0 ratings0% found this document useful (0 votes) 88 views13 pagesJ E Storer 1957
This is some information i collected about foster reactance thearom
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14 IMPEDANCE SYNTH
of (2.4). Hence any rational, positive real function is accept-
able as either an impedance or an admittance
PROBLEMS
2.1. Prove that a pole at ¢ = 0 of a driving-point impedance must be simple and
have a positive residue.
. Which of the following functions are realizable as impedances or admittances?
s+3,
s+l
ststs (o St Bt +85 +5,
© wert @ +1
2,3, What are the conditions on the constants a, b, «, 8, y for the following functions
to be realizable as impedances or admittances?
Oy
yet Foe Fe
you find any way of providing the conditions for (b) from the results of (a)?
Prove that the impedance
yt tes +8
a bas +o
8)
is realizable if H, «, 8, a, b are positive and
va - Vor
What property does the impedance possess if the last expression is a
equality? CHAPTER 3
FOSTER’S REACTANCE SYNTHESIS
of general impedances, it is worthwhile
ynthesis of a circuit containing only two types of
elements, In this chapter, the problem of synthesizing an impedance
Feontaining only inductances and capacitance: a sine reactance,
‘will be considered. The next chapter will treat the case where only
resistances and inductances, or resistances and. fee are present.
The first synthesis of a reactance is due to I his
procedure was to observe that a reactance, Z(jo) = jX(w), can be
ritten in the form
As a prelude to the
first to consider the
Gl
Where waye1 are the zer
ant. He then proceeds
of X(w), wa the poles, and H is a positive con-
to prove that
d
de
X(w) 20 (3.2) “
This fact (Eq. (3.2)] in connection with (3.1) implies the property of
‘separation of poles and zeros.” In other words, the poles and zeros of
(@.1) can be ordered in such a fashion that
0 Sw <1 < we a BB 2
B 2B, 2By
ao aye
co L eee
Fig. 3.2. Foster IT synthesis of reactances.
0, in which case the inductance is absent in the
A circuit of this form (Fig, 3.1) is known as the
included by taking wo
first parallel LC cire
Foster I type.
Foster II synthesis is identical to Foster I synthesis except that the
admittance instead of the impedance is dealt with. Thus,
1 _ (8? + wo?) (8? + 2%)
Zs) ~ sls? + wi) (3? + os
¥(s) = FOSTER’S REACTANCE SYNTHESIS
M
Bes
8 — jor © 8 + jorss
Bars Bhs )
(3.8)
e, again, if the admittance is to be realizable, the Bs... must be real
positive. The cireuit follows directly parallel connection of
onant circuits from (3.8) and is given in Fig. 3.2. The leading shunt
ance is not present if a» = 0.
jX(w), it is seen that
+ Age
s, Foster’s original condition (3.2) is verified as being necessary.
ing (3.9) by w and adding or subtracting it from (3.10), the some-
ronger inequality
a (@.1)
ained.
;, perhaps, worthwhile to illustrate the Foster synthesis with a
ical example. Consider the impedance
24 1)(st +3)
8
4S) = SF + 2s +4)
(3.12)
le this function has its poles and zeros located on the imaginary axis
esses the “separation property,” it is realizable as a reactance.
foster I expansion is of the form
a ‘ts Ass 448 3.13
Zs) tedoteqa (8.13)
¢ problem is to evaluate Ap, As, and Ay. is can be done most
ly by dividing (3.13) by s and setting « hen from (3.12) and
Za) _ 8@+ V@ + 3) _ 24 Ae 44
s@+2)(@+4 « 'rt+2'a+4 18 IMPEDANCE. SYNTHESIS
Equation (3.14) is now a conventional partial fraction expansion. Hence,
iB She oe
Ag = lim [xF(2)] =? 22%? =
i nee oc
lim | [@ + 2)F(@)] = (
Ac= lim [(@ + 4)F@)] = Se fe > 2
=3
and so Z(s) [Eq. (3.12)] can be written as
with the cor The Foster H synthesis
pty ;
Zis\—> t As) st
é L
Sb as
oe
oe
3.4, Foster I synthesis of Eq. (3.12).
Fic
. Foster I synthesis of Hq. (38.12). Fie.
is equally direct. ‘Thus,
tai)
ate)
1 _ s(t
Ze) Bs? + 1G
Bas
fg + Bas
¥(s) =
Again, dividing by s and setting s?
