Chapter 5(Pozar Book)
Matching and Tuning
By
Dr.Kretika Goel
Introduction
•
•
•
•
This chapter marks a turning point, in that we now begin to apply the theory
and techniques of previous chapters to practical problems in microwave
engineering.
The basic idea of impedance matching is illustrated in Figure 5.1, which shows
an impedance matching network placed between a load impedance and a
transmission line.
The matching network is ideally lossless, to avoid unnecessary loss of power,
and is usually designed so that the impedance seen looking into the matching
network is Z0.
Then reflections will be eliminated on the transmission line to the left of the
matching network, although there will usually be multiple reflections between
the matching network and the load. This procedure is sometimes referred to as
tuning.
•
•
•
•
•
Impedance matching or tuning is important for the following reasons:
Maximum power is delivered when the load is matched to the line (assuming
the generator is matched), and power loss in the feed line is minimized.
Impedance matching sensitive receiver components (antenna, low-noise
amplifier, etc.) may improve the signal-to-noise ratio of the system.
Impedance matching in a power distribution network (such as an antenna
array feed network) may reduce amplitude and phase errors.
As long as the load impedance, ZL, has a positive real part, a matching
network can always be found.
Factors that may be important in the selectio of a particular matching
network include the following:
•
•
•
•
Complexity—As with most engineering solutions, the simplest design that
satisfies the required specifications is generally preferable. A simpler matching
network is usually cheaper, smaller, more reliable, and less lossy than a more
complex design.
Bandwidth—Any type of matching network can ideally give a perfect match
(zero reflection) at a single frequency. In many applications, however, it is
desirable to match a load over a band of frequencies. There are several ways of
doing this, with, of course, a corresponding increase in complexity.
Implementation—Depending on the type of transmission line or waveguide being
used, one type of matching network may be preferable to another. For example,
tuning stubs are much easier to implement in waveguide than are multisection
quarter-wave transformers.
Adjustability—In some applications the matching network may require
adjustment to match a variable load impedance. Some types of matching networks
are more amenable than others in this regard.
Why matching or tuning is important?
•To maximize power delivery and minimize power
loss.
•To improve signal to noise ratio as in sensitive
receiver components such as LNA, antenna, etc.
•To reduce amplitude and phase error as in distributed
network such as antenna array.
Zo
Matching
Network
Load Z L
Basic idea of impedance matching
Concept of maximum power transfer
In lump circuit
PL
Zo
Vi
I
ZL VL
Zo
Power deliver at ZL is
2

1
1 2
1  Vi

 Z L
PL  VL I  I Z L  
2
2
2  Z L  Zo 
Power maximum whence ZL = Zo
ZL
continue
In transmission line
The important parameter is reflection coefficient
No reflection whence ZL = Zo , hence
Z L  Zo
 
