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考虑矩阵集
A = { A = [ 0 ( n − 1 ) × 1 1 ( n − 1 ) × ( n − 1 ) − a 0 ⋯ − a n − 1 ] | a j _ ≤ a j ≤ a j ¯ , j = 0 , 1 , 2 , ⋯ , n − 1 } {\displaystyle {\mathcal {A}}=\{{\textbf {A}}={\begin{bmatrix}{\textbf {0}}_{(n-1)\times 1}&&{\textbf {1}}_{(n-1)\times (n-1)}\\-a_{0}&\cdots &-a_{n-1}\end{bmatrix}}|{\underline {a_{j}}}\leq a_{j}\leq {\overline {a_{j}}},j=0,1,2,\cdots ,n-1\}}
集合中的每个矩阵 A {\displaystyle {\mathcal {A}}} 是 Hurwitz 当且仅当存在 P i ∈ m a t h b b S n , i = 1 , 2 , 3 , 4 {\displaystyle {\textbf {P}}_{i}\in mathbb{S}^{n},i=1,2,3,4} ,其中 P i > 0 , i = 1 , 2 , 3 , 4 , {\displaystyle {\textbf {P}}_{i}>0,i=1,2,3,4,} ,使得
P i A i + A i T P i < 0 , i = 1 , 2 , 3 , 4 , {\displaystyle {\textbf {P}}_{i}A_{i}+{\textbf {A}}_{i}^{T}{\textbf {P}}_{i}<0,\quad i=1,2,3,4,}
其中
A i = [ [ 0 ( n − 1 ) × 1 1 ( n − 1 ) × ( n − 1 ) ] a i ] , i = 1 , 2 , 3 , 4 , {\displaystyle {\textbf {A}}_{i}={\begin{bmatrix}{\begin{bmatrix}{\textbf {0}}_{(n-1)\times 1}&{\textbf {1}}_{(n-1)\times (n-1)}\end{bmatrix}}\\{\textbf {a}}_{i}\end{bmatrix}},\quad i=1,2,3,4,}
a 1 = − [ a _ 0 a _ 1 a ¯ 2 a ¯ 3 ⋯ a _ n − 4 a _ n − 3 a ¯ n − 2 a ¯ n − 1 ] , {\displaystyle {\textbf {a}}_{1}=-[{\underline {a}}_{0}\quad {\underline {a}}_{1}\quad {\overline {a}}_{2}\quad {\overline {a}}_{3}\quad \cdots \quad {\underline {a}}_{n-4}\quad {\underline {a}}_{n-3}\quad {\overline {a}}_{n-2}\quad {\overline {a}}_{n-1}],}
a 2 = − [ a _ 0 a ¯ 1 a ¯ 2 a _ 3 ⋯ a _ n − 4 a ¯ n − 3 a ¯ n − 2 a _ n − 1 ] , {\displaystyle {\textbf {a}}_{2}=-[{\underline {a}}_{0}\quad {\overline {a}}_{1}\quad {\overline {a}}_{2}\quad {\underline {a}}_{3}\quad \cdots \quad {\underline {a}}_{n-4}\quad {\overline {a}}_{n-3}\quad {\overline {a}}_{n-2}\quad {\underline {a}}_{n-1}],}
a 3 = − [ a ¯ 0 a _ 1 a _ 2 a ¯ 3 ⋯ a ¯ n − 4 a _ n − 3 a _ n − 2 a ¯ n − 1 ] , {\displaystyle {\textbf {a}}_{3}=-[{\overline {a}}_{0}\quad {\underline {a}}_{1}\quad {\underline {a}}_{2}\quad {\overline {a}}_{3}\quad \cdots \quad {\overline {a}}_{n-4}\quad {\underline {a}}_{n-3}\quad {\underline {a}}_{n-2}\quad {\overline {a}}_{n-1}],}
a 4 = − [ a ¯ 0 a ¯ 1 a _ 2 a _ 3 ⋯ a ¯ n − 4 a ¯ n − 3 a _ n − 2 a _ n − 1 ] . {\displaystyle {\textbf {a}}_{4}=-[{\overline {a}}_{0}\quad {\overline {a}}_{1}\quad {\underline {a}}_{2}\quad {\underline {a}}_{3}\quad \cdots \quad {\overline {a}}_{n-4}\quad {\overline {a}}_{n-3}\quad {\underline {a}}_{n-2}\quad {\underline {a}}_{n-1}].}
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一系列记录和验证 LMI 的参考资料。