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ranyai12

  • 2 years ago

Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))

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  1. ranyai12
    • 2 years ago
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    @Spacelimbus

  2. Spacelimbus
    • 2 years ago
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    \[\Large \frac{1}{s(s+4)^2} \]?

  3. ranyai12
    • 2 years ago
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    yes

  4. Spacelimbus
    • 2 years ago
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    Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]

  5. ranyai12
    • 2 years ago
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    so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2

  6. Spacelimbus
    • 2 years ago
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    If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]

  7. ranyai12
    • 2 years ago
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    I thought you are supposed to multiply A by the values under B and C?

  8. Spacelimbus
    • 2 years ago
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    I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression

  9. Spacelimbus
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    \[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]

  10. Spacelimbus
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    better now it seems

  11. ranyai12
    • 2 years ago
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    wouldnt you get A=0, B=0 and C=-1/4

  12. Spacelimbus
    • 2 years ago
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    Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.

  13. Spacelimbus
    • 2 years ago
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    \[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]

  14. ranyai12
    • 2 years ago
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    could you please explain how you go that

  15. Spacelimbus
    • 2 years ago
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    \[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]

  16. Spacelimbus
    • 2 years ago
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    Yes, I multiplied the above form by s(s+4)^2 (make sure you cancel on the right hand side right) and then you factor likewise terms. When you did that, you can set both sides of the equation equal, they are equal if and only if their coefficients match.

  17. Spacelimbus
    • 2 years ago
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    \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] General setup of partial fraction decomposition, multiply through by \(s(s+4)^2\), it will disappear entirely on the left hand side of the equation and piecewise on the righthandside of the equation.

  18. Spacelimbus
    • 2 years ago
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    Then expand the righthand side and factor likewise terms, set the coefficients equal to the coefficients on the lefthand side to make a valid equation out of it LHS=RHS, in this case a lot of the coefficients have to be zero, because there is only 1 on the LHS.

  19. Spacelimbus
    • 2 years ago
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    you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*

  20. Spacelimbus
    • 2 years ago
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    okay so far?

  21. ranyai12
    • 2 years ago
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    sorry im still confused on how you got the equation

  22. ranyai12
    • 2 years ago
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    can you please show me by the work?

  23. Spacelimbus
    • 2 years ago
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    The idea of partial fraction decomposition is that you find a different way of writing the quotient, so in the end there will be an easier form for it which is especially useful in integral calculus and for LaPlace Transformations. The idea is that you completely factor the denominator and then you split it up, so each new quotient becomes its own denominator

  24. Spacelimbus
    • 2 years ago
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    |dw:1344895643951:dw|

  25. ranyai12
    • 2 years ago
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    i understand that part but after that I get confuesd on how you distributed the variables

  26. Spacelimbus
    • 2 years ago
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    \[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?

  27. ranyai12
    • 2 years ago
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    yea, I'm not sure how you got that

  28. Spacelimbus
    • 2 years ago
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    I multiplied both sides by s(s+4)^2 so the denominator will completely disappear on the LHS, on the right hand side I have: \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]

  29. Spacelimbus
    • 2 years ago
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    you want to solve for A, B, C in the end, but first you distribute all the binomials out and then factor likewise powers together to make a suitable system of equations.

  30. Spacelimbus
    • 2 years ago
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    \[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

  31. Spacelimbus
    • 2 years ago
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    now read the equation like this, on the righthand side you have a quadratic polynomial, but on the left hand side, you can understand it like this: \[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

  32. Spacelimbus
    • 2 years ago
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    okay?

  33. Spacelimbus
    • 2 years ago
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    so in order that the lefthand side is equal to the righthand side, the coefficients A,B,C on the righthand side have to match the coefficients of the same powers of s on the lefthand side.

  34. Spacelimbus
    • 2 years ago
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    \[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]

  35. ranyai12
    • 2 years ago
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    im sorry but i dont understand the LHS

  36. ranyai12
    • 2 years ago
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    I mean RHS

  37. Spacelimbus
    • 2 years ago
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    \[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]

  38. ranyai12
    • 2 years ago
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    where did the numbers come in from

  39. Spacelimbus
    • 2 years ago
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    just a correction, was mistyping it.

  40. ranyai12
    • 2 years ago
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    and why is A added to B

  41. Spacelimbus
    • 2 years ago
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    Well the numbers come from because you expand a polynomial, namely (s+4)^2

  42. Spacelimbus
    • 2 years ago
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    And \[ As^2+Bs^2 = (A+B)s^2\]

  43. Spacelimbus
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    just normal algebra.

  44. ranyai12
    • 2 years ago
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    ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c

  45. Spacelimbus
    • 2 years ago
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    \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] Could you follow to this step? You see that A, B, and C gets multiplied with all that stuff. A with the first entire term, B with the second entire term, and C with the third. To match the equation we have to write the LHS different and then factor terms together, not only the first expression will have an x^2, also the second one with the Bs in front of it!

  46. Spacelimbus
    • 2 years ago
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    \[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]

  47. ranyai12
    • 2 years ago
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    oh ok! thanks! btw do we have to expand?

  48. Spacelimbus
    • 2 years ago
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    yes, in this case I suggest it, there are methods called cover up etc which can be applied to get at least some of the solutions easily, but I wouldn't recommend it.

  49. Spacelimbus
    • 2 years ago
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    For instance if you set \( s=-4\) in the equation above, the first terms will all go to zero and you are left with \[ \Large 1=-4C\]

  50. Spacelimbus
    • 2 years ago
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    but I don't recommend doing this, because this isn't an easy partial fraction decomposition.

