Inverse Laplace Transform of this expression?

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supermiedos
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Homework Statement



Find the Inverse Laplace Transform of [itex]\frac{1}{(s^{2} + 1)^{2}}[/itex]

Homework Equations





The Attempt at a Solution


I tried using partial fractions but it didn't work. It looks like a cosine transform, but I don't know what else to do. Help please :(
 
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supermiedos said:

Homework Statement



Find the Inverse Laplace Transform of [itex]\frac{1}{(s^{2} + 1)^{2}}[/itex]

Homework Equations





The Attempt at a Solution


I tried using partial fractions but it didn't work. It looks like a cosine transform, but I don't know what else to do. Help please :(

Think of it as$$
\frac 1 {s^2+1}\cdot \frac 1 {s^2+1}$$What theorem do you know about the product of transforms?
 
Ohh the convolution theorem?? I'm working on it
 
I got it now. Thank you so much
 
supermiedos said:

Homework Statement



Find the Inverse Laplace Transform of [itex]\frac{1}{(s^{2} + 1)^{2}}[/itex]

Homework Equations





The Attempt at a Solution


I tried using partial fractions but it didn't work. It looks like a cosine transform, but I don't know what else to do. Help please :(

Another way: start with
[tex]g_a(s) = \frac{1}{(s^2+1)(s^2+a^2)}, \: a \neq 1[/tex]
expand in partial fractions, find the inverse Laplace transform ##F_a(t)##, then take the limit as ##a \to 1.##
 
I expanded as you suggested and applied the transform. I got:

[itex]\frac{sin t}{a^{2} - 1}[/itex] + [itex]\frac{sin(at)}{a(1 - a^{2} )}[/itex]

But if I try to take the limit as a goes to 1, they just go to infinity. What am I doing wrong?
 
supermiedos said:
I expanded as you suggested and applied the transform. I got:

[itex]\frac{sin t}{a^{2} - 1}[/itex] + [itex]\frac{sin(at)}{a(1 - a^{2} )}[/itex]

But if I try to take the limit as a goes to 1, they just go to infinity. What am I doing wrong?

Set ##a = 1+\epsilon##, and expand things out until you get something in which you can use l'Hospital's rule to evaluate the limit as ##\epsilon \to 0.## I've done it, and it works!

Or, you can write your result as
[tex]\frac{\sin(at) - a \sin(t)}{a(1-a^2)}[/tex] and then use l'Hospital.
 
Last edited:
Omg you are right. I made the sum of fractions and used L'Hopital rule... I got the same result. That was amazing! I learned a valuable method today, thank you