# How can I find ccL^(-1){(1)/(1+2e^(−s)+e^(−2s))}?

Inverse Laplace transform of $\frac{1}{1+2{e}^{-s}+{e}^{-2s}}$
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honejata1
I guess that you would like to find a function g(t) such that $h\left(s\right)\ast g\left(s\right)=\delta \left(t\right)$ (where ∗ means convolution), where $h\left(t\right)=\delta \left(t\right)+2\delta \left(t-1\right)+\delta \left(t-2\right)$
After taking the Laplace Transform of both parts we get $H\cdot G=\mathcal{L}\left\{\delta \right\}=1$ (where ⋅ means multiplication) . Hence, we need to compute the inverse Laplace transform of
$G\left(s\right)=\frac{1}{H\left(s\right)}=\frac{1}{1+2{e}^{-s}+{e}^{-2s}}=\frac{1}{\left(1+{e}^{-s}{\right)}^{2}}=\sum _{n=0}^{\mathrm{\infty }}\left(n+1\right)\left(-1{\right)}^{n}{e}^{-ns}.$
By recalling that ${\mathcal{L}}^{-1}\left({e}^{-ns}\right)=\delta \left(t-n\right)$, we obtain
$g\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}\left(n+1\right)\left(-1{\right)}^{n}\delta \left(t-n\right).$