I'm trying to design a controller that will achieve reference tracking of the second state for the following system.

$$ \begin{aligned} \dot{x} &=\underbrace{\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]}_A x+\underbrace{\left[\begin{array}{l} 0 \\ 1 \end{array}\right]}_B u \\ y &=\underbrace{\left[\begin{array}{ll} 1 & 0 \end{array}\right]}_C x \end{aligned} $$ where $$x=\left[\begin{array}{l} q \\ \dot{q} \end{array}\right] $$ This problem should be equivalent to tracking a ramp on the first state, hence we require a type-2 system if we are to achieve zero steady-state error. Therefore, we require a double integrator to augment the system with the integrals of the error between the reference and performance output. The error $e$, reference $r$ (step) and performance matric $H = \left[\begin{array}{l} 0 & 1 \end{array}\right]$ are related as $$e=r-Hx$$ If $e=0$, the reference is tracked with zero steady-state error. The system can be augmented by adding states $\int e$ and $\int \int e$, with dynamics $$\dot{w}=\underbrace{\left[\begin{array}{l} 0 & 0 \\ 1 & 0 \end{array}\right]}_Mw+\underbrace{\left[\begin{array}{l} 1 \\ 0 \end{array}\right]}_Ne$$ where $$w=\left[\begin{array}{l} \int e \\ \int \int e \end{array}\right]$$ The augmented system is then given by $$\left[\begin{array}{l} \dot{x} \\ \dot{w} \end{array}\right]=\underbrace{\left[\begin{array}{l} A & 0_{2 \times 2}\\ -NH & M \end{array}\right]}_{A_e}\left[\begin{array}{l} x \\ w \end{array}\right]+\underbrace{\left[\begin{array}{l} B \\ 0_{2 \times 1} \end{array}\right]}_{B_e}u+\left[\begin{array}{l} 0_{2 \times 1} \\ N \end{array}\right]r$$ My question: strangely enough, the pair $(A_e,B_e)$ is uncontrollable for the provided matrices, but it is controllable when I change to $H = \left[\begin{array}{l} 1 & 0 \end{array}\right]$. After designing a controller, it can track a ramp on the state $q$, which implies tracking a step on $\dot{q}$, but I feel that the former $H$ should yield identical behaviour when a step $r$ is applied to track $\dot{q}$. Is there an explanation for this?

  • $\begingroup$ . . . . . . . . $\endgroup$
    – AJN
    Feb 15, 2022 at 15:24
  • $\begingroup$ I ought to be able answer his but have forgotten my learning on matrices, sadly $\endgroup$
    – Rhodie
    Feb 22, 2022 at 19:31
  • $\begingroup$ The goal is the state $x_2$ to track a step input ? $\endgroup$ Feb 23, 2022 at 0:17

1 Answer 1


The uncontrollability makes sense, because the first error integral will be a shifted version of $q$, hence we cannot control them independently.


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