GATE 2022 IN Q.41: solution and answer

GATE 2022 · INQ.41 · 2 marks · MCQ

A proportional-integral-derivative (PID) controller is employed to stably control a plant with transfer function

P (s) = \frac{1}{( s + 1 ) ( s + 2 )}

Now, the proportional gain is increased by a factor of 2, the integral gain is increased by a factor of 3, and the derivative gain is left unchanged. Given that the closed-loop system continues to remain stable with the new gains, the steady-state error in tracking a ramp reference signal ________

(A)
Remains unchanged
(B)
Decreases by a factor of 2
(C)
Decreases by a factor of 3
(D)
Decreases by a factor of 5

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Answer and solution

ANSWER

(C) Decreases by a factor of 3. The ramp error is $e_{ss} = 2 A / K_{i}$ , which depends only on the integral gain, and that gain is tripled.

Official answer: CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Take the standard unity-feedback loop with the PID controller in series with the plant. The controller is $C (s) = K_{p} + \frac{K _{i}}{s} + K_{d} s = \frac{K _{d} s ^{2} + K _{p} s + K _{i}}{s}$ .
Open-loop transfer function: $G (s) = C (s) P (s) = \frac{K _{d} s ^{2} + K _{p} s + K _{i}}{s ( s + 1 ) ( s + 2 )}$ The integral term puts one pole at the origin, so the loop is type 1, provided $K_{i} \neq = 0$ (and $K_{i}$ must be nonzero here because the question multiplies it by 3).
A stable type-1 loop follows a ramp $r (t) = A t$ with a finite steady-state error $e_{ss} = \frac{A}{K _{v}}$ , where $K_{v} = lim_{s \to 0} s G (s)$ is the velocity error constant.
Compute $K_{v}$ : $K_{v} = s \to 0 lim \frac{K _{d} s ^{2} + K _{p} s + K _{i}}{( s + 1 ) ( s + 2 )} = \frac{K _{i}}{( 1 ) ( 2 )} = \frac{K _{i}}{2}$ The $K_{p} s$ and $K_{d} s^{2}$ terms vanish as $s \to 0$ , so only $K_{i}$ is left.
So $e_{ss} = \frac{2 A}{K _{i}}$ . The ramp error depends only on the integral gain, not on $K_{p}$ or $K_{d}$ .
With the new gains $K_{p}^{'} = 2 K_{p}$ , $K_{i}^{'} = 3 K_{i}$ , $K_{d}^{'} = K_{d}$ : $e_{ss}^{'} = \frac{2 A}{3 K _{i}} = \frac{e _{ss}}{3}$ Doubling $K_{p}$ changes the transient response, not the ramp error. The question says the loop stays stable, so the final value theorem still applies and the error really does settle to this value.
The steady-state error in tracking a ramp decreases by a factor of 3, which is option (C).

KEY IDEA

With a PID controller the loop is type 1, and its velocity error constant is $K_{v} = lim_{s \to 0} s C (s) P (s) = K_{i} \cdot P (0)$ . At $s = 0$ the proportional and derivative terms drop out, so the steady-state ramp error $A / K_{v}$ is inversely proportional to $K_{i}$ alone.

WATCH OUT FOR

Thinking every gain increase reduces the error. Some students multiply the factors (2 x 3 = 6, not an option) or add them (2 + 3 = 5, which gives option D). Others take $K_{p}$ as the gain that sets the ramp error and pick option (B). In a PID loop, $K_{p}$ and $K_{d}$ vanish from $lim_{s \to 0} s G (s)$ , so only the factor of 3 on $K_{i}$ matters.

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GATE 2022 IN Q.41: solution and answer

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