network synthesis part i...gabor c. temes & jack w. lapatra, “introduction to circuit synthesis...
TRANSCRIPT
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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/
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OUTLINE
• References
• Definition
• Network Functions
• Realizability Conditions
2Dr. Mohamed Refky
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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
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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
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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
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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
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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 =
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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 =
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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 =
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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 =
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 acircuit.
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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
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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 theorigin.
Network Synthesis
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Realizability Conditions
Dr. Mohamed Refky
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2) For the function
𝑍 𝑠 or 𝑌 𝑠
The power of the numerator and denominator in 𝑠 must differ atmost 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
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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
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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
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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
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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
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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 toput 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 + ⋯
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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
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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 toput it in the form
Network Synthesis
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Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑌 𝑠 = 𝑌1 𝑠 +1
𝑍2 𝑠 +1
𝑌3 𝑠 +1
𝑍4 𝑠 +1
𝑌5 𝑠 + ⋯
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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
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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 toput 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 + ⋯
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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
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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 toput it in the form
Network Synthesis
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Cauer Form (Continued Fraction
Expansion)
Dr. Mohamed Refky
𝑌 𝑠 = 𝑌1 𝑠 +1
𝑍2 𝑠 +1
𝑌3 𝑠 +1
𝑍4 𝑠 +1
𝑌5 𝑠 + ⋯
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Realize the following admittance function in the second Cauer
Form
𝑌 𝑠 =𝑠2 + 4𝑠 + 3
𝑠2 + 2𝑠
Network Synthesis
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Example (6)
Dr. Mohamed Refky