Propagation Mechanism

Transcription

1 Propagation Mechanism ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 1

2 Propagation Mechanism Simplest propagation channel is the free space: Tx free space Rx In a more realistic scenario, there may be dielectric and conducting obstacles (Interacting Objects, IOs) in the medium. If the IOs have a smooth surface waves are reflected and part of the energy is absorbed by the IO and part of the energy penetrates the IO (transmission). If the IOs have a rough surface waves are diffusely scattered. Waves can also be diffracted at the edges of the IOs. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 2

3 Reflection and Transmission Snell s Law Homogeneous plane wave incident on to a dielectric halfspace. Isotropic material relative permeability μ r =1. Complex dielectric constant, δ: conductivity carrier frequency Lossy medium Angle between the plane wave and the halfspace is Θ e Reflection angle: Transmission angle: ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 3

4 Snell s Law If the magnetic field component is paralel with the boundary TM (Transversal Magnetic) case If the electric field component is paralel with the boundary TE (Transversal Electric) case ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 4

5 Reflection and Transmission Reflection and transmission coefficients for: TE waves: TM waves: Brewster Angle: angle at which no reflection occurs in the medium of origin which only occurs for vertical polarization. For air-water interface, the Brewster angle is 53 o for light. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 5

6 Transmission through a Wall (Layered structures) By summing the partial waves, we can calculate: Total transmission coefficient: Total reflection coefficient: Electrical length of the wall T 1, T 2, ρ 1 and ρ 2 : transmission and reflection coef.s of the boundaries. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 6

7 Transmission through a Wall Example: Find T and ρ for a 50-cm-thick brick wall at 4-GHz carrier freq. for perpendicularly incident waves. Θ r = Θ t = 0, f = 4 GHz (λ = 7.5 cm), relative permittivity ε r = 4.44, conductivity σ=0. air-brick interface electrical length of the wall Total transmission and reflection coef.s brick-air interface >1? T gives the amplitude ratio of the fields only. >1 does not violate conservation of energy. (Verify that T 2 + ρ 2 =1?) (Verify that 1+ ρ =T (for TE), TM?) (Power transmitted to the other side?) ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 7

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9 The d -4 Power Law A direct path (Line-of-sight, LOS) and a ground reflection path is present. An interference pattern takes form at the receiver: depending on distance two waves may add constructively or destructively (up to ~ 90 m) For distances greater than received power becomes: loss: d -n (n = 4 for this example) ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 9

10 The d -4 Power Law A very simplified scenario, generally does not fit to practice. 1. Figure shows a decay of d -2 before d break and d -4 after d break, 2. In practice, transition never occurs that sharp at d break, 3. n = 4 is not a universal decay exponent, 1.5 < n < 5.5 is possible. (n = 4 can at best be a mean value for n) 4. In practice, there is a second break point beyond with n > 6, due to the curvature of the earth (LOS is not possible for d>25km) ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 10

11 Diffraction Single Screen or Wedge Screen prevents waves passing to the rhs. There is still «spherical waves» in the «shadow zone» Instead of paralel waves, think about point sources lined up (Huygen s principle). Screen obstruct part of these sources, but the remaining continue to radiate. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 11

12 Diffraction Single Screen or Wedge (Knife-edge) Screen in the middle of height h s. Calculate diffraction coefficient. From the geometry, Fresnel parameter is calculated: Then, Fresnel integral: Then, the total field at the receiver: (calculated numerically!) field strength when there is no screen. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 12

13 Diffraction Single Screen or Wedge (Knife-edge) Diffraction loss due to a knife-edge: (Think about the relation bw. the graph and Fresnel zones (slide 14). ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 13

14 Diffraction Single Screen or Wedge (Knife-edge) Example: Compute the diffraction coef. if d TX = 200 m, d RX = 50 m, h TX = 20 m, h RX = 1.5 m, h s = 40 m, f c = 900 MHz. How much power loss in db? (-34dB) ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 14

15 Fresnel Zones The impact of an obstacle can be assessed qualitatively by the concept of Fresnel Zones. Concentric ellipsoids foci: TX and RX for an ellipsoid: distance «TX ellipsoid RX» is the same for all points on it. Ellipsoids with «TX ellipsoid RX» run length greater than LOS by integer multiple of λ/2 are called Fresnel ellipsoids. λ/2 phase shift between each neighbouring ellipsoid is π rad. ith Fresnel ellipsoid results in a phase shift of «i π» rad. In general, if an obstruction does not block the volume contained within the first Fresnel zone, then the diffraction loss may be neglected. Rule of thumb (for microwave links): as long as 55% of the first Fresnel zone is kept clear, then further Fresnel zone clearance does not significantly alter the diffraction loss. Otherwise, use the calculations at the previous slides. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 15

16 Diffraction by Multiple Screens What happens if there are two or more obstacles causing diffraction? Mathematically difficult problem. There are many models in the literature. Epstein-Petersen method For each screen, calculate the diffraction loss seperately: Place a virtual TX and RX on the tips of the screens to the left and right of the considered screen. Calculate the diff. loss due to a single screen Add (?) all losses together. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 16

17 Diffraction by Multiple Screens Deygout s method: Determine the attenuation bw. TX and RX if only the i th screen is present (for all i) Screen with max. attenuation main screen, call it i ms Compute the attenuation bw. Tx and the tip of the main screen caused by the j th screen (1 j < i ms ) Do the same bw. the main screen and the RX (i ms < j (no. screens)) Add up (?) the losses from all considered screens. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 17

18 Diffraction by Multiple Screens (Deygout s method) Example: 3 screens 20 m apart from each other, and 30, 40 and 25 m high. TX 1st screen: 30m, last screen RX: 100 m. h TX = 1.5m, h RX = 30m. Calculate 900 MHz by Deygout s method. i. Attenuation due to Screen 1: ii. Repeat for all screens: main screen: Screen 2 iii. Calculate the loss bw. Tx and a «virtual RX» at the top of Screen 2 iv. Do the same for Screen 3 (TX on Screen 2) v. Add them up: ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 18 ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 18

19 Scattering A wavefront imping on a «smooth» surface is reflected. A wavefront imping on a «rough» surface is scattered. Height variability is taken as random and is assumed to be smaller than wavelength of the wave. Kirchoff Theory pdf of the surface amplitude (height) is assumed to be Gaussian. Different scattering points do not influence each other, i.e. they do not cast shadow onto each other. Rayleigh roughness effective reflection coef. of rough surface reflection coef. of smooth surface standard deviation of surface height profile angle of incidence ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 19

20 Waveguiding Street canyons, corridors and tunnels due to large buildings. Results in lower propagation exponents d -n (How come n can be < 2?) ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 20

21 Effect of rain and vegetation Attenuation due to rain Presence of raindrops can severely degrade the link perfoamcen Attenuation depends on drop shape/size, rain rage and TX freq. Estimated rain attenuation: attenuation (db/km) rain rate (mm/hr) a, b: depends on drop size and freq. Plants (trees) may cause diffraction and scattering, depends on the speed of wind. Trees cause considerable attenuation when fall into the first Fresnel zone. ELE 492 FUNDAMENTALS OF WIRELESS COMMUNICATIONS 21