Characterization Parameters of Microwave Circuit
Impedances and Admittances :
Consider an arbitrary N-port microwave network
+
V2 , I2
+
V1 , I1
- V2 , I 2
+
+
V1-, I1
+
+
VN , IN




Vn  Vn  Vn
In  In  In
VN- , IN
nN
V n , I n are total voltage and current at port n .
The impedance matrix[z] of the microwave network relates these
voltages and currents is as follows ,
 V1   Z11
V   Z
 2   21
  
 
  
  
  
VN   Z N 1
or
[V]=[Z][I]
Z12
Z 22



ZN2
  
Z 1N   I 1 
Z 2N  I 2 
 

   
 

   

   
 
   Z NN   I N 
Z ij 
Vi
Ij
I k  0 fork j
Z ij is obtained by driving port j with the current I j opening circuit
all other ports and measuring the open – circuit voltage at port i.
Reciprocity principle reveals that
Z ij  Z ji
The admittance matrix [Y] is
 I 1   Y11 Y12
 I  Y
Y22
 2   21

  
 

  
  

  
 I N  Y N 1 Y N 2
   Y1N   V1 
Y 2 N   V2 
 

   
 

   
    
 
   Y NN  VN 
or
[I]=[Y][V] where [y]= [ z ]
1
Property of matrix elements for lossless network
Pave 
1
1
1
Re{[V ]t [ I ]*}  Re{([ I ][ Z ]) t [ I ]*}  Re{[ I ]t [ Z ]t [ I ]*}
2
2
2
1
 ( I 1 Z 11 I 1 *  I 1 Z 12 I 2 *  I 2 Z 21 I 1 * ...)
2

1 N N
 I m Z mn I n *
2 n 1 m 1
Since the net power delivered to the network is zero , we have
Re {Pave }  0
This gives us
Re{ I n Z nm I n }  I n Re{ Z nm }  0
2
*
Re {Z nm }  0
(1)
＠ indicates that :
Re {( I n I m  I m I n )Z mn }  0
*
*
(In Im  Im In )
*
*
Z mn  Z nm
is a real quantity
We therefore have
Re {Z mn }  0
(2)
(1) and (2) show that the elements of impedance matrix are purely
imaginary .
Scattering Matrix
The scattering matrix relates the voltage waves incident on
the ports to those reflected from the ports. We assume that Vn  and
Vn

are the amplitudes of incident and reflected waves at port n.
V1   S11
  
V2   S 21
  
 
  
  
  
VN   S N 1
S12
S 22



SN 2
  
S1N  V1 
 
S 2 N  V2 


   
 

   
    
 
   S NN  VN 
or
[ V  ]=[S][ V  ]
A specific element of the [S] matrix can be determined as

V
Sij  i 
Vj
Vk  0 fork j
S ij is found by driving port j with an incident wave of


voltage V j and measuring the reflected wave amplitude Vi
coming out of port i , while terminating all other ports in
matched loads.
Transformation among [S] , [Z] , and [Y] matrices
We assume that the characteristic impedances Z on of all
the ports are identical . For convenience , we set Z on  1

Vn  Vn  Vn




I n  I n  I n = Vn  Vn

[Z][I]= [Z] [ V  ]-[Z] [ V  ]=[V]= [ V  ]+[ V  ]
([Z]+[U]) [ V  ]=([Z]-[U]) [ V  ]
[U] is the unit matrix
[s]=
[Z ] 
[V  ] [ Z ]  [U ]

[V  ] [ Z ]  [U ]
[U ]  [ S ]
[U ]  [ S ]
[Y ]  [ Z ]1 
[U ]  [ S ]
[U ]  [ S ]
Some properties of [S] matrix
t
(1) For a reciprocal network , [s]= [ s]
1

Vn  (Vn  I n )
2
[ V  ]= 1 ([Z]+[U])[I]
2
1

Vn  (Vn  I n )
2
[ V  ]= 1 ([Z]-[U])[I]
2

[ V  ]= ([Z]-[U]) ([Z]  [U]) -1 [ V ]

