ISRO EC 2025 Q.42: solution and answer

ISRO 2025 · ECQ.42 · 1 mark · MCQ

Let $x (t) = rect (t /2)$ , where $rect (t /2)$ is the rectangular pulse of width 2 centered at $t = 0$ , i.e., $x (t) = 1$ for $∣ t ∣ \leq 1$ and 0 otherwise. What is the Fourier Transform $X (w)$ of $x (t)$ ?

(A)
$X (w) = 2 sinc (w)$
(B)
$X (w) = sinc (w)$
(C)
$X (w) = 2 sinc (w /2)$
(D)
$X (w) = sinc (w /2)$

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Answer and solution

ANSWER

$X (ω) = 2 sinc (ω) = \frac{2 sin ω}{ω}$ , which is option (A).

Official answer: ASource: ISRO's final answer key of 15 May 2026

The pulse is $x (t) = 1$ for $∣ t ∣ \leq 1$ and $0$ elsewhere, so $X (ω) = \int_{- 1}^{1} e^{- j ω t} d t .$
Integrate: $X (ω) = [\frac{e ^{- j ω t}}{- j ω}]_{- 1}^{1} = \frac{e ^{j ω} - e ^{- j ω}}{j ω} .$
Use $e^{j ω} - e^{- j ω} = 2 j sin ω$ : $X (ω) = \frac{2 sin ω}{ω} .$
With the engineering definition $sinc (ω) = \frac{sin ω}{ω}$ , this is $X (ω) = 2 sinc (ω)$ .
Check: $X (0) = \int x (t) d t =$ area of the pulse $= 2$ , and $2 sinc (0) = 2$ , whereas option (B) gives $1$ . The first zero of $X (ω)$ is at $ω = π$ (a pulse of width 2 has zeros at $ω = k π$ ), but $sinc (ω /2) = \frac{sin ( ω /2 )}{ω /2}$ in options (C) and (D) has its first zero at $ω = 2 π$ . Hence option (A).

KEY IDEA

The transform of a rectangular pulse of width $T$ and height 1 is $T \frac{sin ( ω T /2 )}{ω T /2}$ : its value at $ω = 0$ is the pulse area, and its first zero is at $ω = 2 π / T$ . Here $T = 2$ , so the area is 2 and the first zero is at $ω = π$ .

WATCH OUT FOR

Forgetting the height factor (the area) and writing $sinc (ω)$ alone, or confusing the pulse half-width with its width and putting $ω /2$ inside the sinc. A quick check at $ω = 0$ and at the first zero rules these out.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Fourier transform questions, with notes