Partial fraction expansion

Basic overview¶

Our goal is to develop a general procedure for computing the inverse Laplace transform of any rational transfer function. A rational function is a ratio of polynomials. These occur, for example, whenever we derive the transfer function of an LTI system. The basic approach consists of three steps:

1. Start with a rational function of $s$, for example:

1. Perform a partial fraction expansion (PFE) to decompose the function into simpler terms. For the example above, the PFE is:

3. Apply the inverse Laplace transform to each term separately. This produces

In this section, we focus on step 2: the algebraic technique of partial fraction expansion. In the next section, we will turn our attention to step 3.

Properties of polynomials¶

Before discussing partial fraction expansion, we review some important properties of polynomials with real coefficients.

Two more important properties:

1. If a polynomial has real coefficients (all the polynomials we consider will have real coefficients), then complex roots always occur in conjugate pairs. That is, if $a + bj$ is a root, then $a - bj$ is also a root. (Here $j$ is the imaginary unit, satisfying $j^2=-1$).

2. For roots that occur as conjugate pairs, the corresponding factors combine to form a quadratic with real coefficients:

   $$\begin{aligned} \bigl(s - (a + bj)\bigr)\bigl(s - (a - bj)\bigr) &= \bigl((s-a) + bj\bigr)\bigl((s-a) - bj\bigr) \\ &= (s-a)^2 - (bj)^2 \\ & = (s-a)^2 + b^2 \end{aligned}$$

   (5)

   These quadratics are called irreducible quadratics because they have complex roots and cannot be factored further using real numbers. This leads us to the following alternative factorization.

Rational functions¶

A rational function is a ratio of two polynomials:

We have defined a lot of new terms so far! Expand the box below for a summary.

Long division of polynomials¶

Consider a rational function of the form (8). Partial fraction expansion requires the rational function to be strictly proper. If it is not, we can perform polynomial long division to rewrite it as the sum of a polynomial and a strictly proper rational function.

Example A: long division¶

Consider $\displaystyle\frac{s^3 + s - 1}{s^2 + 3s + 2}$.

This is an improper rational function because the degree of the numerator (3) is greater than or equal to the degree of the denominator (2). Polynomial long division is similar to numerical long division. We divide the leading term of the numerator by the leading term of the denominator to get the first term of the quotient, and continue until the degree of the remainder is less than the degree of the denominator.

The quotient is $s - 3$ and the remainder is $8s + 5$. Therefore, we have:

Now we can apply partial fraction expansion to the strictly proper part $\displaystyle\frac{8s + 5}{s^2 + 3s + 2}$.

Partial fraction expansion overview¶

The goal of partial fraction expansion (PFE) is to decompose a proper rational function into a sum of simpler fractions. Each simple fraction will correspond to a term whose inverse Laplace transform we can look up in a table.

Given a strictly proper rational function of the form (8), the first step is to factor the denominator $a(s)$ into its irreducible factors using real-coefficient factorization (Eq. (7)). Each factor contributes a term to the PFE. Here are the details of how each factor type contributes:

The first step in PFE is to write down a general form for the expansion based on the factorization of the denominator. At this stage, we still have unknown variables such as the $a_i$ and $b_i$ in the table above.

Example B: complicated general form¶

Suppose we want to find the PFE of

The first step is to factor the denominator. For complicated examples such as this one, it is easiest to use a somthing like MATLAB’s roots command to find the roots numerically.

The factorization contains a repeated linear factor $(s+1)^2$, a distinct linear factor $(s+3)$, and an irreducible quadratic factor $(s^2 + 2s + 5)$. Therefore, the general form of the PFE is:

The next step is to solve for the unknown coefficients $A,B,C,D,E$ using one of several methods. We will discuss these methods in the following sections.

Systematic method¶

One method for finding the unknown coefficients in a PFE is to multiply both sides of the equation by the denominator and then match coefficients of powers of $s$ on both sides. This is also called the method of undetermined coefficients.

This method works for any combination of factors (distinct linear, repeated linear, irreducible quadratic). However, it can lead to large systems of equations that are tedious to solve.

