FOR 0.125 TO 15.5 GHZ - DTIC · 2011. 5. 13. · report no. faa-rd-80-1 lb an atlas of basic...
133
Report No. FAA-RD-80-1 Lb AN ATLAS OF BASIC TRANSMISSION LOSS FOR 0.125 TO 15.5 GHZ 0 M.E. Johnson and G.D. Gierhart U.S. DEPARTMENT OF COMMERCE NATIONAL TELECOMMUNICATIONS AND INFORMATION ADMINISTRATION BOULDER, COLORADO 80303 ot T N00 AUGUST 1980 Document ;s available to the public through the National Technical Information Service, Springfield, Virginia 22151. Prepared for U.S. DEPARTMENT OF TRANSPORTATION LA. FEDERAL AVIATION ADMINISTRATION Systems Reserch & Development Service Washington, D.C. 20590 So0 8 21. 053
Transcript of FOR 0.125 TO 15.5 GHZ - DTIC · 2011. 5. 13. · report no. faa-rd-80-1 lb an atlas of basic...
Report No. FAA-RD-80-1 Lb
AN ATLAS OFBASIC TRANSMISSION LOSS
FOR 0.125 TO 15.5 GHZ0
M. E. Johnson and G.D. Gierhart
U.S. DEPARTMENT OF COMMERCE
NATIONAL TELECOMMUNICATIONS AND INFORMATION ADMINISTRATION
BOULDER, COLORADO 80303
ot T
N00
AUGUST 1980
Document ;s available to the public through theNational Technical Information Service,
Springfield, Virginia 22151.
Prepared for
U.S. DEPARTMENT OF TRANSPORTATIONLA. FEDERAL AVIATION ADMINISTRATION
Systems Reserch & Development ServiceWashington, D.C. 20590
So0 8 21. 053
NOIC
This document is disseminated under the sponsorship of theDepartment of Transportation in the interest of informationexchange. The United States government assumes no liabilityfor its contents or use thereof.
Technical Report Documentation Page
1. Repo~tNo~ 2.Goernm~ent Accession No. 3. Risi cnt's C010,09 No.
155-I Me-rJA~2~~ 5Re o~a, V te I[An Atlas of Basic Transmission Loss for .125' August 19-80I to 15 5 .l.z 6. Performing Organization Code
-8. P- ming Organization Report No.
r. E. Johnson *.G. D./Gierhart__/'.. .... ~ -m.----• e ni- e,• ,s. . -. -10, Work Uit No. (TRAIS)
U.S. Department of CommerceilNat_ anal Telecam .3cations & InomtinAministrat i,.+ . I1nstitute for Telecommunif.ation Sciences • DOT-FA77WAI-742)_Výi:,: ~325 Broadway Bouilder, CO 80303 k/• ',,,+•a,.
12. Sponsoring Agency Name and Address
U.S. Department of TransportationFederal Aviation Administration .Systems Research and Development service S4 Sponsoring Agency Cod.
lWashington, DC 20591 ARD-4501S. Supplementary Note,
.Performed for the Spectrum Management Branch, ATC Spectrum EngineeringH \ Section.
*16. Abstract poie ta f~. i/i
This report provides an atlas of air/ground and air/air basic transmis-ion loss predictions made with the IF-77 (ITS-FAA-1977) propagation
model- Sets of predictions are provided for each of six frequencies;i.e., 0.125, 0.3, 1.2, 5.1, 9.4, and 15.5 GHz. Each set containscurves for 90 antenna height combinations where one antenna heightvaries from 3 to 30,000 m, and the other varies from 300 to 30,000 m.Each height combination is repeated for three time availabilities:0.05, 0.5, and 0.95. Hence, 1620 transmission loss versus distancecurves are provided. The maxim-um distance used is 1800 km. Propagationfactors and application considerations are discussed. Example problems,are used to illustrate the various application methods provided.
'Note: The graphs in this document have been photo reduced in order tofit a standard paper size. Copies of large size graphs are availablefrom the sponsoring agency upon written request (send letter to theaddress given in item 12 above, attention: ARD-450)....
17. Key Ward, Air/air, air/ground, com- 18. Distribution Statement This document is
puter program, frequency sharing, available to the public through theinterference, navigation aids, pro- National Technical information Ser-agation model, transmission loss. vice, Springfield, VA 21151.
119. Security Classif. (of this report) 20, Security Classif. (of this page) 2A. No. of Pages 21. tic*
tnclassified Unclassified 134
Fo@m DOT F 1700.7 :8-72) Reproduction of completed page authorized
44
' iAM i-.. .
ENGLISH/METRIC CONVERSION FACTORS
LENGTH
From cm m km in ft -i mi
-cm 1 0.01 1xIO"$ 0.3937 0.0328 6.21x106 5.39110-'a 100 1 0.001 39,37 3.281 0.0006 0.0005
km 100,000 1000 1 39370 3281 0.6214 0.5395in 2.340 0.0254 2.54x10"s 1 0.0633 1.SIzlOs 1,37x10s
ft 30.48 0.3048 3.05XI1C" 12 1 1.89x10"• 1.64x1O"4
,d 160,900 1609 1.609 63360 S280 1 0.8618
.. 185.200 1852 1.852 72930 6076 1.151 1
AREA
FromT cm m2 km2 in2 ft 2 M12 hmu2
c92 1 Q..1Q9_ 1xlO-1 0.1550 0.0011 3.86x10"11 S.11XlO-11
02 10.000 1 1xlO-6 15S0 10.76 3.86x107 S.IxlO07
km2 lxtl0q 1x106 1 l.S5w109 1.08x1O 0.3861 0.29'14
in2 6.452 0.0006 6.45x10"10 1 0.0069 2.49x10 1 0 1.88x10-lO
ft 929.0 0.0929 9.29x108 144 1 3.5910"8 2.71x1O"8
.12 2.59x1010 2.59x106 2.590 4.01x109 2.79x10o 1 0.7548vui2 3.4341010 3.434106 3.432 5.31x109 3.70x10? 1.325 1
VOL UME
ro m liter .3i 3 f 3 yd 3 fl. oz. fl. pt. fl. qt. gal.
cm3 1 0.001 lXlO-6 0.0610 33340"5 1.31xlO-6 0.0338 0.0021 0.0010 0.0002
liter 1000 1 0.001 61.02 0.0353 0.0013 33.81 2.113 1.057 0.2642
39 lxlO6 1000 1 61.000 35.31 1.308 33,800 2113 1057 264.2
in3 16.39 0.01653 1.64xl"S 1 0.0006 2.14x!0"5 0.5541 0.0346 2113 0.0043
ft3 28.300 28.32 0.0283 1728 1 0.0370 957.5 59.84 0.0173 7.481yd3 765,000 764.5 0.7646 46700 27 1 25900 1616 807.9 202.0
fl. ot. 29.S7 0. 29$7 2.96X10"s 1.805 0.0010 ý,.87x10"5 1 0.0625 0.0312 0.0078
fl. pt. 473.2 0.4732 O.0005 28.88 0.0167 0.0006 16 1 0.5000 0.1250
fl. qt. 948.4 0.9463 0.0009 S7.75 0.0334 0.0012 32 2 1 0.2500
Sal. 3785 3.785 0.0038 231.0 0.1337 0.0050 128 8 4 1
MASS
To
From * g kg ot lb ton
1 1 0.001 0.0353 0.0022 1.10x1O6
kS 1000 1 35.27 2.205 0.0011
o. 28.35 0.0283 1 0.0625 3.12xlO5
lb 453.6 0. 4S36 16 1 O.0005
ton 907,000 907.2 32,000 2000 1
TEMPERATURE
C a5/9 OF 7-32),
*F . 9A C ) 323
ii
FEDERAL AVIATION ADMINISTRATIONSYSTEMS RESEARCH AND DEVELOPMENT SERVZCE
SPECTRUM MANAGEMENT BRANCH
STATEMENT OF MISSION
The mission of the Spectrum Management Branch is to assist the Depart-ment of State, National Telecommunications and Information Administra-tion, and the Federal Communications Commission in assuring the FAA'sand the nation's aviation interests with sufficient protected electro-magnetic telecommunications resources throughout the world and toproviue for the safe conduct of aeronautical flight by fosteringeffective and efficient use of a natural resource - the electromagne-tic radio frequency spectrum.
This objective is achieved through the following services:
Planning and defending the acquisition and retentionof sufficient radio frequency spectrum to supportthe aeronautical interests of the nation, at homeand abroad, and spectrum standardization for theworld's aviation community.
Providing research, analysis, engineering, andevaluation in the development of spectrum relatedpolicy, planning, standards, criteria, measurement
h equipment, and measurement techniques.
* Conducting electromagnetic compatibility analyses todetermine intra/intersystem viability and designparameters, to assure certification of adequatespectrum to support system operational use andprojected growth patterns, to defend aeronauticalservices spectrum from encroachment by others, andto provide for the efficient use of the aeronauticalspectrum.
Developing automated frequency selection computerprograms/routines to provide frequency planning,frequency assignment, and spectrum analysis capabil-ities in the spectrum supporting the National Air-space System.
Providing spectrum management consultation, assistance,and guidance to all aviation interests, users, andproviders of equipment and services, both naand international.
.I. .............
TABLE OF CONTENTS
Page
TECHNICAL REPORT DOCUMENTATION PAGE. . . . . . . . . . . ..
ENGLISH/METRIC CONVERSION FACTORS. . . . . . . . . . . . . ii
MISSION STATEMENT, FAA SPECTRUM MANAGEMENT BRANCH . . . . . . iii
TABLE OF CONTENTS . . . . . . . . . . . . . . . v. . v
LIST OF FIGUtES. . . . ................ . . . . . . vi
1. INTRODUCTION .... .... .............. . . . . . 1
2. BASIC TRANSMISSION LOSS CALCULATIONS ... ........ 2
3. TRANSMISSION LOSS VERSUS DISTANCE CURVES . . . . . . .. 4
4. APPLICATION CONSIDERATIONS . . . . . . . . . . . . . 90
4.1 Received Signal Level ....... ............. . . . 904.2 Power Density .................. . . . . . . 914.3 Interpolation Between Curves .... ............ .. 924.4 Service Range, Without Interference. .. . . . 944.5 Protection Ratio.............. . . . . . . 954.6 Station Separation ..... .... .96
4.7 Constant Protection Rtio Locus ........... 004.8 Multiple Interference Sources. . . . .... . 103
4.8.1 Simultaneous Independent Source. . . . 1034.8.2 Intermittent Coordinated Sources . . .... 107
4.9 Service Probability ..... ........... . . . . .. 1104.10 Radar Application. . . . . . . . . . . . . . . . . . 114
APPENDIX. LIST OF SYMBOLS ............... . . . . . 117
REFERENCES . . . . . . . . . . . . . . . . . . . . .. . . .. . . 122
v
Vj
LIST OF FIGURES
Figures Page
1 -14 Basic transmission loss versus distance, 125 MHz,2 H1 - 3 m . . . . . . .. . . . . . ... 6
2 H1 - 10 m . . . . . . . . . . . . . . . . . .. . . . 7
3 H1 15 m . . . . . . . . . . . . . . . . . . . . . . 8
4 H1 3 50 m . . . . . . . . .. . . . . . . . . . . 9
5 H1 150 m . . . . . . . . . . . . . . . . . . . . 10
6 H1 - 300 m . . . . .. . . . . . . . . . . . . .. . 17
1 H1 2 500 m . . . . 12
1 H1 -30000 . .......... . .. .. .. . . . . 13
9 H1 5000 m . .. . . . . . . . . . . . . . . . . 14
10 H1 = 0 000 m ................... . 15
11 - 15 00 m. ............................... 16
12 H1 = 0 00 m . . . . . . .................. 1713 H1 . 25 000 m . o. . . . . . . . . . . . . . . .. . . 18
14 H1 = 300 00. m .. .. ....... . . 1915-28 Basic transmission loss versus distance, 300 MHzf
1 5H I 3 m ... . . . . . .. . . . .. .. . . 2 0
16 H 10 m . . . . . . . . 2117 H1 15 m .. . . . . . . . . . . .22
18 H1 30 m . . . . . . . . . . . . . . . . . . . .. . 23
19 H1 150 m . . . . . . . . . . . . . . . . . . . . . . 24
20 HI 300 m. . . . . . . . . . .. . . . . . . .. . . 25
21 H1 500 m . . . . . ... . .. . . ... 26
22 H 1 000 m . *. . . . . . . . . . .. . . . .. 271
23 H1 a 5000 m . . . . . . . . . . . . . . . . 28
24 HI a 10 000 m ........ .............. . . . . . 29
vi
LIST OF FIGURES (continuedl
Figures Pg
25 H 15 000 m . . . .... . . . . . . . 30
26 = 20 000 m . . . . . . . . . . . . . . . . . . . . 31
27 H1 5 00 0 . . . . . . . . . . . . 3228 H = 30 000 m . . . . . . . . ... .. . . .. . . . . 33
29-42 Basic transmission loss versus distance, 1.2 GHz,29 H1 3 m . . . . . . .. . . . . 3430= 10 . . . . . . . . . . 35
31 H1 = 15 m . . . . . . . . . . . .. 36
32 H1 30 m 7. . . . . . 3833 H 15 0 .. . . . . . . . . . . . . . . . .. . . . . 38
34 H1 = 300 m ... . . .... . . . . . . . . . ... . . . . 39
35 H 500 m S m . . . . . .. . . . . . . 40
36 H = 1000 mi. . . . . . . . . . . . . ............... 41
37 H1 = 5000 m . . . . . . . . . . . . 42
38 H1 10 000 m . . . . . . . . . . ... . . ....... .... . . . .. 43
39 H = 15 000 mi . . .. . . . . . . .. . . .. . 44
40 H= 20 000 m...................... 45
41 H 25 000 m. . . . . . . . . . . . . . . 45
42 H1 30 000 7. . . .o 47
43-56 Basic transmission loss versus distance, 5.1 GHz,
43 H1 = 3 m . ... . . . .. . .. .................... . 48
47 H1 = 150 m ................ . . . . . 52
45 = 300 m . . . ....... . . . . . . . . . .. . 53
vii
S ... . ........ .. . . .• . .. . .- - i i• , I . : •?- •--
F r LIST OF FIGURES (continued)Figures Page
49 H1 - 500 M. . . .. . . .. . . . . . . . . . . . . 54
50 H1 - 1000 m . . . . . . . . . . . . . . . . . . . 55
52 H1 " 10 000 m . . . .. . . 0. . . . .. .. . . .. . 57
53 H1 - 15 000 m . . . . . . . . . . . . . . . a . . 5853 H1 - 15 000 m . .... . . . . . .... 58
54 H1 20 000 m . . . . . . . . . . .. . . . . . . . 59
55 H - 25 000 m # . . . . . . .... 60
56 H1 - 30 000 m . . . . . . . . . . . . . . . . . . . . 61
57-70 Basic transmission loss versus distance, 9.4 GHz,
57 H1 - 3 m. . .. . . . .. . .. . . . 62
58 H1 - 10 m ... .. . .. . . . . 63
59 H1 - 15 m . . . . . . .. . . . . .. 64
60 H1 100 i . o . . . . . . . . . . . . . . . . . . 65
61 H1 - 150 m . . . . ............... . . . 66
66 H1 = 1000 m. o.. . . . 6.............................71
63 H -1500 ...... . . . . .............. 68
64 - 1000 m . ... . . . . . . 69
65 H1 -25000 mio . .......... . . . . . . 70
66 H1 1 30 000 m . . . . . . . . . . . . . . . . . . 71
67 H1 - 15 000 m . . . . . . . . . . . . . . . ... 72
68 HI - 20 000 m . . . . . . . 0.. 0. . . . . .. . . . 73
69 H1 - 25 000 m . . . . . . . . . . .. . . . . . 74
70 H1 - 30 000 m . . . . . . . . . . . . . .. . . . 75
71-84 Basic transmission loss versus distance, 15.5 GHz,
71 H1 a 3 m . ..................... . . . . . . . . . . 76
72 H 1 I0 m . . . .0 0 0 . . 0 . . . . . . . . 77
viii
LIST OF FIGURES (continued)
Figures Page
73 H1 - 15 m6 .&. . . .. . . . . . . . 78
74 H1 30 79
"75 H1 150 m . . . . . . . . . . . . . . . . . . . . 80
76 H1 a 300 m . . . . . . . . . . . . . . . . . . . . . 81
77 H1 - 500 m . . . . . . . . . . . . . . . . . . . . . 8278 = 1000 m . . . . . . . . . . . .. . . . . . . . . 83
79 H1 a 5000 m . . . . . . . . . . . . . . . . . . .. . 84
80 H = 10 000 m . . . . . . . . . . . . . . . . . . . 85
81 H - 15 000 m . . . . .. . .. . .. . . . . . 86
82 H1 = 20 000 m . 0 . . . . * . .. . . . . . . . 87
83 H1 = 25 000 m . . . . . . . . . . . . . . . . . 88
84 H1 30 000 m . . . . . . . . . . . . . . . . . . . 89
85 Sketch illustrating interference configuration . . . 97
86 Locus of constant D/U(q)for an adjacent-channelundesired facility . . . . . . . . . . . . . . . .. 101
87 The standard normal deviate Zmo versus serviceprobability Q. . . . . . . . . . . . . . . .. . . . 112
ix
| . .,.,-& } ' -- --- .--...-
AN ATLAS OF BASIC TRANSMISSION i qS FOR 0.125 TO 15.5 GHZ
M. E. Johnson and G. Giernart'
1. INTRODUCTION
Increasing air traffic density and fast, high-flying jets havemade reliable navigation and communication systems for aircraft moreimportant than ever before (9]2. Radio frequencies available for thedevelopment of naw systems and the expansion of present ones to meetfuture demands are limited partly by international agreements. Effi-cient use of these frequencies is therefore imperative to satisfythese demands without increasing the hazards of air travel.
