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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 Ci(s) and Pi(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 xi of A results in a denominator polynomial of
the form Pi(s) = s - xi; each complex conjugate pair of roots xi± yii 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

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

where the second equation holds because every term on the right-hand side
contains a factor (s - xi) except for the term (s - xi)Ci/(s - xi). So, we can
determine Ci by deleting one of the factors (s - xi) of Ai from our rational
function, and evaluating the result at xi.
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 Ci(s)/Pi(s)
in our expansion, where degree(Ci) < d. In this case we can determine the
coefficients of Ci by evaluating Pi(s)B(s)/A(s) at the d points where Pi 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 ana logous 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.

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