Prof. David R. Jackson Dept. of ECE Notes 14 ECE 5317-6351 Microwave Engineering Fall 2011 Network Analysis Multiport Networks 1
Prof. David R. Jackson Dept. of ECE Notes 14 ECE 5317-6351 Microwave Engineering Fall 2011 Network Analysis Multiport Networks 1.
Dec 23, 2015
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1
Prof. David R. JacksonDept. of ECE
Notes 14
ECE 5317-6351 Microwave Engineering
Fall 2011
Network AnalysisMultiport Networks
2
Multiport NetworksA general circuit can be represented by a multi-port network, where the “ports” are defined as access terminals at which we can define voltages and currents.
Examples:
1) One-port network
2) Two-port network
Note: Equal and opposite currents are assumed on the two wires of a port.
R
1I
1V
+
-
+
-
1I
1V
+
-
+
-
+
-
+
-
1I 1I2I2I
1V 1V2V
2V
3
3) N-port Network
To represent multi-port networks we use:
- Z (impedance) parameters- Y (admittance) parameters- h (hybrid) parameters- ABCD parameters
- S (scattering) parameters
Not easily measurable at high frequency
Measurable at high frequency
Multiport Networks (cont.)
+
+
+
+
+
1I
2I
3I
mI
NI
1V
2V
3V
mV
NV
-
-
-
-
-
4
Poynting Theorem (Phasor Domain)
2s f d m eP P j P W W
*
2 2
2 2
1ˆ
2
1 1
2 2
1 12
4 4
i
V S
c
V
c
V
E J dV S n dS
E H dV
j H E dV
The last term is the VARS consumed by the region.
The notation < > denotes time-average.
5
Consider a general one-port network
Complex power delivered to network:
Self Impedance
*1 1
1
2
1
22ˆi S m en dPP E H n j Wd W
V I
s
Average power dissipated in [W]
Average magnetic energy (in [J]) stored inside V
Average electric energy (in [J]) stored inside V
,E H n
S
V1V
1I
+
-
d dP P
m mW W
e eW W
6
** 1 1
1 12
2 21
2
1
2
2
1 1
1
1
1
1
121 12
4 (
221
)
22
di
in
in in
m
in
e
n
i
m e
n
dR jX
PR
I
V I
W WX
V I P
I I I
P j W W
VZ
I
I
I
Define Self Impedance (Zin)
,E H n
S
V1V
1I
+
-
7
We can show that for physically realizable networks the following apply:
*1 1
*1
*
1
in
in in
in
Z Z
V V
I
X
I
R
is an even function of
is an odd function of
Self Impedance (cont.)
Please see the Pozar book for a proof.
in in inZ R jX
Note: Frequency is usually defined as a positive quantity. However, we consider the analytic continuation of the functions into the complex frequency plane.
8
1 11 1 12 2
2 21 1 2
1 11 12 1
2 21 22 2
2 2
V Z I Z I
V Z Z I
V Z Z IV Z I
V Z I Z I
Consider a general 2-port linear network:
In terms of Z-parameters, we have (from superposition)
Impedance (Z) matrix
Two-Port Networks
1 2+
-
+
-1V
1I 2I
2V
9
Elements of Z-Matrix: Z-Parameters(open-circuit parameters )
0k
iij
j I k j
VZ
I
1
1
112
2 0
222
2 0
I
I
VZ
I
VZ
I
Port 2 open circuited Port 1 open circuited
2
2
111
1 0
221
1 0
I
I
VZ
I
VZ
I
1 2+
-
+
-1V
1I 2I
2V
1 11 1 12 2
2 21 1 22 2
V Z I Z I
V Z I Z I
10
Z-Parameters (cont.)
0k
iij
j I k j
VZ
I
+
jI
iV -
We inject a current into port j and measure the voltage (with an ideal voltmeter) at port i. All ports are open-circuited except j.
N-port network
11
Z-parameters are convenient for series connected networks.
1 1 1
2 2 2
A B
A B
I I I
I I I
Series
Z-Parameters (cont.)
A
B
1AI
1BI 2
BI
2AI
+
+
+
+
2AV
2BV
1AV
1BV
1V 2V
2I1I++
--
1 2
1 2
- -
--
1 1 1
2 2 2
1
2
1
2
A B
A B
A A B B
A B
A B
A B
V V V
V V V
Z I Z I
Z Z I
IZ Z
I
IZ Z
I
1 111 11 12 12
2 221 21 22 22
A B A B
A B A B
V IZ Z Z Z
V IZ Z Z Z
12
1 11 12 1
2 21 2
0
2 2
k
iij
j V k j
I Y Y VI Y V
IY
Y Y V
V
I
Admittance (Y) ParametersConsider a 2-port network:
Admittance matrix
Short-circuit parameters
1 21V
1I
+
-
2I
2V+
-
1 11 1 12 2
2 21 1 22 2
I Y V Y V
I Y V Y V
or
13
Y-Parameters (cont.)
