1. Wireless Communication 171004 Arjav A. Bavarva Dept. of Electronics and Communication

2. Diffraction Arjav A. Bavarva Dept. of Electronics and Communication

3. 3 Huygen’s Principal All points on a wave front can be considered as point sources for producing 2 nd ry wavelets 2 nd ry wavelets combine to produce new wave front in the direction of propagation Diffraction arises from propagation of 2 nd ry wave front into shadowed area Field strength of diffracted wave in shadow region =  electric field components of all 2 nd ry wavelets in the space around the obstacle

4. 4 Knife Edge Diffraction Geometry for h t = h r h TXRX hrhr htht d2d2 d1d1 h obs dd

5. Excess Path Length = difference between direct path & diffracted path  =  d – (d 1 +d 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 = d 1 - distance from receiver = d 2 - LOS distance between transmitter & receiver = d = d 1 +d 2 Knife Edge Diffraction Geometry for h t = h r h TXRX hrhr htht d2d2 d1d1 h obs dd  d =  d 1 +  d 2, where,  d i =  = = + – (d 1 +d 2 )

6. 6 Phase Difference between two paths given as 3.54   Assume h > then by substitution and Taylor Series Approximation Knife Edge Diffraction Geometry h t > h r  d2d2 d1d1   h TX RX hrhr htht h obs h’ = = 3.55 =

7. 7 (0.4 rad ≈ 23 o ) x = 0.4 rad  tan(x) = 0.423 tan(x) x when tan x  x   =  +     Equivalent Knife Edge Diffraction Geometry with h r subtracted from all other heights   d2d2 d1d1 TX RX h t -h r  h obs -h r 180-  tan  =   tan  =  

8. 8 Eqn 3.55 for  is often normalized using the dimensionless Fresnel- Kirchoff diffraction parameter, v 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

9. 9 (1) Fresnel Zones used to describe diffraction loss as a function of path difference,  around an obstruction represents successive regions between transmitter and receiver n th 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

10. 10 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 d 1,d 2 >> r n  = = T R slice an ellipsoid with a plane  yields circle with radius r n given as h = r n = =v = then Kirchoff diffraction parameter is given as thus for given r n  v defines an ellipsoid with constant  = n /2

11. 11 Phase Difference,  pertaining to n th 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 1 st Fresnel Zone unblocked to obtain free space transmission conditions 1 st Fresnel Zone is volume enclosed by ellipsoid defined for n = 1 ≤ Δ ≤ n th Fresnel Zone is volume enclosed by ellipsoid defined for n and is defined as  relative to LOS path

12. 12 destructive interference  = /2  d = /2 + d 1 +d 2 For 1 st Fresnel Zone, at a distance d 1 from Tx & d 2 from Rx diffracted wave will have a path length of  d d1d1 d2d2 dd TxRx constructive interference:  d = + d 1 +d 2  = For 2 nd Fresnel Zone

13. 13 Fresnel Zones slice the ellipsoids with a transparent plane between transmitter & receiver – obtain series of concentric circles circles represent loci of 2 nd ry 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 d1d1 d2d2 h Q If an obstruction were present, it could block some of the Fresnel zones

14. 14 Assuming, d 1 & d 2 >> r n  radius of n th Fresnel Zone can be given in terms of n, d 1,d 2, r n =(3.58) radii of concentric circles depends on location between Tx & Rx - maximum radii at d 1 = d 2 (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 n th circle 2 3 /2 3 /2 1  = n /2 n Tx Rx

15. 15 (2) Diffraction Loss caused by blockage of 2 nd ry (diffracted) waves partial energy from 2 nd ry 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)

16. 16 (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 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 = E LOS (free space value) Rx Tx

17. 17 Fresnel zones: ellipsoids with foci at transmit & receive antenna if obstruction does not block the volume contained within 1 st Fresnel zone  then diffraction loss is minimal rule of thumb for LOS uwave: if 55% of 1 st Fresnel zone is clear  further Fresnel zone clearing does not significantly alter diffraction loss d2d2 d1d1  and v are positive, thus h is positive TX RX h  excess path length /2 3 /2 v = e.g.

