i was given this:
$$f(t) = \cases{3 \quad t

Question

i was given this:
$$f(t) = \cases{3 \quad t < 2 \ 1 \quad 2<t<5 \ t \quad 5 < t < 8 \ \frac{t^2}{10} \quad 8<t}$$
i got the equation $$3 - 2u(t-2) + (t-1)u(t-5) + (\frac{t^2}{10} - t) u (t-8)$$
and i got the laplace transform $$\frac 3s - \frac{2e^{-2s}}{s} + e^{-5s} \left[ \frac{1}{s^2} + \frac 4s \right] + e^{-8s} \left[\frac{1}{5s^3} - \frac{1}{s^2} + \frac 8s \right]$$
did i make anything wrong?

lgbasallote · Answer

look at all the beautiful latex lol

lgbasallote · Answer

the final answer is $$\Large \frac 3s - \frac{2e^{-2s}}{s} + e^{-5s} \left[ \frac{1}{s^2} + \frac 4s \right] + e^{-8s} \left[\frac{1}{5s^3} - \frac{1}{s^2} + \frac 8s \right]$$

lgbasallote · Answer

^just in case it cant be read earlier

anonymous · Answer

the first equation (step function) you got seems to be little bit wrong.

lgbasallote · Answer

the 3 - 2u thingy? that's right...i confirmed it with my notes

anonymous · Answer

yes 3-2u thing . let me check it again then.

lgbasallote · Answer

another thing i'd like to ask btw...do i write the final answer as y(s) = or y(t) =?

anonymous · Answer

it should in in the S domain since we are finding Laplace (from t -> s)

lgbasallote · Answer

so y(s)?

anonymous · Answer

F(s)  i guess because f(t) is given !

lgbasallote · Answer

heh same shiz...anyway...

lgbasallote · Answer

why do you think my equation is wrong?

anonymous · Answer

i think it should be  3u(t-2)

lgbasallote · Answer

but the graph would cross the 0 right?

lgbasallote · Answer

|dw:1345468355120:dw|
something like tha?