Question

Find the inverse Laplace transform of the given function:

Answer 1

\[\huge F(s)=\frac{2(s-1)e^{-2s}}{s^2-2s+2}\]

I got as far as to complete the square in the denominator so that

\[F(s)=\frac{2(s-1)e^{-2s}}{(s-1)^2+1}\]I can also see that we're going to end up with a cos somewhere given the general form of F(s)..

Answer 2

@ganeshie8

Answer 3

Would it be: \[\large 2e^t \cos(t) u_2(t)\] or\[\large 2e^{t-2} \cos(t-2)u_2(t)\]

Answer 4

sort out the algebra, (s-1)^2

Answer 5

\[\mathcal{L}(u_a(t)f(t-a)) = e^{-as}F(s)\]

\[\mathcal{L(e^{at}f(t))} = F(s-a)\]

Answer 6

\[F(s)=\frac{2(s-1)e^{-2s}}{(s-1)^2+1} = 2\mathcal{L} ^{-1}\left( e^{-2s} \times \frac{ (s-1) }{ (s-1)^2+1 } \right)\] then use \[\mathcal{L}^{-1}(e^{ct}f(t)) = F(t-c)\]

Answer 7

I like to split it like that, as it's clearer I think

Answer 8

Mine? That's how our prof does it haha

Answer 9

no, mine

Answer 10

Well it's just one more extra step

Answer 11

you have \[2u_2(t)f(t-2)\] then you can find f which is \[e^tcos(t)\]
plug it all in \[2u_2 (t)e^{t-2}cos(t-2)\] actually I'm not entirely sure if we put t-2 in the exponent

Answer 12

I actually just learnt all this about 10 minutes ago doing a problem rofl

Answer 13

Here's what I did...
\[G(s)=\frac{2(s-1)}{(s-1)^2+1} \rightarrow \mathcal{L}^{-1}G(s)=2e^t \cos(t)\]
Then because of the e^(-2s), then t becomes t-2
\[2e^{t-2} \cos(t-2)u_2(t)\]

Answer 14

Yeah that looks good to me, but confirm with @ganeshie8

Answer 15

confirm with wolfram
http://www.wolframalpha.com/input/?i=inverse+laplace+%5Cfrac%7B2%28s-1%29e%5E%7B-2s%7D%7D%7B%28s-1%29%5E2%2B1%7D

look at alternate forms

Answer 16

What's with the thet?

Answer 17

theta*

Answer 18

Oh ok that's there way of saying u2(t)?

Answer 19

the same unit step funciton,

\(\theta(t-a) = u_a(t) \)

Answer 20

Yeah -- I don't understand those alternate forms..

Answer 21

I gotcha

Answer 22

Wolf uses theta as the heaviside/ step function

Answer 23

cos(2-t) mhm

Answer 24

@CShrix
\(u_a(t)\) is a shortcut for \(u(t-a)\)

Answer 25

you get \(u_a(t)\) by translating the unit step function to right by \(a\) units

Answer 26

Ahh gotcha.. Yeah why is it cos(2-t)? Shouldn't it be t-2?

Answer 27

does it matter

Answer 28

I think it does matter.. wouldn't that be a huge difference between -t and t?

Answer 29

The exponent has t-2, but the cos has 2-t..?

Answer 30

recall, cos is symmetric about y axis :
\[\cos(x) = \cos(-x)\]

Answer 31

Pretty much cosx is an even function where sinx is odd

Answer 32

So even the sign of the 2 doesn't matter either?
cos(2-x)=cos(x-2)

Answer 33

Grr haha

Answer 34

\[2-x = -(x-2)\]

happens... :P

Answer 35

asdfasdf

lol

Answer 36

http://www.wolframalpha.com/input/?i=cos%282-x%29%3Dcos%28x-2%29 just to confirm

Answer 37

XD Gotcha now.. Thanks everyone! X)

Answer 38

Thanks for the question :D good practice hehe

Answer 39

You're welcome X) I have an exam on Tuesday and I haven't exactly had a lot of time recently to stay ahead of the game.. :(

Answer 40

Ah darn, my final is about 2 weeks away lol

Answer 41

Same.. We have an exam on Tuesday and then the final the week after LOL

Answer 42

That's brutallll

Answer 43

Hopefully if I study hard enough for this test, I won't have to focus as much on it during finals week! X)

Answer 44

I hear ya, good luck! It's sort of insane how many methods we've learnt in such little time for DE's!

Answer 45

Seriously! Haha.
Best of luck to you too!

Answer 46

you drop Laplace Transforms pretty much as soon as you start questioning why it all works

and i'm not sure that's a bad thing. cos they don't lead anywhere interesting.

Answer 47

fourier transforms are way more challenging, and way more esoteric too.

Answer 48

I hear that, use it in PDEs right

Answer 49

lol, I don't think I'll have to learn Fourier for differential eq's.

Answer 50

Well, ordinary

Answer 51

I might have to take PDE's :'(

Answer 52

Ah =/ My major is Mechanical (probably changing to aerospace!) engineering. So I think this is my last pure-math class

Answer 53

But a pre-req for it here is real analysis...and I haven't done that

Answer 54

Ah, nice one