Inverse Laplace Transform $\frac{p+2}{16((p+2)^2 + 4)}$

Question

How do you work out the inverse laplace transform of

$$\dfrac{p+2}{16((p+2)^2 + 4)}$$

I know the $p+2$ is $e^{-2x}$ but what is the inverse of $(p+2)^2 +4$ ?

Answer 1

First step: Assume you are given $$\frac{s}{s^2+4}$$ Second step: Solve $\mathcal{L}^{-1}\left(\frac{s}{s^2+4}\right)$ by this fact that $\mathcal{L}(\cos at)=\frac{s}{s^2+a^2}$.

Third step: Note that multiplying the function $f(t)$ by $\exp(kt)$, make its L.T. to be $$\mathcal{F}(s)|_{s\to s-k}$$

Answer 2

Hint: Laplace transform of $e^c\cos(ax)$ is given by

$$ {\frac {s-c}{ \left( s-c \right) ^{2}+{a}^{2}}} $$

Answer 3

ok see here.....

you can write : the numerator as : (P - (-2))

and the denominator as: ((P - (-2))^2) + (2)^2

Then you will get something llike this: (p - (-2)) / (((p - (-2))^2) + (2)^2) -------this will be the total fraction!

and now apply the fommula : (e^(at))*(cos(bt)) = (s-a) / (((s-a)^2) - (b^2)) ------- (note that for this problem you have P in place of S and you have to manipulate the signs and apply the formula thats all... )

where, a= (-2) and b= 2. ------according to your problem.

And if we compare with the formula we will get the answer as: (e^(-2t))*(Cos(2t))

Hope the discussion will help you....best of luck.