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Chapter 5 Impedance Matching and TuningChapter 5 Impedance Matching and Tuning
• The matching network is ideally lossless and is usuallyThe matching network is ideally lossless, and is usually designed so that the impedance seen looking into the matching network is Z0.
• Only if Re[ZL] ≠ 0, a matching network can always be found.• The quarter-wave impedance transformer is for read load.q p• Important factors in selecting matching networks:
(1)complexity, (2)bandwidth, (3)implementation, (4)adjustability.( ) p y, ( ) , ( ) p , ( ) j y
1
Review of Smith Chart
zL = ZL / Z0zL ZL / Z0= rL + jxL
2
5.1 Lumped Element Matching (L-networks)p g ( )L-type or L-network matching is the simplest matching network.
(1) zL = ZL/Z0 is outside the r=1(1) zL = ZL/Z0 is inside the r=1 (1) zL ZL/Z0 is outside the r 1 circle (rL < 1).
(2) Z i i i h B h
(1) zL ZL/Z0 is inside the r 1 circle (rL > 1).
(2) Z i h i h B h (2) ZL is series with jB, then shunt with jX.
(2) ZL is shunt with jB, then series with jX.
3
Analytical Solutions (rL>1)
• When rL > 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
01
1inZ jX ZjB
R jX
0 0( )L L
L L L
R jXB XR X Z R Z
0
2 2
(1 )L L L
L
X BX BZ R X
RX R X Z R
00
2 2
L L L L
L L
X R X Z RZ
BR X
inZ
0 01 L
L L
X Z ZXB R BR
4• Note that when rL > 1, the square root in B has real results.
Analytical Solutions (rL<1)
• When rL < 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
0
1 1 1( )in L L
jBZ R j X X Z
0 0
0
( )( )
L L
L L
BZ X X Z RX X BZ R
0Z
0
0
( ) /L LZ R RB
Z
inZ0
0( )L L L
Z
X X R Z R
• Two analytical solutions are physically realizable.
• One of the two solutions may result in smaller reactive elements and
5
One of the two solutions may result in smaller reactive elements, and may have better matching, bandwidth, or better SWR on the line.
Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching
ZL = 200-j100 Ω, Z0 = 100 Ω, j , ,f = 500 MHz, zL = ZL/Z0 = 2-j1, rL = 2 >1 inside the r = 1 circle.
Exact solutions:
2 2R 2 20
02 2
LL L L L
L L
RX R X R ZZ
BR X
3 12 2
2 2 3 1
2.899 10100 2 200 100 100 200 200 100 6 899 10
L LR X
3 1
0 0
200 100 6.899 10
122.471 LX Z ZX
6
122.47L LB R BR
Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching
Th l iThe two solutions:
(1)3
6
2.899 10 0.9228 pF2 500 10
BC
( ) 6
6
p2 500 10
122.47 38.98 nH2 500 10
XL
62 500 10
1 1(2)6
1 1122.47 2 500 10
2 599 pF
CX
3 6
2.599 pF1 1
6 899 10 2 500 10L
B
7
6.899 10 2 500 1046.14 nH
B
Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching
The two solutions:The two solutions:
8
Lumped Elements for Microwave Integrated Circuits (MICs)
9
5.2 Single-Stub TuningShunt‐stub tuning Series‐stub tuning
• No lumped element is required. Convenient for MIC fabrication.• Characteristic impedances of the lines and the stub can be different.• Keep the matching stub as close as possible to the load. • Solution procedure for shunt-stub and series-stub tunings.(1) Locate zL (yL) on the Smith Chart.(1) Locate zL (yL) on the Smith Chart.(2) Move along the const-Γ circle with 2d/λ wavelengths to
