Transcript Notes 20 - Power dividers and couplers part 2

ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 20
Power Dividers and Couplers
Part 2
1
Directional Couplers
Four-Port Networks
Consider a reciprocal 4-port network with matched ports:
0
S
12

S

 
 S13

 S14
S12
S13
0
S23
S23
0
S24
S34
S14 
S24 

S34 

0
There are 6 degrees of freedom
(independent S parameters).
If lossless  [S] is unitary.
The device is called a “directional coupler” if:
S14  S23  0
2
Directional Couplers
Directional coupler:
0
S
 S    12
 S13

0
S12
S13
0
0
0
0
S24
S34
0
S24 

S34 

0
There are now 4
degrees of freedom.
 All ports are matched.
 No power flows to port 4 from in the input port 1 (i.e., port 4 is “isolated”).
 Ports 2 and 3 are isolated (power incident on port 2 does not get to port 3 and vice versa).
3
Directional Couplers
Directional coupler symbol:
Input
Through
1
2
4
3
Isolated
Coupled
 All ports are matched.
 No power flows to port 4 from in the input port 1 (i.e., port 4 is “isolated”).
 Ports 2 and 3 are isolated (power incident on port 2 does not get to port 3 and vice versa).
4
Directional Couplers
0
S
 S    12
 S13

0
S12
S13
0
0
0
0
S24
S34
0
S24 

S34 

0
From the unitary property of the matrix, we have the following results:
S12  S13  1
2
2
(Dot column 1 with its conjugate)
*
S12 S13*  S24 S34
0
(Dot column 2 with the conjugate of column 3)
S13  S24
(Dot column 1 dotted with its conjugate, subtract from column 2 dotted with its conjugate)
S12  S34
(Dot column 2 dotted with its conjugate, subtract from column 3 dotted with its conjugate,
and use the with previous result above)
5
Directional Couplers (cont.)
Choose
S12  S34  
S13   e j , S24   e j
The first equation is really just a choice of reference planes on ports 2 and 3,
which makes these parameters real.
S12  S13  1   2   2  1
2
2
*
S12 S13*  S24 S34
 0   e  j   e j  0

  e j 1  e
j   
       2 n
0
(usually, n = 0)
6
Directional Couplers (cont.)
Two possible choices:
1) Symmetrical coupler
0

 S   
 j

0

j
0
0
0
0
j

(     / 2)
0
j 



0
Example: 90o quadrature hybrid coupler
2) Anti-symmetrical coupler   0;    
0

 S   


0


0
0
-
0
0

0
- 



0
Example: 180o rat-race hybrid coupler
7
Directional Couplers (cont.)
Assume a signal into port 1
 P1- 
dB
RL  -10log     -20log S11
 P1 
Return loss
P1
P1
P2
Input
Ideally + dB
Through
1
2
4
3
Isolated
Directivity
D
dB
 P3 
S

 10log     20log 31  20log
S41
S41
 P4 
Ideally + dB
Isolation
I
dB
 P1 
1
 10log     20log
S41
 P4 
Ideally + dB
P4
Coupling
C
 P1 
1
 10log     20log
 -20log 
P
S
31
 3 
P3
Assume ports 2-4 are matched.
A “hybrid” coupler is one for
which the coupling is 3 dB
(equal power split at the
output ports 2 and 3):
  S31 
dB
Coupled
1
;
2
C dB  3 dB
Note : I dB  DdB  C dB
8
90o (Quadrature) Hybrid Coupler
“A quadrature coupler is one in which the input is split into two signals (usually
with a goal of equal magnitudes) that are 90 degrees apart in phase. Types of
quadrature couplers include branchline couplers (also known as quadrature
hybrid couplers), Lange couplers and overlay couplers.”
Taken from “Microwaves 101”
http://www.microwaves101.com/encyclopedia/Quadrature_couplers.cfm
This is very useful for obtaining circular polarization:
There is a 90o phase difference between ports 2 and 3.
9
Quadrature Hybrid Coupler (cont.)
2
1
90o hybrid
4
3
0 j 1 0


