Review of Smith Chart
Chapter 5 Impedance Matching and Tuning
• The matching network is ideally lossless,
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 qquarter-wave impedance
p
transformer is for read load.
• Important factors in selecting matching networks:
((1)complexity,
)
p
y, (2)bandwidth,
( )
, (3)implementation,
( ) p
, (4)adjustability.
( ) j
y
zL = ZL / Z0
= rL + jxL
1
2
Analytical Solutions (rL>1)
5.1 Lumped
p Element Matchingg ((L-networks))
L-type or L-network matching is the simplest matching network.
• When
rL > 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
Z in  jX 
1
jB 
1
RL  jX L
 Z0
 B( XRL  X L Z 0 )  RL  Z 0

 X (1  BX L )  BZ 0 RL  X L
(1) zL = ZL/Z0 is inside the rr=11
circle (rL > 1).
(2) ZL is
i shunt
h
with
i h jB,
B then
h
series with jX.

XL 

 B 



X

(1) zL = ZL/Z0 is outside the rr=11
circle (rL < 1).
(2) ZL is
i series
i with
i h jB,
B then
h
shunt with jX.
3
RL
Z0
RL 2  X L 2  Z 0 RL
Z in
RL 2  X L 2
1 X L Z0 Z0


B
RL
BRL
• Note that when rL > 1, the square root in B has real results.
4
Analytical Solutions (rL<1)
• When
Example 55.1
1L
L-Section
Section Impedance Matching
rL < 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
ZL = 200-j100
j
Ω,, Z0 = 100 Ω,,
f = 500 MHz, zL = ZL/Z0 = 2-j1,
rL = 2 >1 inside the r = 1 circle.
1
1
1
 jB 

Z in
RL  j ( X  X L ) Z 0
BZ 0 ( X  X L )  Z 0  RL
( X  X L )  BZ 0 RL
Z0

( Z 0  RL ) / RL
 B
Z0


 X   X L  RL ( Z 0  RL )
Exact solutions:
XL 
Z in
B

• Two analytical solutions are physically realizable.
X
• One of the two solutions may result in smaller reactive elements,
elements and
may have better matching, bandwidth, or better SWR on the line.
RL
Z0
RL 2  X L 2  RL Z 0
RL 2  X L 2
3
1
100  2 2002  1002  100  200 2.899  10  



2002  1002
6 899 103  1 
6.899
122.47 
1 X L Z0 Z0




B
RL
BRL 122.47 
5
Example 55.1
1L
L-Section
Section Impedance Matching
6
Example 55.1
1L
L-Section
Section Impedance Matching
The two solutions:
Th two solutions:
The
l i
2.899 103
 0.9228 ppF
 2  500 106
X
122.47
 38.98 nH
L 
 2  500 106
( ) C
(1)
B

1
1

X  122.47  2  500 106
 2.599
2 599 pF
(2) C  
1
1

3
B 6.899
6 899  10  2  500  106
 46.14 nH
L
7
8
Lumped Elements for
Microwave Integrated Circuits (MICs)
5.2 Single-Stub Tuning
Shunt‐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.
(2) Move along the const-Γ circle with 2d/λ wavelengths to
reach the g = 1 (r = 1) circle.
(3) Sh
Shuntt with
ith a susceptance
t
or series
i with
ith a reactance
t
to
t
cancel the imaginary part.
10
9
Shunt-Stub Tuning Analytical Solution
Z in
no stub
Gin 
 Z0
Z L  jZ 0t
Z 0  jZ L t
t  tan  d 
Shunt-Stub Tuning Analytical Solution
If RL  Z0 ,
1 RL  j ( X L  Z 0t )
1

Y0 Z 0  X Lt  jRLt Gin  jBin
t
RL 2t  ( Z 0  X L t )( Z 0t  X L )
RL (1  t 2 )
,
B
Y

in
0
RL 2  ( X L  Z 0t ) 2
RL 2  ( X L  Z 0t ) 2
X L Z 0  ( X L Z 0 ) 2  Z 0 ( RL  Z 0 )( RL Z 0  RL 2  X L 2 )
Z 0 ( RL  Z 0 )
XL  (

