10; 
338 
sorbed by the load and hence is equal to 
een me SES (11) 
(Ra Pp R,)? 
where P, is called the received power. In the same 
way 
A V*Ra 
(Ra ar R,)? 
is the power scattered by the doublet. 
It is easy to show that the maximum power is 
delivered to the load if R, = R,. In this, the matched 
load case, 
(12) 
Ss 
2 2 
el ite Nv 
4R, 4R, 
Now the resistance of the doublet, neglecting its 
low ohmic resistance, is only the radiation resistance 
[equation (9)] and the potential or voltage across 
the terminals is equal to Hol, where £p is the field 
strength of the incident plane wave and L is the 
effective length of the doublet. Inserting these 
quantities into equation (13), 
P. ri P. o> ots o ay 3 
1207 87 
In these equations it has been assumed that the line 
of the doublet has been oriented parallel to the 
electric vector of the incident wave in order to obtain 
maximum power absorption. 
The factor H,?/1207 will be recognized from 
equation (6) as the power per unit area of the inci- 
dent wave. The formula thus says that all the power 
crossing an area 3\7/87 is received, and that all the 
power crossing an equal area is scattered. The 
(13) 
(14) 
area 3\’/87 is therefore called the absorption cross ; 
section or scattering cross section of the matched 
doublet. Since the antenna has been placed parallel 
to the polarization of the incident wave, this is the 
PROPAGATION THROUGH THE STANDARD ATMOSPHERE 
maximum absorption cross section. Moreover, this 
formula holds only when the doublet has been 
matched to its load, and consequently 3\?/87 is the 
maximum absorption cross section. 
It will be noted, however, that 3\?/87 is not the 
maximum scattering cross section. This maximum 
is achieved by shorting out the load, that is, setting 
R, = 0. In this case, 
P,=0 
and 
_ Be 3M (15) 
7 10 DR 
Hence the scattering cross section of the shorted 
(dummy) doublet is four times the scattering cross 
section of the matched load doublet. 
Transmission between Doublets 
in Free Space 
Assume that two doublets, one to function as a 
transmitter and the other as a receiver, a distance 
d>>) apart, are adjusted for maximum power 
transfer. This means that the axes of the doublets 
are parallel and lie in their common equatorial plane 
and that each is matched to its connected circuit. 
Then the power radiated by the transmitting doublet, 
from equation (7), is equal to 
Ed? 
3) 
Py = (16) 
watts, 
and from equation (14) the power delivered to the 
load circuit of the receiving doublet is given by 
Ee 3 
1207 ; 87 
Hence the ratio of the received power (to the load 
circuit) to the output power for maximum power 
\(METERS) 
watts. 
(17) 
d (KILOMETERS) 
Ficure 3. Free-space gain for doublets P:/P; = (8\/87d)? = A%. (Adjusted for maximum power transfer.) 
