184
9. Problems in Chapter 9 - Section 9.5
P9.1
1  W
w W 
 4.5
2 2
4
f  10
 4.5 

  26.57
 10 
 o  tan 1 
(a) Edge illumination is 1. cos 20  o  19.38 dB.
(b) From Eqs. 9.8 with h  0.005 ,  r  2 and the shape function A( )  cos 20  , the problem is
solved but in a complete form. The solved that was interested, the pattern given below was
obtained.
185
The first sidelobe occurs. The first sidelobe occurs at 19  and has a level of about  17.07 dB
relative to the peak.
(c) To calculate the scattered field contribution, use is now made of Eq. 9.11. This field is attached to
the field calculated in (b). The resulting radiation patterns obtained from the total field is shown
below. The nulls have been filled in and the third sidelobe has split into two. The first sidelobe peak
which occurs near 20° (ie. on Field ) has increased its level to -14.9 dB (ie. in increase of about 2.8 dB
) from the earlier figure.
186
On boresight Eqs. 9.8 become (with u=0=v and  b  0  b )
P9.2
where
e  jkr
E (r , ,  )   Eij (r ,  ,  )  
r
i, j
i , j 1
e  jkr
E (r , ,  )   Eij (r , ,  )  
r
i, j


N
1

N
ij
1
i , j 1
e  jkf cos  Gijx cos   Gijy sin  
e  jkf  Gijx sin   Gijy cos  
jkEo ab
exp(  jkh  r ) cos(kh  r ) . The power radiated by an axisymmetric feed
2
with a radix N power pattern is given by
 Eo
PT 
 o (2 N  1)
2
.
The maximum gain is given by
Gmax 

2r 2
2
2
E (r , 0, 0)  E (r , 0, 0)
PT o

ie.
Gmax 
2
2
2 N  1 
Eo
2
N
2
1
 i

 , j 1 ij
Gijx 


G
)

ijy
 .
i , j 1  ij

N
2
1
At   0 , Gijy  Gijx and therefore
Gmax
.
ij
2
 N 1
 kab 




2
N

1
Gijx

 i
  
, j 1  ij

2




.
187
P9.3
The hexagonal reflectarray consists of 37 elements of 0.6  0.6 waveguide with a regular
spacing of 0.65 .
The length of the waveguide sections are chosen to satisfy the reflectarray condition Eq. 9.1
k (  ij  rij  rˆb )  ij  2m
; i  1,...., N x ; j  1,...., N y
Now in this case rˆb  zˆ and as rij are in the x-y plane, the condition becomes
ij  k ij  2m
where  ij is the radial distance from the focus to element ij.
The pattern of the feed is assumed to be cosine to the radix N ie. cos  where N is chosen to give
N
a -6 dB illumination at the centre of the widest located element. Thus
w
 3.5 x 0.8  and
2
188
 2.6 
 c  tan 1    25.017  .
 6 


The most extreme edge element should be illuminate at -6 dB ie. 20 N log 10 cos c  6 . Ie.
N  7.012 . It is assume that the length of each waveguide element is chosen to meet the
reflectarray condition and thereby cancel out the phase variation across the face of the array due to
the feed. Assume each element radiates perfectly and with an amplitude that is the same as the
N
initial excitation ie. cos  c .
The field radiated by the hexagonal array will be approximately
E ( ,  )  E  ( ,  ) AF ' (u , v)
and
E ( ,  )  E  ( , ) AF ' (u, v)
where E  ( ,  ) and E  ( ,  ) are the element patterns of rectangular waveguide, given by Eqs.
4.5 and AF ' (u, v) is formed from Eq. 7.15. Thus
Nr r
AF ( ,  ) 1  
'
n 1
 1   (u
2
2
 v2 )
 exp j kn
6
n
m 1 l 1

u 2  v 2 sec(2l / n  1) / 6


v
sin  tan 1    (m  1) / 3  l / 3n  
u


where
N r = no. of rings in the hexagonal array (here N r  3 )
  is the spacing between the rings; here    0.8 ,
and also cos   1  2 (u 2  v 2 ) , sin   u 2  v 2 and   tan 1 (v / u ) in the global spherical
co-ordinate system. The radiation patterns in the two principal planes are shown below.
189
To compare these patterns with those of a parabolic reflector of the same equivalent diameter,
consider

