GATE 2025 EC Q.26: solution and answer

GATE 2025 · ECQ.26 · 1 mark · MSQ

Consider the unity-negative-feedback system shown in Figure (i) below, where gain $K \geq 0$ . The root locus of this system is shown in Figure (ii) below.

For what value(s) of $K$ will the system in Figure (i) have a pole at $- 1 + j 1$ ?

GATE 2025 EC question 26: Figure (i): block diagram. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$K = 5$
(B)
$K = \frac{1}{5}$
(C)
For no positive value of $K$
(D)
For all positive values of $K$

Answer and solution

ANSWER

(C) For no positive value of $K$ . At $- 1 + j 1$ the angle of $G$ is $- 9 0^{\circ}$ , not $\pm 18 0^{\circ}$ , so the point is not on the root locus. Solving the characteristic equation there needs $K = - j /5$ , which is not real.

Official answer: CSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Read the plot. The open-loop poles are at $- 1, - 2$ and the zeros are at $- 3, - 4$ , so $G (s) = \frac{( s + 3 ) ( s + 4 )}{( s + 1 ) ( s + 2 )}$ . Any positive constant multiplying this can be absorbed into $K$ . The real-axis segments $[- 2, - 1]$ and $[- 4, - 3]$ are the ones with an odd number of poles and zeros to their right, which confirms that this is the ordinary $K \geq 0$ (180°) locus.
A point $s_{0}$ is a closed-loop pole for some $K \geq 0$ only if it lies on the root locus. That requires the angle condition, $∠ G (s_{0}) = \pm 18 0^{\circ}$ (an odd multiple of $18 0^{\circ}$ ). The magnitude condition $K = 1/∣ G (s_{0}) ∣$ gives the gain only after the angle condition holds.
Test $s_{0} = - 1 + j 1$ . Zero vectors: $s_{0} + 3 = 2 + j 1$ has angle $tan^{- 1} (1/2) = 26.5 7^{\circ}$ , and $s_{0} + 4 = 3 + j 1$ has angle $tan^{- 1} (1/3) = 18.4 3^{\circ}$ .
Pole vectors: $s_{0} + 1 = j 1$ has angle $9 0^{\circ}$ , and $s_{0} + 2 = 1 + j 1$ has angle $4 5^{\circ}$ .
$∠ G (- 1 + j 1) = (26.5 7^{\circ} + 18.4 3^{\circ}) - (9 0^{\circ} + 4 5^{\circ}) = 4 5^{\circ} - 13 5^{\circ} = - 9 0^{\circ} \neq = \pm 18 0^{\circ}$ The angle condition fails, so $- 1 + j 1$ is not on the locus. It is not on the $K < 0$ locus either, which would need $0^{\circ}$ . So no real $K$ puts a pole there.
Cross-check with the geometry. The characteristic equation is $(s + 1) (s + 2) + K (s + 3) (s + 4) = 0$ , so $K = - \frac{( s + 1 ) ( s + 2 )}{( s + 3 ) ( s + 4 )}$ . Setting $\frac{d K}{d s} = 0$ gives $2 s^{2} + 10 s + 11 = 0$ , so $s = - 2.5 \pm 0.866$ : breakaway at $- 1.634$ and break-in at $- 3.366$ .
For the complex roots, write the equation as $(1 + K) s^{2} + (3 + 7 K) s + (2 + 12 K) = 0$ . Then $Re (s) = - \frac{3 + 7 K}{2 ( 1 + K )}$ and $∣ s ∣^{2} = \frac{2 + 12 K}{1 + K}$ . Substituting these gives $∣ s + 2.5 ∣^{2} = ∣ s ∣^{2} + 5 Re (s) + 6.25 = \frac{3}{4}$ for every $K$ . The complex part of the locus is therefore the circle centred at $- 2.5$ with radius $3 /2 = 0.866$ .
The distance from $- 1 + j 1$ to the centre is $(1.5)^{2} + 1^{2} = 3.25 = 1.80 \neq = 0.866$ , so the point is off the circle. No $K \geq 0$ puts a closed-loop pole there.
Direct confirmation, which also shows where the wrong options come from. Put $s = - 1 + j 1$ into the characteristic equation: $(s + 1) (s + 2) = j (1 + j) = - 1 + j$ and $(s + 3) (s + 4) = (2 + j) (3 + j) = 5 + 5 j$ . So $K = - \frac{- 1 + j}{5 + 5 j} = \frac{1 - j}{5 ( 1 + j )} = \frac{( 1 - j ) ^{2}}{10} = - \frac{j}{5}$ The gain needed is imaginary, so no real gain works. Its magnitude is $\frac{1}{5}$ , the same value the magnitude condition gives on its own: $∣ G (- 1 + j 1) ∣ = \frac{5 10}{1 \cdot 2} = 5$ , so $K = \frac{1}{5}$ . That is option (B), and it means nothing here because the angle condition has already failed. Option (A), $K = 5$ , is $∣ G ∣$ itself, taken wrongly as the gain. Option (D) is ruled out because the point is never on the locus. Only (C) is correct.

KEY IDEA

A point is a closed-loop pole for some $K \geq 0$ only if it lies on the root locus. That is decided by the angle condition, $∠ G (s) = \pm 18 0^{\circ}$ , not by the magnitude condition. A quicker check is to substitute the point into $1 + K G (s) = 0$ and solve for $K$ . The point is reachable only if $K$ comes out real and non-negative; here $K = - j /5$ . For two real poles and two real zeros, the complex part of the locus is a circle, here centred at $- 2.5$ with radius $0.866$ , and $- 1 + j 1$ is $1.80$ from that centre.

WATCH OUT FOR

Using only the magnitude condition. $∣ G (- 1 + j 1) ∣ = 5$ suggests $K = 1/5$ , which is option (B), or $K = 5$ if the ratio is inverted, which is option (A). Both skip the angle test, which gives $- 9 0^{\circ}$ . Another trap is judging from the drawing: the stretched vertical scale makes the circle look as if it reaches $j 1$ , but its true radius is $0.866$ and it never gets near $- 1 + j 1$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 7 Root locus questions, with notes

GATE 2025 EC Q.26: solution and answer

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