Z L  Zo
 0
The load ZL can be matched as long as ZL not equal to
zero (short-circuit) or infinity (open-circuit)
Factors in selecting matching network
• Complexity: simpler, cheaper, more reliable and
low loss circuit is preferred.
• Bandwidth: match over a desirable bandwidth.
• Implementation: depend on types of transmission
line either cable, stripline, microstripline,
waveguide, lump circuit etc.
• Adjustability:some network may need adjustment
to match a variable load.
Matching with lumped elements
The simplest matching network is an L-section using two reactive elements
jX
Configuration 1
Whence RL>Zo
Zo
jB
ZL
ZL=RL+jXL
jX
Configuration 2
Whence RL<Zo
Zo
jB
ZL
continue
If the load impedance
(normalized) lies in unity
circle, configuration 1 is
used.Otherwise configuration
2 is used.
The reactive elements are
either inductors or capacitors.
So there are 8 possibilities for
matching circuit for various
load impedances.
Matching by lumped elements
are possible for frequency
below 1 GHz or for higher
frequency in integrated
circuit(MIC, MEM).
Configuration 2
Configuration 1
Impedances for serial lumped elements
Serial circuit
Reactance
relationship
values
+ve
-ve
X=2pfL
X=1/(2pfC)
L=X/(2pf)
C=1/(2pfX)
L
C
R
R
Impedances for parallel lumped elements
Parallel circuit
Susceptance
relationship
values
+ve
-ve
B=2pfC
B=1/(2pfL)
C=B/(2pf)
L=1/(2pfB)
L
R
C
R
Lumped elements for microwave integrated circuit
Lossy film
Planar resistor
Chip resistor
Loop inductor
r
Dielectric
Interdigital gap
capacitor
Metal-insulatormetal capacitor
Spiral inductor
r
Chip capacitor
Matching by calculation for configuration 1
jX
Zo
jB
ZL
For matching, the total impedance of L-section plus ZL should equal to Zo,thus
Z o  jX 
1
jB  1 / RL  jX L 
Rearranging and separating into real and imaginary parts gives us
B XRL  X L Z o   RL  Z o
*
X 1  BX L   BZ o RL  X L
**
continue
Solving for X from simultaneous equations (*) and (**) and substitute X in (**)
for B, we obtain
B
X L  RL / Z o RL2  X L2  Z o RL
RL2  X L2
+ve capacitor
-ve inductor
Since RL>Zo, then argument of the second root is always positive, the series
reactance can be found as
Zo
1 X LZo
X  

B
RL
BR L
+ve inductor
-ve capacitor
Note that two solution for B are possible either positive or negative
Matching by calculation for configuration 2
jX
Zo
jB
ZL
For matching, the total impedance of L-section plus ZL should equal to 1/Zo,thus
1
1
 jB 
Zo
RL  j  X  X L 
Rearranging and separating into real and imaginary parts gives us
BZ o  X  X L   Z o  RL
*
 X  X L   BZ o RL
**
continue
Solving for X and B from simultaneous equations (*) and (**) , we obtain
X   R L Z o  R L   X L
B
Z o  R L  / R L
Zo
+ve inductor
-ve capacitor
+ve capacitor
-ve inductor
Since RL<Zo,the argument of the square roots are always positive, again
two solution for X and B are possible either positive or negative
Matching using Smith chart
parallel capacitance
(+ve)
serial inductance
(+ve)
(-ve)
parallel inductance
serial capacitance
(-ve)
Matching using lumped
components
Serial
L
Serial C
Parallel
C
C2
L2
10
C1
50
L1
Parallel
L
C1
L2
50
C2
L!
10
Example
Design an L-section matching network to match a series RC load with an
impedance ZL=200-j100 , to a 100  line, at a frequency of 500 MHz.
Solution
Normalized ZL we
have : ZL= 2-j1
Parallel L
(-j0.7)
Serial C
(-j1.2)
ZL= 2-j1
Serial L
(j1.2)
Parallel C
(+j0.3)
Solution 1
Solution 2
continue
C
L
b
 0.92 pF
2p f Z o
38.8nH
0.92pF
200-j100
x Zo
 38 .8nH
2p f
2.61pF
1
C
 2.61 pF
2p fx Z o
 Zo
L
 46 .1nH
2p f b
46.1nH
ZL=200-j100
Solution 2 seems to be better
matched at higher frequency
reflection coefficient
Reflection coefficient
1.2
1
0.8
solution 1
0.6
solution 2
0.4
0.2
0
0
0.5
freq (GHz)
1
1.5
Single stub-matching
Parallel configuration
Short-stub matching
x
short-stub
ZL
Zo
d
x
open-stub
Open-stub matching
ZL
Zo
d
Example
Design two single–stub shunt tuning networks to match a load ZL=15+j10 to
50 at 2GHz. The load consists of a resistor and inductor in series. Plot the
reflection magnitude from 1 GHz to 3 GHz for each solution.
Solution
•Normalized the load zL=0.3+j0.2
•Construct SWR circle and convert load to admittance
•Then move from load to generator to meet a circle (1+jb) to obtain two
points i.e y1=1-j1.33 and y2=1+j1.33.
•The distance d from the load to stub is obtained either of these two points i.e.
d1= 0.044l and d2=0.387l.
• To improve bandwidth, the stub is chosen as close as possible to the load.
•The tuning requires a stub of j1.33 for y1 and –j1.33 for y2.
•For open circuit-stub i1 =0.147l and i2=0.353l.
Smith Chart
Continue
0.147 l
0.353 l
15
50
15
50
50
50
0.796nH
0.796nH
0.044 l
0.387 l
Solution 1
Solution 2
Convert j0.2 to inductance value, thus
2p  2  10
 0.796nH
9
Use this value to calculate reflection
coefficient
Z L  Zo

