Partial fraction decomposition

Partial fraction decomposition is the technique for writing a rational function $X(s)=N(s)/D(s)$ as a sum of simple fractions, each of which has a known inverse Laplace transform. It is the bread-and-butter skill for inverting Laplace transforms.

The basic recognition we use over and over: $A/(s+\alpha )$ inverts to $A{e}^{-\alpha t}u(t)$ (for the causal ROC $\sigma >-\alpha$). So if we can break $X(s)$ into a sum of terms of this form, we just sum the inverse transforms.

Three cases

Case 1: distinct simple poles

Decompose $X(s)=N(s)/[(s-p_1)(s-p_2)\cdots (s-p_n)]$ as

$X(s)={\sum }_{k=1}^n\frac{A_k}{s-p_k}.$

To find $A_k$, cover up the factor $(s-p_k)$ and evaluate the remaining expression at $s=p_k$:

$A_k={\left[(s-p_k)X(s)\right]}_{s=p_k}.$

This is the cover-up method, and it’s the fastest way to compute residues at simple poles.

Worked example: $G(s)=10s/[(s+3)(s+1)]$.

$A={\left[\frac{10s}{s+1}\right]}_{s=-3}=\frac{-30}{-2}=15,B={\left[\frac{10s}{s+3}\right]}_{s=-1}=\frac{-10}{2}=-5.$

So $G(s)=15/(s+3)-5/(s+1)$, inverse-transforming to $g(t)=(15e^{-3t}-5e^{-t})u(t)$ for the causal ROC.

Case 2: repeated poles

A pole of order $n$ at $s=p$ contributes $n$ terms: $A_1/(s-p)+A_2/(s-p{)}^2+\cdots +A_n/(s-p{)}^n$.

The highest-power coefficient $A_n$ is found by simple cover-up:

$A_n={\left[(s-p{)}^{n}X(s)\right]}_{s=p}.$

Lower-power coefficients require differentiating first:

$A_{n-k}=\frac{1}{k!}\frac{d^k}{d{s}^k}{\left[(s-p{)}^{n}X(s)\right]}_{s=p}.$

Worked example: $G(s)=(s+5)/[s^2(s+2)]$. Double pole at $s=0$, simple pole at $s=-2$.

For $B$ at $s=-2$: $B={\left[(s+5)/s^2\right]}_{s=-2}=3/4$.

For $A_2$ (the highest power at the double pole): $A_2={\left[(s+5)/(s+2)\right]}_{s=0}=5/2$.

For $A_1$ (the lower power): $A_1=(d/ds){\left[(s+5)/(s+2)\right]}_{s=0}=\left[(s+2)-(s+5)\right]/(s+2{)}^2{|}_{s=0}=-3/4$.

So $G(s)=-3/4\cdot 1/s+5/2\cdot 1/s^2+3/4\cdot 1/(s+2)$. Inverse:

$g(t)=\left(-\frac{3}{4}+\frac{5}{2}t+\frac{3}{4}{e}^{-2t}\right)u(t).$

Repeated poles give polynomial-times-exponential terms in time: $t{e}^{-\alpha t}$, $t^2{e}^{-\alpha t}$, etc.

Case 3: complex-conjugate pole pairs

A rational $X(s)$ with real coefficients has complex poles in conjugate pairs $p,p^{\ast }$. Two approaches:

Option A: treat as ordinary partial fractions with complex residues. The residues are conjugates of each other, and the inverse transforms are complex exponentials that recombine to a real damped sinusoid.

Option B: keep the quadratic factor $(s-p)(s-p^{\ast })=(s+\alpha {)}^2+{\omega }_0^2$ (after completing the square) and match to the damped-sinusoid pairs:

$\frac{s+\alpha }{(s+\alpha {)}^2+{\omega }_0^2}\leftrightarrow e^{-\alpha t}\cos ({\omega }_{0}t)u(t),\frac{{\omega }_0}{(s+\alpha {)}^2+{\omega }_0^2}\leftrightarrow e^{-\alpha t}\sin ({\omega }_{0}t)u(t).$

Decompose your quadratic-factor term as a linear combination of these two, then inverse-transform. Option B usually gives cleaner real-valued answers.

Improper rational functions

If the numerator degree is $\ge$ the denominator degree, $X(s)$ is improper, and you have to do polynomial long division first. The quotient is a polynomial in $s$ (which inverse-transforms to a sum of impulses and impulse derivatives), and the remainder gives a proper rational function that you can partial-fraction normally. See Polynomial division for improper rational functions.

What to remember

• For simple poles, cover-up is fast and reliable.
• For repeated poles, the highest-power coefficient is cover-up; lower powers require derivatives.
• For complex-conjugate pairs, complete the square and match to the damped-sinusoid pairs.
• Pole location ↔ time-domain shape: real part is decay rate, imaginary part is oscillation frequency. See s-plane.