GATE 2025 IN Q.50: solution and answer

GATE 2025 · INQ.50 · 2 marks · MSQ

The plant in the feedback control system shown in the figure is $P (s) = \frac{a}{s ^{2} - b ^{2}}$ , where $a > 0$ and $b > 0$ . The type(s) of controller $C (s)$ that CANNOT stabilize the plant is/are

GATE 2025 IN question 50: Unity negative feedback loop. — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
proportional (P) controller
(B)
integral (I) controller
(C)
proportional-integral (PI) controller
(D)
proportional-derivative (PD) controller

Answer and solution

ANSWER

P, I and PI controllers cannot stabilize $P (s) = \frac{a}{s ^{2} - b ^{2}}$ . Only a PD controller with $K_{d} > 0$ and $K_{p} > b^{2} / a$ can. Answer: (A), (B), (C).

Official answer: A, B, CSource: Official answer key, GATE 2025 IN (IIT Roorkee)

Closed-loop characteristic equation for unity negative feedback: $1 + C (s) P (s) = 0 \Rightarrow s^{2} - b^{2} + a C (s) = 0$ (after clearing any denominator of $C (s)$ ). The plant on its own has poles at $s = \pm b$ , so one pole is in the right half-plane and the controller has to move it. Test used throughout: a polynomial can have all its roots in the open left half-plane only if every coefficient is present and all have the same sign. For a second-order polynomial this condition is also sufficient.
P controller, $C (s) = K_{p}$ : $s^{2} + (a K_{p} - b^{2}) = 0$ There is no $s^{1}$ term. If $a K_{p} < b^{2}$ the roots are $\pm b^{2} - a K_{p}$ , one of them real and positive. If $a K_{p} = b^{2}$ there is a double root at $s = 0$ . If $a K_{p} > b^{2}$ the roots are $\pm j a K_{p} - b^{2}$ , which only gives sustained oscillation (marginal stability). No real $K_{p}$ puts both roots in the left half-plane, so P cannot stabilize the plant.
I controller, $C (s) = \frac{K _{i}}{s}$ : multiply through by $s$ : $s^{3} + 0 \cdot s^{2} - b^{2} s + a K_{i} = 0$ The $s^{2}$ term is missing, and the $s$ coefficient is $- b^{2} < 0$ whatever $K_{i}$ is. The necessary condition fails for every $K_{i}$ , so I cannot stabilize the plant.
PI controller, $C (s) = K_{p} + \frac{K _{i}}{s}$ : multiply through by $s$ : $s^{3} + 0 \cdot s^{2} + (a K_{p} - b^{2}) s + a K_{i} = 0$ The $s^{2}$ coefficient is zero for every choice of $K_{p}$ and $K_{i}$ . That coefficient equals minus the sum of the three roots, so the roots always sum to zero and cannot all have negative real parts. (In the Routh array the $s^{2}$ row starts with a zero for the same reason.) So PI cannot stabilize the plant.
PD controller, $C (s) = K_{p} + K_{d} s$ : $s^{2} + a K_{d} s + (a K_{p} - b^{2}) = 0$ A second-order polynomial is stable exactly when all its coefficients are positive. Since $a > 0$ , that holds for $K_{d} > 0$ and $K_{p} > \frac{b ^{2}}{a}$ . For example, with $K_{d} > 0$ and $K_{p} = \frac{2 b ^{2}}{a}$ the equation is $s^{2} + a K_{d} s + b^{2} = 0$ , whose roots are both in the left half-plane. So PD can stabilize the plant.
Why this happens: write $C (s) = \frac{N _{c} ( s )}{D _{c} ( s )}$ , so the characteristic polynomial is $D_{c} (s) (s^{2} - b^{2}) + a N_{c} (s)$ . The plant polynomial $s^{2} - b^{2}$ has no damping ( $s^{1}$ ) term. With P, $N_{c}$ is a constant, so no $s$ term appears. With I or PI, $D_{c} = s$ gives $s^{3} - b^{2} s$ , and $a N_{c} (s)$ has degree at most 1, so nothing can fill the $s^{2}$ slot. Only the derivative term $K_{d} s$ in $N_{c}$ adds the missing $s^{1}$ coefficient.
So the controllers that CANNOT stabilize the plant are P, I and PI: options (A), (B) and (C).

KEY IDEA

Form the closed-loop characteristic polynomial $D_{c} (s) (s^{2} - b^{2}) + a N_{c} (s)$ and apply the necessary condition for stability: every coefficient present and positive. The plant has no $s^{1}$ (damping) term. P cannot add one. An integrator raises the order to three but always leaves the $s^{2}$ coefficient at zero. Only derivative action ( $K_{d} s$ ) supplies the missing term.

WATCH OUT FOR

Treating P control with $a K_{p} > b^{2}$ as stabilizing. It puts the poles at $\pm j a K_{p} - b^{2}$ , which is only marginally stable (sustained oscillation), not asymptotically stable. Another mistake is assuming PI must work because it has two gains to tune: the $s^{2}$ coefficient stays zero whatever $K_{p}$ and $K_{i}$ are. Also watch the word CANNOT, which makes some candidates mark only (D).

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 10 Compensators & PID questions, with notes

GATE 2025 IN Q.50: solution and answer

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