Observable System L Matrix, can a value be negative?

[miniradman]

Homework Statement

An error matrix is in the form, has a characteristic equation:
$CE: s^2 + 120s + 7200 = 0$

A state variable feedback system is described by:
$A_F = \begin{bmatrix}0 & 1 \\-616.8 & -40 \end{bmatrix}$
$B = \begin{bmatrix}0 \\ 1 \end{bmatrix}$
$C = \begin{bmatrix}616.8 & 24 \end{bmatrix}$
$D = \begin{bmatrix}0 \end{bmatrix}$

Find the L error feedback matrix, using a state transformation to convert to observer canonical form:
$L = \begin{bmatrix}l_1 \\ l_2 \end{bmatrix}$

Homework Equations

$T = [O_m P]^-1$
$L_z = T L_x$
$O_m = \begin{bmatrix}C \\CA \end{bmatrix}$

The Attempt at a Solution

From what I understand, observable canonical form entails that:
$P = \begin{bmatrix}1 & 0 \\616.8 & 1 \end{bmatrix}$
$A_e = \begin{bmatrix} -120 & 1 \\ -7200 & 0 \end{bmatrix}$
And equating with the existing system:
$A_e = \begin{bmatrix} -(40+l_1) & 1 \\ -(616.8+l_2) & 0 \end{bmatrix} = \begin{bmatrix} -120 & 1 \\ -7200 & 0 \end{bmatrix}$
This yields the L matrix:
$\begin{bmatrix} l_1\\ l_2 \end{bmatrix} = \begin{bmatrix} 80 \\ 6583,2 \end{bmatrix}$
However this is form is not in observer canonical form, my question is when I use the Observable Matrix with my Transformation matrix:
$O_m = \begin{bmatrix}C \\CA \end{bmatrix}$
Does C need to be in observable canonical form?
$C = \begin{bmatrix} 1 & 0 \end{bmatrix}$
Because my current C matrix is not in that standard form. If so, how do I convert it, or does it fall out in the wash when I apply the P matrix?
$T = [O_m P]^-1$

 

[KataKoniK]

Hello,

Thank you for your question. It appears that you are on the right track with your solution. To answer your question, yes, the C matrix needs to be in observable canonical form in order to use the transformation matrix T. This can be achieved by multiplying the existing C matrix by the P matrix, which will result in the observable canonical form. So your new C matrix would be:

$C = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\616.8 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix}$

Then, you can proceed with your solution using the transformed C matrix. I hope this helps. Let me know if you have any other questions or concerns. Good luck with your work.