GATE 2023 EC Q.47: solution and answer

GATE 2023 · ECQ.47 · 2 marks · MCQ

Consider a discrete-time periodic signal with period $N = 5$ . Let the discrete-time Fourier series (DTFS) representation be $x [n] = \sum_{k = 0}^{4} a_{k} e^{\frac{j k 2 π n}{5}}$ , where $a_{0} = 1$ , $a_{1} = 3 j$ , $a_{2} = 2 j$ , $a_{3} = - 2 j$ and $a_{4} = - 3 j$ . The value of the sum $\sum_{n = 0}^{4} x [n] sin \frac{4 π n}{5}$ is

(A)
$- 10$
(B)
$10$
(C)
$- 2$
(D)
$2$

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Answer and solution

ANSWER

The sum is $- 10$ . This is option (A).

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

DTFS analysis equation: $a_{k} = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j k \frac{2 π}{N} n}$ . With $N = 5$ this gives $\sum_{n = 0}^{4} x [n] e^{- j k \frac{2 π}{5} n} = 5 a_{k}$ .
Euler's formula: $sin \frac{4 π n}{5} = \frac{1}{2 j} (e^{j \frac{4 π n}{5}} - e^{- j \frac{4 π n}{5}})$ .
Identify each exponential with an analysis kernel. Since $e^{- j 2 π n} = 1$ for integer $n$ : $e^{j \frac{4 π n}{5}} = e^{j \frac{4 π n}{5}} e^{- j 2 π n} = e^{- j \frac{2 π ( 3 ) n}{5}}$ , which is the $k = 3$ kernel. The term $e^{- j \frac{4 π n}{5}} = e^{- j \frac{2 π ( 2 ) n}{5}}$ is the $k = 2$ kernel.
So $n = 0 \sum 4 x [n] sin \frac{4 π n}{5} = \frac{1}{2 j} (5 a_{3} - 5 a_{2}) = \frac{5}{2 j} (- 2 j - 2 j) = \frac{5 ( - 4 j )}{2 j} = - 10.$
Cross-check by writing $x [n]$ out. Because $a_{4} = a_{1}^{*}$ and $a_{3} = a_{2}^{*}$ , $x [n]$ is real, and $e^{j \frac{2 π ( 4 ) n}{5}} = e^{- j \frac{2 π n}{5}}$ , $e^{j \frac{2 π ( 3 ) n}{5}} = e^{- j \frac{4 π n}{5}}$ . So $x [n] = 1 + 2 Re (3 j e^{j \frac{2 π n}{5}}) + 2 Re (2 j e^{j \frac{4 π n}{5}}) = 1 - 6 sin \frac{2 π n}{5} - 4 sin \frac{4 π n}{5}$ .
Orthogonality over one period: the constant term and the $- 6 sin \frac{2 π n}{5}$ term both sum to zero against $sin \frac{4 π n}{5}$ . Only the $- 4 sin \frac{4 π n}{5}$ term is left, and $\sum_{n = 0}^{4} sin^{2} \frac{4 π n}{5} = \sum_{n = 0}^{4} \frac{1 - c o s \frac{8 π n}{5}}{2} = \frac{5}{2}$ . That gives $- 4 \times \frac{5}{2} = - 10$ , the same result. Answer: option (A).

KEY IDEA

A sum of the form $\sum_{n = 0}^{N - 1} x [n] e^{- j k \frac{2 π}{N} n}$ is just $N a_{k}$ , so you never need to compute $x [n]$ . Split the sine into two complex exponentials with Euler's formula. Then use $N$ -periodicity in $k$ (here $a_{- 2} = a_{3}$ ) to match each exponential to a coefficient you were given.

WATCH OUT FOR

The usual slip is sending $e^{+ j \frac{4 π n}{5}}$ to $a_{2}$ instead of $a_{3}$ . The analysis kernel carries a minus sign, so $e^{+ j \frac{4 π n}{5}}$ belongs to $k = - 2 \equiv 3$ . Swapping $a_{2}$ and $a_{3}$ gives $+ 10$ , option (B). Two other mistakes are leaving out the factor $N = 5$ from the analysis equation, which gives $\pm 2$ (options C and D), and forgetting the $\frac{1}{2 j}$ from Euler's formula.

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GATE 2023 EC Q.47: solution and answer

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