reach the g = 1 (r = 1) circle.(3) Sh t ith t i ith t t(3) Shunt with a susceptance or series with a reactance to
cancel the imaginary part. 10
Shunt-Stub Tuning Analytical Solution
0 00 tan
( )1 1L L Lin t dno stub
Z jZ t R j X Z tZ ZZ jZ t Y Z X t jR t G jB
0 0 0
220 0
022 2 2
( )( )(1 ) ,
L L L
L L LLin in
in inZ jZ t Y Z X t jR t
R t Z X t Z t XR tG B Y
G jB
022 20 0
2 2 20 0 0
2,( ) ( )
(1 ) ( ) real part
in inL L L L
in L L L
GR X Z t R X Z t
G Y Z R t R X Z t
0 0 0in L L L
00
If , and tan2
LL
XR Z t dZ
0
11 tan , 02 2
LL
X XZd
0
1
2 2
1 tan , 0LL
ZdX X
0
tan , 02 2 LX
Z
11
Shunt-Stub Tuning Analytical Solution0
2 2 20 0 0 0 0
If Z ,
( ) ( )( )L
L L L L L L
R
X Z X Z Z R Z R Z R Xt
0 0 0 0 0
0 0
2 2 2
( ) ( )( )( )
( ) [( ) ]
L L L L L L
L
L
tZ R Z
RX R Z X
00
0
( ) [( ) ]L L L
L
X R Z XZ
R Z
here tant lLet the stub susceptance Bstub (or Bs) = -Bin,
(a) open-stub: and0stubZ jZ t sstubstub jBjBtjYY 0
here tant l
1 1
0 0
1 1tan ( ) tan2 2
o s inl B BY Y
(b) Short-stub: and tjZZstub 0 sstubstub jBjBtjYY 0
1 10 01 1tan ( ) tansl Y Y tan ( ) tan
2 2s inB B
If lo < 0 or ls < 0, λ/2 must be added to have a realistic result. 12
Example 5.2 Shunt Single-Stub Tuning
Match ZL = 15 + j10 Ω to a 50-Ω line. Use a shunt open-stub.Sol: RL ≠ Z0,Sol: RL ≠ Z0,
2 2 20( ) ( )L
L L LRX R Z X 0
0
0
( ) ( )10 397.5
35
L L L
L
Zt
R Z
d
11 1
10 397.5 1 ( tan ) 0.38735 2
1
dt t
l B
1
0
1 ... tan2
10 397 5 1
o inin
l BBY
d
12 2
1
10 397.5 1 tan 0.04435 2
1o in
dt t
l BB
13
1
0
... tan2
o ininB
Y
Example 5.2 Shunt Single-Stub Tuning (Cont’d)Use a shunt open-stub to match ZL = 15 + j10 Ω line.
Solution 1: Solution 2:
14
5.3 Double-Stub Tuning – Analytical Solution
• In single-stub tuning, the length d is not tunable.
• If this causes difficulty in circuit implementation, use double-stub tuning.
0 tanLY jY bY Y G jB
15
00
0 tanL
L L L
L
jY Y G jBY jY b
5.3 Double-Stub Tuning – Analytical Solution2 2 2 2 2
0 1 0
2
0 4 ( ) (1 )
1Lt Y B t B t Y t
Y
20
0 2 2
102 sin
has an upper limit
LYtG Y
t dG
( )Y G j B B 2 2 2
has an upper limit. Given , one can obtain:
(1 )
LGd
Y t G Y G t 1 1
1 02 0
0 1
( )( ) , tan
( )
L L
L L
L L
Y G j B BG j B B Y tY Y t d
Y jt G jB jB
0 01
2 2 2
(1 )
(1 )
L LL
Y t G Y G tB B
tG t G Y G t
0 1
2 0 2 222
2 0 1
( )Re[ ] , Im[ ]
( )1 0
L L
L
Y jt G jB jBY Y Y B
Y B t B ttG Y G
01 0
10
(1 )
1
L L LG t G Y G tB Y
tB
2 0 10 2 2
2 220 1
( ) 0
4 ( )1 1 1
LL L
L
G Y Gt t
t Y B t B ttG Y
101 2
0
1 0
1 tan , ,2
1s
B B B BY
Y
B B B16
0 10 2 2 2
0
( )1 1(1 )
LLG Y
t Y t
1 0tan2
s
B
1 2, ,B B B
Example 5.4 Double-Stub Tuning - PerformanceUse double-stub matching scheme to match ZL = 60 – j80Ω at 2.0 GHz to a 50-Ω line.
17
5.4 The Quarter-Wave Transformer• A single-section transformer may suffice for a narrow-band
impedance matching.• Single-section quarter-wave impedance matching λ= λ0 / 4 at the
desired frequency. (See Chap 2)
• Multisection quarter-wave transformer designs can be synthesized to yield optimum matching characteristics over a broad bandwith.
18
yield optimum matching characteristics over a broad bandwith.
2.5 The Quarter-Wave TransformerReview
A quarter-wave transformer is an impedance matching circuit
2tanR jZ l Z 1 11
12
tan tan
Lin
L Ll
R jZ l ZZ ZZ jR l R
Z Z
0
0
inin
in
Z ZZ Z
19
0 1 0If 0 is required, , then in in LZ Z Z Z R
Estimate the Bandwidth of a Single-Section Impedance Transformera Single Section Impedance Transformer
1 0 LZ Z Z
11
1
, tan tan Lin
L
Z jZ tZ Z tZ jZ t
Z jZ t
11 0 2
0 1 0 1 012
10 1 0 1 01 0
( ) ( )( ) ( )
L
in L LL
Lin L L
Z jZ tZ ZZ Z Z Z Z j Z Z Z tZ jZ t
Z jZ tZ Z Z Z Z j Z Z Z tZ Z
1 01
002 2
1 2 4( ) 4
L
LL
Z jZ tZ ZZ Z
Z Z j t Z Z Z ZZ Z t Z Z
2 20 0 00 02 4( ) 4 1L L LL L
L
Z Z j t Z Z Z ZZ Z t Z ZZ Z
22
0
sec
0LZ Z 0
0
0
When , cos2 2
L
L
L
Z ZZ Z
Z Z
0
0
Set a max , or cos2
Lm m m
L
Z ZZ Z
20
Bandwidth of the Matching Transformer
20
2( / 2 )41 1
m
LZ Z
202 2
0
: 1 sec( )
2
Lm m
m L
m m
Z Zff f f
00 0
0
, ,2 2
2
m mm m
L
ff f ff f
Z Z
0
20
cos1
Lmm
LmZ Z
F ti l b d idth10 or 0.1
0
Fractional bandwidth :2( ) 42m mf ff
f f
4 or 0.25
0 0
01 242 cos Lm
f f
Z Z
2 or 0.5
21
20
2 cos1 Lm
Z Z
Example 5.5Single Section Quarter Wave Transformer BandwidthSingle-Section Quarter-Wave Transformer Bandwidth
ZL = 10 Ω, Z0 = 100 Ω, SWR = 1.2S lSol:
1 0 31.6LZ Z Z
1 0.11m
SWRSWR
max 0 0
min 0 0
11
V V VSWRV V V
01
2
242 cos1
Lm Z Zff Z Z
20 0
1
1
4 0.1 2 31.6 62 cos 2 9%
Lmf Z Z
22
2 cos 2 9%900.99
5.5 Theory of Small Reflections• For applications requiring more bandwidth than that a single quarter-
wave section can provide, multi-section transformers can be used.
(1) Single-Section Transformer
212
121
ZZZZZZ
2
23
2ZZZZZ
L
L
1
12
2121
21
21
ZT
ZZZT
12
1212 1
ZZT
23
5.5 Theory of Small Reflections
(1) Multi-Reflections
2 2 41 12 21 3 12 21 3 2 ...j jT T e T T e
1 12 21 3 12 21 3 2
22 2 41 12 21 3 3 2 3 2(1 ...)j j jT T e e e
2 21 12 21 3 3 2
02 2
nj j n
nj j
T T e e
T T e e
12 21 3 1 31 2 2
3 2 1 31 1
j j
j j
T T e ee e
• If is small, • Γ the reflection from the initial discontinuity between Z1 and Z2 + the
312
1 3je
24
first reflection from the discontinuity between Z2 and ZL
Multisection Transformer
• N commensurate (equal-length) sections of transmission lines. Let ( q g )the total reflection be Γ.
• Assume that all Z increase or decrease monotonically ZL is real• Assume that all Zn increase or decrease monotonically, ZL is real, and “The theory of small reflections” holds.
• It can be validated that
NL ZZZZZZZZ
23
212
101
0
25
NLN ZZZZZZZZ
...,,,23
212
101
0
Multisection Transformer
2 2
2 4 20 1 2
[ ( ) ( ) (if s mmetr]
.
)
..j j j
j N j NjN jN jN
NNe e e
0 1
0 1 /2
[ ( ) ( ) ... (if symmetry]
cos cos 2 ... , 2
)j jjN jN jN
NjN
e e e e e
N N N evene
0 1 1 /2
2cos cos 2 ... cos ,
j
N
eN N N odd
• We can synthesize any desired response as a function of frequency• We can synthesize any desired response as a function of frequency or θ, by properly choosing the Γn’s and enough number of sections N.
• The most commonly used passband responses are:
(1) Binomial or maximally flat response, and(2) Chebyshev or equal-ripple response. 26
H Fi d h C di R fl i ?How to Find the Corresponding Reflection?
