Suppose I have the a system of the type:
$\dot{x}=\begin{bmatrix}
-0.2152 & 3.9030\\
-1.2734& -4.7848
\end{bmatrix}\begin{bmatrix}
0.8\\
0.5
\end{bmatrix}+\begin{bmatrix}
0.5 &-0.5 \\
-1.4& 1.43
\end{bmatrix}\begin{bmatrix}
0\\
1
\end{bmatrix}$
$0=\begin{bmatrix}
-1.7891 & -0.6831
\\
1.7891& 0.6831
\end{bmatrix}\begin{bmatrix}
0.8\\
0.5
\end{bmatrix}+\begin{bmatrix}
1 &0 \\
0& 1
\end{bmatrix}\begin{bmatrix}
0\\
1
\end{bmatrix}$
what considerations can I do about controllability and observability?
I have that the input vector is $u=\begin{bmatrix}
0\\
1
\end{bmatrix}$, so, let's say I do a diagonalization of the matrix $A$, I would have that the eigenvalues, or the singular values, would be on the diagonal, and so the fact that the input has this form , does it have some implications on controllability?
I have already done the tests for controllability and observability, and the system is both controllable and observable.
My doubts come from the fact that with the input of the following form, it means that, if i consider a diaginal structure for the eigenvalues, I have that the first input cannot control the state $x_1$ and the control comes only from the input $u_2$.
Am I doing a wrong reasoning or am I right? Can somebody please help me understand?
Best Answer
First of all controllability is about the existence of an input function that drives the system from some initial state to a desired final state. It doesn't say anything for a specific input, in which case you can already calculate the states.
I believe you want to investigate the case that your input vector-function is of the form $u(t) = \begin{bmatrix}0 \\ 1\end{bmatrix} f(t)$ for some function $f$. In this case you are basically using just one input and ignoring the other one. So it is equivalent of having a $2\times1$ $B$ matrix (that is the second column of your original $B$).
Diagonalizing $A$ is done via a similarity transformation, which doesn't affect controllability/observability. You simply look at the states in a different coordinate system. Also, even if $A$ is diagonal and you have a single input, this input can still affect both states so that the system is still controllable. For example; $$\dot{x}=\begin{bmatrix}1&0\\0&2\end{bmatrix}x+\begin{bmatrix}1\\1\end{bmatrix}u$$ It is a good exercise to calculate $u$ which drives the system from any $x_0$ to any $x_f$ with this small example.