INTRODUCTION

CHAPTER

3 SYNTHESIS OF

ONE-PORT REACTANCE

CIRCUITS

Four simple canonical forms for the realization of one-port reactance functions are the first and second Foster forms and the first and second Cauer forms. These four topologies are canonical in that they can always be realized, and they are also minimal in that they realize the driving point immittance with the least number of elements. The four canonical forms described in this chapter indicate that there is more than one way to synthesize a one-port driving point immittance.

The two Foster forms involve a sequential removal of the poles of an immittance function by first putting it into a partial fraction form. If the pole is at the origin or at infinity the order of the function is reduced by unity. If it is at s = ±jwi the order of the function is reduced by two. The two Cauer canonical forms involve the removal of poles exclusively at the origin or at infinity. It is of note that the zeros of the immittance function are not involved in the construction of one-port networks.

In many instances it may be desirable to extract a pole at a finite frequency from a reactance function that does not coincide with its pole zero diagram. One way this may be achieved is by employing a zero shifting technique. This principle is separately outlined in this chapter.

28

FIRST FOSTER CANONICAL FORM

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 29

The first Foster form realization of a one-port immittance function is obtained by expanding Z(s) by partial fractions and identifying terms in the summation with impedances of simple networks:

(3-l)

The general fonn for Eq. (3-l) is obtained by combining Eqs (2-44) and (2-45) in Chapter 2:

The residues are given by

K0 = sZ(s)l.=o s1. + w�

2K1 = --' Z(s)i.•= -wi s

(3-2)

(3-3)

(3-4)

(3-5)

The firs t term in the partial fraction expansion is recognized as the reactance of a series capacitance C0 having a value

F (3-6)

The second term is recognized as the reactance of a parallel tuned circuit with capacitance and inductance C; and L;:

l C.=-

' 2K; F (3-7)

2K. (3-8) L,=-' H

(JJ� l

respectively. The las t term m this expansion represents the reactance of an inductance with a value

H (3-9)

If the function in Eq. (3-2) has a pole at s = 0 and at s = jw its order is reduced by one if either of these poles is extracted from Z(s). If the function Z(s) has a pair of conjugate poles at s ±jw1 the order is reduced by two if these poles are extracted in the form of a tank circuit. Figure 3-1 gives the schematic diagram of the first Foster form realization of an impedance function.

30 SYNTHESIS OF LUMPED ELEMENT. DISTRIBUTED AND PLANAR FILTERS

I o--f

2K1 w:

I 2K.

2K. w!

ro------------------------------------------------J

Z(s)

FIGURE 3-l

To demonstrate the realization of a one-port reactance function in the first Foster form consider the following reactance fu nction:

Z(s) =

(sz + 1 )(sz + 9)

s(s2 + 4)

Z(s) is a reactance function in that it satisfies the properties of an LC function listed in Chapter 2. Its poles and zeros are interlaced in the manner shown in Fig. 3-2.

The residues of the poles in the partial fraction expansion of Z(s) are given by Eqs (3-3) to (3-5) as

Ko = sZ(s)l.�o = t

sz+4 I 15 2K2 = -- Z(s) •'"' -4 = 4

s

Each residue is positive in keeping with the p.r. condition in Chapter 2. The partial

0 0

FIGURE 3-2 Pole zero diagram or

(s1 + l)(sz + 9) Z(s) = "---,---:-­

s(sl + 4)

.. 0 2 C()

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 31

1 H tF

o-----1flnr------���----�

Z(s)

FIGURE 3-3 First Foster one-port network of

(s2 + l)(s2 + 9) Z(s) +4)

fraction expansion of Z(s) is therefore

) 9/4 15s/4

Z(s = +-- +s s s2 + 4

!fH

Figure 3-3 shows the one-port equivalent circuit of Z(s). It is observed that this equivalent circuit may be obtained by a sequential extraction of the poles of Z(s) in Fig. 3-2.

