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Network Synthesis
Part I
Dr. Mohamed Refky Amin
Electronics and Electrical Communications Engineering Department (EECE)
Cairo University
http://scholar.cu.edu.eg/refky/
OUTLINE
• References
• Definition
• Network Functions
• Realizability Conditions
2Dr. Mohamed Refky
Gabor C. Temes & Jack W. Lapatra, “Introduction to Circuit
Synthesis and Design”, McGraw-Hill Book Company.
M.E. Van Valkenburg, “Introduction to Modern Network
Synthesis”, John Wiley Inc.
References
3Dr. Mohamed Refky
What we have used to do so far is the calculation of the response
of a known circuit to a given excitation.
This is called analysis of circuits.
Network Synthesis
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Definition
Dr. Mohamed Refky
In network synthesis we try to find a new circuit that provides a
required response to a given input excitation
Network Synthesis
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Definition
Dr. Mohamed Refky
Synthesis solutions are not unique
In network synthesis, complex frequency
𝑠 = 𝛿 + 𝑗𝜔
is used to analyze the circuits because it simplifies algebraic work
by including the imaginary part in 𝑠.
Network Synthesis
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Definition
Dr. Mohamed Refky
𝐼 =𝑉
𝑅 + 𝑗𝜔𝐿 +1𝑗𝜔𝐶
𝐼 =𝑉
𝑅 + 𝑠𝐿 +1𝑠𝐶
For a single port network, synthesis may be operated on the
following functions:
Network Synthesis
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One-Port Networks
Dr. Mohamed Refky
𝑉 𝑠
𝐼 𝑠
𝐼 𝑠
𝑉 𝑠
Driving point impedance
𝑍 𝑠 =
Driving point admittance
𝑌 𝑠 =
For a two port network, synthesis may be operated on the
following functions:
Network Synthesis
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Two-Port Networks
Dr. Mohamed Refky
𝑉1 𝑠
𝐼1 𝑠
𝑉2 𝑠
𝐼2 𝑠
Driving point impedance
𝑍11 𝑠 =
Driving point admittance
𝑍22 𝑠 =
For a two port network, synthesis may be operated on the
following functions:
Network Synthesis
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Two-Port Networks
Dr. Mohamed Refky
𝑉1 𝑠
−𝐼2 𝑠
−𝐼2 𝑠
𝑉1 𝑠
Driving point impedance
𝑍12 𝑠 =
Driving point admittance
𝑌21 𝑠 =
For a two port network, synthesis may be operated on the
following functions:
Network Synthesis
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Two-Port Networks
Dr. Mohamed Refky
𝑉2 𝑠
𝑉1 𝑠
−𝐼2 𝑠
𝐼1 𝑠
Driving point impedance
𝐺21 𝑠 =
Driving point admittance
𝛼21 𝑠 =
We will focus on the synthesis of driving point functions for one-
port networks.
The functions used are generally in the form of ratios of
polynomials
𝑍 𝑠 or 𝑌 𝑠
Network Synthesis
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One-Port Networks
Dr. Mohamed Refky
=𝜙 𝑠
𝜓 𝑠=𝛼𝑚𝑠
𝑚 + 𝛼𝑚−1𝑠𝑚−1 +⋯+ 𝛼0
𝛽𝑛𝑠𝑛 + 𝛽𝑛−1𝑠
𝑛−1 +⋯+ 𝛽0
=𝛾𝑚𝛾𝑛
𝑠 − 𝑧1 𝑠 − 𝑧2 … 𝑠 − 𝑧𝑚𝑠 − 𝑝1 𝑠 − 𝑝2 … 𝑠 − 𝑝𝑛
𝑍 𝑠 =1
𝑠𝐶// 𝑅 + 𝑠𝐿
=
1𝑠𝐶
𝑅 + 𝑠𝐿
𝑅 + 𝑠𝐿 +1𝑠𝐶
=𝑅 + 𝑠𝐿
𝑠𝐶𝑅 + 𝑠2𝐿𝐶 + 1
Network Synthesis
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Example (1)
Dr. Mohamed Refky
Find the impedance of the shown circuit
𝑍 𝑠 or 𝑌 𝑠
𝛼’s and 𝛽’s are positive constants
𝑚 is the orders of 𝜙 𝑠 .
