State space model with input derivative

Question

How to include the derivative of an input directly into a state space model?

\begin{equation} \begin{aligned} \dot{\mathbf x}(t) &= \mathbf A \mathbf x(t) + \mathbf B \dot{\mathbf u}(t) \\ \mathbf y(t) &= \mathbf C \mathbf x(t) \end{aligned}\tag{1}\label{eq1} \end{equation}

By definining $\mathbf z(t) \triangleq \mathbf x(t) - \mathbf B \mathbf u(t) $ we can write: \begin{equation} \begin{aligned} \dot{\mathbf z}(t) &= \mathbf A \mathbf z(t) + \mathbf A\mathbf B \mathbf u(t) \\ \mathbf y(t) &= \mathbf C \mathbf z(t) + \mathbf C \mathbf B \mathbf u(t) \end{aligned}\tag{2}\label{eq2} \end{equation}

where the input now enterse the system directly. Systems $\eqref{eq1}$ and $\eqref{eq2}$ appear to deliver the same results when $\mathbf x(0) = \mathbf B\mathbf u(0)$ for $\eqref{eq1}$ and $\mathbf z(0) = 0$ for $\eqref{eq2}$.

Is this a general result?

Answer 1

It is indeed a general result for also independent $x(0)$ and $u(0)$. Consider the solution of the first system: $$\begin{align} x(t) &= e^{At} x(0) + \int_0^t e^{A(t-\tau)} B \dot{u} (\tau) d \tau \\ &= e^{At} x(0) + e^{At} \left[ \left. e^{-A \tau}B u(\tau) \right|_{\tau=0}^t + \int_0^t e^{-A\tau} A B u(\tau) d\tau \right] \\ x(t) - Bu(t) &= e^{At} \left[ x(0) - B u(0) \right] + \int_0^t e^{A(t-\tau)} A B u(\tau) d\tau \end{align}$$ which is the solution to the second equation for $z(0) = x(0) - B u(0)$. It is easy to show that $y(t)$ are also the same for both equations.

Answer 2

Take a look at

I will type it later. For more info see Modern Control Engineering by Ogata and State variables for engineers by DeRusso et al. They provide full details. The example above from Ogata.