Time-Domain Response of Control Systems

Time-domain analysis of first- and second-order systems predicts transient behaviour — rise time, overshoot, settling — the core performance metrics tested heavily in GATE EC (2–3 marks).

Key results

• Standard 2nd-order: T(s) = ωₙ²/(s² + 2ζωₙs + ωₙ²); ζ = damping ratio, ωₙ = natural frequency.
• Damped frequency ω_d = ωₙ√(1−ζ²); poles s = −ζωₙ ± jω_d.
• Peak overshoot M_p = e^{−πζ/√(1−ζ²)} (×100%); peak time t_p = π/ω_d.
• Settling time t_s ≈ 4/(ζωₙ) (2% band), 3/(ζωₙ) (5%). Rise time falls as ωₙ rises.
• Steady-state error via static error constants K_p, K_v, K_a and system type (number of poles at origin).

ζ range	Response
0	undamped (sustained oscillation)
0 < ζ < 1	underdamped (overshoot)
ζ = 1	critically damped (fastest, no overshoot)
ζ > 1	overdamped (slow, no overshoot)

Exam Tricks & Tips

• 🎯 Peak overshoot depends ONLY on ζ: M_p = e^{−πζ/√(1−ζ²)} — memorise ζ=0.5 → ~16%, ζ=0.707 → ~4.3%.
• 🎯 Settling time t_s ≈ 4/(ζωₙ) for the 2% criterion; smaller ζωₙ (real part) → slower settling.
• 🎯 ζ = 0.707 is often optimal (fast, ~4% overshoot, good phase margin).
• 🎯 Steady-state error: type-0 has finite K_p (nonzero step error); type-1 tracks steps with zero error; type-2 tracks ramps.
• 🎯 Increasing ωₙ speeds up the response (smaller rise/peak time) without changing overshoot.
• ❌ Common mistake: using ωₙ instead of ω_d for peak time (t_p = π/ω_d, not π/ωₙ).

Expected exam pattern

1–2 marks: find ζ, ωₙ, overshoot, peak/settling time from a transfer function, or compute steady-state error for a given input and system type.

Quick recap

2nd-order: T=ωₙ²/(s²+2ζωₙs+ωₙ²). ω_d=ωₙ√(1−ζ²). M_p=e^{−πζ/√(1−ζ²)}, t_p=π/ω_d, t_s≈4/ζωₙ. Overshoot depends only on ζ. Steady-state error set by system type & K_p/K_v/K_a.

Cover the answer, recall, then check. 12 cards on time-domain response for GATE EC.

Q1. Write the standard second-order transfer function.
A1. T(s) = ωₙ²/(s² + 2ζωₙs + ωₙ²).

Q2. Give the damped natural frequency.
A2. ω_d = ωₙ√(1 − ζ²).

Q3. Formula for peak overshoot?
A3. M_p = e^{−πζ/√(1−ζ²)} (as a fraction; ×100 for percent).

Q4. Peak overshoot depends on which parameter?
A4. The damping ratio ζ only (not ωₙ).

Q5. Formula for peak time?
A5. t_p = π/ω_d.

Q6. Settling time for the 2% criterion?
A6. t_s ≈ 4/(ζωₙ).

Q7. Classify the response by damping ratio.
A7. ζ=0 undamped, 0<ζ<1 underdamped, ζ=1 critically damped, ζ>1 overdamped.

Q8. Where are the poles of an underdamped second-order system?
A8. s = −ζωₙ ± jω_d (complex conjugates).

Q9. Overshoot at ζ = 0.5?
A9. About 16.3%.

Q10. What is system "type"?
A10. The number of open-loop poles at the origin (integrators); it determines steady-state error to step/ramp/parabola.

Q11. Steady-state error to a step for type-0 and type-1 systems?
A11. Type-0: finite (1/(1+K_p)); type-1 or higher: zero.

Q12. How does increasing ωₙ affect the response?
A12. It speeds it up (smaller rise/peak/settling time) while leaving overshoot (which depends on ζ) unchanged.

How fast, how oscillatory, and how accurately a system responds is read from its time-domain response. For the ubiquitous second-order system, damping ratio ζ and natural frequency ωn set the entire step response — overshoot, settling time, rise time. GATE almost always has a question extracting these specs or a steady-state error from the system type.

Core concept: A standard second-order system's step response is governed by ζ and ωn; transient specs (overshoot, settling, rise time) and steady-state error follow from them and the system type.