If Z(s) is p.r., 1/Z(s) is also p.r. Defining such a new reactance function yields

Z(s) =

s(sl + 4)

(s2 + l)(s2 + 9)

The pole zero diagram for this reactance function is depicted in Fig. 3-4. Its residues are

o�-----•------�oe-----�•�-----------------0 0 2 ro

FIGURE J...4 Pole zero diagram of

s(s2 + 4) Z(s)

(s1 + l)(s1 + 9)

32 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

Z(s)

FIGURE 3-5

IH

First Foster one-port network of

s(s2 + 4) Z(s) = --=------=--

(s2 + l)(s' + 9)

Thus

IF

f.:H

Z(s) = 3s/8 + �s/8

s2 + 1 s2 + 9

It is again observed that the poles of Z(s) in Fig. 3-4 have been extracted in forming the equivalent one-port circuit in Fig. 3-5.

SECOND FOSTER CANONICAL FORM

The second Foster form realization of a one-port immittance function is the dual of the first one except that it involves a partial fraction expansion of an admittance function instead of an impedance one:

(3-10) The general form for the above equation is deduced by duality from Eq. {3-2) as

where

K0 � 2Kis Y(s) = --- + � -2--2 + · · · + K00s s i=ls +wi

K0 = sY(s)is=O sl + w2

2Ki = ----'- Y(s)is>= -ror s

(3-11)

(3�12)

(3-13)

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 33

K -I- Y(s)l 00- tm�-s "'-oo

(3-14)

The term K0/s is recognized as the susceptance of a shunt inductance L0 having a value

1 Lo=­Ko H (3-15)

The second term represents the reactance of a series tank circuit with elements C1 and L1 in shunt with the network:

C-=2K1 I 2 W;

1 L-=---·-I 2K,

F (3-16)

H (3-17)

respectively. The term K00s is recognized as the susceptance of a shunt capacitance given by

F (3-18) The second Foster realization of a one-port immittance function has the form depicted in Fig. 3-6. lt is observed that it is the exact dual of that in Fig. 3-l.

The order of the function in Eq. (3-11) is reduced by one if it has a pole at s = 0 or s = joo removed in the form of shunt inductance or capacitance. lt is reduced by two if it has a pair of conjugate poles at s = ± jw1 extracted in the form of a shunt tank circuit in shunt with the input terminals.

To illustrate the synthesis of a one-port admittance function in the second Foster form, consider the susceptance function obtained by taking the reciprocal of Z(s) used in the first Foster form problem.

Y(s)

2�.-----1 2�.

2K, 2K.

[ K, K�

o-----..1-----'-----_,JL.----_,JL.--w: __ -__ _J w!

Y(s) FIGURE J..{i Second Foster realization of a one-port admittance function.

34 SYNTHESIS ot· LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

o�----�M�------�0�------�M�------------------o 0 2 J

FIGURE J.-7 Pole zero diagram of

s(s2 + 4) Y(s) = --:-----::--­

(s2 + l)(s2 + 9)

00

The pole zero diagram for this susceptance function is depicted in Fig. 3-7. The residues of the poles of Y(s) in Eq. (3-ll) are given with the help of Eqs (3-12) to (3-14) by

s2 + 1 2K1 = -- Z(s)j•'=-1 =i

s

s2 + 9 2K3 = -- Z(s)j,,= -9 = i

s

The partial fraction expansion of Y(s) has the following form:

3s/8 Ss/8 Y(s)=- -+ -­

s2 + 1 s2 + 9

The one-port circuit formed in this way is iilustrated in Fig. 3-8. The tank circuits in this network are recognized as the poles of the susceptance function.

As a final example consider the synthesis of the admittance function below in the second Foster form

Y(s) =

(s2 + 1 )(s2 + 9)

s(s2 + 4)

The pole zero diagram for this function is shown in Fig. 3-9.

o___,.-1----...i" l" I

'F T

6F 0

Y(s)

FIGURE J.-8 Second Foster one-port network of

s(s1 + 4) Y(s) = ---·-··---

(s2 + I )(s2 + 9)

0

FIGURE 3-9 Pole zero diagram of

(s2 + l)(s2 + 9) Y{s)=---­

s(s2 + 4)

Y(s)

FIGURE 3-10

2

IF

Second Faster one-port network of

. (s2 + I )(s2 + 9) Y(s) = ---::----

s(s1 + 4)

SYNTHESIS OF ONE· PORT REACTANCE CIRCUITS 35

00

The partial fraction expansion for this susceptance f unction is readily determined as

( 4/9 l5sj4

y s)= + s +4

The one-port circuit for this admittance is illustrated in Fig. 3-lO.

FIRST CAVER FORM (REMOVAL OF POLES AT INFINITY)

A useful property of a reactance or susceptance function is that it has either a pole or zero at s = joo. An immittance function has a pole at infinity if the degree of the numerator polynomial P(s) is of degree one higher than that of the denominator polynomial Q(s). If Z(s) or Y(s) have no poles at infinity <heir reciprocals do. One ladder network due to Cauer is obtained by a repeated removal of poles at infinity. If such a pole is removed .from an reactance function there is now a zero there. However, the susceptance function obtained by inverting Z(s) has now a pole there which can be again extracted. This process is repeated until the degree of the polynomial is reduced to zero.

J6 SYNTHESIS OF LUMPED ELEMENT, DISTIUBVTED "ND PL"N"R FILTERS

To illustrate this synthesis technique consider the reactance in the example in the previous section:

s4

+ 10s2 +9 Z(s) =

--s3::-+_

4_s_

Z(s) has a pole at infinity since the degree of P(s) is one larger than that of Q(s). The residue of this pole can be obtained by dividing P(s) by Q(s):

Thus

6s2 + 9 Z(s) = s + --

s3 +4s

Z1(s)=s

6s2 + 9 Z2(s)=-­

s3 + 4s

Extracting the pole at infinity from Z(s) yields the equivalent circuit in Fig. 3-11a. Z2(s) has now a zero at infinity instead of a pole there. However, Y2(s) has

a pole there. Forming Y2(s) gives

1 s3 + 4s Y2(s)=--=-----,--­

Z2(s) 6s2 + 4s

The residue of the pole of Y2(s) is now evaluated by dividing P(s) by Q(s):

or

where

s 5s/2 Y2(s)=-+ --

6 6s2 + 9

Y2(s) = Y3(s) + Y4(s)

s Y3(s) = -

6

Y4(s) =

5s/2

6s2 + 9

Extracting the pole of Y2(s) at infinity produces the equivalent circuit in Fig. 3-11b. Y4(s) has now a zero at infinity. Forming Z4(s) = 1/Y4(s)locates a pole there:

1 6s2 + 9 Z4(s)= -- =--

Y4(s) 5s/2

The residue of the new pole at infinity may now be evaluated by dividing P(s) by Q(s):

12s 18 Z4(s)=-+-

5 5s

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 37

� z,

I Z(s)o--:_ _ ___,

(a) After removal of first pole at infinity

IH

z,

r tF

I Z(s)C--L-----1'-------'

(b) After removal of second pole at infinity

z. [ IF

Z(s) >-------

--1.....______,

(c) After removal of third pole at infinity

F1GURE 3-11 First Cauer form of one-port network of

(d) After removal of fourth pole at infinity

s4 + 10s1 + 9 Z(s)=--c----

s3 +4s

38 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

or

where 12s

Z5(s) = ·-

5

18 Z6(s)=-

5s

Figure 3-llc depicts the equivalent circuit of Z(s) with this pole removed. Z6(s) has a zero at infinity but Y6(s) has a pole there which may again be extracted. Forming Y6(s) by inverting Z6(s) gives

5s Y6(s)=-

18

The residue of the pole of Y6(s) is 5/18 by definition. Thus the final equivalent circuit for Z(s) has the form shown in Fig. 3-lld.

