Transcript Microwave Engineering - Universiti Sains Malaysia

Power divider, combiner and
coupler
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Power divider and combiner/coupler
Basic
P2= nP1
P1
P1
divider P =(1-n)P
3
1
combiner
P2
Divide into 4 output
P3=P1+P2
S-parameter for power divider/coupler
Generally
 S11
S    S 21
 S31
S12
S 22
S32
S13 
S 23 
S33 
For reciprocal and lossless network
N

S kiS *ki  1
k 1
S11  S12  S13  1
S21  S22  S23  1
S31  S32  S33  1
N
* 0
S
S
 ki kj
for i  j
k 1
Row 1x row 2
*
*
*
S11S21
 S12S22
 S13S23
0
Row 2x row 3
*
*
*
S21S31
 S22S32
 S23S33
0
Row 1x row 3
*
*
*
S11S31
 S12S32
 S13S33
0
Continue
If all ports are matched properly , then Sii= 0
For Reciprocal
network
 0
S    S12
 S13
S12
0
S 23
For lossless network, must satisfy unitary
condition
S13 
S 23 
0 
2
2
2
 S 23
2
1
S *23 S12  0
S13  S 23
2
1
* S 0
S12
13
S12  S13  1
S12
2
*
S13
S 23  0
Two of (S12, S13, S23) must be zero but it is not consistent. If S12=S13= 0, then
S23 should equal to 1 and the first equation will not equal to 1. This is invalid.
Another alternative for reciprocal network
Only two ports are matched , then for reciprocal network
 0
S   S12
 S13
S12
0
S 23
S13 
S 23 
S 33 
For lossless network, must satisfy unitary
condition
2
2
2
2
2
2
S12  S 23
The two equations show
that |S13|=|S23|
thus S13=S23=0
and |S12|=|S33|=1
These have satisfied all
*
S13
S 23  0
S12  S13  1
*
*
S12
S13  S23
S33  0
1
S13  S 23  S33
2
1
*
*
S23
S12  S33
S13  0
Reciprocal lossless network of two matched
S21 =ej 
 0 e j 0  1
 j

S   e
0
0 
 0
j 
0
e


S12 =ej 
S33 =ej 
3
2
Nonreciprocal network (apply for circulator)
For lossless network, must satisfy unitary
condition
 0
S    S 21
 S31
S12
0
S32
S13 
S 23 
0 
2
2
S12  S13  1
S *31 S32  0
2
2
*
S 21
S 23  0
2
2
*
S 12
S13  0
S 21  S 23  1
S31  S32  1
The above equations must satisfy the following either
or
S12  S23  S31  0
S21  S32  S13  1
S21  S32  S13  0
S12  S23  S31  1
Circulator (nonreciprocal network)
0 0 1 
S   1 0 0
0 1 0
0 1 0 
S   0 0 1
1 0 0
2
1
3
2
1
3
Four port network
Generally
 S11 S12
S
21 S 22

S   
S31 S32

S 41 S 42
For reciprocal and lossless network
N
 SkiS *ki
1
k 1
S11  S12  S13  S14  1
S13
S 23
S33
S 43
N
S14 
S 24 
S34 

S 44 
* 0
S
S
 ki kj
for i  j
k 1
R 1x R 2
*
*
*
*
S11S21
 S12S22
 S13S23
 S14S24
0
R1x R3
*
*
*
*
S11S31
 S12 S32
 S13S33
 S14 S34
0
S21  S 22  S 23  S24  1
R1x R4
*
*
*
*
S11S41
 S12S42
 S13S43
 S14S44
0
S31  S32  S33  S34  1
R 2x R3
*
*
*
*
S21S31
 S22S32
 S23S33
 S24S34
0
S41  S 42  S 43  S44  1
R2x R4
R3x R4
*
*
*
*
S21S41
 S22S42
 S23S43
 S24S44
0
*
*
*
*
S31S41
 S32S42
 S33S43
 S34S44
0
Matched Four port network
Say all ports are matched and symmetrical network, then
 0
S
S    12
S13

S14
S12
0
S 23
S13
S 23
0
S 24
S34
S14 
S 24 
S34 

0 
The unitarity condition become
*
*
S13S23
 S14 S24
0
*
S12  S13  S14  1
*
*
S12 S23
 S14 S34
0
S12  S23  S24  1
*
*
S12 S24
 S13S34
0
**
@
S13  S23  S34  1
S14  S24  S34  1
*
*
S12 S13
 S24 S34
0
*
*
S12 S14
 S23S34
0
*
*
S13S14
 S23S24
0
@@
#
##
To check validity
Multiply eq. * by S24* and eq. ## by S13* , and substract to obtain
*  S 2  S 2 0
S14
 13
14 


