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  1. #1
    ChaosThyre81
    Guest

    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


    Are you talking about the amount of energy the transmitter is using to
    send out the information? Or the actual amount of radio energy it is
    sending out in -dbm that reaches the tower from the phone and vice-versa
    has to be higher?

    M.


    --
    ChaosThyre81



    See More: Cellular Interference




  2. #2
    Klein
    Guest

    Re: Cellular Interference

    On Sat, 19 Mar 2005 03:45:27 +0000, ChaosThyre81
    <[email protected]> wrote:

    >
    >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
    >
    >
    >Are you talking about the amount of energy the transmitter is using to
    >send out the information? Or the actual amount of radio energy it is
    >sending out in -dbm that reaches the tower from the phone and vice-versa
    >has to be higher?


    The received energy per bit is determined by the required signal to
    noise ratio (Eb/No) and the received interference and noise spectral
    density. The required transmitted energy per bit is equal to the
    required received energy per bit plus the path loss. The path loss is
    a function of frequency, antenna height and (most important) distance.
    Also to be included in the link power budget are the receiver antenna
    gain and the transmitter antenna gain. In cellular communications,
    the path loss is usually a function of the fourth power of distance
    (as opposed to second power in free-space transmission).

    Klein



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