[S]= [V  ] =([Z]-[U]) ([Z]  [U]) -1
[V ]
[S] t = {([Z]  [U]) -1}t ([Z]  [U]) t
t
If the network is reciprocal , [Z] is symmetrical and [Z] =[Z]
[S] t = ([Z]  [U]) ([Z]-[U])
-1
Since both ([Z]  [U]) -1 and ([Z]-[U]) are symmetrical , we then get
[S]= [S] t
(2)For a lossless network , [S] is a unitary matrix
The power delivered to the network can be expressed as
Pave 
1
1
Re{[V ]t [ I ]*}  Re{([V  ]t  [V  ]t )([V  ]*  [V  ]* )}
2
2

1
Re{([V  ]t [V  ]*  [V  ]t [V  ]*  [V  ]t [V  ]*  [V  ]t [V  ]* )}
2
1
1
 [V  ]t [V  ]*  [V  ]t [V  ]*
2
2
If the network is lossless , no real power can be delivered to the
network . We then get
[V  ]t [V  ]*  [V  ]t [V  ]*
Note that we have
[V  ]  [ S ][V  ]
This gives us
[V  ]t [V  ]*  [V  ]t [ S ]t [ S ]* [V  ]*
We therefore get
[ S ]t [ S ]*  [U ]
[S ]* ={ {[ S ]t }1 }
Therefore [S] is a unitary matrix and
N
S
k 1
ki
S kj *   ij
for all i=j
Example :
For the delta shown below
 0.100
S
0
0.890
0.8900 

0.200 
determine whether the network is reciprocal or lossless. If a short
circuit
is placed on port 2 . What will be the return loss at port 1.
Solution:
Since [S] is symmetrical , the network is reciprocal .
S11  S 21  0.12  0.82  0.65  1
2
2
The network is not lossless .
V1  S11V1  S12V2  S11V1  S12V2
V2  S 21V1  S 22V2  S 21V1  S 22V2
V2 

S 21
V1
1  S 22
V1
V2
S S

S

S
 S11  12 21
11
12


V1
V1
1  S 22
= 0.1 
( j 0.8)( j 0.8)
 0.633
1  0.2
The return lossless is
R.L.=  20 log   3.97dB
Example :
Find the S parameters of the following 3dB attenuation.
8.56 ohm
8.56 ohm
141.8 ohm
Port 1
Port 2
Solution:
S11 is defined as

V
S11  1
V1
 1 V  0 
2
V2  0
Z in,1  Z 0
Z in,1  Z 0
Where Z in,1 is
Z in,1  8.56 
141.86(8.56  50)
 50
141.86  (8.56  50)
S11 =0= S 22
V2  V2  V1
141.86 //( 8.56  50)
50
8.56  [141.86 //( 8.56  50)] 50  8.56
= 0.707V1  0.707V1

V
S 21  2
V1
 0.707  S12

V2 0
We may use the following circuit to obtain the transmission coefficient
S 21
Zo
+
2V1
8.56
8.56
141.8
V2
Zo
Transmission ABCE matrix
This representation is especially approximate for the configuration of
the Cascade connection of two or more two – port networks.
The ABCD matrix is defined as
V1  AV2  BI 2
I1  CV2  DI 2
I2
I1
A B
C D 


+
V1
Port 1
-
+
V2
Port 2
-
Note that the direction of I 2 is out of the network.
V1   A1
 I   C
 1  1
B1  V2 
D1   I 2 
Consider a cascade connection shown below
I3
I2
I1
A B
C D 


+
V1
-
+
V2
-
The ABCD representation becomes
V1   A1
 I   C
 1  1
B1  V2   A1

D1   I 2  C1
 A B   A1
C D  C

  1
B1   A2
D1  C 2
B1   A2
D1  C2
B2 
D2 
B2  V3 
D2   I 3 
A B
C D 


+
V3
-
Example:
ABCED parameters of a series impedance Z
I2
I1
Z
+
V1
-
+
V2
-
V1  AV2  BI 2
I1  CV2  DI 2
A
C
V1
V2
I1
V2
1
I 2 0
0
I 2 0
B
V1
I2
D
I1
I2
Z
V2 0
1
V2  0
 A B  1 Z 
C D   0 1 

