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Controls and Systems · Chapter 4 of 10 · Intermediate

Transient Response Analysis

How fast, how much overshoot, how long to settle: a second-order system answers all three with just two numbers, the damping ratio and the natural frequency.

Readiness check

This chapter reads response from poles. Tick only what you can do closed-notes.

Recall the standard second-order denominator.
Identify ω_n and ζ from a transfer function.
Evaluate an exponential with a decimal argument.
Read a step-response plot.
Find a time constant from a first-order pole.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit the standard form in Chapter 3.

3 or more weak itemsReview damped vibration in Dynamics and Vibrations.

The core idea

A first-order system is set by one time constant; a second-order system by ω_n and ζ. From those two, overshoot, peak time, and settling time follow by formula.

M_p = e^{−ζπ/√(1−ζ²)}t_p = π/(ω_n√(1−ζ²))t_s ≈ 4/(ζω_n)

A first-order pole at −1/τ gives an exponential approach with time constant τ; the response reaches 63 percent in one τ and effectively settles in four. A second-order system's pole pair −ζω_n ± jω_n√(1−ζ²) sets its character: the real part fixes how fast it decays, the imaginary part how fast it oscillates. The damping ratio ζ controls overshoot, and the natural frequency ω_n scales the speed. With both, the transient specifications are direct formulas, no inversion required.

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

The skill breaks down when: overshoot is expected from an overdamped system, or the natural frequency is confused with the damped frequency.

The concept. An underdamped step response: it rises past the final value to a peak M_p at time t_p, then the oscillation decays and the output settles within a tolerance band.

The skills, taught in order

Five skills cover the first-order time constant, the second-order parameters, the damping cases, and the transient specifications.

4.1 First-order response

A first-order system K/(τs + 1) responds to a step with c(t) = K(1 − e^−t/τ). The time constant τ is the time to reach 63 percent; the response is within 2 percent of final after about 4τ. There is no overshoot.

4.2 Second-order parameters

The standard form ω_n²/(s² + 2ζω_ns + ω_n²) has poles at −ζω_n ± jω_n√(1−ζ²). The natural frequency ω_n sets the speed; the damping ratio ζ sets how oscillatory the response is. The damped frequency is ω_d = ω_n√(1−ζ²).

4.3 The damping cases

ζ = 0 is undamped (sustained oscillation); 0 < ζ < 1 is underdamped (decaying oscillation with overshoot); ζ = 1 is critically damped (fastest without overshoot); ζ > 1 is overdamped (slow, no overshoot). Most designs aim for ζ near 0.5 to 0.7.

Specification	Formula	Set by
Percent overshoot M_p	e^{−ζπ/√(1−ζ²)}	ζ only
Peak time t_p	π/(ω_n√(1−ζ²))	ω_n and ζ
Settling time t_s (2%)	4/(ζω_n)	real part ζω_n

4.4 Transient specifications

Overshoot depends only on ζ, so it fixes the damping. Settling time depends on ζω_n, the distance of the poles from the imaginary axis. Peak time depends on the damped frequency. Together they translate a pole location into a response shape.

4.5 Pole location and response

Moving the pole pair left speeds up settling; moving it down (more imaginary) speeds the oscillation; a constant ζ keeps overshoot fixed along a radial line. Reading response from pole position is the intuition the rest of the course leans on.

Engineering connection: design specifications (overshoot, settling) become target pole regions that root locus and compensators aim for.

Worked example 1: second-order step specifications

A closed-loop system has T(s) = 25/(s² + 6s + 25). Find its natural frequency, damping ratio, percent overshoot, peak time, and 2 percent settling time.

Figure 1. With ζ = 0.6 the system overshoots about 9.5 percent. The poles at −3 ± 4j set the peak time and the settling time directly.

ProblemFind ω_n, ζ, M_p, t_p, and t_s for the system in Figure 1.
Given / findT(s) = 25/(s² + 6s + 25). Find the five quantities.
AssumptionsStandard underdamped second-order system; unit step input.
ModelMatch to ω_n²/(s² + 2ζω_ns + ω_n²), then apply the spec formulas.
Equationsω_n² = 25, 2ζω_n = 6M_p = e^{−ζπ/√(1−ζ²)}, t_p = π/(ω_n√(1−ζ²)), t_s = 4/(ζω_n)
Solveω_n = 5 rad/s, ζ = 6/(2·5) = 0.6. M_p = e^−0.6π/0.8 = e^−2.36 = 9.5%. t_p = π/(5·0.8) = 0.79 s. t_s = 4/(0.6·5) = 1.33 s.
CheckThe poles are −3 ± 4j: the real part 3 gives t_s = 4/3 = 1.33 s, and the imaginary part 4 = ω_d gives t_p = π/4 = 0.79 s. Both agree.
ConclusionTwo numbers produced the full response shape. Overshoot came from ζ alone, while ω_n scaled the timing.

Result. ω_n = 5 rad/s, ζ = 0.6, M_p = 9.5%, t_p = 0.79 s, t_s = 1.33 s.

