Assuming a input-output system
$$u(t) = c \cdot \big(h(t)\circledast m(t) \big)$$

where its impulse response function is:

$$ h(t) = \begin{cases} \frac{A}{\tau}te^{-\frac{t}{\tau}}, &\quad t\geq0 \\ 0, &\quad t<0 \\ \end{cases} $$

Given only these, how can one derive the differential equation below, whose Green's function is the above?:

$$ \ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t) = \frac{A}{\tau}m(t) $$

Would, considering an inverse n-order L differential operator and performing the consequent integrals, be a possible way?

  • $\begingroup$ dunno what role "$c$" plays. it can be (and is) simply folded into $h(t)$. i didn't remove it, but it is clear that $c=1$. $\endgroup$ – robert bristow-johnson Aug 11 '19 at 22:54
  • $\begingroup$ Thanks for the observation. I also guess it is folded into the h(t). I would like to take the chance, although i closed the question, to ask about the second part of the post: How could one derive the differential equation using the fact that h(t) is the Green's function? In this case, we couldn't know the order of the L operator. Would the n consequent integrations and comparison of the result with the h(t) be a logical way of thinking this out? $\endgroup$ – axel Aug 11 '19 at 23:09
  • $\begingroup$ @Alex: You could derive it just the way I've shown it in my answer. The order is implicit in the given $h(t)$. It's inverse Laplace transform will explicitly show the order. $\endgroup$ – Matt L. Aug 12 '19 at 7:02

A simple way to derive the differential equation from the impulse response is to transform the latter to the frequency domain, rewrite the input/output relation, and then transform the resulting equation back to the time domain.

The Laplace transform of the given impulse response $h(t)$ is

$$\begin{align} H(s) &=\frac{A}{\tau}\frac{1}{\left(s+\frac{1}{\tau}\right)^2} \\ &=\frac{A}{\tau}\frac{1}{s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}} \\ &=\frac{U(s)}{M(s)}\tag{1} \end{align}$$

where $U(s)$ and $M(s)$ are the Laplace transforms of the output $u(t)$ and input $m(t)$, respectively.

From $(1)$ we get


Transforming $(2)$ back to the time domain finally gives the desired differential equation



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.