Rewriting state-space to error dynamics in order to achieve tracking a constant

Question

Given the following system $$\begin{equation} \begin{bmatrix} \dot{x_1} \\ \dot{x_2} \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix}}_A\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \underbrace{\begin{bmatrix} 0 \\ 1 \end{bmatrix}}_B u, \end{equation}$$ I which to track $x_1 = x_2 = r$, where $r$ is a piece-wise continuous constant value. This is equivalent to driving $x_1-r$ and $x_2-r$ to $0$. The pair $(A,B)$ is controllable. I rewrote my state-space in terms of error dynamics with states $z_1 = x_1-r$ and $z_2 = x_2-r$. \begin{equation} \label{eq:rewritten} \begin{bmatrix} \dot{z_1} \\ \dot{z_2} \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix}}_A\begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \underbrace{\begin{bmatrix} 0 \\ 1 \end{bmatrix}}_B u, \end{equation} The output is given by \begin{equation} y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \end{equation} Now, if I know the reference $r$, let's say it's 1, how can solve this tracking problem by just using the error dynamics state-space (so no adding integrators)?

Answer 1

Let us consider the control law $u=Kx+v$ where $K$ is the state-feedback gain and $v$ is an exogenous input. This the maximum you can do with your assumptions.

The closed-loop system is given by $$\begin{bmatrix}\dot{x}_1 \\ \dot{x}_2\end{bmatrix}=\begin{bmatrix}0 & 1\\-2+k_1 & 3+k_2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}v,$$

where $K=\begin{bmatrix}k_1 & k_2\end{bmatrix}$.

Now we need two things: 1) The closed-loop system should be asymptotically stable (i.e. $A+BK$ Hurwitz stable), and 2) $v$ should be chosen such that $x_1,x_2$ are both equal to $r$ at equilibrium.

The first objective is achievable since your system is controllable. So, we can safely assume that we have a gain $K$ such that $A+BK$ is Hurwitz stable. We show now that the second objective is not achievable. Let $x_1^*,x_2^*$ be the values of $x_1,x_2$ at equilibrium. Solving for the system of equations, we get

$$\begin{bmatrix}x_1^*\\x_2^*\end{bmatrix}=-\begin{bmatrix}0 & 1\\-2+k_1 & 3+k_2\end{bmatrix}^{-1}\begin{bmatrix}0\\1\end{bmatrix}v.$$

This can be simplified to $$\begin{bmatrix}x_1^*\\x_2^*\end{bmatrix}=\dfrac{1}{2-k_1}\begin{bmatrix}1\\ 0\end{bmatrix}v.$$

Obviously, the second state cannot be set to anything else than 0 at equilibrium. In fact, this can be easily explained since the first state is the integral of the second, so if we want the system to reach an equilibrium then $x_2^*$ must necessarily be zero.

If you just want the first state-to be equal to $r$ at equilibrium, this is fairly easy. You just need to set $v=(2-k_1)r$ and $K$ such that $A+BK$ is Hurwitz stable.