GATE 2022 EC Q.15: solution and answer

GATE 2022 · ECQ.15 · 1 mark · MCQ

The Fourier transform $X (j ω)$ of the signal
$x (t) = \frac{t}{( 1 + t ^{2} ) ^{2}}$
is ________.

(A)
$\frac{π}{2 j} ω e^{- ∣ ω ∣}$
(B)
$\frac{π}{2} ω e^{- ∣ ω ∣}$
(C)
$\frac{π}{2 j} e^{- ∣ ω ∣}$
(D)
$\frac{π}{2} e^{- ∣ ω ∣}$

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Answer and solution

ANSWER

$X (j ω) = \frac{π}{2 j} ω e^{- ∣ ω ∣}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Standard pair: $e^{- ∣ t ∣} ⟷ \frac{2}{1 + ω ^{2}}$ , because $\int_{- \infty}^{\infty} e^{- ∣ t ∣} e^{- j ω t} d t = \frac{1}{1 + j ω} + \frac{1}{1 - j ω} = \frac{2}{1 + ω ^{2}}$ .
Duality ( $g (t) \leftrightarrow G (j ω) \Rightarrow G (t) \leftrightarrow 2 π g (- ω)$ ), with $g (t) = e^{- ∣ t ∣}$ even: $\frac{2}{1 + t ^{2}} \leftrightarrow 2 π e^{- ∣ ω ∣}$ , so $\frac{1}{1 + t ^{2}} \leftrightarrow π e^{- ∣ ω ∣}$ .
Write $x (t)$ as a derivative: $\frac{d}{d t} (\frac{1}{1 + t ^{2}}) = \frac{- 2 t}{( 1 + t ^{2} ) ^{2}}$ , so $x (t) = - \frac{1}{2} \frac{d}{d t} (\frac{1}{1 + t ^{2}})$ .
Time-differentiation property ( $\frac{d}{d t} \leftrightarrow j ω$ ): $X (j ω) = - \frac{1}{2} (j ω) π e^{- ∣ ω ∣} = - \frac{j π}{2} ω e^{- ∣ ω ∣}$
Since $\frac{1}{j} = - j$ : $X (j ω) = \frac{π}{2 j} ω e^{- ∣ ω ∣}$ .
Symmetry check: $x (t)$ is real and odd, so $X (j ω)$ must be purely imaginary and odd in $ω$ . Only option (A) is both: (B) and (D) are real, and (C) is imaginary but even.
Answer: (A).

KEY IDEA

Write $x (t)$ as $- \frac{1}{2} \frac{d}{d t} (\frac{1}{1 + t ^{2}})$ . The transform of $\frac{1}{1 + t ^{2}}$ is $π e^{- ∣ ω ∣}$ , by duality with $e^{- ∣ t ∣} \leftrightarrow \frac{2}{1 + ω ^{2}}$ . Differentiating in time multiplies the transform by $j ω$ . A real, odd signal must have a purely imaginary, odd transform, and that check alone picks out (A).

WATCH OUT FOR

Getting a sign wrong when converting between $j$ and $\frac{1}{j}$ . Since $\frac{1}{j} = - j$ , the result $- \frac{j π}{2} ω e^{- ∣ ω ∣}$ is exactly $\frac{π}{2 j} ω e^{- ∣ ω ∣}$ . If you drop the minus sign from the derivative ( $\frac{d}{d t} \frac{1}{1 + t ^{2}} = \frac{- 2 t}{( 1 + t ^{2} ) ^{2}}$ ), you get the negative of (A) and may still pick (A) for the wrong reason. Other slips: leaving out the $2 π$ in duality (which changes the constant), or dropping the $j$ and choosing the real option (B).

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 22 Fourier transform questions, with notes

GATE 2022 EC Q.15: solution and answer

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