The GATE Grind

GATE 2024 EC – Question 48

Networks, Signals and Systems · Continuous-time Signals · 2 marks · Multiple choice

A continuous time signal $x(t)=2\cos(8\pi t+\pi/3)$ is sampled at a rate of 15 Hz. The sampled signal $x_s(t)$ when passed through an LTI system with impulse response

$$h(t)=\left(\frac{\sin2\pi t}{\pi t}\right)\cos(38\pi t-\pi/2)$$

produces an output $x_o(t)$. The expression for $x_o(t)$ is ______.

  1. $15\sin(38\pi t+\pi/3)$
  2. $15\sin(38\pi t-\pi/3)$
  3. $15\cos(38\pi t-\pi/6)$
  4. $15\cos(38\pi t+\pi/6)$

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Correct answer: (C) $15\cos(38\pi t-\pi/6)$

Explanation

$x(t)$ has components at $\pm4$ Hz, each of amplitude 1 and phase $\pm\pi/3$. Sampling at 15 Hz creates replicas at $\pm4+15k$ Hz, each scaled by $F_s=15$. The filter $h(t)$ is a band-pass at 19 Hz: $\frac{\sin2\pi t}{\pi t}$ has gain 1 for $|f|<1$ Hz, and the modulation splits it into two passbands at $\pm19$ Hz of gain $\frac12e^{\mp j\pi/2}$, i.e. width 18-20 Hz. Only the replica at $19=15+4$ Hz (and its mirror at $-19$) passes. The output is $2\,\mathrm{Re}\{15\cdot\frac12e^{-j\pi/2}e^{j(38\pi t+\pi/3)}\}=15\cos(38\pi t-\pi/6)$.