Question

Inverse Laplace transform

Answer 1

\[F(S) = \frac{e^{-2s}}{s^2 +s - 2}\]

Answer 2

Completed the square \[\frac{e^{-2s}}{s^2 + s +.25 -2 - .25} = \frac{e^{-2s}}{(s + .5)^2 - \frac{9}{4}}\]

Answer 3

And now im stuck

Answer 4

@IrishBoy123

Answer 5

hey tyler

here's the cheat's solution, http://www.wolframalpha.com/input/?i=inverse+laplace+transform++%5Cfrac%7Be%5E%7B-2s%7D%7D%7Bs%5E2+%2Bs+-+2%7D

you've snuck a heaviside in there(!! ie: \(e^{-2s}\)), so i guess you need to work it all about \(t - 2 \) which is where the step up happens. [H, or u, appears as \(\theta\) in math wolfram for reasons i do not get]

i'd need a table of transforms to hand to provide specific input. i assume you have a table?! if not, i can post one.

Answer 6

My table is in my book, just a sec

Answer 7

Laplace transform table

Answer 8

@amistre64

Answer 9

couldnt we factor the bottom and decompose it?

Answer 10

Im not even sure. There's one example in the book and they completed the square on it so that's what im going off of.

Answer 11

(s+2)(s-1) =ss-2-s+2s

Answer 12

e^(-2s) = A(s-1) + B(s+2)

e^(-2) = B(1+2)

B = e^(-2)/3

A = -e^4/3

but then im not sure if that helps or hinders us

Answer 13

Yeah i see what you did but i'm not sure if it helped either.

Answer 14

is there a laplace for 1/(s+a) ?

Answer 15

constants pull out of the transform, since its linear right?

Answer 16

right

Answer 17

theres a 1/s-a

Answer 18

its exp(at)

Answer 19

1/(s-(-2)) and 1/(s-1) then is what we can work with if so

Answer 20

\[A~L^{-1}\{\frac{1}{s-(-2)}+B~L^{-1}\{\frac{1}{s-(1)}\}\]
\[A~e^{-2t}+B~e^t\]
right?

Answer 21

correct

Answer 22

\[-\frac{1}{3}e^{-2t + 4} +\frac{1}{3}e^{t-2}\]

Answer 23

we can factor and such if need be

Answer 24

Just missing the step function part now, the rest looks correct.

Answer 25

at least its looking mostly good lol

Answer 26

lol yes

Answer 27

The answer is \[\frac{1}{3}u_{2}(t)[e^{t-2} - e^{-2(t-2)}]\]. Not sure how the step side function got in there

Answer 28

pretty sure its a result of t being modified.

Answer 29

e^(t-2) is a shift to the right of e^(t)

Answer 30

im notthat uptodate on Laplace stuff, just havent had the experience needed to know anything other than some basics

Answer 31

Oh ok, i wasn't sure if we were suppose to obtain an \[\frac{e^{-2s}}{s}\] some how which is the laplace transform of u_2(t)

Answer 32

http://archive.nathangrigg.net/teaching/diff-eq/laplace-with-heaviside.pdf

Answer 33

awesome, thank you!

Answer 34

good luck :)