Power Dividers and

Directional Couplers
Divider or
coupler

Divider or
coupler

0
S S12
S13
2

Power combining

Power division

S12
0
S 23

S13
S 23

,
2

S S *T

Identity matrix

S12 S13 1 , S12 S 23 1 , S13 S 23 1

S13* S 23 0

*
, 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

13

S 24

*
*
*
S12
S 23 S14* S 34 0 , S14
S12 S 34
S 23 0

S 23 S12 S 34

S14 S 23 which results in a directional coupler

S12 S13 1 , S12 S24 1

2

S13 S34 1 , S24 S34 1

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

0
0
j

j
0
0

0
0 0

0 0

0
j

(The symmetrical Coupler)

(The Antisymmetrical Coupler

3

Input
Isolated

Through or
direct
Coupled

Directional Coupler

P1
Coupling C 10log 20 log dB
P3
Directivity D 10log

P3

20 log
dB
P4
S14

P1
Isolation I 10log 20 log S14 dB
P4
I D C dB

Hybrid couplers are special cases of Directional

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
21 0

0 1

1 0
0 1

0 1

1 0

Directional Couplers
4
3 Pf

Pb
Pi

Pt
1

2
The coupling C is :
Pi
C 10 log
Pf
The directivity D is :
Pf
D 10 log
Pb

For ideal coupler S14 S 23 0

S11 S 22 0
0
S
S 12
S13

S12
0
0
S 24

S13
0
S33
S34

S S *t U
*
S13 S33
0

0
S 24
S34

S 44

*
S 24 S 44
0

S33 0 S 44
0
S
S 12
S13

S12
0
0
S 24

S13
0
0
S34

0
S 24
S34

*
*
S12 S 24
S13 S34
0

*
S12 S13* S 24 S34
0

S12 S 24 S13 S34

S12 S13 S 24 S34
S13 S 24
S12 S34
2

S12 S13 1
S12 S 24 1
S12 C1

0
C
1

jC2

, S13 jC2
C1

jC2

0
jC2

0
C1

0
jC2
C1

Directional Couplers
Two-hole Waveguide Couplers
Port 4 P
b

b4
Kf a1

Kb a1

Port 3

0
-2L

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 directivity D is :
D 20 log
20 log

2Kf
Kr 1 e

Kf
Kr

2 jL

20 log

Kf
K r cos L

20 log sec L

The directivity is the sum of the directivity of the single

aperture plus a directivity associated with the array.
10

Multi-element Couplers
To achieve good directivity over a band of frequencies, couplers
With many apertures may be used.
Isolated

Bb
A

Input

Coupled
B0 A

F0

B1 A

F1 A

n=1

n=0

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 waveguide is :
F Ae

jNd

F
n 0

11

At the plane of the first aperture the total backward wave in the
upper waveguide is :
N

B A Bn e j 2 nd
n 0

The coupling and directivity are :

F
C 2 log -20log
A

F
n 0

B
D 20 log 20 log
F

j 2 nd
B
e
n
n 0

Fn

C 20 log Bn e j 2 nd
n 0

n 0

12

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 ,

Bn 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

C 20 log K f 20 log rn3

n 0

3 j 2 nd
r
: Array factor ( same as the expression for the
ne
n 0

reflection coefficien t from N - section quarter wave transformer.

A maximally flat passband characteri stic can be achieved by choosing
rn3 KCnN , K is constant, CnN 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

2rn cos( N 2n ) d
n 0

M ( N 1) / 2 for N odd
M ( N ) / 2 for N even
Choose S

2r
n 0

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 value 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 ,

contributed 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
Even-mode Odd-mode
pair
pair
a1/2
a1/2

Port 1

b1

a1/2
-a1/2

Z0

Port 4

Z01

Z0
b4

a1
b1 (e o )
,
2
a1
b3 (Te To )
,
2
A B C D

,
A B C D

Z02

Z0

b2

Port 2

Z01

l
Port 3

Z02

Z0

b3

a1
b2 (Te To )
2
a1
b4 (e o )
2
2
T
A B C D
16

Short circuits

Open circuits

Even-mode excitation

j

D
Z
02

0 0
j

1
Z 01

Odd-mode excitation

Z 01
Z 02

jZ 01 1

j

0
Z 02

is for even mode

jZ 01
1

jZ 01 Z 2 Z 2
02
01

Z 01

Z 02

, /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, - 90o 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

