Step Response

The step response of an LTI system is the output when the input is a unit step function u(t). It directly reveals how fast a system reaches steady state, how much it overshoots, and whether it oscillates, making it the standard benchmark in control system design.

Core Concept

Every LTI system has an impulse response h(t). When you feed it a unit step instead of an impulse, the output is the step response s(t). Because the step is the running integral of the impulse, s(t) is simply the running integral of h(t). This relationship lets you extract h(t) from s(t) by differentiation.

For a first-order RC circuit or a first-order control system, the step response follows an exponential: s(t) = (1 - e^(-t/tau)) * u(t). The time constant tau sets how fast the system responds. At t = tau, the response reaches 63.2% of its final value. Engineers specify settling time as approximately 4*tau or 5*tau for practical purposes.

Second-order systems produce step responses that may overshoot and ring before settling. The damping ratio zeta and natural frequency wn together characterize the overshoot percentage and the settling time. Underdamped systems (zeta < 1) oscillate; critically damped (zeta = 1) settle fastest without overshoot; overdamped (zeta > 1) settle slowly without overshoot.

Key Formula

Relationship between step response and impulse response: s(t) = integral from -inf to t of h(tau) d(tau), and h(t) = d/dt [s(t)].

First-order step response: s(t) = (1 - e^(-t/tau)) * u(t). tau = RC for an RC circuit or L/R for an RL circuit.

Second-order step response: s(t) = 1 - (e^(-zeta*wn*t) / sqrt(1-zeta^2)) * sin(wd*t + phi), where wd = wn*sqrt(1-zeta^2) and phi = arccos(zeta).

Problem: Find the step response of H(s) = 1 / (s + 2).

Given: H(s) = 1/(s+2), input X(s) = 1/s (Laplace of unit step)

Step 1 — Output in s-domain:
  Y(s) = H(s) * X(s) = 1/[(s+2)*s]

Step 2 — Partial fractions:
  1/[s*(s+2)] = A/s + B/(s+2)
  A = 1/2 (cover s=0),  B = -1/2 (cover s=-2)
  Y(s) = (1/2)/s - (1/2)/(s+2)

Step 3 — Inverse Laplace:
  y(t) = (1/2 - (1/2)*e^(-2t)) * u(t)
       = (1/2)(1 - e^(-2t)) * u(t)

Final Answer: s(t) = 0.5*(1 - e^(-2t)) u(t)
Time constant = 0.5 s, steady-state value = 0.5

Exam Tip: The steady-state value of the step response equals H(0), which is H(s) evaluated at s=0 (DC gain). This is a one-step check without computing the full inverse Laplace. For GATE, also remember that the step response of an integrator H(s)=1/s is a ramp, not a step; the output grows without bound, confirming the system is unstable in the bounded-input bounded-output sense.

Properties Summary

• Integral relation: s(t) = integral of h(tau)dtau; differentiating s(t) recovers h(t).
• DC gain: steady-state step response = H(s)|_{s=0} for stable systems.
• First-order time constant tau: step reaches 63.2% at t=tau, 98% at t=4*tau.
• Second-order peak overshoot: %OS = 100 * exp(-pi*zeta / sqrt(1-zeta^2)).
• Rise time: time to go from 10% to 90% of final value; inversely related to bandwidth.
• Settling time: approximately 4/(zeta*wn) for a 2% criterion in second-order systems.

Quick Revision

• Step response = running integral of impulse response.
• Impulse response = derivative of step response.
• First-order system time constant tau = RC or L/R.
• 63.2% of final value is reached at exactly one time constant.
• Steady-state value = H(0) for any stable system.
• Underdamped: zeta < 1, oscillates; critically damped: zeta = 1, fastest no-overshoot response.
• Step response in frequency domain: Y(s) = H(s)/s.
• Exam trap: confusing the step response formula with the impulse response; for H(s)=1/(s+a), the impulse response is e^(-at)u(t) but the step response is (1/a)(1-e^(-at))u(t).