GATE 2022 IN Q.11: solution and answer

GATE 2022 · INQ.11 · 1 mark · MCQ

The input $x (t)$ to a system is related to its output $y (t)$ as

\frac{d y ( t )}{d t} + y (t) = 3 x (t - 3) u (t - 3)

Here $u (t)$ represents a unit-step function. The transfer function of this system is ________

(A)
$\frac{e ^{- 3 s}}{s + 3}$
(B)
$\frac{3 e ^{- 3 s}}{s + 1}$
(C)
$\frac{3 e ^{- (s /3)}}{s + 1}$
(D)
$\frac{e ^{- (s /3)}}{s + 3}$

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Answer and solution

ANSWER

The transfer function is $G (s) = \frac{3 e ^{- 3 s}}{s + 1}$ , which is option (B).

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

A transfer function is defined with all initial conditions zero, so take the Laplace transform of both sides with $y (0^{-}) = 0$ .
Differentiation property: $L {\frac{d y}{d t}} = s Y (s) - y (0^{-}) = s Y (s)$ . The left side therefore becomes $s Y (s) + Y (s) = (s + 1) Y (s)$ .
Time-shift property: for a causal input, $L {x (t - 3) u (t - 3)} = e^{- 3 s} X (s)$ . The factor $3$ is a constant gain and passes straight through, so the right side becomes $3 e^{- 3 s} X (s)$ .
Equate the two sides: $(s + 1) Y (s) = 3 e^{- 3 s} X (s)$
Divide through by $X (s)$ and by $(s + 1)$ : $G (s) = \frac{Y ( s )}{X ( s )} = \frac{3 e ^{- 3 s}}{s + 1}$ This is a first-order lag with its pole at $s = - 1$ , a DC gain of $G (0) = 3$ , and a pure transport delay of $3 s$ . That is option (B).

KEY IDEA

A delay of $T$ seconds in time becomes a multiplying factor $e^{- s T}$ in the $s$ -domain, because $L {x (t - T) u (t - T)} = e^{- s T} X (s)$ . The delay changes the exponent only. The gain $3$ and the pole at $s = - 1$ come from the rest of the differential equation, so the answer is a gain of 3, a first-order lag $\frac{1}{s + 1}$ , and a delay factor $e^{- 3 s}$ multiplied together.

WATCH OUT FOR

Two traps match the wrong options. (1) Treating the time shift like time scaling and writing $e^{- s /3}$ . That factor would come from $x (t /3)$ -type manipulations, not from a shift, and it gives options (C) and (D). (2) Letting the input coefficient $3$ move into the denominator and writing $s + 3$ . The $3$ multiplies the input, so it sets the gain. The coefficient of $y (t)$ is $1$ , so the pole stays at $s = - 1$ . That mistake gives options (A) and (D).

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GATE 2022 IN Q.11: solution and answer

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