Page 1
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
Page 2
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 2
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
Page 3
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 3
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
Page 4
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 4
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
Page 5
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 5
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
⎛ ⎝
⎞ ⎠
Page 6
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 6
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
Page 7
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 7
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
Page 8
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 8
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⎧ ⎨ ⎩
Page 9
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 9
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
Page 10
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 10
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
Page 11
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 11
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.
Page 12
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 12
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
Page 13
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 13
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
Page 14
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 14
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
Page 15
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 15
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!
Page 16
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 16
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
Page 17
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 17
See EL675-405.ppt
Rows of Houses on a Hill in San Francisco
Page 18
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 18
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)
Page 19
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 19
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
Page 20
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 20
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
Page 21
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 21
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
Page 22
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 22
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
Page 23
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 23
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
Page 24
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 24
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
Page 25
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 25
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
Page 26
Polytechnic University, Brooklyn, NY
©2002 by H.L. Bertoni 26
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
Page 27
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©2002 by H.L. Bertoni 27
Rows of Trees Next to Buildings
Modify numerical integration to account forPartial transmission through trees
s
d
screen phase
gAttenuatin
screen phase
Absorbing
Page 28
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©2002 by H.L. Bertoni 28
Effect of Trees on Rooftop Fields for a 900MHz Plane Wave Incident at α = 0, 0.5o