Upload
tobias-richards
View
229
Download
1
Tags:
Embed Size (px)
Citation preview
September 2003 ©2003 by H.L. Bertoni 1
VII. Diffraction by an Absorbing Half-Screen
• Kirchhoff-Huygens Approximation for Plane Wave Diffraction by an Edge
• Geometrical Theory of Diffraction• Uniform Theory of Diffraction
September 2003 ©2003 by H.L. Bertoni 2
Plane Wave Illumination of an Absorbing Half-Plane
E x,y,0( ) =jk4π
E0 1+cosα( )e−jkr
rd ′ y d ′ z
−∞
∞
∫0
∞
∫
where
r = x2 + y− ′ y ( )2+ ′ z ( )
2
zx
y
r
dy
dz
€
E in =E0e−jkx
( x, y, 0 )
September 2003 ©2003 by H.L. Bertoni 3
Evaluation of the Integration Over z
€
Let r = x2 +(y−y')2 +(z')2 = ρR2 +(z')2 where ρR = x2 +(y− ′ y )2
In the exponent r ≈ρR +( ′ z )2 2ρR
In the denominator r ≈ρR
Then
1+cosα( )e−jkr
rd ′ z
−∞
∞
∫ ≈1+cosα
ρR
e−jkρR exp−jk
2ρR
( ′ z )2⎡
⎣ ⎢ ⎤
⎦ ⎥ d′ z
−∞
∞
∫
Let u=z'e j π 4 k 2ρR so that dz'= e−j π 4 2ρR k( )du and then
1+cosα( )e−jkr
rd ′ z
−∞
∞
∫ =1+cosα
ρR
e−jkρR e−j π 4 2ρR
ke−u2
du−∞
∞
∫
=1+cosα
ρR
e−jkρR e−j π 4 2πk
π
• The Contribution to the integral come from a small region about z = 0
September 2003 ©2003 by H.L. Bertoni 4
Evaluation of the Integration Over y
E x,y,0( ) =E0
jk4π
1+cosαρR
e−jkρR e−j π 4 2πk
d ′ y 0
∞
∫
=E0
ejπ 4
2k
2π1+cosα
ρR
e−jkρRd ′ y 0
∞
∫
where ρR = x2 +(y− ′ y )2 , cosα =x ρR
We distinguish two cases that are most easily solved
1) Well inside the region y > 0 illuminated by the plane wave.
2) Well inside the shadow region y < 0.R
Xy’
y
R
X y’
y
September 2003 ©2003 by H.L. Bertoni 5
Inside the Illuminated Region y > 0
€
E x,y,0( ) =E0
e j π 4
2k
2π1+cosα
ρR
e−jkρR d ′ y −1+cosα
ρR
e−jkρR d ′ y −∞
0
∫−∞
∞
∫⎧ ⎨ ⎩
⎫ ⎬ ⎭
Contribution to the first integral comes from ′ y near y
Contribution to the second integral comes from ′ y near 0
y
y
Re1+cosα
ρR
e−jkρR⎡
⎣ ⎢ ⎤
⎦ ⎥
Interruptedcancellation
Cancellation ofalternate half cycles
September 2003 ©2003 by H.L. Bertoni 6
Evaluating y Integral for y > 0
€
First Integral: use ρR = x2 +(y−y')2 and expand about y'=y
ρR ≈x+(y− ′ y )2
2x in exponent
ρR ≈x and cosα ≈1 in amplitude
Second Integral: ρR = ρ2 −2y ′ y +( ′ y )2 where ρ≡ x2 +y2 and expand
about y'=0
ρR ≈ρ−y ′ y ρ
in exponent
ρR ≈ρ and cosα ≈x/ρ in amplitude
E x,y,0( ) =E0e j π 4
2k
2π2x
e−jkx e−j k
2x(y− ′ y )2
d ′ y −1+x/ρ
ρe−jkρ e
jkyρ
′ y
d ′ y −∞
0
∫−∞
∞
∫⎧ ⎨ ⎩
⎫ ⎬ ⎭
September 2003 ©2003 by H.L. Bertoni 7
Evaluating y Integral for y > 0 - cont.