2 4 20 1 2
Theory of Samll Reflection: ( ) ...j j j N
Ne e e 0 1 2( )
Binomial Multi-Section Transformer:
N
2
Binomial Multi-Section Transformer:(1 )j NA e
Chebyshev Multi-Section Transformer:
(sec cos )n mAT
27
5.6 Binomial Multisection Matching Transformer• Maximally flat response: the response is as flat as possible near the design
frequency. Also called Binomial matching transformer.• Let 2(1 ) A d tifi i llj NA • Let
2(1 ) Assumed artificially
2 cos
j N
NN
A e
A
0
0
00 2
2N L
NL
Z ZA AZ Z
0
2 2
0
!(1 ) ,( )! !
LN
j N N j n Nn n
n
NA e A C e CN n n
2 4 20 1 2
( ) ...
(0)
j j j NN
N N
e e e
AC C
N! ("enn factorial“)
Properly
( )2
configure the i
N Nn n nNAC C
mpedances to synthesize the needed
N! ( enn factorial )“Four factorial" is written as "4!”
reflections .n28
Binomial Multisection Matching Transformer
nnn xxxZZZ 11 1112lnln1
LNN
LNN
Nn
nnnn
ZCZZCACZ
xx
xZZZ
01
1
1
ln1122ln
11,1
2ln,ln2
NNnC
Ln
nNL
nNnnn
ZZ
ZC
ZZCAC
Z2/
1
001 ln
2222ln
n ZZ 0
• Exact results for Z 1 / Z0 for N = 2 thru 6 are given in Table 5 1• Exact results for Zn-1 / Z0 for N = 2 thru 6 are given in Table 5.1.
29
Binomial Transformer Design Table 5.1
30
Example Binomial Transformer Design
ZL/Z0
N=5
Z1 /Z0 Z2 /Z0 Z3 /Z0 Z4 /Z0 Z5 /Z0
1.0 1.0000 1.0000 1.0000 1.0000 1.0000
1.5 1.0128 1.0790 1.2247 1.3902 1.4810
2.0 1.0220 1.1391 1.4142 1.7558 1.9569
3 0 1 0354 1 2300 1 7321 2 4390 2 89743.0 1.0354 1.2300 1.7321 2.4390 2.8974
4.0 1.0452 1.2995 2.0000 3.0781 3.8270
6.0 1.0596 1.4055 2.4495 4.2689 5.6625
8.0 1.0703 1.4870 2.8284 5.3800 7.4745
10.0 1.0789 1.5541 3.1623 6.4346 9.2687
A 5th d bi i l f f Z 6Z• A 5th order binomial transformer for ZL = 6Z0
31• Also note that 4055.1
2689.46,0596.1
6625.56
E ample Binomial Transformer Design Z /Z <1Example Binomial Transformer Design ZL/Z0 <1
ZL/Z0
N=5
Z1 /Z0 Z2 /Z0 Z3 /Z0 Z4 /Z0 Z5 /Z0
6.0000 1.0596 1.4055 2.4495 4.2689 5.6625
0.1667 0.9438 0.7115 0.4082 0.2343 0.1766
• A 5th order binomial transformer for ZL = Z0 / 6
32
Bandwidth of the Binomial Transformer,cos2 m
NNm A i.e, tolerable max over the passband.m
N/1
m
m A1
21cos
N
mmm
Afff
ff
/1
1
0
0
0 21cos4242)(2
Example 5.6 N = 3, ZL = 2Z0 = 100 Ω, find the BW for Γm = 0.05.Sol: From Table 5.1, the required impedance Zn can be found to be
1 2 354.5 , 70.7 , 91.71 100 502 (0) 0 4167N
Z Z Z
A
11 3
2 (0) 0.41678 100 50
4 1 0.052 cos ( ) 70%
A
f
33
3
02 cos ( ) 70%
2 0.4167f
Binomial Transformer’s Frequency Response
Reflection coefficient magnitude versus frequency for multisectionbinomial matching transformer of Ex. 5.6 with ZL = 2Z0 = 100 Ω and
34
binomial matching transformer of Ex. 5.6 with ZL 2Z0 100 Ω and Γm = 0.05.
Example (2nd Midterm)Design a Butterworth transformer of N = 2 for ZL = 4Z0. Let Γ0, Γ1 and ΓL be
DIYg L 0 0, 1 L
respectively the reflection coefficients at the Z0 – Z1, Z1 – Z2 and Z2 – ZLjunctions, and Γ0 = ΓL. (a) Based on the “theory of small reflection,” find the Γ terms of Γ Γ and θ (b) If Γ = 0 and are required when θ0/ fΓin terms of Γ0, Γ1 and θ. (b) If Γin = 0 and are required when θ= λ / 4, find a and b.
