Network Analysis and Synthesis

Chapter 4
Synthesis of deriving point functions
(one port networks)
 Elementary Synthesis procedures
• The basic philosophy behind the synthesis of
driving-point functions is to break up a
positive real (p.r.) function Z(s) into a sum of
simpler p.r. functions Z1(s), Z2(s) . . . Zn(s).
• Then to synthesize these individual Zi(s) as
elements of the overall network whose dp
impedance is
Z ( s ) Z1 ( s )  Z 2 ( s )  ...  Z n ( s )
 Breaking up process
• One important restriction is that all Zi(s) must be
positive real.
• If we were given all the Zi(s), we could synthesize a
network whose driving point impendance is Z(s) by
simply connecting the Zi(s) in series.
• However, if we were to start from Z(s) alone, how do we
decompose Z(s) into Zi(s)?
an s n  an  1s n  1  ...  a1s  a0 P ( s )
Z ( s)  m m 1

bm s  bm  1s  ...  b1s  b0 Q( s )

• There are four important removal operations(breaking up

process)
 Removing a pole at s=∞ ( n-m=1)
• Consider an impedance function Z(s) having a simple
pole at infinity(n=m+1)

• If Z(s) has a pole at s=∞, we can write Z(s) as

Z ( s ) Ls  R( s )
Z 1 ( s )  Z 2 ( s )

• Z1(s) is an inductor.
• We know Z1(s) is positive real, is Z2(s) positive real?
 Is Z2(s) positive real?
• The poles of Z2(s) are also poles of Z(s), hence, Z2(s)
doesn’t’ have poles on the right hand side of the s
plane and no multiple poles on the jw axis.
– Satisfies the first 2 properties of p.r. functions.
• What about Re(Z2(jw))?
ReZ ( jw)  ReZ1 ( jw)  Z 2 ( jw)  ReZ1 ( jw)  ReZ 2 ( jw) 
ReZ 2 ( jw) 

• Since Z(s) is p.r. Re(Z2(jw))=Re(Z(jw))>0.

• Hence, Z2(s) is p.r.
• Network interpretation of the removal of a pole at infinity
 Removing a pole at s=0( b 0 0 )
an s n  an  1s n  1  ...  a1s  a0 P ( s )
Z ( s)  m m 1

bm s  bm  1s  ...  b1s  b0 Q( s )

• If there is a pole at s=0, we can write Q(s) as

Q( s ) sG ( s )

• Hence, Z(s) becomes

D
Z ( s)   R( s)
s
Z 1 ( s )  Z 2 ( s )

• Z1(s) is a capacitor.
• We know Z1(s) is positive real, is Z2(s) positive
real?
 Is Z2(s) positive real?
• The poles of Z2(s) are also poles of Z(s), hence,
Z2(s) doesn’t’ have poles on the right hand side of
the s plane and no multiple poles on the jw axis.
– Satisfies the first 2 properties of p.r. functions.
• What about Re(Z2(jw))?
ReZ ( jw)  ReZ1 ( jw)  Z 2 ( jw)  ReZ1 ( jw)  ReZ 2 ( jw) 
ReZ 2 ( jw) 

• Since Z(s) is p.r. Re(Z2(jw))=Re(Z(jw))>0.

• Hence, Z2(s) is p.r.
 Removing complex conjugate poles on the jw axis
• If Z(s) has complex conjugate poles on the jw
axis, Z(s) can be expanded into
2 Ks
Z ( s)  2 2
 Z 2 ( s)
s  1
• Note that  2 Kjw 
Re 2  0
2 
 s  1 
• Hence, ReZ ( s )  Re( Z ( s ))  0
2

• Z2(s) is p.r.
• Thus, removal of conjugate imaginary poles from
impedance function Z(s) corresponds to the
removal of the parallel combination of L − C and
from admittance function Y(s) corresponds to
removal of series combination of L − C
 Removing a constant K
• If Re(Z(jw)) is minimum at some point wi and if
Re(Z(jw)) = Ki as shown in the figure

• We can remove a constant k Ki as
Z ( s ) K i  Z 2 ( s )

