Question

inverse laplace of F(s) = ln|(s-2)/(s+2)|

Answer 1

Take the integral as you did in the previous problem you asked for.

Answer 2

So I tried going backwards from the last problem like exactly step by step and it gave me an answer but entirely wrong haha

Answer 3

\[\iota ^-1\int\limits_{s}^{∞} F(s)ds \]

Answer 4

but i dont have f(t) i have F(s)

Answer 5

crap haha

Answer 6

do u know this property ?
\(\huge L^{-1}[F(s)]=\frac{-1}{t}L^{-1}[\frac{d}{ds}F(s)]\)

Answer 7

yes

Answer 8

so we know F(s)

Answer 9

so what did u get d/ds (F(s)) ?

Answer 10

but how do we take LT inverse of it

Answer 11

what did u get d/ds (F(s)) ?

Answer 12

first use. ln A/B = ln A-ln B

Answer 13

4/(s^2 - 4)

Answer 14

okay i think i got it! :)

Answer 15

u get something of this form, A/(s+2) + B/(s-2)

Answer 16

or u can directly use hyperbolic function, i think...

Answer 17

wait no i dont have it

Answer 18

so i got 4/(s^2 - 4) which is 4/1 * 1/(s^2 - (2)^2)

Answer 19

is the answer in terms of hyperbolic function or exponential function ?

Answer 20

because that will decide next step...

Answer 21

ln|(s-2)/(s+2)| = ln (s-2) - ln(s+2)
derivative will be
= 1/s-2 - 1/ s+2
now the inverse LT of this is easy.

Answer 22

pellet son! i 4got laplace transform.

Answer 23

oh okay

Answer 24

so, did u get the answer ?

Answer 25

no :(

Answer 26

what u got ? and whats the actual answer ?

Answer 27

answer: -(2sinh(2t))/t

Answer 28

oh so the 2 is from the 4

Answer 29

yeah, u know the Laplace of sinh t , right ?

Answer 30

-1/t came from the property
4/(s^2-2^2) = 2 *[2/(s^2-2^2)] ---> 2sinh t
got it ?

Answer 31

yeah sure sounds good