GATE 2023 EE Q.13: solution and answer

GATE 2023 · EEQ.13 · 1 mark · MCQ

In the Nyquist plot of the open-loop transfer function
$G (s) H (s) = \frac{3 s + 5}{s - 1}$
corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in $s$ -plane is mapped into a point at

GATE 2023 EE question 13: Single-loop negative feedback block diagram. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
$G (s) H (s) = \infty$
(B)
$G (s) H (s) = 0$
(C)
$G (s) H (s) = 3$
(D)
$G (s) H (s) = - 5$

Answer and solution

ANSWER

The infinite semicircular arc maps to the single point $G (s) H (s) = 3$ . Option (C).

Official answer: CSource: Official answer key, GATE 2023 EE (IIT Kanpur)

The infinite arc of the Nyquist contour: the large semicircle is $s = R e^{j θ}$ with $R \to \infty$ and $θ$ running from $+ 9 0^{\circ}$ to $- 9 0^{\circ}$ (clockwise). Its image is $lim_{∣ s ∣ \to \infty} G (s) H (s)$ .
Compare the degrees: the numerator and the denominator are both degree 1, so $G (s) H (s)$ is proper but not strictly proper. The limit is therefore a finite, nonzero constant, which rules out $0$ and $\infty$ .
Divide the numerator and denominator by $s$ : $∣ s ∣ \to \infty lim \frac{3 s + 5}{s - 1} = ∣ s ∣ \to \infty lim \frac{3 + 5/ s}{1 - 1/ s} = \frac{3}{1} = 3$
Both $5/ s$ and $1/ s$ go to zero as $R \to \infty$ whatever the value of $θ$ , so every point on the arc tends to the same value. The whole infinite arc collapses to the single point $G (s) H (s) = 3$ , the ratio of the leading coefficients.
Why not $- 5$ : $- 5$ is $G (0) H (0) = \frac{5}{- 1}$ . That is the image of the point $s = 0$ on the $j ω$ axis, not of the arc at infinity.

KEY IDEA

On the Nyquist contour, the arc at infinity maps to $lim_{∣ s ∣ \to \infty} G (s) H (s)$ . If the numerator and denominator have the same degree, that limit is the ratio of their leading coefficients, so the whole arc shrinks to one finite point on the real axis. If the function is strictly proper, the point is the origin. Here the ratio is $3/1 = 3$ .

WATCH OUT FOR

The usual mistake is to pick $0$ out of the habit that "the infinite arc maps to the origin". That is only true for strictly proper functions, and this one has equal degrees. The other trap is $- 5$ , which comes from putting $s = 0$ into the function. That gives the start of the $j ω$ -axis image, not the image of the arc at infinity.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 11 Nyquist questions, with notes

GATE 2023 EE Q.13: solution and answer

Answer and solution

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