Potential radio frequency intarference problems must be recog-nized and dealt with effectively. Information contained in this
report will resolve many such problems. Curves are included that maybe used to estimate gross transmission characteristics of electro-
magnetic radiation at frequencies ranging from 0.125 to 15.5 GHz forantenna elevations applicable to air/ground and air/air transmissions.
Propagation of radio frequency energy at VHF/UHF/SHF is affectedby the troposphere, specifically by variations in the refractiveindex of the atmosphere. The terrain along and in the vicinity ofthe great-circle path between transmitter and receiver also plays animportant part. A 1977 propagation model (IF-77) developed by theInstitute for Telecormuni.ations (ITS) for the Federal Aviation Admin-istration (FAA) that includes allowances for terrain and atmosphericeffects was used to develop the basic transmission loss curves forthis report, and it is discussed in Section 2. Similar curves werepreviously developed (4, 51, but the IF-77 model is an improvementover the methods used previously and new predictions are appropriate.In particular, previous curves were for a time availability of 0.5(median) only, and those provided here apply to time availabilities
of 0.05, 0.5, or 0.95.
'The authors are with the Institute for Telecommunication Sciences,National Telecommunications and Information Administration, U.S.Department of Commerce, Boulder, Colorado 80303.,
References are listed alphabetically by author at the end of thereport so that reference numbers do not a.pear sequentially in thetext.
-~ - . - - - - -- - - - -.. .-... ....
Transmission loss [17, Sec. 2] is the ratio, expressed in deci-bels, of power radiated to the power that would be available at thereceiving antenna terminals if there were no circuit losses exceptthose associated with the radiation resistance of the receivingantenna. When both antennas are considered isotropic and loss-free,transmission loss is referred to as a basic transmission loss. Long-term median values of basic transmirdion loss are considered equiva-lent to the median level of hourly medtans when there are many hourly
medians.Basic transmission loss, Lb, versus distance curves with time
availabilities, q, of 0.05, 0.5, and 0.95 are given in Section .
The Lb (0.05) curves may be used to estimate Lb values for an unde-sired interfering transmitter that are exceeded during 95% (10'%-5%)of the time. Median (50%) propagation conditions may be estihiatedwith the Lb(0.5) curves. Curves of Lb(0. 9 5 ) may be used to estimatethe service range for a desired transmitter at which service would beavailable for 95% of the time in the absence of interference. Adiscussion of application consideration, complete with illustrative
examples, is provided in Section 4.Except where otherwise indicated, all equations provided here
are dimensionally consistent; e.g., all lengths in a particularequation are expressed in the same units. Calculations are made inthe computer programs with all lengths expressed in kilon•tars.Braces are used around parameter dimensions when particular units arecalled for or when a potential dimension difficulty exists. A listof symbols is provided in the Appendix.
2. BASIC TRANSMISSION LOSS CALCULATIONSDuring 1960-1973, the transmission loss prediction methods
developed at the National Bureau of Standards [17] were extended andincorporated into an air/ground propagation model. This 1973 model(ITS-FAA-1973 or IF-73) was developed by ITS for the FAA [6] and hassince evolved into the IF-77 model, which is applicable to air/ground,air/air, ground/satellite, and air/satellite paths [7, 10, 11]. Itcan also be used for ground/ground paths that are line-of-sight,smooth earth, or have a common horizon. These methods are based on aconsiderable amount of experimental data, and extensive comparisonsof predictions with data have been made [12, 14].
2
In performing these calculations with IF-77, a smooth (terrain
parameter Ah=O) earth with an effective earth-radius factor k of 4/3
(surface refractivity Ns=301 N-units) was used along with compensa-
tion for the excessive ray bending associated with the 4/3 model at
high altitudes. Constants for average ground, horizontal polariza-
tion, isotropic antennas, and long-term power fading statistics for a
continental temperate climate were also used. Although these param-
eters may be considered either reasonable or near worst-case formany applications, the curves should be used with caution if conditions
differ drastically from those assumed.For example, use of 5arface roughness, vertical polarization and
nonisotropic antenna patterns could result in less signal level vari-ability because of a lower reflection coefficient for the earth'ssurface. Thus, the assumed parameters would oiten be expected to
provide worst-case values. However, the smooth earth curves may notprovide worst-case condition situations that involve irregular terrain.
With the excepticn of a region "near" the radio horizon, valuesof median basic transmission loss for "within-the-horizon" paths were
obtained by adding the attenuation due to atmospheric absorption (indecibels) to the transmission loss corresponding to free-space condi-tions. Within the region just inside the radio horizon, values ofthe transmission loss were calculated using geometric optics, toaccount for the phasor addition of the direct ray and a ray reflectedfrom the surface of the earth. Segments of curves resulting fromthese two methods were joined to form a curve that shows median basictransmission loss as increasing monotonically with distance. Transi-tion between the line-of-sight and diffraction regions was made usinga straiaht line. to connect a diffraction value at the radio horizonwith a point just inside the line-of-sight region where the ray opticsformulation is valid [6, Sec. A.4.2; 7, Sec. 71.
As indicated above, the two-ray interference model was not usedexclusively for within-the-horizon median calculations, because thelobing structure obtained from it for short paths is highly dependent
on surface characteristics (roughness as well as electrical constants),atmospheric conditions (the effective earth radius is variable intime), and antenna characteristics (polarization, orientation, andgain pattern). Such curves would often be more misleading thanuseful; i.e., the detailed structure of the lobing is highly dependent
3
on parameters that are difficult to determine with sufficient pre-
cision. However, the lobing structure is given statistical considera-
tion in the calculation of variability.
For time availabilities other than 0.5 (median), the basic
transmission loss, Lb, curves do not always increase monotonicallywith distance. This occurs because variability changes with distance
can sometimes overcome the median level changes. Variability includes
contributions from both hourly-median or long-term power fading andwithin-the-hour or short-term phase interference fading. Bothsurface reflection and tropospheric multipath are included in the
short-term fading formulation used. However, surface reflection
(i.e., lobing statistics) is not allowed tc contribute to short-termvariability in the region just inside the radio horizon (i.e., horizon
side of horizon lobe). This results in a decrease in short-termvariability in the horizon lobe region which can show up as a decreasein Lb (0.95) values. A decrease in short-term variability near theradio horizon has been observed in experimental air/ground propagation
data [1].
3. TRANSMISSION LOSS VERSUS DISTANCE CURVES
Figures 1 through 84 at the end of this section show Lb Versusdistance for frequencies of 0.125, 0.3, 1.2, 5.1, 9.4, and 15.5 GHz;
terminal 1 antenna elevations, Hi, of 3, 10, 15, 30, 150, 300, 500,
1000, 5000, 10 000, 15 000, 20 000, 25 000, and 30 000 m; and terminal2 antenna heights, H2 , that range from 300 to 30 000 m using thevalues just given for H1 . These figures are cataloged by increasing
'1 within each frequency group. Copies of large size graphs are avail-
able from the sponsoring agency upon written request (see p. i).Each figure contains those curves for which H2 values are equal
to or greater than H1 * Each figure consists of three curve setswhere the upper, middle, and lower sets provide Lb(0.05), Lb(O.5),
and Lb(0. 9 5 ), respectively. These correspond to time availabilities,
q, of 0.05, 0.5, and 0.95. For example, Lb(0. 9 5 ) = 200 dB, means
that the basic transmission loss would be 200 dB or less during 95%
of the time, and Lb(.5)= 190 dB means that the median basic trans-
mission loss is 190 dB.
4
.. .. .. ..... ..........,n v
A free space, Lbf' curve is included for each set, and it was
calculated using the H1 of the set and H2=30,000 m. At zero distance
it is the difference in antenna heights that determines L bf so the
Lbf. curves shown are for the highest L bf height combination. However,
for path distances somewh~at greater than the altitude difference, Lb
values are nearly identical.
The IF-77 propagation model (Sec. 2) used to calculate these
curves includes allowances for a large number of factors that affect
propagation (e.g., long-term power fading, surface reflection multi-
path, tropospheric multipath, etc.) and blends calculations made for
the line-of-sight, diffraction, and scatter regions together intransition regions. These complications and the use of simple linear
interpolation to obtain curves from the points actually calculated
result in curves that are occasionally bumpy and have some discontin-uities. For the most part, these are minor, but some are severeenough to encourage manual smoothing of the computer generated plots.Since such smoothing would tend to obscure places where problems with
the IF-77 may exist and would require extensive drafting, it was notdone. Cormmients concerning some of these bumps have been added to
Figure 68.
5
*,. A
* --- , I5%
41)( 60 --0 1001 -- 60 10
I ~ ~ ~ ~ . .~ .. .....- - -
A D F W
Antnn heightas 12dicated
E ___ ___ ___ ___
H0=
4- - 100 m0 0 0 00 120 10 60 1
D 00________ 10000__m
U
2~ 0" 0( 80 100 1200 1400 1600 1800 300mlOG - -95%
1~C -Continenltal
INC tepperate climato,N 01 (k-4/3),
.=oth earth C'.h=O) ,horizontal polarization,average goun,isotropic. antennas.
20.^ dOC 600 800 1000 1200 1400 1600 1800OiSlance in km
Figure 1. Basic transmission loss versus distance; F -125 M5Hz, H = m
A
~Kt~w125MHz
0
:00 400 600 Son~ 1000l I2zo 140C' boo 1800
C:
Antenna heights______ __________ _____dn indicated.
Cf 100Cm
D 500 m
F 15000 mi
15 10 00130000 m10 95%
ISO__ _____- HtinontalA D\ N 1:3%)raWr climlvte,
N .01 (k-4/3),U oth earth (&w-0),
horizontal polarization,awiraae ground,±8o~oLic .antAfnias.
................................FrweSpace
2 400 8u0 B00 1000 1100 1400 1600 1800
DiStance in k~mFicre 2. Basic transmission loss versus distanice, F 125 MHz, H1 10 m.
5%Q..
zoo__ 125 MHz
1C. 200 400 600 80Ow ll 1200 10 s I
_____ _ ___ 0 %
'4E 10000 m
H 1500 m
A 3C00 mB 25000 m
E 30000 m
2ý10 400 600 8oo 1000 1200 1400 1600 18500
isOcriztinetal ~aiain
average groun~d,isotropic antainnas.
.. ................................................... Freas ae
-cc 600 S00 1000 ZOO0 1400 1600 I800
Distance in kmFicure 3. Basic transmission loss versus distance; F -125 MHz, Hi 1 15 m.
________ 125 MHz
z-i 4 4 600 800 1000 1200 idfO 1600 1900
11550
N t 300 (k4/
200~ ~~~~ 1000 m0 0 0 0 40 10 0
Dpac in00 mm
Figur 4. ' Ba 0i ra Smisono s 1 esu isa0e0 1200 1400 -RO 10 00
(k-4/3).-
___ 125 MHz
Boo 10 00 80 000 1200 1400 1600 1600
A _ _ _F_\
4 ______Antenna heightsas indicated.
22
A 300 m3 SO0 mC 1000m
_______ D__ __ 5000 mE 10000 mF 15000 m
U ~G 20000Gm____________H 25000 m
200 400 600 8100 l000 1203 1400 260 1 300 m
Ootinentaltallprate clilate,
:1;301 (k-4/3),moth oasth (*h-),
'A±sontal. polarization~,avrage 9r,istropic antennas.
.... FreespVOW
2CO 400 600 Sao 2000 1200 1400 1600 11100 2Distance in km
Figure 5. Basic transmission loss vf.r*,u, tance: P 125 MHz, H 1 -150 m.
10
7 -777
___ 125 MHz
2 cle 4flC 600 g00 l0o0 a100 140 I00 IS *lF
4 * IIms height
______as
ROD
~0 JO 400~00 500 Q00 tOO 00 w iso I 6W2
A 3w 0a 900~
3 10003)
Fas eIWO a-eM
- -5MD
400~~ -0.o 00 19 141 w w M
..... .... . .............. ...
150 Cb*IrM*(
Distanceink
riqure 6. Basic transmission lose versus distance$ P 1 25 Nil. Nl 30 200.
4 -... .. . .
- 7 400 V00 500 1000 1200 _...., 400.," [00 oo
A
Antenna heights
an indiicated.
3. 500 m
H 2
.~~~ ~~~~~ 100 M(0 4i ~ 6C 80 00 10 40 A0' 10
.'~~~'C'~~ 2000 m0 0 00 20 40 60 10
G. 250
___ 125 MHz
I" 4ý O 5C I 000) 1200 14OC 00 1600110
50
an indcated
C 10000 m
w _ _ _ A__ _ _ _ _ _ _ _ 15000 M
E 20000 m
G 30000 m
100 200O 400 600 g00 1000 1100 1400 1600 1500
iso Continental
Aý~ -301F (k-4/3),_ _1's Jo earth (ah-0),
horizontal polarization,
.... Free SPeCS
'0 00 600 Soo 1000 1100 1400 1000 180Distance in km
CFigure 8. Basic transmission loss versus distanocy F -125 MHz, H 1 -1000 m.
13
lot __ __ 125 MHz
-i(00 'In0 6(1 00 so 000 1100 14n0 I0 bo 800
o0 _ _0_%
"is0
as inicaetd.
_____~~~~~~ _ _ _ _ _ __ _ _ _ _ _ 115000 M
215 A10000 m
B 10000nC 15000 m
0 20 40 60 SC 000 200 140 60 80 20000 in
F4 -301 0 mk4/
iomtirocantaim
Dirizonta Inlriakmn
Figure ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ iotoi 9.Bsctanmsinlosvru dsaoiF 2 11: 00 m.
14.. re pc
*2
250 -- .........
4 60 50 10 20 4C .0 10
50
c ........ .... ._......._ .......
Antenna hohihtsas iniicatm.
E2H 10000 m~
___ ____ ___C 20000 m
100 400 600 s00 1000 1200 1400 1600 1800
1 0 Continentalj tosperate cliznate,
N 30 1 (k-4/3).______.Yxth' earth (AhO0),
horizontal polari motion,
2 400 600 goo 1000 ZOO0 1400 1600 8500
Dislance in km
rigure 1U. Basic transmission loss versus distance; F -125 M~Iz, 11 10 000 m.
15
5%
125 MHz
0C- ----------
InI
Lq ZO:A 15000 P1J
D3 20000 mn-'C 25000 m
D3 30000 Mn
0antinen*tal
ISO ~taiiprate climate,
]psi __ ____________ roth earth (ah-O),horizontal polarization,average 9-4m,isotropic antennas.
.............................Free Space
I F F
4 ~ :^ *30 B00 1000 ZROO 1400 1600 11110
D stance in kmn
F±;urs 11. Basic trans?,ission loss versus distance:- 125 MHL, H,- 15 000 m.
16
".. . .. . .. 5%
A a,'C
, ! -. . _" 125 MHz
0I
"" 4"C E•1C0 00 B 000 1200 1400 1600 B00
IC
i • ~ ................... . ......... .......... .... . . . . . . . .
IA
____ Antenna heiahts'• . -- - as indicatcd.
_____ . ' IIl - 20000 m
A 20000 mB 25000 mC 30000 m
407 6,0 800 1O00 1200 1400 1600 1800,oo .... 95%
.. ........ ..... ............ ... . . .. .......... . . . . . ......
Continentalt.erate clim~te,NS31(k-4/3),
_earth (Ah-0),horizontal polarization,average grord,isotropic antennas.
.................. Free Space
II
ŽO0 400 000 goo 000 1100 1400 1600 t800
Ds,$1ance in kfn
-:.:re 12. 3asic tran.s-i'sion loss -/ersus distance; F - 125 MHz, I1 - 20 000 m.
17
vi,,
*.. . .t".. ........ .. .....
_ _ __125 MHz
C
6:0 oe. :000 120 1400 1600 i-0
........ ...'.= ... .. . .......... . ............
V, Antenna hoi ahts
I7 -as indicatmd.
El
' '--- 1 ".. . 25000 m
A 25000 000m
I ' 0000 m
..4 - - 6 . 00 Soo 1000 1200 1400 1600 1800
00. T 95%
S... .. ... .- . ..... .. .. ..... ... ......- - ---- ---- --.. .
-5k continentaltmperate climte,
I N -301 (k-4/3),Azth earth (,h-0o),
horizontal polariz.:tio)n,av-ge ground;,
- - -isotropic antennas.
.1 .Free Space2;t%
. . . , 40 0 Goo 1000 1200 1400 ISO' I1I110
Distance in km
±io.re 1.. Basic transnission lois versus distance; F r 125 MHz, H1 - 25 On m.
18
'5 .. •. i.. ____. _ .~.I ..* .~ , . .. ....... .. .. .. .. . . _, ___ . ____. _ _,_.-
:.;fr1
* - - *- -..... _ 5
A"5
1-- -125 hHz
sitC
0
I, jAntenna heightsas £rlicated.
E
_ _ I.__ __
Hi 3000 m 1
"" A 30000 m
° 01 O" 400 600 800 .000.1200 140. 1600 8.00
.. . .... .... .. .................. ................... ................................... .......
Contine~ntaltAWx,ratr' clinlh,,N --301 (k• 4/I),
I - .• __•m n tJi cvrth (All=0),S•,xrJ wxm~nt | iolar izatin,,
aver.Kpq qIrowr6,
isotLrpeic antennas.2?0C . . -- -
........ r Space
ZZSIi4
"2 : ,,,C 600 ,00 1000 1O00 1400 160', 1W1,.
