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Network Analysis and Synthesis
Chapter 6
Synthesis of two port networks
5.1 Introduction
• In this chapter we will discuss 2 of the most
widely used two port synthesis methods
– Coefficient matching and
– Darlington method (insertion loss method)
5.2 Coefficient matching
• It is one of the most simple and effective
method for two port synthesis.
• In this method,
– We compare the transfer function we want to
synthesize with the transfer function of a circuit we
already know.
– If they have the same form, we try to match the
coefficients of the two functions. (Remember that
the coefficients of the second transfer function is a
function of the circuit elements.)
– Once we match the coefficients, we know the
values of the elements in the circuit.
Example 1
• The voltage transfer function of the following
circuit is
R1-source impedance
R2-load impedance
• If we are required to synthesize the following
voltage transfer function.
1
T ( s)  K 2
s  2s  1
• Comparing the two equations and assuming
equal termination and normalizing the
resistors to 1Ω, R1=R2=1Ω
1
1 1 1

  2
LC 2 C L
L 2
and
C 2
Hence, the elemental values of the normalized transfer function are
R 1  R2  1,
L 2
and
C 2
Example 2
• Consider the following network
• The voltage transfer function is
K
T (s) 
LC1C2 R p




 1

1  
1
1
1
  s
s 3  s 2 




 R2C2 R1C1   L 1  1  C1C2 R1 R2  LC1C2 R p
 C C 

2 
  1

R2
RR
K
and
Rp  1 2
R1  R2
R1  R2
• If we were to synthesize the following voltage
transfer function assuming equal termination
with 1Ω resistor.
1
s 3  2s 2  2s  1
By compairing the two equations and assuming equal terminati on
1
1
1
1
1
Rp  , K  ,
 1, 
2
2
2 LC1C2 R p
C1 C2
T ( s) 
1 1
1 
1
   
2
L  C1 C2  C1C2
Solving
C1  C2  1 f , L  2h
Pros and Cons of Coefficient matching
• Pros
– Simple
– Effective
• Cons
– When the order of the transfer function increases,
the number of simultaneous equations we have to
solve for increases.
– Doesn’t demonstrate sophisticated network
design methods.
5.3 Insertion Loss (Darlington Method)
• A low pass characteristics can be obtained by
using RC, RL or LC networks.
• Low cost, low sensitivity to component variations
and simplicity of design make LC two port
networks the most widely used filter networks.
• Here the network is assumed to be doubly
terminated. (A valid assumption in almost all
cases.)
• A powerful method for designing doubly
terminated LC two port networks is the
Darlington method.
• It is one of the most effective method of
realizing a two port network: Insertion loss
method (Darlington method).
• In Darlington method of filter design,
– The specifications of the insertion loss of the filter is
converted to the reflection coefficients (related to the
maximum power that can be delivered by the source
vs. the actual delivered power to the load) of the
filter.
– From the reflection coefficients the driving point
impedance of the terminated networks is obtained.
– Then this driving-point impedance is developed into
resistively terminated LC ladder network.
Procedures of Darlington synthesis
• The derivation of Darlington method is
complicated, hence, we will just discuss the
procedure for using the Darlington method.
• Procedure
1. From F(s) obtain the reflection coefficient p(s)
4 R1 R2
p( s)
 ( s )  ( s)  1 
F (s) 
2
q( s)
R1  R2 
4 R1 R2
1
2
R1  R2 
 (s) 
p1 ( s )
q1 ( s )
o The zeros of q1(s) are the left plane zeroes of q(s).
o The zeros of p(s) are equally distributed between the
zeros of ρ (s) and ρ (-s), with restrictions that conjugate
zeros must be together.
2. From ρ(s) determine the normalized Z(s).
Z ( s) 
1   ( s)
1   ( s)
or
Z ( s) 
1   ( s)
1   ( s)
3. Expand Z(s) into continued fraction expansion
about infinity and obtain the ladder. The two
impedances defined above lead to 2 (dual)
ladders, one terminated with R2 and the other
1/R2.
Example 3
• Synthesize the following voltage transfer function
using Insertion loss method
• Solution:
1
F (s) 
,
6
1 s
R1  R2  1
– To find the reflection coefficient
4 R1 R2
s6
 ( s )  ( s)  1 
F ( s)  6
2
s 1
R1  R2 
– The zeros of q(s) are z1=1, z2=-1, z3=-0.5+j0.866,
z4=-0.5-j0.866, z5=0.5+j0.866, z6=0.5-j0.866
– Hence, poles of ρ(s) are z2=-1, z3=-0.5+j0.866, z4=0.5-j0.866
– The zeros of p(s) are 6 multiple poles at s=0.
– Hence, the zeros of ρ(s) are 3 multiple poles at
s=0.
– Hence
3
 ( s) 
s
s  1s  0.5  j 0.866s  0.5  j 0.866
s3
 ( s)  3
s  2s 2  2s  1
– The driving point impedance is
1   ( s ) 2s 3  2s 2  2s  1
Z ( s) 

