0.20 Partial fraction expansion

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:

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.

Heaviside "cover-up" method

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 $D(x)$ :

where $a_1$ $a_p$ are the roots of $D(x)$ . We can then rewrite $G(x)$ as a sum of partial fractions:

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:

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

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 ${}_r$ .

Repeated roots

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

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:

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:

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

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 ${}_1$ :

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

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.