17 Broadband Anti-Reflection Coating Design
Transcript of 17 Broadband Anti-Reflection Coating Design
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Broadband Anti-reflectionCoating Design
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Quarter Wave Transformer (1)
4g
Z0 Z01
0
2
01 ZR
ZZ
L
in
)(realRL
001 ZRZ L
Require the load is resistive.
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Quarter Wave Transformer (2)
Narrow bandwidth
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Theory of Small Reflection (1)
d
T21T12
Z1 Z2 1212
1
ZZ
ZZ
12
21
2
121
2
1
ZZ
Z
T
21
1
212
2
1
ZZ
Z
T
je
je
je
j
e
je
L
LZ
312T
3
12T 3
2
11
2
23
ZZ
ZZ
L
L
dkz
21T
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Theory of Small Reflection (2)
...422
32112
2
321121 jj eTTeTT
n
n
jj
eeTT
0
2
32
2
321121
j
j
eeTT
2
32
2
3211211
1
11212112 1,1,Since TT
j
j
ee
231
2
3111
1
111
j
j
e
e2
31
2
31
1
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Theory of Small Reflection (3)
When the differences between Z1 & Z2, and Z2 & ZL are small,
1|&|3| are small.
If |13|
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Multi-section Transformer (1)
,01
010ZZ
ZZ
Z0 Z1 Z2 ZN
ZL
0 1 2N
,1
1
nn
nnn
ZZ
ZZ
NL
NLNZZ
ZZ
N
n
jnn
jNN
jj eeee0
2242
210 ...
For small reflections:
Note: at center frequency.2
n
nn
n
n
nn
nnnZ
ZZxxxZ
Z
ZZ
ZZ
11
1
1 ,)1ln(since,ln2
1
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Multi-section Transformer (2)
...2210 NjNjjNjNjN eeeee
In designing wide-band transformers, we force the
reflection coefficients to be symmetrical, i.e.
Note, this does not imply that theZns are symmetrical.
,...,, 22110 NNN
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Multi-section Transformer (3)
When N is odd,
When N is even,
Can synthesize any desired reflection coefficient response as a
function of frequency by properly choosing ns and usingenough sections (N).
2
102
1...2cos...2coscos2 Nn
jN nNNNe
cos...2cos...2coscos22
110 Nn
jN nNNNe
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Specifications on Bandwidth
Design specs: center frequency f0 , bandwidth f ,maximum allowable reflection coefficient m
Fractional bandwidth:
mm
f
f 42
2/
2
00
0
24 f
f
m
or
f
m
m
0 2/0
m 2
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Binomial Design (1)
satisfies the above 2 conditions
1-1,2,...Nnfor0
2when0
2
n
n
d
d
NjeA 21
Njjj eeeA
NNA cos2||
Require:
We can let
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Binomial Design (2)
Design specs: center frequency , bandwidth ,
maximum allowable reflection coefficient
Need to determine: (1) order N(2) all Zns.
00
20 ZZ
ZZ
A L
LN
mN
L
LmZZZZ cos
0
0
mNN
mm A cos2
mf
0f
Letf=0 or
At m
0
02ZZ
ZZA
L
LN
Step 1: determine N
)ln(cos
ln0
0
m
L
LmZZ
ZZ
N
Round up to the next integer!
(MatLab: ceil)
Always overdesign a little bit.
0
24 f
fm
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Binomial Design (3)
!!
!where,1
0
22
nnN
NCeCAeA Nn
N
n
jnN
n
Nj
),1,:(Note 110 NCCCCCAC N
N
NNN
nN
N
n
N
nn
Step 2: determine Zn
N
n
jn
ne0
2
MatLab: nchoosek(N,n)
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.
0
)1(
0
0 ln22Z
Z
ZZ
ZZA LN
L
LN
n
nn
n
n
nn
nnn
ZZZxxx
ZZ
ZZZZ
11
1
1 ,)1ln(since,ln21
neZZ nn
2
1 use iteratively
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Binomial Design (5)
00 ,
x
zik ki ak 1
y
rkr kak 1
tkt k ak 2
i
r t10 , 20 , 10 , N N ,0
10 , N
1d 2d 1Nd Nd
Design a multi-section binomial transformer for 45o TE incidence
from air to a thermal plastic antenna radome material with r=3 atcenter frequency of 5 GHz. Require SWR = 1.01 from 4GHz to
6GHz. All the materials are non-magnetic.
