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?

  • 2
    \$\begingroup\$ While Laplace Transforms are used in EE, given that this reads as pretty much a pure math question, I'm wondering if it would be more at home on math.se. \$\endgroup\$
    – SSilk
    May 3, 2023 at 18:06
  • \$\begingroup\$ It could be; I posted it here because I was studying this while studying control theory, but you're right it's more abstract than that. Should I move it? \$\endgroup\$
    – Mauro
    May 3, 2023 at 21:00

1 Answer 1


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\$.

  • \$\begingroup\$ I edited the title, thanks. Why would \$H(t)\$ give the same result? I understand the answer you're giving, but I thought that between \$H(t)\$ and \$H(t - t_0)\$ there is a difference, hence the question: namely, I thought the latter is 0 also between t = 0 and \$t_0\$, while the former isn't. Do you mean they are the same if you understand the "switching" really happens at \$t_0\$? \$\endgroup\$
    – Mauro
    May 3, 2023 at 21:08
  • \$\begingroup\$ Do the LT of \$f(t-t_0)H(t)\$ and then of \$f(t-t_0)H(t-t_0)\$ for any "causal" function \$f(t)\$ and see for your self. If the function is not causal then you are correct there will be a difference. \$\endgroup\$
    – RussellH
    May 3, 2023 at 21:49
  • \$\begingroup\$ What about if I have more than one exponential? I was trying to plot the inverse transform of \$Y(s)=4(e^{-s}/s^2-e^{-3s}/s^2)\$ (that should be \$y(t)=4(H(t-1)(t-1)-H(t-3)(t-3))\$), but since the right form is that with \$H(t-a)\$ and I have two different exponentials I'm not sure how I should plot it: the first starts climbing at t = 1, the second at t = 3, so there is an initial ramp then it levels at y = 8? \$\endgroup\$
    – Mauro
    May 4, 2023 at 19:14
  • \$\begingroup\$ @Mauro yes! that is correct \$\endgroup\$
    – RussellH
    May 4, 2023 at 21:29

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