Partial fraction decomposition

Partial fraction decomposition breaks rational functions into simpler ones we can manipulate. This often includes taking the inverse Laplace transform and integrating. We break functions into those we can integrate into $\ln (x)$ or use substitution methods.

In MATLAB, we can use partfrac(exp, var), where var is optional.

General idea

We integrate something of the form: $\int \frac{P(x)}{Q(x)}dx$. Where the degree of the numerator is greater than or equal to the degree of the denominator, we perform long division to reduce it:

$$\frac{x^2+3x+5}{x+1}=x+2+\frac{3}{x+1}$$

This is easier to integrate. We have a convenient theorem:

For any polynomial of degree $n\ge 3$, it has at least one real root.

So where the degree of the denominator is greater than the numerator, we can always reduce into either linear factors or a quadratic factor or a combination. The quadratic factor will have no real roots. We can always find numbers $A$ and $B$ to make this split decomposition work:

$$\frac{11x+64}{(x+5)(x+8)}=\frac{A}{x+5}+\frac{B}{x+8}$$

For a quadratic term in the denominator, we use a numerator term of $Ax+B$.

For repeated terms in the denominator, we “step up” with the split terms’ denominator exponents.

$$\frac{2x^5-3x^2+2}{(x^2+1{)}^2(x-5{)}^3(2x-8{)}^4}=\frac{Ax+B}{(x^2+1{)}^1}+\frac{Cx+D}{(x^2+1{)}^2}+\frac{E}{(x-5{)}^1}+\dots$$

Cover-up method

For functions where the denominator doesn’t have repeated roots, we can use a special technique called the cover-up method. We can just pick a convenient value for one of the roots and solve for the coefficient.

For example, for:

$$Q(s)=\frac{3}{(s+4)(s+1)}=\frac{k_1}{s+4}+\frac{k_2}{s+1}$$

We can then assume $s=-1$ (to solve for $k_2$), and $s=-4$ (to solve for $k_1$). This is pretty fast! We can generalise:

$$k_i=(s-p_i)F(s){|}_{s=p_i}$$

Complex poles

For any polynomial in the denominator, we know that if it has a complex root, the conjugate will be another root, i.e.:

$$F(s)=\cdots +\frac{k}{s+\alpha -j\beta }+\frac{k^{\ast }}{s+\alpha +j\beta }$$

See also

• If there’s an irreducible quadratic term (i.e., no real roots) in the denominator, we can complete the square to still be able to integrate or find the inverse Laplace transform.