chapter diffraction

7/28/2019 Chapter Diffraction

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3.7 Diffraction

allows RF signals to propagate to obstructed (shadowed) regions

- over the horizon (around curved surface of earth)

- behind obstructions

received field strength rapidly decreases as receiver moves into

obstructed region

diffraction field often has sufficient strength to produce useful signal

Segments

3.7.1 Fresnel Zone Geometry


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Huygens Principal

all points on a wavefront can be considered aspoint sources for

producing 2ndry wavelets

2ndry wavelets combine to produce new wavefront in the direction

of propagation

diffraction arises frompropagation of 2ndry wavefront into

shadowed area

field strength of diffracted wave in shadow region = electric field

components of all 2ndry wavelets in the space around the obstacle

slit knife edge


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Excess Path Length = difference between direct path & diffracted path

= d (d1+d2)

3.7.1 Fresnel Zone Geometry

consider a transmitter-receiver pair in free space

let obstruction of effective height h &width protrude to page

- distance from transmitter = d1- distance from receiver = d2- LOS distance between transmitter & receiver = d = d1+d2

Knife Edge Diffraction Geometry forht= hr

hTX RX

hrht

d2d1hobs

dd = d1+ d2, where ,

22

idh di =

2

1

2 dh = 222 dh + (d1+d2)


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Phase Difference between two paths given as

3.54

21

212

2 dd

ddh

Assume h hr

d2d1

hTX

RX

hrht hobs

h

21

212

2

22

dd

ddh

= 3.55=

21

212 2

2 dd

ddh


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(0.4 rad 23o

)

x = 0.4 radtan(x) = 0.423

tan(x)

x

when tan x x= +

21

21

21 dd

ddh

d

h

d

h

Equivalent Knife Edge Diffraction Geometry with hrsubtracted from

all other heights

d2d1

TX

RXht-hr

hobs-hr

180-

tan = 1

d

h

tan = 2d

h


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Fresnel-Kirchoffdiffraction parameter, v

)(2)(2

21

21

21

21

dddd

ddddh

v = (3.56)

when is inunits of radians is given as

= 2

2v

(3.57)

from equations 3.54-3.57, the phase difference, between LOS &

diffracted path is function ofobstructions height & position

transmitters & receiversheight & position

simplify geometry by reducing all heights to minimum height


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(1) Fresnel Zones

used to describe diffraction loss as a function ofpath difference,

around an obstruction

represents successive regions between transmitter and receivernthregion = region where path length of secondary waves is n/2

greater than total LOS path length

regions form a series of ellipsoids with foci at Tx & Rx

/2 + d

1.5+ d

d

+ d

at 1 GHz= 0.3m


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Construct circles on the axis ofTx-Rx such that = n/2, for given integern

radii of circles depends on location of normal plane between Tx and Rx

given n, the set of points where = n/2 defines a family ofellipsoids

assuming d1,d2 >> rn

=

21

212

22 dd

ddhn

T

R

slice an ellipsoid with a plane yields circle with radius rn given as

h = rn =21

21

dd

ddn

= n2v =

21

21

21

21

21

21 22

dd

dd

dd

ddn

dd

ddh

then Kirchoffdiffraction parameter is given as

thus for given rnvdefines an ellipsoid with constant = n/2


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(2) Diffraction Loss caused by blockage of 2ndry (diffracted) waves

partial energy from 2ndry waves is diffracted around an obstacle

obstruction blocks energy from some of the Fresnel zones

only portion of transmitted energy reaches receiver

received energy = vector sum of contributions from all unobstructed

Fresnel zones

depends on geometry of obstruction

phase of secondary (diffracted) E-field

Obstacles may block transmission pathscausing diffraction loss

construct family ofellipsoids between TX & RX to represent

Fresnel zones

join allpoints for which excesspath delay is multiple of/2

compare geometry of obstacle with Fresnel zones to determine

diffraction loss (or gain)


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Fresnel zones: ellipsoids with foci at transmit & receive antenna

if obstruction does not block the volume contained within 1st Fresnel

zone then diffraction loss is minimal

rule of thumb forLOS uwave:if 55% of 1st Fresnel zone is clear further Fresnel zone clearingdoes not significantly alter diffraction loss

d2d1

and v are positive, thus h is positive

TX RXh

excess path length

/2

3/2

)(2)(2

21

21

21

21

dddd

ddddh

v =e.g.