%
Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain
S 23  S12

2
2
 S34   0

$
Both equations % and $ will be satisfy if S14 = S23 = 0 . This means
that no coupling between port 1 and 4 , and between port 2 and 3 as
happening in most directional couplers.
Directional coupler
If all ports matched , symmetry and S14=S23=0 to be satisfied
 0
S
S    12
S13

 0
S12
0
0
S 24
0 
S 24 
S34 

S34 0 
S13
0
0
The equations reduce to 6 equations
** S12  S13  1
* S12  S 24  1
** S13  S34  1
* S24  S34  1
*
*
S12 S24
 S13S34
0
*
*
S12 S13
 S24 S34
0
By comparing these equations yield S13  S 24
By comparing equations * and ** yield
S12  S34
Continue
Simplified by choosing S12= S34= ; S13=e j  and S24=  e j

Where  +  = p + 2np
2 cases
1. Symmetry Coupler  =  = p/2
2. Antisymmetry Coupler  =0 , =p
Both satisfy 2 +2 =1
 0  j
 0 0
S   
j 0 0

 0 j 
0 
 0
S   
 0

0 
0
j 


0 
 0 
0   
0  

 0
Physical interpretation
|S12 | 2 = power deliver to port 2= 2 =1- 2
1
Input
2
Through
4
Isolated
3
Coupled
|S13 | 2 = coupling factor = 2
Characterization of coupler
Coupling= C= 10 log P1  20 log  dB
Directivity= D= 10 log
P3
P3

 20 log
P4
S14
Isolation = I= 10 log P1  20 log S14
P4
I = D + C dB
dB
dB
For ideal case
|S14|=0
Practical coupler
Hybrid 3 dB couplers
 =  = p/2
=  = 1 / 2
0 1 j 0 

j 
1 1 0 0

S  
2 j 0 0 1 


j 1
0
 0
Magic -T and Rat-race couplers
 =0 , =p
=  = 1 / 2
0

1 1
S  
2 1

0
0
0 0  1
0 0 1

1 1 0 
1
1
T-junction power divider
E-plane T
H-plane T
Microstrip T
T-model
Lossy line
Z1
Vo jB
Lossless line
Z2
Yin
1
1
Yin  jB 

Z1 Z 2
1
1
Yin 

Z1 Z 2
If Zo = 50,then for equally
divided power, Z1 = Z2=100
Example
• If source impedance equal to 50 ohm and the
power to be divided into 2:1 ratio. Determine Z1
and Z2
1 Vo2 1
P1 
 Pin
2 Z1 3
Z1  3Zo  150
1 Vo2 2
P2 
 Pin
2 Z2 3
3Zo
Z2 
 75
2
1 Vo2
Pin 
2 Zo
Zo  Z1 // Z2  50
Resistive divider
Zo/3
Zo/3
P1
Zo
V1
Zo
P2
V2
V
Zo/3
V3
Zo
P3
Zo
Z
 Zo
3
Z o 2Z o
Zin 

 Zo
3
3
2Zo / 3
2
V
V1  V
Zo / 3  2Zo / 3
3
V2  V3 
Zo
3
1
V  V V
Zo  Zo / 3
4
2
1 V12
Pin 
2 Zo
P2  P3 
2


