GATE 2024 IN Q.13: solution and answer

GATE 2024 · INQ.13 · 1 mark · MCQ

A discrete-time sequence is given by $x [n] = [1, 2, 3, 4]$ for $0 \leq n \leq 3$ . The zero lag auto-correlation value of $x [n]$ is

(A)
1
(B)
10
(C)
20
(D)
30

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Answer and solution

ANSWER

The zero-lag autocorrelation equals the energy of the sequence: $R_{xx} [0] = 1^{2} + 2^{2} + 3^{2} + 4^{2} = 30$ . The answer is option (D).

Official answer: DSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

For a real energy sequence, the autocorrelation at lag $l$ is $R_{xx} [l] = n = - \infty \sum \infty x [n] x [n - l]$
Set the lag to zero, $l = 0$ : $R_{xx} [0] = n = - \infty \sum \infty x^{2} [n]$ This is exactly the energy of the sequence.
The sequence is non-zero only for $0 \leq n \leq 3$ , so the sum has four terms: $R_{xx} [0] = 1^{2} + 2^{2} + 3^{2} + 4^{2} = 1 + 4 + 9 + 16 = 30$
The zero-lag autocorrelation is therefore $30$ , which is option (D).

KEY IDEA

At zero lag the sequence is multiplied by an unshifted copy of itself, so $R_{xx} [0] = \sum_{n} x^{2} [n]$ , which is the signal's energy. It is also the largest value the autocorrelation takes at any lag: $R_{xx} [0] \geq ∣ R_{xx} [l] ∣$ .

WATCH OUT FOR

Adding the samples instead of their squares gives $1 + 2 + 3 + 4 = 10$ , option (B), which is a trap. Dividing by the number of samples (average power, $30/4 = 7.5$ ) is also wrong here: the question asks for the autocorrelation of an energy signal, not a time-averaged one.

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GATE 2024 IN Q.13: solution and answer

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