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.

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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

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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)

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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

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Path Criterion

A radio wave takes the path of minimum time. In free space (no diffraction or reflection) this will be a straight line.

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• 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

⎛

⎝ ⎜

⎞

⎠ ⎟

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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

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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

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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

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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

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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

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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.

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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

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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

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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

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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]

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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

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Path Models

• Free-Space

• Partially Obstructed

• Largely Obstructed

• Totally Obstructed

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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

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Fresnel Zones• Elliptical zones of radiated energy between

transmitted and receiver

Wikipedia - http://en.wikipedia.org/wiki/Fresnel_zone

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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%

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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

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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.

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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

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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.

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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.

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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

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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.

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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.

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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.

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Diffraction Model

http://www.mike-willis.com/Tutorial/PF7.htm

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Obstruction Loss Model

http://www.mike-willis.com/Tutorial/PF7.htmHorizontal axis is

Obstruction of Fresnel Zone

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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

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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[ ]

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Obstructed Fresnel Zones in Path

From Lewis Girod, Graph attributed to Rappaport, Wireless Communications: Principles and Practice, Prentice Hall, 1996, p 97

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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

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Fresnel Loss Magnitude

F() =12

12+ S2 +C2 −S−C⎡⎣ ⎤⎦

⎧⎨⎩

⎫⎬⎭

12

Square root of F() * F* ()

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Logarithmic Fresnel Loss

F() dB =10Log1012

12+ S2 +C2 −S−C( )

⎡⎣⎢

⎤⎦⎥

⎧⎨⎩

⎫⎬⎭

This loss should be added to the free-space propagation loss in dB.

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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:

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Fn =nλd4

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LOS Distance Attenuation

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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]

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Radiation Path

• Black line on terrain at lower right is path

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Close-up of Radiation Path

• Point A (Left) is on a hilltop

• Point B (Right) is in a developed area

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Path and Cross-Section

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Discussion of Programming

Discussion in class ...

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