GATE 2024 EC Q.14: solution and answer

GATE 2024 · ECQ.14 · 1 mark · MCQ

A digital communication system transmits through a noiseless bandlimited channel $[- W W]$ . The received signal $z (t)$ at the output of the receiving filter is given by $z (t) = \sum_{n} b [n] x (t - n T)$ , where $b [n]$ are the symbols and $x (t)$ is the overall system response to a single symbol. The received signal is sampled at $t = m T$ . The Fourier transform of $x (t)$ is $X (f)$ . The Nyquist condition that $X (f)$ must satisfy for zero intersymbol interference at the receiver is ______.

(A)
$\sum_{m = - \infty}^{\infty} X (f + \frac{m}{T}) = T$
(B)
$\sum_{m = - \infty}^{\infty} X (f + \frac{m}{T}) = \frac{1}{T}$
(C)
$\sum_{m = - \infty}^{\infty} X (f + m T) = T$
(D)
$\sum_{m = - \infty}^{\infty} X (f + m T) = \frac{1}{T}$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

Option A: $\sum_{m = - \infty}^{\infty} X (f + \frac{m}{T}) = T$ , the Nyquist zero-ISI criterion with the pulse normalised so that $x (0) = 1$ .

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Sample the output at $t = m T$ : $z (m T) = \sum_{n} b [n] x ((m - n) T) = b [m] x (0) + \sum_{n \neq = m} b [n] x ((m - n) T)$ .
The second sum is the intersymbol interference. It vanishes for arbitrary symbols only if $x (k T) = 0$ for every integer $k \neq = 0$ . Taking the usual normalisation $x (0) = 1$ , the time-domain Nyquist criterion is $x (k T) = δ [k]$ .
Form the ideally sampled signal $x_{s} (t) = \sum_{k} x (k T) δ (t - k T) = δ (t)$ . Its Fourier transform is $X_{s} (f) = 1$ for all $f$ .
Impulse sampling at spacing $T$ replicates the (two-sided) spectrum at multiples of the sampling rate $1/ T$ : $X_{s} (f) = \frac{1}{T} m = - \infty \sum \infty X (f - \frac{m}{T}) .$
Set $X_{s} (f) = 1$ : $\frac{1}{T} \sum_{m} X (f - \frac{m}{T}) = 1$ . Since $m$ runs over all integers, replacing $m$ by $- m$ gives $m = - \infty \sum \infty X (f + \frac{m}{T}) = T .$
The shifts are $m / T$ in frequency (multiples of the symbol rate), not $m T$ , which rules out C and D. With $x (0) = 1$ the right-hand side is $T$ , not $1/ T$ , which rules out B. The answer is option A.

KEY IDEA

Zero ISI means $x (k T) = δ [k]$ . Sampling at interval $T$ repeats the spectrum every $1/ T$ Hz and scales it by $1/ T$ . The sampled pulse is a single impulse, whose transform is 1, so the shifted copies $X (f + m / T)$ must add up to the constant $T$ .

WATCH OUT FOR

Shifting the copies by $m T$ instead of $m / T$ : the spectral copies sit at multiples of the symbol rate $1/ T$ . Another slip is forgetting the $1/ T$ factor in the sampling theorem, which leads to $1/ T$ on the right-hand side (option B) instead of $T$ . The right-hand side assumes $x (0) = 1$ ; with a general $x (0)$ it would be $T x (0)$ .

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 15 Digital communication questions, with notes

GATE 2024 EC Q.14: solution and answer

Answer and solution

More Digital communication questions