Ill

\

ll1ll||1|1|||

I!
H

I
|

HI

Hll

1|

 

NETWORK SYNTi-iESl-S. PROBLEMS AND
{EXAMPLES ILLUSTMTING THE
USE 01" THEOREMS

Thesis fox flu Doom of M. S.
MKHIGAN STATE COLLEGE

Ncah Herbert Kramer
3949.

This is to certify that the

thesis entitled

Network Synthesis. Problems and
Examples Illustrating The

Use of Theorems.
presented by

Noah Herbert Kramer

has been accepted towards fulfillment
of the requirements for

 

Me So degree ill—L.

 

NE'I“..’ORZ‘Z SYNTHESIS. PliOifiLEMS AND EXAMPLES

ILLUSTRA'I‘ING THE 33 OF 111303316".

"Y

NOAH KERBERT KhAqu

M

A TEES IS

Submitted to the School of Graduate Studies of Vichigan
State College of Agriculture and Applied Science
in partial fulfillment of the requirements
for the degree of

wasrsfi or SCIENC~

[.

A CKNOW LEDGEL‘ZLZN T

The author wishes to eXpress his thanks
to Doctor J. A. Strelzoff for his help in the
develOpment of this thesis, and for his patience

in reading the manuscript.

No Ho lXI'alTieI’

( OJ
5') 9“” ._
(5.1 1 h)" ‘

TH [SIS

Noah Herbert Kramer

(ERR (\
w \ /
//K¢¢/ - €77.27 C-

{\wammkk‘l V‘Q

ABSTRACT

An R-C Low- Pass Filter whose Bower Insertion
Ratio approximates a prescribed function of fre-
quency is synthesized.

The Filter is in the form of a symmetrical
lattice with prescribed, but equal, terminating
resistors.

The problem of approximation is first consi-
dered.

The realization of a lattice that approximates
a trapezoidal pattern is then shown.

A few realization conditions, peculiar to this
thesis are then discussed and some avenues for

further study are made available.

Wet

I. Introduction.

The purpose of this thesis is to
present some examples of some theorems on
Network Synthesis.

Most of the theorems will be taken from
the book 'Network Analysis and Feedback Amplifier
Design' by Doctor Hendrik w. Bode.l

\JGenerous use has been made of two books

by Doctor Ernst A. Guillemin,2:5 particularly the
second volume of 'Communication Networks'.

The amount of help derived from the last
named book is not sufficiently noted by references
alone, for the book was used to get alternate
approaches to problems presented by Dr. Bode.

After a short discussion to explain and
define terms, the thesis will follow a standard
pattern.

OOCOIOOOOOOOOO‘OOOOO00.0.0.0....0.00COOOOOOOIOOOOOOO

l Bode, H.W. Network Analysis and Feedback Amplifier
Design. 1945 New York: D. Van Nostrand.

2 Guillemin, Ernst A. Communication Networks
Vol. II 1935 New Yerk: John Wiley

5 Guillemin, Ernst A. Communication Networks
Vol. I 1955 New Ybrk: John Wiley

There will be presented a portion of a theorem
from.Dr. Bode's book. This may or may not be
paraphrased.

Then a demonstrating example will be formed,
and the necessary work will be done to show the
mechanism of the theorem.

If a conclusion is deemed advisable, it too

will be presented.

2. The problem of Synthesis.

The study of circuits is most generally a
problem of analysis. This is simply; given a
circuit, find its behavior.

In this thesis the problem.is not analysis,
but synthesis. Given a desired behavior of a
circuit, design a network that will have that
behavior characteristic.

The latter problem.may lead to circuits
that are not readily formed with present laboratory
equipment. The analysis problem has generally
a unique solution, while the synthesis problem

seldom leads to a single result.

5. The 'p' notation.

It is necessary to find some symbolic
notation to describe the circuit that is under

4 provides such a

study. Operational Calculus
handy method. In Operational Calculus the Operator
d/dt is noted as 'p' and the operator ‘fdt is
written as l/p. Thus the common expression
Ldi/dt + (l/c)jidt can be written as (Lle/cpfi.

