References : Network ynthesis Gabor C. Temes & Jack W. Lapatra, Introduction to Circuit ynthesis and Design, McGraw-Hill Book Company. M.E. Van Valkenburg, Introduction to Modern Network ynthesis, John Wiley Inc. Definitions : Analysis : Known Excitation Known Circuit Response? Calculation of the response of a known circuit to a given excitation.
ynthesis : Known Excitation Circuit? Required Response Finding a new circuit which provides a required response to a given input excitation. N.B.ynthesis solutions are not unique. Using the Complex Frequency : The complex frequency δ j ω is used to analyse the circuits because : It simplifies algebraic work by including the imaginary part in. Ex : I I R jωl V V R L C jωc becomes :
It simplifies circuit synthesis. Network Functions : ynthesis Could be operated on different network functions. Also, the network may be a single port network or a two-port one. ingle Port Networks : For a single port network, synthesis may be operated on the following functions : I V - N V Driving Point Impedance I I Driving Point Admittance Y V
Two-Port Networks : N V - I V - Driving Point Impedance Driving Point Impedance I V I V I Transfer Admittance Y Transfer Impedance V I I V Current Transfer Ratio Voltage Transfer Ratio G I I V V α
Note : This chapter is concerned only with synthesis of driving point functions for single port networks. The functions used are generally in the form of ratios of polynomials : F φ ψ Example : R L C C R R L L C
LC CR L R LC L R L R C P P H
Poles and eros of Network Functions : F Ψ Φ m m m m o n n n n o b b b b a a a a F Where :a s and b s are positive constants and n&m are the orders of Φ and Ψ. are factorized: and If Ψ Φ m n o o P P P b a F cale Factor H b a F are the poles of,p, P,P F are the zeros of,,, Where: o o m n
Conditions of Realizability of Driving Point Immitances D.P.I. : The function must be a Positive Real one P.R., i.e. Re al{ F } 0 for Re al{ } 0 This condition means that the power flows from the source to the circuit and may be achieved if : ithe poles of the function are negative or, if complex, they have a negative real part. This condition makes the circuit stable. iithe poles on the jω axis must be simple poles. iiif must not have multiple zeros or poles at the origin.
For : F Φ Ψ F a b o o n m a b n m a b n m a b n m The power of the numerator and denominator in must differ at most by ± i.e. m n ± This is because the function must be reduced to one of the elements : R, L, c or G, C, L If these conditions are satisfied, several methods of synthesis are available. The following are the basic methods :
First Foster Form st F.F. : Partial Fractions for Example : ynthesize the function : olution :
B A 0 A B
Ω F Ω F st F.F. econd Foster Form nd F.F. : Partial Fractions for Y Example : ynthesize the function : 7 Y
olution : Y 7 7 6 Y A B Y C A B C
C A : of Coefficiects B 0 : Coefficients of C : Coefficients of 0,, 0 A C B Y Y 9 6
Y F 6H H F 9 nd F.F. Continued Fraction Expansion Cauer Forms: st Cauer Form Descending orders for Descending orders for Y nd Cauer Form Ascending orders for Ascending orders for Y
Example : Realize the following function in the st Cauer form : olution :
½ ½ 6
Which results in the following ladder network : H H ½F 6 F st Cauer Form These steps may be done as follows : ½ 6 ½ 6 ½ ½ --
H H ½F 6 F st Cauer Form Example : Realize the following admittance function in the first Cauer Form : Y
olution : ½ ½ ½ 6 ½ ½ ----- - - N.B. Whenever a minus sign appears in the division, the process stops and the circuit is not realizable in a passive form.
Y 6 st Cauer Form Example : Realize the following function in the nd Cauer form :
olution : nd Cauer Form
Example : Give the second Cauer realization of the admittance function : Y
Y Y
nd Cauer Form ynthesis with LC Elements : Applying partial fractions to an LC impedance or admittance function will produce the following possible parts : F N D K K o K ω Kn ω n
Properties of the driving point D.P. immittance : F is the ratio of an even N and an odd D or vice versa. The degrees of N and D differ by exactly. At 0, there is either a zero k o 0 or a pole K >0. At there is either a zero K 0 or a pole K >0. F has only simple poles and zeros located alternatively on the jω axis. 6The residues at all poles are real and positive.