0k
iij
j V k j
IY
V
jV
iI
+-
We apply a voltage across port j and measure the current (with an ideal current meter) at port i. All ports are short-circuited except j.
N-port network
14
1 1 1
2 2 2
111 11 12 12
221 21 22 22
A B
A B
A B A B
A B A B
I I I
I I I
VY Y Y Y
VY Y Y Y
1 1 1
2 2 2
A B
A B
V V V
V V V
Parallel
1AI
1AV
1I
1V A
B1BV
2V
2I
2BI
2AI
2BV
2AV
+ +
+ +
+ +1
1
2
2
1BI
- -
--
- -
Admittance (Y) Parameters
Y-parameters are convenient for parallel connected networks.
15
1Y Z
Admittance (Y) Parameters
Relation between [Z] and [Y] matrices:
V Z I
I Y V
V Z Y V
Z Y V
Z Y U Identity Matrix
Hence
Therefore
16
Reciprocal Networks
If a network does not contain non-reciprocal devices or materials* (i.e. ferrites, or active devices), then the network is “reciprocal.”
ij ji ij jiZ Z Y Y
Z Y
and are symmetric
* A reciprocal material is one that has reciprocal permittivity and permeability tensors. A reciprocal device is one that is made from reciprocal materials
Note: The inverse of a symmetric matrix is symmetric.
Example of a nonreciprocal material: a biased ferrite
(This is very useful for making isolators and circulators.)
17
Reciprocal MaterialsD E
B Η
x xx xy xz x
y yx yy yz y
z zx zy zz z
D E
D E
D E
x xx xy xz x
y yx yy yz y
z zx zy zz z
B
B
B
Ferrite: 0
0
0
0 0 1
j
j
is not symmetric!
Reciprocal: ,ij ji ij ji
18
We can show that the equivalent circuits for reciprocal 2-port networks are:
T-equivalent
Pi-equivalent
11 21Z Z 22 21Z Z
21Z
21Y
11 21Y Y22 21Y Y
Reciprocal Networks (cont.)
19
ABCD-ParametersThere are defined only for 2-port networks.
1 2'
1 2
V VA B
I IC D
' '2 2
2 2
1 1
2 20 0
1 1' '2 20 0
I I
V V
V IA C
V V
V IB D
I I
1 2 1I
1V2V
'2I
'2 2I I
20
Cascaded Networks
'
'
21 1
1 1 2
1
1
2
2
AAA
A A
BA
B
B
A B
B
VV VABCD
I I I
VABCD
I
VABCD ABCD
I
1 2'
1 2
ABV VABCD
I I
1 2 1 2
1AI '
2AI
1AV
2AV
1BV 2
BV
1I'2I
1V 2V
1BI
'
2BI
A B
A nice property of the ABCD matrix is that it is easy to use with cascaded networks: you simply multiply the ABCD matrices together.
21
At high frequencies, Z, Y, h & ABCD parameters are difficult (if not impossible) to measure.
o V and I are not uniquely definedo Even if defined, V and I are extremely difficult
to measure (particularly I).o Required open and short-circuit conditions are
often difficult to achieve.
Scattering (S) parameters are often the best representation for multi-port networks at high frequency.
Scattering Parameters
22
S-parameters are definedassuming transmission lines are connected to each port.
On each transmission line:
0 0
0 0
i i i iz zi i i i
i i i ii
i
i i i
ii
iV zV z V e V e
V z V zI z
Z Z
V z
1, 2i
1 2
1z 2z
1a
1b 2b
2a01 1,Z 02 2,Z
Scattering Parameters (cont.)
Local coordinates
0
0
i i i i i
i i i i i
a z V z Z
b z V z Z
Incoming wave function
Outgoing wave function
23
For a One-Port Network
01
01
1
1
1
1
11
0
0
0
0
L
Z
Z
V
V
S
b
a
1l
1a
1b01Z
L
1 1
11 1
0 0
0
Lb a
S a
0
0
i i i i i
i i i i i
a z V z Z
b z V z Z
Incoming wave function
Outgoing wave function
For a one-port network, S11 is defined to be the same as L.
