DIFFRACTION BY TERRAIN 467 
part of the figure. The height ho is conveniently 
determined from a profile of the transmission path 
obtained from a topographic map. If the heights 
hy, he, and h of transmitter, receiver, and obstacle, 
respectively, above a given reference level such as 
sea level are given, we have 
_ dha + dah 
dy + dy 
where the signs have been chosen so that ho is nega- 
tive when the receiver is in the shadow of the ridge 
and positive when it is in the illuminated region. 
Now, by equation (38), 
v= in? Ge 
1 
In these equations, give the angles a; and a2 the same 
sign as fo and give v- the same sign that fo has in 
equation (12). 
The ratio of the field to the free-space field at tHe 
receiver is now given by | H/E) | = 2(v), defined by 
equation (7) and plotted in Figure 4. In Figure 9, 
this ratio is given in decibels as a function of the 
quantity z= —h/Vdd (all lengths in meters). 
The successive curves in Figure 9 correspond to 
different values of the ratio d,/d2 or d:/d; (choose 
whichever one is the smaller). Only the field below 
the line of sight is shown. . 
—h, (12) 
(G22 Hs (eee as) 3) 
FIELD IN SHADOW BEHIND DIFFRACTING RIDGE 
AN KG 
NANNY 
HSS 
CLL ETI DSS CNN 
015 0.2 0.3 0.4 
DB BELOW FREE SPACE 
fo 
a 
AN 
40 
Ficure 9. Field in shadow behind a diffracting ridge. 
Figure 10. Diffraction field above the diffracting edge. 
Field Near the Line of Sight 
The fact that just above the line of sight the field 
increases above its free-space value may sometimes 
be used to obtain a favorable site (Figure 10). The 
maximum value of the field is about 1.17 times the 
free-space value (Figure 4), equivalent to 1.36 db. 
On the other hand, there are advantages in avoiding 
a line TR that is too close to grazing the top of 
an intervening obstacle, as this will substantially 
reduce the signal. At the line of sight, the signal is 
6 db below free space. In order to get approximately 
the free-space value of the field, the crest of the 
obstacle should be sufficiently below the line TR 
so that v > 0.8 where v is given by equation (13), 
ho being the clearance between the line TR and the 
obstacle. In cases where the heights and distances 
are not quite certain, it is therefore preferable to 
select a higher and definitely unobstructed site 
rather than to try to utilize the small gain that might 
possibly be had from the diffraction field. 
Diffraction with Reflecting Ground 
When the ground near the transmitter or receiver 
is smooth and reflecting, the diffraction problem 
becomes very complicated. It can be solved by the 
method of images on assuming that the radiation 
reflected on the transmitter side of the obstacle 
issues from an image transmitter and that the radia- 
tion reflected on the receiver side is incident upon 
4 R 
T 
oe x 
Ficure 11. Diffraction of direct and reflected rays. 
an image receiver (Figure 11). The total field at the 
receiver may be written 
E = FE, — BE, — Ey + Es, 
where each term on the right-hand side is of the 
form of equation (6), #, corresponding to the direct 
radiation, H, to the radiation from the image trans- 
mitter to the receiver, H; to the radiation from the 
transmitter to the image receiver, and H, to the 
radiation from one image to the other. These four 
terms differ in the value of v assigned to each of 
them; the effective height ho computed by equation 
