Chapter 5

Synthesis of 1-port
(Driving Point)
Networks
Outline
5.1. Elementary synthesis procedures

5.2. Driving point (dp) function synthesis using partial fraction method

5.3. Driving point (dp) function synthesis using continued fraction method

Reference: F.F. Kuo, "Network Analysis and Synthesis" Chapter 10-11

V.K. Aatre, "Network Theory and Filter Design“: Chapter 9

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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)
• One important restriction during breaking up 𝑍(𝑠) is that all 𝑍𝑖 𝑠 must be
positive real.
• If all the 𝑍𝑖 𝑠 were given , one can synthesize a network whose driving
point impendence 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 ?
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Removing a pole at 𝒔 = 𝟎
• Suppose 𝑍(𝑠) is given as an s n + an −1s n −1 + ... + a1s + a0 N ( s )
Z (s) = m −1
=
bm s + bm −1s + ... + b1s + b0 D ( s )
m

• If there is a pole at 𝑠 = 0 (i.e. 𝑏0 = 0) , we can write 𝐷(𝑠) as

D( s ) = sG ( s )
• Dividing 𝑁(𝑠) by 𝐷(𝑠) , 𝑍(𝑠) becomes
D
Z ( s) = + R( s) D  0
s
• 𝑍1 𝑠 is a capacitor. = Z1 ( s ) + Z 2 ( s )
• We know 𝑍1 𝑠 is positive real, is 𝑍2 𝑠 positive real?
i. 𝑍2 𝑠 must have no poles on the RHS of s-plane
ii. 𝑍2 𝑠 may have only simple poles on the 𝑗𝜔 axis with real and positive residues
iii. 𝑅𝑒 𝑍2 𝑗𝜔 ≥ 0 for all 𝜔.
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Removing a pole at 𝒔 = 𝟎 …
• Examining the three criteria
✓Poles of 𝑍2 𝑠 are also poles of 𝑍 𝑠
✓Using the same argument, a simple partial fraction expansion doesn't affect the
residues of the other poles
✓When 𝑠 = 𝑗𝜔
D
Re Z ( j ) = Re Z1 ( j ) + Z 2 ( j ) = Re Z1 ( j ) + Re Z 2 ( j ) = Re   + Re Z 2 ( j )
 j 
Re Z ( j ) = Re Z 2 ( j )  0

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Removing a pole at 𝒔 = ∞
• If 𝑍(𝑠) has a pole at 𝑠 = ∞, (𝑛 − 𝑚 = 1), we can write 𝑍(𝑠) as
Z ( s ) = Ls + R ( s )
= Z1 ( s ) + Z 2 ( s )
• Using a similar argument as previous case we can show that 𝑍2 𝑠 is p.r.
• 𝑍1 𝑠 is an inductor.

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Removing complex conjugate poles on the 𝒋𝝎 axis.
• If 𝑍(𝑠) has complex conjugate poles (𝑠 = ±𝑗𝜔1 ) on the 𝑗𝜔 axis, 𝑍(𝑠) can be
expanded into 2 Ks
Z ( s) = + Z 2 ( s)
s + 1
2 2

 2 Ks   j 2 K 
• Note: Re  2 2
= Re  2 2
=0
 s + 1  s = j  − + 1 

• Hence, ReZ 2 ( s ) = ReZ ( s )  0

• 𝑍2 𝑠 is p.r.

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Removing a constant 𝑲
• If Re(𝑍(𝑗𝜔)) is minimum at some point 𝜔𝑖 and if Re 𝑍 𝑗𝜔𝑖 = 𝐾𝑖 as shown
in the figure
• We can remove that a costant 𝐾 ≤ 𝐾𝑖 from Re 𝑍 𝑗𝜔
Z (s) = Ki + Z 2 (s)
so that the remainder 𝑍2 𝑠 is p.r.
• This is essentially removing a resistor.

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Synthesis
• 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.

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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) = +
• 2/𝑠 is a capacitor s s+3
• 𝑠/(𝑠 + 3) is a parallel connection of a resistor and an inductor.

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Example 1…
• 2/𝑠 is a capacitor with 𝐶 = 1/2.
• 𝑠/(𝑠 + 3) is a resistor 𝑅 = 1 connected in parallel with an inductor 𝐿 = 1/3.