A similar procedure indicates that the equivalent circuit in Fig. 3-12

corresponds to the susceptance function below:

s4 + l0s2 + 9 Y(s) = -- ---­

s3 + 4s

The realization of a one-port reactance or susceptance function in the first Cauer form involves the extraction of a pole at infinity by long division, inverting the remainder and dividing again to remove the next pole there (and so on). It is therefore concluded that it is possible to synthesize an LC ladder network by a

continued fraction expansion of the reactance or susceptance function. This method may be illustrated by repeating the synthesis of Z(s) in Fig. 3-ll:

s4 + 10s2 + 9 Z(s)=--­

s3 + 4s

Forming a continued fraction expansion of Z(s) indicates that

S3 + 4s)s4 + 10s2 + 9(s-> z

s4 + 4s2

6s2 + 9)s3 + 4s(s/6-> y

3 3s s +-

2

5s -)6s2 + 9(12s/ 5-> z

2 6s

5s 9) 2- (5s/18-> y

5s

2

0 (3-19)

[ 1H

[ �H

(d)

(a)

�H

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 39

ooo-

::1 F Y,

(b)

MF Y,

(c)

{�H

FIGURE 3-12: First Cauer form of one-port network of

s• + 10s2 + 9 Y(s)=-�

--

s' +4s

40 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

The first element of the ladder network is an impedance s, the next one is a shunt admittance s/6, the third element of the ladder network is a series impedance 12s/6 and the last element of the network is a shunt admittance 5s/18. These elements of the continued fraction expansion are in keeping with those of the ladder network in Fig. 3-11.

If Z(s) is replaced by Y(s) in the above example the first element of the ladder network would be a shunt element, as indicated in Fig. 3-12.

SECOND CAUER FORM (REMOVAL OF POLES AT THE ORIGIN)

Another useful property of an LC function is that it has always a pole or zero at the origin. If Z(s) or Y(s) has no pole there its reciprocal does. The second Cauer ladder synthesis of an immittance function consists of the successive extraction of such poles. If a pole is removed at the origin there is now a zero there. However, inverting the function introduces a new pole there which may once more be

extracted. This process is repeated until the degree of the function is reduced to zero. Synthesis of a ladder network in the second Cauer form proceeds as in the

case of the first Cauer form except that it is necessary to arrange both P(s) and Q(s) in ascending order before division to identify the residue of the pole. This will now be illustrated for the reactance function used in the earlier example:

s4 + 10s2 + 9 Z(s) = --

s3=---+

_4_s_

Rearranging P(s) and Q(s) in ascending order and forming a continued fraction expansion of Z(s) yields

4s + s3)9 + 10s2 + s4(9/4s--+ z 9s2

9+-4

15s3 31s2 -)- +s4(961/15s-+z

31 4 31s2

31

15s3 s4)- (15/31s--+ y

31 15s3

31

0

The ladder network obtained in this way is depicted in Fig. 3-13.

(3-20)

tF

o------1

[ tF

o------1

[

SYNTHESIS OF ONE-PORT REACTI\NCE CIRCUITS 41

[ z,

Z(s)

>----

�---­

(a)

flH z,

(b)

/AF

z,

(c)

/ftF ;F

o-----J I-----r--11---,

FIGURE 3-13 Second Cauer form of one-port network of

s4 + !Os2

+ 9 Z(s)= -----

s3 + 4s

42 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

ZERO SHIFTING AND POLE REMOVAL

The poles of the input immittance of a two-port reactance network terminated in a resistive termination correspond to the transmission zero (attenuation poles) of the circuit. Much of modern filter synthesis therefore involves extracting the poles of an LCR rational function in such a way that the ensuing circuit has the proper topology and exhibits the required attenuation poles of the transfer function. If all the poles either reside at the origin or at infinity then a first or second Cauer synthesis procedure will directly display them without further ado. If the attenuation poles of the two-port network lie at finite frequencies, as in the case of a transfer function that has equal ripple in the pass- and stopbands, then some other procedure is necessary. Scrutiny of this problem suggests that whereas it is not possible to directly extract a pole at a specified frequency, it may in fact be done indirectly by zero shifting followed by inversion in order to interchange the

positions of the poles and zeros of the immittance function. This technique may be understood by noting that the extraction of a pole in one or other of the two Cauer forms moves a zero to the origin or infinity to replace whichever pole has been extracted, since one or the other must always appear there in a rational reactance function. In fact, all the zeros of the rational function are shifted by the removal of a pole, although the other poles are left unperturbed. A procedure based on zero shifting and inversion is one common technique utilized in practice for the removal of poles at finite frequencies.

