GATE 2025 BT Q.43: solution and answer

GATE 2025 · BTQ.43 · 2 marks · MCQ

The output $y (t)$ of a first-order process is governed by the following differential equation

τ_{p} \frac{d y}{d t} + y = K_{p} f (t)

where $τ_{p}$ is a non-zero time constant, $K_{p}$ is the gain and $f (t)$ is the input with $f (0) = 0$ .

Assume $y (0) = 0$ . The transfer function for this process is (consider $s$ as the independent variable in the Laplace domain)

(A)
$\frac{K _{p}}{τ _{p} s + 1}$
(B)
$\frac{τ _{p}}{K _{p} s + 1}$
(C)
$\frac{τ _{p}}{K _{p} ( s + 1 )}$
(D)
$\frac{K _{p}}{τ _{p} ( s + 1 )}$

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Answer and solution

ANSWER

(A) $\frac{K _{p}}{τ _{p} s + 1}$

Official answer: ASource: Official answer key, GATE 2025 BT (IIT Roorkee)

Symbols: $τ_{p}$ is the process time constant (units of time), $K_{p}$ is the steady-state gain, $Y (s) = L {y (t)}$ and $F (s) = L {f (t)}$ .
Laplace transform of a derivative: $L {\frac{d y}{d t}} = s Y (s) - y (0)$ . Because $y (0) = 0$ , this is just $s Y (s)$ . Since $y (0) = 0$ and $f (0) = 0$ , both variables already start from zero, so they are already deviation variables.
Transform both sides of the model: $τ_{p} s Y (s) + Y (s) = K_{p} F (s)$ .
Factor out $Y (s)$ : $Y (s) (τ_{p} s + 1) = K_{p} F (s)$ .
The transfer function is output over input with zero initial conditions: $G (s) = \frac{Y ( s )}{F ( s )} = \frac{K _{p}}{τ _{p} s + 1}$
Check against the options. The steady-state gain must be $G (0) = K_{p}$ (by the final-value theorem, a unit step in $f$ settles $y$ at $K_{p}$ ), and the pole must be at $s = - 1/ τ_{p}$ . B gives $G (0) = τ_{p}$ and C gives $G (0) = τ_{p} / K_{p}$ , so both are wrong. D, $\frac{K _{p}}{τ _{p} ( s + 1 )}$ , gives $G (0) = K_{p} / τ_{p}$ and a pole at $s = - 1$ . Only A meets both conditions, so the answer is (A).

KEY IDEA

With zero initial conditions, take the Laplace transform term by term ( $d y / d t \to s Y (s)$ ) and form $G (s) = Y (s) / F (s)$ . A first-order process always has the standard form $K_{p} / (τ_{p} s + 1)$ : the gain is $G (0) = K_{p}$ and the single pole is at $s = - 1/ τ_{p}$ .

WATCH OUT FOR

Picking D, $K_{p} / [τ_{p} (s + 1)]$ , by dividing the whole equation by $τ_{p}$ and then factoring wrongly. That form changes the steady-state gain to $K_{p} / τ_{p}$ and moves the pole to $s = - 1$ . It is also dimensionally inconsistent, because $s$ has units of $time^{- 1}$ and cannot be added to a pure number $1$ unless it is multiplied by $τ_{p}$ .

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GATE 2025 BT Q.43: solution and answer

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