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I have to solve the following exercise. Let $$ \dot x(t) = \begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}x(t)+\begin{bmatrix}2\\0\end{bmatrix}u(t); $$ $$ \qquad\qquad y(t)=\begin{bmatrix}1&0\end{bmatrix}x(t);\quad x(0)=\begin{bmatrix}1\\1\end{bmatrix};\ \ \ t\geq0. $$ be a continuous-time system where $y(t)$ is the output variable and $u(t)$ is the input variable. The requests of the exercise are the following:

  • determine, if possible, a control input which moves the state system from the state $x(0)$ at $t=0$ to the zero state at $t=1$;
  • determine, if possible, a control input, with support in $[0, 1]$, such that, for $t\geq1$ the output (in free evolution) is zero but the state (still in free evolution) is not.

For the first point I have no problem. I have verified the complete reachability of the system and I compute the desidered control input with the formula $$ u(\tau)=B^Te^{A^T(1-\tau)}\Gamma^{-1}(1)\left(\begin{bmatrix}0\\0\end{bmatrix}-e^{A}x(0)\right),\quad\tau\in[0, 1], $$ where $B=\begin{bmatrix}2\\0\end{bmatrix}$ and $\Gamma$ is the reachability gramian. I have more problems for the second point. I cannot know how to interpret "free evolution" because, if the exercise means the free response and the free state of the system, it is not necessary to determine a control input for them, because they do not depend on the input. Some hints?

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  • $\begingroup$ I believe it means free evaluation ($u(t)=0$) for $t > 1$. $\endgroup$ – obareey Jul 19 '17 at 10:38
  • $\begingroup$ Ok, and so it is what I'm thinking. But, if $u(t)=0$ for $t>1$ the request of the exercise loses meaning because it is not necessary to determine a control input (because we already know that it is zero). Is this the conclusion? Thank you $\endgroup$ – Jeji Jul 19 '17 at 13:17
  • $\begingroup$ @Jeji check if the system is observable. After this you will see what the question is trying to ask. $\endgroup$ – Kwin van der Veen Jul 19 '17 at 13:53
  • $\begingroup$ I check it and the system is not completely observable, but I still do not understand what the exercise is trying to ask me..(Perhaps the question is if is it possible to determine $x(0)$ such that the free response and the free state are zero and non-zero respectively?) $\endgroup$ – Jeji Jul 19 '17 at 14:04
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It will help if we re-write the system as

$$ {{\dot x}_1} = 2u\qquad {{\dot x}_2} = {x_1}\qquad y = {x_1} $$

So clearly we can only influence the first state, but since it influences the second state we are controllable as you have indeed verified. On the other hand we can only observe the first state as well. So I guess the answer depends if the second question is in addition to the first or not.

If it is in addition, then for the second question think of bringing $x_1$ and $x_2$ to the origin at time $t=1$ and then turning off the control. If $x_1$ were the position of a particle, what this means is that at time $t=1$ we have managed to park it at the origin ($ x_1 = 0$) with zero velocity ($\dot x_1 = 2u$ and $u=0$ for $t \geq 1$). Is it possible that $x_2$ however is nonzero? We are already at the origin for both states at $t=1$ and now all the dynamics are turned off.

However, if the question was a stand-alone question, then think of a control that brings only the output to zero, that is state $x_1=0$ at time $t=1$ and then you turn control off. Or in other words drive the system from $(1,1)$ to $(0,1)$ using your control law. Then $\dot x_2 = x_1$ so indeed the second state will stop evolving after $t \geq 1$ but the state (of the whole system) is not really zero.

Here is a simulation:

enter image description here

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  • $\begingroup$ Ok, now all is clear. Thank you! $\endgroup$ – Jeji Jul 20 '17 at 16:28

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