review noise concepts direct transmission - friis formula
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Direct Transmission, Part One
1 Review noise concepts
2 Direct transmission - Friis formula
3 Atmospheric gas attenuation
4 Total attenuation on a path
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 1 / 46
II. Direct transmission
Applies to situations where reflection and refraction are neglected
Scattering may occur, but we will only consider it to reduce signalstrength
Usually appropriate for satellite and radar systems
It is easy to predict received power under these conditions - Friisformula
We will also consider attenuation due to atmospheric gases and rainwith direct transmission
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 2 / 46
For an isotropic antenna transmitting power PT , the radiated flux densityat distance R is
S(R) =PTυT
4πR2(1)
For a directive antenna, we need to include the gain
S(R, θT , φT ) =PTυT
4πR2DT (θT , φT ) =
PT GT (θT , φT )
4πR2(2)
For a polarization and impedance matched receiving antenna in the farfield,
PR =PT GT (θT , φT )GR(θR , φR)λ2
(4π)2R2(3)
This is the Friis transmission formula - very useful for system calculations
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 3 / 46
Excepting for the ohmic losses in the antennas, included in υT and υR , weneglected all other losses in the previous equation. We can include them by
PR =PT GT (θT , φT )GR(θR , φR)λ2
(4π)2R2TgasTrainTimpTpolTcoupling
where
Tgas = exp
(−∫
pathαgas dl
)(4)
Train = exp
(−∫
pathαrain dl
)(5)
These are due to the fact that flux density will be proportional to e−αx ina scattering medium, but α may change along the path
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 4 / 46
It is often easier to work with this equation in dB as
PR,dbW = PT ,dbW − Lsys,db (6)
where the system loss Lsys,db is given by
Lsys,db = 20 log10 Rkm + 20 log10 fMHz + 32.44 + Lgas,db + Lrain,db + Lpol ,db
+Limp,db + Lcoup,db − GTdb − GRdb (7)
and
Ldb = −10 log10 T (8)
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 5 / 46
This gives
Lgas,db =
∫path
αgas,db dl (9)
and
Lrain,db =
∫path
αrain,db dl (10)
where αgas,db and αrain,db are the “specific attenuation” due to gas andrain in dB/km
αgas,db = 4.34 αgas
αrain,db = 4.34 αrain (11)
Thus to get the total loss along a path, we need to integrate the specificattenuations over the path
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 6 / 46
Practical considerations:
Consider a 1 MWatt transmitter (PT = 106 Watts)
Consider a sensitive receiver with 100 ohm input impedance that candetect 1 µvolt
PR,min = V 2/R =(10−6
)2/100 = 10−14
Maximum tolerable system loss: 10−14/106 = 10−20 or 200 dB
Thus, system losses greater than 200 dB are not very practical formost systems
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 7 / 46
Another sample problem: Communicating with Galileo
Approximate distance to Galileo: = 628× 106 km
Let’s use a 2 GHz system: fMHz = 2, 000
Neglecting system and atmospheric losses we get
Lsys,db = 20 log10 Rkm + 20 log10 fMHz + 32.44− GTdb − GRdb
= 175 + 66 + 32.44− GTdb − GRdb
= 273.44− GTdb − GRdb (12)
Assume Galileo has a 10 dBi gain (high gain antenna never opened)and transmits 20 W, then
PR = 13− 273.44 + 10 + GRdb (13)
It’s pretty clear we need a high gain antenna on the ground and a goodreceiver with low F - JPL quotes 10−20 watts as being received!
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 8 / 46
III. Atmospheric gas attenuation
Gas molecules in the atmosphere are able to absorb microwave power- it is re-emitted as thermal noise
Effects not very significant below 10 GHz
Most significant around molecular resonance frequencies, in particularoxygen and water vapor; ITU-R Recommendation P. 676-7.
