Transcript Outline - 正修科技大學

Chapter 4
Microwave Network Analysis
Equivalent Voltage and Current
 For non-TEM lines, the quantities of voltage, current, and impedance
are nor unique, and are difficult to measured. Following
considerations can provide useful result:
1) Voltage and current are defined only for a particular mode, and
are defined so that the voltage is proportional to the transverse
electric field, and the current is proportional to transverse magnetic
field.
2) The product of equivalent voltage and current equals to the power
flow of the mode.
3) The ratio of the voltage to the current for a single traveling wave
should be equal to the characteristic impedance of the line. This
impedance is usually selected as equal to the wave impedance of the
line.
The Concept of Impedance
 Various types of impedance:
1) =(/)1/2 =intrinsic impedance of medium. This impedance is
dependent only on the material parameters of medium, and is equal
to the wave impedance of plane wave.
2) Zw=Et /Ht=1/Yw =wave impedance, e.g. ZTEM, ZTM, ZTE. It may
depend on the type of line or guide, the material, and the operating
frequency.
3) Z0 =(L /C)1/2 =1/Y0 =characteristic impedance. It is the ratio of
voltage to current. The characteristic impedance is unique definition
for TEM mode but not for TM or TE modes.
 The real and imaginary parts of impedance and reflection coefficient
are even and odd in 0 respectively.
Impedance and Admittance Matrices
 The terminal plane (e.g. tN) is important in providing a phase
reference for the voltage and current phasors.
 At the nth terminal (reference) plane, the relations are given as:
V   Z I ;
Z ij 
Vij
I ij
|Vk 0 for
I   Y V ;
Yij 
I ij
Vij
|Vk 0 for
Z is impedancemat rix
k j
Y is admit tancemat rix
k j
Z   Y 1
 Reciprocal Networks
 If the arbitrary network is reciprocal ( no active devices, ferrites, or
plasmas), [Y] and [Z] are symmetric matrices .
Zij  Z ji
Yij  Y ji
Z   Z T
Y   Y T
 Lossless Networks
 If the arbitrary network is lossless, then the net real power delivered
to the network must be zero. Besides
Re{Zij }  0,
Re{Yij }  0, for any i, j
Example4.1: Find the Z parameters of the two-port network?
Solution
Z11 
V1
|I 2 0  Z A  Z C
I1
Because t henet workis reciprocal
V2
Z 22 
|I1 0  Z B  Z C
I2
By volt agedivision
V1
V2 Z C
Z12  |I1 0 
 ZC
I2
I 2 Z B  ZC
The Scattering Matrices
 The scattering parameter Sij is the transmission coefficient from j port
to port i when all other ports are terminated in matched loads.
Vi 
Sij   |V  0 for
Vj K
 [Z] or [Y]  [S]
S   [ Z ]  [U ] ;
k j
[ Z ]  [U ]
[U ] is unit or ident it ymat rix
1
0

.

0
 [S]
0
1
.
0
.

. 0

0 1
.
 [ Z]
[U ]  [ S ]
Z  
[U ]  [ S ]
Example4.2: Find the S parameters of the 3 dB attenuator
circuit?
Solution

1
 V2 0
1
V
S11 
|
V
Z  Z0
  |V  0 
|Z0 on port 2
2
Z  Z0
1
1
in
1
in
A matched 3B attenuator with a
50 Ω Characteristic impedance
8.56  [141.8(8.56  50)]
Z 
 50
141.8  8.56  50
So S11  0. From symmetry feature, S22  0.
1
in
V2
S21   |V  0
V1 2
41.44
50
V  V2  V1 (
)(
)  0.707V1
41.44  8.56 50  8.56
 S21  S12  0.707

2
 Reciprocal Networks
 Lossless Networks
 [S] is symmetric matrix
 No real power delivers to network.Besides
S   S 
t
N
*
S
S
 ki kj  1; for i  j
k 1
N
*
S
S
 ki kj  0; for i  j
k 1
Example4.3: Determine if the network is reciprocal, and lossless? If
port 2 terminated with a matched load, what is the return loss at port 1?
If port 2terminated with a short circuit, what is the return loss seen at
port 1?
Solution
 0.150 0.85  45
[S ]  

