GATE 2019 BT Q.41: solution and answer

GATE 2019 · BTQ.41 · 2 marks · MCQ

Which one of the following statements is CORRECT about proportional controllers?

(A)
The initial change in control output signal is relatively slow
(B)
The initial corrective action is greater for larger error
(C)
They have no offset
(D)
There is no corrective action if the error is a constant

See the question in the paper

Answer and solution

ANSWER

(B) The initial corrective action is greater for larger error. A proportional controller's output change is $K_{c} e$ : it responds at once, in proportion to the error, and leaves a steady-state offset.

Official answer: BSource: Official answer key, GATE 2019 BT (IIT Madras)

Proportional (P) control law: $c (t) = \overset{c}{ˉ} + K_{c} e (t)$ , where $e (t) = y_{s p} (t) - y_{m} (t)$ is the error (set point $y_{s p}$ minus measured value $y_{m}$ ), $K_{c}$ is the controller gain and $\overset{c}{ˉ}$ is the bias, the controller output when $e = 0$ . In deviation form $c^{'} (t) = c (t) - \overset{c}{ˉ} = K_{c} e (t)$ , so the transfer function is $G_{c} (s) = K_{c}$ : a pure gain with no dynamics.
(A) Because $G_{c} (s) = K_{c}$ has no lag, the output changes the moment the error appears. A step error $e = M$ gives a step change in output of $K_{c} M$ at $t = 0^{+}$ . The initial change is immediate, not slow, so (A) is false.
(B) The corrective action is $Δ c = K_{c} e$ , directly proportional to the error. A larger initial error gives a proportionally larger initial correction, so (B) is true.
(C) Offset check: take a first-order process $G_{p} (s) = \frac{K _{p}}{τ s + 1}$ (process gain $K_{p}$ , time constant $τ$ ) in a unity-feedback loop with P control and a unit step in set point. The closed-loop response is $Y (s) = \frac{K _{c} K _{p}}{τ s + 1 + K _{c} K _{p}} \cdot \frac{1}{s}$ . By the final-value theorem, $y (\infty) = lim_{s \to 0} s Y (s) = \frac{K _{c} K _{p}}{1 + K _{c} K _{p}}$ , so the offset is $1 - y (\infty) = \frac{1}{1 + K _{c} K _{p}} \neq = 0$ for any finite $K_{c}$ . P control leaves a steady-state offset, so (C) is false.
Physical reason for the offset: to hold the process at a new steady state, the controller output must stay away from $\overset{c}{ˉ}$ . Since $c - \overset{c}{ˉ} = K_{c} e$ , that requires a nonzero error. Removing the offset needs integral action.
(D) With a constant error $e = M$ , the P output stays at $\overset{c}{ˉ} + K_{c} M$ , which is a steady, nonzero corrective action. Giving zero action for a constant error is how derivative control behaves ( $τ_{D} \frac{d e}{d t} = 0$ when $e$ is constant, where $τ_{D}$ is the derivative time), not proportional control. So (D) is false.
The only correct statement is (B).

KEY IDEA

A P controller is a pure gain, $c - \overset{c}{ˉ} = K_{c} e$ . It acts at once and in proportion to the error, so a bigger error gets a bigger correction. It still acts on a constant error, and it cannot remove offset, because holding a new steady state needs a nonzero error.

WATCH OUT FOR

Mixing up the three control modes. Zero action for a constant error describes derivative control (D), and eliminating offset describes integral control (C). A slow initial response belongs to lag-type or integral action (A). None of these describes proportional control.

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GATE 2019 BT Q.41: solution and answer

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