I was trying to understand the inverse transform of \$Y(s)=F(s)e^{-as}\$, but on different sources I found three different answers (H being the step): \$f(t-a)\$, \$H(t)f(t-a)\$ and \$H(t-a)f(t-a)\$.
The first and the second I guess could be the same, given usually the analysis starts at t = 0, but which inverse transform is the right one?
The Laplace Transform on the interval \$0<t<\infty\$ is common used in electronics. Certainly causal signals \$u(t)=0; t<t_0\$.
We all get lazy when writing equations because often the context is understood.
For example: if the Laplace transform of \$f(t)=t\$, then all the values for t<0 get ignored. What we take the transform of is \$f(t)H(t)\$ where \$H(t)\$ is the unit step function (also called the switching function).
So then the inverse Laplace transform \$L^{-1}[G(s)] = g(t)H(t)\$. The multiplication by H(t) is often not written but understood to be there.
When a causal function is shifted to the right by \$t_0\$, the values of the function from 0 to \$t_0\$ is zero. So in this case the "switching" happens at \$t_0\$ instead of 0, so \$H(t-t_0)\$ is more descriptive but \$H(t)\$ will give the same result.$$L^-{1}[F(s)e^{-t_0 s}]=f(t-t_0)H(t-t_0)\tag{equ 1}$$is the correct and most descriptive form. However the other forms are correct if it is "understood that they mean equ 1
Solutions for the Laplace transform are valid for \$t \ge 0\$.
