Question

Got another Inverse Laplace Transformation I'd like to get some assistance with.

Please and thank you!

Answer 1

Find Inverse Laplace { (s+1) / (4s^2 + 1) }

Below is my work

Answer 2

\[L^{-1} = \frac{s+1}{4s^{2}+1} = L ^{-1} ( \frac{ 1 }{ 4s } + \frac{ 1 }{ 4s ^{2} + 1} )\]

Answer 3

\[\frac{1}{4} + L^{-1}( \frac{1}{4} \frac{1}{s^{2}+\frac{1}{4}} ) = \frac{1}{4} +\frac{1}{2} \frac{\frac{1}{2}}{s^{2} + (\frac{1}{2})^{2}}) = \frac{1}{4} + \frac{1}{2}\sin(\frac{1}{2}t)\]

Answer 4

Plugging in the reverse laplace equation into wolfram we get the following

\[L^{-1} = \frac{1}{4} ( 2\sin(\frac{t}{2}) + \cos(\frac{t}{2}) )\]

Answer 5

Fan and medal for correction of my work.

Answer 6

Not sure about the last step. Why do you need L^-1? is it not that it is just 1/4 + (1/2) sin (t/2)?

Answer 7

Because I need to find the inverse transform of the rest of the equation so:
\[\frac{1}{4} * L^{-1}\frac{1}{2}\frac{\frac{1}{2}}{s^{2}+\frac{1}{2}^{2}} \]

Is the laplace of \[L(coskt) = \frac{s}{s^{2}+k^{2}}\]

So\[k= \frac{1}{2}\]

Answer 8

You are given F(s) , hence \(L^{-1} (F(s)) = f(t) \)
And your work gave \(f(t) = \dfrac{1}{4} +\dfrac{1}{2} sin(t/2)\)
and we are done. What I don't get is why do you take L^(-1) = 1/4 (2sin(t/2) +cos (t/2))

Answer 9

Ugh I apologize my syntax it's suppose to be a + not a * Not add not multiply

Answer 10

Oh no for that, it was the given solution via Wolfram.

I didn't understand how they got to it and that's why I'm thinking my answer is incorrect.

Answer 11

post the wolfram's link, please

Answer 12

Sure thing:

http://www.wolframalpha.com/input/?i=Inverse+Laplace+ {%28s%2B1%29%2F%28%284s^2%29%2B1%29}

Answer 13

My plugin: Inverse Laplace {(s+1)/((4s^2)+1)}

Answer 14

ok, I need look up at my book :)

Answer 15

Sure this may help, how I broke it into two different laplace equations

\[L^{-1}(\frac{s+1}{4s^{2}+1}) = L^{-1}(\frac{1}{4s} + \frac{1}{4s^{2}+1}) = L^{-1}(\frac{1}{4s}) + L^{-1}(\frac{1}{4s^{2}+1})\]

Given:

\[L(1) = \frac{1}{s}\]

\[L(sinkt) = \frac{k}{s^{2}+k^{2}}\]

Answer 16

I know how to get that step :)

Answer 17

oh, yeah, that is the formula :)!!

Answer 18

LOL sorry, it just seems to straight forward Im not sure how they got anything different.

I know for imaginary roots with auxilary equations we come up with both sin and cos

But how did it come about this way?

Answer 19

think of this
\(\dfrac{s+1}{4s^2 +1}=\dfrac{s}{4s^2+1}+\dfrac{1}{4s^2+1}\)

the first term is \(L^{-1}\dfrac{s}{4(s^2+(1/2)^2)}= (1/4) cos (t/2)\)

the second term is \(L^{-1} \dfrac{1}{4(s^2+(1/2)^2)}=1/2 sin (1/2 t)\)

Answer 20

Crap... I'm sorry.

I made a HUGE booboo right from the get go..

I split the fractions incorrectly and instead of

\[\frac{s}{4s^{2}+1}\]

I got

\[\frac{1}{4s}\]

Answer 21

Thank you helping me realize that. I spotted it as soon as I saw your equation

Answer 22

:)