[Math] Why Laplace Transform of 1 and u(t) are the same.

laplace transformstep function

I looked at the Laplace Transform Table on the Internet and I saw that Laplace Transform of step function is equal to laplace transform of 1 .Is it because the region of convergence of the Laplace transform of 1 or what?

Secondly Why does the Laplace transform of u(t) not have impulsive term?

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Best Answer

Because $u(t)= \begin{cases} 1,& \text{if } t\geq 0\\ 0, & \text{otherwise} \end{cases}$
The laplace transform of a function $f$: $$F(s):=\int_0^\infty f(t)e^{-st}\mathrm{d}t$$ And in the domain $t \in [0,\infty)$ both the $f(t)=1$ and $u(t)$ equals to $1$.

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