Double integrator

The differential equations which represent a double integrator are:

${\displaystyle {\ddot {q}}=u(t)}$

${\displaystyle y=q(t)}$

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

${\displaystyle {\dot {\textbf {x}}}(t)={\frac {d{\textbf {x}}}{dt}}={\begin{bmatrix}{\dot {q}}\\{\ddot {q}}\\\end{bmatrix}}}$

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

The normalized state space model of a double integrator takes the form

${\displaystyle {\dot {\textbf {x}}}(t)={\begin{bmatrix}0&1\\0&0\\\end{bmatrix}}{\textbf {x}}(t)+{\begin{bmatrix}0\\1\end{bmatrix}}{\textbf {u}}(t)}$

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

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

Taking the Laplace transform of the state space input-output equation, we see that the transfer function of the double integrator is given by

${\displaystyle {\frac {Y(s)}{U(s)}}={\frac {1}{s^{2}}}.}$

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