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The analog signal m(t) is given below m(t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt, the Nyquist sampling rate will be

Correct Answer: 1/300

Explanation:

m (t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt


Nyquist sampling freq fs ≤ 2fm where fm is highest frequency component in given signal and highest fm in 3rd part


2pfmt = 300 pt


fm = 150 Hz


fs = 2 x 150 p 300 Hz


.


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