Gin  Y0  Z 0 RL (1  t 2 )  RL 2  ( X L  Z 0t ) 2 real part
RL 2
) [( RL  Z 0 ) 2  X L 2 ]
Z0
RL  Z 0
here t  tan  l
Let the stub susceptance Bstub (or Bs) = -Bin,
(a) open-stub: Z stub   jZ 0 t and Ystub   jY0t  jBstub  jBs
X
If RL  Z 0 , and t  tan  d   L
2Z 0
lo

 X 
1
tan 1   L  , X L  0

 2Z 0 
d  2
 
 1 
X L 
1 


tan



 , X L  0
 2
 2Z 0  
 


B 
B
1
1
tan 1 ( s )  
tan 1  in 
Y0
2
2
 Y0 
(b) Short-stub: and
Z stub  jZ 0t
Ystub   jY0 t  jBstub  jBs
ls
11

1
tan 1 (
Y0
)
1
Y 
tan 1  0 

2
Bs
2
 Bin 
If lo < 0 or ls < 0, λ/2 must be added to have a realistic result.
12
Example 5.2 Shunt Single-Stub Tuning
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:
Match ZL = 15 + j10 Ω to a 50-Ω line. Use a shunt open-stub.
Sol: RL ≠ Z0,
XL  (
t
RL 2
) ( RL  Z 0 ) 2  X L 2 
Z0 
RL  Z 0

10  397.5
35
10  397.5
d
1
 
(  tan 1 t1 )  0.387
35
 2
B 
l
1
 Bin  ...  o  
tan 1  in 

2
 Y0 
t1  
10  397
397.5
5
d
1
 
tan 1 t2  0.044
35
 2
B 
l
1
 Bin  ...  o  
tan 1  in 

2
 Y0 
t2  
13
14
5.3 Double-Stub Tuning – Analytical Solution
5.3 Double-Stub Tuning – Analytical Solution
• In single-stub tuning, the length d is not tunable.
0  4t 2 (Y0  BL t  B1t ) 2  Y0 2 (1  t 2 ) 2
Y0
1 t2

2
2t
sin 2  d
GL has an upper limit.
limit
0  GL  Y0
• If this causes difficulty in circuit implementation, use double-stub tuning.
Given d , one can obtain:
Y1  GL  j ( BL  B1 )
Y2  Y0
GL  j ( BL  B1  Y0t )
, t  tan  d
Y0  jt (GL  jBL  jB1 )
Re[Y2 ]  Y0 , Im[Y2 ]   B2
YL  Y0
YL  jjY0 tan  b
 GL  jBL
Y0  jYL tan  b
15
(Y  BL t  B1t ) 2
1 t2
GL 2  Y0 2 GL  0
0
t
t2
4t 2 (Y0  BL t  B1t ) 2
1  t 2 
GL  Y0 2 1  1 
t 
Y0 (1  t 2 ) 2
B1   BL 
Y0  (1  t 2 )GLY0  GL 2t 2
t
GL  (1  t )GLY0  GL 2t 2
2
B1  Y0
0


1
2
 
1
s


  2
t


B
tan 1   , B  B1 , B2
Y
 0
Y 
tan 1  0  , B  B1 , B2
B
16
Example 5.4 Double-Stub Tuning - Performance
5.4 The Quarter-Wave Transformer
Use double-stub matching scheme to match ZL = 60 – j80Ω at 2.0 GHz to a
50-Ω line.
• 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.
17
Review
18
Estimate the Bandwidth of
a Single
Single-Section
Section Impedance Transformer
2.5 The Quarter-Wave Transformer
A quarter-wave transformer is an impedance matching circuit
Z1  Z 0 Z L
Z in  Z1
Z L  jZ1t
, t  tan    tan 
Z1  jZ Lt
Z  jZ1t
Z1 L
 Z0
Z in  Z 0
Z ( Z  Z 0 )  j ( Z12  Z 0 Z L )t
Z1  jZ Lt


 1 L
Z in  Z 0 Z Z L  jZ1t  Z
Z1 ( Z L  Z 0 )  j ( Z12  Z 0 Z L )t
1
0
Z1  jZ Lt
R  jZ1 tan  l
Z in  Z1 L
Z1  jRL tan  l
in 

