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ragman

  • 3 years ago

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

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  1. Skaematik
    • 3 years ago
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    note

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

  3. Skaematik
    • 3 years ago
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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

  4. ragman
    • 3 years ago
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    oh ok sounds good thanks

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

  6. lgbasallote
    • 3 years ago
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    ??

  7. lgbasallote
    • 3 years ago
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    @ragman you there?

  8. ragman
    • 3 years ago
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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 )

  9. ragman
    • 3 years ago
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    @lgbasallote

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

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

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

  13. ragman
    • 3 years ago
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    yup perfect

  14. hartnn
    • 3 years ago
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    can u do partial fractions ??

  15. ragman
    • 3 years ago
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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

  16. lgbasallote
    • 3 years ago
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    where's C?

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

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

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

  20. ragman
    • 3 years ago
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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

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

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

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

  24. lgbasallote
    • 3 years ago
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    i think it will change your A value too

  25. lgbasallote
    • 3 years ago
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    so you best check

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

  27. lgbasallote
    • 3 years ago
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    nope nevermind..it stays the same

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

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

  30. hartnn
    • 3 years ago
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    yup, same.Good work.

  31. lgbasallote
    • 3 years ago
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    A =-3?

  32. lgbasallote
    • 3 years ago
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    yeah that's right

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

  34. lgbasallote
    • 3 years ago
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    everything's right

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

  36. hartnn
    • 3 years ago
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    yup, and use inverse laplace formulas.

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

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

  39. ragman
    • 3 years ago
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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

  40. lgbasallote
    • 3 years ago
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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?

  41. lgbasallote
    • 3 years ago
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    uhhh wait... wrong formula for last one

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

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

  44. lgbasallote
    • 3 years ago
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    n is the exponent

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

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

  47. ragman
    • 3 years ago
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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 ?

  48. hartnn
    • 3 years ago
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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?

  49. ragman
    • 3 years ago
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    yup thats what i got thanks

  50. hartnn
    • 3 years ago
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    welcome :)

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