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  • Question

  • 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


  • Options
  • A. 1/100
  • B. 1/200
  • C. 1/300
  • D. 1/600

  • 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

    The analog signal m(t) is given below m(t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt, the Nyquist s.

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