Standing wave ratio SWR =m
m
1
1
1SWR
1SWRor
m
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Binomial Design (6)
0
0
02
0
2
220
2
00
2
0
22
0
2
0
22
0
sin
2
sin1
sin
1
sin
sinsin
f
c
Zk
Z
ZZ
kZ
kkkkkk
n
irnzn
gn
i
n
rn
irn
nn
irnzn
n
irnirnxrnzn
Details in binomial.m
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Chebyshev Design (1)
Ripple in passband!
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Chebyshev Design (2)
Chebyshev function
nTn cos)(cos
)2cos(cos)1cos(cos2 nnnsince
)(cos)(coscos2)(cos 21 nnn TTT
Let cosx
we have
)()(2)( 21 xTxxTxT nnn
xxxT
xxTxxT
xT
34)(
12)()(
1)(
3
3
2
2
1
0
If n is even, Tn (x) is even.
If n is odd, Tn (x) is odd.
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Chebyshev Design (3)
)coscos()( 1xnxTn
)coshcosh()( 1xnxTn
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Chebyshev Design (4)
)cos
cos(
...2cos...2coscos2 10
mN
jN
n
jN
TAe
nNNNe
mm
0 2/0
m 2
|||)(| Amm
obviously
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Chebyshev Design (5)
Key: expansion of )cos(sec)cos
cos(
mN
m
N TT in terms of cos(n
)()(2)( 21 xTxxTxT nnn
)cos()]cos(cos2[
)cos()cos(cos2)cos(
21
21
pTpTp
pTpTppT
nn
nnn
Letmp sec
from
)1cos()1cos(coscos2 mmmsince
)]cos(cos2)cos( 1 pTpS nn Let
This means the coefficient of cosm ( ) inTn-1 will add to
the coefficients of cos(m+1)and cos(m1) in Sn.
MatLab program: Chebycoeff.m
)2cos()1()cos(
,cos)cos(,1)cos(
22
2
10
pppT
ppTpT
for0m
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Chebyshev Design (6)
Design specs: center frequency , bandwidth ,
maximum allowable reflection coefficient
Need to determine: (1) order N(2) all Zns.
00
)(sec0 ZZ
ZZ
AT L
L
mN
mf
0f
Letf=0 or
Step 1: determine N
Round up to the next integer!
Always overdesign a little bit.
||Am )coshcosh()( 1xnxTn
0
01)(secZZ
ZZT
L
L
m
mN
0
01 1
))(seccoshcosh( ZZ
ZZ
NL
L
m
m
)(seccosh
1cosh
1
0
01
m
L
L
m ZZ
ZZ
N
0
24 f
fm
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Chebyshev Design (7)
)cos(sec
...2cos...2coscos2 10
mN
jN
n
jN
TAe
nNNNe
,...2,1,0
2 )2cos()cos(secn
nNmN nNDT Assume
2
2nNnNnAD
,...,, 22110 NNN
Step 2: determine Zn
If N is even 02
ADN
,
1
1
nn
nnn
ZZ
ZZ
From
n
nnn ZZ
1
11 use iteratively
n
nnn ZZ
1
11or
0
0
1
))(seccoshcosh(
1
ZZ
ZZ
N
A
L
L
m
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Chebyshev Design (8)
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 thespecs are satisfied.
(1) If N is odd:
1
1
1 1
1
nn
nn ZZUse for n = 1, , (N-1)/2
n
nnn ZZ
1
11Use for n = N, N-1, , (N+3)/2
2
3
2
1
2
1 NNN ZZZAlthough this will grantee
2
1
2
1 NN
the value will deviate a little from the designed value.
cos...2cos...2coscos22
110 Nn
jN nNNNe
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Chebyshev Design (9)
(2) If N is even:
1
11
1
1
n
nnn ZZUse for n = 1, , N/2
n
nnn ZZ
1
11Use for n = N, N-1, , N/2+1
This will never use2N whose actual value will deviate from the designed value.
2
102
1...2cos...2coscos2 Nn
jN nNNNe
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Chebyshev Design (10)
m
Find actual fractional bandwidth (due to overdesign)
m
f
f 42
0
0
01 1))(seccoshcosh(ZZ
ZZN
L
L
m
m
0
01 1cosh1
coshsecZZ
ZZ
N L
L
m
m
From the definition of Zn, we can find rn and then find kzn atf0.Then from kzndn = find dn.
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Tapered Line
since
dzZ
Z
dz
deL dzzkj
z
z
0
0
)(2
ln2
1)( 0