chapter 4 microwave network analysis. equivalent voltage and current for non-tem lines, the...
TRANSCRIPT
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:
1
for 0
for 0
|
matrix admittance is ;
|
matrix impedance is ;
YZ
V
IY
YVYI
I
VZ
ZIZV
jkVij
ijij
jkVij
ijij
k
k
Reciprocal Networks If the arbitrary network is reciprocal ( no active devices,
ferrites, or plasmas), [Y] and [Z] are symmetric matrices . Tjiij
Tjiij
YYYY
ZZ ZZ
Lossless Networks
If the arbitrary network is lossless, then the net real power delivered to the network must be zero. Besides
jiY
Z
ij
ij
,any for ,0}Re{
,0}Re{
Example4.1: Find the Z parameters of the two-port network?Solution
CCB
CI
CBI
CAI
ZZZ
Z
I
V
I
VZ
ZZI
VZ
ZZI
VZ
2
20
2
112
02
222
01
111
1
1
2
|
division By voltage
|
reciprocal isnetwork theBecause
|
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.
jkVj
iij
KV
VS
for 0|
[Z] or [Y] [S]
10.0
0..
.10
0.01
matrixidentity or unit is ][
;][][
][][
U
UZ
UZS
[S] [Z]
][][
][][
SU
SUZ
Example4.2: Find the S parameters of the 3 dB attenuator circuit?
SolutionA matched 3B attenuator with a 50 Ω Characteristic impedance
02 2
2
111 0
11 on port 210 01 0
1
11 22
221 0
1
2 2 1
| | |
8.56 [141.8(8.56 50)]50
141.8 8.56 50So 0. From symmetry feature, 0.
|
41.44 50( )( ) 0.7041.44 8.56 50 8.56
inZV V
in
in
V
V Z ZS
V Z Z
Z
S S
VS
V
V V V
1
21 12
7
0.707
V
S S
Reciprocal Networks No real power delivers to
network.Besides tSS
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
02.04585.0
4585.0015.0][S
Lossless Networks
jiSS
jiSS
N
kkjki
N
kkjki
for ;0
for ;1
1
*
1
*
[S] is symmetric matrix
Since [S] is not symmetric, the network is not reciprocal.
So the network is not lossless.
1745.0)85.0()15.0( 22221
211 SS
When port 2 terminated with a matched load, =S11=0.15.dB 5.16)15.0log(20log20 RL
When port 2 terminated with a short circuit,
dB 9.6)452.0log(20log20 So
452.02.01
)4585.0)(4585.0(15.0
1
result above theusing and by equationfirst theDividing
1
gives equation second The
22
211211
1
21211
1
1
1
22
2112
2221212221212
2121112121111
RL
S
SSS
V
VSS
V
V
V
S
SVV
VSVSVSVSV
VSVSVSVSV
Summary
Reciprocal Networks (symmetric)
No active elements, no anisotropic material
ttt SSYYZZ
Lossless Networks
No resistive material, no radiation
Unitary
matries , , of partsImaginary
0} Re{0}Re{
*
USS
SZY
YZ
T
Example4.4: Determine if the network is reciprocal, and lossless ?Solution
A matched 3B attenuator with a 50 Ω Characteristic impedance
36.1508.141
8.14136.150
8.14156.88.141
8.1418.14156.8Z
From the result of example 4.2
10
01
5.00
05.0][][
0707.0
707.00][
*SS
S
T
Since the network is reciprocal but not lossless, [S] should be symmetric but not unitary.
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
2 2
1
2
1
111 0 0
1
112 0
2 2
2 121 0
1 1
2
222 0
2 2
| | 0 ( 0)
0| 0
0|
0 1|
0 0 [ ] 1
Since the network is neither reciprocal nor lossless,
gV V g
gs
Vds
mV m
mds
Vds
mds
IIY I
V V
IY
V I R
I g VY g
V V
VgI RY
V V R
Yg R
[ ] should be neither symmetric nor imaginary.Y
0Z0Z
D
S
GPort 1Port 2
dsRm gsg V
gsV
Port 1 Port 2
D
SS
G
unitary.nor symmetricneither be should ][
lossless,nor reciprocalneither isnetwork theSince
2
01][
/1
/1
0)/1)(0(
0)/1)(0(
))((
))((
2/1
2
0)/1)(0(
2
))((
2
00)/1)(0(
02
))((
2
1/1
/1
0)/1)(0(
0)/1)(0(
))((
))((
0
0
0
0
0
0
0
0
00
00
2112220110
211222011022
0
0
000
0
2112220110
02121
002112220110
01212
0
0
00
00
2112220110
211222011011
S
ZR
ZR
RZ
RZgS
ZR
ZR
RY
RY
gRYY
gRYY
YYYYYY
YYYYYYS
RZ
RZg
RY
g
gRYY
Yg
YYYYYY
YYS
gRYYYYYYYY
YYS
RY
RY
gRYY
gRYY
YYYYYY
YYYYYYS
ds
ds
ds
dsm
ds
ds
ds
ds
mds
mds
ds
dsm
ds
m
mds
m
mds
ds
ds
mds
mds
Problem 1Problem 1:Determine if the inductance networks are reciprocal, and lossless ?
