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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).

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

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

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