0
.
85

45

0
.
2

0



Since [S] is not symmetric, the network is not reciprocal.
2
2
S11  S21  (0.15)2  (0.85)2  0.745  1
So the network is not lossless.
When port 2 terminated with a matched load, =S11=0.15.
RL  20log   20log(0.15)  16.5 dB
When port 2 terminated with a short circuit,
V1  S11V1  S12V2  S11V1  S12V2
V2  S21V1  S22V2  S21V1  S22V2
T hesecondequation gives
V2
 V1
S21
1  S22
Dividing thefirst equation by V1 and using theabove result
V1
V2
S S
    S11  S12   S11  12 21
1  S22
V1
V1
(0.85  45)(0.8545)
 0.15 
1  0.2
 0.452
So
RL  20log   20log(0.452)  6.9 dB
Summary
 Reciprocal Networks (symmetric)
No active elements, no anisotropic material
Z   Z t
Y   Y t
S  St
 Lossless Networks
No resistive material, no radiation
Re{Z }  0
Re{Y }  0
 Imaginarypart sof Y , Z , S mat ries
S T S *  U 
 Unit ary
Example4.4: Determine if the network is reciprocal, and lossless ?
Solution
From the result of example 4.2
0.707
 0
[S ]  
0 
0.707
0.5 0  1 0
T
*
 [S ] [S ]  



0
0
.
5
0
1

 

A matched 3B attenuator with a
50 Ω Characteristic impedance
Since the network is reciprocal
but not lossless, [S] should be
symmetric but not unitary.
141.8  150.36 141.8 
8.56  141.8
Z   



141
.
8
8
.
56

141
.
8
141
.
8
150
.
36

 

Since the network is reciprocal
but not lossless, [Z] should be
symmetric but not imaginary.
Example4.5: Determine if the network is reciprocal, and lossless ?
Solution
Ig
I1
Y11  |V2 0 
|V2 0  0
V1
Vgs
Y12 
(
I g  0)
Y22 
I2
|V1 0 
V2
V2
Rds
G
Z0
Z0
I
g V 0
Y21  2 |V2 0  m 1
 gm
V1
V1
V2
Port 2
Port 1
I1
0
|V1 0 
0
V2
I 2  Rds
gm  0 
D
S
G

1
Rds
Port 1 
Vgs
D
g mVgs
Port 2
Rds

0 
0
S

[Y ]  
1
 g m
Rds 
 Since the network is neither reciprocal nor lossless,
[Y ] should be neither symmetric nor imaginary.
S
S11 
(Y0  Y11)(Y0  Y22 )  Y12Y21 (Y0  0)(Y0  1 / Rds )  0  g m Y0  1 / Rds


1
(Y0  Y11)(Y0  Y22 )  Y12Y21 (Y0  0)(Y0  1 / Rds )  0  g m Y0  1 / Rds
S12 
 2Y12Y0
 20

0
(Y0  Y11)(Y0  Y22 )  Y12Y21 (Y0  0)(Y0  1 / Rds )  0  g m
S21 
 2Y21Y0
(Y0  Y11)(Y0  Y22 )  Y12Y21

 2  g mY0
 2  gm
Z R

 2  g m 0 ds
(Y0  0)(Y0  1 / Rds )  0  g m Y0  1 / Rds
Z 0  Rds
(Y0  Y11)(Y0  Y22 )  Y12Y21
S22 
(Y0  Y11)(Y0  Y22 )  Y12Y21
(Y0  0)(Y0  1 / Rds )  0  g m Y0  1 / Rds Rds  Z 0



(Y0  0)(Y0  1 / Rds )  0  g m Y0  1 / Rds Rds  Z 0
1
0 

Rds  Z 0 
 [ S ]    2  g Z 0 Rds
m

Z 0  Rds Rds  Z 0 

 Since thenetworkis neitherreciprocalnor lossless,
[ S ] should be neithersymmetricnor unitary.
 Problem 1:Determine if the inductance networks are
reciprocal, and lossless ?
I1
L1
L2
L3
Port 1 V1
I2
V2 Port 2
 Problem 2:Determine if the capacitance networks are
reciprocal, and lossless ?
I1
Port 1 V1
I2
C3
C1
C2
V2 Port 2
Three Port Network
 S11 S12
[ S ]   S21 S22

 S31 S32
 Reciprocal Network
S12  S21
S13  S31
S23  S32
 Matching at all ports
S11  0
S22  0
S33  0
 Lossless Network
[ S ] is unitary 
S13 
S23

S33 
[ S ]T [ S ]*  [U ]
Condition A: S12  S13  1
2
Condition B: S13* S23  0
2
S12  S23  1
2
*
S23
S12  0
2
S13  S23  1
2
2
S12* S13  0
From condition B, two of the three parameters, S12 , S13 , S23 must be zero.
This can not satisfy condition A.
 A three port network can not be lossless, reciprocal,
and match at all pors simultaneously.
Applications
 A counter-clockwise circulator
0 0 1 
[S ]  1 0 0


0 1 0
 [S] is unitary and matched at all ports, but not symmetric.
Therefore, circulator is lossless and matched, but not
reciprocal.
 Power splitters
 0

