The GATE Grind

GATE 2021 EC – Question 14

Networks, Signals and Systems · Continuous-time Signals · 1 mark · Multiple choice

Consider a real-valued base-band signal $x(t)$, band limited to 10 kHz. The Nyquist rate for the signal $y(t)=x(t)\,x\left(1+\frac t2\right)$ is

  1. 15 kHz
  2. 30 kHz
  3. 60 kHz
  4. 20 kHz

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Correct answer: (B) 30 kHz

Explanation

$x\left(1+\frac t2\right)$ is $x(t)$ shifted and stretched in time by a factor of 2, so its spectrum is compressed and it is band limited to $\frac{10}{2}=5$ kHz. Multiplication in time convolves the spectra, so $y(t)$ is band limited to $10+5=15$ kHz. The Nyquist rate is $2\times15=30$ kHz.