Factoring in Inverse Laplace Transformations

In summary, the conversation discusses a problem with breaking up a fraction using partial fractions. The expression given in the conversation is incorrect and Maple 11 provides the correct solution. The concept of partial fractions is a standard tool in calculus and can be easily researched or found in a textbook.
  • #1
xzibition8612
142
0

Homework Statement



see attachment

Homework Equations





The Attempt at a Solution


I don't get the part where it says "Solving this equation for X(s), we obtain ..."
Specifically jumping from 3/s(s2+3s+5) to (3/5)(1/s)-(3/5)[(s+2)/(s2+2s+5)].

How did the problem break up this fraction into separate subtracting fractions?? Is there a trick or something? I'm completely lost here. Thanks.
 

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  • #2
xzibition8612 said:

Homework Statement



see attachment

Homework Equations





The Attempt at a Solution


I don't get the part where it says "Solving this equation for X(s), we obtain ..."
Specifically jumping from 3/s(s2+3s+5) to (3/5)(1/s)-(3/5)[(s+2)/(s2+2s+5)].

How did the problem break up this fraction into separate subtracting fractions?? Is there a trick or something? I'm completely lost here. Thanks.

It is just partial fractions. However, the expression given above is wrong: Maple 11 gets:

gs:=3/s/(s^2+3*s+5):lprint(%);
3/s/(s^2+3*s+5)

(Maple uses notation A/B/C to stand for A/(B*C), etc, and uses A/B*C to mean (A/B)*C, etc.)

g1:=convert(gs,parfrac,s): lprint(%);
3/5/s + 1/5*(-3*s-9)/(s^2+3*s+5)

So, Maple gets -(3/5)(s+3) instead of -(3/5)(s+2) in the numerator of the second term. You can easily check that (s+2) gives the wrong answer when you combine the two terms.

BTW: "partial fractions" is a standard tool in calculus, often used in computing integrals. Just Google 'partial fractions' if you have not seen this material before, or consult a calculus textbook.
 
  • #3
i get it thx
 

Related to Factoring in Inverse Laplace Transformations

What is inverse Laplace factoring?

Inverse Laplace factoring is a mathematical technique used to find the original function from its Laplace transform. It is the reverse of Laplace factoring, where a function is transformed into its Laplace transform.

Why is inverse Laplace factoring important?

Inverse Laplace factoring is important because it allows us to find the original function from its Laplace transform, which is useful in solving differential equations and other mathematical problems. It also helps in understanding the behavior of a system in the time domain.

What are the steps involved in inverse Laplace factoring?

The steps involved in inverse Laplace factoring include identifying the Laplace transform of the function, using partial fraction decomposition to write the Laplace transform in terms of simpler functions, looking up the inverse Laplace transform in a table or using algebraic manipulations to find the original function.

What are the common challenges in inverse Laplace factoring?

One of the common challenges in inverse Laplace factoring is identifying the Laplace transform of a function. This requires knowledge of Laplace transform properties and techniques for finding Laplace transforms. Another challenge is finding the inverse Laplace transform of a function that cannot be easily looked up in a table, which may require advanced algebraic manipulations.

How is inverse Laplace factoring used in real-world applications?

Inverse Laplace factoring is used in a variety of real-world applications, such as in engineering, physics, and economics. It is used to solve differential equations that model real-world systems, such as electrical circuits, heat transfer, and population growth. It is also used in signal processing and control systems to analyze and design systems in the time domain.

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