18. 18 h = 0   and v =0 TX RX d2d2 d1d1 d2d2 d1d1  and v are negative  h is negative h TX RX  v =

19. 19 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

20. 20 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 2 nd ry Huygen’s sources in the plane above the knife edge Knife Edge Diffraction Geometry, R located in shadowed region Huygens 2 nd dry source d2d2 d1d1 T R h’

21. 21 Electric field strength, E d 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 E 0 = Free Space Field Strength in the absence of both ground reflections & knife edge diffraction (3.59)= F(v) =

22. 22 G d (dB) = Diffraction Gain due to knife edge presence relative to E 0 G d (dB) = 20 log|F(v)| (3.60) G d (dB) -3 -2 -1 0 1 2 3 4 5 Graphical Evaluation 5 0 -5 -10 -20 -30 v Fresnel diffraction parameter v

23. 23 Table for G d (dB) [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 0 v G d (dB)

24. 24 Example: = 0.333 (f c = 900MHz), d 1 = 1km, d 2 = 1km, For h = 25m, 0 and -25m. For each of these cases, identify the fresnel zone within which the tip of obstruction lies. 2. diffraction loss from graph is G d (dB)  -22dB from table G d (dB)  20 log (0.225/2.74) = - 21.7dB v = = 2.74 1. Fresnel Diffraction Parameter 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 1 st 3 Fresnel zones Compute Diffraction Loss at h = 25m

25. 25 e.g. Let: = 0.333 (f c = 900MHz), d 1 = 1km, d 2 = 1km 2. diffraction loss from graph is G d (dB)  1dB v == -2.74 1. Fresnel Diffraction Parameter 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 1 st 3 Fresnel zones diffraction losses are negligible since obstruction is below LOS path Compute Diffraction Loss at h = -25m

26. 26 f = 900MHz  = 0.333m  = tan -1 (75-25/10000) = 0.287 o  = tan -1 (75/2000) = 2.15 o  =  +  = 2.43 o = 0.0424 radians Example: Find diffraction loss v = from graph, G d (dB) = -25.5 dB = find h if G d (dB) = 6dB for G d (dB) = 6dB  v ≈ 0 then  = 0 and  = -  and h/2000 = 25/12000  h = 4.16m  2km10km  T R 25m  75m 2km10km 100m T 25m 50m R  =0 2km10km  T R 25m  h

27. 27 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

28. 28 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

29. 29 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 h c for given incident angle  i h c =(3.62) Let h = maximum protuberance – minimum protuberance if h < h c  surface is considered smooth if h > h c  surface is considered rough h

30. 30 stone – dielectric properties  r = 7.51  = 0.028  = 0.95 rough stone parameters h = 12.7cm  h = 2.54  h = standard deviation of surface height about mean surface height

31. 31 For h > h c  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 I 0 is Bessel Function of 1 st kind and 0 order  s = (3.64)

32. 32 Reflection Coefficient of Rough Surfaces (1)  polarization (vertical antenna polarization) 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 ideal smooth surface Gaussian Rough Surface Gaussian Rough Surface (Bessel) Measured Data forstone wall h = 12.7cm,  h = 2.54

33. 33 (2) || polarization (horizontal antenna polarization) 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 ideal smooth surface Gaussian Rough Surface Gaussian Rough Surface (Bessel) Measured Data forstone wall h = 12.7cm,  h = 2.54

34. 34 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 = m 2 RCS = power density of radio wave incident upon scattering object power density of signal scattered in direction of the receiver

35. 35 d T = distance of transmitter from the scattering object d R = distance of receiver from the scattering object assumes object is in the far field of transmitter & receiver P r (dBm) = P t (dBm) + G t (dBi) + 20 log( ) + RCS [dB m 2 ] – 30 log(4  ) -20 log d T - 20log d R 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:

36. 36 RCS can be approximated by surface area of scattering object (m 2 ) measured in dB relative to 1m 2 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  m 2 [Sei91] for medium and large buildings, 5-10km 14.1 dB  m 2 < RCS < 55.7 dB  m 2

37. References T. S. Rappaport, “Wireless Communication”, Prentice hall