F i-j re .4. lss! ".rans.sission toss versur distance: r 12s mIwz, II 3f) 000 m.
19 -
':-- 2•- ; -- " " ' " ...... .. -
___300 MH z
200~_ __7 __ý __
250 20 400 600 Go0 1000 1200 1400 Ifi00 3800
Antenna hei~ifts______ _____ ______an irKdicatea.
H2
__ ___ _A 300 mB S00 M
ABC~~ 10 0m~ ~ii~aD 5000rat IlmtE(10000 M
..00 .BoL 1000 1200 2400 21600 1 BOO'
:A191I 15. O'\ i trtmoinlssvru itnc;F-3 , rat climate
(k4/)
IIF
C'.C. 20 400 00 00 00 200 1400 1600 bo0o
Antenna heights___ ___ ___ ___ ___ as indicated.
BH1 10 tr.
12__ __ __ ____ __A 300-it
B 500m11
_______~ ~~~ 200 60 0 1ft00 10 10 0
1 300 -301
2zc: 400 60'0 800 1000 1200 140 1600 5600;7Ca~c 95%M
...ur .. 6.. Ba.. c .r n mi s a .o..es.....c l 3 0 M~ , H - 10 m
............. 2.. 1
4- minna
l~f 5%
___ __ ___300 MHz
-20"1 400 600 800 100 1200 1400 3600 1500
III Antenna heights______ _____ ____as indicated.
C _____ _____H1 15 M
H2
A 300 m8 500rm
____~~ 1000 m__ _ __ _ ____________________~~~ ~~~~ 5000________ m________ _________
-~~~~~~~ 10000_ m~'~ ~ C ~ ~ 3C0 30
N 31 300003)m
........................ ... ................................. reSac
4 : '-.~.~--~ ,. ..............-. _ _
1001
S-...... .......... _ ___
II____ ___ __ ___ ___ __ __ _ ___ ___ _ I 30000M z
20-0 400 600 $00 000 1200 1400 1600 1000
Ap0 EF~ t~ease indicated.
~~o 500~~ 1000 m20 40 160 10
F~~~~~~~ure~~~~~~~ 1000i rasiso os essdstne 00t~i 0 m.
1223
~... ....3A .~ ....j ..... .....______________ al
________300 MHz
Zon 400 600 1100 1000 1200 1400 1600 111S
A 7,C 0 E
Ante'nna howihts
.xj\ as indicated.
22t -.--. 7.A 300 i
-IB 500 mIC 1000ra
____D 51000!'m
c,20000 m
1_ 00000
t: Continental
AC 0 E -. teq~rate climiate,
horizontal polarization,average ground,isotropic antennas.
........................................Free Spacm
soO'C 1000 1200 1400 1600 too(
0Dsance -n Om
~:ze19. Basic transmission loss versus distance; F - 300 MHz, I =1501
'2'~~ ~ ... - _________________24_
SIt
..2
-300 Mrz
NZ.
A00C
1 0 00 400 600 800 1000 120o 1400 1600 80
- -30 (k-Iwo)50% erh(A-)
Antenna heigehtse
25B -0 M____
0 20~ 400 600 80 1000 M0 0 0 0
tranamiusion~~~~~ lo15neditne 30Mz 000 m
250
___ ______ ___ ___300 MH z
0
10 200 40 -0 600 B00 1000 1200 1400 2600 1SQO
50%
Antnn heights
______ ____ ____________ as indicated.
Hi 500 m
22'A 500 m
C5000 inU10000 Mn
____ ___ ____E 15000 ni
F 20000 MG25000 mn
U H 30000 mn
*4 0 201) 400 6011 Bo0 000 1200 1400 16Cl 1800
S 95%
Oontainentaltemrieratc climate,
E~ F-301 (k-4/3),s=cth earth, (Ah-0),horizonital polarizat-0n,aveage ground,isotropi antennas.
..............................................................Free Space
21: 400 600 B00 1000 1200 1400 1600 1800
D~slance in km
Figure 21. Basic transmission loss versus distance; r - 300 MHz, H 500 m.
26
-4A. 0 D.E F a
___ __ __ ___300 MHz
CO 00 400 60^' 600 100 Z OO 1400 1600 1 800
50%
Anterna ýwights_____ _____ ______ as jirdcated.
H1 1000 m
H 2A~ 1000 m
7 ~ ~~~~~ 1000 60 0 00 20 404 60 40
AG 300 m 30
2. 40 100C)244U ~ ..
600 goo2 ~ 0 1200 1400 1600 14800
L.~~odm .4 *.a~ . A
I?'.00-"-
___ 300 MHz
21O 40 ca Soo Iwo 1200 1400 '0 Soo OC0
:"as indicated.
____ ____H 1 5000 m
22!__ ___ A 50000m
D20000mE 25000 zoF 30000 m
2 ~ 4ý) 6'0 800) '000 1200 1400 1600 1500
Ootinentalteq~rate climate,
ýA~~ -3 D01. (k-4/3).a=oth earth (Ah=0),horizontal polarization,average qmtind,
.............................Freem Space
6' 00 1000 1200 1400C M0n (180
:),sance in k10
Ficurc, 23. Basic transmission loss versus distance; F -300 M4Hz, 5001I
28
00 5000%
E30000Mz
20 400 V6 100 1000 1000 1OO 400 1~0 Go- 800
tn ~ureAitenm hegt
as indicated.d
itrpc 20otenam
.*---~---~~.________________E 300 20 0 m~,si
0 24.0 60 Boosi tr'S 1s000 loss versu distnce le 3 0MI. n .
295
. -~~~~. ....~- .. - --.-
1~50%C __......_.. ......
___
C)
V - ____.............._
I~~ 1 5000
I~ 030000 m~
4"', 6 00 1OF 000 1 OO 10 60 10
4' 40 bO 4095%
Conltinentaltelpe-ate climate,
N=0(k-4/3),sR=th earth (*Ah=0),horizontal polarization,average ground,isotropic ant.annas,
- . O) I0C 12~00 14Z00 I MLCC 1?1500
a- kC'~ .rn ~vau. s.~ ic~a~ transmission loss versus distance, F' 100 MhIz, 11 15 00 M.
30 .
.. ... ...... .... ..... .. •• • + ++ , _ • • •?++•. ' . .• • • •+ • J• :-:S• • • ,'+ ' • ••" •
-- -- -Ir" °." - rT7l , Tf
..........
20o1 300MHz
0 200 400 400 am 1000 1200 1400 1600 10*0
AMem aýgtsI as iWicated.
E
- 4
,I A 200o 0 ,_____-- - ______I -____ ____ ____ __ ..__ .... ).
"f' I " "2-O"- -;2•l 25000 m
SiC 301)00 rp
Soo 1000 1200 1400 '600 500:;• + i "95% +
Cbntinental
- - utlooth 11a smt(k-4/.#h,
is,)trq *a, 1,F flý ID.
V" C . r0 110¢, 1400 1600 [so0c
rizx.r 26. Basic transmission lose versus distancel F - 300 MHz, H, - 20 0'0 m.
31
F' ... ...... . -........ ... ....... .... ... ..... .. .... ......... ..' ..-. .. . .. . . .. .". .-. -
___ __ _ _ __ ______300 MHzI.
-5 PADC 400 600 Soc 1rC 2OO 1400 1600 [Bo
1012;-- -- ___0%___ __
'_ _
4 _____ ______ ______ ______Antennal 1 ghts-- I - ________as ilmlicated
*. I I I
I 1 =25000 m.
" SO0 00O0 1000! :200 41400 1600 .
.;2'. . .. "-'t---2A 5 09 al
4-0B 60000 m
S1 WA - 0
SContinentalGteperate climate,
Ns=301 (k=4/3),A~~J smooth earth 1Ah=o) ,""-. .. -- . horizontal polarization,
____average ground,isotropic antennas.
..... . .Free pac
k m_ ___. 1" - :". 6.: S^" 4140' I200 1400 1600 l500 O
Figure 2-. Basi- t.'ansmission loss veisms distance; F = 300 MHZ, HI = 25 000 m.
32
Sil l 2 .. . . . . ...
- _ _ _ _ _ _ _ _ _ _*r* * ~ _ _ _ _ _ ~ '1* M~
I°' -- _ ____--5%
.... .. ........... ...... .. ..
I _
_001 300MHz200 - -,_______ ________ ___
- I
200 400 600 goo 1000 1200 1400 1600 1800
50%
- l •"1..............- .. .... ....... ... I... .
Antenna heights
S. ._ _ _ _ _ _ as indicated .
1= 30000 r
12
I5- -A 30000 r
I I
- 2 400 600 800 I000 1200 iInn W600 1800
-, ,-95%
: .. ... .� -Contilnental
- temperate climate,-•A• Ns-301 (k-4/3),
smooth earth (Ah=0),
"",' •horizontal polarization,a-age ground,I I \isotropic antennas.
... ... ... . .. .........Free Space
I<
"o n ,, 0 0 1200 1400 1600 I80
sla,:e in km
Ficure 26. Basic transmission loss versus distance; F - 300 MHz, H 30 000 m.
33
5%F
[ 1200 MHz.1-
* . 4C 61no Boo I2n00 - 100 1400 1600 [Soo
At 0 G\IA~OICAntenna heights
_____as indicated.
aI I 13 m
B 500 rC00mD 5000 ri
____E 10000 mnF 15000 mG 20000 m11 25000 m
ŽC~~~~~ 4C ffQ- 10 8C 130000 m210. 4c (ý O 000 10 1400 1600 185%
{.777Continental
3ý 1 temperate climate,
I I N =01 (k-4/3),'J 1C PX G\'H I j _ _ _a=th earth (Ah=O),' S C _ _ _ _ _ _ _ _ _ - - - - h o r i z o n t a l p o ~l a r i z a t i o n ,
I average ground,
I I isotropic anteirm',s.
..... Free Space
oo 1 1 ~ SC 00 200 1400 1600 1800I
rFijre 2ý. Basic transmission loss versus distance; F =1.2 GHz, H11 3 m.
34
_ _ _5%
A IC Doo" 0z
___ ___1200 MHz
O, i 60) 800 10 1200 1400 100 ]s800
. Z . . .......-- 0
", '\ as indicated.
3=10 m
x H2_______ -- A 300nm•.. B 500m"•" .•.• •',C 1000 m
G 20000 mH 25000 m
: "5• """1 30000 m" ""0-0 nn- 10000 1200 1400 1600 1800
______ -- Continental
I~ N301 Mk-4/3) ,
s il e u isotropic antennas.
. Free Space
an I C1) 10 400 1600 o80
31)~ J. la~ tansinission loss versus distance: F =1.2 rHz, ill = 10 m.
35
________________1200 MHz
dc C,' VC' 800 2000 122, C400 I600 1800
2--150%
I ~Antenina heiqhts1__ as indicated.
.. ~~~~I ,,- __ __ 115 m
A5000 m* ~ ~~~ 5 .___ 000 mi
F' 15000 mG 20000 mn
-. ~1 25. 3000 m
I - V f59 %
I - - -___ -- Continent~al. ...... tenprate cliniate,
I N 301 (k-4/3),___j~wthearth (ti=O) ,
horizonta1 polarization,I--- a-erage qrcwKI,
ac I co 1 '0 200 2400I 2600 280
- ~'0 ' 3as i -fans.-ission loss versus distance; F =1.2 '111zm.
-. ~~~7J7Z~i7777736
r..... 5%
_ 1200 MHz
; 0" bOOý 80 I So 1000 120n 140n 1600 i0
I 1 I
G'Antenna heithtts
_ ,- --. -- _as itdicated.
,.., *-------_*_.i,-4------... ..* .... __•__ ____00_
o3 m
" -�-"A [00 m
D 5000 m
2- .A* 1- -.. .. ' - A 300r0 nr"F 150000m
G 20000 m
"I" i~1 30000 m1.. 400 61 1' g00 1000 IZOO 1400 . 600 ]00
X .I I I
-> - T .... .teme~rate clinate,N=01 (k-4/3),
=?oth earth (&h=0) Ihori zxital polarization,
"F !average ground,\i 1 \ iSOtipic antennas.
K \--\....... Free Space
_____[____.....
""a. ýT ,C00 I0 10 1600 1800
P'c k M
F izure 32. 9asi: transmission loss versus distance; F = 1.2 -Hz, H1 = 30 m.
37
'1.
5%
__ __ _ 1200 MHz
0
1, 0 zoo 400 600 B00 1000 1200 1400 IOCO 160C
1T1 50%
0 _
A¶1 I Antenna heights.. ........... __ as i~ndicated.
0 IH 150 mn
H2
~~~~~~~ 300 60 0 mO10 40 60 10:00B 500 __ _ _
C 00
D 00
2t E 000.
0C. 4 00 1000 Bo ILO 1200 1400 1600 Moo0
2aart climate,
S. . . . . . ....... F e 1C
i------ -------
125
1200 MHz
4 600 800 1000 1200 1400 I100 1800
T°[ F -- .- .. T .. .. -50%
E ... 7AY1I. 'I.... ... . . sA:OP D: E' H Antenna hoiqhts
A 300 inB500 Mc 1000 m
D 000 in• •E 10000 50n"F 15000 m
G2000 In
___ _ _ ____ _ __ I1 30000m2- --. 40.L0 o 1000 1200 1400 1600 )
S- . ooContinental
. .. . temperate climate,- N =01 (k'ý4/3),
. . . . . ......... "................ .... earth (Mh=0),
.. i' -horizontal polarization,average groutn,
isotrop'ic Olt4'nm.s4.
....... S" w w',;I+It+
"t o+1. 01, ," 140f-t
- 4. ý iiSJ trans7-.ssion los,' v(,r•rus distianc(.; F" 1.2 "ill,, II j In.
39
V.....
___ ___1200 MHz
Z(,( 400 600 FIO 1 no0 1200 1400 is I 0%
A50%
___ ____ ___
2 IC,
2V' 2
D 1 0000 m
~~~'v~~~ 2000 m0 0 0 0 40 60 I_G 25000 0m
*teroperate cliffatc-,N 31(k-4/3).
eaoiir'oh ('h 0) ,- -- )wri7.Ontal I1)larizatiofl,
All D F G wcraqc qrnunid,k isotTTpic antennaR.
.... IPree Space~
7-.zre Ras;.,:..-ansf'i3:;on loss versus distanice; F 1.2 GlIZ, H1
40
___ ___ 1200 MHz
-3' 400 go0 800 000 1200 1400 1600 I500
Z...................... .. .... 2 Antenna heights------ --- as indcicated.
..- ,-----*--- -+-. =1000 M
12=
______ A 1000 m13 500D MC 10000 M0 15000m
-- ~E 20000 m
-. 400 600 $D0 1000 1200 1400 16o'f 1800
____ __{i~ 9 5%
Continental
I ___________________________smo~oth earth (Ah-0),
- horizontal polarizat ion,average ground~,isotropic antennas.
...................................................... Free Sparo
SOO 1000 1200 1400 100 1 8.'a
~.r 3 ~aiCtransmissionl loss versus distance; F =1.2 0,11z, 11, 1000 Tn.
41
Y .. yT.7TZ7U777i7'A'
___ _ _ __ ___1200 MHz
J___ _ _ _
I---
2m( 400 p.00 Elio 000c 1200 1400 16010 I*C.0
_T ___ 50
Antenna heiqhts____________ ______ as indicated.
AI 5000 m
C 1500r
A 25000 rF 30000 m
400 600 son 1000 1200 1400 1600 t 100 %
I _95%
...... Crntinental
N =301 (k-4/3),__________ ______ Joth earth I(Ah-L),
hoirizontal polariLation,A E C 4 EF a-eage ground,
ipotropic antennas.
.......................................................Free Sp~aCe
6 -7::' Sol 1000 1200 1400 1600 8900
>stance r) km
Fi;-ure 37. Basic transmission loss versus distance; F 1.2 GINz, H1 5000 m.
.12
5%
___ __ __ ___ 1200 MHz
0
.. .... ................
1 Antenna heightsas indicated.
B 15000 m~C 20000 mD 25000 msE 30000 ms
40
1 600 800 1000 ]zoo 140C 1600 1800
Continental
* tue~rate clifrue,
____________s~ot~h earth (Ah-~0),
horizontal polarization,avrage ground,
isotropic antennaks.
.... Free Space
VIC 10 100010 ZO 1400 1600 1800
Fiz2ýra 38. Basic tý:ansmission lcss versus distance- F =1.2 GOHs, I-: 10 000 mi.
S. - ~ ~ .. .- ___ ____ ___4-3
_______________1200MHz
00______ ___ ____10
20Clý 400 600 b00 1000 1200 1400 1600 150%c
o - ___ ____ ____
-4 ,.)___ AB ! ~Antenna ku.,gts................. _ as indicattd.
:1IH 1 15000 m
______ ____ IH 2 -
IA ISUOO mB20000 miC25000 mnD30000 In
Z:I 400 600 800 100 iaoo 20 1400 1600 10
N301 (k-4/3),tNprt i..ae____ __ ___s&3th earth (&PMO),horizontal polarization,
avareg ground,iinotr ic antennas.
...................................................... Free Space
40C' GO 30CI 1000 1200 1400 1600 1600
D~slarce in km
Ficore 39. Basic transmission loss versus distance; F - 1.2 GH7,, H~ 1 15 000 M.
