GATE 2022 EE Q.49: solution and answer

GATE 2022 · EEQ.49 · 2 marks · MCQ

The discrete time Fourier series representation of a signal $x [n]$ with period $N$ is written as $x [n] = \sum_{k = 0}^{N - 1} a_{k} e^{j (2 k nπ / N)}$ . A discrete time periodic signal with period $N = 3$ , has the non-zero Fourier series coefficients: $a_{- 3} = 2$ and $a_{4} = 1$ . The signal is

(A)
$2 + 2 e^{- (j \frac{2 π}{6} n)} cos (\frac{2 π}{6} n)$
(B)
$1 + 2 e^{(j \frac{2 π}{6} n)} cos (\frac{2 π}{6} n)$
(C)
$1 + 2 e^{(j \frac{2 π}{3} n)} cos (\frac{2 π}{6} n)$
(D)
$2 + 2 e^{(j \frac{2 π}{6} n)} cos (\frac{2 π}{6} n)$

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Answer and solution

ANSWER

The signal is $x [n] = 2 + e^{j \frac{2 π}{3} n} = 1 + 2 e^{j \frac{2 π}{6} n} cos (\frac{2 π}{6} n)$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

DTFS coefficients repeat with period $N$ : $a_{k} = a_{k + N}$ . With $N = 3$ : $a_{- 3} = a_{0} = 2$ and $a_{4} = a_{1} = 1$ . No other coefficient is non-zero, so $a_{2} = 0$ in the fundamental range $k = 0, 1, 2$ .
Synthesis equation over one period, $k = 0, 1, 2$ , with $ω_{0} = \frac{2 π}{3}$ : $x [n] = a_{0} + a_{1} e^{j \frac{2 π}{3} n} + a_{2} e^{j \frac{4 π}{3} n} = 2 + e^{j \frac{2 π}{3} n}$
Rewrite to match the options. Split the exponent in half, so that $e^{j \frac{2 π}{3} n} = e^{j \frac{2 π}{6} n} e^{j \frac{2 π}{6} n}$ and $1 = e^{j \frac{2 π}{6} n} e^{- j \frac{2 π}{6} n}$ : $1 + e^{j \frac{2 π}{3} n} = e^{j \frac{2 π}{6} n} (e^{- j \frac{2 π}{6} n} + e^{j \frac{2 π}{6} n}) = 2 e^{j \frac{2 π}{6} n} cos (\frac{2 π}{6} n)$
Result: $x [n] = 1 + (1 + e^{j \frac{2 π}{3} n}) = 1 + 2 e^{j \frac{2 π}{6} n} cos (\frac{2 π}{6} n)$ which is option (B).
Check the other options with the same identity: (D) expands to $2 + 1 + e^{j \frac{2 π}{3} n} = 3 + e^{j \frac{2 π}{3} n}$ and (A) to $3 + e^{- j \frac{2 π}{3} n}$ . Both give $4$ at $n = 0$ , but $x [0] = 3$ . (C) expands to $1 + e^{j π n} + e^{j \frac{π}{3} n}$ , which also gives $3$ at $n = 0$ . At $n = 1$ , though, it gives $e^{j \frac{π}{3}}$ , while $x [1] = 2 + e^{j \frac{2 π}{3}} = \frac{3}{2} + j \frac{3}{2}$ . (C) is not even periodic with period $3$ . Only (B) matches.

KEY IDEA

DTFS coefficients are periodic in $k$ with period $N$ , so an index outside $0$ to $N - 1$ maps back into that range: $a_{- 3} = a_{0}$ and $a_{4} = a_{1}$ . The half-angle factorisation $1 + e^{j θ} = 2 e^{j θ /2} cos (θ /2)$ then turns $2 + e^{j \frac{2 π}{3} n}$ into the form used in the options.

WATCH OUT FOR

Treating $a_{- 3}$ and $a_{4}$ as harmonics at $- 3 ω_{0}$ and $4 ω_{0}$ that are separate from $a_{0}$ and $a_{1}$ . A common slip is to add $a_{- 3} + a_{4}$ as the DC term, or to read $2 e^{j \frac{2 π}{6} n} cos (\frac{2 π}{6} n)$ as one term instead of $1 + e^{j \frac{2 π}{3} n}$ . Either way the constant comes out as $2$ , which leads to option (D). Expand each option back into exponentials and check it at one or two values of $n$ .

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GATE 2022 EE Q.49: solution and answer

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