How Does Partial Fraction Decomposition Apply to Laplace Transforms?

[reece]

Basically I don't know how F(s) can be split up to below.

F(s) = $$\frac{1}{s^{2}(s-2)}$$

= $$\frac{1}{4}$$ ( - $$\frac{1}{s}$$ - 2 $$\frac{1}{s^{2}} + \frac{1}{s-2}$$ )

I thought it would be 1/s^2 - 1 / s-2

How does this work? Please explain.
thanks

 

[rezwan]

hey

try using partial fraction:

1/s^2(s-2)=A/s+B/s^2+C/(s-2)
then multiply both side of the identity by s^2(s-2)
then group the terms: s^2,s and constants(terms without s)
finally equate both sides of the identity. and then calculate A,B C
and then put in the term A/s+B/s^2+C/(s-2)
then you will find the expected form.

thanks

rezwan

 

[tacman]

The Laplace splitting fraction method is a technique used to simplify and solve complex Laplace transforms. It involves breaking down the original transform into smaller, simpler transforms that can then be solved individually. In this case, the original transform F(s) = \frac{1}{s^{2}(s-2)} is being split into three smaller transforms: - \frac{1}{s}, -2 \frac{1}{s^{2}}, and \frac{1}{s-2}.

To understand how this works, let's first look at the denominator of F(s), which is s^{2}(s-2). This can be factored as s^{2}(s-2) = s^{2}-2s. Using partial fraction decomposition, we can rewrite this as \frac{1}{s^{2}(s-2)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s-2}, where A, B, and C are constants.

Now, using the properties of Laplace transforms, we can break down the original transform F(s) into the three smaller transforms: F(s) = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s-2} = A \mathcal{L} \{1\} + B \mathcal{L} \{t\} + C \mathcal{L} \{e^{2t}\}. This is where the - \frac{1}{s}, -2 \frac{1}{s^{2}}, and \frac{1}{s-2} terms come from.

So, to answer your question, the Laplace splitting fraction method works by breaking down the original transform into smaller transforms that can then be solved individually. By doing this, we can simplify the process and solve for the Laplace transform of the original function. I hope this explanation helps!