24
Scattering matrix
1 11 1 12 2
2 21 1 22
1 11 12 1
2 21 22 2
2
0 0 0
0 0
0
0
0
0
0
b S a S a
b S a S
b S S ab S a
b S S
a
a
For a Two-Port Network
1 2
1z 2z
1a
1b 2b
2a01 1,Z 02 2,Z
25
Scattering Parameters
Output is matched
Input is matched
input reflection coef. w/ output matched
reverse transmission coef. w/ input matched
forward transmission coef. w/ output matched
output reflection coef. w/ input matched
1 11 1 12 2
2 21 1 22 2
0 0 0
0 0 0
b S a S a
b S a S a
2
1
2
1
111
1 0
112
2 0
221
1 0
222
2 0
0
0
0
0
0
0
0
0
a
a
a
a
bS
a
bS
a
bS
a
bS
a
Output is matched
Input is matched
26
Scattering Parameters (cont.)
ja
ib
For a general multiport network:
All ports except j are semi-infinite (or matched)
0
0
0k
iij
j a k j
bS
a
Port j
Port i
Semi-infinite
N-port network
27
Scattering Parameters (cont.)
1
2
3
2a
2b
3a
3b
1a
1b
Illustration of a three-port network
28
For reciprocal networks, the S-matrix is symmetric.
ij jiS S i j
Scattering Parameters (cont.)
Note: If all lines entering the network have the same characteristic impedance, then
0
0
0k
iij
j V k j
VS
V
29
Why are the wave functions (a and b) defined as they are?
Note:
*
2
0
01 10 Re 0 0
2 2i
i i ii
VP V I
Z
0
2
0 0
10 0
2
i i i
i i
a V Z
P a
Scattering Parameters (cont.)
1 2
1z 2z
1a
1b 2b
2a01 1,Z 02 2,Z
(assuming lossless lines)
30
Similarly,
2
0
20100
2
1
2i
i ii
VP b
Z
Also,
Scattering Parameters (cont.)
2 2 2
2 2 2
0
1 10
21 1
2
0
2
02
i i
i
i i
i
i
i
li i i i i
li i i i i
li i i
li i i
P
V
l a l a e
l V e
V l
l b
e
P l b e
V
31
Example
1a
1b
2a
2b0Z 0Z
Z2V1V
1z 2z
Find the S parameters for a series impedance Z.
Note that two different coordinate systems are being used here!
32
Example (cont.)
02 2
1 1 011
1 1 0
0
00
0
0
0 0
0 0in
ina a
b V Z ZS
a V Z Z
Z Z Z
Z Z Z
1102
ZS
Z Z
By symmetry:
S11 Calculation:
1a
1b 2b0Z 0Z
Z2V1V
inZ
1z 2z
Semi-infinite
22 11S S
33
2
2
221
1 0
2
1 0
0
0
0
0
a
a
bS
a
V
V
2 2 2
02 1
0
1 1 0 11
02 2 1 0 11
0
0 0 0
0 0
0 1
0 0 1
a V V
ZV V
Z Z
V a Z S
ZV V a Z S
Z Z
Example (cont.)S21 Calculation:
1a
1b 2b0Z 0Z
Z2V1V
inZ
1z 2z
Semi-infinite
1 1 00 0V a Z
34
01 0 11
021
1 0
0 0 0 011
0 0 0 0 0
0 1
0
2 21 1
2 2
Za Z S
Z ZS
a Z
Z Z Z Z Z ZS
Z Z Z Z Z Z Z Z Z Z
021
0
2
2
ZS
Z Z
Hence
Example (cont.)
1a
1b 2b0Z 0Z
Z2V1V
inZ
1z 2z
Semi-infinite
12 21S S
35
Example
0Z 0Z0 ,s sZ
L
z 2z1z
Find the S parameters for a length L of transmission line.
Note that three different coordinate systems are being used here!
36
Example (cont.)
2
2
2
0022
0
111
0 01
( )in a
in aa
Z ZS
b
Z ZS
a
by symmetry
2
2
0 00 0 20
0 0
1tan
tan 1
s
s
j LLs s
in s s j Las s L
eZ jZ LZ Z Z
Z jZ L e
0 0
0 0
sL
s
Z Z
Z Z
0Z 0Z0 ,s sZ
L
z 2z1z
1 0V 2 0V+ +
- -
L
+-
V z
Semi-infinite
S11 Calculation:
37
Example (cont.)
