Question

find the inverse laplace of F(s) = e^(-3s)/(s^2).

Answer 1

So this relates to the unit step function. Ummm, I'm trying to remember the formula...\[\large \mathcal L^{-1}\left[e^{-as}F(s)\right]\quad =\quad f(t-a)\;u(t-a)\]

Answer 2

yes

Answer 3

but it's u(t - a)f(t-a)

Answer 4

How is that different from what I wrote? :O

Answer 5

Oh based on the fact that I wrote the exponential first? :) fair enough!

Answer 6

yeah haha

Answer 7

Understand how to solve your problem now or is still confusing? :O

Answer 8

still confusing haha

Answer 9

so i plug in and get u(t - 3)f(t - 3)

Answer 10

\[\large \mathcal L^{-1}\left[e^{-as}\cdot F(s)\right]\quad =\quad \large \mathcal L^{-1}\left[e^{-3s}\cdot\frac{1}{s^2}\right]\]
So we just need to think back to what the inverse laplace of 1/s^2 is.

Answer 11

okay

Answer 12

so F(s) = 1/s^2 * e^(-3s)

Answer 13

No, F(s) is just the 1/s^2 part.
See how I tried to match them up in the last equation?
e^-3s is our e^-as part of the equation.

Answer 14

Remember inverse laplace of 1/s^2? :D Look back at your chart if not!! Nice easy one!

Answer 15

okay so its t

Answer 16

Yesss good :D

So now the exponential term is applying some type of shift.
Hmm I wish I had a better understanding of this so I could explain it more thoroughly lol.\[\large t \qquad \rightarrow \qquad (t-3)\;u(t-3)\]

Answer 17

okay so why isnt it u(t - 3)f(t - 3)

Answer 18

oh cause its just t

Answer 19

We were able to find f(t),\[\large f(t)=t\]Our final answer (because of what the exponential is doing to the function) should look like this,\[\large f(t-a)\;u(t-a)\]

\[\text{If}\; f(t)=t,\; \text{then}\; f(t-a)=t-a\]

Answer 20

Hah yah D: grr you too fast lol

Answer 21

haha okay so then where do i go from there, integration?

Answer 22

Umm that IS your final answer.
Unless your teacher wants it written without the unit step, in which case you might be able to go a step further as write the function as a piece-wise.
Hmm.

Answer 23

oh yeah thats it SWEET

Answer 24

THANK YOU KINDLY ♥

Answer 25

Welcome :3