1 55 0 .. ... .. .. ... ..... .. .5 %
. ... .j. __ __ _ ~1200 MHz
2 t
-•o B' '0• 1C O €O I00C I ZO on I4t.C' 1800BI ---- F-.
,,4. . . . . . . . .'.... O S ,,-... . OO- '.. .. ... -i .. . ,C . .IU
, --- .-- i.-- I I i "
: s ....... ... ... ...'I i I \ Antenn heights
I ~------- as inkiicatedi.
< i "', u11420000 m
I I '-. -. 27-o
"- • -B 23 000 m
I I c 30000 i
I 1!
I I II
___A:. i- i Li__BOO' ,0" 1200 140 1I600 8
-. I ii'
- -- . . . .... . ... Continental
S. temperate climate,NI N=301 (k-4/3),
S smooth earth (Y'lb0),S --- -- ---- ... horizontal polarization,
A 1C average ground,lA isotropic antennas.
- ........ F Space
S' I• . . - '
' ) '. ~ ' I !". ': ..-
Fio'lre 40. Basic .:,'ansmission loss vers-is distance; 1 1.2 GnZ, I1 = 20 0Of)0 .
45
, ,t .
• -.. " . •. ,i,-.ZII7I,, ,
..... .
2 1 ~5%
* II ~ ~~ ~ ~~ .. -........ _______
t_?_-'-i
z~o ! N"1200 MHz
" 200 400 600 8C0 100I 0 1200 I4C6v, 3600 'S 1,o " 50%
"...... ..... ........ .... ............ ..Antenna heights
I '_ _ as indicated.
H 2A 25000m
B 30000 rr,
"" '. .600 80 0 000 0 03 400 I20600 1800605%
....... ... ..... .ti.ntal. .tperate climate,J3 =301 (k-4/3),
_ I th earth (Ah=O),___- _ __" "__ -horizontal polarization,
A.B average ground,isotropic antennas.
.. . ........ Free Space
i" S0(C 1000 1200 1400 60C. '800"
":stance, n k m
Fioure 4.. Basic transmission loss versus distance; F - 1.2 GtIz, H. 25 000 m.
46
I.
150 - __15%
. ~ ~ ~ ~ .......................
_____ ____-
20o--•- 1200 MHz
0
200 400 600 Boo 1000 I200 1400 1600 1800
S • .............. -- ...... . .. I . . . .
A i !Antenna heights
as indicate:,•
""____; I_________
I ___ ____ ____
•' : %'\HI 30000 m
I (k=--/H2=
:•- _ _ - 4 -A 30000 m
.2: 4 -- 600 m00 1o00 ]oýO 1400 16100 180
S.... .. . ..-:!-•i ' . . . tem e ate clima e,
42. o...loss... ... horizontas polarization,
I I isotropic antennas.'
i , \ ,....... Free Spac
6- 1301" 7 ,^' J,O Ino mO 200 ! 400 I fOO I 800
S 1 3rýe n Pm
Fi7--•re 42. ý-asi: tr-ansmission. loss versus distance; F =1.2 G;Hz, If1 30 000 m.
47
-,
_ _o_ __ _ _ --- -- 5%
.. . .. .. . . . . . . . . ... ,......... . . .. ............... .......
_ _ _ _5100MHz
"400 600 800 000 1200 1400 600 0 %
Antenna heights.... ...... as indicated.
SC tE.ý Fý Gý 3m
M H2 =
___A 300m22 •."B 500m
C 1000CmD 5000 m
_ _, _ _ _E 10000 mF 15000 mG 20000 m
.i.J..........11 25___ I3000 m- ., 0 401 600 800 R000 200 400 300
Si , )9 5 %
- I
Continental
temperate climate,
N301 Mk-4/3),"""th earth (Ah-0),
horizontal polarization,,average ground,
A'BC D E F G H I '�otropic antennas.
,,r....... F e Space
"KY.,x•.,• " "\,,'
.5
BO : E 8CCO 1000 1200 1400 1600 ;800
-a'ace P)0 km
Fijure 43. Basic transmission loss versus distance; F . 5.1 GHz, H] = 3 m.
48
100 0 0 80 10 O 10 60 fO
~. ..........5100MHz
NIN
I -• -\-:
" 200 400 6O B0oo '000 '200 1400 6r.00 IFO0
" C000050%
A~ntenna heights
. ; D1 G =1o mH 1 0
2H
*I - .- J A 300 m
C 000DmD 5000Dm
F 15000 mnI;.20000 m*I4H 25000 m
220 10 30000m" "CC i F0Or'C, 0 800 1000 1200 1400 1600 1800
95%
__ - ---- .. ContinentalIteiperate climate,.. "
N =301 (k=4/3),
Smooth ear'h -Ž, ,- ____-____--- ]horizontal p.larization,
,' N• I I average ground,"ABIC• D �iisio tropic I o o , antennas.
' Free Space.
SI i__7_______1400___00_1_0094.
2~~~0 4". -* 6O: •F" :•: IZO I+00 1600 l BO
T1~.re 44. Oas c sranms;o 0 !~oss ",rsUs distances F = 5.1 GHz, H1= 10 m.
49
, - - , ,. .' :.*.,,..,.I..S,, . .. • " ' •,+ . " ; "AL
- -C
........ .....•'" ~ ~ ~ .. ...'' , H. . ..... ...... ........ " .......
5100MHz
I0( _______ oo •o ,o0o , 4o 0 . ___.Lgo .600_._226 I2550
I II=i15_Soo " _ _ , \ _ , \ Is
2
100 _____ 400 0 0 1000 Q
F G5000m2000000
B 5000 m~~~C 1•.I".'•S2000 m;
250 D, •• -- H 25000 m
.. 1 30000 m200 40C 600 Boo 1000 I200 14no IG 0 I 1800
95%
;4 a 31 (=/ )- - ----- '-- .. ... ... C~ontinental
. . . . . . .. ........ ........... ....... nEot.2 earth (Ah"O)
I ~average grou1nd,A* is Ijotropic antennas.
- A. .. Spam
ec'. 1000 1200 140 1600 1800
s f a' e Ir0 krm
F-_z-.re 45. lasic transmission loss versus distance; F • 5.1 GHz, FI = 15 m.
50
........ s- ý .ý -- ,-,:...'.•r,- -"
I -C
- - -- - -- _ •.
0 5100MHz
I'Ik
'~~ C 00 100 2020_ 140 I6o £0 800
J ' f '50%
-, '1-T --.--- - - -, .N - ... .
___ Stonna hojaljitsasinjicatedI.
GIIH 'N'3s -
0 0 A00
B 500. m ii
0 25000 a
.- : ' , " -, \ .. , • , 1 5 0 0 rm
"... - -1t-- - 2 30 0 0 0 m
12(H 4000 mO 10• . .. . . . . . . . . . . . . 5. . ,' ! _ I I O XO 12• 4 o ., i 6 0r , 1 3 1,I 0 0 0 0 m 4V 95%.._•_._tt
N
I , HContinentalrate climate,N 3•O1 (k=4/3),
.-. .�----- ~ 4 spooth earth (Ch=0),-orizontal polarization,
A FGHI E I average grou'-d,S", -I isotropic antennas.
....... Free Space
I I 14,•. :\ f ". -, . l I /.................. ..... -. ,--=--..•---4C.---.'J L. .r l
- 8% 1>" '• 200 14C00 so6ni I O
4;re 4I. Has tI-' -i.s•ns iasior, loss versus distance; " = 5.1 (,Uz, = 30 m.
51
100 5%
__oo___ 5100MHz
200______10
0
!0 OC ? 400 qu O0 Poo 0o0 1200 1400 1600 50%50%
125¢
..Antenna heights
as indicated.
zoo - H 150n m
A 300 In
z•= • E1000 5000 m
EG 10000 m
I . ~ 1..""LŽK'KFŽSJF 15000 InS:, 2 • . H 25000 1
- -0000 11• - 4: .2., 400 60n 800o 1000 1200 1400 1600 1800
SI ' I95% 1 0 (
I _ _ _ _ _
:* "- .., - _! C-ontinenta
-. fr'rizontal polarization,average ground,
A'i 0 E'.F C)H)I isotrop)ic antennas.20C I - ________ ________.. .. . . . . . . . . . ...... Free Space
-"C',, I000 1200 I400 1600 800
'n km
.-- u 4- .. , as1 : :ransmission loss versus distance; F = 5.1 GlIz, 1 = 159 n.
52
.. , - . ..,,.,._•. ,• ' ,",,.L• .. .. ... ., . '/'- -• '" .. . . . ., ._ ,
--- -- r-----------
-- 5%
.' .......... ... . ..
2- 5100MHz
-2 0 400 600 Boo 1000 1200 1400 1600 180050
50%
...... Antenna heights
A C asF I indicated.
A 300 mSB 500 m
C 1000 m____.. .. D 5000 m
E 10000 mnr' 15000 Pm
U G 20000 mŽ 0 00 0H 25000 m
-- _ _ _600 800 1000 1200 1400 16no 18nn I 30000 m
'°" !H ---- --- ........ .. 95
-H---- -----+-- .- 4continentaltemerate cl1imate,
" " • -- • .. .. Ns=301 (k=4/3).-| ........... th earth (h=)
horizontal polarization,average ground,
D ~E FO if__H I_ isotropic antennas.
* I I
1 .......LFree Space
1200 1400 1600 I800
"ur . 5asic t:-ansmission loss versus distancel F 5.1 GHz, H = 300 ni.
53
i0o .5%
-- ,s ........ ..... .... ....... . .......... ... . ... ......... ° ... .
+D • 5100MHz
.
tooC - 50%
1A
I2, \ __- _o m
0
A C50E0 HU ________ l0000 I
SF 2 0 0 0 0 m
S~G 25000 mo H 3000 0 m0 200 400 600 Bo0 1000 1200 1400 1600 BO
IU'• IV
. 0 .a..... cl..te,
N_=301 (k-,4/3),
--• ............ ................... ' .... ........ . ... A!W . ... .. .. i... .... ...
SJcot~ear~ (•-C),______ ______ ____ ---- horiw~ta1 polarization,
its ... ....... tnr Inic a u teda.
200 C 4'C 6. 800 1000 I20 O 1400 I 800
C.s'arce nr kFigure 49. Basic t.-ansmission loss versus distance: F - 5.1 GHz, H1- 500 m.
54
* .. l .i,-.
H 2
1 00 5%
" " I ' • ........... .......... .... °. . ..... .... . .. .. . . . .. .. ..
___ 5100MHz
0 iOO H0 O 0 20 10
0%
.-... Antenna heights
'_ _an indicated.
aC___________________H, - 1000 m
"H2A 1000 m
11 5000 m-JiC 10000 mD 15000 m
250 E 20000 mP 25000 m
- G 30000 m
. '00 40o 600 GoO 1000 1 OO I IO I95%
Cotinental• ~tani~ata climate,.. ................................ (k-4/3),S.... ............. earth (Ah,,O),
horizcntal polarization,
average grun,A .• i•uimaropic antemas.
.... .Fres Space
'V % 'a - • ,'.7
j. i InX Po 10 14cr. 16,00 1 lB.r
.-re 51. Basic °arfnsmission loss versus distance: F - 5.1 Gtiz, IlI - Pno0 m.
55
oi4
II
R0C~ 5100MHz
?0 400 600 goo 1000 1200 11100 600o 1800
50%
Antenna heights-- as indicated.
Aý Dý EýIll = 5000 -1
A' 10000
C 15000D 20000 in
F 30000 m
400 60() o8 100 0 100Z 1400 1600 800O
Continental
N 31(k=4/3),-. - .. ~ othearth (Ah=O),-- - - -- ~ - horizontal pOlaxiýt iton,
~, \ avrage ground,
Alla D'~FIsotropic antes.
F.....................iree Space
- . .~. 80 00 1Ž0 4 1 I00 1800
S 'a-e -n kM
::e 3. aSic i:.-ansmission loss versus distance; F =5.1 Gl~z, Ill 5000 M.
56
_5%
___ ___ 5100 MHz
00 2')0 400 S00 goo 1000 1ZOO 1400 1600 1900
_______50%
. ...... ....... .. .Antei"n helqhtsI'S as ind~icated.
_______ _______ ______ _______ ______ _______ l~H 10000 111
A 10000B 1500IC 2000L0 InD 25000E E30000~
60 go -0 O 1400 1600 1 CI %
Continental
teuperate cllzmte,
____________________skoth earth (Ah-0),
.... Free. Space
-c~ 1~0 200 14W0 1.00 IoC
, km
10 010M.:re 32. Basict. insmission lose versus distance; F 5.1 GHz, H1 00) o
57
I.-
_ _ _ _~~- ---- ---- ----- I ~
At 13 C 5100 MHz
2 1" 4on 600 800 1000 1200 1400 I600 100r° - - 5 0 %
Antenna Ix-irihts................... as indicatcd .
A I { 15000
C 25,00 ,i[ D 30'0t'00 1-
u[
25,)oo 400 600 800 1000 1200 1400 1600 1500
S ... ... .. . .... ... .......... ... . . . Contin ntaltemperate climate,. -301 (k,.4/3),.. th. earth (Ah=O),
--- """l_-horizontal polarization,
average ground,isotropic antennas.
............... .............Free Spac
-. --..-.---------.-.- - --....... --
*.. . ...-- - - - -.. -- .. .. . -" --" " -' -'•" ,•,- - -.... - ,-....-
- V' I ?Or- 400 r b
."as 'c :ransr.i!slor. loss vorsus distance; F " 5.1 (1l, I11 15 000) m.
58
..................-* ----- *---- - -w
!0 ,o5%
____________51lot MHz
200 0O 40C 600 600 1000) 1200 1I100 1600 t R0)•
I; 5~0% l
Mt .ahoiolhts
____________- __ _____aa indicatcnl,
II
Ii = 00005a
A 2000a
2520001
y 0 200 40C 600 to0 1000 2?00 1400 1600 1 sm.
Conteim hnioh
--. . . . . ..2Y.oth eath (.h-.)
.......... .......... .I as i"nd"cated.)
[i' iL'r • • •\ ., . •-- ~~~horizonatalarei, ticru, anenspolarization
" " I •' I( ~ 1 1400 2600 I00
.a..e 54. 2 : -•nsj-is on loss versus distance; P- 5.1 t1, , I•1= 20 100 m
59 .
2A.0 40C~ 600 Soo 100 1- 1400 160 MCI
-5%
_ _ 5 100M 'z
0 -- 2?0 400 600 B00 1000 1200 1400 600 1800
I50
. ............ ......k ....... . ..... ................. . .. Ante .n height3
• - "'.• asindicat(K.•:
_ _ _ _ __If O =25000 122
A 25000-)B 30000L
-2,¢ 4C. 600 Bo0 1000 22(0 1400 I C00 2800
S' i / 9 5 %
SContinentaltemperate climte,N =\ : (k=4/3),
______ ____ _____[ ____ soth earth (Ah=()),horizontal polarization,i averawe grond,isotropic antennas.
........ Free Space
I ZOO 1400 IbOO 1800
Fi.7jre 45. -Sasic trnsmission loss versus distance; r 5.1 GlIz, I11 25 000 m.
60
.-.- , , -A,
201 5%
k\ - ___- ___
i.,rj~....
..
•...'-'•
1''
""A 5100MHz
0,.i ",, 200 400 600 a 00 1000 1200 o400 1600 1800
550
............... ... ntenlna heŽights. . ................... n indicatedl.
H1 30000 m
2_______ _______ A 30000 i
250I
I tI- - -
""i.". Zoo 400 600 Boo 100 IOOO 00o 1400 [Goo Is 0
"-' .. temperate climate,-.,, • -.............. eti ............r.... .. _ ,. ............. ... . . .... ........ . ..... ..... F pavrage grourK],
- Z 0 . 1200 1400 lp'(W. I "Id..
:..• _. . . . . h- ...
p7:•re "i€.. ji .: t 'ans:,ission loss versus distance; V' 5.1 fI2z, 1Il = 30 OW) in.
61
'I ,, _ _ , ,I• . .-. ,.'• i ";,..
'°------ - 5%
... ... .... .... ....
_ _._ _9400 MHz
101- 4GO 600 8o 100 0 200 1oo 400 1600 ]a 00100 ~.--I50%
1 25
0
150
MAntenna hiejqhts?I ....... ............. - ... . as indicated.
= \ ' \ 1,= 3
225; A 300im
C 1000 mFD 5000 m
2t .: , _ E 10000 m
G 20000 mH I 25000 in
2400 26001 30000 mC, . 4'O 600 800 I000 zo0 1400 1600 1 0$• •"" ! f95%
mi
, I, 'Continental
teqperate climate,"NS=301 (k=4/3),
.... .. s oth earth (Ah=O),"horizontal polarization, onaverage ground1,4
Aý C Di Eisotropic antennas.
.................... ........ Free Space
4:" . ^C 800 1000 1200 1400 1600 0 BOO
S $1a-ýe in km
"'Z-re 5-. 3-3S.c -ransnission loss versus distance; F 9.4 07111, I1 3 i.
62
.5 • ., . .. ... ....
Fo l-.9400 MHz
100 , 2", 400 600 800 1000 1200 1400 1600 i BO O
I IH
t2
" C IA 300 m
B 500 m
1C 000 mD 5000 mE 10000 m
F15000250 m0G 20000 mHl 25000 m
F -0 00 1 30000 m100 400 14600 1600 1800,: • • "95%
Continental"temperate clitate,
. . . .. .. N=301 (k=4/3),. O' earth (Ah-O),
"•-" " "' ' --':.- ... Ihorizontal polarization,
average ground,
A~C~ I~ Iisotropic antennas.