ZERO SHIFTING BY POLE REMOVAL

The technique of zero shifting and inversion to enable the extraction of poles at finite frequencies will now be demonstrated by way of an example for the reactance function associated with the pole zero diagram in Fig. 3-9:

s4 + 10s2 + 9 Y(s)=---=--­

s(s2 + 4)

This susceptance function has a pole at the origin and at infinity which may be removed in a first and second Cauer manner; the remainder functions are

Y1(s)= Y(s)-s

9 Y1(s) = Y(s)--

4s

respectively. Evaluating these two quantities gives

6s2 + 9 Y1(s)= ---

s(s2 + 4)

s(s2 + 31/4) Y1(s) = -------

s2 + 4

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 43

X 0 )( 0 >< 0 2

��------�0+---,x�-----------o o VI 2

FIGURE 3-14 Pole zero diagrams for

s4+10s2+9 Y(s)=---=--­

s(s2 + 4)

and for same after removal of pole at infinity and at origin.

In the first remainder a zero has been shifted to infinity to replace the pole that has been removed there; in the second one a zero has been moved to the origin to replace the pole that has been extracted. In both instances the other poles of the function have been left unperturbed as asserted. Since the numerator polynomials are also altered all the zeros have in fact been perturbed. Scrutiny of some typical pole zero diagrams suggests that the zeros move in every instance towards the pole being removed; if it is at infinity the zero shifts to infinity and if it is removed at the origin it moves to the origin. Figures 3-14 and 3-15 depict the two situations here. Indeed, the two Cauer forms have realized, in this manner, all the poles of the reactance function at either the origin or at infinity. Since the poles and zeros of reactance functions are interlaced on the jw axis it is not possible to shift a zero past an adjacent pole; it is therefore always necessary to partially remove the pole nearest to the zero that must be shifted.

ZERO SHIFTING BY PARTIAL POLE REMOVAL

The extraction of a pole at a specified frequency with the aid of zero shifting or partial pole removal will now be illustrated by way of an example

s4+ 10s2 +9 Y(s) = -----=---

s(s2 + 4)

This admittance has a pole at infinity but has no pole or zero at s2 = - 10 (say). A pole may, however, be extracted from it by first placing a zero there by a partial removal of the nearest pole which happens in this'case to lie at infinity such that

Y(s)- ksl,z= -to= 0

44 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED A.ND PLA.NAR ALTERS

This equation may now be solved for the residue k of the pole which must be partially extracted:

s4 + 10s2 +91 k= =to s2(s2+4) •>=-Io

Extracting this partial pole at infinity in the form of a shunt capacitor from Y(s)gives

( ) 17s4 + 188s2 + 180 yl s = ---�---20s(s2 + 4)

Y1 (s) has still a pole at s equal to infinity but it now also has a zero at s2 = - 10,as is readily verified. Forming Z1(s) gives a pole there which may be removed:

Z1(s) = -� = 20s(s2 + 4)Y1(s) 17s4 + 188s2 + 180

The pole of Z1(s) at s2 = -10 may now be extracted by expanding it in partial fractions:

21(s) = 15s/19 + 250sf38

s2 + 10 17s2 + 18A pair of conjugate poles may now be removed at s2 = -10 in the form of a parallel LC circuit in series with the network giving a remainder reactance

250s/38 Z2(s)=----17s2 + 18

This reactance has a zero at infinity but its reciprocal has a pole there:

1 646s 684 Y2(s) = Z2(s)

= 250 +

250sThe remaining susceptance may now be realized by inspection by removing a pole at both the origin and at infinity in the form of a shunt capacitor and series inductor.

