-
Notifications
You must be signed in to change notification settings - Fork 4
/
Copy pathStep_12.m
44 lines (39 loc) · 2.05 KB
/
Step_12.m
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
% Step 12: If information on the terrain clearance angle at a
% receiving/mobile antenna adjacent to land is available, correct the field
% strength for terrain clearance angle at the receiver/mobile using the
% method given in Annex 5, § 11.
% 11 Terrain clearance angle correction
% For land paths, and when the receiving/mobile antenna is on a land
% section of a mixed path, if more precision is required for predicting the
% field strength for reception conditions in specific areas, e.g. in a
% small reception area, a correction may be made based on a terrain
% clearance angle. The terrain clearance angle ?tca is given by:
% ?tca =? degrees (30)
% where
% ? is the elevation angle of the line from the receiving/mobile antenna
% which just clears all terrain obstructions in the direction of the
% transmitter/base antenna over a distance of up to 16 km but not going
% beyond the transmitting/base antenna.
% The calculation of ? should not take Earth curvature into account. ?tca
% should be limited such that it is not less than +0.55° or more
% than +40.0°.
% Where the relevant terrain clearance angle information is available, the
% correction to be added to the field strength is calculated using:
% Correction = J (??) – J (?) dB (30a)
% where J(?) is given by equation (12a):
% ??=0.036 sqrt f (30b)
% ? =0.065 ?tca sqrt f (30c)
% ?tca : terrain clearance angle (degrees)
% f : required frequency (MHz).
% It should be noted that the land field-strength curves take account of
% losses due to typical shielding of the receiving/mobile antenna by gently
% rolling terrain. Thus the terrain clearance angle corrections are zero at
% a small positive angle typical of receiving/mobile antenna positions.
% Figure 28 illustrates the terrain clearance angle correction for the
% nominal frequencies.
function e = Step_12(f,tca)
v = 0.036*sqrt(f); %equ'n (30b)
J = 6.9+20*log10(sqrt((v-0.1)^2+1)+v-0.1); %equ'n (12a)
v = 0.065*tca*sqrt(f); %equ'n (30c)
e = J - 6.9+20*log10(sqrt((v-0.1)^2+1)+v-0.1); %equ'n (12a)
end