@+2)@+4) _ Br, _Bs
8@+D@+3) r+i at
x,
Ga) =
and hence
By = lim ((@ + D)G())
Bs = lim [(@@ + 3)G@)1 = M6
3
B,, = lim (@(@)] = 14
The resulting cireuit is given in Fig. 34.
It is to be noted that the Foster synthesis procedure yields canonic
circuits, ie., cireuits synthesized with the least number of elements.
Thus in the numerical example above there are five parameters: two
poles, two zeros, and the constant multiplier. The circuits of Figs. 3.3
and 3.4 each contain five elements. YNTHESIS 19
FOS! ‘ANCE
It is instructive to view these Foster synthesis procedures as a process
of pole remo As an illustration, consider the numerical example
dealt with previously. This impedance (Eq. (3.12)] can be written in
the form
* + 2)(s? + 4)
y 2:
= Hs) - hy =
It is apparent from (3.15) that the poles at +j 1/2 have been removed
F icom Zs). The circuit corresponding to (
1
OOO Zy\s)
} —-!
Zs 4
Fra, 3.5. Single-pole removal from Z(s) [Eq. (3.15a)).
- Obviously, if the poles at 0, +72 were removed from Z,(s) the result-
ing circuit would be identical with the Foster I synthe:
previously. However, Z:(s) can also be developed on
basis, i.e., a Foster II expansion. Thus,
ee a) ae
~ 6 +2) 6
r +2
ith the resulting circuit that of Fig. 3.6, which is different from those
3 and 3.4). It is now apparent that a large
3 toe
ai
6, Generalized Foster synthesis of Z(s) [Eq. (3.12)].
imber of distinct circuits, yielding the same reactance, can be obtained
p removing poles from the impedance, then poles from the remaining
ihittance, then poles from the remaining impedance, ete. In Fig. 3.7,
ere are indicated nine more circuits which synthesize the reactance
(8.12). The notation below each circuit indicates the location of
p pole and whether it was removed from an admittance or impedance.
‘the basis of this notation, the circuit of Fig. 3.3 would be labeled 20 IMPEDANCE SYNTHESIS
OF, 21, 4; that of Fig. 3.4, 1A, 3A, A; and of Fig. 3.6, 21, » A, 24.
It is suggested that the reader derive some of the circuits of Fig. 3.7 for
himself as exercises.
It is apparent that all 12 of these Foster-type circuits (Figs. 3.3, 3.4,
and 3.7) are canonic. Lest the impression be given that such an array
a
TSO
1 te
F
ft
1, 0A, O1, 0A, OF 3A, 01, 31 1A, 01,34
(a) (e) fy
2 4 12
If SOTO
a 7 Bw ee bok A
Sp es
~A, 1,3, OF A, 01, 0A, A ~A,~I, @A, #1, 7A
(g) (h) @
Fic. 3.7. Generalized Foster circuits for the reactance Z(s) = ae SG +2. {
of networks is exhaustive, it must be pointed out that they are all of the
form of generalized ladders, i.e., of the form of Fig. 3.8. If other types
of configurations are used, many more canonic circuits are usually pos-
sible. Thus, the impedance of (3.12) might be synthesized in one (01
more) ways with networks of the form of Fig. 3.9. However, finding
values of the elements in such an array is likely to be very difficult.
For many network topologies, this is impossible without resorting t
numerical techniques. It is an unsolved problem as to how many differ
ent canonie networks there are which synthesize a given reactance,
Consider now the networks 7 and d of Fig. 3.7 which have particularly FOSTER’S REACTANCE SYNTHESIS 21
simple Indder structures. ‘They are termed Cauer I and IT, respectively,
ter their originator.? Since the driving-point impedance of a ladder
© structure such as Fig, 3.8 is given by the continued fraction
Z=4447———_—
1 i :
e Baty I
Ee Zi" ds +
© a standard synthetic-division procedure can be used to determine the
coefficients in this continued fraction. Hence the element values of
. . >
qi 3 3s
22 Z4, Ze
— an
> Fie, 3.8, Ladder network. Fi, 3.9. Other possible network topol-
ogies for the reactance [Bq. (3.12)].
‘the networks (Figs. 3.7i and 3.74) can be obtained in this manner. ‘Thus,
forming a synthetic division on (3.12) yields
3/8
Bs! + 326? + 24 [s® + Os? + Bs
P 5° + 4st + 3s ds
; ~ 288 F Bs [5 3:
ind the continued-fraction expansion is 22 IMPEDANCE SYNTHESIS
from which the element values of the circuit of Fig, 3.7i are obtained
directly. Similarly, doing this synthetic division in terms of 1/s yields
8 6
P+ a1
ats
49/118
and the continued-fraction expansion is
1
es
AC oer ae
+o.