Since iji is the voltage across the induct-
ance L, it follows that p is equivalent to jm, where
w is the angular velocity 2wf, and j is the familiar
operator -1 .

This gives another form of expression
(Lp+l/cp)i = iji + -j/wci.

Guillemino uses )\ instead of p.

4. The 'p' equations.

The p equations can be easily formed in an
analysis problem. Either the mesh or the nodal
equations can be written for the network, in terms
of p. Then the problem would call for the solution
of an unknown current or voltage. This can be done

by Cramer's Rule.5’6

4 Carter, G.W. The Simple Calculation of Electric
Transients, 1945 New Ybrk: MacMillan

° Ibid 2

This rule gives a result which is the quotient of

two polynomials in p. The form is %%E% .
P

This result can lead to several continuations.

If only a steady-state solution is required
p is replaced by jm and the fraction is then evalu-
ated.

If both transient and steady-state are req-
uired, Operational Calculus, Laplace Transform, or
even classical Differential Equations can be used.

All this is the analysis problem. Now consider
the synthesis problem.

Consider the case where the circuit is to be-

have in such a way that the driving point impedance

can be written as 5%2% = Z .
BP

With that as a starting point, the problem is
to synthesize a circuit with the above driving point
impedance.

5 Doherty & Keller. Mathematics of Modern Engineering.
Vol. 1 1956 p63 New York: John Wiley & Sons.

6 Pipes, Louis A. Applied Mathematics for Engineers
and Physicists. p86, 152 New Ybrk: McGraw-Hill.

The problem might have arisen from a very commonplace
engineering experience; a complex circuit might have
been designed and it's driving point impedance found
to be 21.

The circuit designer wishes to find a circuit
that has the same 21, but is composed of different
elements and possibly fewer elements. Now he is faced

with a problem in synthesis.
5. The polynomial in 'p'.

A few interesting facts can be learned by obser-
ving the character of the polynomial in p. Assume

the polynomial is expressing a current function. Then
- - 3 - h
there can be written: 1 = &( := C(p'p') (P3%T%_LEHEI% .
s p (p-p )(p-p p-p

The primed p represent the poles of the current, and the

 

unprimed p represent the zeros. Each pole and zero
can be either real or complex, depending on the form
of the polynomial.

If Operational calculus were to be used to
continue the solution of the current the result would
be a series of increasing and decreasing exponentials

with both real and complex powers.

If the poles were not simple, the result might
be an extremely complicated function of time.

A knowledge of the roots of the eXpression
will give some indication of the form of the

solution.
6. Plot of 'p'.

A small plot will give a picture of p.
Assume p to be in the denominator.‘ Note: since

p = jm, an imaginary p will represent a negative

real frequency.

 

 

 

fig. 1

U

-E.
3 \D
_ a:
§ 21
at \J '
u. 2‘“:
X": ”'2

l—w
1'3 :01,
fig £91. #realp
¢3 P3
‘50 $0
a. 2
2 o
E a.
x X
Ht LIL

-¢~ imaginary p

 

7. Physically Realizable Networks.

This thesis is concerned with only physically
realizable networks. They are defined as, 'networks
of real elements, having no modes of free vibration
that increase indefinitely with time.'° Thus the
forms eat elwt, ejmtt are not allowed.

From physical considerations, Dr. Bode has
made a list of the requirements of the zeros of

B(p) which is the same as the poles of %%E% ,
P

These are the requirements for a physically real-

izable netwak.°°
8. Network Criteria.

1. Zeros and poles are either real or conjugate
complex pairs.

2. Real and imaginary components are respectively
even and odd functions of frequency on the real
frequency axis.

5. No zeros in the right half plane.

4. Zeros on the real frequency axis are all simple.

5. Real component of driving point impedance cannot

be negative at real frequency.

6. Passive power must not be more than that consumed
by generator and its conjugate.

7. A driving point impedance that meets the above
list, when it has no zeros on the real frequency
axis, is known as the ' minimum susceptance type.
If the network has no Poles on the real frequency
axis, the network is called the ' minimum reactance

type'. These two types are not mutually exclusive.

