GATE 2020 IN Q.35: solution and answer

GATE 2020 · INQ.35 · 2 marks · NAT

Let $g [n] = {10 n = 0 n = \pm 1, \pm 2, \pm 3, \dots$ and $h [n] = {10 n = 0, 3, 6, 9, \dots otherwise$ .

Consider $y [n] = h [n] \otimes g [n]$ , where $\otimes$ denotes the convolution operator. The value of $y [2]$ is ____

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Answer and solution

ANSWER

$y [2] = 0$

Official answer: 0Source: Official answer key, GATE 2020 IN (IIT Delhi)

Identify $g [n]$ : it equals 1 at $n = 0$ and 0 at every other integer, so $g [n] = δ [n]$ , the unit impulse.
Write the convolution sum: $y [n] = \sum_{k = - \infty}^{\infty} h [k] g [n - k] = \sum_{k = - \infty}^{\infty} h [k] δ [n - k]$ .
Apply the sifting property. $δ [n - k]$ is non-zero only when $k = n$ , so only one term of the sum survives and $y [n] = h [n]$ . Convolving with the unit impulse leaves any sequence unchanged.
Evaluate at $n = 2$ : $y [2] = h [2]$ .
$h [n] = 1$ only for $n \in {0, 3, 6, 9, \dots}$ , the non-negative multiples of 3. Since $2$ is not a multiple of 3, $h [2] = 0$ .
Therefore $y [2] = h [2] = 0$

KEY IDEA

$g [n]$ is the unit impulse $δ [n]$ , and convolving any sequence with $δ [n]$ gives the sequence back: $h [n] \otimes δ [n] = h [n]$ . So $y [2]$ is simply $h [2]$ , which is 0 because 2 is not a multiple of 3.

WATCH OUT FOR

Not noticing that $g [n]$ is just $δ [n]$ and working through the convolution sum term by term, which risks errors with the flipped index. A second mistake is misreading $h [n]$ as 1 at every $n$ up to 9, or as periodic in both directions, and answering 1. It is non-zero only at $n = 0, 3, 6, 9, \dots$

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 31 LTI & convolution questions, with notes

GATE 2020 IN Q.35: solution and answer

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