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XIE et al.: CONTROL SYNTHESIS OF DISCRETE-TIME T–S FUZZY SYSTEMS 3
the following equality holds true:
P =
f
j
∈K
g
j
,
j∈{1,2,...,m+1}
⎧
⎨
⎩
⎛
⎝
m+1
j=1
α
f
j
j
⎞
⎠
⎛
⎝
m+1
j=1
g
j
!
π
f
j
⎞
⎠
P
⎫
⎬
⎭
(5)
where α
f
j
j
=
r
i=1
α
f
j
i
ji
.
Proof: Considering the fact that
r
i=1
α
ji
= 1, one gets
P =
⎛
⎝
m+1
j=1
α
j1
+···+α
jr
g
j
⎞
⎠
P
=
f
j
∈K
g
j
,
j∈{1,2,...,m+1}
⎧
⎨
⎩
⎛
⎝
m+1
j=1
α
f
j
j
⎞
⎠
⎛
⎝
m+1
j=1
φ
j
P
⎞
⎠
⎫
⎬
⎭
(6)
where φ
j
= C
f
j
1
g
j
C
f
j
2
g
j
−f
j
1
···C
f
j
r
g
j
−f
j
1
−f
j
2
−···−f
j
r−1
.
Employing the fact that g
j
= f
j
1
+ f
j
2
+···+f
j
r
, one has
C
f
j
1
g
j
C
f
j
2
g
j
−f
j
1
···C
f
j
r
g
j
−f
j
1
−f
j
2
−···−f
j
r−1
=
g
j
!
π
f
j
. (7)
Then, from (6) and (7), one gets
P =
f
j
∈K
g
j
,
j∈{1,2,...,m+1}
⎧
⎨
⎩
⎛
⎝
m+1
j=1
α
f
j
j
⎞
⎠
⎛
⎝
m+1
j=1
g
j
!
π
f
j
⎞
⎠
P
⎫
⎬
⎭
. (8)
From (8), it is clear to find that the equality (5)
holds true.
Remark 1: Under the framework of multiinstant homoge-
nous matrix polynomials, the relaxation quality of obtained
stabilization conditions could be further improved, which
presents a multipolynomial dependence on the normalized
fuzzy weighting functions at the current and past instants of
time with some prescribed degrees. However, there always
exist some terms that are constant or with lower degrees,
which are difficult to be integrated into the framework of
multiinstant homogenous matrix polynomials. Indeed, the
result of Lemma 2 will play an important role in the main
proof, i.e., transform those terms into multiinstant homogenous
matrix polynomials with prescribed degrees. As an illustrative
example, with r = 2 variables and m = 1, g
1
= g
2
= 2, one
has P =
f
1
∈K(2)
f
2
∈K(2)
{(
2
j=1
α
f
j
j
)(
2
j=1
g
j
!/π( f
j
))P}=
α
20
1
α
20
2
P + 2α
20
1
α
11
2
P + α
20
1
α
02
2
P + 2α
11
1
α
20
2
P + 4α
11
1
α
11
2
P+
2α
11
1
α
02
2
P + α
02
1
α
20
2
P + 2α
02
1
α
11
2
P + α
02
1
α
02
2
P.
Lemma 3: For some positive integers g
j
with j =
1, 2,...,m + 1 and g
m
≥ 2, the two equalities (9) and (10),
as shown at the bottom of this page, always always hold
true where one has
⎧
⎨
⎩
R
ii
f
1
f
2
···f
m−1
(
f
m
−2e
i
)
f
m+1
= 0, for f
m
− 2e
i
< 0
R
ij
f
1
f
2
···f
m−1
(
f
m
−e
i
−e
j
)
f
m+1
= 0, for f
m
− e
i
− e
j
< 0.
(11)
Proof: First, let us write out an arbitrary term of the left-
hand side of the equality (9) given as follows:
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
R
ii
f
1
f
2
···f
m−1
( f
m
−2e
i
)f
m+1
(12)
where f
j
∈ K(g
j
), j ∈{1, 2,...,m + 1}; 1 ≤ i ≤ r.
If f
m
− 2e
i
< 0, then the term of (12) is zero by using
condition (11). If f
m
−2e
i
≥ 0, then f
m
−2e
i
∈ K(g
m
−2), i.e., it
implies that the term (12) is also one term of the right-hand
side of the equality (9).
Secondly, let us write out an arbitrary term of the right-hand
side of the equality (9) as follows:
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
× h
2
i
R
ii
f
1
f
2
···f
m+1
(13)
where f
j
∈ K(g
j
), j = 1, 2,...,m−1; f
m
∈ K(g
m
−2); f
m+1
∈
K(g
m+1
); 1 ≤ i ≤ r.
Since f
m
∈ K(g
m
− 2), one has f
m
+ 2e
i
∈ K(g
m
). Thus,
the term (13) can be represented as follows:
⎛
⎝
m−1
j=1
h
f
j
(t − m + j)
⎞
⎠
h
f
m
+2e
i
h
f
m+1
(t + 1)
×
R
ii
f
1
f
2
···f
m−1
( f
m
+2e
i
−2e
i
)f
m+1
. (14)
Then, if f
m
+ 2e
i
= p with p ∈ K(g
m
) is
satisfied, one has h
f
m
+2e
i
(z(t))R
ii
f
1
f
2
···f
m−1
( f
m
+2e
i
−2e
i
)f
m+1
=
1≤i≤r;f
j
∈K(g
j
),
j∈{1,2,...,m+1}
⎧
⎪
⎪
⎨
⎪
⎪
⎩
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
×
R
ii
f
1
f
2
···f
m−1
(
f
m
−2e
i
)
f
m+1
⎫
⎪
⎪
⎬
⎪
⎪
⎭
=
f
m
∈K(g
m
−2);f
j
∈K(g
j
),
j∈{1,2,...,m+1},j=m;1≤i≤r
⎧
⎨
⎩
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
× h
2
i
R
ii
f
1
f
2
···f
m+1
⎫
⎬
⎭
(9)
1≤i<j≤r;f
j
∈K(g
j
),
j∈{1,2,...,m+1}
⎧
⎨
⎩
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
×
R
ij
f
1
f
2
···f
m−1
( f
m
−e
i
−e
j
)f
m+1
⎫
⎬
⎭
=
f
m
∈K(g
m
−2);f
j
∈K(g
j
),
j∈{1,2,...,m+1},j=m;1≤i<j≤r
⎧
⎨
⎩
⎛
⎝
m+1
j=1
h
f
j
(t − m + j)
⎞
⎠
× h
i
× h
j
R
ij
f
1
f
2
···f
m+1
⎫
⎬
⎭
(10)
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