Broadband Anti-reflection
Coating Design
UCF
Quarter Wave Transformer (1)
g
Z0
4
Z01
Z 012
Z in 
 Z0
RL
 Z 01  RL Z 0
Require the load is resistive.
RL (real)
UCF
Quarter Wave Transformer (2)
Narrow bandwidth
Theory of Small Reflection (1)
UCF
d
Z1
1
T21
T12
1
T21
 j
e
T12 2
T12
kzd  
ZL
Z2
L
e  j
e  j
e  j
e  j
3
3
3
Z 2  Z1
1 
Z 2  Z1
2  1
ZL  Z2
3 
ZL  Z2
T21  1  1
2Z 2

Z1  Z 2
T12  1  2
2 Z1

Z1  Z 2
UCF
Theory of Small Reflection (2)
  1  T12T213 e 2 j  T12T2132 2 e 4 j  ...


 1  T12T213 e  2 j  2 3 e  2 j
n

n 0
 1  T12T213 e
 2 j
1
1  2 3 e  2 j
Since 2  1 , T21  1  1 , T12  1  1
  1  1  1 1  1 3 e
1  3 e 2 j

1  13 e  2 j
 2 j
1
1  13 e  2 j
UCF
Theory of Small Reflection (3)
When the differences between Z1 & Z2, and Z2 & ZL are small,
Γ1|&|Γ3| are small.
If |Γ1Γ3|<<1,

  1  3 e 2 j
 The total reflection is dominated by the reflection from the
initial discontinuity (Γ1) and the first reflection from the
second discontinuity (Γ3e-2jϑ)
 e-2jϑ accounts for the round-trip phase delay
Multi-section Transformer (1)
UCF

Z0
0


Z1
Z2
1
ZN
2
N
Z1  Z 0
Z n 1  Z n
Z  ZN
, n 
0 
, N  L
Z1  Z 0
Z n 1  Z n
ZL  ZN
For small reflections:
   0  1e
Note:  
 2 j

2
 2 e
 4 j
 ...  N e
at center frequency.
 2 jN
N
  n e  2 jn
n 0
Z n 1  Z n
Z n 1  Z n 1  Z n 1 
, since ln(1   x )   x ,  x 
 ln 
n 
Zn
Z n 1  Z n 2  Z n 
ZL
UCF
Multi-section Transformer (2)
In designing wide-band transformers, we force the
reflection coefficients to be symmetrical, i.e.
0  N , 1  N 1 , 2  N  2 ,...
Note, this does not imply that the Zns are symmetrical.
 
 
 
   e  jN 0 e jN  e  jN  1 e j  N 2   e  j  N 2   ...
Multi-section Transformer (3)
UCF
When N is odd,
   2e
 jN


0 cos N  1 cos N  2  ...  n cos N  2n   ...   N 1 cos  
2


When N is even,
   2e
 jN

1 
0 cos N  1 cos N  2  ...  n cos N  2n   ...  N 
2 2

• Can synthesize any desired reflection coefficient response as a
function of frequency  by properly choosing Γns and using
enough sections (N).
UCF
Specifications on Bandwidth
Design specs: center frequency f0 , bandwidth f ,
maximum allowable reflection coefficient m
 f
0   / 2
    2m
4
f
    2 m


 2 m
f0
0
 /2

f 


or  m   2 
4
f0 
0 
m
Fractional bandwidth:
 m

UCF
Binomial Design (1)
Require: 





0
when

2
 n
d 
 0 for n  1,2,...N - 1
 n
 d   
2

We can let

   A 1  e  2 j
    Ae
j
e

N
 j
e 
 j N
| A | 2 N cos N 
satisfies the above 2 conditions
Binomial Design (2)
UCF
Design specs: center frequency f 0 , bandwidth f ,
maximum allowable reflection coefficient m
Need to determine: (1) order N
(2) all Zns.
Step 1: determine N
ZL  Z0
N Z L  Z0
N
 A2
Let f=0 or  0  2 A 
Z L  Z0
ZL  Z0
At m
m
  m   2 N
A cos  m
N
Z L  Z0
 m 
cos N  m
Z L  Z0

Z  Z0
ln  m L
ZL  Z0

N
ln(cos  m )




m

f
  2 
4
f0



Round up to the next integer!
(MatLab: ceil)
Always overdesign a little bit.
Binomial Design (3)
UCF
Step 2: determine Zn

    A 1  e
 2 j

N
A2
N
N
 ZL 
Z L  Z0
 ( N 1)
ln 
2
Z L  Z0
 Z0 
 A C e
n0
N
n
 2 jn 
, where C
N
N
n
N!