0 00 0
0 011 22
0 00 0
0 0
tan
tan
tan
tan
s ss
s s
s ss
s s
Z jZ LZ Z
Z jZ LS S
Z jZ LZ Z
Z jZ L
20 0 0 11 220
0s in aZ Z Z Z S S
Note: If
0Z 0Z0 ,s sZ
L
z 2z1z
1 0V 2 0V+ +
- -
L
+-
V z
Hence
38
2 2
2221
1
0
1 0 00
0
0a a
bS
a
V Z
V Z
1 1 110 0 1V V S
0 0
0 0
sL
s
Z Z
Z Z
Example (cont.)
0Z 0Z0 ,s sZ
L
z 2z1z
1 0V 2 0V+ +
- -
L
+-
V z
11
11
00
1
VV
S
We now try to put the numerator of the S21 equation in terms of V1 (0).
Hence, for the denominator of the S21 equation we have
S21 Calculation:
Semi-infinite
39
21
1
2
2
0 0 1
00
1
0 1s
s s
s s
sj z j z
j L j LL
j L j LL
L
V
V z V e e
V L V e e
VV
e e
Example (cont.)
0Z 0Z0 ,s sZ
L
z 2z1z
1 0V 2 0V+ +
- -
L
+-
V z
2 20 0 0 0 1 LV V V V
Next, use
1
2 2
00 1
1s sLj L j L
L
VV
e e
Hence, we have
40
2
2 1121 2
1 0
0 1 1
0 1
s
s
j LL
j LLa
V S eS
V e
Example (cont.)
0Z 0Z0 ,s sZ
L
z 2z1z
1 0V 2 0V+ +
- -
L
+-
V z
Therefore, we have
1121 122
1 1
1
s
s
j LL
j LL
S eS S
e
by symmetry
11
11
00
1
VV
S
1
2 2
00 1
1s sLj L j L
L
VV
e e
so
41
Special cases:
0 0 11 22
21 12
) 0, 0s
s L
j L
a Z Z S S
S S e
0
0
s
s
j L
j L
eS
e
20 11 220
21
2)
2 2
0
1 1s
g gs
g
in a
j L
b L L
Z Z S S
e S
0 1
1 0S
0 ,s sZ
L
Example (cont.)
42
02 0111
02 01
01 0222 11
02 01
Z ZS
Z Z
Z ZS S
Z Z
2
2
2
221
1
02
1
01 0
0
0
0
0 0
a
a
b
a
V
Z
VS
Z
Example
Find the S parameters for a step-impedance discontinuity.
01Z 02Z
43
22 2
2 2 1 1 1100 00 0 0 0 1
aa aV V V V S
01 0221 12
01 02
2Z Z
S SZ Z
Example (cont.)
1V2V
++
01Z 02Z --Because of continuity of the voltage
across the junction, we have:
2 2
2
02 02
1 1
01 010
1
0
2
11
1
0 10
0 0
a a
V
Z Z
V V
Z Z
V
S
S
02 0111
02 01
02
02 01
1 1
2
Z ZS
Z Z
Z
Z Z
0121 11
02
1Z
S SZ
Hence
so
Semi-infiniteS21 Calculation:
44
For reciprocal networks, the S-matrix is symmetric.
For lossless networks, the S-matrix is unitary.
Identity matrix
TS S
* *T TS S S S U
* *
1 1
( , )N N
Tik kj ki kj ij
k k
i j S S S S
Take element1 ;
0;ij
i j
i j
Properties of the S Matrix
Equivalently,
* 1TS S
Notation: † *H T
S S S
† 1S S
so
A B U
B A U
Note :
If
then
N-port network
45
Example:
11 12 13
21 22 23
31 32 33
S S S
S S S S
S S S
* * *11 11 21 21 31 31
* * *12 12 22 22 32 32
* * *13 13 23 23 33 33
* * *11 12 21 22 31 32
* * *11 13 21 23 31 33
* * *12 13 22 23 32 33
1
1
1
0
0
0
S S S S S S
S S S S S S
S S S S S S
S S S S S S
S S S S S S
S S S S S S
Unitary
Properties of the S Matrix (cont.)
The column vectors form an orthogonal set.
The rows also form orthogonal sets (see the note on the previous slide).
46
Example
50
02 2
0 02
0 02
j j
jS
j
12
350
S
Not unitary Not lossless
1) Find the input impedance looking into port 1 when ports 2 and 3 are terminated in 50 [] loads.
2) Find the input impedance looking into port 1 when port 2 is terminated in a 75 [] load and port 3 is terminated in a 50 [] load.
(For example, column 2 doted with the conjugate of column three is not zero.
47
11 11 1
1
0 50[ ]in in
bS Z
a
1 If ports 2 and 3 are terminated in 50 [Ω]:
Example (cont.)