1 
  26.268
4f /D
In the reflector the half-cone angle is  c '  tan 1 
The same axisymmetric feed will be used. The radix of this pattern is N  6.34 . The resulting pattern
is shown below. The pattern is broader than the array pattern, which would correspond to lower
edge illumination for the array, which would be close to 0 dB.
190
P9.4 The main way of increasing the bandwidth of a reflectarray is to maintain the reflectarray
condition over a broader range of frequencies.
This can be done by:
-
loading the patches in such an array with slots or loops;
placing the array elements above a surface with properties that changes appropriately with
frequency;
employing structures consisting of multiple layers;
ridge waveguide elements;
aperture coupled elements;
true-time delay lines; and
reflectarray elements with cell-size <  / 2 to enable elements to be placed closer together.
This list is by no means exhaustive. New approaches are described in the literature all the time.
Whatever, the approach adopted, maintaining the reflectarray condition is the paramount concern.
191
P9.5
The profile of the lens is given by Eq. 9.18, which is
z  z ( x)  0.0001x 3  0.0019 x 2  0.1688x  60.066
.
(9.5.1)
The unit normal to this surface is given by
nˆ  
f
f
where
f  z ( x)  z  0.0001x 3  0.0019 x 2  0.1688 x  60.066  z
 z  0.0001x 3  0.0019 x 2  0.1688x  60.066
f  xˆ
f
f
 zˆ
  xˆ 0.0003x 2  0.0038 x  0.1688  zˆ
x
z


Therefore
nˆ 

0.0003x

 0.0038 x  0.1688  1
xˆ 0.0003x 2  0.0038 x  0.1688  zˆ
2
2
The total electric path length is
L  r1  n ( x 2  x1 ) 2  ( z 2  z1 ) 2  r3
where r1  d sec1
and
(P9.5.2)
r3  ( x3  x2 ) 2  ( z3  z 2 ) 2  (h    z 2 ) sec  3 .
192
By Snell’s law n sin  2  sin 1 .
 sin 1 

 n 
ie.  2  sin 1 
The refracted ray co-ordinates at the interfaces of the lens and the outer medium (air) is found by
solving for the root of Eq. P9.5.1 and also
z  z1  ( x  x1 ) cot  2 .
Here z1  0
x1  d tan 1 .
and
The vector form of Snell’s law is
(s R  ns I )  nˆ  0
where s I  xˆ( x2  x1 )  zˆ( z 2  z1 ) .
Thus ( s Rx n z  s Rz n x )  n( s Ix n z  s Iz n x )  0
ie. s Rx n z  s Rz n x  n( s Ix n z  s Iz n x )
.
Also sin  3 '  n sin  2 ' and cos  2 '  nˆ  s I  n x s Ix  n z s Iz and
 3 ' 3  nˆ  zˆ 
1
0.0003x
2

2
 0.0038 x  0.1688  1
.
ie.
 3   3 '
1
0.0003x
2

2
 0.0038 x  0.1688  1
From these equations, the total path can be calculated from Eq. 9.5.2.
193
A plot of the path length is shown below as a function of the feed ray launch angle 1 . The results
show clearly that the lens shape was synthesized for an illuminating source with a half-cone angle of
45 .
194
P9.6
From conservation of power from the source to the aperture, we use Eq. 9.17 to give
Pa (t )

