Maximum Frequency of a Signal

Discussion in 'Homework Help' started by jegues, Feb 1, 2012.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    If I have a signal,

    f(t) = 2048 + 700cos(2\pi31.25t) - 1100sin(2\pi125t)

    How do I go about finding the maximum frequency?

    What do they really mean by maximum frequency anyways?

    (I haven't taken any signal processing courses yet so my understanding between these distinctions and how to solve for them is still quite minimal)

    Thanks again!
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Is sampling the context? If so, the greater frequency component is 125Hz, and thus your sampler must have at least twice that.
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I'm not quite sure, but I think it is within a sampling context.

    My professor hinted that it should somehow relate to the maximum slope.

    The question was stated as,

     
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
  5. holnis

    Member

    Nov 25, 2011
    50
    4
    The frequencies present in the signal above are

    F1=31.25
    F2=125

    Thus Fmax = 125 Hz and according to the sampling theorem:

    Fs > 2Fmax = 250 hz

    The Nyquist rate is Fn = 2Fmax. ==> Fn = 250 Hz.
     
Loading...