Inverse Laplace Transform of this expression?

In summary, the student attempted to solve a cosine transform equation using partial fractions, but it didn't work. He then tried expanding the equation in partial fractions and using l'Hospital's rule to get a result, but the limit went to infinity. After figuring out that a=1+\epsilon, he solved the equation using that information.
  • #1
supermiedos
63
0

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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  • #2
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?
 
  • #3
Ohh the convolution theorem?? I'm working on it
 
  • #4
I got it now. Thank you so much
 
  • #5
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.##
 
  • #6
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?
 
  • #7
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:
  • #8
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
 

Related to Inverse Laplace Transform of this expression?

1. What is the inverse Laplace transform?

The inverse Laplace transform is a mathematical operation that allows us to find the original function from its Laplace transform. It is the reverse process of the Laplace transform and is often used in engineering and physics to solve differential equations.

2. What is the formula for finding the inverse Laplace transform?

The formula for the inverse Laplace transform is given by the Bromwich integral, which is ∫c-i∞c+i∞F(s)estds, where F(s) is the Laplace transform of the function and c is a constant chosen such that the integral converges.

3. How do you find the inverse Laplace transform of a rational function?

To find the inverse Laplace transform of a rational function, we can use partial fraction decomposition to break the function into simpler fractions. We can then use tables, formulas, or the properties of the Laplace transform to find the inverse transform of each term and combine them to get the final answer.

4. Can the inverse Laplace transform be used for all functions?

No, the inverse Laplace transform can only be used for functions that have a Laplace transform. This means that the function must be piecewise continuous on [0,∞), of exponential order, and have a finite number of discontinuities and poles in the complex plane.

5. Are there any applications of the inverse Laplace transform?

Yes, the inverse Laplace transform has several applications in various fields such as engineering, physics, and economics. It is commonly used to solve differential equations, analyze control systems, and model real-world phenomena. It is also used in signal processing and circuit analysis.

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