rap ch3 large scale 2 part3-1

8/4/2019 Rap Ch3 Large Scale 2 Part3-1

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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



2

Huygen’s Principal

• all points on a wavefront can be considered as point sources for

producing 2ndry wavelets

• 2ndry wavelets combine to produce new wavefront in the direction

of propagation

• diffraction arises from propagation 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 – (d 1+d 2)

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 = d 1 - distance from receiver = d 2- LOS distance between transmitter & receiver = d = d 1+d 2

Knife Edge Diffraction Geometry for ht = hr

h

TX RX

hr ht

d 2d 1hobs

d d = d 1+ d 2, where , 22

id h d i =

2

1

2 d h = 2

2

2 d h + – (d 1+d 2)



4

Phase Difference between two paths given as

3.54

21

212

2 d d

d d h

Assume h << d 1 , h << d 2 and h >> then by substitution and Taylor

Series Approximation

Knife Edge Diffraction Geometry ht > hr

d 2d 1

h TX

RX

hr

ht hobs

h’

21

212

2

22

d d

d d h

= 3.55=

21

212 2

2 d d

d d h



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(0.4 rad ≈ 23o

)

x = 0.4 rad tan(x) = 0.423

tan(x)

x

when tan x x = +

21

21

21 d d

d d h

d

h

d

h

Equivalent Knife Edge Diffraction Geometry with hr subtracted from

all other heights

d 2d 1

TX

RXht -hr

hobs-hr

180-

tan = 1

d

h

tan = 2d

h



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Eqn 3.55 for is often normalized using the dimensionless Fresnel-

Kirchoff diffraction parameter, v

)(2 )(2

21

21

21

21

d d d d

d d d d h

v = (3.56)

when is in units of radians is given as

= 2

2v

(3.57)

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

diffracted path is function of • obstruction’s height & position

• transmitters & receivers height & position

simplify geometry by reducing all heights to minimum height



(1) Fresnel Zones

• used to describe diffraction loss as a function of path difference,

around an obstruction

• represents successive regions between transmitter and receiver• nth region = 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

λ /2 + d

1.5λ + d

d

λ + d

at 1 GHzλ = 0.3m



Construct circles on the axis of Tx-Rx such that = n /2, for given integer n

• 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 of ellipsoids

• assuming d 1 ,d 2 >> r n

=

21

212

22 d d

d d hn

T

R

slice an ellipsoid with a plane yields circle with radius r n given as

h = r n =21

21

d d

d d n

= n2v =

21

21

21

21

21

21 22

d d

d d

d d

d d n

d d

d d h

then Kirchoff diffraction parameter is given as

thus for given r n v defines an ellipsoid with constant = n /2



Phase Difference, pertaining to nth Fresnel Zone is

(n-1) ≤ ≤ n

• contribution to the electric field at Rx from successive Fresnel Zones

tend to be in phase opposition destructive interference

• generally must keep 1st Fresnel Zone unblocked to obtain free space

transmission conditions

• 1st Fresnel Zone is volume enclosed by ellipsoid defined for n = 1

2

1 n

2

n≤ Δ ≤

nth Fresnel Zone is volume enclosed by ellipsoid defined for n and is defined

as relative to LOS path



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destructive interference

• = /2

•d = /2 + d 1+d 2

For 1st Fresnel Zone, at a distance d 1 from Tx & d 2 from Rx

• diffracted wave will have a path length of d

d 1 d 2

d

Tx Rx

constructive interference:

• d = + d 1+d 2

• =

For 2nd

Fresnel Zone



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Fresnel Zones

• slice the ellipsoids with a transparent plane between transmitter &

receiver – obtain series of concentric circles

• circles represent loci of 2ndry wavelets that propagate to receiversuch that total path length increases by /2 for each successive circle