Resonance frequencies: water - 22.3, 183.3, 325.4 GHz, oxygen - 60,118.7 GHz
Density of resonances increases greatly at higher frequencies
Specific attenuation depends on atmospheric composition whichvaries with altitude
Pressure broadening changes line shapes at different altitudes
Tool available atwww.rcru.rl.ac.uk/show.php?page=njt/ITU/ITU676-6
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 9 / 46
10 100 10000.001
0.01
0.1
1
10
100
1000
Frequency (GHz)
Spe
c. a
tten
(dB
/km
)
OxygenWater VaporBoth
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 10 / 46
10 100 10000.001
0.01
0.1
1
10
100
1000
Frequency (GHz)
Spe
c. a
tten
(dB
/km
)
0 km10 km
50 55 60 65 700.001
0.01
0.1
1
10
Frequency (GHz)
Spe
c. a
tten
(dB
/km
)
0 km10 km20 km
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 11 / 46
IV. Total attenuation on a path
Horizontal paths are easy since atmospheric gases are usually uniformhorizontally
Lgas,db =
∫αgas,db dl = αgas,dbL (14)
Vertical or slant paths require integration over path
Typical models for atmospheric variations are like
ρ = ρ0 exp (−h/hs,w ) (15)
or we can use measurements of atmospheric composition
Results are often presented as an “equivalent” L by which αgas,db atthe ground can be multiplied to get through the total atmosphere
We will just rely on the figure for total vertical attenuation throughthe atmosphere
Higher frequencies have many resonances
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 12 / 46
1 10 100 10000.001
0.01
0.1
1
10
100
1000
Frequency (GHz)
Zen
ith a
tten
(dB
)
0 km5 km10 km
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 13 / 46
Attenuation on slant paths can be obtained from the correspondingvertical path through
Lgas,slant(h1, h2) =
∫ s2
s1
αgas,db(s) ds =
∫ h2
h1
αgas,db(h) csc θ dh
= Lgas,vert csc θ (16)
For attenuations at two different angles,
Lg (θ1)
Lg (θ2)=
csc θ1csc θ2
=sin θ2sin θ1
(17)
Note these relations do not apply for θ < 10 degrees due to Earthcurvature and refraction
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 14 / 46
dh dS
θ
Ground Distance
Altitude h
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 15 / 46
Direct Transmission, Part Two
1 Rain attenuation
2 Specific attenuation for rain
3 Rain statistical information
4 Probability of outage predictions
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 16 / 46
I. Rain attenuation
Rain also can cause attenuation along atmospheric paths, usuallynegligible for f < 5 GHz
Can cause serious outages as frequency increases
A statistical approach is necessary for discussing rain fades - averagenumber of hours per year to obtain given attenuation
Statistics will vary from location to location!
We will therefore need two types of information: one for attenuationon a path given rain rate (deterministic) and one for probability ofrain rate (statistical)
Systems can be designed with a “rain margin” to allow continuedfunction even during rain events
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 17 / 46
0.1 1 10 100 10000.1
1
10
100
Hours per year rain attenuation is exceeded
Rai
n at
tenu
atio
n (d
B)
Latitude 40°, R0.01
=49 mm/hr
14 GHz, θ=20°
35 GHz, θ=20°
14 GHz, θ=90°
35 GHz, θ=90°
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 18 / 46
II. Specific attenuation for rain
As with atmospheric gases, we can talk about a specific attenuationfor rain, αrain,db in dB/km
Specific attenuation must be integrated over path to get totalattenuation - this is more difficult for rain
We can use theory to derive a specific attenuation for rain through
Drop size distribution n(r)- number density per drop diameterMie scattering theory for a sphere to obtain amount of power absorbedand scattered out of incident beam for one sphere - “extinction crosssection”Integration over size distribution to get total power lossThis is power lost in a differential volume, leads to a differentialequation that shows flux density ∝ e−αx
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 19 / 46
It turns out that both theory and experiment lead to a simpler formfor rain specific attenuation as
αrain,db = krRarr (18)
kr and ar are tabluated coefficients and Rr is the rain rate in mm/hr
We will use the plots in the book that give kr and ar for bothhorizontal and vertical polarizations
These coefficients are functions of frequency since drop scattering andaborption efficiency changes with frequency
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 20 / 46
1 10 100 1000
10−4
10−3
10−2
10−1
100
Frequency (GHz)
k r coe
ffici
ent
HV
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 21 / 46
1 10 100 10000.6
0.8
1
1.2
1.4
1.6
Frequency (GHz)
a r coe
ffici
ent
HV
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 22 / 46
1 10 100 10000.01
0.1
1
10
100
Frequency (GHz)
Rai
n S
peci
fic A
ttenu
atio
n (d
B/k
m)
H pol, 10 mm/hrH pol, 50 mm/hrH pol, 100 mm/hrV pol
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 23 / 46
III. Rain statistical information
We now have an equation relating rain specific attenuation to rainrate Rr
If we have statistical information on Rr for a given location, we canthen obtain statistical information on α through α = krR
arr
Several sources available for this information:
Local measurements: always best but usually not availableNational Weather Service or National Oceanic and AtmosphericAdministration in USOther global models
Note that “tails” of the distribution are most important because it isthe heaviest rains that give the most serious attenuation
The ITU-R model we will use for predicting rain attenuation focusesinstead of the rain rate exceeded 0.01 percent of the time, Rr ,0.01
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 24 / 46
20
25
30
35 40
45
50
55 60
65
70
75
80
25
30
35
50
40
5515 60
85
45
85
6565
15
15
15
50
7570
35
90
20
70
25
25
20
40
25
120oW 108oW 96oW 84oW 72oW
25oN
30oN
35oN
40oN
45oN
50oN
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 25 / 46
IV. Probability of outage predictions
So far we have models for rain specific attenuation and some info onprobability of rain rates
We still don’t have total attenuations though because we haven’tintegrated over paths
Difficult to do this precisely because rain properties vary horizontallyand vertically!
