GATE 2026 IN Q.43: solution and answer

GATE 2026 · INQ.43 · 2 marks · MCQ

Given two signals: $x (t) = u (t - 2) - u (t - 3)$ , and $y (t) = e^{- 4 t} u (t)$ .

If $z (t) = x (t) * y (t)$ , then $Z (s)$ is ________.

(A)
$\frac{e ^{- 3 s} [ e ^{s} - 1 ]}{s ( s + 4 )}$
(B)
$\frac{e ^{- s} [ e ^{- 3 s} - 1 ]}{( s + 3 ) ( s + 4 )}$
(C)
$\frac{e ^{- 3 s} [ 1 - e ^{s} ]}{( s + 3 ) ( s + 4 )}$
(D)
$\frac{e ^{- 3 s} [ 1 - e ^{s} ]}{s ( s + 4 )}$

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Answer and solution

ANSWER

$Z (s) = \frac{e ^{- 3 s} [ e ^{s} - 1 ]}{s ( s + 4 )}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

Convolution property: convolution in time becomes multiplication in $s$ , so $Z (s) = X (s) Y (s)$ .
Time-shift property with $L {u (t)} = \frac{1}{s}$ : $L {u (t - t_{0})} = \frac{e ^{- s t_{0}}}{s}$ , so $X (s) = \frac{e ^{- 2 s}}{s} - \frac{e ^{- 3 s}}{s} = \frac{e ^{- 2 s} - e ^{- 3 s}}{s} .$
Standard pair: $L {e^{- 4 t} u (t)} = \frac{1}{s + 4}$ , so $Y (s) = \frac{1}{s + 4}$ .
Multiply: $Z (s) = \frac{e ^{- 2 s} - e ^{- 3 s}}{s ( s + 4 )} .$
Take out $e^{- 3 s}$ as a common factor: $e^{- 2 s} - e^{- 3 s} = e^{- 3 s} (e^{s} - 1)$ , so $Z (s) = \frac{e ^{- 3 s} [ e ^{s} - 1 ]}{s ( s + 4 )} .$
Checking the other options: the denominator must be $s (s + 4)$ , because the rectangular pulse brings the factor $\frac{1}{s}$ . That rules out (B) and (C). (D) is exactly $- Z (s)$ , since $1 - e^{s} = - (e^{s} - 1)$ . The answer is (A).

KEY IDEA

Convolution in time becomes multiplication in $s$ . The pulse $u (t - 2) - u (t - 3)$ transforms to $\frac{e ^{- 2 s} - e ^{- 3 s}}{s}$ by the time-shift property, and $e^{- 4 t} u (t)$ transforms to $\frac{1}{s + 4}$ . Taking out $e^{- 3 s}$ as a common factor puts the product in the printed form. Quick check: $x (t) \geq 0$ and $y (t) \geq 0$ , so $z (t) \geq 0$ and its area $Z (0)$ must be positive. The pulse has area $1$ and $y$ has area $\frac{1}{4}$ , so $Z (0) = \frac{1}{4}$ . Option (A) tends to $+ \frac{1}{4}$ as $s \to 0$ and (D) tends to $- \frac{1}{4}$ .

WATCH OUT FOR

Getting the sign wrong when taking out the common factor. $e^{- 2 s} - e^{- 3 s}$ equals $e^{- 3 s} (e^{s} - 1)$ , not $e^{- 3 s} (1 - e^{s})$ , and the wrong sign leads to (D). A second mistake is to shift the $\frac{1}{s + 4}$ factor and get something like $(s + 3)$ in the denominator, as in (B) and (C). Only the step functions pick up $e^{- s t_{0}}$ . The exponential is not shifted, so its transform stays $\frac{1}{s + 4}$ .

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 8 Laplace questions, with notes

GATE 2026 IN Q.43: solution and answer

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