2
$\begingroup$

I'm Trying to help in this question which involves Algebraic Ricatti equation. Honestly to say I never met this equation before.

I'm struggling to understand the conditions stated in the limitations of the dare algorithm in matlab. For convinience I quote it here:

The $(A, B)$ pair must be stabilizable (that is, all eigenvalues of $A$ outside the unit disk must be controllable). In addition, the associated symplectic pencil must have no eigenvalue on the unit circle. Sufficient conditions for this to hold are $(Q, A)$ detectable when $S = 0$ and $R > 0$, or $$\begin{bmatrix} Q & S \\ S^T & R \end{bmatrix}>0$$

All $A,B,S$ are matrices ($S=0$ in my case ), so I don't understand how a matrix can be greater then zero? Is this is special notations for positive definite matrix?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is just a notation for positive definite matrix. Several works use this $>$ instead of $\succ$. In the context of control theory both notations mean the same thing.

$\endgroup$
3
  • $\begingroup$ are you familiar with $dare$ by mistake? I tried to create random $R,Q$ with eigenvalues $>0$ together with controllable $(A, B)$ but $dare$ still fails $70\%$ of times which make me craze:( Where am I wrong? $\endgroup$ Commented Dec 11, 2015 at 17:27
  • $\begingroup$ Try making sure they are symmetric. It could help the solver. $\endgroup$
    – AndyOak
    Commented Dec 14, 2015 at 15:22
  • $\begingroup$ look at the code under the link in the question please. It missing the part where I make sure the eigenvalues are positive, but for this part I do something very stupid-simple like stay in a loop until it so. $\endgroup$ Commented Dec 14, 2015 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .