How do you calculate overshoot for a second order system?

Peak overshoot percentage is M_p = e^(−πζ/√(1−ζ²)) × 100%, where ζ is the damping ratio. This applies only to underdamped systems (ζ < 1). For example, ζ = 0.5 gives 16.3% overshoot, ζ = 0.707 gives 4.3%, and ζ = 1.0 gives 0% (critically damped, no overshoot).

What is a unit step input?

A unit step input is a test signal that equals 0 for t < 0 and 1 for t ≥ 0, representing an instantaneous switch from off to on. Its Laplace transform is 1/s. Applying a step input and measuring the output (step response) is the standard method for characterizing system dynamics, revealing rise time, overshoot, settling time, and steady-state behavior.

What is the difference between underdamped and overdamped step response?

Underdamped (ζ 1) responses approach the final value monotonically without oscillation but take longer to settle. Critically damped (ζ = 1) provides the optimal balance — no oscillation with the fastest possible settling time.

Unit Step Response

Q: What is a unit step input?

A unit step input is a test signal that equals 0 for t < 0 and 1 for t ≥ 0, representing an instantaneous switch from off to on. Its Laplace transform is 1/s. Applying a step input and measuring the output (step response) is the standard method for characterizing system dynamics, revealing rise time, overshoot, settling time, and steady-state behavior.

Q: What is the difference between underdamped and overdamped step response?

Underdamped (ζ 1) responses approach the final value monotonically without oscillation but take longer to settle. Critically damped (ζ = 1) provides the optimal balance — no oscillation with the fastest possible settling time.

Quick Answer

The unit step response of a second-order system with transfer function H(s) = ω_n²/(s² + 2ζω_ns + ω_n²) depends on the damping ratio ζ: underdamped (ζ < 1) produces oscillatory response with overshoot ≈ e^(−πζ/√(1−ζ²)) × 100%, critically damped (ζ = 1) gives the fastest non-oscillatory response, and overdamped (ζ > 1) yields sluggish exponential approach. For ζ = 0.5 and ω_n = 10 rad/s, peak overshoot is 16.3% with settling time approximately 4/(ζω_n) = 0.8 seconds.

Calculate this instantly on LAPLACE Calculator →

What Is the Unit Step Response of a Second Order System?

The unit step response is the output y(t) of a system when driven by a unit step input u(t) with zero initial conditions. For the standard second-order transfer function H(s) = ω_n²/(s² + 2ζω_ns + ω_n²), the step response in the s-domain is Y(s) = ω_n²/[s(s² + 2ζω_ns + ω_n²)]. The time-domain response depends entirely on two parameters: the natural frequency ω_n (which sets the time scale) and the damping ratio ζ (which determines the shape). This second-order model describes a vast range of physical systems including mass-spring-damper mechanical systems, RLC electrical circuits, servo motor positioning, and accelerometer dynamics. Computing these step responses via Laplace transforms is streamlined at www.lapcalc.com.

Key Formulas

Underdamped Response (ζ < 1): Oscillations and Overshoot

When ζ < 1, the system poles are complex conjugates at s = −ζω_n ± jω_n√(1−ζ²), producing the oscillatory step response y(t) = 1 − (e^(−ζω_nt)/√(1−ζ²))·sin(ω_d·t + φ), where ω_d = ω_n√(1−ζ²) is the damped natural frequency and φ = arccos(ζ). Peak overshoot occurs at t_p = π/ω_d with magnitude M_p = e^(−πζ/√(1−ζ²)) × 100%. For ζ = 0.1, overshoot is 72.9%; for ζ = 0.3, it's 37.2%; for ζ = 0.5, it's 16.3%; and for ζ = 0.707, it's 4.3%. Most engineering systems are designed with 0.4 ≤ ζ ≤ 0.8 to balance speed against overshoot. The 2% settling time is approximately t_s ≈ 4/(ζω_n), and rise time (10%–90%) is approximately t_r ≈ (1.8)/ω_n for ζ = 0.5.

Compute unit step response Instantly

Get step-by-step solutions with AI-powered explanations. Free for basic computations.

Open Calculator

Critically Damped (ζ = 1) and Overdamped (ζ > 1) Responses

At critical damping (ζ = 1), both poles are real and equal at s = −ω_n, producing y(t) = 1 − (1 + ω_n·t)·e^(−ω_nt) with zero overshoot and the fastest possible non-oscillatory response. This is the optimal damping for many instrumentation and measurement systems (e.g., galvanometers, seismometers) where overshoot would cause measurement errors. For ζ > 1 (overdamped), the poles are two distinct real values s = −ζω_n ± ω_n√(ζ²−1), producing y(t) = 1 − A₁·e^(s₁t) − A₂·e^(s₂t) where A₁ and A₂ depend on pole locations. The overdamped response approaches steady state more slowly than critically damped, with no practical advantage — it represents excessive damping that slows the system unnecessarily. Door closers are a common example of intentionally overdamped systems.

Step Response Performance Specifications

Engineers characterize second-order step responses using four key metrics. Rise time t_r (time from 10% to 90% of final value) measures speed of response — faster rise requires higher ω_n. Peak overshoot M_p (percent by which response exceeds final value) depends only on ζ, not ω_n. Settling time t_s (time for response to stay within ±2% or ±5% of final value) equals approximately 4/(ζω_n) for the 2% criterion. Steady-state error e_ss indicates the final offset from the desired value — a unity-DC-gain second-order system has zero steady-state error for step inputs. These specifications often conflict: reducing rise time typically increases overshoot, requiring design trade-offs. PID controllers and lead-lag compensators are designed to satisfy multiple specifications simultaneously.

Computing Second Order Step Response with Laplace Transforms

To compute the step response analytically, take Y(s) = H(s)/s = ω_n²/[s(s² + 2ζω_ns + ω_n²)], perform partial fraction decomposition, and apply inverse Laplace transforms to each term. For the underdamped case, the partial fractions yield Y(s) = 1/s − (s + 2ζω_n)/(s² + 2ζω_ns + ω_n²). Completing the square in the denominator and using the Laplace pairs for e⁻ᵃᵗcos(ωt) and e⁻ᵃᵗsin(ωt) produces the time-domain expression. MATLAB computes this numerically with step(tf([wn^2],[1 2*z*wn wn^2])), while Python uses scipy.signal.step(). For exact symbolic computation and step-by-step partial fraction decomposition, the LAPLACE Calculator at www.lapcalc.com handles all three damping cases automatically, making it invaluable for control systems coursework and design verification.

Frequently Asked Questions

The step response of a second-order system is its output y(t) when a unit step input is applied with zero initial conditions. Its shape is determined by the damping ratio ζ: oscillatory with overshoot for ζ < 1 (underdamped), fastest non-oscillatory for ζ = 1 (critically damped), and sluggish exponential for ζ > 1 (overdamped). The natural frequency ω_n scales the time axis.

Master Your Engineering Math

Join thousands of students and engineers using LAPLACE Calculator for instant, step-by-step solutions.

Start Calculating Free →