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anonymous

  • one year ago

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

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  1. Luigi0210
    • one year ago
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    Could you rewrite that using the equation tool?

  2. anonymous
    • one year ago
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    y u dont understand?

  3. anonymous
    • one year ago
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    \[(2s \times e^{-s})/(s-1)^2\] ?

  4. anonymous
    • one year ago
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    yes this is right.

  5. Luigi0210
    • one year ago
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    Have you tried anything so far?

  6. anonymous
    • one year ago
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    yes but im little bit confuse.

  7. Luigi0210
    • one year ago
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    Show us what you've tried so far and we'll see if we can help

  8. anonymous
    • one year ago
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    ok my ans is 2exp^t-1.H(t-1) but the book ans is different.

  9. Luigi0210
    • one year ago
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    What's the book answer?

  10. anonymous
    • one year ago
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    2t.exp^t-1H(t-1).

  11. Luigi0210
    • one year ago
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    Did you do this with partial fractions?

  12. anonymous
    • one year ago
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    no.i use unit step function.

  13. hartnn
    • one year ago
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    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?

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

  15. hartnn
    • one year ago
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    because next steps are quite trivial !

  16. anonymous
    • one year ago
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    the book ans is 2t.exp^t-1H(t-1).

  17. anonymous
    • one year ago
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    i need this answer.

  18. hartnn
    • one year ago
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    lets see how we can get there :) you understood the properties i applied? Have the inverse laplace transform table with you ?

  19. anonymous
    • one year ago
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    yes i have.

  20. hartnn
    • one year ago
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    so from 2s exp^-s/(s-1)^2 we came to 2 (s+1)/s^2 by applying 2 properties. you understood both ?

  21. hartnn
    • one year ago
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    for better clarity, do you want me to give you the steps till that step?

  22. anonymous
    • one year ago
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    yes give me details plz

  23. hartnn
    • one year ago
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    give me a moment to type, by that time, please think of next steps :)

  24. anonymous
    • one year ago
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    ok.thank u.

  25. hartnn
    • one year ago
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    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 :)

  26. hartnn
    • one year ago
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    \(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)

  27. anonymous
    • one year ago
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    ok thank u so much.and which softwaer u used to write this things clearly.

  28. anonymous
    • one year ago
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    last thing is tell me the the laplace of exp^t-1 is what??

  29. hartnn
    • one year ago
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    used \(Latex \), which is supported by this site. you can use it too :)

  30. anonymous
    • one year ago
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    can i download it?/

  31. hartnn
    • one year ago
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    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

  32. hartnn
    • one year ago
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    and for the latex you can just start typing in `\( ..\)` try it type `\(Hi\)` and see what u get

  33. anonymous
    • one year ago
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    ok

  34. hartnn
    • one year ago
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    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?

  35. anonymous
    • one year ago
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    sorry exp ^t-x laplace??

  36. hartnn
    • one year ago
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    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?

  37. anonymous
    • one year ago
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    i dont understand why u right 2(s+1) in numerater?

  38. hartnn
    • one year ago
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    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 <<Note how s-1 on left became s>>

  39. hartnn
    • one year ago
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    so applying that property to your function, the numerator s became (s+1)

  40. hartnn
    • one year ago
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    correction : If you want to remove e^at from inverse laplace , replace 's' by 's+a' in your function <<Note how s-a on left became s>>

  41. anonymous
    • one year ago
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    can u right the correct ans which i post in above comments??

  42. hartnn
    • one year ago
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    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...

  43. anonymous
    • one year ago
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    ok thank u so much to give me precious time.nice to meet u.

  44. hartnn
    • one year ago
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    ah, you too :)

  45. sidsiddhartha
    • one year ago
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    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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