𝑛 are the orders of 𝜓 𝑠 .
Network Synthesis
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One-Port Networks
Dr. Mohamed Refky
=𝜙 𝑠
𝜓 𝑠=𝛼𝑚𝑠
𝑚 + 𝛼𝑚−1𝑠𝑚−1 +⋯+ 𝛼0
𝛽𝑛𝑠𝑛 + 𝛽𝑛−1𝑠
𝑛−1 +⋯+ 𝛽0
𝑍 𝑠 or 𝑌 𝑠
𝑧1, 𝑧2, …, 𝑧𝑚 are the zeros of 𝑍 𝑠 or 𝑌 𝑠
𝑝1, 𝑝2, …, 𝑝𝑛 are the poles of 𝑍 𝑠 or 𝑌 𝑠
Network Synthesis
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One-Port Networks
Dr. Mohamed Refky
=𝜙 𝑠
𝜓 𝑠=𝛼𝑚𝑠
𝑚 + 𝛼𝑚−1𝑠𝑚−1 +⋯+ 𝛼0
𝛼𝑛𝑠𝑛 + 𝛽𝑛−1𝑠
𝑛−1 +⋯+ 𝛽0
𝛾𝑚𝛾𝑛
𝑠 − 𝑧1 𝑠 − 𝑧2 … 𝑠 − 𝑧𝑚𝑠 − 𝑝1 𝑠 − 𝑝2 … 𝑠 − 𝑝𝑛
𝛾𝑚𝛾𝑛
is the scale factor
For series impedances
𝑍 𝑠 = 𝑍1 𝑠 + 𝑍2 𝑠
For parallel impedances
𝑍 𝑠 =1
𝑌1 𝑠 + 𝑌2 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
=1
1𝑍1 𝑠
+1
𝑍2 𝑠
For series impedances
𝑌 𝑠 =1
𝑍1 𝑠 + 𝑍2 𝑠
For parallel impedances
𝑌 𝑠 = 𝑌1 𝑠 + 𝑌2 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
=1
𝑍1 𝑠+
1
𝑍1 𝑠
For a combination of series and
parallel impedances
𝑍 𝑠 = 𝑍1 𝑠 + 𝑍𝑝 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
= 𝑍1 𝑠 +1
1𝑍2 𝑠
+1
𝑍3 𝑠
= 𝑍1 𝑠 +1
𝑌2 𝑠 + 𝑌3 𝑠
For a combination of series and
parallel impedances
𝑍 𝑠 =1
𝑌1 𝑠 + 𝑌2 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
=1
1𝑍1 𝑠
+1
𝑍2 𝑠 + 𝑍3 𝑠
For a combination of series and
parallel impedances
𝑌 𝑠 =1
𝑍1 𝑠 + 𝑍𝑝 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
=1
𝑍1 𝑠 +1
𝑌2 𝑠 + 𝑌3 𝑠
=1
𝑍1 𝑠 +1
1𝑍2 𝑠
+1
𝑍3 𝑠
For a combination of series and
parallel impedances
𝑌 𝑠 = 𝑌1 𝑠 + 𝑌𝑠 𝑠
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
=1
𝑍1 𝑠+
1
𝑍2 𝑠 + 𝑍3 𝑠
𝑍 𝑠 = 𝑍1 𝑠 +1
𝑌2 𝑠 +1
𝑍3 𝑠 +1
𝑌4 𝑠 +1
𝑍5 𝑠 + ⋯
𝑌 𝑠 = 𝑌1 𝑠 +1
𝑍2 𝑠 +1
𝑌3 𝑠 +1
𝑍4 𝑠 +1
𝑌5 𝑠 + ⋯
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
𝑍 𝑠 =1
𝑠𝐶 +1
𝑅 + 𝑠𝐿
𝑍 𝑠 =𝑠𝐿 + 𝑅
𝑠2𝐿𝐶 + 𝑠𝑅𝐶 + 1
Network Synthesis
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Realization of a Function
Dr. Mohamed Refky
𝑍 𝑠 =1
𝑠𝐶 +1
𝑅 + 𝑠𝐿
𝑍 𝑠 =𝑠𝐿 + 𝑅
𝑠2𝐿𝐶 + 𝑠𝑅𝐶 + 1
In network synthesis, we try to find a way to convert the function
(𝑍 𝑠 or 𝑌 𝑠 ) into a form that is easier to be realized into a
circuit.