-1  j 0 0 1 
S  
2 1 0 0 j 


0
1
j
0


 The quadrature hybrid is a lossless 4-port (the S matrix is unitary ).
 All four ports are matched.
 The device is reciprocal (the S matrix is symmetric.)
 Port 4 is isolated from port 1 and ports 2 and 3 are isolated from each other.
10
Quadrature Hybrid Coupler (cont.)
The quadrature hybrid is usually used as a splitter:
2
1
90o hybrid
4
+90o out of phase -90o out of phase
3
The signal from port 1 splits evenly between ports 2 and 3, with a 90o phase
difference.
S21  jS31
Can be used to produce right-handed circular polarization.
The signal from port 4 splits evenly between ports 2 and 3, with a -90o phase
difference.
S24   jS34
Can be used to produce left-handed circular polarization.
11
Quadrature Hybrid Coupler (cont.)
(Branch-line coupler)
A microstrip realization is shown here.
Z0
1
g
Z0
4
Z0
Z0
4
2
g 4
Z0
2
2
Z0
plane of symmetry
(POS)
Z0
3
0 j 1 0


-1  j 0 0 1 
S  
2 1 0 0 j 


0
1
j
0


Z0
Note: We only need to study what
happens when we excite port 1, since
the structure is symmetric.
We use even/odd mode analysis
(exciting ports 1 and 4) to figure out
what happens when we excite port 1.
12
Quadrature Hybrid Coupler (cont.)
Even Analysis
V
V1e
Z0
Z0
Z0
2
V2e
V1e Z0
jY0
Z0
2
g 4
V2e
Z0
jY0
Z0
l s  g 8
OC
OC
OC
OC
V
V  Z0
OC
g 4
Z0
V4e
Z0
2
Y0  1/ Z0
V3e  V2 e
V3e
Z0
Ys  jY0 tan   sls 
 jY0 tan  / 4 
 jY0
V4 e  V1e
13
Quadrature Hybrid Coupler (cont.)
Odd Analysis
V
V1o
Z0
Z0
Z0
2
V  Z0
V2o
- jY0
Z0
2
g 4
V2o
Z0
- jY0
Z0
ls  g 8
SC
SC
SC
SC
V 
V1o Z0
OC
g 4
Z0
V4o
Z0
2
Y0  1/ Z0
V3o
Z0
Ys   jY0 cot   sls 
  jY0 cot  / 4 
V  -V
o
3
o
2
  jY0
V4o  -V1o
14
Quadrature Hybrid Coupler (cont.)
Z0
Z0
Consider the
general case:
.
Y
Z0
2
g 4
 + for even 
Y   jY0 

 - for odd 
Y
2-port
In general:
 ABCD Y
 1 0


Y 1
Shunt load on line
 ABCD 
4

 0

j 2
 Z
 0
jZ 0 
2


0 


cos   

 ABCDline   
 j / Z line sin  
0 

Quarter-wave line

jZ 0line sin    


D  cos    
Here : Z 0line  Z 0 / 2
  /2

 ABCD   ABCDY  ABCD  ABCDY
4
15
Quadrature Hybrid Coupler (cont.)
Hence we have

0

 1 0 
ABCD


 

Y 1   j 2
 Z
 0
jZ 0 
2  1

 Y
0 

 jZ 0Y

 1 0  2

 j 2
Y
1


 Z
 0
jZ 0 
2


0 

jZ 0Y


2


 jZ 0Y 2 j 2


Z0
 2
jZ 0 
2 

jZ 0Y 

2 
0
1
j  + for even 
Y 
Z 0  - for odd 
16
Quadrature Hybrid Coupler (cont.)
Continuing with the algebra, we have

 j 
jZ


0 
 Z0 
1 
 ABCD   
2
2


j
j2
 jZ 

 0  Z 0  Z 0
j  j

1 

 1  j2
2  j   
  Z0  Z0

 1
1 

 1 

j
2  
  Z 0 




 j 
jZ 0   
 Z 0  
jZ 0


j   j 


jZ 0
jZ 0 

1

17
Quadrature Hybrid Coupler (cont.)
Hence we have
 1
1 
ABCD


0e
 1 

2 j 
  Z 0 
jZ 0 

1

Convert this to S parameters (use Table 4.2 in Pozar):
 S 0e

 0

 1- j
 2
1- j 
2 

0 

18
Quadrature Hybrid Coupler (cont.)
Z0
Z0
V1
2
Z0
V2
V1
g
V1V4
V2 -
-
4
V4
Z0
V1S11  
V1
a2  a3  a4  0
Hence
S11  0
Z0

Z0
g 4
Z0
2
V3-
V3
Z0
V1- e  V1-o V1- e  V1-o
1  V1- e V1-o 
S11 

    



