Heaviside expansion theorem

Remarks

Since the roots are simple, the partial fraction decomposition of $P/Q$ is $\sum_k c_k / (s - a_k)$. Multiplying by $s - a_k$ and letting $s \to a_k$ gives $c_k = P(a_k) / Q'(a_k)$, because $Q(s)/(s - a_k) \to Q'(a_k)$. Each term $c_k/(s - a_k)$ is the Laplace transform of $c_k e^{a_k t}$ on $\operatorname{Re} s > \operatorname{Re} a_k$.

Example: $P = 1$, $Q(s) = s^2 + 1$, with roots $\pm i$ and $Q'(s) = 2s$. The theorem gives $f(t) = \frac{e^{it}}{2i} + \frac{e^{-it}}{-2i} = \sin t$, the inverse Laplace transform of $1/(s^2 + 1)$.

In Heaviside's form, the response to a unit step is $\varphi(p) H$ with $\varphi = P/Q$, $\deg P \le \deg Q$ and $Q(0) \ne 0$. If the roots $a_k$ of $Q$ are simple, the theorem applied to $P(s)/(s Q(s))$ gives, for $t \ge 0$, $$\varphi(p) H = \frac{P(0)}{Q(0)} + \sum_k \frac{P(a_k)}{a_k Q'(a_k)} e^{a_k t}.$$

The hypothesis of simple roots is needed: $1/s^2$ is the transform of $t$, which is not a combination of exponentials. For repeated roots, the residue formula gives terms $t^j e^{a_k t}$ instead.

Going deeper

For a student

The coefficient $P(a_k)/Q'(a_k)$ is the residue of $P/Q$ at $a_k$, and the theorem is the special case of the residue form of the inverse Laplace transform, $f(t) = \sum_k \operatorname{Res}_{s = a_k} \left( e^{st} P(s)/Q(s) \right)$. It is the main tool for solving linear constant-coefficient systems by the Heaviside operational calculus or by the Laplace transform, because their transfer functions are rational. The exponents $a_k$ are the natural frequencies of the system, and the theorem expresses the response as a superposition of its modes.