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# 3.7 Diffraction PowerPoint PPT Presentation

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

3.7 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

• obstructed region

• diffraction field often has sufficient strength to produce useful signal

Segments

3.7.1 Fresnel Zone Geometry

slit

knife edge

• 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

• field strength of diffracted wave in shadow region =  electric field

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

d = d1+ d2, where ,

di =

h

d

=

+

– (d1+d2)

TX

RX

d1

d2

ht

hobs

hr

Knife Edge Diffraction Geometry for ht = hr

• 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

Excess Path Length = difference between direct path & diffracted path

 = d – (d1+d2)

 

3.54

3.55

 =

=

h

h’

TX

d1

d2

RX

hobs

ht

hr

Knife Edge Diffraction Geometry ht > hr

Assume h << d1 , h <<d2 and h >> then by substitution and Taylor

Series Approximation

Phase Difference between two paths given as

tan(x)

TX

hobs-hr

ht-hr

RX

d1

d2

x

tan  =  

tan  =  

 

x = 0.4 rad  tan(x) = 0.423

Equivalent Knife Edge Diffraction Geometry with hrsubtracted from all other heights

180-

when tan x  x   =  + 

v =

(3.56)

when  is inunits of radians  is given as

 =

(3.57)

Eqn 3.55 for  is often normalized using the dimensionless Fresnel-Kirchoffdiffraction parameter, v

• from equations 3.54-3.57   , the phase difference, between LOS & diffracted path is function of

• obstruction’s height & position

• transmitters & receiversheight & position

• simplify geometry by reducing all heights to minimum height

d

λ/2 + d

λ + d

1.5λ + d

• (1) Fresnel Zones

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

• around an obstruction

• represents successive regions between transmitter and receiver

• nthregion = 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

at 1 GHz λ = 0.3m

R

=

T

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

h = rn =

then Kirchoffdiffraction parameter is given as

v =

=

thus for given rnvdefines an ellipsoid with constant = n/2

• 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 d1,d2 >> rn

nthFresnel Zone is volume enclosed by ellipsoid defined for n andis defined

as  relative to LOS path

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

Phase Difference, pertaining to nthFresnel Zone is

≤ Δ≤

(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

d

d2

d1

• destructive interference

•  = /2

• d = /2 + d1+d2

Tx

Rx

• constructive interference:

• d =  + d1+d2

• = 

For 2nd Fresnel Zone

• For 1st Fresnel Zone, at a distance d1from Tx & d2 from Rx

• diffracted wave will have a path length of d

Q

R

h

d2

O

T

d1

• Fresnel Zones

• slice the ellipsoids with a transparent plane between transmitter &

• receiver – obtain series of concentric circles

• circles represent loci of2ndry wavelets that propagate to receiver

• such that total path length increases by /2 for each successive circle

• effectively produces alternativelyconstructive &destructive

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

• zones

n

 =n/2

1

/2

2

rn =

(3.58)

3

3/2

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

Rx

Tx

Assuming, d1& d2 >> rn radius of nth Fresnel Zone can be given in terms of n, d1,d2, 

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

• - maximum radii at d1 = d2 (midpoint), becomes smaller as plane

• moves towards receiver or transmitter

• - shadowing is sensitive to obstruction’s position and frequency

• (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)

Rx

Tx

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 = ELOS (free space value)

• (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

e.g.

v =

excess path length

/2

3/2

RX

TX

h

d1

d2

 and v are positive, thus h is positive

• 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 clearing

• does not significantly alter diffraction loss

v =

RX

TX

h = 0   and v =0

RX

TX

d1

d2

h

d1

d2

 and v are negative h is negative

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

Huygens 2nddry

source

h’

T

d1

R

d2

Knife Edge Diffraction Geometry, R located in shadowed region

• 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

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

• sources in the plane above the knife edge

= F(v) =

(3.59)

Electric field strength, Ed 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

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

• reflections & knife edge diffraction

Gd(dB) = Diffraction Gaindue to knife edge presence relative to E0

• Gd(dB)= 20 log|F(v)| (3.60)

5

0

-5

-10

-15

-20

-25

-30

v

Graphical Evaluation

Gd(dB)

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

Gd(dB)

v

0

 -1

20 log(0.5-0.62v)

[-1,0]

20 log(0.5 e- 0.95v)

[0,1]

20 log(0.4-(0.1184-(0.38-0.1v)2)1/2)

[1, 2.4]

20 log(0.225/v)

> 2.4

Table for Gd(dB)

= 2.74

v =

3. path length difference between LOS & diffracted rays

 

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

Compute Diffraction Loss at h = 25m

1. Fresnel Diffraction Parameter

• 2. diffraction loss

• from graph is Gd(dB) -22dB

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

• 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

1. Fresnel Diffraction Parameter

v =

= -2.74

3. path length difference between LOS & diffracted rays

 

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

Compute Diffraction Loss at h = -25m

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

• 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

T

R

50m

100m

25m

10km

2km

T

v =

75m

25m

R

=

10km

2km

from graph, Gd(dB)= -25.5 dB

find h if Gd(dB)= 6dB

T

=0

h

• forGd(dB)= 6dB  v≈ 0

• then  = 0 and  = - 

• and h/2000 = 25/12000  h = 4.16m

25m

R

10km

2km

find diffraction loss

f = 900MHz   = 0.333m

 = tan-1(75-25/10000) = 0.287o

 = tan-1(75/2000) = 2.15o

 = +  = 2.43o = 0.0424 radians

• 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 Scattering

• RF waves impinge on rough surface reflected energy diffuses in all directions

• e.g. lamp posts, trees  random multipath components

• actual received signal in mobile environment often stronger than

• predicted by diffraction & reflection models alone

h

• Let h =maximum protuberance – minimum protuberance

• if h < hc  surface is considered smooth

• if h > hc  surface is considered rough

hc =

(3.62)

• Reflective Surfaces

• flat surfaceshas 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

h = standard deviation of surface height about mean surface height

• stone – dielectric properties

• r = 7.51

•  = 0.028

•  = 0.95

• rough stone parameters

• h = 12.7cm

• h = 2.54

rough= s 

(3.65)

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

• local mean, find s as:

s =

(3.63)

• (ii) Boithias modified scattering coefficient has better correlation

• with empirical data

s =

(3.64)

I0 is Bessel Function of 1st kind and 0 order

For h > hc reflected E-fields can be solved for rough surfaces using modified reflection coefficient

||

1.0

0.8

0.6

0.4

0.2

0.0

0 10 20 30 40 50 60 70 80 90

angle of incidence

Reflection Coefficient of Rough Surfaces

(1)  polarization (vertical antenna polarization)

• ideal smooth surface

• Gaussian Rough Surface

• Gaussian Rough Surface (Bessel)

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

| |

1.0

0.8

0.6

0.4

0.2

0.0

angle of incidence

0 10 20 30 40 50 60 70 80 90

Reflection Coefficient of Rough Surfaces

(2) || polarization (horizontal antenna polarization)

• ideal smooth surface

• Gaussian Rough Surface

• Gaussian Rough Surface (Bessel)

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

power densityof signal scattered in direction of the receiver

RCS =

power density of radio wave incident upon scattering object

• units = m2

• determine signal strength by analysis using

• - geometric diffraction theory

• - physical optics

3.8.1 Radar Cross Section Model (RCS)

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

•  accurately predict scattered signal strengths

• scattering in far field region

• describes propagation of wave traveling in free space that

• impinges on distant scattering object

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

– 30 log(4) -20 log dT - 20log dR

• 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

• 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 = dB m2

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

• 14.1 dB  m2 < RCS < 55.7 dB  m2