° ibid 1 p. 120
°° ibid 1 p. 125

-8-

9. Multi-Resonant Circuits.

The circuit designer might have designed a
circuit composed of many meshes. These meshes might
have any combination of R, L, and C. He then looks
at his network and asks if it can be reduced to a
more simple form. Economy dictates that he uses the
least number of elements.

He can easily evaluate the order of the denominator
of the driving point impedance function. The procedure

is as follows:

1. Number each mesh.

2. Follow the contour of each mesh and assign a value
to that mesh.
Value is zero if there is neither L or C in the mesh.
Value is one if there is either L or C in the mesh.
Value is two if there is both L and C in the mesh.

5. Check off one L and C in the mesh if they appear
there. Avoid checking off an element that appears
as a mutual, if this is possible.

4. Add up the total number of mesh values.
This total is the order of the determinant of the

driving point impedance denominator.

OOOOOOCCOOOOOOOOOOOOOO00....O....OOOOOOOOOOOOOOOOOOOOOO

° ibid 2 p. 187

Here is an example illustrating this technique:

fig. 2

 

 

 

 

 

 

 

 

 

 

 

6
5'
L
mesh # Value
1 2
2 2
5 2
4 l
5 1
$3. 9.
Total... ....8

-10-

10. Multi-Resonance; Non-Dissapative Cases.

A circuit composed of only pure L and C is non-
dissapative. A plot of its reactance versus time will
show a series of alternating zeros and poles.

There are four basic cases. They are dist-

inguished by their behavior at zero and infinite

frequency.
(a) Zero at O cps ----- zero at infinity.
(b) Zero at O cps ----- pole at infinity.
(c) Pole at O cps ----- zero at infinity.
(d) Pole at O cps ----- pole at infinity.

Their plots appear below.
fig. 3

—_-

av ‘49

 

 

—- qb—_—

(<7 (d)

 

-_ _-—

-—-‘_‘--

‘—~—-—-
I\

-11-

11. Canonic Forms.

The plot of reactance versus time gives the
basic character of the circuit. The original circuit
can be reduced to a canonic form. These forms will
have the same reactance curve as the original function,
and will contain the least number of elements. A
given circuit can be represented by a very large
number of canonic forms so economics will determine

which one to use.
12. Preparation of the Problem.

Assume that a multi-resonant circuit has been
designed. The driving point impedance can be written
by elementary methods and is called Z = g(p .

Or perhaps the problem might have arisen that a
circuit is required to have a given distribution of
zeros and poles. These are problems in synthesis
and these problems are solved by resorting to
canonic forms.

The two most common methods will be shown.

-12..

13. Foster's Reactance Method.

Foster7 has devised a method to draw the canonic
form.

The required information is the location of the
zeros and poles, and the reactance at any real freq-

uency.

From.g(p) it is easy'tO‘write the form

2 = H(p2-wf) (pa-mg) (pa-w§)p
(pa-mg) (pa-mg) (pa-mg)

The roots of the numerator are the zeros and the

roots of the denominator are the poles.

The most general forms of the solution are

as follows:

”WEB—Cit

In the four basic cases mentioned before their

fig. 4

 

 

solutions will appear as:

7 Foster, R.M. A Reactance Theorem. B.S.T.J.
April 1924. pp259-267.

-15-

First Type:

 

 

fig. 5

 

~ Jul W_‘—"'
I.__.

 

 

Second Type
! .L
(a) m I lr
_ _____ J
i_ l‘ oh_
(.H__rr11nr\____4p__—_.
<y__ITBKTW #4}
.L_ ______ J

 

 

 

 

 

 

-14-

 

 

 

 

(01f

 

 

 

An example of the use of Foster's Theorem is given:

Problem: Design a circuit with this characteristic:

zeros at w = 5,5,7, The reactance curve:

poles at w = 2,4,6, (//[ //fi A/// ////.
I
X = j12.5 at w = 0.5 :///' ///' C/
i I .

