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04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 1
Geographic Information ProcessingRadio WavePropagation
Line-of-Sight Propagation in cross-section and relief - Hupfer et al.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 2
Radio Wave Transmission• Electromagnetic energy in orthogonal fields
– Electric field energy– Magnetic field energy
• Radiate from source antenna– Power spread over spherical wavefront– Power/Area decreases with distance as area of
sphere increases and as losses dissipate energy– Energy can be focused in desired direction
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 3
Radio Wave Reception
• Receiver responds to vector sum of arriving waves (out-of-phase waves interfere)
• Subject to receiver sensitivity limits
• Antennas with directional gain can focus energy from particular direction(s)
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 4
Engineering Problem
• Specify transmitter power and antenna gain in direction of receiver
• Specify receiver sensitivity and the operable corresponding antenna gain
• Determine path loss over topography
• Compare received power with sensitivity limit and modify conditions as necessary
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 5
Path Criterion
A radio wave takes the path of minimum time. In free space (no diffraction or reflection) this will be a straight line.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 6
• Reduction in Output/Input ratio of amplitude of radio wave
• Usually measured in decibels (dB)
– Power Attenuation:
– Voltage Attenuation:
Attenuation
€
Attenuation =10log10
PowerOut
PowerIn
⎛
⎝ ⎜
⎞
⎠ ⎟
€
Attenuation = 20log10
VoltageOut
VoltageIn
⎛
⎝ ⎜
⎞
⎠ ⎟
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 7
Energy in Electric Field
• Where:– w = Energy Density [Joules/m2]– E = electric field intensity [V/m] ε = dielectric permittivity of free space
€
w = 12 εE 2
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 8
Energy in Electric Field
• Where:– p = Average Power Density [Watts/m2]– E = Electric field intensity [V/m]
– Ro = Impedance of free space [~377 Ohms]
€
p =E 2
2R0
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 9
Radio Wave Attenuation Factors• Distance
• Diffraction from objects in propagation path
• Reflection from objects near path
• Conduction/reflection/refraction by various atmospheric effects
• Scattering by objects and atmospheric components
04/19/23 © 2009 Raymond P. Jefferis III Lect 00 - 10
Transmission Losses
Transmitted electromagnetic energy is lost on its way to a receiving station due to a number of factors, including:
– Antenna efficiency – Path loss– Antenna aperture gain – Atmospheric loss– Path loss – Diffraction loss
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 11
Issues with Reflection
• Out-of-Phase waves cancel primary wave– Various reflecting surfaces => different arrivals– Random arrival phasea produce noise floor
• Digital symbols– Inter-symbol interference– Data rate must be limited to allow each symbol
to extinguish itself before next
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 12
Transmitter Power• Pt = 10 Log10 PmW [dBm]
• Example: 5 Watts = 5000 mWPt [dBm] = 10 Log10 5000 = 37 dBm
• Example: 40 Watts = 40000 mWPt = 10 Log10 40000 = 46 dBm
• For propagation loss calculations, dBm units are more convenient than power.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 13
Antenna GainAe = effective antenna apertureG = 4πAe/λ2 (Antenna Gain)d = antenna diameterλ = wavelengthη= aperture efficiency
Ae =ηAπ(d / 2)2
G=4πλ2 Ae
G=ηA
πdλ
⎛⎝⎜
⎞⎠⎟
2
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 14
Path Losses• Effective Aperture (transmit or receive):
Ae = η A• Effective Radiated Power:
EIRP = PtGt = Ptη taAt
where,Gt = 4πAet/λ2 Gr = 4π Aer/λ2
• Path Loss (for path length R):Lp = (4π R/λ2
• Received Power:Pr = EIRP*Gr/Lp
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 15
Receiver Sensitivity• Usually specified in microvolts on 50-Ohm
input connector
• Can be converted to power by:
• For typical sensitivity of 0.18 microVolts:[Watts]€
P =V 2
R
€
Pr =0.18∗10−6
( )2
50= 6.5∗10−10
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 16
Receiver Sensitivity [dBm]
• dBm => dB milliWatts
• Pr [dBm] = 20 log10[Pr/10-3]where Pr is given in Watts
• Converting the receiver sensitivity to dBm,
• For proper reception, the transmitted signal must not fall below this level.€
Pr[dBm] =10log10[6.5∗10−10 /10−3] = −62 [dBm]
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 17
Received Power - dB Model
• (Pratt & Bostian, Eq. 4.11)Pr = EIRP +Gr - Lp - La - Lt - Lr [dBW]– EIRP => Effective radiated power
– Gr => Receiving antenna gain
– Lp => Path loss
– La => Atmospheric attenuation loss
– Lt => Transmitting antenna losses
– Lr => Receiving antenna losses
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 18
Path Models
• Free-Space
• Partially Obstructed
• Largely Obstructed
• Totally Obstructed
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 19
Free-Space Model
• No obstructions in or “near” pathNote: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path.
