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I have to compute a matrix $P$ that defines an Linear Matrix Inequality region in the following way $$L_P = \{ s\in \mathbb{C}|\begin{pmatrix}I \\sI \end{pmatrix}^* P\begin{pmatrix}I \\sI \end{pmatrix}\prec 0\} $$ for the constraint $|\Re (s)|> |\Im (s)|$.

I have used the conic region matrix, knowing that for $\Re (s) \tan(\theta)> |\Im (s)|$ we get the matrix $$\begin{pmatrix}0 & 0&\sin(\theta) &\cos(\theta) \\0 & 0&-\cos(\theta)&\sin(\theta) \\\sin(\theta) & -\cos(\theta)&0 &0\\\cos(\theta) & \sin(\theta)&0 &0 \end{pmatrix}$$ and that the inequality $|\Re (s)|> |\Im (s)|$ can be split into $\Re (s)\tan(\pi/4)> |\Im (s)|$, $\Re (s)\tan(-\pi/4)> |\Im (s)|$.

So I get two different matrices $P_1$,$P_2$, but the problem requires to get only one. How can I put them together? Is it correct to build a matrix $$\begin{pmatrix}P_1 &0 \\0&P_2 \end{pmatrix}$$ and use it as solution?

cholo14
  • 451

1 Answers1

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The problem is discussed in Scherer, Definition 3.7, where the matrix $P$ is partitioned as

$P = \begin{bmatrix} Q & S \\ S^T & R\end{bmatrix}$.

As you obtained two different $P$ each of them satisfying different stability regions, the final $P$ is a block matrix structured as

$ P = \begin{bmatrix} Q_1 & 0 &S_1 & 0 \\ 0 & Q_2 & 0 &S_2\\ S_1^T & 0 & R_1 & 0 \\ 0 & S^T_2 & 0 & R_2 \\ \end{bmatrix}.$

Betelgeuse
  • 484