Question

Find the inverse laplace of e^(2 - 2s)/(s-1).

Answer 1

split the terms into two. that might help identify what you have.

Answer 2

\[e^2(\frac{ e^{-2s} }{ s-1 })\]

Answer 3

no but its e^(2 - 2s)

Answer 4

so it would be e^(2)/e^(2s)

Answer 5

\[e^2(e^{-2s}) = e^{2-2s} \]

Answer 6

oh okay hmmm

Answer 7

You're basically rearranging your given into some form that you can recognize and apply the laplace formulas to. Hope that helps.

Answer 8

ha its more complicated than that but thanks anyways :)

Answer 9

then there's partial fractions.

Answer 10

i just need to find the inverse laplace of e^(2 - 2s)/(s - 1)

Answer 11

no theres not haha

Answer 12

im trying to remember. damit. lol

Answer 13

cause i got the first part using the shift function u(t - a)f(t - a) and it worked perfectly

Answer 14

oh i got it!!!

Answer 15

aren't exponentials usually left behind in the time domain?

Answer 16

not in real time

Answer 17

what did it work out to?

Answer 18

f(t) = e^(t - 1)u(t-1) - e^(2)*e^(t - 2)*u(t-2)

Answer 19

e^(t-1)u(t-1) comes from the part i already solved - e^(-s)/(s-1)

Answer 20

you gone dun it

Answer 21

too bad i dont understand what it means haha