How Do You Select Sigma for Different Regions in Inverse Laplace Transforms?

[roam]

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 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]

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?

 

[roam]

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]

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

I was talking about the part $1/(s^2+4)$ that you were having trouble with; not the entire expression.

What I meant is exactly what I said: that the final answer (to that part) will involve the trigonometric functions $\sin$ and/or $\cos$. The thing you start with, $1/(s^2+4)$ is real, so the answer $f(t)$ that you end up with should be real as well.