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.

$$(h * g)(t) = \int_0^t \ \tau \ \frac{\sin{\omega (t - \tau)}}{\omega} \ d\tau = \frac{1}{\omega^2} (t - \frac{\sin{\omega t}}{\omega})$$


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Updated on March 20, 2021


  • user122415
    user122415 over 2 years

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    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
      Kevin Driscoll almost 10 years
      You 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
      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
      user122415 almost 10 years
      jajaja 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
      Disintegrating By Parts over 9 years
      You 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
      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.