CALCULATION OF RADIO GAIN 415 
For one-way transmission the radio gain is 
10 log = = 20 log A + 10 log (GG). (182) 
1 
For two-way transmission the radar gain is 
10 log = = 40 log A + 10 log (G,G2) 
’ +7.5+10loge—20log. (183) 
Typel The heights and distance apart of the 
transmitter and receiver antennas and the wave- 
length are known. The radio gain is to be found. 
An early-warning set has a horizontal antenna, 
located 118 meters above sea level. A receiver is 
located in an airplane 1,520 meters above sea level, 
at a distance of 300 km. The wavelength is 3 meters. 
The gain of the radar antenna is 96 db and its power 
output 100 kw. (a) The power received by the air- 
plane receiver, assuming a gain of 10 db, is to be 
found. (b) The power returned to the radar by the 
airplane, assuming that the airplane has a radar 
cross section ¢ of 40 square meters, is to be found. 
One-way: From Figure 2, d, = 205 km. Hence 
the receiver may be assumed well within the diffrac- 
tion region. 
From Figure 40, sd = 9.3, with f(6)=1. 
From Figure 41, eh: = 12.3, with g(6) = 1. 
From Figure 37, 20 log A = — 213. 
To convert, 20 log A to 20 log A by equation (175), 
we need 20 log hi and 20 log g;, which are given 
by Figure 41: 
20 log hy = 41.8, 
20 log g: = 1.5. 
Hence 
20 log 4 = —170, 
and by equation (182), 
10 log = 70-96-10 = Ge 
1 
Since P; is 105 watts, 
Pa S108 1 S10 
Radar: Substituting in equation (183), the radar 
gain is given by 
10 log (P2/Px1) 
—340 + 2(96) + 7.5 + 16 — 9.5, 
—134 db 
or 
P, = Py x 107 
The power output P; is 10° watts, so that the max- 
imum received power 
P, = 10°5* w. 
The minimum detectable power of the set is given as 
1.6 X 10° = 10°*° watt, so that under the given 
conditions the power returned by the target would 
be slightly below the threshold of detection. 
Type Il. Gain versus receiver (or target) height 
is to be found for given distance, given wavelength, 
and given transmitter height: A radar has an antenna 
height of hi = 30 meters, a wavelength of \ = 1.5 
From Figure 41, 
meters and a distance from a receiver (or target) of 
d = 100 km. Assuming a receiver antenna gain 
Gz = 1, the variation of P:/P; at the receiver with 
receiver height is to be found. Also, assuming a 
target of cross section ¢ = 50 sq meters, the varia-~ 
tion of P2/P; at the radar receiver with target height 
is to be found. The radar antenna has a gain of 
13.5 db. 
One-way: 
sd = 3.9 (from Figure 40). 
From Figure 37, for the fixed value of sd = 3.9, we 
find a correspondence between values of 20 log A and 
Che, listed in Table 5 below. By means of Figure 41 
or equation (159), ehe is changed to h, and by means 
of equation (176), A to. A. From Figure 41, it is seen 
that 20 log hi = 29.5 and 20 log g: = 0. To change A 
to P:/Pi, the transmitter gain of 13.5 db and the 
receiver gain of 0 db must be taken into account, 
according to equation (182). The result is given in 
Table 5. The values of 10 log (P:/P1) are plotted in 
Figure 25, together with the results found with the 
same data in text onp 397 for the optical-inter- 
ference region. 
TABLE 5* 
ho Radio Radar 
Meters | ehe | 20log 4 |20logA| Gain Gain 
in db in db 
63 0.8 —190 —160.5 —147 —273 
142 1.8 —180 —150.5 —137 —253 
259 3.3 —170 —140.5 —127 —233 
417 5.3 —160 —124.5 —117 —208 
* See also Table 1 and Figure 25. 
Radar: 
P. 
10 log = 40 log A + 27 + 7.5 + 10 log « 
z — 20 log 1.5, 
= 40log A + 27+ 7.5+ 17 — 3.5, 
= 40log A + 48. 
Type III. Gain versus distance is to be found, 
with antenna heights and wavelength given: Using 
the same data given in text on p.398 the gain as a 
function of distance in the diffraction region is to be 
found. The result has been plotted in Figure 26. 
The polarization is horizontal. 
hy = 30 meters G = 22.4 (13.5 db) 
he = 1000 meters G2 (one-way) = 1 (0 db) 
d = 1.5 meters G2 (radar) = 22.4 (13.5 db) 
co = 10 squure meters 
ehe SS 1a, 
Referring to Figure 37, we find a correspondence 
between A and sd. Values of A are to be assumed. 
To change sd to d, use Figure 40. To change A to A, 
use equation (176). From Figure 41, 
20 log 30 = 29.5, 
20 log g: = 0, 
20log A = 20log A + 29.5 + 0. 
