Power Dividers and
Directional Couplers
Divider or
coupler
Divider or
coupler
0
S    S12
 S13
Power combining
Power division
S12
0
S 23
S13 
S 23 

0 
,
SS *T
 Identity matrix 
S12  S13  1 , S12  S 23  1 , S13  S 23  1
2
S13* S 23  0
2
2
2
2
2
*
, S 23
S12  0 , S12* S13  0
 At least two of the three parameters (S12 , S13 , S 23 ) must be zero.
A three port cannot be lossless, reciprocal , and matched at all ports.
1
Four-Port Network (Directional Couplers)
Assume all ports are matched
 0 S12 S13 S14 
S

0
S
S
23
24 
S    12
 S13 S 23 0 S 34 


S
S
S
0
24
34
 14

*
*
*
Lossless  S13
S 23  S14* S 24  0 , S14
S13  S 24
S 23  0
*
14
S
S
2
13
 S 24
2
 0
*
*
*
S12
S 23  S14* S 34  0 , S14
S12  S 34
S 23  0 

S 23 S12  S 34
2
2
 0
S14  S 23 which results in a directiona l coupler
2
S12  S13  1 , S12  S24  1
2
2
2
2
S13  S34  1 , S24  S34  1
2
2
2
2
S13  S24   , S12  S34  
S12  S34   , S13  e j , S24  e j
*
*
S12
S13  S24
S34  0        2n
2   2 1

0

S   
 j

0
0

S   


0
j
0
0
0
0
j



0
0
0
0
 
0
j 
 (The symmetrica l Coupler)


0
0

 (The Antisymmet rical Coupler


0
3
Input
Isolated
1
4
2
3
Through or
direct
Coupled
Directional Coupler
P1
Coupling  C  10log  20 log  dB
P3
Directivit y  D  10log
P3

 20 log
dB
P4
S14
P1
Isolation  I  10log  20 log S14 dB
P4
I  D  C dB
4
Hybrid couplers are special cases of Direction al
coupler, where the coupling factor is 3 dB.
1
 
. The quadrature hybrid has a 900 phase shift between
2
ports 2 and 3 when fed at port 1.
 0 1 j 0


1 1 0 0 j 
S  
2  j 0 0 1


0
j
1
0


The magic T hybrid or rat - race has a 180o phase differnce between
ports 2 and 3 when fed at port 4.
0 1

1 1 0
S  
2 1 0

0  1
1 0
0  1

0 1

1 0
5
Directional Couplers
4
3 Pf
Pb
Pi
Pt
1
2
The coupling C is :
Pi
C  10 log
Pf
The directivit y D is :
Pf
D  10 log
Pb
6
For ideal coupler S14  S 23  0
S11  S 22  0
0
S
S    12
 S13

0
S12
S13
0
0
0
S 33
S 24
S 34
S S *t  U
*
S13 S 33
0
0 
S 24 
S 34 

S 44 
*
S 24 S 44
0
,
S 33  0  S 44
0
S
S    12
 S13

0
S12
0
S13
0
0
0
S 24
S 34
0 
S 24 
S 34 

0 
7
*
*
S12 S 24
 S13S 34
0
,
*
S12 S13*  S 24 S 34
0
S12 S 24  S13 S 34
S12 S13  S 24 S 34
S13  S 24
S12  S 34
S12  S13  1
2
2
S12  S 24  1
2
S12  C1
 0
C
S    1
 jC2

