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anonymous

  • 5 years ago

L^-1[(s^2)+2s+1]/[(s^2)+2s+4]=?

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  1. anonymous
    • 5 years ago
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    I don't know how much detail you need, but,\[\frac{s^2+2s+1}{s^2+2s+4}=\frac{(s^2+2s+(1+3))-3}{s^2+2s+4}=\frac{(s^2+2s+4)-3}{s^2+2s+4}\]\[=1-\frac{3}{s^2+2s+4}=1-\frac{3}{(s+2)^2}\]

  2. anonymous
    • 5 years ago
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    The inverse Laplace of the quotient is then the sum of the inverse Laplaces. The inverse Laplace of 1 is the delta function, and the inverse Laplace of the second is 2e^(-2t)t

  3. anonymous
    • 5 years ago
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    \[\delta (t)-3e^{-2t}t\]

  4. anonymous
    • 5 years ago
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    Sorry, scrap that - falling asleep...the second fraction is wrong, though the first part of the answer is right - the delta function. The second fraction is\[\frac{3}{(s+1)^2+3}=\sqrt{3}.\frac{\sqrt{3}}{(s-(-1))^2+(\sqrt{3})^2}\]which is in a form you can read off a table of inverted Laplace transforms as\[\sqrt{3}e^{-t}\sin \sqrt{3}t\]

  5. anonymous
    • 5 years ago
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    So your inverse Laplace is really\[\delta (t) -\sqrt{3}e^{-t}\sin \sqrt{3}t\]

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