• effectively produces alternatively constructive & destructive

interference to received signal

T

R

O

d

1

d

2

h

Q

• If an obstruction were present, it could block some of the Fresnel

zones



12

Assuming, d 1 & d 2 >> r n radius of nth Fresnel Zone can be given

in terms of n, d 1 ,d 2 ,

21

21

d d

d d n

r n = (3.58)

• radii of concentric circles depends on location between Tx & Rx

- maximum radii at d 1 = d 2 (midpoint), becomes smaller as plane

moves towards receiver or transmitter

- shadowing is sensitive to obstruction’s position and frequency

Excess Total Path Length, for each ray passing through nth circle

2

3 /23

/21

=n /2 n

Tx

Rx



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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

• Fresnel Zones indicate phase of secondary (diffracted) E-field

Obstacles may block transmission paths – causing diffraction loss

• construct family of ellipsoids between TX & RX to represent

Fresnel zones

• join all points for which excess path delay is multiple of /2

• compare geometry of obstacle with Fresnel zones to determine

diffraction loss (or gain)



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(ii) as screen height increases E will vary up & down as screen blocks more

Fresnel zones below LOS path

amplitude of oscillation increases until just in line with Tx and Rx field strength = ½ of unobstructed field strength

Diffraction Losses

Place ideal, perfectly straight screen between Tx and Rx

(i) if top of screen is well below LOS path screen will have little effect

- the Electric field at Rx = E LOS (free space value)

Rx

Tx



15

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 for LOS uwave:if 55% of 1st Fresnel zone is clear further Fresnel zone clearingdoes not significantly alter diffraction loss

d 2d 1

and v are positive, thus h is positive

TX RX h

excess path length

/2

3 /2

)(2 )(2

21

21

21

21

d d d d

d d d d h

v =e.g.



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h = 0 and v =0

TX RX d 2

d 1

d 2d 1

and v are negativeh is negative

h

TX RX

)(

2

)(2

21

21

21

21

d d

d d

d d

d d h

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

K if d M d l i l d l h id i i h i d f



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Knife-edge Model - simplest model that provides insight into order of magnitude for diffraction loss

• useful for shadowing caused by 1 object treat object as a knife edge

• diffraction losses estimated using classical Fresnel solution for field

behind a knife edge

Consider receiver at R located in shadowed region (diffraction zone)

• E- field strength at R = vector sum of all fields due to 2ndry Huygen’s

sources in the plane above the knife edge

Knife Edge Diffraction Geometry, R located in shadowed region

Huygens 2nddrysource

d 2d 1

T

R

h’



19

Electric field strength, E d of knife-edge diffracted wave is given by:

F(v) = Complex Fresnel integral• v = Fresnel-Kirchoff diffraction parameter

• typically evaluated using tables or graphs for given values of v

E 0 = Free Space Field Strength in the absence of both ground

reflections & knife edge diffraction

(3.59)= F(v) =

dt

t j j

v

2exp2

1 2

0 E

E d



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Gd (dB) = Diffraction Gain due to knife edge presence relative to E 0

• Gd (dB) = 20 log|F(v)| (3.60)

G d

( d B )

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

Graphical Evaluation

5

0-5

-10

-15

-20-25

-30 v



21

Table for Gd (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)

L 0 333 (f 900MH ) d 1k d 1k h 25



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e.g. Let: = 0.333 (f c = 900MHz), d1 = 1km, d2 = 1km, h = 25m