2
1
l

2
Z

RL
Z in  Z 0
Z in  Z 0
Z L  Z0
Z L  Z0
  

Z L  Z 0  j 2t Z 0 Z L
( Z L  Z 0 ) 2  4t 2 Z 0 Z L
When  
If in  0 is required, Z in  Z 0 , then Z1  Z 0 RL

2
,
 
Z L  Z0
2 Z L Z0
Set a max    m , or  m 
19
1
1
4Z L Z 0
 Z L  Z0 
2
sec 2 
cos 
Z L  Z0
2 Z L Z0
cos  m
20
Example 5.5
Single Section Quarter
Single-Section
Quarter-Wave
Wave Transformer Bandwidth
Bandwidth of the Matching Transformer
  2( / 2   m )
ZL = 10 Ω, Z0 = 100 Ω, SWR = 1.2
S l
Sol:
4Z 0 Z L
1
 1
sec 2  m
2
m
(Z 0  Z L )2
m :
 f
 fm

, m 
,
2 f0
cos  m 
2 f0
m
1  m
2
fm 
2 m

Z1  Z 0 Z L  31.6
f0
2 Z0 Z L
Z0  Z L
10 or 0.1
Fractional
F
ti l bandwidth
b d idth :
4
f 2( f 0  f m )

 2 m
f0
f0

 2
2 Z0 Z L
4  1   m
cos 
2

 1   m Z 0  Z L

SWR  1
 0.1
SWR  1
Vmax V0  V0 1  


SWR 
Vmin V0  V0 1  
m 
 
2 Z0 Z L  
4
f
m

 2   cos 1 
2 Z Z
f0



1


0
L 
m



4
6
 0.1 2  31.6  
 2   cos 1 
  2   9%



 0.99 90  
4 or 0.25


 
2 or 0.5
21
22
5.5 Theory of Small Reflections
5.5 Theory of Small Reflections
• For applications requiring more bandwidth than that a single quarterwave section can provide, multi-section transformers can be used.
(1) Multi-Reflections

(1) Single-Section Transformer
1 
Z 2  Z1
 2
Z 2  Z1
3 
ZL  Z2
ZL  Z2
 1  T12T213e  j 2  T12T2132  2 e  j 4  ...
 1  T12T213e  j 2 (1  3 2 e  j 2   3 2  e  j 4  ...)
2

 1  T12T213e  j 2   3 2  e  j 2 n
n
n 0
T21  1  1 
2Z
2
Z2
Z 2  Z1
T12  1  2 
2 Z1
Z 2  Z1
 1 
 j 2
T12T213e
1  3e  j 2

1  3 2 e  j 2 1  13e  j 2
2 j
• If 13 is small,   1  3e
• Γ  the reflection from the initial discontinuity between Z1 and Z2 + the
first reflection from the discontinuity between Z2 and ZL
23
24
Multisection Transformer
Multisection Transformer
   0  1e  j 2   2 e  j 4  ...   N e  j 2 N
• N commensurate ((equal-length)
q
g ) sections of transmission lines. Let
the total reflection be Γ.
 e  jN [ 0 (e jN  e  jN )  1 (e
e
 j  N  2 
)  ...]
(if ssymmetry
mmetr )
  0 cos N  1 cos  N  2    ...   N /2  ,
N  even

