Network
Network
Network
where Y Z 1
Reciprocal and Lossless Networks
Reciprocal networks usually contain nonreciprocal media
such as ferrites or plasma, or active devices. We can show
that the impedance and admittance matrices are symmetrical,
so that.
Zij Z ji or Yij Yji
I1 I2
V1 ZC V2
Solution
Port 2 open-circuited Similarly we can show that
V1 V2
Z11 Z A ZC Z 21 ZC
I1 I 2 0
I1 I 2 0
Z B ZC
V1 V2 ZC ZC V2
Z12 ZC Z 22 Z B ZC
I2 I1 0
I 2 Z B ZC Z B ZC I2 I1 0
S-parameters
Port 1 Port 2
Microwave device
Vt Vr
Vi Vi
S-parameters
Voltage of traveling wave away from port 1 is
Vr1 V
Vb1 Vi1 t 2 Vi 2
Vi1 Vi 2
Voltage of Voltage of
Reflected wave Transmitted wave
From port 1 From port 2
Logarithmic form
b1 S11 S12 a1 S11=20 log(1)
S-matrix b S a S22=20 log(2)
2 21 S 22 2 S12=20 log(12)
S21=20 log(21)
S-parameters
Vt 2
Vr1 S12
S11 Vi 2 Vr 2 0
Vi1 Vr 2 0
Vt1 Vr 2
S 21 S 22
Vi1 Vr1 0
Vi 2 Vr1 0
Vt 2
S21 V1 Vo 141.8 V2
Vi 2 V
r 2 0
From the fact that S11=S22=0 , we know that Vr1=0 when port 2 is
matched, and that Vi2=0. Therefore Vi1= V1 and Vt2=V2
Z 2 // Z 3 Z o Zo
Vt 2 V2 V1 Vo
Z 2 // Z 3 Z1 Z 3 Z o Z3 Z o
41.44 50
V1 0.707V1
41.44 8.56 50 8.56
S
0 0.707
Therefore S12 = S21 = 0.707
0.707 0
Lossless network
For lossless n-network , total input power = total output power. Thus
n n
ai ai* bi bi* Where a and b are the amplitude of the signal
i 1 i 1
In summation form
n 1 for i j
*
SkiSkj
0
for i j
k 1
Conversion of Z to S and S to Z
S Z U 1Z U
Z U S 1U S
where 1 0 . 0
0 . . .
U
. . 1 .
0 . . 1
Reciprocal and symmetrical network
Since the [U] is diagonal , thus U U t
S S t
Example
A certain two-port network is measured and the following scattering
matrix is obtained:
0.10o 0.890o
S o
0.890 o
0.20
From the data , determine whether the network is reciprocal or lossless. If
a short circuit is placed on port 2, what will be the resulting return loss at
port 1?
Solution
Since [S] is symmetry, the network is reciprocal. To be lossless, the S
parameters must satisfy
For i=j
n 1 for i j
SkiSkj 0 for i j
* |S |2 + |S |2 = (0.1)2 + (0.8)2 = 0.65
11 12
b1 b2
S11 S12 S11
S12 S 21
0.1
j 0.8 j 0.8
0.633
a1 a1 1 S 22 1 0.2
I 2 V2 V1 V2 I1 I 2 In matrix form
V V1
A 1 for port 2 open circuit B for port 2 short circuit
V2 I2 V
I 2 0 2 0
V1 V1
A for port 2 open circuit B for port 2 short circuit
V2 I 2 0
I2 V
2 0
I I1
C 1 for port 2 open circuit D for port 2 short circuit
V2 I2 V
I 2 0 2 0
I1 = - I2 = 0 hence C= 0 I1 = - I2 hence D= 1
Z3
V1 V2
V1
A for port 2 open circuit
V2 I 2 0 therefore
then V2
Z3 V1 Z1 Z 3 Z1
V1 A 1
Z1 Z 3 V2 Z3 Z3
Continue
V1
B for port 2 short circuit Z1 I2
I2 V
2 0
Analysis
I 2 I1
Therefore
I1 1
V2 I 2 Z3 I1Z3 C
V2 Z 3
Continue
I1
D for port 2 short circuit Z1 I2
I2 V I1
2 0
Z2 VZ2
Z3
I1 is divided into Z2 and Z3, thus
Z3
I2 I1
Z 2 Z3 Full matrix
Hence Z1 Z1Z 2
1 Z Z1 Z 2
Z3
D
I1 Z
1 2 2
I2 Z3 1 1
Z2
Z 3 Z3
ABCD for transmission line
I1 I2
Input V1 Transmission line Zo g V2
z = - z =0
V f e j t e g z Vb e j t eg z
V ( z ) V f e j t e g z Vb e j t eg z Vf Vb
Zo
If Ib
I ( z)
1
Zo
V f e j t e g z Vb e j t eg z
f and b represent forward and backward propagation voltage and current
Amplitudes. The time varying term can be dropped in further analysis.