Tis * 1
6
Ls
and hence the circuit of Fig. 2
Now, the question obviously arises, if one were actually building a
reactance, which one of the many Foster-type circuits should be chosen?
Considering the point of view of economy, as well as the fact that induet-
ances are more lossy than capacitors, the circuit containing the least
total inductance is probably the best choice, On this basis for the .
impedance of (3.12), the Foster I form (Fig. 3.3) and the network of
Fig. 3.7c are the best, with the Cauer II form (Fig. 3.7d) running a close
third. Since the Cauer forms are easiest to derive, and usually one
of them has nearly the minimum inductance possible, they are normally
chosen as the network to synthesize a given reactance. FOSTER’S REACTANCE SYNTHESIS 23
In conclusion, it is desirable to point out that the Caner forms serve
another very useful purpose. In later chapters, it will be necessary
to decide whether a given function is realizable as a reactance. The
Caner development provides a simple and direct way of doing this.
‘Thus, for example, suppose it were desired to find out whether the function
s+ 4s? + 3s + 2s? + 2s
Stsitsitse +l
Zs) =
is realizable as a reactance. ‘The process of locating its poles and zeros
is a tedious job. However, if it is a reactance, it has a Cauer develop-
ment. Thus, one can perform the synthetic division for its continued-
fraction expansion; i.e.,
3
stot ott st + 1s) Fast + 3s? 4 2s? + 2s
e+ st+ st+ stt 8
: s+ +s
8/3
Bet + Bs + 5? $ 5 5
ae
e+ e+
9s
Be +255 + s+ 8
3s? + Gs* + Gs? + 9s
= 48° — 5st — Bs
Since negative numbers occur, the process can at once be terminated as
this function obviously cannot be a reactanee.
PROBLEMS
3.1, Prove that the conditions (3.1) and (3.2) are sufficient for a function to be
_ realizable as a reactance.
3.2. For a reactance synthesized by a eanonie network, prove that the number
of inductances and capacitances cannot differ by more than 1. On the basis of
the dependence of the reactance at s = 0 and 5 = «, give conditions for deciding
whether the number of inductances exceeds, is equal to, or is less than the number of,
capacitances.
F 88, Synthesize the following reactances in at least four ways:
_. (2 + DG? + 100),
a s(s? + 99)
_s(s +2)
(a) Z(s) (b) Z(s)
3.4, Are either of the following functions realizable as reactances?
bg) 2 tot tos St 18s" + 8st + 668
2st F 19st + 4a? +3 238 + 278" +
FF 38 CHAPTER 7
HURWITZ POLYNOMIALS
In the succeeding chapter on Darlington’s synthesis procedure, and in
later chapters dealing with the synthesis of transfer impedances, some
results will be needed concerning a class of polynomials known as Hurwitz
lynomials. It is the purpose of this chapter to present proofs of these
roperties and some of their applications to impedances.
A Hurwitz polynomial Q(s) is defined as a polynomial that possesses
the following properties:
1. Q(s) is real for s real.
2. Q(s) has its zeros located in the left, half plane or on the imaginary
t is apparent from the definition above that an impedance is the ratio
two Hurwitz polynomials. An alternate definition, and”a very useful
perty of Hurwitz polynomials, is expressed by the following theorem:
A necessary and sufficient condition for a polynomial to be Hurwitz any
t the function defined by the ratio of its odd to even parts is realizable as a
re reactance,
Thus, if the Hurwitz polynomial Q(s) is explicitly
x
Q8) = Y gus” = gels) + als)
a
even and odd parts are
Ge) = Zgan8™ — Gols) = Tyanyrs?et!
e above theorem then states that the function qA8)/qe(8) is realizable
a reactance if, and only if, the polynomial Q(s) is Hurwitz.
proof of this property can be constructed as follows: Consider a
tion defined by
: ges) qe(s) 7
Ye) = £2) 2 _ s)_ a
© = Oe) = ae) + G0) ee
(s) is Hurwitz, then the function ¥(s) is positive real, and hence a
izable admittance, since
[. It is real for s real.