(a2 = a3 = 0)
2 0 1
2
3
2a
2b
3a
3b3 0
50S 1a
1b1in
1 11 1 12 2 13 3b S a S a S a
48
22
2
75 50 1
75 50 5
a
b
2) If port 2 is terminated in 75 [Ω] and port 3 in 50 [Ω]:
Example (cont.)
2
1
5
1
2
3
2a
2b
3a
3b3 0
50S 1a
1b1in
49
31 21 11 12 13
1 1 1
212 2 12 2 21
1
1 1
5 102 2
in
b j jS
ab aS S S
a a a
S Sa
2 2 2a b
1
1
1 44.55[1
501
]inin
in
Z
Example (cont.)
11
2 2
3 3
02 2
0 02
0 02
ab
b a
j j
j
jb a
322 1 21 22 23
1 1
/aa
b a S S Sa a
2
1
5
1
2
3
2a
2b
3a
3b3 0
50S 1a
1b1in
50
Transfer (T) Matrix
For cascaded 2-port networks:
2
1 11 12 2
1 21 22 2
2
a T T b
b T
b
T
T
a
a
22
21 21
11 11 2212
21 21
1 S
S ST
S S SS
S S
A1Aa 2
Aa2Ba
1Ba
1Bb
2Bb
2Ab1
Ab B1 12 2
T Matrix:
21
22 22
212 12
1122 22
1T
T TS
T TT
T T
(Derivation omitted)
51
1 2
1 2
1
1
2 1
2 1
21
1
1
1 2
A AA
A A
AA
A
A
A B
A
A
A B
B
B
BB
B
a bT
b a
aT
b
b a
a
a
b
a
b
bT
b aT
But
Transfer (T) Matrix (cont.)
Hence
The T matrix of a cascaded set of networks is the product of the T matrices.
ABT
52
Conversion Between Parameters
53
ExampleDerive Sij from the Z parameters.
(The result is given inside row 1, column 2, of the previous table.)
11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2
011 1
0
inin
in
Z ZS
Z Z
11 21 21 22 21 0||inZ Z Z Z Z Z Z
S11 Calculation:
54
Example (cont.)
11 21 21 22 21 0
21 22 0 2111 21
22 0
11 21 22 0 21 22 0 21
22 0
211 22 11 0 21 22 21 0 21 22 21 0 21
22 0
211 22 11 0 21
22 0
||inZ Z Z Z Z Z Z
Z Z Z ZZ Z
Z Z
Z Z Z Z Z Z Z Z
Z Z
Z Z Z Z Z Z Z Z Z Z Z Z Z
Z Z
Z Z Z Z Z
Z Z
11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2
55
Example (cont.)
211 0 22 21
22 0in
Z Z Z ZZ
Z Z
11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2
011
0
in
in
Z ZS
Z Z
211 0 22 21 0 0 22
11 211 0 22 21 0 0 22
Z Z Z Z Z Z ZS
Z Z Z Z Z Z Z
so
56
Example (cont.)11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2
211 0 22 21 0 0 22
11 211 0 22 21 0 0 22
2 211 0 11 22 21 0 0 22
2 211 0 11 22 21 0 0 22
20 22 11 0 21
20 22 11 0 21
Z Z Z Z Z Z ZS
Z Z Z Z Z Z Z
Z Z Z Z Z Z Z Z
Z Z Z Z Z Z Z Z
Z Z Z Z Z
Z Z Z Z Z
57
Example (cont.)11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2
20 22 11 0 21
11 20 22 11 0 21
Z Z Z Z ZS
Z Z Z Z Z
Note: to get S22, simply let Z11 Z22 in the previous result.
20 11 22 0 21
22 20 11 22 0 21
Z Z Z Z ZS
Z Z Z Z Z
58
Example (cont.)
11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2cV 2 0V 1 0V
1 110 1V S
Assume 1 0 1 [V]V
21 22 21 01
11 21 21 22 21 0
||0
||c
Z Z Z ZV V
Z Z Z Z Z Z
Use voltage divider equation twice:
0
222 21 0
0 c
ZV V
Z Z Z
21 2 20 0S V V
S21 Calculation:
59
Example (cont.)
Hence
21 22 21 0 021 11
22 21 011 21 21 22 21 0
||1
||
Z Z Z Z ZS S
Z Z ZZ Z Z Z Z Z
11 21Z Z 22 21Z Z
21Z
Semi infinite
0Z 0Z11S
1 2cV 2 0V 1 0V
After simplifying, we should get the result in the table.
(You are welcome to check it!)
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