Download Microwave Engineering

Network Parameters
Impedance and Admittance matrices
For n ports network we can relate the voltages and currents by impedance
and admittance matrices
Impedance matrix
V1   Z11
V   Z12
 2 
 .  .
 .  .
  
Vn   Z1n
Admittance matrix
Z 21 . . Z n1   I1 
Z 22 . . Z n 2   I 2 
.
. .
.  . 
 
.
. .
.  . 
Z 2n . . Z nn   I n 
where
 I1  Y11 Y21
 I  Y12 Y22
 2 
.
 .  .
 .  .
.

 
 I n  Y1n Y2n
Y   Z 1
. . Yn1  V1 
. . Yn 2  V2 
. . .  . 
 
. . .  . 
. . Ynn  Vn 
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
Lossless networks can be shown that Zij or Yij are imaginary
Refer to text book Pozar pg193-195
Example
Find the Z parameters of the two-port T –network as shown below
ZA
ZB
I1
I2
V1
V2
ZC
Solution
Port 2 open-circuited
Z11 
V1
I1
 Z A  ZC
Z12
I1 0
Z 21 
I 2 0
Port 1 open-circuited
V1

I2
Similarly we can show that
V2
I1
 ZC
I 2 0
This is an example of reciprocal network!!
ZC
V2 ZC

 Z B  ZC 
 ZC
I 2 Z B  ZC
Z B  ZC
Z 22 
V2
I2
 Z B  ZC
I1 0
S-parameters
Port 1
Port 2
Microwave device
Input signal
reflected signal
transmitted signal
Vi1
Vr1
Vt2
Transmission and reflection coefficients
 
Vt
Vi
 
Vr
Vi
Vi2
Vr2
Vt1
S-parameters
Voltage of traveling wave away from port 1 is
Vb1 
V
Vr1
Vi1  t 2 Vi 2
Vi1
Vi 2
Voltage of
Reflected wave
From port 1
Voltage of
Transmitted wave
From port 2
Voltage of transmitted wave away from port 2 is
Vb 2
Vt1
Vr 2

Vi1 
Vi 2
Vi1
Vi 2
Vt1
Vt 2 
Vr1
, 12 
, 21  V
Let Vb1= b1 , Vi1=a1 , Vi2=a2 ,  1
i1
Vi1
Vi 2
Vr 2
and  2 
Then we can rewrite
Vi 2
S-parameters
b1  1 a1  12a2
Hence
In matrix form
S-matrix
b2   21 a1   2a2
 b1   1 12   a1 
b   
 a 

2  2 
 2   21
 b1   S11 S12   a1 
b   S
 a 
S
22   2 
 2   21
•S11and S22 are a measure
of reflected signal at port
1 and port 2 respectively
•S21 is a measure of gain or
loss of a signal from port 1
to port 2.
•S12 ia a measure of gain or
loss of a signal from port 2
to port 1.
Logarithmic form
S11=20 log(1)
S22=20 log(2)
S12=20 log(12)
S21=20 log(21)
S-parameters
S11 
S12 
Vr1
Vi1
Vt 2
Vi 2
Vr 2 0
Vr 2 0
Vr2=0 means port 2 is matched
S 21 
Vt1
Vi1
S 22 
Vr1 0
Vr1=0 means port 1 is matched
Vr 2
Vi 2
Vr1 0
Multi-port network
Port 5
Port 1
Port 4
network
 b1   S11 S12
b   S
S 22
 2   21
b3    S31 S32
b   S
S 42
 4   41
b5   S51 S52
S13
S14
S 23
S 24
S33
S34
S 43
S53
S 44
S54
S15   a1 
S 25  a2 
S35   a3 
 
S 45  a4 
S55   a5 
Example
Below is a matched 3 dB attenuator. Find the S-parameter of the circuit.
8.56 
8.56 
141.8 
Z1=Z2= 8.56  and Z3= 141.8

