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3. Problems in Chapter 3 - Section 3.9
P3.1
(a) Consider an aperture A in a conducting ground plane. The equivalence principle says that two
having the same value within a a given region are equivalent in that region. Consider an ewquivalent
problem to the one shown in the figure above. Suppose thbe original fields E, H exist for z  0 but
there is a null field for z  0 . To support the external field surface currents must occur on z  0 .
In addition place a perfect electric conductor over z  0 , on top of which we place a magnetic
current M s . Further, for z  0 , we specify the same field as for the original problem.
Since the tangential components of the aperture field E a are zero just behind the magnetic
conductor then M s  E  zˆ z 0  E a  zˆ . This is illustrated below
Now image the magnetic currents in the ground plane. Thus
13
Hence the equivalent currents are M s  2E a  ẑ A and J s  0
.
Returning to the original problem, the electric field radiated by an aperture containing fields
E a , H a is
E(r , ,  ) 
jk
4


e  jkR
A R  M   o J  Rˆ Rˆ dS '
where M  E  ẑ A and J  zˆ  H A
M  Ms
and
E(r , ,  ) 
(b)
jk
2
. In the equivalent problem shown above
J  0 . Hence
e  jkR
A R zˆ  E a  Rˆ dS '
.
The distance to the field point P is R  r  r'
. When r  r '
R  r  rˆ  r' . In the amplitude of the integrand a greater degree of approximation is used R  r
and Rˆ  rˆ while in the phase function let R  r  rˆ  r' . Hence
E
jk e  jkr
rˆ  zˆ  N 
2 r
where N 

A
.
E a exp( jkrˆ  r ' ) dS '
.
rˆ  zˆ  N  rˆ  Nzˆ  rˆ  zˆ N where
rˆ  xˆ sin  cos   yˆ sin  sin   zˆ cos
. Therefore
rˆ  N   sin  N x cos   N y sin   and rˆ  zˆ   cos . Now converting from rectangular to
spherical co-ordinates using the relations in Appendix A, we find that
rˆ  zˆ  N   ˆ N x cos   N y sin    ˆ N x sin   N y cos  
Hence
.
14
Er  0
E 
jk e  jkr
N x cos   N y sin  
2 r
E 
jk e  jkr
cos   N x sin   N y cos  
2 r
As given in eqns. 3.26
.
15
P3.2
For a coaxial transmission line terminated in an infinite ground plane, the fields in the aperture due
to the presence of the TEM mode are
E   ˆ
V
1  j z
e
 r ln( b / a) 
and
H
1
o
zˆ  E
where ˆ  xˆ cos   yˆ sin  , V is the voltage between the conductors and  r is the dielectric
constant of the material between the conductors. Assuming that the reflection coefficient at the
aperture is  , the aperture electric field is approximately given by E a  1    Ez 0 . The
radiated field can be calculated using the results of P3.1, the current on the aperture surface is
M s  2 1   E  zˆ z 0
and the field is obtained from the transform function
2V 1   
N
d '  d ' exp  jk ' cos(   ' )xˆ cos  ' yˆ sin  ' .
 r ln( b / a) 0
a
2
b
The integrals over  ' can be completed in closed form my means of the identities
2
cos  '
cos  
 d ' sin  '  exp jZ cos(   ' )  2j J (Z )sin  
1
0
d
J 0 (Z )   J1 (Z )
dx
where J n (Z ) is a Bessel function of order n. Therefore,
16
4j V 1   
Nx  
cos   d ' J 1 ( w sin  )
 r ln( b / a)
a
b

