On 24 ene, 02:29, "Kaimbridge" wrote:
I've seen the different conversions for mW, mV, mV/m, dBm, mostly based
on 50 ohms.
What I'm interested in is converting a given field strength mV/m, at a
given distance for a particular frequency, to the equivalent (decayed)
baseline wattage.
For instance, in section 73.318, the FCC defines the FM blanket
overload boundary as 562 mV/m (115 dBu), having a distance equalling
.394 * KW^.5 (ERP), where (I believe) ".394" is supposed to be
50000^.5/562.
So, assuming flat terrain with "raw reception" (meaning no antenna and
line gain/loss), what would the equivalent ERP wattage be (i.e., if you
took a field strength meter and held it next to a transmitter, what
would the wattage be to produce 562 mV/m?)?
Since wattge can be found from mV/m, what would the equation for
decayed wattage (mW_d) be, given an ERP (KW_erp), frequency (MHz) and
distance (Km)? Instead of 50, should 2*Pi*60 Hz (~=~ 377) be used for
ohms?
~Kaimbridge~
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Hello,
The formula you should use (or propagation graph) to calculate the ERP
(or input wattage), based on field strength measurement, depends on
many factors like: Frequency, height of transmitter and receiver,
ground properties, antenna gain, etc.
The reason for that is that at certain distance from the transmitting
antenna, the ground reflected wave interferes with the direct wave.
Whether the reflected wave will result in destructive or constructive
interference depends on the path length difference and ground
reflection coefficient. At large difference, there is always
destructive interference (this causes that in reality for VHF and low
UHF, field strength is inversely proportional to distance squared).
Maybe you can give some more details about antenna type, transmitter
height, distance, frequency, etc. One formula to calculate the path
loss that incorporates ground reflection is the "two-ray propagation
formula". It comes in two versions one without sinusoidal functions
(the simple one) and the complete one (with goniometric functions) that
also gives reasonable results when the antenna is more close to the
antenna.
At large distance other phenomena become dominant (for example over the
horizon diffraction).
Hope this will help you,
Wim
PA3DJS