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Determining Frequencies that Exist in a Signal

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Using trig identities from a calculus book or other, write out the results to the following modulations. State which frequencies exist in the signal s(t).

    2. Relevant equations
    a) s(t) = Acos(2∏f₁t) * Bcos(2∏f₂t)


    3. The attempt at a solution
    cos(s) * cos(t) = cos (s + t)/2 + cos (s – t)/2

    s(t) = ((AB)/2)*(cos(2∏t(f1 + f2)) + cos(2∏t(f1 - f2)))

    What I don't quite understand is the part that asks to state which frequencies exist in the signal. I understand that in modulation, half of the signal is shifted to the right by f2, then the other half of the signal is shifted to the left by f2. Which would make the answers f2 and -f2.

    According to the answer it should be f1, f2, |f2-f1|, f1+f2.
     
    Last edited: Jul 29, 2012
  2. jcsd
  3. Jul 30, 2012 #2

    CWatters

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    Your maths looks right.

    s(t) = ((AB)/2)*(cos(2∏t(f1 + f2)) + cos(2∏t(f1 - f2)))

    is similar to the sum of two waves..

    = Cos(2∏Fusbt) + Cos(2∏Flsbt)

    when

    Fusb = upper side band frequency = f1+f2
    Flsb = lower side band frequency = f1-f2
     
  4. Jul 30, 2012 #3

    CWatters

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    In radio signal f2 would be the speech/music and f1 the RF carrier. Since speech and music are not pure tones the result is a band of frequencies either side of the carrier as shown in the diagram

    http://en.wikipedia.org/wiki/Sideband
     
  5. Aug 1, 2012 #4

    rude man

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    The answer is wrong. f1 and f2 do not exist in the product, just the sums and differences, just as you have derived.

    In ordinary amplitude modulation (think radio broadcast AM), , they do, but you have what is called "double-sideband, suppressed-carrier" modulation. Think of f1 as the carrier and f2 as the modulation, then you can see why it's called what it is.
     
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