Partial fraction decomposition rewrites one fraction of polynomials as a sum of smaller fractions, each with a simple denominator. It is the step that makes a rational function easy to integrate, and easy to take back from the Laplace or z domain. I'll start with why we want the split, then walk through one example step by step. After that we look at the forms the terms can take, and at two more examples that cover the other cases.
- Why do we split a fraction into partial fractions ?
- Example 1 - Two distinct linear factors
- Which terms go into the decomposition ?
- Example 2 - A repeated factor
- Example 3 - A numerator of higher degree
Why do we split a fraction into partial fractions ?
A fraction such as (3x + 2) / (x2 + x) is hard to work with as a whole. Nobody has a table entry for it. But each piece of the split has a table entry, so the split turns a problem we cannot look up into several problems we can.
Integration is the first use. After the split in Example 1, the integral of (3x + 2) / (x2 + x) becomes the integral of 2/x plus the integral of 1/(x + 1). Each of these is a logarithm, so the result is 2 ln|x| + ln|x + 1| + C.
The inverse Laplace transform is the second use, and it is the one that engineers meet most often. A transfer function or a response in the s domain is usually a ratio of two polynomials. Take F(s) = (3s + 2) / (s(s + 1)), which is Example 1 with x renamed to s. The split gives 2/s + 1/(s + 1). The table gives 2 for 2/s and e-t for 1/(s + 1), so f(t) = 2 + e-t for t >= 0. The inverse z transform uses the same idea in discrete time.
Each partial fraction has a known integral and a known inverse transform : the split is what connects a rational function to those tables.Each factor of the denominator gives its own terms : in a Laplace problem, each term is one mode of the time response, such as a constant or a decaying exponential.
Example 1 - Two distinct linear factors
This example uses the cover-up method, which is the fastest method when the denominator factors into different linear terms. The idea is to multiply both sides by one factor, and then to choose the value of x that makes every other term vanish.
Step 1 :
Factor the denominator first. Here x2 + x = x(x + 1), which has two different roots, x = 0 and x = -1.
Step 2 :
Write one unknown fraction for each factor. The numerators A and B are constants, because each denominator is linear.
Step 3 :
Multiply both sides by x. The factor x cancels on the left and in the A term, but it stays in the B term.
Step 4 :
Set x = 0. The B term is multiplied by 0 and vanishes, so A is left alone and A = 2.
Step 5 :
Go back to the equation of Step 2 and multiply both sides by (x + 1) this time. Now the factor (x + 1) cancels on the left and in the B term.
Step 6 :
Set x = -1. The A term is multiplied by 0 and vanishes, and the left side becomes 3(-1) + 2 over -1, which is 1. So B = 1.
Answer :
Always check the answer by adding the fractions back together. Over the common denominator x(x + 1), the numerator is 2(x + 1) + x = 3x + 2, which is the original numerator.
The two multiplications and substitutions can be done in one move. To find A, cover the factor x in the original fraction and put x = 0 into what is left: (3 x 0 + 2) / (0 + 1) = 2. To find B, cover (x + 1) and put x = -1: (3 x (-1) + 2) / (-1) = 1. This short form is why the method is called cover-up. It is also known as the Heaviside method.
Each distinct linear factor gives one constant : the value of x that makes that factor zero isolates its constant.Multiply by one factor at a time : Step 5 starts again from the equation of Step 2, not from the result of Step 3.Check by recombining : one line of algebra confirms the answer.
Which terms go into the decomposition ?
Example 1 had the simplest kind of denominator. Before we try other examples, we need the rule that decides which unknown terms to write down. The rule depends only on the factors of the denominator, and on one check of the degrees.
The check comes first. The method needs a proper fraction, where the degree of the numerator is lower than the degree of the denominator. If it is not lower, divide the polynomials first. The quotient is a polynomial, and only the remainder fraction is split. Then factor the denominator completely over the real numbers, and write the terms from the table below.
Factor in the denominator |
Terms to write |
distinct linear factor (x - a) |
A / (x - a) |
repeated linear factor (x - a)k |
A1 / (x - a) + A2 / (x - a)2 + ... + Ak / (x - a)k |
quadratic factor x2 + px + q with no real roots |
(Bx + C) / (x2 + px + q) |
A quadratic factor gets a linear numerator Bx + C, not a constant. Here is a short example. The fraction (x2 + x + 2) / (x(x2 + 1)) has the form A/x + (Bx + C)/(x2 + 1). Cover-up at x = 0 gives A = 2. Comparing the coefficients of x2 and x then gives B = -1 and C = 1. So the result is 2/x + (1 - x)/(x2 + 1).
The number of unknowns equals the degree of the denominator : a repeated factor (x - a)k needs k constants, and a quadratic factor needs two.Cover-up finds only some of the constants directly : the others come from comparing coefficients, or from putting any other convenient value of x into the equation.
Example 2 - A repeated factor
A repeated factor needs one term for each power. The cover-up method still finds two of the three constants here, and one extra substitution finds the last one.
Let's split (3x + 1) / ((x - 1)(x + 1)2). The table gives the form A/(x - 1) + B/(x + 1) + C/(x + 1)2.
To find A, cover (x - 1) and put x = 1 into what is left: (3 + 1) / (22) = 1. To find C, cover the whole (x + 1)2 and put x = -1: (-3 + 1) / (-1 - 1) = 1. B has no factor that can be covered alone. So put x = 0 into both sides. The left side is 1 / ((-1)(1)) = -1, and the right side is -A + B + C = -1 + B + 1. This gives B = -1.
So the result is 1/(x - 1) - 1/(x + 1) + 1/(x + 1)2. Recombining over the common denominator gives (x + 1)2 - (x - 1)(x + 1) + (x - 1) = 3x + 1, which confirms it.
Cover-up works for the highest power of a repeated factor : covering (x + 1)2 gave C, the constant over (x + 1)2.The lower powers need one more equation : any value of x that is not a root gives it, and x = 0 is usually the easiest.
Example 3 - A numerator of higher degree
When the numerator degree is equal to or higher than the denominator degree, a sum of partial fractions alone cannot equal the fraction. Every partial fraction goes to zero for large x, but such a fraction does not. So the polynomial division must come first.
Let's split (x3 + 2) / (x2 + x). The numerator has degree 3 and the denominator has degree 2. Division gives x3 + 2 = (x2 + x)(x - 1) + (x + 2). So the fraction equals x - 1 + (x + 2) / (x(x + 1)).
Now only the proper remainder fraction is split. Cover-up at x = 0 gives 2 / 1 = 2, and cover-up at x = -1 gives 1 / (-1) = -1. The final result is x - 1 + 2/x - 1/(x + 1).
Keep the polynomial part in view when you use the result. In an integral it adds x2/2 - x to the logarithms. In a Laplace problem, a polynomial part such as s - 1 in F(s) turns into impulse terms in f(t), so it cannot be dropped either.
Divide first when the fraction is not proper : the quotient x - 1 stays as a polynomial part, and only the remainder is split.A computer algebra system can check the result : for example, apart() in Python sympy returns x - 1 - 1/(x + 1) + 2/x for this fraction.