6341 notes 29 High Frequency Scattering by Cylinder

Report
ECE 6341
Spring 2014
Prof. David R. Jackson
ECE Dept.
Notes 29
1
High-Frequency Scattering by Cylinder
y
E
i
a
ka
1
x
PEC cylinder
H
i
E  zˆ e  jkx
i
 1   jkx
ˆ
H  y e
 
i
Assume
2
Physical Optics
Physical Optics Approximation
E
i
H
i
Lit
Normal
Dark
Dark region
(not seen by incident
plane wave)
J s  0 on dark region
3
Physical Optics (cont.)
Physical Optics Approximation
Lit
Dark
Locally, the reflection acts like
plane-wave reflection from a
flat surface.
  1 (PEC)
Normal
J s  nˆ  H
Lit region:


  nˆ   H   2 nˆ  H
ˆn  H  nˆ  H i  H r
 nˆ  H
i
i
i
4
Physical Optics (cont.)
Physical Optics Approximation
E
i
H
i
Lit
Lit region:
Dark region:
Dark
i
ˆ
J s  2n  H
Js  0
5
High-Frequency Scattering by Cylinder (cont.)
Physical Optics Approximation
 2 nˆ  H i ,
Lit region
Js  
0,
Dark region
Lit region:

2
  
3
2
y
a
PEC
x
lit
6
High-Frequency Scattering by Cylinder (cont.)
  1   jkx 
J s  2 ˆ  H  2  xˆ cos    yˆ sin     yˆ   e 
  

i
 2 
 zˆ   cos   e  jkx
 
 2 
J s  zˆ   cos  e jka cos
or
 

2
  
3
2
7
High-Frequency Scattering by Cylinder (cont.)
Scattered field:
Consider a z-directed
line source at the origin:
y
I
x
Az 
I
4j
H 0(2)  k  
 2

2
(2)
Ez 

k
A


j

A


I
H

 z
z
0 k  
j  z
4

1
8
High-Frequency Scattering by Cylinder (cont.)
Far field:

2  jk 
Ez ~ 
I
e e
4
 k
Next,
For consider
current atthe
line source to be
located at (x´, y´):
j

4
, include phase shift terms
j k xk
 = phase term  e
k x  k cos 
k y  k sin 
Hence
e

x
y y

x  a cos  
y  a sin  
jka cos  
9
High-Frequency Scattering by Cylinder (cont.)


2 or  jk  4
Ez ~ 
I
e e 
4
 k
j
Hence, letting
e
jka cos  
I  dI  J sz  ad
Hence
dEzs  
 dI
4
2  jk  j 4 jka cos  '
e e e
k
or or

j
2
s
 jk 
4 jka cos  


dEz  
e e e
 J sz   a d 
4
 k

10
High-Frequency Scattering by Cylinder (cont.)
or

j






2
2
 jka cos  
 jk 
4 jka cos  


dEzs  
cos

e
ad

e
e
e
 

4   
  k
Hence
Integrating,

3 /2
j

2
j  ka cos  cos 
s
 jk 
4
Ez 
e e a  cos  e
d 
 /2
2  k 
For the integral
11
High-Frequency Scattering by Cylinder (cont.)
This may be written as


2  jk  4
E 
e e aI
2  k 
j
s
z
where
Hence
I 
3 /2
 /2
cos  e
j  ka  cos  cos  
d 
I      f  x  e jg  x dx
b
Compare with
a
For the integral
Hence, we can identify
f    cos  
g    cos      cos  
  ka
12
High-Frequency Scattering by Cylinder (cont.)
Find
the stationary-phase point (SPP):
SPP
g   cos     cos
 
sin       sin    0
sin        sin  
g  0  0
or
0
0
0
A
  0  0  2 n
  2 n  g     0
(No SPP. Assume   2 n)
0
sin A   sin B
A   B  2 n
or
A  B    2 n
B
  0  0    2 n
 

0     n
2
2
13
High-Frequency Scattering by Cylinder (cont.)
We require the
(b)restriction that
  3 
0   , 
2 2 
From the previous slide,
 

0     n
since
2 2
Also,
0

2

choose
Hence, choose n = -1 :
 

0  
2 2
14
Geometrical Optics
The
specular
point of reflection is the point at which the ray reflects
Specular
point
off and travels to the observation point.
Observation point
We can show that

Specular point
y
s 
s  0
x
15
Geometrical Optics (cont.)

Specular point 
Proof
s  0

    2

   2   s

   2s
y




s 
  
 s    
2 2
x
16
High-Frequency Scattering by Cylinder (cont.)
Note that there is always a stationary-phase point, for all
Then
observation angles (except  = 0).
y
 

0  
2 2
x
17
High-Frequency Scattering by Cylinder (cont.)
Next, calculate the g function at the stationary-phase point:
At SPP:
g   cos     cos
 


g 0  cos   0  cos 0

   
  
 cos         cos   
 2 2 
2 2

  
  
 cos     cos   
2 2
2 2
 
 
 
 sin    sin    2sin  
2
2
2
18
High-Frequency Scattering by Cylinder (cont.)
Next, calculate the second derivative of the g function:
At SPP:
g   cos     cos
 


g  0   cos   0  cos 0
 
  g 0
 
 2sin  
2
Note:
 
g 0  0
19
High-Frequency Scattering by Cylinder (cont.)
Recall:
jg x
Hence theI integral
 f ( x )ise  0 
0
, g   x0   0

j
2
e 4
 g   x0 
, g   x0   0


f    cos  
0 
g    cos      cos  
 
g 0  2sin  
2
 
g  0  2sin  
2
2

 
 
  ka
2
Hence
 

j
2
    j ka 2sin 2 
I ~ cos    e
e 4

2 2
ka 2sin
2
20
High-Frequency Scattering by Cylinder (cont.)


I ~ cos    e
2 2
Use
or
 
j  ka 2sin 
2
2
ka 2sin

e
j

4
2
  
 
cos     sin  
2 2
2
Then we have

 
I ~
sin  e
ka
 2
 
j  ka  2sin    j 
 2
4
e
21
High-Frequency Scattering by Cylinder (cont.)
Recall

j

2
Ezs 
e jk  e 4 a I
2  k 
Then
Therefore
 


j
ka
2sin




j 



2




 2
Ezs ~
e jk  e 4 a  
e 4
 sin   e
2  k 
2
 ka 

j
or

 /   k
or
a
    jk 
E ~
sin  e e
2
2
s
z
 
j  ka  2sin  
 2
22
High-Frequency Scattering by Cylinder (cont.)
Then
a
    jk 
E ~
sin   e e
2
2
s
z
 
j  ka 2sin 
2
Radiation pattern of cylinder (scattered field)
y
y
E
i
a
ka
1
x
x
PEC cylinder
H
i
23
High-Frequency Scattering by Cylinder (cont.)
In the backscattered direction ( = ):
Then
a  jk  j 2 ka 
E ~
e e
2
s
z
Echo width (monostatic RCS):
 E0
We 
 2

2
s 2
z
 1  E



  2  2

We  lim
Note: E0 = 1 in our case.
 
2 E
E0
s 2
z
2
24
High-Frequency Scattering by Cylinder (cont.)
We  2 E
Then
s 2
z
a  jk  j 2 ka 
E ~
e e
2
s
z
Hence
 a
We  2 
 2
We   a
2

   a

(circumference of lit region)
25

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