Chapter 4 Mobile Radio Propagation: Large-Scale Path Loss...ECE 8708 Wireless Communications :...

Yimin Zhang, Villanova University 1

ECE 8708 Wireless Communications : Propagation – Large-Scale Path Loss

Chapter 4Chapter 4

Mobile Radio Propagation: Mobile Radio Propagation: LargeLarge--Scale Path LossScale Path Loss

Yimin Zhang, Ph.D.Department of Electrical & Computer Engineering

Villanova Universityhttp://yiminzhang.com/ECE8708



OutlinesOutlines

Propagation Model - Attenuations

- Free Space

- Reflection

- Diffraction

- Scattering

Log-Normal Shadowing

Practical Link Budget Design



Propagation ModelsPropagation Models



Mobile Radio PropagationMobile Radio Propagation

• Mobile radio channel is an important controlling factor in wireless communication systems.

• Transmission path between transmitter and receiver can vary in complexity.

• Wired channels are stationary and predictable, whereas radio channels are extremely random and have complex models.

• Modeling of radio channels is done in statistical fashion based on measurements for each individual communication system or frequency spectrum.




• Large scale propagation models - To predict the average received signal strength over large T-R separation distances (several hundreds or thousands of meters).

Typically, the local average received power is computed by averaging signal measurements over a measurement track of 5 to 40 wavelengths.

• Small scale propagation models (or fading models) - To characterize the rapid fluctuations of the receiver signal strength over very short distances ( a few wavelengths) or short time durations (on the order of seconds).

In small scale fading, the received signal power may vary by as much as three to four orders of magnitude (30 to 40 dB).



Electrical FieldThe electric field is expressed as a vector E

and its magnitude is given by

The unit of electrical field is volts/meter.

Electrical Field

zyx EzEyExE ++=

222zyx EEEEE ++==



The standard unit of power is Watt, but dBm is more commonly used.

P (dBm) = 10 log10 [ P (mW)]

P (mW) P (dBm)

10 101 0

10-1 -1010-2 -20

Minimum usable signal strength to be received at a base station is typically between -90 dBm -100 dBm.

Units of Received Signal StrengthUnits of Received Signal Strength



• Free Space Propagation: Transmitter and receiver have a clear, unobstructed LOS path between them.

• Reflection: From the surface of the earth and from buildings and walls. Usually dimensions of reflecting object are much greater than wavelength.

• Diffraction: Bending of electromagnetic waves around sharp edges such as, sharp towers or peaks.

• Scattering: Due to objects in the medium that are small compared to wavelength and the number of objects is many (e.g., foliage, street signs, lamp posts, rain, shower).




Radiating Power to Electric FieldRadiating Power to Electric Field




)/(32

0

0 12

cos cdtj

cr

cedj

cdc

LiE −

+= ω

ωπεθ

Electric and magnetic fields launched from electric current i0

)/(3

2

220

0

4sin cdtj

c

c cedj

cdc

dj

cLiE −−

++= ω

θ ωω

πεθ

)/(2

0

4sin cdtjc ce

dc

dj

cLiH −

+= ω

φω

πθ

0=== θφ HHE r

In the near field (d is small), 1/d2 and 1/d3 terms may dominate.



When d is large, 1/d2 and 1/d3 terms becomes negligible, and only

survive.

Far field regions: regions far away from the transmitter satisfying

d >> df = 2D2/λ (df is called the Fraunhofer distance)

Additionally, to be in the far-field region, df must also satisfy

df >> λ and df >> D

Hereafter, we only consider propagation at far-field regions.


)/(2

0

0

4sin cdtjc ce

dj

cLiE −−

= ω

θω

πεθ

)/(0

4sin cdtjc ce

dj

cLiH −

= ω

φω

πθ



Effective isotropic radiated power (EIRP)

EIRP = Pt Gt

where Pt : Transmitter powerGt : Transmitter antenna gain

EIRP represents the maximum radiated power available from a transmitter in the direction of maximum antenna gain, compared to an isotropic radiator.

