Analytic Solution for Shunt Stub
00
00
022
0
00
20
20
002
20
2
2
0
00
;2
;/])[(
/1satisfy tochosen is
])([
))((
)(
)1(
1Let
tan where;)(
)(
is load thefrom lengtha down impedance The
1 impedance Load
ZRZ
X
ZRZR
ZXRZRX
t
ZYGd
tZXRZ
tZXtXZtRB
tZXR
tRG
jBGZ
Y
dttjXRjZ
tjZjXRZZ
dZ
jXRY
Z
LL
LL
LLLL
LL
LLL
LL
L
LL
LL
LLL
L
0for ])(tan[2
1
0for )(tan2
1
stub circuited-shorta for while
0for ])(tan[2
1
0for )(tan2
1
stub circuited-open anfor Then, .satisfy tochosen is
0for )tan(2
1
0for tan2
1
therefore,2
tantan
0
1
0
1
0
1
0
1
1
1
BY
B
BY
Bl
BY
B
BY
Bl
BBl
tt
ttd
ddt
O
O
S
Problem 1Problem 1: Repeat example 5.5 using analytic solution.
Example5.6: Design two single-stub (open circuit) series tuning networks to match this load ZL = 100+j 80 to a 50 line, at a frequency of 2 GHz?
1. The normalized load impedance ZL= 2-j1.6. 2. SWR circle intersects the 1+jx circle at both points z1 = 1.0-j1.33 z2 = 1.0+j1.33. Reading WTG can obtain: d1= 0.328-0.208=0.12 d2= 0.672-0.208=0.463.3. The stub length for tuning z1 to 1 requires l1 = 0.397,and for tuning z1 to 1 needs l2 = 0.103.
Solution
1. ZL = 100+j 80 at 2 GHz can find R= 100,L=6.37nH.
Analytic Solution for Series Stub
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00
022
0
00
20
20
002
20
2
2
0
00
;2
;/])[(
/1satisfy tochosen is
])([
))((
)(
)1(
1Let
tan where;)(
)(
is load thefrom lengtha down admittance The
1 admittance Load
YGY
B
YGYG
YBGYGB
t
YZRd
tYBGY
tYBtBYtGX
tYBG
tGR
jXRY
Z
dttjBGjY
tjYjBGYY
dY
jBGZ
Y
LL
LL
LLLL
LL
LLL
LL
L
LL
LL
LLL
L
0for ])(tan[2
1
0for )(tan2
1
stub circuited-opena for while
0for ])(tan[2
1
0for )(tan2
1
stub circuited-short anfor Then, .satisfy tochosen is
0for )tan(2
1
0for tan2
1
therefore,2
tantan
0
1
0
1
0
1
0
1
1
1
BY
B
BY
Bl
XZ
X
XZ
Xl
BBl
tt
ttd
ddt
O
S
S
Problem 2Problem 2: Repeat example 5.6 using analytic solution.
Double-Stub Matching
Variable length of length d between load and stub to have adjustable tuning between load and the first stub.
Shunt stubs are easier to implement in practice than series stubs.
In practice, stub spacing is chosen as /8 or 3/8 and far away 0 or /2 to reduce frequency sensitive.
Original circuit
Equivalent circuit
adjustable tuning
Disadvantage is the double-stub tuner cannot match all load impedances. The shaded region forms a forbidden range of load admittances.
Two possible solutions
b1,b2 and b1’,b2’ with the same distance d.
Example5.7: Design a double-stub (open circuit) shunt tuning networks to match this load ZL = 60-j 80 to a 50 line, at a frequency of 2 GHz?
1. The normalized load impedance YL= 0.3+j0.4 (ZL= 1.2-j1.6). 2. Rotating /8 toward the load (WTL) to construct 1+jb circle can find two values of first stub b1 = 1.314 b’1 = -0.114. 3. Rotating /8 toward the generator (WTG) can obtain y2= 1-j3.38 y’2= 1+j1.38.