 2e  jjN 
  0 cos N  1 cos  N  2    ...   N 1 /2 cos   , N  odd
• Assume that all Zn increase or decrease monotonically,
monotonically ZL is real
real,
and “The theory of small reflections” holds.
• It can be validated that
0 
j  N  2 
Z1  Z 0
Z  Z2
Z  ZN
Z  Z1
, 1  2
, 2  3
, ... N  L
Z1  Z 0
Z 2  Z1
Z3  Z 2
ZL  ZN
25
• 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
5.6 Binomial Multisection Matching Transformer
H to Fi
How
Find
d the
h C
Corresponding
di R
Reflection?
fl i ?
• Maximally flat response: the response is as flat as possible near the design
frequency. Also called Binomial matching transformer.
• Let     A(1  e 2 j ) N  Assumed
A
d artificially
tifi i ll
Theory of Samll Reflection:
    2 N A cos 
( )   0  1e  j 2   2 e  j 4  ...   N e  j 2 N
  0  2N A 
Binomial Multi-Section Transformer:
N
  0
Z L  Z0
 A N
2
Z L  Z0
N
    A(1  e 2 j ) N
    A(1  e 2 j ) N  A CnN e  j 2 n , CnN 
n 0
  0  1e
Chebyshev Multi-Section Transformer:
 j 2
N!
( N  n))!n !
  2 e  j 4  ...   N e  j 2 N
((0)) N
N! (("enn
enn factorial
factorial“))
Cn
N
“Four factorial" is written as "4!”
2
Properly configure the impedances to synthesize the needed
 n  ACnN 
    ATn (sec  m cos  )
reflections  n .
27
28
Binomial Multisection Matching Transformer
n 
ln
Z n 1  Z n 1 Z n 1
x 1
 ln
, ln x  2
,
Z n 1  Z n 2
Zn
x 1
x  1  1
Z n 1
Z  Z0
1
1
Z
 N CnN ln L
 2n  2 ACnN  N 1 CnN L
Zn
2
Z L  Z0 2
Z0
Z n 1  Z L 
 
Z n  Z 0 
• Exact
Binomial Transformer Design Table 5.1
C nN / 2 N
results for Zn-11 / Z0 for N = 2 thru 6 are given in Table 5.1.
51
29
30
Example Binomial Transformer Design
ZL/Z0
Z1 /Z0
Z2 /Z0
N=5
Z3 /Z0
Z4 /Z0
Z5 /Z0
1.0
1.5
2.0
30
3.0
4.0
6.0
8.0
10.0
1.0000
1.0128
1.0220
1 0354
1.0354
1.0452
1.0596
1.0703
1.0789
1.0000
1.0790
1.1391
1 2300
1.2300
1.2995
1.4055
1.4870
1.5541
1.0000
1.2247
1.4142
1 7321
1.7321
2.0000
2.4495
2.8284
3.1623
1.0000
1.3902
1.7558
2 4390
2.4390
3.0781
4.2689
5.3800
6.4346
1.0000
1.4810
1.9569
2 8974
2.8974
3.8270
5.6625
7.4745
9.2687
E ample Binomial Transformer Design ZL/Z0 <1
Example
N=5
6
 1.0596 ,
5.6625
6
 1.4055
4.2689
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
• A 5th order
d binomial
bi
i l transformer
f
for
f ZL = 6Z0
• Also note that
ZL/Z0
31
32
Bandwidth of the Binomial Transformer
Binomial Transformer’s Frequency Response
m  2 N A cos N  m , i.e, tolerable max m over the passband.
 1   1/ N 
 m  cos   m  
2 A
   
1
1 
f 2( f 0  f m )
4
4

 2  m  2  cos 1   m 
2 A
f0
f0


  
1/ N




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
Z1  54.5 , Z 2  70.7 , Z3  91.7
1 100  50
 0.4167
0 4167
8 100  50
 1 0.05 1 
f
4
 2  cos 1  (
) 3   70%
f0

 2 0.4167 


A  2 N (0) 
DIY
Reflection coefficient magnitude versus frequency for multisection
binomial matching transformer of Ex. 5.6 with ZL = 2Z0 = 100 Ω and
Γm = 0.05.
33
Example (2nd Midterm)
Design
g a Butterworth transformer of N = 2 for ZL = 4Z0. Let Γ0, Γ1 and ΓL be
respectively the reflection coefficients at the Z0 – Z1, Z1 – Z2 and Z2 – ZL
junctions, and Γ0 = ΓL. (a) Based on the “theory of small reflection,” find the
Γin terms of Γ0, Γ1 and θ.
θ (b) If Γin = 0 and  in / f  0 are required when θ
= λ / 4, find a and b.
in  0  1e  j 2  2 e  j 4  20 cos 2  1
 
20 cos 2    1  0  1  20
2

 in
 40 sin 2  0

a 1 4  b
 4a  ab  4  b  4a  4  ab  b  ab  4

a 1 4  b
ba
a 1
1
4
2
 ab  a 2  b  a  2(ab  a 2  b  a)  a 3  a 2  a  4  0
ba
a 1
3
3
1 4
4  1 / 9 35
 1 1
r     12  
 1.9273 , q 

