GATE 2026 IN Q.59: solution and answer

GATE 2026 · INQ.59 · 2 marks · NAT

A linear time invariant (LTI) system has a transfer function

G (s) = \frac{10 ( s + 1 )}{s ( s ^{2} + 2 s + 5 )}

The system is placed in unity negative feedback configuration.

For a unit ramp reference, the steady state error is ________.

(rounded off to two decimal places)

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Answer and solution

ANSWER

The steady-state error to a unit ramp is $e_{ss} = 1/ K_{v} = 1/2 = 0.50$ .

Official answer: 0.49 to 0.51Source: Official answer key, GATE 2026 IN (IIT Guwahati)

Check closed-loop stability first, because the final value theorem only applies to a stable loop. The characteristic equation is $1 + G (s) = 0$ : $s (s^{2} + 2 s + 5) + 10 (s + 1) = s^{3} + 2 s^{2} + 15 s + 10 = 0$
Routh array: row $s^{3}$ : $1, 15$ ; row $s^{2}$ : $2, 10$ ; row $s^{1}$ : $\frac{2 \cdot 15 - 1 \cdot 10}{2} = 10$ ; row $s^{0}$ : $10$ . The first column $1, 2, 10, 10$ has no sign change, so the closed loop is stable.
Find the system type. $G (s)$ has one pole at the origin, so the loop is type 1. A type-1 loop follows a ramp with a finite, non-zero steady-state error.
Velocity error constant $K_{v}$ : $K_{v} = s \to 0 lim s G (s) = s \to 0 lim \frac{10 ( s + 1 )}{s ^{2} + 2 s + 5} = \frac{10 \cdot 1}{5} = 2$
Steady-state error for a unit ramp $R (s) = 1/ s^{2}$ : $e_{ss} = s \to 0 lim \frac{s R ( s )}{1 + G ( s )} = s \to 0 lim \frac{1}{s + s G ( s )} = \frac{1}{K _{v}} = \frac{1}{2} = 0.50$

KEY IDEA

In a unity-feedback loop, the ramp error depends only on the velocity constant: $e_{ss} = 1/ K_{v}$ with $K_{v} = lim_{s \to 0} s G (s)$ . That formula is only valid once you have confirmed the closed loop is stable, which here takes a quick Routh check on $s^{3} + 2 s^{2} + 15 s + 10$ .

WATCH OUT FOR

One slip is to cancel the $s$ wrongly and use the zero at $s = - 1$ or the numerator gain alone, giving $K_{v} = 10$ and $e_{ss} = 0.1$ . The correct value is $K_{v} = 10 \cdot 1/5 = 2$ , because the constant term $5$ of $s^{2} + 2 s + 5$ divides the gain. Another mistake is to skip the stability check. If the loop were unstable, $1/ K_{v}$ would give a meaningless number.

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GATE 2026 IN Q.59: solution and answer

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