GATE 2018 EC Q.39: solution and answer

GATE 2018 · ECQ.39 · 2 marks · NAT

The input $4 sinc (2 t)$ is fed to a Hilbert transformer to obtain $y (t)$ , as shown in the figure below:

[Figure]

Here $sinc (x) = \frac{sin ( π x )}{π x}$ . The value (accurate to two decimal places) of $\int_{- \infty}^{\infty} ∣ y (t) ∣^{2} d t$ is ______.

GATE 2018 EC question 39: A single block labelled 'Hilbert Transform'. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

$\int_{- \infty}^{\infty} ∣ y (t) ∣^{2} d t = 8.00$

Official answer: 8Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Input spectrum. With the normalised sinc, $sinc (t) \leftrightarrow rect (f)$ , and time scaling gives $sinc (a t) \leftrightarrow \frac{1}{a} rect (\frac{f}{a})$ for $a > 0$ . With $a = 2$ : $X (f) = 4 \cdot \frac{1}{2} rect (\frac{f}{2}) = {2, 0, ∣ f ∣ < 1 ∣ f ∣ > 1$
Hilbert transformer. Its frequency response is $H (f) = - j sgn (f)$ , so $∣ H (f) ∣ = 1$ for every $f \neq = 0$ . So $Y (f) = - j sgn (f) X (f)$ , and $∣ Y (f) ∣^{2} = ∣ X (f) ∣^{2}$ everywhere except at the single point $f = 0$ . A single point adds nothing to an integral.
Parseval's theorem: $\int_{- \infty}^{\infty} ∣ y (t) ∣^{2} d t = \int_{- \infty}^{\infty} ∣ Y (f) ∣^{2} df = \int_{- 1}^{1} 2^{2} df = 4 \times 2 = 8$
Check in the time domain. Since $\int_{- \infty}^{\infty} sinc^{2} (a t) d t = \frac{1}{a}$ , the input energy is $16 \cdot \frac{1}{2} = 8$ . The Hilbert transform does not change the energy, so $E_{y} = 8.00$ .

KEY IDEA

A Hilbert transformer is an all-pass filter, $H (f) = - j sgn (f)$ with $∣ H (f) ∣ = 1$ . It only shifts phase by $\mp 9 0^{\circ}$ , so the output energy equals the input energy. Find the input energy with Parseval: $4 sinc (2 t)$ has a flat spectrum of height $2$ over $∣ f ∣ < 1$ , so the energy is $2^{2} \times 2 = 8$ .

WATCH OUT FOR

(1) Reading $sinc (2 t)$ as $\frac{s i n 2 t}{2 t}$ instead of the stated normalised form $\frac{s i n 2 π t}{2 π t}$ . That gives $16 \cdot \frac{π}{2} = 8 π \approx 25.13$ . (2) Taking the scaled spectrum as $rect (f /2)$ without the $\frac{1}{2}$ factor, which gives 32. (3) Treating the Hilbert transformer like the analytic-signal operation, which removes negative frequencies, and halving the energy to 4. The Hilbert transformer keeps every frequency and changes only its phase.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 22 Fourier transform questions, with notes

GATE 2018 EC Q.39: solution and answer

Answer and solution

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