GATE 2026 EC Q.13: solution and answer

GATE 2026 · ECQ.13 · 1 mark · MCQ

The Laplace Transform of the signal $x (t) = u (t - 2) * (t u (t))$ is given by which of the following expressions?

[" $*$ " represents convolution operator]

(A)
$\frac{e ^{- 2 s}}{s ^{2} ( s - 2 )}$
(B)
$\frac{e ^{- 2 (s - 2)}}{s ^{3}}$
(C)
$\frac{s e ^{- 2 s}}{( s - 2 ) ^{2}}$
(D)
$\frac{e ^{- 2 s}}{s ^{3}}$

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Answer and solution

ANSWER

$X (s) = \frac{e ^{- 2 s}}{s ^{3}}$ for $Re (s) > 0$ , so the answer is option (D).

Official answer: DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Convolution property. Convolution in time becomes multiplication in $s$ : $X (s) = L {u (t - 2)} \cdot L {t u (t)}$
Time-shifting property. $L {u (t)} = \frac{1}{s}$ for $Re (s) > 0$ . Delaying a signal by 2 multiplies its transform by $e^{- 2 s}$ , so $L {u (t - 2)} = \frac{e ^{- 2 s}}{s}, Re (s) > 0$
Standard pair (ramp). $L {t u (t)} = \frac{1}{s ^{2}}, Re (s) > 0$
Multiply. Both transforms exist for $Re (s) > 0$ , so the product is valid there: $X (s) = \frac{e ^{- 2 s}}{s} \cdot \frac{1}{s ^{2}} = \frac{e ^{- 2 s}}{s ^{3}}, Re (s) > 0$
Check in time. Convolving with $u (t)$ integrates a signal from $0$ to $t$ , so $u (t) * t u (t) = \frac{t ^{2}}{2} u (t)$ . With the step delayed by 2, the result is delayed by 2: $x (t) = \frac{( t - 2 ) ^{2}}{2} u (t - 2)$ . Since $L {t^{n} u (t)} = \frac{n !}{s ^{n + 1}}$ , we get $L {\frac{t ^{2}}{2} u (t)} = \frac{1}{s ^{3}}$ . The delay multiplies this by $e^{- 2 s}$ , which gives $\frac{e ^{- 2 s}}{s ^{3}}$ , option (D).

KEY IDEA

Convolution in time becomes a product in $s$ , and a time delay of $t_{0}$ becomes the factor $e^{- s t_{0}}$ . Here $\frac{e ^{- 2 s}}{s} \cdot \frac{1}{s ^{2}} = \frac{e ^{- 2 s}}{s ^{3}}$ . In the time domain, convolving with a delayed step integrates the ramp once and delays the result: $x (t) = \frac{( t - 2 ) ^{2}}{2} u (t - 2)$ .

WATCH OUT FOR

Mixing up the two shifting properties. A delay in time, $x (t - 2)$ , multiplies the transform by $e^{- 2 s}$ . A shift in $s$ , such as $(s - 2)$ or $e^{- 2 (s - 2)}$ , comes from multiplying by $e^{2 t}$ , and this signal has no such factor. That confusion produces options (A), (B) and (C). A second slip is to multiply the two signals in time instead of their transforms.

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

GATE 2026 EC Q.13: solution and answer

Answer and solution

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