Solution
Vr1
S11 
Vi1 V
r 2 0
Zin  Z o

Zin  Z o
By assuming the output port is terminated by Zo = 50 ,
then
Z  Z  Z //( Z  Z )
in
1
3
2
o
 8.56  141.8(8.56  50) /(141.8  8.56  50)  50 
S11 
50  50
0
50  50
Because of symmetry , then S22=0
Continue
8.56 
S21 
Vt 2
Vi 2 V
r 2 0
V1
Vo
8.56 
141.8 
V2
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 

  Vo 

Vt 2  V2  V1
 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 
Therefore S12 = S21 = 0.707
S   
0
0.707
0 
0.707
Lossless network
For lossless n-network , total input power = total output power. Thus
n

i 1
ai ai*
n
  bibi*
Where a and b are the amplitude of the signal
i 1
at a* = bt b*
=at St S* a*
Putting in matrix form
Thus
at
(I –
St
S*
)a*
=0
In summation form
n

k 1
*
SkiSkj

1
0
for i  j 

for i  j 
This implies that
Note that bt=atSt and b*=S*a*
St
S*
=I
Called unitary
matrix
Conversion of Z to S and S to Z
S   Z  U 1Z  U 
Z   U  S 1U  S 
where
1 0 . 0
0 . . . 

U   
. . 1 .


0 . . 1 
Reciprocal and symmetrical network
Since the [U] is diagonal , thus
U   U  t
For reciprocal network
Z   Z t
Thus it can be shown that
S   S  t
Since [Z] is symmetry
Example
A certain two-port network is measured and the following scattering
matrix is obtained:
 0.10o 0.890o 
S   
o
o 
0
.
8

90
0
.
2

0


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 
2 + |S |2 = (0.1)2 + (0.8)2 = 0.65
*
|S
|
11
12
 SkiSkj  0 for i  j 

k 1
Since the summation is not equal to 1, thus
it is not a lossless network.
continue
Reflected power at port 1 when port 2 is shorted can be calculated as follow
and the fact that a2= -b2 for port 2 being short circuited, thus
b1=S11a1 + S12a2 = S11a1 - S12b2
(1)
b2=S21a1 + S22a2 = S21a1 - S22b2
(2)
From (2) we have
S 21
b2 
a1
1  S 22
Short at port 2
(3)
Dividing (1) by a1 and substitute the result in (3) ,we have

b1
b2
S12 S 21
j 0.8 j 0.8
   S11  S12  S11 
 0.1 
 0.633
a1
a1
1  S 22
1  0.2
Return loss
 20 log   20 log 0.633  3.97 dB
a2
-a2=b2
ABCD parameters
I1
V1
I2
Network
V2
Voltages and currents in a general circuit
I 2  V2  V1
V2  I1  I 2
This can be written as
V1  V2  I 2
I1  V2  I 2
Or
V1  AV2  BI 2
I1  CV2  DI 2
A –ve sign is included in the definition of D
In matrix form
V1   A B   V2 
 I   C D  I 
  2
 1 
Given V1 and I1, V2
and I2 can be
determined if ABDC
matrix is known.
Cascaded network
I1a
V1a
I2a
a
V1a   Aa
 I   C
 1a   a
I1b
I2b
b
V2a V1b
V1b   Ab
 I   C
 1b   b
Ba   V2a 
Da   I 2a 
V2b
Bb   V2b 
Db   I 2b 
However V2a=V1b and –I2a=I1b then
V1a   Aa
 I   C
 1a   a
Ba   Ab
Da  Cb
Bb   V2b 
Db   I 2b 
Or just convert to one matrix
V1a   A B   V2b 
 I   C D  I 
  2b 
 1a  
The main use of ABCD matrices are
for chaining circuit elements together
Where
 A B   Aa
C D  C