continue
At the input z = -
V1 V () V f e g
Vb e g (1) I1 I ()
1
Zo
V f eg Vb e g (2)
At the output z = 0
V1
A for port 2 open circuit
V2 I 2 0
g g (e x e x )
Vo (e e ) cosh( x)
A cosh( g ) 2
2Vo
(e x e x )
sinh( x)
V1 2
B for port 2 short circuit
I2 V
2 0
For V2 = 0 , Eq. (3) implies –Vf= Vb = Vo . From Eq. (1) and (4) we have
Z oVo (eg e g )
B Z o sinh( g )
2Vo
continue
I1
C for port 2 open circuit
V2 I 2 0
For I2=0 , Eq. (4) implies Vf = Vb = Vo . From Eq.(2) and (3) we have
Vo (eg e g ) sinh( g )
C
2Vo Z o Zo
I1
D for port 2 short circuit
I2 V
2 0
For V2=0 , Eq. (3) implies Vf = -Vb = Vo . From Eq.(2) and (4) we have
Z oVo (eg e g )
D cosh( g )
2 Z oVo
continue
Note that
The complete matrix is therefore
Z
1 Z
Series impedance 0 1
Z Shunt impedance 1 0
1
Z 1
Table of ABCD network
Z1 Z1 Z1Z 2
1 Z Z1 Z 2
Z2
Z3
2
Z3 T-network 1 1
Z2
Z 3 Z3
Z3
Z3
1 Z 3
Z1 Z2
Z2 p-network
1 1 Z Z
3 1 3
Z1 Z 2 Z1Z 2 Z1
n 0
1 Ideal transformer
0 n
n:1
Short transmission line
cos(k ) jZo sin( k )
Lossless transmission line ABCD tline sin( k )
j cos(k )
Zo
If << l then cos(k ) ~ 1 and sin (k ) ~ k then
1 jZo k
ABCD tlineshort 1
j k 1
Zo
Embedded short transmission line
Z1 Transmission line Z1
1 0 1 jZo k 1 0
ABCD embed 1 1 1 1
Z j Z k 1 1
1 o 1
Z
Solving, we have
jZo k
1 jZo k
Z1
ABCD embed
2 jZo k j k 1 jZo k
Z1 Z12 Zo Z1
Comparison with p-network
Z3
1 Z3
Z2
ABCDp net
1 1 Z3 1
Z3
Z1 Z 2 Z1Z 2 Z1
jZo k
1 jZ o k
Z1
ABCD embed
2
o jjZ k k jZ k
1 o
Z1 Z12 Zo Z1
Shunt
k
If Zo << Z1 then the series impedance Z j
Zo
This is a capacitance which is given by C
Z oc
Equivalent circuits
Zo ZoL Zo
Zo
Zo >> Z1 L
c
Zo Zoc Zo
Zo << Z1
C
Z oc
Transmission line parameters
It is interesting that the characteristic impedance and propagation constant of a
transmission line can be determined from ABCD matrix as follows
B
Zo
C
g cosh A ln A A2 1
1 1 1
Conversion S to ABCD
For conversion of ABCD to S-parameter
1 S1,1
. 1 S 2,2 S1,2 .S 2,1 Z .1 S1,1
. 1 S 2,2 S1,2 .S 2,1
1
A( S ) . 1
2.S 2,1 .1 S1,1
. 1 S 2,2 S1,2 .S 2,1 1 S1,1 . 1 S 2,2 S1,2 .S 2,1
Z o
Odd and Even Mode Analysis
Usually use for analyzing a symmetrical four port network
(1) Excitation •Equal ,in phase excitation – even mode
•Equal ,out of phase excitation – odd mode
Line of
symmetry
3 4
Since the network is symmetry, Instead of 4 ports , we can only analyze 2 port
continue
We just analyze for 2 transmission lines with characteristic Ze and Zo
respectively. Similarly the propagation coefficients be and bo respectively.