In practice, effective radiated power (ERP) is more commonly used

ERP = EIRP /1.64 or ERP (dB) = EIRP (dB) – 2.15

ERP represents the maximum radiated power available from a transmitter in the direction of maximum antenna gain, compared to a half-wave dipole antenna.


half-wave dipole antenna radiation pattern



Free Space PropagationFree Space Propagation

)W/m(44

EIRP 22

22 ηππE

dGP

dP tt

d ===

Power flux density:

where η=120π=377 (Ω) is the intrinsic impedance of free space.




ant

ant

ant

antr R

VR

VdP4

)2/()(22

==

It the receiver antenna is modeled as a matched resistive load to the receiver, then the receiver antenna will induce an voltage into the receiver which is half of the open circuit voltage at the antenna.



: effective aperture (of the receiver antenna)

Pr : Received powerD : Max dimension of transmitting antennaGr : Receiver antenna gainL : System loss factor (L ≥ 1, transmission lines etc,

but not due to propagation)λ= c / f = 3 • 108 / f : Wavelength

(units – f : Hz, c = 3 • 108 : meters/sec, λ: meters)


LdGGPAPdP rtt

edr 22

2

)4()(

πλ

==

T Rd

πλ

4

2r

eGA =



Path loss: signal attenuation as a positive quantity measured in dB, is defined as the difference (in dB) between the effective transmitted power and the received power, and may or may not include the effect of antenna gains.

(when antenna gains are included)

(when antenna gains are excluded)


−== 22

2

)4(log10log10)dB(PL

dGG

PP rt

r

t

πλ

−== 22

2

)4(log10log10)dB(PL

dPP

r

t

πλ



For free space propagation, the path loss exponent is 2.

or


2

00 )()(

−

=

dddPdP rr

−

=

−=

0

0

00

log20)W(001.0)W()(log10

log20)dBm()()dBm()(

dddP

dddPdP

r

rr

For simplicity of computations, the reference distance d0 for practical system using low-gain antennas in the 1-2 GHz region is typically chosen to be 1 m in indoor environment and 100 m or 1 km in outdoor environments.



Program: Given a transmitter produces 50W of power. If this power is applied to a unity gain antenna with 900 MHz carrier frequency, find the received power at a free space distance of 100 m from the antenna.

What is the received power at 10 km? Assume unity gain for the receiver antenna.

Program: Given a transmitter produces 50W of power. If this power is applied to a unity gain antenna with 900 MHz carrier frequency, find the received power at a free space distance of 100 m from the antenna.

What is the received power at 10 km? Assume unity gain for the receiver antenna.

ExampleExample

Solution: fc = 900 MHz λ = (3 • 108) / (900 • 106) = 0.333 m; Pt = 50 W; Gt = 1; Gr = 1; L = 1;

At d = 100 m

or

At d = 10 km

Solution: fc = 900 MHz λ = (3 • 108) / (900 • 106) = 0.333 m; Pt = 50 W; Gt = 1; Gr = 1; L = 1;

At d = 100 m

or

At d = 10 km

)mW(105.3)W(105.31100)4(

333.01150)4(

3622

2

22

2−− ×=×=

×××××

==ππ

λLd

GGPP rttr

)dBm(5.24))mW(log(10)dBm( −== rr PP

)mW(105.3)W(105.3 710 −− ×=×=rP )dBm(5.64)dBm( −=rP



Electric Properties of Material Bodies

Permittivity ε F/m Farads/m

Permeability µ H/m Henries/m

Conductivity σ S/m Siemens/m

For lossless dielectrics, the permittivity is real and can be expressed as ε = ε0 εr, where ε0 = 8.85 • 10-12 F/m is the permittivity of free space, and εr is relative permittivity (dielectric constant)

For lossy dielectric materials, ε = ε0 εr - j ε’, where

ReflectionReflection

fπσε

2'=




For lossydielectrics, σ may be sensitive to frequency.

For lossydielectrics, σ may be sensitive to frequency.

For a good conductor (f<σ/εrε0), terms εrand σ are generally insensitive to the operating frequency.

For a good conductor (f<σ/εrε0), terms εrand σ are generally insensitive to the operating frequency.




111 ,, σµε

222 ,, σµε

Dielectric 1

Dielectric 2

When a radio wave propagating in one medium impinges upon another medium having different electric properties, this wave is partially reflected and partially transmitted.

If the second medium is a perfect conductor, then all incident energy is reflected back into the first medium without loss of energy.