V V
2V
2V
V 
1 e

S11  S11o   0  0

2
By symmetry:
S11  S22  S33  S44  0
19
Quadrature Hybrid Coupler (cont.)
Z0
Z0
V1
2
Z0
V2
V1
g
V1V4
-
Z0
V2 S21  
V1
V2 -
Z0

a2  a3  a4  0
g 4
4
V4
Z0
Z0
2
V3-
V3
Z0
V2 - e  V2 -o V2 - e  V2 -o 1 e
o
S21  

  S21  S21



V V
2V
2
1  -1- j   1- j  
 

2  2   2  
-j

2
By symmetry and reciprocity:
-j
S21  S12  S43  S34 
2
20
Quadrature Hybrid Coupler (cont.)
Z0
Z0
V1
2
Z0
V2
V1
g
V1V4
-
Z0
V3S31  
V1
V2 -
Z0
a2  a3  a4  0
g 4
4
V4

Z0
Z0
2
V3-
V3
Z0
V3- e  V3-o V3- e  V3-o V2 - e  V2 -o 1 e
o
S31  


  S21 - S21




V V
2V
2V
2
1  -1- j   1- j  
 
2  2   2  
-1

2
By symmetry and reciprocity:
-1
S31  S13  S24  S42 
2
21
Quadrature Hybrid Coupler (cont.)
Z0
Z0
V1
2
Z0
V2
V1
V1V4 -
Z0
V4 S41  
V1

a2  a3  a4  0
V2 -
g
Z0
4
V4
Z0
g 4
Z0
2
V3-
V3
Z0
V4- e  V4-o V4- e  V4-o V1- e  V1-o
1 e
S41 



S11 - S11o   0





V V
2V
2V
2
By symmetry and reciprocity:
S41  S14  S23  S32  0
22
Quadrature Hybrid Coupler (cont.)
Summary
Z0
1
g
Z0
4
Z0
Z0
4
2
Z0
g 4
Z0
0 j 1 0


-1  j 0 0 1 
S  
2 1 0 0 j 


0
1
j
0


2
Z0
2
3
Z0
The input power to port 1
divides evenly between ports 2
and 3, with ports 2 and 3 being
90o out of phase.
23
180o Hybrid Coupler (Rat-Race Ring Hybrid)
“Applications of rat-race couplers are numerous, and include
mixers and phase shifters. The rat-race gets its name from its
circular shape, shown below.”
Taken from “Microwaves 101”
http://www.microwaves101.com/encyclopedia/ratrace_couplers.cfm
Figure 7.42 of Pozar
Photograph of a
microstrip ring hybrid.
Courtesy of M. D. Abouzahra, MIT
Lincoln Laboratory
24
Rat-Race Ring Hybrid (cont.)
2
1
180o hybrid
4
0 1

- j 1 0
S  
2 1 0

0 -1
3
1 0
0 -1

0 1

1 0
 The rat race is a lossless 4-port (the S matrix is unitary).
 All four ports are matched.
 The device is reciprocal (the S matrix is symmetric).
 Port 4 is isolated from port 1 and ports 2 and 3 are isolated from each other.
25
Rat-Race Ring Hybrid (cont.)
The rat race can be used as a splitter:

2
1
In phase
180o hybrid

4
180o out of phase
3
The signal from the “sum port”  (port 1) splits evenly between ports 2 and 3,
in phase.
S21  S31
 The signal from the “difference port”  (port 4) splits evenly between ports 1
and 2, 180o out of phase.
S24  S34
26
Rat-Race Ring Hybrid (cont.)
The rat race can be used as a combiner:
V1  V2  S12

Signal 1 (V1)
180o hybrid

V1 V2  S42
2
1
4
3
Signal 2 (V2)
The signal from the sum port  (port 1) is the sum of the input signals 1
and 2.
S12  S13
 The signal from the difference port  (port 4) is the difference of the
input signals 1 and 2.
S42  S43
27
Rat-Race Ring Hybrid (cont.)
A microstrip realization is shown here.
2
g
1
4
Z0
Z0
2Z 0
g 4
Z0
3
0 1

