Study notes

Qwestrum Engineering360 · Chemical Engineering · Process Dynamics & Control

First and Second Order Responses

A first-order system responds to a step with an exponential rise reaching 63.2% at one time constant; a second-order system adds a damping ratio ζ that decides whether the response is overdamped, critically damped or oscillatory.

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.

$63.2% of the final value is reached at t = \tau$ (first order)
ζ < 1 oscillatory, ζ = 1 critically damped, ζ > 1 overdamped
K is the steady-state gain; τ measures the response speed

Topic details

Introduction

This Coughanowr topic characterises the standard dynamic responses that most processes approximate. You interpret first-order step, ramp and impulse responses to extract gain and time constant, and analyse second-order behaviour — overshoot, decay ratio and period — through the natural frequency and damping ratio.

Key relations & formulas

G(s) = \frac{K}{(\tau s + 1)}

(first-order, time constant τ)

G(s) = \frac{K}{(\tau^{2} s^{2} + 2 ζ \tau s + 1)}

(second-order)

y(t) = K M (1 - e^(-\frac{t}{\tau}))

(first-order step response of magnitude M)

Notation and sign conventions

Relation 1 —

G

G(s) = \frac{K}{(\tau s + 1)}

(first-order, time constant τ)

Write this relation with symbols exactly as in Process Systems Analysis & Control — Coughanowr & LeBlanc before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Relation 2 —

G

G(s) = \frac{K}{(\tau^{2} s^{2} + 2 ζ \tau s + 1)}

(second-order)

Relation 3 —

y

y(t) = K M (1 - e^(-\frac{t}{\tau}))

(first-order step response of magnitude M)

Concept in depth

The first-order lag is the workhorse model: it rises smoothly and monotonically to a new steady state, reaching 63.2% of the change in one time constant and about 98% in four. Two first-order systems in series, or an inertia-plus-resistance system, give second-order dynamics where the damping ratio ζ governs the character. An overdamped system (ζ > 1) is sluggish but smooth; critically damped (ζ = 1) is the fastest non-oscillating response; underdamped (ζ < 1) overshoots and rings, with lower ζ meaning more overshoot and slower settling. These prototypes let you predict behaviour before tuning a controller.

Assumptions and validity limits

State assumptions explicitly before using any relation for first and second order responses — 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 Process Dynamics & Control 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 Process Dynamics & Control 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 first and second order responses.
4. Use equation 1:

G

.
5. Use equation 2:

G

.
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

First and Second Order Responses appears in DCS and plant automation. In Indian chemical curricula this topic is tested because it connects theory to dynamic models and loop tuning.

GATE and semester exams often combine first and second order responses with earlier units — revise prerequisites before attempting mixed problems.

Industry interview panels sometimes ask: "Where did you use first and second order responses?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

Students misread the time constant from a step response (it is the 63.2% point, not the settling time), confuse steady-state gain with dynamic gain, and mislabel damping regimes. Treating an oscillatory response as first-order is a conceptual error.

Quick revision checklist

Before attempting first and second order responses problems, confirm you can:

63.2% of the final value is reached at t = \tau

(first order)
2. ζ < 1 oscillatory, ζ = 1 critically damped, ζ > 1 overdamped
3. K is the steady-state gain; τ measures the response speed

Revise the solved examples in Process Systems Analysis & Control — Coughanowr & LeBlanc 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.

First-order response value

Problem

A first-order process (K = 2, τ = 5 s) receives a unit step. Find the output at t = 5 s and t = 10 s.

Solution

y(5) = 2(1 − e⁻¹) = 2×0.632 = 1.26; y(10) = 2(1 − e⁻²) = 2×0.865 = 1.73. It approaches the final value 2 asymptotically.

Conceptual check — First and Second Order Responses

Problem

In a Process Dynamics & Control semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of first and second order responses." What should a complete answer include?

Exams & GATE

Coughanowr Ch. 5 — read τ and K from a step-test graph.

📖 Standard books (India)

Process Systems Analysis & Control — Coughanowr & LeBlanc
Read: Syllabus unit
Dynamic modelling and control loops

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📚Key formulas & points