P ( )
x1 (t )
x12 (t )  d 2
z 

t  (z  d ) 
t 

z 

 2 (t ) ( z  d )  t 
t 

where z (t ) is the lens profile, t is the radial distance, and  (t ) 
is given by Eq. 9.18 and is
z (t )  0.0001t 3  0.0019t 2  0.1688t  60.066 .
Therefore
z
 0.0003t 2  0.0038t  0.1688
t
In addition
x1  t  ( z 2  0) tan  2  d tan 
Therefore
tan  
1
t  z 2 tan  2 
d
Also n sin  2  sin 
 sin  
and sin  2  sin 1 

 n 
.
z(t )  d 2  t 2 . The lens profile
195
The aperture field is given by Eq. 9.15 as
E a  xˆA( )e  jkL
Pa (t )
P( )
where A( ) is the power pattern of the feed and L(t )  L1  L2  L3 is the total path length from
the focus to the aperture plane. In this instance, A( )  cos N  . Choose N  3 to give an edge
illumination of -7.7 dB . Therefore
E a  xˆ cos 3  e  jkL
Pa (t )
P ( )
.
The aperture field for the lens profile given by Eq. 9.18 is shown below.
196
P9.7
The field correlation theorem gives (Eq. 3.62)
Prec 
2
1 Sr
2 Re  (E a  H a* )  nˆ dS
(E a  H b )  nˆ dS r

Sa

.
Suppose the incident field is ‘b’ then

Prec 1

Pb
4
(E a  H b )  nˆ dS r
Sr
2
.
PT Pb
where PT 


1
Re  (E a  H a* )  nˆ dS is the power radiated by the feed. Now suppose
Sa
2
(E b , H b )  (E a , H a ) . Then
Prec

Pb

Sr
(E a  H a )  nˆ dS r
2
T
4P
2
 r
2
.
This is now the power from the feed that is reflected from the surface S r . r is the reflection
coefficient where
r  e  j o

Sr
(E a  H a )  nˆ dS r
2 PT
where  o is the electrical phase reference.
197
Suppose the feed radiation has an axisymmetric pattern

e  jk
Ea 
A( ) ˆ cos   ˆ sin 

1
Ha 
o

ˆ  E a
2
1

Re  d sin  A( ) 
Sa
2 o
o
PT 


0
d sin  A( ) .
2
Consider the numerator where the unit vector n̂ is normal to the surface S r and the integration is
done over that surface. Now

g
   ˆ
g


nˆ  

 2  2
 ˆ


 2  2 

where   (n  1) f /( n cos  1) , n is the refractive index and
 

(n  1) f

n sin 
 (n cos  1) 2
.
Now dS r  sin    2  2 d d
and E a  H a 
ˆ E a  E a 
1
o
Therefore
e  j o
2 o PT
 E
e  j o

2 o PT
 E
r 

Sr
 e  j
 o PT

a
 E a ˆ  nˆ  dS r
a
 E a  sin  dd
Sr
c
o

0

d A( ) 2 sin  exp  jk 2  ( )
Finally
c
r  e
 j o

0
d A( ) 2 sin  exp  jk 2  ( )


0
d sin  A( )
2
.
198
In the case of a feed with a pattern that is cosine to the radix N ie. A( )  cos N  , the
denominator is


0
d sin  A( )  1 /( 2 N  1) and let u c  cos c . Thus
2
c

 
r  e  j o (2 N  1)  du u 2 N exp  j

0
 (nu  1) 
where   2kf (n  1) . For example, if N=3, n=2.5 and f=10 cm the reflection coefficient is shown
below as a function of frequency.
Magnitude, dB
Phase, deg.
199
P9.8 Based on a HPBW of about 5°, the desired radius of the Luneburg lens is
R  1.12

HPBW
.
After some minor adjustments around this estimate, a radius of R  6.3  was selected. With a
refractive index of n=-2 and a cosine feed pattern radix of m=1, a gain of 31.77 dBi was achieved
with an efficiency of 95.9%. The radiation pattern has a HPBW of 5° and the pattern is shown below.
200
Ways of achieving the desired variation of refractive index from n  2 at the centre to n  1 on the
outside is to follow the Luneburg refractive index equation in a piecewise fashion as illustrated
below.
201
P9.9
The ray path in a Maxwell lens is given by
r ( ,  )  R  sin  cot  

sin  cot  2  1

(a) If    / 2 . Consider the positive root first. Then

r ( / 2,  )  R  cot   cot 2   1

R
cos   1  R tan 
sin 
2

Similarly, if    / 2 and choose the negative root ie.