 0
2
, S13  jC2
C1
jC2
0
0
jC2
0
0
C1
0 
jC2 
C1 

0 
8
Directional Couplers
Two-hole Waveguide Couplers
b4 Port 4 Pb
Port 3
0
Kf a1
-2L
Kb
a1
L
Port 1
a1 0
Pf
Kf a1
-L
Kf a1
-L
b3
b2
a1 -L
Pi
Port 2
Pt
Kf and Kr
are the forward
and reverse aperture coupling coeffiecients
9
The coupling C is :
C  -20 log 2 K f
The directivit y D is :
D  20 log
 20 log
2Kf
Kr 1 e
Kf
Kr
 2 jL
 20 log
Kf
K r cos L
 20 log sec L
The directivit y is the sum of the directivit y of the single
aperture plus a directivit y associated with the array.
10
Multi-element Couplers
To achieve good directivity over a band of frequencies, couplers
With many apertures may be used.
Coupled
Isolated
Bb
A
Input
B0 A
F0
A
B1A
F1 A
n=1
n=0
d
Bf
Through
Let the aperture coupling in the forward direction be Cn
And the reverse coupling in the reverse direction be Dn
At the position of the Nth aperture the total forward wave in the
upper wave guide is :
F  Ae
 jNd
N
F
n 0
n
11
At the plane of the first aperture the total backward wave in the
upper wave guide is :
N
B  A Bn e  j 2 nd
n 0
The coupling and directivit y are :
F
C  2  log
 -20log
A
N
F
n 0
N
B
D  20 log
 20 log
F
 j 2 nd
B
e
 n
n 0
N
 Fn
N
 C  20 log  Bn e  j 2 nd
n 0
n 0
12
n
Assume that the apertures are round holes with identical position relative
to the edge of the guide, with rn being the radius of the nth aperture.
Fn  K f rn3 ,
B n  K b rn3 , K f , K b are constant for the forward and
backward coupling coefficien ts that are the same for all aperture.
N
D  C  20 log K b  20 log  rn3e  j 2 nd  C  20 log K b  20 log S dB
n 0
N
C  20 log K f  20 log  rn3
n 0
S
N
3  j 2 nd
r
: Array factor ( same as the expression for the
 ne
n 0
reflection coefficien t from N - section quarter wa ve transform er.
A maximally flat passband characteri stic can be achieved by choosing
rn3  KCnN , K is constant,
C nN is binomial coefficien ts.
13
Chebyshev Response
In order to obtain an equal-ripple characteristics in the pass band
The array factor F is made proportional to Chebyshev polynomial.
Choose r0  rN , r1  rN 1 ,... then
3  2 jn
n
S r e
M
  2rn cos( N  2n ) d
n 0
M  ( N  1) / 2 for N odd
M  ( N ) / 2 for N even
Choose S 
M
 2r
n 0
n
cos( N  2n )  K TN (sec mcos  )
  d and sec  m is the value at the upper and lower edges of
the passband. Choose K such that it gives the desired coupling C
in the center of the band where   /2.
14
D  C - 20log K b  20 log S  0
TN (sec m)
Kf
 -20log
 20 log
dB
Kb
TN (sec mcos  )
The smallest v alue of D will occur when TN (sec mcos  )  1,
since TN (sec m)  TN (sec mcos  ,
Kf
Kb
is very small , the minimum value of D min in the passband ,
contribute d by the array factor occurs when
TN (sec  m cos  )  1.
Dmin  20 log TN (sec  m )
dB
Given D min , then  m can be determined (bandwidth )
15
Branch-Line Coupler
Z0
Even-mode Odd-mode
pair
pair
a1/2
a1/2
Z0
b2
Port 1
Port 2
b1
Z01
a1/2
-a1/2
Z02
Port 4
l
Z01
l
Port 3
Z0
Z02
Z0
b3
b4
a1
b1  (e  o )
,
2
a1
b3  (Te  To )
,
2
A  B C  D

,
A  B C  D
a1
b2  (Te  To )
2
a1
b4  (e  o )
2
2
T
A  B C  D
16
Open circuits
Even-mode excitation
A