Lots of models available; ITU-R studies provide recommendations onuse
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 26 / 46
All of the total path attenuation models are based on either
Lrain,db =
∫path
αrain,db(l) dl ≈ αrain,db(0)Leff (19)
where αrain,db(0) is the specific attenuation at sea level and Leff is an“effective” path length, or
Lrain,db =
∫path
αrain,db(l) dl ≈ aRbeff L (20)
where Reff is an “effective” rain rate. Empirical methods are usually usedto find these quantities.
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 27 / 46
ITU-R Model for Earth to Space Paths
ITU-R Recommendation P. 618-9 provides an empirical method forpredicting LE on Earth to space paths
Stated as being accurate to within 20 percent for most locations atfrequencies up to 55 GHz
Inputs: Rr ,0.01, station height in km, hS , satellite elevation angle θ,ground station latitude φ, rain height hR , kr and ar coefficients
Output: Effective path length LE to be used with rain rate Rr ,0.01 toobtain attenuation exceeded 0.01 percent of the time
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 28 / 46
10 20 30 40 502
4
6
8
Frequency (GHz)
L E (
km)
(a) θ=90 deg, φ=40 deg
20 mm/hr50 mm/hr100 mm/hr
0 20 40 60 800
10
20
30
40
50
Elevation angle (deg)
L E (
km)
(b) 35 GHz, φ=40 deg
20 mm/hr50 mm/hr100 mm/hrL
S
0 50 1000
2
4
6
8
Station Latitude (deg)
L E (
km)
(c) 35 GHz, θ=90 deg
20 mm/hr50 mm/hr100 mm/hr
0 50 1003
4
5
6
7
Rain Rate (mm/hr)
L E (
km)
(d) θ=90 deg, φ=40 deg
20 GHz35 GHz50 GHz
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 29 / 46
ITU-R Model for Earth to Earth Paths
ITU-R Recommendation P. 530-12 provides an empirical method forpredicting LE on Earth to Earth paths
LE =d
1 + d/d0
where d is the path distance in km and
d0 = 35e−0.015Rr,0.01 Rr ,0.01 ≤ 100
d0 = 7.81 Rr ,0.01 > 100 (21)
with Rr ,0.01 in mm/hr.The quantity d0 is an average horizontal extent of the rain, and decreasesfrom 35 km at low rain rates to 7.81 km at very high rain rates.
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 30 / 46
Probability Scaling
The Preceeding models determine the attenuation exceeded 0.01 percentof the time:
Lrain,0.01 =(krR
arr ,0.01
)LE
A “probability scaling” approach is recommended to obtain attenuationsexceeded p percent of the time. For Earth-to-space paths
Lrain,p = Lrain,0.01
( p
0.01
)−[0.655+0.033 ln(p)−0.045 ln(Lrain,0.01)−β(1−p) sin θ]
where
β = 0 p ≥ 1 or |φ| ≥ 36◦
β = −0.005 (|φ| − 36◦) p < 1 and |φ| < 36◦ and θ ≥ 25◦
β = −0.005 (|φ| − 36◦) + 1.8− 4.25 sin θ otherwise
This approach is specified as accurate for p ranging from 0.001 to 5percent (see equations in book for Earth-to-Earth paths).
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 31 / 46
Use of rain attenuation models enables us to ascertain how manyhours of fades of a given number of dB to expect per year
Systems can be designed to have enough S/N ratio to function evenduring these fades
This is not always possible with practical satellite systems however
Problem increases with frequency, so satcom and radar systemsoperating at f > 10 GHz can expect significant outage times due torain
Program for the ITU-R model available via course website
Frequency scaling approaches also available to scale attenuationsobserved at one frequency into another at the same location
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 32 / 46
Direct Transmission, Part Three
1 Rain attenuation example
2 Site diversity improvements
3 Scintillations on atmospheric paths
4 Look angles to geostationary satellites
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 33 / 46
I. Rain attenuation example
12 GHz Earth to satellite link with the ground station location nearColumbus, Ohio (latitude ≈ 40◦, station height 0 km). The satellite isobserved at range 38,000 km from the ground station at elevation angle40 degrees. Suppose the satellite transmitter transmits 120 Watts ofpower in a 24 MHz bandwidth using circular polarization, and has anantenna gain of 46 dBi. Suppose also that the receiver has a noise figureof 6 dB and the receive antenna gain is 30 dBi.