The theory

Beginner — the standard second-order system

Closed-loop form: T(s) = ωn²/(s² + 2ζωn·s + ωn²). The poles are at −ζωn ± jωn√(1−ζ²). Response by ζ:

• ζ = 0: undamped, pure oscillation.
• 0 < ζ < 1: underdamped, decaying oscillation (most control systems).
• ζ = 1: critically damped, fastest without overshoot.
• ζ > 1: overdamped, sluggish.

Intermediate — transient specifications (underdamped step)

• Damped frequency: ωd = ωn√(1 − ζ²).
• Peak (percent) overshoot: %OS = 100·exp(−πζ/√(1−ζ²)) — depends only on ζ.
• Peak time: tp = π/ωd.
• Settling time (2%): ts ≈ 4/(ζωn); (5%): ts ≈ 3/(ζωn).
• Rise time: decreases with ωn.
  So ζ sets the overshoot, ωn sets the speed.

Advanced — steady-state error and system type

For unity feedback, steady-state error to standard inputs depends on the system type (number of poles at the origin, i.e., integrators in the open loop G(s)):

• Position constant Kp = lim(s→0)G(s); step error ess = 1/(1+Kp).
• Velocity constant Kv = lim(s→0)sG(s); ramp error ess = 1/Kv.
• Acceleration constant Ka = lim(s→0)s²G(s); parabola error ess = 1/Ka.
  A Type-0 system has finite Kp (nonzero step error); Type-1 has zero step error but finite ramp error; Type-2 tracks ramps with zero error. Adding integrators reduces steady-state error but hurts stability.

Worked example

A second-order system has ωn = 10 rad/s and ζ = 0.5. Find %overshoot, peak time, and 2% settling time.
%OS = 100·exp(−πζ/√(1−ζ²)) = 100·exp(−π·0.5/√0.75) = 100·exp(−1.814) = 16.3%.
ωd = ωn√(1−ζ²) = 10·√0.75 = 8.66 rad/s → tp = π/ωd = π/8.66 = 0.363 s.
ts (2%) = 4/(ζωn) = 4/(0.5·10) = 0.8 s.

GATE relevance

Very high yield: extract ζ, ωn from a transfer function; compute %OS, tp, ts, tr; steady-state error from system type and error constants (Kp, Kv, Ka); and effect of adding a pole/zero or integrator. NAT and MCQ both common.

Exam tricks and shortcuts

• %OS depends only on ζ; ts ≈ 4/(ζωn) (2%); tp = π/ωd.
• ζ sets overshoot, ωn sets speed — remember which knob does what.
• Steady-state error: Type-0 → step error 1/(1+Kp); Type-1 → ramp error 1/Kv; Type-2 → parabola error 1/Ka.
• Mnemonic: "Zeta for overshoot, omega-n for speed."

Key formulas (second-order, standard form)

• T(s) = ω_n²/(s² + 2ζω_n s + ω_n²).
• Damped frequency: ω_d = ω_n√(1 − ζ²).
• Peak overshoot: M_p = e^(−ζπ/√(1−ζ²)) × 100%.
• Rise time, peak time: t_p = π/ω_d; settling time t_s ≈ 4/(ζω_n) (2%).
• Steady-state error from static error constants K_p, K_v, K_a.

The damping ratio ζ controls overshoot and settling; ω_n sets the response speed.

Worked Example

Problem: A unity-feedback second-order system has closed-loop transfer function T(s) = 25/(s² + 6s + 25). For a unit-step input, find the natural frequency, damping ratio, percentage peak overshoot, peak time, and 2% settling time.

Solution:
Compare the denominator with s² + 2ζωₙs + ωₙ²:
ωₙ² = 25 ⇒ ωₙ = 5 rad/s.
2ζωₙ = 6 ⇒ ζ = 6/(2×5) = 0.6 (underdamped).

Damped natural frequency:
ω_d = ωₙ√(1 − ζ²) = 5√(1 − 0.36) = 5 × 0.8 = 4 rad/s.

Percentage peak overshoot:
%M_p = 100 · e^(−ζπ/√(1−ζ²)) = 100 · e^(−0.6π/0.8) = 100 · e^(−2.356) ≈ 9.5%.

Peak time (time to first peak):
t_p = π/ω_d = π/4 ≈ 0.785 s.

Settling time (2% criterion):
t_s = 4/(ζωₙ) = 4/(0.6 × 5) = 4/3 ≈ 1.33 s.

Answer: ωₙ = 5 rad/s, ζ = 0.6, %M_p ≈ 9.5%, t_p ≈ 0.785 s, t_s ≈ 1.33 s.