polytechnic university, brooklyn, ny ©2002 by h.l. bertoni1 xi. influence of terrain and vegetation...

$: Polytechnic University, Brooklyn, NY ©2002 by H.L. Bertoni1 XI. Influence of Terrain and Vegetation Terrain Diffraction over bare, wedge shaped hills Diffraction$
Polytechnic University, Brooklyn, NY

©2002 by H.L. Bertoni 1

XI. Influence of Terrain and Vegetation

• Terrain Diffraction over bare, wedge shaped hills

Diffraction of wedge shaped hills with houses

Diffraction over rounded hills with houses

• Vegetation Effective propagation constant in trees

Forest with a uniform canopy of trees

Rows of trees next to rows of houses



Influence of Terrain on Path Loss

Adapting theoretical results to various terrain conditions.

€

Location :

A - Use local angle αA to compute Q

B - Account for diffraction loss at top of hill, use local angle

αB to compute Q

C - Use wedge diffration or creeping ray description of the fields

going over the hill

effh A

BC

θBαAα



Diffraction Loss Over Bare, Wedge Shaped Hill

PG = PG0{ }R1 +R2

R1R2

DT φ, ′ φ ( )2⎧

⎨ ⎩

⎫ ⎬ ⎭

Q2 gPB( ){ } PG2{ }

where

PG0 =λ

4π R1 +R2( )

⎡

⎣ ⎢ ⎤

⎦ ⎥

2

and

QgPB( ) =3.502gPB −3.327gPB2 +0.962gPB

3

with gPB =sinαB

d

λ

R ≈R1 +R2

1R

2R

Bα

φ

φ′

( )π−n2



Heuristic Wedge Diffraction Coefficient for Impedance Boundary Conditions

UTD diffraction coefficient

DT φ, ′ φ ( ) =D1w +D2

w +ΓnD3w +Γ0D4

w

For TE or TM polarization, plane wave reflection coefficients

Γ0 =ΓE,H

π2

⎛ ⎝

⎞ ⎠ − ′ φ

⎡ ⎣ ⎢

⎤ ⎦ ⎥ and Γn =ΓE,H φ− n−

12

⎛ ⎝

⎞ ⎠ π

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Partial Coefficients

D1,2w =

−12n 2πk

cotπ ± φ− ′ φ ( )

2n⎡

⎣ ⎢ ⎤

⎦ ⎥ F kLa± φ− ′ φ ( )[ ]

D3,4w =

−1

2n 2πkcot

π ± φ+ ′ φ ( )2n

⎡

⎣ ⎢ ⎤

⎦ ⎥ F kLa± φ+ ′ φ ( )[ ]

where F⋅[] is the transition function

’ (2-n)π

π/2-’ -(n-1/2)π



Transition Function for Wedge Diffraction

Argument : L =R1R2

R1 +R2

and a± β( ) =2cos2 nπN± −β2

⎛ ⎝

⎞ ⎠

where N± are the integers closest to β ±π( ) 2πn

For broad wedge 1<n<1.5( ) and shallow diffraction

π <φ- ′ φ <1.5π( ) and π <φ + ′ φ <2π( )

N+ φ− ′ φ ( ) =1 ; N− φ− ′ φ ( ) =0

N+ φ+ ′ φ ( ) =1 ; N− φ+ ′ φ ( ) =0

and

a+ φ− ′ φ ( ) =2cos2 nπ −φ− ′ φ

2⎛ ⎝

⎞ ⎠ ; a− φ− ′ φ ( ) =2cos2

φ− ′ φ 2

⎛ ⎝

⎞ ⎠

a+ φ+ ′ φ ( ) =2cos2 nπ −φ + ′ φ

2⎛ ⎝

⎞ ⎠ ; a− φ + ′ φ ( ) =2cos2

φ+ ′ φ 2

⎛ ⎝

⎞ ⎠



Example of Wedge Diffraction at 900 MHz

€

Wedge angle = 180o −2×5.711o =168.578o 0.9365π( )

n= 2π −Wedge angle( ) π =1.0635

Half width of transition zone is the same as the Fresnel width

Wf = λR1R2

R1 +R2

≈131000×1000

2000=12.9m

500 500 500 500

10 50 90

12.9

5.71o

188o

2.29o 2.29o

168.6o

5.71o

3.42o



Example – Arguments of Transition Functions

k =2π λ =6π

L =R1R2

R1 +R2

=1000×1000

2000=500

′ φ =3.420o(0.0190π rad) ; φ =188.002o(1.0445π rad)

