I need to find a function $g(t)$ such that $h(t)*g(t)=\delta(t)$, where $$ h(t) = \delta(t) + 2\delta(t-1) + \delta(t-2) \;. $$ I have found $$ H(s) = 1 + 2\mathrm{e}^{-s} + \mathrm{e}^{-2s} $$ and $$ H(s)G(s)=1 $$ and thus $$ G(s) = \dfrac{1}{1 + 2e^{-s} + \mathrm{e}^{-2s}} \;, $$ but I am unsure how to find the inverse Laplace transform of $G(s)$. I have reason to believe that it may involve an infinite series, but I can't find any examples similar to this.
How can I find $\mathcal{L}^{-1}\left\{\dfrac{1}{1 + 2e^{-s} + \mathrm{e}^{-2s}}\right\}$ ?
I guess that you would like to find a function $g(t)$ such that $h(s)*g(s)=\delta(t)$ (where $*$ means convolution), where $h(t) = \delta(t) + 2\delta(t-1) + \delta(t-2)$.
After taking the Laplace Transform of both parts we get $H\cdot G=\mathcal{L}\{\delta\}=1$ (where $\cdot$ means multiplication) . Hence, we need to compute the inverse Laplace transform of $$G(s)=\frac{1}{H(s)}=\frac{1}{1 + 2e^{-s} + e^{-2s}}=\frac{1}{(1 + e^{-s})^2} =\sum_{n=0}^{\infty}(n+1)(-1)^ne^{-ns}.$$ By recalling that $\mathcal{L}^{-1}(e^{-ns})=\delta(t-n)$, we obtain $$g(t)=\sum_{n=0}^{\infty}(n+1)(-1)^n\delta(t-n). $$