C
Short circuits
Odd-mode excitation
jZ 01   1
0  0
 j

j

1 

0 
  Z 01
  Z 02
Z 01



j
Z
01 

Z 02


 1
Z 01 
1 



 jZ 01  Z 2  Z 2   Z 
02 
02 
 01

B   1j
  
D  Z
02

0
1  is for even mode

,   /4
17
If we choose the Impedance Z02  Z 01 / 2
1
Then e  0 , Te 
(1  j ) for even mode
2
1
o  0 , To 
(1  j ) for odd mode
2
b1  0 port 1 is matched
j
b2  
(half power, - 90 o phase shift from port 1 to 2)
2
1
b3  
(half power, - 180 o phase shift from port 1 to 3)
2
b 4  0 (no power to port 4)
18
Parallel-Coupled Lines Directional Coupler
Input
Port 1
L
Isolated Port 4
Coupled Port 3
a1/2
Input
Port 1
Direct Port 2
L
Direct Port 2
H wall
a1/2
Coupled Port 3
Isolated Port 4
Even Mode excitation
19
Z0
I3
3
I1
4
Z0e
Z0o
I2
2
1
Z0
Z0
I4
Z0
V
V1 V1e  V1o
Z in 
 e
I1
I 1  I 1o
Z 0  jZ 0 e tan 
Z Z
Z 0 e  jZ0 tan 
e
in
e
0
,
Z 0  jZ0o tan 
Z Z
Z 0o  jZ 0 tan 
o
in
o
0
o
e
Z
Z
V1o  V o in
, V1e  V e in
Z in  Z o
Z in  Z o
V
I  o
Z in  Z o
o
1
V
, I  e
Z in  Z o
e
1
20
Z ino ( Z ine  Z o )  Z ine ( Z ino  Z o )
Z in 
Z ine  Z ino  2 Z o
Let Z ein Z ino  Z oe Z 0o  Z o2 , then Z in  Z o
j ( Z 0 e  Z 0o ) tan 
V3  V
2 Z 0  j ( Z 0 e  Z 0o ) tan 
jC tan 
V3  V
1  C 2  j tan 
V2  V  V  V
e
2
o
2
,
,
Z 0 e  Z 0o
Defin e C 
Z 0 e  Z 0o
V4  V4e  V4o  V2e  V2o  0
1 C2
1  C 2 cos   j sin 
V3
V2
For    2 Then
C ,
  j 1 C2
V
V
1 C
1 C
Z 0e  Z o
, Z 0o  Z o
1 C
1 C
21
Hybrid Junctions Magic T
0
S
S    12
 S12

0
S12
0
0
S 24
0 1
1 0
S   
1 0

0  1
S12
0
0
 S 24
0 
S 24 

 S 24 

0 
1 0
0  1

0 1

1 0
22
Microstrip Hybrid Ring
Even
Even
1/2
Odd
1/2
Vb
+
2
a
3
Odd
+
Vb1/2
/4
-1/2
or -Vb+
/4
/4
Va+ 1
Zc
4 V+
a
or -Va+
Z1
3/4
a
The hybrid ring (rat-race)
Ports 1 and 3 are uncoupled, ports 2 and 4 are uncoupled
23
1
Va+
Yc jB1
Y1
Vb+
jB2 Yc
Equivalent circuit for
one half of Hybrid ring
For even excitation :
V1   S11oc S12oc  Va 
     oc
oc    
V2   S 21 S 22  Vb 
For odd excitation :
V1   S11sc
     sc
V2   S 21
S12sc 
sc 
S 22 
Va 
 
Vb 
,
,
V4   S11oc
     0c
V3   S 21
V4   S11sc
     sc
V3   S 21
S12oc 
oc 
S 22 
S12sc 
sc 
S 22 
Va 
 
Vb 
 Va 
 
 Vb 
24
Use superposit ion V1  2Va , V2  2Vb
V3  V4  0
V1  1  S11oc  S11sc
     oc
sc
2
V
S

