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

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

consider a transmitter-receiver pair in free space 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 = Knife Edge Diffraction Geometry for ht = hr h TX RX hr ht d2 d1 hobs d  = + – (d1+d2)

Knife Edge Diffraction Geometry ht > hr Assume h << d1 , h << d2 and h >>  then by substitution and Taylor Series Approximation 3.54   Phase Difference between two paths given as  = 3.55 = Knife Edge Diffraction Geometry ht > hr  d2 d1   h TX RX hr ht hobs h’

Equivalent Knife Edge Diffraction Geometry with hr subtracted from all other heights   d2 d1 TX RX ht-hr  hobs-hr 180- when tan x  x   =  +  tan  =   (0.4 rad ≈ 23o ) x = 0.4 rad  tan(x) = 0.423 tan(x) x tan  =    

Eqn 3.55 for  is often normalized using the dimensionless Fresnel-Kirchoff diffraction parameter, v (3.56) when  is in units of radians  is given as  = (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 GHz λ = 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 d1,d2 >> rn T R  = slice an ellipsoid with a plane  yields circle with radius rn given as h = rn = = v = then Kirchoff diffraction parameter is given as thus for given rn  v defines an ellipsoid with constant  = n/2

1st Fresnel Zone is volume enclosed by ellipsoid defined for n = 1 ≤ Δ ≤ nth Fresnel Zone is volume enclosed by ellipsoid defined for n and is defined as  relative to LOS path 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

For 1st Fresnel Zone, at a distance d1 from Tx & d2 from Rx diffracted wave will have a path length of d d1 d2 d Tx Rx  destructive interference  = /2 d = /2 + d1+d2 constructive interference: d =  + d1+d2  =  For 2nd Fresnel Zone

slice the ellipsoids with a transparent plane between transmitter & 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 T R O d1 d2 h Q If an obstruction were present, it could block some of the Fresnel zones

radii of concentric circles depends on location between Tx & Rx Assuming, d1 & d2 >> rn  radius of nth Fresnel Zone can be given in terms of n, d1,d2,  rn = (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  2 3/2 3 /2 1  =n/2 n Tx Rx

(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)

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

 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 = e.g. d2 d1  and v are positive, thus h is positive TX RX h  excess path length /2  3/2

v = h = 0   and v =0 TX RX d2 d1 d2 d1  and v are negative h is negative h TX RX 

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

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 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 2nddry source d2 d1 T R h’

Electric field strength, Ed of knife-edge diffracted wave is given by: (3.59) = F(v) = 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 Gain due to knife edge presence relative to E0 Gd(dB) = 20 log|F(v)| (3.60) Graphical Evaluation 5 -5 -10 -15 -20 -25 -30 v Gd(dB) -3 -2 -1 0 1 2 3 4 5

Table for Gd(dB) Gd(dB) v [0,1] 20 log(0.5 e- 0.95v) [-1,0] 20 log(0.5-0.62v) > 2.4 20 log(0.225/v) [1, 2.4] 20 log(0.4-(0.1184-(0.38-0.1v)2)1/2)  -1 v Gd(dB)

e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m Compute Diffraction Loss at h = 25m 1. Fresnel Diffraction Parameter v = = 2.74 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   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

e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m Compute Diffraction Loss at h = -25m v = = -2.74 1. Fresnel Diffraction Parameter 2. diffraction loss from graph is Gd(dB)  1dB 3. path length difference between LOS & diffracted rays   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

find diffraction loss  2km 10km  T R 25m  75m 100m 50m f = 900MHz   = 0.333m  = tan-1(75-25/10000) = 0.287o  = tan-1(75/2000) = 2.15o  = +  = 2.43o = 0.0424 radians v = from graph, Gd(dB) = -25.5 dB = find h if Gd(dB) = 6dB for Gd(dB) = 6dB  v ≈ 0 then  = 0 and  = -  and h/2000 = 25/12000  h = 4.16m =0 2km 10km  T R 25m  h

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 provides additional RF energy at receiver actual received signal in mobile environment often stronger than predicted by diffraction & reflection models alone

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 = (3.62) Let h = maximum protuberance – minimum protuberance if h < hc  surface is considered smooth if h > hc  surface is considered rough h

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

For h > hc  reflected E-fields can be solved for rough surfaces using modified reflection coefficient rough = s  (3.65) s = (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 I0 is Bessel Function of 1st kind and 0 order s = (3.64)

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 

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 

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 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

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