Attenuator
Let R1 and R2 be the
normalized resistances
R1
R1
R2
A

C
B  1 R1 
  0 1 
D 

The input reflection

R1
R12 
1
2 R1 
 1 R1  1 R




R
R
1
2
2 
1




1  0 1   1
R1 

1
 R2

 R2
R2 
coefficien t must be  0 or Z in  1,
1  R12
A  B  C  D , then R 2 
2 R1
The insertion loss :
2
LT  10 log C  D , 
1  R1 
LT  20 log 

1

R
1

Find the values of R1 , R2 . The terminati on is 50 ,
Multiply each resistor by 50.
Directional Couplers
Multihole Waveguide Couplers
b2
Kr a1
Kr a1
Port 3
Port 2
-2bL
Kf a1
-bL
0
Kf a1
-bL
L
Port 1
a1 0
Kf and Kr
b3
b4
a1 -bL
Port 4
are the forward
and reverse aperture coupling coeffiecients
Parallel-Coupled Lines Directional Coupler
Input
Port 1
L
Isolated Port 2
Coupled Port 3
a1/2
Input
Port 1
Direct Port 4
L
Direct Port 4
H wall
a1/2
Coupled Port 3
Isolated Port 2
Even Mode excitation
L
Input
a /2
Port 11
Direct Port 4
Coupled Port 3
-a1/2
Odd-mode excitation
The ABCD of a
Lossless Transmission Line
l
Z0k,b
E wall
Isolated Port 2
 cos bl
 j sin bl

 Z 0k
jZ0k sin bl 

cos bl 

Use the even and odd impedance , from the ABCD we can
Obtain the reflection and transmission coefficients for
even and odd cases.
For a perfect input match b1=0, which requires that:
e  0  0  Z oe Z 0o  1 or Z 0 e Z 0o  Z 02
b2  0 (i.e infinite directivit y)
ja1 ( Z 0 e  Z 0o ) sin b
a1
b3  ( e  o ) 
2 2 cos b  j (( Z 0 e  Z 0o ) sin b
a1
2a1
b4  (Te  To ) 
2 2 cos b  j (( Z 0 e  Z 0o ) sin b
The coupling k c  b3 / a1 is maximum when    / 4. At this
Z 0 e  Z 0o
1
frequency k c 
, Kc  2
Z 0 e  Z 0o
kc
 Z 0 e  Z 0o 

Maximum Coupling is 10 log 
 Z 0 e  Z 0o 
Z 0e  Z 0
1  kc
and
1  kc
Z 0o  Z 0
1  kc
1  kc
2
Branch-Line Coupler
Z0
Even-mode Odd-mode
pair
pair
a1/2
a1/2
Z0
b3
Port 2
Port 3
b2
Z01
-a1/2
a1/2
Z02
Port 1
l
Z01
l
Port 4
Z0
Z02
Z0
b4
b1
a1
b1  ( e  o )
,
2
a1
b3  (Te  To )
,
2
A  B C  D

,
A  B C  D
a1
b2  ( e  o )
2
a1
b4  (Te  To )
2
2
T
A  B C  D
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



jZ 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
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