GATE 2024 EC Q.13: solution and answer

GATE 2024 · ECQ.13 · 1 mark · MCQ

In the feedback control system shown in the figure below $G (s) = \frac{6}{s ( s + 1 ) ( s + 2 )}$ .

$R (s)$ , $Y (s)$ , and $E (s)$ are the Laplace transforms of $r (t)$ , $y (t)$ , and $e (t)$ , respectively. If the input $r (t)$ is a unit step function, then __________.

GATE 2024 EC question 13: Unity negative feedback loop. — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
$lim_{t \to \infty} e (t) = 0$
(B)
$lim_{t \to \infty} e (t) = \frac{1}{3}$
(C)
$lim_{t \to \infty} e (t) = \frac{1}{4}$
(D)
$lim_{t \to \infty} e (t)$ does not exist, $e (t)$ is oscillatory

Answer and solution

ANSWER

(D) The limit of e(t) as t goes to infinity does not exist, because e(t) oscillates. The closed-loop poles are s = -3 and s = ±j√2, so the error keeps an undamped sinusoid at √2 rad/s.

Official answer: DSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Check that the closed loop is stable before using the final value theorem. With unity negative feedback the characteristic equation is $1 + G (s) = 0$ , so $s (s + 1) (s + 2) + 6 = 0$ : $s^{3} + 3 s^{2} + 2 s + 6 = 0$
Routh array. Row $s^{3}$ : $1, 2$ . Row $s^{2}$ : $3, 6$ . Row $s^{1}$ : $\frac{3 \cdot 2 - 1 \cdot 6}{3} = 0$ . The whole $s^{1}$ row is zero, which means some roots sit symmetrically about the origin. The first-column entries that exist ( $1$ , $3$ ) do not change sign, so no root lies in the right half-plane.
Take the auxiliary polynomial from the $s^{2}$ row: $3 s^{2} + 6 = 0 \Rightarrow s = \pm j 2$ . Factoring confirms it: $s^{3} + 3 s^{2} + 2 s + 6 = s^{2} (s + 3) + 2 (s + 3) = (s + 3) (s^{2} + 2)$ . The closed-loop poles are $- 3$ and $\pm j 2$ , so the system is marginally stable.
For unity feedback the error transform is $E (s) = \frac{R ( s )}{1 + G ( s )} = \frac{1}{s} \cdot \frac{s ( s + 1 ) ( s + 2 )}{( s + 3 ) ( s ^{2} + 2 )} = \frac{( s + 1 ) ( s + 2 )}{( s + 3 ) ( s ^{2} + 2 )}$
$s E (s)$ has poles at $\pm j 2$ , on the $j ω$ -axis. The final value theorem needs every pole of $s E (s)$ to be in the open left half-plane, so it cannot be used here. The numerator does not cancel these poles: at $s = j 2$ it equals $(1 + j 2) (2 + j 2) = 3 j 2 \neq = 0$ .
Partial fractions: $E (s) = \frac{2/11}{s + 3} + \frac{\frac{9}{11} s + \frac{6}{11}}{s ^{2} + 2}$ so $e (t) = \frac{2}{11} e^{- 3 t} + \frac{9}{11} cos (2 t) + \frac{6}{11 2} sin (2 t), t \geq 0$
The exponential term dies out. The sinusoid at $ω = 2 rad/s$ never decays, and its amplitude is $\frac{3}{11}$ . So $e (t)$ keeps oscillating and $lim_{t \to \infty} e (t)$ does not exist. This is option (D).

KEY IDEA

The final value theorem only applies when every pole of $s E (s)$ is in the open left half-plane. So check closed-loop stability first. Here $1 + G (s) = (s + 3) (s^{2} + 2)$ , which puts a pair of poles on the $j ω$ -axis at $\pm j 2$ . The loop is marginally stable, so the error ends up as a steady sinusoid instead of settling to a limit.

WATCH OUT FOR

The common mistake is to apply the final value theorem without checking stability. Blindly, $lim_{s \to 0} s E (s) = \frac{( 1 ) ( 2 )}{( 3 ) ( 2 )} = \frac{1}{3}$ , which gives the trap option (B). A second trap: noticing $G (s)$ is type 1 and writing $e_{ss} = 0$ for a step, which gives option (A). Both results assume a stable closed loop, and this loop is not asymptotically stable.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 22 Time response questions, with notes

GATE 2024 EC Q.13: solution and answer

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