# A state-space representation of an integro-differential equation implies a false statement

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.

(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.

• The problem is that the transform of a product isn't the product of the transforms – Federico Dec 7 '18 at 18:58
• You end up with the convolution of $\delta$ and $x$, which is $x$. Mystery solved – Federico Dec 7 '18 at 18:59
• @Federico I can't follow. Can you please elaborate? – user8396743 Dec 7 '18 at 19:07
• 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 – Federico Dec 7 '18 at 19:09
• 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. – user8396743 Dec 8 '18 at 8:46

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.