Consider the following function:
$$
f(t) = \begin{cases} \cos(t) &\mbox{if } 0 \leq t \le π \\
\sin(t) & \mbox{if } π \leq t \le 2π \\
e^t & \mbox{if } t > 2\pi \end{cases}
$$
A way to approach the solution would be using the definition, that is:
$$\int_{0}^{π} \cos(t)e^{-st} dt + \int_{π}^{2π} \sin(t) e^{-st}dt + \int_{2π}^{+\infty} e^{-t}e^{-st} dt$$
then solve the integrals and find the solution.
My question is:
Can we use basic laplace transforms on a cased function and if not why?
With the term basic laplace transforms I mean
paradigms such as:
$$ L[\cos(at)] = \frac{s}{s^2+a^2}$$
$$ L[\sin(at)] = \frac{a}{s^2+a^2}$$
$$ L[e^{at}]= \frac{1}{s-a}$$
In that case the Laplace transform of f(t) would be the following:
$$ L[f(t)] = 2*\frac {s^2}{s^2+1} + \frac {1}{s – 1} $$ but it just doesn't feel right.
Best Answer
$$\begin{align*} f(t)&=\begin{cases}\cos t&\text{for }0\le t\le\pi\\\sin t&\text{for }\pi<t\le2\pi\\e^t&\text{for }t>2\pi\end{cases}\\[1ex] &=\cos t(u(t)-u(t-\pi))+\sin t(u(t-\pi)-u(t-2\pi))+e^tu(t-2\pi) \end{align*}$$
where $u(t)$ is the unit step (Heaviside step) function defined by
$$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}$$
It's easy to show
$$\mathcal L_s\{f(t-c)u(t-c)\}=e^{-cs}\mathcal L_s\{f(t)\}\quad\quad(*)$$
where $\mathcal L_s\{f(t)\}$ is the Laplace transform of $f(t)$. We can expand $\mathcal L_s\{f(t)\}$ as
$$\mathcal L_s\left\{\cos t\,u(t)+\cos(t-\pi)u(t-\pi)-\sin(t-\pi)u(t-\pi)+\sin(t-2\pi)u(t-2\pi)+e^{2\pi}e^{t-2\pi}u(t-2\pi)\right\}$$
since $-\cos t=\cos(t-\pi)$, $\sin t=-\sin(t-\pi)$, and $\sin(t-2\pi)=\sin t$. Then use $(*)$ to simplify this accordingly.