# Thread: Dirac Delta and Laplace Transform

1. ## Dirac Delta and Laplace Transform

Hi guys, my professor recently gave us this problem:

$\displaystyle y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ 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...

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

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.

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

$\displaystyle y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ 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...

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

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

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

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

Where $\displaystyle u(t)=\begin{cases}0, \text{ if } t < 0 \\ 1, \text{ if } t \ge 0 \end{cases}$ the Heaviside function.

3. You have $\displaystyle \displaystyle Y(s) = \frac{1}{s^2 + 4}\sum_{k = 1}^{\infty}e^{-sk\pi} = \sum_{k = 1}^{\infty}e^{-sk\pi}\left(\frac{1}{s^2 + 4}\right)$.

Now recall that $\displaystyle \displaystyle \mathcal{L} ^{-1}\left[e^{-as}F(s)\right] = f(t-a)H(t-a)$, and the laplace transform of a sum is the sum of the laplace transforms.

4. Oh! Nice idea guys! I don't know why I missed that

Thanks both of you