Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by \(x(t) = 20\cos 500\pi t\).
DSSSB AE - 2023,
Answer: C
Write the signal as \(x(t) = 20\cos(2\pi f t)\) with \(2\pi f = 500\pi\), so the frequency is
\[ f = 250\ \text{Hz} \]
To avoid aliasing the sampling rate must be at least twice the highest frequency (the Nyquist rate).
\[ f_s = 2f = 500\ \text{Hz} \]
The value 250 Hz is the signal frequency itself, which is only the Nyquist frequency, and 125 Hz and 375 Hz are not related to it. The answer is 500 Hz, option C.
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