Worked example 2: a first-order response

A first-order system has G(s) = 10/(s + 5). Find its DC gain, time constant, and 2 percent settling time, and the final value of its unit-step response.

Figure 2. A first-order system rises smoothly with no overshoot, reaching 63 percent of its final value in one time constant and essentially settling in four.

ProblemFind the DC gain, τ, t_s, and final value for the system in Figure 2.
Given / findG(s) = 10/(s + 5), unit step input. Find DC gain, τ, t_s, c(∞).
AssumptionsFirst-order, stable system; unit step.
ModelWrite G in the form K/(τs + 1): the pole gives τ, G(0) gives the DC gain, and t_s ≈ 4τ.
EquationsG = 10/(s + 5) = 2/(0.2s + 1)τ = 1/5, c(∞) = K · 1
SolveDC gain = G(0) = 10/5 = 2. τ = 1/5 = 0.2 s. t_s ≈ 4τ = 0.8 s. Final value of the unit-step response = K = 2.
CheckThe final value theorem gives lim_s→0 s · [10/(s(s+5))] = 10/5 = 2, matching the DC gain. No overshoot, as a single real pole requires.
ConclusionOne pole, one time constant: a first-order system is fully described by τ and its DC gain, with a clean exponential response.

Result. DC gain 2, τ = 0.2 s, t_s ≈ 0.8 s, final value 2.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Expecting overshoot when overdamped	Overshoot predicted for ζ ≥ 1	"Is ζ less than 1?"	Overshoot occurs only for an underdamped system.
Confusing ω_n and ω_d	Peak time off	"Which frequency sets the oscillation?"	Use the damped frequency ω_d = ω_n√(1−ζ²) for t_p.
Settling time from ω_n alone	t_s too short or long	"What is the real part of the pole?"	t_s ≈ 4/(ζω_n), the distance from the imaginary axis.
Reading τ from the wrong pole	First-order timing wrong	"Is the denominator in τs + 1 form?"	Normalise so τ is the reciprocal of the pole magnitude.

Practice ladder

Level 1 · Direct skill

A first-order system has G = 4/(s + 8). Find its time constant and settling time.

Show answer

τ = 1/8 = 0.125 s; t_s ≈ 4τ = 0.5 s. DC gain is 4/8 = 0.5.

Level 2 · Mixed concept

A second-order system has ω_n = 10 rad/s and ζ = 0.5. Find the percent overshoot and settling time.

Show answer

M_p = e^{−0.5π/√0.75} = e^−1.814 = 16.3%. t_s = 4/(0.5·10) = 0.8 s.

Level 3 · Independent problem

A system has poles at −2 ± 2j. Find ω_n, ζ, and the percent overshoot.

Show answer

ω_n = √(2² + 2²) = 2.83 rad/s; ζ = 2/2.83 = 0.707. M_p = e^{−0.707π/√0.5} = e^−π = 4.3%. A classic well-damped design.

Level 4 · Transfer to real engineering

Find a real system with a known step response (a motor, a thermal loop) and estimate its ω_n and ζ, or its τ, from the response shape.

What good work looks like

Overshoot read off to get ζ, peak or settling time to get ω_n (or τ for first order), with a sanity check against the measured curve.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used ζ for overshoot and ζω_n for settling time."

"Give me four pole pairs; I will find ω_n and ζ for each."

"What is the overshoot?" Applying the spec formulas yourself is the skill.

"Describe this response." Reading it from the poles is the point.

Portfolio task

Take a real second-order system, find ω_n and ζ, predict overshoot and settling time, and compare with a measured or simulated step response.

Must include: ω_n and ζ, predicted M_p and t_s, and a comparison with the actual response.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What sets a first-order response?

The time constant τ; 63 percent at one τ, essentially settled by 4τ, with no overshoot.

2. What does overshoot depend on?

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

3. Give the settling time.

t_s ≈ 4/(ζω_n), set by the real part of the poles.

4. Name the four damping cases.

Undamped (ζ = 0), underdamped (0 < ζ < 1), critically damped (ζ = 1), overdamped (ζ > 1).

5. Where are the second-order poles?

At −ζω_n ± jω_n√(1−ζ²).

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

+1 dayRe-derive the specs of a second-order system from a blank page.

+3 daysFind the response shape for two new pole pairs.

+7 daysCarry response into steady-state error, Chapter 5.

+30 daysReuse pole-to-response intuition in root locus design.

Textbook mapping

Item	Mapping
Primary source	Ogata, Modern Control Engineering, Chapter 5 (Transient and Steady-State Response Analysis)
Cross-reference	Nise, Ch. 4 · Dynamics and Vibrations
Core topics	4.1 First-order response · 4.2 Second-order parameters · 4.3 Damping cases · 4.4 Transient specs · 4.5 Pole location and response
Engineering connection	Specs become target pole regions for design.
Read next	Chapter 5: Steady-State Error.