Question

Use the residue computation formula derived in Problem 38 to determine quickly the partial fraction expansion for$F\left(s\right)=\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}$

Short answer

$\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=-\frac{1}{2s}+\frac{1}{s-1}-\frac{1}{2\left(s+2\right)}$

Step-by-step solution

Define Inverse Laplace transform

Given a function , if there is a function that is continuous on

and satisfies,then we say that is the inverse Laplace transform of and employ the notation

Non-repeated Linear Factors

If can be factored into a product of distinct linear factors,

where the 's are all distinct real numbers, then the partial fraction expansion has the form

where the 's are real numbers. There are various ways of determining the constants . In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If is a factor of and is the highest power of that divides , then the portion of the partial fraction expansion of that corresponds to the term is

where the 's are real numbers.

Quadratic Factors

If is a quadratic factor of that cannot be reduced to linear factors with real coefficients and is the highest power of that divides , then the portion of the partial fraction expansion that corresponds to is

it is more convenient to express in the form when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

Find the factor of the denominator 

There we have

$P\left(s\right)=2s+1,Q\left(s\right)=s\left(s-1\right)\left(s+2\right)$

Using partial fractions we get

$F\left(s\right)=\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{2\left(s+2\right)}$

and using the equation from Problem 28 we obtain

$$\begin{gathered} A=\underset{s\to 0}{\lim }sF\left(s\right) \\ =\frac{2s+1}{\left(s-1\right)\left(s+2\right)} \\ =-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} B=\underset{s\to 1}{\lim }sF\left(s\right) \\ =\underset{s\to 1}{\lim }\frac{2s+1}{s\left(s+2\right)} \\ =1 \end{gathered}$$

$$\begin{gathered} C=\underset{s\to -2}{\lim }sF\left(s\right) \\ =\underset{s\to -2}{\lim }\frac{2s+1}{s\left(s-2\right)} \\ =-\frac{1}{2} \end{gathered}$$

Therefore

$\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=-\frac{1}{2s}+\frac{1}{s-1}-\frac{1}{2\left(s+2\right)}$