GATE 2025 IN Q.54: solution and answer

GATE 2025 · INQ.54 · 2 marks · NAT

Let $X (e^{j ω})$ represent the discrete time Fourier transform of a 4-length sequence $x [n]$ , where $x [0] = 1$ , $x [1] = 2$ , $x [2] = 2$ , and $x [3] = 4$ . $X (e^{j ω})$ is sampled at $ω_{k} = \frac{2 π k}{3}$ to generate a periodic sequence in $k$ with period 3, where $k$ represents an integer. Let $y [n]$ represent another sequence such that its discrete Fourier transform $Y [k]$ is given as $Y [k] = X (e^{j ω_{k}})$ for $0 \leq k \leq 2$ . The value of $y [0]$ is ______ (in integer).

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Answer and solution

ANSWER

$y [0] = x [0] + x [3] = 5$

Official answer: 5Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Sample the DTFT at $N = 3$ equally spaced frequencies $ω_{k} = \frac{2 π k}{3}$ : $Y [k] = X (e^{j 2 π k /3}) = n = 0 \sum 3 x [n] e^{- j 2 π k n /3}, k = 0, 1, 2.$
Take the 3-point IDFT and swap the order of summation: $y [n] = \frac{1}{3} k = 0 \sum 2 Y [k] e^{j 2 π k n /3} = m = 0 \sum 3 x [m] [\frac{1}{3} k = 0 \sum 2 e^{j 2 π k (n - m) /3}], 0 \leq n \leq 2.$
The bracketed sum is $1$ when $n - m$ is a multiple of $3$ and $0$ otherwise. So sampling in frequency aliases in time: the length-4 sequence wraps modulo 3, $y [n] = r \sum x [n + 3 r], 0 \leq n \leq 2.$
For $n = 0$ , the samples with $m \equiv 0 (mod 3)$ are $m = 0$ and $m = 3$ . (Equivalently, $e^{- j 2 π k \cdot 3/3} = e^{- j 2 π k} = 1$ , so $x [3]$ enters $Y [k]$ exactly as a sample at $n = 0$ would.) Hence $y [0] = x [0] + x [3] = 1 + 4 = 5,$ and $y [1] = x [1] = 2$ , $y [2] = x [2] = 2$ .
Numerical check. With $W = e^{- j 2 π /3}$ we have $W^{3} = 1$ and $W + W^{2} = - 1$ . So $Y [0] = 1 + 2 + 2 + 4 = 9$ and $Y [1] = 1 + 2 W + 2 W^{2} + 4 W^{3} = 5 + 2 (W + W^{2}) = 3$ . In the same way $Y [2] = 1 + 2 W^{2} + 2 W^{4} + 4 W^{6} = 5 + 2 (W^{2} + W) = 3$ . Then $y [0] = \frac{1}{3} (9 + 3 + 3) = 5$ .

KEY IDEA

Taking $N$ samples of a DTFT and inverting them with an $N$ -point IDFT gives the time-aliased sequence $y [n] = \sum_{r} x [n + r N]$ . Here the sequence has 4 samples and $N = 3$ , so $x [3]$ folds back onto $n = 0$ .

WATCH OUT FOR

Answering $y [0] = x [0] = 1$ treats the 3-point DFT as though it simply cut off $x [3]$ . That is wrong because $N = 3$ is less than the sequence length 4, so the time-domain aliasing puts $x [3]$ on top of $x [0]$ . A second slip is to take $y [0]$ as $Y [0] = 9$ , the sum of all the samples. $Y [0]$ is a frequency-domain value; $y [0]$ is the average of the three $Y [k]$ .

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