Z L  Zo
1
refl. coeff.
L
0.2  50
Reflection coefficient
0.8
0.6
Solution 1
0.4
Solution 2
0.2
0
1
1.5
2
f (GHz)
2.5
3
Formulas for calculation
Let ZL=RL+jXL, then the impedance at distance d from the load is
Zd  Zo
Z L  jZ o tan  d
( R  jX L )  jZ o tan  d
 Zo L
Z o  jZ L tan  d
Z o  j ( RL  jX L ) tan  d
1
Yd  G  jB 
Zd
Admittance at this point is
Thus, by substituting Zd and separating real and imaginary, we have
G
B

RL 1  tan 2  d

RL2   X L  Z o tan  d 2
RL 2 tan  d  Z o  X L tan  d  X L  Z o tan  d 

Z o RL2   X L  Z o tan  d 2

continue
Equating G = Yo = 1/Zo,and t  tan  d , thus we have


1
RL 1  t 2
 2
Z o RL   X L  Z o t 2


RL2   X L  Z ot 2  Z o RL 1  t 2  0
Z o ( RL  Z o )t 2  2 X L Z ot  ( RL Z o  RL2  X L2 )  0
Solving
t


X L  RL Z o  RL 2  X L2 / Z o
 XL
t
2Z o
 RL  Z o 
for RL  Z o
for RL  Z o
continue
The two principle solution are
 1
1
tan
t
d  2p

l  1 p  tan 1 t
 2p

for t  0
 for t  0
To find the stub length,
1
1  B 

tan  
l 2p
 Yo 
for open  stub
1
1  Yo 

tan  
l 2p
B
for short  stub


Single stub-matching
Serial configuration
Zo
Zo
ZL
Short-stub matching
i
d
short-stub
Zo
Open-stub matching
i
Zo
openstub
d
ZL
Example
Match a load impedance of ZL=100+j80 to a 50 W line using a single series
open-stub.Assuming the load consists of resistor and inductor in series at
2GHz. Plot the reflection coefficient from 1 GHz to 3 GHz.
Solution
•Normalized the load zL=2+j1.6
•Construct SWR circle
•Then move from load to generator to obtain two points on unity circle(1+jx)
z1=1-j1.33 and z2=1+j1.33.
•The distance d from the load to stub is obtained either of these two points i.e.
d1= 0.120l and d2=0.463l.
• To improve bandwidth, the stub is chosen as close as possible to the load.
•The tuning requires a stub of j1.33 for z1 and –j1.33 for z2.
•For open circuit-stub i1 =0.397l and i2=0.103l.
Smith Chart
Continue
0.397 l
50
0.103 l
0.120 l
50
0.463 l
100
100
50
50
50
50
6.37nH
6.37nH
Solution 1
Solution 1
Convert j1.6 to inductance value, thus
2p  2  10
 6.37nH
9
Use this value to calculate reflection
coefficient
Z L  Zo

Z L  Zo
1
refl. coeff.
L
1.6  50
Reflection coefficient
0.8
0.6
Solution 1
0.4
Solution 2
0.2
0
1
1.5
2
f (GHz)
2.5
3
Formulas for calculation
Let YL=GL+jBL, then the load admittance at distance d from the load is
YL  jYo tan  d
(G L  jBL )  jYo t
Yd  Yo
 Yo
Yo  jYL tan  d
Go  j (G L  jBL )t
Where t = tan d
1
Z d  R  jX 
Yd
Impedance at this point is
Thus, by substituting Yd and separating real and imaginary, we have
R