Speed of propagation

(at free space 3x108m/s)

Intrinsic impedance

(at free space 120π Ω)

iiiv

εµ1

=

i

ii ε

µη =

Et

Ei Er



(vertical polarization) (horizontal polarization)

Reflection Reflection -- PolarizationsPolarizations



Relationship between angles

Snell’s Law

When µ1=µ2

Ei Er

Et

θi θr

θi = θr

Reflection Reflection –– Angle RelationshipsAngle Relationships

)90sin()90sin( 2211 ti θεµθεµ −=−

ri θθ =

θt)90sin()90sin( 21 ti θεθε −=−



Reflection coefficient

Transmission coefficient

The value of Γ depends on polarization.

ΓEE

i

r =

ΓTEE

i

t +== 1

it

it

i

r

EEΓ

θηθηθηθη

sinsinsinsin

12

12|| +

−==

ti

ti

i

r

EEΓ

θηθηθηθη

sinsinsinsin

12

12

+−

==⊥

Reflection Reflection –– Magnitude RelationshipsMagnitude Relationships

Ei Er

Et

θi θr

θi = θr

θt

E-field in the plane of incident

E-field normal to the plane of incident



When the first medium is free space, and µ1=µ2

irir

irir

irr

i

irr

i

irt

irt

i

r

t

i

r

t

it

it

i

r

EEΓ

θεθε

θεθε

θεε

θ

θεε

θ

θεθθεθ

εµθ

εεµθ

εµθ

εεµθ

θηθηθηθη

2

2

2

2

0000

0000

12

12||

cossin

cossin

sincos1

sincos1

sinsinsinsin

sinsin

sinsin

sinsinsinsin

−+

−+−=

+−

−−=

+

−=

+

−

=+−

==


)90sin()90sin( tri θεθ −=−r

itt ε

θθθ2

2 cos1cos1sin −=−=

iri

iriΓθεθ

θεθ2

2

cossin

cossin

−+

−−=⊥



Parallel / Perpendicular /vertical polarization horizontal polarization

θi = θr θi = θrEi = Er Ei = –Er

Reflection from Perfect ConductorReflection from Perfect Conductor

Ei Er

θi θrθi = θr



Ei

θi θ0

ELOS

Er=Eg

ETOT = ELOS +Eg

R (receiver)

hr

ht

T (transmitter)

d

Ground Reflection Ground Reflection –– TwoTwo--Ray ModelRay Model




−=

cdtj

ddEtdE cωexp),( 00

General E-field expression in free space

Line-of-sight (LOS) wave

where d0 is a reference distance at the far-field.

−=

cdtj

ddEtdE c

'exp'

),'( 00LOS ω

Ground reflection wave

−=

cdtj

ddEΓtdE c

"exp"

),"( 00g ω




Assume horizontal E-field and perfect ground reflection, Γ= –1.

The combined E-field becomes

−−

−=+=

cdtj

ddE

cdtj

ddEtdEtdEtdE cc

"exp"

'exp'

),"(),'(),( 0000gLOSTOT ωω

The path difference between the LOS and reflected paths is

)2/()(/)(1)( 22222 dhhddhhddhh rtrtrt ++≈++=++

When d >> ht+hr,

2222 )()( dhhdhh rtrt +−−++=∆

)2/()(/)(1)( 22222 dhhddhhddhh rtrtrt −+≈−+=+−

dhh

dhh

dhh rtrtrt 2

2)(

2)( 22

=−

−+

≈∆




At some time, say at t=d”/c

"exp

')",( 0000

TOT ddE

cj

ddE

cdtdE c −

∆

==ω

Also note that, when d >> ht+hr,

Denote , then )/( cc ∆=∆ ωθ

ddE

ddE

ddE 000000

"'≈≈

( )

=−=

++−=

+−=

∆∆

∆∆∆

∆∆

2sin2cos22

sin1cos2cos

sin1cos)(

0000

2200

2200TOT

θθ

θθθ

θθ

ddE

ddE

ddE

ddEdE θ

θθθ2

22

sin21sincos2cos

:Ref

−=

−=




When is small ( )2/∆θ

220000

TOT22

2sin2)(

dkhh

ddE

ddEdE rt =≈

= ∆

λπθ

rad3.0)2/(2/ <∆=∆ ccωθ

cdhh

crtcc ωωθθ

≈∆

=≈

∆∆

222sin λλ

π rtrt hhhhd 203

20:athappensIt

≈>

Conclusion

λrthhd 20

>For

4

22

dhhGGPP rt

rttr =

)log20log20log10log10(log40)dB( rtrtr hhGGdP +++−=

For example, λ = 0.15m, ht = 50m, hr = 1.5m d = 10,000m =10km

For example, λ = 0.15m, ht = 50m, hr = 1.5m d = 10,000m =10km




http://home.earthlink.net/~loganscott53/Two_Ray_Propagation.htm



ExampleExample



FresnelFresnel Zone GeometryZone Geometry

hhTT

d1d1 d2d2

RR

When d1 , d2 >> h, h >> λ, the excess path length (difference between the direct path and the diffracted path) is

