# Canonical Forms

In order to see if we are in one of these canonical forms, just write the 
equations from the block diagram, and find the associated $S(A, B, C, D)$.

> [!TIP]
> In order to find a rough diagram, use 
> [Horner Factorization](MODERN-CONTROL.md/#horner-factorization) to find 
> $a_i$ values. Then put all the $b_i$ to the right integrator by shifting them
> as many left places, starting from the rightmost, for the number of 
> associated $s$

## Control Canonical Form
It is in such forms when: 
$$
A = \begin{bmatrix}
- a_1 & -a_2 & -a_3 & \dots & -a_{n-1} &-a_n\\
1 & 0 & 0 & \dots  & 0 &  0\\
0 & 1 & 0 & \dots & 0 & 0\\
\dots & \dots & \dots & \dots  & \dots \\
0 & 0 & 0 & \dots & 1 & 0
\end{bmatrix}

B = \begin{bmatrix}
1 \\ 0 \\ \dots \\ \dots \\ 0
\end{bmatrix}

C = \begin{bmatrix}
b_1 & b_2 & \dots & b_n
\end{bmatrix}

D = \begin{bmatrix}
0
\end{bmatrix}
$$

## Modal Canonical Forms
> [!CAUTION]
> This form is the most difficult to find, as this varies drastically in cases 
> of double roots
>
$$
A = \begin{bmatrix}
- a_1 & 0 & 0 & \dots & 0\\
0 & -a_2 & 0 & \dots & 0\\
0 & 0 & -a_3 & \dots & 0\\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & 0 & -a_n
\end{bmatrix}

B = \begin{bmatrix}
1 \\ 1 \\ \dots \\ \dots \\ 1
\end{bmatrix}

C = \begin{bmatrix}
b_1 & b_2 & \dots & b_n
\end{bmatrix}

D = \begin{bmatrix}
0
\end{bmatrix}
$$

## Observable Canonical Form
<!--TODO: Correct here -->

[^reference-input-pole-allocation]: [MIT | 06 January 2025 | pg. 2](https://ocw.mit.edu/courses/16-30-feedback-control-systems-fall-2010/c553561f63feaa6173e31994f45f0c60_MIT16_30F10_lec11.pdf)