For the third situation, noting that
( P 12 0 − εP 11) TP−1( P
11
12 0
− εP 11) ≥ 0
(20)
we have
PT
12
P−1 P
+ ε P
0
11
12 0
≥ ε PT 12
0
12 0
− ε 2 P 11
(21)
(21) implies
(
(
(
PT
1) T + ε 1) − ε 2 1)
12 P−1 P
P
P
(22)
11
12 ≥ εP 12
12
11
Output Feedback Control of Discrete-time LTI Systems: Scaling LMI Approaches
145
Hence we complete the proof.
Remark 1. If ε ≡ 0 is set , then Theorem 1 recovers the result stated in (Bara & Boutayeb, 2006). We
shall note that ε actually plays an important role in the scaling LMI formulation in Theorem 1. If ε ≡ 0 ,
Theorem 1 implies AT 22 P 22 A 22 − P 22 < 0 and P 22 > 0 , i.e., the system matrix A 22 must be Schur stable, which obviously is an unnecessary condition and limits the application of this LMI formulation.
However, with the aid of ε, we relax this constraint. A searching routine, such as fminsearch (simplex
search method) in Matlab ©, can be applied to the following optimization problem (for a fixed ε, we have
an LMI problem):
min λI, s. t. Φ(Θ) < λI
(23)
ε, P, R
The conservatism of Theorem 1 lies in these relaxations (15) or (16) on (5). To further relax the
conservatism, we may choose a diagonal matrix
= diag{ ε 1, ..., εm}, εi ≥ 0 , instead of the single
scalar ε. For example,
PT
+
12 P−1 P
P
11
12 ≥ PT
12
12 −
P 11
(24)
Then we shall search the optimal value over multiple scalars for (23).
Remark 2. In (Bara & Boutayeb, 2006), a different variable replacement is given:
P 2 = P 22 − PT
12 P−1 P
11
12
(25)
in (8). However, it is easily proved that these two transformations actually are equivalent. In fact, in
(8), we have P 11 > 0 and P 2 > 0 since P > 0 . Based on (17), we have
⎡
⎤
⎣ AT 0 0 A − Λ
0 P
0
∗ ⎦
2
< 0
(26)
P 11 KC + [ P 11 P 12] A − P 11
where
Λ0 = P 11
P 12
= P
(27)
PT P
P−1 P
12
2 + PT
12 11
12
Hence, for the above three situations, we have an alternative condition, which is stated in the
following lemma.
Theorem 2. The discrete-time system (1)-(2) is stabilized by (3) if there exist P 11 > 0 , P 2 > 0 , P 12
and R with P defined in (27), such that
⎧
⎨ Υ(Λ1) < 0, m = n − m
⎩ Υ(Λ2) < 0, m < n − m
(28)
Υ(Λ3) < 0, m > n − m
where ε ∈ R ,
⎡
⎤
Υ(Λ
⎣ AT 0 0 A − Λ i ∗ ⎦
i) =
0 P 2
,
RC + [ P 11 P 12] A − P 11
Λ1 = P 11
P 12
,
PT P
12
2 − ε 2 P 11 + εP 12 + εPT
12
146
Discrete Time Systems
⎡
⎤
P 11
P 12
Λ
⎣
⎦
2 =
,
PT P
+ ε[ PT 0] − ε 2 P 11 0
12
2 + ε
P 12
0
12
0 I
P
Λ
11
P 12
3 =
(
(
(
.
PT P
1) T + εP 1) − ε 2 P 1)
12
2 + εP 12
12
11
Furthermore, a static output controller gain is given by (13).
Proof: We only consider the first case. Replacing P 2 and R by P 22 and K using (25) and (13), we
can derive that (28) is a sufficient condition for (5) with the P defined in (8).
3.2 Co with full row-rank
When Co is full row rank, there exists a nonsingular matrix Tc such that CoT−1
o
= [ Il 0].
Applying a similarity transformation to the system (1)-(2), the closed-loop system (4) is stable
if and only if
˜
Ac = A + BKC is stable
where A = Tc AoT−1
c
, B = TcBo and C = CoT−1
c
= [ Il 0].
