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7/21/2019 Synthesis Two Ele
http://slidepdf.com/reader/full/synthesis-two-ele 1/20
Synthesis of one-port networks
with two kinds of elements
11.1 Properties of L-C immittance functions
11.2 Synthesis of L-C driving-point immittances
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Contents
Properties of L-C immittance functions Properties
Examples of immittance and non-immittance
functions
Synthesis of L-C driving-point immittance Synthesis of L-C circuit
Examples of synthesis
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Property 1. L-C immittance function
1. ZLC (s or !LC (s is the ratio of odd to even or even to odd
polynomials.
Consider the impedance Z(s of passive one-port net"or#.
($ is even % is odd
&s "e #no"' "hen the input current is I ' the average po"er
dissipated y one-port net"or# is )ero*
&verage Po"er = =0
(for pure reactive net"or#
1 1
2 2
( ) ( )( )( ) ( )
M s N s Z s M s N s
+=
+
21Re[ ( )]
2 Z j I ω
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1 2 1 2( ) ( ) ( ) ( ) 0 M j M j N j N jω ω ω ω − =
1 2 1 2
2 2
2 2
( ) ( ) ( ) ( )( ) 0
( ) ( )
M s M s N s N s EvZ s
M s N s
−= =
+
1 20 M N = = OR 2 10 M N = =
1 1
2 2
( ) , ( ) M N
Z s Z s N M
= =
Z(s) or Y(s) is the ratio of een to odd or odd to een!!
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Property ". L-C immittance function
+.,he poles and )eros are simple and lie on the
axis.
Since oth $ and % are ur"it)' they have only
imaginary roots' and it follo"s that the poles and )eros
of Z(s or !(s are on the imaginary axis.
Consider the example
jω
1 1
2 2
( ) , ( ) M N
Z s Z s
N M
= =
4 2
4 2 0
5 3
5 3 1
( ) a s a s a
Z sb s b s b s
+ +=
+ +
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Ex highest order of the numerator * +n - highest order of the
denominator can either e +n-1 (simple pole at s/ or
the order can e +n01 (simple )ero at s/ .
mpedance function cannot hae multiple poles or #eros on the
axis.
$n order for the impedance to e positive
real the coefficients must e real and
positive.
4 2
4 2 0
5 3
5 3 1
( ) a s a s a
Z sb s b s b s
+ +=
+ +
jω
∞
,he highest po"ers of the numerator and the denominator
polynomials can differ y' at most' unity.
∞
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Property %. L-C immittance function
%.&he poles and #eros interlace on the a'is. jω
0 +1
+2
+3
+4
ω
(w)
i*hest power+ "n -, ne't hi*hest power must e "n-"
&hey cannot e missin* term. nless/
if s025
5 1 0b s b s+ = 1/ 4 ( 2 1) / 41
5
( ) i k
k
b s e
b
π −
=
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2 2 2 2 2 2
1 3
2 2 2 2 2 2
2 4
( )( )...( )...( )
( )( )...( )...
i
j
K s s s Z s
s s s s
ω ω ω
ω ω ω
+ + +=
+ + +
3e can write a *eneral L-C impedance or admittance as
0 2 4
2 2 2 2
2 4
2 2( ) ...
K K s K s Z s K s
s s sω ω ∞
= + + + +
+ +
Since these poles are all on the 2" axis' the residues must e real andpositive in order for Z(s to e positive real .
S04w Z(4w)04(w) (no real part)
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2 20 2 2
2 2 2
2
( )( )...
( )
K K dX K
d
ω ω ω
ω ω ω ω ∞
+= + + +
+
( )0
dX
d
ω
ω
≥
Since all the residues 3i are positive' it is eay
to see that for an L-C function
Ex)
2 2
3
2 2 2 2
2 4
( )( )
( )( )
Ks s Z s
s s
ω
ω ω
+=
+ +
2 2
3
2 2 2 2
2 4
( )( )
( )( )
K jX j
ω ω ω ω
ω ω ω ω
− += +
− + − +
w
X(w)
0w2
w3
w4
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Properties 5 and 6. L-C immittance
function
,he highest po"ers of the numerator and
denominator must differ y unity4 the lo"est
po"ers also differ y unity.
