LECTURE 17
§ Modes of Propagation
Waves produced by an antenna are subject to the solutions of
Maxwell's equations.
The boundary value problem is shown here:
The methods of solution are beyond the scope of this class. But the
results will be summarized here.
For the far-zone fields ( r   ) , there are three types of waves:
1) Space-waves (spherical plane waves)
The ground acts as a reflector with certain losses, depending on the
frequency band.
2) Sky-waves
Space-waves reflected and refracted by the ionosphere back to the
earth.
Within certain frequency bands, ionosphere acts like a plasma.
3) Surface waves (ground waves)
A special type of wave where the power is confined near the
surface
Each of the three types of waves is for particular frequency bands
due to the nature of ground and ionosphere material characteristics.
VLF: 3-30 KHz
LF : 30-300 KHz
MF : 300-3000 KHz
HF : 3-30 MHz
VHF: 30-300 MHz
UHF: 300-3000 MHz
SHF: 3-30 GHz
Surface-Wave
for short distance, <100 Km
Sky-Wave
long range, hundreds Km
Space-Wave
shorted range (<10 Km)
path loss is too much for
a longer range
SHF  Microwaves
§ Space-wave propagation
Suitable for VHF, UHF and Microwaves, particularly important in
current information technology.
Free-space (spatial) loss
If S=incident power density at a receiver,
PR ( power received)  Ae  S ,
where Ae : effective antenna aperture.
Also, from antenna theory, max antenna gain (assuming that the
receiver is at power peak)
4
GR  2 Ae

If radiation from an antenna is uniform (never in the real case)
P  total radiated power
S T 2
4 r
for non-uniform (anistropic) radiation:
P
S  T 2 GT
4 r
Friis formula:
2
P
PR  S
GT  T 2 GT GR 2
4
4 r
2
  
PR
Power received
 
 GT GR 

PT
Power out of a transmit ter
 4 r 
LF  spatial loss
P
 4 r   4 r 
LF  T GT GR  
f
 
PR

c

 

2
2
Note: for a fixed distance
LF  if f 
Implication for 1-2GHz:
f is high (109 )  LF is high
Cells (with local hubs) are needed  cellular phone
If we express LF in dB, we have:
[eq 4-22]
 4 R

LF ( dB)  10 log
f  106 
5 MHz
 3  10

R in Km for distance
c = 3 105 Km/sec
2
 40 
LF (dB)  20 log
  20 log R  20 log f MHz
 3 
2
LF (dB)  32.44  20 log R  20 log f MHz
distance increase LF 

frequency increase LF 
Total path loss: L  LF  Lex
Example
If a system's max allowable loss is 148.3 dB, what is the max range
the system can use at 900MHz?
(ignore excess loss)
148.3  32.44  20 log R  20 log 900
 20 log R  56.77

R  689Km.