Inverse Laplace Transform

  • #1
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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. :confused:

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.
 

Answers and Replies

  • #2
Ray Vickson
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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. :confused:

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
[tex] \frac{1}{s^2+4} = \frac{1}{(s+2i)(s-2i)} [/tex]
in partial fractions. You will end up with a trigonometric function as your answer.
 
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  • #3
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Expand
[tex] \frac{1}{s^2+4} = \frac{1}{(s+2i)(s-2i)} [/tex]
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?
 
  • #4
Ray Vickson
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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?
 
  • #5
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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".
 
  • #6
Ray Vickson
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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.
 
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