E x,y,0( ) =E0ejπ 4
2k
2π2x
e−jkx e−j
k2x
(y− ′ y )2
d ′ y −1+x/ ρ
ρe−jkρ e
jkyρ
′ y d ′ y
−∞
0
∫−∞
∞
∫⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪
Let −u =ejπ 4 k (2x)(y−y' ) =1
jky ρe
jkyρ
′ y ⎤
⎦ ⎥ ⎥ −∞
0
du=ejπ 4 k (2x)dy' =ρ jky( )
e−j
k2x
(y− ′ y )2
d ′ y =-∞
∞
∫ e−jπ 4 2xk
e−u2du=e−jπ 4 2x
k−∞
∞
∫ π
September 2003 ©2003 by H.L. Bertoni 8
Evaluating y Integral for y > 0 - cont.
E x,y,0( ) =E0ejπ 4
2k
2π2x
e−jkx e−jπ 4 2πxk
⎛
⎝ ⎜ ⎞
⎠ ⎟ −
1+x/ ρρ
e−jkρ ρjky
⎛
⎝ ⎜ ⎞
⎠ ⎟
⎧ ⎨ ⎩
⎫ ⎬ ⎭
=E0e−jkx+E0e
−jπ 4 e−jkρ
ρD θ( )
where y ρ =sinθ, x ρ =cosθ
and
D θ( ) =−12πk
1+cosθ2sinθ
The first term is the incident plane wave.
The second term is a cylindrical wave
propagating away from the edge
with pattern function D θ( ).
Y
x
Incident plane wave
September 2003 ©2003 by H.L. Bertoni 9
Inside the Shadow Region y < 0
€
E x,y,0( ) =E0
e j π 4
2k
2π1+cosα
ρR
e−jkρR d ′ y 0
∞
∫
Contribution to the integral comes from ′ y near 0
yy
Re1+cosα
ρR
e−jkρR⎡
⎣ ⎢ ⎤
⎦ ⎥
Interruptedcancellation
Cancellation ofalternate half cycles
September 2003 ©2003 by H.L. Bertoni 10
Evaluating y Integral for y < 0
€
ρR ≈ x2 +(y− ′ y )2 = ρ2 −2yy'+(y')2 ≈ ρ2 −2yy' where ρ≡ x2 +y2
In the exponent use ρR ≈ρ−y ′ y ρ
in exponent
In amplitude use ρR ≈ρ and cosα ≈x/ρ=cosθ
E x,y,0( )=E0
ej π 4
2k
2π1+x/ρ
ρe−jkρ e
jkyρ
′ y
d ′ y 0
∞
∫
Since ejk
yρ
′ y d ′ y
0
∞
∫ =ρjky
exp jkyρ
y'⎛
⎝ ⎜ ⎞
⎠
⎤
⎦ ⎥ 0
∞
=−1
jkyρ=
e−j π 2
−ky ρ( )
Then E x,y,0( )=E0e−jπ /4 e−jkρ
ρD θ( )
Where D θ( )=12
k2π
1+x/ρ−ky ρ( )
=−1
2πk
1+cosθ2sinθ
This is a cylindrical wave propagating away from the edge with pattern function D(θ)
September 2003 ©2003 by H.L. Bertoni 11
Geometrical Theory of Diffraction (GTD)Total field
EGTD =EG0U θ( )+ED
Incident plane wave
EG0 =E0e−jkx
Diffracted wave
ED =E0e−jπ / 4 e−jkρ
ρD θ( )
D θ( ) =−12πk
1+cosθ2sinθ
GTD is not valid near the shadow boundary θ =0 where D θ( ) =∞.
()
y
x
Shadowboundary
DiffractedCylindrical wave
IncidentPlanewave
September 2003 ©2003 by H.L. Bertoni 12
GTD Valid Outside Transition Region
Fresnelzone
Shadowboundary
Incidentplanewave
y
x
€
Example: If x =10 m and f =1GHz, λx = 3=1.7 m.