1) The function must be a Positive Real (PR)
𝑅𝑒𝑎𝑙 𝑍 𝑠 or 𝑌 𝑠 ≥ 0 for 𝑅𝑒𝑎𝑙 𝑠 ≥ 0
This condition means that the power flows from the source to the
circuit
Network Synthesis
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Realizability Conditions
Dr. Mohamed Refky
𝑍1 𝑠 = 𝑅 + 𝑠𝑋 𝑍2 𝑠 = −𝑅 + 𝑠𝑋
𝑌1 𝑠 = 𝐺 +1
𝑠𝑋𝑌2 𝑠 = −𝐺 +
1
𝑠𝑋
1) The function must be a Positive Real (PR)
𝑅𝑒𝑎𝑙 𝑍 𝑠 or 𝑌 𝑠 ≥ 0 for 𝑅𝑒𝑎𝑙 𝑠 ≥ 0
This condition means that the power flows from the source to the
circuit
The poles of the function are negative or, if complex, they have
a negative real part. This condition makes the circuit stable.
The poles on the 𝑗𝜔 axis must be simple poles.
𝑍 𝑠 or 𝑌 𝑠 must not have multiple zeros or poles at the
origin.
Network Synthesis
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Realizability Conditions
Dr. Mohamed Refky
2) For the function
𝑍 𝑠 or 𝑌 𝑠
The power of the numerator and denominator in 𝑠 must differ at
most by ±1.
This is because the function must be reduced to one of the
elements
Network Synthesis
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Realizability Conditions
Dr. Mohamed Refky
=𝜙 𝑠
𝜓 𝑠=𝛼𝑚𝑠
𝑚 + 𝛼𝑚−1𝑠𝑚−1 +⋯+ 𝛼0
𝛽𝑛𝑠𝑛 + 𝛽𝑛−1𝑠
𝑛−1 +⋯+ 𝛽0
𝑅, 𝑠𝐿,1
𝑠𝐶𝐺,
1
𝑠𝐿, 𝑠𝐶or
In first foster form, partial fraction is used to factorized 𝑍 𝑠
Network Synthesis
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First Foster Form
Dr. Mohamed Refky
𝑍 𝑠 =𝑘1
𝑎1𝑠 + 𝑏1+
𝑘2𝑎2𝑠 + 𝑏2
𝑍 𝑠 =𝛼1𝑠 + 𝛼0
𝛽2𝑠2 + 𝛽2𝑠 + 𝛽0
Use the first foster form to synthesize the function
𝑍 𝑠 =𝑠2 + 4𝑠 + 3
𝑠2 + 2𝑠
Network Synthesis
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Example (1)
Dr. Mohamed Refky
In second foster form, partial fraction is used to factorized 𝑌 𝑠
Network Synthesis
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Second Foster Form
Dr. Mohamed Refky
𝑌 𝑠 =𝑘1
𝑎1𝑠 + 𝑏1+
𝑘2𝑎2𝑠 + 𝑏2
𝑌 𝑠 =𝛼1𝑠 + 𝛼0
𝛽2𝑠2 + 𝛽2𝑠 + 𝛽0
Use the second foster form to synthesize the function
𝑌 𝑠 =4𝑠4 + 7𝑠2 + 1
𝑠 2𝑠2 + 3
Network Synthesis
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Example (2)
