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
Answer: C
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