N  n ! n!
MatLab: nchoosek(N,n)
     n e  2 jn
n0
 n  AC nN (Note : C nN  C NNn , C 0N  1, C1N  C NN1  N )
Z n 1  Z n 1  Z n 1 
Z n 1  Z n


, since ln(1   x )   x ,  x 
n 
 ln 

Z n 1  Z n 2  Z n 
Zn
Z n 1  Z n e 2 n
use iteratively
The essence using this approximation can make the design to be consistent.
We obtain N+1 n ’s for n = 0,1,…N. But we need to find only N n ’s for n = 1,2,…N.
Binomial Design (4)
UCF
n
n
p 0
p 0
ln Z n 1  ln Z n  2n  ln Z 0  2 p  ln Z 0  2 A C pN
 ZL 
 ln Z N 1  ln Z 0  2 A C  ln Z 0  2 A2  ln Z 0  ln   ln Z L
m 0
 Z0 
 ZL 
 ( N 1)

A2
ln 
 Z N 1  Z L (consistent!)
 Z0 
N
N
m
N
From the definition of Zn, we can find rn and then find kzn at f0.
Then from kzn dn = find dn.
Find actual fractional bandwidth (due to overdesign)
1  ZL
m  ln
2  Z0

 cos N  m    cos 1  m ln Z L  
m
Z 

2

 0  
4 m
f

 2
f0

1/ N
Binomial Design (5)
UCF
k r  k 1a k r
0 , 0
θr
x
y
0 ,1
0 , 2

 0 ,  N 1
0 , N
θi
k t  k 2 a kt
θt
 0 ,  N 1
k i  k1a k i
d1
d2
d N 1
dN
Design a multi-section binomial transformer for 45o TE incidence
from air to a thermal plastic antenna radome material with r =3 at
center frequency of 5 GHz. Require SWR = 1.01 from 4GHz to
6GHz. All the materials are non-magnetic.
1  m
SWR  1
or  m 
Standing wave ratio SWR =
1  m
SWR  1
z
Binomial Design (6)
UCF
k zn  k 02  rn  k x2  k 02  rn  k 02 sin 2  i  k 0  rn  sin 2  i
Zn 
Zn 
 gn
 0
k zn
Zn
0


0
 rn  sin 2  i
1
 rn  sin  i
2
  rn
0
2


 0 Z n
2
k zn
 rn  sin  i
0 
c
f0
Details in binomial.m
1
 2  sin 2  i
Zn
UCF
Chebyshev Design (1)
Ripple in passband!
UCF
Chebyshev Design (2)
Chebyshev function
Tn (cos  )  cos n 
since 2 cos  cos( n  1)  cos n   cos( n  2 )
 Tn (cos  )  2 cos  Tn 1 (cos  )  Tn  2 (cos  )
Let x  cos 
we have T0 ( x )  1
T1 ( x )  x
T2 ( x )  2 x 2  1
If n is even, Tn (x) is even.
If n is odd, Tn (x) is odd.
T3 ( x )  4 x 3  3 x

Tn ( x )  2 xT n 1 ( x )  Tn  2 ( x )
UCF
Chebyshev Design (3)
Tn ( x )  cos( n cos 1 x )
T n ( x )  cosh( n cosh 1 x )
Chebyshev Design (4)
UCF
   2e  jN 0 cos N  1 cos N  2  ...  n cos N  2n  ...
 Ae
 jN
cos
TN (
)
cos m
0 
m
0   / 2
    2m
 m
obviously  m  |  ( m ) | | A |