4j V 1   
cos  J 0 ( wb)  J 0 ( wa)
 r ln( b / a)
where w  k sin  . Similarly,
Ny 
4j V 1   
sin  J 0 ( wb)  J 0 ( wa)
 r ln( b / a)
.
The far-zone electric field components are
2V 1    e  jkr
J 0 (wb)  J 0 (wa),
E (r , ,  )  
 r ln( b / a) r
E (r , ,  )  0  E r (r , ,  ) .
This field amplitude is independent of the azimuth direction  due to symmetry of the geometry and
the polarization varies about the z-direction. When   0 and    , along the axis the field goes to
zero. When the transmission line is small in terms of wavelengths, kb, ka  0 as is the case for
conventional coaxial cables then J 0 ( Z )  1  Z 2 / 4 and
2
2
V 1    e  jkr
2 b  a 

k sin   
E (r , ,  ) 
 r ln( b / a) r
 2 
,
which shows the radiated field is maximum and normal to the infinite conducting plate.
Consider the example of a transmission line with a slightly larger dimensions, kb  1 and ka  0.5 ,
filled with a dielectric with  r  2.5 and   0.3 , the radiated field in the far-field is shown below
In 3-D the amplitude is doughnut-shaped.
17
P3.3
Given the aperture field
E a  xˆ Eo exp(  jx) ; x  a ;
Also H a 
1
o
y  b.
zˆ  E a .
The Fourier transform function of the aperture field is
a
b
N x   dx' exp( j 2ux' jx' )  dy' exp( j 2uy' ) .
b
a
The integrals can be evaluated as follows:

b
b
dy' exp( j 2uy' )  2bS (2vb)
a
 dx' exp j(2u   ) x'  2aS (2u   )a.
a
Therefore
N x  4abS (2u   )aS (2vb) and also
Ly 
1
o
Nx
We see that the beam maximum has shifted from u  0 to u   / 2 , v  0 . That is
1

sin  o   / 2 and o  0
  

 2 
ie.  o  sin 1 
.
For example, when ka  4  2a /  , the beam maximum varies with
a, radians
0
/12
/8
/4
/3
a as follows:
o, deg.
0
3.75
5.61
11.32
15.2
Note that the only affect of the phase slope over the aperture field is a shift in the beam maximum,
the radiation pattern remains the same.
18
P3.4
This time the aperture field has a quadratic phase factor
E a  xˆ Eo exp(  jx 2 ) ; x  a ;
where x 2  1 . Also H a 
1
o
y b
zˆ  E a .
The Fourier transform function of the aperture field is
a
b
N x   dx' exp( j 2ux' jx' 2 )  dy' exp( j 2uy' ) .
b
a
The integrals can be evaluated as follows:

b
b
dy' exp( j 2uy' )  2bS (2vb)
 dx' exp  j(2ux'x' )  2aS 2ua  j 
a
2
a
a
a
dx' x' 2 exp( j 2ux' ) .
The last integral is evaluated by repeated use of integration by parts. Finally

a
a
dx' x' 2 exp( j 2ux' )  2a 3 S (2ua) 
a
cos(2ua)  S (2ua)
(u ) 2
a
2
2
 dx' exp( j 2ux' jx' )  2aS (2ua) (1  j a ) 
a
and
j
cos(2ua)  S (2ua) . Finally
2(u ) 2

j
cos(2ua)  S (2ua)
N x (u, v)  4abEo S (2vb)S (2ua) (1  j a 2 ) 
2
2(u )


For moderately large apertures, N x gives the main lobe and the first few sidelobes. Thus in the Eplane (   0,  ), v  0 and

j
cos(2ua)  S (2ua)
N x (u,0)  4abEo S (2ua) (1  j a 2 ) 
2
2(u)


When   0 , there is no phase error then
N x (u,0)  4abEo S (2ua) .
Next consider when u  0 . If   0 then
N x (0,0)  4abEo .
If a 2 is small on axis
.
19


j 2ab

N x (0,0)  Eo 4ab (1  j a 2 )  Lim
cos(2ua)  S (2ua) .
2
u 0 (u )


sin x
x2
x2
For small arguments S ( x) 
and cos x  1 
. Therefore
 1
2
x
6
1
 cos( xa)  S ( x) 
Lim 
 .
2

x 0
3
x


This results in
N x (0,0)  4abEo (1  j
 a2
3
)  4abEo exp(  ja 2 / 3) .
A small quadratic phase produces a phase shift on axis but maximum gain is unaffected.
Next consider the pattern nulls. In the error-free case these occur in the v  0 plane when
S ( 2ua )  0 ie. when u  n / 2a ; n  1, 2,  . In these ‘null’ angles when there is quadratic phase
error
N x (n / 2a,0)   j
ie.
 8a 3b
(1) n Eo .
2
(n)
N x (n / 2a,0)  
8a 3b
Eo .
(n) 2
The approximate power level at the n-th null relative to the peak level at u  0  v is
 2a 2
PDn  
2
 (n)
2

 .

