communications link analysis - hacettepe universitytoker/ele492/2. propagationmechanism.pdf ·...

$: Communications Link Analysis - Hacettepe Universitytoker/ELE492/2. PropagationMechanism.pdf · Diffraction loss due to a knife-edge: (Think about the relation bw. the graph and Fresnel$
PropagationMechanism

ELE 492 – FUNDAMENTALS OF WIRELESS COMMUNICATIONS 1Spring 2017

Propagation Mechanism

ELE 492 – FUNDAMENTALS OF WIRELESS COMMUNICATIONS 2

Simplest propagation channel is the free space: Tx → free space → Rx

In a more realistic scenario, there may be dielectric and conducting obstacles (InteractingObjects, IOs) in the medium.

If the IOs have a smooth surface → waves are reflected and part of the energy is absorbed bythe 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.

Spring 2017

Reflection and Transmission – Snell’s Law Homogeneous plane wave incident on to a dielectric halfspace. Isotropic material

relative permeability μr=1.

Complex dielectric constant, δ:

Angle between the plane wave and the halfspace is Θe

Reflection angle:

Transmission angle:


conductivity

carrier frequency

Lossy medium

Spring 2017

Snell’s LawIf 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


Reflection and Transmission


Reflection and transmission coefficients for:

TE waves:

TM waves:

Brewster Angle: angle at which no reflectionoccurs in the medium of origin which onlyoccurs for vertical polarization.For air-water interface, the Brewster angleis 53o for light.

Spring 2017

Transmission through a Wall (Layered structures)

By summing the partial waves, we can calculate:

Total transmission coefficient:

Total reflection coefficient:

T1, T2, ρ1 and ρ2: transmission and reflection coef.s of the boundaries.


Electrical length of the wall

Spring 2017

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. Θr = Θt = 0, f = 4 GHz (λ = 7.5 cm), relative permittivity εr = 4.44, conductivity σ=0.

air-brick interface electrical length of the wall

Total transmission and reflection coef.s

brick-air interface

(Verify that |T|2+|ρ|2=1?)

(Verify that 1+ ρ =T (for TE), TM?)

(Power transmitted to the other side?)


>1 ?T gives the amplitude ratio of the fields only.>1 does not violate conservation of energy.

Spring 2017

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:


10 10 10 10 10loss: d-n (n = 4 for this example)

Spring 2017

10 10 10 10 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 dbreak and d-4 after dbreak,

2. In practice, transition never occurs that sharp at dbreak,

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).


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.


Diffraction – Single Screen or Wedge (Knife-edge)

Screen in the middle of height hs. Calculate diffraction coefficient.

From the geometry, Fresnel parameter is calculated:

Then, Fresnel integral:

Then, the total field at the receiver:


field strength whenthere is no screen.

(calculated numerically!)

Spring 2017


Diffraction loss due to a knife-edge:

(Think about the relation bw. the graph and Fresnel zones (slide 14).



Example: Compute the diffraction coef. if dTX = 200 m, dRX = 50 m, hTX = 20 m, hRX = 1.5 m, hs = 40 m, fc = 900 MHz.


How much power loss in dB? (-34dB)

Spring 2017

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.


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.


Diffraction by Multiple Screens Deygout’s method: Determine the attenuation bw. TX and RX if only the ith screen is present (for all i)

Screen with max. attenuation → main screen, call it ims

Compute the attenuation bw. Tx and the tip of the main screen caused by the jth screen (1 ≤ j < ims)

Do the same bw. the main screen and the RX (ims < j ≤ (no. screens))

Add up (?) the losses from all considered screens.


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. hTX = 1.5m, hRX = 30m. Calculate attenuation @ 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:

Diffraction by Multiple Screens (Deygout’s method)



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. theydo not cast shadow onto each other.


effective reflectioncoef. of rough surface

reflection coef. of smooth surface

standard deviation of surface height profile

angle of incidence

Rayleigh roughness

Spring 2017

Waveguiding Street canyons, corridors and tunnels due to large buildings.

Results in lower propagation exponents d-n

(How come n can be < 2?)


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:

Plants (trees) may cause diffraction and scattering, depends on the speed of wind.

Trees cause considerable attenuation when fall into the first Fresnel zone.


attenuation (dB/km)rain rate (mm/hr)

a, b: depends on drop size and freq.

Spring 2017

communications link analysis - hacettepe universitytoker/ele492/2. propagationmechanism.pdf ·...

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