2. diffraction loss

• from graph is Gd(dB) -22dB

• from table Gd(dB) 20 log (0.225/2.74) = - 21.7dB

)10(333.0

)2000(225 )(26

21

21 d d

d d h

v = = 2.74

1. Fresnel Diffraction Parameter

3. path length difference between LOS & diffracted rays

m

d d

d d h625.0

10

2000

2

25

2

6

2

21

212

4. Fresnel zone at tip of obstruction (h=25)

• solve for n such that = n /2

• n = 2· 0.625/0.333 = 3.75

• tip of the obstruction completely blocks 1st

3 Fresnel zones

Compute Diffraction Loss at h = 25m

L 0 333 (f 900MH ) d 1k d 1k h 25



23

e.g. Let: = 0.333 (f c = 900MHz), d1 = 1km, d2 = 1km, h = 25m

2. diffraction loss from graph is Gd(dB) 1dB

)10(333.0)2000(225 )(26

21

21 d d d d h

v = = -2.74

1. Fresnel Diffraction Parameter

3. path length difference between LOS & diffracted rays

md d

d d h625.0

10

2000

2

25

2 6

2

21

212

4. Fresnel zone at tip of the obstruction (h = -25)• solve for n such that = n /2

• n = 2· 0.625/0.333 = 3.75

• tip of the obstruction completely blocks 1st 3 Fresnel zones

• diffraction losses are negligible since obstruction is below LOS path

Compute Diffraction Loss at h = -25m

fi d diff i l



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f = 900MHz = 0.333m

= tan-1(75-25/10000) = 0.287o

= tan-1(75/2000) = 2.15o

= + = 2.43o = 0.0424 radians

find diffraction loss

)(

2

21

21

d d

d d

v =

from graph, Gd (dB) = -25.5 dB

24.4)12000(333.0)2000)(10000(20424.0 =

find h if Gd (dB) = 6dB

• for Gd (dB) = 6dB v ≈ 0

• then = 0 and = -

• and h /2000 = 25/12000 h = 4.16m

2km 10km

T

R25m

75m

2km 10km

100m

T

25m50m

R

=0

2km 10km

T

R25m

h



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3.7.3 Multiple Knife Edge Diffraction

• with more than one obstruction compute total diffraction loss

(1) replace multiple obstacles with one equivalent obstacle

• use single knife edge model

• oversimplifies problem

• often produces overly optimistic estimates of received signal

strength

(2) wave theory solution for field behind 2 knife edges in series

• Extensions beyond 2 knife edges becomes formidable• Several models simplify and estimate losses from multiple obstacles

3 8 S tt i



26

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



27

Reflective Surfaces

• flat surfaces has dimensions >>

• rough surface often induces specular reflections

• 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

• if h < hc surface is considered smooth • if h > hc surface is considered rough

h



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stone – dielectric properties• r = 7.51• = 0.028• = 0.95

rough stone parameters• h = 12.7cm• h = 2.54

h = standard deviation of surface height about mean surface height



29

For h > 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

I 0

is Bessel Function of 1st kind and 0 order

2

0

2sin

8sin

8expl

I ihih

s = (3.64)

R fl ti C ffi i t f R h S f



30

Reflection Coefficient of Rough Surfaces

(1) polarization (vertical antenna polarization)

1.0

0.8

0.6

0.4

0.2

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

R fl ti C ffi i t f R h S f



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(2) || polarization (horizontal antenna polarization)

1.0

0.8

0.6

0.4

0.2

0.00 10 20 30 40 50 60 70 80 90

| |

angle of incidence

Reflection Coefficient of Rough Surfaces

• ideal smooth surface

• Gaussian Rough Surface

• Gaussian Rough Surface (Bessel)

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

3 8 1 Radar Cross Section Model (RCS)



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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 density of signal scattered in direction of the receiver

Urban Mobile Radio



33

• d T = distance of transmitter from the scattering object

• d R = distance of receiver from the scattering object

• assumes object is in the far field of transmitter & receiver

Pr (dBm) = Pt (dBm) + Gt (dBi) + 20 log() + RCS [dB m2] – 30 log(4) -20 log d T - 20log d R

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 1m2 reference

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

• useful in predicting received power which scatters off large

objects (buildings)

• units = dB m2

• [Sei91] for medium and large buildings, 5-10km

14.1 dB m2 < RCS < 55.7 dB m2

rap ch3 large scale 2 part3-1

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