The depth of the first null ( n  1) is calculated from the above formula is:
a2 (radians)
0
/36
/16
/8
/4
Null Depth
(dB)
0.00
-35.05
-28.00
-21.98
-15.96
The row highlighted is the case that was requested.
Finally, consider the first sidelobe level. The peak occurs when S (2ua )  0.2173 ie. -13.26dB
when 2ua  4.494 or u  2.247 / a . At the sidelobe peak
20



N x (2.247 / a, 0)  2abEo 2(0.2173)(1  ja 2 )  j
(0.2167  0.2173)
2
(2.247 / a)


Neglecting the second term in the square braces as it is small
N x (2.247 / a, 0)  4abEo 0.2173(1  ja 2 )
ie.
N x (2.247 / a, 0)  4ab Eo 0.2173 1  (a 2 ) 2
 4ab Eo 0.2173
.
Hence the first sidelobe peak increases only slightly when with small quadratic phase error.
In summary, the main effect on the pattern of a small quadratic phase error is a filling in of the
pattern nulls.
.
21
P3.5
Given the aperture electric field
E a  xˆ Eo exp j( x) ; x  a ;
Also H a 
1
o
y b
zˆ  E a . The approach adopted is a more general one than the previous problems.
The far-field is determined from
a
N x  Eo  dy'  dx' exp  jk sin  ( x' cos   y' sin  )exp  j( x' ) .
b
b
a
Let U  k sin  cos  and V  k sin  sin  then
a
N x  Eo  dy' exp( jVy' )  dx' exp( jUx' ) exp  j( x' )
b
b
a
where

b
b
dy' exp( jVy' )  2bS (Vb) .
Suppose the maximum deviation of (x ) across the aperture is
( x)  ( x)  ( x)  M
;x a
as used by Cheng (1955) where M is a finite number and (x) is the average value. The factor
exp(  j( x)) could be taken outside the integral since its value is constant. Also
exp  j( x)  cos   j sin 
 1
1
2  j 
2
Hence the second integral becomes
 1
2
adx' exp( jUx' ) exp j( x' )  adx' exp( jUx' )1  2    adx' exp( jUx' ) exp(  j)
a
.
The first term becomes
a
a
22
1
 1
2
2
adx' exp( jUx' )1  2    adx' exp( jUx' )  2  adx' exp( jUx' )
a
a
a
 1
2
 2aS (Ua)1   
 2

2
 

 2aS (Ua)1 
M 2 
2


The second integral becomes
a
a
a
a
 dx' exp( jUx' ) exp(  j)   j  dx'  exp( jUx' )  0
Combining all results so far we obtain
 2 2
N x  4abEo S (Ua) S (Vb)1 
M 
2


.
The electric field in the far-zone is
E 
jk e  jkr
(1  cos  ) N x cos 
4 r
and
E  
jk e  jkr
(1  cos  ) N x sin  .
4 r
The gain on axis is
Gmax 
4r 2 Pr
PT
where PT 
1
Pr 
2 o
2
 k  1
  2 Nx
 2  r
1
ie. Pr 
2 o
Therefore
1
2
Eo 4ab and
2 o
2
 k  Eo


2
 2  r
2
2
4ab 2 1  
2
2
M2 .
.
23
2
G max
2
4r 2  k  E o
2 2
2
4ab  1  M


2 o  2  r 2
2

1
2
E o 4ab
2 o

k2

4ab 1  
2
2
M2 
4

2
.
4ab 1  
2
2
M2
This compares with a uniformily illuminated rectangular aperture
Go 
4A