X = ij(wz'm3) (we-g3) (”2'05)

 

 

Evaluate H
(ma-mg) (ma-mg) (mg-wg) w=0.5
312.5 =_1.5H(.25-9)(.25-252(.25-49) 5

(.25-4)(.25-16)(.25-36)

H=5=L2h=L8

 

 

 

_ - Y _
Ck _ 038030)]: LR "' l/wflck
02 = 3L4’161(4'56) = .01825 L“=lOQ/4(l.6) = 15.58
5(4-9)(4-25)(4-49)
04 = 'L15’4)(16'55) =.o2510 L4=lOO/16(2.5)= 2.704
5(16-9)(l6-25)(16-49)
_ -g56-4)(36-18) = = - =
Ce — 5(36_97(36_257T55:49) .03315 L6 lOQ/o6(5.5) .858

The final network will look like this:

Lg [2 [y 16
{TYYL_.__

F—- «—4+~— I
1‘. Cr ‘4

-15-

The alternative Foster solution.

L. L. L5? L7

05—]. (35]- 07 r]:

Lk = wfi(wfi-9)(o§-2s)(w§-49)5
(mg-4)(wfi-16)<w§-ss)(mg-mg)

= = = 9 25 49 5) _
L1 (w 0) .a2. 4 16 56 - 23.92

5L9)(9-251(9-49)

L5 = 9-4(9-16)(9-36)

= 30.5

5(25)(25-9)(25-49)

(25-4)(25-18)(25-36) = 25:1

L5 =

L - 5(49)(49-9)(49-25)

.. = 12.18
(49-4)(49-l6)(49-35)

-15-

_ - mZ
Lk — _%——a
w -wk
ck = 1/wfiL
Cl = 0
CS =.OO564
C5=.OOl73
C.7 =.00167

14. Cauer‘s method.

Cauer8° has a method similar to that of Foster°°, to
reduce a non-dissapative network to a canonic form.

The loop equation of the network is written.
The driving point impedance is then written as a
fraction of two polynomials in p.

There are two possible solutions The terms can
be arranged in either ascending or descending powers
of p.

This choice will determine flhe form of the
solution. The two forms are equivalent, but one may
be the wiser choice from.the standpoint of economics.

After the terms are placed in order the method

of continued fractions is used to bring about a solution.

The two basic forms are:

 

 

 

 

8 Cauer, W. DieVerwirklichung von Wechselstromwider-
standen vorgeschriebener Frequenzahabhangigkeit.
Archiv f. Elektrotechnik, vol. 17. P.555, 1927

° Ibid 5 p.198
°° Ibid 7

-17-

Cauer Method Example

2x10'6fd.

Given: L1 2xlo'5h cl

C} ‘L 67']— L5 05X1O-5h CS 4x10'6fd.

‘43-? 4/? FIND: Canonic Forms.
1

O (L1+L3)P + (l/cl+l/C5)(l/P)
Lsp + l/CBP -L5p + l/Csp

-L3p + l/Csp (L1+L5)p + (1/c1 + 1/c3)1/p
250k _5 1 250k 3
z' .__ .5x10'3p+ r (2.5x10 )pmwsom - .SXIO'Sp-I- p

2.5x10-5p + 750k
p

 

 

 

= 1x10'§p2+750+125,000x106 = 10'p4+750p3+125x109

2.5x10-3p+ ZEQE 2.5x10“5p5+750x103p
p

Arrange in ascending order.
2 = 125x109+7sop3+1x10‘5p4
750,ooop+2.5x10-5p5

Continued Fraction Expansion

+.....
750,000p = 2.53:10'31)a /1.25x10§ + 750p‘ + 10‘6P4

1.25x109 + 416p8
5325' + 10‘5p4

 

167x103 + 554p“ + 10'§p4 167x105
-
P 750,000p + 2.5x10 5P5 P

Invert the remainder and continue the expansion.

-18-

3
a 4 2.25x10 / +
354p + lop / 750K103p + 2.5XlO-3p o o o o o o o o o o

750x105p + 2.25x10'3p3
.25x10'5p5

 

 

Result 2-25X103 + .25x10'325 -—§ 2.25x105
P 534p8 + 10'6p4 p

Invert remainder and continue division.

/F___________1§in;Q§_£n +..............
.25x10‘3p3 554p2 + lo-6p4

 

 

 

 

534p2 + o
10'5p
1540x105 + 10-624 _~, 1540x105
p .25x10-5p5 p
Invert remainder and divide.
250[: if, 250
10’6P4 / .25x10-5p5 ““9 p

The successive results represent a series of Z,Y, in

order.