21

212

2 ddddh +

≈∆

The corresponding phase difference is

21

212

222

ddddh +

≈∆

=λπ

λπφ




21

21

ddddnrn +

=λ is the radius corresponding to the nth Fresnel zone,

which has nλ/2 path difference, or nπ phase difference to the LOS.

A rule of thumb is that as long as 55% (many materials say 60%) of the first Fresnel zone is kept clear, the diffraction loss will be minimal.



http://gbppr.dyndns.org:8080/fresnel.main.cgi





Observations:

• rn is dependent of the wavelength (or frequency).

• If d1 + d2 is fixed, rntakes smaller value when the position is closer to either end.

21

21

ddddnrn +

=λ



KnifeKnife--Edge Diffraction GeometryEdge Diffraction Geometry• Diffraction allows radio signals to propagate around the curved

surface or propagate behind obstructions.

• Based on Huygen’s principle of wave propagation.



KnifeKnife--Edge Diffraction GeometryEdge Diffraction Geometry



KnifeKnife--edge Diffraction Geometryedge Diffraction Geometry

The field strength at point R is a vector sum of the fields due to all of the secondary Huygen’s sources in the plane above the knife edge.




Approximate diffraction gain expressions

21

21 )(2ddddhv

λ+

=




21

21 )(2ddddhv

λ+

=




• When there are multiple obstructions, the problem becomes much more complicated.

• A simple approach is to use a single equivalent obstacle.



ExampleExample



• When a radio wave impinges on a rough surface, the reflected energy is spread out or diffused in all directions. (e.g., foliage).

• A surface is considered smooth if its minimum to maximum protuberance h is less than the critical height.

• The surface is considered rough if the protuberance is greater than hc.

ScatteringScattering

ich

θλ

sin8=

rough surface

ch

smooth surface

ch



• Modified reflection coefficient for rough surface

• Scattering loss factorby Ament

modified by Boithias


ΓΓ sρ=rough

−=

2sin8expλ

θπσρ ihs

−=

22 sin8sin8expλ

θπσλ

θπσρ iho

ihs I

σh: standard deviation of the surface height

I0: Bessel function of the first kind and zero order

σh: standard deviation of the surface height

I0: Bessel function of the first kind and zero order



Radar Cross Section (RCS) ModelRadar Cross Section (RCS) Model

object scattering on theincident waveradio ofdensity Power receiver ofdirection in wavescattered ofdensity Power RCS =

For a medium and large size building located 5-10 km away, RCS values is in the range of 14.1 to 55.7 dBm2.



Radar Cross Section (RCS) ModelRadar Cross Section (RCS) Model

RT

TTR

ddGPP

log20log20)4log(30]m RCS[dB)20log((dBi)(dBm)(dBm) 2

−−−+++=

πλ

223

2

)4(RCS

RT

TTR dd

GPPπ

λ ⋅=

PT = Transmitted power (mW)GT = Gain of transmitting antenna dT = Distance of scattering object from transmitterdR = Distance of scattering object from receiver



• Most radio propagation models are derived using a combination ofanalytical and empirical models.

• Empirical approach is based on fitting curves or analytical expressions that recreate a set of measured data.

• Advantages: Takes into account all propagation factors, both known and unknown.

• Disadvantages:New models need to be measured for different environment or frequency.

Practical Link BudgetPractical Link Budget



• Over many years, some classical propagation models have been developed, which are used to predict large-scale coverage for mobile communication system design.

LongLong--Distance Path ModelDistance Path Model

)()(/at lossPath 000 dPLdPPd T ==

)()(/at lossPath dPLdPPd T ==

n

dd

dPLdPL

=

00 )()(

+=

00 log10[dB])([dB])(

ddndPLdPL



LongLong--DistanceDistance Path ModelPath Model



• Long-distance path loss gives only the average value of path loss.• Surrounding environment may be vastly different at two locations

having the same T–R separation d.• More accurate model includes a random variable to account for

change in environment.

