3
$\begingroup$

I have been asked to show that controllability and observability are not affected by replacing $A$ with $(A+αI)$. And also to show that this is not necessarily true for stabilizability.

I have thought about different approaches:

Popov-Belevitch-Hautus test

If the system is controllable must be true that $$Rank [(A - \lambda I) B] = n$$

If we replace $A$ with $(A+αI)$ then $$Rank [((A +αI) - \lambda I) B] $$ $$Rank [(A - (\lambda -α)I) B] $$ $$Rank [(A - gI) B] $$ where $g = (\lambda -α) \in C$

$(A - gI)$ has already $rank = n$ for all $g \in C$ except for those which are eigenvalues for $A$.

For those values, we need to prove that the concatenation of $B$ will guarantee the rank to be n.

The issue is that since B is the same we can't guarantee it.

Controllability Matrix

Given an LTI if and only if the control matrix $C$ has full column rank, then the system is controllable. $$ C = [B ,AB, A²B ... A^{n-1} B]$$ but then altering the diagonal of $A$ can affect the rank of C, or if not, how come?

Controllability Gramian

I tried to plug $(A+αI)$ in the integral, but doing that then I don't know how to prove that the matrix $W$ is still nonsingular for any t > 0.

$\endgroup$

1 Answer 1

3
$\begingroup$

Controllability: there does not exist a left-eigenvector for $A$ that is orthogonal for $B$.

For your problem, for any $x$ such that $xA = \lambda x$, $xB\neq0$.

Since the eigenvectors of $A$ are also the eigenvecors of $A+\alpha I$, i.e., $x(A+I\alpha) = x(\lambda+\alpha)$, you have $xB\neq0$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.