r ( / 2,  )  R cot   cot 2   1

R
cos   1   R tan 
sin 
2

Thus r ( / 2,  )   R tan

2
.
(b) The path length through the lens is
o
L( o ,  )  

 r ( ,  ) 

d n(r ) r ( ,  )  
  
2
2
where n(r ) is the refractive index. For a Maxwell lens of radius R , it is given by
n( r ) 
2
r
1  
R
2
 , radians
 ( ) , radians
1.6
493.316
2.
493.316
2.5
493.316
3.
493.316
There does not appear to be closed form solution to the path length integral, and so the path length
is calculated numerically. The electrical phase over the path with launch angle
( )   L o ,   .
 is given by
202
(c) The rays crossing the vertical axis at   / 2 do so in a normal direction and this is indicated by
r /    / 2   . To estimate the aperture field for a given  variation, for conservation of
power Pa (  ) d  P( ) d where Pa (  ) is the power in the aperture and P( ) is the power
launched at angle  . At the aperture there is a significant mismatch as it is half-way the spherical
lens. The power transmission coefficient is
T ( ) 
2  r ( )
.
1   r ( )
Thus the half-way power is P ' ( )  T P ( ) and if this is the aperture then
2
Pa ' (  )  T P ( )
2
d
d
.
The amplitude distribution will be
Aa (  )  Pa ' (  ) 

d
d
T P ( )
2
d
d
 2 n(  ) 

P ( ) 
 1  n(  ) 
Now
d 1

 R sec 2
d 2
2
Aa (  )  2 cos

2
d 2

 cos 2
d R
2
or
P ( ) n(  )
R 1  n(  )
By means of the refractive index equation for the Maxwell lens
n(  )
2R 2

1  n(  ) 3R 2   2
Therefore
Aa (  )  2 cos

2
P ( )
2R
3R   2
2
203
P9.10 At the aperture of a hemispherical Maxwell lens, the refractive index varies with distance
from the centre as
n( r ) 
2
r
1  
R
2
where R is the lens radius. Therefore, simple quarter-wave matching will not be as effective as with
homogeneous materials. For the latter, a perfect match at a centre frequency, say f o , is achieved
between a top layer with refractive index n L and air with n  1if a quarter wave layer of material
with refractive index nI  nL is interposed.
(a) Suppose the choice of n L is based on achieving an average power match. The reflected power
from the lens is
Pt
2

Pinc
where  
1  n(  )  2  R 2

1  n(  )  2  3R 2
1 R
d  (  )
R 0
1 R
 2  R2
  d 2
R 0
  3R 2
.
av 
Let   3R tan 
av 
.
where d  3R sec 2  d and  o  tan 1 (1 / 3)
1 o
3R 2 tan 2   R 2
2
d

3
R
sec

R 0
3R 2 tan 2   3R 2
o
 3  d tan 2   1 / 3
0
o

 3  d 1 / 3  tan 2 
0

1

 3   o  1  sec  o 
3

That is
av  0.57 . Choose n L from
204
av  0.57 
nL  1
.
nL  1
ie. nL  3.65 and therefore nI  1.912 (or  r  3.654 ).
(b) There are several ways of improving the matching obtained beyond simple averages. Two are:
(i) Fashion the lens from discrete shell and choose the values of the refractive index to minimize
aperture mismatch. One recent approach is to sue gradient refractive index metamaterials (eg. Xu et
al., 2014).
(ii) Instead of a single layer of material on the aperture, two or more layers can be used beneficially
to minimize mismatch over a band of frequencies. An example of this approach taken by Qi et al.
(2013).
205
P9.11
Assume initially a plane wave of amplitude A is incident in the negative z-direction onto a ground
plane where the reflection coefficient is 1  1exp( j1 ) . The reflected wave is of the form
A1 exp(  jkz) . This wave impinges on surface 2, which has a reflection coefficient
2  1exp( j2 ) with a magnitude close to 1 but it allows a small part of the power to be
transmitted. The wave reflected from surface 2 is A1 exp(  jkd)2 . This reflected wave once again
is incident on surface 1 and arrives at the surface in the form A1 exp(  jkd)2 exp(  jkd) . And so
the cycle continues. One complete cycle corresponds to a  2 phase-shift. To maximize the signal
over a complete cycle, the phase should be
(1  2 )  2kd  m2
;
m  1,2,
which is the Fabry–Pérot relation Eq. 9.26.
206
P9.12
In region 1, the electric field is expressed
E x1  A1e  jk1z  B1e jk1z
and the associated magnetic field is
H y1  
1
1
A e
 jk1 z
1

