GATE 2025 EC Q.16: solution and answer

GATE 2025 · ECQ.16 · 1 mark · MCQ

The Nyquist plot of a system is given in the figure below. Let $ω_{P}$ , $ω_{Q}$ , $ω_{R}$ , and $ω_{S}$ be the positive frequencies at the points $P$ , $Q$ , $R$ , and $S$ , respectively.

Which one of the following statements is TRUE?

GATE 2025 EC question 16: Nyquist plot in the complex plane. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$ω_{S}$ is the gain crossover frequency and $ω_{P}$ is the phase crossover frequency
(B)
$ω_{Q}$ is the gain crossover frequency and $ω_{R}$ is the phase crossover frequency
(C)
$ω_{Q}$ is the gain crossover frequency and $ω_{S}$ is the phase crossover frequency
(D)
$ω_{S}$ is the gain crossover frequency and $ω_{Q}$ is the phase crossover frequency

Answer and solution

ANSWER

(D): $ω_{S}$ is the gain crossover frequency, because the locus crosses the unit circle at $S$ . $ω_{Q}$ is the phase crossover frequency, because the locus crosses the negative real axis at $Q$ .

Official answer: DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Definitions. The gain crossover frequency $ω_{g c}$ is where $∣ G (j ω) H (j ω) ∣ = 1$ . On a Nyquist plot, that is where the positive-frequency locus crosses the unit circle. The phase crossover frequency $ω_{p c}$ is where $∠ G (j ω) H (j ω) = - 18 0^{\circ}$ , i.e. where the locus crosses the negative real axis.
Find the positive-frequency branch. Follow the arrows from $ω = 0^{+}$ . The locus rises through $S$ , then $R$ , then $Q$ , then goes round the upper half of the small loop through $P$ and ends at the origin at $ω = + \infty$ . The large arc on the right (from $ω = 0^{-}$ to $ω = 0^{+}$ ) and the branch from the origin ( $ω = - \infty$ ) through the lower half of the loop and $Q$ up to $ω = 0^{-}$ are not positive frequencies. Along the positive branch the phase keeps falling: about $- 12 5^{\circ}$ at $S$ , $- 13 5^{\circ}$ at $R$ , $- 18 0^{\circ}$ at $Q$ and $- 22 5^{\circ}$ at $P$ .
Point $S$ gives the gain crossover. $S$ lies on the dash-dot unit circle, so $∣ G (j ω_{S}) H (j ω_{S}) ∣ = 1$ and $ω_{S} = ω_{g c}$ . The positive branch meets the unit circle only once: it is outside the circle before $S$ and stays inside after it. So this is the only gain crossover. Note that $S$ is not on the $- 13 5^{\circ}$ ray. That ray crosses the locus at $R$ , so the phase at $S$ is not $- 13 5^{\circ}$ . It is less lagging, roughly $- 12 5^{\circ}$ as drawn.
Point $Q$ gives the phase crossover. $Q$ lies on the negative real axis, so $∠ G (j ω_{Q}) H (j ω_{Q}) = - 18 0^{\circ}$ and $ω_{Q} = ω_{p c}$ . The positive branch crosses the negative real axis only here, because the upper half of the loop stays above the axis until it reaches the origin. At $Q$ the magnitude is about $0.45 \neq = 1$ , so $Q$ is not a gain crossover.
Rule out $R$ and $P$ . $R$ is on the $- 13 5^{\circ}$ ray inside the circle: $∣ G H ∣ < 1$ and $∠ G H = - 13 5^{\circ}$ . $P$ is on the $13 5^{\circ}$ ray inside the circle. The phase has already passed $- 18 0^{\circ}$ at $Q$ and is still falling, so that angle means $∠ G H = - 22 5^{\circ}$ , and again $∣ G H ∣ < 1$ . Neither point meets either crossover condition. That rules out (A), which makes $ω_{P}$ the phase crossover, and (B), which makes $ω_{R}$ the phase crossover. (B) and (C) also fail because they make $ω_{Q}$ the gain crossover, and (C) makes $ω_{S}$ the phase crossover.
Answer. $ω_{S}$ is the gain crossover frequency and $ω_{Q}$ is the phase crossover frequency, which is option (D). As a check, the gain margin is $1/∣ G (j ω_{Q}) H (j ω_{Q}) ∣ \approx 1/0.45 \approx 2.2 > 1$ . The phase margin is $18 0^{\circ} + ∠ G (j ω_{S}) H (j ω_{S})$ , roughly $5 5^{\circ}$ as drawn. The $4 5^{\circ}$ marks are not the phase margin. They mark the rays through $R$ and $P$ , and those points are distractors.

KEY IDEA

On a Nyquist plot, read each crossover from where the positive-frequency locus is, not from the angle marks. It crosses the unit circle at the gain crossover ( $∣ G H ∣ = 1$ ) and the negative real axis at the phase crossover ( $∠ G H = - 18 0^{\circ}$ ).

WATCH OUT FOR

Swapping the two definitions, which gives (C). Another error is to treat $R$ as special because it sits on the marked $4 5^{\circ}$ ray. It looks like a $4 5^{\circ}$ phase margin, but $R$ has $∣ G H ∣ < 1$ , so it is neither crossover. $P$ is on the other $4 5^{\circ}$ ray at phase $- 22 5^{\circ}$ , not $- 18 0^{\circ}$ . Some students also assume $S$ is on the $- 13 5^{\circ}$ ray and write the phase margin as $4 5^{\circ}$ . In the figure $S$ is below that ray, so its phase is less lagging than $- 13 5^{\circ}$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 27 Bode & margins questions, with notes

GATE 2025 EC Q.16: solution and answer

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