Given that the signal to noise ratio required for the system to function is10 dB, how many hours per year (on average) will the system fail?
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 34 / 46
Compute SNR using Friis formula neglecting atmospheric gas and rainattenuation:
PR,dbW = PT ,dbW + GTdb + GRdb − 20 log10 Rkm − 20 log10 fMHz − 32.44
= 20.8 + 46 + 30− 91.6− 81.6− 32.44 = −108.8 dbW
Thermal noise power level at the receiver (assuming a 290 Kelvin externalnoise temperature) is
PN,Watts = FkBT0B
= 4(1.38× 10−23
)(290)
(24× 106
)= 3.84× 10−13 W
or −124.1 dBW.The SNR neglecting rain and gas attenuation is 15.3 dB, so that 5.3 dBmargin is achieved over the 10 dB signal-to-noise threshold required foroperation.Gas attenuation < 0.1 dB so can be neglected
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 35 / 46
From rain rate map, Rr ,0.01 in Columbus around 45 mm/hr.kr around (0.024,0.024), ar around (1.12,1.18) for h and v polarizationsfrom figuresCombine using equation for circular pol to get kr = 0.024, ar = 1.15.ITU-R Earth-to-space model gives LE = 3.93 km so
Lrain,0.01 = 3.93×(0.024× 451.15
)= 7.5 dB (22)
Use probability scaling to get other percentages; note scaling is hard toinvert so may need an iterative or graphical method to find whatpercentage gives 5.3 dB attenuationGraph on the next page shows 0.025 percent, or around 2.2 hours per year
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 36 / 46
0.001 0.01 0.1 10.5
1
2
5
10
20
Percent of time rain attenuation is exceeded
Rai
n at
tenu
atio
n (d
B)
12 GHz, Latitude 40°, R0.01
=45 mm/hr
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 37 / 46
II. Site diversity improvements
Serious rain attenuation usually only happens over a small horizontalextent - heavy rain cells are finite
Thus if we have two antennas separated by a wide distance it isunlikely that both would have serious rain fades at the same time
This is a “diversity” concept - use multiple antennas to avoid fades
Signals from antennas can be combined either by switching toantenna with highest power or always using both in a varying ratio
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 38 / 46
Two quantities defined: diversity gain and diversity improvement
Dependence on distance: measurements done at OSU showsaturating exponential curve
An empirical model was developed from this data
Models are necessary because buying and placing another antennamay not be easy!
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 39 / 46
0.001 0.01 0.110
100
Percent of time rain attenuation is exceeded
Rai
n at
tenu
atio
n (d
B)
One antennaTwo antennas
I
Gd
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 40 / 46
0 5 10 15 20 25 30 350
1
2
3
4
5
6
Distance between antennas (km)
Div
ersi
ty G
ain
(dB
)
14 GHz, θ=20°
2 dB6 dB10 dB
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 41 / 46
III. Scintillations on atmospheric paths
As we know, the atmosphere is not a homogeneous medium, andsmall variations cause variations in propagated signals
All radio paths will observe some small fluctuations
These should not be confused with system effects!
Scintillations are usually larger at lower elevation angles because thereis more path length in which the atmosphere can vary
May appear to be amplitude or angle of arrival variations dependingon scan rate
Twinkling of stars is a scintillation phenomenon! Important problemfor optical astronomy
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 42 / 46
IV. Look angles to geostationary satellites
Satellites orbiting at 35, 900 km have an orbital period equal to therotation of the Earth - thus they stay fixed in the sky
This makes ground station design simple - no tracking required
Note this is pretty far away however; it remains to be seen how thesesystems will compare with LEO satellite for PCS systems now underdevelopment
It is fairly straight forward to derive the angles (∆,Az) at which aground antenna should be oriented given the station latitude θ′,longitude φ′, and sub-satellite longitude φ0
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 43 / 46
Station in Northern hemisphere: azimuth is Az = π + β for a satellite westof the station, Az = π − β for a satellite east of the station.
cos δ = cos θ′ cos(φ′ − φ0
)(23)
cosβ = tan θ′ cot δ (24)
To find the elevation angle use the law of cosines
R2 = R2ε + (Rε + RS )2 − 2Rε (Rε + RS ) cos δ (25)
and
CB = R sin ∆ (26)
= CO − OB = (Rε + RS ) cos δ − Rε (27)
so that
sin ∆ =(Rε + RS ) cos δ − Rε
R(28)
Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 44 / 46
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