  51. ranyai12
    • 2 years ago
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    so then you get As^2+8As+16A+Bs^2+4Bs+Cs

  52. Spacelimbus
    • 2 years ago
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    yes, that's what you get. If you read your equation careful enough you will see that there are multiple s^2 and s in it, you want to collect them and then match them to the left hand side, this is a valid technique.

  53. ranyai12
    • 2 years ago
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    ok now I understand that part! thank you! after that you solve for the variable right

  54. Spacelimbus
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    \[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \] See, the right hand side, you multiply A+B times s^2, if A+B are 0, then the right hand side becomes 0s^2, so it is equal to the left hand side and we are happy.

  55. Spacelimbus
    • 2 years ago
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    yes, match all the coefficients with the the coefficients on the LHS

  56. ranyai12
    • 2 years ago
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    would you get C=1

  57. Spacelimbus
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    \[ A+B=0 \\ 8A+4B+C=0 \\ 16A=1\] See how I got that 3x3 System?

  58. ranyai12
    • 2 years ago
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    isnt it 16A=1

  59. Spacelimbus
    • 2 years ago
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    Sorry, C is not zero no.

  60. Spacelimbus
    • 2 years ago
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    yes, I got it right, confused myself with my silly notes (-:

  61. Spacelimbus
    • 2 years ago
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    I get the following (concentrating hard now) \[A= \frac{1}{16}, \ B=- \frac{1}{16}, \ C= -\frac{1}{4} \]

  62. ranyai12
    • 2 years ago
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    so A=1/16 B=-1/16 C=-1/4

  63. Spacelimbus
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    yes

  64. ranyai12
    • 2 years ago
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    awesome! wat do we do now?

  65. Spacelimbus
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    So we have just proved that the left hand side can also be written as \[\Large \frac{1}{16s}- \frac{1}{16(s+1)}-\frac{1}{4(s+1)^2}\]

  66. ranyai12
    • 2 years ago
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    isnt it s+4

  67. Spacelimbus
    • 2 years ago
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    of course it is, I just typed it wrong. (-:

  68. Spacelimbus
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    \[\Large \frac{1}{16s}- \frac{1}{16(s+4)}-\frac{1}{4(s+4)^2} \]

  69. ranyai12
    • 2 years ago
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    ok awesome! whats the next step

  70. Spacelimbus
    • 2 years ago
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    well you know that the Laplace is a linear operator, so you can factor out constants and then grab a cheat sheet to see if you can find inverse laplace transforms

  71. ranyai12
    • 2 years ago
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    ok but the there are two constaNTS

  72. Spacelimbus
    • 2 years ago
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    For the first one \[ \Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)\]

  73. ranyai12
    • 2 years ago
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    F(s)=(1/16)[(1/s)-(1/(s+4)]-(1/4)[1/(s+4)^2)]

  74. Spacelimbus
    • 2 years ago
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    The inverse of 1/s = 1

  75. Spacelimbus
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    http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

  76. ranyai12
    • 2 years ago
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    1/16[e^0-e^-4t]

  77. Spacelimbus
    • 2 years ago
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    yes looks good.

  78. ranyai12
    • 2 years ago
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    what do i do about 1/(s+4)^2

  79. Spacelimbus
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    you forgot the last one though.

  80. Spacelimbus
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    \[\Large t^ne^{at}= \frac{n!}{(s-a)^{n+1}} \] with \(n=1, 2, 3 ...\) in our case n=1

  81. Spacelimbus
    • 2 years ago
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    got that?

  82. ranyai12
    • 2 years ago
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    kind of

  83. Spacelimbus
    • 2 years ago
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    You have to use a cheat sheet for these kind of problems, those formulas have been derived once, (some of them get ridiculously complicated, because of the long integration by parts problems)

  84. ranyai12
    • 2 years ago
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    for 1/(s+4)-->n!/(s^(n+1))

  85. Spacelimbus
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    I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

  86. ranyai12
    • 2 years ago
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    i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n

  87. Spacelimbus
    • 2 years ago
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    \[\Large e^{at}= \frac{1}{s-a} \] In our case \( a=-4\)

  88. Spacelimbus
    • 2 years ago
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    non related formulas.

  89. ranyai12
    • 2 years ago
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    but we have s+4 not s-4

  90. ranyai12
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    oh nvr mind i confused t^n and e^at

  91. Spacelimbus
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    \[ \Large t^1e^{-4t}= \frac{1!}{(s+4)^{2}} \]

  92. Spacelimbus
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    this is what I get for my substitution.

  93. ranyai12
    • 2 years ago
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    ok so I ended up with (1/16)(1)-(1/16)[1/(s+4)]-(1/4)[1/(s+4)^5)]

  94. Spacelimbus
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    No, that's not what you are doing.

  95. Spacelimbus
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    You will use this for differential equations, so you compute the inverse, the inverse is a function.

  96. ranyai12
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    what do you do?

  97. Spacelimbus
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    \[\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t} \]

  98. Spacelimbus
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    That is what I get, if I didn't make any careless mistakes.

  99. ranyai12
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    so you dont use the transform equation

  100. Spacelimbus
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    look at the cheat sheet again, right you have the Laplace Transform, and on the other side you have the inverse, you want to compute the inverse.

  101. ranyai12
    • 2 years ago
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    oh ok

  102. Spacelimbus
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    \[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace- \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]

  103. Spacelimbus
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    see that? I didn't carry it out entirely, but that above is a Laplace Transform, whenever you have something of such a form, you want to compute its inverse.

  104. Spacelimbus
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    The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.

  105. Spacelimbus
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    It basically turns differential equation problems into Algebra problems.

  106. ranyai12
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    oh i get it thank you sooo much!

  107. Spacelimbus
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    welcome

  108. ranyai12
    • 2 years ago
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    :)

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