1

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

Segments3.7.1 Fresnel Zone Geometry

2

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

3

Excess Path Length = difference between direct path & diffracted path

= d – (d1+d2)

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

Knife Edge Diffraction Geometry for ht = hr

hTX RX

hrht

d2d1

hobs

d d = d1+ d2, where , 22

idh di =

21

2 dh = 22

2 dh + – (d1+d2)

4

Phase Difference between two paths given as

3.54

21

212

2 dd

ddh

Assume h << d1 , h << d2 and h >> then by substitution and Taylor

Series Approximation

Knife Edge Diffraction Geometry ht > hr

d2

d1

hTX

RX

hr

hthobs

h’

21

212

2

22

dd

ddh

= 3.55=

21

212 2

2 dd

ddh

5

(0.4 rad ≈ 23o )

x = 0.4 rad tan(x) = 0.423

tan(x)

x

when tan x x = +

21

21

21 dd

ddh

d

h

d

h

Equivalent Knife Edge Diffraction Geometry with hr subtracted from all other heights

d2d1

TX

RX ht-hr

hobs-hr

180-

tan = 1d

h

tan = 2d

h

6

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

)(

2

)(2

21

21

21

21

dd

dd

dd

ddh

v = (3.56)

when is in units of radians is given as

= 2

2v

(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

7

(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

8

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

=

21

212

22 dd

ddhn

T

R

slice an ellipsoid with a plane yields circle with radius rn given as

h = rn = 21

21

dd

ddn

= n2v =

21

21

21

21

21

21 22

dd

dd

dd

ddn

dd

ddh

then Kirchoff diffraction parameter is given as

thus for given rn v defines an ellipsoid with constant = n/2

9

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

• 1st Fresnel Zone is volume enclosed by ellipsoid defined for n = 1

2

1 n

2

n≤ Δ ≤

nth Fresnel Zone is volume enclosed by ellipsoid defined for n and is defined as relative to LOS path

10

destructive interference• = /2d = /2 + d1+d2

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

constructive interference: • d = + d1+d2 • =

For 2nd Fresnel Zone

11

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

hQ

• If an obstruction were present, it could block some of the Fresnel zones

12

Assuming, d1 & d2 >> rn radius of nth Fresnel Zone can be given in terms of n, d1,d2,

21

21

dd

ddn

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

23/23

/21 =n/2 n

Tx

Rx

13

(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)

14

(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 = ELOS (free space value)

Rx

Tx

15

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 clearingdoes not significantly alter diffraction loss

d2d1

and v are positive, thus h is positive

TX RX h

excess path length/2

3/2

)(

2

)(2

21

21

21

21

dd

dd

dd

ddh

v = e.g.

16

h = 0 and v =0

TX RX d2

d1

d2d1

and v are negative h is negative

hTX RX

)(

2

)(2

21

21

21

21

dd

dd

dd

ddh

v =

17

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

18

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

TR

h’

19

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

(3.59)= F(v) =

dttjj

v

2

exp2

1 2

0E

Ed

20

Gd(dB) = Diffraction Gain due to knife edge presence relative to E0

• Gd(dB) = 20 log|F(v)| (3.60)

Gd(

dB)

-3 -2 -1 0 1 2 3 4 5

Graphical Evaluation

50

-5

-10-15-20-25

-30 v

21

Table for Gd(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

vGd(dB)

22

e.g. Let: = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m

2. diffraction loss• from graph is Gd(dB) -22dB

• from table Gd(dB) 20 log (0.225/2.74) = - 21.7dB

)10(333.0

)2000(225

)(26

21

21 dd

ddh

v = = 2.74

1. Fresnel Diffraction Parameter

3. path length difference between LOS & diffracted rays

mdd

ddh625.0

10

2000

2

25

2 6

2

21

212

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

Compute Diffraction Loss at h = 25m

23

e.g. Let: = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m

2. diffraction loss from graph is Gd(dB) 1dB

)10(333.0

)2000(225

)(26

21

21 dd

ddh

v = = -2.74

1. Fresnel Diffraction Parameter

3. path length difference between LOS & diffracted rays

mdd

ddh625.0

10

2000

2

25

2 6

2

21

212

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

Compute Diffraction Loss at h = -25m

24

f = 900MHz = 0.333m

= tan-1(75-25/10000) = 0.287o

= tan-1(75/2000) = 2.15o

= + = 2.43o = 0.0424 radians

find diffraction loss

)(

2

21

21

dd

dd

v =

from graph, Gd(dB) = -25.5 dB

24.4)12000(333.0

)2000)(10000(20424.0 =

find h if Gd(dB) = 6dB

• for Gd(dB) = 6dB v ≈ 0

• then = 0 and = - • and h/2000 = 25/12000 h = 4.16m

2km10km

T

R 25m 75m

2km10km

100m

T

25m50mR

=0

2km10km

T

R 25m h

25

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

26

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

27

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 = i

sin8 (3.62)

Let h = maximum protuberance – minimum protuberance

• if h < hc surface is considered smooth

• if h > hc surface is considered roughh

28

stone – dielectric properties• r = 7.51• = 0.028• = 0.95

rough stone parameters• h = 12.7cmh = 2.54

h = standard deviation of surface height about mean surface height

29

For h > hc reflected E-fields can be solved for rough surfaces using

modified reflection coefficient

rough = s (3.65)

s =

2sin

exp

ih (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

2

0

2sin

8sin

8expl

I ihih

s = (3.64)

30

Reflection Coefficient of Rough Surfaces

(1) polarization (vertical antenna polarization)

1.00.80.60.40.20.0

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

31

(2) || polarization (horizontal antenna polarization)

1.00.80.60.40.20.0

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

32

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

33

• 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

Pr(dBm) = Pt (dBm) + Gt(dBi) + 20 log() + RCS [dB m2]

– 30 log(4) -20 log dT - 20log dR

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:

34

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