W = Cl = 6x10 fd

 

p
5 -
212§£19_,= L1 = .00445x10 5h.
p
6
l:§%§;9_ = 02 = .746x10-5rd.
2:0 = L2 = .004 h.
Thus the Cauer form is:
C,‘ ‘ 6; fig. 10
4, A;

-19..

The other alternative

2 = 10'6p4 + 75099 + 125x109

2.5x10-3p5 + 750,000p

44x1 '51) + -------

2.51:10'5293 + 750.000p /10-5p4 + 750p T125x10g
10-6p4 + 500p8 + 0

 

4505*“?‘125x10g‘
.4x10-5p=L1p

a 9 5.55x10-6p + -------
450p + 125x10 /2.5x10-5p8 + 750,000p
2.5x10'3p5 + 695,000p

 

8.1x10-Qp
55.000p /450p‘ + 125x10g

450pa

 

125x109
o 44X10-6
125x109 /‘“‘3§;fi$$;“‘JL'

Network: 1., [2.

55,000p 6
5.55X10- p=Clp
8.1x10-5p=L2p

.44x10’5p=c2p
fig. 11
L1 = .4 mh
L2 = 8.1 mh
cl = 5.55uf

Note in this particular problem, the second form

gives results more easily constructed than the first

fo n. This is not a general rule.

-20..

15. Minimum Resistance Function.
Theorem°:

A passive immittance will continue to meet
the requirements for physical realizability in pass-
ive networks, if it is diminished by any real con-
stant, as long as the real component of the resultant
empression does not become negative at any real
frequency.

To use this theorem.it is necessary to be able
to convert a p expression into some eXpression
containing m.

Take the p expression, replace p by jm, and
then separate the real and imaginary parts. Call
the real part R'.

From.R' can be subtracted any real number, as
long as the remainder R" is never negative for any
real frequency. R' will reach a minimum value at
some frequency, and as long as this value is subtracted
from the original R', the new expression R" will
never be negative.

Illustration:

 

Separate function of p into real and imaginary
components. Differentiate the real part and find

the minimum.value. Subtract this value from the

original expression.
O...OOOOOOOOOOOOOOOOOO0.0.00.00...00....0.00.0000...

° Ibid p. 172
-21...

Illustration: Minimum Resistance Function.

 

 

 

 

 

 

 

 

Given.a passive network FIND
3+ +2 Minimum Resistance
Z =‘§—:R———— Function.
..... 2-1§2:é-----------__--__-_----_-_-_-----___-_-_
a
p = jw 00000000 Z = w +jw+2 rationalize
-5w”+3jm+4
Z = -w8+iw+2(-5w2+4-51w) ...... 5w4-llma+8+l(-2w3-2w)
(-5mz+4+5jm)(-5w‘+4-350) (4-5w ) +9w
Separate real from.the imaginary. Z = R+jX
R = 5m4-llw‘ +8 X = -2m5 -2m
(4-5w2)“+9w‘ (4-58”)“+9m“
Let X = ma
(4-521)a +9x (4-5x)2 +9x 25x3-51x+16

g; = o = (25x3-51x+18)(10x-11)-(5x“-11x+8)(50x-31)

0 = 120x“-240x+72 = 10x2-20x+6 ..... x = 20t1@063215
X. = 1.65, .5675 20

 

Choose the first root.

x 1.65 ....R 6 .2-50. + 6 _ EI:7'— 0,105

This is the minimum at we = 1.63. By inspection we see
that the alternate root gives a higher valued solution.

The minimum.resistance function is then
2-0.105 = 23(1-.525)+p(1-.5151+2g1-.420)
5p2 t 5p + 4

z' =

= .481;2 + 0.69p + 1.58
5p2 + 3p + 4

-22-

Illustration: Minimum Conductance Function
The same procedure can be followed to find min-

imum conductance with an admittance function.

