# Inverse Laplace Transform

## Homework Statement

Given the Laplace transform

$$F_L(s) = \frac{1}{(s+2)(s^2+4)},$$

by using the complex inversion formula compute the inverse Laplace transform, ##f(t),## for the following regions of convergence:

(i) ##Re(s)<-2;##
(ii) ##-2<Re(s)<0;##
(iii) ##Re(s)>0.##

## Homework Equations

Inverse Laplace transform relationship:

$$f(t) = \frac{1}{j2\pi} \int^{\sigma + j \infty}_{\sigma-j\infty} F_L (s) \exp(st) \ ds \tag{1}$$

Where ##s=\sigma + j \omega,## and ##\sigma## must be chosen to lie within the region of absolute convergence of ##F_L.##

## The Attempt at a Solution

So, using equation (1), how do I exactly choose the values of ##\sigma## for each case? I am very confused about this part.

I tried to solve this without the complex inversion formula (just to see what the solution has to look like). I started out by expanding using partial fractions as:

$$F_L(s) = \frac{1}{(s+2)(s^2+4)} = \frac{1}{8(s+2)} + \frac{1}{8(s^2 +4)}$$

There is a pole at ##s=-2## due to the first term. The first term has the form ##1/(s-a),## so its transform can be written as ##\frac{1}{8} e^{-2t}.## However I am unable to proceed further because I don't see in Laplace transform tables what the transform of the form ##1/(s^2 +a)## looks like.

Any help would be appreciated.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Given the Laplace transform

$$F_L(s) = \frac{1}{(s+2)(s^2+4)},$$

by using the complex inversion formula compute the inverse Laplace transform, ##f(t),## for the following regions of convergence:

(i) ##Re(s)<-2;##
(ii) ##-2<Re(s)<0;##
(iii) ##Re(s)>0.##

## Homework Equations

Inverse Laplace transform relationship:

$$f(t) = \frac{1}{j2\pi} \int^{\sigma + j \infty}_{\sigma-j\infty} F_L (s) \exp(st) \ ds \tag{1}$$

Where ##s=\sigma + j \omega,## and ##\sigma## must be chosen to lie within the region of absolute convergence of ##F_L.##

## The Attempt at a Solution

So, using equation (1), how do I exactly choose the values of ##\sigma## for each case? I am very confused about this part.

I tried to solve this without the complex inversion formula (just to see what the solution has to look like). I started out by expanding using partial fractions as:

$$F_L(s) = \frac{1}{(s+2)(s^2+4)} = \frac{1}{8(s+2)} + \frac{1}{8(s^2 +4)}$$

There is a pole at ##s=-2## due to the first term. The first term has the form ##1/(s-a),## so its transform can be written as ##\frac{1}{8} e^{-2t}.## However I am unable to proceed further because I don't see in Laplace transform tables what the transform of the form ##1/(s^2 +a)## looks like.

Any help would be appreciated.

Expand
$$\frac{1}{s^2+4} = \frac{1}{(s+2i)(s-2i)}$$
in partial fractions. You will end up with a trigonometric function as your answer.

roam
Expand
$$\frac{1}{s^2+4} = \frac{1}{(s+2i)(s-2i)}$$
in partial fractions. You will end up with a trigonometric function as your answer.

Thank you very much for this hint. I expanded the fraction and got:

$$\frac{1}{8} \left( \frac{1}{s+2} + \frac{j}{4(s+2j)} - \frac{j}{4(s-2j)} \right)$$

now that we reduced the expression into a sum of standard forms, the only transform applicable in this case is ##\frac{1}{s-a} \iff e^{at}.## Therefore we obtain

$$f(t) = \frac{1}{8} \left( e^{-2t} + \frac{j}{4} \left( e^{-2jt} + e^{2jt} \right) \right).$$

Is this correct? When you say you end up with a trigonometric function, do you mean that I should express the exponentials in the last equation as sines and cosines?

Ray Vickson
Homework Helper
Dearly Missed
Thank you very much for this hint. I expanded the fraction and got:

$$\frac{1}{8} \left( \frac{1}{s+2} + \frac{j}{4(s+2j)} - \frac{j}{4(s-2j)} \right)$$

now that we reduced the expression into a sum of standard forms, the only transform applicable in this case is ##\frac{1}{s-a} \iff e^{at}.## Therefore we obtain

$$f(t) = \frac{1}{8} \left( e^{-2t} + \frac{j}{4} \left( e^{-2jt} + e^{2jt} \right) \right).$$

Is this correct? When you say you end up with a trigonometric function, do you mean that I should express the exponentials in the last equation as sines and cosines?

Why don't you try it for yourself, to see what you get?

Why don't you try it for yourself, to see what you get?

I already have. I was asking you to clarify what you meant by "ending up with a trigonometric function as your answer".

Ray Vickson