I would like to convert the equation $\ddot{y}+\int_0^t y(\tau)d\tau=0$ to state-space representation. Below, I present my attempt, which seems to be contradicting, and then ask my question at the end.
Coversion
Let $x_1=y$ and $x_2=\dot{y}$. Also, let $x=\left[\begin{array}{c}x_1\\x_2\end{array}\right]$ and so:
$$\dot{x}=
\left[\begin{array}{c}x_2\\-\int_0^t x_1 d\tau \end{array}\right]
$$
Take Laplace transform, assuming 0 initial conditions:
$$
sX=\left[\begin{array}{c}X_2\\ -\frac{X_1}{s} \end{array}\right]=
\left[\begin{array}{cc}0 & 1\\-\frac{1}{s} & 0\end{array}\right]X
$$
Inverse Laplace transform:
$$
\dot{x}=
\left[\begin{array}{cc}0 & \delta(t)\\-1 & 0\end{array}\right]x
$$
where $\delta(t)$ is the delta-dirac function (infinity at 0, and 0 elsewhere).
Question
The last equation implies $\dot{y}=\delta(t)\dot{y}$ and this implies $1=\delta(t)$, a false statement.
Please let me know the error in my logic.
Comments
(1) I know I can model the original equation using another state assignments without running into such problem of contradicting statements. For example, I can use the states $x_1=\int_0^t y(\tau)d\tau$, $x_2=y$, and $x_3=\dot{y}$. This state assignment will not result into a problem like the former one. However, this assignment results in 3-by-3 system, whereas the former results in 2-by-2 system.
(2) I also know I can differentiate the original ODE to get rid of the integral, but this will also result in a 3-by-3 system.
(3) The bottomline here: The main objective of this question is to uncover the error in my first attempt that used Laplace transform.
Best Answer
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.