Question

\[\begin{array} \text{Find} \; \mathcal L \lbrace A(t) \rbrace \; \text{where} \; A(t) &=0 \quad 02\end{array}\]

Answer 1

is it just \[\huge \left[-\frac{te^{-st}}{s} - \frac{e^{-st}}{s^2}\right|_1^2\]

Answer 2

I dont understand

Answer 3

am i right @UnkleRhaukus ?

Answer 4

i cant remember ,
can you show all working

Answer 5

\[\Huge \int\limits_{0}^{1}e^{-st}(0)+\int\limits_{1}^{2}te^{-st}+\int\limits_{2}^{\infty}e^{-st}(0)\]

Answer 6

it was a paiecewise function .the first and the third integrals are zero.
your work looks fine.

Answer 7

^yes that's what i did

Answer 8

so is it right?

Answer 9

yes it is. just apply the limits.

Answer 10

ahh thank you thank you!

Answer 11

yw:)

Answer 12

is that laplace transformation ?

Answer 13

yes

Answer 14

i got \[\huge -\frac{2e^{-2s}}{s} - \frac{e^{-2s}}{s^2} + \frac{2e^{-s}}{s}\]
is that right?

Answer 15

oh , I studied that yesterday in class and didn't understand what my teacher has wrote ~.~
I really think I must retake calculus 1

Answer 16

i dont get it either hah lol

Answer 17

the laplace is in the frequency domain,
where as the original function is in the time domain

Answer 18

....so am i right? o.O

Answer 19

@lgbasallote its ok .just if you want to take LCM s^2 and simplify it.

and do not worry you will get it i am here :P

Answer 20

but i am right so far right?

Answer 21

right

Answer 22

thanks!!

Answer 23

i just wanted to make sure that you applied limits correctly i am getting .
\[\Large \frac{e^{-s}}{s}-\frac{2e^{-2s}}{s}+\frac{e^{-s}}{s^2}-\frac{e^{-2s}}{s^2}\]

Answer 24

@lgbasallote i hope you can simplify it more.

Answer 25

ahh yes i made a careless mistake thanks