# Impulse Response Function-Green's Function

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?

• 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 $c=1$. Commented Aug 11, 2019 at 22:54
• 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?
– axel
Commented Aug 11, 2019 at 23:09
• @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. Commented Aug 12, 2019 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

$$U(s)\left[s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}\right]=\frac{A}{\tau}M(s)\tag{2}$$

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

$$\ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t)=\frac{A}{\tau}m(t)\tag{3}$$