Signals and Systems

Q1:
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

A 1/100

B 1/200

C 1/300

D 1/600

ANS:C - 1/300

m (t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt Nyquist sampling freq f_s ≤ 2f_m where f_m is highest frequency component in given signal and highest f_m in 3^rd part 2pf_mt = 300 pt f_m = 150 Hz f_s = 2 x 150 p 300 Hz

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