0/ fin
2cos2 104
22
10 ee jjin
202
2cos2 0110
10210
in
4444444
11
02sin4 0
abbababababb
aa
in
359/141141
0434
31)(2
112
412322
aaaabaababaabaa
abab
ba
2751.0,5707.1
4321.08135
99/14,9273.1
271
27112
94
61
3 233 23
rqrrqr
qr
The “exact” solution in Table 5.1 is a = 1.4142 and b =2.8285. The error is from
358435.2/4
4067.12751.05707.191
ab
a the approximation used in the “theory of small reflection.”
5.7 Chebyshev Multisection Matching Transformers
Chebyshev polynomials: Recurrence formula:)()(
1)(
1
0
xxTxT
34)(24
33 xxxT
2)()(2)( 21 nxTxxTxT nnn
12)( 22 xxT 188)( 24
4 xxxT
Tn(x)345
n=1
2
n=1
xpassband
36
p
5.7 Chebyshev Multisection Matching Transformers(1) mapped to passband
Let x = cosθ, it can be shown that (2) t id th b d
1)(,1 xTx n
nTn cos)(cos 1)(1 T(2) outside the passband
(3) In general, 1)coscos()( 1 xxnxTn
1)coshcosh( 1 xxn
1)(,1 xTx n
1)coshcosh( xxn
Tn(x)345
n( )
1
2
n=1
xpassband
37
passband
Magnitude of Chebyshev Polynomials
|Tn(x)|345
22
n=1
xx
passband
38
passband
Chebyshev Responses
39
Mapping of Passband and Stopband
Define is mapped to passbandthe upper passband edgethe lower passband edge
1cos/cos xx m,1 xm1 x the lower passband edge ,1 xm
1
1
( ) cos [cos (cos / cos )] (sec cos )(sec cos ) sec cos
n m n mT x n TT
1
22
3
(sec cos ) sec cos
(sec cos ) sec (1 cos 2 ) 1
(sec cos ) sec (cos3 3cos ) 3sec cos
m m
m m
TTT
40
3
4 24
(sec cos ) sec (cos3 3cos ) 3sec cos
(sec cos ) sec (cos 4 4cos 2 3) 4sec (1 cos 2 ) 1m m m
m m m
TT
Design of Chebyshev Transformer
jN NNN ]22[2)(
LmN
L
njN
ZZZZ
TAAT
ZZZZ
nNNNe
)(sec1)(sec)0(
...]2cos...2coscos[2)(
00
10
mmNm
LmNL
ATAZZTZZ
)sec(cos)(sec 00
L
L
mm
L
L
mL
LmN
f
ZZZZ
NZZZZ
ZZZZ
AT 1cosh1coshsec,11)(sec
0
01
0
0
0
0
41m
ff 420
Table 5.2 Chebyshev Transformer Design
42
Example 5.6Design a Chebyshev transformer of N=3 for ZL = 2Z0 = 100 Ωwith Γm = 0.05.
3j Sol: 30 1
33
( ) 2 [ cos3 cos ]
(sec cos )
j
jm
eAe T
3sec (cos3 3cos ) 3 sec cos
0.05m m
m
A AA
1 01 1sec cosh cosh
m
Lm
Z ZN Z Z
0
11 20cosh cosh 1.4083 3
m LN Z Z
30 0 3
3 3
2 sec 0.0698mA
4312 3A 3
1 2(sec sec ) 0.1037m m
Example 5.6Design a Chebyshev transformer of N=3
50.5711 0
01
ZZ
81.70111
112
0
ZZ
20.87111
223
1
ZZ
%10142
1
7.44
2
om
m
ff
0mf
44
Example: “Chebyshev轉換器特性之深入探討電腦模與實驗量p測”清大物理 碩士論文 張靜宜
0chebyshev4-section
measured
-20
ion(
dB)
simulation
-40Ref
lect
i
10 11 12-60Load/
Transformer
ModeFrequency(GHz)
Ch b h 四階轉換器電腦模擬與實驗結果之比較
terminationMode
converter
45
Chebyshev四階轉換器電腦模擬與實驗結果之比較
gx XÇw Éy VtÑA H Ñ
46