Understanding Laplace Transform of f(t)

[georgeh]

i have f(t) defined piece-wise and continous..
f(t) = 0, t < 2pi
t-pi , pi <=t<2pi
0 , t >=2pi

i have so far g(t)=U_pi*f(t-pi)-U_2pi*f(t-pi)
if i do the laplace,
i get e^-pis/s^2-e^-2pis/s^2
in the book, they have
e^-pi*s/s^2 -e^-2pi*s/s^2 (1+pi*s)
I am not sure how they got the factor of (1+pi*s)
..

 

[FrogPad]

Are you sure $f(t)$ is correct. You have:
$$f(t) = \left\{ \begin{array}{l} 0, \,\, t<2\pi \\ t-\pi, \,\, \pi \leq t < 2\pi \\ 0, \,\, t \geq 2\pi$$So when $t<2\pi$ and $\pi \leq t < 2\pi \\$ it equals $0$ and $t-\pi$. I'm assuming you mean $f(t) = 0|t<\pi$ ?

I got (assuming $f(t)$ is wrong):

$$\frac{e^{-\pi s}}{s^2} - \frac{\pi e^{-2\pi s}}{s} - \frac{e^{-2\pi s}}{s^2}$$

 

[georgeh]

sorry, what i meant was
on the first interval, t < pi, not two pi.