GATE 2026 EC Q.40: solution and answer

GATE 2026 · ECQ.40 · 2 marks · MCQ

For the control system shown in the Figure, the transfer function of a plant, $G (s) = \frac{1}{( s + 1 ) ( s + 2 )}$ is connected in cascade with a compensator $C (s) = K (s + α)$ , where $K$ and $α$ are positive real valued constants.

Which of the following pairs $(K, α)$ represent the correct values for the closed loop system to have poles at $(- 3 \pm j 5)$ ?

GATE 2026 EC question 40: Unity negative feedback block diagram. — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
2, 3
(B)
3, 4
(C)
2, 4
(D)
3, 3

Answer and solution

ANSWER

$(K, α) = (3, 4)$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

With unity negative feedback and $C (s)$ in cascade with $G (s)$ , the open-loop transfer function is $C (s) G (s) = \frac{K ( s + α )}{( s + 1 ) ( s + 2 )}$
The closed-loop poles are the roots of $1 + C (s) G (s) = 0$ : $(s + 1) (s + 2) + K (s + α) = 0 \Rightarrow s^{2} + (3 + K) s + (2 + K α) = 0$
The characteristic polynomial the poles $- 3 \pm j 5$ call for is $(s + 3)^{2} + (5)^{2} = s^{2} + 6 s + 14$
Match the $s$ coefficients: $3 + K = 6 \Rightarrow K = 3$ .
Match the constant terms: $2 + K α = 14 \Rightarrow 3 α = 12 \Rightarrow α = 4$ .
Check: $s^{2} + 6 s + 14 = 0$ gives $s = \frac{- 6 \pm 36 - 56}{2} = - 3 \pm j 5$ , as required. Both $K$ and $α$ are positive, so $(K, α) = (3, 4)$ , which is option (B).

KEY IDEA

For a second-order closed loop, the complex poles $- σ \pm j ω_{d}$ fix the characteristic polynomial exactly as $s^{2} + 2 σ s + (σ^{2} + ω_{d}^{2})$ . The compensator gives two free parameters, one for each coefficient. Here $K$ alone sets the $s$ coefficient ( $3 + K$ ), and once $K$ is known, $α$ sets the constant term ( $2 + K α$ ).

WATCH OUT FOR

Taking the constant term of the desired polynomial as $3^{2} - 5 = 4$ or as $9$ , when it is $3^{2} + (5)^{2} = 14$ . The other mistake is forgetting that $K$ multiplies $α$ in the constant term, which gives $2 + α = 14$ . Every distractor is close. Option (D), $(3, 3)$ , gets $K$ right but gives $s^{2} + 6 s + 11$ , whose poles are $- 3 \pm j 2$ . Options (A) and (C) have $K = 2$ , which gives an $s$ coefficient of 5 and a real part of $- 2.5$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 10 Compensators & PID questions, with notes

GATE 2026 EC Q.40: solution and answer

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