3.7 Diffraction

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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

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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)= 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

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
• 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