Figure 3-15 illustrates the development of the circuit. This circuit has one pole at s2 = -10 and a pair of poles at infmity. It is noted that the network in Fig. 3-15 is no longer canonical in that it requires one more element for its realization than was necessary to synthesize the same rational function using either the two Foster or Cauer forms. Indeed, whenever a pole at either the origin or at infinity is partially removed, one additional element is required. If an internal pole is partially removed, two additional elements are required. Thus, where possible, partial pole removals usually only involve poles at the origin or at infinity.

If it is desired to remove a pole at s2 = - ! (say) from the rational function in Eq. (3-1) a partial removal of a pole at the origin is required and the first step takes the following form:

Y1(s)= Y(s)-�1 =0s ,.0=-t

)( 0 )( 0 2

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 45

0 )(

0----------�x�--o�----�x 0 2 ./¥

F1GURE 3-15 Pole zero diagrams for

s4+ l0s2 +9 Y(1) = --:--­

s(s2 + 4)

1howing zero shifting to remove pole at s2 = -10.

As another example, the following susceptance associated with a degree 3 elliptic filter is to be realized with a pair of complex poles at s2 = - 1.742 292:

s(2s2 + 3.68633) Y(s)- ---=----- 2.272 67s2 + 3.007 52

The first step in the realization of this circuit is to move a zero to s2 = -1.742 292 by partially removing a pole at infinity:

The residue is

Y1 (s) = Y(s) - ksls'= -3.035 58

k= 2s2 + 3.686 33

I = 0.6129 2.272 61s2 + 3.007 52 sl=-3.035 58

and the first element is a shunt capacitor

C1 = 0.6129 H

. and the remainder susceptance is

s(s2 + 3.035 58) Y1 (s) = -2.-27_:_

2_6_7--,s2,-+_3 ___ ()()_7_5_ 2

The required complex poles may now be extracted by forming Z 1 (s) and either expanding it in partial fractions or evaluating the required residue. Employing the latter approach gives

s2 + 3.035 58 2k = Z1(s)ls'= _3_03558 = 2.11177

s

46 SYNTHESIS OF LUMPED ELEMENT, DISTRIBUTED AND PLANAR FILTERS

r=-Z(s)

'---------'

� /oF Y1(s)

I Z(:--_..___

--:_ __ __.

Y,(s)

r=� Z(s)

L---------'

FIGURE 3-16 Schematic diagram for

2s3 + 3.68633s Y(s)- -,--c--:-:--:----

2.27267s2 + 3.00752

SYNTHESIS OF ONE-PORT REACTANCE CIRCUITS 47

The reactance assoc iated with the pair of complex conjugate poles at s2 -3.035 58,

2.111 77s

+ 3 .035 58

may be realized as a parallel LC network in series with the circuit with elements

C2 =0.4735 F

L2 = 0.6955 H

The remainder reactance is

or

Y3(s) has a pole at infinity:

Z3(s) = Z1(s) Z2(s)

( 1.63 184

Z3s)= --­

s

Y3(s) = 0.6129s

which may be realized as a shunt element given by

C3 = o.6129 F

The required circuit is given in Fig. 11-llb in Chapter 11.

PROBLEMS

J-1 Obtain the two Fosler canonical realizations for each of the following two immittance functions:

s' + 8s

+9

(s2 + 1 )(s2 + 3)(s2 + 5)

s(s2 + 2)(s1 + 4)

J.l Synthesize the two Cauer forms for each of the functions in Prob. 3-1.

J-3 Deduce the upper and lower bounds on the coefficient Q in order for the following function to be a positive real one:

s3 + Qs

s4 + 10s1 + 9

BffiLIOGRAPHY

Cauer, W ., 'The realization of impedance with prescribed frequency dependence', Arch. Eleclrolech., Vol. 15, pp. 355--388, 1926.

Foster, R. M., 'A reactance theorem', Bell System Tech. J., No. 3, pp. 259-267, 1924.