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Example 2
• Synthesis the following p.r. function
7s + 2
Y (s) =
• Solution: 2s + 4
• Note that there are no poles on 𝑠 = 0 or 𝑠 = ∞ or 𝑗𝜔 axis.
• Lets find the minimum of Re{𝑌 𝑗𝜔 }

 7 j + 2   (2 + j 7 )(4 − j 2 ) 
Re(Y ( j ) ) = Re  = Re 
 2 j + 4   16 + 4 
2

8 + 14 2
=
16 + 4 2

• The minimum of Re 𝑌 𝑗𝜔 = 1/2 at 𝜔 = 0.

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Example 2…
• Removing the constant
1
Y ( s ) = Y1 ( s ) + Y2 ( s ) = + Y2 ( s )
2
7s + 2 1 3s
Y2 ( s ) = Y ( s ) − Y1 ( s ) = − =
2s + 4 2 s + 2
1 3s
Y (s) = +
2 s+2 3s
• 𝑌(𝑠) is a ½ conductance in parallel with Y2 ( s ) =
s+2

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Exercise
• Synthesize the following p.r. function.

6s + 3s + 3s + 1
3 2
Z ( s) =
6s + 3s
3

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Synthesis of one port networks with two kinds of elements
• In this section we will focus on the synthesis of networks with only two of
R,L,C elements.
• Thus the networks can be either L-C, R-C or R-L.
• The driving point impedance/admittance of these kinds of networks have
special properties that makes them easy to synthesize.
• First study the properties of the particular type of network then synthesize it.

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L-C imittance 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.)
• Consider an impedance 𝑍(𝑠) of an L-C driving point impedance Z(s)

M 1 ( s ) + N1 ( s )
Z (s) =
M 2 (s) + N 2 (s)
M1 and M2 even parts
N1 and N2 odd parts

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L-C imittance functions …
• The average power dissipated by the network is
1
Average Power = Re ( Z ( j ) ) I ( j ) = 0
2

2
 Re ( Z ( j ) ) = 0
M 1 ( s ) M 2 ( s ) − N1 ( s ) N 2 ( s )
=
M 22 (s) − N 22 (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)
odd even
Z (s) = or Z ( s) =
even odd
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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 (with
positive residues) 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.
5. Either a zero or a pole at origin or infinity.
6. 𝑅𝑒 𝐹 𝑗𝜔 = 0 for all 𝜔.
7. The slope of 𝑋 𝜔 is strictly increasing.
• There are two kinds of network realization types for two element networks.
• Foster and Cauer
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Examples
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.
5. Either a zero or a pole at origin or
infinity.

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Foster synthesis
• Uses decomposition of the given 𝐹(𝑠) into simpler two element
impedances/admittances.
• For an L-C network with system function 𝐹(𝑠), it can be written as
• Partial fraction expansion
K 0 = sF ( s ) s =0
K0 2 Ki s
F (s) = + K s + 2 + ... F (s)
s s + i 2
K =
s s →
• This is because 𝐹(𝑠) has poles on the 𝑗𝜔 axis only.
Ki =
( s 2
+ i )
2

F (s)
2s
s =−i 2

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Foster synthesis…
•c

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Foster synthesis…
• Using the above partial fraction decomposition, we can realize 𝐹(𝑠) as

For a driving point

impedance
Foster I
(Foster Series)

For a driving point

admittance
Foster II
(Foster Parallel)

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Example 3
• Synthesize 𝐹 𝑠 as driving point impedance.

F ( s) =
(
2 s2 +1 s2 + 9)( )
Solution: (
s s2 + 4 )
• 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

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Example 3…
• For driving point impedance
s 9 15
Z ( s) = 2s + + 2 2 2

s s +4

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Exercise
• Synthesize 𝐹 𝑠 as driving point admittance.

s (s + 2 )(s + 4 )
2 2

F (s) = 2
(s + 1)(s + 3)
2

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Cauer synthesis
• Uses continued fraction expansion method.
• It is based on removing pole at 𝑠 = ∞ (Cauer I) or 𝑠 = 0 (Cauer II).

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 inverse it and remove the pole at 𝑠 = ∞.

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Cauer I synthesis…
Case 1: pole at 𝑠 = ∞
• In this case, 𝐹(𝑠) can be written as
N3 (s)
F (s) = K  s + ,
M 2 (s)
Order of M 2 ( s ) = Order of N 3 ( s ) + 1 → Zero at infinity
M 2 (s)
Thus has a pole at infinity.
N3 (s)
Hence,
1 1
F (s) = K  s + = K s +
M 2 (s) 1
K1s +
N3 (s) K 2 s + ...
• This expansion can easily be realized using a ladder network as shown
in the figure.
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Cauer I synthesis…
• Case 2: zero at 𝑠 = ∞
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.

