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9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 1 Geographic Information Processing Radio Wave Propagation Line-of-Sight Propagation in cross-section.

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Presentation on theme: "9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 1 Geographic Information Processing Radio Wave Propagation Line-of-Sight Propagation in cross-section."— Presentation transcript:

1 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 1 Geographic Information Processing Radio Wave Propagation Line-of-Sight Propagation in cross-section and relief - Hupfer et al.

2 9/21/2015© 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

3 9/21/2015© 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)

4 9/21/2015© 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

5 9/21/2015© 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.

6 9/21/2015© 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

7 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 7 Energy in Electric Field Where: –w = Energy Density [Joules/m 2 ] –E = electric field intensity [V/m] –  = dielectric permittivity of free space

8 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 8 Energy in Electric Field Where: –p = Average Power Density [Watts/m 2 ] –E = Electric field intensity [V/m] –R o = Impedance of free space [~377 Ohms]

9 9/21/2015© 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

10 9/21/2015© 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

11 9/21/2015© 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

12 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 12 Transmitter Power P t = 10 Log 10 P mW [dBm] Example: 5 Watts = 5000 mW P t [dBm] = 10 Log 10 5000 = 37 dBm Example: 40 Watts = 40000 mW P t = 10 Log 10 40000 = 46 dBm For propagation loss calculations, dBm units are more convenient than power.

13 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 13 Antenna Gain A e = effective antenna aperture G = 4  A e /  2 (Antenna Gain) d = antenna diameter λ = wavelength  = aperture efficiency

14 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 14 Path Losses Effective Aperture (transmit or receive): A e =  A Effective Radiated Power: EIRP = P t G t = P t  ta A t where, G t = 4  A et /  2 G r = 4  A er /  2 Path Loss (for path length R): L p = (4  R/  2 Received Power: P r = EIRP*G r /L p

15 9/21/2015© 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]

16 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 16 Receiver Sensitivity [dBm] dBm => dB milliWatts P r [dBm] = 20 log 10 [P r /10 -3 ] where P r is given in Watts Converting the receiver sensitivity to dBm, For proper reception, the transmitted signal must not fall below this level. [dBm]

17 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 17 Received Power - dB Model (Pratt & Bostian, Eq. 4.11) P r = EIRP +G r - L p - L a - L t - L r [dBW] –EIRP => Effective radiated power –G r => Receiving antenna gain –L p => Path loss –L a => Atmospheric attenuation loss –L t => Transmitting antenna losses –L r => Receiving antenna losses

18 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 18 Path Models Free-Space Partially Obstructed Largely Obstructed Totally Obstructed

19 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 19 Free-Space Model No obstructions in or “near” path Note: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path. Obstructions must be outside first “Fresnel Zone” - See next slide

20 9/21/2015© 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

21 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 21 Partially Obstructed Path Model Obstructions in, but not occluding, path Note: Path is elliptical (Fresnel) volume surrounding line-of-sight ray path. Obstructions inside first “Fresnel Zone” but not by more than 40%

22 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 22 Highly Obstructed Path Model Obstructions in, but not occluding, path Note: 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

23 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 23 Fully Obstructed Path Model Obstructions occluding, path Note: 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.

24 9/21/2015© 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

25 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 25 Loss Calculation #1 Let f = 146 MHz, d = 10 km Loss dB = 32.4 + 20 log 10 (146) + 20log 10 (10) Loss dB = 32.4 + 43.3 + 20 = 95.7 dB The receiver signal strength for 40-Watts is: P r [dBm] = 46 - 95.7 = - 49.7 dB Conclusion: 10 km free-space signal path is okay at this frequency.

26 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 26 Loss Calculation #2 Let f = 146 MHz, d = 100 km Loss dB = 32.4 + 20 log 10 (146) + 20log 10 (100) Loss dB = 32.4 + 43.3 + 40 = 115.7 dB The receiver signal strength for 40-Watts is: P r [dBm] = 46 - 115.7 = - 69.7 dB Conclusion: 100 km free-space signal path is too long for reliable reception at this frequency.

27 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 27 1300 MHz Digital Data Sensitivity

28 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 28 Loss Calculation #3 Let f = 1300 MHz, d = 20 km Loss dB = 32.4 + 20 log 10 (1300) + 20log 10 (20) Loss dB = 32.4 + 63.3 + 26 = 120.7 dB The receiver signal strength for 10-Watts is: P r [dBm] = 40 - 120.7 = - 80.7 dB Conclusion: 20 km free-space signal path is too long at this frequency.

29 9/21/2015© 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.

30 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 30 Ex. #3 with 16 dB Antenna Gain Let f = 146 MHz, d = 100 km Loss dB = 32.4 + 20 log 10 (146) + 20log 10 (100) Loss dB = 32.4 + 43.3 + 40 = 115.7 dB The receiver signal strength for 40-Watts is: P r [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.

31 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 31 Diffraction Model http://www.mike-willis.com/Tutorial/PF7.htm

32 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 32 Obstruction Loss Model http://www.mike-willis.com/Tutorial/PF7.htm Horizontal axis is

33 Obstruction of Fresnel Zone 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 33 Building roof at edge of 1st Fresnel zone

34 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 9/21/2015© 2009 Raymond P. Jefferis III Lect 06 - 34

35 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 35 Obstruction Loss Calculations http://www.mike-willis.com/Tutorial/PF7.htm Use value of , enter previous graph, and read loss in dB, or calculate knife-edge loss J(  )as:

36 9/21/2015© 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

37 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 37 More Exact Fresnel Loss Fresnel loss, where C = Fresnel cosine integral S = Fresnel sine integral

38 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 38 Fresnel Loss Magnitude Square root of F(  ) * F * (  )

39 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 39 Logarithmic Fresnel Loss This loss should be added to the free-space propagation loss in dB.

40 9/21/2015© 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.

41 Fresnel Radius(f) For d 1 = d 2 = d the Fresnel radius at the midpoint becomes: 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 41

42 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 42 LOS Distance Attenuation

43 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 43 Path Shaded by LOS Attenuation 1300 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)

44 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 terrain 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 44

45 Example Point A Antenna antlat = 40.057880 N antlon = 75.598214 W anthgt = 40.0 [meters] Point B Fire truck trklat = 40.123831 N trklon = 75.513662 W trkhgt = 2.5 [meters] 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 45

46 Radiation Path Black line on terrain at lower right is path 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 46

47 Close-up of Radiation Path Point A (Left) is on a hilltop Point B (Right) is in a developed area 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 47

48 Path and Cross-Section 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 48

49 Discussion of Programming Discussion in class... 9/21/2015© 2009 Raymond P. Jefferis III Lect 09 - 49


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