ragman
find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))
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Skaematik
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note
ragman
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'note' sorry this is my first time using this what do you mean by that?
Skaematik
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Im just bookmarking this question so that I can return to it later. :D
Id want to know the answer too
ragman
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oh ok sounds good thanks
lgbasallote
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your question is \[\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]
lgbasallote
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??
lgbasallote
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@ragman you there?
ragman
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hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )
ragman
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@lgbasallote
lgbasallote
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uhh s^2 + 22 + 5??
ragman
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dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)
lgbasallote
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so..
\[\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]
ragman
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yup perfect
hartnn
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can u do partial fractions ??
ragman
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@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
lgbasallote
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where's C?
hartnn
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those are correct, so u need only C right ?
where u took C above s+1
lgbasallote
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your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]
lgbasallote
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so you have \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]
ragman
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oh 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
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nopes only C, if it were s^2+1 u would have taken Cx+D
hartnn
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so i think u can get C now, right?
ragman
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oh i see now and yeah i believe i can get C now trying it right now
lgbasallote
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i think it will change your A value too
lgbasallote
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so you best check
hartnn
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great tell us what u get C as....we all will verify that.
A is correct.
lgbasallote
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nope nevermind..it stays the same
ragman
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ok sounds good got B=5 finding A now
ragman
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A=-3 and C=4 is that what you guys got?
hartnn
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yup, same.Good work.
lgbasallote
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A =-3?
lgbasallote
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yeah that's right
ragman
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ok cool i was about to write out how i got it lol
lgbasallote
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everything's right
ragman
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and now i just plug back all the variables right
hartnn
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yup, and use inverse laplace formulas.
lgbasallote
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yep find \[\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}\]
lgbasallote
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uhh that should be 5/s^2
ragman
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ok 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
lgbasallote
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here'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?
lgbasallote
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uhhh wait... wrong formula for last one
lgbasallote
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\[\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}\]
better
ragman
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yes thats very helpful but im getting confused with what n should be in the middle equations
lgbasallote
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n is the exponent
ragman
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so it would be 2?
lgbasallote
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for example:
\[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]
ragman
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thats 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
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yes thats true, so u will have 5t^1= 5t ....isn't it ?
u can verify that u should get final answer as 5t-3+4e^(-t)
did u get that?
ragman
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yup thats what i got thanks
hartnn
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welcome :)