anonymous
  • anonymous
find the inverse laplace of 2s exp^-s/(s-1)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Luigi0210
  • Luigi0210
Could you rewrite that using the equation tool?
anonymous
  • anonymous
y u dont understand?
anonymous
  • anonymous
\[(2s \times e^{-s})/(s-1)^2\] ?

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anonymous
  • anonymous
yes this is right.
Luigi0210
  • Luigi0210
Have you tried anything so far?
anonymous
  • anonymous
yes but im little bit confuse.
Luigi0210
  • Luigi0210
Show us what you've tried so far and we'll see if we can help
anonymous
  • anonymous
ok my ans is 2exp^t-1.H(t-1) but the book ans is different.
Luigi0210
  • Luigi0210
What's the book answer?
anonymous
  • anonymous
2t.exp^t-1H(t-1).
Luigi0210
  • Luigi0210
Did you do this with partial fractions?
anonymous
  • anonymous
no.i use unit step function.
hartnn
  • 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
  • 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
  • hartnn
because next steps are quite trivial !
anonymous
  • anonymous
the book ans is 2t.exp^t-1H(t-1).
anonymous
  • anonymous
i need this answer.
hartnn
  • hartnn
lets see how we can get there :) you understood the properties i applied? Have the inverse laplace transform table with you ?
anonymous
  • anonymous
yes i have.
hartnn
  • 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
  • hartnn
for better clarity, do you want me to give you the steps till that step?
anonymous
  • anonymous
yes give me details plz
hartnn
  • hartnn
give me a moment to type, by that time, please think of next steps :)
anonymous
  • anonymous
ok.thank u.
hartnn
  • 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
  • hartnn
\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)
anonymous
  • anonymous
ok thank u so much.and which softwaer u used to write this things clearly.
anonymous
  • anonymous
last thing is tell me the the laplace of exp^t-1 is what??
hartnn
  • hartnn
used \(Latex \), which is supported by this site. you can use it too :)
anonymous
  • anonymous
can i download it?/
hartnn
  • 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
  • hartnn
and for the latex you can just start typing in `\( ..\)` try it type `\(Hi\)` and see what u get
anonymous
  • anonymous
ok
hartnn
  • 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
  • anonymous
sorry exp ^t-x laplace??
hartnn
  • 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
  • anonymous
i dont understand why u right 2(s+1) in numerater?
hartnn
  • 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
  • hartnn
so applying that property to your function, the numerator s became (s+1)
hartnn
  • hartnn
correction : If you want to remove e^at from inverse laplace , replace 's' by 's+a' in your function <>
anonymous
  • anonymous
can u right the correct ans which i post in above comments??
hartnn
  • 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
  • anonymous
ok thank u so much to give me precious time.nice to meet u.
hartnn
  • hartnn
ah, you too :)
sidsiddhartha
  • 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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