Please check my work. I need to find the inverse Laplace Transform for the following function.
$$\mathcal{L}\{f(t)\}=\frac{e^{-s}}{s^2+\pi^2}$$
My solution:
Recognizing the exponential factor as a t-shift we pull it to the side. We can also add a $\pi$ factor to the numerator and denominator to reveal a transform of a sine function…
$$\mathcal{L}\{f(t)\}=e^{-s}\cdot \frac{1}{\pi}\frac{\pi}{s^2+\pi^2}$$
sprinkle in a little pattern matching to get…
$$\mathcal{L^{-1}}\{f(t)\}=\frac{1}{\pi}\cdot sin(\pi (t-1))\cdot u(t-1)$$
but…
$$sin(\pi (t-1))=-sin(\pi t)$$
hence…
$$f(t)=-\frac{1}{\pi}\cdot sin(\pi t)\cdot u(t-1)$$
Best Answer
Instead of trying to manipulate forms (and pattern matching), we can use the inverse the Laplace transform, $$ \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds = \sum\text{Res} $$ Your $F(s)$ has two poles at $s=\pm i\pi$. \begin{align} \mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{s(t-1)}}{s^2+\pi^2}ds\\ &=\lim_{s\to i\pi}(s-i\pi)\frac{e^{s(t-1)}}{s^2+\pi^2}+\lim_{s\to -i\pi}(s-i\pi)\frac{e^{s(t-1)}}{s^2+\pi^2} \end{align} but before we determine the limits we have to think about when $e^{s(t-1)}$ converges since $e^t\to\infty$ as $t\to\infty$. That is, we require $s(t-1)<0$ or $t<1$. We can achieve this from the shifted Unit step function $$ \mathcal{U}(t-1)=\begin{cases} 0, &t<1\\ 1, & t>1 \end{cases} $$ From our limits, we obtain $$ \mathcal{L}^{-1}\{F(s)\} = \frac{e^{i\pi(t-1)}}{2i\pi}\mathcal{U}(t-1)+\frac{e^{-i\pi(t-1)}}{-2i\pi}\mathcal{U}(t-1)=\frac{\sin[\pi(t-1)]\mathcal{U}(t-1)}{\pi} $$