2

4
2
4ab .
Therefore
Gmax  Go 1 
2
2
M2 ,
which gives an upper bound on the gain. Clearly when   0
Gmax  Go as required.
24
P3.6
Once again as in P3.5, assume a uniform aperture field. This time simply
E a  x̂ Eo
and H a 
1
o
zˆ  E a . At this stage we could use the results of P3.5 and set   0 and obtain the
required result for a rectangular aperture. However, proceeding with the problem, the far-field is
determined from
a
N x  Eo  d '  d '  ' exp  jw ' cos(   ' )
2
0
a
where w  k sin  and a  D / 2 . Then
a
N x  2E o  d '  ' J 0 ( w ' )
a
 J ( wa) 
 (a ) E o 2 0

 ( wa) 
.
2
The far-fields are obtained from
E 
jk e  jkr
(1  cos  ) N x cos 
4 r
and
jk e  jkr
E  
(1  cos  ) N x sin 
4 r
.
2
On axis   0 and N x  (a ) Eo and therefore
E 
jka2 e  jkr
Eo
2
r
; 0
(x-direction)
The radiated power is
Pr 
1
2 o
 k 2a4  2
 2  Eo .
 4r 
The total input power with uniform illumination is
PT 
1
2
E o (a 2 ) .
2 o
The maximum gain is defined as
25
4r 2 Pr
Go 
PT
1  k 2a4 

 Eo
2 o  4r 2 
1
2
E o (a 2 )
2 o
4r 2

 (ka) 2
That is
Go  (ka) 2
 2a 


  
 D 


  
2
2
as was required.
26
P3.7
A circular aperture of radius a in an infinite ground plane radiates fields
jk e  jkr
E 
N x cos 
2 r
and
E  
jk e  jkr
cos  N x sin 
2 r
where
 J ( wa) 
N x  (a 2 ) Eo 2 0

 ( wa) 
in which w  k sin  . Also the radiated magnetic field is
H
1
o
rˆ  E .
The same aperture in free-space radiates the electric field
E 
jk e  jkr
(1  cos  ) N x cos 
4 r
and
E  
jk e  jkr
(1  cos  ) N x sin 
4 r
.
27
with all variables the same as in the infinite ground plane case. The main differences in the
expressions for the field components are as follows:
Field component
Free-space
E
 1  cos  


2


 1  cos  


2


E
Ground plane
1
cos
These factors on the field are sometimes referred to as Huygen’s factors or obliquity factors.
28
P3.8
3.27
K
D/2
The field radiated by a uniformly illuminated rectangular aperture of width D is from eqn.
dx' exp( j 2ux' )  KDS (uD)
D / 2
where S ( x)  sin( x) / x and K is a constant for fixed azimuth angle. In this instance let   0 and
u  k sin  .
If the aperture has a cosine taper over D , which goes to zero at the edge of the aperture, by eqn.
3.28 the field dependence is
2D
 x' 
dx' cos  exp( j 2ux' )  K
C (uD)
D / 2

D
K
D/2


where C ( x)  cos( x) / 1  (2 x /  ) 2 .
In the case of a uniformly illuminated circular aperture of diameter D , the transform for the field is
(eqn. 3.29)
2
K  d '
0
D/2
 d ' exp  jk sin   ' cos(   ' )  2K
0
D/2
 d ' J
0
(k sin   ' )
0
D
 K  
2
2
.
 J 0 ( wD / 2) 
2 ( wD / 2) 


The aim of this problem is to compare the half-power beamwidth and the first sidelobe level in one
plane for the three cases, namely uniform rectangular, cosine-taper on rectangular and uniform
circular. This can be done through evaluating each of the functions S (x ) , C (x ) and 2 J 0 ( x) / x . The
following results are obtained as listed in the table below.
Aperture type
Rectangular uniform illum.
Half HPBW
0.44