 

ABCD parameters of some fundamental circuit configurations
Example :
Transformation of impedance matrix into ABCD transmission matrix
The impedance matrix parameters are given as
V1  Z11I1  Z12 I 2
V2  Z 21I1  Z 22 I 2
V1  AV2  BI 2
I1  CV2  DI 2
We have that
A
B
V1
V2
V1
I2
 Z 11
C

I 1 Z 11 Z 11

I 1 Z 21 Z 21

I 1 Z11  Z 12 I 2
I2
I 2 0
V2 0
 Z11
V2 0
I1
I2
 Z 12
V2 0
Z 22
Z Z  Z 12 Z 21
 Z 12  11 22
Z 21
Z 21
I1
V2

I 2 0
I
D 1
I2
I1
1

I 1 Z 21 Z 21
I 2 Z 22

V2  0
I2
Z 21

Z 22
Z 21
AD-BC= Z11 Z 22  ( Z11Z 22  Z12 Z 21
Z 21 Z 21

Z 21
1
)
Z 21
Z 12
Z 21
If the network is reciprocal Z12  Z 21 , we have
AD-BC=1
Transformations among Z,Y,S and ABCD parameters
Network Analyzer Calibration : using thru – reflect – line
S parameters are ratios of complex voltage amplitudes. The primary
reference plane is generally at some point within the analyzer itself.
We consider the following schematic.
a1
a2
Device
under
test(DUT)
Error box
[S]
b1
A1 B1
C1 D1
Mesurement plane
for port 1
A’B’
C’ D’
Reference plane
for device port 2
Reference plane
for device port 1
Error box
[S]
A2 B2
C2 D2
b2
Mesurement plane
for port 2
A calibration procedure is used to characterize the error boxes
before the measurement of DUT.
Consider the following error box
a1
a2
Error box
[S]
b1
A1 B1
C1 D1
L
in
L 
ZL
b2
Z L  Z0
Z L  Z0
b1  S11a1  S12a2  S11a1  S12b2 L
b2  S21a1  S22a2  S21a1  S22b2 L
b2 (1  S22 L )  S21a1
b2 
S 21a1
(1  S 22 L )
b1  S11a1  S12
in 
S 21a1L
(1  S 22 L )
b1
S S 
 S11  12 21 L
a1
1  S 22 L
Assume that the error box is reciprocal ,i.e.,
in  S11 
Z12  Z 21
S122 L
1  S 22 L
To calibrate the error box means to find scattering parameters S ij of
the error box. We consider the following three situations
(1) the load is short – circuited , L  1
2
in,s
S
 S11  12
1  S 22
… (a)
(2) the load is open– circuited , L  1
2
in,o  S11 
S12
1  S 22
…(b)
(3) the load is matched , L  0
in  S11
…(c)
By solving (a) (b) and (c) , we obtain S11 , S12 and S 22 .We
then get the scattering parameters S ij of the error box.
The work with a cascade of three two – port networks , we
convert the S parameters of error boxes to the corresponding ABCD
parameters . As a result , the overall measurement transmission matrix
is
 Am
C
 m
Bm   A1

Dm  C1
B1   A B    A2
D1  C  D  C 2
B2 
D2 
1
The ABCD transmission matrix of DUT is
 A B    A1
C  D   C

  1
1
B1   Am
D1  C m
Bm   A2
Dm  C 2
B2 
D2 
Finally , we convert the ABCD transmission matrix of DUT into
the S parameters of DUT.
Two – port power gain Calculation by using Scattering Parameters
Consider the following two – port network
Zs I1
+
+V1
V1
V1
-
Vs
I2
[S]
+
V2
(Z0)
V2
Zl
>
>
>
s
+
V2
>-
in
l
out
We want to find the following three power gains
Power Gain=G= Pl P the ratio of power dissipated in the load to
in
the power delivered to the input of two – port network.(independent of
Zs )
Available Gain G A  Pavn P the ratio of maximum power available to the
avs
two - port
network to the maximum power available from the source .(dependent
on
Zs
, but not
Zl
)
Transducer power Gain= GT

