Laplace transform and region of convergence

Why it matters

Sensors, signal-conditioning circuits and control loops are described by differential equations, and the Laplace transform turns those into algebra. It also handles signals and systems the Fourier transform cannot (growing exponentials, unstable plants), and the region of convergence (ROC) tells you at a glance whether a system is causal and stable.

Key ideas

Definition. The bilateral Laplace transform weights the signal by e^(−st) with complex s = σ + jω. The real part σ adds an exponential damping e^(−σt), which makes the integral converge for many signals whose Fourier transform does not exist. On the line σ = 0, X(s) reduces to the CTFT X(jω) — provided that line is inside the ROC.

Region of convergence. The ROC is the set of s for which the integral converges. It depends only on σ = Re(s), so it is always a vertical strip, a half-plane or the whole plane, and it never contains a pole. The same expression X(s) can belong to different signals; the ROC is part of the answer.

• Right-sided signal (zero before some time, e.g. causal): ROC is to the right of the rightmost pole, Re(s) > σmax.
• Left-sided signal (zero after some time): ROC is to the left of the leftmost pole.
• Two-sided signal: ROC is a strip between two poles (or does not exist).
• Finite-duration, absolutely integrable signal: ROC is the entire s-plane.

Causality and stability from the ROC (for an LTI system with transfer function H(s))

• Causal ⇔ ROC is a right half-plane (for rational H(s)).
• BIBO stable ⇔ ROC includes the jω-axis.
• Causal and stable ⇔ all poles have negative real parts (ROC is right of all poles and includes the jω-axis).
• A system can be stable but non-causal: poles at −1 and +2 with the strip −1 < Re(s) < 2.

Unilateral transform. For causal signals and initial-value problems, the one-sided transform (lower limit 0⁻) is used. Its differentiation property brings in initial conditions automatically, which is why it is the tool for circuits and control systems.

Inverse transform. For rational X(s), use partial fractions, then match each term with a standard pair, choosing right- or left-sided according to the ROC. Repeated poles give terms like t·e^(−at); complex poles give damped sinusoids.

Poles and time behaviour (causal signals): a pole at s = −a gives e^(−at); a pair at −σ ± jωd gives e^(−σt) cos(ωd t + φ); poles on the jω-axis give undamped oscillation or a step; right-half-plane poles give growth.

Formulas

• X(s) = ∫ from −∞ to ∞ of x(t) e^(−st) dt — bilateral transform; s = σ + jω.
• X(s) = ∫ from 0⁻ to ∞ of x(t) e^(−st) dt — unilateral transform.
• e^(−at)u(t) ↔ 1/(s + a), ROC Re(s) > −a.
• −e^(−at)u(−t) ↔ 1/(s + a), ROC Re(s) < −a.
• δ(t) ↔ 1 (all s); u(t) ↔ 1/s, ROC Re(s) > 0; t·u(t) ↔ 1/s².
• tⁿ e^(−at) u(t) ↔ n!/(s + a)^(n+1).
• e^(−at) cos(ω0t) u(t) ↔ (s + a)/((s + a)² + ω0²), e^(−at) sin(ω0t) u(t) ↔ ω0/((s + a)² + ω0²).
• x(t − t0) ↔ e^(−st0) X(s); e^(−at) x(t) ↔ X(s + a); x * h ↔ X(s)H(s).
• dx/dt ↔ sX(s) − x(0⁻) (unilateral); ∫ from 0⁻ to t of x ↔ X(s)/s.
• x(0⁺) = lim(s→∞) sX(s) — initial value theorem (X(s) strictly proper).
• x(∞) = lim(s→0) sX(s) — final value theorem, valid only if all poles of sX(s) are in the left half-plane.

Symbols: s complex frequency (s⁻¹); σ neper frequency (s⁻¹); ω angular frequency (rad/s); a decay rate (s⁻¹); t0 delay (s); X(s) has the unit of x·s.

Worked examples

Example 1 (standard). Find the causal x(t) whose transform is X(s) = (s + 3)/(s² + 3s + 2), and check it with the initial- and final-value theorems.

1. Factor: s² + 3s + 2 = (s + 1)(s + 2). Write X(s) = A/(s + 1) + B/(s + 2).
2. A = (s + 3)/(s + 2) at s = −1 = 2/1 = 2. B = (s + 3)/(s + 1) at s = −2 = 1/(−1) = −1.
3. Causal ⇒ ROC Re(s) > −1: x(t) = (2e^(−t) − e^(−2t)) u(t).
4. Initial value: lim(s→∞) s(s + 3)/(s² + 3s + 2) = 1, and x(0⁺) = 2 − 1 = 1 ✓.
5. Final value: poles of sX(s) are at −1, −2 (left half-plane), so x(∞) = lim(s→0) sX(s) = 0 ✓.
6. x(t) = (2e^(−t) − e^(−2t)) u(t), with x(0⁺) = 1 and x(∞) = 0.

Example 2 (GATE level). H(s) = 1/((s + 1)(s − 2)). List the possible ROCs, say which is stable, and find h(t) for the stable one.

1. Poles at s = −1 and s = +2. Possible ROCs: Re(s) > 2 (causal, unstable), Re(s) < −1 (anti-causal, unstable), −1 < Re(s) < 2 (two-sided).
2. Only the strip −1 < Re(s) < 2 contains the jω-axis ⇒ stable but non-causal.
3. Partial fractions: H(s) = (−1/3)/(s + 1) + (1/3)/(s − 2) (check: at s = −1, 1/(−1 − 2) = −1/3; at s = 2, 1/(2 + 1) = 1/3).
4. Pole at −1 is to the left of the strip ⇒ right-sided term: (−1/3) e^(−t) u(t).
5. Pole at +2 is to the right of the strip ⇒ left-sided term: (1/3) × [−e^(2t) u(−t)].
6. h(t) = −(1/3) e^(−t) u(t) − (1/3) e^(2t) u(−t) — both terms decay away from t = 0, so ∫|h| = 1/3 + 1/6 = 0.5 is finite.

Common mistakes

• Giving X(s) without its ROC, or assuming every ROC is Re(s) > something.
• Using the final value theorem on a signal with poles on the jω-axis or in the right half-plane (e.g. x(t) = sin t), which gives a meaningless number.
• Forgetting the minus sign in the left-sided pair: −e^(−at)u(−t) ↔ 1/(s + a).
• Applying the initial value theorem to an improper X(s) (degree of numerator ≥ denominator) — split off the impulse part first.
• Equating "poles in the left half-plane" with stability for a non-causal system; the real test is whether the ROC contains the jω-axis.
• Dropping the initial condition in sX(s) − x(0⁻) when solving a differential equation.

For GATE IN

Questions ask you to pick the correct ROC for a given signal, decide causality and stability from pole locations and an ROC, find an inverse transform by partial fractions, or use the initial- and final-value theorems (including recognising when the final value theorem does not apply). Practise two-sided signals and the three-ROC case for two real poles.

Quick check

1. What is the ROC of x(t) = e^(−2t)u(t) + e^(−5t)u(t)?
2. What is the Laplace transform of e^(3t)u(−t), with ROC?
3. X(s) = 10/(s(s + 5)) — final value of x(t)?
4. Can the ROC of a finite-duration pulse contain a pole?
5. A causal system has a pole at s = 0.5. Is it stable? Answers: 1. Re(s) > −2. 2. −1/(s − 3), Re(s) < 3. 3. lim s·10/(s(s + 5)) = 2. 4. No — its ROC is the whole s-plane, and its transform has no finite poles. 5. No — causal with a right-half-plane pole is unstable.