I have the following system:
$\ddot{y_1}=-y_1+\alpha y_2+u_1$
$\ddot{y_2}=-y_2+\alpha y_1-2u_2$
I am trying to answer 4 questions:
- For what values of $\alpha$ is the system controllable
- For what values of $\alpha$ is the system controllable from $u_1$ alone
- For what values of $\alpha$ is the system controllable from $u_2$ alone
- For what values of $\alpha$ is the system controllable if $u_1=u_2$
My workings:
First, let's convert it into state space form by setting
$x_1=y_1, x_2=y_2,x_3=\dot{y_1},x_4=\dot{y_2}$
We get
$\begin{bmatrix}\dot{x_1}\\\dot{x_2}\\\dot{x_3}\\\dot{x_4}\\\end{bmatrix}=\begin{bmatrix}0&0&1&0\\0&0&0&1\\-1&\alpha&0&0\\\alpha&-1&0&0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix}+\begin{bmatrix}0&0\\0&0\\1&0\\0&-2\end{bmatrix}\begin{bmatrix}u_1\\u_2\\\end{bmatrix}$.
To answer 1) We check controllability by checking the rank of $[B,AB,A^2B,A^3B]$ with $B=\begin{bmatrix}0&0\\0&0\\1&0\\0&-2\end{bmatrix}$. This matrix appears to always be full rank.
To answer 2) We check controllability by checking the rank of $[B,AB,A^2B,A^3B]$ with $B=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$. This matrix appears to be full rank only when $\alpha\ne0$.
To answer 3) We check controllability by checking the rank of $[B,AB,A^2B,A^3B]$ with $B=\begin{bmatrix}0\\0\\0\\-2\end{bmatrix}$ This matrix appears to be full rank only when $\alpha\ne0$.
To answer 4) We check controllability by checking the rank of $[B,AB,A^2B,A^3B]$ with $B=\begin{bmatrix}0\\0\\1\\-2\end{bmatrix}$ This matrix appears to be full rank only when $\alpha\ne0$.
Are the answers really this trivial or am I making a mistake somewhere?
Best Answer
I have given a detailed solution for all the 4 parts using Kalman's test for controllability of a linear control system:
(a) When we work with two controls $u_1$ and $u_2$:
First we form the controllability matrix as $$ Q = \left[ \matrix{B & A B & A^2 B & A^3 B \cr} \right] $$ and calculate the rank of $Q$.
A simple calculation yields $$ Q = \left[ \begin{array}{cccccccc} 0 & 0 & 1 & 0 & 0 & 0 & -1 & -2 \alpha \\[2mm] 0 & 0 & 0 & -2 & 0 & 0 & \alpha & 2 \\[2mm] 1 & 0 & 0 & 0 & -1 & -2 \alpha & 0 & 0 \\[2mm] 0 & -2 & 0 & 0 & \alpha & 2 & 0 & 0 \\[2mm] \end{array} \right] $$
Since the first columns of $Q$ are linearly independent,
$$ \mbox{rank}(Q) = 4 $$
Thus, we conclude that the linear system is controllable for all values of $\alpha$.
(b) When we work with one control only $u_1$.
In this case, we take $$ B = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} \right] $$
The controllability matrix $Q$ reduces to $$ Q = \left[ \matrix{B & A B & A^2 B & A^3 B \cr} \right] = \left[ \begin{array}{cccc} 0 & 1 & 0 & -1\\ 0 & 0 & 0 & \alpha \\ 1 & 0 & -1 & 0 \\ 0 & 0 & \alpha & 0 \\ \end{array} \right] $$
When $\alpha \neq 0$, the controllability matrix has full rank and the system is completely controllable.
When $\alpha = 0$, the controllability matrix has rank 3 and the system is not controllable.
(c) When we work with one control only $u_2$.
In this case, we take $$ B = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ -2 \\ \end{array} \right] $$
The controllability matrix $Q$ reduces to $$ Q = \left[ \matrix{B & A B & A^2 B & A^3 B \cr} \right] = \left[ \begin{array}{cccc} 0 & 0 & 0 & -2 \alpha \\ 0 & -2 & 0 & 2 \\ 0 & 0 & -2 \alpha & 0 \\ -2 & 0 & 2 & 0 \\ \end{array} \right] $$
When $\alpha \neq 0$, the controllability matrix has full rank and the system is completely controllable.
When $\alpha = 0$, the controllability matrix has rank 3 and the system is not controllable.
(d) When we work with $u_1 = u_2$.
In this case, we take $$ B = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ -2 \\ \end{array} \right] $$
The controllability matrix $Q$ reduces to $$ Q = \left[ \matrix{B & A B & A^2 B & A^3 B \cr} \right] = \left[ \begin{array}{cccc} 0 & 1 & 0 & -2 \alpha - 1 \\ 0 & -2 & 0 & \alpha + 2 \\ 1 & 0 & -2 \alpha - 1 & 0 \\ -2 & 0 & \alpha + 2 & 0 \\ \end{array} \right] $$
When $\alpha = 0$, the matrix $Q$ has rank 2 as it has only two linearly independent columns. In this case, the system is not controllable.
When $\alpha \neq 0$, the matrix $Q$ has full rank and the system is completely controllable.