Partial pole placement by LQ regulators an inverse problem approach

Partial Pole Placement by LQ Regulators:An Inverse Problem Approach Kenji Sugimoto Abstract—This paper gives a necessary and sufficient condition under which a state feedback control law

Partial Pole Placement by LQ Regulators:

An Inverse Problem Approach

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.

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

specifica-tion describing transient responses is closed-loop pole configuraspecifica-tion

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  # 0 for a

control weightingR = 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

adjustingQ 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 do placen 0 m poles exactly at specified points

by a finite LQ regulator, wheren 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]

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.

II PRELIMINARIES Consider the system

wherex 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

J(u) = 1

0 (xTQx + uTRu) dt (2) withQ  0 and R = I, then the optimal control is given by a state feedback law

whereX is a minimal solution of the Riccati equation

XA + ATX 0 XBBTX + Q = 0: (4) This paper aims at findingK, which is optimal in this sense for some weightingQ  0 and which will place n 0 m poles exactly at any specified points

Notation: For a constant matrixA, AT denotes its transpose For

a rational matrixW (s), W (s): = WT(0s) We frequently use the notation

C D : = C(sI 0 A)

01B + D:

III POLE PLACEMENT

We will use the following right coprime factorization by polyno-mial matrices [12]

(sI 0 A)01B = P (s)M(s)01: (5)

In the actual calculation, however, we will not have to compute

P (s) 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 feedbacku = 0Kx induces another right coprime factorization (sI 0 A + BK)01B = 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 aK such that

M(s) + KP (s) = (sI + 8)E(s) (7) for a given constant matrix8 and a polynomial matrix E(s) Namely,

we factorize the denominator polynomial matrix into the two factors

We will then design the factorE(s) to place poles exactly and sI +8

to guarantee LQ optimality

Lemma 1: If (7) holds, then

for someL such that LB = I

0018–9286/98$10.00  1998 IEEE

Proof: Post-multiplying (7) byM(s)01, we have

(sI + 8)E(s)M(s)01= I + K(sI 0 A)01B: (9)

Hence, we have

sE(s)M(s)01! I (s ! 1): (10)

SinceE(s)M(s)01is strictly proper, the existence ofL in (8) follows

from the structure theorem of Wolovich [12] Then, from (8) and (5)

we have

sE(s)M(s)01= sL(sI 0 A)01B

= LB + LA(sI 0 A)01B: (11) ThusLB = I holds by (10)

Lemma 2: AssumeLB = I Then the gain K satisfies (7) with

(8) iff

Proof—Necessity: Note that from (5) we have

(sI 0 A)P (s) = BM(s):

Premultiplying this withL, we have

sLP (s) = LAP (s) + M(s) (13) becauseLB = 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 = AA1 A2

with block element matrices of compatible sizes (Otherwise we can

use a suitable similarity transformation on the system.) Then,LB = I

iff

Now take

T = LI 0

and compute

T (A 0 BK)T01= A10 A2L1 A2

Hence, them closed-loop poles are placed as the eigenvalues of 08,

and the remainingn 0 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

IV LQ OPTIMALITY Now let us find a condition on8 under which K in (12) is LQ

optimal for some weighting Q  0

Define the return difference matrix

W (s): = I + K(sI 0 A)01B: (18) Then it is well known that the condition

W (s)W(s) > I for all s = j! (19)

gives a solution to the inverse problem as follows:K is LQ optimal

for some (unknown) weightingQ  0 if (19) holds [3], [6] The

following theorem guarantees optimality ofK by using this fact

Theorem: For (14), consider the feedbacku = 0Kx in (12) with

L given by (15) Then, (19) holds iff

8T8 > 2(s) for all s = j! (20) and

where

2(s): =

0CT

LCL 0AT

L 0ST

L

(22)

and

AL BL

CL DL = LI 0

1 I A 0LI1 0I

SL= CLAL+ DT

LCL

RL= DT

LDL+ (CLBL)T+ CLBL:

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

W (s) = (M(s) + KP (s))M(s)01

= (sI + 8)E(s)M(s)01: (23) Condition (19) is then equivalent to

(sI + 8) (sI + 8) > fM(s)E(s)01g M(s)E(s)01

Let us compute the right-hand side In view of (11) and LB = I,

we have

M(s)E(s)01= s A B

01

= s A 0 BLA B

(25)

V (s): = A 0 BLA (A 0 BLA)B

(26)

By using (14) and T in (16)

T (A 0 BLA)T01= A0L B0L : Equation (26) is hence reduced to

V (s) = AL BL

CL DL

Substitute (25) and (27) into (24) Then s(8T0 8) + 8T8 > s(DL0 DT

L) + V (s)V (s) + AL ALBL

CL CLBL

+ AL ALBL

CL CLBL

= s(DL0 DT

L) + 2(s) for all s = j!: (28)

Note that this is equivalent to (19) in the present case Now assume

this condition ThenK 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 = L1A2+ A4+ 8 = DL+ 8: (29)

Thus we have (21) Furthermore

8T0 8 = DL0 DT

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 that2(s) is

a para-Hermite proper rational matrix IfA10 A2L1is stable, then

2(s) is in RL1 Hence, there exists a real number such that

I  2(s) for all s = j!: (31) Then, (20) holds if8T8 = I + Q for any positive definite Q This

is equivalent to

XTDL+ DT

LX 0 XTX 0 DT

LDL+ I + Q = 0 (32) whereX: = 8 + DL X is symmetric iff (21) holds The design

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

A Design Algorithm

1) DesignL1placingn 0 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

3) TakeQ > 0 such that 0DT

LDL+ I + Q > 0 and compute a positive definite solutionX of (32) (X exists in this case.)

4) Calculate 8 = X 0 DL Then, the desired gainK 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

A0=

:006 0:999 0:0578 :0369 :0092 0:0012

B0=

20 0

0 10

:

We first make a similarity transformation

A: = V01A0V; B: = V01B0

V : = diag(1; 1; 1; 1; 20; 10)

so that (14) holds According to the eigenstructure specification in

[5] and [8], we have

L1= 0:0064 0:0305:0566 :0154 0:2393 :0031:0822 :0008

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) forQ = 0 is

X = 8:55 :115:115 15:9 and hence

8 = X 0 DL= 17:590:38 17:59 ::38 Then we can readily obtain the gainK by (12), and our gain in the original coordinates is K0: = KV01 This gain satisfies (19) and

hence is LQ optimal for some weighting

Now the eigenvalues ofA00 B0K0are 04:00; 0:63 6 2:42j; 0:05; 017:59 6 :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

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