 B1e jk1z .
In region 1, which is free-space, k1  k and 1   o . In region 2
E x 2  A2 e  jk2 z  B2 e jk2 z
H y2  
1
2
A e
 jk2 z
2
 B2 e jk2 z

.
Finally, in region 3, which is also free-space, there is no reflected wave and therefore
E x 3  A3 e  jkz
H y3  
1
o
A3 e  jkz
.
The boundary conditions at z=0 and z   require continuity of the tangential field components.
Thus, at z=0
E x1  E x 2
z 0
H y1  H y 2
z 0
(P9.12.1)
(P9.12.2)
207
and at z  
E x 2  E x3
(P9.12.3)
z 
H y 2  H y3
(P9.12.4)
z 
This gives four equations in four unknowns ( A1 is the initial amplitude). The final result is of the form
A3  f (k , k 2 , o , 2 , d , ,  r ) A1 , which is still in the form of a plane wave possibly with a
significantly reduced amplitude compared with A1 depending on the top surface material (  r ), the
separation distance d and thickness  . Now to find this functional relation,
(1) gives A1  B1  A2  B2
(2) gives
1
o
 A1  B1  
1
2
(P9.12.5)
 A2  B2 
(P9.12.6)
(3) gives A2 e  jk2  B2 e jk2  A3 e  jk
And (4)
1
2
A e
 jk2
2

 B2 e jk2 
1
o
(P9.12.7)
A3 e  jk
(P9.12.8)
Take the ratio of Eqs. P9.12.5 and P9.12.6 results in
1  A1  B1  1  A2  B2 

 o  A1  B1   2  A2  B2 
Let R1 
B1
A1
and
R2 
B2
so that
A2
1 1  R1  1 1  R2 

 o 1  R1   2 1  R2 
(P9.12.9)
Also, the ratio of Eqs. P9.12.7 and P9.12.8 give
1 1  R2 exp( 2 jk2 )  1
.

 2 1  R2 exp( 2 jk2   o
Upon re-arranging this gives
  o
R2  exp( 2 jk2 ) 2
 2  o



(P9.12.10)
208
Now Eq. P9.12.9 can be arranged into the form
R1 
 2 1  R2    o 1  R2 
.
 2 1  R2    o 1  R2 
(P9.12.11)
Now from Eq. P9.12.5
(1  R1 ) A1  (1  R2 ) A2
and from Eq. P9.12.7
A3  (1  R2 e 2 jk2 ) A2 e  j ( k2  k )
ie. A3  (1  R2 e
As well (1  R2 e
2 jk2
2 jk2
 1  R1   j ( k2 k )
 A1e
)
 1  R2 
  o
)  1   2
2  o

2 2
 
 2  o
and from (11)
1  R1 
2 2 1  R2 
 2 1  R2    o 1  R2 
.
Therefore
 2 2
A3  
2  o


2 2 1  R2 

 A1 exp  j (k 2  k ) 
  2 1  R2    o 1  R2  
.
The two terms on the right side in large curved braces correspond along with the exponential factor
correspond to transmission coefficients. The latter shows the difference in phase across the
interfaces at z=0 and z   . If    , the term
 2 2 

2 2 



2 2 1  R2 
o 
 2

 2 1  R2    o 1  R2 
 2 2 
 2 0
   o 
 2 
2  o 
2  o




2 22
 22   o2
o
o
2 22
2 2
2
2

;
;



2
2
 r 2 o
2  o 1   r 2
 r2 2  o 1   r2
If  r 2 is close to unity, the transmission coefficients approach unity and, therefore, A3  A1 .