D

Rectangular cosine illum.
0.6
Circular - uniform
illum.
0.52
D

D
1st sidelobe level
 13.2dB
 23dB
 17.5dB
Location of
sidelobe
1.44
1.89
1.64

D

D

D
1st null
1 .0
1 .5

D

D
1.22

D
29
P3.9
Assume the link antennas have cirular apertures and are uniformly illuminated. Also the aperture
field is x-directed. Therefore, E a  xˆ and H a  yˆ /  o .
2
 D  J ( w)
N x  2   1
w
2
2  D  J 1 ( w)
Ly 
 
o  2  w
2
where w  k sin  .
In the far-field
jk e  jkr
E 
N x cos    o cos  L y cos 
4 r


.
jk e  jkr
 D  J ( w)

(1  cos  ) cos  2   1
4 r
w
2
2
Similarly
E 
jk e  jkr
 cos  N x sin    o L y sin 
4 r


.
jk e  jkr
 D  J ( w)

(1  cos  ) sin  2   1
4 r
w
2
2
The total radiated power per unit solid angle is
2
2
  D  2   J 1 ( w)  2
1
1  k 
2
2
Pr 
E 


 (1  cos  ) 2    
2 o
2 o  4r 
  2    w 
.
A study of the geometry of the path shows that the impact of curvature of the ray-path will be small.
30
The cosine law gives
d 2  Re  h1   Re  h2   2Re  h1 Re  h2 cos
2
2
Assume h1  h2 and Re  h1 , h2
d 2  2 Re2  2 Re2 cos   2 Re2 (1  cos  )
 4 Re2 sin 2

2
That is
sin

2

d
2 Re
.
Also
Re  h3  Re  h1  cos
Therefore
h3  h1 cos

2
 h1 1  sin 2

2
 d
 h1 1  
 2 Re



2

2
.
.
31
That is h3  h1 . For the specified geometry shown below
 70 

  0.2674
15000


 b  tan 1 
. This is now substituted into the normalized power pattern
of the circular aperture with diameter D . That is
 J ( ) 
Pr  (1  cos  )  1

  
2
2
where  
D
sin 

. A variety of aperture diameters were tried when    b . It has been found
that an aperture with diameter of  148 would ensure that the beam out to 6dB would be clear of
blockage. The calculations are given below.
Beamwidth Circular Aperture
Enter diam eter):
(
D
Enter test angle (deg):
P(  )
P0
(1
cos (  ) ) 

P 6 
20 log
180
P0
P6( 6 )  6.027
h


15000tan
6 
180
 h  70.006
6
0.2674
J1(   D  sin (  .0001) )
(   D  sin (  .0001) )
P( 0 )
P6( 6 )
148
32
P3.10
In the Fresnel region, the radiated electric field is given by
jk e  jkR  jk \ 2 / R
rˆ  M s  dS '
E
e
4 R A
where R  r  r' and
rˆ  M s  2rˆ  zˆ  E a 
 2 cos  E a
.
Assume E a  x̂Eo where Eo is a constant. Therefore
a
2
jk e  jkR 2
d '  d '  ' 2 cos  E o e  jk ' / R

0
4 R 0
 jkR
a
j
e
 xˆ E o
2  d '  ' cos  exp(  jk ' 2 / R)
0

R
 ka 2   jkR
e
 xˆ 2 j E o cos  sin 
 4R 
E  xˆ
The radiated power per unit solid angle is
Pr 
 ka 2
1
1
2
E 
4 R 2 E o2 sin 2 
2 o
2 o
 4R
E2
 o
2 o
 A   sin  
 

    
2
where   ka 2 / 4 R and A  a 2 .



33
The total radiated power is
PT 
E o2
A
2 o
and therefore the gain at R is
4A  sin  
G 2 

   
4A
2
 2 S ( )
2
.

