Double integrator

In systems and control theory, the double integrator is a canonical example of a second-order control system. It models the dynamics of a simple mass in one-dimensional space under the effect of a time-varying force input $$\textbf{u}$$.

Differential equations
The differential equations which represent a double integrator are:
 * $$\ddot{q} = u(t)$$
 * $$y = q(t)$$

where both $$q(t), u(t) \in \mathbb{R} $$ Let us now represent this in state space form with the vector $$\textbf{x(t)} = \begin{bmatrix} q\\ \dot{q}\\ \end{bmatrix}$$


 * $$ \dot{\textbf{x}}(t)= \frac{d\textbf{x}}{dt} = \begin{bmatrix}

\dot{q}\\ \ddot{q}\\ \end{bmatrix} $$

In this representation, it is clear that the control input $$\textbf{u}$$ is the second derivative of the output $$\textbf{x}$$. In the scalar form, the control input is the second derivative of the output $$q$$.

State space representation
The normalized state space model of a double integrator takes the form
 * $$\dot{\textbf{x}}(t) = \begin{bmatrix}

0& 1\\                              0& 0\\                             \end{bmatrix}\textbf{x}(t) + \begin{bmatrix} 0\\ 1\end{bmatrix}\textbf{u}(t)$$
 * $$ \textbf{y}(t) = \begin{bmatrix} 1& 0\end{bmatrix}\textbf{x}(t).$$

According to this model, the input $$\textbf{u}$$ is the second derivative of the output $$\textbf{y}$$, hence the name double integrator.

Transfer function representation
Taking the Laplace transform of the state space input-output equation, we see that the transfer function of the double integrator is given by
 * $$\frac{Y(s)}{U(s)} = \frac{1}{s^2}.$$

Using the differential equations dependent on $$ q(t), y(t), u(t)$$ and $$\textbf{x(t)}$$, and the state space representation: