GATE 2022 EC Q.55: solution and answer

GATE 2022 · ECQ.55 · 2 marks · NAT

For a vector $\overset{x}{ˉ} = [x [0], x [1], \dots, x [7]]$ , the 8-point discrete Fourier transform (DFT) is denoted by $\overset{ˉ}{X} = D F T (\overset{x}{ˉ}) = [X [0], X [1], \dots, X [7]]$ , where

X [k] = n = 0 \sum 7 x [n] exp (- j \frac{2 π}{8} nk) .

Here, $j = - 1$ . If $\overset{x}{ˉ} = [1, 0, 0, 0, 2, 0, 0, 0]$ and $\overset{y}{ˉ} = D F T (D F T (\overset{x}{ˉ}))$ , then the value of $y [0]$ is ________ (rounded off to one decimal place).

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$y [0] = 8.0$ .

Official answer: 7.9 to 8.1Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Double-DFT property. Applying the $N$ -point DFT twice gives the input scaled by $N$ and circularly time-reversed: $D F T (D F T (x)) [n] = N x [(- n)_{N}]$ This holds because $k = 0 \sum N - 1 X [k] e^{- j \frac{2 π}{N} nk} = m = 0 \sum N - 1 x [m] k = 0 \sum N - 1 e^{- j \frac{2 π}{N} (m + n) k}$ and the inner sum is $N$ when $m + n \equiv 0 (mod N)$ and $0$ otherwise.
At $n = 0$ the reversal has no effect, because $(- 0)_{8} = 0$ . So $y [0] = 8 x [0] = 8 \times 1 = 8$
Direct check. The only non-zero samples are $x [0] = 1$ and $x [4] = 2$ , and $e^{- j \frac{2 π}{8} (4) k} = e^{- j π k} = (- 1)^{k}$ . So $X [k] = 1 + 2 (- 1)^{k}$
That gives $\overset{ˉ}{X} = [3, - 1, 3, - 1, 3, - 1, 3, - 1]$ .
Second DFT at index 0. At index 0 every exponential equals 1, so the DFT there is just the sum of the samples: $y [0] = k = 0 \sum 7 X [k] = 4 (3) + 4 (- 1) = 8$
Both methods give the same result: $y [0] = 8.0$ .

KEY IDEA

Taking the $N$ -point DFT twice gives $N x [(- n)_{N}]$ , the input scaled by $N$ and circularly reversed. Index 0 does not move under that reversal, so $y [0] = N x [0]$ . Also, the DFT at index 0 is simply the sum of the input samples.

WATCH OUT FOR

Two errors are common. The first is to assume that $D F T (D F T (x))$ gives back $x$ , which leads to the answer 1. That confuses the forward DFT with the inverse DFT, which has the opposite sign in the exponent and a $1/ N$ factor. The second is to forget the factor $N$ and use the time-reversal alone. Some students also add $x [4]$ in as well, getting 3 or 24, when $x [4]$ actually lands at $y [4] = 16$ and not at $y [0]$ .

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 11 DTFT & DFT questions, with notes

GATE 2022 EC Q.55: solution and answer

Answer and solution

More DTFT & DFT questions