Example : Find the two Foster Form realizations of the function : 9 olution : Imag. j j j 0 -j -j -j -plane Real
Foster : Ko K K K 9 K o K 9 0 0 9 9
9 6 9
nd Foster Form : 9 Y 9 K K Y 8 9 Y K 8 9 9 9 Y K
9 8 8 Y ] 7 8 [ ] 8 8 [ Y
Example : Find the first and second Cauer realizations for the impedance function given by : 9 olution : st Cauer realization : 0 9
0 9 [6] 6 9 [] 6 9 6 [8] 9 ----- 6 8
nd Cauer realization : 9 0
9 6 9 0 9 9 6 96 60 -----
RC Immittances : In order to realize an immittance function with RC elements, the partial fraction expansion must be in the form : Or Y K K K o K n Ki i δ o i n Ki i δ i Properties of RC Immittances RC Impedances All poles and zeros are simple and lie on the δ negative real axis of the -plane. RC admittances All poles and zeros are simple and lie on the δ negative real axis of the -plane. Poles and zeros alternate on the δ axis. Poles and zeros alternate on the δ axis.
The first critical freq. the smallest in absolute value on the δ axis is a pole. It may be at 0. The first critical freq. the smallest in absolute value on the δ axis is a zero. It may be at 0. The last critical freq. is a zero which may be at infinity. The last critical freq. is a pole which may be at infinity. The residues at the poles of are real and positive. The residues at the poles of Y are real and negative. The residues of {Y} are real and positive. Example of a zero at infinity :
0 0 Example of a pole at infinity : Y 0 Y
Example : Find the Foster realizations of the impedance function : olution : jω -plane δ - - - 0
K K K o K K o 0 K
Y Y Which is unrealizable.
Y Y Y Y
Example : Find two Cauer realizations for the function given by :
olution: st Cauer Form : 8 6 8 6 6 8 6 6
8
nd Cauer Form : 6 8 6 8 6 0
0
RL Immittances : RL immittances have the partial fraction expansions : Or : Y K K K K o o n Ki i δ i n Ki i δ i Properties of RL Immittances RL impedances All poles and zeros are simple and lie on the δ real negative axis of the -plane. RL admittances All poles and zeros are simple and lie on the δ real negative axis of the -plane. Poles and zeros alternate on the δ axis. Poles and zeros alternate on the δ axis.
The first critical frequency on the δ axis is a zero. It may be at 0. The first critical frequency on the δ axis is a pole. It may be at 0. The last critical frequency is a pole which may be at infinity. The last critical frequency is a zero which may be at infinity. The residues of the poles of are real and negative. The residues of are real and positive. The residues of the poles of are real and positive. Example : Realize the immittance function given by :
olution : jω - - - 0 δ The first critical frequency is a zero on the δ axis. The network is RL. st Foster Form : K K
K K
The realized circuit will be :
nd Foster Form : Y Y Ko K K
K K o Y Y 0 K Y 6 Y 6 6
Example : Find the nd Cauer realization of the impedance function given by : 6 olution : jω -6 - - - 0 δ Poles and zeros alternate on the δ axis and the first critical frequency is a zero. The network is RL.
8 8 8 6 8 7 9 7 7 8
98 7 8 7 8 8 8 _ 6 7 98 8
caling : Example : Using magnitude and frequency scaling, synthesize two possible networks for the impedance function : 00 00 00 000
Magnitude caling : 00 Frequency caling : 00 jω - - - First critical frequency is a pole at 0. The network is RC. 0 δ
st Foster Form : K Ko K K K o 0 K
00 { 00 } 00 00 00 { 00 00 00
nd Foster Form : Y Y Y K K Y
Y 6] [ ] [ 6] [ ] [ 00 Y 00 6 00 00 00
Y 00 00 00 00 6 00 00
Example : Find all possible circuits to realize the following admittance function : Y olution : st Foster Form : K K o K
H F ¼ H F st Foster Realization
nd Foster Form : Y K K Y Y Y 6
st Cauer Form for :
6 ---- 6
st Cauer Form for Y : Y Unrealizable
nd Cauer Form for :
----
6nd Cauer Form for Y : Y
9 8 Unrealizable