GATE 2020 EC Q.29: solution and answer

GATE 2020 · ECQ.29 · 2 marks · MCQ

A finite duration discrete-time signal $x [n]$ is obtained by sampling the continuous-time signal $x (t) = cos (200 π t)$ at sampling instants $t = n /400$ , $n = 0, 1, \dots, 7$ . The 8-point discrete Fourier transform (DFT) of $x [n]$ is defined as

X [k] = n = 0 \sum 7 x [n] e^{- j \frac{π k n}{4}}, k = 0, 1, \dots, 7.

Which one of the following statements is TRUE?

(A)
All $X [k]$ are non-zero.
(B)
Only $X [4]$ is non-zero.
(C)
Only $X [2]$ and $X [6]$ are non-zero.
(D)
Only $X [3]$ and $X [5]$ are non-zero.

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Answer and solution

ANSWER

Only $X [2]$ and $X [6]$ are non-zero (each equals 4), so option (C) is TRUE.

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

Sample the signal. With $t = n /400$ : $x [n] = cos (200 π \cdot \frac{n}{400}) = cos (\frac{π n}{2})$ , so for $n = 0, \dots, 7$ the samples are $1, 0, - 1, 0, 1, 0, - 1, 0$ .
Write it at a DFT bin frequency. The DFT kernel is $e^{- j \frac{π k n}{4}} = e^{- j \frac{2 π}{8} k n}$ . Since $\frac{π}{2} = \frac{2 π \cdot 2}{8}$ , $x [n] = cos (\frac{2 π \cdot 2 n}{8})$ : exactly 2 whole cycles fit in the 8 samples, so the tone sits exactly on bin $k_{0} = 2$ and nothing leaks into other bins. (Equivalently: $f_{s} = 400 Hz$ , bin spacing $f_{s} / N = 50 Hz$ , and $100 Hz$ falls exactly on bin 2.)
Euler's form. $x [n] = \frac{1}{2} e^{j \frac{2 π}{8} 2 n} + \frac{1}{2} e^{- j \frac{2 π}{8} 2 n} = \frac{1}{2} e^{j \frac{2 π}{8} 2 n} + \frac{1}{2} e^{j \frac{2 π}{8} 6 n}$ , since $e^{j \frac{2 π}{8} 8 n} = 1$ , i.e. $- 2 \equiv 6 (mod 8)$ .
Orthogonality of DFT exponentials. $n = 0 \sum N - 1 e^{j \frac{2 π}{N} (m - k) n} = N$ if $k \equiv m (mod N)$ , and $0$ otherwise.
Result. Substituting, $X [k] = \frac{1}{2} n = 0 \sum 7 e^{j \frac{2 π}{8} (2 - k) n} + \frac{1}{2} n = 0 \sum 7 e^{j \frac{2 π}{8} (6 - k) n} = 4 δ [k - 2] + 4 δ [k - 6]$ , so $X [2] = X [6] = 4$ and every other $X [k] = 0$ .
Direct check on the bins that decide the options. Only even $n$ contribute, with $x [0] = x [4] = 1$ , $x [2] = x [6] = - 1$ . $X [0] = 1 - 1 + 1 - 1 = 0$ (rules out A). $X [4] = \sum x [n] (- 1)^{n} = 1 - 1 + 1 - 1 = 0$ (rules out B). $X [2] = \sum x [n] (- j)^{n} = 1 \cdot 1 + (- 1) (- 1) + 1 \cdot 1 + (- 1) (- 1) = 4$ and $X [6] = \sum x [n] (j)^{n} = 1 + 1 + 1 + 1 = 4$ (rules out D, which says only $X [3]$ and $X [5]$ are non-zero). Only $X [2]$ and $X [6]$ are non-zero, which is option (C).

KEY IDEA

Sampling at $400 Hz$ gives $x [n] = cos (π n /2) = cos (\frac{2 π \cdot 2 n}{8})$ , which completes exactly 2 cycles in the 8-point window. A cosine sitting exactly on a DFT bin $k_{0}$ puts all its energy into bins $k_{0}$ and $N - k_{0}$ , each with value $N /2$ . Here that means $X [2] = X [6] = 4$ and every other bin is zero, with no leakage.

WATCH OUT FOR

Picking bin 4 by treating $π /2$ as "half of $π$ " and so half of $N = 8$ , instead of writing $π /2 = 2 π k_{0} /8$ , which gives $k_{0} = 2$ . The other common mistake is forgetting the negative-frequency half: the $e^{- j \frac{2 π}{8} 2 n}$ term lands at $k = 8 - 2 = 6$ , not at $k = - 2$ . Expecting leakage into every bin (option A) is also wrong here, because an integer number of cycles fits in the window.

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GATE 2020 EC Q.29: solution and answer

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