.4 ' ,i , !..... Fre s-pace\,
0: "00 00 1400 1600 1800
SS'are ,n km
i 7-.re 2-. !!as i-' "r ~n ission loss versus distance; P = 9.4 .Ilz, il 1 20 m.
63
', --- tJ
F . .w '""
' 1 "4 ... m
----- - - --
S/•• .....~ .... . ... ..... e i h '
V~ A F G H I
0 i A9400 MH7
21
6 b0 000 2 1400 i0
2 . D-5 0050 %
E'
a in00 a.AC2G00
'1 _ _.-A 300 r-
S •B 5000 mC 1000 mD 5000 InE 10000 mF 15000 moG 20000 mo
UH 25000 moC 20, 400 600 son* 1000 1200 1400 I 13 00,10(1 160 __81__
I95%
'11251 --------- 4 - 1__
Continental. temperate climate,
. . . ...... .. ... . . " N s = 3 0 1 (k = 4 / 3 ) ,I' ........ .......... s ooth earth (Ah=O),.horizontal polarization,
' average groumd,
A 11 isotropic antennas.AC' DI F Q I __ _ __ _
.. ...... Free Space
2zt1--~--
goo ' 300 1000 1200 1400 I 600 1800
D ,Stance in km
Fiure 5-1. Basic transmission loss versus distance; F 9.4 GHz, Ii1 = 15 m.
64
-SFWi -
__ _ _ 11 5%.. .r, ..... .... .... . ... ... .
dFý HI ý -
\\- - -9400 MHz
400 600 80so I WOO 12 1400 160 t B ioo.
a .-... 50%
0
,.. .... . .Antenna heights... ... ....... . ... as indicated .
If, 30 r.
H12=
____A 300 mn' 500i*mC 1000 mD 5000 m
._ __ _UI; 10000 mF 15000 mG 20000 mII 25000 mn
______ ______1 30000 mn
2• 400 600 Boo 1000 1200 1400 1600 Is0o
95%
S•: • IContinental
" "o-erate climate,• .. N ,'301(k=4/3),' ~~ ........ ........... • c~ earth 1Ah-o),
...... • ................• ... ... . ...... .horizental polarization,
isotropic antennas.
.. F.. . 'Fice
6-1C B0o 1 000 1000 0 1600 1800
S sta•Ce in kM
F' •re 6$. Basi: transmission loss versus distance; " - 9.4 GlIz, Ii1 = 30 m.
65
47 ,7.....*. .. ,==. -•i 7 -
~5%
\ý- D ' F' fHI 9400 MHz
-- ,50%
-I.I* I:
:253 -. ._____ ___ ____ _______
Antenna heicthts. ....... . . ............... . ... ................... a s in d ic a te d .
A 0 E H I I I = 150 m
12A.1 /5 300 r"
1B 500:
C 1000D 50W1"
1 000.)-_____ -- F 15000
G 20000II 25000 r1 30000m
00 60(C Boo 1000 1200 1400 1600 1800
.. i195%
Continental" " "" .i ' terri::erate climate,
Ns=301 (k=4/3),." " ............ ...... .............. ............ smo th earth (Ah=O),
horizontal polarization,average ground,isotropic antennas.
.... ....... Free Space
"-\
2so 4I-• &( 800 0000 00 1400 1600 1800
,S lfa-ce in km
- .6. i: rans.ission loss versus distance; F = 9.4 G;z, H1 = 150 n.
C6
.. ,,__._.___._.... . ,__J o7 7 ___.___"._.... ._:__..___
00 _ __ _ __ -__ __ .5%
\_ ___1 9400 MHz
10
25 2
E A CmH 300 m
A 3000 m
1' 15000 mn
D 25000 mn
20(l 400 600 800 1000 1200 1400 16ovn I 3000Oi
2 - - 195
=el~rate 4i1iznate,............. N 2s01 (k-4/3),
. .... .... .......... .Rc th earth (&Pw0) ,.... -.---- -.. horizontal polarization,
average grownd,Iisotropic a ntennas.
D1slance in ItI
1~e 2. ý-sic trnmission loss versus distance; F - 9.4 GHZ, Ill= 300 In.
67
---i - - ... ... .. ..... .
o_, ,9400 MHz
! .. , 4,1"- irJ,0 Bo o 0 00 1200 1400 1600 1800- ---- 50%
In. Antenna hEiahtsS................. as indicated.
E C -NG_ HI 500..,
H2 =w f -- ____ A'B 1000 mC 5000 r,.
I D10000 in4 E 15000 m
F 20000 miG 25000 mr.
- I
-2= ----- _____ -- ~~nA.5: . - Continental
temperate climrate,N -301(k-4/3),. ....... ...... earth ( h-O),
. horizontal polarization,
- Iaverage groun,isotrop)ic antennas.
................................Free Space
" 7 42-T 60O0 Boo 1000 I200 1400 1600 1800
D-stance n km
Fi'-:re 63. La.ýiic transmsssion loss versus distance; F - 9.4 Gflz, H1 = 500 m.
6G3
*.. ... ... .. . 1' a rm. -
_-_ _ . . . . 9 400 M H z
- oi. ,o'o o ,o ,o ,o 0 °o0
20?C 400I 600 800 1000 1200 1400 1600 800[ 50%
0
0 "
15C ___ ___ --___ ____ __
Antenna h~eights_ i".....................as indicated.
A IC IE C 1000oo m
" ,: ___ - -:_. . A 1000 m•"IrB 5000
C 110000
D 15000 m
__F 25000 m
2M____ 400 Goo Soo 1000 1200 1400 1600 180095%
S. . . . ,C on t i nen taltaVurate clinate,
.. ................... . (k-4/3),. . ..... .. .. th ( "-0),
...........................horizontal polarization,
l ist i cz i antennas.
..... Free Space
2 - . " A 800oo 1000 i 0o o 0 146 0 1800
D,slance in km
e 4. "asic transmission Icss versus distance; F - 9.4 GHZ, HI 1 1000 m.
69 Si1 i~ A - .. .
5%
____________9400 MHz
400 60C DOO Oc 2 400 6100 0s
112
___ _ __ __ _ _ __ __ __A 5000 r
__________________ _____B 10000
4D 20000 80
Continental Flrzto
average ground,3ý_ 0D ýEýisotropic antennas.
............................................................roe Space
60 800 1g00 Ž0o ]o 1400 600 ]BooD sIoa n ce i n km
Fizure ý5. 2asic 1''-irmission loss versus distance, F 9.4 Gunz, 1I1 5000 m.
'70r
- -Moo,
___ ___9400 MHz
11' 00 60 OC. 1000 1200 Il0o 1600 1 R0T 1 50%
Antenna heights........... as ind~icated.
EBýC E 1l1 10000 m
12______________ ______ ______A 10000
B 15000 r!C 20000 mD 25i000m
- ___ ___ ___E 30000 mn
40C £00 800 1000 1200 1400 1600 1800
-- ~--continenltaltie~rate climate,
................................ th earth (Ah-OL.,..... . ..... ...... horizontal polarization,
average ground,isotropic antennas.
............................Frree Spacmý
410 £00 Boo 1000 1200 1400 160" 1800
DiStance in km
*i~r 6. Bas~ic t.rasmirission loss versus distance; F =9.4 Gliz, 11~ 10 0010 mn.
71
1~~ K I15%
A B Cl 9400 MHz
4 600 Soo 000o 100 t 0 1600 18)
7A B' G - ___ __ 15000 r.
______ - IA %1,00 7!
-~~~~ 2'5) _ __ _ __
400 60"____ 800 1000 1200 1400 1600 1800
i__ 5c__I_( Continental
taiperate clbijate,
~~~. ........................ Ismoheree Space)
........................... FreSpc
75 - - 9400 MHz
U1
Wn 25T'-
0
I'[ 60 100 000 1200 140 60 0
The1, shwnis for the H,-20 000 mnend H ' 30 000 Is case 17, see. 81. 50%At zerol distance the ray length used to determine L, is the differ-
C ence in antenna heights so that the height combination selected gives
.25va lues near zero distance that are larger than tose associated
0 wth othe antenna heights.
IlThe L.,,(0.5) curve is the same as the 'fcurve.
U) IIIAntenna heights
----- - I-a- indicated.
C L (C' 5) to be greater than 6 hsH 00
zo becom!.ng less than by an excessive--
25IVAB the aircraft become further apart, the-____ A 20000 niI
oxygn an watr vaor lyersand tmoC 30000 11ph.eric absorption increases [6, sec. A.4.5).
Vo VThis transition region connects the far
o radio horizon by' a straight line [7, sec. 'I.
1000~ ZOO 400 600 g00 1000 120 40 10 0
U) ~ Vimie transition region connects the dilfraction region with the 95%forward scatter region by using a straight line [6, eqn. 145).In this case the diffraction and scatter lines do not cross and
!25 .,the line is drawn between the radio horizon point and the firstCO T valid scatter point.
SIV VII Forw d scatter region (7, sec. 51.
....... H .30 . -4/3),
--... ?rizor.tal polarizaticn,average growndisotropic antennas.
VII0 s h A B IV IIIV11Rteaircraft become further apart, Free Space
the ray between them passes throughmore of the wster vapor layer and
225 tropobpheric inultipath increases, but_this increase in variability is not
slo*-wed to affect the Jtv(0,5) curve(6, sec. A.71.
2 I02c"' 400 600 G00 1000 1 Ml 1400 1000 1600
. ..__ .... .. ..__. ... ....
I |I .. .
•-- 9400 MHz
I II
0
' ntenna heights
.. .......... .. as indica t i..
2~C2S 2' ___BI_ _=__ooo_
SkA 25000 mB 30000 m
0' K "
"" 200 400 600 800 I000 1200 1400 1600 1800
0,oo 9 5%
I B 1= 5005
15e Continenta'
terrate climate,S.......... I.N.... -4/3) ,
• ............ ............. .............. ....... . .......... horizontalea tpolarization,(a O ,I "3• ='----- '•average ground,
S....... Free Space
225 30
20 11 60,e o Soo 1000 12o0 1400 ,600 1800
-j• --- - t -- - - - W ', -. ,
I 4
. ......... .. .. . ....... .i ................. • . X
o.. 9400 MHz
InI
I - _ _j
£- .. .__]
_____1400 1006000
600 800 1000 1 zo O C0 i600 1800:"Oo , 50%
Anten-i heights
............... ........... . a s ind ic a ted .
, II= 30000 m
200-
____ _____ _____ _____ _____ ______A 30000r
'; I24
200 40C 600 80 000 1 00 1400 1 600 1 00
Ioo---___ II- =____ -
_________ - - - - - Continentalterrperat~e climate,N 01 (k--4/3),
I -- ~ - - . . horizontal polarization," Iaverage grcnd,isc1"C opic antennas.
A\
. . . . . . . ........ Free Spao
". , " 600 800 1200 1400 1,00 I 100
__ ~--5%
150~ ~~ -. . . .. ... .. .... ..... .
zoo___ _ _ 15500 MHz
0 200 400 600 b00 1000 1200 1400 1600 1800[ 50%
0
•, 5° ---1 -i____________ ___ ___
SAntenna heights
S ,.......... ..... . ... as indicated.
Al D - F I H =3 m
2O2HS~H 2 =
I225 A 300 m
•4 cB 500 m
D 5000 mE 10000 m
___ _ __ ] r 15000 m3G 20000 mo1100I 25000 m
2 0 400 600 000 1000 1200 1400 1600 1BC.. 1 0
"I•19_I_ 95%
I -t j t Continentalt i teffperate climate,
N .301 (k-4/3) ,
,...... .. . ..- J. earth (Ah-O),S -. horizontal polarization,
aw Bwmgrourmi,_ _ i.:: _ ic antennas.
I ....... Free Space
"- -
\*\,' 0 i
2 5 "0 " 0, 6j0 0r., "< I? -"• I-..' I1300
iI.• ." ......................
200 'j15 500 MHz
I0 200 400 600 80() 1000 1200 1400 1600 _180050%
1250
i _ __.__,_ Antenna heights
.___ ...... _ as indicated.S............. .....E aIA1C 11 G II - 10 m
S200 ,
22 A 300 m1B 500 mC 1000 m
-D 5000 mk•tE 10000 m
F 15000 mG 20000 n
, H 25000 mI 130000 m
"" 100 ( 2o 400 600 600 1000 1200 1400 1600 3800
0 95%I
I5O Cotinental
temperate climate,()-4/3) ,
. . ...... .................. . ..... e a r th (A h = 0 ) ,IT ........ .... .. .. ... ...... .. I....... ho izo tal polarization,
a-eage growrd,isotropic aptetnas.
20C____
Al C: 1 F (3 I...... Free SpaceA I 1
22-5
C 200 400 610 800 '000 1200 1400 1600 1380O
Dstance in km
7 .-- --- -- --- - - - - - ---
I ;' I ,, + .. . ...... . .. 1. 5 0 0 M.z
A 0 F I0 1 15 500 MHz
H 1-[ I, [ \]
, ý ýG D,,G1'ý
o ISOK 2o0 400 ŽŽ00 . 2O 000 1202 00 0O i .j0 50% •
1H 2
C 1000
15 ' Antenna heights
C lOC m2=
I D 5000 m_,,IE 10000 m2
F 15000 mG 20000 mII 25000 m
. ____ _ __"1_ I 30000 ri-- C0. 400 600 800 1000 1200 '.00 1600 1800
1 0(- 95%125
150 •'..... iContinental
t temperate climate,.......... NS-301 (k=4 3)..... .......... .......... wT ooth earth (Ah=O),....... ......I'!i -horizontal pDlarization,
average ground5,isotropic anltenlnas.
200 ___ ___
AýC! ..... Free Spno
25 400 600 0oo 1000 1200 1400 1600 1 o0Sance t n km r0 0 8
Figure 73. Basic transmission loss versus distance; F = 15.5 GIIz, Il= 1
7.'/
A 0 F~il15530MHzI0
1oo 2; 400 600 800 1000 1200 1400 1600 1800
0;
015%
Antenna heiqhts__............. .. . . .as . .i licated.
BHi = 30 m
S200
_______A 300 m"B 500 mC 1000 m
__ __ _ _ _ _ _ 10000 mF 15000 mG 20000 m
L- _H 25000 m i
______ ,0.1 400 600 800 1000 1200 1400 160" 8s00*G 95%A I
1_50_____ _..... . Continentaltemperate climate,Ns.301 (k-4/3),S.............. o'•th ea th ( 0),
1 ,............. ..... ..... horizontal polarization,................................................. average giound,
isotropic antennas.
A 1 B o ~ ilFree Space
"2"Z:, 40f" 600 800 1 00 Iom 1400 1 00o BOO
's Iarce i kM
Figure 74. Basic transmission loss versus disti.nce; F = 15.5 GHZ, H1 = 30 m.
_'
5 15500 MHz
C
0
ISOS
Anterind height~s
--j____ as indicated.
I A5 1 D1150 m
€A 300 mB 500 m
I.I
D .. H 1500 m4
-- - E 10000 mnP15000 In
G 20000 mH 25001 m3•30000 m
?00 400 600 00 i000 1200 1400 Ifi0o__'" '°°° I9 5%
I Z_
50 ,Continentaltemperate climnate,
Ns-301 (k-4/3),.....................- iX th earth (Ah-O),
IT -s horizonital polaization
.............................................................. avarag. ground,
A, OCý E (' 600 C0060 S
OSIance in km
Figure 75. sasic transmission loss versus distance; p = 15.5 GHz, H = 150 m.
-- .-..... 5%
1 0I 5
S EF15 500 MHz
20C
"°"10 -,
,.,,•'°' t ] 50 %800 0 6 O 1000 20 ZO 400 L600 1800O
0
. . Antenna heights
"".I .. .... . .. . .. . a indicated.
~~1 - __ H 300 MSCI•i. ______, _• oo
; 100 DmSA 300 m
B 500DmC 1000 MS \D 5000 m
E 10000 m
F 15000 mG 20000 m
""IH 25000 mI 30000 m
"200 400 600 Boo 1000 2I'00 1400 1600 1800
SSo I Continentaltemperate climate,
"N1=301 (k=4/3),
smooth earth (A=0)horizontal Ix~larization,average qrcidisotropic antennas.
A! 110 El F 3 1.... Free Space
_ - 4'• 61"0 800 1000 1200 1400 1600 Ic00
L. ,s!an'ne ,n km
FiGure 76. ti ransmission loss versus distance; F = 15.5 nHz, H 1 300 m.
155%
A •I 0 "•e. • .......~~ ~~........ .......... .... 4....... ..... ..... ". 5 5 0 I~
r - --O --
I~c
AZOO D E CJI H 15 500 MHz
0
."~~ .............. ..... .............. ..... .... ....AH 500 In
ZJ.
22 B 1000 MS, 1 0 0 0 0 M
E. 15000 M
u H 30000 in
I 00 20 40 ,OO •0 Soo 1amIZO0 10 G'. 95%
ISO 0o(,ntinental
terriper-ate clirrate,
,, (k-4/3),.............. eart.h (A1-0) o
isotropic anter-s-as,
2 't . , 600 IlOO 1000 I 200 1400 1600O 1800""is ta n c e i n km_.ic-ure 7-. Basic transmission loss versus distance; F - 15.5 GHZ, N
*&
_ _ _ -5%
F15 500 MHz
OC 40 1 1 R 50%22. 1 .
0I' ____5___0___ --- __ ___
'P. Antenna heiqhtsas indicated.
