GATE 2022 EC Q.60: solution and answer

GATE 2022 · ECQ.60 · 2 marks · NAT

The block diagram of a closed-loop control system is shown in the figure. $R (s)$ , $Y (s)$ , and $D (s)$ are the Laplace transforms of the time-domain signals $r (t)$ , $y (t)$ , and $d (t)$ , respectively. Let the error signal be defined as $e (t) = r (t) - y (t)$ . Assuming the reference input $r (t) = 0$ for all $t$ , the steady-state error $e (\infty)$ , due to a unit step disturbance $d (t)$ , is ________ (rounded off to two decimal places).

GATE 2022 EC question 60: Unity-feedback loop. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

The steady-state error due to a unit step disturbance is $e (\infty) = - 0.10$ .

Official answer: -0.11 to -0.09Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

With $r (t) = 0$ , the error is $E (s) = R (s) - Y (s) = - Y (s)$ . Let the controller gain be $G_{c} = 10$ and the plant be $G_{p} (s) = \frac{1}{s ( s + 10 )}$ .
The plant input is the sum of the controller output and the disturbance: $Y (s) = G_{p} (s) [D (s) + 10 E (s)] = G_{p} (s) [D (s) - 10 Y (s)]$
Collecting $Y (s)$ gives the disturbance-to-output transfer function: $\frac{Y ( s )}{D ( s )} = \frac{G _{p}}{1 + 10 G _{p}} = \frac{1}{s ( s + 10 ) + 10} = \frac{1}{s ^{2} + 10 s + 10}$
Stability check before using the final value theorem: $s^{2} + 10 s + 10 = 0$ gives $s = - 5 \pm 15 \approx - 1.13, - 8.87$ . Both closed-loop poles are in the left half-plane (for a second-order polynomial, all coefficients positive is sufficient), so the final value theorem applies.
Final value theorem with $D (s) = \frac{1}{s}$ : $y (\infty) = s \to 0 lim s \cdot \frac{1}{s} \cdot \frac{1}{s ^{2} + 10 s + 10} = \frac{1}{10} = 0.1$
Hence $e (\infty) = r (\infty) - y (\infty) = 0 - 0.1 = - 0.10$ .

KEY IDEA

Treat the disturbance as the input and find $Y (s) / D (s) = \frac{G _{p}}{1 + G _{c} G _{p}}$ . Whether a step disturbance is rejected depends on integrators placed BEFORE the point where the disturbance enters, not on the system type as a whole. Here the only integrator is in the plant, after the disturbance, so the step is not rejected: the output settles at $1/ G_{c} = 1/10$ , and because $r = 0$ the error is the negative of that, $e (\infty) = - y (\infty) = - 0.1$ .

WATCH OUT FOR

Two errors are common. The first is reporting $+ 0.10$ , which forgets that $e = r - y = - y$ when $r = 0$ . The second is answering $0$ on the grounds that the loop is type 1. The integrator sits after the disturbance entry point, so it gives zero error for a step REFERENCE but cannot reject a step DISTURBANCE. Only an integrator in the controller, before the disturbance enters, would drive this error to zero.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 22 Time response questions, with notes

GATE 2022 EC Q.60: solution and answer

Answer and solution

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