1
/
2
V
1
1
2
Zo
1
 Pin
4
Wilkinson Power Divider
/4
/4
70.7
50
/2 Z o
100
70.7
Zo
50
50
2Zo
Zo
/2 Z o
Zo
For even mode
2
Z
Z ine  T
2
Therefore
ZT  2 Z o
For Zin =Zo=50 
ZT  2 50  70.7
And shunt resistor R =2 Zo = 100
Analysis (even and odd mode)
1
Z
2
+V2
4
r/2
Z
r/2
Port 2
Vg2
Port 1
2
1
4
Port 3
+V3
Vg3
For simplicity all values are
normalized to line characteristic
impedance , I.e Zo = 50 .
For even mode Vg2 = Vg3 and
for odd mode Vg2 = -Vg3. Since
the circuit is symmetrical , we
can treat separately two
bisection circuit for even and
odd modes as shown in the next
slide. By superposition of these
two modes , we can find S parameter of the circuit. The
excitation is effectively Vg2=4V
and Vg3= 0V.
Even mode
Vg2=Vg3= 2V
Port 2 1
4
Port 1
O.C
If
Z 2
Z 2
+V2e
Z
e
2 +V1
Looking at port 2
Zine= Z2/2
Therefore for matching
r/2
2V
Note:
Z 2  Z in Z out
O.C
then V2e= V since Zine=1 (the circuit acting like voltage divider)
To determine V2e , using transmission line equation V(x) = V+ (e-jx + Ge+jx) , thus
Reflection at port 1, refer to Z  2
V2e  V ( )  jV  (1  G)  V
4
V1e  V (0)  jV  (1  G)  jV
G 1
G 1
G
Then
2 2
2 2
V1e   jV 2
is
Odd mode
Vg2= - Vg3= 2V
Port 2 1
Z
o
2 +V1
+V2o
4
r/2
Port 1
At port 2, V1o =0 (short) ,
/4 transformer will be
looking as open circuit ,
o
2V thus Zin = r/2 . We choose
r =2 for matching. Hence
V2o= 1V (looking as a
voltage divider)
S-parameters
S11= 0
(matched Zin=1 at port 1)
S22 = S33 = 0 (matched at ports 2 and 3 both even and odd modes)
V1e  V1o
 j/ 2
S12 = S21 =
e
o
V2  V2
S13 = S31 =  j / 2
S23 = S32 = 0 ( short or open at bisection , I.e no
coupling)
Example
Design an equal-split Wilkinson power divider for a 50 W system
impedance at frequency fo
The quarterwave-transformer characteristic is
Z  2 Z o  70.7
R  2Z o  100
The quarterwave-transformer length is

o
4 r
Wilkinson splitter/combiner
application
Power Amplifier
matching
netw orks
/4
/4
70.7
70.7
50
100
70.7
Splitter
100
70.7
combiner
50
Unequal power Wilkinson
Divider
2
1
Z03  Zo
Z02
R2=Zo/K
Zo
R
K3
Z02  K 2Z03  Zo K (1  K 2 )
Z03
3
1 K 2
R3=Zo/K
1

R  Zo  K  
K

P3 Power at port  3
K 

P2 Power at port  2
2
Parad and Moynihan power divider
Zo4
Zo2
Zo
Zo
Zo1
1
Zo3
R
Z05
Z02  Zo K
3/ 4
Zo
3
1/ 4


Z01  Zo 
2
1 K 
K
2
Z 04  Z o K
1  K 
2 1/ 4

1 K 
Z03  Zo
2 1/ 4
K5/ 4
Zo
Z05 
K
1

R  Zo  K  
K

P3 Power at port  3
K 

P2 Power at port  2
2
Cohn power divider



1


2






VSWR at port 1
VSWR at port 2 and port 3
Isolation between port 2 and 3
Center frequency fo
Frequency range (f2/f1)
= 1.106
= 1.021
= 27.3 dB
= (f1 + f2)/2
=2
3
Couplers
2
2
Yse
 1  Ysh
Branch line coupler
Yo
Yo
Yse
E1
E2
/4
Ysh
Yo
E3
2Ysh

2
2
E 2 1  Ysh
 Yse
/4
x dB coupling
Ysh
Yo
E3
E3
 10 x
E1
E12  E 22  E32
2
or
20 
E  E 
1  2    3 
E  E 
 1  1
2
Couplers
3 dB Branch line coupler
E2  E3
2
2
Yse
 1  Ysh
2
Ysh  1
input
Zo
Zo / 2
/4
Zo
isolate
Zo
Z sh  50
Yse  1.414
Zo
/4
Zo / 2
Zo  50
Output
3dB
Z se  35.5
Zo
Zo
Output
3dB 90o out of phase
Couplers
9 dB Branch line coupler
E3
 109 20   0.355
E1
2
 E2 
1     0.355 2
 E1 
 E2 
   1  0.355 2  0.935
 E1 
 E3  0.355
  
