Inverse Laplace Transformations

In summary, to find the inverse Laplace transform of the given functions, use partial fractions to write the problems in a form similar to that of the inverse Laplace transform table. Then, use the linearity of the Laplace transform to solve for the constants and simplify the fractions. The answers for the given problems are 3. \frac{2}{5}e^t-\frac{2}{5}e^{-4} and 7. 2e^{t}cos(t)+3e^{t}sin(t). It may be helpful to refer to a different table of inverse Laplace transformations for easier reading.
  • #1
jimmypoopins
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0

Homework Statement


Find the inverse Laplace transform of the given functions:
3. [tex]\frac{2}{s^2+3s-4}[/tex]

7. [tex]\frac{2s+1}{s^2-2s+2}[/tex]

Homework Equations


Inverse Laplace Transform Table

The Attempt at a Solution


on 3. i made the denominator look like [tex](s+4)(s-1)[/tex] but i got lost from there. i couldn't find anything on the table resembling the equation.

on 7. i completed the square and got [tex]\frac{2s+1}{(s-1)^2+1}[/tex], which resembles [tex]\frac{s-a}{(s-a)^2+b^2}[/tex] on the table, yet i can't seem to make the numerator look like [tex](s-a)[/tex].

if it helps anyone, the answers are 3. [tex]\frac{2}{5}e^t-\frac{2}{5}e^{-4}[/tex] and 7. [tex]2e^{t}cos(t)+3e^{t}sin(t)[/tex].

thanks in advance.edit: p.s. if anyone has a better table of inverse laplace transformations, the one i posted is really hard to read. I'm using the one in my book but it might be useful, if you need to refer to one in a table when trying to help, to refer to a table you're more comfortable reading.
 
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  • #2
For the first one, use partial fractions to write the problem as A/(s-1)+B/(s+4) (find the constants A and B). Break up the second fraction as well. Write it as 2(s-1)/((s-1)^2+1) +C/((s-1)^2+1) (find the constant C). Use the linearity of the laplace transform.
 
Last edited:
  • #3
Dick said:
For the first one, use partial fractions to write the problem as A/(s-1)+B/(s+4) (find the constants A and B). Break up the second fraction as well. Write it as 2(s-1)/((s-1)^2+1) +C/((s-1)^2+1) (find the constant C). Use the linearity of the laplace transform.

thank you very much. I'm really bad with partial fractions, so i guess my brain sort of tried to not think of that method. i guess i'll touch up on them because this chapter is filled with inverse laplace transformations.
 

Related to Inverse Laplace Transformations

1. What is an inverse Laplace transformation?

An inverse Laplace transformation is a mathematical operation that converts a function from the Laplace domain back into the time domain. It is the reverse process of a Laplace transformation.

2. Why do we need to perform inverse Laplace transformations?

Inverse Laplace transformations are necessary for solving differential equations in the time domain. They allow us to find the original function from its Laplace transform, making it easier to analyze and understand the behavior of dynamic systems.

3. What is the difference between a Laplace transform and an inverse Laplace transform?

A Laplace transform converts a function from the time domain to the Laplace domain, while an inverse Laplace transformation does the opposite - it converts a function from the Laplace domain back to the time domain.

4. How do you perform an inverse Laplace transformation?

To perform an inverse Laplace transformation, you need to use a table of Laplace transforms or algebraic techniques such as partial fraction decomposition and the residue theorem. The method used depends on the complexity of the function.

5. What are some real-world applications of inverse Laplace transformations?

Inverse Laplace transformations have a wide range of applications in fields such as engineering, physics, and economics. They are used to analyze and model dynamic systems, such as electrical circuits, mechanical systems, and economic systems. They are also used in signal processing to reconstruct signals from their frequency domain representation.

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