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ragman Group Title

find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))

  • 2 years ago
  • 2 years ago

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  1. Skaematik Group Title
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    note

    • 2 years ago
  2. ragman Group Title
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    'note' sorry this is my first time using this what do you mean by that?

    • 2 years ago
  3. Skaematik Group Title
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    Im just bookmarking this question so that I can return to it later. :D Id want to know the answer too

    • 2 years ago
  4. ragman Group Title
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    oh ok sounds good thanks

    • 2 years ago
  5. lgbasallote Group Title
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    your question is \[\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]

    • 2 years ago
  6. lgbasallote Group Title
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    ??

    • 2 years ago
  7. lgbasallote Group Title
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    @ragman you there?

    • 2 years ago
  8. ragman Group Title
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    hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )

    • 2 years ago
  9. ragman Group Title
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    @lgbasallote

    • 2 years ago
  10. lgbasallote Group Title
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    uhh s^2 + 22 + 5??

    • 2 years ago
  11. ragman Group Title
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    dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)

    • 2 years ago
  12. lgbasallote Group Title
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    so.. \[\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]

    • 2 years ago
  13. ragman Group Title
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    yup perfect

    • 2 years ago
  14. hartnn Group Title
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    can u do partial fractions ??

    • 2 years ago
  15. ragman Group Title
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    @hartnn yeah thats what i was doing but for some reason i could figure out the other two variable maybe im solving it incorrectly but i got -3 for A and 5 for B

    • 2 years ago
  16. lgbasallote Group Title
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    where's C?

    • 2 years ago
  17. hartnn Group Title
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    those are correct, so u need only C right ? where u took C above s+1

    • 2 years ago
  18. lgbasallote Group Title
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    your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]

    • 2 years ago
  19. lgbasallote Group Title
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    so you have \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]

    • 2 years ago
  20. ragman Group Title
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    oh ok so thats where i messed up i thought that where C is it should be (Cs+D) so i was trying to solve for both of those variables

    • 2 years ago
  21. hartnn Group Title
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    nopes only C, if it were s^2+1 u would have taken Cx+D

    • 2 years ago
  22. hartnn Group Title
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    so i think u can get C now, right?

    • 2 years ago
  23. ragman Group Title
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    oh i see now and yeah i believe i can get C now trying it right now

    • 2 years ago
  24. lgbasallote Group Title
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    i think it will change your A value too

    • 2 years ago
  25. lgbasallote Group Title
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    so you best check

    • 2 years ago
  26. hartnn Group Title
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    great tell us what u get C as....we all will verify that. A is correct.

    • 2 years ago
  27. lgbasallote Group Title
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    nope nevermind..it stays the same

    • 2 years ago
  28. ragman Group Title
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    ok sounds good got B=5 finding A now

    • 2 years ago
  29. ragman Group Title
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    A=-3 and C=4 is that what you guys got?

    • 2 years ago
  30. hartnn Group Title
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    yup, same.Good work.

    • 2 years ago
  31. lgbasallote Group Title
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    A =-3?

    • 2 years ago
  32. lgbasallote Group Title
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    yeah that's right

    • 2 years ago
  33. ragman Group Title
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    ok cool i was about to write out how i got it lol

    • 2 years ago
  34. lgbasallote Group Title
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    everything's right

    • 2 years ago
  35. ragman Group Title
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    and now i just plug back all the variables right

    • 2 years ago
  36. hartnn Group Title
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    yup, and use inverse laplace formulas.

    • 2 years ago
  37. lgbasallote Group Title
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    yep find \[\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}\]

    • 2 years ago
  38. lgbasallote Group Title
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    uhh that should be 5/s^2

    • 2 years ago
  39. ragman Group Title
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    ok sorry im just learning this but how do you go about doing that im trying to find the L^-1 formula but my notes arent really helping

    • 2 years ago
  40. lgbasallote Group Title
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    here's a hint: \[\huge \mathcal L \{ a \} = \frac as\] \[\huge \mathcal L \{ t^n\} = \frac{n!}{s^{n+1}}\] \[\huge \mathcal L \{e^{at} \} = \frac{a}{s -a}\] does that help?

    • 2 years ago
  41. lgbasallote Group Title
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    uhhh wait... wrong formula for last one

    • 2 years ago
  42. lgbasallote Group Title
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    \[\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}\] better

    • 2 years ago
  43. ragman Group Title
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    yes thats very helpful but im getting confused with what n should be in the middle equations

    • 2 years ago
  44. lgbasallote Group Title
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    n is the exponent

    • 2 years ago
  45. ragman Group Title
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    so it would be 2?

    • 2 years ago
  46. lgbasallote Group Title
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    for example: \[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]

    • 2 years ago
  47. ragman Group Title
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    thats whats a lil confusing to me because on the denominator it would have to be n=1 so that it would equal s^2 but on the numerator 1! equals 1 right ?

    • 2 years ago
  48. hartnn Group Title
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    yes thats true, so u will have 5t^1= 5t ....isn't it ? u can verify that u should get final answer as 5t-3+4e^(-t) did u get that?

    • 2 years ago
  49. ragman Group Title
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    yup thats what i got thanks

    • 2 years ago
  50. hartnn Group Title
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    welcome :)

    • 2 years ago
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