O. Kilic EE 542

TOPIC 8:Propagation Mechanisms

EE 542Fall 2008

O. Kilic EE 542

References

Saunders, S. R. Antennas and Propagation for Wireless Communication Systems, Wiley

Internet, Google search under Fresnel, Knife Edge, Propagation

O. Kilic EE 542

Introduction

We will discuss how the basic parameters of antennas can be used together with an understanding of propagation mechanisms.

The objective is to calculate the range of a wireless communication system.

We will introduce approximate models which are of idealized nature for simplicity.

O. Kilic EE 542

Path Loss

The path loss between a pair of antennas is the ratio of the transmitted power to the received power, usually expressed in dB.

It includes all of the possible elements of loss associated with interactions between the propagating wave and any objects between the Tx and Rx antennas.

In order to define the path properly, the losses and gains in the system must be considered.

O. Kilic EE 542

Elements of a Wireless Communication System

T T R T T RR

T R

P D D P G GP

L LL L= =

All gains G and losses L are expressed as power ratios, and powers expressed as watts.

** Note that the book definition is wrong!!

O. Kilic EE 542

Effective Isotropic Transmitted Power (EIRP)

( )

;

10log 10log

T TT T TI

T

R T T RRI R

R

TI T T R

RI R

TIdB

RI

RdB R

P DEIRP P G P

L

P P G GP P

G L

P P G GL

P P

PL L

P

P EIRP G L dBW

= =

=

= = = =

= +

The advantage of expressing powers in terms of EIRP is that the path loss, L can be expressed independently of system paramettransmitted and received EIRP.

ers by defining it as the ratio of

O. Kilic EE 542

A Note on DecibelsUnit Reference Power Application

dBW 1 W Absolute power

dBm 1 mW Absolute powerP [dBW] = P [dBm] - 30

dB any Gain or loss of a network

dBi Power radiated by an isotropic reference antenna

Gain of an antenna

dBd Power radiated by a half-wave dipole

Gain of an antenna0 dBd = 2.15 dBi

O. Kilic EE 542

Example 1

A base station transmits a power of 1 W with a gain of 12 dBd in the direction of a mobile receiver, which has a gain of 0 dBd. The mobile receiver has a sensitivity of -104 dBm.

a) Determine the effective isotropic radiated power

b) Determine the maximum acceptable path loss

O. Kilic EE 542

Solution

0 14.15 2.15 ( 134)148.3 dB

T T R RL P G G P= + + = + + =

Quantity Value in original units

Value in consistent units

PT 1 W 0 dBW

GT 12 dBd 14.15 dBi

GR 0 dBd 2.15 dBi

PR -104 dBm -134 dBW

1.415

0 14.15 14.15 dBW

10 26 W

T TEIRP P G

EIRP

= += + =

= =a)

b)

O. Kilic EE 542

Propagation Modeling

The main goal of propagation modeling is to predict the path loss L as accurately as possible, so that

The range of a radio system can be determined accurately before installation.

O. Kilic EE 542

Why consider propagation?1.Could my system operate correctly?

Wanted signal intensity/ range/ coverage? Accounting for

Required quality Required distance/ area/ volume Required geographic/ climatic region Required time period

2. Could my system suffer unacceptable interference from other systems?3. Could my system produce such interference to other systems?

Strength of unwanted interfering signals? Received by the system at hand Radiated by the system at hand

Physical area of interference? Time period of interference? Degradation in quality of wanted signals?

O. Kilic EE 542

Simplest possible case Single unobstructed path Valid approximation for satellite communications The transmitter uses power PT. At distance d, what is the (average) received power PR? Friis free space equation (H.T. Friis, 1946):

Simplest Case: LOS: Free space propagation

O. Kilic EE 542

Friis Equation

In Friis equation:

= Wavelength ( = c/f) f = Frequencyc = Speed of lightGT =Transmitter antenna gainGR = Receiver antenna gaind = Distance

Recall that antenna gain measures the ability of the antenna to focus in a particular direction.

2

4R T T RP PG G

R

= We derived before in antenna discussions

* The dependence on arises from the effective aperture of an isotropic antenna.