- j 1 0
S  
2 1 0

0 -1
1 0
0 -1

0 1

1 0
3g / 4
g / 4
Z0
4
28
180o Hybrid Coupler (Magic T)
A waveguide realization of a 180o hybrid coupler is shown here, called a “Magic T.”
“Magic T”
Please see p. 361 of Pozar.
IEEE Microwave Theory and Techniques Society
0 1

- j 1 0
S  
2 1 0

0 -1
1 0
0 -1

0 1

1 0
Note the logo!
29
Rat-Race Ring Hybrid (cont.)
2
g
1
4
1
Z0
Z0
2Z 0
V
Z0
2Z 0
g 4
Z0
V3
3g / 4
g / 4
Z0
4
Layout
V2
2Z 0
2Z 0
3
2
g / 4
V1
Plane of
symmmetry
Z0
g
3g
4
4
g / 4
V4
4
3
Z0
2Z 0
Z0
Schematic
30
Rat-Race Ring Hybrid (cont.)
Even Analysis
1
Z0
Z0
2Z 0
V
V1 e
OC

V
V3 e
2Z 0
Z0
g
3g
8
8
2
V2 e
Z0
jY0
2

jY0
2
2Z 0
OC
OC
OC
2Z 0
3
Y0  1 / Z 0
Y0 s  Y0 / 2
Ys1  jY0 s1 tan   s ls1 


 j Y0 / 2 tan  / 4 
V4 e
 jY0 / 2
4
Ys 2  jY0 s 2 tan   sls 2 


 j Y0 / 2 tan  3 / 4 
  jY0 / 2
31
Rat-Race Ring Hybrid (cont.)
Odd Analysis
1
2Z 0
Z0
V
g
V1-o
8
Z0
2
V2-o
3g

jY0
2
Z0
jY0
2
8
SC
SC
SC
SC
Ys1   jY0 s1 cot   s ls 


  j Y0 / 2 cot  / 4 
-V 
3
2Z 0
Z0
-0
4
V
V3-o
4
Z0
Z0
Y0  1/ Z0
  jY0 / 2
Ys1   jY0 s 2 cot   sls 


  j Y0 / 2 cot  3 / 4 
 jY0 / 2
32
Rat-Race Ring Hybrid (cont.)
Proceeding as for the 90o coupler, we have:
 1
 ABCD 0e    jY0
 2
0  0

1
1  j
  2 Z 0
j 2Z0   1

jY0
0 
 
2
 1
   jY0

 2
0  1

1
1  j
  2 Z 0
j 2Z0 

0 

 1

 2
j Z
0

j 2Z0 


1 

0

1

33
Rat-Race Ring Hybrid (cont.)
Converting from the ABCD matrix to the S matrix, we have
 1

 ABCD 0e   2
j Z
 0
j 2Z0 


1 

Table 4.2 in Pozar
 S 0e
- j  1

1
2
1
1
34
Rat-Race Ring Hybrid (cont.)
For the S parameters coming from port 1 excitation, we then have:
V1S11  
V1
a2  a3  a4  0
S11  S33  0
V1- e  V1-o
1 e
o
S11 

S

S


11
11
2V 
2
1 -j
j 
 


2 2
2
0
(symmetry)
V2 S21  
V1
a2  a3  a4  0
S21  S12  S34  S43  - j
2
(symmetry and reciprocity )
V2 - e  V2 -o
1 e
o
S21 

S

S

21
21 

2V
2
1 -j
-j 
 


2 2
2
-j

2
35
Rat-Race Ring Hybrid (cont.)
V3S31  
V1
a2  a3  a4  0
S31  S13  (symmetry )
j
2
V3- e  V3-o
1 e
o
S31 

S
S

11
11 
2V 
2
1 -j
j 
 

2 2
2
-j

2
Similarly, exciting port 2, and using symmetry and reciprocity, we have the
following results:
S22  S44  0
S23  S32  S14  S41  0
S24  S42  j
2
36
Rat-Race Ring Hybrid (cont.)
Summary
2
g
0 1

- j 1 0
S  
2 1 0

0 -1
1 0
0 -1

0 1

1 0
1
4
Z0
Z0
2Z 0
g 4
Z0
3
3g / 4
g / 4
Z0
4
37