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Inverse laplace tables, where am I going wrong?

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the inverse Laplace transform of [tex]\frac{2s+1}{s^2-2s+2}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]
    \begin{align*}
    L^{-1}\{ \frac{2s+1}{s^2-2s+2} \} &= 2 L^{-1}\{ \frac{s}{(s-1)^2+1} \} + L^{-1}\{ \frac{1}{(s-1)^2+1} \} \\
    &= 2 L^{-1}\{ \frac{s}{(s-1)^2+1} \} + e^t sin(t) \\
    &= L^{-1}\{ \frac{2s+1-1}{(s-1)^2+1} \} + e^t sin(t) \\
    &= L^{-1}\{ \frac{2s-1}{(s-1)^2+1} \} + L^{-1}\{ \frac{1}{(s-1)^2+1} \} + e^t sin(t) \\
    &= 2e^t cos(t) + e^t sin(t) + L^{-1}\{ \frac{1}{(s-1)^2+1} \} \\
    &= 2e^t cos(t) + e^t sin(t) + e^t sin(t)
    \end{align*}
    [/tex]

    My book has an answer of [itex]2e^t cos(t) + 3e^t sin(t)[/itex]. Am I making a mistake in the algebra?
     
    Last edited: Sep 16, 2011
  2. jcsd
  3. Sep 16, 2011 #2
    Instead of 2s+1-1 put 2s+2-2 then you will habe 2(s-1)+2 and the you will have the answer
     
  4. Sep 16, 2011 #3
    Ok I see how that works, but what about my answer is wrong? What algebra mistake am I making?
     
  5. Sep 16, 2011 #4
    Just realized I had a typo in the original post. :cry: Lines 5 and 6 I typed "sin" instead of "cos." It's all changed now and my question still stands.
     
  6. Sep 16, 2011 #5
    When you write (2s-1)/(s-1)^2+1, it wont give you cos because you should have nA/A^2+1 then you will have a cos so you should have n(s-1)/(s-1)^2+1 which n=2 here. Since you will have 2(s-1)=2s-2 and to make it equal to 2s in line 2 you should write like 2s-2+2 the separate to 2s-2 and 2 then you will have the answer
     
  7. Sep 16, 2011 #6
    Oh, I think you must be doing it differently? My table has:

    [tex]e^{at} sin(bt) = \frac{b}{(s-a)^2+b^2}, s>a [/tex]
    [tex]e^{at} cos(bt) = \frac{s-a}{(s-a)^2+b^2}, s>a[/tex]

    I'm matching it to that. Should I be doing it a "longer" way? I still don't see how my original method is wrong. 2s+1-1=2s, just as 2s+2-2=2s and 2(s+1)-2=2s.
     
    Last edited by a moderator: Sep 16, 2011
  8. Sep 16, 2011 #7
    (2s + 1)/(s^2 - 2s + 2) = (2s + 1)/[(s - 1)^2 + 1] = [2(s - 1) + 3]/[(s - 1)^2 + 1]

    Which equals 2*(s - 1)/[(s - 1)^2 + 1] + 3*1/[(s - 1)^2 + 1].

    Use the appropriate formulas from the table of Laplace transforms.
     
  9. Sep 16, 2011 #8
    I still don't see it but thank you for your time. :redface:
     
  10. Sep 16, 2011 #9
    Thats it for cos you should have s-a in numerator and and s-a is s-1 when you write it like 2s-1 it is not s-a it is 2s-a and for solving it you should have 2(s-a) not 2s-a. Right?
     
  11. Sep 16, 2011 #10
    I understand why your method works.

    Why won't the method I used in the first post work? Is it not algebraically sound and does it not fit the form given in the Inverse Laplace Transform tables? What am I doing wrong?
     
  12. Sep 16, 2011 #11
    Essentially all you have to do is make the denominator and the numerator look similar. The easiest way to do it is to complete the square.

    You need (s - 1) in the denominator and in the numerator to match the formulas in the table of Laplace transforms.
     
  13. Sep 16, 2011 #12
    I must be missing something major because I still don't get it. I think all of my steps match the formulas in my table.
     
  14. Sep 16, 2011 #13
    Have a look at the 4th line. (2s - 1)/[(s - 1)^2 + 1] ≠ 2*[(s - 1 )/(s - 1)^2 + 1].

    L^-1{2* (s - 1)/[(s - 1)^2 + 1]} = 2e^t*cos(t).
     
  15. Sep 16, 2011 #14
    Thank you, that was exactly what I needed to see -- where I made the algebra mistake.[URL]http://smiliesftw.com/x/bowdown.gif[/URL]
     
    Last edited by a moderator: Apr 26, 2017
  16. Sep 16, 2011 #15
    Next time use the method I described above it is much easier.
     
  17. Sep 16, 2011 #16
    Ok, another quick question. Say I have this:

    [tex]L^{-1} \{ \frac{1}{(s+1)^2+4} \}[/tex]

    This is very close to

    [tex]e^{at} sin(bt) = \frac{b}{(s-a)^2+b^2}, s>a[/tex]

    So to get the b equal I would multiply the transform by 2?

    [tex]2L^{-1}= \{ \frac{1}{(s+1)^2+4} \} \rightarrow f(t) = 2e^{-t} sin(2t)[/tex]

    But my book says the answer is [tex]f(t) = \frac{1}{2}e^{-t} sin(2t)[/tex]

    Am I making another dumb mistake? Why is the book multiplying by 1/2 and not 2?

    edit: Nevermind, I see it now.

    Thank you everyone for your help.
     
    Last edited: Sep 16, 2011
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