O. Kilic EE 542

Free Space Path Loss

( )

2

2 2

( )

4

4 4

420log

32.4 20log 20log

= = = =

= = + +

R T T R

T T RF

R

F dB

km MHzF dB

P PG GR

PG G R RfLP c

RL

L R f

The free space path loss increases by 6 dB for each doubling either in frequency or distance. For most practical applications the total path loss will be in excess of the free space loss.

O. Kilic EE 542

Friis Equation Limitations

The Friis equation breaks down for small d. It is valid only in the far field region. Far field threshold is df = 2D2/ Define a reference point d0 >> df and measure (or predict) PR(d0).

Then:

where n is the path loss exponent and for free space equals to 2. But for other medium it is usually larger.

00

( ) ( )n

R R

dP d P dd

=

O. Kilic EE 542

Example 2

The communication system described in Example 1 is operated under free space propagation conditions at 900 MHz. Determine its maximum range.

O. Kilic EE 542

Solution

( )

( )

32.4 20log 20log

32.4 20loglog

20148.3 32.4 20log900 2.84

20693 km

= + + = =

km MHzF dB

MHzF dB

L R f

L fR

R

This is impractically large. In practice, other factors will reduce the range substantially, so more reasonable loss factors need to be considered. Free space path loss serves as a first cut minimum loss for a given range.

O. Kilic EE 542

Example 3

A satellite at a distance of 40,000 km from a point on the earths surface radiates a power of 2 W from an antenna with a gain of 17 dB in the direction of the observer. Find the radiated power density at the receiving point, and the power received by an antenna with an effective area of 10 m2.

O. Kilic EE 542

Solution

( ),

-15 222 7

1.710

2

-14

2 50=4.97 10 W/m

4 4 4 10

where

10 10 50

143 dBW/m

4.97 10 W

133 dBW

T dB

dB

T T

G

T

dB

R e

R

P GS

R

G

S

P A S

P

= =

= = =

=

= = =

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Example 4

Consider the satellite in Example 3 and assume it operates at a frequency of 11 GHz. The receive antenna has a gain of 52.3 dB. Find the received power.

O. Kilic EE 542

Solution

R

F

R

(dBW)

10log2 17 20 dBW

G 52.3 dB

4L 20log 205.3 dB

P 20 52.3 205.3 133 dBW

R R F

T T

P EIRP G L

EIRP P G

R

= + = + = + =

= = =

= + =

Same as in Example 3

O. Kilic EE 542

ObservationRemember the relation between effective area and gain

of a receive antenna:

2

4 eR

AG

=

In Example 2, Ae = 10 m2.

2 28

9

( )

4 10 40168862.2

3 1011 10

10log 52.3 dB

R

R dB R

Gcf

G G

= = = = =

Therefore the receive antennas in both examples are equivalent.

O. Kilic EE 542

Propagation Modes

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Effects on Propagation

Atmospheric

Terrain

zAbsorptionzRefractionzDuctingzRain Scattering

zReflectionzDiffractionzTerrain Scattering

O. Kilic EE 542

Propagation Mechanisms

Usually there is no line-of-sight path. Need to consider other mechanisms: Reflection Diffraction Scattering

Reflection is fairly easy to model, diffraction and scattering are harder.

O. Kilic EE 542

Reflection/Refraction Phenomena

Dense air

Thin air

Electromagnetic waves travel at different speeds in different media.

Velocity of light waves is more in media of lower refractive index.

This causes the waves to bend from normal to surface, when it

travels from medium of higher refractive index to lower.

O. Kilic EE 542

Reflection

Occurs when the wave encounters an object with large dimensions.The wave is partially reflected and partially transmitted (refracted). Proportions and angles depend on the materials and the surface Incident and reflected angles are equal: i = r. Refracted wave follows Snell's law:

n1 sin 1 = n2 sin 2

where n is the refractive index, r rn =

O. Kilic EE 542

Definitions: Planes of Incidence and the Interface and the polarizationsPerpendicular (horizontal) polarization sticks out of or into the plane of incidence.

Plane of the interface (here the yz plane) (perpendicular to page)

Plane of incidence(here the xy plane) is the plane that contains the incident and reflected k-vectors.

ni

nt

ikG

rkG

tkG

i r

t

Ei Er

Et

Interface

x

y

zParallel (vertical)polarization lies parallel to the plane of incidence.

