Calculate the inverse Laplace transform of $\frac{1}{1-e^{-s}}$

Question

During my signals and systems class i came across this and i have to find it's inverse Laplace transform. I don't know how.

$$\mathcal{L}^{-1} \Big\{ \frac{1}{1-e^{-s}} \Big\} = \ ?$$

Any help appreciated. Thanks in advance.

Answer 1

Utilizing the property \begin{align} \mathcal{L}\{f(t)\} = \frac{\int_{0}^{T} e^{- s u} \, f(u) \, du}{1 - e^{-s T}} \end{align} where $f(t+T) = f(t)$ then \begin{align} \mathcal{L}^{-1}\{\frac{1}{1- e^{-s}}\} = \frac{\int_{0}^{1} e^{-su} \, \delta(u) \, du}{1- e^{-s}} \end{align} for which $f(t) = \delta(t)$ with the periodic property $f(t+1) = f(t)$. This leads to $$f(t) = \sum_{n=0}^{\infty} \delta(t-n)$$ Alternatively \begin{align} \mathcal{L}^{-1}\left\{\frac{1}{1-e^{-s}}\right\} = \sum_{n=0}^{\infty} \mathcal{L}^{-1}\{e^{-ns}\} = \sum_{n=0}^{\infty} \delta(t-n) \end{align}