Describes particle fraction expansion using the Heaviside method. Decompose rational functions of polynomials in Laplace or Z-transforms.

Oliver Heaviside (1850-1925), was an English electrical engineer, mathematician and physicist who among many things adapted complex numbers to the study of electrical circuits. He introduced a method to decompose rational function of polynomials as they occur when using the Laplace transform to solve differential equations.\(\)

Whenever the denominator of a rational function can be factored into distinct linear factors, the fraction can be expressed as the sum of partial fractions.

A rational function \(\eqref{eq:factors}\) with \(N\) distinct poles \(r_k\), can be expressed as a summation of simple fractions

Written out as

The constants \(c_k\) can be obtained by dividing out the \((x-r_k)\) factor in equation \(\eqref{eq:distinctpoles}\) and evaluating at \(x=r_k\)

## Non-distinct poles

If a pole \(r_k\) is not distinct, the partial fraction expansion *for that multiple pole* becomes a bit more involved. When the pole occurs \(q\) times as in \((x-r_k)^q\), that pole expands to a summation of decreasing powers of \((x-r_k)\)

where the constants \(d_i\) follow to satisfy

In practice, it is easier to first find \(d_q\) by dividing out the highest power \((x-r_k)^q\) in equation \(\eqref{eq:factors}\) and evaluating at \(x=r_k\)

The remaining \(d_k\) constants are found by substituting known constants and matching like powers.

## For example

his is one of those cases where examples might shed some light on the topic.

### Example 1

Consider the rational polynomial

The denominator is a second-order polynomial. The roots of any quadratic equation \(ax^2+bx+c=0\) are

$$ r_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \nonumber $$

The two roots of the denominator polynomial in equation \(\eqref{eq:example}\) are

This gives equation \(\eqref{eq:example}\) with a factorized denominator

According to \(\eqref{eq:nondistict}\), this can be expressed as partial fractions

Find the constant \(c_1\) using using Heaviside’s cover up method: multiply all terms by \((1-x)^2\) and then evaluate for \(1-x=0\) so that only the term \(c_1\) is left on the right

Given the value of \(c_1\), constant \(c_2\) can be found by substituting any numerical value (other than \(0\) in equation \(\eqref{eq:heavyside2}\)

The sum of partial fractions follows as

### Example 2

Consider the rational polynomial where the order of the numerator is the same as that of the denominator

Make the degree of the numerator \(1\) less than that of the denominator by dividing both sides by \(x\). This makes \(\frac{f(x)}{z}\) is a strictly proper rational function. According to Heaviside, this can be expressed as partial fractions

Multiply both sides of equation \(\eqref{eq:example2a}\) by \(x\) so that only \(c_0\) is left on the right; then evaluate for \(x=0\)

Use Heaviside’s cover up method to find the constant \(c_3\). (multiply both sides with \((x-2)^3\) so that only \(c_1\) is left on the right; then evaluate for \(x=2\)

Bring all the terms into a common denominator

Substitute the known values \(c_1\) and \(c_2\), expand and regroup

Match corresponding powers of \(x\)

Given \(c_4\), we only need one equation to solve for \(c_3\)

Substitute the values of \(c_{1\ldots4}\) in \(\eqref{eq:example2a}\)

Multiplying both sides by \(x\) gives the sum of partial fractions