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  • Laplace Transform : sin^2(wt) question

    Brad Grillakis

    Brad Grillakis

    @brad-grillakis-aE4GRM
    Updated: Oct 26, 2011
    Views: 1.7K
    Hello! I am working on my Laplace Transform homework and I am trying to work out the steps on how my professor got this answer but I am not understanding it. The question is:

    Q: What is the inverse Laplace transform of {32 / (s(s[SUP]2[/SUP] + 64))} ?
    A: sin[SUP]2[/SUP](4t)

    The Laplace Look Up Table formula is :

    Signal                                Transform
sin[SUP]2[/SUP](w[SUB]0[/SUB]t)                          2w[SUB]0[/SUB][SUP]2[/SUP] / (s(s[SUP]2[/SUP] + 4w[SUB]0[/SUB][SUP]2[/SUP]))
    When I am breaking down this function and I am using the transform equation I can only break down the 64 (since 8[SUP]2[/SUP] = 64) but that 4 right next to it and that 32 is throwing me for a loop.

    I really appreciate it if someone can help me understand how the answer is sin[SUP]2[/SUP](4t)
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  • Sagar07

    Member • Oct 27, 2011

    Didn't get your question exactly.......
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  • Brad Grillakis

    Member • Oct 27, 2011

    Sagar07
    Didn't get your question exactly.......
    My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)

    The 2 formula I gave is from the Laplace Transform Table.
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  • ritesh05

    Member • Oct 27, 2011

    Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is
    [​IMG]
    so after using this formula you will get -.5(cos8t-1)=-.5(1-2sin^(4t)-1)=sin^2(4t)...............{cos2t=1-2sin^2(t)}
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  • rashidsohail

    Member • Nov 1, 2011

    i dont understand that
    Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is
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  • simplycoder

    Member • Jan 1, 2012

    Brad Grillakis
    My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)

    The 2 formula I gave is from the Laplace Transform Table.
    This can be solved by partial fractions. All you need is proper substitution.

    Substitute it as (A/s)+(Bs+C)/(s^2+64).

    A=0.5,B=-0.5,C=0.

    so you get 0.5/s- 0.5/(s^2+64)

    Laplace inverse of the first term is 1, and that of second term is cos(8t).
    So you get 0.5*(1-cos8t).By trignometry, cos8t=1-2sin^2 (4t).
    So you get 0.5*(1-(1-2sin^2(4t))).
    =0.5*2sin^2(4t).
    =sin^2(4t).//
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  • silverscorpion

    Member • Jan 2, 2012

    Write the fraction 32/(s(s2+64) as follows..

    32/(s(s2+64) = (2*4[sup]2[/sup])/(s*(s[sup]2[/sup] + 4*4[sup]2[/sup]))

    So, you get w[sub]0[/sub] to be 4. Hence, the inverse Laplace transform is, from the formula you have given, sin[sup]2[/sup] 4t
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