 0.38
 E2  0.935
Let say we choose Ysh  0.8
2Ysh
2
1  Ysh
 Yse2

2  0.8
1  0.8
2
 Yse2
 0.38
1.6
Yse 
 0.36  1.962
0.38
Z0  50
Zsh  50 / 0.8  62.5
Zse  50 /1.962  25.5
Note: Practically upto 9dB coupling
Couplers
Hybrid-ring coupler
1/2
4
isolated
Ge
Output in-phase
/4
3
1
/4
Go
2
Te
/8
2 /8 2
2
OC
OC
1/2
/4
/4
1
/4
2
To
2
/8
/8
2
2
1
Input
/4
•Can be used as splitter , 1 as input and 2 and 3
2
as two output. Port is match with 50 ohm.
Output in-phase •Can be used as combiner , 2 and 3 as input
and 1 as output.Port 4 is matched with 50 ohm.
Analysis
The amplitude of scattered wave
B1 
1
1
Ge  Go
2
2
B3 
1
1
Ge  Go
2
2
1
1
B2  Te  To
2
2
1
1
B4  Te  To
2
2
Couple lines analysis
w
b
w
w
r
s
b
s
Planar
Stacked
w
d
r
s
w
r
Coupled microstrip
The coupled lines are usually assumed to operate in TEM mode.
The electrical characteristics can be determined from effective
capacitances between lines and velocity of propagation.
Equivalent circuits
Odd mode
Even mode
+V
+V
+V
E-w all
2C12 2C12
H-w all
C11
-V
C22
C11
C22
C11 and C22 are the capacitances between conductors and the ground
respectively. For symmetrical coupled line C11=C22 . C12 is the
capacitance between two strip of conductors in the absence of ground. In
even mode , there is no current flows between two strip conductors , thus
C12 is effectively open-circuited.
Even mode
Continue
The resulting capacitance Ce = C11 = C22
Therefore, the line characteristic impedance
Z oe 
L

Ce
Odd mode
The resulting capacitance Co = C11 + 2 C12 = C22 + 2 C12
Therefore, the line characteristic impedance
Z oo 
1
 Co
LCe
Ce

1
 Ce
Planar coupled stripline
Refer to Fig 7.29 in Pozar , Microwave Engineering
Stacked coupled stripline
w >> s and w >> b
C11 
 r  oW

 r  oW
b  s  / 2 b  s  / 2

4b r  oW
b2  s2
F /m
 
C12  r oW F / m
s
4b r  oW
Ce  C11 
F /m
2
2
b s
 2b
1
Co  C11  2C12  2 r  o w
  F / m
 b2  s 2 s 
  1  r  o o
1
b2  s 2
Z oe 
 Zo
 Ce
4bw  r
1
1
Z oo 
 Zo
 Co
2w  r 2b / b 2  s 2  1 / s
 
 
Coupled microstripline
Refer to Fig 7.30 in Pozar , Microwave Engineering
Design of Coupled line Couplers
Coupling
3
4
Isolated
(can be matched)
w
s
Layout
w
/4
input
output
2
1
wc
Zo
3
+V3
Schematic circuit
2V
Zo

I3
1
+V1
I1
Zoe
Zoo
I4
4
+V4
I2
2
+V2
Zo
Zo
Even and odd modes analysis
Even
I1e = I 3e
e
e
e
e
V4 = V 2
3
+V3e
+
V
_
Same
excitation
voltage
I4e = I 2e
V1 = V 3
I3e
Zo
+
V
_
(99)
Zo
I1e
Zoe
1
+V1e
I4e
4
+V4e
I2e
2
+V2e
Zo
Zo
Odd
I3o
Zo
_
V
+
+
V
_
3
+V3o
Zo
1
+V1o
I1o
Zoo
I4o
4
+V4o
I2o
2
+V2o
Zo
I1o = -I3o
I4o =- I2o
Zo
V1o = -V3o
V4o = -V2o
Reverse
excitation
voltage
(100)
Analysis
V
Z in  1
I1
V1e  V1o

I1e  I1o
(101)
From transmission line equation , we have
e Z
Z in
oe
o
Z in
 Z oo
Z o  jZoe tan
Z oe  jZo tan
Z o  jZoo tan
Z oo  jZo tan
By voltage division
V1o  V
(102)
V1e  V
(103)
where
Zo = load for transmission line
 = electrical length of the line
Zoe or Zoo = characteristic impedance of
the line
I1o 
I1e 
o
Z in
o
Z in
 Zo
(104)
e
Z in
e
Z in
 Zo
(105)
V
o Z
Z in
o
V
e Z
Z in
o
(106)
(107)
continue
Substituting eqs. (104) - (107) into eq. (101) yeilds






