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

The analog signal () is given below () = 4 cos 100 + 8 sin 200 + cos

1.	The analog signal () is given below () = 4 cos 100 + 8 sin 200 + cos 300 , the Nyquist sampling rate will be
Answer» 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 .