Question

Use the residue formulas derived in Problems 38 and 42 to determine the partial fraction expansion for

$F\left(s\right)=\frac{6s^2+28}{\left(s^2-2s+5\right)(s+2)}$

Short answer

$\frac{6s^2+28}{\left(s^2-2s+5\right)(s+2)}=\frac{2(s-1)+6}{{(s-1)}^2+2^2}+\frac{4}{s+2}$

Step-by-step solution

Define Inverse Laplace transform

Given a function $F\left(s\right)$, if there is a function$f\left(t\right)$that is continuous on $[0,\infty )$and satisfies $\mathcal{L}\left\{f\right\}=F$,then we say that $f\left(t\right)$is the inverse Laplace transform of $F\left(s\right)$and employ the notation $f={\mathcal{L}}^{-1}\left\{F\right\}$.

Non-repeated Linear Factors

If Q(s) can be factored into a product of distinct linear factors,

$Q\left(s\right)=\left(s-r_1\right)\left(s-r_2\right)\cdots \left(s-r_n\right),$

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

$\frac{P\left(s\right)}{Q\left(s\right)}=\frac{A_1}{s-r_1}+\frac{A_2}{s-r_2}+\cdots +\frac{A_n}{s-r_n},$

where the $A_i$'s are real numbers. There are various ways of determining the constants $A_1,\dots ,A_n$In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If $s-r$is a factor of $Q\left(s\right)$and $(s-r{)}^m$ is the highest power of s - r

that divides $Q\left(s\right)$, then the portion of the partial fraction expansion of

$P\left(s\right)/Q\left(s\right)$that corresponds to the term $(s-r{)}^m$is

$\frac{A_1}{s-r}+\frac{A_2}{{(s-r)}^2}+\cdots +\frac{A_m}{{(s-r)}^m},$

where the $A_i$'s are real numbers.

Quadratic Factors

If $(s-\alpha {)}^2+{\beta }^2$is a quadratic factor of $Q\left(s\right)$ that cannot be reduced to linear factors with real coefficients and $m$ is the highest power of

$(s-\alpha {)}^2+{\beta }^2$that divides $Q\left(s\right)$then the portion of the partial fraction expansion that corresponds to $(s-\alpha {)}^2+{\beta }^2$is

$\frac{C_{1}s+D_1}{{(s-\alpha )}^2+{\beta }^2}+\frac{C_{2}s+D_2}{{\left[{(s-\alpha )}^2+{\beta }^2\right]}^2}+\cdots +\frac{C_{m}s+D_m}{{\left[{(s-\alpha )}^2+{\beta }^2\right]}^m}.$

it is more convenient to express $C_{i}s+D_i$in the form $A_i(s-\alpha )+\beta B_i$when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_1(s-\alpha )+\mathbf{\beta }{B}_1}{{(s-\alpha )}^2+{\mathbf{\beta }}^2}+\frac{A_2(s-\alpha )+\mathbf{\beta }{B}_2}{{\left[{(s-\alpha )}^2+{\mathbf{\beta }}^2\right]}^2}+\cdots +\frac{{\mathbf{A}}_m(s-\alpha )+\mathbf{\beta }{B}_m}{{\left[{(s-\alpha )}^2+{\mathbf{\beta }}^2\right]}^m}.$

Find the factor of the denominator

There we have

$s^2-2s+5=(s-1{)}^2+4=(s-1{)}^2+2^2$

so $\alpha =1,\beta =2$and

$F\left(s\right)=\frac{6s^2+28}{\left(s^2-2s+5\right)(s+2)}=\frac{A(s-1)+2B}{{(s-1)}^2+2^2}+\frac{C}{s+2}$

Using the equation from Problem 42 we obtain

$$\begin{gathered} 2B+4Ai=\underset{S\to 1+2i}{\lim }\frac{\left(s^2-2s+5\right)\left(6s^2+28\right)}{\left(s^2-2s+5\right)(s+2)} \\ =\underset{S\to 1+2i}{\lim }\frac{6s^2+28}{s+2} \\ =\frac{10+24i}{3+2i}\frac{3-2i}{3-2i} \\ =\frac{30-20i+72i+48}{9+4} \\ =\frac{78+52i}{13} \\ =6+4i \end{gathered}$$

from where it follows

$\left\{\begin{array}{l}2B=6\\ 2A=4\end{array}\right.\Rightarrow \left\{\begin{array}{l}A=2\\ B=3\end{array}\right.$

Using the equation from Problem 38 we obtain

$$\begin{gathered} C=\underset{S\to -2}{\lim }\frac{(s+2)\left(6s^2+28\right)}{\left(s^2-2s+5\right)(s+2)} \\ C=\underset{t\to -2}{\lim }\frac{6s^2+28}{s^2-2s+5}=4 \end{gathered}$$

Finally we get

$\frac{6s^2+28}{\left(s^2-2s+5\right)(s+2)}=\frac{2(s-1)+6}{{(s-1)}^2+2^2}+\frac{4}{s+2}$