The gain in the far-field is Go 
4A

2
. Therefore, G  Go S ( ) .
2
The relative gain is required at R  2a 2 /  . At this distance G  0.81Go and at four times this
distance at R  8a 2 /  , G  0.99 Go .
34
P3.11
Suppose the aperture is subdivided into N annuli A  A1  A2   AN . At the radial limits of the
n-th annulus the path lengths are separated by  / 2 . Thus
rn  R  n / 2 at the lower rim of the annulus and rn1  R  (n  1) / 2 at the upper edge. The
radiated electric field at P is
E
jk e  jkR
F dS '
4 A R
where F is a generalised vector current which is uniform over the aperture.
Let R  r  r' where
R  r  r'

R  n / 22   ' 2
 n 
 R  nR      ' 2
 2 
2
2
n   ' 
 R 1
 
R R
 R
 '2
R

n
2
Thus for the n-th annulus
2
.
35
En 
jk e  jkR
exp( jkn / 2)  exp( jk ' 2 / R)F dS '
4 R
An
.
Let the n-th integral be given by
 exp( jk '
2
/ R)F dS '  vˆ Fn  Fn 1  / 2
An
where v̂ is a field direction and Fn and Fn1 are contributions from the rims of the annulus. Then
the contribution from An is
En 
jk e  jkR
vˆ exp( jkn / 2) Fn  Fn 1  / 2 .
4 R
The total field at P is this
N
E   En 
n 1
jk e  jkR N
vˆ  (1) n Fn  Fn 1  / 2
4 R
n 1
jk e  jkR

vˆ  F1  (1) N FN 1 / 2
4 R


.
Therefore, the contributions to the integral cancel except for the first and the last. Clearly, the field
decays as 1 / R with increasing distance. However, the amplitude goes through a series of maxima
and minima because as R increases, N decreases . A description of this phenomena with Fresenel
zones can be found in the references (Silver, 1946), (Born & Wolf, 1959).
36
P3.12
Define the following functions:

E-field pattern function A( ) occurs at   0 and

E-field pattern function B ( ) occurs at   90 .
A suitable field function is E(r , ,  ) 


e  jkr ˆ
 A( ) cos   ˆ B( ) sin  .
r
Now the co-polar direction is defined p  ˆ cos   ˆ sin  and the cross-polar direction
q  ˆ sin   ˆ cos  . Therefore
E p  E  pˆ 
e  jkr
A( ) cos 2   B( ) sin 2 
r

e  jkr
 A( )  B( )sin  cos 
E q  E  qˆ 
r

and
.
When    / 4 , cos   1 / 2 . Therefore
e  jkr 1
 A( )  B( )
Ep 
r 2
Eq 
e  jkr 1
 A( )  B( )
r 2
and
.
That is the co-polar pattern is the average of the E- and H-plane patterns and the cross-polar
patterns is obtained from the difference of the two principal plane patterns.
37
P3.13
Consider the integral
I (h)   f ( z ) e jh ( z ) dz
C
where C is a contour in the complex z-plane, h is a large positive parameter and
( z )   cos(  z )
. Assume there a single stationary point of the exponential at z  z s
Expand  (z ) and f (z ) in power series at this point. First consider  (z ) expanded as
( z )  ( z s ) 
' ( z s )
' ' ( z s )
(z  zs ) 
(z  zs )2  
1!
2!
d ( z )
, etc. The stationary point occurs at ' ( z )  0 . Thus,
dz
' ( z )  sin(   z )  0 . That is z  z s   .
where  ' ( z ) 
At this point we can find ( z s )  1 and ' ' ( z s )  1 . Let us now make a change of variable so
that s  z  z s . Therefore, the phase function can be approximated as
1
 ( s)  1  s 2
2
Note that
.
ds
 1 and also the change of variable will be used in the integral. In addition, the
dz
contour is varied to along the real axis around the stationary point. Thus

I ( h)   f ( s  z s )

where G ( s )  f ( s  z s )

dz jh ( s )
e
ds   G ( s) e jh ( s ) ds .

ds
dz
 f (s  z s ) .
ds
Expand G (s ) in a power series also about s  0 . Thus
G( s)  G ( 0) (0) 
G (1) (0)
G ( 2) (0) 2
s
s 
1!
2!
where G ( n ) (0)  d n G ( s ) / ds n | s 0
I ( h) ~ e
 jh
. Therefore

