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Homework Help: Laplace Transforms

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Okay, i'm pretty lost in my differential equations class and I'm not sure how to transform the functions

    2. Relevant equations

    f(t)= 1 / (s+2)^2
    f(t)= 1 / (s+3)^3
    f(t)= 1 / (s^2-3^2)

    They seem pretty simple, but i have no clue if whether i'm suppose to use 2 formulas or separate them?


    I'd look at the solutions in the back of my book, but they're even problems :(
    Thanks, I'd appreciate any help!
     
    Last edited: Jun 27, 2010
  2. jcsd
  3. Jun 27, 2010 #2

    Zryn

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    Are you familiar with Laplace transform tables or do you need to show calculations, as per:

    [tex]\int e^{-st} f(t) dt[/tex]
     
  4. Jun 27, 2010 #3
    Yes, i am familiar with the tables, but I'm not sure how to use them, lol.
    I don't need the calculations, just the solution. Thanks again.
     
  5. Jun 27, 2010 #4

    Zryn

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    So looking at the tables, you have a list of time domain functions and a list of equivalent Laplace domain functions. It doesnt matter which you start with, you can work from either side back to the other.

    Your question gives you Laplace domain functions and wants the time domain equivalent.

    Can you transform [tex]\frac{1}{s}[/tex] and [tex]\frac{1}{s^{2}}[/tex] into their time domain equivalents?
     
  6. Jun 27, 2010 #5
    Yes I can, 1 and t
     
  7. Jun 27, 2010 #6

    Zryn

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    Yep, spot on.

    How about [tex]\frac{2}{s^{3}}[/tex] and [tex]\frac{1}{s+2}[/tex] ?
     
  8. Jun 27, 2010 #7
    Hm, I'm not sure about 2/s^3 but 1/s+2 is e^-2t, right?
     
  9. Jun 27, 2010 #8

    Zryn

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    Ah, I see the problem then.

    There are a billion and one different Laplace transform tables, some more comprehensive than others.

    If for example, you took a look at:

    http://math.fullerton.edu/mathews/c2003/laplacetransform/LaplaceTransformMod/Images/Table.12.2.jpg [Broken]

    Check out transform #2 and #5.

    * Yes [tex]\frac{1}{s+2}[/tex] becomes [tex]e^{-2t}[/tex] by the way, how about, [tex]\frac{2}{s^{3}}[/tex] ?
     
    Last edited by a moderator: May 4, 2017
  10. Jun 27, 2010 #9
    ^is it t^2?
    Ah, I'm still not really sure, could you please show me the solution to
    f(t)= 1 / (s+2)^2
    That will probably make it much clearer to me. Thanks
     
  11. Jun 27, 2010 #10

    Zryn

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    Using #5 on that table:

    [tex]\frac{n!}{(s-a)^{n+1}}[/tex] becomes [tex]t^{n}[/tex][tex]e^{at}[/tex]

    Therefore:

    [tex]\frac{1!}{(s- -2)^{1+1}}[/tex] turns into ...

    * Yes, [tex]\frac{2}{s^{3}}[/tex] turns into [tex]t^{2}[/tex]
     
  12. Jun 27, 2010 #11
    te^-2t?
     
  13. Jun 27, 2010 #12
    So f(t)= 1 / (s+3)^3 would be te^-3t?
     
  14. Jun 27, 2010 #13

    Zryn

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    Bingo!

    Now the second and third equation of yours require a small amount of manipulation. You may notice that the second equation is almost the same as the first, except you would need a 2 as the numerator for the transform to work.

    When this happens, its just a matter of being tricksy.

    Keep in mind that[tex]\frac{1}{2}[/tex]*2 = 1. Make the numerator of your equation a 2 (multiply by 2) instead of a 1 so the transform works, then multiply the answer after you finish transforming by [tex]\frac{1}{2}[/tex] so that everything works out.

    What answer do you get?

    * You type faster than me, and you're fairly close but to get that answer you cant do the transform with the 1 as the numerator. Much like with 2/s^3 = t^2, you cant have 1/s^3 = t^2, you need to make (2*)1/s^3=(1/2*)t^2.
     
