Solutions to Practice Problems
Practice Problem 25.1
What is the longest line-of-sight communication range between a transmitter whose transmitting antenna is 350 feet high and
a receiver whose receiving antenna is 25 feet high?
d  2h1  2h2  2  350   2  25  33.5 miles
Distance is 33.5 miles
Practice Problem 25.2
Your cell phone transmits at a power level of 500 mW, with an antenna gain of 2.0 dB. The cell tower has an antenna gain of
8.0 dB, and is a distance of 5 miles away. For LTE, you’re transmitting at 700 MHz. Will your signal make it to the tower
and will it have sufficient power to “close the link” and allow you communicate? Or will you suffer the fate of a cellular
“dead zone”? (note: 1 mile = 1.609 km, and consider −105 dBm as the minimum power required to be able to “close the
link”)
Calculate received power:
Pr 
2
PG
t t Gr 
 4 d 
Gt = 100.2 = 1.58
Pt = 0.5 W
2
Gr = 100.8 = 6.31
d = 5 mi (1.609 km/mi) = 8.045 km = 8045 m
λ = c / f = 3x108 / 700x106 = 0.429 m
Pr 
2
PG
t t Gr 
 4 d 
2

(0.5W)(1.58)(6.31)(0.429m) 2
 4
8045m 
2
 8.98 1011 W
Convert this to dBm:
 8.98 1011 W 
Pr  dBm  10log10 
  70.47 dBm
 0.001W 
This received power is greater than the minimum power required to close the link, so communication takes
place.