GL 1  t 2

GL2  BL  Yo t 2
X 
GL 2t  Yo  BL t BL  Yot 

Z o GL2  BL  Yot 2

continue
Equating R = Zo = 1/Yo , thus we have


1
GL 1  t 2
 2
Yo GL  BL  Yot 2


GL2  BL  Yot 2  YoGL 1  t 2  0
Yo (GL  Yo )t 2  2 BLYot  (GLYo  GL2  BL2 )  0
Solving
t
t


BL  G L Yo  G L 2  BL2 / Yo
GL  Yo 
 BL
2Yo
for G L  Yo
for G L  Yo
continue
The two principle solution are
 1
1
tan
t
d  2p

l  1 p  tan 1 t
 2p

for t  0
 for t  0
To find the stub length,
1
1  X 


tan 
l 2p
 Zo 
for short  stub
1
1  Z o 

tan 

l 2p
 X 
for open  stub


Double-stub matching
S2
S1
Open or short
stubs
Zo
ZL
d
x
The advantage of this technique is the position of stubs
( d and x) are fixed. The matching are done by changing
the length of stubs. The disadvantage of this technique
is not all impedances can be matched.
x
S1
d
admittance cannot be
matched
B’
2
S
B
A= load admittance
A’=admittance of A at
stub 2
B= Adjust of stub S2 to
bring to the S2 circle
B’=Admittance of B at S1
Then by adjusting stub 1
will bring the admittance
to the center of the chart
A
A’
x
Example
Design a double-stub shunt tuner to match a load ZL=60-j80 to a 50  line.
The stubs are to be short circuited stubs, and are spaced l/8 apart. The load
consists of a series resistor and capacitor that match at 2 GHz. Plot the
reflection coefficient magnitude versus frequency from 1 GHz to 3GHz.
Solution
•Plot normalized load zL=1.2 –j 1.6  and convert to admittance we have
yL= 0.3 +j0.4.
•Construct a rotated 1+ j b circle which is a distance d=l/8 from a 1+jb
circle. Get two possible points on the rotated 1+jb circle, y1 and y1’ by
adding susceptance of the first stubs. In this case we take x=0(match
section immediate after the load . I.e b1=1.314 and b1’=-0.114.The length
of stub will be i1=0.396l and i1’=0.232l.
•Now transform both point onto 1+jb circle along SWR circles.This
bring two solution y2 =1-j3.38 and y2’=1+j1.38.
•Then the second stub should be b2= 3.38 and b2’=-1.38. The length of
stub will be i2=0.454l and i2’=0.100l.
continue
Rotated 1+jb
circle
l/8
y1
b1
y2’
yL
y1’ b1’
b2’
b2
y2
continue
Reflection coefficient
Reflection Coefficient vs frequency
1
0.8
0.6
Solution 1
0.4
Solution 2
0.2
0
1
1.5
2
2.5
3
Frequency(GHz)
The shorter the stubs the wider will be the bandwidth
Formulas for calculation
Let YL=GL+jBL, then the load admittance at distance x from the load is
YL  jYo tan  x
YL  Yo
 G L'  jBL'
Yo  jYL tan  x
x is distance between stub
and load
'
Admittance at first stub

Y1  G L'  j BL'  B1

After transforming to stub 2 , we have
Y2  Yo
GL '  j ( BL'  B1  Yo t )


Yo  j GL'  jBL'  jB1 t
Where t = tan d
continue
At this point the ral part of Y2 =Yo ,thus
GL '2 GL 'Yo
1 t2
t
2

Yo  BL ' t  B1t 2
t
2
0
Solving


4t 2 Yo  BL ' t  B1t 2 
1 t2 
GL '  Yo
1 1
2 

'
2 2
2t
Y
1

t
o




Since GL’ is real, the quantity in square root must be nonnegative, thus
0
4t Yo  BL ' t  B1t 
2
2


