GATE 2020 EC Q.11: solution and answer

GATE 2020 · ECQ.11 · 1 mark · MCQ

The pole-zero map of a rational function $G (s)$ is shown below. When the closed contour $Γ$ is mapped into the $G (s)$ -plane, then the mapping encircles

GATE 2020 EC question 11: An s-plane with a horizontal axis labelled Re and a vertical axis labelled Im. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
the origin of the $G (s)$ -plane once in the counter-clockwise direction.
(B)
the origin of the $G (s)$ -plane once in the clockwise direction.
(C)
the point $- 1 + j 0$ of the $G (s)$ -plane once in the counter-clockwise direction.
(D)
the point $- 1 + j 0$ of the $G (s)$ -plane once in the clockwise direction.

Answer and solution

ANSWER

The image of $Γ$ encircles the origin of the $G (s)$ -plane once in the clockwise direction. The answer is option (B).

Official answer: BSource: Official answer key, GATE 2020 EC (IIT Delhi)

Direction of $Γ$ . The only arrowhead on the curve is on the upper-left arc, pointing up and to the right along the curve. Going up the left side and then across the top is clockwise. The second arrow, at the lower left, starts at the label $Γ$ and points across the curve rather than along it. It only marks which curve is $Γ$ and says nothing about direction.
Count what $Γ$ encloses. Zeros inside are the real-axis zero just inside the left edge and the complex-conjugate pair in the left half-plane, so $Z = 3$ . Poles inside are the two real-axis poles on either side of the origin, so $P = 2$ . The far-left zero, the real pole just beyond the right edge and the right-half-plane complex poles all lie outside $Γ$ . No pole or zero lies on $Γ$ , so $G (Γ)$ is a closed curve that never passes through the origin.
Principle of the argument. Write $G (s) = K \frac{\prod _{i} ( s - z _{i} )}{\prod _{k} ( s - p _{k} )}$ . Then $∠ G (s) = ∠ K + \sum_{i} ∠ (s - z_{i}) - \sum_{k} ∠ (s - p_{k})$ . As $s$ goes once round $Γ$ , the vector $s - a$ makes one full turn if $a$ is inside $Γ$ . If $a$ is outside, the vector ends at the angle it started from, so its net change is zero. Only the enclosed poles and zeros change $∠ G (s)$ .
Angles are measured counter-clockwise positive, and $Γ$ is clockwise, so each enclosed factor turns by $- 2 π$ . Zeros enter $∠ G$ with a plus sign and poles with a minus sign: $Δ∠ G = Z (- 2 π) - P (- 2 π) = - 2 π (Z - P) = - 2 π (3 - 2) = - 2 π$
Net encirclement. A net angle change of $- 2 π$ means the image point $G (s)$ goes round the origin of the $G (s)$ -plane once clockwise, i.e. $N = Z - P = 1$ clockwise encirclement. This is option (B).
Eliminate the rest. Option (A) names the right point but the wrong direction. You get (A) if you take $Γ$ as counter-clockwise or use $P - Z$ for clockwise encirclements. For (C) and (D): an encirclement of $- 1 + j 0$ by $G (Γ)$ is an encirclement of the origin by $1 + G (Γ)$ . That count is the number of zeros of $1 + G (s)$ inside $Γ$ minus the number of its poles there. The poles of $1 + G$ are the poles of $G$ . Its zeros depend on the gain $K$ , and a pole-zero map does not give $K$ . So the figure settles neither (C) nor (D), and (B) is the only option it determines.

KEY IDEA

By the principle of the argument, when $s$ goes once clockwise round a closed contour, the image $G (s)$ goes clockwise round the origin $N = Z - P$ times. Here $Z$ and $P$ count only the zeros and poles of $G$ inside the contour. $Γ$ is clockwise and encloses 3 zeros and 2 poles, so $N = 1$ clockwise. A pole-zero map tells you about encirclements of the origin only. Encirclements of $- 1 + j 0$ depend on the zeros of $1 + G (s)$ , which change with the gain.

WATCH OUT FOR

One mistake is to count every pole and zero on the map instead of only those inside $Γ$ . Another is to read the arrow beside the label $Γ$ as a direction arrow. A third is to use $P - Z$ where the rule for clockwise encirclements needs $Z - P$ . The last two turn the correct answer (B) into (A). Candidates who go straight to Nyquist habits also pick (C) or (D), because they expect every encirclement question to be about $- 1 + j 0$ . That point can't be decided from a pole-zero map with no gain given.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 11 Nyquist questions, with notes

GATE 2020 EC Q.11: solution and answer

Answer and solution

More Nyquist questions