o Ze  Z  Ze Zo  Z
o Ze  Z 2
Z in
2
Z
in
o
in in
o Z 
in in
o
Z in 
o
e
o
e
o
Z in
 Z in
 2Z o
Z in
 Z in
 2Z o
Let
(108)
Zo  Zoo Zoe
Therefore eqs. (102) and (103) become
e Z
Z in
oe
Z oo  j Z oe tan
Z oe  j Z oo tan
(109)
o Z
Z in
oo
Z oe  j Z oo tan
Z oo  j Z oe tan
(110)
For matching we may consider the second term of eq. (108) will be zero , I.e
o Ze  Z 2  0
Zin
in
o
or
o e
Zin
Zin  Zoo Zoe  Zo 2
and (108) reduces to Zin=Zo
continue
Since Zin = Zo , then by voltage division V1 = V. The voltage at port 3, by
substitute (99), (100) , (104) and (105) is then
o
 Ze

Z in
e
o
e
o
in

V3  V3  V3  V1  V1  V 

e
o
 Zin  Z o Zin  Z o 
(111)
Substitute (109) and (110) into (111)
o
Z in
Z o  jZoo tan

o
Z in
 Z o 2Z o  j Z oe  Z oo  tan
Then (111) reduces to
V3  V
e
Z in
e
Z in
 Zo

Z o  jZoe tan
2Z o  j Z oe  Z oo  tan
j Z oe  Z oo  tan
2 Z o  j Z oe  Z oo  tan
(112)
continue
We define coupling as
and
C
1 C 2 
Z oe  Z oo
Z oe  Z oo
2Z o
Z oe  Z oo
Then V3 / V , from ( 112) will become

Z oe  Z oo 
j
tan

Z oe  Z oo 
jC tan
V3  V
V
Z  Z oo  tan
2Z o
1  C 2  j tan
 j oe
Z oe  Z oo  Z oe  Z oo 
Similarly
V4  V4e  V4o  V2e  V2o  0
V2  V2e  V2o  V
1 C 2
1  C 2 cos  j sin
V1=V
Practical couple line coupler
V3 is maximum when  = p/2 , 3p/2, ...
Thus for quarterwave length coupler  = p/2 , the eqs V2 and V3 reduce to
V1=V
V2  V
V3  V
1 C 2
1  C 2 cosp / 2  j sinp / 2
jC tanp / 2
1  C 2  j tanp / 2
V4  0
V
1 C 2
V
  jV 1  C 2
j
jC()
1  C 2  j
V
jC
1 C 2
j

 VC
Z oe  Z o
1 C
1 C
1 C
Z oo  Z o
1 C
Example
Design a 20 dB single-section coupled line coupler in stripline with a 0.158 cm
ground plane spacing , dielectric constant of 2. 56, a characteristic impedance
of 50  , and a center frequency of 3 GHz.
Coupling factor is C = 10-20/20 = 0.1
Characteristic impedance of even
and odd mode are
Z oe  50
1  0.1
 55.28
1  0.1
1  0.1
Z oo  50
 45.23
1  0.1
Then multiplied by
r
 r Zoe  88.4
 r Zoo  72.4
From fig 7.29 , we have
w/b=0.72 , s/b =0.34. These
give us
w=0.72b=0.114cm
s= 0.34b = 0.054cm
Multisection Coupled line coupler (broadband)
Coupled

V3


C2
C1
C3


....
CN-2
Isolated


V4
CN-1 C
N
V1
V2
input
Through
For single section , whence C<<1 , then
V3

V1
V2

V1
jC tan
2
1  C  j tan

jC tan
 jC sin e  j
1  j tan
1 C 2
1  C 2 cos  j sin
 e  j
V4=0
and For = p/ 2 then V3/V1= C
and V2/V1 = -j
Analysis
Result for cascading the couplers to form a multi section coupler is

 
1

V3  jC1 sin e  j V1  jC2 sin e  j V1e 2 j 
...  jC sin e  j V e 2 j ( N 1)

N
For symmetry C1=CN , C2= CN-1 ,
etc
V3  jV1 sin e  j C1 1  e 2 j ( N 1)  C2 e 2 j  e 2 j ( N 2)
...  CM e  j ( N 1) 
 