Dr. Mohamed Refky
𝑌 𝑠 =4𝑠4 + 7𝑠2 + 1
2𝑠3 + 3𝑠
= 2𝑠 +𝑠2 + 1
𝑠 2𝑠2 + 3
First Cauer Form of 𝑍 𝑠 starts with
𝑍 𝑠 =𝛼𝑚𝑠
𝑚 + 𝛼𝑚−1𝑠𝑚−1 +⋯+ 𝛼0
𝛽𝑛𝑠𝑛 + 𝛽𝑛−1𝑠
𝑛−1 +⋯+ 𝛽0
then Continued Fraction Expansion (CFE) is applied to 𝑍 𝑠 to
put it in the form
Network Synthesis
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Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑍 𝑠 = 𝑍1 𝑠 +1
𝑌2 𝑠 +1
𝑍3 𝑠 +1
𝑌4 𝑠 +1
𝑍5 𝑠 + ⋯
Realize the following function in the first Cauer form
𝑍 𝑠 =𝑠4 + 4𝑠2 + 3
𝑠3 + 2𝑠
Network Synthesis
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Example (3)
Dr. Mohamed Refky
First Cauer Form of 𝑌 𝑠 starts with
𝑌 𝑠 =𝛽𝑛𝑠
𝑛 + 𝛽𝑛−1𝑠𝑛−1 +⋯+ 𝛽0
𝛼𝑚𝑠𝑚 + 𝛼𝑚−1𝑠
𝑚−1 +⋯+ 𝛼0
then Continued Fraction Expansion (CFE) is applied to 𝑌 𝑠 to
put it in the form
Network Synthesis
33
Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑌 𝑠 = 𝑌1 𝑠 +1
𝑍2 𝑠 +1
𝑌3 𝑠 +1
𝑍4 𝑠 +1
𝑌5 𝑠 + ⋯
Realize the following admittance function in the first Cauer Form
𝑌 𝑠 =𝑠2 + 4𝑠 + 3
𝑠2 + 2𝑠
Network Synthesis
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Example (4)
Dr. Mohamed Refky
Secound Cauer Form of 𝑍 𝑠 starts with
𝑍 𝑠 =𝛼0 +⋯+ 𝛼𝑚−1𝑠
𝑚−1 + 𝛼𝑚𝑠𝑚
𝛽0 +⋯+ 𝛽𝑛−1𝑠𝑛−1 + 𝛽𝑛𝑠
𝑛
then Continued Fraction Expansion (CFE) is applied to 𝑍 𝑠 to
put it in the form
Network Synthesis
35
Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑍 𝑠 = 𝑍1 𝑠 +1
𝑌2 𝑠 +1
𝑍3 𝑠 +1
𝑌4 𝑠 +1
𝑍5 𝑠 + ⋯
Realize the following admittance function in the second Cauer
Form
𝑍 𝑠 =𝑠4 + 4𝑠2 + 3
𝑠3 + 2𝑠
Network Synthesis
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Example (5)
Dr. Mohamed Refky
Secound Cauer Form of 𝑌 𝑠 starts with
𝑌 𝑠 =𝛽0 +⋯+ 𝛽𝑛−1𝑠
𝑛−1 + 𝛽𝑛𝑠𝑛
𝛼0 +⋯+ 𝛼𝑚−1𝑠𝑚−1 + 𝛼𝑚𝑠
𝑚
then Continued Fraction Expansion (CFE) is applied to 𝑌 𝑠 to
put it in the form
Network Synthesis
37
Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑌 𝑠 = 𝑌1 𝑠 +1
𝑍2 𝑠 +1
𝑌3 𝑠 +1
𝑍4 𝑠 +1
𝑌5 𝑠 + ⋯
Realize the following admittance function in the second Cauer
Form
𝑌 𝑠 =𝑠2 + 4𝑠 + 3
𝑠2 + 2𝑠
Network Synthesis
38
Example (6)
Dr. Mohamed Refky