# Controllability for second order coupled system

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:

1. For what values of $$\alpha$$ is the system controllable
2. For what values of $$\alpha$$ is the system controllable from $$u_1$$ alone
3. For what values of $$\alpha$$ is the system controllable from $$u_2$$ alone
4. 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?

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.

• Isn't this what I did? Apr 5 at 4:45
• Kindly compare the controllability matrix you have given in the question and the controllability matrix in my answer. I spent some time to calculate this matrix correctly! Apr 5 at 4:50
• You are right. Let me edit my answer and correct it. Thank you! Apr 5 at 5:43
• Thank you as well for validating my answers! Apr 5 at 5:44
• Yes, looks fine! Apr 5 at 5:57