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
1
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
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 = 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)
d =  d1+  d2, where , di = h 2  di2
 = h  d + h  d – (d1+d2)
2
2
1
2
2
2
d
h
TX
ht
d1
d2
hobs
RX
hr
Knife Edge Diffraction Geometry for ht = hr
3
Assume h << d1 , h << d2 and h >>  then by substitution and Taylor
Series Approximation
h 2  d1  d 2 
 


2  d1d 2 
3.54
Phase Difference between two paths given as
=

2  h 2 d1  d 2 


 =  h 2  2d1  d 2  


  2 d1d 2 
2  d1d 2 
2
TX
ht

d1
h
3.55

h’
d2
hobs

RX
hr
Knife Edge Diffraction Geometry ht > hr
4
Equivalent Knife Edge Diffraction Geometry with hr subtracted from
all other heights

TX
ht-hr

180-
hobs-hr
d1
when tan x  x   =  + 
h
tan  =  
d1
h
tan  =  
d2

d2
RX
tan(x)
x = 0.4 rad  tan(x) = 0.423
(0.4 rad ≈ 23o )
 d1  d 2 
h h

  h

d1 d 2
 d1d 2 
x
5
Eqn 3.55 for  is often normalized using the dimensionless FresnelKirchoff diffraction parameter, v
v= h
2(d1  d 2 )
2d1d 2
 
d1d 2
 (d1  d 2 )
(3.56)
when  is in units of radians  is given as
=

2
v2
(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
6
(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
d
at 1 GHz λ = 0.3m
λ/2 + d
λ+d
1.5λ + d
7
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
R
n  h 2 d1  d 2 

=
2  2 d1d 2 
T
slice an ellipsoid with a plane  yields circle with radius rn given as
h = rn =
nd1d 2
d1  d 2
then Kirchoff diffraction parameter is given as
2d1  d 2 
nd1d 2 2d1  d 2 

v= h
d1d 2
d1  d 2 d1d 2
=
2n
thus for given rn  v defines an ellipsoid with constant  = n/2
8
nth Fresnel Zone is volume enclosed by ellipsoid defined for n and is defined
as  relative to LOS path
n  1
2
n
≤Δ≤
2
• 1st Fresnel Zone is volume enclosed by ellipsoid defined for n = 1
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
9
For 1st Fresnel Zone, at a distance d1 from Tx & d2 from Rx
• diffracted wave will have a path length of d
d
Tx
d2
d1
destructive interference
•  = /2
•d = /2 + d1+d2
Rx

For 2nd Fresnel Zone
constructive interference:
• d =  + d1+d2
•=
10
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 receiver
such that total path length increases by /2 for each successive circle
• effectively produces alternatively constructive & destructive
interference to received signal
Q
h
O
T
R
d
2
d
1
• If an obstruction were present, it could block some of the Fresnel
zones
11
Assuming, d1 & d2 >> rn  radius of nth Fresnel Zone can be given
in terms of n, d1,d2, 
rn =
nd1d 2
d1  d 2
(3.58)
• 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
Excess Total Path Length,  for each ray passing through nth circle
n
1
2
3
 =n/2
/2

3/2
Rx
Tx
12
(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)
13
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
Rx
Tx
14
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
e.g.
v= h
2(d1  d 2 )
2d1d 2
 
d1d 2
 (d1  d 2 )

TX
RX
h
d1
d2
excess path length
/2

3/2
 and v are positive, thus h is positive
15
v=
h
2(d1  d 2 )
2d1d 2
 
d1d 2
 (d1  d 2 )
d1
d2
TX
RX
h = 0   and v =0
RX
TX
d1
h
d2

 and v are negative h is negative
16
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
17
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
Huygens 2nddry
source
T
h’
d1
d2
R
Knife Edge Diffraction Geometry, R located in shadowed region
18
Electric field strength, Ed of knife-edge diffracted wave is given by:
Ed
= F(v) =
E0
1  j   exp   jt 2 dt
2

v


2


(3.59)
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
19
Gd(dB) = Diffraction Gain due to knife edge presence relative to E0
• Gd(dB) = 20 log|F(v)|
(3.60)
Gd(dB)
Graphical Evaluation
5
0
-5
-10
-15
-20
-25
-30
-3 -2 -1
0 1
2
3
4
5 v
20
Table for Gd(dB)
Gd(dB)
0
20 log(0.5-0.62v)
20 log(0.5 e- 0.95v)
20 log(0.4-(0.1184-(0.38-0.1v)2)1/2)
20 log(0.225/v)
v
 -1
[-1,0]
[0,1]
[1, 2.4]
> 2.4
21
e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m
Compute Diffraction Loss at h = 25m
1. Fresnel Diffraction Parameter
v= h
2(d1  d 2 )
2(2000)
 25
= 2.74
d1d 2
0.333(106 )
2. diffraction loss
• from graph is Gd(dB)  -22dB
• from table Gd(dB)  20 log (0.225/2.74) = - 21.7dB
3. path length difference between LOS & diffracted rays
h 2  d1  d 2  252  2000 

 
 
 6   0.625m
2  d1d 2  2  10 
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
22
e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m
Compute Diffraction Loss at h = -25m
1. Fresnel Diffraction Parameter
v= h
2(d1  d 2 )
2(2000)
 25
6 = -2.74
d1d 2
0.333(10 )
2. diffraction loss from graph is Gd(dB)  1dB
3. path length difference between LOS & diffracted rays
h 2  d1  d 2   252  2000 


 

  0.625m
2  d1d 2 
2  106 
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
23
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
2d1d 2

v=
 (d1  d 2 )
2(10000)(2000)
0
.
0424
 4.24
=
0.333(12000)
T
50m
R
25m
100m
10km
2km

T

25m
75m
10km

R
2km
from graph, Gd(dB) = -25.5 dB
find h if Gd(dB) = 6dB
T

• for Gd(dB) = 6dB  v ≈ 0
25m

• then  = 0 and  = - 
10km
• and h/2000 = 25/12000  h = 4.16m
h
=0
R
2km
24
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
25
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
26
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 =

8 sin i
(3.62)
Let h = maximum protuberance – minimum protuberance
• if h < hc  surface is considered smooth
h
• if h > hc  surface is considered rough
27
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
28
For h > hc  reflected E-fields can be solved for rough surfaces using
modified reflection coefficient
rough = s 
(3.65)
(i) Ament, assume h is a Gaussian distributed random variable with a
local mean, find s as:
   h sin i 2 
 
s = exp 

 

(3.63)
(ii) Boithias modified scattering coefficient has better correlation
with empirical data
   h sin i 2     h sin i 2 
 
s = exp  8    I 0  8
l
   
 
 
(3.64)
I0 is Bessel Function of 1st kind and 0 order
29
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
0 10 20 30 40 50 60 70 80 90
angle of incidence
30
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
| |
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
31
3.8.1 Radar Cross Section Model (RCS)
• if a large distant objects causes scattering & its location is known
 accurately predict scattered signal strengths
power density of 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
32
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 that
impinges on distant scattering object
• wave is reradiated in direction of receiver by:
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 field of transmitter & receiver
33
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
34