GATE 2026 IN Q.60: solution and answer

GATE 2026 · INQ.60 · 2 marks · NAT

The system $G (s) = \frac{K}{( s + 1 ) ^{4}}$ has a gain margin of 20 dB. The value of $K$ is ______. (rounded off to two decimal places)

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Answer and solution

ANSWER

K = 0.40

Official answer: 0.39 to 0.41Source: Official answer key, GATE 2026 IN (IIT Guwahati)

Phase of the open-loop transfer function. All four poles sit at $s = - 1$ , so $∠ G (j ω) = - 4 tan^{- 1} ω$ This falls steadily from $0^{\circ}$ at $ω = 0$ towards $- 36 0^{\circ}$ as $ω \to \infty$ , so it passes through $- 18 0^{\circ}$ exactly once.
Phase-crossover frequency (where the phase is $- 18 0^{\circ}$ ): $4 tan^{- 1} ω_{p c} = 18 0^{\circ} \Rightarrow tan^{- 1} ω_{p c} = 4 5^{\circ} \Rightarrow ω_{p c} = 1 rad/s$
Magnitude at $ω_{p c}$ : $∣ G (j ω_{p c}) ∣ = \frac{K}{( ω _{p c}^{2} + 1 ) ^{4}} = \frac{K}{( 2 ) ^{4}} = \frac{K}{4}$
Use the definition of gain margin: $GM = - 20 lo g_{10} ∣ G (j ω_{p c}) ∣ = 20 dB \Rightarrow ∣ G (j ω_{p c}) ∣ = 1 0^{- 1} = 0.1$
Solve for $K$ : $\frac{K}{4} = 0.1 \Rightarrow K = 0.40$

KEY IDEA

Gain margin is read at the phase-crossover frequency. With four identical poles at $s = - 1$ , each pole adds $- 4 5^{\circ}$ at $ω = 1 rad/s$ , so the total phase is $- 18 0^{\circ}$ there. A 20 dB gain margin means $∣ G (j ω_{p c}) ∣ = 0.1$ , which gives $K /4 = 0.1$ .

WATCH OUT FOR

Setting $∣ G (j ω_{p c}) ∣ = 10$ instead of $0.1$ (sign slip in $GM = - 20 lo g_{10} ∣ G ∣$ ), which gives $K = 40$ . Another slip is writing $(2)^{4}$ as $2$ instead of $4$ , which gives $K = 0.2$ . Some students also take the gain-crossover frequency ( $∣ G ∣ = 1$ ) in place of the phase-crossover frequency.

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GATE 2026 IN Q.60: solution and answer

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