GATE 2023 IN Q.48: solution and answer

GATE 2023 · INQ.48 · 2 marks · NAT

Consider a standard negative feedback configuration with $G (s) = \frac{1}{( s - 2 ) ( s - 3 )}$ and the controller $C (s) = K_{P} + \frac{K _{I}}{s} + K_{D} s$ . The root-locus of $G (s) C (s)$ is presented in the figure below. The gain $C (j ω) = 2$ at $ω = 1$ rad/s. The value of $K_{D}$ is ______ (rounded off to one decimal place).

GATE 2023 IN question 48: Root-locus plot. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

$K_{D} = 1.0$ . The root locus forces a double zero at $s = - 1$ , so $C (s) = \frac{K _{D} ( s + 1 ) ^{2}}{s}$ , $C (j 1) = 2 K_{D} = 2$ , and $K_{D} = 1.0$ .

Official answer: 0.9 to 1.1Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Write the PID as one fraction: $C (s) = \frac{K _{D} s ^{2} + K _{P} s + K _{I}}{s}$ . So $G (s) C (s) = \frac{K _{D} s ^{2} + K _{P} s + K _{I}}{s ( s - 2 ) ( s - 3 )}$ has poles at $0, 2, 3$ and at most two zeros, the roots of the PID numerator.
Real-axis rule: a point on the real axis is on the locus if the total number of real poles and zeros to its right is odd. For a point in $(- 1, 0)$ there are 3 poles to the right, which is odd, so the segment from $0$ to $- 1$ is on the locus. The plot also shows the whole real axis left of $- 1$ on the locus. If $- 1$ were a single zero (either $K_{D} = 0$ , or a second zero further left), a point just left of $- 1$ would see $3 + 1 = 4$ poles and zeros to its right. That is even, so the point would not be on the locus, which contradicts the figure. Only one zero marker is drawn, so the second zero must also sit at $- 1$ : the PID has a double zero at $s = - 1$ , and left of $- 1$ the count is $3 + 2 = 5$ (odd).
Branch check: 3 poles and 2 finite zeros give $3 - 2 = 1$ asymptote, at $18 0^{\circ}$ . This matches the single branch running to $- \infty$ on the plot. The pole at $0$ ends on one of the zeros at $- 1$ , and the branch coming right from the break-in point ends on the other.
Breakaway check with $K = - \frac{s ( s - 2 ) ( s - 3 )}{( s + 1 ) ^{2}}$ . Setting $\frac{d K}{d s} = 0$ and cancelling the factor $(s + 1)$ gives $s^{3} + 3 s^{2} - 16 s + 6 = 0,$ with roots $s \approx 2.48$ , $s \approx - 5.89$ and $s \approx 0.41$ . The root at $0.41$ is not on the locus, since only 2 poles lie to its right. The other two are the breakaway point near $2.5$ and the break-in point near $- 5.9$ seen on the plot, which confirms the double zero at $- 1$ .
So $K_{D} s^{2} + K_{P} s + K_{I} = K_{D} (s + 1)^{2} = K_{D} s^{2} + 2 K_{D} s + K_{D}$ . That gives $K_{P} = 2 K_{D}$ and $K_{I} = K_{D}$ , and $C (s) = \frac{K _{D} ( s + 1 ) ^{2}}{s} .$ The drawn locus is the positive-gain locus, and its gain is $K_{D}$ , so $K_{D} > 0$ .
Evaluate at $ω = 1$ : $C (j 1) = \frac{K _{D} ( 1 + j ) ^{2}}{j} = \frac{K _{D} \cdot 2 j}{j} = 2 K_{D} .$ Checking term by term: $C (j 1) = K_{P} + j (K_{D} - K_{I}) = 2 K_{D}$ . $C (j 1)$ is real and positive, so reading 'gain $= 2$ ' either literally or as a magnitude gives the same equation.
Set $2 K_{D} = 2$ , so $K_{D} = 1.0$ (and hence $K_{P} = 2$ , $K_{I} = 1$ ).

KEY IDEA

Read the controller's zeros off the root locus. A PID adds a pole at the origin and two zeros. The whole real axis left of $- 1$ is on the locus, which by the odd-count rule is only possible if both zeros sit at $- 1$ . So $C (s) = K_{D} (s + 1)^{2} / s$ , and at $ω = 1$ the proportional, integral and derivative terms combine into the real number $2 K_{D}$ .

WATCH OUT FOR

Taking the single 'o' marker as a single zero, $C (s) = K_{P} (s + 1) / s$ . That leaves $K_{D}$ undetermined (or wrongly zero), and it contradicts the plot, because a single zero at $- 1$ would take the real axis left of $- 1$ off the locus. Overlapping zeros are drawn as one marker, so count them with the real-axis rule, not by eye. A second slip is using $∣ j ω + 1∣$ instead of $∣ j ω + 1 ∣^{2}$ in the magnitude, which gives $K_{D} = 2$ .

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 10 Compensators & PID questions, with notes

GATE 2023 IN Q.48: solution and answer

Answer and solution

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