H 1000 m
H2 =
225 A 100
B 5000 MC 1000ci mD 15000 m
E 20000 m
F 25000 mC~ 30000 m
"" 2CC 400 600 800 1000 I200 1400 1600 100
ilr r f 95%
12's ____
15, lCntint ntalteaperate clinate,
N301 (k-4/3),. . .th earth (Ah=0),
- horizontal polarization,S...aveage qrou~nd,
isotropic antennas.
2C
.....F Free Space
2: 4 600 Soo 100coo100 10 1SC
-s lance in k m100 40 60 80
7~r 3. %asic t;ransmission 10.3s versus distance: F -15.5 GHz, If 10100m
I • 'i. - _5 %
A C DEF 15 500 MHz
22,
0
I- - 50%
12O
0
l~t.
@1 Antenam heightsas indicate.
H 5000 M
i 0€ 'OO tO0 O0 lCO 1010 1 00100000 I rn
S5000M
B°° 100 m
SC 1 S000 m
D 20000 m2- -. _ E 25000 m
F 30000 n
0 4 0 0 6 0 0 $ Go 1 0 0 0 1 2 0 0 1 1 6 >05
12,
I O -".. _- -tinentaltemprate climate,N -301 (k-4/3),
.......... t~hearth ýb-0),------- horizontal po1larization,.. ............ average ground,isotropic antennas.
....F.e Space
"Vý 40C, 600 Bo0 1000 1 ZOCI 1400 C00 " I00C IS tance in km
rigure 79. Ba~ic transmission loss versus distance; F - 15.5 GHS, H1 = 5000 m.
RAA
1.- -�...
.. . 5 %
__ _ __ C 15500MHz
201)
2 0
0-50%
100 400 600 SCO 1000 1200 1400 1ROO Boo
• '5 aAntenna heirhtsS~as ird ".cared."'• •'•' I" " •" '"'"'"'• •.. ......" .......... '............ .... '. . . . .' : . .. 1 " 1 0 0 mI
H, -0 100
H2
A 10000 mB2 B 1500n iSC 20000 m
D 25000 nE 30000 m
25fu
200 400 00 •0oo 1000 zno 1400 60 j0o
I95%,2 o I
150 Continentalteterate climate,
01 (k-4/3),................ th earth ( Oh-0),
isotropic antennas.
.... .. . Free ... .............. ..............................'ri Spoate
fin 4So o00 60 100 200 1400 1600 I100
1-.~r > Basic tranw jon loss versus distance: F - 15.5 (Hz, H 10 000 in.
ii.
_ ,
" ... , .., .... .... ...
ýAI 15 500 MHz
2216
-0 0 - _ _
200 400 600 goo 1000 i 200 1403 1600 10co0
oICCo - -\\ 50
50%
0
ISO
I •Antenn' heightsI p Ias indicated.
H 15000 m
m ~H2ll• ~~~A 15000 \\ _B0Om
C 25000 m"-J InD 30000 m
i10 200 400 600 oo0 i00 200 1400 i60 0 00"-, _ -9 5%
1 0 • Continentaltuerate climate,
........... . 01 (-4/3),=clth earth (Ah-0),.horizotal polarizatioi.,•i ....... ................ ... ....... ... ...... average ground ,
.0o• •imotrapic an--ms.
....Free Sp&Le
l:c OC ( kCo too 1000 100 1400 1600 1100D iteoC* in km
rigq-re 81. Basic tronsmission loss versus distancei F" - 15.5 rfl, P. 15 )IO n
~ri.................... .
-- - A 1 C -0_ 15 500MI-z
10,- 200 400 600 too 1000 1200 1400 1600 91~0
S 50%C
125
! • •Antenna heights
"7 .i ..- as indicated.
Q~ZO ~ ] H1 20000 m
13 11 2 =
•;,'A 20000 mi 1B 25000 m
C 30000 m
'I
2C-:) 400 600 too 1000 1200 1400 If0oo I 0 toS~95%
- I
-Continental
tatperate climate,H M(k-4/3),
_____....... I___ ____ ____ ____u th PartJ1 (Ah-0),~~~. ......... ........ ............ .. ,,)1:5 1 • • ..... ] .............. ............ J...... ... i ...... horizontal polarization,
-avrage ground,isotropic antennas.
....... Free Spac2111
t ,
. " 600 600 1000 1ZOO 1400 1600 150Distance in km
ric'.re S2. Basic transaission lcss versus distance; F - 15.S GHz, HI - 20 000 m.
87It.
- -- - --- -
I r•,~ ~ ~ ~ ~ ~ ~ ~~~~....... .................... ... .. :-,..., .............. ...
zoo__ 1_ .500 MHz
o 1
"I cc 200 400 600 G00 1000 1200 2400 4 600 T MC
_,0%
OIt
0
i ISO•,
AntA- . heiohts
".......... ...... .. .... . as indicated .
200. H 25DO0 mI'00
•z•.A 2500m
B 30000 m
S 200 600 800 2000 2200 2400 2600 1900II
2,o !95%
1 25
N301 (ki'4/3),
t.erierate climate, ,
,. ... ..... ...... .- earth (Mh-, ),. ........ .... ... ...... . ...... . ......... horixontal polarization,
"" ave e gr4d
isotropic antennas.
, ..... Fre P
" " z: 0 -o 600 So0 1000 I200 2400 1600 1200
Dislance in km
.Fiqjre 83. Basic transmission loss versus distdnce; F * 15.5 rHz, HI = 2 oo00 2.
88
• -• .... ......... ......_ ,• , ;• . ,.,i
~OC
..... .... ...............
,Ao,5500M0z
ZOO
0
F-50%0 z
= Antennia heights_---eas indicated.
1 A H 1 30000 m
200 112 =
us .fA 30000 m
• 1 _ .-I I
U .o400 600 Boo I OC, 1200 1400 1600 1 00Io• 95%
temperate climate,01(k,-4/3),
.. . . . . .. .. .. . . .. . . . ..:: . ... . . ....... ... . ....... . . . --. - h o r i z o n t a l p o l a r i z a t i o n ,......... ..... .......... a e.agie ground, '
-------- •isotropic Antennas.
200
2t:
"400 600 Boo 1000 0120 1400 1600 1800
Oistance in kni
7i:.:,re 34. zasic transmiission loss versus distance: P 15.5 GHz, 11 30 00n In,
0 89
Yk_
4. APPLICATION CONSIDERATIONS
Some application considerations for the curves provided in Sec-
tion 3 are given in thi~s section along with some sample problems.
While use of these curves may provide rapid estimates of service
limitations imposed by radio wave propagation for many problems, use
of the computer programs built around IF-77 propagation model will
provide more accurate answers and may also make a considerable reduc-
tion in the manual computation effort required to obtain some results.
Atlas users are encouraged to obtain the "Applications Guide" [11]
for the IF-77 program set since it contains a multitude of sample
application problems along with the information needed to obtain
computer generated curves for- specific applications-from ITS.
4.1 Received Signal Level
In addition to Lb , estimates of received signal level, RSL, from
a desired facility, D, require information on the equivalent isotrop-
ically radiated power, EIRP, receiving antenna gain, Gr and receivingantenna system line losses, L ; i.e.,
D(q) = EIRP + Gr Lr L (q)()
where the D(q) is the power level that is available at the receiver
input for a fraction q of the time. Compatible decibel type units
are used in (1); e.g., units of dBW for EIRP would result in dBW units
for D. The transmitting antenna gain included in the EIRP and Gr (in
dBi) should be applicable to the direction of propagation since the
full main beam antenna gains may not be realized.
Received Signal Level (RSL)
Problem 1: Estimate the RSL available for at least 95% of the trans-
mission time for an aircraft at 10,300 m above mean sea level (insi)
and 400 kmn from th~e transmitting facility. Assume that a surface
elevation of 300 in, a transmitting antenna height of 15 mn, a radiated
power of 10 dBW, an effective transmi~tting antenna gain of 6 dBi, an
effective receiving antenna gain of 2 dBi, a receiving system line
loss of 1 dB, and a frequency of 125 MHz are applicable.
Solution: The lower graph of Figure 3 is applicable to this problem,
and it gives L (0.9 5) =147 dB for H1 15 m and H2 10 300 -300
90/
10 000 m. This value and the other parameters of the problem state-
ment may be used in (1) to calculate D(0.95); i.ei.,
D(0.95) - EIRP + Gr - Lr - Lb(0. 9 5 ;
D(0.95) 10 + 6 + 2 - 1 - 147
or
D(0.95) - -130 dBW = -100 dBm
4.2 Power Density
Calculation of power density, SR, at the receiving antenna is
similar to the calculation of D(q) except that the recil•rocal of theeffective receiving area of an isotropic antenna (A• in 0,B-sq m)s is
used >I place of Gr - L, in (1); i.e.,
SR(q) = EIRP - AI - Lb(q) . (2)
Compatible decibel type units are used in (2); e.g., units of dBm forEIRP would result in dB-m/sq m units for SR. Frequency (f) is used
to ca..culate A from
AI[dB-sq m] = 38.54 - 20 log f[MHz] , (3)
where log is the common (base 10) logarithm.
Power DensityProblem 2: Estimate the median power density available at a groundreceiving facility from an aircraft at 10 500 m-msl and 400 km away.
Assume that a surface elevation of 500 m, an aircraft transmitting
EIRP of 45 dBm, a ground facility antenna height of 10 m, and a fre-
quency of 300 MHz are applicable.
3 The notation used for the units of these quantities is intended toimply that they are decibel-type quantities obtained by taking 10 logof a quantity with the units indicated after dB-; e.g., AI[dB-sq ml =10 log{X2 [sq m]/47r)} (where X[m] is wavelength). Equations used inthis report are dimensionally consistent. Where difficulties withunits could occur, brackets are used to indicate proper units.
/9
911
Solution: The middle graph of Figure 16 is applicable to this prob-
lem, and it gives Lb(0.5) = 147 dB for H10 m and H2 = 10 500 -500b ~ 1210 000 M. This value and the other parameters of the problem
statement may be used in (3) and (2) to calculate SR(0.5); i.e.,
Ai[dB-sq m] = 38.54 - 20 log f(MHz] ,
AI = 38.54 - 20 log(300) ,
AI = -11.00 dB-sq m,
SR(0.5) = EIRP - AI - Lb(0.5),
S(0.5) -= 45 - (-11) - 147
and
SR(0.5) = -91 dB-m/sq m = -121 dB-W/sq mi
4.3 Interpolation Between Curves
Interpolation between the curves supplied in Section 3 may be
required for some problems. When an adequate estimate can not be
made by visual means, a simple logarithmic interpolation method is
expected to yield satisfactory results for most cases; i.e.,
Lb = L bl + (Lb 2 -Lbl) log(x/x1 ) (log (x 2 /x 1 )
where L for x is calculated based on L and x values taken from theb b
curves at points 1 and 2 (Lbl, 2 and x1 , 2 ). In (4), the parameter x
can be either antenna height or frequency.
Interpolation for HeightProblem 3: Estimate the Lb(0. 9 5 ) for an aircraft at 10 000 m above
the surface and 400 km from the transmitting facility by interpola-
tion between curves for altitudes of 5000 and 15 000 m. Assume that
a facility antenna height of 15 m and a frequency of 125 MHz are
applicable.
92
14..
Solution: Curves from the lower graph of Figure 3 are applicable;
i.e., at x = 5000 m, Lbl(0. 9 5 ) = 187 dB, and at x2 = 15 000 m,
Lb 2 (0. 9 5 ) = 133 dB. These values are used in (4) to obtain Lb(0. 9 5 )
for x = 10 000 m; i.e.,
[Lb 2 (0. 9 5 ) - Lbl(0.95)]Lb(0. 9 5 ) = Lbi(0. 9 5 ) + log(x/xl)log (x2/xI1)
[= 1 133-187] log(10000/5000)
log (15000/5000)
and
L (0.95) 153 dBb
This value is 6 dB greater than the 147 dB level that would have beenobtained by using the 10 000 m curve directly which is about 10% of
the 54 dB interpolation range. Hence, values obtained by interpola-tion should be used with caution.
Interpolation for Frequency
Problem 4: Estimate the Lb (0.5) for a frequency of 300 MHz by inter-
polation between curves for 125 and 1200 MHz. Assume that an aircraft
altitude of 10 000 m, a facility antenna height of 10 m, and a dis-
tance of 400 km are applicable.
Solution: Curves from the middle graph of Figure 2 and 30 are appli-
cable; i.e., for x = 125 MHz, Figure 2 gives Lbl(0.5) = 144 dB, and
for x2 = 1200 MHz, Figure 30 gives Lb2 (0.5) = 152 dB. These values
are used in (4) to obtain Lb(0.5) for x = 300 MHz; i.e.,
[Lb2(0.5) - L bl(0.5)] log(x/x 1 )
Lb(0.5) = Lbl(0.5) + log (x 2 /xI)
(152-144) log(30!)/125)
Lb (0.5) 144 + log(1200/125)
and
Lb(0.5) = 147 dB
93
-- - --.- ---
This value is the same as the value that would have been obtained by
using the 300 MHz cuirve of Figure 16 directly. Hence, (4) can yield
good results.
4.4 Service Range, Without Interference
The curves of Section 3 can be used directly to estimate servicerange when service is not limited by interference. Stich service
requirements specified in terms of D(q) or SR can be translated to a
Lb level by rearrangement of (1) ox (2); i.e.,
Lb(q) =EIRP + Gr Lr -D(q) , (5)
or
Lb(q) - EIRP SR(q) - A, (6)
Service Range, Power Density
Problem 5: In the absence of interference, determine the maximum gap-
less service range available to an air 'aft at 10 000 m above the
surface when a power density of -106.6 dB-W/sq m with a time avail-ability of 95% is used to define satisfactory service. Assume that aground facility antenna height of 15 m, an EIRP of 20 dBW, and a
frequency of 125 MHz is applicable to this problem.
Solution: The Lb value that corresponds to the edge of the servicerange is determined from (3) and (6); i.e.,
AI[dB-sq ml = 38.54 - 20 log f[MHzj ,
AI = 38.54 - 20 log(125) != -3.40 dB-sq m
Lb(0. 9 5) = EIRP - SR(0. 9 5 ) - A dB
L (0.95) = 20 -(-106.6) - (-3.4) dB
b
and
Lb(0. 9 5 ) = 130 dB
The lower graph of Figure 3 is applicable to this problem. However,
the appropriate curve gives distance values of 213 and 261 km for an
Lb(0. 9 5 ) of 130 dB. Hence, the maximum gap-less range foL SR(0.95)
= -106.6 dB-W/sq m at 10 000 m is 213 km.94
4.5 Protection Ratio
The protection ratio or the desired-to-undesired signal ratio
(D/U) exceeded at the receiver for at least 95% of the time, D/U(0.95),
can be estimated using the curves of this report as follows:
D/T(0.5) +Gt+GrLb(0.50 - Pt+Gt+Gr-Lb (0.5) (7)
Y(q) Lb(0.5) - Lb(q) , (8)
DU
In (9), P 's are radiated powers and Gt,r 's are transmitting andt r
receiving antenna gains--all must be in consistent decibel type
units. Bracket subscripts indicate that all items within the brackets
are associated with either the desired (subscript D) or undesired
(subscript U) facility.
Additional variabilities could easily be included in (9) for
such things as antenna gain when variabilities for them can bedetermined. Continuous (100%) or simultaneous channel utilization
is implicit in the D/U(0.95) formulation provided above so that theimpact of intermittent transmitter operation must be considered
separately (Sec. 4.8.2).
Protection Ratio
Problem 6: Determine D/U(0.5) and D/U(0.95) for an aircraft receiving
125 MHz transmission at 10 000 m above the surface from a desired
facility 300 km away and an undesired facility 600 km away. Assume
that identical equipment parameters (PtIs and G 's) for the facil-ities, facility antenna heights of 10 m, and simultaneous facility
operation are applicable.
Solution: Equations (7) through (10) and values read from the curves
of Figure 2 are used to solve this problem as follows:
95
----------- _ ___ __ __
[y(0.95))D = [Lb(o.5) - Lb(0. 9 5 )ID = 126.0 - 130.0 = -4.0 dB
[Y(0.05)HU a (Lb(O.5) - Lb(O.05)lU M 176.8 - 167.1 - 9.7 dB
Y (0.95) - -[ (0.95)]2 + [Y(O.o5) ]2 - - --4.o)2 + (9.7)2 -10.5 dB,
D/U(0.5) - [Pt+Gt+Gr-Lb(0.5),D- [Pt+Gt+Gr-Lb(0.5)] U
D/U(0.5) ((Pt+Gt+Gr]D- (Pt+Gt+Gr]u) + [Lb(0.5)]U -[Lb(0.5D
D/U(0.5) - 0 + 176.8 - 126.0 = 50.8 dR
and
D/U(0.95) = D/U(0.5) + YDU(0. 9 5 ) = 50.8 + (-10.5) 40.3 dB
For this problem D/U(0.5) exceeds D/U(0.95) by 10.5 dB.
4.6 Station Separation
The geometry for station separation calculations is illustrated
in Figure 85. Station separation, S, is the sum of the desired
facility to aircraft distance, dD, and the undesired facility to
aircraft distance, du; i.e.,
S = dD +du , (11)
where all terms of (11) are in the same units of length. While
facility separation, Sf, is only equal to station separation when
the aircraft is over the great-circle path connecting the facilities,
many of the problems for which the use of this atlas is anticipated
invclve the S = Sf case where the station separation required to
protect an aircraft located at a "protection point" over the great-
circle path must be determined. Even this simple case can involve
tedious calculations because of the iterative procedure required
when (10) is used. Other situations may be compounded by such
things as facility antenna gain changes with azimuth, and serious
96
41-)
.44
4-4
0
Q)'44
4.1
:3n
-o wa+ -v F4
Q - 4-i
If 4-)
U)
4.-)
DOO
0. Lf4
0U
I-9
v0)
97
consideration should be given to using special predictions made usingone of the computer programs built around the IF-77 model [111.