S
21
 2
 21
S12oc  S12sc 
oc
sc 
S 22  S 22 
V1 
 
V2 
V4  1  S11oc  S11sc S12oc  S12sc  V1 
     oc
sc
oc
sc    
2
 S 21  S 21 S 22  S 22  V2 
V3 
1 oc
1
1 oc
1
sc
sc
S11  ( S11  S11 )  ( e  o ) , S12  ( S12  S12 )  (Te  To )
2
2
2
2
1 oc
1 oc
sc
sc
S 21  ( S12  S12 ) , S 22  ( S 22  S 22
)
2
2
1 oc
1
1 oc
1
sc
sc
S 31  ( S12  S12 )  ( e  o ) , S32  ( S 22  S 22 )  (Te  To )
2
2
2
2
1 oc
1 oc
sc
S 41  ( S11  S11 ) , S 42  ( S12  S12sc )
2
2
25
S 44  S11 , S33  S 22 , S34  S 21
Port 2
1/2
2
Te
/4
/8
1
2
2 e
Port 2
1/2
Port 1 To
/8
3/8
2
O.C
Even Mode
/4
2
1
2 o
2
3/8
2
S.C
Odd Mode
O.C
S.C
 1 j 2
 1
A B
A B

 , 
C D   

1 

e  j 2
C D  o  j 2
j
j
j
j
e 
, Te 
, o 
, To 
2
2
2
2
S11  o i.e. port 1 is matched.
S 21 
j
2
,
S31 
j
2
Port 1
j 2

1 
, S31  0
26
0
1
S    j 
20

 1
1 0  1
0 1 0 
1 0 1

0 1 0
27
Power Dividers
P2
Z2
P1
Z1
Z3
A lossless three port juntion
P3
1
1
 2
 2
P1  Y2 V1  Y3 V1
2
2
P2  P1 , P3  (1   ) P1
Y1  Y2  Y3 for impedance matching
Y2


Y3 1  
.
28
This type of losselss power divider wi ll not have matched
outputs ports .
S 23  0, no islolation between th e output ports.
It is desired to have S 23  0 , the reflected power at port 2 does
not couple into port 3.
If we connect resistor between port 2 and port 3, the reflected power
from the output port is absorbed by by the resistor. When the output
ports are terminate d in correct load impedance, there is no current in the
resistor.
Z /4
2
Zc
R
Z3
ZL2
ZL3
29
P3  K 2 P2 and maintain zero current in R
when port 2 and 3 are terminate d in the matched load Z L2 and Z L 3
Then V2  V3 . In order to obtain the power ratio we require :
2
K V
 2
2
Z L2

 2
3
V
Z L3
 K 2 Z L3  Z L 2
For matched load at port 1 Yin  Yc
Z L 2 Z L3
Yin  2  2  Yc
Z2
Z3
( K 2 Z 32  Z 22 ) Z L 3
Z 22 Z 32

Zc
At port 1 :
Yin,2
Z L2
 2 ,
Z2
Yin,3
Z L3
 2 ,
Z3
30
1 2
2 1
 2
V1 Yin,3  K
V1 Yin, 2 
2
2
Z2  K 2Z3
At any frequency Yin,2  Y2
Yin,3
YL 2  jY2 tan 
Y2  jYL 2 tan 
YL 3  jY3 tan 
 Y3
 K 2Yin, 2
Y3  jYL 3 tan 
VL 2 I L 2 Z L 2
I L2 Z L2

 2
1
VL3 I L3 Z L3 K I L 2 Z L3
There is no current in R at any frequency as long
as port 2 and 3 are teminated in their matched load
impedances .
31
Yin,2  Yin,3  (1  K 2 )Yin, 2
Yc  Yin
Yc2  (1  K 2 ) 2 Y22
in 
 2
Yc  Yin Y c(1  K 2 ) 2 Y22  2 j (1  K 2 )Yc Y2 tan 
Choose Yc2  (1  K 2 ) 2 Y22  in  0
In order to analyze the coupling between 2 and 3, terminate 1
in a load impedance Yc
Port 2 and 3 are uncoupled if G  Y23
32
 I 2  Y22 Y23  V2 
 I   Y
 V 
Y
 3   23 33   3 
I3
Y23 
V2 V 0
3
G
I3
V3
-Y23
-Y33+Y23 -Y22+Y23
z3
I2
V2
I3
Yc
sc
Y3 ,
Equivalent circuit with port
2 and 3 excited
z2
V2
Y2 ,
Equivalent circuit between
Port 2&3 with port 3 sc and
R removed
33
2Y2Y3
Y23 
Yc (1  cos 2 )  j (Y2  Y3 ) sin 2
Y2Y3
When   /2 Y23 
Yc
Z2Z3
R
,
Zc
Z L2
K 2R
 2
K 1
Z 2  K RZ c
,
Z L3
1
, Z3 
K
RZ c
R
 2
K 1
34
Passive Microwave Devices
Attenuators
R1
Zc
Vg
R1
R2
Zc
R2
T section
R2 ( R1  Z c )
Rin  R1 
R1  R2  Z c
R1
R1
 section
For Rin  Z c  R1 ( R1  2 R2 )  Z c2
VTH

