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shafqat_uet Group Title

Can anyone find inverse Laplace of this (s^2-pi^2)/(s^2+pi^2)^2

  • one year ago
  • one year ago

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  1. shafqat_uet Group Title
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    \[\frac{(s^{2}-\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]

    • one year ago
  2. itsmylife Group Title
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    \[\frac{ (s- \pi) (s+ \pi)}{ (s+ \pi) (s+ \pi) }\] now you can do it ;)

    • one year ago
  3. Callisto Group Title
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    The denominator is not quite right?

    • one year ago
  4. shafqat_uet Group Title
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    thanks itsmylife

    • one year ago
  5. itsmylife Group Title
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    i think its done , you gotta separate terms now thats all ;) your welcome @shafqat_uet

    • one year ago
  6. itsmylife Group Title
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    holda thers a mistake

    • one year ago
  7. shafqat_uet Group Title
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    in denomirator you made a mistake

    • one year ago
  8. itsmylife Group Title
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    \[\frac{ (s - \pi)(s + \pi) }{ (s^2 + \pi^2)(s^2+ \pi^2) } \] this looks pretty easier but m stuck now :(

    • one year ago
  9. Callisto Group Title
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    \[\frac{(s^{2}-\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]\[=\frac{(s^{2})}{(s ^{2}+\pi ^{2})^{2}} - \frac{\pi^2}{{(s ^{2}+\pi ^{2})^{2}}}\] \[=(\frac{s^2}{(s ^{2}+\pi ^{2})})(\frac{1}{(s ^{2}+\pi ^{2})}) - \frac{\pi^2}{{(s ^{2}+\pi ^{2})}}\times \frac{1}{{(s ^{2}+\pi ^{2})}}\]

    • one year ago
  10. shafqat_uet Group Title
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    @itsmylife good try but I have also tried like this. Is it possible using derivative or integral of the laplace

    • one year ago
  11. itsmylife Group Title
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    no i guess wat @Callisto has done is quite easier , you gotta take inverse now

    • one year ago
  12. Callisto Group Title
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    Sorry, last step: \[=(\frac{s}{(s ^{2}+\pi ^{2})})(\frac{s}{(s ^{2}+\pi ^{2})}) - \frac{\pi}{{(s ^{2}+\pi ^{2})}}\times \frac{\pi}{{(s ^{2}+\pi ^{2})}}\]

    • one year ago
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