φ- ′ φ 2

=0.5128π ; φ+ ′ φ

2=0.5318π

kLa+ φ− ′ φ ( ) =6000π cos2 1.0635π −0.5128π( ) =154.2π

kLa− φ− ′ φ ( ) =6000π cos2 0.5128π( ) =8.74π

kLa+ φ+ ′ φ ( ) =6000π cos2 1.0635π −0.5318π( ) =59.4π

kLa− φ+ ′ φ ( ) =6000π cos2 0.5318π( ) =59.4π

Since all arguments >> π , F ⋅[] =1 in all terms



Example – Reflection and Partial Diffraction Coefficients

Angle of incidence : π 2− ′ φ = 0.5−0.0190( )π =0.481π 86.58o( )

φ- n-12( )π = 1.0445−0.5634( )π =0.481π

For Vertical Polarization and εr =15 :

sinθT =115

sin86.58o and θT =14.9o

ΓH =εr cosθ −cosθT

εr cosθ +cosθT

=−0.614

Partial diffraction coefficients : φ- ′ φ =1.0255π, φ + ′ φ =1.0635π( )

D1,2w =

−12×1.06352π ×6π

cotπ ±1.0255π2×1.0635

⎡ ⎣

⎤ ⎦

=0.286

1.146⎧ ⎨ ⎩

D3,4w =

−12×1.06352π ×6π

cotπ ±1.0635π2×1.0635

⎡ ⎣

⎤ ⎦

=0.459

0.459⎧ ⎨ ⎩



Example – Diffraction Coefficient

D =D1w +D2

w +Γ D3w +D4

w( )

=0.286+1.146+ −0.614( ) 0.459+0.459( ) =0.868

Using Felsen coefficient for angle

θ =π − φ− ′ φ ( ) =π 1−1.0445+0.0190[ ]

= −0.0255π −4.59o( )

D=−12πk

1θ

+1

2π −θ⎧ ⎨ ⎩

⎫ ⎬ ⎭

=−1

2π ×6π1

−0.0255π+

12.0255π

⎧ ⎨ ⎩

⎫ ⎬ ⎭

=1.161



Example – Path Gain and Path Loss

PG =λ

4π R1 +R2( )

⎡

⎣ ⎢ ⎤

⎦ ⎥

2R1 +R2

R1R2

D 2⎡

⎣ ⎢ ⎤

⎦ ⎥

=λ

4π

⎛ ⎝

⎞ ⎠

2D 2

R1R2 R1 +R2( )=

13

4π⎛ ⎝

⎞ ⎠

2 0.868( )2

1000×1000×2000

=2.65×10−13

PL =130−10log2.65=125.8 dB

For antenna in free space

PG0 =λ

4π R1 +R2( )

⎡

⎣ ⎢ ⎤

⎦ ⎥

2

=1.76×10−10

PL0 =100−10log1.76=97.5 dB



Wedge Shaped Hills Covered With Houses

Path gain is the sum of the free space path gain, the total excess gain due to the buildings, and the gain for diffraction to mobile.

Compute the total excess gain by replacing the buildings by absorbing half-screens and use numerical integration to go from one screen to the next.

Use line source at the transmitter location for the initial field.



Example – Numerical Evaluation of Roof Top Fields for Houses on Wedge Shaped Hill

MHzf

md

900

50

==

€

For line source radiation

e−jkR

kR

-250



Example – Path Gain for Point Source Excitation from Line Source Results

For line source

Excess Path Gain=20logELS

1 kR

⎡

⎣ ⎢ ⎤

⎦ ⎥ =20logELS +10logkR( )