UCF
Chebyshev Design (5)
Key: expansion of TN (
Let p  sec  m
cos
)  TN (sec m cos ) in terms of cos(n
cos m
T0 ( p cos  )  1, T1 ( p cos  )  p cos  ,
T 2 ( p cos  )  ( p 2  1)  p 2 cos( 2 ) 
from Tn ( x )  2 xT n 1 ( x )  Tn  2 ( x )
Tn ( p cos  )  2 p cos  Tn 1 ( p cos  )  Tn  2 ( p cos  )
 p[ 2 cos  Tn 1 ( p cos  )]  Tn  2 ( p cos  )
Let S n ( p cos  )  2 cos Tn 1 ( p cos  )]
since 2 cos  cos m   cos( m  1)  cos( m  1) for
This means the coefficient of cosm (m  0) in Tn-1 will add to
the coefficients of cos(m+1) and cos(m1) in Sn.
MatLab program: Chebycoeff.m
UCF
Chebyshev Design (6)
Design specs: center frequency f 0 , bandwidth f ,
maximum allowable reflection coefficient m
Need to determine: (1) order N
(2) all Zns.

Step 1: determine N
m 
Let f=0 or  0  ATN (sec m )  Z L  Z 0
1 Z L  Z0
TN (sec m ) 
m Z L  Z 0
Z L  Z0

f 
 2 

4
f0 
m | A |
T n ( x )  cosh( n cosh 1 x )
1 Z L  Z0
cosh( N cosh (sec m )) 
m Z L  Z 0
 1 Z L  Z0 
1

cosh 
m Z L  Z 0 
Round up to the next integer!

N
Always overdesign a little bit.
cosh 1 (sec m )
1
UCF
Chebyshev Design (7)
Step 2: determine Zn
Z L  Z0
1
A
cosh( N cosh 1 (sec m )) Z L  Z 0
   2e  jN 0 cos N  1 cos N  2  ...  n cos N  2n   ...
 Ae  jN TN (sec m cos )
0  N , 1  N 1 , 2  N  2 ,...
Assume TN (sec m cos ) 
 n  N n 
D
n 0,1, 2,...
ADN 2n
2
N 2n
cos(N  2n)
If N is even N  AD0
2
Z n 1  Z n Z  Z 1  n or Z  Z 1  n
From n 
use iteratively
n
n 1
n
, n 1


1
1  n
n
Z n 1  Z n
Chebyshev Design (8)
UCF
We obtain N+1 n ’s for n = 0,1,…N. But we need to find only N n ’s for n = 1,2,…N.
There may always be a little error in the design. Since the whole methodology is
developed from small reflection approximation, we need to check using rigorous
impedance transformation formula or chain matrix approach and find whether the
specs are satisfied.
(1) If N is odd:
1  n 1
Use Z n  Z n 1
for n = 1, …, (N-1)/2
1  n 1
1  n
Use Z n  Z n 1
for n = N, N-1, …, (N+3)/2
1  n
Although this will grantee N 1  N 1
Z N 1  Z N 1 Z N 3
2
2
2
   2e
 jN
2
2
the value will deviate a little from the designed value.


0 cos N  1 cos N  2  ...  n cos N  2n   ...   N 1 cos  
2


Chebyshev Design (9)
UCF
(2) If N is even:
1  n 1
Use Z n  Z n 1
for n = 1, …, N/2
1  n 1
1  n
Use Z n  Z n 1
for n = N, N-1, …, N/2+1
1  n
This will never use N whose actual value will deviate from the designed value.
2
   2e
 jN

1 
0 cos N  1 cos N  2  ...  n cos N  2n   ...  N 
2 2

UCF
Chebyshev Design (10)
From the definition of Zn, we can find rn and then find kzn at f0.
Then from kzn dn = find dn.
Find actual fractional bandwidth (due to overdesign)
1 Z L  Z0
cosh( N cosh (sec m )) 
m Z L  Z 0
1
1
 1 Z L  Z0
1
 sec m  cosh  cosh 
 N
 m Z L  Z 0
4 m
f

 2
f0





 m
Tapered Line
UCF
since
 2 j  k z ( z ) dz d
 Z 
1
0
( )   e
ln dz
2 0
dz  Z 0 
L
z