# Inverse Laplace transform of $\frac{1}{1 + 2e^{-s} +e^{-2s}}$

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 $\mathcal{L}^{-1}\left\{\dfrac{1}{1 + 2e^{-s} + \mathrm{e}^{-2s}}\right\}$ ?

• Sorry, I didn't realize how similar the asterisk looks to the convolution symbol in Tex. – Matt Jun 17 '17 at 7:03
• I don't understand the edits made here: math.stackexchange.com/posts/2325824/revisions – Matt Jun 17 '17 at 7:11
• The Laplace transform of $\delta(t-1)$ is $e^{-s}$ and not $2^{-s}$. Do you agree? BTW if you are new here: math.stackexchange.com/tour – Robert Z Jun 17 '17 at 7:19
• I agree. However, the laplace transform of $2\delta(t-1)$ is $2e^{-s}$, right? I do see that I was missing an "e" in one of those. – Matt Jun 17 '17 at 7:22
• Sorry. My fault. – Robert Z Jun 17 '17 at 7:23

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 $$g(t)=\sum_{n=0}^{\infty}(n+1)(-1)^n\delta(t-n).$$
• @Matt Take the derivative of $1/(1-y)=\sum_{n=0}^{\infty}y^n$, let $x=-y$ and shift the index. – Robert Z Jun 17 '17 at 6:55
• @Matt Is my guessing about your question correct? That is that you would like to find a function $g(t)$ such that the convolution of $h(s)$ and $g(s)$ is $1$. In case could you please edit your question? – Robert Z Jun 17 '17 at 6:58