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.


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).


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.


(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.

  • $\begingroup$ The problem is that the transform of a product isn't the product of the transforms $\endgroup$
    – Federico
    Dec 7, 2018 at 18:58
  • $\begingroup$ You end up with the convolution of $\delta$ and $x$, which is $x$. Mystery solved $\endgroup$
    – Federico
    Dec 7, 2018 at 18:59
  • $\begingroup$ @Federico I can't follow. Can you please elaborate? $\endgroup$ Dec 7, 2018 at 19:07
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    $\begingroup$ @Federico Please answer the question. I will accept it. It is important to have the answer visible, just in case someone else runs into the same problem. $\endgroup$ Dec 7, 2018 at 19:20
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    $\begingroup$ I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $\det (A-\lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form. $\endgroup$ Dec 8, 2018 at 8:46

1 Answer 1


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.


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