  15. Jun 27, 2010 #14
    I'm sorry, I'm not really sure of what you mean?
     
  16. Jun 27, 2010 #15

    Zryn

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    No problems!

    When you do the intgral of 2x, you bring the 2 out the front of the integral sign, integrate x, then multiply the answer by 2, to get x^2. You can confirm this by then differentiating x^2 to get 2x.

    When you do Laplace transforms, you can do the same thing.

    If you have 2/s, you actually have 2* 1/s. 1/s Transforms from Laplace domain to time domain to 1, and then you multiply by 2 for a final answer of 2.

    If you had 4/s^3 you actually have 2 * 2/s^3. You transformed 2/s^3 earlier to t^2, so you then multiply the transformed answer by 2 and the final answer is 2t^2.

    If you have 1/(s+3)^3, you actually have 1/2 * 2/(s+3)^3. You need the 2 in the numerator to be able to do the transform, which then becomes t^2e^-3t, multiplying by the 1/2 gives a final answer of 1/2*t^2e^-3t.

    Does that make more sense?
     
  17. Jun 27, 2010 #16
    Sort of, but I'm still kind of lost..
    What is the solution to f(t)= 1 / (s+3)^3?
    I'll understand if I see the solution.
     
  18. Jun 27, 2010 #17

    Zryn

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    Based on Line 6 in my previous post. Repetition (+Bold) for emphesis!

    If you have 1/(s+3)^3, you actually have 1/2 * 2 * 1/(s+3)^3 (since 1/2 * 2 = 1, and you can just about always multiply something by 1 if youre in the mood to do so). You need the 2 in the numerator to be able to do the transform, and you take the 1/2 out the front of the transform (like you would take 2 out the front of the integral example I mentioned earlier) which then becomes t^2e^-3t, multiplying by the 1/2 gives a final answer of 1/2*t^2e^-3t.

    * Theres only two problems with learning Laplace theory, the first is knowing how to use a transform table, and the second (and much harder) is being able to see what you need to do to an equation to get it into the form of one of the transforms on the transform table. If you can do both of those you will be able to solve all of your class problems however! Just takes a bit of practice.
     
    Last edited: Jun 27, 2010
  19. Jun 27, 2010 #18
    So is it 2t^2e^-3t?
     
  20. Jun 27, 2010 #19

    Zryn

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    Not quite.

    Starting on the time domain side of the table, what does te^-3t transform into? (using transform #5 on the laplace table). Suffer through my continuous questions if you can, theres a point and a goal we're heading towards ;)
     
  21. Jun 27, 2010 #20
    1/(s+3)^2?
     
  22. Jun 27, 2010 #21

    Zryn

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    Right!

    Now, how about t^2e^-3t?
     
  23. Jun 27, 2010 #22
    2/(s+3)^3?
     
  24. Jun 27, 2010 #23

    Zryn

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    Right again!

    Now, you may notice that the equation we are working on is 1/(s+3)^3, which does not match the form of any of the transforms on our Laplace table, but we do know that t^2e^-3t transforms into 2/(s+3)^3, which is really really close. In fact, its so close, its only out by a multiplication of x, where x = 1/2 in this case.

    2/(s+3)^3 * x = 1/(s+3)^3

    What we can do is say that if we multiply the Laplace transform by x to get a transform we can work on, we then multiply the time domain answer by 1/x to get the correct result. Theres no doubt some commutation based mathematical proof somewhere if you're keen, but thats the process which happens.

    So, our function 1/(s+3)^3 * x (where x = 2) becomes 2/(s+3)^3 which transforms into t^2e^-3t. We then multiply this time domain function by 1/x, so the final answer is 1/2*t^2e^-3t.

    *This method works from Laplace to Time, and vice versa. You can practice by seeing what 1/4*t^3e^-5t transforms into!

    Hows that sound? (You need to do this process to solve your third equation, and probably most other equations in the future, so you will get lots of practice)
     
  25. Jun 28, 2010 #24

    Mark44

    Staff: Mentor

    Per the rules of this forum
    "Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made."

    If you haven't looked at the rules, they are here: https://www.physicsforums.com/showthread.php?t=5374
     
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