0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The problem is that the transform of a product isn't the product of the transforms $\endgroup$ – Federico Dec 7 '18 at 18:58
  • $\begingroup$ You end up with the convolution of $\delta$ and $x$, which is $x$. Mystery solved $\endgroup$ – Federico Dec 7 '18 at 18:59
  • $\begingroup$ @Federico I can't follow. Can you please elaborate? $\endgroup$ – user8396743 Dec 7 '18 at 19:07
  • $\begingroup$ 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 $\endgroup$ – Federico Dec 7 '18 at 19:09
  • 1
    $\begingroup$ 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. $\endgroup$ – user8396743 Dec 8 '18 at 8:46
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.