Example A: systematic method¶

Let’s find the PFE of the strictly proper part from the long division of Example A:

Finding a common denominator, we can write:

These must be the same for all possible values of $s$, so we can equate the numerators:

This leads to the system of equations:

Solving this system, we find $A = -3$ and $B = 11$. Therefore:

Combining with the polynomial part from Eq. (11), we have our complete PFE:

Example C: irreducible quadratic¶

Now let’s find the PFE for

The quadratic $s^2 - 2s + 10$ has complex roots (discriminant $= 4 - 40 < 0$), so we keep it as an irreducible quadratic:

Multiplying by the denominator:

Matching coefficients:

Solving, we fine $A = 1$, $B = -1$, $C = 3$. Therefore:

Example B: complicated case¶

The systematic method works fine for small problems, but can become tedious for larger problems. Consider finding the PFE for Example B, whose general form is given by Eq. (14):

As you might be thinking, this is going to be a lot of algebra. Combining the right-hand side over a common denominator gives the following numerator:

Matching this with the numerator on the left-hand side of Eq. (26), we must solve

Again, this is rather unpleasant to solve by hand. Skipping the algebra, we find that the solution is $A = 1$, $B = -2$, $C = 2$, $D = 2$, $E = -2$. Therefore,

Clearly this method can become tedious for larger problems. In the next sections, we will explore some shortcuts that work for special cases.

Cover-up method¶

The cover-up method, also called Heaviside’s method, is a shortcut for finding the coefficients in a partial fraction expansion when the denominator factors into distinct linear factors. It allows us to find each coefficient directly without solving a system of equations.

The method is most effective for distinct linear factors. For example, consider a case where the denominator factors as

where all the $r_i$ are distinct real numbers. Then, we expect the PFE to look like:

The cover-up method provides a way to find each $A_i$ directly. Here are the steps.

1. Multiply both sides by $(s - r_i)$. This factor cancels (“covers up”) the corresponding denominator term, and we obtain:

   $$\begin{aligned} &\frac{b(s)}{(s - r_1) \cdots (s - r_{i-1})(s - r_{i+1}) \cdots (s - r_n)} &= A_i + \sum_{k \neq i} \frac{A_k (s - r_i)}{s - r_k} \end{aligned}$$

   (32)

2. Now set $s=r_i$ on both sides. The right-hand side simplifies to $A_i$ because all other terms contain $(s - r_i)$ in the numerator, which becomes zero. We are left with:

   $$A_i = \frac{b(r_i)}{\prod_{k \neq i}(r_i - r_k)}$$

   (33)

3. Repeat for each $i = 1, 2, \ldots, n$ to find all $A_i$.

Let’s see this method in action with some examples.

Example D: cover-up method¶

Let’s find the PFE for

First, factor the denominator: $s^3 + 6s^2 + 11s + 6 = (s+1)(s+2)(s+3)$.

Therefore, the PFE general form is

Using the cover-up method:

Therefore, the final PFE for this example is

This is much faster than the systematic method, which would have required more algebraic manipulation and also solving a system of 3 equations in 3 unknowns.

Limitations of cover-up¶

The cover-up method can always be used to find the coefficients associated with distinct linear factors, so we should use it whenever possible. However, if the expression also contains irreducible quadratic factors or repeated factors, we must resort to other methods to find the remaining coefficients. Two options are:

• use the systematic method to solve for the remaining coefficients.

• choose specific values of $s$ to create additional equations.

We will illustrate these approaches through examples.

Example C: irreducible cover-up¶

Let’s return to Example C, which we originally solved using the systematic method.

As before, we write the general form of the PFE:

Using the cover-up method, we can find $A$ directly:

To find $B$ and $C$, we can apply the systematic method. Substituting $A=1$ and finding a common denominator on the right-hand side gives:

This system of equations is easy to solve; we obtain $B = -1$ and $C = 3$. Therefore, the complete PFE is:

This is the same solution we found earlier using the systematic method alone, but we saved some work by using the cover-up method to find $A$ directly. This allowed us to reduce the size of the system of equations we had to solve.

Example B: complicated cover-up¶

Let’s revisit Example B, whose general form is given by Eq. (14):

We can use cover-up to find $C$ since $(s+3)$ is a distinct linear factor. We cannot use cover-up to $(s+1)$ because it is repeated. Multiplying by $(s+1)$ and setting $s=-1$ will lead to division by zero! However, cover-up will work on the largest power $(s+1)^2$. So we can use cover-up to find $B$.