Problem 2Problem 2:Determine if the capacitance networks are reciprocal, and lossless ?
1L 2L
3LPort 1 Port 2
1I 2I
1V 2V
Port 1 Port 2
1I 2I
1V 2V
3C
1C 2C
Three Port Network Reciprocal Network
333231
232221
131211
][
SSS
SSS
SSS
S322331132112 SSSSSS
Matching at all ports
0 0 0 332211 SSS Lossless Network
*
2 2 2 2 2 2
12 13 12 23 13 23
* * *13 23 23 12 12 13
[ ] is unitary [ ] [ ] [ ]
Condition A: 1 1 1
Condition B: 0 0 0
From condition B, two of the three
TS S S U
S S S S S S
S S S S S S
12 13 23 parameters, , , must be zero.
This can not satisfy condition A.
A three port network can not be lossless, reciprocal,
and match at all pors simultaneously.
S S S
[S] is unitary and matched at all ports, but not symmetric. Therefore, circulator is lossless and matched, but not reciprocal.
A counter-clockwise circulator
Power splitters
010
001
100
][S
Applications
002
1
002
12
12
10
][S
[S] is symmetric and matched at all ports, but not unitary. Therefore, circulator is reciprocal and matched, but not lossless.
Port 3
Port 2Port 1
Splitters
Twice the electric length represents that the wave travels twice over this length upon incidence and reflection.
A Shift in Reference Planes
mj
m
nnn
SeS
l
n
2'
Generalized Scattering
Parameters
jkVij
jiij
kZV
ZVS
,00
0|
If the characteristic impedances of a multi-port network are different,
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 non-identical characteristic impedances, and is defined as following:
jkaj
iij
i
ii
j
ij
ka
bS
iZ
Vb
jZ
Va
for 0
0
0
|
,...2,1;
,..2,1;
i
jidentijjka
ij
jiidentnonij
Z
ZS
ZV
ZVS
k
0
0,for 0
0
0, |
Port 1 Port 2A Two-Port
Network
1a 2a
1b 2b
01Z 02Z
The Transmission (ABCD) Matrix
ABCD matrix has the advantage of cascade connection of multiple two-port networks.
2
2
1
1
I
V
DC
BA
I
V
2
2
22
22
11
11
1
1
I
V
DC
BA
DC
BA
I
V
Table 4-2 Conversions between two-port network parameters
1 BCAD
Example4.6: Find the S parameters of network?
Solution
For reciprocal network, [Z] is is symmetric. Hence, Z12=Z21
From Table 4-1
00022
00
0
0
0
0
1
01
0
0
1
01
1
01
2cos
2sin
2sin
2cos
1
01
CZjYZCj
jZCZ
CjjY
jZ
Cj
CjjY
jZ
Cj
DC
BA
Port 1 Port 2
4
C C0Z 0Z0,Z
4
C C0,Z
From Table 4-2
20
220
20
22
00
0022
20
220
001221
20
220
20
22
020
220
020
220
00
0011
22
22
2
2
22
)()(
)()(
ZCjjCZ
ZCj
DCZZBA
DCZZBAS
ZCjjCZ
DCZZBASS
ZCjjCZ
ZCj
CZZCjjjCZ
CZZCjjjCZ
DCZZBA
DCZZBAS
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.
2
2
2221
1211
1
1
V
V
TT
TT
V
V
2
2'
22'
21
'12
'11
2221
1211
1
1
V
V
TT
TT
TT
TT
V
V
21
221112
21
11
21
22
21
2221
1211
1
S
SSS
S
SS
S
STT
TT
Port 1 Port 2 T Two-Port
Network
1V 2V
1V 2V
Port 1 Port 2 T Two-Port
Network
T Two-Port
Network
1V
1V
'1V
'1V
2V
2V
'2V
'2V
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 Circuit for Two-Port Networks
Equivalent circuits for some common microstrip discontinuities. (a) Open-ended. (b) Gap. (c) Change in width. (d) T-junction.