'
2 2
Yo 1  t
1
'
Simplified to 0  G L  Yo
1 t2
t2

Yo
sin 2  d
continue
So susceptances of stubs are
B1   BL '
and
B2 
 Yo
1  t 
Yo 
2
' 2 2
G L ' Yo  G L t
t
2
' 2 2
1  t G L ' Yo  G L t  G L ' Yo


GL ' t
To find the stub length ,
1
1  B 

tan  
l 2p
 Yo 
for open  stub
1
1  Yo 

tan  
l 2p
B
for short  stub


B either B1 or B2
continue
The two principle solution are
 1
1
tan
t
d  2p

l  1 p  tan 1 t
 2p

for t  0
 for t  0
To find the stub length,
1
1  X 


tan 
l 2p
 Zo 
for short  stub
1
1  Z o 

tan 

l 2p
 X 
for open  stub


Graphical method
Suitable for load impedance laying inside the unity circle.
• Plot ZL normalised to 50 ohm
•Find bisector and point intersect the chart axis (I.e A)
•Draw circle at center A touching the center of the chart
•Obtain R1 and calculated new characteristic impedance of the
line
ZT  R1 Z o
•Renormalised the ZL to ZT, thus we have ZL2 . Then plot on the
chart.
•Obtain x for the length of the matching section by moving
towards generator.
Graphical method
x
ZL2
ZL
R1
Unity circle
A
R2
Transmission line transformer
l
Quarter-wave
transformer
Z1
Z o  Z1Z 2
Z2
Matching at one frequency .For other frequency the impedance at the
input of matching section will be
Z in  Z o
Z 2  jZ o t
Z o  jZ 2t
At matched frequency tan d = tan (p/2)
Where t = tan d
continue
Reflection coefficient


Z in  Z1 Z o Z 2  Z1   jt Z o2  Z1Z 2


Z in  Z1 Z o Z 2  Z1   jt Z o2  Z1Z 2
Since

  2 p2   m
Zo2=Z1Z2, this reduces to

Z 2  Z1 

Z 2  Z1   j 2t Z1Z 2
 


m
1
1  4Z Z /Z  Z  sec  
2
1 2



Z 2  Z1
2 Z1Z 2
cos 
2
2
12
1
for  near p / 2
0
m p pm
2
p
continue
Fractional bandwidth is given by
 

2
Z
Z
f 2 f o  f m 
4
1 2 
m

 2  cos 1 
fo
fo
p
 1   2 Z 2  Z1 
m


Reflection coeff. vs frequency
Ref. Coeff.
1
0.8
10:01
0.6
4:01
0.4
2:01
0.2
0
0
0.5
1
1.5
2
f/fo
Reflection coefficient with different ratio of Z1:Z2
Example
Design a single section quarter-wave matching transformer to match a 10 
load to a 50  line at fo =3GHz. Determine the percentage bandwidth for
which the SWR < 1.5
For a pure resistive load we can just calculate directly
Z 0  Z1Z 2  10  50  22 .36 
and the length of transformer is l/4
Then determine reflection coefficient for SWR=1.5
m 
SWR  1 1.5  1

 0 .2
SWR  1 1.5  1
Hence fractional bandwidth is




2
Z
Z
m
f
4
4
0 .2
2 50  50 
1 2 
 2  cos 1 
 2  cos 1 
 0.29
 1  0.2 2 50  10 
fo
p
p
 1   2 Z 2  Z1 
m




or 29%
Multisection transformer
Zo 
o
Z1 

Two types of transformer
Z2 

Z3
Zn N

ZL
•Binomial multisection
•Chebyshev multisection
Reflection coefficient for multisection can be written as