 0.4321
81
6 9
9
 27 27
3
r  q 3  r 2  1.5707 ,
3
r  q 3  r 2  0.2751
1
a   1.5707  0.2751  1.4067
9
b  4 / a  2.8435
The “exact” solution in Table 5.1 is a =
1.4142 and b =2.8285. The error is from
the approximation used in the “theory of
35
small reflection.”
34
5.7 Chebyshev Multisection Matching Transformers
Chebyshev polynomials:
T0 ( x)  1
T1 ( x)  x
Recurrence formula:
T3 ( x)  4 x 3  3x
Tn ( x)  2 xTn 1 ( x)  Tn  2 ( x)
4
2
T2 ( x)  2 x 2  1 T4 ( x)  8 x  8 x  1
Tn(x)
n2
54 3
2
n=1
x
ppassband
36
5.7 Chebyshev Multisection Matching Transformers
Magnitude of Chebyshev Polynomials
(1) x  1, Tn ( x)  1 mapped to passband
Let x = cosθ, it can be shown that Tn (cos  )  cos n
(2) x  1, Tn ( x)  1 outside
t id the
th passband
b d
(3) In general, Tn ( x)  cos(n cos 1 x)
x 1
 cosh(n cosh 1 x)
Tn((x))
|Tn(x)|
54 3
2
x 1
54 3
2
n=1
n=1
1
x
x
passband
passband
37
Chebyshev Responses
38
Mapping of Passband and Stopband
Define x  cos  / cos  m  x  1
is mapped to passband
 m  x  1,
the upper passband edge
the lower passband edge
   m  x  1,
Tn ( x)  cos n[cos 1 (cos  / cos  m )]  Tn (sec  m cos  )
T1 (sec  m cos  )  sec  m cos 
T2 (sec  m cos  )  sec2  m (1  cos 2 )  1
T3 (sec  m cos  )  sec3  m (cos 3  3cos  )  3sec  m cos 
39
T4 (sec  m cos  )  sec4  m (cos 4  4 cos 2  3)  4sec 2  m (1  cos 2 )  1 40
Design of Chebyshev Transformer
Table 5.2 Chebyshev Transformer Design
( )  2e  jN [0 cos N  1 cos N  2   ...  n cos N  2n   ...]]
 ( 0) 
Z L  Z0
1
Z L  Z0
 ATN (sec  m )  A 
Z L  Z0
TN (sec  m ) Z L  Z 0
m  A TN (cos  m sec  m )  A
TN (sec  m ) 
1
 1 Z L  Z0
1 Z L  Z0
1 Z L  Z0

, sec  m  cosh  cosh 1 
A Z L  Z 0 m Z L  Z 0
 N
 m Z L  Z 0
4
f
 2 m
f0





41
42
Example 5.6
Example 5.6
Design a Chebyshev transformer of N=3 for ZL = 2Z0 = 100 Ω
with Γm = 0.05.
 j 3
Sol: ( )  2e [ 0 cos 3  1 cos  ]
Design a Chebyshev transformer of N=3
 Ae  j 3 T3 (sec  m cos  )
 A sec3  m (cos 3  3cos  )  3 A sec  m cos 
A   m  0.05
1
 1 Z L  Z0
sec  m  cosh  cosh 1 
 m Z L  Z0
 N

 
 
Z1  Z 0
1  0
 57.50
1  0
Z 2  Z1
1  1
 70.81
1  1
Z3  Z 2
1  2
 87.20
1  2
4
f
 2 m
f0

1
 20  
 cosh  cosh 1     1.408
 3 
3
2 0  A sec3  m
  0  0.0698  3
 m  44.7 o
 101%
21  3A(sec3  m  sec  m )  1  0.1037   2
43
44
Example:
p “Chebyshev轉換器特性之深入探討電腦模與實驗量
測”清大物理 碩士論文 張靜宜
0
chebyshev
4-section
Reflectiion(dB)
measured
simulation
g{x XÇw Éy V{tÑAÑ H
-20
-40
Transformer
Mode
converter
Load/
termination
-60
10
11
Frequency(GHz)
12
Ch b h 四階轉換器電腦模擬與實驗結果之比較
Chebyshev
45
46