Incident medium

Transmitting medium

O. Kilic EE 542

Reflection - Terrain Effect

Direct beam

Reflected beam

z Radio waves are reflected by ground, bulidings water sheet over lakes, rivers, and sea etc.

z During Reflection , the waves suffer a loss, defined by reflection coefficient.

z At receiver, energy arrives from direct and reflected path ( causing multipath).

z If the two waves are in phase, there is an enhancement of the signal.z If the two waves are out of phase , a cancellation occurs ( disrupting

transmission).

O. Kilic EE 542

Parallel(Vertical)polarization

Perpendicular(Horizontal)polarization

-1( )

Note that book has error

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Reflection Coefficient for Earth

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Reflection and Transmission Coefficients for Dry Ground

f = 100 MHz

Brewster angle

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Reflection and Transmission Coefficients for Wet Ground

f = 100 MHz

Pseudo Brewsterangle

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Some Properties of Reflection Notice that the reflection coefficient for the

vertically polarized (parallel) wave, v decreases and goes to zero at one angle Brewster angle.

For highly conducting medium, v never quite goes to zero but is minimal pseudo Brewster angle.

Notice that at grazing incidence; i.e. = 90o, the reflection coefficient for both polarizations have unit amplitude. (h = v = -1)

O. Kilic EE 542

Reflection Coefficient for Earth

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Reflection Coefficient (h and v pol)

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Reflection from a Flat Ground

hT

hR

D

DT DR

direct

reflectedi

Free space model no longer applies.

O. Kilic EE 542

Modification of Friis Equation for Planar Earth

For distances less than tens of km, the earths curvature can be neglected and earths surface can be assumed flat.

Typically, D >> hT, hR. Then is very small:

0, 90 90o oi = This is known as grazing incidence, and || = 1

2

2T R

R T T R

h hP PG GD

=

O. Kilic EE 542

Planar Earth Approximation Derivation

12

12

22 2

2

22 2

2

2;

( )( ) 12

( )( ) 12

,2 2 4;

dir o

jref o

ref dir

T Rref T R

T Rdir T R

T R

T R T Rref dir

E E u

E E e u L

L L L

h hL D h h DD

h hL D h h DD

where D h hh h h hL L L LD D

== =

= + = + + + = + +

= = = =

GG

O. Kilic EE 542

Planar Earth Approximation Derivation

4

4

22

22

Assume grazing angle :1

1

22 sin

24 sin ;4

24 sin4

= = =

= + = = = = =

G

G G G

G

T R

T R

h v

h hjD

ref o

h hjD

R dir ref o

T RR o

T RR dir dir T T R

T RR T T R

E E e u

E E E uE e

h hE ED

h hP P P PG G Friis eqnD R

h hP PG GR

2

2

2 2, sin ( ) .

T RT T R

T R

h hPG GD D

where h h D and is used

O. Kilic EE 542

Plane Earth Loss2

2T R

R T T R

h hP PG GD

=

( ) ( )( ) 10log

( ) 10log( ) 10log( ) 20log 20log 40log( )

R

T

P T R T R

PL dBP

L dB G G h h D

= = + + +

Differs from free space loss: No frequency dependence (as a result of the assumption hT, hR

O. Kilic EE 542

Earths Curvature Effects

When the separation between the Transmit and Receive antennas are large, one needs to account for the Earths curvature.

Question: What is the maximum range that Tx and Rx antennas can see each other over a smooth Earth?

O. Kilic EE 542

Earths Curvature Effects

a

hThR

dT dR

Maximum range is at the horizon; i.e. the radio ray is tangent to the Earth.

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Max. Range for Spherical Eartha = Earths radius = 6370 km >> hT, hR

( )

( )

2 2

max

2 1 2 12 4

1 2

, 2

2

= + = + +

= + = +

T TT T T T

TT T

R R

T R T R

h hd a h a ah aha a

h d ahaSimilarly d ah

d d d a h h

a

dT

hThR

dR

O. Kilic EE 542

Modification for Plane Earth

Earths curvature affects the total received field calculations in the flat earth model.

The effective height of the antennas over the curved earth need to be modified to use the plane earth model.