  a
Ba   Ab
Da  Cb
Bb 
Db 
Determination of ABCD parameters
V1  AV2  BI 2
I1  CV2  DI 2
Because A is independent of B, to determine A put I2 equal to zero and determine
the voltage gain V1/V2=A of the circuit. In this case port 2 must be open circuit.
V
A 1
V2
I
C 1
V2
for port 2 open circuit
B
for port 2 short circuit
2 0
I 2 0
for port 2 open circuit
I 2 0
V1
 I2 V
D
I1
 I2 V
2 0
for port 2 short circuit
ABCD matrix for series impedance
I1
I2
Z
V2
V1
A
V1
V2
for port 2 open circuit
B
I 2 0
for port 2 short circuit
2 0
V1= - I2 Z hence B= Z
V1= V2 hence A=1
I
C 1
V2
V1
 I2 V
for port 2 open circuit
D
I 2 0
I1 = - I2 = 0 hence C= 0
The full ABCD matrix can be written
I1
 I2 V
for port 2 short circuit
2 0
I1 = - I2 hence D= 1
1 Z 
0 1 


ABCD for T impedance network
I1
Z1
Z3
V1
A
V1
V2
then
Z2
I2
V2
for port 2 open circuit
therefore
I 2 0
V2 
Z3
V1
Z1  Z 3
V1 Z1  Z 3
Z1
A

 1
V2
Z3
Z3
Continue
B
V1
 I2 V
Z1
for port 2 short circuit
I2
2 0
Solving for voltage in Z2
VZ 2
Z 2 Z3
Z 2  Z3

V
Z 2 Z3 1
Z1 
Z 2  Z3
But
VZ2   I 2 Z 2
Z3
Z2
Hence
B
V1
ZZ
 Z 2  Z1  1 2
 I2
Z3
VZ2
Continue
I
C 1
V2
I1
for port 2 open circuit
Z1
I2
I 2 0
Z3
Analysis
 I 2  I1
V2   I 2 Z3  I1Z3
Therefore
C
I1
1

V2 Z 3
V2
Continue
D
I1
 I2 V
for port 2 short circuit
2 0
I1
Hence
D
I1
Z
 1 2
 I2
Z3
I2
Z3
I1 is divided into Z2 and Z3, thus
 Z3
I2 
I1
Z 2  Z3
Z1
Z2
VZ2
Full matrix
 Z1
1  Z
2

 1
 Z 3
Z1Z 2 
Z1  Z 2 
Z3 

Z2

1

Z3
ABCD for transmission line
I1
Input V1
I2
Transmission line
z = -
Zo
g
V2
z =0
V f e j t e g z  Vb e j t eg z
For transmission line
V ( z )  V f e j t e g z  Vb e j t eg z
I ( z) 

1
V f e j t e g z  Vb e j t eg z
Zo
Zo 

Vf
If

Vb
Ib
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
V f eg   Vb e g 
Zo

(2)
At the output z = 0
V2  V (0)  V f  Vb
(3)
Now find A,B,C and D using the above 4 equations
A
V1
V2

1
I 2  I (0) 
V f  Vb
Zo
for port 2 open circuit
I 2 0
For I2 =0 Eq.( 4 ) gives Vf= Vb=Vo giving

(4)
continue
From Eq. (1) and (3) we have
A
B
g
Vo (e
V1
 I2 V
e
2Vo
g 
)
 cosh( g )
for port 2 short circuit
Note that
(e x  e  x )
cosh( x) 
2
(e x  e  x )
sinh( x) 
2
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
C
I1
V2
for port 2 open circuit
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
D
I1
 I2 V
for port 2 short circuit
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
The complete matrix is therefore
cosh(g ) Z o sinh( g )
 sinh( g )

cosh(g ) 

 Zo

When the transmission line is lossless this reduces to
 cos( k )
 sin( k )
j
 Zo
jZo sin( k )

cos( k ) 

