# 0.20 Partial fraction expansion

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This module describes the method of partial fraction expansion, in which a ratio of polynomials can be split into a sum of small polynomials. TheHeaviside cover-up method is discussed in detail with examples. Finding a partial fraction expansion in matlab is also discussed.

Splitting up a ratio of large polynomials into a sum of ratios of small polynomials can be a useful tool,especially for many problems involving Laplace-like transforms. This technique is known as partial fractionexpansion. Here's an example of one ratio being split into a sum of three simpler ratios:

$\frac{8x^{2}+3x-21}{x^{3}-7x-6}=\frac{1}{x+2}+\frac{3}{x-3}+\frac{4}{x+1}$

There are several methods for expanding a rational function via partial fractions. These include the method of clearingfractions, the Heaviside "cover-up" method, and different combinations of these two. For many cases, the Heaviside"cover-up" method is the easiest, and is therefore the method that we will introduce here. For a more complete discussion,see Signal Processing and Linear Systems by B.P. Lathi, Berkeley-Cambridge Press, 1998, pp-24-33. Someof the material below is based upon this book.

## No repeated roots

Let's say we have a proper function $G(x)=\frac{N(x)}{D(x)}$ (by proper we mean that the degree $m$ of the numerator $N(x)$ is less than the degree $p$ of denominator $D(x)$ ). In this section we assume that there are no repeatedroots of the polynomial $D(x)$ .

The first step is to factor the denominator $D(x)$ :

$G(x)=\frac{N(x)}{(x-{a}_{1})(x-{a}_{2})(x-{a}_{p})}$

where ${a}_{1}$ ${a}_{p}$ are the roots of $D(x)$ . We can then rewrite $G(x)$ as a sum of partial fractions:

$G(x)=\frac{{}_{1}}{x-{a}_{1}}+\frac{{}_{2}}{x-{a}_{2}}++\frac{{}_{p}}{x-{a}_{p}}$

where ${a}_{1}$ ${a}_{p}$ are constants. Now, to complete the process, we must determine the values of these  coefficients. Let's look at how to find ${}_{1}$ . If we multiply both sides of the equation of G(x) as a sum of partial fractions by $x-{a}_{1}$ and then let $x={a}_{1}$ , all of the terms on the right-hand side will go to zeroexcept for ${}_{1}$ . Therefore, we'll be left over with:

${}_{1}=(x, , (x-{a}_{1})G(x))$

We can easily generalize this to a solution for any one of the unknown coefficients:

${}_{r}=(x, , (x-{a}_{r})G(x))$

This method is called the "cover-up" method because multiplying both sides by $x-{a}_{r}$ can be thought of as simply using one's finger to cover up this term in the denominator of $G(x)$ . With a finger over the term that would be canceled bythe multiplication, you can plug in the value $x={a}_{r}$ and find the solution for ${}_{r}$ .

In this example, we'll work through the partial fraction expansion of the ratio of polynomials introduced above.Before doing a partial fraction expansion, you must make sure that the ratio you are expanding is proper. If itis not, you should do long division to turn it into the sum of a proper fraction and a polynomial. Once this isdone, the first step is to factor the denominator of the function:

$\frac{8x^{2}+3x-21}{x^{3}-7x-6}=\frac{8x^{2}+3x-21}{(x+2)(x-3)(x+1)}$

Now, we set this factored function equal to a sum of smaller fractions, each of which has one of thefactored terms for a denominator.

$\frac{8x^{2}+3x-21}{(x+2)(x-3)(x+1)}=\frac{{}_{1}}{x+2}+\frac{{}_{2}}{x-3}+\frac{{}_{3}}{x+1}$

To find the alpha terms, we just cover up the corresponding denominator terms in $G(x)$ and plug in the root associated with the alpha:

${}_{1}=(x, , (x+2)G(x))=(x, , \frac{8x^{2}+3x-21}{(x-3)(x+1)})=1$
${}_{2}=(x, , (x-3)G(x))=(x, , \frac{8x^{2}+3x-21}{(x+2)(x+1)})=3$
${}_{3}=(x, , (x+3)G(x))=(x, , \frac{8x^{2}+3x-21}{(x+2)(x-3)})=4$

We now have our completed partial fraction expansion:

$\frac{8x^{2}+3x-21}{(x+2)(x-3)(x+1)}=\frac{1}{x+2}+\frac{3}{x-3}+\frac{4}{x+1}$

## Repeated roots

When the function $G(x)$ has a repeated root in its denominator, as in

$G(x)=\frac{N(x)}{(x-b)^{r}(x-{a}_{1})(x-{a}_{2})(x-{a}_{j})}$

Somewhat more special care must be taken to find the partial fraction expansion. The non-repeated terms areexpanded as before, but for the repeated root, an extra fraction is added for each instance of the repeatedroot:

$G(x)=\frac{{}_{0}}{(x-b)^{r}}+\frac{{}_{1}}{(x-b)^{(r-1)}}++\frac{{}_{r-1}}{x-b}+\frac{{}_{1}}{x-{a}_{1}}+\frac{{}_{2}}{x-{a}_{2}}++\frac{{}_{j}}{x-{a}_{j}}$

All of the alpha constants can be found using the non-repeated roots method above. Finding the betacoefficients (which are due to the repeated root) has the same Heaviside feel to it, except that this time wewill add a twist by using the derivative to eliminate some unwanted terms.

Starting off directly with the cover-up method, we can find ${}_{0}$ . By multiplying both sides by $(x-b)^{r}$ , we'll get:

$(x-b)^{r}G(x)={}_{0}+{}_{1}(x-b)++{}_{r-1}(x-b)^{(r-1)}+{}_{1}\frac{(x-b)^{r}}{x-{a}_{1}}+{}_{2}\frac{(x-b)^{r}}{x-{a}_{2}}++{}_{j}\frac{(x-b)^{r}}{x-{a}_{j}}$

Now that we have "covered up" the $(x-b)^{r}$ term in the denominator of $G(x)$ , we plug in $x=b$ to each side; this cancels every term on the right-hand side except for ${}_{0}$ , leaving the formula

${}_{0}=(x, , (x-b)^{r}G(x))$

To find the other values of the beta coefficients, we can take advantage of the derivative. By taking the derivativeof the equation after cover-up (with respect to $x$ the right-hand side becomes ${}_{1}$ plus terms containing an $x-b$ in the numerator. Again, plugging in $x=b$ eliminates everything on the right-hand side except for ${}_{1}$ , leaving us with a formula for ${}_{1}$ :

${}_{1}=(x, , \frac{d (x-b)^{r}G(x)}{d x}})$

Generalizing over this pattern, we can continue to take derivatives to find the other beta terms. The solutionfor all beta terms is

${}_{k}=(x, , \frac{1}{k!}\frac{d^{k}(x-b)^{r}G(x)}{dx^{k}})$

To check if you've done the partial fraction expansion correctly, just add all of the partialfractions together to see if their sum equals the original ratio of polynomials.

## Finding partial fractions in matlab

Matlab can be a useful tool in finding partial fraction expansions when the ratios become too unwieldy to expand byhand. It can handle symbolic variables. For example, if you type syms s , $s$ will be treated as a symbolic variable. You can then use it as such when you make function assignments.

If you've done this and have then made a function, say $H(s)$ , which is a ratio of two polynomials in the symbolic variable $s$ , there are two ways to get the partial fraction expansion of it.A trick way is to say diff(int(H)) . When you use these functions together, Matlab gives back $H$ expanded into partial fractions. There's also a more formal way to do it using the residue command. Type help residue in Matlab for details.

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