Station separation required to obtain a specified D/U(0.95)level at an aircraft location (protection point) fixed relative tothe desired station may be estimated using the atlas curves as follows:
(1) Determine [Lb(0. 9 5)] D from the appropriate curve andcalculate [Lb(0.0 5)H using
[Lb(0.05)] = D/U(0.95) - [P +G +G -L (0.95)] + [P+G
b U1 t t r- b 9 1D + [t+G+rlU, (12)
(2) Determine the first estimate of dU, dul, byreading the appropriate cLrve at the [Lb(O.0 5 )]UIlevel determined in step 1. If d is multivalued,U1the largest value should be taken as dul. Thisestimate for dU is somewhat larger than the dUactually required and can be taken as a worst case
value.(3) Determine the second estimate of dU, du 2 , by taking
the value calculated using (12) as a median value(i.e., [Lb(0. 5 )]U 2 = [Lb(0.05H)I) and reading the
appropriate median curve to obtain a value for
d u2. This estimate for dU is somewhat smaller thanthe d actually required and can be taken as a best
Ucase value.
(4) Values obtained for du1 and du 2 in the previous
step bracket du; i.e.,
dU2 : dU ! dul1
The maximum amount that dU can be decreased byadditional calculations from the worst case (dul)value is dul-du2 , and if this difference is small
enough, the iterative calculations required toimprove the initial estimate are not needed. Whena better estimate is required, dui values between
d and d are selected, and the D/U(0.95)iU2 Ul
associated with them calculated using (10) . A
98
bracketing procedure based on previously obtained D/U(0.95) i values can be used to select the next dUivalue. The process is continued until the D/U(0.95)iobtained is sufficiently close to the requiredD/U(0.95). Then station separation is calculated
from (11).
Station Separation
Problem 7: Determine the station separation necessary to obtainD/U(0.95) = 20 dB for a protectiLn point at 10 000 m and 400 km fromthe desired station. Assume that identical equipment parameter (Pt's
wtand G s) for the facilities, a frequency of 125 MHz, facilityt, rantenna heights of 10 m, and simultaneous facility operation areapplicable.Solution: The procedure given above and the curves of Figure 2 areused to solve this problem as summarized below.
(1) [Lb(0. 9 5 )]D = 150 dB
[Lb(0.05)1UI = D/U(0.95) - [P +G +G -L (0.95)]D + [Pt+Gt+Gru
b 1t t r b 9- 1 [tG+l
[Lb(0.05)]UI = D/U(0.95) + [Lb(0. 9 5 ]D + ([Pt+Gt+Gr'u- [Pt+Gt+Grr]
[Lb(0.05)]Ul = 20 + 150 + 0 = 170 dB
(2) For [Lb(0.05)]Ul = 170, dui = 627 km.(3) For [Lb(0.5)]U2 170, d U2 = 466 km.
(4) Since 466 < dU < 627 kin, calculate [D/U(0.95)] 3 for
du 3 4 dU2 + (dul-dU2 )/2
du 3 = 466 + (627-466)/2 = 546 km
that is,
99
.44
[y(0.95)] D= [Lb(0.5) - Lb(0. 9 5 )] D = 143.8 - 150 = -6.2 dB
y(0.05] = [Lb(0.5) - Lb(0.05]3 = 177.9 - 161.9 = 16 dB ,
YDU3 =-/[Y(0.95)]2 + [Y(0.05)]2 3 = - (-6.2)2 + 162 = -17.2 dB
[D/U(0.5)] 3 = [Lb(0.5)]U3 - [Lb(0.5)]D + [Pt+Gt+Gr]D - [Pt+Gt+Gr]
[D/U(0.5)] 3 = 177.9 - 143.8 + 0 = 34.1 dB
and
[D/U(0.95)] 3 = [D/U(0.5)] 3 + YDU3(0. 9 5 ) = 34.1 + (-17.2) - 16.9
Therefore, d 546 < d < 627 = dul; and a new trial is needed;
i.e., dU4 = 546 + (627-546)/.2 = 586 km. Values resulting from the new
trial are as follows:
(Y(0.05]U4 := 179.7 - 165.3 = 14.4
DU4 .2 (14.4)2 = -15.7
[D/U(0.5)] 4 = 179.7 - 143.8 + 0 = 35.9 ,
and
[D/U(0.95)] 4 = 35.9 + (-15.7) = 20.2 or 20 dB
Hence, du = d = 586 km will give D/U(0.95) = 20 dB for the conditionsU4
of the problem. Note that d was high by 41 km (7%) while du2 was
low by 120 km (20%).
4.7 Constant Protection Ratio Locus
A method of approximating the locus of a constant protection
ratio as a circle enclosing the undesired facility has been developed
[8, pp. 104-1071 and is illustrated in Figure 86. This method is
useful in estimating the region around an adjacent-channel undesired
100
- -------------. ---- ------
UNDESIRED DESIREDSTATION- STATION
I !I
du - dD .
Figure 86. Locus of constant D/U(q) for an adjacent-channelundesired facility.
facility in which service is unsatisfactory. For a given aircraft
altitude, the circle has radius R given by
SBR(13)
where S is calculated using (11) and the ratio B is calculated from
B dD/dU . (14)
The circle is centered on the undesired facility side of an exten-
sion to the line connecting the ground facilities at a distance
C S (B+l) (15)C=2 (B2 -1)
from the point half-way between the facilities. Generally, service
can be regarded as being unsatisfactory within the circle, even
10
101
though some locations having satisfactory service may exist above
the undesired facility because of the vertical pattern of its antenna.The two basic assumptions associated with this method are as
follows:(a) The geometry of Figure 86 can be treated as plane
geometry; i.e., the earth is assumed to be flat,and slant range projections, dDU, onto the hori-zontal plane are approximately equal to the actualranges, rD,U" This asstmption is reasonable ifdD > d > aircraft altitude/260 for distances in
kilometers and aircraft altitude in meters.
(b) The protection ratio, D/U(q), is assumed to be pro-portional to the logarithm of the ratio of the dis-tances; i.e., D/U(q) = M log (dD/du), where M is aconstant.
This method is only approximate. The second assumption, (b), isviolated by lobing in the transmission loss versus distance curve,which may occur at any constant altitude because of ground reflections
or by facility antenna patterns that are not omnidirectional in thehorizontal plane. Assumption (b) is also violated by a change in the
slope of the transmission loss versus distance curve which, as anexample, occurs in the vicinity of the radio horizon. However, theuse of the smallest d corresponding to a particular station spacing
Dand a particular protection ratio avoids the ambiguity due to lobing.
Furthermore, in most applications service is limited by noise ratherthan by interference if ranges beyond the radio horizon of the desired
station are encountered.
Protection Ratio Locus
Problem 8: Find the locus of the D/U(0.95) = -40 dB contour about anadjacent-channel undesired facility at an altitude of 5000 m. Assumethat the D/U(0.95) = -40 dB location on the great circle connectingthe facility has been previously determined so that dD = 200 km andIDdu = 40 km.
Solution: This problem is solved by using the dD and dU provided in
the problem statement in equations (11) and (13) through (15); i.e.,
102
102 _ _ _ _
S... ..... ... ...... •.. .• J I ."-- :
S =d + d= 200 + 40 =240 km,
B d dD/dU = 200/40 = 5,
R = B/(B2 -1) = (240) (5)/(52_1) = 50 km, and23(B2+1) (240) (52+1)
C 3 (B •(5240) = 130 kin.
Hence, at 5000 m harmful adjacent-channel interference (D/U(0.95) L
-40 dB) occurs about the undesired facility in the region defined bya 50 km radius circle centered at 10 km (C-S/2) on the far side ofthe undesired facility along the line connecting the facilities.
4.8 Multiple Interference Sources
Two types of interference from multiple sources will be con-sidered. The first type involves the case where the interfering orundesired power results from incoherent statistically independent
multiple sources that all operate continuously. These are called"simultaneous independent sources" and are discussed in Section 4.8.1.
The other type is the case where the undesired power results
from multiple sources that are operated intermittently in a coordi-nated fashion such that only one transmits at a time, and periods withno transmissions are possible. These are called "intermittent coordi-nated sources" and are discussed in Section 4.8.2.
4.8.1 Simultaneous Independent SourcesA method for estimating D/U(0.95) when the undesired power is
made up of contributions from incoherent statistically independentmultiple sources is provided in this section. It is the "log-normal"method developed by Norton et al. [16] and previously used for thispurpose [8, Sec. 3.2]. The method may be summarized as follows:
(1) For each ith facility of the n undesired facilitiesinvolved, determine its median received power
[U(0.5)] and standard deviation a. in using
[U(0.5)]i = [Pt + Gt + Gr + Lr - Lb(0"5)]i (16)
103
[Y(0.05)] = [Lb(0.05) - Lb(0.5)]i (17)i bb
and
ai = [Y(0.05)]i/1.6448 (18)
where the variables involved are as previously
defined. However, all radiated powers, Pt's, must
be in the same decibel type unit such as dBW.(2) For each ith undesired facility, determine the mean
power, ai, and variance, Pi, in nondecibel or watt
type units using
=i exp[O.5(ai/c) 2 + [U(0.5)]i/c] (19)
and
i= a (exp[(ai/cC) 2 1) , (20)
where
c = 10 log e = 4.3429 . (21)
If [U(0.50)Ii was in dBW, ai will be in watts and
Pi in watts squared.
(3) Determine [Y(O.05)]U and U(0.5) for the resulting
total undesired power from the n undesired facili-
ties using
Cru = chn(l + [jp i/( (ai) 2 (22)
[Y(O.05)U = 1.6448 U, (23)
and
U(.5) = c[in(•ci) - 0.5( ucr/)2] , (24)
where in is the natural (base e) logarithm. Note
that a is in dB and U(0.5) is in the same decibel
type units used for the [U(0.5)]i's.
104
* .***
(4) Determine D/U(0.95) using
D(0.5) =P + Gt + Gr - L- Lb(0"50)]D (25)
D/U(0.5) = D(O.5) - U(0.5) , (26)
with (9) and (10).
Multiple Simultaneous Sources
Problem 9: Find D/U(0.95) when the undesired power is made up of
contributions from four undesired facilities that may be treated asmultiple simultaneous sources. Assume that the facilities havereceived power parameter,; as explained below.
S[U (0.50) ] i[Y(0.05) ]i
dBW ::B
1 95 0
2 -100 53 -105 10
4 -110 15
5 -115 20
D(0.50)=-65 dBW [Y(O. 9 5)]D=-5 dB
Solution: The step-by-step procedure as given above is followed;
i.e.,
(l)-(2) the first two steps result in the values tabu-
lated below which also include the summation terms
required in step 3.
105
-------- --- --
i i ai Iii
dB W W2
1 0 3.1616 x 101 0
2 3.0399 1.2773 x 10-10 1.0315 x lo-20
3 6.0798 8.4229 x 10-11 4.3264 x 10- 2 0
4 9.1196 9.0657 x 10-11 6.6761 x 10-19
5 12.1595 1.5927 x 10-10 6.4363 x 10-17
TOTALS 7.780 x 10-10 6.508 x 10-17
(3) = c/'n(1 + [n Ili/ q ai) 2 ])
CU 4.3429 n(l + [(6.508 x 10-1)/(7.780 x 10-) ]) = 9.402 dB
[Y(0.05)]U = 1.6448 U = (1.6448)(9.402) = 15.46 dB
U(0.5) c(ln(A ci) - 0.5(au/C) 2]
and
U(0.5) - 4.3429 11n(7. "2 x 10- 10 -0.5(9.402/4.3429)1 = -101.27 dBW,
(4)
D/U(0.50) D(0.50) - U(0.50) = -65 -(-101.27) 36.27 dB ,
YDU /[Y(.95)]D + [Y(0.05)] 2 _ -(5)2 + (15.46)2 -16.25 dB
and
D/U(0.95) = D/U(0.5) + YDU = 36.27 + (-16.25) = 20 dB
106
4.8.2 Intermittent Coordinated Sources
A method is provided in this section for estimating D/U(0.95)
when the undesired power is made up of contributions from multiple
sources that are operated intermittently in a coordinated fashion
such that only one transmits at a time, and periods with no transmis-
sions are possible. This method is applicable to an air traffic
control (ATC) situation where the undesired signals come from another
sector, but it estimates a D/U(0.95) value instead of a "probabilityof interference" [131. Users in the interfering or undesired sector
are assumed to transmit one at a time, and D/U(0.95) is taken over
the time that the desired facility transmits; i.e., channel utiliza-
tion for the desired facility is neglected by assuming D/U valueswhen no desired power is available are not of interest. The methodis summarized as follows:
(1) Determine D(0.5) and [Y( 0 . 9 5 )]D using (25) and (8)
along with transmission loss values read from theappropriate curves.
(2) Divide the airspace of the undesired sector into
volumes and characterize each by single values of
[U(0.05)]i, [U(0.5)]i, and [U(0.95)]i where theseare calculated as in (16) from Lb values read from
the appropriate curves. Estimate the channelutilization expected for aircraft transmitting from
each volume under the conditions of full sector
utilization. Channel utilization, Uc, is the
percent of time that transmissions from aircraft
within the volume takes place. Pair each power
level with a level utilization, UL' where
0.8 Uc for (U(0.5)]i
UL = (27)
0.1 Uc for fU(0.05)]i and [U(0.95)]i
(3) Repeat step 2 for each undesired facility of the
interfering sector.
107
St i , i
(4) Reorder the "level-UL pairs determined in steps 2
and 3 by decreasing power levels and calculate an
accumulated utilization, UA, that gives channel
utilization percentage for levels equal to or greater
than a particular tabulated value; i.e.,
UA([U(q)]j)= UL i for all [U(q)]i ?. [U(q)] • (28)
(5) Using the tabulation of step 4, estimate levels for
U(0.05) and U(0.5). Then calculate [Y(O.05)JU from
[Y(0.05)] U =: U(0.05) - U(0.5) (29)
and determine D/U(0.95) using (9) and (10).
Multiple Simultaneous Sources
Problem 10: Find D/U(0.95) when the undesired power is made up of
contributions irom .Arcraft and a facility in an interfering sector.
Assume that steps 1 ..hrough 3 of the procedure given above have
resulted in the following:
(1) Values obtained for step 1 are D(0.5) = -95 dBW and
[Y(0.95)]D 3 dB.
(2) (3) Values obtained for steps 2 and 3 are as shown
below.
Uc [U(0.05)]i UUL [U(0"5)]i U, [U(0.95)].i ULSdBW % dBW % dBW %
5 .100 0,.5 -105 4 -110 0.5
5 -103 0.5 -108 (4 -113 0.5
10 -106 1 -j12 8 -116 1
5 -109 0.5 -115 4 -119 0.5
40 -148 4 -160 32 -1.72 4
108_________
Si 4.
-- • -I ' iit - --
In the tabulation above, transmissions with Uc
40% are from the undesired facility, and the
remaining utilization (20%) is from four different
aircraft volumes. Note that the total uUiliza-
tion for the undesired sector is 65% so that there
are no transmissions from the undesired sector
during 35% of the time.Solution: In thi3 case, the solution starts with step 4 of the pro-
vided procedure.
(4) Reorder the values obtained in steps 2 and 3 and
calculate UA s using (28) as shown.
Level UL UA
Tj, A
dBW % %
-100 0.5 0.5-103 0.5 1-105 4 5-*- U(0.05) = -105 dBW-106 1 6-108 4 10-109 0.5 10.5-110 0.5 11-112 8 19-113 0.5 19.5-115 4 23.5-116 1 24.5-119 0.5 25-148 4 29.. Interpolate to-160 32 61 get U(0.5)-172 4 65None 35 100
(5) Use the step (4) tabulation to obtain U(0.05) and
U(0.5). Since a UA = 5% value is given in thetabulation, U(0.05) is obtained directly as U(0.05)
= -105 dBW. However, U(0.5) is determined using
linear interpolation between -148 and -160 dBW; i.e.,
U•0.5) = (160-148) (61-50) -160 -156 dBW(61-29)
109
mw"• -- t'". .... . ~- - ,- , . _ _i, . ,:, , •
Then determine D/U(0.95) using (26), (29), (9),and (10); i.e.,
D/U(0.5) - D(0.5) - U(0.5) = -95 -(-156) = 61 dB
(Y(O.05)]U - U(0.05) - U(0.5) = -105 -(-156) - 51 dB
Y (0.95)D2 + [Y(0 2 =. ..2/(-3T + 512= -51 dBYDU(.5 =-Y095'D .
and
D/U(0.95) D/U(0.5) + YDU(0. 9 5 ) = 61 + (-51) = 10 dB
4.9 Service Probability
Semi-empirical methods of predicting the time variability, loca-tion-to-location variability, and prediction uncertainty of transmis-
sion loss for tropospheric communication circuits have been developed 1and incorporated in the concept of service probability [2; 3; 15,
Annex I; 17, Annex V]. Although a comprehensive discussion of thesemethods is beyond the scope of this report, a simple method of using
the service probability concept with the atlas curves is given inthis section.I
Se, vice probability is taken here as the probability that satis-
factory service will be obtained where satisfactory service is definedby a required time availability for a critical signal or protectionratio level. (This assumes that the transmitter is always on, and thetransmitter down time is not considered in this analysis.) It may
be interpreted as bias or margin used in addition to time availability.A lix designed so that the critical signal level is available when
the transmission loss for the path corresponds to Lb(O. 9 5 ) couldhave a time availability of 0.95 and a service probability of 0.5; i.e.,
the probability that a particular link will provide a time availabilityof 0.95 is 0.5. Variation from this design may be interpreted as achange of time availability and/or service probability (8, Fig. 33].
