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System Dynamics · Module 8 of 10

Second-Order System Response

Two poles, two parameters. The natural frequency sets the speed and the damping ratio sets the overshoot, and from them you predict the rise, the ring, and the settle of almost any machine.

Readiness check

This module reads the second-order response. Tick only what you can do closed-notes.

Recall ω_n and ζ from a second-order denominator.
Recall the damped frequency ω_d = ω_n√(1 − ζ²).
Evaluate an exponential and a square root.
Distinguish underdamped from overdamped.
Recall that overshoot depends on damping.

0 or 1 weak itemsContinue with this module.

2 weak itemsRevisit damping ratio in Module 2.

3 or more weak itemsReview second-order vibration in Dynamics and Vibrations, Module 10.

The core idea

A second-order system is fixed by its natural frequency and damping ratio. For an underdamped system those two numbers set the overshoot, the peak time, and the settling time of the step response.

G(s) = ω_n²/(s² + 2ζω_ns + ω_n²)M_p = e^{−ζπ/√(1−ζ²)}t_p = π/ω_d, t_s ≈ 4/(ζω_n)

A second-order system has two poles and a standard transfer function ω_n²/(s² + 2ζω_ns + ω_n²). The damping ratio ζ decides the character: overdamped (ζ > 1) and critically damped (ζ = 1) responses rise without overshoot, while underdamped (ζ < 1) responses overshoot and ring before settling. For the underdamped case the step response is fully described by a few formulas. The percent overshoot M_p = e^{−ζπ/√(1−ζ²)} depends on damping alone. The peak time t_p = π/ω_d and the settling time t_s ≈ 4/(ζω_n) (to 2 percent) depend on both parameters through the damped frequency ω_d = ω_n√(1 − ζ²) and the decay rate ζω_n. Two numbers, read from the model, predict the entire transient.

The skill works when: you extract ω_n and ζ, then apply the overshoot, peak, and settling formulas.

The skill breaks down when: overshoot is expected for ζ ≥ 1, or settling time is built from ω_n alone without the damping.

The concept. The underdamped step response rises past its final value to a peak, overshoots, and rings down. The overshoot is set by damping; the peak and settling times by the natural frequency and damping together.

The skills, taught in order

Five skills cover the standard form, the damping regimes, and the three response metrics.

8.1 The second-order standard form

The canonical transfer function is ω_n²/(s² + 2ζω_ns + ω_n²). Matching a model's denominator to this form reads off ω_n and ζ, the only two numbers needed to characterise the response.

8.2 The damping regimes

The damping ratio sorts the response: overdamped (ζ > 1) is slow with two real poles, critically damped (ζ = 1) is the fastest non-overshooting case, and underdamped (ζ < 1) overshoots and rings with a complex pole pair. Most machines and instruments are underdamped.

8.3 Percent overshoot

For an underdamped system the peak rises a fraction M_p = e^{−ζπ/√(1−ζ²)} above the final value. It depends on the damping ratio alone, so overshoot directly measures ζ: about 5 percent at ζ = 0.7, around 16 percent at ζ = 0.5.

Metric	Formula	Depends on
Overshoot M_p	e^{−ζπ/√(1−ζ²)}	ζ only
Damped frequency ω_d	ω_n√(1 − ζ²)	ω_n, ζ
Peak time t_p	π/ω_d	ω_n, ζ
Settling time t_s (2%)	4/(ζω_n)	ω_n, ζ

The standard second-order metrics. Overshoot fixes the damping; the times depend on both parameters.

8.4 Peak time and damped frequency

The response first peaks at t_p = π/ω_d, half a period of the damped oscillation ω_d = ω_n√(1 − ζ²). A higher natural frequency reaches the peak sooner, and heavier damping lowers ω_d slightly, lengthening it.

8.5 Settling time

The response settles within 2 percent of its final value after about t_s ≈ 4/(ζω_n), set by the decay rate ζω_n, the real part of the poles. Moving the poles further left, by more damping or a higher frequency, settles the system faster.

Engineering connection: these metrics become the specifications a controller must meet in Control Systems, where ζ and ω_n are placed by design.

Worked example 1: step metrics from omega and zeta

A second-order system has ω_n = 5 rad/s and ζ = 0.3. Find the percent overshoot, the damped frequency, the peak time, and the 2 percent settling time.

Figure 1. With ζ = 0.3 the lightly damped system overshoots by 37 percent and rings several times before settling. The peak comes early, but full settling takes nearly 2.7 seconds.

ProblemFind M_p, ω_d, t_p, and t_s for the system in Figure 1.
Given / findω_n = 5 rad/s, ζ = 0.3. Find M_p, ω_d, t_p, t_s.
AssumptionsStandard underdamped second-order system; step input.
ModelApply the overshoot, damped-frequency, peak-time, and settling-time formulas.
EquationsM_p = e^{−ζπ/√(1−ζ²)}ω_d = ω_n√(1 − ζ²)t_p = π/ω_d, t_s = 4/(ζω_n)
SolveM_p = e^{−0.3π/√0.91} = e^−0.988 = 0.372 = 37.2%. ω_d = 5√0.91 = 4.77 rad/s. t_p = π/4.77 = 0.659 s. t_s = 4/(0.3 × 5) = 4/1.5 = 2.67 s.
Checkζ = 0.3 is light damping, so a large overshoot near 37 percent is expected. The settling time, set by ζω_n = 1.5, is much longer than the peak time, as light damping requires.
ConclusionTwo numbers predict the whole transient: a 37 percent overshoot, a peak at 0.66 s, and settling at 2.67 s. Lower damping would overshoot more and settle later.

