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

  • $\begingroup$ Sorry, I didn't realize how similar the asterisk looks to the convolution symbol in Tex. $\endgroup$ – Matt Jun 17 '17 at 7:03
  • $\begingroup$ I don't understand the edits made here: math.stackexchange.com/posts/2325824/revisions $\endgroup$ – Matt Jun 17 '17 at 7:11
  • $\begingroup$ 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 $\endgroup$ – Robert Z Jun 17 '17 at 7:19
  • $\begingroup$ 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. $\endgroup$ – Matt Jun 17 '17 at 7:22
  • $\begingroup$ Sorry. My fault. $\endgroup$ – 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 $$g(t)=\sum_{n=0}^{\infty}(n+1)(-1)^n\delta(t-n). $$

  • $\begingroup$ Thank you. I think the big part I'm missing, though, is that last step where the fraction becomes a summation. How is that expanded? $\endgroup$ – Matt Jun 17 '17 at 6:51
  • $\begingroup$ @Matt Take the derivative of $1/(1-y)=\sum_{n=0}^{\infty}y^n$, let $x=-y$ and shift the index. $\endgroup$ – Robert Z Jun 17 '17 at 6:55
  • $\begingroup$ @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? $\endgroup$ – Robert Z Jun 17 '17 at 6:58
  • $\begingroup$ Apologies. I have corrected the confusing symbols in my question. I am looking to find g(t) to "undo" h(t) in the signal path. $\endgroup$ – Matt Jun 17 '17 at 7:04
  • $\begingroup$ @Anonymous Yes, corrected. Thanks! $\endgroup$ – Robert Z Jun 17 '17 at 7:25

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