For sake of simplicity, consider the super classic LTI system \begin{equation*} \begin{cases} & \dot{{x}}(t) =A {x}(t)+B {u}(t) , \quad \quad \text{for } 0\le t \le t_f \\ & x(0)=x_i \end{cases} \end{equation*}

I interpret this setting as a certain known dynamic $\dot{{x}}(t) =A {x}(t)$ describing a real process that we want to control by injecting the control term $B {u}(t)$. Now, assume that the system is controllable and say that we want to steer it from $x(0)=x_i$ to $x(t_f)=t_f$ with a control $u(t)$ that acts from $t=0$ to $t=t_f$.

By definition, since the system is controllable we can always find such a control $u$ but my question is: what happens after $t_f$? Since I interpret $\dot{{x}}(t) =A {x}(t)$ as a dynamical model that describes a real process, I also assume that such a dynamic will persist after $t_f$. If $x_f$ is not an equilibrium point of the dynamic, the system would revert to an equilibrium (assuming that $A$ is stable) after $t_f$. As you can notice, I am assuming $u(t)=0$ for $t>t_f$.

In other words: I bring the system to $x(t_f)=x_f$ with the control $u$ but after $t_f$ it seems that I am left with the system \begin{equation*} \begin{cases} & \dot{{x}}(t) =A {x}(t), \quad \quad \text{for } t > t_f \\ & x(t_f)=x_f \end{cases} \end{equation*} which would converge to an equilibrium, possibily different from $x_f$. Was the effect of my control policy to bring the system to $x_f$ for an instant only? Should one consider an infinite-horizon in this case?

  • $\begingroup$ You are right. But the answer to your question depends on the answer to "what is the desired behavior of the underlying real process"? $\endgroup$
    – Arastas
    Jun 1, 2020 at 13:41
  • $\begingroup$ Well, I guess one might want to keep a system at $x_f$, or sufficiently close to it, for a certain period of time. $\endgroup$ Jun 1, 2020 at 13:55
  • $\begingroup$ How about solving the linear system $B \bar{u} = - A x_f$ for $\bar{u}$? $\endgroup$ Jun 3, 2020 at 12:19
  • 1
    $\begingroup$ Then you cannot put $u=0$. Typically, we say that $u(t)=u_0+u_1(t)$, where $u_0$ is the steady-state constant control that makes $x_f$ an equilibrium, and $u_1(t)$ is the control signal that drives $x$ to $x_f$. Then you can change coordinates as $e=x-x_f$ that yields $\dot{e} = Ae + Bu_1 + Ax_f + Bu_0$ and under the condition $Ax_f + Bu_0=0$ you have $\dot{e}=Ae+Bu_1$, which you want to stabilize at $e=0$. $\endgroup$
    – Arastas
    Jun 3, 2020 at 12:20
  • $\begingroup$ How can I practically design $u_0$? I guess for $u_1(t)$ I could use, for example, the minimum energy control that drives the system to $x_f$. $\endgroup$ Jun 3, 2020 at 12:31

1 Answer 1


First of all notice that controllability is an open-loop property of the system, i.e. there is no feedback involved. This property is used to check our absolute limits of what we can and cannot do to control the system. It gives a geometric limit of where we can reach in the phase space if we don't have any limitations on the control signal. In reality, the calculated control signal is not used at all.

However, it turns out that in LTI systems controllability is equivalent to existence of a static state feedback controller that assigns system eigenvalues arbitrarily in the complex plane, which is actually useful for real-world applications.

To summarize, controllability is very important as a concept, but it is not directly used for control signal calculations. And it isn't interested in what happens after $t_f$.

  • $\begingroup$ This is interesting. How can one derive such a static state feedback controller? $\endgroup$ Jun 3, 2020 at 12:21
  • $\begingroup$ For single input systems such a controller is unique and you can use the so-called Ackermann's formula (en.wikipedia.org/wiki/Ackermann%27s_formula). For multi input systems there are many different approaches but simplest one is the dyadic approach, where you can select a fan-out vector $f$ and convert the system to single input such as $(A,Bf)$. In general this problem is called eigenstructure assignment. $\endgroup$
    – obareey
    Jun 3, 2020 at 12:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .