5. Impedance Matching and Tuning
• Apply the theory and techniques of the
previous chapters to practical problems in
microwave engineering.
• Impedance matching is the 1st topic.
Figure 5.1 (p. 223)
A lossless network matching an arbitrary load
impedance to a transmission line.
1
• Impedance matching or tuning is important
since
– Maximum power is delivered when the load is
matched to the line, and power loss in the feed line
is minimized.
– Impedance matching sensitive receiver
components improves the signal-to-noise ratio of
the system.
– Impedance matching in a power distribution
network will reduce the amplitude and phase errors.
2
• Important factors in the selection of matching
network.
–
–
–
–
Complexity
Bandwidth
Implementation
Ajdustability
3
5.1 Matching with Lumped Elements
• L-section is the simplest type of matching
network.
• 2 possible configurations
Figure 5.2 (p. 223)
L-section matching networks.
(a) Network for zL inside the 1 + jx circle.
(b) Network for zL outside the 1 + jx circle.
4
Analytic Solution
• For Fig. 5. 2a, let ZL=RL+jXL. For zL to be
inside the 1+jx circle, RL>Z0. For a match,
1
Z 0  jX 
jB  1/( RL  jX L )
B( XRL  X L Z 0 )  RL  Z 0
X (1  BX L )  BZ 0 RL  X L
• Removing X 
X L  RL / Z 0 RL2  X L2  Z 0 RL
B
RL2  X L2
Z0
1 X L Z0
X 

B
RL
BRL
5
• For Fig.5.2b, RL<Z0.
1
1
 jB 
Z0
RL  j ( X  X L )
BZ 0 ( X  X L )  Z 0  RL
( X  X L )  BZ 0 RL
X   RL ( Z 0  RL )  X L
( Z 0  RL ) / RL
B
Z0
6
Smith Chart Solutions
• Ex 5.1
7
Figure 5.3b (p. 227)
(b) The two possible Lsection matching circuits.
(c) Reflection coefficient
magnitudes versus
frequency for the
matching circuits of (b).
8
Figure on page 228.
9
5.2 Single Stub Tuning
Figure 5.4 (p. 229)
Single-stub tuning
circuits.
(a) Shunt stub.
(b) Series stub.
10
• 2 adjustable parameters
– d: from the load to the stub position.
– B or X provided by the shunt or series stub.
• For the shunt-stub case,
– Select d so that Y seen looking into the line at d
from the load is Y0+jB
– Then the stub susceptance is chosen as –jB.
• For the series-stub case,
– Select d so that Z seen looking into the line at d
from the load is Z0+jX
– Then the stub reactance is chosen as –jX.
11
Shunt Stubs
• Ex 5.2 Single-Stub Shunt Tuning
ZL=60-j80
Figure 5.5a (p.
230)
Solution to
Example 5.2.
(a) Smith chart for
the shunt-stub
tuners.
12
Figure 5.5b (p. 231)
(b) The two shunt-stub tuning solutions. (c) Reflection
coefficient magnitudes versus frequency for the tuning
circuits of (b).
13
• To derive formulas for d and l, let ZL= 1/YL=
RL+ jXL. Z  Z 0 ( RL  jX L )  jZ 0 tan  d
Z 0  j ( RL  jX L ) tan  d
Y  G  jB 
1
Z
RL (1  tan 2  d )
where G  2
RL  ( X L  Z 0 tan  d ) 2
RL2 tan  d  ( Z 0  X L tan  d )( X L  Z 0 tan  d )
B
Z 0 [ RL2  ( X L  Z 0 tan  d ) 2 ]
• Now d is chosen so that G = Y0=1/Z0,
Z 0 ( RL  Z 0 ) tan 2  d  2 X L Z 0 tan  d  ( RL Z 0  RL2  X L2 )  0
X L  RL [( Z 0  RL ) 2  X L2 ] / Z 0
tan  d 
, for RL  Z 0
RL  Z 0
14
• If RL = Z0, then tanβd = -XL/2Z0. 2 principal
solutions are
 1
XL 
XL
1 

d  2

 1
 2

tan  
0
 for 2Z 0
 2Z 0 

XL 
XL
1 
0
   tan  
  for 2Z0
 2Z 0  

• To find the required stub length, BS = -B.
for open stub
l0
1
1
1  BS 
1  B 


tan    
tan  
2
2
 Y0 
 Y0 
1
1
1  Y0 
1  Y0 

tan   
tan  

2
B
 BS  2
l0
for short stub
15
Series Stubs
• Ex 5.3 Single Stub Series Tuning
ZL = 100+j80
Figure 5.6a (p. 233)
Solution to Example
5.3.
(a) Smith chart for
the series-stub tuners.
16
Figure 5.6b (p. 232)
(b) The two seriesstub tuning solutions.
(c) Reflection
coefficient
magnitudes versus
frequency for the
tuning circuits of (b).
17
• To derive formulas for d and l, let YL= 1/ZL=
GL+ jBL. Y  Y0 (GL  jBL )  jY0 tan  d
Y0  j (GL  jBL ) tan  d
Z  R  jX 
1
Y
GL (1  tan 2  d )
where R  2
GL  ( BL  Y0 tan  d ) 2
GL2 tan  d  (Y0  BL tan  d )( BL  Y0 tan  d )
X
Y0 [GL2  ( BL  Y0 tan  d ) 2 ]
• Now d is chosen so that R = Z0=1/Y0,
Y0 (GL  Y0 ) tan 2  d  2 BLY0 tan  d  (GLY0  GL2  BL2 )  0
BL  GL [(Y0  GL ) 2  BL2 ] / Y0
tan  d 
, for GL  Y0
GL  Y0
18
• If GL = Y0, then tanβd = -BL/2Y0. 2 principal
solutions are
 1
BL 
BL
1 

d  2

 1
 2

tan  
0
 for 2Y0
 2Y0 

BL  
BL
1 
0
   tan  
  for 2Y0
 2Y0  

• To find the required stub length, XS = -X.
for short stub
l0
1
1
1  X S 
1  X 


tan 
tan  

2
2
 Z0 
 Z0 
1
1
1  Z 0 
1  Z 0 

tan 
tan  


2
X 
 X S  2
l0
for open stub
19
5.3 Double-Stub Tuning
• If an adjustable tuner was desired, single-tuner
would probably pose some difficulty.
Smith Chart Solution
• yL  add jb1 (on the rotated 1+jb circle) 
rotate by d thru SWR circle (WTG)  y1 
add jb2  Matched
• Avoid the forbidden region.
20
Figure 5.7 (p.
236)
Double-stub
tuning.
(a) Original
circuit with the
load an arbitrary
distance from the
first stub.
(b) Equivalentcircuit with load
at the first stub.
21
Figure 5.8 (p.
236)
Smith chart
diagram for the
operation of a
double-stub
tuner.
22
Ex. 5.4 ZL = 60-j80
Open stubs, d = λ/8
Figure 5.9a (p.
238)
Solution to
Example 5.4.
(a) Smith chart
for the doublestub tuners.
23
Figure 5.9b (p. 239)
(b) The two double-stub tuning solutions.
(c) Reflection coefficient magnitudes versus frequency
for the tuning circuits of (b).
24
Analytic Solution
• To the left of the first stub in Fig. 5.7b,
Y1 = GL + j(BL+B1) where YL = GL + jBL
• To the right of the 2nd stub,
GL  j ( BL  B1  Y0t )
Y2  Y0
where t  tan  d
Y0  jt (GL  jBL  jB1 )
• At this point, Re{Y2} = Y0
2
2
(
Y

B
t

B
t
)
1

t
L
1
GL2  GLY0 2  0
0
2
t
t
4t 2 (Y0  BL t  B1t ) 2
1 t2 
GL  Y0
1 1
2 
2t 
Y02 (1  t 2 ) 2



25
• Since GL is real,
4t 2 (Y0  BLt  B1t ) 2
0
1
2
2 2
Y0 (1  t )
Y0
1 t2
0  GL  Y0 2  2
t
sin  d
• After d has been fixed, the 1st stub susceptance
can be determined as
Y0  (1  t 2 )GLY0  GL2t 2
B1   BL 
t
• The 2nd stub susceptance can be found from
the negative of the imaginary part of (5.18)
26
• B2 =
Y0 Y0GL (1  t 2 )  GL2t 2  GLY0
GLt
• The open-circuited stub length is
1
1  B 

tan  
 2
 Y0 
l0
• The short-circuited stub length is
1
1  Y0 

tan  

2
B
l0
27
5.4 The Quarter-Wave Transformer
• Single-section transformer for narrow band
impedance match.
• Multisection quarter-wave transformer designs
for a desired frequency band.
• One drawback is that this can only match a
real load impedance.
• For single-section,
Z1  Z0 Z L
28
Figure 5.10 (p. 241)
A single-section quarter-wave matching
transformer.   0 4 at the design
frequency f0.
29
• The input impedance seen looking into the
matching section is
Z L  jZ1t
Z in  Z1
Z1  jZ L t
where t = tanβl = tanθ, θ = π/2 at f0.
• The reflection coefficient
Zin  Z 0 Z1 ( Z L  Z 0 )  jt ( Z12  Z 0 Z L )


Zin  Z 0 Z1 ( Z L  Z 0 )  jt ( Z12  Z 0 Z L )
• Since Z12 = Z0ZL,
Z L  Z0

Z L  Z 0  j 2t Z 0 Z L
30
• The reflection coefficient magnitude is
 


Z L  Z0
 Z
2
1
L

 Z 0  /( Z L  Z 0 )   4t Z 0 Z L /( Z L  Z 0 ) 
2
1  4Z Z
L

1/ 2
 Z L  Z 0   4t Z 0 Z L 


2
1  4Z Z
L
2
2
2
1
2
2
2
1/ 2

/( Z L  Z 0 ) sec  
2
0

/( Z L  Z 0 )    4t Z 0 Z L /( Z L  Z 0 ) 
1
2
0
1/ 2
1/ 2
31
• Now assume f ≈ f0, then l ≈ λ0/4 and θ ≈ π/2.
Then sec2 θ >> 1. 

Z L  Z0
2 Z0 Z L
cos  , for  near  / 2
32
• We can define the bandwidth of the matching
transformer as   2     

2
 2 Z0 Z L

1
 1 
sec  m 
2
 Z L  Z0

m


m


2
m
2 Z0 Z L
or cos  m 
1   2m Z L  Z 0
• For TEM line,
• At θ = θm,
fm 
2 f v p
f
  l 

v p 4 f0 2 f0
2 m f 0

33
• The fractional bandwidth is
fm
4 m
f 2( f 0  f m )

 22
 2
f0
f0
f0

 
2 Z0 Z L
m
 2  cos 

 1   2m Z L  Z 0
4
1



• Ex. 5.5 Quarter-Wave Transformer
Bandwidth
ZL = 10, Z0 = 50, f0= 3 GHz, SWR ≤ 1.5
34
Figure 5.12 (p. 243)
Reflection coefficient magnitude versus frequency for a
single-section quarter-wave matching transformer with 35
various load mismatches.
5.5 The Theory of Small Reflection
Single-Section Transformer
Z 2  Z1
Z L  Z2
1 
,  2  1 ,  3 
,
Z 2  Z1
Z L  Z2
2Z 2
2 Z1
T21  1  1 
, T12  1   2 
Z 2  Z1
Z 2  Z1
  1  T12T213e 2 j  T12T21 32 2e 4 j 
T12T213e 2 j
 1 
1   2 3e 2 j
2 j
1  3e

1  13e 2 j
13
1

1   3e 2 j
36
Figure 5.13 (p. 244)
Partial reflections and transmissions on a single-section
37
matching transformer.
Multisection Transformer
Z1  Z 0
Z n 1  Z n
ZL  ZN
0 
, n 
, N 
,
Z1  Z 0
Z n 1  Z n
ZL  ZN
( )   0  1e
2 j
  2e
4 j

 Ne
2 jN
• Assume the transformer is symmetrical,
 0   N , 1   N 1,  2   N 2, etc.
38
( )  e jN { 0 [e jN  e  jN ]  1[e j ( N 1)  e  j ( N 1) ] 
}
• If N is odd, the last term is ( N 1) / 2 (e j  e j )
while N is even,  N / 2
( )  2e  jN [ 0 cos N 
 1 cos( N  2) 
1
  N / 2 ], for N even,
2
( )  2e  jN [ 0 cos N 
 1 cos( N  2) 
1
  ( N 1) / 2 cos  ], for N odd,
2
39
5.6 Binomial Multisection Matching Transformer
• The response is as flat as possible near the
design frequency.  maximally flat
• This type of response is designed, for an Nsection transformer, by setting the first N-1
derivatives of |Γ(θ)| to 0 at f0.
• Such a response can be obtained if we let
( )  A(1  e j 2 ) N
( )  A e
 j
j
e e
 2 A cos 
N
2 N
N
40
• Note that |Γ(θ)| = 0 for θ=π/2, (dn |Γ(θ)|/dθn ) =
0 at θ=π/2 for n = 1, 2, …, N-1.
• By letting f  0,
Z L  Z0
(0)  2 A 
Z L  Z0
A2
N
N
Z L  Z0
Z L  Z0
N
( )  A(1  e j 2 ) N  A CnN e2 jn ,
n 0
N!
where C 
( N  n)!n !
N
n
N
( )  A C e
n 0
N 2 jn
n
 0  1e
2 j
  2e
4 j

 Ne
2 jN
41
 n  ACnN
• Γn must be chosen as
• Since we assumed that Γn are small, ln x ≈ 2(x1)/(x+1),
Z n 1  Z n
n 
Z n 1  Z n
Z n 1
ln
Zn
1 Z n 1
ln
2
Zn
Z L  Z0 N
2 n  2 AC  2(2 )
Cn
Z L  Z0
N
n
N
ZL
2 C ln
Z0
N
N
n
• Numerically solve for the characteristic
impedance  Table 5.1
42
• The bandwidth of the binomial transformer
 m  2 N | A | cos N  m
1/ N





1
 m  cos 1   m  
 2  | A |  
m
f 2( f 0  f m )

 24
f0
f0

1/ N


4
1 1   m 
 2  cos  
 

 2  | A |  
• Ex. 5.6 Binomial Transformer Design
43
Figure 5.15 (p. 250)
Reflection coefficient magnitude versus frequency for
multisection binomial matching transformers of
Example 5.6 Z = 50Ω and Z = 100Ω.
44
5.7 Chebyshev Multisection Matching
Transformer
Chebyshev Polynomial
• The first 4 polynomials are
T1 ( x)  x,
T2 ( x)  2 x  1,
2
T3 ( x)  4 x 3  3 x,
T4 ( x)  8 x 4  8 x 2  1.
• Higher-order polynomials can be found using
Tn ( x)  2 xTn1 ( x)  Tn2 ( x)
45
Figure 5.16 (p. 251)
The first four Chebyshev polynomials Tn(x).
46
• Properties
– For -1≤x ≤1, |Tn(x)|≤1  Oscillate between ±1 
Equal ripple property.
– For |x| > 1, |Tn(x)|>1  Outside the passband
– For |x| > 1, |Tn(x)| increases faster with x as n
increases.
• Now let x = cosθ for |x| < 1. The Chebyshev
polynomials can be expressed as
Tn (cos  )  cos n
More generally,
Tn ( x)  cos(n cos 1 x)
1
Tn ( x)  cosh(n cosh x)
for | x | 1,
for | x | 1.
47
• We need to map θm to x=1 and π- θm to x = -1.
For this,
 1  cos   
cos 
Tn (
)  cos n (sec  m cos  )  cos n cos 

cos  m
 cos  m  

• Therefore,
T1 (sec  m cos  )  sec  m cos  ,
T2 (sec  m cos  )  sec 2  m (1  2 cos  )  1,
T3 (sec  m cos  )  sec  m (cos 3  3cos  )  3sec  m cos  ,
3
T4 (sec  m cos  )  sec 4  m (cos 4  4 cos 2  3)
4sec 2  m (cos 2  1)  1.
48
Design of Chebyshev Transformers
• Using (5.46)
( )  2e jN [0 cos N 
 N cos( N  2n) 
 1 cos( N  2) 
]
 Ae jN TN (sec  m cos  )
• Letting θ = 0,
Z L  Z0
(0) 
 ATN (sec  m )
Z L  Z0
Z L  Z0
1
A
Z L  Z 0 TN (sec  m )
49
• If the maximum allowable reflection
coefficient magnitude in the passband is Γm,
1 Z L  Z0
1 Z L  Z0
TN (sec  m ) 

A Z L  Z0 m Z L  Z0
1
ZL
ln
2 m
Z0
 cosh( N cosh 1 (sec  m ))
1
 1 Z L  Z0 
1
sec  m  cosh  cosh 
 

 N
  m Z L  Z 0  
1
 1
Z L 
1
cosh  cosh 
ln
 

Z 0  
 N
 2 m
50
• Once θm is known,
m
f
 24
f0

Ex 5.7 Chebyshev Transformer Design
Γm = 0.05, Z0 = 50, ZL = 100
Use Table 5.2
51
Figure 5.17 (p. 255)
Reflection coefficient magnitude versus frequency for
the multisection matching transformers of Example 5.7. 52
Figure 5.18 (p. 256)
A tapered transmission line matching
section and the model for an incremental
length of tapered line. (a) The tapered
transmission line matching section.
53
Figure 5.19 (p. 257)
A matching section with an exponential
impedance taper. (a) Variation of impedance.
(b) Resulting reflection coefficient
54
triangular taper
for d(In Z/Z0/dz.
(a) Variation of
impedance. (b)
Resulting
reflection
coefficient
magnitude
response.
55
(a) Impedance
variations for the
triangular,
exponential, and
Klopfenstein tapers.
(b) Resulting
reflection coefficient
magnitude versus
frequency for the
tapers of (a).
56
with passive and
lossless networks
(ω0 is the center
frequency of the
matching
bandwidth). (a)
Parallel RC.
(b) Series RC. (c)
Parallel RL.
(d) Series RL.
57
Figure 5.23 (p. 263)
Illustrating the Bode-Fano criterion. (a) A
possible reflection coefficient response. (b)
Nonrealizable and realizable reflection
58