Discussion
The Nyquist rate of the given analogue signal is:
\(X(t) = 3\cos 50\pi t + 10\sin 300\pi t\)
Answer: C
The frequencies in the signal are found from the angular frequencies: \(50\pi\) rad/s is 25 Hz and \(300\pi\) rad/s is 150 Hz.
\[ f_{max} = \frac{300\pi}{2\pi} = 150\ \text{Hz} \]
The Nyquist rate is twice the highest frequency in the signal.
\[ f_N = 2 \times 150 = 300\ \text{Hz} \]
The value 150 Hz is the highest frequency itself and 50 Hz is twice the lower component, and 200 Hz does not come from either of them. The answer is 300 Hz, option C.
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