1   ( s)
2s 2  2s  1
– Using continued fraction expansion
Example 4
• Synthesize the following voltage transfer
function using Insertion loss method with
R1=R2=1Ω
1
F (s) 
1  s8
• Solution:
– To find the reflection coefficient
4 R1 R2
s8
 ( s )  ( s)  1 
F ( s)  8
2
s 1
R1  R2 
– The zeros of p(s) are 8 multiple poles at s=0,
hence the zeros of ρ(s) are 4 multiple poles at s=0.
(Refer at the end for a more detailed explanation on how to get
roots of polynomials of the form sn+a)
– The zeros of q(s) are evenly distributed on the unit
circle on the s plane.
– The angle between the two zeros is
0
360

 450
8
– Since no zero on real axis or jw plane
and because the zeros of q(s)
have to be conjugate complex,
the angle of one of the roots
from the real axis should be equal
for two conjugate roots.
• The zeros of q(s) are then
cos 22.50  j sin 22.50 , cos 67.50  j sin 67.50 , cos 67.50  j sin 67.50 , cos 22.50  j sin 22.50
cos 22.50  j sin 22.50 , cos 67.50  j sin 67.50 , cos 67.50  j sin 67.50 , cos 22.50  j sin 22.50
• The roots of ρ(s) are the left hand zeros of q(s)
 cos 67.50  j sin 67.50 , cos 22.50  j sin 22.50
 cos 67.50  j sin 67.50 , cos 22.50  j sin 22.50
Computing the cosine and sine of 22.50 and 67.50
p1, 2  0.924  j 0.383 and
p3, 4  0.383  j 0.924
• ρ(s) becomes
s4
 ( s) 
s  0.924  j 0.383s  0.924  j 0.383s  0.383  j 0.924s  0.383  j 0.924
s4
 ( s)  4
s  2.163s 3  3.414 s 2  2.613s  1
• The driving point impedance becomes:
Z ( s) 
1   ( s)
1   ( s)
2s 4  2.163s 3  3.414s 2  2.613s  1
Z ( s) 
2.163s 3  3.414s 2  2.613s  1
• Taking the continuous fraction expansion
• Hence, the network becomes
Finding roots of polynomials of the
form sn+a
• The roots will be located on the circle with
radius of a1/n on the s plane.
• Separation between them is
360 0

n
• Check if the polynomial has root at s=a1/n or
s=-a1/n or s=ja1/n or s=-ja1/n.
• Case 1: root at one of these
– Start at that root and plot
the roots with separation of θ.
• Case 2: no root at those locations
– Because the roots have to be conjugate complex,
the angle one of the roots makes with the real axis
will be equal to the negative angle of the
conjugate root.
– Hence, one of them will be
r1  cos

2
 j sin

2
– Start from this root and plot the rest
with separation angle of θ.
Example
• What are the roots of s8+1
• Separation between the roots is
3600

 450
8
• Root on one of the axis? No
• Hence one of the root is
450
450
r1  cos
 j sin
2
2
• Plot the rest starting from this one.
Example
• s6-1
• The separation angle is
3600

 600
6
• Root at one of the axis? Yes at r1=1.
• Hence, plot the rest with separation of 600
starting from r1=1.
• The roots are then
r1, 2  1, r3, 4  cos 600  j sin 600 , r5,6   cos 600  j sin 600