GATE 2025 IN Q.59: solution and answer

GATE 2025 · INQ.59 · 2 marks · NAT

A feedback control system is shown in the figure.

The maximum allowable value of $n$ such that the output $y (t)$ , due to any step disturbance signal $d (t)$ , becomes zero at steady-state, is ________ (in integer).

GATE 2025 IN question 59: Block diagram with one summing junction, both inputs marked +. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

Applying the final value theorem formally, as the question intends, gives n = 1. Strictly, though, the closed loop (characteristic polynomial s^3 + s^2 + 1) has two right-half-plane poles, so y(t) grows without bound for every n and the condition cannot be met.

Official answer: marks to all candidatesSource: Official answer key, GATE 2025 IN (IIT Roorkee)

Loop equation. The junction adds $D (s) / s^{n}$ and $- \frac{1}{s ^{2}} Y (s)$ , and its output passes through $\frac{1}{s + 1}$ : $Y (s) = \frac{1}{s + 1} [\frac{D ( s )}{s ^{n}} - \frac{1}{s ^{2}} Y (s)]$ The $- 1/ s^{2}$ block feeding a $+$ input is ordinary negative feedback with $H (s) = 1/ s^{2}$ .
Disturbance-to-output transfer function. Collect the $Y (s)$ terms and multiply numerator and denominator by $s^{2} (s + 1)$ : $\frac{Y ( s )}{D ( s )} = \frac{\frac{1}{s ^{n} ( s + 1 )}}{1 + \frac{1}{s ^{2} ( s + 1 )}} = \frac{s ^{2 - n}}{s ^{3} + s ^{2} + 1}$
Check stability before using the final value theorem. The characteristic polynomial is $s^{3} + s^{2} + 0 \cdot s + 1$ . Its $s$ coefficient is zero, so it fails the necessary condition for stability. Routh array: $s^{3} s^{2} s^{1} s^{0} : 10 : 11 : \frac{1 \cdot 0 - 1 \cdot 1}{1} = - 1 : 1$ The first column $1, 1, - 1, 1$ changes sign twice, so there are two right-half-plane poles. The roots are $s \approx - 1.466$ and $s \approx 0.233 \pm j 0.793$ .
What this means for $y (t)$ . A step disturbance $D (s) = A / s$ ( $A \neq = 0$ ) gives $Y (s) = \frac{A s ^{1 - n}}{s ^{3} + s ^{2} + 1}$ The numerator is just a power of $s$ , so it can never cancel the poles at $0.233 \pm j 0.793$ . So $y (t)$ always contains a term that grows like $e^{0.233 t} sin (0.793 t + ϕ)$ with a nonzero coefficient. Strictly, $y (t)$ grows without bound for every $n$ , and no value of $n$ makes it settle to zero. The question is flawed as printed.
The method the question intends (final value theorem applied formally). $y_{ss} = s \to 0 lim s Y (s) = s \to 0 lim \frac{A s ^{2 - n}}{s ^{3} + s ^{2} + 1} = s \to 0 lim A s^{2 - n}$ because the denominator tends to $1$ . This limit is zero only if $2 - n > 0$ , that is $n < 2$ .
Test the boundary values formally. $n = 1$ : $lim A s = 0$ . $n = 2$ : $lim A = A \neq = 0$ . $n \geq 3$ : $A s^{2 - n} \to \infty$ . Every $n \leq 1$ gives zero, so the largest allowed integer is $n = 1$ .
Conclusion. The intended answer, from applying the final value theorem formally, is $n = 1$ . Strictly speaking, the closed loop $s^{3} + s^{2} + 1$ is unstable, so the theorem does not apply and no $n$ satisfies the condition. The official key is MTA, so neither reading can lose marks.

KEY IDEA

Rejecting a step disturbance needs the disturbance-to-output transfer function to have at least one zero at the origin. Here the loop's feedback $1/ s^{2}$ puts $s^{2}$ in the numerator of $Y / D$ , and the disturbance path's $1/ s^{n}$ removes $n$ of those zeros. That leaves $s^{2 - n}$ , and a step (one more $1/ s$ ) is rejected only if $2 - n \geq 1$ . A second rule matters just as much: the final value theorem only holds for a stable closed loop, so do a Routh check first. Here it fails, because $s^{3} + s^{2} + 1$ has no $s$ term.

WATCH OUT FOR

(1) Using the final value theorem without checking stability. The characteristic polynomial $s^{3} + s^{2} + 1$ has a missing coefficient, so the loop is unstable and the theorem's result is only formal. (2) Answering $n = 2$ by asking only for a finite steady state ( $2 - n \geq 0$ ) rather than a zero one ( $2 - n > 0$ ). (3) Reading the $- 1/ s^{2}$ block going into a $+$ junction as positive feedback. That gives $s^{3} + s^{2} - 1$ , which is also unstable, so it changes the algebra but not the formal answer.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 22 Time response questions, with notes

GATE 2025 IN Q.59: solution and answer

Answer and solution

More Time response questions