Note that
g    jk
Where
= attenuation
k=wave propagation
constant
Lossless line
=0
cosh( jk )  cos( k)
sinh( jk )  j sin( k)
Table of ABCD network
Transmission line
Z
Series impedance
Z
Shunt impedance
cosh(g ) Z o sinh( g )
 sinh( g )

cosh(
g

)


Z
o


1 Z 
0 1 


 1 0
1 
 Z 1
Table of ABCD network
Z1
Z2
T-network
Z3
Z3
Z1
Z2
p-network
 Z1
1  Z
2

 1
 Z 3
Z3


1

Z
3


Z2


Z
Z
1
1
 
 3 1 3 
 Z1 Z 2 Z1Z 2
Z1 
n 0 
 1
0 n 
n:1
Z1Z 2 
Z3 

Z2

1

Z3
Z1  Z 2 
Ideal transformer
Short transmission line
Lossless transmission line
ABCD tline
 cos(k )
  sin( k )
j
 Zo
If  << l then cos(k ) ~ 1 and sin (k  ) ~ k 
ABCD tlineshort
jZo sin( k )

cos(k ) 

then
 1
 1
j k
 Zo
jZo k 

1 

Embedded short transmission line
Z1
ABCD embed
Solving, we have
ABCD embed
Transmission line
 1 0  1
  1 1  1
Z
j Z k
 1  o
Z1
jZo k   1 0
 1

1
1 

Z
 1 
jZo k 


1
jZo k  

Z1


 2  jZo k   j k  1  jZo k  
 Z1
Zo
Z1 
Z12
Comparison with p-network
ABCDp net
ABCD embed
Z3

1


Z2

 1  1  Z3
 Z1 Z 2 Z1Z 2

Z3 

Z3 
1
Z1 
jZo k 


1

jZ
k

o


Z1


jZ
k

jZ
k

2
k

o
  o j

1

 Z1
Zo
Z1 
Z12
It is interesting to note that if we substitute in ABCD matrix in p-network,
Z2=Z1 and Z3=jZok we see that the difference is in C element where we
have extra term i.e
k
j
Zo
So the transmission line
Zok 
k
Both are almost same if

exhibit a p-network
2
Zo
Z1
Comparison with series and shunt
Series
If Zo >> Z1 then the series impedance Z  jZo k 
This is an inductance which is given by
L
Zo 
c
Where c is a velocity of light
Shunt
If Zo << Z1 then the series impedance
Z j
This is a capacitance which is given by
C
k
Zo

Z oc
Equivalent circuits

Zo
ZoL
Zo
Zo >> Z1
L
Zo 
c
C

Z oc

Zo
Zoc
Zo << Z1
Zo
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
1
1 
1
g  cosh  A  ln  A  A2  1 


 
Conversion S to ABCD
For conversion of ABCD to S-parameter
 S11
S
 21
 Z o A  B  Z o2C  Z o D

S12 
1
2Z o  AD  BC 



S 22  Z o A  B  Z o2C  Z o D 
2Z o
 Z o A  B  Z o2C  Z o D 
For conversion of S to ABCD-parameter
Z o 1  S11 1  S 22   S12 S 21 
 1  S11 1  S 22   S12 S 21
A B
1 

C D  2S  1 1  S11 1  S 22   S12 S 21 



1

S
1

S

S
S
11
22
12 21 


21 Z
 o

Zo is a characteristic impedance of the transmission line connected to the
ABCD network, usually 50 ohm.
MathCAD functions for conversion
For conversion of ABCD to S-parameter
S ( A) 
2.Z .A1,1. A2,2  A1,2 . A2,1 
Z . A1,1  A1,2  Z .Z . A2,1  Z . A2,2

1

2.Z
 Z . A1,1  A1,2  Z .Z . A2,1  Z . A2,2 
Z . A1,1  A1,2  Z .Z . A2.1  Z . A2,2 
For conversion of S to ABCD-parameter
. 1  S 2,2   S1,2 .S 2,1
Z .1  S1,1 
. 1  S 2,2   S1,2 .S 2,1 
 1  S1,1 
1 

