706 IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL 43, NO 5, MAY 1998 Partial Pole Placement by LQ Regulators: An Inverse Problem Approach II PRELIMINARIES Consider the system _ = Ax + Bu x Kenji Sugimoto Abstract—This paper gives a necessary and sufficient condition under which a state feedback control law places part of the closed-loop poles exactly at specified points and, at the same time, is linear quadratic optimal for some quadratic weightings This is made possible by means of a solution to the inverse problem of optimal control A design example is given to illustrate the result Index Terms— Inverse problem of optimal control, LQ control, pole placement where x and u are n- and m-dimensional vector signals, respectively Throughout the paper, we assume that (A; B ) is controllable and B is of full column rank If, for the above system, we are given the performance index with Q and R feedback law (xT Qx + uT Ru) dt (2) = I , then the optimal control is given by a state u Manuscript received June 28, 1996 The author is with Department of Aerospace Engineering, Graduate School of Engineering, Nagoya University, Furo-cho, Chikusa-ku, Nagoya, 464-01 Japan (e-mail: sugimoto@suzu.nuae.nagoya-u.ac.jp) Publisher Item Identifier S 0018-9286(98)01442-1 ( )= J u I INTRODUCTION Linear quadratic (LQ) regulation is widely used in designing feedback systems It is, however, often very difficult to select suitable quadratic weightings of a performance index A more direct specification describing transient responses is closed-loop pole configuration Hence, relationships between the weight selection and pole placement have extensively been studied; see [1], [2], [4], [5], and [7]–[11], just to name a few References [1], [2], [4], [10], etc have studied successive methods and algorithms shifting the entire set of poles or a pair of complex conjugate poles or a single real pole by LQ regulators, but this is restrictive as a design method In [7], [9], and [11], an LQ regulator is designed so that all closed-loop poles are placed inside a given region The idea of this regional pole placement is interesting, particularly from a robustness point of view In the present paper, however, we aim at placing some dominant poles at specified points rather than all poles in one region This is because in many cases, some poles mainly affect the response, provided that the remaining poles lie far enough along the real negative half-line In this context, it is well known [3], [8] that as # for a control weighting R = I with a fixed state weighting Q 0, some of the closed-loop poles tend to the invariant zeros of the system (A; B; Q1=2 ), while the others tend to infinity (cheap control) This method enables us to achieve partial pole placement quite freely by adjusting Q However, this is done only asymptotically, not exactly It seems to be a common belief that poles can never be placed in such a freedom without resorting to high-gain LQ regulators In this paper, we place n m poles exactly at specified points by a finite LQ regulator, where n and m are the dimensions of the state and the input vectors, respectively We give a necessary and sufficient condition under which a state feedback attains this partial pole placement, while at the same time it is LQ optimal for some weightings This is made possible by using a solution to the inverse problem of LQ optimal control (for a detailed study of this problem, see [3], [6], and references therein) Designing LQ regulators based on solutions of the inverse problem was originally proposed by Fujii [5] He placed the closed-loop poles asymptotically, as in [3] and [8] (1) = 0Kx; K = BT X (3) where X is a minimal solution of the Riccati equation XA + AT X XBB T X + Q = 0: (4) This paper aims at finding K , which is optimal in this sense for some weighting Q and which will place n m poles exactly at any specified points Notation: For a constant matrix A, AT denotes its transpose For a rational matrix W (s), W (s): = W T (0s) We frequently use the notation A B C D : = C (sI A)01 B + D: III POLE PLACEMENT We will use the following right coprime factorization by polynomial matrices [12] (sI A)01 B = P (s)M (s)01 : (5) In the actual calculation, however, we will not have to compute ( ) and M (s) explicitly The state-space data (A; B ) will be used directly Given the fractional representation (5), it is well known that any state feedback u = 0Kx induces another right coprime factorization P s (sI A + BK )01 B = P (s)(M (s) + KP (s))01 (6) which means that the closed-loop properties are characterized in terms of the denominator polynomial matrix M (s) + KP (s) Our first objective is to find a K such that ( ) + KP (s) = (sI + 8)E (s) M s (7) for a given constant matrix and a polynomial matrix E (s) Namely, we factorize the denominator polynomial matrix into the two factors We will then design the factor E (s) to place poles exactly and sI +8 to guarantee LQ optimality Lemma 1: If (7) holds, then ( ) = LP (s) E s for some L such that LB 0018–9286/98$10.00 1998 IEEE = I (8) IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL 43, NO 5, MAY 1998 M (s)01 , we have (sI + 8)E (s)M (s) = I + K (sI A)01B: Proof: Post-multiplying (7) by (9) 707 L Theorem: For (14), consider the feedback given by (15) Then, (19) holds iff Hence, we have sE (s)M (s)01 ! I (s ! 1): (10) Since E (s)M (s) is strictly proper, the existence of L in (8) follows from the structure theorem of Wolovich [12] Then, from (8) and (5) we have sE (s)M (s)01 = sL(sI A)01 B = LB + LA(sI A)01 B: Thus LB = I holds by (10) Lemma 2: Assume LB = I Then the gain K satisfies (7) (8) iff K = LA + 8L: Proof—Necessity: Note that from (5) we have (sI A)P (s) = BM (s): Premultiplying this with because (13) LB = I Now assume (7) and (8) Then M (s) + KP (s) = sLP (s) + 8LP (s) = M (s) + (LA + 8L)P (s) by (13) Hence (12) holds because of controllability Sufficiency is readily shown by direct calculation Let us consider the role of (12) in the state space With no loss of generality, we assume that A= B = I0 A1 A2 ; A3 A (20) + DL is symmetric (21) where AL T CL 0AT C L L 2(s):= (14) with block element matrices of compatible sizes (Otherwise we can use a suitable similarity transformation on the system.) Then, LB = I iff and W (s) = (M (s) + KP (s))M (s)01 = (sI + 8)E (s)M (s)01: (sI + 8) (sI + 8) > fM (s)E (s)01g M (s)E (s)01 for all s = j!: Let us compute the right-hand side In view of (11) and we have = LI1 I0 (16) =s A2 : T (A BK )T 01 = A1 00A2 L1 08 (17) and compute W (s): = I + K (sI A)01 B: Then it is well known that the condition W (s)W (s) > I for all s = j! V (s): = By using (14) and T LB (24) = I, B 01 A LA I A BLA B = sI V (s) I (A BLA)B : LAB 0LA A BLA LA (25) (26) in (16) T (A BLA)T 01 = A0L B0L : Equation (26) is hence reduced to IV LQ OPTIMALITY Now let us find a condition on under which optimal for some weighting Q Define the return difference matrix (23) Condition (19) is then equivalent to (15) Hence, the m closed-loop poles are placed as the eigenvalues of 08, and the remaining n m poles are specified by L1 ; they are given as a solution of the pole placement problem for the pair (A1 ; A2 ) In the remainder of the paper, we assume the canonical form (14) for simplicity RL Remark: As opposed to a weaker condition with “>” replaced by “,” (19) is not necessary for optimality, yet fairly close to it; see [6] Hence, there is little loss of generality in requiring this condition Proof of the Theorem: Substituting (5) into (18) and using (7), we have L = (L1 I ): T (22) AL BL = I A I CL DL L1 I 0L1 I SL = CL AL + DLT CL RL = DLT DL + (CL BL )T + CL BL : M (s)E (s)01 = s Now take BL 0SLT BLT SL (12) L, we have sLP (s) = LAP (s) + M (s) 8T > 2(s) for all s = j! and (11) with u = 0Kx in (12) with K in (12) is LQ V (s) = AL BL CL DL : (18) Substitute (25) and (27) into (24) Then (19) s(8T 8) + 8T > s(DL DLT ) + V (s)V (s) AL AL BL AL gives a solution to the inverse problem as follows: K is LQ optimal for some (unknown) weighting Q if (19) holds [3], [6] The following theorem guarantees optimality of K by using this fact + + (27) AL BL CL CL BL CL CL BL T = s(DL DL ) + 2(s) for all s = j!: (28) 708 IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL 43, NO 5, MAY 1998 Note that this is equivalent to (19) in the present case Now assume this condition Then K is optimal for some weighting [3], [6] Since K is written as (3) for some Riccati solution X , KB must be symmetric In view of (12), (14), and (15) KB = L1 A2 + A4 + = DL + 8: and hence (30) and hence (28) reduces to (20) Conversely, assume (20) and (21) Then we have (28) directly, and hence (19) holds Now we are ready to state our design method Note that 2(s) is a para-Hermite proper rational matrix If A1 A2 L1 is stable, then 2(s) is in RL Hence, there exists a real number such that I 2(s) for all s = j!: (31) Then, (20) holds if 8T = I + Q for any positive definite Q This is equivalent to X T DL + DLT X X T X DLT DL + I + Q = X : = + DL X is symmetric iff (21) holds The (32) design where problem is thus reduced to finding a stabilizing solution to this degenerate Riccati equation for possibly sign-indefinite coefficients These observations lead to the following 1) Design L1 placing n m poles at specified points in the open left half-plane; see Section III 2) Compute a minimal satisfying (31) via, say, a bisection method T DL + I + Q > and compute a 3) Take Q > such that 0DL positive definite solution X of (32) (X exists in this case.) 4) Calculate = X DL Then, the desired gain K is given by (12) and (15) V AN EXAMPLE To illustrate our design method, we adopt the same flight control problem for the F-4 fighter as treated in [5] and [8] The system is described by the state-space matrices :387 012:9 :952 6:05 0:174 4:31 01:76 0:416 0:999 0:0578 :0369 :0092 0:0012 0 0 0 0 010: 0 0 05: : We first make a similarity transformation A: = V 01 A0 V; B : = V 01 B0 V : = diag(1; 1; 1; 1; 20; 10) so that (14) holds According to the eigenstructure specification in [5] and [8], we have L1 = :59 :38 = X DL = 17 0:38 17:59 : Then we can readily obtain the gain K by (12), and our gain in the original coordinates is K0 : = KV 01 This gain satisfies (19) and hence is LQ optimal for some weighting Now the eigenvalues of A0 B0 K0 are 04:00; 0:63 2:42j; 0:05; 017:59 :38j: We observe that the four poles are placed at the points specified in [5] and [8] exactly The remaining two poles are not so large in magnitude, compared with [5] and [8], which means that our gain is not very high gain Note that the accuracy of pole placement largely affects transient responses, especially when some are close to the origin, as in the above example Note also that if we place all poles in one specific region, we will have entirely different responses REFERENCES A Design Algorithm 0:746 :024 : 006 A0 = 1: 0 0 0 B0 = 00 00 20 0 10 :55 :115 X = 8:115 15:9 (29) Thus we have (21) Furthermore 8T = DL DLT which coincides with F1 in the expression [5, eq (6.1)] with coordinate changes taken into account By means of a numerical search, we obtain = 309:4 satisfying (31) A solution to the Riccati equation (32) for Q = is 0:0064 0:0305 :0822 :0008 :0566 :0154 0:2393 :0031 [1] A T Alexandridis and G D Galanos, “Optimal pole-placement for linear multi-input controllable systems,” IEEE Trans Circuits Syst., vol CAS-34, pp 1602–1604, 1987 [2] M H Amin, “Optimal pole shifting for continuous multivariable linear systems,” Int J Contr., vol 41, pp 701–707, 1985 [3] B D O Anderson and J B Moore, Optimal Control: Linear Quadratic Methods Englewood Cliffs, NJ: Prentice Hall, 1990 [4] W L Eastman and J A Bossi, “Design of linear quadratic regulators with assigned eigenstructure,” Int J Contr., vol 39, pp 731–742, 1984 [5] T Fujii, “A new approach to the LQ design from the viewpoint of the inverse regulator problem,” IEEE Trans Automat Contr., vol AC-32, pp 995–1004, 1987 [6] T Fujii and M Narazaki, “A complete optimality condition in the inverse problem of optimal control,” SIAM J Contr Optim., vol 22, pp 327–341, 1984 [7] K Furuta and S B Kim, “Pole assignment in a specified disk,” IEEE Trans Automat Contr., vol AC-32, pp 423–427, 1987 [8] C A Harvey and G Stein, “Quadratic weights for asymptotic regulator properties,” IEEE Trans Automat Contr., vol AC-23, pp 378–387, 1978 [9] S O R Moheimani and I R Petersen, “Quadratic guaranteed cost control with robust pole placement in a disk,” Proc Inst Elec Eng Contr Theory Appl., vol 143, pp 37–43, 1996 [10] K Sugimoto and Y Yamamoto, “On successive pole assignment by linear quadratic optimal feedbacks,” Lin Alg Its Appl., vol 122–124, pp 697–724, 1989 [11] B Wittenmark, R J Evans, and Y C Soh, “Constrained Poleplacement using Transformation and LQ-design,” Automatica, vol 23, pp 767–769, 1987 [12] W A Wolovich, Linear Multivariable Systems New York: SpringerVerlag, 1974 ... Hence, the m closed-loop poles are placed as the eigenvalues of 08, and the remaining n m poles are specified by L1 ; they are given as a solution of the pole placement problem for the pair (A1... (12) is LQ V (s) = AL BL CL DL : (18) Substitute (25) and (27) into (24) Then (19) s(8T 8) + 8T > s(DL DLT ) + V (s)V (s) AL AL BL AL gives a solution to the inverse problem as follows: K is LQ optimal... eigenstructure,” Int J Contr., vol 39, pp 731–742, 1984 [5] T Fujii, “A new approach to the LQ design from the viewpoint of the inverse regulator problem, ” IEEE Trans Automat Contr., vol AC-32, pp 995–1004,