• Obstructions must be outside first “Fresnel Zone” - See next slide
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 20
Fresnel Zones• Elliptical zones of radiated energy between
transmitted and receiver
Wikipedia - http://en.wikipedia.org/wiki/Fresnel_zone
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 21
Partially Obstructed Path Model
• Obstructions in, but not occluding, pathNote: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path.
• Obstructions inside first “Fresnel Zone” but not by more than 40%
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 22
Highly Obstructed Path Model
• Obstructions in, but not occluding, pathNote: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path.
• Obstructions inside first “Fresnel Zone” and occlude it by more than 40% but not completely
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 23
Fully Obstructed Path Model
• Obstructions occluding, pathNote: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path.
• Obstructions inside first “Fresnel Zone” occlude it completely. Energy confined to higher-order Fresnel zones.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 24
Free-Space Path Loss Model
• Free space loss [Watts]:
• Free space loss [dB]:32.4 + 20 Log f [MHz] + 20 Log d [Km]
– f is the radio frequency [MHz]– d is the distance [km] between the transmitting and
receiving antennas
€
Pr(d) = KPt
d2
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 25
Loss Calculation #1
• Let f = 146 MHz, d = 10 kmLossdB = 32.4 + 20 log10(146) + 20log10(10)LossdB = 32.4 + 43.3 + 20 = 95.7 dB
• The receiver signal strength for 40-Watts is:Pr [dBm] = 46 - 95.7 = - 49.7 dB
• Conclusion: 10 km free-space signal path is okay at this frequency.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 26
Loss Calculation #2
• Let f = 146 MHz, d = 100 kmLossdB = 32.4 + 20 log10(146) + 20log10(100)LossdB = 32.4 + 43.3 + 40 = 115.7 dB
• The receiver signal strength for 40-Watts is:Pr [dBm] = 46 - 115.7 = - 69.7 dB
• Conclusion: 100 km free-space signal path is too long for reliable reception at this frequency.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 27
1300 MHz Digital Data Sensitivity
€
Pr =1.58∗10−6( )
2
50= 50∗10−10
€
Pr[dBm] =10log10[50.0∗10−10 /10−3] = −53
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 28
Loss Calculation #3
• Let f = 1300 MHz, d = 20 kmLossdB = 32.4 + 20 log10(1300) + 20log10(20)LossdB = 32.4 + 63.3 + 26 = 120.7 dB
• The receiver signal strength for 10-Watts is:Pr [dBm] = 40 - 120.7 = - 80.7 dB
• Conclusion: 20 km free-space signal path is too long at this frequency.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 29
Antenna Gain
• Antenna radiation patterns direct energy, resulting in gain in certain directions.