A( S ) 
. 1 
. 1  S 2,2   S1,2 .S 2,1  1  S1,1 
. 1  S 2,2   S1,2 .S 2,1 
2.S 2,1  .1  S1,1 
 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
(2) Draw intersection line for symmetry and apply
•short circuit for odd mode
•Open circuit for even mode
(3) Also can apply EM analysis of structure
•Tangential E field zero – odd mode
•Tangential H field zero – even mode
(4) Single excitation at one port= even mode + odd mode
Example 1
Edge coupled line
1
2
Line of
symmetry
4
3
The matrix contains the odd and even parts
 S11ev  S11od

1  S 21ev  S 21od
S  
2  S31ev  S31od

 S 41ev  S 41od
S12ev  S12od
S13ev  S13od
S 22ev  S 22od
S32ev  S32od
S 42ev  S 42od
S 23ev  S 23od
S33ev  S33od
S 43ev  S 43od
S14ev  S14od 
S 24ev  S 24od 
S34ev  S34od 

S 44ev  S 44od 
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
ABCD tline
 cos( b )
  sin( b )
j
Z

jZ sin( b )

cos( b ) 

Using conversion
 Z o A  B  Z o2C  Z o D

2Z o  AD  BC 
S  


2Z o
 Z o A  B  Z o2C  Z o D 
Z o A  B  Z o2C  Z o D 
1
continue

 Z 2  Z o2 

 j sin b 

Z
1



S  

 Z 2  Z o2 

2Z o
2Z cos b   j sin b 
 Z2  


Taking b  
p
2

2Z o


 Z 2  Z o2 

j sin b 


Z


(equivalent to quarter-wavelength transmission line)
Then
 Z 2  Z o2
S   2 2 
Z  Z o  j 2ZZ o
1
 j 2ZZ o 

Z 2  Z o2 
continue
S13
S23
Odd + even
S11 S12
S21 S22
Convert to
S31
S11 S12
S21 S22
S11 S12
S21 S22
S11 S12
S21 S22
S11 S12
S21 S22
S14
S24
S34
S33
S44
S41
2-port network
matrix
S42
S32
4-port network
matrix
S43
continue
Follow symmetrical properties
ev+ od
S11 S12
S21 S22
S13 S14
S23 S24
ev- od
ev- od
S31 S32
S41 S42
S33 S34
S43 S44
ev+ od
Assuming bev = bod = p
2
Then
S 41  S14  S32
jZo  Z ev
Z od 
 S 23  
 2
2
2

2  Z ev  Z o Z od  Z o2 
2


(
Z
Z

Z
jZo  ev od
o ) ( Z od  Z ev ) 

2
2
2  ( Z ev
 Z o2 ) ( Z od
 Z o2 ) 
For perfect isolation (I.e S41=S14=S32=S23=0 ),we choose Zev and Zod such that
Zev Zod=Zo2.
continue
ev+ od
S11 S12
S21 S22
S13 S14
S23 S24
ev- od
ev- od
S31 S32
S41 S42
S33 S34
S43 S44
ev+ od
Similarly we have
S11  S 22
2
2
 Z o2 Z od
 Z o2 
1  Z ev
 S33  S 44 
 2
2
2

2  Z ev  Z o Z od  Z o2 
2 2

Z ev
Z od  Z o4
1 


2
2
2  ( Z ev
 Z o2 )( Z od
 Z o2 ) 
Equal to zero if Zev Zod=Zo2.
continue
ev+ od
S11 S12
S21 S22
S13 S14
S23 S24
ev- od
ev- od
S31 S32
S41 S42
S33 S34
S43 S44
ev+ od
We have
S31  S13  S 24  S 42
2
2
 Z o2 Z od
 Z o2 
1  Z ev

