Inverse Transforms and Initial-Value Problems

Partial fractions for distinct, repeated and complex poles, residues, differential equations with initial conditions, and zero-input and zero-state responses

Cambridge — making Heaviside respectable

Thomas Bromwich came top in the Cambridge mathematics examinations of 1895 — Senior Wrangler — and lectured at St John's College. Today he is best known for justifying Oliver Heaviside's operator calculus (Thomas John I'Anson Bromwich), and the general formula for inverting a Laplace transform, a line integral in the complex plane, carries his name (Inverse Laplace transform). For the transforms engineers meet most — ratios of polynomials in

s

— the integral collapses into something far friendlier: one simple term for each root of the denominator. That is partial fractions, and it turns differential equations with initial conditions into a page of algebra.

Systems built from resistors, capacitors, inductors, masses and springs have rational transforms,

X(s) = N(s)/D(s)

. The roots of

D(s)

are the poles, and each one contributes a term of known shape. If the degree of

N

is not below that of

D

, divide first; what remains is a proper fraction.

💡Partial fractions

Distinct poles. $X(s) = \sum_k \dfrac{r_k}{s - p_k}$ , with residues $r_k = (s - p_k)X(s)\big|_{s = p_k}$ , and $x(t) = \sum_k r_k e^{p_k t}$ for $t \ge 0$ .
A repeated pole of order $m$ contributes $\dfrac{r_1}{s - p} + \dots + \dfrac{r_m}{(s - p)^m}$ , and $\dfrac{1}{(s - p)^j} \leftrightarrow \dfrac{t^{j-1}e^{pt}}{(j - 1)!}$ .
A complex pair $\sigma \pm j\omega$ has conjugate residues $r$ and $r^*$ , and together they give a real oscillation, $2|r|\,e^{\sigma t}\cos(\omega t + \angle r)$ .

📐Three cases

$\dfrac{s + 3}{(s + 1)(s + 2)} = \dfrac{2}{s + 1} - \dfrac{1}{s + 2}$ , so $x(t) = 2e^{-t} - e^{-2t}$ .
$\dfrac{s^2 + 1}{(s + 1)^2(s + 3)} = \dfrac{-3/2}{s + 1} + \dfrac{1}{(s + 1)^2} + \dfrac{5/2}{s + 3}$ , so $x(t) = -\tfrac32 e^{-t} + t\,e^{-t} + \tfrac52 e^{-3t}$ .
$\dfrac{5}{s(s^2 + 2s + 5)} = \dfrac1s - \dfrac{(s + 1) + 1}{(s + 1)^2 + 4}$ , so $x(t) = 1 - e^{-t}\cos 2t - \tfrac12 e^{-t}\sin 2t$ .

SciPy's residue and SymPy's apart produce exactly these coefficients.

Derivation

A circuit switched on with stored energy

In a series RLC circuit driven by

v_{in}(t)

, the capacitor voltage obeys

\ddot v + \frac{R}{L}\dot v + \frac{1}{LC}v = \frac{1}{LC}v_{in} .

Transform both sides, using the derivative rule with the initial voltage

v(0^-)

and its rate

\dot v(0^-) = i(0^-)/C

s^2V - s\,v(0^-) - \dot v(0^-) + \frac RL\big(sV - v(0^-)\big) + \frac{1}{LC}V = \frac{1}{LC}V_{in}(s) .

Solve for

V(s)

V(s) = \underbrace{\frac{(s + R/L)\,v(0^-) + \dot v(0^-)}{s^2 + \frac RL s + \frac{1}{LC}}}_{\text{initial conditions}} + \underbrace{\frac{V_{in}(s)/LC}{s^2 + \frac RL s + \frac{1}{LC}}}_{\text{input}} .

The initial conditions arrive as extra terms in the numerator, and the denominator — the same in both parts — fixes the shape of everything.

📐Five volts on a charged capacitor

Take

R = 1\ \Omega

L = 1

C = 0.25

F: the denominator is

s^2 + s + 4

, an underdamped pair with

\omega_n = 2

rad/s and

\zeta = 0.25

. Switch a 5 V source onto the circuit at

t = 0

while the capacitor holds 1 V and no current flows. Then

V(s) = \frac{s + 1}{s^2 + s + 4} + \frac{20}{s\,(s^2 + s + 4)} .

t = 1

s the first part contributes

-0.0706

V and the second

5.3532

v(1) = 5.2826

V, which solve_ivp and lsim reproduce to four decimals. The final-value theorem gives 5 V, as it must.

💡Zero-input and zero-state responses

Every linear system's response splits in two: the zero-input response, what the stored energy does with no input, and the zero-state response, what the input does to a system starting from rest. Convolution with

h(t)

(Unit 2) gives only the second. The unilateral transform gives both at once — which is why circuits, structures and vehicles switched on in mid-motion are solved with it.

Impulse responses from transfer functions. With zero initial conditions,

Y(s) = H(s)X(s)

, and the impulse response is the inverse transform of

H(s)

. For

H(s) = 4/(s^2 + s + 4)

, completing the square gives

h(t) = \frac{8}{\sqrt{15}}\,e^{-t/2}\sin\!\big(\tfrac{\sqrt{15}}{2}t\big)

, which peaks at 1.42 after 0.68 s — the same curve scipy.signal.impulse computes numerically.

[F]or it, nothing would be uncertain and the future, as the past, would be present to its eyes.

– Pierre-Simon Laplace, of an intelligence that knew every force and every position, A Philosophical Essay on Probabilities (1814), quoted on Wikiquote

🕰Bromwich's integral

The inverse transform is the line integral

x(t) = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty} X(s)\,e^{st}\,ds

along a vertical line inside the region of convergence — known as the Bromwich integral, the Fourier–Mellin integral, or Mellin's inverse formula (Laplace transform). For a rational

X(s)

, closing the contour turns it into a sum of residues of

X(s)e^{st}

at the poles: partial fractions, by another name.

⚠0⁻ or 0⁺?

The unilateral transform uses the state just before the switch, at

t = 0^-

. If the input contains an impulse, the state can jump at

t = 0

, and

x(0^+)

differs from

x(0^-)

. Use the values before the event — the transform accounts for the jump by itself.

💡Where you will use this

Pole positions read straight off these expansions (Lesson 7-3), and closed-loop responses are inverted the same way (Lesson 7-4). The z-transform's residuez (Lesson 8-2) is the discrete twin of residue, and state-space simulation with initial states is what flight and orbit simulators do at every step.

📚Further reading

Oppenheim and Willsky, Signals and Systems, sections 9.3 and 9.9 — the inverse transform and the unilateral transform for differential equations.
Inverse Laplace transform on Wikipedia — Bromwich's formula and Post's inversion.

↓But there is a problem

Every term in every expansion was

e^{p t}

for some pole

p

. Poles in the left half-plane meant decay, poles on the axis meant oscillation that never stops — and a pole in the right half-plane would mean growth without limit. On 7 November 1940, in a moderate wind, a new suspension bridge in Washington State showed the whole world what that looks like.