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Laplace Transform question

  1. Apr 6, 2008 #1
    [SOLVED] Laplace Transform question

    1. The problem statement, all variables and given/known data

    Use the Laplace transform to solve the given IVp
    [tex]
    {\frac {d^{2}}{d{t}^{2}}}y \left( t \right) +2\,{\frac {d}{dt}}y
    \left( t \right) +5\,y \left( t \right) ={e^{-t}}\sin \left( 2\,t
    \right)
    [/tex]
    [tex]y(0)=1, y'(0)=-1[/tex]


    2. Relevant equations
    [tex]L(f) = \int^{\infty}_{0}e^{-st}f(t)dt[/tex]
    [tex]
    L \left( {e^{-t}}\sin \left( 2\,t \right) \right) =2\, \left(
    \left( s+1 \right) ^{2}+4 \right) ^{-1}
    [/tex]

    [tex] y'' = Ys^{2}-y(0)s-y'(0)[/tex]
    [tex] y' = Ys - y(0)[/tex]
    3. The attempt at a solution
    I first converted the problem into frequency space like so:
    [tex] Y(s^{2}+2s+5) = 2\, \left(
    \left( s+1 \right) ^{2}+4 \right) ^{-1} +2s+3[/tex]
    [tex]Y={\frac {2\,{s}^{3}+7\,{s}^{2}+16\,s+17}{ \left( {s}^{2}+2\,s+5
    \right) ^{2}}}[/tex]

    Then I used Partial Fractions:
    [tex]
    Y={\frac {As+B}{{s}^{2}+2\,s+5}}+{\frac {Cs+D}{ \left( {s}^{2}+2\,s+5
    \right) ^{2}}}
    [/tex]
    Solving the 4 equations that result I got the following values for the constants:
    [tex] A = 2[/tex]
    [tex] B = 3[/tex]
    [tex] C = 0[/tex]
    [tex] D = 2[/tex]
    Putting these constants back into my initial partial fractions equation i get:
    [tex]Y={\frac {2\,s+3}{{s}^{2}+2\,s+5}}+2\, \left( {s}^{2}+2\,s+5 \right) ^
    {-2}[/tex]
    [tex]
    Y=2\,{\frac {s+1}{ \left( s+1 \right) ^{2}+4}}+ \left( \left( s+1
    \right) ^{2}+4 \right) ^{-1}+2\, \left( {s}^{2}+2\,s+5 \right) ^{-2}
    [/tex]
    The first of these two I can find the inverse Laplaces for, but the last I cant figure out what to do with. This is where I am stuck at:
    [tex]
    y=2\,{e^{-t}}\cos \left( 2\,t \right) +1/2\,{e^{-t}}\sin \left( 2\,t
    \right) +\mbox {{\tt `\#msup(mi("L"),mo("`}}L^{-1}\mbox {{\tt `"))`}}
    \left( 2\, \left( {s}^{2}+2\,s+5 \right) ^{-2} \right) [/tex]

    I cannot figure out what to do with the last term -- I dont know how to put it into a form which i can find the inverse laplace transform for. If someone could please tell me what to do with that I would appreciate it greatly.
     
  2. jcsd
  3. Apr 6, 2008 #2
    Does anyone have any clue at all how I can break up the last term so that I can find an inverse for it? I cant expand it any further with partial fractions and there is nothing else I can think of. Once again, any help at all would be greatly appreciated.
     
  4. Apr 6, 2008 #3
  5. Apr 7, 2008 #4
    Multiplication in the s-domain is convolution in the time domain.

    You already have the factorisation for the polynomial, [tex]s^2+2s+5[/tex].
    [tex]L^{-1}\{2\left(\frac{1}{(s+1)^2+4}\right)\left(\frac{1}{(s+1)^2+4}\}\right)[/tex]
    =
    [tex]2(e^{-t}sin(2t)*e^{-t}sin(2t))[/tex]
     
    Last edited: Apr 7, 2008
  6. Apr 7, 2008 #5
    Thank you for all the help. I read about the convolutions, but this section precedes convolution in the book and so we must solve it without that knowledge. Actually that gives the wrong answer also which is kind of weird. So, i'm still not sure. If anyone has any other suggestions I would appreciate it greatly if you shared them with me!
     
  7. Apr 8, 2008 #6
    Where did you get 2s + 3 on the right of the equation?

    [tex]Y(s^{2}+2s+5) = 2\, \left(
    \left( s+1 \right) ^{2}+4 \right) ^{-1} +2s+3[/tex]

    Given the initial conditions you gave, I get:
    [tex]Y(s^{2}+2s+5) = \frac{2}{
    (s+1)^{2}+4} +s[/tex]

    Before we go further, do you agree?
     
  8. Apr 13, 2008 #7

    exk

    User Avatar

    I get the following.

    Take Laplace transform of the function with your initial conditions:
    [tex]Ys^{2}-s+1+2(Ys-1)+5Y=\frac{2}{(s+1)^{2}+4}[/tex]

    Rewriting a little we get:
    [tex]
    Y(s^{2}+2s+5) = \frac{2}{
    (s+1)^{2}+4} +s+1
    [/tex]

    Divide both sides by [tex](s+1)^{2}+4[/tex]

    We get this:

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

    Now you need to take the inverse transform: let [tex]g(t)=e^{-t}sin(2t)[/tex] and [tex]g(t)*g(t)=\int^{t}_{0}g(z)g(t-z)dz[/tex] the convolution of g(t) with itself.

    [tex]y(t)=(\frac{1}{2})(g(t)*g(t))+e^{-t}cos(2t)[/tex]

    Take the convolution (I used a calculator, masochism is is out of style ;) ) and you get:

    [tex]y(t) = \frac{-e^{-t}(2tcos(2t)-sin(2t))}{8}+e^{-t}cos(2t)[/tex]

    Checking on calculator I get:
    [tex]
    y(0)=1, y'(0)=-1
    [/tex]

    and plugging y(t) into the initial equation I get the desired result.

    I am not sure how you could solve this without using a convolution because I am not sure what the inverse laplace transform of [tex]
    (\frac{1}{2})\frac{4}{
    ((s+1)^{2}+4)^{2}} [/tex] is otherwise.
     
  9. Apr 13, 2008 #8
    Thank you for all the help exk and Eidos! Yes i havent been able to do it any other way than using convolution integrals either. That is weird that this question is presented in the book before that section. Once again, thank you.
     
  10. Apr 13, 2008 #9
    What calculator do you have that can do convolution? :!!)
     
  11. Apr 13, 2008 #10

    exk

    User Avatar

    Just a rusty old TI-89, convolution is just an integral ;)
     
  12. Apr 13, 2008 #11
    I've only ever heard about those mythical beasts (TI-89), Im used to using non-graphical, non-programmable calculators since those are the only permitted ones in our tests.

    I must see about this calculator...
     
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