Isolated Port 4

Coupled Port 3
a1/2
Input
Port 1

Direct Port 2

Direct Port 2
H wall

a1/2
Coupled Port 3

Isolated Port 4
Even Mode excitation
19

Z0

I3
3

Z0e

I1

Z0o

I2

Z0

Z0

I4

Z0

V1 V1e V1o
Z in
e
I1
I 1 I 1o
Z 0 jZ 0 e tan
Z Z
Z 0 e jZ 0 tan
e
in

e
0

Z 0 jZ 0o 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
Define 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

S12
0
0
S 24

0 1
1 0
S
1 0

0 1

S12
0
0
S 24

0
S 24

S 24

1 0
0 1

0 1

1 0

22

Microstrip Hybrid Ring

Even
1/2

Odd
1/2

Vb+

Even

Odd

+
Vb1/2

-1/2

or -Vb+

Va+ 1
Zc

4V+
a
or -Va+

Z1

a
The hybrid ring (rat-race)
Ports 1 and 3 are uncoupled, ports 2 and 4 are uncoupled
23

1
Va+

Yc jB1

Y1

jB2 Yc

Vb+

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 Va

sc
S 22 Vb

V4

S11oc
0c
V3 S 21
V4

S11sc
sc
V3 S 21

S12oc Va

oc
S 22 Vb
S12sc Va

sc
S 22 Vb

24

Use superposition V1 2Va , V2 2Vb

V3 V4 0
oc
sc
V1
1 S11 S11
oc
sc
2
V
S

S
21
2
21

S12oc S12sc V1
oc
sc

S 22 S 22 V2

oc
sc
oc
sc

V4
1 S11 S11 S12 S12 V1
oc
sc
oc
sc

2
V3
S 21 S 21 S 22 S 22 V2
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
S 21 ( S12 S12 ) , S 22 ( S 22 S 22sc )
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

Port 2
1/2
Port 1 To

S.C
Odd Mode

O.C

2
3

S.C

1 j 2
1
A B
A B
,

C D

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

O.C
Even Mode

j 2

, S31 0
26

 0
1
j
S
2 0

1 0 1
0 1 0
1 0 1

0 1 0

27

Power Dividers
P2

Z2
P1

Z1
Z3

P3

A lossless three port juntion

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 will not have matched

outputs ports .
S 23 0, no islolation between the 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 terminated in correct load impedance, there is no current in the
resistor.
Z
Zc

R
Z3

ZL2

ZL3
29

P3 K 2 P2 and maintain zero current in R

when port 2 and 3 are terminated 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

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

V3 0

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

Vg

8Z c

Vg

8Z c

R2

Power attenuation, R in Z c
R1 R2 Z c
1 K
2K
R1
Zc
,
R2
Zc
2
1 K
1 K
For 3 - dB attenuator R 1 8.58, and R 2 141.4.
K
2

PIN diode can be used in parallel with R 2 in the - section configuration

to switch it in and out of the circuit

36

OC

Phase shifters

( 2 1 ) / v p ( 2 1 ) /

Bias Current Input

input

output
l1

Bias
Circuit
Ground

Length l2

Bias
Circuit
Ground

Bias Current Input

oc
Incremental-line-type phase shifter
37

l=
V+1

V+2

V+3

V+4

V-1

V-2

V-3

V-4

Bias
Current
Input

jB

jB

OC

A phase shifter using switched reactive elements

38

Transmission matrix of a normalized shunt suceptance jB:

B
1 j B
j
2
2
A1
_
_

j B 1 j B

2
2
Transmission matrix of a section of transmission line of
Electrical length :

e j
A2
0

0
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 tanthen : 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

B
B j / 2
j / 2
1 j e

2
4

T14

j (1 jB ) 1
When B 1,

T14

B
j 1 jB

(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

d1

A phase shifter using open circuited stubs spaced

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
AB
b4 jZ 01a1

Z 01
1 Z 01

, Te 1 Z 01 jZ 01 and T0 1 Z 01 jZ 01
and

Coupling 10 log
2
1 Z 01

b3 1 Z 01 a1
, since P3 b32 and P1 a12

43