Similarly to Section 3.1, we can also divide this problem into three situations: l = n − l,
l < n − l and l > n − l. We use the condition (6) here and partition Q as Q =
Q 11 Q 12 ,
QT 12 Q 22
where Q 11 ∈ R l× l.
Theorem 3. The discrete-time system (1)-(2) is stabilized by (3) if there exist Q > 0 and R, such that
⎧
⎨ Γ( ¯Θ1) < 0, l = n − l
Γ( ¯Θ
⎩
2) < 0, l < n − l
(29)
Γ( ¯Θ3) < 0, l > n − l
where ε ∈ R ,
Γ( ¯Θ i) =
A ¯
Θ iAT − Q
∗
(
,
A[ Q 11 Q 12] T + BR) T − Q 11
¯
Θ1 = 0
0
,
0 Q 22 + ε 2 Q 11 − εQ 12 − εQT 12
⎡
⎤
0
0
¯
Θ
⎣
⎦
2 =
,
0 Q 22 + ε 2 Q 11 0 − ε Q 12 − ε[ QT
0 0
0
12 0]
¯
Θ3 = 0
0
(
(
(
,
0 Q
1)
1) T
1)
22 + ε 2 Q
− εQ
− εQ
11
12
12
(
(
Q 1) and Q 1) are properly dimensioned partitions of Q
11
12
11 and Q 12 . Furthermore, a static output
feedback controller gain is given by
K = RQ−1
(30)
11
Output Feedback Control of Discrete-time LTI Systems: Scaling LMI Approaches
147
Proof: We only prove the first case l = n − l, since the others are similar. Noting that
( BKC) Q( BKC) T = BKQ 11 KTB and BKCQ = BK[ Q 11 Q 12], (6) is equivalent to ( A[ Q 11 Q 12] T + BKQ 11) Q−1( A[ Q
11
11 Q 12] T + BKQ 11) T
−
Q
(31)
A
11
Q−1[ Q
QT
11
11 Q 12] AT + AQAT − Q < 0
12
Using the fact that
Q − Q 11 Q−1[ Q
QT
11
11 Q 12] =
0
0
12
0 QT 12 Q−1 Q
11
12
we infer that stability of the close-loop system is equivalent to the existing of a Q > 0 such
that
A ¯
Θ0 AT − Q
∗
(
< 0
(32)
A[ Q 11 Q 12] + BKQ 11) T − Q 11
where
¯
Θ0 = 0
0
0 Q 22 − QT Q−1 Q
12
11
12
Since
( Q 12 − εQ 11) TQ−1( Q
11
12 − εQ 11) ≥ 0
(33)
or equivalently,
QT
+ ε
12 Q−1 Q
Q
11
12 ≥ εQT
12
12 − ε 2 Q 11
(34)
It follows that (29) implies (32). Hence we complete the proof.
Remark 3. How to compare the conditions in Theorem 3 and Theorem 1 remains a difficult problem.
In the next section, we only give some experiential results based on numerical simulations, which give
some suggestions on the dependence of the results with respect to m and l.
3.3 Transformation-dependent LMIs
The result in this subsection builds a connection between the sets L, K c, K o, ˜
K c and ˜
K o, which
are defined as follows. Without causing confusion, we omit the subscript o for Ao, Bo and Co
in this subsection.
L = { K ∈ R m× l : ¯ A stable}
(35)
i.e., the set of all admissible output feedback matrix gains;
K c = { Kc ∈ R m× n : A + BKc stable}
(36)
i.e., the set of all admissible state feedback matrix gains;
K o = { Ko ∈ R n× l : A + KoC stable}
(37)
i.e., the set of all admissible observer matrix gains. Based on Lemma 1, we can easily formulate
the LMI solution for sets K c and K o. In fact, they are equivalent to following two sets
respectively:
˜
K c = { Kc = Wc 2 W−1 ∈ R m× n : ( W
c 1
c 1, Wc 2) ∈ W c}
(38)
148
Discrete Time Systems
and
W c = { Wc 1 ∈ R n× n, Wc 2 ∈ R m× n : Wc 1 > 0, Ψ c < 0}
(39)
where Ψ c =
− Wc 1
AWc 1 + BWc 2 .
