Easy laplace transform question

Discussion in 'Homework Help' started by ihaveaquestion, Oct 4, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    The laplace transform of f(t) = (2t+4)u(t) = 2/s^2 + 4/s

    I can see that the laplace of f(t) = (2t) = (1/2)*(1/(s/2)^2) = 2/s^2

    but I'm stumped at what I do with the 4... I see nothing like it on the laplace transform table... matlab gives the laplace of (t+4), a broken down version, to be 1/s^2 + 4/s. Again I see the 1/s^2 part from the t, but I don't se how they get 4/s for the 4 part... the laplace of u(t) is 1/t, is this why? Some explanation would be appreciated, thanks.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The Laplace transform of a constant k is simply k/s. So the LT of a constant 4 is 4/s as you have indicated.

    For the purposes of performing the Laplace transform, the Heavyside Function u(t) constrains the value of the time domain function it acts upon such that, for t<0 the function is zero and for t>=0 the function is the multiplicand of u(t). Presumably this is something that a rigorous application of the underlying mathematics demands - maybe it's all a bit academic for us ordinary folk.
     
  3. ihaveaquestion

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    May 1, 2009
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  4. t_n_k

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    Laplace Transform of sin(3t) is 3/(s^2+9) - not what you have written.

    Laplace Transform of cos(3t) is s/(s^2+9) - again not what you wrote.

    Where did you get those forms you wrote - from a table of transfrm pairs?


    Laplace Transform of f(t)=6sin(3t)+8cos(3t) is

    F(s) = 6*3/(s^2+9) + 8*s/(s^2+9)
    = 18/(s^2+9) +8s/(s^2+9)
    = (8s+18)/(s^2+9)
     
  5. ihaveaquestion

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    May 1, 2009
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  6. t_n_k

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    Mar 6, 2009
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    You made a fairly simple error near the end. Should be like this .....

    1/2 X (s-2+s+6)/(s^2+4s-12) = 1/2X(2s+4)/(s^2+4s-12)=(s+2)/(s^2+4s-12)
     
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