find the inverse laplace of 2s exp^-s/(s-1)^2

- anonymous

find the inverse laplace of 2s exp^-s/(s-1)^2

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- schrodinger

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- Luigi0210

Could you rewrite that using the equation tool?

- anonymous

y u dont understand?

- anonymous

\[(2s \times e^{-s})/(s-1)^2\]
?

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## More answers

- anonymous

yes this is right.

- Luigi0210

Have you tried anything so far?

- anonymous

yes but im little bit confuse.

- Luigi0210

Show us what you've tried so far and we'll see if we can help

- anonymous

ok my ans is 2exp^t-1.H(t-1) but the book ans is different.

- Luigi0210

What's the book answer?

- anonymous

2t.exp^t-1H(t-1).

- Luigi0210

Did you do this with partial fractions?

- anonymous

no.i use unit step function.

- hartnn

After applying time shifting, you'll need to take inverse laplace of
2s/(s-1)^2
how did you find it?
you say you got 2e^(t) for this?

- hartnn

lets see :)
Applying frequency shifting,
\(\Large L^{-1}[ 2s/ (s-1)^2] = e^t L^{-1} [2(s+1)/s^2 ]\)
clear till here ?

- hartnn

because next steps are quite trivial !

- anonymous

the book ans is 2t.exp^t-1H(t-1).

- anonymous

i need this answer.

- hartnn

lets see how we can get there :)
you understood the properties i applied?
Have the inverse laplace transform table with you ?

- anonymous

yes i have.

- hartnn

so from 2s exp^-s/(s-1)^2
we came to 2 (s+1)/s^2 by applying 2 properties.
you understood both ?

- hartnn

for better clarity, do you want me to give you the steps till that step?

- anonymous

yes give me details plz

- hartnn

give me a moment to type,
by that time, please think of next steps :)

- anonymous

ok.thank u.

- hartnn

Since \(\Large L^{-1} [e^{-as}F(s)] = f(t-a) H(t-a) \)
we first find
\(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] \)
and then use the above property
Now
\(\Large L^{-1}f(s-a) = e^{at}f(t) \)
so
\(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] = e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}] \)
see whether you get these steps first :)

- hartnn

\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)

- anonymous

ok thank u so much.and which softwaer u used to write this things clearly.

- anonymous

last thing is tell me the the laplace of exp^t-1 is what??

- hartnn

used \(Latex \), which is supported by this site.
you can use it too :)

- anonymous

can i download it?/

- hartnn

why would you need that ?
we're here :
\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)
you'll get the same answer which is in you book, be patient lol
we're just half way done

- hartnn

and for the latex
you can just start typing in `\( ..\)`
try it
type `\(Hi\)`
and see what u get

- anonymous

ok

- hartnn

so we need to solve
\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)
`\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)`
any ideas how to proceed?

- anonymous

sorry exp ^t-x laplace??

- hartnn

sorry, didn't get your question.
we're solving an inverse laplace question,
and I just asked you whether you know how we can proceed with
\( L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)
are we on a same page?

- anonymous

i dont understand why u right 2(s+1) in numerater?

- hartnn

glad that you ask doubts :)
\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)
This is a standard property
what does it say? If you want to remove e^at from inverse laplace ,
replace 's' by 's+1' in your function <>

- hartnn

so applying that property to your function,
the numerator s became (s+1)

- hartnn

correction :
If you want to remove e^at from inverse laplace ,
replace 's' by 's+a' in your function <>

- anonymous

can u right the correct ans which i post in above comments??

- hartnn

if I just post all the steps, you'll understand none of them.
I don't suppose you want that...
thats why I am posting few steps and asking you whether you got them...
if this approach doesn't suit you, I am really not the right person to help you...

- anonymous

ok thank u so much to give me precious time.nice to meet u.

- hartnn

ah, you too :)

- sidsiddhartha

here
\[F(s)=\frac{ 2s.e^{-s} }{ (s-1)^2 }\\s~In ~laplace~domain~means~\frac{ d }{ dt }~In ~time~domain\\f(t)=2.\frac{ d }{ dt }L^{-1}[\frac{ e^{-s} }{ (s-1)^2 }]=2.\frac{ d }{ dt }[(t-1)e^{t-1}H(t-1)]\\used~frequency~shifting~In~last~step\\f(t)=2[(t-1)e^{t-1}+e^{t-1}]H(t-1)=2t.e^{t-1}H(t-1)\]
easy this way :) @Yasirist

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