Network Parameters

1
Network Parameters
2
Impedance and Admittance matrices
For n ports network we can relate the voltages
and currents by impedance and admittance matrices
Impedance matrix
Admittance matrix
where
3
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.
Lossless networks can be shown that Zij or Yij
are imaginary
Refer to text book Pozar pg193-195
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Example
Find the Z parameters of the two-port T network
as shown below
I1
I2
V1
V2
Solution
Similarly we can show that
Port 2 open-circuited
This is an example of reciprocal network!!
Port 1 open-circuited
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S-parameters
Microwave device
Port 1
Port 2
Vi2
Vi1
Input signal
Vr2
Vr1
reflected signal
Vt1
transmitted signal
Vt2
Transmission and reflection coefficients
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S-parameters
Voltage of traveling wave away from port 1 is
Voltage of Reflected wave From port 1
Voltage of Transmitted wave From port 2
Voltage of transmitted wave away from port 2 is
Let Vb1 b1 , Vi1a1 , Vi2a2 ,
Then we can rewrite
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S-parameters

• 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.

Hence
In matrix form
Logarithmic form S1120 log(r1) S2220
log(r2) S1220 log(t12) S2120 log(t21)
S-matrix
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S-parameters
Vr20 means port 2 is matched
Vr10 means port 1 is matched
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Multi-port network
Port 5
network
Port 1
Port 4
Port 2
Port 3
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Example
Below is a matched 3 dB attenuator. Find the
S-parameter of the circuit.
Z1Z2 8.56 W and Z3 141.8 W
Solution
By assuming the output port is terminated by Zo
50 W, then
Because of symmetry , then S220
11
Continue
V1
V2
Vo
From the fact that S11S220 , we know that Vr10
when port 2 is matched, and that Vi20. Therefore
Vi1 V1 and Vt2V2
Therefore S12 S21 0.707
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Lossless network
For lossless n-network , total input power
total output power. Thus
Where a and b are the amplitude of the signal
Putting in matrix form at a bt
b at St S a
Note that btatSt and bSa
Called unitary matrix
Thus at (I St S )a 0
This implies that St S I
In summation form
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Conversion of Z to S and S to Z
where
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Reciprocal and symmetrical network
Since the U is diagonal , thus
For reciprocal network
Since Z is symmetry
Thus it can be shown that
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Example
A certain two-port network is measured and the
following scattering matrix is obtained 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 ij
S112 S122 (0.1)2 (0.8)2 0.65
Since the summation is not equal to 1, thus it
is not a lossless network.
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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
(1)
b1S11a1 S12a2 S11a1 - S12b2
Short at port 2
a2
(2)
b2S21a1 S22a2 S21a1 - S22b2
-a2b2
From (2) we have
(3)
Dividing (1) by a1 and substitute the result in
(3) ,we have
Return loss
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ABCD parameters
I1
I2
Network
V2
V1
Voltages and currents in a general circuit
In matrix form Given V1 and I1, V2 and I2 can
be determined if ABDC matrix is known.
This can be written as
Or
A ve sign is included in the definition of D
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Cascaded network
I1a
I2a
I1b
I2b
b
a
V1a
V2a
V1b
V2b
However V2aV1b and I2aI1b then
The main use of ABCD matrices are for chaining
circuit elements together
Or just convert to one matrix
Where
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Determination of ABCD parameters
Because A is independent of B, to determine A put
I2 equal to zero and determine the voltage gain
V1/V2A of the circuit. In this case port 2 must
be open circuit.
for port 2 open circuit
for port 2 short circuit
for port 2 short circuit
for port 2 open circuit
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ABCD matrix for series impedance
I2
I1
Z
V2
V1
for port 2 open circuit
for port 2 short circuit
V1 - I2 Z hence B Z
V1 V2 hence A1
for port 2 short circuit
for port 2 open circuit
I1 - I2 0 hence C 0
I1 - I2 hence D 1
The full ABCD matrix can be written
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ABCD for T impedance network
I1
I2
Z1
Z2
Z3
V1
V2
for port 2 open circuit
therefore
then
22
Continue
Z1
for port 2 short circuit
I2
VZ2
Z2
Solving for voltage in Z2
Z3
Hence
But
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Continue
Z1
I1
I2
for port 2 open circuit
V2
Z3
Analysis
Therefore
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Continue
for port 2 short circuit
Z1
I2
I1
VZ2
Z2
Z3
I1 is divided into Z2 and Z3, thus
Full matrix
Hence
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ABCD for transmission line
I1
I2
V1
V2
Input
Zo
g
Transmission line
z 0
z -
For transmission line
f and b represent forward and backward
propagation voltage and current Amplitudes. The
time varying term can be dropped in further
analysis.
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continue
At the input z -
(2)
(1)
At the output z 0
(4)
(3)
Now find A,B,C and D using the above 4 equations
for port 2 open circuit
For I2 0 Eq.( 4 ) gives Vf VbVo giving
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continue
Note that
From Eq. (1) and (3) we have
for port 2 short circuit
For V2 0 , Eq. (3) implies Vf Vb Vo . From
Eq. (1) and (4) we have
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continue
for port 2 open circuit
For I20 , Eq. (4) implies Vf Vb Vo . From
Eq.(2) and (3) we have
for port 2 short circuit
For V20 , Eq. (3) implies Vf -Vb Vo . From
Eq.(2) and (4) we have
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continue
Note that
The complete matrix is therefore
Where a attenuation kwave propagation
constant
When the transmission line is lossless this
reduces to
Lossless line a 0
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Table of ABCD network
Transmission line
Z
Series impedance
Shunt impedance
Z
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Table of ABCD network
Z1
Z2
T-network
Z3
Z3
Z1
Z2
p-network
Ideal transformer
n1
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Short transmission line
Lossless transmission line
If ltlt l then cos(k ) 1 and sin (k
) k then
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Embedded short transmission line
Z1
Z1
Transmission line
Solving, we have
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Comparison with p-network
It is interesting to note that if we substitute
in ABCD matrix in p-network, Z2Z1 and Z3jZok
we see that the difference is in C element where
we have extra term i.e
So the transmission line exhibit a p-network
Both are almost same if
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Comparison with series and shunt
Series
If Zo gtgt Z1 then the series impedance
This is an inductance which is given by
Where c is a velocity of light
Shunt
If Zo ltlt Z1 then the series impedance
This is a capacitance which is given by
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Equivalent circuits
ZoL
Zo
Zo
Zo gtgt Z1
Zoc
Zo
Zo
Zo ltlt Z1
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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
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Conversion S to ABCD
For conversion of ABCD to S-parameter
For conversion of S to ABCD-parameter
Zo is a characteristic impedance of the
transmission line connected to the ABCD network,
usually 50 ohm.
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MathCAD functions for conversion
For conversion of ABCD to S-parameter
For conversion of S to ABCD-parameter
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Odd and Even Mode Analysis
Usually use for analyzing a symmetrical four port
network

