User Manual

229
MN024-08
r
min
= { 10 · exp [ (7 - 5.25) / 10 ] · 0.226} / (4·π·50) } · exp (-1/2) = 0.023
m
Example 3. Let’s suppose to have a Low Power TFAM90/20 connected to an
omnidirectional antenna through a 20-metre length ½” cable (no splitters
used). Let’s suppose that the antenna Gain is 7 dB and that our Britecell
system distributes two GSM900 carriers and one UMTS2100 carrier.
Moreover, let’s assume that the maximum allowed electrical field strength is:
E = 6 V m
(typical Italian reference level for exposure to time-varying electric and
magnetic fields). The corresponding value of the maximum allowed power
density is:
S = E
2
/377 = 0.1 W/m
2
By reading the Britecell bulletin, we get that the output power at the TFAM
antenna port is 14 dBm/carrier (=0.025 W) for a 2-carrier GSM900 MHz
distribution, and 17 dBm (0.05 W) for 1 WCDMA carrier.
The 900 MHz and 2100 MHz output powers at the remote unit ports are:
P
900MHz,TFAx
= 0.025W+0.025W=0.05W (for 900MHz signals)
P
2100MHz,TFAx
= 0.05W (for 2100MHz signals)
Let’s assume that the ½” cable losses are 0.07 dB/m in the 900 MHz band and
0.18 dB/m in the 2100 MHz band; the total losses between the TFAM90/20
output port and the antenna input ports can therefore be estimated as follows:
L
900MHz
= 0.07 (dB/m) x 10 (m) = 0.7 dB on 900MHz signals
L
2100MHz
=0.18 (dB/m) x 10 (m) =1.8 dB on 2100MHz signals
The term “10 exp (G-L/10) P” which appears inside the relation 6.1 should
therefore be calculated apart for each frequence, and then added in order to
calculate the composite contribution:
P
900MHz, ant
= 10 exp[(7-0.7)/10]· 0.05 = 0.213 W
P
2100MHz,ant
= 10 exp[(7-1.8)/10]·0.05 = 0.165 W
P
composite
= P
900MHz, ant
+ P
2100MHz,ant
= 0.378W
By dividing the total power through (4·π·S) and taking the square root
according to the relation 6.1, we therefore get the the following minimum
safety distances from the antenna:
r
min
= { P
composite
/(4·π·0.1)} · exp (-1/2) = 0.54 m