Pl
Pavs
the ratio of power delivered to the
load to the power available from the source.(dependent on both Z s and
Zl )
The reflection coefficient looking from the network toward the load is
l 
Z l  Z 0 V2

Z l  Z 0 V2
The reflection coefficient looking from the network toward the source
is
s 
Z s  Z 0 V1

Z s  Z 0 V1
We next want to find in and out
V1
in  
V1
, out
V2
 
V2
V1  S11V1  S12V2  S11V1  S12 lV2
V2  S 21V1  S 22V2  S 21V1  S 22 lV2
V2 (1  S 22 l )  S 21V1
V1  S11V1 
in 
S12 S 21l 
V1
1  S 22 l
S S  Z  Z0
V1
 S11  12 21 l  in

V1
1  S 22 l Z in  Z 0
By the same token , we obtain
out  S 22 
S12 S 21s
1  S11s
The voltage
V1  Vs
V1
is
Z in
 V1  V1  V1 (1  in )
Z s  Z in
Z in  Z 0
1  in
1  in
1  in
1  in
Vs
 V1 (1  in )
1  in
Zs  Z0
1  in
Z0
V1  Vs
Z0
Z 0 (1  in )  Z s (1  in )
We assume that
V1  Vs
Z 0 =1
1
1
 Vs
1  in  Z s  Z s in
(1  Z s )  in ( Z s  1)
V 1  s
1
 s
Z s  1 2 1  s in
1  in
Zs 1
1
1 Zs
 Vs
Z 1 1 Zs 1 Zs 1
1
1
1
(1  s )  (1  s ) 

2
2
Zs 1 2
Zs 1
Zs 1
where
The average power delivered to the network is
 2
V
1  s
1 V1
2
2
Pin 
(1  in )  s
(1  in )
2
2 Z0
8 1  s in
2
2
The power delivered to the load is
Pl 
Pl 
V2
2
2
2Z 0
V1
V2 
(1  l )
2
S 21 (1  l )
2
2
1  S 22 l
2Z 0
2

Vs
2
8
S 21V1
(1  S 22 l )
S 21 (1  l )(1  s )
2
2
2
1  S 22 l 1  s in
2
2
The power gain G is
S 21 (1  l )
2
G=
Pl
Pin

2
1  S 22 l (1  in )
2
2
*
The maximum power available to the network occurs when in = s
We then get Pavs  Pin 
in  s *

VS
2
1  s
2
8Z 0 (1  s 2 )
Similarly , the maximum power available to the load occurs when
l = out* . Therefore , we have
Pavs  Pin
l  out*
2
VS

S 21 (1  out )(1  s )
2
2
8Z 0 1  S 22 out * 2 1  s in 2
1  11s (1  out )
2
1  s in
2
l  out*

2
2
1  S22out *
2
We get
Pavn 
S 21 1  s
2
VS
2
2
8Z 0 1  S11s 2 (1  out 2 )
The available power gain G A is
S 21 (1  s )
2
GA 
Pavn
Pavs

2
1  S11s (1  out )
2
2
Transducer power Gain GT is
S 21 (1  s )(1  l )
2
GT 
Pl
Pavs

2
2
1  S 22 l 1  s in
2
2
The matched (or maximum) transducer power gain occurs when
s  l  0
GT ,max  S 21
2
The unilater power gain is
S 21 (1  s )(1  l )
2
GT ,u  GT
in  S11, S12 0

2
1  S11S 1  S 22 l
2
2
2
Chain Scattering Matrix
a1  T11 T12   b2 
 b   T
 
 1   21 T22  a2 
chain scattering parameters
scattering transfer parameters
T parameters
*Relationship between the S and T parameters
T11
T
 21
and
 1
T12   S 21

T22   S11
 S 21
 S11
S
 21
S 22 
S 21 

S11S 22 
S12 
S 21 

 T21
S12   T11

S 22   1
 T11
T21T12 
T11 

T12 

T11 
T22 