—— 
COVERAGE DIAGRAMS 439 
where 
d _——I 
S| 
£ p = 1, and the effect of antenna directivity is 
ieglected, K = D, and 
v= 
as VGido | ~ D)? + 4D sin? (22) 
dr ov 2 
The following discussion illustrates how one con- 
tour of a coverage diagram, corresponding to a 
particular value of radio gain, may be plotted on 
Figure 4 or its equivalent, Figure 7. The result is a 
curve similar to Figure 2, but plotted in u,v coordi- 
nates instead of he and d. See also Figures 16 to 39. 
For illustration, let the transmitter gain, G, = 1 and 
let the radio gain be such a value that do = dp/dp = 2. 
Further, let \ = 0.1 meter and h: = 20 meters. 
From Figure 15 it is seen that r = 1,030d/hi?/? = 1.2. 
Select one of the curves for, sin? (2/2) in Figure 12, 
Chapter 5, say sin? (0/2) = 1, for which Q = 7, 37, 
5, etc. These values correspond to tips of the lobes 
for which n = 1, 3, 5, etc., since, for perfect reflec- 
tion, ® = nz by equation (116) in Chapter 5. 
Next select values of K = D and note the corre- 
sponding value of the radical 
Ja — Ky + 4K sin? 
given by Figure 12 in Chapter 5. Equation (22) 
then gives the value of v. These quantities together 
with & = nr may conveniently be tabulated, as in 
Table 2. Corresponding values of D and v are plotted 
as crosses on Figure 7. The line drawn through these 
points is the locus of the tips of the lobes. The 
D-v LOC] AND LOBES 
Table 2. Values of v and F# for sin?(/2)=1, d,=2. 
! 
EY), 5 (r = 1.2) 
D |Af — D+ 4D sin? SI uation R=nr| Lobe 
(Figure 12, Chapter 5) | (22)] 4 ~ ™? \maxima 
2 He ae 1 1.2 |Ist lobe 
0 ie a8 Bin eesiou(2dilobe 
0.6 1 af 5 | 6.0 |8dlobe 
08 i: a6 7 | 8.4 (4th lobe 
0.95 1.95 3.9 
1.0 2.0 4.0 
actual position of each lobe tip is then marked with a 
circle where the corresponding value of R crosses 
the locus in Figure 7. 
Additional lobe points are located by choosing 
some other value of sin? (Q/2), say sin? (0/2) = 0.7. 
Each value of sin? (2/2) now gives two points on each 
lobe, one on the upper branch and the other on the 
lower. Again choose values of K = D, obtain the 
corresponding values of the radical 
Ale = Ky + 4K sin? | 
from Figure 12 in Chapter 5, and calculate new 
values for v. The D-» values are plotted as crosses 
on Figure 7 and the line through them is the locus of 
points for which sin? (Q/2) = 0.7 (see Table 3). 
For sin? (9/2) = 0.7, sin (9/2) = +0.836, 0/2 
= 0.3157 or 0.6857. Then 2 = nx = 0.632 + 2kr 
Ue 
fa 
\s a \ 
DIVERGENCE D 
— — — PATH DIFFERENCE 
Ficure 7. D-v loci and lobes. 
