Transcript Wireless Communication Arjav A. Bavarva Dept. of Electronics and Communication Diffraction Arjav A. Bavarva Dept. of Electronics and Communication Huygen’s Principal • All points on a wave front.

Wireless
Communication
171004
Arjav A. Bavarva
Dept. of Electronics and Communication
Diffraction
Arjav A. Bavarva
Dept. of Electronics and Communication
Huygen’s Principal
• All points on a wave front can be considered as point sources for
producing 2nd ry wavelets
• 2ndry wavelets combine to produce new wave front in the direction
of propagation
• Diffraction arises from propagation of 2ndry wave front into
shadowed area
• Field strength of diffracted wave in shadow region =  electric field
components of all 2ndry wavelets in the space around the obstacle
3
d
h
TX
RX
d1
ht
d2
hobs
hr
Knife Edge Diffraction Geometry for ht = hr
4
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
d
h
2
2
TX
ht
d1
d2
hobs
RX
hr
Knife Edge Diffraction Geometry for ht = hr
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
= 
2  h 2 d1  d 2


  2 d1d 2
TX
ht

d1

=

h
 2d1  d 2  

h 2 
2  d1d 2 

3.55

h’
d2
hobs

RX
hr
Knife Edge Diffraction Geometry ht > hr
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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
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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
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(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
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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
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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
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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
•=
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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 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
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Assuming, d1 & d2 >> rn  radius of nth Fresnel Zone can be given
in terms of n, d1,d2, 
nd1d 2
rn =
(3.58)
d1  d 2
• 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
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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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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
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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 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
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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
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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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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
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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
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Gd(dB) = Diffraction Gain due to knife edge presence relative to E0
• Gd(dB) = 20 log|F(v)|
(3.60)
Graphical Evaluation
Gd(dB)
5
0
-5
-10
-20
-30
-3
-2
-1
0
1
2
3
4
5
v
Fresnel diffraction parameter v
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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
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Example:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, For h = 25m, 0
and -25m. For each of these cases, identify the fresnel zone within which
the tip of obstruction lies.
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
h2  d1  d 2  252  2000 

 
 
 6   0.625m
2  d1d2  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
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e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km
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
h2  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
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Example: 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
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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
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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 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
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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
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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
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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
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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
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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
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
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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
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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
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References
T. S. Rappaport, “Wireless Communication”, Prentice hall