G ( n ) (0) n
 h 
s exp  j s 2 
 ds 
n!
 2 
n 0

.
We now make use of the following identity (Felsen & Marcuvitz, 1973, p. 384)
38




ds s n exp  s 2

 (n  1) / 2
  ( n 1) / 2 ; n even


 0 ;
n odd
.
Therefore
G ( 2 n ) (0) (2n  1) / 2
( 2 n 1) / 2
n  0 ( 2n) ! 
h
 j 
2


I (h) ~ e  jh 
.
We re-express the following as
1..3.5....(2n  1)
(2n  1)!!
where
(2n  1) / 2  n  12   12 
 
n
2
2n
n!! n.(n  2)!!
and
1
h

 j 
2


( 2 n 1) / 2
1
  j / 2 h 
e

2

( 2 n 1) / 2
 e j ( n 1 / 2 ) / 2
(2) ( n 1 / 2 )

h ( n 1 / 2)
2 j / 4 2 n e jn / 2
e
h
hn
so
that
I ( h) ~

e  jh
2 e
h
e  jh

n 0
2 e
where a n 
G ( 2 n ) (0) 2 n e jn / 2 (2n  1)!!

hn
2n
n  0 ( 2n) !

an
h
h
j / 4
j / 4
n
G ( 2 n ) (0) jn / 2
e
(2n  1)!!
( 2 n) !
.
G ( 0)  G(0)  f ( ) and therefore a0  2 e j / 4 f ( ) . To obtain higher-order terms, we follow
Felsen & Marcuvitz (1973, p. 431). Let G ( s )  f ( z ) w( s ) where w( s ) 
G (s ) results in
G (1) (s)  f ' ( z) w(0) 2  f ( z )w(1) (0)
G ( 2) (s)  f ' ' ( z ) w(0) 3  3 f ' ( z ) w(0)w(1) (0)  f ( z )w( 2) (0) .
By means of L’Hôpital’s rule
w(0) 
2
 (zs )
( 2)
dz
. Taking derivatives of
ds
39
2  ( 3) ( z s )
w (0) 
3  ( 2) ( z ) 2
s
(1)
2
(1)

  ( 4) ( z s )
 ( 3) ( z s )
3 
2
 w (0) 

4
2
(1)
w (0)  

w
(
0
)

w
(
0
)
w
(
0
)





( 2)
w(0)  2 
4
 ( z s )  4!




( 2)
As a result
G( s)  2 f ( z )
and
1


G ( 2) ( s )  2 2  f ( )  f ' ' ( )
4


Therefore
a0  2 e j / 4 f ( )
and
1


a1  2e j / 2  f ( )  f ' ' ( ) 
4


.
Finally, to second –order, the integral is asymptotically given by
I (h) ~ e  jh
2j
h

j 
1

 f ( )  2h  f ' ' ( )  4 f ( ) 



.
.
40
P3.14
Assume a uniformly illuminated ature with M s  2 nˆ  xˆE o  and J s  0 .
The electric field radiated in the near-field by this aperture is given by
E


jk e  jkR ˆ
R  M s dS '
4 A R
where R  r  r' with
r'  xˆ ' cos  ' yˆ ' sin  '
r  r xˆ sin  cos   yˆ sin  sin    zˆr cos  .
, z' 0 and
Thus
R  r  r'
 xˆ r sin  cos    ' cos  '  yˆ r sin  sin    ' sin  '  zˆr cos  
R̂ 
.
R
R

R r  r'
Rˆ  M s  2 Rˆ   yˆE o 

2E o
 xˆr cos   zˆ r sin  cos    ' cos  ' 
r  r'
.
The x-directed component of the far-field will be evaluated. This is
Ex  
jk
cos 
2

2
0
a
d '  d '  ' r
0
exp(  jkR)
r  r'
2
.
In this example 2a  50 . The radiated field given by this component is evaluated at (a) r  10 ;
(b) r  100 ; and (c) r  5,000 . The far-field distance is Ro  2 D 2 /  . This corresponds to the
41
distance Ro  2 D 2 /   5,000 . Plots of the radiation patterns at the three distances (a) to (c) are
shown below.
42
P3.15
The triangular aperture field distribution in the x-direction and uniform in the y-direction results in
the aperture field
 2