Wc 1 AT + WT
c 2 BT
− Wc 1
˜
K o = { Ko = W−1 W
o 1
o 2 ∈ R n× l : ( Wo 1, Wo 2) ∈ W o}
(40)
and
W o = { Wo 1 ∈ R n× n, Wo 2 ∈ R n× l : Wo 1 > 0, Ψ o < 0}
(41)
where Ψ o =
− W 1 o
Wo 1 A + Wo 2 C .
ATWo 1 + CTWT
−
o 2
W 1 o
Lemma 2. L = ∅ if and only if
1. ¯
K c = K c { Kc : KcYc = 0, Yc = N ( C)} = ∅ ; or
2. ¯
K o = K o { Kc : YoKo = 0, Yo = N ( B )} = ∅ .
In the affirmative case, any K ∈ L can be rewritten as
1. K = KcQCT( CQCT)−1 ; or
2. K = ( BT PB)−1 BT PKo.
where Q > 0 and P > 0 are arbitrarily chosen.
Proof: The first statement has been proved in Geromel et al. (1996). For complement, we
give the proof of the second statement. The necessity is obvious since Ko = BK. Now we
prove the sufficiency, i.e., given Ko ∈ ¯
K o, there exists a K, such that the constraint Ko = BK
is solvable. Note that for ∀ P > 0, Θ o =
BT P
is full rank, where Y
YT
o = N ( BT ). In fact,
o
rank(Θ oYo) = rank( BTPYo ) ≥ n − m. Multiplying Θ
I
o at the both side of Ko = BK we have
n− m
BTPKo = BTPBL
YT
o Ko
0
Since BTPB is invertible, we have K = ( BTPB)−1 BTPK 0. Hence, we can derive the result.
Lemma 3. L = ∅ if and only if there exists Ec ∈ R n×( n− l) or Eo ∈ R n×( n− m) , such that one of the following conditions holds:
1. rank( Tc =
C
) = n and C( E
ET
c) = ∅ ; or
c
2. rank( To = B Eo ) = n and O( Eo) = ∅ .
where
C( Ec) = W c {( Wc 1, Wc 2) : CWc 1 Ec = 0, Wc 2 Ec = 0}
O( Eo) = W o {( Wo 1, Wo 2) : BTWo 1 Eo = 0, EToWo 2 = 0}
In the affirmative case, any K ∈ L can be rewritten as
1. K = Wc 2 CT( CWc 1 CT)−1 ; or
2. K = ( BTWo 1 B)−1 BTWo 2 .
Output Feedback Control of Discrete-time LTI Systems: Scaling LMI Approaches
149
Proof: We only prove the statement 2, since the statement 1 is similar. For the necessity, if there
exist K ∈ L, then it shall satisfy Lemma 1. Now we let
Wo 1 = P, Wo 2 = PBK
Choose Eo = P−1 Yo, Yo = N ( BT). It is known that B Eo is full rank. Then we have
BTWo 1 E = BTYo = 0, ETWo 2 = YT
o BK = 0
For sufficiency, we assume there exists Eo such that the statement 2) is satisfied. Notice that
Wo 1 > 0 and the item Wo 2 in Ψ o can be rewritten as Wo 1 W−1 W
o 1
o 2.
W−1 W
o 1
o 2 = To ( TT
o Wo 1 To )−1 TT
o Wo 2 = B( BTWo 1 B)−1 BT Wo 2
(42)
since To is invertible and BTWo 1 E = 0, ETWo 2 = 0. Hence, W−1 W
o 1
o 2 can be factorized as
BK, where K = ( BTWo 1 B)−1 BTWo 2. Now we can derive (5) from the fact Ψ o < 0. Thus we
complete the proof.
Remark 4. For a given To,
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