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Carsons rule, which bits go where?

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data
    calculate the frequency deviation using an approximate technique


    2. Relevant equations
    input analogue signal is 1.2kHz
    min bandwidth to avoid over-sampling is 3.6kHz
    signal transmitted over 1.1km with a 12kHz bandwidth on a FM carrier
    Carrier freq is in the GSM reverse band at 875MHz


    3. The attempt at a solution

    Bandwidth requirement (CBR) =2(Δf + fm)
    0.5CBR - fm= Δf = freq deviation

    does CBR=12kHz and fm=875MHz?
    Don't know what goes where
     
  2. jcsd
  3. Dec 6, 2012 #2
    CBR is the total bw estimate of the FM signal (12kHz)

    fm is the bandwidth of the message signal

    Δf is peak frequency deviation of the instantaneous FM frequency and depends on the amplitude of the message signal.
     
    Last edited: Dec 6, 2012
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