Finding laplace inverse of $\frac{1}{(p+2)^2(p^2+\omega^2)}$

Question

I am finding the laplace inverse of the below :- $$\frac{1}{(p+2)^2(p^2+\omega^2)}$$

I can apply convolution theorem by taking $$f(p)=\frac{1}{(p+2)^2}, g(p)=\frac{1} {(p^2+\omega^2)}$$ and calculating the function which yield the above functions on laplace transform i.e. essentially , $$F(t)=e^{-2t}t,G(t)=\frac{\sin{\omega}t}{\omega}$$ then solve by routine integration .

But this seems a little cumbersome . Is there a better way to solve the above ?

Thanks for reading !

Answer 1

The inverse Laplace transformation is define as $f(t)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}F(p)e^{pt}dp$, where integral is taken in the complex plane along some line $(a-i\infty,a+i\infty)$, where $a$ is some real number (greater than the real part of all singularities of $F(p)$) https://en.wikipedia.org/wiki/Inverse_Laplace_transform

We can close the contour, integrating along a part of the circle of big radius $R$. Integral along this circle $\to0$ at $R\to\infty$ if $|F(p)|\to 0$ at $|p|\to\infty$ ; this is the case for $F(p)=\frac{1}{(p+2)^2(p^2+\omega^2)}$

$$f(t)=\oint_C=\frac{2\pi{i}}{2\pi{i}}\sum_{\text{all poles}}Res\frac{e^{pt}}{(p+2)^2(p^2+\omega^2)}=\frac{e^{i\omega t}}{2i\omega(2+i\omega)^2}+\frac{e^{-i\omega t}}{-2i\omega(2-i\omega)^2}+\frac{d}{dp}\bigl(\frac{e^{pt}}{p^2+\omega^2}\bigr)|_{p=-2}$$ $$=\frac{1}{\omega}\Im\frac{e^{i\omega t}}{(2+i\omega)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$ $$=\frac{(4-\omega^2)\sin \omega t-4\omega\cos\omega t}{\omega(4+\omega^2)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$

Please, check.

Answer 2

Hint:

Use Partial Fraction Decomposition

$$\dfrac1{(p^2+w^2)(p+2)^2}=\dfrac a{p+2}+\dfrac b{(p+2)^2}+\dfrac{cp+d}{p^2+w^2}$$

Multiply both sides by $(p^2+w^2)(p+2)^2$

and use the coefficients of different exponents of $p$ to find $a,b,c,d$

Finally use the standard formulae