[S ]   1
2
1

2
1
2
0
0

2
0 

0 

1
Port 2
Port 1
Splitters
 [S] is symmetric and matched at all ports, but not unitary.
Therefore, circulator is reciprocal and matched, but not
lossless.
Port 3
 A Shift in Reference Planes
 n   n ln
Sm'  e 2 jn Sm
 Twice the electric length
represents that the wave
travels twice over this length
upon incidence and reflection.
 Generalized Scattering
Parameters
 If the characteristic impedances
of a multi-port network are
different,
Sij 
Vi  Z 0 j
V j
Z 0i
|V  0,k  j
k
Generalized Scattering Matrices
 The scattering parameter Sij defined earlier was based on the
assumption that all ports have the same characteristic impedances
( usually Z0=50). However, there are many cases where this may
not apply and each port has a non-identical characteristic impedance.
 A generalized scattering matrix can be applied for network with nonidentical characteristic impedances, and is defined as following:
Vi 
aj 
; j  1,2,..
Z0 j
Vi 
bi 
; i  1,2,...
Z 0i
bi
Sij 
|ak 0 for
aj
Sij,nonident 
k j
Vi  Z 0 j
V j
Z 0i
a1
Z01
Port 1
a2
A Two-Port
Network
b1
|ak 0 for k  j  Sij,ident
Z02
Port 2
b2
Z0 j
Z 0i
The Transmission (ABCD) Matrix
V1   A B  V2 
 I   C D   I 
 2 
 1 
V1   A1 B1   A2
 I   C D  C
1 2
 1  1
B2  V2 
D2   I 2 
 ABCD matrix has the advantage of cascade connection of multiple
two-port networks.
 Table 4-2 Conversions between two-port network parameters
 For reciprocal network, [Z] is is symmetric. Hence, Z12=Z21

AD  BC  1
Example4.6: Find the S parameters of network?
Solution
From Table 4-1
A B
C D 



4
Port 1
Z0



cos
jZ
sin
0
0 
0
 1
 1
2
2

  jC 1



j

C
1

  jY0 sin

cos

2
2 

jZ 0   1
0  0
0
 1

  jC 1
  jY
0
j

C
1

 0


C
 CZ 0
jZ 0 


2 2


j

C
Z

jY


CZ
0
0
0

Port 2
Z0 , 
C
C
Z0


4

Z0 , 

C
From Table 4-2
S11 
A  B Z 0  CZ 0  D
A  B Z 0  CZ 0  D
 CZ 0  j  ( j  j 2C 2 Z 02 )  (CZ 0 )

 CZ 0  j  ( j  j 2C 2 Z 02 )  (CZ 0 )
j 2C 2 Z 02

 2CZ 0  2 j  j 2C 2 Z 02
S21  S12 

2
A  B Z 0  CZ 0  D
2
 2CZ 0  2 j  j 2C 2 Z 02
S22 
 A  B Z 0  CZ 0  D
A  B Z 0  CZ 0  D
j 2C 2 Z 02

 2CZ 0  2 j  j 2C 2 Z 02
The Transmission [T] Matrix
 At low frequencies, ABCD matrix is defined in terms of net voltages
and currents. When at high frequencies, T matrix defined in terms of
incident and reflected waves will become very useful to evaluate
cascade networks.
V1  T11 T12  V2 
   
  
T
T
V1   21 22  V2 
V1
Port 1
V1
V2
T  Two-Port
Port 2 Port 1
Network
V1
V2
T11
T
 21
V1  T11 T12  T11' T12'  V2 
   
 '
'   
T
T
V1   21 22  T21 T22  V2 
 1
T12   S21


T22   S11
 S21
V1
T  Two-Port
V2 V ' 
1
Network
V2' 
T  Two-Port
Network
V2
S22 

S21 

S11S22 
S12 
S21 
V1' 
Port 2
V2' 
Equivalent Circuit for Two-Port Networks
A coax-to-microstrip
transition and equivalent
circuit representations.
(a) Geometry of the transition.
(b) Representation of the
transition by a “black box.”
(c) A possible equivalent
circuit for the transition.
 Equivalent circuits for some common microstrip discontinuities.
(a) Open-ended. (b) Gap. (c) Change in width. (d) T-junction.
 Equivalent circuits for a reciprocal two-port network.
(a) T equivalent
V1   Z11 Z12   I1 
V    Z
I 
Z
22   2 
 2   21
 I1  Y11 Y12  V1 
 I   Y
 V 
Y
 2   21 22   2 
(b)  equivalent
Example4.7: Find the network as equivalent T and  model at 1GHz?

