anonymous
  • anonymous
How can I find the inverse Laplace Transform of s/(s+2)^4?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1447569784837:dw|
ganeshie8
  • ganeshie8
try using \[\large \mathcal{L}^{-1}\{ F(s-a)\} = e^{at}\mathcal{L}^{-1} \{F(s)\}\]
IrishBoy123
  • IrishBoy123
if you can't find a simple solution in a table of LPT's, use convolution and simpler standard solutions eg you can re-write it as \[\dfrac{s}{(s+2)^2}. \dfrac{1}{(s+2)^2}\] that's \[\mathcal{L}^{-1} \{ \mathcal{L}\{e^{-2t}(1-2t) \} . \mathcal{L}\{te^{-2t} \} \}\]

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anonymous
  • anonymous
I am still unclear.
anonymous
  • anonymous
|dw:1447617794696:dw| What i did was split the s-1 into two functions.
IrishBoy123
  • IrishBoy123
hi mickey. so what you really have is: \[\dfrac{s-1}{(s+2)^4} [??]\\ = \dfrac{s}{(s+2)^4} - \dfrac{1}{(s+2)^4} \\ \color{blue}{= \dfrac{s+2-3}{(s+2)^4}} \\ \color{red}{=\dfrac{1}{(s+2)^3} - \dfrac{3}{(s+2)^4}}\] so we go back to ganeshie's original suggestion 🤕 http://www.wolframalpha.com/input/?i=inverse+laplace+transform+%7B%281%29%2F%28s%2B2%29%5E3%7D \(-\) http://www.wolframalpha.com/input/?i=inverse+laplace+transform+%7B%283%29%2F%28s%2B2%29%5E4%7D i dunno. i think the convolution is simpler. does this get you the right answer to your problem?
anonymous
  • anonymous
Hello, I had posted this same problem in another thread because i accidentally closed it. It all makes sense. http://www.wolframalpha.com/input/?i=inverse+laplace+transform+%7B%28s-1%29%2F%28s%2B2%29%5E4%7D

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