A.H. Systems PAM-0118P Operation Manual - Page 10

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A.H. Systems Model PAM-0118P Pre-Amplifier
TYPICAL CONVERSION FORMULAS
The constant in the above equation is derived as follows. Power is related to voltage according to
Ohm's law. The Log
Note, the resistance of 50 used above reflects that RF systems are matched to 50
systems use decibels referenced from 1 mW, the corresponding voltage increase for every 1 mW
The logarithmic form of Ohm's law shown above is provided to describe why the log of the
The constant in this equation is derived following similar logic. First, consider the pointing vector
which relates the power density (W/m
Where is the free space characteristic impedance equal to 120
decibels and using the appropriate conversion factor to convert dBW/m
density and dBV/m to dBV/m for the electric field, the constant becomes 115.8.
Where AF is the antenna factor of the antenna being used, provided by the antenna manufacturer or
To derive the constant for the above equation, simply convert the characteristic impedance of free
The derivation for the constant in the above equation comes from the decibel equivalent of the factor
 A.H. Systems inc. – May 2014
REV D
function is used for relative (dB) scales, so applying the logarithmic function to
10
Ohm's law, simplifying, and scaling by ten (for significant figures) yields:
10Log
10
power increase can be calculated with another form of Ohm's law:
V = (PR)
Given a resistance of 50and a power of 1 mW
20Log
corresponding voltage is multiplied by 20.
dBmW/m
a calibration that was performed within the last year.
Not much to this one; just plug away!
dBA/m = dBV/m – 51.5
space to decibels, as shown below.
As above, simply plug away.
dBW/m
A simple relation to calculate decibel-Watts per square meter.
dBmW/m
of 1000 used to convert W to mW and vice versa, as shown below.
dBmW = dBV – 107
P = V
/ R
2
[P] = 20Log
[V] – 10Log
10
= 0.223 V = 223000 V
0.5
[223000 V] = 107 dB
10
= dBV/m – 115.8
2
) to the electric field strength (V/m) by the following equation.
2
P=|E|
/
2
dBV/m = dBV + AF
V/m = 10
{[(dBuV/m)-120]/20}
20Log
[120] = 51.5
10
A/m = 10
{[(dBuA/m)-120]/20}
= 10Log
[V/m – A/m]
2
10
= dBW/m
+ 30
2
2
[50
]
10
. Transforming this equation to
to dBmW/m
2
. Since RF
for power
2
10