2
1
2221
1211
2
1
I
I
ZZ
ZZ
V
V
2
1
2221
1211
2
1
V
V
YY
YY
I
I
Equivalent circuits for a reciprocal two-port network. (a) T equivalent (b) equivalent
Example4.7: Find the network as equivalent T and model at 1GHz?Solution
From Table 4-1
0100
1000
2cos
1002sin
2sin100
2cos
24
2
jj
j
j
DC
BA
l
From Table 4-2
0
1001
0
22
2112
11
C
DZ
jC
ZZ
C
AZ
0100
1000
j
jZ
100
100
100
12
1222
1211
jZ
jZZ
jZZ
Port 1 Port 2
4
1 100Z
0 50Z 0 50Z
From Table 4-2
pF5915.110012
1
nL915.1510012
CCG
LLG
0100
1000
j
jY
100
100
100
12
1222
1211
jY
jYY
jYY
0
1001
0
22
2112
11
B
AY
jB
YY
B
DY
nL915.151001
12
1
pF5915.1100112
LLG
CCG
Equivalent T model
Equivalent model
15.915nH
Port 1 Port 21.592pF
15.915nH
100j
Port 1 Port 2
100j
100j
15.915nH
Port 1 Port 21.592pF
1
100
j
Port 1 Port 2
1.592pF
1
100
j 1
100
j
Example4.8: Find the equivalent model of microstrip-line inductor?Solution
From Table 4-1
lljY
ljZl
DC
BA
cossin
sincos
0
0
From Table 4-2
w
h
r
l
ljYljY
ljYljYY
cotcsc
csccot
00
00
ljYljZ
l
B
AY
ljYljZB
YY
ljYljZ
l
B
DY
cotsin
cos
cscsin
11
cotsin
cos
00
22
00
2112
00
11
2tan
2cos
2sin2
)2
sin2()cos1(
sin
csccot
0
20
0
001211
ljY
ll
ljY
ll
jY
ljYljYYY
Equivalent model
Port 1 Port 2
L
C
10 cscjY l
10 tan
2
ljY
Port 1 Port 2
C
10 tan
2
ljY
ljYY
ljY
ljYljYYY
csc2
tan
csccot
012
0
001222
0
0 00
0 0 0 00
0
0 0
0 0 0 0 0
Inductance of microstrip-line
1sin
csc
;(H)
Parasitic capacitance
tan2 2
2 2 ;(F)2
l
eff eff
l
eff eff
j L jZ l jZ ljY l
Z l Z lZ lL
c
l lj C jY jY
l lY Y Y l
Cc
Example4.9: Find the equivalent T model of microstrip-line capacitor?Solution
From Table 4-1
lljY
ljZl
DC
BA
cossin
sincos
0
0
From Table 4-2
w
h
r
l
ljZljY
l
C
DZ
ljZljYC
ZZ
ljZljY
l
C
AZ
cotsin
cos
cscsin
11
cotsin
cos
00
22
00
2112
00
11
ljZljZ
ljZljZZ
cotcsc
csccot
00
00
2tan
2cos
2sin2
)2
sin2()cos1(
sin
csccot
0
20
0
001211
ljZ
ll
ljZ
ll
jZ
ljZljZZZ
Equivalent T model
Port 1 Port 2
0 tan2
ljZ
Port 1 Port 2
L L
C
0 tan2
ljZ
0 cscjZ l
ljZZ
ljZ
ljZljZZZ
csc2
tan
csccot
012
0
001222
(H);222
2tan
2tan
inductance Parasitic
(F);sin
csc1
line-microstrip of eCapacitanc
00000
00
0
00
00
0
0
0
0
c
lZlZlZ
lZ
L
ljZLj
cZ
l
Z
l
Z
l
Z
lC
ljZCj
effeffl
effeffl
Problem3Problem3: Design a 6GHz attenuator ?
(Hint: -20logS21=6 S21=0.501 )
0501.0
501.00S
Problem4Problem4: Design a 6nH microstrip-line inductor on a 1.6mm thick FR4 substrate. The width of line is 0.25mm. Find the length (l ) and parasitic capacitance?
Problem5Problem5: 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?
6nH0Z
0Z
l
w=0.25mm
h=1.6mm
r=4.5
2pF0Z 0Z
w=5mm
h=1.6mm
r=4.5
l