• Z2(s) is p.r.
• This is essentially removing
a resistor.
 Constructing
• Assume that using one of the removal
processes discussed we expanded Z(s) into
Z1(s) and Z2(s).
• We connect Z1(s) and Z2(s) in series as shown
on the figure.
 Example 1
• Synthesize the following p.r. function
s 2  2s  6
Z (s) 
s ( s  3)
• Solution:
– Note that we have a pole at s=0. Lets remove it
A Bs  C
Z (s)  
s s 3
A 2, B 1, C 0
2 s
Z (s)  
s s 3
– Note that 2/s is a capacitor, while s/(s+3) is a
parallel connection of a resistor and an inductor.
 Construct the n/w
• 2/s is a capacitor with C=1/2.
• While s/(s+3) is a R=1 connected in parallel
with an inductor L=1/3.
 Example 2
• Synthesis the following p.r. function
7s  2
Y ( s) 
2s  4
• Solution
– Note that there are no poles on s=0 or s=∞ or jw
axis.
– Lets find the minimum of Re(Y(jw))
 7 jw  2   2  j 7 w4  j 4 w
ReY ( jw)  Re  Re 2 
 4 jw  4   16  16 w 
8  28w2

16  16w2
4  14 w2

8  8w 2
 Ex 2…
• Note that minimum of Re(Y(jw))=1/2.
• Lets remove it
1 3s
Y (s)  
2 s2
3s
• ½ is a conductance in parallel with Y2(s)= s2
• Note that Y2(s) is a conductance 1/3 in series
with an inductor 3/2.
 Ex…….
• Synthesize the following p.r. function.
6 s 3  3s 2  3s  1
Z (s) 
6 s 3  3s
• The real part of the function is constant, equal to unity.
Removing a constant of 1Ω,

• The reciprocal of is an admittance

• has a pole at s=∞ and the pole is removed by finding partial

expansion of
• Removing a term with the pole at s=∞ give a capacitor
of 2 farads in parallel with .
• is now obtained as:
=

• The reciprocal of is:

• = 3s + which is clearly an inductor of 3H in series with
capacitor of 1F

• The final network is:

 Synthesis of one port networks with two
kinds of elements
• In this section we will focus on the synthesis of
networks with only L-C, R-C or R-L elements.
• The deriving point impedance/admittance of
these kinds of networks have special
properties that makes them easy to
synthesize.
 1. L-C immittance functions
• These networks have only inductors and
capacitors.
• Hence, the average power consumed in these
kind of networks is zero. (Because an inductor
and a capacitor don’t dissipate energy.)
• If we have an L-C deriving point impedance
Z(s)
M 1 ( s )  N1 ( s ) M1 and M2 even parts
Z ( s) 
M 2 ( s)  N 2 ( s) N1 and N2 odd parts
 L-C immittance functions…
• The average power dissipated by the network
is 1
Average Power  ReZ ( jw)  I ( jw) 0
2

2
 ReZ ( jw)  0
M 1 ( s ) M 2 ( s )  N1 ( s ) N 2 ( s )
 2 2
M 2 ( s)  N 2 ( s)
 M 1 ( s ) M 2 ( s )  N1 ( s ) N 2 ( s ) 0
 M 1 ( s ) 0  N 2 ( s ) or M 2 ( s ) 0  N1 ( s )
N1 ( s ) M 1 (s)
 Z (s)  or Z (s) 
M 2 (s) N 2 ( s)
even odd
Z ( s)  or Z (s) 
odd even
 Properties of L-C function
1. The driving point impedance/admittance of an
L-C network is even/odd or odd/even.
2. Both are Hurwitz, hence only simple imaginary
zeros and poles on the jω axis.
3. Poles and zeros interlace on the jω axis.
4. Highest power of the numerator and
denominator may only differ by 1; the lowest
powers also differ by unity.
5. Either a zero or a pole at origin and infinity.
Example
 Synthesis of L-C networks
• There are two kinds of network realization
types for two element only networks.
– Foster and
– Cauer
• We can write a general L-C impedance or
admittance as
 Foster synthesis
• Uses decomposition of the given F(s) into simpler two or
more element impedances/admittances.
• The partial fraction expansion of LC function is
expressed in general terms as:
K0 2K s
F (s)   K  s  2 i 2  ...
s s  i
• K - must be positive and real
i

– The synthesis is accomplished directly from the partial fraction

expansion by associating the individual terms in the
expression with network elements
 Foster synthesis
• The partial fraction expansion method is based up
on the elementary synthesis procedures of
removing poles on jω-axis.
• This is because F(s) has poles on the jω axis only.
• Networks synthesized by partial fraction expansions
are sometimes called Foster-type networks.
• The impedance form is sometimes called Foster
Series network(Foster I realization) and the
admittance form is a Foster parallel network(Foster
II realization)
• Using the above decomposition, we can
realize F(s) as

