GATE 2023 EE Q.50: solution and answer

GATE 2023 · EEQ.50 · 2 marks · NAT

The discrete-time Fourier transform of a signal $x [n]$ is $X (Ω) = (1 + cos Ω) e^{- j Ω}$ . Consider that $x_{p} [n]$ is a periodic signal of period $N = 5$ such that

x_{p} [n] = x [n], for n = 0, 1, 2 = 0, for n = 3, 4

Note that $x_{p} [n] = \sum_{k = 0}^{N - 1} a_{k} e^{j \frac{2 π}{N} k n}$ . The magnitude of the Fourier series coefficient $a_{3}$ is ________ (Round off to 3 decimal places).

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Answer and solution

ANSWER

$∣ a_{3} ∣ = \frac{1}{5} (1 + cos \frac{6 π}{5}) = \frac{3 - 5}{20} \approx 0.038$ .

Official answer: 0.037 to 0.039Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Expand the DTFT into exponentials (Euler). $X (Ω) = e^{- j Ω} + \frac{1}{2} (e^{j Ω} + e^{- j Ω}) e^{- j Ω} = \frac{1}{2} + e^{- j Ω} + \frac{1}{2} e^{- j 2Ω}$
Read off $x [n]$ by comparing with $X (Ω) = \sum_{n} x [n] e^{- j Ω n}$ . $x [0] = \frac{1}{2}$ , $x [1] = 1$ , $x [2] = \frac{1}{2}$ , and $x [n] = 0$ for every other $n$ . So $x [n]$ lies entirely inside $0 \leq n \leq 2$ , and one period of $x_{p} [n]$ ( $n = 0, \dots, 4$ ) is exactly $x [n]$ followed by the zeros at $n = 3, 4$ .
DTFS coefficients are samples of the DTFT. With the synthesis form given in the question, the analysis equation is $a_{k} = \frac{1}{N} \sum_{n = 0}^{N - 1} x_{p} [n] e^{- j \frac{2 π}{N} k n}$ . Since one period equals $x [n]$ and $x [n]$ is zero outside that period, the sum is the DTFT evaluated at $Ω = \frac{2 π k}{N}$ : $a_{k} = \frac{1}{N} X (Ω)_{Ω = \frac{2 π k}{N}}$
Evaluate at $k = 3$ , $N = 5$ . $a_{3} = \frac{1}{5} (1 + cos \frac{6 π}{5}) e^{- j \frac{6 π}{5}}$
Take the magnitude. Since $∣ e^{- j θ} ∣ = 1$ and $1 + cos θ \geq 0$ , $∣ a_{3} ∣ = \frac{1}{5} (1 + cos \frac{6 π}{5})$ . Here $cos \frac{6 π}{5} = - cos \frac{π}{5} = - \frac{1 + 5}{4} = - 0.80902$ , so $∣ a_{3} ∣ = \frac{1 - 0.80902}{5} = \frac{3 - 5}{20} = 0.03820$
Check by direct summation (optional). $5 a_{3} = \frac{1}{2} + e^{- j \frac{6 π}{5}} + \frac{1}{2} e^{- j \frac{12 π}{5}} = (0.5 - 0.80902 + 0.15451) + j (0.58779 - 0.47553) = - 0.15451 + j 0.11226$ , whose magnitude is $0.19098$ , giving $∣ a_{3} ∣ = 0.03820$ again.
Round to 3 decimal places: $∣ a_{3} ∣ \approx 0.038$ .

KEY IDEA

When one period of a periodic signal is a finite-length signal $x [n]$ padded with zeros, its DTFS coefficients are the DTFT of $x [n]$ sampled at $N$ equally spaced frequencies and divided by $N$ : $a_{k} = \frac{1}{N} X (\frac{2 π k}{N})$ . Here $x [n] = {\frac{1}{2}, 1, \frac{1}{2}}$ fits inside $0 \leq n \leq 4$ , so nothing overlaps (no time-domain aliasing) and $a_{3} = \frac{1}{5} X (\frac{6 π}{5})$ .

WATCH OUT FOR

Forgetting the $\frac{1}{N}$ factor, which gives $∣ X (6 π /5) ∣ = 0.191$ instead of $0.038$ . Another mistake is using $cos (6 π /5) = + 0.809$ (the angle $21 6^{\circ}$ is in the third quadrant, so its cosine is negative), which gives $0.362$ . Students also sometimes sample at $Ω = 2 π k$ or $π k / N$ instead of $2 π k / N$ .

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GATE 2023 EE Q.50: solution and answer

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