Solutions to End of Chapter Problems
1. Is diffraction harmful or advantageous in radio communications? Explain.
Diffraction is helpful because it allows communication around objects.
2. What are the three modes that an electromagnetic wave can travel from a source to a destination?
Ground wave, sky wave and space wave.
3. What is the term used for an electromagnetic wave that propagates by line-of-sight?
Space wave.
4. A ship-to-ship marine-band VHF radio operates at 156 MHz and is limited to a maximum of 25 watts. The signal
propagates via space propagation, so it is limited in range to direct line-of-sight. A Coast Guard transmitting station on shore
has a monopole antenna that is 350 feet tall.
(a) If a ship is 35 miles (56,315 m) away from the CG station, how high must the ship’s monopole antenna be mounted
to ensure reception?
d=
2h1 + 2h2 , so 35 =
2 ( 350 ) + 2 ( h2 )
h2 =
36.5 ft
Height is 36.5 feet
(b) Using the Friis Free-Space equation, calculate the received power at the ship.
Pt = 25 W
Gt = 1.45
Gr = 1.45
d = 56,315 m
λ = c / f = 3x108 / 156x106 = 1.92 m
2
PG
(25W)(1.45)(1.45)(1.92m) 2
t t Gr λ
=
Pr =
2
2
( 4π d )
( 4π 56,315m )
= 3.87 ×10−10 W
Received power is 3.87 x10-10 W
(c) If someone is standing in a life raft with a hand-held VHF radio (assume antenna height of 6’), what is the maximum
range from which they could contact the ship in part a?
d = 2h1 + 2h2 = 2 ( 36.5 ) + 2 ( 6 ) =12 miles
Distance is 12 miles
5. In a certain communication link, the transmit power is 5 W and the path loss is 100 dB. What is the received power in
mW?
PLdB (d ) =
Pt ( dBm ) − Pr ( dBm ) (d ) so Pr ( dBm ) (d ) =
Pt ( dB ) − PLdB (d )
 Pt mW 
 5000 mW 
=
Pt ( dBm ) 10
=
log10  ( )  10 log=
10 
 37 dBm
 1 mW 
 1 mW 
−63 dBm
Pr ( dBm ) (d ) =
37 dBm − 100 dB =
Received power is −63 dBm
−7
Thus the power is 5 × 10 mW
6. Use the log-normal model to solve for the distance of transmission (d) given the following parameters. Use the Friis FreeSpace equation to determine the path loss at d0 = 1 m.
n = 2.7, f = 900 MHz, Pt = 10 dBm, Pr = −70 dBm, Gt = 1.64, Gr = 5 dB, PLdB(d) = 110 dB.
Pt(mW) = 101 = 10 mW
Pt(dBm) = 10 dBm
d0 = 1 m
Gt = 1.64
Gr = 100.5 = 3.16
λ = c / f = 3x108 / 900x106 = 0.333 m
Friis equation: compute received power in mW at d0=1m:
Convert received power at 1 m to dBm:
=
Pr (1m)
2
PG
(10 mW)(1.64)(3.16)(0.333m) 2
t t Gr λ
=
2
2
( 4π d )
( 4π 1 m )
= 3.64 ×10−2 mW
 3.64 ×10−2 mW 
Pr (1m) in dBm = 10 log10 
 = −14.4 dBm
1 mW


Compute PLdB(1 m):
Write the equation for PLdB(d):
Plug in known values:
Solve for distance d:
Distance is 1.479 km