Laplace transform of proper rational function

  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
##\frac{2s + 1}{((s - 1)(s - 1)} = \frac{A(s - 1) + B(s - 1)}{(s - 1)(s - 1)}##
For this problem (b),
1713580812875.png

The solution is,
1713580841965.png

However, I don't understand how they got their partial fractions here (Going from step 1 to 2).

My attempt to convert into partial fractions is:

##\frac{2s + 1}{(s - 1)(s - 1)} = \frac{A(s - 1) + B(s - 1)}{(s - 1)(s - 1)}##

Thus,

##2s + 1 = A(s - 1) + B(s - 1)##
##2s + 1 = (A + B)s - A - B##

##2 = A + B##
##1 = - A - B##
##-1 = A + B##

However, ##2 ≠ -1##

Does someone please know how they got their partial fractions expression in step 2?

Thanks!
 
Physics news on Phys.org
  • #2
I disagree with their first step. ##\frac {2s+1}{s^2-2s+1} = \frac {2s}{(s-1)^2} + \frac{1}{(s-1)^2}##.
I don't know why they are adding 1 to the denominators.
CORRECTION: The text in the OP is a typo and the actual problem has denominator ##s^2-2s+2##. Thanks, @Orodruin .
 
Last edited:
  • Love
Likes member 731016
  • #3
FactChecker said:
I disagree with their first step. ##\frac {2s+1}{s^2-2s+1} = \frac {2s}{(s-1)^2} + \frac{1}{(s-1)^2}##.
I don't know why they are adding 1 to the denominators.
Because the actual problem has the denominator ##s^2 - 2s + \color{Red}{2}## … the 1 in the solution’s first expression is a typo.
 
  • Like
  • Love
Likes member 731016 and FactChecker
  • #4
Apart from that, your partial fraction ansatz is equivalent to assuming
$$
2s + 1 = (A+B)(s-1)
$$
This obviously will not work. You should check how to do partial fractions when the denominator is a square (or higher power). Even if it will not be useful here.
 
  • Love
Likes member 731016
  • #5
Orodruin said:
is a typo
Typos seem quite overabundant in the solutions of the OP’s threads. If this is the main mode of learning, they may want to consider a different source.
 
  • Love
Likes member 731016
  • #6
There are no tricky partial fractions involved in this problem. Every step is a simple rearrangement of the fractions.
 
  • Love
Likes member 731016
  • #7
I mean, you can do partial fractions, but the roots of the denominator are complex and the approach in the solution is much more straightforward.
 
  • Love
Likes member 731016

FAQ: Laplace transform of proper rational function

What is a proper rational function?

A proper rational function is a fraction where the degree of the numerator is less than the degree of the denominator. For example, the function \( \frac{2x + 1}{x^2 + 3x + 2} \) is a proper rational function because the degree of the numerator (1) is less than the degree of the denominator (2).

How do you compute the Laplace transform of a proper rational function?

To compute the Laplace transform of a proper rational function \( \frac{P(s)}{Q(s)} \), where \( P(s) \) and \( Q(s) \) are polynomials, you can use the definition of the Laplace transform: \( \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt \). For rational functions, it's often easier to perform partial fraction decomposition on \( \frac{P(s)}{Q(s)} \) and then apply the Laplace transform to each term individually.

What is the significance of the poles in the Laplace transform?

The poles of the Laplace transform, which are the values of \( s \) that make the denominator zero, are significant because they determine the stability and behavior of the system being analyzed. The location of the poles in the complex plane indicates whether the system is stable, marginally stable, or unstable. Specifically, poles in the left half-plane correspond to stable systems, while poles in the right half-plane correspond to unstable systems.

Can improper rational functions be transformed using the Laplace transform?

Yes, improper rational functions, where the degree of the numerator is greater than or equal to the degree of the denominator, can also be transformed using the Laplace transform. However, before applying the Laplace transform, you should perform polynomial long division to rewrite the improper rational function as a sum of a polynomial and a proper rational function. You can then take the Laplace transform of each part separately.

What are common applications of the Laplace transform in engineering?

The Laplace transform is widely used in engineering for analyzing linear time-invariant (LTI) systems, particularly in control theory and signal processing. It helps in solving differential equations, designing control systems, and analyzing system stability. Additionally, it is used in circuit analysis to determine the behavior of electrical circuits in the s-domain, making it easier to analyze transient and steady-state responses.

Back
Top