For a driving point

impedance (Foster I)

For a driving point

admittance(Foster II)

• If Z(s) has no pole at the origin then capacitor is not present in the network.
Similarly, if there is no pole at ∞, inductor is not present in the network.
• If Y (s) has no pole at the origin then inductor is not present. Similarly, if there is
no pole at infinity, capacitor is not present.
 Example
F ( s) 
 
2 s 2 1 s 2  9 
• Synthesize   as driving point
s s2  4
impedance and admittance using foster
synthesis method
Solution:
– Decompose F(s) into simpler forms
K0 2 K1 s
F (s)   K  s  2
s s 4
9 15
K  2, K 0  , K1 
2 4
• For driving point impedance(Foster I)
K0 2 K1 s
F (s)   K  s  2
s s 4
9 15
K  2, K 0  , K1 
2 4

• Do for driving point admittance(Foster II)

• Synthesize driving point admittance using
foster methods (Foster II)
 Z(s) has a pole at the origin then capacitor will present
in the network. Similarly, there is pole at ∞, thus
inductor will present in the network.
Y(s) has no pole at the origin then inductor will not
present in the network. Similarly, there is no pole at ∞,
thus capacitor will not present in the network.
 Cauer synthesis
• Uses continued fraction expansion method.
• It is based on removing pole at s=∞ or s=0
N1 ( s ) M 1 ( s)
Z ( s)  or Z ( s) 
M 2 (s) N 2 (s)
• Since the degree of the numerator and
denominator differ by only 1, there is either a
pole at s=∞ or a zero at s=∞.
– If a pole at s=∞, then we remove it.
– If a zero at s=∞, first we invert it and remove the
pole at s=∞.
• Case 1: pole at s=∞(n=m+1)
– In this case, F(s) can be written as
N 3 (s)
F ( s ) K  s  ,
M 2 (s)
Order of M 2 ( s ) Order of N 3 ( s )  1
Hence,
1
F ( s ) K  s 
M 2 ( s)
N 3 (s)
1
K  s 
1
K1 s 
K 2 s  ...
• If F(s) is impedance function, this expansion
can easily be realized as

– series arms are inductors

– Shunt arms are capacitors
• Case 2: zero at s=∞(n=m-1)
1
– In this case G ( s)  will have a pole at s=∞.
F ( s)

– We synthesize G(s) using the procedure in the

previous step.
– Remember that if F(s) is an impedance function,
G(s) will be an admittance function and vice versa.
– Since we always remove a pole at s=∞ by
inverting remainder and dividing, we conclude
that we can synthesize an L-C ladder network by a
continued fraction expansion. The quotients
represent the poles at s=∞, which we remove, and
we invert the remainder successively until the
remainder is zero.
• If the initial function is an impedance, the first
quotient must necessarily be an impedance.

• When the first function is an admittance, the

first the first quotient is admittance
• Example: Using Cauer realization(Cauer I)
synthesize 5 3
2 s  12 s  16 s
Z ( s) 
s 4  4s 2  3

Solution:
– This is an impedance function.
– We have a pole at s=∞, hence, we should remove it.
 3
4s 10s
Z (s) 2s  4
s  4s
We see that the quotients of CFE give the elements of
ladder network. Because the CFE always inverts each
remainder and divides, the successive quotients
alternate between Z and Y and Z again
 Cauer I realization of Z(s)

• If we were given Y(s) instead our realization

would be?
 Cauer II (Alternate realization) :
Removing a Pole or a zero @ s=0
• Since the lowest degree of numerator and
denominator of an L-C immittance must differ
by unity, it follows that there must be a zero or
a pole at s=0.
– Arrange both numerator and denominator in
ascending order
– Divide the lowest power of the denominator in to
the lowest power of the numerator,
– Then invert the remainder and divide again
Synthesize :
 R-C driving point impedance/ R-L
admittance
• R-C impedance and R-L admittance driving point
functions have the same properties.
• By replacing the inductor in LC by a resistor an R-C
driving point impedance or R-L driving point admittance,
it can be written as
K 0 sF ( s ) s 0
K0 Ki
F (s)   K    ... K  lim F ( s)
s  i
s 
s
K i s   i  F ( s ) s  
• Where i

