Here are the two problem statements I'm trying to understand:

  • Reachability. The reachability problem is to “find the set of all the final states $x(T)$ reachable starting from a given initial state $x(t_0)$”.

  • Controllability. The controllability problem is “to find the set of all the initial states $x(t_0)$ controllable to a given final state $x(T)$”.

In my view, both problems are the Same. I do not really get the difference especially given that the reachability and controllability gramians are exactly the same. Also, I want to understand why reachability always implies controllability but controllability does not imply reachability (unless $A$ in full rank).

  • $\begingroup$ You are only considering LTI systems? $\endgroup$ – Kwin van der Veen Dec 8 '18 at 9:17

They are not the same problem, but equivalent for linear continuous systems. Also, reachability and controllability gramians are slightly different. To understand the differences let's start with a general linear continuous time-varying system. $$ \dot{x}=A(t)x(t)+B(t)u(t) $$

Its solution can be given as $$ x(t) = \phi(t,t_0) x(t_0) + \int_{t_0}^t \phi(t,\tau) B(\tau) u(\tau) d\tau $$ where $\phi(\cdot,\cdot)$ is the state transition matrix.

Now, let's say we want to "reach" to the state $x(t_1)=x_1$ at time $t_1$ for a given $x(t_0)=x_0$. Then, we can use the input function $$ u(t) = B^T(t) \phi^T(t_1, t) W_r^{-1}(t_1,t_0) \left(x_1 - \phi(t_1,t_0) x_0 \right) $$ where $$W_r(t_1,t_0) := \int_{t_0}^{t_1} \phi(t_1, \eta) B(\eta) B^T(\eta) \phi^T(t_1, \eta) d\eta$$

Note that if the reachability gramian is full-rank, we can reach any state we want from any initial condition, hence full reachability. If it does not have a full-rank you can still show that the reachable subspace at time $t_1$ is $$\begin{align*}\mathcal{R}(t_0;t_1) &= \operatorname{Im} \int_{t_0}^{t_1} \phi(t_1,\tau) B(\tau) d\tau \\ &= \operatorname{Im} W_r(t_1,t_0)\end{align*}$$

For controllability, suppose your final state is given as $x(t_1)=x_1$ and you want to find which initial states can reach this final state. Then using the properties of the state transition matrix, $$x_0 = \phi^{-1}(t_1,t_0) x_1 - \int_{t_0}^{t_1} \phi(t_0, \tau) B(\tau) u(\tau) d\tau$$ which is now essentially the same problem with reachability, but backwards in time. So, the controllable subspace is $$\begin{align*}\mathcal{C}(t_0;t_1) &= \operatorname{Im} \int_{t_0}^{t_1} \phi(t_0,\tau) B(\tau) d\tau \\ &= \operatorname{Im} W_c(t_1,t_0)\end{align*}$$ where $$W_c(t_1,t_0) := \int_{t_0}^{t_1} \phi(t_0, \eta) B(\eta) B^T(\eta) \phi^T(t_0, \eta) d\eta$$

Discrete time case is more interesting, because reachability and controllability is not equivalent in this case as you pointed out. The reason is the state transition matrix (which is $A^k$ for discrete LTI case) might not be invertible (we cannot always go backwards in time) as it is in continuous time case. But the thought process is the same.

To summarize,

  • For full reachability in linear continuous systems: $\operatorname{Im} W_r(t_1,t_0) = \mathbb{R}^n$
  • For full controllability in linear continuous systems: $\mathbb{R}^n = \operatorname{Im} \phi(t_1,t_0) \subseteq \operatorname{Im} W_c(t_1,t_0)$
  • For full reachability in LTI discrete systems: $\operatorname{Im} \sum_{i=0}^{k-1} A^i B = \mathbb{R}^n$
  • For full controllability in LTI discrete systems: $\operatorname{Im}A^n \subseteq \operatorname{Im} \sum_{i=0}^{n-1} A^i B$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.