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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?

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1 Answer 1

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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.

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  • $\begingroup$ Isn't this what I did? $\endgroup$
    – Programmer
    Apr 5 at 4:45
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    $\begingroup$ 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! $\endgroup$
    – Dr. Sundar
    Apr 5 at 4:50
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    $\begingroup$ You are right. Let me edit my answer and correct it. Thank you! $\endgroup$
    – Dr. Sundar
    Apr 5 at 5:43
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    $\begingroup$ Thank you as well for validating my answers! $\endgroup$
    – Programmer
    Apr 5 at 5:44
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    $\begingroup$ Yes, looks fine! $\endgroup$
    – Dr. Sundar
    Apr 5 at 5:57

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