# Dirac Delta and Laplace Transform

• Mar 27th 2011, 05:53 PM
kenndrylen
Dirac Delta and Laplace Transform
Hi guys, my professor recently gave us this problem:

$y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...

$Y(s)=\frac{1}{s^2+4}\sum_{k=1}^{\infty}e^{-sk\pi}=\cfrac{1}{(e^{s\pi}-1)(s^2+4)}$.

However I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.
• Mar 27th 2011, 07:42 PM
TheEmptySet
Quote:

Originally Posted by kenndrylen
Hi guys, my professor recently gave us this problem:

$y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...

$Y(s)=\frac{1}{s^2+4}\sum_{k=1}^{\infty}e^{-sk\pi}=\cfrac{1}{(e^{s\pi}-1)(s^2+4)}$.

However I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.

Here is an idea instead of summing the series just invert the series term by term

$\displaystyle Y(s)=\sum_{k=1}^{\infty}\frac{e^{-s\pi k}}{s^2+4}$

$\displaystyle \mathcal{L}^{-1}\left( \sum_{k=1}^{\infty}\frac{e^{-s\pi k}}{s^2+4}\right)=\sum_{k=1}^{\infty}\mathcal{L}^{-1}\left( \frac{e^{-s \pi k}}{s^2+4}\right)=\sum_{k=1}^{\infty}\sin\left(t-k\pi \right)u(t-k\pi)$

Where $u(t)=\begin{cases}0, \text{ if } t < 0 \\ 1, \text{ if } t \ge 0 \end{cases}$ the Heaviside function.
• Mar 27th 2011, 07:45 PM
Prove It
You have $\displaystyle Y(s) = \frac{1}{s^2 + 4}\sum_{k = 1}^{\infty}e^{-sk\pi} = \sum_{k = 1}^{\infty}e^{-sk\pi}\left(\frac{1}{s^2 + 4}\right)$.

Now recall that $\displaystyle \mathcal{L} ^{-1}\left[e^{-as}F(s)\right] = f(t-a)H(t-a)$, and the laplace transform of a sum is the sum of the laplace transforms.
• Mar 27th 2011, 08:12 PM
kenndrylen
Oh! Nice idea guys! I don't know why I missed that (Giggle)

Thanks both of you (Nod)