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Page 18 of 23 Tan et al. Complex Eng Syst 2023;3:6 I http://dx.doi.org/10.20517/ces.2023.10

In addition, the other scalars are given as follows

∑
31 T T
Ξ = ( ) B − 2(1 + )R − S 1 − S 2 − S 3 − S 4 ,

=1
33 41
Ξ = −2(4 + )R + 2S 1 − 2S 2 + 2S 3 − 2S 4 , Ξ = S 1 + S 2 − S 3 − S 4 ,
43 44
Ξ = −2(2 − )R − S 1 + S 2 + S 3 − S 4 , Ξ = −Q − 4(2 − )R ,
51 53 54 55
˜
Ξ = S 3 + S 4 , Ξ = −S 3 + S 4 + 3(2 − )R , Ξ = 3(2 − )R , Ξ = −3(2 − )R ,
61 62 66
Ξ = 3(1 + )R , Ξ = S 2 + 3(1 + )R , Ξ = −3(1 + )R ,
71 1 T T 77 +6,1 T T +6, +6
Ξ = B , Ξ = − 0 1 Ω , Ξ = B , Ξ = − 0 Ω ,

[ ]
11 T
Γ = G 0 0 0 0 0 0 · · · 0 ,
12 −1 −1 −1 −1

Γ = −( − 1) { , · · · , , ≠ , · · · , 1 },
1
1
[ ]
21 ∑ 1

Γ = A 0 B =1 0 0 0 B · · · B ,
31 [ ]
Γ = A 0 0 0 0 0 0 · · · 0 ,
[ ]
41 ∑

Γ = 0 0 B =1 0 0 0 0 · · · 0 ,
51 [ 1 ]
Γ = 0 0 0 0 0 0 B · · · 0 ,
61 [ ]
Γ = 0 0 0 0 0 0 0 · · · B ,
71 [ ]
Γ = G 0 0 0 0 0 0 · · · 0 ,
ˇ −1 −1 −1
R = −( − 1) {(1 + 2 + 3 + · · · + , +3 ) Φ(R 1 ), · · · , (1 + 2 + 3 + · · · + , +3 ) Φ(R ), ≠ ,
−1
· · · , (1 + 2 + 3 + · · · + , +3 ) Φ(R )},
2
Φ(R ) = −2 + , = {1, · · · , ( ≠ ), · · · , },

 0 0 S 1 + S 3 −S 1 + S 3 −S 3 0 0 · · · 0 
 
 
81  0 0 S 2 + S 4 −S 2 + S 4 −S 4 0 0 · · · 0
Γ =   ,
S 1 + S 3 0 −S 1 + S 3 0 0 −S 3 0 · · · 0
 
 S 2 + S 4 0 −S 2 + S 4 0 0 −S 4 0 · · · 0 
 
ˆ −1 −1 −1 −1
R = {−(1 − ) R , −3(1 − ) R , − R , −3 R }.

Furthermore, the memory controller gain matrix is = .
−1

−1 −1
Proof: Define = P ,thenpre-multiplyingandpost-multiplying (15),(17),(20) with { , · · · , , R

−1
−1
−1
−1
−1
R , R , R , R , R , , , , } respectively, and define = , R = R , Q =

Q , S 1 = S 1 , S 2 = S 2 , S 3 = S 3 , S 4 = S 4 , Q = Q , Ω =
Ω . According to Schur complement and Lemma 2, and considering (ℎ) ∈ [ (ℎ), (ℎ)],

(31), (32), (35), (36), (40) can be obtained. This completes the proof.
Remark 4. For a given dwell time ℎ, the TR (ℎ) can be regarded as a linear combination of its upper and
lower bounds, namely (ℎ) = 1 (ℎ) + 2 (ℎ), where 1 + 2 = 1 and 1 > 0, 2 > 0. We can change

1 and 2 to get the corresponding value of (ℎ).
Remark 5. Theorem 2 presents a solution algorithm in terms of linear matrix inequality to obtain the con-
troller gains. However, the transition probability in Case 3 is completely unknown. We introduce (ℎ) to