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Math Help - Laplace transform

  1. #1
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    Laplace transform

    Q1) L {4cost sin2t}

    Q2) L { f 0- infinite(the big f sign) e pwt-3t ( cos4t - t^2)dt}

    Q3) Given iR + 1/C f 0- infinite(the big f sign) i dt + Ldi/dt = 4, where R,C,L are constants. show that if i(0) = 0, then

    L {i} = 4C/LCs^2+RCs+1


    Ans:
    1) (4/(s-1)^2+4) +(4/s+1)^2 +4

    2) 1/s [ (s+3/(s+3)^2 + 16) - 2/(s+3)^2]

    3) I(s) = 4s/ (s+3)^2(s^2+4)


    Help... i just started learning laplace transform. i am struggling because i am like direct to 2yr to my school and i have never done anything related before.. anyone can refer me to any good online materials that teaches the basics(if any)....also provide me with the workings/expl for the above question. Thanks for helping.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathsnerd View Post
    Q1) L {4cost sin2t}
    Use the trig identity:

    sin(A+B) - sin(A-B) = 2 sin(A)cos(B)

    to reduce this to:

    L {4 cos(t) sin(2t)} = L 4[sin(3t) - sin(t)] = 2 [Lsin(3t) + Lsin(t)]

    Then look up the Laplace transform of sin(at) in a table to complete
    the task.

    2 [Lsin(3t) + Lsin(t)] = 2[3/(s^2+9) + 1/(s^2+1)]

    ................ .....=[8 (s^2 + 3) / ((s^2 + 1)(s^2 + 9))]

    Q2) L { f 0- infinite(the big f sign) e pwt-3t ( cos4t - t^2)dt}

    Q3) Given iR + 1/C f 0- infinite(the big f sign) i dt + Ldi/dt = 4, where R,C,L are constants. show that if i(0) = 0, then

    L {i} = 4C/LCs^2+RCs+1
    You will have to improve the way you are typing these questions to make them easier
    for us to understand, Q2 and 3 are incomprehensible as they are

    Ans:
    1) (4/(s-1)^2+4) +(4/s+1)^2 +4

    2) 1/s [ (s+3/(s+3)^2 + 16) - 2/(s+3)^2]

    3) I(s) = 4s/ (s+3)^2(s^2+4)


    Help... i just started learning laplace transform. i am struggling because i am like direct to 2yr to my school and i have never done anything related before.. anyone can refer me to any good online materials that teaches the basics(if any)....also provide me with the workings/expl for the above question. Thanks for helping.
    Look here (but I suspect you are lacking a lot of background that
    you need to follow this, you might need to revise single variable
    calculus first).

    RonL
    Last edited by CaptainBlack; April 20th 2007 at 11:37 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Use the trig identity:

    sin(A+B) - sin(A-B) = 2 sin(A)cos(B)

    to reduce this to:

    L {4 cos(t) sin(2t)} = L 4[sin(3t) - sin(t)] = 2 [Lsin(3t) + Lsin(t)]

    Then look up the Laplace transform of sin(at) in a table to complete
    the task.

    2 [Lsin(3t) + Lsin(t)] = 2[3/(s^2+9) + 1/(s^2+1)]

    ................ .....=[8 (s^2 + 3) / ((s^2 + 1)(s^2 + 9))]
    You can have a bit of fun with this using the online Fourier-Laplace calculator here.
    This gives for the LT of 4 cos(t) sin(2t):

    -2i (s-2i)/[(s-2i)^2+1] + 2i (s+2i)/[(s+2i)^2+1]

    which you can have a lot of fun proving is the same as what I gave in the
    previous post.

    RonL
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