Using the Inverse Laplace Transform for Step and Impulse Response

You have a transfer function G(s) that describes how your control system behaves. You need y(t), the actual time-domain output, when the input is a step or an impulse. The operation that gets you there is the inverse Laplace transform step response calculation. You will learn what each term in the answer means and how the location of G(s)’s poles in the s-plane determines whether your system oscillates, decays, or flies apart.

The impulse response is the simplest case. An impulse input, a Dirac delta δ(t), has a Laplace transform of exactly 1. So the output Y(s) = G(s) × 1 = G(s). The time-domain impulse response is the inverse Laplace transform of the transfer function itself: y(t) = ℒ⁻¹{G(s)}.

If G(s) = 4/(s+2), the impulse response is 4e^{-2t} for t ≥ 0. The pole at s = -2 tells you the exponential decay rate. A pole further left, say s = -10, would give a faster decay. A pole on the imaginary axis, s = jω, would produce sustained oscillation. This direct link between pole location and the shape of the response is why control engineers spend so much time looking at pole-zero plots.

A unit step input u(t) has a Laplace transform of 1/s. The output Y(s) is therefore G(s)/s. The step response is y(t) = ℒ⁻¹{G(s)/s}.

Using the same G(s) = 4/(s+2), Y(s) = 4/(s(s+2)). Partial fraction decomposition gives 2/s - 2/(s+2). Inverting each term yields y(t) = 2(1 - e^{-2t}). The response starts at 0 and approaches a steady state of 2. That steady-state value can be verified using the final value theorem: lim_{s→0} s·[4/(s(s+2))] = 4/2 = 2. The step response is essentially the integral of the impulse response, which is why it rises to a constant rather than returning to zero.

Every pole of G(s) contributes one term to y(t). The type of term is determined entirely by the pole’s location and multiplicity. A table is the most direct way to see this mapping, but the prose version works too.

Real Poles

If a is negative, the term decays. If a is positive, it grows without bound, the system is unstable.

Complex Conjugate Poles

The damping ratio ζ controls the decay envelope. Lower ζ means more oscillation and a longer settling time.ζ > 1 gives two real poles and a sluggish, non-oscillatory response.

Right Half-Plane Poles

Any pole with a positive real part, or any complex pair with a positive real part, makes the response grow exponentially. This is an unstable system. A pole on the imaginary axis (real part zero) produces sustained oscillation. The final value theorem fails here because the limit as t→∞ does not exist.

Take a simple first-order system: G(s) = 5/(s+3).

Impulse response: Y(s) = G(s) = 5/(s+3). The inverse is y(t) = 5e^{-3t}. The time constant is 1/3 ≈ 0.33 seconds. After 3 time constants (1 second), the response has decayed to about 5% of its initial value.

Step response: Y(s) = G(s)/s = 5/(s(s+3)). Decompose: 5/(s(s+3)) = A/s + B/(s+3). Multiply through: 5 = A(s+3) + Bs. Set s=0 to get A = 5/3. Set s=-3 to get B = -5/3.The steady-state value is 5/3 ≈ 1.67. The time constant is again 1/3 seconds. A common mistake is to forget the factor of 1/s in the step response and invert G(s) directly.

The standard second-order system in control engineering is written as G(s) = ω_n²/(s² + 2ζω_n s + ω_n²). The parameters ζ (damping ratio) and ω_n (natural frequency) completely define the behaviour. The formulas for overshoot, peak time, and settling time are derived from the inverse Laplace transform of this form. Ogata 'Modern Control Engineering' 5th edition, Chapter 7, Section 7-2, Table 7-1, and Nise 'Control Systems Engineering' 6th edition, Chapter 4, Section 4-3, Table 4-1, both provide the standard relationships. The percent overshoot for a unit step is 100e^{-ζπ/√(1-ζ²)}. The peak time is π/(ω_n√(1-ζ²)). The 2% settling time is approximately 4/(ζω_n).

Worked Example: Underdamped System

G(s) = 16/(s² + 3.2s + 16). Compare to the standard form: ω_n² = 16, so ω_n = 4 rad/s. 2ζω_n = 3.2, so ζ = 3.2/(2×4) = 0.4. This is underdamped. The damped natural frequency ω_d = 4√(1-0.16) = 4×0.917 = 3.67 rad/s. The step response will show a 25.4% overshoot (from the formula) and a 2% settling time of 4/(0.4×4) = 2.5 seconds.

Worked Example: Overdamped System

G(s) = 25/(s² + 12s + 25). Here ω_n = 5 rad/s. 2ζω_n = 12, so ζ = 12/(2×5) = 1.2. This is overdamped. The poles are real: s = -6 ± √(36-25) = -6 ± √11 = -2.68 and -9.32. The step response contains two decaying exponentials and no oscillation. The slower pole at s = -2.68 dominates; the faster one at s = -9.32 decays quickly. The 2% settling time is roughly 4/2.68 ≈ 1.5 seconds, which is slower than what a critically damped design at the same ω_n would achieve.

Before you trust your inverse Laplace transform result, verify the steady-state value using the final value theorem. The theorem states that if all poles of sF(s) have negative real parts (except possibly a single pole at s=0), then lim_{t→∞} f(t) = lim_{s→0} sF(s). Ogata gives the condition as all poles of sF(s) having negative real parts except for a single pole at s=0. Nise states that sF(s) must be analytic in the right half-plane and on the imaginary axis, except for a simple pole at the origin.

For the step response example G(s) = 5/(s+3), sF(s) = 5/(s+3). The pole is at s = -3, in the left half-plane. The limit as s→0 is 5/3, matching the steady state from the inverse transform. This check catches algebraic errors in partial fraction decomposition. A mismatch means you made a mistake. Do not apply the theorem when the system has poles on the imaginary axis, an undamped oscillator, for example, because the theorem will return a finite value when the true response oscillates forever.