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Nyquist rate

  1. Apr 21, 2012 #1
    Hello, I have the following Fourier complex signal:
    v(t) = 2/πsin(500πt)+1/2sin(1000πt)+1/3sin(1500πt)

    I need to find the minimum sampling rate for this signal.
    So, according to the Nyquist Theorem, the sampling rate must be twice the highest frequency component contained in the original signal.

    So I assume my sampling rate would be twice the third harmonic (highest frequency component).

    Now, because the format of a sine wave is expressed as 'sin(2πft)', I assume each component within brackets represents '2πft'. So for instance:
    (500πt) = 2πft, so f = 250
    (1000πt) = 2πft, so f = 500
    (1500πt) = 2πft, so f = 750

    Therefore, my Nyquist sampling rate would need to be twice the highest frequency component which would mean a sampling rate of 2 * 750 = 1500 samples per second ?

    Any confirmation of whether I am correct here or not would be great.

    Thanks kindly.
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 22, 2012 #2


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    Staff: Mentor

    Looks right! :smile:
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