Here is the question:
F(t) = {
0, t<4
5 sin(pi * t) 4<=t<5
0, t>=5
How do I find the Laplace transformation of this function?
My answer is (5pi/(s^2+pi^2))(e^(-4s)-e^(-5s)) but it's wrong..
help plz!!
Another more useful expression of the f(t) is...
$\displaystyle f(t) = 5\cdot \{\sin \pi t\cdot \mathcal{U}(t-4) - \sin \pi t\cdot \mathcal{U}(t-5)\}$
... where $\displaystyle \mathcal{U} (*)$ is the 'Haeviside step function'. Now You have to compute the LT of (1) 'term by term' in appropriate way...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The definition of the Laplace transform of a function $\displaystyle f(t)$ is $\displaystyle L[f(t)]=\int_{0}^{\infty}f(t)e^{-st}\,\,dt$
For this particular function we have:
$\displaystyle L[f(t)]=\int_{0}^{4}0e^{-st}\,\,dt + \int_{4}^{5}5sin(\pi t)e^{-st}\,\,dt + \int_{5}^{\infty}0e^{-st}\,\,dt$
The first and third integral terms above are equal to $\displaystyle 0$, so the Laplace transform would be $\displaystyle \int_{4}^{5}5sin(\pi t)e^{-st}\,\,dt$
Now, using integration by parts, I got the answer:
$\displaystyle L[f(t)]=\frac{5\pi (e^{4s} + e^{5s})}{e^{9s}(s^{2} + \pi^{2})}$
I kinda rushed through the integration, so check for yourself what the integral equates to.