Find the Laplace function of a piecewise function

Question

$$f(t)=\left\{\begin{array}{ll}t^2&\text{if}\ 0\leq t\lt 1\\2t-1&\text{if}\ 1\leq t\lt5\\9&\text{if}\ t\geq5\end{array}\right.$$

Please help, I don't understand the steps.

Answer 1

Well, you know the definition of the Laplace transform:

$$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}=\int_0^\infty\text{f}\left(t\right)\exp\left(-\text{s}t\right)\space\text{d}t\tag1$$

So, for your function we get:

$$\text{F}\left(\text{s}\right)=\int_0^1t^2\exp\left(-\text{s}t\right)\space\text{d}t+\int_1^5\left(2t-1\right)\exp\left(-\text{s}t\right)\space\text{d}t+\int_5^\infty9\exp\left(-\text{s}t\right)\space\text{d}t\tag2$$

Answer 2

Rewrite your function as: $$f(t)=t^2(u(t)-u(t-1))+(2t-1)(u(t-1)-u(t-5))+9u(t-5)$$ $$f(t)=t^2u(t)-(t-1)^2u(t-1)-2(t-5)u(t-5)$$ And apply the following property of the Laplace Transform: $$\mathcal {L}\{u(t-c)f(t-c)\}=e^{-cs}F(s)$$