Qwestrum Engineering360 · Electrical & Electronics · Control Systems
Time Response Analysis
The time response of a standard second-order system is set by the damping ratio ζ and natural frequency ω_n; ζ controls overshoot and settling while ω_n controls the speed of response.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.
Key formulas & points
Skim these first — then read the full notes below.
- Underdamped: oscillatory; overdamped: no overshoot
- Steady-state error e_ss from type number and input
- Pole location in s-plane determines stability and speed
Topic details
Introduction
Most control specifications reduce to a standard second-order form s² + 2ζω_n s + ω_n². From ζ and ω_n you obtain every transient metric: peak overshoot M_p = e^(−ζπ/√(1−ζ²)), peak time t_p = π/ω_d, and settling time t_s ≈ 4/ζω_n.
Scope in B.Tech and GATE syllabus
The damped natural frequency ω_d = ω_n√(1−ζ²). Underdamped systems (0 < ζ < 1) oscillate; critically damped (ζ = 1) reach steady state fastest without overshoot; overdamped (ζ > 1) are sluggish.
Key relations & formulas
Formulas (Indian textbook notation)
Formulas (Indian textbook notation)
- Peak time t_{p} = \frac{\pi}{\omega_{d}}; M_{p} = e^(-ζ\frac{\pi}{\sqrt}{1-ζ^{2}})
(2% criterion)
Notation and sign conventions
Relation 1 —
Formulas (Indian textbook notation)
Write this relation with symbols exactly as in Control Systems Engineering — Nagarath & Gopal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Peak time t_{p} = \frac{\pi}{\omega_{d}}; M_{p} = e^
Formulas (Indian textbook notation)
- Peak time t_{p} = \frac{\pi}{\omega_{d}}; M_{p} = e^(-ζ\frac{\pi}{\sqrt}{1-ζ^{2}})
Write this relation with symbols exactly as in Control Systems Engineering — Nagarath & Gopal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
(2% criterion)
Write this relation with symbols exactly as in Control Systems Engineering — Nagarath & Gopal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Fundamentals and definitions
Reading ζ and ω_n from a characteristic equation: match s² + as + b to the standard form, so ω_n = √b and ζ = a/(2ω_n).
Governing relations in practice
Steady-state error depends on the system type (number of poles at the origin) and the input: a type-0 system has finite error to a step, a type-1 system has zero step error but finite ramp error given by 1/K_v.
Design and analysis considerations
Pole location on the s-plane visualises the response: distance from the origin sets speed (ω_n), angle from the imaginary axis sets damping (ζ = cos of that angle).
Assumptions and validity limits
State assumptions explicitly before using any relation for time response analysis — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Control Systems viva and GATE descriptive questions, listing valid assumptions often earns separate marks.
Step-by-step problem approach
1. Read the question and list given data with SI units (common in Control Systems papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to time response analysis.
4. Use equation 1:
5. Use equation 2:
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to time response analysis.
4. Use equation 1:
.
5. Use equation 2:
Peak time t_{p} = \frac{\pi}{\omega_{d}}; M_{p} = e^
.6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
Applications & exam relevance
Time Response Analysis appears in process plants and automation. In Indian electrical curricula this topic is tested because it connects theory to modelling, stability, and controller design.
GATE and semester exams often combine time response analysis with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use time response analysis?" — answer with a lab, mini-project, or plant visit example if possible.
Common mistakes in exams
• Using ω_n instead of ω_d in the peak-time formula
• Expressing overshoot as a fraction when a percentage is asked (or vice versa)
• Forgetting the settling time depends on the tolerance band (4/ζω_n for 2%, 3/ζω_n for 5%)
• Misidentifying system type when finding steady-state error
• Expressing overshoot as a fraction when a percentage is asked (or vice versa)
• Forgetting the settling time depends on the tolerance band (4/ζω_n for 2%, 3/ζω_n for 5%)
• Misidentifying system type when finding steady-state error
Quick revision checklist
Before attempting time response analysis problems, confirm you can:
1. Underdamped: oscillatory; overdamped: no overshoot
2. Steady-state error e_ss from type number and input
3. Pole location in s-plane determines stability and speed
2. Steady-state error e_ss from type number and input
3. Pole location in s-plane determines stability and speed
Revise the solved examples in Control Systems Engineering — Nagarath & Gopal and one previous-year GATE or university paper for this unit.
Worked examples
Try the problem first — open the solution when you are ready to check.
Damping ratio and overshoot
Problem
A second-order system has the characteristic equation s² + 6s + 25 = 0. Find ω_n, ζ and the percentage peak overshoot.
Solution
ω_n = √25 = 5 rad/s.
2ζω_n = 6 → ζ = 6/(2×5) = 0.6.
M_p = e^(−ζπ/√(1−ζ²)) = e^(−0.6π/√(1−0.36)) = e^(−1.885/0.8).
= e^(−2.356) = 0.0948 → 9.5% overshoot.
2ζω_n = 6 → ζ = 6/(2×5) = 0.6.
M_p = e^(−ζπ/√(1−ζ²)) = e^(−0.6π/√(1−0.36)) = e^(−1.885/0.8).
= e^(−2.356) = 0.0948 → 9.5% overshoot.
Conceptual check — Time Response Analysis
Problem
In a Control Systems semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of time response analysis." What should a complete answer include?
Exams & GATE
Nagarath & Gopal — find ζ, ω_n from characteristic equation.
📖 Standard books (India)
Control Systems Engineering — Nagarath & Gopal
Read: Syllabus unit
Transfer functions, stability, and PID
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