For point source

PG =10logλ

4πR⎛ ⎝

⎞ ⎠

2

+Excess Path Gain

=10logλ

4πR⎛ ⎝

⎞ ⎠

2

kR⎡

⎣ ⎢ ⎤

⎦ ⎥ +20logELS

=10logλ

8πR⎛ ⎝

⎞ ⎠

+20logELS

At RB =4000m, R=4250m and PG =10log1/3

8π ×4250⎛ ⎝

⎞ ⎠

−67 =−122.1 dB

At RC =1000m, R =1250m and PG =10log1/3

8π ×1250⎛ ⎝

⎞ ⎠

−78 =−127.7 dB



Diffraction Over an Idealized Wedge Shaped Hill with Houses – Analytic Approach

At rooftops beyond the hill

PGB =λ

4π

⎛ ⎝

⎞ ⎠

2 D θB( )2

R1RB R1 +RB( )Q gP1( )Q gP2( )Q gPB( ){ }

2

At rooftops on backside of hill

PGC =λ4π

⎛ ⎝

⎞ ⎠

2 D θC( )2

R1RC R1 +RC( )

Q gP1( )NC

⎧ ⎨ ⎩

⎫ ⎬ ⎭

2

1α2α

1RBR

CR

Bθ

Cθ

Bα



Comparison of Analytic and Numerical Approaches for House at RB=4000m

€

Since Tx is at the same elevation as the highest rooftop

α1 =θC =tan−1 114−152000

⎛ ⎝

⎞ ⎠ =0.04946 rad and αB =θB =tan−1 114−7

4000⎛ ⎝

⎞ ⎠ =0.02674 rad

α2 =θC −θB =0.02272 rad

Also gP =sinα λ d and Q gP( )=3.502gP −3.327gP2 +0.9629gP

3

gP1 =0.606 , Q gP1( )=1.117 while gP 2 =0.278 , Q gP2( )=0.738 and gP3 =0.327 , Q gP3( ) =0.823

Transition parameter S=2kR1RB

R1 +RB

sin2 θB 2( ) =8.985>π so that F (S) ≈1

Thus D θB( )2=

12πk

1θB

−1

2π +θB

⎡

⎣ ⎢ ⎤

⎦ ⎥

2

=11.709

PGB =λ4π

⎛ ⎝

⎞ ⎠

2 D θB( )2

R1RB R1 +RB( )Q gP1( )Q gP 2( )Q gP3( ){ }

2=8.92×10−13

PG( )dB=−120.5dB compared to −122.1 dB Good agreement!



Comparison of Analytic Approaches for House at RC = 1000m

€

NC =100050

=20

D θC( )2=3.398

PGC =λ4π

⎛ ⎝

⎞ ⎠

2 D θC( )2

R1RC R1 +RC( )

Q gP1( )NC

⎧ ⎨ ⎩

⎫ ⎬ ⎭

2

=2.39×10−14

PGC( )dB=−136.2dB compared to −127.7dB

Analytic method is too pesimistic



See EL675-405.ppt

Rows of Houses on a Hill in San Francisco



Diffraction Past Houses on a Cylindrical Hill

-1000 0 1000 2000 3000 4000 5000Screen Placement (m)

0

10

20

30

40

50

60

Screen Height (m) Tx

-1000 0 1000 2000 3000 4000 5000Screen Position (m)-120-110-100-90-80-70-60-50-40-30

Field Strength (dB)



Geometry for Finding Path Gain in the Presence of Cylindrical Hill

At housesbeyond the hill

At houses onthe backside of hill

R

1R1P

2PBα

2R

HR

θ̂

Tangent Points

R

1R1P

HR

θ̂

Tangent Point



Diffraction Coefficient for Cylindrical Hills

0 5000 10000 15000 20000 25000Hill Radius (m)

0.40

0.50

0.60

0.70

0.80

0.90

1.00

1.10

D

900 MHz

1800 MHz

H

0 5000 10000 15000 20000 25000Hill Radius (m)

3.0

4.0

5.0

6.0

7.0

8.0

9.0

D1

900 MHz

1800 MHz



Path Gain at Rooftops of Houses on Cylindrical Hills

€

At houses beyond the hill

PGB =λ4π

⎛ ⎝

⎞ ⎠

2D1

2e−2ψ ˆ θ

kR1R2RQ gPB( ){ }

2

where the attuation constant is

ψ =2.02πRH

λ⎛ ⎝

⎞ ⎠

13

−1.04dλ

At houses on the backside of hill

PGC =λ4π

⎛ ⎝

⎞ ⎠

2DH

2 e−2ψ ˆ θ

R1R

0 5000 10000 15000 20000 25000Hill Radius (m)