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h = 0and v =0

TX RXd2d1

d2d1

and v are negativeh is negative

h

TX RX

)(

2

)(2

21

21

21

21

dd

dd

dd

ddh

v =


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3.7.2 Knife Edge Diffraction Model

Diffraction Losses

estimating attenuation caused by diffraction over obstacles is

essential for predicting field strength in a given service area

generally not possible to estimate losses precisely

theoretical approximations typically corrected with empirical

measurements

Computing Diffraction Losses

for simple terrain expressions have been derived

for complex terrain computing diffraction losses is complex


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Electric field strength,Edof knife-edge diffracted wave is given by:

F(v) = Complex Fresnel integralv = Fresnel-Kirchoff diffraction parameter

typically evaluated using tables or graphs for given values ofv

E0 = Free Space Field Strength in the absence of both ground

reflections & knife edge diffraction

(3.59)=F(v) =

dt

tjj

v

2exp2

1 2

0E

Ed


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Gd(dB) = Diffraction Gaindue to knife edge presence relative toE0

Gd(dB)= 20 log|F(v)| (3.60)

G

d(dB)

-3 -2 -1 0 1 2 3 4 5

Graphical Evaluation

5

0-5

-10

-15

-20-25

-30 v


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Table forGd(dB)

[0,1]20 log(0.5 e- 0.95v)[-1,0]20 log(0.5-0.62v)

> 2.420 log(0.225/v)[1, 2.4]20 log(0.4-(0.1184-(0.38-0.1v)

2

)1/2

)

-10

vGd(dB)


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3.8 Scattering

RF waves impinge on rough surface reflected energy diffuses in all

directions

e.g. lamp posts, trees random multipath components

provides additional RF energy at receiver

actual received signal in mobile environment often stronger than

predicted by diffraction & reflection models alone


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Reflective Surfaces

flat surfaces has dimensions >>

rough surface often induces specularreflections

surface roughness often tested using Rayleigh fading criterion

- define critical height for surface protuberances hc for given

incident angle i

hc =i

sin8(3.62)

Let h =maximum protuberance

minimum protuberance ifh < hc surface is considered smooth ifh > hc surface is considered rough

h


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Forh > hc reflected E-fields can be solved for rough surfaces using

modified reflection coefficient

rough= s (3.65)

s =

2

sin

exp

ih(3.63)

(i) Ament, assume h is a Gaussian distributed random variable with a

local mean, find s as:

(ii)Boithias modified scattering coefficient has better correlation

with empirical data

I0 is Bessel Function of1

st

kindand 0 order

2

0

2sin

8sin

8expl

I ihih

s = (3.64)


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Reflection Coefficient of Rough Surfaces

(1) polarization (vertical antenna polarization)

1.0

0.8

0.6

0.40.2

0.00 10 20 30 40 50 60 70 80 90

||

angle of incidence

ideal smooth surface

Gaussian Rough Surface Gaussian Rough Surface (Bessel)

Measured Data forstone wall h = 12.7cm, h = 2.54


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3.8.1 Radar Cross Section Model (RCS)

if a large distant objects causes scattering & its location is known

accurately predict scattered signal strengths

determine signal strength by analysis using

- geometric diffraction theory

- physical optics

units = m2

RCS =power density of radio wave incident upon scattering object

power densityofsignal scattered in direction of the receiver


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where

Pris the received signal power

Ptis the transmit power

Gt is the transmit antenna gainr1 is the transmitter-to-target range

sb is the target bistatic RCS

r2 is the target-to-receiver range

Gris the receive antenna gain

lis the radar wavelength

22

2

2

2

1

3

2

)4( rr

GGP

P

rtt

r

U b M bil R di


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dT=distance of transmitter from the scattering object

dR=distance of receiver from the scattering object

assumes object is in the far fieldof transmitter & receiver

Pr(dBm) =Pt(dBm) + Gt(dBi) + 20 log() +RCS[dB m2]

30 log(4) -20 log dT - 20log dR

Urban Mobile Radio

Bistatic Radar Equation used to find received power from

scattering in far field region

describes propagation of wave traveling in free space thatimpinges on distant scattering object

wave is reradiated in direction of receiver by:


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RCS can be approximated by surface area of scattering object (m2)

measured in dB relative to 1m2reference

may be applied to far-field of both transmitter and receiver

useful in predicting received power which scatters off large

objects (buildings)

units = dBm2

for medium and large buildings, 5-10km

14.1 dB m2

chapter diffraction

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