,here must e either a )ero or a pole at the
origin and infinity.
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Summary of properties
1. or !LC (s is the ratio of odd to even or even toodd polynomials.
+. ,he poles and )eros are simple and lie on the 2" axis
5. ,he poles and )eros interlace on the 2" axis.
6. ,he highest po"ers of the numerator and denominatormust differ y unity4 the lo"est po"ers also differ y unity.
7. ,here must e either a )ero or a pole at the origin andinfinity.
( ) LC Z s
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Examples
2
2 2
( 4)( )
( 1)( 3)
Ks s Z s
s s
+=
+ +
5 3
4 2
4 5( )
3 6
s s s Z s
s s
+ +=
+
2 2
2 2
( 1)( 9)( )
( 2)( 10)
K s s Z s
s s
+ +=
+ +
2 2
2
2( 1)( 9)( )
( 4)
s s Z s
s s
+ +=
+
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11." Synthesis of L-C 7riin* point
immittances
L-C immittance is a positive real function "ith
poles and )eros on the 2" axis only.
,he synthesis is accomplished directly from the partial
fraction .
8(s is impedance - then the term *
capacitor of 193 farads
the 3(infinites is an inductance of 3(infinite henrys.
0 2 4
2 2 2 22 4
2 2( ) ...
K K s K s Z s K s
s s sω ω ∞
= + + + +
+ +
0 / K s
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s a parallel tan# capacitance and inductance.
'
8or Z(s) partial fraction
2 22 /( )i i K s s ω +
22 /
i i K ω 1/ 2 i K
2 2
2 2
9 152( 1)( 9) 2 2( ) 2
( 4) 4
s s s
Z s s s s s s
+ += = + +
+ +
2/15f
2/9 f 2h 15/8 h
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8or Y(s) partial fraction
In admittance
2 2
2 2 2 2
1 3
( 2)( 4) 2 2
( ) ( 1)( 3) 3 1
s s s s s
Y s s s s s s
+ += = + +
+ + + +
(!)1f
3/2 f
2/3 h2h
1/" f
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9nother methodolo*y
:sing property 6 ;,he highest po"ers if numerator and
denominator must differ y unity4 the lo"est po"ers also
differ y unity.< ,herefore' there is al"ays a )ero or a pole at s/infinite .
suppose Z(s numerator*+n 'denominator*+n-1
this net"or# has pole at infinite. - "e can remove this
pole y removing an impedance L 1 s
Z+ (s / Z(s - L 1 s
=egree of denominator * +n-1 numerator*+n-+ Z+ (s has )ero at s/infinite.
!+ (s /19 Z+ (s ' !5 (s / !+ (s - C+ s
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This infinite term removing process continue until
the remainder is zero.
Each time e remove the pole! e remove aninductor or capacitor depending upon hether the
function is an impedance or an admittance.
"inal synthesized is a ladder hose series arms
are inductors and shunt arms are capacitors.
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5 3
4 2
2 12 16( )
4 3
s s s Z s
s s
+ +=
+ +
5 3 3
2 4 2 4 2
2 12 16 4 10( ) 2
4 3 4 3
s s s s s Z s s
s s s s
+ + += − =
+ + + +
24 2
3 3 3
3 34 3 1 2( )
4 10 4 4 10
s s s
Y s s s s s s
++ +
= − =
+ +
4 32
8 2
( ) 333
2
s
Z s Z s s
= − =
+2h
2h
1/4 f 3/4 f
2/3 h
8/3 h
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,his circuit (Ladder called as Cauer
ecause Cauer discovered the continues fraction
method.
>ithout going into the proof of the statement m
in can e said that oth the 8oster and Cauer
form gice the minimum numer of elements for aspecified L-C net"or#.