What will be the graph of inverse Laplace transform of s/(s+1)

Question

[1]Inverse laplace transform of $\frac{s}{s+1}$ will be $-e^{-t}u(t)+\delta(t)$.How to draw its graph?

Answer 1

Well, that is true:

$$\text{f}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{1+\text{s}}\right]_{\left(t\right)}=\delta\left(t\right)-e^{-t}u(t)\tag1$$

So, for the dirac delta function:

$$\delta\left(t\right):=\begin{cases}+\infty\space\space\space\space\space\space\space\space t=0\\\\0\space\space\space\space\space\space\space\space\space\space\space\space\space t\ne0\end{cases}\tag2 $$

So, when $t>0$:

$$\text{f}\left(t\right)=\delta\left(t\right)-e^{-t}=-e^{-t}\tag3$$

Answer 2

Dirac delta $\delta(t)$ is often plotted as a right arrow at the origin, and zero elsewhere. Overall, the plot of

$$\color{red}{-e^{-t}u(t)}+\color{blue}{\delta(t)}$$ is something like the following: