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Qwestrum Engineering360 · Electrical & Electronics · Control Systems

Root Locus Technique

The root locus plots how the closed-loop poles migrate in the s-plane as the loop gain K varies from zero to infinity, revealing at a glance the range of K for stability and the achievable damping.
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.

  • Root locus shows closed-loop pole movement vs gain K
  • Asymptotes for n > m: angles (2k+1)180°/(n−m)
  • Dominant poles approximate second-order response

Topic details

Introduction

Root locus starts at the open-loop poles (K = 0) and ends at the open-loop zeros or infinity (K → ∞). A point lies on the locus if it satisfies the angle condition (net angle = odd multiple of 180°); the gain at any point is found from the magnitude condition.

Scope in B.Tech and GATE syllabus

The design value is immediate: you can pick the K that places the dominant poles at a desired ζ, or find the K at which the locus crosses into the right half-plane (stability limit).

Key relations & formulas

Formulas (Indian textbook notation)

  • Anglecondition:Σ∠(s−zi)−Σ∠(s−pj)=±180§K0§Angle condition: Σ∠(s-z_{i}) - Σ∠(s-p_{j}) = ±180^{§K0§}

Formulas (Indian textbook notation)

  • Magnitudecondition:K=Π∣s−pj∣/Π∣s−zi∣Magnitude condition: K = Π|s-p_{j}|/Π|s-z_{i}|

Formulas (Indian textbook notation)

  • BreakawaypointsfromdKds=0onrealaxisBreakaway points from \frac{dK}{ds} = 0 on real axis

Notation and sign conventions

Relation 1 —
Anglecondition:Σ∠Angle condition: Σ∠

Formulas (Indian textbook notation)

  • Anglecondition:Σ∠(s−zi)−Σ∠(s−pj)=±180§K0§Angle condition: Σ∠(s-z_{i}) - Σ∠(s-p_{j}) = ±180^{§K0§}
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 —
Magnitudecondition:K=Π∣s−pj∣/Π∣s−zi∣Magnitude condition: K = Π|s-p_{j}|/Π|s-z_{i}|

Formulas (Indian textbook notation)

  • Magnitudecondition:K=Π∣s−pj∣/Π∣s−zi∣Magnitude condition: K = Π|s-p_{j}|/Π|s-z_{i}|
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 —
BreakawaypointsfromdKds=0onrealaxisBreakaway points from \frac{dK}{ds} = 0 on real axis

Formulas (Indian textbook notation)

  • BreakawaypointsfromdKds=0onrealaxisBreakaway points from \frac{dK}{ds} = 0 on real axis
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

Construction rules: the locus on the real axis lies to the left of an odd count of real poles and zeros; branches equal the number of poles n; (n−m) branches go to infinity along asymptotes at angles (2k+1)180°/(n−m) meeting the real axis at the centroid σ = (Σpoles − Σzeros)/(n−m).

Governing relations in practice

Breakaway/break-in points, where branches leave or join the real axis, are found from dK/ds = 0. The imaginary-axis crossing (found via Routh) gives the critical gain for marginal stability.

Design and analysis considerations

Dominant poles — the closest complex pair to the imaginary axis — let you approximate the higher-order response by an equivalent second-order system.

Assumptions and validity limits

State assumptions explicitly before using any relation for root locus technique — 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 root locus technique.
4. Use equation 1:
Anglecondition:Σ∠Angle condition: Σ∠
.
5. Use equation 2:
Magnitudecondition:K=Π∣s−pj∣/Π∣s−zi∣Magnitude condition: K = Π|s-p_{j}|/Π|s-z_{i}|
.
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

Root Locus Technique 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 root locus technique with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use root locus technique?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Using the wrong real-axis segment (must be left of an odd count of poles+zeros)
• Computing asymptote angles with n instead of (n−m)
• Forgetting to verify breakaway points give real positive K
• Ignoring the effect of a nearby zero when assuming dominant-pole behaviour

Quick revision checklist

Before attempting root locus technique problems, confirm you can:
1. Root locus shows closed-loop pole movement vs gain K
2. Asymptotes for n > m: angles (2k+1)180°/(n−m)
3. Dominant poles approximate second-order response
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.

Asymptotes and centroid

Problem

For an open-loop transfer function G(s)H(s) = K/[s(s+2)(s+4)], find the number of asymptotes, their angles and centroid.

Solution

Poles at 0, −2, −4 (n = 3); no finite zeros (m = 0).
Number of asymptotes = n − m = 3.
Angles = (2k+1)180°/3 = 60°, 180°, 300°.
Centroid σ = (Σpoles − Σzeros)/(n−m) = (0 − 2 − 4)/3 = −6/3 = −2.

Conceptual check — Root Locus Technique

Problem

In a Control Systems semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of root locus technique." What should a complete answer include?

Exams & GATE

Nagarath & Gopal — sketch root locus with jω-axis crossing.

📖 Standard books (India)

  • Control Systems Engineering — Nagarath & Gopal

    Read: Syllabus unit

    Transfer functions, stability, and PID

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