     A 1  e

 j2  N
At center frequency   
N






2
A cos   
Or its magnitude
p
2
(i.e  =l/4)
 2  0
p
N
Binomial transformer
For binomial expansion the reflection coefficient can be written as
     A(1  e
 j2  N
)
N
 A
n 0
C nN e  j 2 n 
where
CnN 
N!
N  n !n!
Note CnN=CN-nN, C0N=1 and C1N=N=CN-1N
Let
n  AC nN
n 
then
Z n 1  Z n 1 Z n 1
 ln
Z n 1  Z n 2
Zn
Solving for    0 to obtain value of A
Z L  Z0
 0   2 A 
Z L  Zo
N
Hence
n 
or
A  2 N
Z L  Z0
Z L  Zo
Z  Zo N
1 Z n 1
Z
ln
 AC nN  2 ( N 1) L
C n  2 ( N 1) C nN ln L
2
Zn
Z L  Zo
Zo
continue
For bandwidth
m  2 N A cos N  m
Therefore
1/ N 
 
1  1  m 
 
 m  cos
 2  A  


The fractional bandwidth, thus
1/ N 
 
4 m
f 2 f o  f m 
4
1  1  m 
 

 2
 2  cos
 2  A  
fo
fo
p
p


Example
Design a three section binomial transformer to match 50 load to a 100 line
Solution
N=3 , ZL=50 , Z0=100 then
C03 
But
3!
1
3!0!
C13 
3!
3
2!1!
C 23 
Z L  Z0
1
Z
 N 1 ln L  0.0433
Z L  Zo 2
Zo
3!
3
1!2!
1 Z n 1
ln
 AC nN
2
Zn
Z1
 2 AC 03  2( 0.0433 )1
Z0
Z
ln 2  2 AC 13  2( 0.0433 )3
Z1
Z
ln 3  2 AC 23  2( 0.0433 )3
Z2
ln
A  2 N
Z1=91.7 
Z2=70.7

Z3=54.5 
Chebyshev transformer
Chebyshev polinomial in general
Tn ( x)  cos( n cos 1 x)
for
x 1
Tn ( x)  cosh( n cosh 1 x)
for
x 1
Useful forms of Chebyshev polinomial are
T1 (sec  m cos  )  sec  m cos 
T2 (sec  m cos  )  sec 2  m 1  cos 2   1
T3 (sec  m cos  )  sec 3  m cos 3  3 cos    3 sec  m cos 
T4 (sec  m cos  )  sec 4  m cos 4  4 cos 2  3
 4 sec 2  m cos 2  1  1
continue
For N section,Chebyshev expansion the reflection coefficient can be written as
    2e  jN o cos N  1 cos( N  2)  ....  n cos( N  2n) 
 Ae  jN TN sec  m cos  
Solving for      0 to obtain value of A
Z  Z0
 0   AT N sec  m   L
Z L  Zo
or
A
Z L  Z0
1
Z L  Z o TN sec  m 
For maximum reflection coefficient magnitude
A  m
Therefore
Z L  Z0 1
1
ZL
TN sec  m  

ln
Z L  Z o m 2m Z o
or
1
1
Z L 
1 

sec  m  cosh  cosh 
ln

Z o  
 N
 2m
Example
Design a three section Chebyshev transformer to match a 100 load to a 50 line.
Taking m=0.05.
Solution
1
1
1
Z L 
1
100  
1 
1 


sec  m  cosh  cosh
ln
ln
 
  cosh  cosh 


Z o  
50  
 2  0.05
 N
3
 2m
 1.408
   Ae  jN T3 sec  m cos    2e  jN o cos 3  1 cos  
*
T3 (sec  m cos  )  sec 3  m cos 3  3 cos    3 sec  m cos 
**
Equating * and ** and A=m=0.05
cos
2o  A sec 3  m   0.05 sec 3 1.408
o  0.0698

continue


21  3 A sec 3  m  sec  m  3  0.05 sec 3 1.408  sec(1.408)
cos
1  0.1037
For symmetrical
But
3  0  0.0698
1  Z n 1 

n  ln
2  Zn 
Start n=0
Then
and
2  1  0.1037
ln Z n1  2n  ln Z n
ln Z1  2o  ln Z o  20.0698  ln 50  4.051
Z1  57 .5 
n=1
ln Z 2  21  ln Z1  20.1037   ln 57 .5  4.259
Z 2  70 .7 
n=2
ln Z 3  22  ln Z 2  20.1037  ln 70.7  4.466
Z 3  87.0 
