# [SOLVED]modes observable and modes controllable

#### dwsmith

##### Well-known member
How do you find observable and controllable modes?
$\mathcal{L}\Big\{\big(\mathbf{A} - s\mathbf{I}\big)^{-1}\Big\} = \begin{bmatrix} -e^{-t} - te^{-t} + \frac{1}{2}t^2e^{-t} & te^{-t} & -te^{-t} + \frac{1}{2}t^2e^{-t}\\ -te^{-t} & -e^{-t} & te^{-t}\\ te^{-t} - \frac{1}{2}t^2e^{-t} & te^{-t} & -e^{-t} + te^{-t} - \frac{1}{2}t^2e^{-t} \end{bmatrix}\\$
and then
$X(t) = -e^{-t}\mathbb{I} - te^{-t} \begin{bmatrix} 1 & -1 & -1\\ 1 & 0 & -1\\ 1 & -1 & -1 \end{bmatrix} - \frac{t^2}{2}e^{-t} \begin{bmatrix} -1 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 1 \end{bmatrix} = e^{-t}\mathbb{I} + te^{-t} \begin{bmatrix} 1 & -1 & -1\\ 1 & 0 & -1\\ 1 & -1 & -1 \end{bmatrix} + \frac{t^2}{2}e^{-t} \begin{bmatrix} -1 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 1 \end{bmatrix}.$
I read that the loss of observability or controllability then the residue is zero then the pole doesn't show up in the transfer function.

How can this be used or can it?

Also,
\begin{align}
\mathbf{A} &=
\begin{bmatrix}
0 & -1 & -1\\
1 & -1 & -1\\
1 & -1 & -2
\end{bmatrix}\\
\mathbf{B} &=
\begin{bmatrix}
1 & 0\\
0 & 1\\
0 & 0
\end{bmatrix}\\
\mathbf{U} &=
\begin{bmatrix}
u_1\\
u_2
\end{bmatrix}\\
\mathbf{T} &= (\mathbf{A} - s\mathbb{I})^{-1}\mathbf{B}\\
&=
-\frac{1}{(s+1)^3}
\begin{bmatrix}
s^2 + 3s + 1 & -(s + 1)\\
s+1 & (s+1)^2\\
s & -(s+1)
\end{bmatrix}\\
\mathbf{X}(s) & = \mathbf{T}\mathbf{U}
\end{align}

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