47 48 IMPEDANCE SYNTHESIS
__[ae(je)}?___
(qe(Ge)]? = [gol For
3. Since Q(s) is Hurwitz, its poles are confined to the left half plane.
There are no poles on the imaginary axis as any imaginary zeros of Q(s)
are also zeros of 9.(s)
ince Y(s) is a positive real function, Z(s) = 1/¥(s) can be synthesized
as an impedance. Note that, for (7.1),
2. Re [¥(ja)] =
= Re gelde) + Golde) _ 79
Re [Z(ja)] = Re 8 aie) = 1 (7.2)
Hence the impedance 4,(s), defined by
tg ad — Gols) 7
Za) = Hs) —1 = yoy — 1 = (7.3)
is realizable since it merely represents the removal of the minimum value
of the resistance from Z(s), in a fashion identical to the start of the Brune
procedure. Thus, the function Z,(s) in (7.3) has been shown to be
realizable; moreover, it is a pure reactance (its resistance is zero). Com-
bining this result [Eq. (7.3)] with (7.1) demonstrates the fact that if Q(s)
is Hurwitz, q.(s)/q-(s) is a realizable reactance. A proof of sufficiency
can be obtained by reversing the procedure.
One consequence of the above proof is that the even and odd parts of
Hurwitz polynomials have their zeros located on the imaginary axis. A
direct demonstration of this without using the idea of positive real fune-
tions can be quite difficult. It also should be pointed out that the fact
that q.(s)/q-(s) is a realizable reactance if, and only if, Q(s) is Hurwita
means that a simple test as to whether a given polynomial is Hurwitz
can be made by carrying out the synthetic-division procedure disc
connection with the Cauer reactance forms (Chap. 3).
connection with the Foster preamble, this synthetic division will also
locate any imaginary zeros Q(s) might have.'
y Next, consider a realizable impedance
PO) _ pels) + pols) (7.42)
QW) ~ als) + mls)
where the subscripts ¢ and 0 denote the even and odd parts of the poly-
nomials. It will now be shown that all four functions
Pols) Gols) Pol 8) Go 8)
pas) als) ls) ala) ee)
! This process of determining whether a polynomial is Hurwitz, i.e., whether it has
all its zeros located in the left half plane, is equivalent to the Routh test for stability
of mechanical systems. (EB. J. Routh, “Dynamies of a System of Rigid Bodies,” 3d
ed., Macmillan, London, 1877.)
Zs) = HURWITZ POLYNOMIALS 49
_ are themselves realizable reactances. That the first two are reatesplel
~ reactances are direct consequences of the fact that P
(74a) must be Hurwitz polynomials as Z(s) is a positive real faeaa
To prove that the function p,( ) is a realizable reactance, it is
necessary to show that its poles lie on the imaginary axis and have positive
residues. That they lie on the imaginary axis is a consequence of the
fact that Q(s) is Hurwitz, and hence all the zeros of q.(s) are imaginary.
To prove that the residues are positive, the original impedance of (7.4a)
can be written in the form
: Bels)/qels) + pol s)/qe(s) 7
Ce) V+ gols)/qe(s) )
= Let je, be one of the poles of p,(s)/q-(s). Evaluating (7.5) at ja, yields
a [ts 4 Pals |
Ze) = lim ds Lacs)" gel)
sie, d qo(s)
& [: i #¢|
= Pe( Jerr) /qe(Ger») + Dol jerr)/ qa jan)
Gol Jor») /qi(Jon)
(7.6)
‘where the primes denote differentiation with respect tos. It is to be noted
in (7.6) that p,(je,)/qi(Je,) is purely imaginary, while p,(jo,)/q!(jo,) and
go(Jo,)/q5 (Jw) are real. Taking the real part of (7.6) yields
Pol Jerr) /Gi( Jere)
pe ql jeas)/ qe Jeo) Hae
which must be positive since Z(s) is realizable. It is to be noted that
o(jes»)/q, (Jor) is itself positive, since it is the residue of the pole at jo
the realizable reactance q.(s)/g.(s). Hence, from (7.6),
PolJeo) =0
GG
hich states that the residue of p.(s)/g.(s) at the pole jw is positive.
Hence p,(s)/qo(s) is a realizable reactance. A similar proof can be
ried out to show that q,(s)/p.(s) is a realizable reactance.
In conclusion, it is worthwhile to.consider how one might determine
ether or not a given function Z(s) is a realizable impedance. There
at least three possible methods, namely:
1. Determine whether or not Z(s), or 1/Z(s), is a positive real funetion.
©2. Determine whether or not Z(s), or 1/Z(s), satisties conditions (2.4).
»3. Use a synthesis procedure to determine whether or not Z(s), or
/4(s), can be synthesized,
