GATE 2023 IN Q.36: solution and answer

GATE 2023 · INQ.36 · 2 marks · MCQ

The impulse response of an LTI system is $h (t) = δ (t) + 0.5 δ (t - 4)$ , where $δ (t)$ is the continuous-time unit impulse signal. If the input signal $x (t) = cos (\frac{7 π}{4} t)$ , the output is ______.

(A)
$0.5 cos (\frac{7 π}{4} t)$
(B)
$1.5 cos (\frac{7 π}{4} t)$
(C)
$0.5 sin (\frac{7 π}{4} t)$
(D)
$1.5 sin (\frac{7 π}{4} t)$

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Answer and solution

ANSWER

The output is $y (t) = 0.5 cos (\frac{7 π}{4} t)$ , option (A).

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

Convolving a signal with $δ (t - t_{0})$ delays it by $t_{0}$ : $x (t) * δ (t - t_{0}) = x (t - t_{0})$ . Convolution is linear, so $y (t) = x (t) * h (t) = x (t) + 0.5 x (t - 4) .$
Substitute the input: $y (t) = cos (\frac{7 π}{4} t) + 0.5 cos (\frac{7 π}{4} (t - 4)) = cos (\frac{7 π}{4} t) + 0.5 cos (\frac{7 π}{4} t - 7 π)$
$7 π$ is an odd multiple of $π$ , so $cos (θ - 7 π) = cos (θ - π) = - cos θ$ .
So $y (t) = cos (\frac{7 π}{4} t) - 0.5 cos (\frac{7 π}{4} t) = 0.5 cos (\frac{7 π}{4} t) .$
Check with the frequency response (a sinusoid is an eigenfunction of an LTI system): $H (j ω) = 1 + 0.5 e^{- j 4 ω}$ . At $ω = \frac{7 π}{4}$ we get $4 ω = 7 π$ and $e^{- j 7 π} = cos 7 π - j sin 7 π = - 1$ , so $H = 1 - 0.5 = 0.5∠ 0^{\circ}$ . The output is the input scaled by $0.5$ with no phase shift, $0.5 cos (\frac{7 π}{4} t)$ : option (A).

KEY IDEA

An impulse at $t_{0}$ in $h (t)$ adds a copy of the input delayed by $t_{0}$ . Here the delay of 4 s is a phase lag of $\frac{7 π}{4} \times 4 = 7 π$ rad, an odd multiple of $π$ . That turns the delayed copy into $- 0.5 cos (\frac{7 π}{4} t)$ , which cancels half of the direct term. Put another way, $∣ H (j ω) ∣ = ∣1 + 0.5 e^{- j 4 ω} ∣$ is at its minimum of $0.5$ , with zero phase, at this frequency.

WATCH OUT FOR

Adding the two impulse strengths without accounting for the delay ( $1 + 0.5 = 1.5$ ) gives option (B). That is only right if the phase lag $4 ω$ is an even multiple of $π$ . A second trap is thinking the delay turns cosine into sine (options C and D). A phase shift of $7 π$ only flips the sign; a sine would need an odd multiple of $π /2$ .

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 31 LTI & convolution questions, with notes

GATE 2023 IN Q.36: solution and answer

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