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  1. #1
    ChaosThyre81
    Guest

    Klein Wrote: (My Question is on the bottom)

    The above kinds of things were never reported with analog cell phones
    because these phones use an FM waveform which is at a constant
    strength, i.e., no pulsations. When the first GSM and TDMA phones
    came into the consumer marketplace, such things began to be reported.
    This happens because the fundamental waveform of these standards is a
    pulsating waveform that has either a 1/8 or 1/3 duty cycle, depending
    on which standard it is (also iDEN does this.) In the case of the US
    TDMA (IS-136) standard, the pulsations occur at a 67 Hz rate. The
    digital speech signals are gathered up over a 20 msec time interval
    and then transmitted in two bursts of 3.33 msec. The rest of the time
    (13.3 msec) other cell phones will transmit. Because the transmitter
    is only on 1/3 of the time, it must transmit at 3 times the power
    level that would be needed if continuous non-TDMA waveforms were being
    used. In the case of GSM, the duty cycle is 1/8 and the burst rate is
    a little over 200 Hz. Of course, the GSM signal has a peak power 8
    times higher than the average power.


    Hello, I'm curious about something:

    "The rest of the time
    (13.3 msec) other cell phones will transmit. Because the transmitter
    is only on 1/3 of the time, it must transmit at 3 times the power
    level that would be needed if continuous non-TDMA waveforms were being
    used. In the case of GSM, the duty cycle is 1/8 and the burst rate is
    a little over 200 Hz. Of course, the GSM signal has a peak power 8
    times higher than the average power."

    Why does the transmitter have to transmit at 3 times the power because
    it's only on a 3rd of the time?

    Thanks!

    M.


    --
    ChaosThyre81



    See More: Cellular Interference




  2. #2
    Klein
    Guest

    Re: Cellular Interference

    On Wed, 16 Mar 2005 19:42:03 +0000, ChaosThyre81
    <[email protected]> wrote:

    >
    >Klein Wrote: (My Question is on the bottom)
    >
    >The above kinds of things were never reported with analog cell phones
    >because these phones use an FM waveform which is at a constant
    >strength, i.e., no pulsations. When the first GSM and TDMA phones
    >came into the consumer marketplace, such things began to be reported.
    >This happens because the fundamental waveform of these standards is a
    >pulsating waveform that has either a 1/8 or 1/3 duty cycle, depending
    >on which standard it is (also iDEN does this.) In the case of the US
    >TDMA (IS-136) standard, the pulsations occur at a 67 Hz rate. The
    >digital speech signals are gathered up over a 20 msec time interval
    >and then transmitted in two bursts of 3.33 msec. The rest of the time
    >(13.3 msec) other cell phones will transmit. Because the transmitter
    >is only on 1/3 of the time, it must transmit at 3 times the power
    >level that would be needed if continuous non-TDMA waveforms were being
    >used. In the case of GSM, the duty cycle is 1/8 and the burst rate is
    >a little over 200 Hz. Of course, the GSM signal has a peak power 8
    >times higher than the average power.
    >
    >
    >Hello, I'm curious about something:
    >
    >"The rest of the time
    >(13.3 msec) other cell phones will transmit. Because the transmitter
    >is only on 1/3 of the time, it must transmit at 3 times the power
    >level that would be needed if continuous non-TDMA waveforms were being
    >used. In the case of GSM, the duty cycle is 1/8 and the burst rate is
    >a little over 200 Hz. Of course, the GSM signal has a peak power 8
    >times higher than the average power."
    >
    >Why does the transmitter have to transmit at 3 times the power because
    >it's only on a 3rd of the time?


    Transmission of 1 bit of information over a certain distance requires
    a certain transmission energy. Energy is the product of power and
    time. If you wish to transmit one bit in t seconds, then a certain
    amount of power will be required. If you wish to transmit one bit in
    t/2 seconds, then twice the power will be required in order for the
    energy to remain the same. Another way to look at is that the average
    power required for a given data rate is independent of the time taken
    to do the transmission. If you decrease the transmission interval,
    the peak power must increase proportionally in order to keep the
    average power the same.

    "There ain't no free lunch."
    Klein



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