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Control Systems is one of the most rewarding subjects in Part B1 (Electrical) of the GATE Robotics and Automation (RA) 2027 paper, because a handful of standard results — the second-order response parameters, the Routh–Hurwitz test, and the static error constants — unlock the majority of the questions. This guide reviews the transfer function, transient response, stability, steady-state error and frequency response, gives you the formula set the exam actually tests, and then works through five GATE-style problems in full.

Transfer functions & block diagrams

A linear time-invariant system is described by its transfer function G(s) = C(s)/R(s), the Laplace-domain ratio of output to input at zero initial conditions. Systems are combined by block-diagram reduction: blocks in series multiply, blocks in parallel add, and a feedback loop closes to the standard form. For a negative-feedback loop with forward path G and feedback H, the closed-loop transfer function is G/(1 + GH).

C(s)/R(s) = G(s) / (1 + G(s)H(s))  •  open-loop = G(s)H(s)

Second-order transient response

The standard second-order system C/R = ωn² / (s² + 2ζωns + ωn²) is the single most examined model. Reading the natural frequency ωn and damping ratio ζ straight from the denominator gives every transient specification: peak overshoot, peak time, rise time and settling time.

t c(t) 1 peak (Mp) underdamped step response (0 < ζ < 1)

Figure 1. Underdamped second-order step response: overshoot and settling are fixed by ζ and ωn.

Mp = e−ζπ/√(1−ζ²)  •  tp = π/(ωn√(1−ζ²))  •  ts ≈ 4/(ζωn) (2%)

Stability & the Routh criterion

A system is stable if every pole lies in the left half of the s-plane. The Routh–Hurwitz criterion tests this without solving the characteristic equation: build the Routh array and require that all entries in the first column have the same sign. Any sign change signals right-half-plane roots, and a row of zeros signals marginal stability. Routh is the go-to tool for finding the range of a gain K that keeps a system stable.

Steady-state error

The steady-state error to a standard input follows from the static error constants, which depend on the type of the system (the number of poles at the origin in the open-loop transfer function). The position constant Kp governs the step response, the velocity constant Kv the ramp, and the acceleration constant Ka the parabola.

Kp = lims→0 G(s)H(s)  •  Kv = lims→0 sG(s)H(s)  •  ess,ramp = 1/Kv

Frequency response

Frequency-domain analysis uses the Bode plot (magnitude and phase versus frequency) and the polar/Nyquist plot to judge relative stability through the gain margin and phase margin. The gain crossover frequency is where |G(jω)| = 1 (0 dB), and the phase crossover is where the phase reaches −180°. Larger margins mean a more robustly stable, better-damped system.

Worked examples (GATE-style)

Example 1 — damping ratio and overshoot

A closed-loop system has C/R = 25 / (s² + 6s + 25). Find ωn, ζ and the percentage peak overshoot.

Solution. Comparing with ωn²/(s² + 2ζωns + ωn²): ωn² = 25 ⇒ ωn = 5 rad/s. And 2ζωn = 6 ⇒ ζ = 6/10 = 0.6.

Mp = e−0.6π/√(1−0.36) = e−2.356 = 0.0948 = 9.48%.

Example 2 — peak time and settling time

For the same system (ωn = 5, ζ = 0.6), find the peak time and the 2% settling time.

Solution. The damped frequency is ωd = ωn√(1−ζ²) = 5×0.8 = 4 rad/s.

tp = π/ωd = π/4 = 0.785 s.

ts = 4/(ζωn) = 4/3 = 1.33 s.

Example 3 — range of K for stability (Routh)

A unity-feedback system has G(s) = K / [s(s+1)(s+2)]. Find the range of K for stability.

Solution. Characteristic equation: s(s+1)(s+2) + K = s³ + 3s² + 2s + K = 0. Routh array first column: s³: 1; s²: 3; s¹: (3×2 − K)/3 = (6−K)/3; s&sup0;: K.

For stability all must be positive: (6−K)/3 > 0 and K > 0, giving 0 < K < 6. (At K = 6 the system is marginally stable.)

Example 4 — steady-state error to a ramp

A unity-feedback system has G(s) = 20 / [s(s+5)]. Find the steady-state error for a unit-ramp input.

Solution. This is a type-1 system, so the velocity constant governs the ramp error. Kv = lims→0 sG(s) = 20/5 = 4.

ess = 1/Kv = 0.25.

Example 5 — first-order time constant

A first-order system has closed-loop transfer function C/R = 10/(s + 2). Find the time constant and the 2% settling time.

Solution. Writing it as K/(τs + 1): divide through by 2 to get 5/(0.5s + 1), so τ = 0.5 s.

The 2% settling time is 4τ = 4×0.5 = 2 s (and the DC gain is 10/2 = 5).

Key formulas

FORMULA SHEET

Closed loop:   C/R = G/(1 + GH)
Standard 2nd order:   ωn²/(s² + 2ζωns + ωn²)
Overshoot:   Mp = e−ζπ/√(1−ζ²)
Peak time:   tp = π/(ωn√(1−ζ²))   |   Settling: ts = 4/(ζωn)
Error constants:   Kp, Kv = lim sG H,   ess,ramp = 1/Kv
Stability:   Routh first column all same sign

Common mistakes to avoid

  • Reading ζ without matching the standard form. Always identify ωn from the constant term first, then use 2ζωn = coefficient of s.
  • Using the wrong error constant. Step error uses Kp, ramp uses Kv, parabola uses Ka — and the system type decides which are finite.
  • Forgetting the pole at the origin when counting the system type; s in the denominator makes it type 1.
  • Sign errors in the Routh array — a single wrong subtraction flips the stability conclusion.
  • Confusing gain margin and phase margin, or reading them at the wrong crossover frequency.

GATE ROBOTICS & AUTOMATION 2027

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Frequently asked questions

What is the damping ratio and why does it matter?

The damping ratio ζ measures how oscillatory a second-order system is. ζ < 1 is underdamped (overshoots), ζ = 1 is critically damped, and ζ > 1 is overdamped. It directly sets the percentage overshoot and, with ωn, the settling time.

When is a system marginally stable?

A system is marginally stable when it has poles exactly on the imaginary axis — in a Routh array this shows up as a complete row of zeros. It oscillates with constant amplitude rather than decaying or growing.

How does system type affect steady-state error?

The system type is the number of open-loop poles at the origin. A higher type reduces steady-state error: a type-1 system has zero step error but a finite ramp error, while a type-0 system has a finite step error.

This solved-problems guide is part of the complete GATE RA 2027 Syllabus overview. For the full topic breakdown, see the Control Systems syllabus guide.

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