GATE 2023 EE Q.18: solution and answer

GATE 2023 · EEQ.18 · 1 mark · MCQ

The Fourier transform $X (ω)$ of the signal $x (t)$ is given by

X (ω) = 1, for ∣ ω ∣ < W_{0} = 0, for ∣ ω ∣ > W_{0}

Which one of the following statements is true?

(A)
$x (t)$ tends to be an impulse as $W_{0} \to \infty$ .
(B)
$x (0)$ decreases as $W_{0}$ increases.
(C)
At $t = \frac{π}{2 W _{0}}$ , $x (t) = - \frac{1}{π}$
(D)
At $t = \frac{π}{2 W _{0}}$ , $x (t) = \frac{1}{π}$

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Answer and solution

ANSWER

(A) is correct. As $W_{0} \to \infty$ , $x (t)$ tends to an impulse. Here $x (t) = \frac{sin ( W _{0} t )}{π t}$ , and as the spectrum widens to $X (ω) = 1$ for all $ω$ it becomes the transform of $δ (t)$ .

Official answer: ASource: Official answer key, GATE 2023 EE (IIT Kanpur)

Inverse Fourier transform of the rectangular (ideal low-pass) spectrum: $x (t) = \frac{1}{2 π} \int_{- W_{0}}^{W_{0}} e^{j ω t} d ω = \frac{1}{2 π} \cdot \frac{e ^{j W_{0} t} - e ^{- j W_{0} t}}{j t} = \frac{sin ( W _{0} t )}{π t}$
Value at the origin (area under the spectrum divided by $2 π$ ): $x (0) = \frac{1}{2 π} \int_{- W_{0}}^{W_{0}} 1 d ω = \frac{W _{0}}{π}$ This grows as $W_{0}$ increases, so (B) is false.
At $t = \frac{π}{2 W _{0}}$ : $x (\frac{π}{2 W _{0}}) = \frac{sin ( π /2 )}{π \cdot \frac{π}{2 W _{0}}} = \frac{2 W _{0}}{π ^{2}}$ This value is positive and depends on $W_{0}$ . It is never $- \frac{1}{π}$ , and it equals $\frac{1}{π}$ only for the one value $W_{0} = \frac{π}{2}$ . The question leaves $W_{0}$ general, so (C) and (D) are both false.
As $W_{0} \to \infty$ , $X (ω) \to 1$ for every $ω$ . Since $δ (t) \leftrightarrow 1$ , $x (t) \to δ (t)$ . You can see this in the sinc itself: its peak $x (0) = W_{0} / π$ grows without bound, its main lobe (between the zeros at $t = \pm π / W_{0}$ ) has width $2 π / W_{0}$ , which shrinks to zero, and its total area stays $\int_{- \infty}^{\infty} x (t) d t = X (0) = 1$ . So (A) is true.
Only statement (A) is true.

KEY IDEA

Time and bandwidth trade off. A rectangular spectrum of half-width $W_{0}$ gives $x (t) = sin (W_{0} t) / (π t)$ , whose peak is $W_{0} / π$ , main-lobe width is $2 π / W_{0}$ and area is $X (0) = 1$ . As the band widens without limit, the pulse gets taller and narrower while its area stays 1, which is an impulse. Equivalently, a flat spectrum $X (ω) = 1$ is the transform of $δ (t)$ .

WATCH OUT FOR

Forgetting the $1/ (2 π)$ factor, or taking $x (t) = sin (W_{0} t) / t$ or $sin (W_{0} t) / (π t) \cdot 2$ . At $t = π / (2 W_{0})$ that leads students to a $W_{0}$ -free value like $\pm 1/ π$ , so they pick (C) or (D). The correct value, $2 W_{0} / π^{2}$ , depends on $W_{0}$ . A second trap is thinking $x (0)$ falls as $W_{0}$ grows because 'the sinc gets narrower'. The pulse does narrow, but its peak $W_{0} / π$ rises.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 22 Fourier transform questions, with notes

GATE 2023 EE Q.18: solution and answer

Answer and solution

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