLEDGEMENT
This work was supported by the National Science Council of the Republic of China under Grant NSC 89-2213-E-146-008.
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