- 256 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about '3.7 Diffraction' - farica

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

- 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

knife edge

- 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

di =

h

d

=

+

– (d1+d2)

TX

RX

d1

d2

ht

hobs

hr

Knife Edge Diffraction Geometry for ht = hr

- 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)

3.54

3.55

=

=

h

h’

TX

d1

d2

RX

hobs

ht

hr

Knife Edge Diffraction Geometry ht > hr

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

Series Approximation

Phase Difference between two paths given as

TX

hobs-hr

ht-hr

RX

d1

d2

x

tan =

tan =

x = 0.4 rad tan(x) = 0.423

(0.4 rad ≈ 23o )

Equivalent Knife Edge Diffraction Geometry with hrsubtracted from all other heights

180-

when tan x x = +

(3.56)

when is inunits of radians is given as

=

(3.57)

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

- from equations 3.54-3.57 , the phase difference, between LOS & diffracted path is function of
- obstruction’s height & position
- transmitters & receiversheight & position
- simplify geometry by reducing all heights to minimum height

λ/2 + d

λ + d

1.5λ + d

- (1) Fresnel Zones
- used to describe diffraction loss as a function of path difference,
- around an obstruction
- represents successive regions between transmitter and receiver
- nthregion = 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

at 1 GHz λ = 0.3m

=

T

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

h = rn =

then Kirchoffdiffraction parameter is given as

v =

=

thus for given rnvdefines an ellipsoid with constant = n/2

- 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

nthFresnel Zone is volume enclosed by ellipsoid defined for n andis defined

as relative to LOS path

- 1st Fresnel Zone is volume enclosed by ellipsoid defined forn = 1

Phase Difference, pertaining to nthFresnel Zone is

≤ Δ≤

(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

d2

d1

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

Tx

Rx

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

For 2nd Fresnel Zone

- For 1st Fresnel Zone, at a distance d1from Tx & d2 from Rx
- diffracted wave will have a path length of d

R

h

d2

O

T

d1

- Fresnel Zones
- slice the ellipsoids with a transparent plane between transmitter &
- receiver – obtain series of concentric circles
- circles represent loci of2ndry wavelets that propagate to receiver
- such that total path length increases by /2 for each successive circle
- effectively produces alternativelyconstructive &destructive
- interference to received signal

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

=n/2

1

/2

2

rn =

(3.58)

3

3/2

Excess Total Path Length, for each ray passing through nth circle

Rx

Tx

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

- 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

(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)

Tx

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

v =

excess path length

/2

3/2

RX

TX

h

d1

d2

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

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

source

h’

T

d1

R

d2

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

(3.59)

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

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

- Gd(dB)= 20 log|F(v)| (3.60)

5

0

-5

-10

-15

-20

-25

-30

v

Graphical Evaluation

Gd(dB)

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

v

0

-1

20 log(0.5-0.62v)

[-1,0]

20 log(0.5 e- 0.95v)

[0,1]

20 log(0.4-(0.1184-(0.38-0.1v)2)1/2)

[1, 2.4]

20 log(0.225/v)

> 2.4

Table for Gd(dB)

v =

3. path length difference between LOS & diffracted rays

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

Compute Diffraction Loss at h = 25m

1. Fresnel Diffraction Parameter

- 2. diffraction loss
- from graph is Gd(dB) -22dB
- from table Gd(dB) 20 log (0.225/2.74) = - 21.7dB

- 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

1. Fresnel Diffraction Parameter

v =

= -2.74

3. path length difference between LOS & diffracted rays

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

Compute Diffraction Loss at h = -25m

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

- 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

R

50m

100m

25m

10km

2km

T

v =

75m

25m

R

=

10km

2km

from graph, Gd(dB)= -25.5 dB

find h if Gd(dB)= 6dB

T

=0

h

- forGd(dB)= 6dB v≈ 0
- then = 0 and = -
- and h/2000 = 25/12000 h = 4.16m

25m

R

10km

2km

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

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

- 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

- Let h =maximum protuberance – minimum protuberance
- if h < hc surface is considered smooth
- if h > hc surface is considered rough

hc =

(3.62)

- Reflective Surfaces
- flat surfaceshas 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

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

(3.65)

- (i) Ament, assume h is a Gaussian distributed random variable with a
- local mean, find s as:

s =

(3.63)

- (ii) Boithias modified scattering coefficient has better correlation
- with empirical data

s =

(3.64)

I0 is Bessel Function of 1st kind and 0 order

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

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

(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

angle of incidence

0 10 20 30 40 50 60 70 80 90

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

power densityof 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

3.8.1 Radar Cross Section Model (RCS)

- if a large distant objects causes scattering & its location is known
- accurately predict scattered signal strengths

- 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 fieldof transmitter & receiver

RCS can be approximated by surface area of scattering object (m2)

- measured in dB relative to 1m2reference
- 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

Download Presentation

Connecting to Server..