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Can anyone find inverse Laplace of this (s^2pi^2)/(s^2+pi^2)^2
 one year ago
 one year ago
Can anyone find inverse Laplace of this (s^2pi^2)/(s^2+pi^2)^2
 one year ago
 one year ago

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shafqat_uetBest ResponseYou've already chosen the best response.0
\[\frac{(s^{2}\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]
 one year ago

itsmylifeBest ResponseYou've already chosen the best response.0
\[\frac{ (s \pi) (s+ \pi)}{ (s+ \pi) (s+ \pi) }\] now you can do it ;)
 one year ago

CallistoBest ResponseYou've already chosen the best response.4
The denominator is not quite right?
 one year ago

itsmylifeBest ResponseYou've already chosen the best response.0
i think its done , you gotta separate terms now thats all ;) your welcome @shafqat_uet
 one year ago

itsmylifeBest ResponseYou've already chosen the best response.0
holda thers a mistake
 one year ago

shafqat_uetBest ResponseYou've already chosen the best response.0
in denomirator you made a mistake
 one year ago

itsmylifeBest ResponseYou've already chosen the best response.0
\[\frac{ (s  \pi)(s + \pi) }{ (s^2 + \pi^2)(s^2+ \pi^2) } \] this looks pretty easier but m stuck now :(
 one year ago

CallistoBest ResponseYou've already chosen the best response.4
\[\frac{(s^{2}\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]\[=\frac{(s^{2})}{(s ^{2}+\pi ^{2})^{2}}  \frac{\pi^2}{{(s ^{2}+\pi ^{2})^{2}}}\] \[=(\frac{s^2}{(s ^{2}+\pi ^{2})})(\frac{1}{(s ^{2}+\pi ^{2})})  \frac{\pi^2}{{(s ^{2}+\pi ^{2})}}\times \frac{1}{{(s ^{2}+\pi ^{2})}}\]
 one year ago

shafqat_uetBest ResponseYou've already chosen the best response.0
@itsmylife good try but I have also tried like this. Is it possible using derivative or integral of the laplace
 one year ago

itsmylifeBest ResponseYou've already chosen the best response.0
no i guess wat @Callisto has done is quite easier , you gotta take inverse now
 one year ago

CallistoBest ResponseYou've already chosen the best response.4
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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