Solution
4
From Table 4-1
2  
  l 
 
 4 2


cos
 A B 
2

C D  

  j sin 
2  100

Port 2
Z1  100
Z 0  50
Z 0  50
Port 1

j100sin   0
j100
2 

  j
0 

cos
  100
2 
From Table 4-2
A
Z11   0
C
1
Z12  Z 21   100 j
C
D
Z 22   0
C
 j100
 0
Z   


j
100
0


 Z11  Z12  j100
Z 22  Z12  j100
 Z12   j100
Equivalent T model
2  1G  L  100  L  15.915nL
1
 100  C  1.5915pF
2  1G  C
From Table 4-2
 0
Y    j

 100
D
Y11   0
B
 Y11  Y12 
1 j
Y12  Y21 

100
B
Y22  Y12 
A
Y22   0
B
 Y12   j
j100
Port 1
j

100
0 

100
2  1G  C  1
 C  1.5915pF
100
1
1
 L  15.915nL
100
2  1G  L
Port 2
 j100
15.915nH 15.915nH
Port 2
Port 1
1.592pF
j
100
j
100
j100
Equivalent  model
 j 1

100
Port 1
j 1

100
Port 2
j 1

100
15.915nH
Port 1
1.592pF
1.592pF
Port 2
Example4.8: Find the equivalent  model of microstrip-line inductor?
Solution
From Table 4-1
 A B   cos l
C D   jY sin l

  0
l
jZ0 sin l 
cos l 
From Table 4-2
D
cos l
Y11  
  jY0 cot l
B jZ 0 sin l
w
r
h
1
1
Y12  Y21 

 jY0 csc l
B
jZ 0 sin l
 Y11  Y12   jY0 cot l  jY0 csc l
A
cos l
Y22  
  jY0 cot l
2 l
B jZ 0 sin l
jY0 (2 sin
)
jY0
2

(1  cos l ) 
l
l
sin l
jY0 csc l 
 jY0 cot l
2 sin cos
Y   
2
2

jY
csc

l

jY
cot

l
0
 0

l
 jY0 tan
2
Y22  Y12   jY0 cot l  jY0 csc l
 jY0 tan
Equivalent  model
l
 jY0 csc  l 1
2
 Y12   jY0 csc l
jY0 tan
Inductance of microstrip-line
l
2
 1
jY0 tan
l 0
1
j L 
 jZ 0 sin  l  jZ 0  l
 jY0 csc  l
L
Z0 l


Z 0l 0 0 eff


Z 0l  eff
c
jC  jY0 tan
 jY0
l
2
 1

;(H)
L
Port 1
Parasitic capacitance
 l l 0
Port 2
Port 1
l
2
2
l
l
Y0
Y0  0 0 eff Y l 
0
eff
2  2
C

;(F)


2c
C
Port 2
C
Example4.9: Find the equivalent T model of microstrip-line capacitor?
Solution
From Table 4-1
 A B   cos l
C D   jY sin l

  0
l
jZ0 sin l 
cos l 
w
From Table 4-2
A
cos l
Z11  
  jZ 0 cot l
C jY0 sin l
r
h
1
1

  jZ 0 csc l
C jY0 sin l
 Z11  Z12   jZ 0 cot l  jZ 0 csc l
D
cos l
Z 22  
  jZ 0 cot l
2 l
jZ 0 (2 sin
)
C jY0 sin l
jZ 0
2

(1  cos l ) 
l
l
sin l
  jZ0 cot l  jZ0 csc l 
2 sin cos
Z   
2
2


jZ
csc

l

jZ
cot

l
0
0


l
 jZ 0 tan
2
Z12  Z 21 
Z 22  Z12   jZ 0 cot l  jZ 0 csc l
 jZ 0 tan
Equivalent T model
l
2
Z12   jZ 0 csc l
jZ 0 tan
Port 1
Capacitance of microstrip- line
l
2

jZ 0 tan
2

Port 2
 jZ 0 csc  l
1
  jZ 0 csc l
jC
l 0 0 eff
l  eff
sin l l 0 l
C



; (F)
Z 0
Z 0
Z 0
Z 0c
l

L
P arasiticinductance
Port 1
C
l
jL  jZ 0 tan
2
l
Z 0 tan l 0
l Z 0l 0 0 eff Z 0l  eff
2
L
 Z0


; (H)

2
2
2c
L
Port 2
 Problem3: Design a 6GHz attenuator ?
(Hint: -20logS21=6  S21=0.501
0.501
 0
S   
 )
0
.
501
0


 Problem4: Design a 6nH microstrip-line inductor on a
1.6mm thick FR4 substrate. The width of line is 0.25mm.
l
Find the length (l ) and parasitic capacitance?
Z0
6nH
Z0
w=0.25mm
r=4.5
h=1.6mm
 Problem5: Design a 2pF microstrip-line capacitor on a
1.6mm thick FR4 substrate. The width of line is 5mm. Find
the length (l ) and parasitic inductance?
l
Z0
w=5mm
2pF Z 0
r=4.5
h=1.6mm