What is the inverse laplace transform of $\frac{s}{s+1}$?
My work was: $$ X(s)=\frac{s}{s+1}\\ X(s)=s\frac{1}{s+1}\\ x(t)=\frac{d}{dt}e^{-t}=-e^{-t} $$
My only issue is that when I check my answer with wolfram alpha, it says that the inverse laplace transform of $\frac{s}{s+1}$ is actually $-e^{-t}+\delta(t)$. What is the correct way to find the inverse transform?
$\endgroup$2 Answers
$\begingroup$HINT: $$\frac s{s+1}=1-\frac1{s+1}$$
We know, $\mathcal{L}(e^{at})=\frac1{s-a}$
and we can prove $\mathcal{L}(\delta(t-b))=e^{-bs}$, put $b=0$
$\endgroup$ $\begingroup$Simple way to solve this problem sorry about the formatting
\begin{align} &=\frac s{s+a}\\ &=\frac{(s+a)-a}{s+a}\\ &=\frac{s+a}{s+a}-\frac a{s+a}\\ &=1-\frac a{s+a}\\ &=δ(t)-\mathcal(ae^{-at}) \end{align}
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