# The Laplace Transform of Piecewise Function

Write the following as an unit step function and find the Laplace transform.

$f(t)=\begin{cases}{t}&0 \leq t < 3\\ 3&3 \leq t < 4\\ 11-2t& 4 \leq t < 5.5 \\ 0&t \geq 5.5 \end{cases}$

Workings:

$f(t) = t - u(t-3)(t) + u(t-3)(3) - u(t-4)(3) + u(t-4)(11-2t) - u(t-5.5)(11-2t)$

$f(t) = t - u(t-3)t+ 3u(t-3) - 3u(t-4) + u(t-4)(11-2t) - u(t-5.5)(11-2t)$

$\mathcal{L}\{f(t)\} = \mathcal{L}\{t\} - \mathcal{L}\{u(t-3)t\} + 3\mathcal{L}\{u(t-3)\} - 3\mathcal{L}\{u(t-4)\} + \mathcal{L}\{u(t-4)(11-2t)\} - \mathcal{L}\{u(t-5.5)(11-2t)\}$

$\mathcal{L}\{f(t)\} = \frac{1}{s^2} - \mathcal{L}\{u(t-3)t\} + \frac{3e^{-3s}}{s} - \frac{3e^{-4s}}{s} + + \mathcal{L}\{u(t-4)(11-2t)\} - \mathcal{L}\{u(t-5.5)(11-2t)\}$

Now I'm not sure what to do. Any help will be appreciated.

• Just Laplace transform each of the four pieces rather than trying to write out the whole function first – danimal Mar 26 '15 at 14:23
• Okay. $\mathcal{L}\{t\} = \frac{1}{s^2}$ $\mathcal{L}\{t(u(t-3)\} = \frac{1}{s^2} \frac{e^{-3s}}{s}$ $3\mathcal{L}\{u(t-3)\} = \frac{3e^{-3s}}{s}$ $3\mathcal{L}\{u(t-4)\} = \frac{3e^{-4s}}{s}$ $\mathcal{L}\{u(t-4)(11-2t)\} = \frac{e^{-4s}}{s}\frac{2e^{-11s}}{s}$ $\mathcal{L}\{u(t-5.5)(11-2t)\} = \frac{e^{-5.5s}}{s}\frac{2e^{-11s}}{s}$ – TillermansTea Mar 26 '15 at 14:51

Note that if $a>0$, $$\mathcal{L}(u(t-a)f(t)) = \int_0^{\infty} e^{-st} u(t-a)f(t) \, dt = \int_a^{\infty} e^{-st} f(t) \, dt,$$ since $u(t-a)$ is $0$ for $t<a$. Then setting $x+a=t$, we have $$\mathcal{L}(u(t-a)f(t)) = e^{-as}\int_0^{\infty} e^{-sx} f(a+x) \, dx,$$ if you prefer.