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Discussion :: Signals and Systems

  1. 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

    2. A.
    1/100
    B.
    1/200
    C.
    1/300
    D.
    1/600

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    Answer : Option C

    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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