INFINITE EIGENVALUE ASSIGNMENT BY AN OUTPUT FEEDBACK FOR SINGULAR SYSTEMS
T
ADEUSZKACZOREK
∗∗
Institute of Control and Industrial Electronics Warsaw University of Technology ul. Koszykowa 75, 00–662 Warszawa, Poland
e-mail:
kaczorek@isep.pw.edu.plThe problem of an infinite eigenvalue assignment by an output feedback is considered. Necessary and sufficient conditions for the existence of a solution are established. A procedure for the computation of the output-feedback gain matrix is given and illustrated with a numerical example.
Keywords: infinite eigenvalue assignment, feedback, singular system
1. Introduction
It is well known (Dai, 1989; Kaliath, 1980; Wonham, 1979; Kaczorek, 1993; Kuˇcera, 1981) that if the pair (A, B) of a standard linear system ˙x = Ax + Bu is controllable then there exists a state-feedback gain ma- trix K such that det[I
ns − A + BK] = p(s), where p(s) = s
n+ a
n−1s
n−1+ · · · + a
1s + a
0is a given ar- bitrary n-th order polynomial. By changing K we may modify arbitrarily only the coefficients a
0, a
1, . . . , a
n−1but we are not able to change the degree n of the poly- nomial which is determined by the matrix I
ns. In sin- gular linear systems we are also able to change the de- gree of the closed-loop characteristic polynomials by a suitable choice of the state-feedback matrix K. The problem of finding a state-feedback matrix K such that det[Es − A + BK] = α 6= 0 (α is independent of s) was considered in (Kaczorek, 2003; Chu and Ho, 1999). The infinite eigenvalue assignment problem by a feedback is very important in the design of perfect observers (Kaczo- rek, 2000; 2002; 2003).
In this paper the problem of an infinite eigenvalue as- signment by an output feedback is formulated and solved.
This is an extension of the method given in (Kaczo- rek, 2003) for an output feedback case. Necessary and sufficient conditions for the existence of a solution to the problem will be established and a procedure for the computation of an output-feedback gain matrix will be presented.
2. Problem Formulation
Let R
n×mbe the set of real n × m matrices and R
n:=
R
n×1. Consider the continuous-time linear system E ˙ x = Ax + Bu, y = Cx, (1) where ˙ x = dx/dt and x ∈ R
n, u ∈ R
mand y ∈ R
pare respectively the semistate, input and output vectors.
Moreover, E, A ∈ R
n×n, B ∈ R
n×m, C ∈ R
p×n. The system (1) is called singular if det E = 0 and it is called standard when det E 6= 0.
It is assumed that rank E = r < n, rank B = m, rank C = p and the pair (E, A) is regular, i.e.
det[Es − A] 6= 0 (2)
for some s ∈ C (the field of complex numbers). Let us consider the system (1) with the output feedback
u = v − F y, (3)
where v ∈ R
mis a new input and F ∈ R
m×pis a gain matrix. From (1) and (3) we have
E ˙ x = (A − BF C)x + Bv. (4)
Problem 1. Given matrices E, A, B, C of (1) and a nonzero scalar α (independent of s), find an F ∈ R
m×psuch that
det[Es − A + BF C] = α. (5)
In this paper necessary and sufficient conditions for the existence of a solution to Problem 1 will be established and a procedure for the computation of F will be pro- posed.
3. Problem Solution
From the equality
Es − A + BF C = [Es − A, B]
"
I
nF C
#
= [I
n, BF ]
"
Es − A C
# (6)
and (5) it follows that Problem 1 has a solution only if
rank [Es − A, B] = n (7)
and
rank
"
Es − A C
#
= n (8)
for all finite s ∈ C. The problem will be solved using the following two-step procedure:
Step 1. (Subproblem 1). Given E, A, B of (1) and a scalar α, find a matrix K = F C such that
det[Es − A + BK] = α. (9)
Step 2. (Subproblem 2). Given C and K depending on some free parameters k
1, k
2, . . . , k
l(found in Step 1), find a matrix F satisfying the equation
K = F C. (10)
The solution of Subproblem 1 is based on the following lemma (Chu and Ho, 1999; Kaczorek, 2003):
Lemma 1. If the condition (2) is satisfied, then there exist orthogonal matrices U and V such that
U [Es − A]V =
"
E
1s − A
1∗ 0 E
0s − A
0# ,
U B =
"
B
10
#
, (11a)
where E
1, A
1∈ R
n1×n1, E
0, A
0∈ R
n0×n0, B
1∈ R
n1×m, the subsystem (E
1, A
1, B
1) is completely con- trollable, the pair (E
0, A
0) is regular, E
1is upper trian- gular and ‘∗’ denotes an unimportant matrix. Moreover,
the matrices E
1, A
1and B
1are of the form
E
1s − A
1=
E
11s − A
11E
12s − A
12−A
21E
22s − A
220 −A
32. . . .
0 0
· · · E
1,k−1s − A
1,k−1E
1ks − A
1k· · · E
2,k−1s − A
2,k−1E
2ks − A
2k· · · E
3,k−1s − A
3,k−1E
3ks − A
3k. . . .
· · · 0 −A
k,k−1E
kks − A
kk
,
B
1=
B
110 .. . 0
, (11b)
where E
ij, A
ij∈ R
n¯iׯnj, i, j = 1, . . . , k, B
11∈ R
n¯i×m, P
ni=1