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Modification for Plane Earth

a

hR

hT

hT

hR

hT

hR

2 2

2 2

2

2

12

2''

R R

R R

TT

T T T

R R R

h d a a

d da aa a

dha

h h hh h h

= + = +

== =

effective height of antennas

dT

dR

O. Kilic EE 542

Ground Roughness

So far a smooth reflecting surface was assumed. This results in specular reflection at the point

where the transmitted wave hits the Earths surface.

When the surface is rough, the specularreflection assumption is no longer valid.

This results in diffuse reflection. Furthermore, random nature of the surface results in unpredictable situations.

O. Kilic EE 542

Ground Roughness Encounters with a rough surface is more

appropriately described as a scattering mechanism rather than reflection.

This implies that waves reflect back in multiple directions.

Only a small fraction of the incident energy will be scattered in the direction of the receiver.

Therefore, the reflected term form the ground may make a negligible contribution to the received signal.

O. Kilic EE 542

O. Kilic EE 542

What is rough?

A surface that might be considered rough at certain frequencies and incidence angles may be smooth at others.

Rayleigh criterion is a measure for roughness.

Calculate the range of path differences from the scattered waves from the surface.

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What is rough?2

2 cos( )i

l

l h

= =

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Rayleigh Criterion

0, ( . . ) .If i e h thesurfaceappears smooth

,If reflected rays cancel each other =

Smooth Surface:

Extreme Roughness:

Practical Definition:

( )2

8cos i

rough surface

h

Function of wavelength and incidence angle.At grazing incidence surfaces seem smoother.

O. Kilic EE 542

O. Kilic EE 542

Practical Definition of rough

For typical mobile applications, the incident angle is almost 90o; i.e. the station height

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Random Roughness

The Rayleigh criterion for a randomly rough surface is expressed by the standard deviation:

4 cos( )s iC =

< 0.1 smooth surface, specular reflection> 10 highly diffuse, reflected term negligibleC

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Equivalent Reflection Coefficient

Roughness factor, f(s)2

24 cos( )1

22( )( )

s iC

s

eq s s

f e eR f

= ==

O. Kilic EE 542

Roughness Factor

Roughness for various values of /

O. Kilic EE 542

Irregular Terrain, Diffraction

Mobile radio systems have a variety of applications over a variety of coverage areas.

Obstacles rather than interfaces may be present in the line of sight.

Propagation behind edges or beyond the horizon is of interest. Diffraction

O. Kilic EE 542

Obstacles and Irregular Terrain

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Interaction of em waves with a barrier - Diffraction

Not real

diffraction

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Fresnel Zone

a

a

ellipsoid circle

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Fresnel ellipsoids and Fresnel zones In studying radiowave propagation between two points A and B, the intervening space can be subdivided by a family of ellipsoids, known as Fresnel ellipsoids, all having their focal points at A and B such that any point M on one ellipsoid satisfies the relation:

2

AB MBAM +=+ n (1) where n is a whole number characterizing the ellipsoid and n = 1 corresponds to the first Fresnel ellipsoid, etc., and is the wavelength. As a practical rule, propagation is assumed to occur in line-of-sight, i.e. with negligible diffraction phenomena if there is no obstacle within the first Fresnel ellipsoid. The radius of an ellipsoid at a point between the transmitter and the receiver is given by the following formula:

2/1

21

21

+

=ddddnRn (2)

or, in practical units:

2/1

21

21)(

550

+= fdd

ddnRn (3)

where f is the frequency (MHz) and d1 and d2 are the distances (km) between transmitter and receiver at the point where the ellipsoid radius (m) is calculated.

O. Kilic EE 542

Fresnel EllipsoidsAn obstruction is generally considered to be significant when it impinges into the first Fresnel zone

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Fresnel Circles

R1

R2TX RX

r

o ro

diffracted waves

L2L1

R

L1 is constant as long as the tip of the vector traces the circle with radius R1. L2 is constant as long as the tip of the vector tracesthe circle with radius R2. Visualize that it is a cone.

TXR

1

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Fresnel Ellipses

TX RX

n rn

o ro

Rn

0 0 0

0 2

L r

L r l L L n

= += + =

Definition of ellipse: loci of points with a constant distance between two fixed points. The fixed points are called focal points.

