[BC] Total Power Of An AM Directional Array

[Paul B. Walker, Jr.]

What is the dumb quesiton fee charge?

Paul



On 4/4/07, Thomas G. Osenkowsky <tosenkowsky at prodigy.net> wrote:
>
> The total power in a DA system would be base current
> squared times base operating resistance. There will be
> phasing/coupling system and ground losses. The FCC
> allows an increase in transmitter power to compensate
> for these. Part of the base current will be inclusive of
> the ground losses.
>
> Some people mistakenly believe all the base currents should
> add up to equal the common point current. This is NOT so!
> Also, some measure the base operating impedance with an
> OIB and then do the I squared R calculation. The OIB has
> an insertion effect which varies with each tower in each array.
> This is not an accurate means of determining total radiated
> power.
>
> Some DA's exhibit 'gain' or a higher RMS than their counterpart
> non-D radiator. These designs, however, usually have degraded
> bandwidth (high RSS:RMS ratio) and low operating impedances.
> Some patterns have a 'negative tower' which returns power to the
> common point. A negative tower is NOT one with a lagging phase
> angle.
>
> Yours was not a dumb question, therefore you will not be charged
> a dumb question fee.
>
> Tom Osenkowsky, CPBE
>
> > In an AM multiple tower directional array, does the power of
> > all towers as determined by "P equals I squared times R" add
> > up to the authorized power of the station?
> > For example, for a one thousand watt station with four towers
> > would the power of each tower add up to one thousand watts?
> > Thanks to all who have answered my occasional questions to
> > satisfy my curiosity about radio, even if the questions
> > appear to be dumb ones.
> > Tom
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-- 
Paul B. Walker, Jr.
www.walkerbroadcasting.com
walkerbroadcasting at gmail.com