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Finding the laplace of a Heaviside system

  1. Nov 15, 2011 #1
    I need to use Laplace transforms to find the solution to this system of Heaviside functions but I'm not sure where to start because the two different x's in the system are confusing me.

    Should I start by taking the laplace transforms of both sides where the laplace of H(t-1) = e-s/s

    These are the functions:

    dx1/dt = x2 + 2 - H(t-1)

    dx2/dt = -x1 + 1 - H(t-1)

    with x1(0) = 1 and x2(0) = 0


    Thanks.
     
  2. jcsd
  3. Nov 15, 2011 #2

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    Welcome to PF, jamie_18! :smile:

    Yes, you should start taking the laplace transforms on both sides.
    This will give you 2 equations in the s-domain.
    Then solve the system of equations...
     
  4. Nov 15, 2011 #3
    Hello, this is the same person as the account jamie_18 which is getting deleted, I forgot password so had to create a new account lol...

    Anyways,

    I did the laplace for the two functions and got to

    s2[itex]\overline{x}[/itex]1 - sx1(0) - [itex]\dot{x}[/itex]1(0) = [itex]\overline{x}[/itex]2 + 2/s - e-s/s

    and

    s2[itex]\overline{x}[/itex]2 - sx2(0) - [itex]\dot{x}[/itex]2(0) = -[itex]\overline{x}[/itex]1 + 1/s - e-s/s

    How would I go about solving this system?

    Thanks
     
  5. Nov 15, 2011 #4

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    You appear to have the wrong laplacian for [itex]dx_1 \over dt[/itex].
    It should be [itex]s \overline x_1 -x_1(0)[/itex].

    But otherwise it looks good!

    Furthermore, you can fill in the boundary conditions.

    Then you will have 2 equations in [itex]\overline x_1[/itex] and [itex]\overline x_2[/itex].
    Do you know how to solve that?
     
  6. Nov 15, 2011 #5
    I was accidently using the laplace for the second derivative, not very smart of me lol..

    I got the new laplace transforms: (with the boundary conditions applied)

    s[itex]\overline{x}[/itex]1 - 1 = [itex]\overline{x}[/itex]2 + 2/s - e-s/s

    and

    s[itex]\overline{x}[/itex]2 = -[itex]\overline{x}[/itex]1 + 1/s - e-s/s

    I modified the first laplace function to get it into the form

    [itex]\overline{x}[/itex]1 = 1/s + [itex]\overline{x}[/itex]2/s = 2/s2 - e-s/s2

    I then subbed this into the second laplace equation... and got:

    [itex]\overline{x}[/itex]2 = -1/s2 - [itex]\overline{x}[/itex]2/s2 - 2/s3 + e-s/s3 + 1/s2 - e-s/s2

    I am not sure if this is the correct way to go by solving these equations...


    Thanks
     
  7. Nov 15, 2011 #6

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    Seems fine by me.
    But you did not solve completely for [itex]\overline x_2[/itex] yet...
     
  8. Nov 15, 2011 #7
    okay.. I've solved through to find:

    [itex]\overline{x}[/itex]2(s2 +1) = -2/s + e-s/s - e-s

    which gives

    [itex]\overline{x}[/itex]2 = -2/s(s-i)(s+i) = e-s/s(s-i)(s+i) + e-s/(s-i)(s+i)

    How would I do the partial fractions with imaginary numbers?

    Thanks
     
  9. Nov 15, 2011 #8

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    You need to apply the inverse Laplace transform.
    How do you usually do that?

    I'm afraid it's not useful to do partial fractions with imaginary numbers for that.
     
  10. Nov 15, 2011 #9
    I would do this for:

    [itex]\overline{x}[/itex] = a/(s+b)
    x = ae-bt

    and for [itex]\overline{x}[/itex] = e-as/bs
    x = 1/b(H(t-a)

    and for [itex]\overline{x}[/itex] = e-as/(s+b)
    x = e-b(t-a)H(t-a)


    But I'm not sure how I would find the inverse of a function with more than just a basic (s+a), etc, on the bottom...

    Thanks
     
  11. Nov 15, 2011 #10

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    Hmm, do you have the Laplace transforms for cos(t) and for sin(t)?
     
  12. Nov 15, 2011 #11
    The laplace of sin(at) is a/a2+s2

    and cos(at) is s/a2+s2

    Does this mean the e-s/(s2+12) becomes sint*e-s?

    If I wrote the function out as being [itex]\overline{x}[/itex]2 = (-2/s)*(1/s2+1) + (e-s/s)*(1/s2+1) + (e-s)*(1/s2+1) would that make it easier to solve?

    Thanks
     
  13. Nov 15, 2011 #12

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    It's a step in the right direction!

    But no, it does not mean that e-s/(s2+12) becomes sint*e-s.
    For starters, after transformation there should not be an "s" anymore...

    What is the Laplace transform of f(t-a)?
     
  14. Nov 15, 2011 #13
    Yes that is true... the s should dissapear.

    The laplace of f(t-a) is e-st/s ... I think..

    I tried inversing it by splitting it up and got:

    x = -2sint + H(t-1)sint + (???)sint..

    I'm not sure what the inverse of e-s is (the ???) without the s underneath it..

    Thanks.
     
  15. Nov 15, 2011 #14

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    Hmm, the Laplace transform of f(t-a)H(t-a) is: [itex]e^{-as} \overline f(s)[/itex].

    So to find the inverse Laplace transform from something with [itex]e^{-as}[/itex] in it, you need to shift in the time-domain.

    What are you using for your Laplace transforms? :confused:
    You seem to be missing some crucial information...
     
  16. Nov 15, 2011 #15
    I have been using the laplace transform table on wikipedia.

    How would I go about shifting it in the time domain?

    Have I managed to get the rest of it right?

    I will be back on tomorrow to try and finish the question.

    Thanks a lot for the help tonight :smile:
     
  17. Nov 15, 2011 #16

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    Okay, let me ask this then:

    What is the Laplace transform of sin(t-1)?
     
  18. Nov 16, 2011 #17
    Would the laplace of sin(t-1) be 12/s2+12 .. ?

    This does not appear in the table so its probably wrong..

    thanks
     
  19. Nov 16, 2011 #18

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  20. Nov 16, 2011 #19
    Would e-s become f(t-1)H(t-1)...
     
  21. Nov 16, 2011 #20

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    No, e-sF(s) becomes f(t-1)H(t-1).
     
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