Simplifying rational functions

This document describes how to use synthetic division and partial fraction
expansion to reduce a rational function to its canonical form. Synthetic division
and partial fraction expansion are implemented in Matlab's residue function,
which is a good way to experiment with them.

1 Partial fractions

Suppose we have a rational function

We would like to represent it in a simpler form. It turns out that any rational
function can be decomposed in the partial fraction expansion

Here K(s) is a polynomial, and C_i(s) and P_i(s) are \simple" polynomials. K(s)
is called the direct term; it is necessary only if n≥ m, and if it exists it has
degree (n - m).

The denominator polynomials P _i depend on the roots of A, which are also
called poles of the rational function. (Roots of B are called zeros of the rational
function.) Each isolated root x_i of A results in a denominator polynomial of
the form P_i(s) = s - x_i; each complex conjugate pair of roots x_i± y_ii gives a
denominator of the form Multiple roots result in
terms of higher degree; for example a real root xi with multiplicity k gives a
denominator

To determine the direct term we can use synthetic division (see below). So
for now let us assume n < m. In this case we can use the Heaviside method
(also called the cover-up method) to determine the coefficient polynomials C_i.
The simplest case is an isolated root x_i of A(s). In this case, C_i is a constant,
and we have

where the second equation holds because every term on the right-hand side
contains a factor (s - x_i) except for the term (s - x_i)C_i/(s - x_i). So, we can
determine C_i by deleting one of the factors (s - x_i) of A_i from our rational
function, and evaluating the result at x_i.
For example, suppose we have

We will then have a term in our expansion

To determine a, we evaluate 1/(s² + 1) at s = 2. This tells us that a = 1/5, so
our term is

This way of determining coe cients gives the method its name: we \covered
up" the factor 1/(s - 2) of B(s)/A(s) and evaluated the remaining expression
at s = 2.

If our denominator has a repeated root or a complex conjugate pair of roots
(or even a repeated conjugate pair), then we will have a factor P _i(s) in the de-
nominator which has degree d > 1. This factor will result in a term C_i(s)/P_i(s)
in our expansion, where degree(C_i) < d. In this case we can determine the
coefficients of C_i by evaluating P_i(s)B(s)/A(s) at the d points where P_i is zero;
this will result in d equations in the d unknown coefficients.

For example, consider again the rational function

The factor (s² + 1) leads to a term in our expansion

To determine a and b, we evaluate 1/(s-2) at the two points at which (s² +1)
is 0, namely ± i. This gets us two equations,


Solving these equations gives a = -1/5 and b = -2/5; combining the new term
with our previous result tells us that our final expansion is

2 Synthetic division

We are given a rational function B(s)/A(s) with numerator degree n and de-
nominator degree m. If n≥ m, we can pull out a quotient term K(s), leaving
a remainder term R(s) with degree(R) < m, so that

The process is analogous to long division, and is called synthetic division. We
will illustrate it by example: suppose we start with

We are looking for K(s) and R(s), with degree(R) < 2, so that

B(s) = K(s)A(s) + R(s)

To get the highest- order term of B(s) (namely s³) right, we can see that we
have to multiply A(s) by s. If we set K₁(s) = s, we have

R₁(s) = B(s) - K₁(s)A(s) = (s³ - s² + s + 1) - (s³ - 4s² + 3s) = 3s² - 2s + 1

This gets us partway to our goal: R₁(s) has a smaller degree than B(s) did,
but not small enough. But, we can repeat the process: to get rid of the leading
term of R₁(s) (namely 3s²), we can multiply A(s) by 3. Setting K₂(s) = s + 3,
we have

R₂(s) = B(s)-K₂(s)A(s) = (s³ -s² +s+1)-(s³ -4s² +3s)-(3s² -12s+9)

Cancelling terms gives R₂(s) = 10s-8, which has sufficiently low degree, so we
can take R = R₂ and K = K₂.