λx
€
GTD is not valid in
the transition region
September 2003 ©2003 by H.L. Bertoni 13
Example of Shadowing at Building Corners
Building
y
x
FromTransmitter
2 m
For f =1 GHz at y =−2 m
in shadow region
ρ = 102 +22 =10.2
θ =arctanyx
⎛ ⎝
⎞ ⎠
=−11.3°
k =2πf / c=20.9
€
ED =E0
1
10.2
1
2π ×20.9
1+cos11.32sin11.3
=0.138E0
−20log(0.138)=17 dB
Signal at 2 m into the shdow region is 17 dB below incident signal.
September 2003 ©2003 by H.L. Bertoni 14
Uniform Theory of Diffraction for Small y
€
ρR = x2 +y2 −2y ′ y +( ′ y )2
In exponent use Taylor series expansion in ′ y up to second order
ρR ≈ρ−yρ
( ′ y )+x2
2ρ3 ( ′ y )2 where ρ = x2 +y2
In amplitude ρR =ρ, cosα ≈x/ρ=cosθ
End point integrals for ED :
(y>0) ED =E0
e jπ /4
2k
2π−
1+cosθ
ρe−jkρ⎡
⎣ ⎢ ⎤
⎦ ⎥ exp−jk
x2
2ρ3 ( ′ y )2 −yρ
( ′ y )⎡
⎣ ⎢ ⎤
⎦ ⎥ ⎧ ⎨ ⎩
⎫ ⎬ ⎭ d ′ y
−∞
0
∫
=E0ejπ /4
2k
2π−
1+cosθρ
e−jkρ⎡
⎣ ⎢ ⎤
⎦ ⎥ exp−jk
x2
2ρ3 ′ y 2+
y
ρ′ y
⎡
⎣ ⎢ ⎤
⎦ ⎥ ⎧ ⎨ ⎩
⎫ ⎬ ⎭ d ′ y
0
∞
∫
(y<0) ED =E0
e jπ /4
2k
2π1+cosθ
ρe−jkρ⎡
⎣ ⎢ ⎤
⎦ ⎥ exp −jk
x2
2ρ3 ( ′ y )2 +yρ
′ y ⎡
⎣ ⎢ ⎤
⎦ ⎥ ⎧ ⎨ ⎩
⎫ ⎬ ⎭ d ′ y
0
∞
∫
September 2003 ©2003 by H.L. Bertoni 15
Evaluating End Point Integrals
€
For both y>0 and y <0
ED =E0
e jπ /4
2k
2π−sgn(y)
1+cosθρ
e−jkρ⎡
⎣ ⎢ ⎤
⎦ ⎥ exp0
∞
∫ −jkx2
2ρ3 y'2+
y
ρy'
⎡
⎣ ⎢ ⎤
⎦ ⎥ ⎧ ⎨ ⎩
⎫ ⎬ ⎭ dy'
Complete the square of the exponent
kx2
2ρ3 y'2+
yρ
y'⎡
⎣ ⎢ ⎤
⎦ ⎥ =
xρ
k2ρ
y'+kρ2
yx
⎡
⎣ ⎢ ⎤
⎦ ⎥
2
−kρ2
yx
⎛
⎝ ⎜ ⎞
⎠ ⎟
2
Define s=kρ2
y
x
⎛
⎝ ⎜ ⎞
⎠ ⎟
2
so that - jkx2
2ρ3 y'2+
yρ
y'⎡
⎣ ⎢ ⎤
⎦ ⎥ =−j
xρ
k2ρ
y'+ s⎡
⎣ ⎢ ⎤
⎦ ⎥
2
+js
September 2003 ©2003 by H.L. Bertoni 16
Evaluating End Point Integrals - cont.
