Decomposing Fractions Using Laplace Transformations

In summary, the method for separating a fraction with multiple terms in the denominator involves multiplying each term in the denominator by a variable and setting up equations to solve for those variables. The correct answer for the given fraction is A = 2s - 4, B = 2, and C = -4.
  • #1
leopard
125
0
Separating a fraction

I don't remember what this method is named in English, but I want to write the fraction

[tex]\frac{1}{(s^2 + 1)(s-3)(s+2)}[/tex]

in the form

[tex]\frac{A}{s^2 + 1} + \frac{B}{s-3} + \frac{C}{s+2}[/tex]

I multiply A with (s-3)(s+2), B with (s^2 + 1)(s+2) and C with (s^2 + 1)(s-3), to get

[tex]As^2 - As - 6A + Bs^3 + 2Bs^2 + Bs + 2B + Cs^3 - 3Cs^2 + Cs - 3C = 1[/tex]

and the equations

(1) B + C = 0
(2) A + 2B - 3C = 0
(3) -A + B + C = 0
(4) -6A + 2B - 3C = 1

(1) and (3) gives A = 0
(2) then gives B = C = 0

This is obviously wrong. The correct answer is

A = 2s - 4
B = 2
C = -4

according to my book. There must be something I don't understand here...
 
Last edited:
Physics news on Phys.org
  • #2
For the fraction containing the denominator of (s^2+1), the numerator should be As+B.
 

FAQ: Decomposing Fractions Using Laplace Transformations

What is a Laplace transformation?

A Laplace transformation is a mathematical tool used in calculus to transform a function of time into a function of complex frequency. It is often used to simplify differential equations and solve them algebraically.

What is the purpose of using a Laplace transformation?

The purpose of using a Laplace transformation is to transform a function from the time domain to the frequency domain. This allows for easier analysis and solution of differential equations, as well as simplifying certain mathematical operations.

How is a Laplace transformation calculated?

A Laplace transformation is calculated by integrating a function with respect to time, multiplied by a complex exponential function. The integral is evaluated over a specific range of time, usually from 0 to infinity.

What are the advantages of using Laplace transformations?

Laplace transformations have several advantages, including simplifying differential equations, allowing for algebraic solutions, and providing a way to analyze systems in the frequency domain. They also allow for the use of complex numbers, which can make calculations easier and more accurate.

What are some practical applications of Laplace transformations?

Laplace transformations have many practical applications in physics, engineering, and other sciences. They are commonly used in electrical circuit analysis, control systems, signal processing, and fluid dynamics, among others.

Similar threads

Back
Top