I'm trying to find the Laplace transform of $7 e^{-3t} u(t-3)$, where $u$ is the heaviside step function. However, we've never really gone through what the Laplace transform of the heaviside step function actually is, so I'm a little confused as to how this would work out.

(This is a question from a previous exam paper, I'm just studying for my exam in a few days. I didn't want to tag homework, because it isn't exactly homework, and it's not something we've covered in lectures, it's more of a knowledge question).

  • $\begingroup$ The step function only restricts the domain of integration. $\endgroup$ – Jas Ter Oct 31 '13 at 8:14

You can compute $\mathcal{L}\left\{7e^{-3t}u(t-3)\right\}$ straight from the integral definition (it's actually rather straightforward in this case), but I would also suggest that you learn how to compute these using the appropriate shift formulas; I'll at least derive one of them for you.

For starters, let's compute $\mathcal{L}\{f(t-a)u(t-a)\}$, where

$$u(t-a) = \begin{cases} 1 & t\geq a\\ 0 & t<a\end{cases}$$

Then it follows that

$$\begin{aligned}\mathcal{L}\{u(t-a)f(t-a)\} &= \int_0^{\infty} e^{-st} f(t-a)u(t-a)\,dt \\ &= \int_a^{\infty} e^{-st}f(t-a)\,dt \\ &= \int_0^{\infty} e^{-s(\tau+a)}f(\tau)\,d\tau\quad(\text{by taking $t-a=\tau$}) \\ &= e^{-as}\int_0^{\infty}e^{-s\tau}f(\tau)\,d\tau\\ &= e^{-as}F(s)\end{aligned}$$

where $F(s) = \mathcal{L}\{f(t)\}$. The above result is one of the shift formulas.

So to compute $\mathcal{L}\left\{7e^{-3t}u(t-3)\right\}$, you'll need to rewrite your function in the form $f(t-3)u(t-3)$; i.e. $$7e^{-3t}u(t-3) = 7e^{-3(t-3+3)}u(t-3) = 7e^{-9}e^{-3(t-3)}u(t-3)$$ Thus, it now follows that $$\begin{aligned}\mathcal{L}\{7e^{-3t}u(t-3)\} &= 7e^{-9}\mathcal{L}\{e^{-3(t-3)}u(t-3)\}\\ &= 7e^{-9}\left(e^{-3s}\mathcal{L}\{e^{-3t}\}\right)\\ &=\frac{7e^{-9}e^{-3s}}{s+3} \\ &= \frac{7e^{-3(s+3)}}{s+3}\end{aligned}$$

There's another way of computing $\mathcal{L}\left\{7e^{-3t}u(t-3)\right\}$. You can couple this with another type of shift formula; in particular,

$$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$$

and hence (without much difficulty), we see that

$$\mathcal{L}\{e^{bt}f(t-a)u(t-a)\} = e^{-a(s-b)}F(s-b)$$

Taking $f(t)=1$ in your problem leaves us with

$$\begin{aligned}\mathcal{L}\{7e^{-3t}u(t-3)\} &= 7e^{-3(s+3)}\left.\mathcal{L}\{1\}\right|_{s\to s+3}\\ &= \frac{7e^{-3(s+3)}}{s+3}\end{aligned}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.