• Equal ,in phase excitation even mode
• Equal ,out of phase excitation odd mode

(1) Excitation

• (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
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Example 1
Edge coupled line
The matrix contains the odd and even parts
Since the network is symmetry, Instead of 4 ports
, we can only analyze 2 port
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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
Using conversion
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continue
Taking
(equivalent to quarter-wavelength transmission
line)
Then
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continue
S13
S14
S23
S24
Odd even
Convert to
S34
S33
S31
S44
S41
S42
S32
S43
4-port network matrix
2-port network matrix
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continue
Follow symmetrical properties
ev- od
ev od
ev- od
ev od
Assuming bev bod Then
For perfect isolation (I.e S41S14S32S230 ),we
choose Zev and Zod such that Zev ZodZo2.
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continue
ev- od
ev od
ev- od
ev od
Similarly we have
Equal to zero if Zev ZodZo2.
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continue
ev- od
ev od
ev- od
ev od
We have
if Zev ZodZo2.
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continue
ev- od
ev od
ev- od
ev od
if Zev ZodZo2.
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continue
This S-parameter must satisfy network
characteristic
(1) Power conservation
Reflected power
transmitted power to port 2
transmitted power to port 4
transmitted power to port 3
Since S11 and S410 , then
(2) And quadrature condition
50
continue
For 3 dB coupler
or
Rewrite we have
In practice Zev gt Zod so
However the limitation for coupled edge
bev and bod are not pure TEM thus not equal
(Gap size ) also
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A l/4 branch line coupler
Odd
Symmetrical line
Even
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Analysis
Stub odd (short circuit)
Stub even (open circuit)
The ABCD matrices for the two networks may then
found
Transmission line
stub
stub
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continue
Convert to S
For perfect isolation we require
Thus
or
From previous definition
54
continue
Substituting into S-parameter gives us
and
Therefore for full four port
And
55
continue
For power conservation and quadrature conditions
to be met
Equal split S
or
And
If Zo 50 W then Z2 35.4 W