1 1
, ,... Capaictors for R - C impedance and inductor for R - L admittance
K0 Ki
Ki
K, ,... Represent resistors
i
 Properties of R-C impedance or R-L
admittance functions
1. Poles and zeros lie on the negative real axis.
2. The singularity nearest to(or at) origin must
be a pole and a zero near(or at) infinity.
3. The residues of the poles must be positive
and real.
4. Poles and zeros must alternate on the
negative real axis.
Ex:
Synthesis of R-C impedance or R-L admittance
• Foster
– In foster realization we decompose the function into
simple imittances according to the poles. That is we
write F(s) as
K0 K1
F (s)   K    ...
s s  i
 K0 K1
F (s)   K    ...
s s  i

• For R-L admittance

• For R-L admittance

 Example
3( s  2)( s  4)
F (s) 
• Synthesize as R-C impedance
s ( s  3)
and R-L admittance in foster realization.
Solution:
– Note that the singularity near origin is a pole and a
zero near infinity.
– The poles and zeros alternate
– We can expand F(s) as F ( s) 8  1  3
s s 3
– R-C impedance
• R-L admittance
• Cauer realization
– Cauer realization uses continued fraction expansion.
– For R-C impedance and R-L admittance we remove a
resistor first.
– Then invert and remove a capacitor
– Then invert and remove a resistor . . .
 Example
3( s  2)( s  4)
F ( s) 
• Synthesize using Cauer I realization
s ( s  3)
as R-C impedance and R-L admittance.
Solution:
– Note that the singularity near origin is a pole.
– The singularity near infinity is a zero.
– The zeros and the poles alternate.
F(s) is R-L impedance
or R-C admittance
– Note that the power of the numerator and denominator
is equal, hence, we remove the resistor first.
For R-C impedance For R-L admittance
 R-L impedance/R-C admittance
• R-L impedance deriving point function and R-C
admittance deriving point function have the
same property.
• If F(s) is R-L impedance or R-C admittance, it
can be written as
Ki s
F ( s ) K  s  K 0   ...
s  i
1 1
, ,... Inductors for R - L impedance and Capacitors for R - C admittance
K Ki
Ki
K0 , ,... Represent resistors
i
 Properties of R-L impedance/R-C
admittance
1. Poles and zeros are located on the negative
real axis and they alternate.
2. The nearest singularity near origin(or at) is
zero. The singularity near infinity(or at) is a
pole.
3. The residues of the poles must be real and
negative.
• Because the residues are negative, we can’t use
standard decomposition method to synthesize.
EX
 Synthesis of R-L impedance and R-C
admittance
• Foster
– If F(s) is R-L impedance d.p or R-C admittance d.p
function. We can write it as
Ki s
F ( s ) K  s  K 0   ...
s  i
– Because of the third property of R-L impedance/R-
C admittance d.p. functions, we can’t decompose
F(s) into synthesizable components with the way
we were using till now.
– We have to find a new way where the residues
wont be negative.
 Foster…
• If we divide F(s) by s, we get
F (s) K 0 Ki
  K   ...
s s s  i
• Note that this is a standard R-C impedance d.p.
function, hence, the residues of the poles of
F(s)/s will be positive.
• Once we find Ki and σi we multiply by s and
draw the foster realization.
 Example
2( s  1)( s  3)
F ( s) 
• Synthesize as R-L
( s  2)( s  6)
impedance and R-C admittance using Foster
realization.
Solution:
– Note that the singularity near origin is a zero.
– The singularity near infinity is a pole.
– The zeros and the poles alternate.
F(s) is R-L impedance
or R-C admittance
• We divide F(s) by s.
F ( s) ( s  1)( s  3)

s s( s  2)( s  6)
1 1 5
 2 4  4
s s 2 s 6

Then multiplyin g by s
1 s 5 s
F ( s)  1  4  4
2 s 2 s 6
• R-L impedance

• R-C admittance
/.
• Cauer realization
– Using continued fractional expansion
– We first remove R0. To do this we use fractional
expansion method by focusing on removing the
lowest term first.
– We write N(s) and M(s) starting with the lowest
term first.
 Example
2( s  1)( s  3)
F ( s) 
• Synthesize as R-L
( s  2)( s  6)
impedance and R-C admittance using Cauer
realization.
• Solution: P( s)
F (s) 
Q( s)
– We write P(s) and M(s) as
P ( s ) 6  8s  2 s 2
Q( s ) 12  8s  s 2
Cauer realization…
Synthesis of certain R-L-C functions
• Under certain conditions, R-L-C driving point
functions may be synthesized with the use of
either partial fractions or continued fractions.
• For Example below function is neither L-C, R-C,
nor R-L.
Z

– The function can be synthesized by continued

fraction expansion