Inverse laplace transform of a hard partial fraction, $1/[s^2(s^2+\omega^2)]$
Let $F(s) = H(s) G(s)$ with $H(s) = \frac{1}{s^2}$ and $G(s) = \frac{1}{s^2 + \omega^2}$
By the convolution theorem $\mathcal{L}^{1}\big(H(s) G(s)\big) = (h * g)(t) = \int_0^t \ h(\tau) g(t\tau) \ d\tau$
From a table, $h(\tau) = \mathcal{L}^{1}( 1/s^2 ) = \tau \ u(\tau)$ where $u(\tau)$ is the unit step function. $g(\tau) = \mathcal{L}^{1}\big( 1/(\omega^2+ s^2 )\big) = \frac{\sin{\omega \tau}}{\omega} u(\tau)$.
$$(h * g)(t) = \int_0^t \ \tau \ u(\tau) \frac{\sin{\omega (t  \tau)}}{\omega} u(t\tau) \ d\tau$$
Since $\tau$ is always positive $u(\tau) = 1$ on this domain. Similarly, since $t > \tau$ on this domain, $u(t\tau) = 1$. The resulting integral can be done easily.
$$(h * g)(t) = \int_0^t \ \tau \ \frac{\sin{\omega (t  \tau)}}{\omega} \ d\tau = \frac{1}{\omega^2} (t  \frac{\sin{\omega t}}{\omega})$$
Related videos on Youtube
user122415
Updated on March 20, 2021Comments

user122415 over 2 years
So the question is find the inverse of $\dfrac{1}{s^2(s^2+\omega^2)}$. And here is the solution. I have no idea why its done this way. I would think to take a partial fraction of the form $\dfrac{a}{s}+\dfrac{b}{s^2}+\dfrac{c}{s^2+\omega^2}$ and then solve that, but that seems too complicated. This textbook solution looks like some brute force factorization specific to this problem only. Is there a general concept or way to tackle these kinds of problems thanks.

Kevin Driscoll almost 10 yearsYou know the inverse transform of $\frac{1}{s^2}$ and the inverse transform of $\frac{1}{s^2 + \omega^2}$. Since they are multiplied here I would try using the Convolution theorem.

Amzoti almost 10 years@user122415: It is often helpful to do a partial fraction expansion on $F(s)$ such that you can get the result into forms that allow you to use a Laplace Transform table. Do the partial fraction expansion first and then the inverse transform of that result.

user122415 almost 10 yearsjajaja factor it into something that's on the laplace table, then it becomes easy. ok i just didn't realize that. thx a lot kevin. u 2 amz

Disintegrating By Parts over 9 yearsYou don't need the $1/s$ term because only powers of $s^{2}$ are found on the left. Another way to see this: let $z=s^{2}$ on the left, do the partial fraction decomposition in $z$ and substitute $z=s^{2}$ back in at the end.

Did over 9 years"I would think to take a partial fraction of the form a/s+b/s2+c/s2+ω2 and then solve that" This is exactly what the authors are doing.
