GATE 2018 EC Q.42: solution and answer

GATE 2018 · ECQ.42 · 2 marks · NAT

The figure below shows the Bode magnitude and phase plots of a stable transfer function $G (s) = \frac{n _{0}}{s ^{3} + d _{2} s ^{2} + d _{1} s + d _{0}}$ .

Consider the negative unity feedback configuration with gain $k$ in the feedforward path. The closed loop is stable for $k < k_{0}$ . The maximum value of $k_{0}$ is ______.

GATE 2018 EC question 42: Left: Bode plots sharing a common frequency axis. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

$k_{0} = 0.1$ . At the phase crossover frequency $∣ G (j ω_{p c}) ∣ = 20 dB = 10$ , so the closed loop stays stable only while $10 k < 1$ .

Official answer: 0.1Source: Official answer key, GATE 2018 EC (IIT Guwahati)

The open-loop transfer function is $k G (s)$ (with $k > 0$ ). The question says $G (s)$ is stable, so $k G (s)$ has $P = 0$ poles in the right half-plane. By the Nyquist criterion the closed loop is stable exactly when the polar plot of $k G (j ω)$ does not encircle the point $- 1$ .
Phase crossover frequency $ω_{p c}$ : this is where $∠ G (j ω) = - 18 0^{\circ}$ . On the phase plot, the curve starts at $0^{\circ}$ and falls monotonically towards $- 27 0^{\circ}$ , so it crosses $- 18 0^{\circ}$ only once, at the vertical dashed line. That means the polar plot crosses the negative real axis only once, at $ω_{p c}$ .
Now read the magnitude on the same vertical dashed line: $∣ G (j ω_{p c}) ∣ = 20 dB$ , so $∣ G (j ω_{p c}) ∣ = 1 0^{20/20} = 10.$
So $G$ alone (with $k = 1$ ) crosses the negative real axis at $- 10$ , which is outside $- 1$ , and the loop would be unstable ( $GM = - 20 dB$ ). The gain $k$ scales the whole polar plot, and it has to bring this crossing inside $- 1$ .
With gain $k$ , the negative-real-axis crossing sits at $- 10 k$ . There is no encirclement of $- 1$ as long as $k ∣ G (j ω_{p c}) ∣ < 1 \Rightarrow 10 k < 1 \Rightarrow k < 0.1$
So $k_{0} = \frac{1}{∣ G ( j ω _{p c} ) ∣} = 0.1$ . The $36 dB$ DC level and the resonant peak do not matter here: they occur at frequencies where the phase is still above $- 18 0^{\circ}$ , so they never touch the negative real axis.

KEY IDEA

If the open loop is stable ( $P = 0$ ) and the phase crosses $- 18 0^{\circ}$ exactly once, the largest stable gain is the gain margin as a plain ratio: $k_{0} = 1/∣ G (j ω_{p c}) ∣$ . Read the magnitude at the phase crossover frequency, not at the gain crossover and not at the peak.

WATCH OUT FOR

There are three common slips. (1) Reading the wrong magnitude: the 36 dB DC level (giving $k_{0} = 1 0^{- 36/20} \approx 0.016$ ), the resonant peak, or the 0 dB gain crossover, instead of the magnitude where the phase is $- 18 0^{\circ}$ . (2) Converting 20 dB to 20 instead of 10: magnitude in dB is $20 lo g_{10} ∣ G ∣$ . (3) Forgetting the reciprocal and answering $k_{0} = 10$ . The magnitude at $ω_{p c}$ is already greater than 1, so $k$ must shrink the loop gain, not boost it.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 27 Bode & margins questions, with notes

GATE 2018 EC Q.42: solution and answer

Answer and solution

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