Solution
4. The susceptance of the second stubs should be b2 = 3.38 b’2 = -1.38.5. The lengyh of the open-circuited stubs are found as l1 = 0.146, l2 = 0.204,or l1 = 0.482, l2 = 0.350.6.ZL = 60-j 80 at 2 GHz can find R= 60, C=0.995pF.
Analytic Solution for Double Stub
1 1
1
1 02 0
0 1
The first stub with admittance
( )
where is the load, is the susceptance of the first stub.
The second stub with admittance
( ); where tan ,
( )
L L
L L L
L L
L L
Y G j B B
Y G jB B
G j B B Y tY Y t d Y
Y j G jB jB t
00
2 0
2 2 20 1
0 2 2 2 20
0
1 (1)
Let Re{ } Re{ } can derive
1 4 ( )[1 1 ] (2)
2 (1 )
Since is real, it gives the range of for matching a given .
1 0
LL
L L
L
Z
Y Y
t t Y B t B tG Y
t Y t
G G d
tG Y
20
2 2sin
Y
t d
2 2 20 0
1
2 2 20 0 0
2
1
0
After has been fixed, both stubs can be found as
(1 ) from (2)
(1 ) from (1)
For an open-circuited stub
1tan ( )
2
while for a shor
L LL
L L L
L
O
d
Y t G Y G tB B
t
Y t G Y G t G YB
G t
l B
Y
1 0
1 2
t-circuited stub
1tan ( )
2where or
Sl Y
BB B B
Problem 3Problem 3: Repeat example 5.7 using analytic solution.
Quarter-Wave transformer
It can only match a real load impedance.
The length l= /4 at design frequency f0.
The important characteristics
]2
1[cos
42 BandwidthFraction
d tolertatemaximunfor )2
(2
)sec2
(11
0
0
2
1
0
2
0
02
01
ZZ
ZZ
f
f
ZZ
ZZ
ZZZ
L
L
m
m
mm
mL
L
m
L
Example5.8: Design a quarter-wave matching transformer to match a 10 load to a 50 line? Determine the percent bandwidth for SWR1.5?Solution
%2929.0
]5010
50102
)2.0(1
2.0[cos
42
]2
1[cos
42
2.015.1
15.1
1
1
36.221050
2
1
0
0
2
1
0
01
ZZ
ZZ
f
f
SWR
SWR
ZZZ
L
L
m
m
m
L
Binomial Multi-section MatchingThe passband response of a binomial matching transformer is
optimum to have as flat as possible near the design frequency, and is known as maximally flat.
The important characteristics
])(2
1[cos
42
42
)(2 bandwidth Fraction
2 ;cos2t coefficien reflection Maximun
)4
( 2
at , !)!(
! , 2 where
;)1()( response Binomial
11
0
0
0
00
0
0
22
Nmm
mmNN
m
Nn
L
LN
Nnn
N
n
jnNn
Nj
Af
ff
f
f
A
lfnnN
NC
ZZ
ZZA
CAeCAeA
Binomial Transformer Design
If ZL<Z0, the results should be reversed with Z1 starting at the end.
00
01 ln2)2(22lnZ
ZCC
ZZ
ZZ
Z
Z LNn
NNn
L
LNn
n
n
Example5.9: Design a three-section binomial transformer to match a 50 load to a 100 line? and calculate the bandwidth for m=0.05?Solution
5.5400.4ln2lnln:2
7.7026.4ln2lnln:1
7.91518.4ln2lnln:0
3!2!1
!3 3
!1!2
!3 1
!0!3
!3
%707.0])0433.0
05.0(
2
1[cos
42
])(2
1[cos
42
0433.0ln2
1 2
3For
10
3223
20
3112
10
3001
32
31
30
311
11
0
01N
0
0
ZZ
ZCZZn
ZZ
ZCZZn
ZZ
ZCZZn
CCC
Af
f
Z
Z
ZZ
ZZA
N
LN
LN
LN
Nm
L
L
LN
Using table design for N=3 and ZL/Z0=2(reverse) can find coefficient as 1.8337, 1.4142, and 1.0907.
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