GATE 2022 EE Q.38: solution and answer

GATE 2022 · EEQ.38 · 2 marks · MCQ

The open loop transfer function of a unity gain negative feedback system is given as

G (s) = \frac{1}{s ( s + 1 )} .

The Nyquist contour in the $s$ -plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point $(- 1 + j 0)$ by the Nyquist plot of $G (s)$ , corresponding to the Nyquist contour, is denoted as $N$ . Then $N$ equals to

GATE 2022 EE question 38: The s-plane has a horizontal axis labelled Re(s) with an arrow pointing right, and a vertical axis labelled Im(s) with an… — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
0
(B)
1
(C)
2
(D)
3

Answer and solution

ANSWER

The Nyquist plot goes round $- 1 + j 0$ exactly once, counter-clockwise, so $N = 1$ : option (B).

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Open-loop poles inside the contour. $G (s)$ has poles at $s = 0$ and $s = - 1$ . The small indentation bulges into the left half plane, so the pole at the origin lies inside the contour. The indentation passes to the right of $- 1$ , so that pole lies outside. Therefore $P = 1$ .
Closed-loop poles inside the contour. Setting $1 + G (s) = 0$ gives $s^{2} + s + 1 = 0$ , so $s = - \frac{1}{2} \pm j \frac{3}{2}$ . Both roots are in the left half plane at distance $1$ from the origin, well outside the small neighbourhood of the origin. Therefore $Z = 0$ .
Nyquist criterion. The contour is traversed clockwise. Count encirclements of $- 1 + j 0$ as positive when they are clockwise. Then $N = Z - P = 0 - 1 = - 1,$ which means one encirclement, counter-clockwise.
Check by mapping, segment $ω : 0^{+} \to \infty$ . $G (j ω) = \frac{1}{- ω ^{2} + j ω} = - \frac{1}{1 + ω ^{2}} - j \frac{1}{ω ( 1 + ω ^{2} )} .$ The plot starts at $- j \infty$ along the asymptote $Re = - 1$ and ends at the origin. For every $ω > 0$ the real part lies strictly between $- 1$ and $0$ and the imaginary part is negative, so this branch never crosses the negative real axis.
Small left-side arc. On the arc $s = ε e^{j θ}$ , $θ$ goes $27 0^{\circ} \to 18 0^{\circ} \to 9 0^{\circ}$ , which is clockwise about the origin. There $G \approx \frac{1}{s} = \frac{1}{ε} e^{- j θ}$ , and its angle goes $+ 9 0^{\circ} \to 18 0^{\circ} \to 27 0^{\circ}$ . Check at the ends: $G (- j ε) \approx + j / ε$ , $G (- ε) \approx - 1/ ε$ , $G (+ j ε) \approx - j / ε$ . So the arc maps to an infinite-radius arc traced counter-clockwise from $+ j \infty$ through $- \infty$ to $- j \infty$ . It passes to the left of $- 1$ .
Close the plot. The $ω < 0$ half is the mirror image of the $ω > 0$ branch: it runs from the origin up to $+ j \infty$ . The large $s$ -plane semicircle maps to the origin. In order, the closed curve goes origin $\to + j \infty \to - \infty \to - j \infty \to$ origin. Seen from $- 1$ , the angle increases by a full $36 0^{\circ}$ , so the curve goes round $- 1 + j 0$ exactly once, counter-clockwise. This agrees with $N = - 1$ from the criterion. The number of encirclements is $1$ , which is option (B).

KEY IDEA

Where the contour is indented decides $P$ . Here the indentation goes into the left half plane, so the pole at the origin is inside the contour and $P = 1$ . The closed loop is stable, so $Z = 0$ . Then $N = Z - P = - 1$ : one counter-clockwise encirclement of $- 1$ . On the plot, the small left-side arc maps to an infinite arc swept counter-clockwise on the left, and that arc wraps round $- 1$ .

WATCH OUT FOR

Using the usual textbook contour, which indents to the right of the origin. That leaves the origin pole outside, gives $P = 0$ , and leads to $N = 0$ (option A). With the indentation drawn to the left, $P = 1$ . The small arc then maps to an infinite arc on the left, passing through $- \infty$ , instead of one on the right, and it encircles $- 1$ once. A second trap is the sign: $N = - 1$ means one counter-clockwise encirclement. The number of encirclements is still $1$ .

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 11 Nyquist questions, with notes

GATE 2022 EE Q.38: solution and answer

Answer and solution

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