Find Laplace inverse

Question

Let $${{{{x^{\ast}(s) = \left( \frac{1}{(s+\mu_1 + \mu_2) (s + \hat{\lambda}_2) (s + \lambda_1 +\lambda_2 )}\right)}}}}$$ be the laplace transform in question, where $\mu_1,\mu_2, \lambda_1,\lambda_2, \hat{\lambda}_2$ are just real parameters.

How can i get its inverse by means of convolution, $x(t) = \mathcal{L}^{-1} (x^{\ast}(s)) $?

I know that the convolution is associative, but how can i get a solution of such problem?

Thanks

Answer 1

You right, first thing here is a convolution $$ x(t) = \mathcal{L}^{-1} \left( \frac{1}{(s+\mu_1 + \mu_2) (s + \hat{\lambda}_2) (s + \lambda_1 +\lambda_2 )}\right)= \mathcal{L}^{-1} \left( \frac{1}{s+\mu_1 + \mu_2 }\right)*\mathcal{L}^{-1} \left( \frac{1}{ s + \hat{\lambda}_2 }\right) *\mathcal{L}^{-1} \left( \frac{1}{s + \lambda_1 +\lambda_2 }\right) = f(t)*g(t)*w(t) $$ where $f(t)= \mathcal{L}^{-1}\left( \frac{1}{s+\mu_1 + \mu_2 }\right)$, $g(t)=\mathcal{L}^{-1} \left( \frac{1}{ s + \hat{\lambda}_2 }\right)$ and $w(t)= \mathcal{L}^{-1} \left( \frac{1}{s + \lambda_1 +\lambda_2 }\right)$

Just to note, since convolution is associative and commutative, you can evaluate it in any order (which is why no brackets were needed above), $$x(t)= (f(t)*g(t))*w(t) = f(t)*(g(t)*w(t)) = (g(t)*f(t))*w(t) = f(t)*(w(t)*g(t)) = (f(t)*w(t))*g(t)) $$ Now use the formula $(f*g)(t) =\int_0^t f(\tau)g(t-\tau)d\tau$ and finish the transform.

Answer 2

By the residue theorem:

$$ \frac{1}{(x-a)(x-b)(x-c)}=\frac{1}{(a-b)(a-c)(x-a)}+\frac{1}{(b-a)(b-c)(x-b)}+\frac{1}{(c-a)(c-b)(x-c)}$$ and $\mathcal{L}^{-1}\left(\frac{1}{x-c}\right)=e^{cs}$.

Answer 3

Hint:

You can proceed by direct convolution, using

$$\mathcal{L}(e^{ax})=\frac1{x-a},$$and

$$e^{ax}*e^{bx}=\int_{t=0}^x e^{a(x-t)}e^{bt}dt=\left.e^{ax}\frac{e^{(b-a)t}}{b-a}\right|_{t=0}^x=\frac{e^{bx}-e^{ax}}{b-a}.$$