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anonymous
 3 years ago
find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))
anonymous
 3 years ago
find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0'note' sorry this is my first time using this what do you mean by that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im just bookmarking this question so that I can return to it later. :D Id want to know the answer too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok sounds good thanks

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your question is \[\huge \mathcal L ^{1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so.. \[\huge \mathcal L ^{1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2can u do partial fractions ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@hartnn yeah thats what i was doing but for some reason i could figure out the other two variable maybe im solving it incorrectly but i got 3 for A and 5 for B

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2those are correct, so u need only C right ? where u took C above s+1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you have \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok so thats where i messed up i thought that where C is it should be (Cs+D) so i was trying to solve for both of those variables

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2nopes only C, if it were s^2+1 u would have taken Cx+D

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2so i think u can get C now, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see now and yeah i believe i can get C now trying it right now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think it will change your A value too

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2great tell us what u get C as....we all will verify that. A is correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope nevermind..it stays the same

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok sounds good got B=5 finding A now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A=3 and C=4 is that what you guys got?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok cool i was about to write out how i got it lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and now i just plug back all the variables right

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2yup, and use inverse laplace formulas.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep find \[\huge \mathcal L ^{1} \left \{ \frac 3s + \frac 5s + \frac 4{s+1} \right \}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhh that should be 5/s^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok sorry im just learning this but how do you go about doing that im trying to find the L^1 formula but my notes arent really helping

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here's a hint: \[\huge \mathcal L \{ a \} = \frac as\] \[\huge \mathcal L \{ t^n\} = \frac{n!}{s^{n+1}}\] \[\huge \mathcal L \{e^{at} \} = \frac{a}{s a}\] does that help?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhhh wait... wrong formula for last one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \mathcal L \{e^{at} \} = \frac{1}{sa}\] better

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes thats very helpful but im getting confused with what n should be in the middle equations

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for example: \[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats whats a lil confusing to me because on the denominator it would have to be n=1 so that it would equal s^2 but on the numerator 1! equals 1 right ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2yes thats true, so u will have 5t^1= 5t ....isn't it ? u can verify that u should get final answer as 5t3+4e^(t) did u get that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup thats what i got thanks
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