E a  xˆ  x E1  E o   E o 
 a

;x 
a
2
.
Assume M s  2 nˆ  E a and J s  0 . In that case by eqns. 3.26 the far-fields are given by
E 
jk e  jkr
N x cos   N y sin 
2 r
E 
jk e  jkr
cos   N x sin   N y cos  where
2 r



and

N   Ea ( x' , y' ) exp( jk sin  ( x' cos   y' sin  )) dx' dy'
A
,
which is evaluated at z' 0 . In this case N y  0 and
Nx  
a/2
a / 2
 2

dy '  x E1  Eo   Eo  exp( jw( x' cos   y ' sin  ))
b / 2
 a

dx 
b/2
where w  k sin  . The integral with respect to y’ is given by
b/2
1
b / 2 dy' exp( jwy' sin  )  jw sin  exp( jw sin  ) .
b / 2
b/2
43
That is
 b

dy ' exp( jwy' sin  )  bS  w sin  
b / 2
 2


b/2
where S ( x)  sin( x) / x . Thus
 b
 a/2
 2

N x  bS  w sin    dx'  x' E1  Eo   Eo  exp( jwx' cos  ) .

a
/
2
 2

 a

Consider the first term under the integral ie.

a/2
dx' x' exp( jKx' )  
a / 2
a/2
0
dx' x' exp( jKx' )  
0
a / 2
dx' x' exp( jKx' ) .
The integral has a general form namely
 dx x exp( x ) 
exp( x )
2
x  1
.
Therefore
a/2

a/2
0
exp( jKx' )
 jKx'1
dx' x' exp( jKx' ) 
( jK ) 2
0
.
 a
 a
exp  jK 
exp  jK 
 2 a 
 2  1

jK
2
( K ) 2
( K 2 )
Similarly
a/2
exp( jKx' )
 jKx'1
dx
'
x
'
exp(
jKx
'
)

a / 2
( jK ) 2
0
0
a
a


exp   jK 
exp   jK 
1
2 a
2




  
2
2
jK
( K )
( K )
 2
.
Therefore

a/2
dx' x' exp( jKx' ) 
a / 2
That is
a   a
a  1

exp  jK   exp   jK   2

2( jK )   2 
2  K

  a
a  2

exp  jK 2   exp   jK 2   K 2



 
44

a/2
dx' x' exp( jKx' ) 
a / 2
a2
2
2
 a 2
 a 2 a
S  K   2 cos K   2 
2
 2 K
 2 K
 a 2
S K   2
 2 K
  a 
cos K 2   1
 
 
a2  a  4
 a
 S  K   2 sin 2  K 
2  2 K
 4
2
a2   a  1  a  
 S  K   S  K  
2   2  2  4  
Therefore
2
  a
 b
 
 1  a
 
 a

N x  ab S w sin   E1  Eo  S  w cos    S  w cos     Eo S  w cos   .
 2
 
 2  4
 
 2

  2
The results for two special cases:
(a) When E1  E o the illumination is uniform and
 b

 a

N x  ab S  w sin   Eo S  w cos   which was obtained previously.
 2

 2

(b) When E1  0 the illumination at the edge goes to zero
 b
E
N x  ab S  w sin   o
 2
 2
 a

S  w cos  
 4

2
From these transform functions we evaluate the far-field from
jk e  jkr
E 
N x cos 
2 r
and
E  
jk e  jkr
cos  N x sin 
2 r
.
For example, consider a 2 x2 aperture with uniform and triangular illumination. The edge
illumination of the latter is -20dB. The radiation patterns in the E- and H-planes for uniform and
triangular illuminations are shown below.
Uniform illumination case ( E1  1):
45
2.5
0
10
20
EEdB(  PII )
EHdB(  PII ) 30
40
50
60 60
0
0
20
40
60
80
100
90
60
80
100
90

Triangular illumination case ( E1  0.1 ).
2.5
0
10
20
EEdB(  PII )
EHdB(  PII ) 30
40
50
60 60
0
0
20
40