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Example 4
• Using Cauer-I realization synthesize

2s 5 + 12s 3 + 16s
Z ( s) =
s + 4s + 3
4 2

Solution:
• This is an impedance function.
• We have a pole at 𝑠 = ∞, hence, we should remove it.
• The continued fraction expansion is:

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Example 4…

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Cauer II synthesis
• Since the lowest degrees of numerator and denominator of an L-C
admittance must differ by unity, it follows that there must be a zero or a pole
at 𝑠 = 0.
• If we follow the same procedure we have just outlined, and remove
successively poles at 𝑠 = 0, we will have an alternate realization in a ladder
structure.
• To do this by continued fractions, we arrange both numerator and
denominator in ascending order and divide the lowest power of the
denominator into the lowest power of the numerator; then we invert the
remainder and divide again.

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Example 5
• Using Cauer-II realization synthesize

Z (s) =
(s 2
+ 1)(s 2 + 3)
• Solution: s (s 2 + 2 )

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R-C driving point impedance/ R-L admittance
• R-C impedance and R-L admittance driving point functions have the same
properties.
• If 𝐹(𝑠) is and R-C driving point impedance or R-L driving point admittance,
it can be written as
K0 K1
F (s) = + K + + ...
s s +i
• Where
1 1
, ,... Capaictors for R - C impedance and inductor for R - L admittance
K0 Ki
Ki
K , ,... Represent resistors
i

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Properties: R-C impedance/ R-L admittance
1. Poles and zeros lie on the negative real axis.
2. The critical frequency nearest to origin must be a pole and a zero near
infinity.
3. The residues of the poles must be positive and real.
4. Poles and zeros must alternate on the negative real axis.
5. The slope of 𝑍𝑅𝐶 (𝜎) and 𝑌𝑅𝐿 (𝜎) is strictly decreasing.

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Synthesis: R-C impedance/ 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 Ki
F (s) = + K + + ...
s s +i

K 0 = sF ( s ) s =0
K  = lim F ( s)
s →

K i = ( s +  i ) F ( s ) s =−
i

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Synthesis …

• For R-C impedance

• For R-L admittance

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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 . . .

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Example 6
• Synthesize 𝐹(𝑠) using Foster and Cauer realization as R-C impedance and R-
L admittance.
3( s + 2)(s + 4)
F ( s) =
s( s + 2)
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-C impedance or R-L admittance

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Example 6…
• Partial fraction expansion of 𝐹(𝑠) yields

3( s + 2)( s + 4)
F (s) =
s ( s + 2)
For R-C impedance
8 1
= + +3
s s+3

For R-L admittance

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Example 6…
• 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

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Exercise
• Synthesize 𝐹(𝑠) of the previous example using Foster realization as R-C
impedance and R-L admittance.

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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

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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 critical frequency near origin is zero. The critical frequency
near infinity is a pole.
3. The residues of the poles must be real and negative.
4. Because the residues are negative, we can’t use standard decomposition
method to synthesize.
5. The slope of 𝑍𝑅𝐿 (𝜎) and 𝑌𝑅𝐶 (𝜎) is strictly increasing.

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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.
• If we divide F(s) by s, we get
F (s) K 0 Ki
= + K + + ...
s s s +i
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Synthesis…
• 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.

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Example 7
• Synthesize 𝐹(𝑠) as R-L impedance and R-C admittance using Foster
realization.
2( s + 1)(s + 3)
F ( s) =
( s + 2)(s + 6)
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

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Example 7…
• We divide 𝐹(𝑠) by 𝑠. For R-L impedance

F ( s ) s ( s + 1)( s + 3)
=
s s ( s + 2)( s + 6)
1 1 5
= 2+ 4 + 4
s s+2 s+6
For R-C admittance
Then multiplyin g by s
1 s 5 s
F (s) = 1 + 4 + 4
2 s+2 s+6

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Cauer realization
• Using continued fractional expansion
• We first remove 𝑅0 . To do this we use fractional expansion method by
focusing on removing the lowest 𝑠 term first.
• We write 𝑁(𝑠) and 𝐷(𝑠) starting with the lowest term first.

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Example 8
• Synthesize 𝐹(𝑠) as R-L impedance and R-C admittance using Cauer
realization.
2( s + 1)(s + 3)
F ( s) =
( s + 2)(s + 6)
Solution:

6 + 8s + 2 s2

F (s) =
12 + 8s + s 2

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Example 8…
• R-L impedance

• R-C admittance

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Reading Assignment
• Synthesis of RLC networks
• F.F. Kuo, "Network Analysis and Synthesis" Chapter 11

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