Question

Find the laplace transform of the function:
f(t)=0 for t <1; f(t)=t for t >= 1

Answer 1

Write it in terms of the unit step function.$$f(t)=u(t-1)t$$Can you do it now?

Answer 2

http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

$$\mathcal{L}\{f(t)\}=\mathcal{L}\{u(t-1)t\}\\F(s)=e^{-s}\frac1{s^2}$$

Answer 3

Thanks for the help. The final answer you gave was wrong but the initial step helped a lot. Seemed to have blanked on that property.

\[\mathcal{L}\{u(t-1)t^1\} = (-1)^1F^{(1)}(x) = -\frac{d}{dt}\frac{e^{-s}}{s} = e^{-s}(s^{-1}+s^{-2})\]

This agrees with the final answer in the book. Thank you for the help.

Answer 4

hm? how'd you get that?

Answer 5

oops nvm

Answer 6

The book calls it 'Differentiation of Transforms.

L(-t*f(t)) = F'(S)
IF you carry it out then you get
\[L\{t^n*f(t)\} = (-1)^n F^{(n)}(s)\]