LogLog--Normal ShadowingNormal Shadowing

Xσ : Zero mean Gaussian random variable (dB)σ : Standard deviation (dB)

σσ XddndPLXdPLdPL +

+=+=

00 log10[dB])([dB])([dB])(

[dB])(-[dBm])([dBm])( dPLdPdP tr =



• Values of n and σare computed from measured data, using linear regression such that the difference between measured and estimated path losses is minimized in a mean square error sense over a wide range of measurement locations and T–R separations.




Some results of Q(z) and erf are listed in pages 647 and 649.

Q FunctionQ Function

∫∞

−=

−=

z

zerfdxxzQ2

121

2exp

21)(

2

π

)(1)( zQzQ −=−

5.0)0( =Q

Matlab has erf function.

Q function or error function (erf) can be used to determine the probability that the received signal will exceed (or fall below) a particular level.



LogLog--Normal Shadowing ModelNormal Shadowing Model

−=>

σγγ )(])(Pr[ dPQdP r

r

−=<

σγγ )(])(Pr[ dPQdP r

r

The probability that the received signal level (in dB power unit) will exceed a certain value γ can be calculated from the cumulative density function as

The probability that the received signal level will be below γ can be calculated from



Pr (x≥x0)

x0 xm

Gaussian Probability Density FunctionGaussian Probability Density Function

−−= 2

2

2)(exp

21)(

σπσmxxP



Gaussian Gaussian pdfpdf--Q Function RelationQ Function Relation

−−= 2

2

2)(exp

21)(

σπσmxxP

dxmxxxPx

r

−−=> ∫

∞

2

2

0 2)(exp

21)(

0σπσ

σmxy −

=Let

)(2

exp21 0

20

0

zQmxQdyymxxPmx

r =

−

=

−=

−

> ∫∞

− σπσσ

σmxz −

= 0 where



Area AArea A

• Given a circular coverage area of radius R...

• There is a desired received signal level γ

• We are interested in computing U(γ), the percentage of useful service area (i.e., the percentage of area with a received power PR ≥ γ)

r R

Percentage of Coverage AreaPercentage of Coverage Area

θγπ

γπ

γπ

rdrdrPR

dArPR

U r

R

r ])(Pr[1])(Pr[1)(2

0 022 <=<= ∫ ∫∫

−−=

−=>

2)(

21

21)(])(Pr[

σγ

σγγ dPerfdPQdP rr

r




++−−−=

+−−−=

−−=

−=>

2))]/log(10)/log(10)(([

21

21

2))]/log(10)(([

21

21

2)(

21

21)(])(Pr[

00

00

σγ

σγ

σγ

σγγ

RrndRndPLPerf

drndPLPerf

dPerfdPQdP

t

t

rrr

2log10,

2)/log(10)(Let 00

σσenbdRndPLPγa t =

++−=

−

−

−

+−=

+⋅−= ∫

baberf

babaerf

drRrbaerfr

RU

R

1121exp)(121

ln121)(

2

02γ



ExampleExample



• Longley Rice model: point-to-point communication systems (40MHz–100MHz)

• Okumara model: widely used in urban areas (150 MHz – 300 MHz)

• Hata model: graphical path loss (150 MHz – 1500 MHz)

Outdoor Propagation ModelsOutdoor Propagation Models

Radio transmission in a mobile communications system often takesplace over irregular terrain.

The terrain profile of a particular area needs to be taken into account for estimating the path loss.

Most of the models are based on systematic interpretation of measurement data obtained in the service area.

Commonly used outdoor propagation models:



OutlinesOutlines

Propagation Model - Attenuations

-Free Space

- Near-Field vs. Far-Field

- Reflection

- Two-Ray Model

- Diffraction

- Fresnel Zone

- Scattering

Log-Normal Shadowing

Practical Link Budget Design

- Outdoor Propagation Models

Chapter 4 Mobile Radio Propagation: Large-Scale Path Loss...ECE 8708 Wireless Communications :...

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Transcript of Chapter 4 Mobile Radio Propagation: Large-Scale Path Loss...ECE 8708 Wireless Communications :...