• Antenna gain [dB] adds to the reference propagation gain (subtracts from propagation loss) in certain directions.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 30
Ex. #3 with 16 dB Antenna Gain
• Let f = 146 MHz, d = 100 kmLossdB = 32.4 + 20 log10(146) + 20log10(100)LossdB = 32.4 + 43.3 + 40 = 115.7 dB
• The receiver signal strength for 40-Watts is:Pr [dBm] = 16 + 46 - 115.7 = - 53.7 dB
• Conclusion: 100 km free-space signal path is okay for reliable reception at this frequency, if trans-mitter and receiver each have 8 dB antenna gain.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 31
Diffraction Model
http://www.mike-willis.com/Tutorial/PF7.htm
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 32
Obstruction Loss Model
http://www.mike-willis.com/Tutorial/PF7.htmHorizontal axis is
Obstruction of Fresnel Zone
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 33
Building roof at edge of 1st Fresnel zone
Conclusions
• A building reaching to edge of the 1st Fresnel zone produces 6 dB loss (loss of ¾ of radiated power)
• Obstruction of entire 1st Fresnel zone would be a significant loss to a communication system
04/19/23 © 2009 Raymond P. Jefferis III Lect 06 - 34
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 35
Obstruction Loss Calculations
http://www.mike-willis.com/Tutorial/PF7.htm
€
υ =±h2 d1 + d2( )
λd1d2
Use value of , enter previous graph, and read loss in dB, or calculate knife-edge loss J()as:
€
J(ν ) = 20log10 F(ν ) ≈ 6.9 + 20log10 ν 2 +1 + ν[ ]
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 36
Obstructed Fresnel Zones in Path
From Lewis Girod, Graph attributed to Rappaport, Wireless Communications: Principles and Practice, Prentice Hall, 1996, p 97
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 37
More Exact Fresnel Loss
F() =12−121+ j( ) C()− jS()[ ]
Fresnel loss, whereC = Fresnel cosine integral
S = Fresnel sine integral
C(x) = cosπ2t2⎛
⎝⎜⎞⎠⎟
0
x
∫ dt
S(x) = sinπ2t2⎛
⎝⎜⎞⎠⎟
0
x
∫ dt
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 38
Fresnel Loss Magnitude
F() =12
12+ S2 +C2 −S−C⎡⎣ ⎤⎦
⎧⎨⎩
⎫⎬⎭
12
Square root of F() * F* ()
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 39
Logarithmic Fresnel Loss
F() dB =10Log1012
12+ S2 +C2 −S−C( )
⎡⎣⎢
⎤⎦⎥
⎧⎨⎩
⎫⎬⎭
This loss should be added to the free-space propagation loss in dB.
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 40
Practical Problem
Calculate the path loss between two points on a topographic map having a free-space path between them with no obstructions near the first Fresnel zone.
Fresnel Radius(f)
• For d1 = d2 = d the Fresnel radius at the midpoint becomes:
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 41
Fn =nλd4
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 42
LOS Distance Attenuation
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 43
Path Shaded by LOS Attenuation1300 GHz radio path over rough terrain, showing First Fresnel zone. Terrain is tinted according to the Line-of-Sight propagation losses.(Blue is max. loss)
Line-of-Sight Path
• Get starting point and its antenna height
• Get destination point with antenna height
• Draw air path between these points
• Calculate and plot terrain below air path
• Calculate radiation-terrain clearance on path
• Identify interfering terrain
• Add transmission loss from terrain04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 44
Example
• Point A Antennaantlat = 40.057880 Nantlon = 75.598214 Wanthgt = 40.0 [meters]
• Point B Fire trucktrklat = 40.123831 Ntrklon = 75.513662 Wtrkhgt = 2.5 [meters]
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 45
Radiation Path
• Black line on terrain at lower right is path
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 46
Close-up of Radiation Path
• Point A (Left) is on a hilltop
• Point B (Right) is in a developed area
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 47
Path and Cross-Section
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 48
Discussion of Programming
Discussion in class ...
04/19/23 © 2009 Raymond P. Jefferis III Lect 09 - 49