Partial pole placement by LQ regulators an inverse problem approach

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  0 and R

feedback law

0

(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: ).
Publisher Item Identifier S 0018-9286(98)01442-1.

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  # 0 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 do place n 0 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 0 XBB T X + Q = 0:

(4)

This paper aims at finding K , which is optimal in this sense for some
weighting Q  0 and which will place n 0 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 0 A)01 B + D:

III. POLE PLACEMENT
We will use the following right coprime factorization by polynomial matrices [12]

(sI 0 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 0 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 8 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
0
1
(sI + 8)E (s)M (s) = I + K (sI 0 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)
0
1
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 0 A)01 B
= LB + LA(sI 0 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 0 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 4

(20)

8 + DL is symmetric

(21)

where

AL
0
T CL 0AT
0
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 0 BK )T 01 = A1 00A2 L1 08

(17)

and compute

W (s): = I + K (sI 0 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 0 BLA

B

= sI 0 V (s)
I
(A 0 BLA)B
:
LAB

0LA
A 0 BLA
LA

(25)

(26)

in (16)

T (A 0 BLA)T 01 = A0L B0L :
Equation (26) is hence reduced to

IV. LQ OPTIMALITY

Now let us find a condition on 8 under which
optimal for some weighting Q  0.
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 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.

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 0 A I 0
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 8 > 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 0 8) + 8T 8 > s(DL 0 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  0 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 0 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 + 8 = 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 0 A2 L1 is stable, then
2(s) is in RL . Hence, there exists a real number  such that

1

I  2(s) for all s = j!:
(31)
Then, (20) holds if 8T 8 = I + Q for any positive definite Q. This

is equivalent to

X T DL + DLT X 0 X T X 0 DLT DL + I + Q = 0
X : = 8 + 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 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.
T DL + I + Q > 0 and compute a
3) Take Q > 0 such that 0DL
positive definite solution X of (32). (X exists in this case.)
4) Calculate 8 = X 0 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
0 :952 6:05
0:174 4:31
0 01:76 0:416
0:999 0:0578 :0369 :0092 0:0012
0
0
0
0
0
0
0
0 010:

0
0
0
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
8 = X 0 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 0 B0 K0 are

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.
REFERENCES

A. Design Algorithm

0:746
:024
:
006
A0 =
1:
0
0
0 0
0 0
B0 = 00 00
20 0
0 10

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

(29)

Thus we have (21). Furthermore

8T 0 8 = DL 0 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 = 0 is

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

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