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 $(p+2)^2 +4$ ?

Okay. Now combine (and multiply with the constant $\frac{1}{16}$). — Daniel Fischer, Jan 12 '14 at 18:36

score 2 · Answer 1 · answered Jan 12 '14 at 18:38

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 $$\mathcal{F}(s)|_{s\to s-k}$$

score 1 · Answer 2 · answered Jan 12 '14 at 18:38

1

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

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

answered Jan 12 '14 at 18:38

Mhenni Benghorbal

48,099

@B.S.: You are absolutely right. Thanks for the comment. – Mhenni Benghorbal Jan 12 '14 at 18:43
Thanks for the answer, but how do I find the inverse of $$\dfrac{p+2}{(p+2)^2 + 4} $$ if I did not know it would turn into $e^c\cos(ax)$ as in the exam we will only be given the formula for $cos(ax)$ – Draxler292 Jan 12 '14 at 19:23
@Draxler292: You need to use the tables and some properties of the Laplace transform that you have been taught. Another approach is to use complex variable techniques. – Mhenni Benghorbal Jan 12 '14 at 19:57

score 0 · Answer 3 · answered Jul 26 '18 at 11:13

Use: $\color{red}{L^{-1}\{F(s-a)\}=e^{at}f(t)}$ $$L^{-1}\left\{\frac{1}{16}\cdot \frac{s}{s^2+2^2}\right\}=\frac{1}{16}\cos \left(2t\right)$$ So $$L^{-1}\left\{\frac{s+2}{16\left(\left(s+2\right)^2+4\right)}\right\}=e^{-2t}\frac{1}{16}\cos \left(2t\right)$$

score 0 · Answer 4 · answered Jan 12 '14 at 20:52

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.

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

4 Answers4