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Math Help - Laplace/ heaviside help

  1. #1
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    Laplace/ heaviside help

    My problem is regarding this function:

    H(t-pi/2) * cos(3t)

    and Im suppose to write the Laplace transformation of this product using the theorem:
     <br /> <br />
L[H(t-c) f(t-c)] = e^{-cs}*L(f)<br />

    I understand the principle behind this problem, and I know the answer is  e^{-pi*s/2} * L(cos(3t) in terms of t-pi/6). Its just I can't figure out how to rewrite cos(3t) in terms of t-pi/2. I tried to do cos (2t + t), use the addition property, and end up getting stuck with a 1-4sin(t)^2 that I dont know what to do with. Can someone help me out?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mike10003 View Post
    My problem is regarding this function:

    H(t-pi/2) * cos(3t)

    and Im suppose to write the Laplace transformation of this product using the theorem:
     <br /> <br />
L[H(t-c) f(t-c)] = e^{-cs}*L(f)<br />

    I understand the principle behind this problem, and I know the answer is  e^{-pi*s/2} * L(cos(3t) in terms of t-pi/6). Its just I can't figure out how to rewrite cos(3t) in terms of t-pi/2. I tried to do cos (2t + t), use the addition property, and end up getting stuck with a 1-4sin(t)^2 that I dont know what to do with. Can someone help me out?
    Recall that \sin\!\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta.

    So \cos\!\left(3t\right)=-\sin\!\left(3t-\tfrac{\pi}{2}\right)

    Now note that \sin\!\left(3t-\tfrac{\pi}{2}\right)=\sin\!\left(3t-\tfrac{3\pi}{2}+\pi\right)=-\sin\!\left(3t-\tfrac{3\pi}{2}\right)=-\sin\!\left(3\left[t-\tfrac{\pi}{2}\right]\right)

    Therefore \cos(3t)=\sin\!\left(3\left[t-\tfrac{\pi}{2}\right]\right)=\left.\sin(3t)\right|_{t\to t-\frac{\pi}{2}}

    Does this make sense?
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  3. #3
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    Yes that is a great help! Didn't think to use those angle identities.

    So the other problem related to this is:

    H(t-pi/6)*sin(2t)

    Using your guide as a reference, I did this
    sin(2t)=cos(pi/2-2t)
    =cos(-2t+pi/2)
    =cos(-2t+pi/3 + pi/6)
    =cos(-2(t-pi/6)+pi/6))

    Here I wasnt sure what to do, so I split it up using the addition formula.

    =cos(-2(t-pi/6))cos(pi/6) - sin(-2(t-pi/6))sin(pi/6)
    =\sqrt(3)/2*cos(-2(t-pi/6)) - 1/2*sin(-2(t-pi/6))

    Then converting to laplace, I get this

    =\sqrt3 /2*(s/s^{2}+4) - 1/2*(-2/s^{2}+4)
    =(\sqrt3 s+2)/(2*(s^{2}+4))
    Then multiply that by the L of the heaviside, e^{-pi*s/6}

    However, the answer in the back of the book is
    (1/2)e^{-pi*s/6}*(4+\sqrt3 s^{2})/(s^{2}+4)

    Am I doing something wrong? I dont know where they got a s^2 in the numerator, since the laplace of cos is s/(s^2+a^2) and sin is a^2/(s^2+a^2).
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