Hi. I have to find the Laplace transform for:
$\displaystyle f(t)=\begin{Bmatrix}{ 0}&\mbox{ if }& t<0\\ \displaystyle\frac{t}{a}-4n & \mbox{if}& 4na<t<(4n+1)a\\ -\displaystyle\frac{t}{a}+4n+2 & \mbox{if}& (4n+1)a<t<(4n+2)a \\ 0 & \mbox{if}& (4n+2)a<t<(4n+4)a \end{matrix},a>0,n=0,1,2,3,...$

So I considered the period $\displaystyle T=(4n+4)a$, but I've noted the detail that for $\displaystyle n\neq{0}$ the function isn't defined on the interval $\displaystyle 0<t<4na$. Anyway, I get to this solution:

$\displaystyle F(s)=\displaystyle\frac{1}{1-e^{-s(4n+4)a}} \left [ \displaystyle\frac{2}{a}\displaystyle\frac{e^{-s(4n+1)a}}{s^2}}+ \displaystyle\frac{1}{a}\displaystyle\frac{e^{-s4na}}{s^2}}+ \displaystyle\frac{(4n-1)}{s}e^{-s(4n+2)a}+ \displaystyle\frac{1}{a}\displaystyle\frac{e^{-s(4n+2)a}}{s^2}} \right ]$

I don't know if its right, and I don't pretend anyone to work with this horrible function, but perhaps someone smarter than me by simple eye check can tell me if it could be right or not.

So, what you say?

Bye there.