R2
 
 R1  R2  Z c

V g

35
The power delivered to the load is :
2

R2
1 VTH
PL 
Z c  
2 2Z c
 R1  R2  Z c

R2

K 
 R1  R2  Z c
1 K
R1 
Zc
1 K
For 3 - dB attenuator
2



2
2
2
Vg
 Vg
2

K
8Z c
 8Z c
2
Power attenuatio n, R in  Z c 
2K
,
R2 
Zc
2
1 K
R 1  8.58, and R 2  141.4.
PIN diode can be used in parallel with R 2 in the  - section configurat ion
to switch it in and out of the circuit
36
OC
Phase shifters
  ( 2   1 ) / v p  ( 2   1 )  / 
/4
Bias Current Input
/4
input
output
l1
/4
Bias
Circuit
Ground
/4
Length l2
Bias
Circuit
Ground
/4
Bias Current Input
/4
oc
Incremental-line-type phase shifter
37
Bl=
V+1
V+2
V+3
V+4
V-1
V-2
V-3
V-4
/4
Bias
Current
Input
jB
jB
/4
OC
A phase shifter using switched reactive elements
38
Transmission matrix of a normalized shunt suceptance jB:
_
_


B 
1  j B
j
2 
A1    _2
_

  j B 1 j B

2
2 
Transmission matrix of a section of transmission line of
Electrical length :
e j
A2   
0
0 
 j 
e 
39
Relationship among wave amplitudes:
V3 
V1 
V2 
V4 
    A1     A1 A2     A1 A2 A3   
V1 
V2 
V4 
V3 
 A11 A12  V4 

  
A
A
22  V4 
 21
Choose V-4 = 0, then V+4 = V+1 /A11 Thus T14 = 1/ A11
1
2
_
_



2
B  j B  j 

T14  1  j  e 
e 
2
4




If tan  2/B then : T14  e j  e  j (  ) ,
T14  1
  (   )      2
For B  2,   /4 ,   /2
For B  1  1.107,   53.14 0
40
Choose the two stubs such that jB2   jB1



2
B  j / 2 B  j  / 2 

T14  1  j  e

e

2
4




  j (1  jB ) 1
_
When B  1,
2
T14
_
2
1

B 
  j 1  jB 

2


2
1
 (1  B 2 ) 1  1
Phase of T14   / 2  tan 1 B   / 2  B
The change in phase when B  B1 and B  -B1is 2B1
A value of B1 as large as 0.2    .4  22.92 0  large mismatch.
This phase shifter is limited o small phase shifts between states in order
to keep the input VSWR small
41
/4
P
d1
P’
d
A phase shifter using open circuited stubs spaced /4
Apart. P and P’ are switched into the circuit when the
Diodes are off and on, respectively.
42
b1  0 port 1 is matched , b2  0 port 2 is decoupled
e  o  0  A  B  C  D , A  D for both even and odd modes,
then B  C or : Z 02 
1
T
AB
b4   jZ 01a1
Z 01
1  Z 01
2
, Te   1  Z 01  jZ 01 and T0  1  Z 01  jZ 01
2
and
2
b3   1  Z 01 a1
 1 

Coupling  10 log 
2 
 1  Z 01 
2
, since P3  b32 and P1  a12
43