(200)
 2 jV1 sin e  jN C1 cosN  1  C2 cosN  3
1
...  CM 
2
V3
C

At center frequency
o
Where M= (N+1)/2
V1  p / 2

Example
Design a three-section 20 dB coupler with binomial response (maximally
flat), a system impedance 50  , and a center frequency of 3 GHz .
Solution
For maximally flat response for three section (N=3) coupler, we require
dn
d
n
C ( )
0
for n  1,2
(201)
 p / 2
From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have
C
V3
1


 2 sin C1 cos 2  C2 
V1
2


 C1sin3  sin   C2 sin  C1 sin3  (C2  C1) sin
(202)
Continue
Apply (201)
dC
 3C1 cos3  (C2  C1 ) cos 
0
d
p / 2
d 2C
d
2
  9C1 sin3  (C2  C1 ) sin 
 10C1  C2  0
p / 2
Midband Co= 20 dB at  =p/2. Thus C= 10-20/20=0.1
From (202), we
C= C2 - 2C1= 0.1
Solving © and © © gives us
©©
C1= C3 = 0.0125 (symmetry) and C2 = 0.125
©
continue
Using even and odd mode analysis, we have
1 C
1  0.0125
Z oe1  Z oe3  Z o
 50
 50.63
1 C
1  0.0125
Z oo1  Z oo3  Z o
1  0.0125
 49.38
1  0.0125
1 C
1  0.125
Z oe 2  Z o
 50
 56.69
1 C
1  0.125
Z oo 2  Z o
1  0.125
 44.1
1  0.125
continue
Let say , r = 10 and d =0.7878mm
Plot points on graph Fig. 7.30
Z oe1  Z oe3  50.63
Z oo1  Z oo3  49.38
We have , w/d = 1.0 and s/d = 2.5 , thus
w = d = 0.7878mm and
s = 2.5d = 1.9695mm
Similarly we plot points
Z oe2  56.69
For section 1 and 3
Z oo2  44.1
We have , w/d = 0.95 and s/d = 1.1 , thus
w = 0.95d = 0.748mm and
s =1.1d = 0.8666mm
For section 2
Couplers
1
2
Lange Coupler
Evolution of Lange
coupler
1= input
2=output
3=coupling
4=isolated
3
4
1
2
4
3
3
1
2
4
4
1
s
w
s
s
w
s
w
w
w
3
2
Analysis
Equivalent circuit
Simplified circuit
o
90
1
1
Cm
2
Cex
2
Cex
3
4
Ze4 Zo4
1
C
C
4
Ze4
where
Zo4
3
1
Cex
CexC m
Cin  Cex 
Cex  C m
Cm
2
Cm
Cin
C
2
3
Cm
Cin
4
Cex
Continue/ 4 wire coupler
Even mode
All Cm capacitance will be at same potential, thus the total capacitance is
Ce4  Cex  Cin
(300)
Odd mode
All Cm capacitance will be considered, thus the total capacitance is
Co4  Cex  Cin  6Cm
(301)
Even and Odd mode characteristic impedance
Z e4 
1
 Ce 4
Z o4 
1
 Co 4
(302)   velocityin transmission line
continue
Now consider isolated pairs. It’s equivalent circuit is same as two wire line ,
thus it’s even and odd mode capacitance is
Ce  Cex
Co  Cex  2Cm
Substitute these into (300) and (301) ,
we have
CexC m
Cin  Cex 
Cex  C m
And in terms of impedance refer
to (302)
C e 3C e  C o 
Ce 4 
Ce  Co
Z e4 
Z oo  Z oe
Z oe
3Z oo  Z oe
Co 3Co  Ce 
Co 4 
Ce  Co
Z o4 
Z oo  Z oe
Z oo
3Z oe  Z oo
continue
Characteristic impedance of the line is
Z oe Z oo Z oo  Z oe 2
Z o  Z e4 Z o4 
3Zoo  Zoe 3Zoe  Z oo 
Coupling


2 Z2
Z e4  Z o4
3 Z oe
oo
C

2
2
Z e4  Z o4 3 Z oe
 Z oo
 2Z oe Z oo


The desired characteristic impedance in terms of coupling is
Z oe 
Z oo 
4C  3  9  8C 2
2C 1  C  / 1  C 
4C  3  9  8C 2
2C 1  C  / 1  C 
Zo
Zo
VHF/UHF Hybrid power splitter
50
output
50
input
1
5
T1
6
7
3
2
T2
100
C
4
8
50
output
Guanella power divider
(VHF/UHF)
I2
Rg
I1
V2
V1
Vg
RL
V2
I1
I2