40

50

60

70

80

90

100

110

120

ψ

d=50m

d=100m

James



Influence of Trees

Canopy versus trunk

For elevated base station, canopy is important For mobile to mobile links, trunk give dominant effect over short links

Leaves and branches

Scatter and absorb wave energy Mean field dominates over short distances For short distances, attenuation ≤ 20 dB

Waves propagate as exp[-j(k+)L] = ’ - j” is the change in phase constant and attenuation constant

depends on polarization and direction of propagation At longer distances incoherent field dominates

Isolated trees vs. small group of trees vs. forests



and for Horizontal Propagation Through Canopy

′ ′′

€

κ"= attnuation in nepers/m

α = attenuation dB/m

α =8.7κ"

At f =1 GHz

α =8.7×0.11=0.96 dB/m

Wavenumber in the trees

kT =k+κ =k εr

so that

εr =k+κ( )

2

k2 ≈1+2κk

Polarizability

χ =εr −1≈2κ k

At f =1 GHz

χ =2(0.30−j0.11) (20π /3)

=0.019−j0.010



Propagation to Mobile Inside the Forest

Forest

BSh

THmh

R

α−=θ o90 αθT ≈θC

€

PG=λ

4πR⎛ ⎝

⎞ ⎠

2

T2exp2 HT −hm( )kIm εr −sin2θ[ ]{ }

Provided HT −hm( )kIm εr −sin2θ[ ] ≤3

Note that at 200 MHz and θ→ 90o,

kIm εr −sin2θ[ ] =kIm εr −1[ ] ≈kIm2κk

⎡

⎣ ⎢ ⎤

⎦ ⎥ =Im 2κk[ ]

=Im 20.05−j0.09( )4π /3=−0.471

For this case, HT −hm ≤3 0.471=6.4m

€

Wavenumber in

vertical direction

kT cosθT

=k εr cosθT

=k εr(1−sin2θT )

=k εr −sin2θ



Approximations for Mobile Inside Forest

€

For small angle α

cosθ =sinα << εr −1 = 2κ /k and sinθ=cosα ≈1

Since εr sinθT =sinθ , then cosθT = 1−sin2θT = 1−sin2θ

εr

≈ 1−1εr

For vertical E polarization

T =2 εr cosθ

εr cosθ+cosθT

≈2 εr sinα

εr sinα +1εr

εr −1≈

2εr sinαεr −1

≈2 hBS −HT( )R 2κ /k

(The same approximate T is found for horizontal E polarization.)

PG=λ

4π⎛ ⎝

⎞ ⎠

24R4

hBS −HT( )2

2κ /kexp2 HT −hm( )kIm 2κ /k[ ]{ }

PG/PG0 =4R2

hBS −HT( )2

2κ /kexp2 HT −hm( )kIm 2κ /k[ ]{ }

If hBS −HT =10m, HT −hm =5m and R=1000m, then at f =200MHz,

PG /PG0( )dB =−64.6 dB



Propagation to a Mobile in a Clearing

€

Total field incident on the edge is Ein +ΓEin =(1+Γ)Ein =TEin. Therefore

PG=λ

4πR⎛ ⎝

⎞ ⎠

2

T2D

2 1ρ

≈λ

4πR⎛ ⎝

⎞ ⎠

22 hBS −HT( )R 2κ /k

2

D2 1ρ

Use the transmission coefficient T into forest and Felsen diffraction coefficient

Note that TD is approximately the heuristic diffraction coefficient for a dielectric edge.

Effective height of the edge may be less than the tree height at low frequencies.

Forest

BSh

TH

mh

R

α

x



Rows of Trees Next to Buildings

Modify numerical integration to account forPartial transmission through trees

s

d

screen phase

gAttenuatin

screen phase

Absorbing



Effect of Trees on Rooftop Fields for a 900MHz Plane Wave Incident at α = 0, 0.5o

polytechnic university, brooklyn, ny ©2002 by h.l. bertoni1 xi. influence of terrain and vegetation...

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