€
Then
ED =E0
ejπ /4
2k
2π−sgn(y)
1+cosθ
ρe−jkρ⎡
⎣ ⎢ ⎤
⎦ ⎥ e js exp−j
xρ
k2ρ
′ y + s⎛
⎝ ⎜ ⎞
⎠ ⎟
2⎡
⎣ ⎢
⎤
⎦ ⎥ d ′ y
0
∞
∫
Using variable u=xρ
k2ρ
′ y + s so that d ′ y =ρx
2ρk
du=s
y ρ( )
2k
du
ED =E0
ejπ /4
j2kk
2π−sgn(y)
1+cosθy ρ( ) ρ
e−jkρ⎡
⎣ ⎢
⎤
⎦ ⎥ 2j se js exp−ju2
( )dus
∞
∫⎡
⎣ ⎢ ⎤
⎦ ⎥
or
ED =E0e−jπ /4
2πk−
1+cosθ2sinθ
e−jkρ
ρ
⎡
⎣ ⎢ ⎤
⎦ ⎥ 2j sejs exp−ju2
( )dus
∞
∫⎡
⎣ ⎢ ⎤
⎦ ⎥
September 2003 ©2003 by H.L. Bertoni 17
Evaluating End Point Integrals - cont.
€
Thus
ED =E0e−jπ /4
2πk−1+cosθ2sinθ
⎡ ⎣
⎤ ⎦ e−jkρ
ρF s( )=E0e
−jπ /4 e−jkρ
ρD θ( )F s( )[ ]
where F s( )=2j se js e−ju2
dus
∞
∫
is a transition function that eliminates the singularity at θ=0.
Since s=kρ2
y
x
⎛
⎝ ⎜ ⎞
⎠ ⎟
2
=kρ2
tanθ( )2 so that s>>1 when θ is large enough.
September 2003 ©2003 by H.L. Bertoni 18
Approximation for F(s)
€
Rational Approximation F s( ) = 2πs f 2s/π( )+jg 2s/π( ){ }
Here f ξ( ) ≈1+0.926ξ
2+1.792ξ +3.104ξ2
gξ( ) ≈1
2+4.142ξ +3.492ξ2 +6.670ξ3
where ξ = 2sπ
For s→ 0, F s( ) ≈ 2πse jπ /4
2=
πkρ2
y
xe jπ /4
September 2003 ©2003 by H.L. Bertoni 19
Field at the Shadow Boundary
€
For y→ 0, ρ ≈x, cosθ ≈1, sinθ ≈y x so that D(θ) ≈-1
2πk y x( )
Then ED ≈E0e−jπ /4 −1
2πk1
y x⎡
⎣ ⎢ ⎤
⎦ ⎥ ⎧ ⎨ ⎩
⎫ ⎬ ⎭
e−jkx
xπ kx
2yx
ejπ /4⎧ ⎨ ⎩
⎫ ⎬ ⎭
≈12
E0e−jkx −y
y
⎛
⎝ ⎜ ⎞
⎠ ⎟
For y=0+
E =E0e−jkx+ED =E0e
−jkx 1−12
⎛ ⎝
⎞ ⎠ =
12
E0e−jkx
For y=0−
E =ED =12
E0e−jkx
Total field is continuous at y=0 and is 12
incident field, or 6 dB lower
than the fiel in the absence of the edge.
September 2003 ©2003 by H.L. Bertoni 20
Value of F(s) for Large s
€
Rational Approximation F s( ) = 2πs f 2s/π( )+jg 2s/π( ){ }
When s→ ∞, f ξ( ) ≈1+0.926ξ
2+1.792ξ +3.104ξ2 →0.3ξ
=0.3π2s
gξ( ) ≈1
2+4.142ξ +3.492ξ2 +6.670ξ3 → 0
F(s)→ 2πs 0.3π2s
⎛
⎝ ⎜ ⎞
⎠ ⎟ =0.3π ≈1
On the edge of the transition region y = λx, s=kρ2
yx
⎛ ⎝
⎞ ⎠
2
=kλ2
ρx
=π
and ξ = 2. Since f ( 2)=0.215, g( 2)=0.0297 and
F(π)= 2π(π) 0.215+j0.0297[ ] ≈0.937
Outside the transition region F(s) ≈1
September 2003 ©2003 by H.L. Bertoni 21
Variation of Field Near Shadow Boundary
f =900 MHz
12
Width of Fresnel zone
λx =303
=3.16
Y
x
x= 30 m
-10 -5 0 5 10-30
-25
-20
-15
-10
-5
0
5
Rec
eive
d S
igna
l(dB
)
y(meters) WF-WF