# Laplace transform of $u(t)$

I have been trying to teach myself about Laplace Transformations before I take Differential Equations in the spring and I ran into a problem that I'm not sure how to solve.
If an undamped system is given a force that consists of two piecewise functions, like $f(t) = u(t) - u(t-1)$, do you take the laplace transform of each function separately, i.e. $\mathscr{L}[f(t)] = \mathscr{L}[u(t)] - \mathscr{L}[u(t-1)]$?

Or do you need to manipulate $u(t)$ so it has the same $a$ as $u(t-1)$ before you find the transform, assuming $u(t-1)$ is in the form $u(t-a)$.

If it is the first case, $\mathscr{L}[u(t)] = \frac{1}{s}$, correct?

I'm aware that this could potentially be a trivial issue but I just want to confirm that I have the right approach before I move on.

Thanks!

• Take the LT of each separately. Commented Oct 20, 2017 at 15:01
• The LT is simply $\int_0^1 e^{-st}\,dt$. Commented Oct 20, 2017 at 15:03
• In this case it is straightforward to compute the Laplace transform explicitly, but in general, since it is a linear operator, we have ${\cal L} (f+g) = {\cal L}f + {\cal L} g$. The only time that you might not do this is in the unlikely case that if the separate $f,g$ do not have a Laplace transform (because of 'wild' behaviour) but the combination does. Commented Oct 20, 2017 at 15:06
• Thanks! I appreciate the help. Commented Oct 20, 2017 at 15:06

$$\mathcal L [\alpha f(x)+\beta g(x)]=\alpha\mathcal L [ f(x)]+\beta\mathcal L[ g(x)]$$ for $\alpha, \beta \in \mathbb R$