# Laplace inverse transform of an exponential multiplying another function

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?

• 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. May 3, 2023 at 18:06
• 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? May 3, 2023 at 21:00

The Laplace Transform on the interval $$\0 is common used in electronics. Certainly causal signals $$\u(t)=0; t.

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\$$.

• 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 $t_0$? May 3, 2023 at 21:08
• 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. May 3, 2023 at 21:49
• 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? May 4, 2023 at 19:14
• @Mauro yes! that is correct May 4, 2023 at 21:29