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Math Help - Inverse laplace transform, help needed!

  1. #1
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    Smile Inverse laplace transform, help needed!

    Hi. I'm trying to find the inverse laplace transform of the following:

    <br />
\frac{(s+2)}{(s+1)*(s^2+4s+5)}<br />

    Problem is - I'm not sure where to begin. Should I complete the square for
    (s^2 +4s+5)? And if so, where do I go next from that?

    Any help would be greatly appreciated.
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  2. #2
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    Start by separating the fraction into partial fractions.
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  3. #3
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    Quote Originally Posted by whipflip15 View Post
    Start by separating the fraction into partial fractions.
    Thanks for replying. I tried that earlier, but was unsure about something:

    <br />
(s+2) = A(s^2+4s+5) + B(s+1)<br />

    Setting s=-1 produces A=\frac{1}{2} ,

    ...but what about B? The solution's complex, so how do I deal with that?
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  4. #4
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    Quote Originally Posted by dan569 View Post
    Thanks for replying. I tried that earlier, but was unsure about something:

    <br />
(s+2) = A(s^2+4s+5) + B(s+1)<br />

    Setting s=-1 produces A=\frac{1}{2} ,

    ...but what about B? The solution's complex, so how do I deal with that?
    Since the quadratic factor is irreducible over real numbers, the correct partial fraction form to use is \frac{A}{s+1} + \frac{Bs + C}{s^2 + 4s + 5}.

    When getting the inverse Laplace Transform of the second term you will need to complete the square.
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  5. #5
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    Lightbulb

    Thanks, I think I've got it.

    y(t) = \frac{1}{2}[e^{-t} + (sin(t) - cos(t)).e^{-2t}]
    Last edited by dan569; December 9th 2008 at 03:53 AM. Reason: added solution
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  6. #6
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    Apologies. The actual wording of the question is:

    ...find the inverse laplace transform of \frac{(s+2)}{(s+1)(s^2 +4s+5)}

    Hence, show that it is equal to \int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\
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  7. #7
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    Quote Originally Posted by dan569 View Post
    Apologies. The actual wording of the question is:

    ...find the inverse laplace transform of \frac{(s+2)}{(s+1)(s^2 +4s+5)}

    Hence, show that it is equal to \int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\
    LT^{-1}\left[ \frac{(s+2)}{(s+1)(s^2 +4s+5)}\right] = LT^{-1}\left[ \frac{1}{s+1} \cdot \frac{s+2}{(s+2)^2 + 1}\right]


    = LT^{-1}\left[ \frac{1}{s+1}\right] * LT^{-1} \left[\frac{s+2}{(s+2)^2 + 1}\right]


    = e^{-t} * ( e^{-2t} \cos t )


    where the second and third lines follow from the usual theorems and standard results.
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  8. #8
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    Quote Originally Posted by dan569 View Post
    Apologies. The actual wording of the question is:

    ...find the inverse laplace transform of \frac{(s+2)}{(s+1)(s^2 +4s+5)}

    Hence, show that it is equal to \int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\
    Or perhaps all you're expected to do is integrate \int_0^t e^{-(t-u)}.e^{-2u}.cos(u) du\ and show that it's equal to \frac{1}{2}[e^{-t} + (sin(t) - cos(t)).e^{-2t}] ....?
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  9. #9
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    Thanks for your reply.

    I imagine it's just the first option, but I've put down both, just in case. Thanks again for your time.
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