 

Given: Find:
a
Y = 52+3P+4 Minimum Conductance Function
p + p + 2

Y = -5wa + film + 4 -5mz+3jm+4 (2-m2-jw)

 

 

 

 

 

 

 

 

-w‘ + 3w + 2 m4 -50” + 4
= -10w2=61w+8+5w4 ~3195 -4w2+51w5+5w8-4jw Let x=m3
w4 -5w3 + 4
Y _ 50 -1183 +8 +1(2w5+2w)
w4 -5w‘ + 4
G = 5Xa-11X +8 B = 2Q5 +20
x2-5x+4 m4 ~5ma +4
'%E = 0 = (x2-3x+4)(1Ox-11) - (5x3-11x+8)(2x-5)
.o = -4x8 +24x-2O Root x=l
le c = 5-11+8 = 2 = 1
l -5 +4 '5
Y!=Y-l=§£8+3p+4 ‘1:
p8 +p + 2
8
Minimum Conductance Function =té2 +2P +2
p3 +p +2

-23-

15. Minimum Reactance Function.

Theorem°

A passive will continue to meet the require-
ments of physical realizability in passive networks
if it is diminished by the reactance or susceptance
of one of its real frequency poles.

This is the same type problem as the minimum

resistance problem.

 

 

 

 

 

 

 

 

Illustration:
Given: FIND:
B
Z = 2234i2il Minimum.Reactance Expression.
p3+pa+p+l
z = ZR? fP +1 .... = 223 12 +1 roots, =1
p5 +p2 +p +1 (P+1)(p3+1) P=t3
A0 = lim z(p-pQ) = lim.pz-p3 2 p: =-1
p po 9 p0 2PO
A = 11m.(p2-1)g2p3+p+1) = -2+ +1 _ 1
0 z 23(1137 "5
p j 23(p+l)(p +1)
In an anti-resonant circuit D = 2Ao L = -220
P
D=l L=1 °
2 g 8
Z = E +(Z') = ZPiip+l Z = 2p fP+l _ E
l+p (1+P)TI+P 5 (p+1)(p8+l) l+p

 

z' — zpfinilgnfzn _ .p“+1 1

- (P+l><p +1) " <p“+1)<p+1) = FT

2 =._l_ +.__E_. 1::rvm
p+1 p2+l F—- I Fig. 12

.OCCOOCOOOOOOOOOOOOOOOOOOO0.0000......OOOOOOOOOOOOOOOOO.

° Ibid 1 p. 175.

 

16. Brune’s Method.

It may be required to find a circuit that is
represented by a p expression. As long as the ex-
pression represents a passive network, this can be
done by the method of Bruneg. He has shown that any
impedance satisfying the requirements to be passive
can be represented by a physical network.

To do this, use is made of the technique
developed to find minimum resistance, conductance and
reactance functions.

The idea is to remove the poles of impedance
until a simple resistance is reached.

The procedure is:
with a p expression given, change it to a minimum
resistance function. Then remove the poles one by
one. Remove a pole, subtract it from the minimum
resistance function, and keep doing this till there
remains only a pure resistance.

The process may be laborious.

9 Brune, 0. Synthesis of a Finite Two Terminal
Network Whose Driving Point Impedance
is a Prescribed Function of Frequency.

Journal of Math. and Phys., vol. 10, pp.191-255.

-25-

Illustration: Brune Method.

 

Given: FIND:
a .
Z = 4P2 +4p + 2 A Network to represent
5p + 3p + 4 this expression.

From previous work the minimum resistance occurred
when w8=l.65. The minimum resistance function was

found to be

2 = °43P2+459P+1.58 m2 = 1.65 w = 1.275
5p2 + 5p + 4

= -.48(1.55)+L.591);1.275)+1.58 _ .7971 .882
-5(1.65) + j5(1.275) +4 ’ =4.IE%35?§5

.211Z-90° a pure negative L or a pure C.