Treat the odd and even mode lines as uniform lossless lines. Taking ABCD
matrix for a line , length l, characteristic impedance Z and propagation
constant b,thus
Taking b
p
(equivalent to quarter-wavelength transmission line)
2
Then
Z 2 Z o2 j 2ZZ o
S 2 2 1
Z Z o j 2ZZ o Z 2 Z o2
continue
S13 S14
S23
S24
S11 S12 S11 S12
Odd + even Convert to S21 S22 S21 S22 S34
S11 S12
S11 S12 S11 S12
S21 S22 S33
S31 S21 S22 S21 S22
S44
S41 S42 S32 S43
( Z
jZo ev odZ Z 2
o ) ( Z od Z ev )
2 ( Z ev
2
Z o2 ) ( Z od 2
Z o2 )
For perfect isolation (I.e S41=S14=S32=S23=0 ),we choose Zev and Zod such that
Zev Zod=Zo2.
continue
S11 S12 S13 S14 ev- od
ev+ od
S21 S22 S23 S24
Similarly we have
1 Z ev
2
Z o2 Z od2
Z o2
S11 S 22 S33 S 44 2
2 Z ev Z o Z od Z o2
2 2
1 2 2
Z ev Z od Z o4
2 ( Z ev
2
Z o2 )( Z od
2
Z o2 )
We have
1 Z ev
2
Z o2 Z od 2
Z o2
S31 S13 S 24 S 42 2
2 Z ev Z o Z od Z o2
2 2
( Z 2
Z 2
) Z 2
2 ev od o
( Z Z )( Z Z )
2 2 2
ev o od o
Z ev Z
od
Z Z if Zev Zod=Zo2.
ev od
continue
S11 S12 S13 S14 ev- od
ev+ od
S21 S22 S23 S24
jZo Z ev Z od
S 21 S12 S34 S 43 2
2 Z ev Z o Z od Z o2
2 2
1
jZo if Zev Zod=Zo2.
Z Z
ev od
continue
This S-parameter must satisfy network characteristic:
(1) Power conservation
2 2 2 2
S11 S21 S31 S41 1
Reflected
power transmitted transmitted transmitted
power to power to power to
port 2 port 3 port 4
Since S11 and S41=0 , then
2 2
S21 S31 1
S11 p
(2) And quadrature condition Arg
S 21 2
continue
For 3 dB coupler
Z ev Z
1
2
Z ev Z od
od 1 or Z Z 2
Z Z 2 ev od
ev od
Rewrite we have
Z ev 1 ( 2 )
3 2 2
Z od 1 ( 2 )
Z ev
In practice Zev > Zod so 3 2 2 5.83
Z od
However the limitation for coupled edge
Z ev
2 (Gap size ) also bev and bod are not pure TEM
Z od thus not equal
A l/4 branch line coupler
Odd 90o
1 Z2 2
90o
1 Z2 2
Z1
Z1
45o 45o
90o 90o
Z1
Z1
90o 1 Z2 2
4 Z2 3
Z1
Z1
45o 45o
O/C O/C
Analysis
p
Stub odd (short circuit) X s ,od Z1 tan Z1
4
p
Stub even (open circuit) X s ,ev Z1 cot Z1
4
The ABCD matrices for the two networks may then found :
Z2
1 0 0 jZ2 1 0
Xs
jZ2
ABCD 1 1 j 1
jX Z 0 1 j jZ2 Z2
s 2 jX s 2
Z2 X s X s
Transmission
stub stub
line
continue
Convert to S
Z o A B Z o2C Z o D 2Z o AD BC
S 1
Z o A B Z o2C Z o D 2Z o Z o A B Z o2C Z o D
Z o2 Z 2 Z o2
jZ 2 j 2
j 2Z o
1 X s Z2
2Z o Z 2 Z o2 Z 2 Z o2 Z o2 Z 2 Z o2
jZ 2 j 2
j 2Z o jZ 2 j 2
j
Xs Xs Z 2 Xs Z 2
Z o2 Z 2 Z o2 Xs
Zo Z2
Z1
S11 jZ2 j j 0 or From
X s2 Z2 Z o2 Z 22 previous
definition
continue
Substituting into S-parameter gives us
0 Zo 0 Zo
S odd
1
Z and S even 2 2 1
Z o Z 2 jZ2 o
2 2 0 Z
Z o Z 2 jZ2 o 0
Equal split S
Z 1 Zo
S 21 2 or Z2
Zo 2 2
And Zo
Zo
Zo Z2 2
X s Z1 Zo
Z o2 Z 22 Z
2
Z o2 o
2