GATE 2022 BT Q.41: solution and answer

GATE 2022 · BTQ.41 · 2 marks · MCQ

Which of the following statements are CORRECT for a controller?

P. In a proportional controller, a control action is proportional to the error

Q. In an integral controller, a control action is proportional to the derivative of the error

R. There is no "offset" in the response of the closed-loop first-order process with a proportional controller

S. There is no "offset" in the response of the closed-loop first-order process with a proportional-integral controller

(A)
P and Q only
(B)
P and R only
(C)
P and S only
(D)
Q and S only

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Answer and solution

ANSWER

(C) P and S only. A proportional controller's action is proportional to the error (P). Integral action is proportional to the integral of the error, not its derivative, so Q is false. A P controller on a first-order process leaves an offset of $1/ (1 + K_{c} K_{p})$ , so R is false. A PI controller removes the offset, so S is true.

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

P: true. Proportional control: $c (t) = c_{s} + K_{c} ε (t)$ , where $c (t)$ is the controller output, $c_{s}$ its bias (the output when the error is zero), $ε = y_{s p} - y$ the error between set point $y_{s p}$ and measured output $y$ , and $K_{c}$ the controller gain. The action is proportional to the error.
Q: false. Integral control: $c (t) = c_{s} + \frac{K _{c}}{τ _{I}} \int_{0}^{t} ε d t$ , with $τ_{I}$ the integral (reset) time. The action is proportional to the integral of the error. Derivative action is the one proportional to $d ε / d t$ .
Now test R and S. Take a first-order process $G_{p} = \frac{K _{p}}{τ _{p} s + 1}$ (process gain $K_{p}$ , time constant $τ_{p}$ ) in a unity-feedback loop with controller $G_{c}$ , and apply a unit step in set point: $Y_{s p} (s) = 1/ s$ . The closed-loop response is $\frac{Y}{Y _{s p}} = \frac{G _{c} G _{p}}{1 + G _{c} G _{p}}$ . Take $K_{c} K_{p} > 0$ (the controller acts in the correct direction), so the loop is stable and the final value theorem applies.
R: false. With a P controller, $G_{c} = K_{c}$ , so $\frac{Y}{Y _{s p}} = \frac{K _{c} K _{p}}{τ _{p} s + 1 + K _{c} K _{p}}$ .
Final value theorem: $y (\infty) = lim_{s \to 0} s \cdot \frac{1}{s} \cdot \frac{K _{c} K _{p}}{τ _{p} s + 1 + K _{c} K _{p}} = \frac{K _{c} K _{p}}{1 + K _{c} K _{p}} < 1$ . So the offset is $1 - y (\infty) = \frac{1}{1 + K _{c} K _{p}} \neq = 0$ for any finite $K_{c}$ . A P controller always leaves an offset.
S: true. With a PI controller, $G_{c} = K_{c} (1 + \frac{1}{τ _{I} s}) = \frac{K _{c} ( τ _{I} s + 1 )}{τ _{I} s}$ . The loop now contains a pure integrator ( $1/ s$ ), and $\frac{Y}{Y _{s p}} = \frac{K _{c} K _{p} ( τ _{I} s + 1 )}{τ _{I} s ( τ _{p} s + 1 ) + K _{c} K _{p} ( τ _{I} s + 1 )} .$ The characteristic equation $τ_{I} τ_{p} s^{2} + τ_{I} (1 + K_{c} K_{p}) s + K_{c} K_{p} = 0$ has all coefficients positive, so this second-order loop is stable.
Final value theorem: $y (\infty) = lim_{s \to 0} s \cdot \frac{1}{s} \cdot \frac{K _{c} K _{p} ( τ _{I} s + 1 )}{τ _{I} s ( τ _{p} s + 1 ) + K _{c} K _{p} ( τ _{I} s + 1 )} = \frac{K _{c} K _{p}}{K _{c} K _{p}} = 1$ . The offset is zero, because the integral term keeps changing the output for as long as any error remains.
Only P and S are correct, which is option (C).

KEY IDEA

Offset is set by the loop's gain at $s = 0$ . With P control, $G_{c} G_{p} (0) = K_{c} K_{p}$ is finite, so $y (\infty) = K_{c} K_{p} / (1 + K_{c} K_{p}) < 1$ . Integral action puts a $1/ s$ in the loop, so $G_{c} G_{p} \to \infty$ as $s \to 0$ , $y (\infty) = 1$ , and the offset disappears.

WATCH OUT FOR

Mixing up integral and derivative action. Q describes derivative control ( $\propto d ε / d t$ ), not integral control. The other common slip is thinking a large enough $K_{c}$ removes the P-controller offset. It only shrinks it, as $1/ (1 + K_{c} K_{p})$ , and never reaches zero at any finite gain.

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GATE 2022 BT Q.41: solution and answer

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