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Laplace transformation problem

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    L{4u(t-ர)cos2t}

    2. Relevant equations

    I have used UNIT STEP FUNCTION but could not get the result

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2010 #2

    vela

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    What did you get? Show your work, not just your final result from your attempt.
     
  4. Sep 8, 2010 #3
    L{4u(t-ர)cos2t
    f(t) = cos2t
    F(s) = s/ s^2+4
    therefore

    4s e^-ர(cosர + sinர)/s^2 +4 Ans
    ________________________________
    that is my work
     
  5. Sep 8, 2010 #4

    gabbagabbahey

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    Where is the factor the of [itex]4e^{-\tau}(\sin\tau+\cos\tau)[/itex] coming from? What rule are you trying to apply here?
     
  6. Sep 9, 2010 #5
    here is my solve, if I have done correctly.
    laplace.gif
     
  7. Sep 10, 2010 #6
    Please expert help me.
     
  8. Sep 10, 2010 #7

    vela

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    Unfortunately, I find your handwriting a bit hard to read. Could you type out your final answer?
     
  9. Sep 10, 2010 #8

    gabbagabbahey

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    It looks good to me (although you have a very strange way of writing [itex]\pi[/itex]). On a side note, since trig functions are [itex]2\pi[/itex] periodic, you should have immediately recognized that [itex]\cos(2t+2\pi)=\cos(2t)[/itex] without appealing to a trig identity.
     
  10. Sep 10, 2010 #9

    vela

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    I think the exponential term in the front incorrectly has an a in it rather than s.
     
  11. Sep 10, 2010 #10
    here is my final Answer
    4e^(-as) s/(s^(2) + 4)

    what do you think.........
     
  12. Sep 11, 2010 #11

    vela

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    The variable a shouldn't be in the final answer.
     
  13. Sep 14, 2010 #12
    can you giving me some hints?
     
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