GATE 2022 EC Q.22: solution and answer

GATE 2022 · ECQ.22 · 1 mark · MCQ

The root-locus plot of a closed-loop system with unity negative feedback and transfer function $K G (s)$ in the forward path is shown in the figure. Note that $K$ is varied from 0 to $\infty$ .

Select the transfer function $G (s)$ that results in the root-locus plot of the closed-loop system as shown in the figure.

GATE 2022 EC question 22: s-plane with horizontal axis 'Re s' and vertical axis 'Im s'. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$G (s) = \frac{1}{( s + 1 ) ^{5}}$
(B)
$G (s) = \frac{1}{s ^{5} + 1}$
(C)
$G (s) = \frac{s - 1}{( s + 1 ) ^{6}}$
(D)
$G (s) = \frac{s + 1}{s ^{6} + 1}$

Answer and solution

ANSWER

The answer is (A): $G (s) = \frac{1}{( s + 1 ) ^{5}}$ . A fifth-order pole at $s = - 1$ with no zeros gives five straight branches leaving $- 1$ at $\pm 3 6^{\circ}, \pm 10 8^{\circ}, 18 0^{\circ}$ .

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Read the plot. Every root-locus branch starts at an open-loop pole (at $K = 0$ ), and there is one branch for each pole. The figure has five branches, all leaving the single point $s = - 1$ . So $G (s)$ has $n = 5$ poles, and all five are at $s = - 1$ : a pole of order 5. No branch ends at a finite point, so there is no finite zero and $m = 0$ .
Count the asymptotes. The five straight branches are $7 2^{\circ} = \frac{36 0 ^{\circ}}{5}$ apart, at $\pm 3 6^{\circ}, \pm 10 8^{\circ}, 18 0^{\circ}$ . The asymptote-angle rule $θ_{k} = \frac{( 2 k + 1 ) 18 0 ^{\circ}}{n - m}$ with $n - m = 5$ gives exactly these angles for $k = 0, 1, 2, 3, 4$ .
Centroid check. For option A, $σ_{A} = \frac{\sum p - \sum z}{n - m} = \frac{5 ( - 1 ) - 0}{5} = - 1$ . That is the point where the branches meet.
Real-axis rule. For $(s + 1)^{5}$ , any real point to the left of $- 1$ has 5 poles to its right. That is an odd count, so the point is on the locus. A point between $- 1$ and $0$ has no poles or zeros to its right, so it is not. This matches the figure, where the only real-axis branch runs left from $- 1$ .
Check the branches exactly. $1 + \frac{K}{( s + 1 ) ^{5}} = 0 \Rightarrow (s + 1)^{5} = - K = K e^{j (2 k + 1) 18 0^{\circ}} \Rightarrow s = - 1 + K^{1/5} e^{j (2 k + 1) 3 6^{\circ}}, k = 0, 1, 2, 3, 4$ As $K$ goes from $0$ to $\infty$ , these are five straight rays from $- 1$ at $3 6^{\circ}, 10 8^{\circ}, 18 0^{\circ}, 25 2^{\circ}, 32 4^{\circ}$ . That is exactly the plot, so A is correct.
Rule out B. The poles of $s^{5} + 1$ are $e^{j (2 k + 1) 3 6^{\circ}}$ , five points spread around the unit circle centred on the origin. Only one of them is at $- 1$ . The five branches would start from five different points, not from one point.
Rule out C and D. Both have a sixth-degree denominator, so $n = 6$ and there would be six branches, but the figure shows five. There are further mismatches. In C the centroid is $σ_{A} = \frac{6 ( - 1 ) - ( + 1 )}{5} = - 1.4 \neq = - 1$ , and the zero at $+ 1$ would put the segment $(- 1, 1)$ of the real axis on the locus, with a branch ending at $s = + 1$ . In D the poles of $s^{6} + 1$ are spread around the unit circle, and the zero at $- 1$ means a branch would end at $- 1$ rather than leave it. The figure shows none of these.

KEY IDEA

If every open-loop pole is stacked at one point and there are no zeros, $1 + \frac{K}{( s + a ) ^{n}} = 0$ becomes $(s + a)^{n} = - K$ . The whole locus is then $n$ straight rays from $s = - a$ at angles $\frac{( 2 k + 1 ) 18 0 ^{\circ}}{n}$ . Count the rays to get $n$ , and read the common point to get $a$ .

WATCH OUT FOR

Picking (B). The poles of $s^{5} + 1$ lie at exactly the angles $\pm 3 6^{\circ}, \pm 10 8^{\circ}, 18 0^{\circ}$ , the same angles as the rays in the figure, which makes it look right. But those are five separate starting points on the unit circle around the origin, not five directions leaving $s = - 1$ . The branches of (B) would start at five different places, while every branch in the figure starts at $- 1$ . A second slip is to overlook that (C) and (D) have six poles, which means six branches, when the figure shows five.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 7 Root locus questions, with notes

GATE 2022 EC Q.22: solution and answer

Answer and solution

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