GATE 2023 EC Q.48: solution and answer

GATE 2023 · ECQ.48 · 2 marks · MCQ

Let an input $x [n]$ having discrete time Fourier transform $X (e^{j Ω}) = 1 - e^{- j Ω} + 2 e^{- 3 j Ω}$ be passed through an LTI system. The frequency response of the LTI system is $H (e^{j Ω}) = 1 - \frac{1}{2} e^{- j 2Ω}$ . The output $y [n]$ of the system is

(A)
$δ [n] + δ [n - 1] - \frac{1}{2} δ [n - 2] - \frac{5}{2} δ [n - 3] + δ [n - 5]$
(B)
$δ [n] - δ [n - 1] - \frac{1}{2} δ [n - 2] - \frac{5}{2} δ [n - 3] + δ [n - 5]$
(C)
$δ [n] - δ [n - 1] - \frac{1}{2} δ [n - 2] + \frac{5}{2} δ [n - 3] - δ [n - 5]$
(D)
$δ [n] + δ [n - 1] + \frac{1}{2} δ [n - 2] + \frac{5}{2} δ [n - 3] + δ [n - 5]$

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Answer and solution

ANSWER

$y [n] = δ [n] - δ [n - 1] - \frac{1}{2} δ [n - 2] + \frac{5}{2} δ [n - 3] - δ [n - 5]$ , which is option (C).

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

Convolution property of the DTFT: for an LTI system, $Y (e^{j Ω}) = X (e^{j Ω}) H (e^{j Ω})$ .
Write $u = e^{- j Ω}$ , so $X = 1 - u + 2 u^{3}$ and $H = 1 - \frac{1}{2} u^{2}$ . By the time-shift property, a term $a u^{k} = a e^{- j k Ω}$ is the DTFT of $a δ [n - k]$ .
Multiply the two polynomials term by term: $Y = (1 - u + 2 u^{3}) (1 - \frac{1}{2} u^{2}) = 1 - u + 2 u^{3} - \frac{1}{2} u^{2} + \frac{1}{2} u^{3} - u^{5}$
Collect like powers. The only overlap is at $u^{3}$ , where $2 + \frac{1}{2} = \frac{5}{2}$ : $Y (e^{j Ω}) = 1 - e^{- j Ω} - \frac{1}{2} e^{- j 2Ω} + \frac{5}{2} e^{- j 3Ω} - e^{- j 5Ω}$
Take the inverse DTFT term by term: $y [n] = δ [n] - δ [n - 1] - \frac{1}{2} δ [n - 2] + \frac{5}{2} δ [n - 3] - δ [n - 5]$
Check the signs against the options. Only option (C) has $+ \frac{5}{2}$ on $δ [n - 3]$ and $- 1$ on $δ [n - 5]$ (A, B and D all have $+ δ [n - 5]$ ).

KEY IDEA

When both $X (e^{j Ω})$ and $H (e^{j Ω})$ are finite sums of powers of $e^{- j Ω}$ , the output transform is just their polynomial product, and each $e^{- j k Ω}$ turns back into $δ [n - k]$ . You can check it in the time domain too: $h [n] = δ [n] - \frac{1}{2} δ [n - 2]$ , so $y [n] = x [n] - \frac{1}{2} x [n - 2]$ with $x [n] = δ [n] - δ [n - 1] + 2 δ [n - 3]$ . That gives the same five terms.

WATCH OUT FOR

Sign slips in the cross terms. The product $(- e^{- j Ω}) (- \frac{1}{2} e^{- j 2Ω}) = + \frac{1}{2} e^{- j 3Ω}$ adds to $2 e^{- j 3Ω}$ to give $+ \frac{5}{2}$ , not $- \frac{5}{2}$ . Also, $(2 e^{- j 3Ω}) (- \frac{1}{2} e^{- j 2Ω}) = - e^{- j 5Ω}$ is negative. Getting either sign wrong leads to option (A) or (B).

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 11 DTFT & DFT questions, with notes

GATE 2023 EC Q.48: solution and answer

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