Now we have reduced the number of unknowns from five $(A,B,C,D,E)$ to three $(A,D,E)$. Substituting the values we found, our expression becomes:

We could use the systematic method here and solve a system of 3 equations in 3 unknowns, but this still requires considerable algebra. Instead, we can plug in specific values of $s$ and create additional equations.

Clearing fractions, simplifying, and rearranging these equations gives:

This system is much easier to solve than the original one. Solving, we find $A = 1$, $D = 2$, and $E = -2$. Therefore, the complete PFE is:

This is the same solution we found earlier using the systematic method alone, but we saved a lot of work by using the cover-up method to solve for some of the coefficients directly and by choosing specific values of $s$ to create a smaller system of equations.

A carefully chosen value of $s$ (usually $s=0$ or $s=\pm 1$) can often lead to simple equations to solve for unknown coefficients, which can be easier than using the systematic method.

Complex roots method¶

Although the cover-up method is not directly applicable to irreducible quadratic factors, it is still possible to use cover-up by working with complex roots. This method is less common, and not advisable for hand calculation because it involves complex arithmetic, but it is worth knowing about because it is the method MATLAB uses.

Example C: complex cover-up¶

Let’s revisit Example C. Consider finding the PFE for

The quadratic $s^2 - 2s + 10$ has complex roots $s = 1 \pm 3j$, so we can factor it as

With complex cover-up, we treat these as distinct linear factors:

We can use the cover-up method to find all three coefficients. For the complex ones, this involves substituting complex numbers.

Therefore, our complex PFE is:

We can check that this is equivalent to the earlier result in Eq. (25) we found using the systematic method:

Limitations of complex cover-up¶

The complex cover-up method handles irreducible quadratic factors, but it still cannot be used for repeated factors. In such cases, as before, we must resort to other methods to find the remaining coefficients.

Using MATLAB¶

For complicated expressions, the MATLAB residue command can compute the partial fraction expansion numerically:

[r, p, k] = residue(num, den)

where num and den are vectors of polynomial coefficients (in descending powers) for the numerator and denominator, respectively. The outputs are:

• r : vector of residues

• p : vector of poles (in the same order as the residues)

• k : quotient polynomial (non-zero only if improper)

when there are repeated poles, they will be listed multiple times in p, with corresponding residues in r, arranged in increasing powers of the repeated factor.

Example A: MATLAB solution¶

Let’s return to Example A:

We can input the numerator and denominator coefficients directly into MATLAB. Note that zero coefficients must be included to account for missing powers!

num = [1, 0, 1, -1];
den = [1, 3, 2];
[r, p, k] = residue(num, den);

Result:

r =
    11
    -3
p =
    -2
    -1
k =
     1    -3

We therefore have residues at $\{11, -3\}$ corresponding to poles at $\{-2, -1\}$, and a quotient polynomial $s - 3$. The final PFE is therefore:

This is the same result we found using the systematic method in Eq. (20).

Example B: MATLAB solution¶

Let’s return to our most complicated example, Example B:

num = [5, 20, 30, 20, -11];
den = [1, 7, 22, 42, 41, 15];
[r, p, k] = residue(num, den);

Result:

r =
   2.0000 + 0.0000i
   1.0000 + 1.0000i
   1.0000 - 1.0000i
   1.0000 + 0.0000i
  -2.0000 + 0.0000i
p =
  -3.0000 + 0.0000i
  -1.0000 + 2.0000i
  -1.0000 - 2.0000i
  -1.0000 + 0.0000i
  -1.0000 + 0.0000i
k =
     []

Note that MATLAB returns complex residues and poles this time. It also returns extra decimals, indicating that these are approximate roots calculated numerically. We have a real pole at $-3$, complex conjugate poles at $-1 \pm 2j$, and a repeated pole at $-1$. The residues of the repeated pole are $1$ and $-2$, listed in increasing powers of the repeated factor. The $k$ vector is empty because the original $G(s)$ was strictly proper. The resulting PFE is:

We can combine the complex conjugate terms to obtain the real PFE:

This coincides with our result from the systematic method and the cover-up method.

Summary¶

In the next section, we will develop the inverse Laplace transform pairs needed to convert each term of the PFE back to the time domain.

Test your knowledge¶