Fresnel ellipse

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What it means The Fresnel circles are defined such that between two adjacent circles the

path difference between Tx and Rx is half a wavelength (n+rn)-(o-ro)=n/2

This defines the radius of each Fresnel circle:

Between each adjacent ring is a 180o phase difference; i.e. each adjacent ring is out of phase.

As the rays follow rings further outside, their power is less due to increased path length.

So the overall contribution oscillates as one moves along each ring: Power levels can be visualized as a function of Fresnel circles:

+ P1 > - P2 > + P3 > - P4 > .R1 < R2 < R3 < R4 < .

0 0 0 01 1

0 0 0 0

; ;n nn r rR R R R nr r

= = =+ +

subt

ract

add

subt

ract Power levels oscillate

O. Kilic EE 542

Knife Edge Diffraction Loss

Avoid obstructing the first Fresnel circle!

TX RX

L.O.S

h = -R1

z

RXL.O.S

z

TX h=0

ObstructionBelow LOS

Obstructionabove LOS

1

2h R

O. Kilic EE 542

Some Observations on KE Diffraction Loss

At v=0 grazing incidence, Loss = 6dB At v = -0.8 h/R1 = 0.56 56% of the

first Fresnel zone is clear. 6dB loss is avoided.

V = 0, grazing angle V = -0.8

R1Tx TxRx Rx

* In practice avoid obstruction to the first Fresnel zone.

O. Kilic EE 542

How do we use these circles and ellipses?

Given a terrain profile between Tx and Rx, Fresnel ellipsoids enable the designer to visualize the degree of obstructions along the path.

O. Kilic EE 542

Significance of Fresnel Zones

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Non-LOS propagation

When the 1st Fresnel zone is obstructed, non LOS propagation occurs.

An obstruction may lie to the side, above, or below the path. Examples: buildings, trees, bridges, cliffs, etc.

Obstructions that do not enter in the 1st Fresnel zone can be ignored. Often one ignores obstructions up to of the zone.

O. Kilic EE 542

Ideal Knife Edge Diffraction; i.eSharp Obstruction

hd1 d2

-

Note that h is the height of the obstruction; i.e. h>0 obstruction above LOSh

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Knife Edge Diffraction

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Knife Edge Diffraction - ExampleFind the radius of the first Fresnel zone for a wireless system operating

at 900 MHz when an obstacle is 5 km away from the transmitter along the LOS path, which is 6km long.

What is the acceptable height for the obstruction if the system would allow for 6dB loss.

5 km 1 km

O. Kilic EE 542

Solution0

0

60 0

1 60 0

56 5 1

19003

5 101 16.673 10

kmr km

cf MHz mf

rn R mr

== == = =

= = = =+

a)

11 2

20 2

2 23.62

Loss dB vRh v R R m

= = = = =

R1 R2

b) From the KE loss graph

h

O. Kilic EE 542

Ground Reflections and KE Diffraction

The reflections can be represented equivalently by image theory.Four possible scenarios need to be added:

dT dR

I

m

a

g

e

t

o

w

e

r

I

m

a

g

e

t

o

w

e

r

O. Kilic EE 542

Contribution 1

dT dR

Contribution 2

dT dR

I

m

a

g

e

t

o

w

e

r

I

m

a

g

e

t

o

w

e

r

dT dR

dT dR

I

m

a

g

e

t

o

w

e

r

I

m

a

g

e

t

o

w

e

r

Contribution 3

Contribution 4

O. Kilic EE 542

Diffraction Around Real Objects

Most real-world obstructions are large in comparison to the wavelength - i.e. not knife edges. It is possible to solve the equations for idealized cases. Solutions for many objects and including reflection effects, loss from trees etc. rapidly become impractical. Therefore path loss prediction models are used. There are several models in general. These involve adding additional loss to the knife edge loss. Where there are multiple knife edges further models are available, the one currently used by the ITU-R for terrestrial links is a modified version of the Deygoutmodel.

O. Kilic EE 542

Atmospheric Effects

The lower part of the atmosphere (troposphere) is a region in which temperature decreases with height.

Tropopause separates troposphere from stratosphere, in which the air temperature seems to be constant with height.