Selection of time availability, q, and service probability, Q,requirements for a particular situation should be determined by the
user. In general, higher values will mean more reliability at a
higher cost where cost may include a requirement for more frequencies j
110110
to provide a particular service. Except when Q=0.5, the inclusion ofservice probability complicates ---he prediction process.
Experience at the FAA indicates that predictions made with q-0.95and Q=0.5 are adequate to define service volumes for air navigationaids when known equipment degradation factors are accounted for byadjusting prediction parameters, such as transmitted power, andcaution is used to avoid situations where conditions such as terraindiffer significantly from those assumed in the predictions.
The D/U(0.95) that is available with probability Q or D/U(0.95,Q)
may be estimated by using Figure 87, which gives the standard normaldeviation Zin as a function of Q and the method that follows.
(1) Estimate the extra standard error, ae' that isneeded to account for uncertainties other thanthe path-to-path variance, a' [17, eqn. V40],
such as those associated with equipment performanceparameters for the desired facility. Then solve
for ar(q, Q) by using parameters applicable to thedesired facility, q=0.95, and the following:
c2 (q) = 12.73 + 0.12 y2 (q) (30)
0 (q) = a 2 (q)+ 2 (31)
ce c e
ce 2 (q) if Lb(q) + Zmo(Q) e< Lbf - 6
C2 (q,Q) , (32)
C2 otherwisee
where Lbf is the free space basic transmi.ssionloss obtained from the appropriate atlas curve,and the test in (32) is intended to prevernt the
use of power levels that are in excess of freespace values by an unrealistic amount. The
resulting o 2 (q,Q) will be called [a2 (0.95, Q)]D"(2) Repeat step 1 with parameters applicable to the
undesired facility, q=0.05 and Q set to Q-l;
e.g., if Q=0.9, ta!t-- Q as 0.1 for this step. Theresulting a 2 (q,Q) will be called [a2 (0.05, Q- 1 )]U*
1i1
- a1E-
M:=T4, '1' t 4
::T - C - I'4f 4
00
I 4
(3) Use the results of steps 1 and 2 along with aD/U(q) determined using one of the methods pre-
viously discussed in the following:
aDU(0. 9 5 , Q) = /[(2 (0.95, Q)]D + 02 (0.05, Q-l)]U (33)
D/U(0.95, Q) = D/U(0.95,0.5) -z (Q) U(0.95, Q) . (34)mo ODU(. 5
Service ProbabilityProblem 11: Determine D/U(0.95, 0.9) when D/U(0.95,0.5) = 20.2 dB
and other parameters are as follows:
Desired Facility Undesired Facility
C= 2 dB a = 3 dBe e
Lbf = 126.8 dB Lbf = 129.9 dB
Lb(0. 9 5 ) = 150 dB Lb(0.05) = 465.3 dB
Y(0.95) = -6.2 dB Y(0.05) = 14.4 dB
Solution:
(1) For the desired facility,
CY o(0.95) = 12.73 + 0.12 Y2 (0.95) = 12.73 + (0.12)(-6.2)2 = 17.34 dB2
2ce2(0.95) = c 2 ( 0 .95) + 0e2 = 17.34 + (2)2 = 21.34 dB 2
Lb(0. 9 5 ) + Zmo (0.9 e = 150 + (1.3)(2) = 152.6 dB
L -6 =126.8- 6 =120.8 dBbf
therefore,
Lb(o 95) + Zmo (0.9) Oe Lbf 6
so that
[a2(0.95, 0 .9)]D Oce 2 (0.95) = 21.34 dB 2
113
(2) For the undesired facility,
o 2(0.05) = 12.73 + 0.12 y 2 V0.35) = 12.73 + (0.12)(14.4)2 = 37.61 dB2 ,
C ce2 (0.05) c 2 (0.05) + ae2 3 37.61. + (3)2 = 46.61 dB2
Lb(0.05) + (1-0.9) (Y 165.3 + (-1.3)(3) 161.4 dB
Lbf - 6 129.9 - 6 = 123.9 dB
therefore,
Lb(0.05) + z (1 0.9) _ L bf 6
so that
[2 (0.05, 0U.1)] = U c (0.05) = 46.61 dB2U ce
aDU(0.95, 0.9) = %/I[2 (0.95, 0 .9) 1 D + [a2(0.05, 0.1))u
ODU(0. 9 5 , 0.9) = 21.34 + 46.61 = 8.2 dB
D/U(0.95, 0.9) = D/U(0.95) - Zmo(0.9) aDU(0.95,)
and
D/U(0.95, 0.9) = 20.2 - (1.3)(8.2) = 9.5 dB
4.10 Radar Application
Basic transmission loss for radar applications, LbR(0. 9 5 ), can
be estimated from Lb(0.5) and Y(0.95) values determined for the
radar-to-target path by using the appropriate curves and (8) along
with A from (3), and the distribution of expected radar cross-
section values, A (q)[dB - sq m]), as follows:
Y (q) = A (q) -A (0.5) (35)
114
(0.95) =- /2[Y(0.95)12 + [Y (0.95)]2 (36)
and
LbR(0. 9 5 ) = 2 Lb(0.5) + AI - A (0.5) - YR(0.95) (37)
Here, A,(q) is the radar cross section available or exceeded for a
fraction q of the time and is expressed as decibels greater than
1 sq m; i.e., 10 log (sq m cross section).
Radar cross section "... can be expressed as 4w times the power
delivered per unit solid angle in the direction of the receiver
divided by the power per unit area incident at the target" where
"...4Ar enters from the definition of solid angle" [18, p. 8]. It may
also be taken as "...the area which would intercept sufficient power
out of the transmitted field to produce the given echo by isotropic
reradiation" [18, p. 9].
The radar cross section of a particular object is dependent upon
such things as its shape, orientation, and composition along with the
frequency and polarization of the radar. A more comprehensive dis-
cussion is beyond the scope of this report, and interested readers
are referred to the Radar Cross Section Handbook [18] for additional
information.
Basic Transmission Loss for Radar
Problem 12: Determine LbR'O. 9 5) for parameters specified as follows:
A (0.5) = 13 dB-sq m (20 sq m cross section)
A (0.95) = 0 dB-sq m (1 sq m cross section)
f = 9.4 GHz = 9400 MHz
Lb(0.5) = 169.6 dB
and
Y(0.95) = -7.5 dB
Solution: By using (3), (35), (36), and (37), LbR(O. 9 5 ) is calculated
as follows:
115
, - . .. .... 9, . • tJ.S % [ , .
A = 38.54 - 20 log f[MHz] ,
A= 38.54 -20 og(9400) -40.9 dB-sq m
Y (0.95) = a0(0.95) - A (0.5) = 0 - 13 = -13 dB-sq m
YR(0.95) = -/2[Y(0.95)]2 + [Y (0.95)]2
R(095) -/2(-7.5 + (-13)2 = -16.8 dB
LbR(0. 9 5 ) 2 Lb(O.5) A, A(0.5)
and
LbR(0. 9 5 ) = 2(169.6) + (-40.9) - 13 - (-16.8) = 302 dB
116
APPENDIX. LIST OF SYMBOLS
This list includes most of the abbreviations, acronyms, and sym-
bols used in this report. Many are similar to those previously usedin other reports [8, 11, 17]. The units given for symbols in thisList are those required by or resulting from equations as ',von in
this report. Except where otherwise indicated, equations are dimen-sionally consistent so that appropriate units can be selected by the
user.In the following list, the English alphabet precedes thie Greek
alphabet, letters precede numbers, and lowei-case letters precedeupper-case letters. Miscellaneous symbols and notations are givenafter the alphabetical items.
ATC Air traffic control.AI Effective receiving area [dB-sq m] of an isotropic
antenna from (3).
A (q) The radar cross section available or exceeded for afraction q of the time. It is expressed as decibelsgreater than 1 sq m; i.e., 10 log (sq m cross section).
B A ratio shown in Figure 86.
c 10 log e = 4.3429.
C A distance shown in Figure 86.
dB Decibels, 10 log (dimensionless ratio of powers).
dBi Antenna gain in decibels greater than isotropic.
dBm Power in decibels greater than 1 milliwatt.
dBW Power in decibels greater than 1 watt.
dB-sq m Effective antenna aperature area in decibels greaterthan 1 square meter.
dB-W/sq m Power density as decibels greater than 1 watt persquare meter,
d Desired facility to aircraft distance as shown inFigure 85.
d Undesired facility to aircraft distance as shown inFigure 85.
dul A "worst case" d used in Section 4.5.
117
Sdu2 A 4best case" dU used in Section 4.5.
D(q) The RSL from the desired facility that is available orexceeded for at least a fraction q of the time.
D/U(0.95) The desired-to-undesired signal ratio [dB] that isavailable or exceeded for 95% of the time, from (10).
D/U(0.95,Q) The D/U(0.95) available with probability Q, from (34).
e 2.718281828
EIRP Equivalent isotropically radiated power (Sec. 4.1) incecibel type units; e.g., dBW or dBm.
f Frequency [MHz], used in (3).
FAA Federal Aviation Administration.
GHz Gigahertz (109 Hz).
G Gain [dBi] of receiving or transmitting antenna inapplicable direction (Sec. 4.1).
H Terminal 1 or 2 antenna elevation [m] above the earth'ssurface.
i Used to indicate ith item.
IF-73 The ITS-FAA-1973 propagation model [6].
IF-77 The ITS-FAA-1977 propagation model [11].
ITS Institute for Telecommunication Sciences.
A subscript used in (28) to designate a specific mem-ber of a set of power levels.
k Effective earth radius factor; i.e., effective radius/actual radius [11, Sec. 4.1, refractivity discussion;17, Sec. 4].
km Kilometers (10i m).
ln Natural (base e) logarithm.
log Common (base 10) logarithm.
LBasic transmission loss in decibels [17, Sec. 2.4].
L L for free space propagation conditions [17, Sec. 2.4].bf b
LbR Lb for radar-target-radar path as calculated from (37).
Lbl, 2 L values for points 1 or 2 between which a value' interpolated using (4).
118
__ _ _ __ :
Lb(q) An Lb level not exceeded for a fraction q of the time.
Lr Line loss [dB] of receiving antenna system (Sec. 4.1).
m Meters.
M A constant used in Section 4.7.
n Number of undesired signal sources, see Section 4.8.1.
N Minimum monthly surface refractivity in N-units [11,Sec. 4.1, refractivity discussion].
N-units Units of refractivity corresponding to (refractiveindex-l) x 106.
P Radiated power in decibel type units (dBm, dBW), usedin (7).
q Time availability (Sec. 1).
Q Service probability, see Section 4.9.
rD A slant range shown in Figure 86.
rU A slant range shown in Figure 86.
R A radius shown in Figure 86.
RSL Received signal level.
S Station separation shown in Figure 85.
Sf Facility separation shown in Figure 85.
S (q) Power density at the receiving antenna [dB-W/sq mlthat is available or exceeded for a fraction of time
q, from (2)
UHF Ultra-High Frequency (300 to 3000 MHz).
UA Accumulated utilization [%], from (28).
U Channel utilization M%], see Section 4.8.2.
UL Level utilization [%J, from (27).
VHF Very High Frequency (30 to 300 MHz).
x Variable (height or frequency) value for which aninterpolated Lb value from (4) is desired.
x Variable (height or frequency) values for points 1 or1 2 between which interpolation using (4) is desired.
119
Y(q) Normalized (about median) variability [dB] of Lb fortime availability q. Calculated using (8).
(Y(O.O5)] U Normalized (about median) variability for the 5%(q-0.05) interference level from the undesired source(s).Calculated using (8), (23), or (29).
YDU (0.95) Normalized (about median) variability [dB] of D/U forq=0.95 from (9).
YR (0.95) Normalized variability for a radar-target-radar pathwith q=0.95. Calculated from (36).
Y (q) Normalized variability for radar cross section, from(35).
zmo(Q) Value of the standard normal deviate for a particularQ, from Figure 87.
a i Mean power (watt type units] for ith undesired facility,from (19).
Ah Terrain parameter used to characterize terrain [11,Table 7, Fig. 53; 15, Sec. 2.21.
X Wavelength.
Variance of power [watts 2 type units] for ith unde-sired facility, from (20).
T 3.141592654.
Used to indicate a summation over n undesired sources,see (22).
a 2 (q,Q) Total standard error for an Lb for a particular q andQ, from (32).
2 (q) Prediction errors or path-to-path variance, from (30).
c2 (q) Combined standard error variance, from (31).
a (0.95,Q) Standard error for a D/U(0.95) prediction for a partic-
ular value of Q, from (33).
S2 Standard error variance used to account for uncertan-e ties associated with equipment performance parameters,see (31).
o. Standard deviation [dB] for ith undesired facility,1 from (18).
Standard deviation [dB) for contributions from bimul-taneous independent undesired sources, from (22).
120 /
J" 1 *-
D Used to indicate that the bracketed items are associ-ated with the desired facility.
Used to indicate that the bracketed items are associ-ated with the ith undesired facility or volume.
U Used to indicate that the bracketed items are associ-Uated with the undesired facility.
121
REFERENCES
(11 Ames, L. A., P. Newman, and T. F. Rogers (1955), VHF troposphericoverwater measurements far beyond the radio horizon, Proc. IRE43, No. 10, 1369-1373.
(21 Barsis, A. P., K. A. Norton, P. L. Rice, and P. A. Elder (1961),Performance predictions for single tropospheric communicationlinks and for several links in tandem, NBS Tech. Note 102 (NTIS,AD 288 532).'
(31 Barsis, A. P., K. A. Norton, and P. L. Rice (1962), Predictingthe performance of tropospheric communication links, singly andin tandem, IRE Trans. Commun. Systems CS-10, No. 1, pp. 2-22.
[4) CCIR (1978), VHF, UHF, and SHF propagation curves for the aero-nautical mobile service, Rec. 528, XIV Plenary Assembly, Kyoto(Intl. Telecomm. Union, Geneva; NTIS, PB 298025).'
(5] Gierhart, G. D., and M. E. Johnson (1969), Transmission loss
atlas for select aeronautical service bands from 0.125 to 15.5GHz, ESSA Tech. Rept. ERL 111-ITS 79 (GPO).2
(6] Gierhart, G. D., and M. E. Johnson (1973), Computer programs forair/ground propagation and interference analysis, 0.1 to 20 GHz,DOT Rept. FAA-RD-/3-103 (NTIS, AD 770 335).1
(71 Gierhart, G. D., and M. E. Johnson (1978), Propagation model(0.1 to 20 GHz) extensions for 1977 computer programs, DOT Rept.FAA-RD-77-129 (NTIS, ADA 055605).'
(8] Gierhart, G. D., R. W. Hubbard, and D. V. Glen (1970), Electro-space planning and engineering for the air traffic environment,DOT Rept. FAA-RD-70-71 (NTIS, AD 718 447).)
(9] IEEE (1970), Special issue on air traffic control, Proc. IEEE58, No. 3.
[101 Johnson, M. E., and G. D. Gierhart (1978), Aerospace propagationprediction capabilities associated with the IF-77 model, AGARDConference Proceedings No. 238, Paper 48.
[11] Johnson, M. E., and G. D. Gierhart (1978), Applications guidefor propagation and interference analysis computer prog7rams (0.1to 20 GHz), DOT Rept. FAA-RD.-77-60 (NTIS, ADA 053242).
(12] Johnson, M. E., and G. D. Gierhart (1979), Comparison of measureddata with IF-77 propagation model predictions, DOT Rept. FAA-RD-79-9 (NTIS, ADA-076508).
[131 Juroshek, J. R. (1975), Probabilistic interference predictionmodel-computer programs, DOT Rept. FAA-RD-75-33 (NTIS, ADA011453).
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(14] Longley, A. G., R. K. Reasoner, and V. L. Fuller (1971), Measuredand predicted long-term distributions of tropospheric trans-mission loss, OT Telecomm. Res. and Eng. Rept. OT/TRER 16 (NTIS,COM-75-11205).'
[15] Longley, A. G., and P. L. Rice (1968), Prediction of troposphericradio transmission loss over irregular terrain, a computermethod-1968, ESSA Tech. Rept. ERL 79-ITS 67 (NTIS, AD 676 874).'
(16] Norton, K. A., H. Staras, and M. Blum (1952), A statisticalapproach to the problem of multiple radio interference to FMand television service, IRE Trans. Ant. Prop. PGAP-1, pp. 43-49.
[17] Rice, P. L., A. G. Longley, K. A. Norton, and A. P. Barsis (1967),Transmission Loss predictions for tropospheric communicationscircuits, NBS Tech. Note 101, I and II revised (NTIS, AD 687820and 687821).-
(18] Ruck, G. T., D. E. Barrick, W. D. Stuart, and C. K. Krichbaum(1970), Radar Cross Section Handbook, 1 and 2 (Plenum Press, NewYork).
lCopies of these reports are sold by the National Technical Info~ina-tion Service, Operations Division, Springfield, VA 22161. Order byindicated accession number.2Copies of these reports were sold by the Superintendent of Documeits,U.S. Government Printing Office, Washington, DC 20402, and may stillbe available.
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