Result. M_p = 37.2%, ω_d = 4.77 rad/s, t_p = 0.66 s, t_s = 2.67 s.

Worked example 2: metrics from a physical system

A mass-spring-damper has m = 2 kg, k = 200 N/m, and c = 12 N·s/m. Find the natural frequency, damping ratio, damped frequency, and percent overshoot.

Figure 2. The physical parameters give ω_n and ζ directly. With the same ζ = 0.3 as Example 1, the overshoot is identical, but the higher natural frequency makes everything happen twice as fast.

ProblemFind ω_n, ζ, ω_d, and M_p for the system in Figure 2.
Given / findm = 2 kg, k = 200 N/m, c = 12 N·s/m. Find ω_n, ζ, ω_d, M_p.
AssumptionsStandard mass-spring-damper; underdamped step response.
Modelω_n = √(k/m), ζ = c/(2√(km)), then ω_d and M_p from the standard formulas.
Equationsω_n = √(k/m), ζ = c/(2√(k m))ω_d = ω_n√(1 − ζ²)M_p = e^{−ζπ/√(1−ζ²)}
Solveω_n = √(200/2) = √100 = 10 rad/s. ζ = 12/(2√400) = 12/40 = 0.3. ω_d = 10√0.91 = 9.54 rad/s. M_p = e^−0.988 = 37.2%.
Checkζ = 0.3 matches Example 1, so the overshoot is identical at 37.2 percent, confirming overshoot depends on ζ alone. The doubled ω_n halves the peak and settling times.
ConclusionThe physical parameters map straight to ω_n and ζ, and from there to the full response. Same damping gives same overshoot; higher frequency gives a faster response.

Result. ω_n = 10 rad/s, ζ = 0.3, ω_d = 9.54 rad/s, M_p = 37.2%.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Overshoot for ζ ≥ 1	A peak drawn on an overdamped response	"Is ζ below 1?"	Only underdamped systems overshoot.
Settling from ω_n alone	t_s wrong by the damping factor	"Did I use ζω_n?"	t_s ≈ 4/(ζω_n), the decay rate.
Overshoot tied to ω_n	Overshoot changes with frequency	"Does M_p depend on ω_n?"	Overshoot depends on ζ only.
ω_d equal to ω_n	Peak time computed with ω_n	"Did I reduce ω_n by √(1−ζ²)?"	Use ω_d = ω_n√(1 − ζ²) for the peak time.

Practice ladder

Level 1 · Direct skill

A system has ω_n = 8 rad/s and ζ = 0.5. Find the damped frequency.

Show answer

ω_d = 8√(1 − 0.25) = 8√0.75 = 8 × 0.866 = 6.93 rad/s.

Level 2 · Mixed concept

For ζ = 0.5, find the percent overshoot and the 2 percent settling time if ω_n = 8 rad/s.

Show answer

M_p = e^{−0.5π/√0.75} = e^−1.814 = 0.163 = 16.3%. t_s = 4/(0.5 × 8) = 1.0 s.

Level 3 · Independent problem

A system has G(s) = 36/(s² + 4.8s + 36). Find ω_n, ζ, and the overshoot.

Show answer

ω_n = √36 = 6 rad/s. 2ζω_n = 4.8, so ζ = 4.8/12 = 0.4. M_p = e^{−0.4π/√0.84} = e^−1.371 = 0.254 = 25.4%.

Transfer task | Real engineering

A positioning system must overshoot no more than 5 percent and settle within 0.5 s. Translate these into requirements on ζ and ω_n.

What good work looks like

5 percent overshoot needs ζ ≈ 0.7. Then t_s = 4/(ζω_n) ≤ 0.5 requires ζω_n ≥ 8, so ω_n ≥ 8/0.7 ≈ 11.4 rad/s. The spec becomes a target pole region.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used ζ alone for the overshoot."

"Give me three second-order systems; I will find ω_n and ζ for each."

"Compute the metrics for me." Extracting ω_n and ζ first is the skill.

"Why does it overshoot?" Reasoning from the damping ratio is the point.

Portfolio task

Take a real underdamped system, find ω_n and ζ, predict its overshoot and settling time, then compare with a measured or simulated step response.

Must include: ω_n, ζ, a predicted M_p and t_s, and a comparison.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the second-order standard form.

G(s) = ω_n²/(s² + 2ζω_ns + ω_n²).

2. What does the overshoot depend on?

The damping ratio ζ alone: M_p = e^{−ζπ/√(1−ζ²)}.

3. Give the peak time.

t_p = π/ω_d, with ω_d = ω_n√(1 − ζ²).

4. Give the 2 percent settling time.

t_s ≈ 4/(ζω_n).

5. Which regimes do not overshoot?

Critically damped (ζ = 1) and overdamped (ζ > 1).

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the metrics from ω_n and ζ on a blank page.

+3 daysFind the metrics of three new systems.

+7 daysRecast the model in state space, Module 9.

+30 daysReuse ζ and ω_n as control specifications.

Textbook mapping

This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.

Topic in this module	Where to read more
Second-order standard form and damping	Karnopp, Margolis & Rosenberg, Chapter 8
Overshoot, peak time, settling time	Karnopp, Margolis & Rosenberg, Chapter 8
Transient response specifications	Karnopp, Margolis & Rosenberg, Chapter 9

Chapter numbers refer to the 5th edition. The second-order metrics are standard, so any recent edition will align closely.