O. Kilic EE 542

Atmospheric Effects

At frequencies > 30 MHz, three effects dominate the propagation of waves: localized refractive index fluctuations any abrupt changes in refractive index with

height ducting due to change in rate of decrease in

refractive index with height

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O. Kilic EE 542

O. Kilic EE 542

O. Kilic EE 542

Refraction How it Affects Wireless Communication

Radio beam propagated in free space follows a straight line

Radio beam propagated through earth's atmosphere becomes curved

Refraction causes loss of line of sight for the receiver

Refraction will also give some insight into fading phenomenon

O. Kilic EE 542

O. Kilic EE 542

RefractionK-factor

K = effective earth radius true earth radius

true earth

effective earth

k - factor is the scaling factor that helps to quantify the curvature of the radio beam

z True earth radius ( a ) = 6370 kmz Effective earth radius for a given atmospheric condition is the radius of

the fictitious earth which allows the microwave beam, to be drawn as a straight line .

O. Kilic EE 542

Ductingz Atmospheric refraction under certain conditions causes the microwave

beam to be trapped in an atmospheric ductz Ducting occurs by low-altitude, high density atmospheric layersz Ducting can also result into Multipath.

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Atmospheric attenuation

Starts becoming relevant above about 5 GHz Depends primarily, but not exclusively on

water vapour content of the atmosphere Varies according to location, altitude, path

elevation angle etc. Can add to system noise as well as

attenuating desired signal Precipatation has a significant effect

O. Kilic EE 542

Rain Scattering

z Scattering of Microwaves by rain is very important above 10 Ghz.

z The rain droplets size becomes appreciable in comparison to wavelength.

z These droplets cause scattering of microwave energy .

z The main effect of scattering is heavy attenuation in the path.

z The loss of horizontally polarized wave is higher than that for vertically polarized.

z Rain drops can also cause depolarization of microwave beam.

z Scattering and depolarization occur simultaneously, so effect is not additive.

O. Kilic EE 542

Ionospheric propagation

Most relevant up to about 30 MHz Many modes of propagation: a

complicated topic. Sporadic E can be important up to

about 70 MHz. (ITU-R P.534) Highly variable

O. Kilic EE 542

0676-0

H O2

H O2

102

10

10 1

10 2

1

10 3

2

5

5

2

5

2

5

2

5

2

FIGURE 5Specific attenuation due to atmospheric gases

S

p

e

c

i

f

i

c

a

t

t

e

n

u

a

t

i

o

n

(

d

B

/

k

m

)

352 52 2102101

Dry airDry airTotal

Frequency, f (GHz)

Pressure: 1 013 hPaTemperature: 15 CWater vapour: 7.5 g/m3

TOPIC 8:Propagation MechanismsReferencesIntroductionPath LossElements of a Wireless Communication SystemEffective Isotropic Transmitted Power (EIRP)A Note on DecibelsExample 1SolutionPropagation ModelingWhy consider propagation?Simplest Case: LOS: Free space propagationFriis EquationFree Space Path LossFriis Equation LimitationsExample 2SolutionExample 3SolutionExample 4SolutionObservationPropagation ModesPropagation MechanismsReflectionReflection Coefficient for EarthReflection and Transmission Coefficients for Dry GroundReflection and Transmission Coefficients for Wet GroundSome Properties of ReflectionReflection Coefficient for EarthReflection Coefficient (h and v pol)Reflection from a Flat GroundModification of Friis Equation for Planar EarthPlanar Earth Approximation DerivationPlanar Earth Approximation DerivationPlane Earth LossEarths Curvature EffectsEarths Curvature EffectsMax. Range for Spherical EarthModification for Plane EarthModification for Plane EarthGround RoughnessGround RoughnessWhat is rough?What is rough?Rayleigh CriterionPractical Definition of roughRandom RoughnessEquivalent Reflection CoefficientRoughness FactorIrregular Terrain, DiffractionObstacles and Irregular TerrainInteraction of em waves with a barrier - DiffractionFresnel ZoneFresnel EllipsoidsFresnel CirclesWhat it meansKnife Edge Diffraction LossSome Observations on KE Diffraction LossHow do we use these circles and ellipses?Significance of Fresnel ZonesNon-LOS propagationIdeal Knife Edge Diffraction; i.e Sharp ObstructionKnife Edge DiffractionKnife Edge Diffraction - ExampleSolutionGround Reflections and KE DiffractionDiffraction Around Real ObjectsAtmospheric EffectsAtmospheric Effects