 2
2
2

2  Z ev  Z o Z od  Z o2 
2
2
2


(
Z

Z
)
Z
ev
od
o

 2
2
2
2
 ( Z  Z )( Z  Z ) 
o
od
o 
 ev
 Z ev  Z 
od 

Z Z 
od 
 ev
if Zev Zod=Zo2.
continue
ev+ od
S11 S12
S21 S22
S13 S14
S23 S24
ev- od
ev- od
S31 S32
S41 S42
S33 S34
S43 S44
ev+ od
S 21  S12  S34
jZo  Z ev
Z od 
 S 43  
 2
2
2

2  Z ev  Z o Z od  Z o2 

1
  jZo 
Z Z
od
 ev




if Zev Zod=Zo2.
continue
This S-parameter must satisfy network characteristic:
(1) Power conservation
2
2
S11  S21
Reflected
power
2
2
 S31  S41  1
transmitted
power to
port 2
transmitted
power to
port 3
transmitted
power to
port 4
Since S11 and S41=0 , then
2
2
S21  S31  1
(2) And quadrature condition
 S11 
p
  
Arg 
2
 S 21 
continue
For 3 dB coupler
 Z ev  Z
od

Z Z
od
 ev
2

 1

2

or
 Z ev  Z
od

Z Z
od
 ev

 1

2

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
Z od
(Gap size )
also bev and bod are not pure TEM
thus not equal
A l/4 branch line coupler
Odd
90o
1
Z2
2
90o
1
Z2
2
Z1
Z1
45o
Z1
90o
Z1
90o
Symmetrical line
Even
1
90o
4
Z2
45o
90o
Z2
2
3
Z1
Z1
45o
45o
O/C
O/C
Analysis
Stub odd (short circuit)
Stub even (open circuit)
X s ,od
p 
 Z1 tan    Z1
4
X s ,ev
p 
  Z1 cot    Z1
4
The ABCD matrices for the two networks may then found :
 1 0  0
ABCD   1 1  j
 jX
 Z
 s  2
stub
jZ2   1
 1
0 
  jX s
Transmission
line
stub
 Z2
0 
Xs

1  j jZ2
   2
 Z2 X s

jZ2 

Z2 
X s 
continue
Convert to S
 Z o A  B  Z o2C  Z o D

2Z o  AD  BC 
S  


2Z o
 Z o A  B  Z o2C  Z o D 
Z o A  B  Z o2C  Z o D 
1

Z o2 Z 2
Z o2
j
 jZ 2  j
2
Z2
1
X

s

2Z o Z 2
Z o2 Z 2
Z o2 
2Z o
 jZ 2  j
j

2
Xs
Z 2 
Xs

2Z o


Z o2 Z 2
Z o2 
jZ 2  j
j

2
Z 2 
Xs
For perfect isolation we require
S11ev  S11od  S11ev  S11od  0
S11  jZ2  j
Z o2 Z 2
X s2
Z o2
j
0
Z2
or
Thus
Xs  
S11ev  S11od  0
Zo Z2
Z o2
 Z 22
 Z1
From
previous
definition
continue
Substituting into S-parameter gives us
S odd
0

Z
2
2
Z o  Z 2  jZ2  o
1
Zo 
0 
and
0
S even   2 2
Z
Z o  Z 2  jZ2  o
Therefore for full four port
1
Z
S 21  S12  S 43  S34  S 21ev  S 21od    j 2
2
Zo
S 41  S14  S32
1
Z 22
 S 23  S 21ev  S 21od    1  2
2
Zo
S11  S22  S33  S44  0
And S31  S13  S42  S24  0
1
Zo 
0 
continue
For power conservation and quadrature conditions to be met
Equal split S
S 21
Z
1
 2 
Zo
2
or
And
X s  Z1 
Zo Z2
Z o2
 Z 22
Z2 

Zo
Zo
2
Z 
Z o2   o 
 2
If Zo= 50  then Z2 = 35.4 
2
Zo
2
 Zo