 

.165h L = -.165h

At w = 1.275 - L = .211/1.275

Subtract from.Z'.
2.. =.48p”+.88p+1.58
5p2+5p+4

s
This can be factored. = 2'4gp'538§;-E9Z)QP +1.54)

-.,165h = .825p3+.975p?+1.55 +1.58

 

 

Invert and consider Y"

2
Y” = 5 +5p+4 = .
3744I.558p+.597)(p'+i.64) P°1e at P i 1 28

A = 1im PI:P§X = 1.64 5 +5°(1.28 +4 = 1.55
° p - p0 2P0 2.56Jl.397+35112287T.55§TT§.EZ)

l/L = 2A0 L = .525h c = 2Ao/p: = .510/1.54 = 1.89
This is a series resonant circuit in shunt

 

 

Fig. 15

-__JTII\- it

L:.323H (“.5099

 

-25-

The admittance of this shunt element is:
gglgg’ Subtract this from'Y" and get yv1:
064+p2

Y' I i = 5L8+5P+4 -5010
2.44(p“+1.64)(5.58p+.597) 1°64+P

= 5p“+5p+4-2.55p‘-5p _ 1
2.44(1.84+p“)(.558p+.597) -2.44 .558p+.597

Z"' .338p+.397 ........ Lp+R
L = .338 R = .397
Now that all these computations are finished, the

network can be drawn.

 

WI

00105 “016511
.323h .338h
-- .1.9 fd .397

 

 

The Brune treatment of an impedance function has
been shown. The same treatment can be used to

handle an admittance function.

 

Given: FIND:
_ 5 3 +5 +2
Y _ P:2+pt§ Brune network.

Minimum conductance when w=1 =1

Minimum conductance expression

 

 

 

 

a
Y‘= §E_$§2$§ This was shown previously.
At co=l ---- Y' 5—17.33 :2 $231 = 21:20;
Treat as a negative L ... w=l ... l/L = -2 L=-.5h
Z" = 111—3 +2 -- o5p -_- n“-m+2+2p5+p‘fla

2p2fip+l 2p” +p +1

= 95+pa+9+1 = (p”+1)(p+1)

2pz+p+1 2pa+p+l
Y" = 3222.1. A0 = limm‘mfi Z

(p2+1)(P+l) p——gp° 2po

A0 = -2 + +1 _ l
2(I+353 2'

In an admittance circuit series resonance

2
l/L=2Ao 0:15.;
:1 =1

Eliminate a series resonant circuit.

-28-

ym=.§2;:atl__ 2.4.2311... 1 1
P +1)(P+1) (p2+l)(p”+l)(p+15 p+l " p+l
This is a series R, L circuit in shunt with the rest

of the circuit.

 

 

 

 

W i
"' 0 5p 1p
1 l
l/p .T 113
Fig. 15

The Brune method is more laborious in the
impedance case. This is not a general rule, but
is due to the choice of the original function.
Another function might have been chosen which.would
tend to give a result most easily when the impedance

approach were chosen.

-29..

17. Negative Resistance.

The concept of negative resistancelo: 11
is introduced to facilitate the change from a
passive network to an active network. The negative
resistance can be considered as a source of power.
One method used to realize this element is to
use the negative s10pe found on a portion of the
Tetrode Tube Characteristic.
3 From B to D the slope

1 of the curve is negative.

Fig. 16

 

e
The use of negative resistance allows the

use of two other circuit elements. They are
negative inductance and negative capacitance.

In the general form, a T network is set up.

The shunt element of the T is a negative resistance.
The other two elements are equal in magnitude to the
shunt element, but opposite in sign. The T is term-

inated with Z". The input impedance is Z: --

..........OOOOOOOOOOOOOCOOO00...........OOOOOOOOOOOO

10 Everitt, W.L. Communication Engineering.
Second Edition, p. 509 McGraw-Hill Co. 1937.

ll Emery, W.L. Ultra-High Frequency Engineering.
p. 162 Macmillan Co. 1942.

° Ibid p. 188.

-50-

 

 

 

 

 

N
II

R + -R(§+Zz = R _ Rz-RZ

Rz-R2 -RZ -R“
z '12“

This is the case where Z is a resistance.

 

In the case where Z is pure L.

2' = R+ ‘R(R+1@Ll. = 1wLR-R?:ijR _ -R8 wL _ -33
jdL ij " -w “ 353

Where ZH is pure Capacitance
2: = R+ -R§R+1ijC2 = R -R' .. R
1 ij ij 105 -33
l/ij l/ij
Thus any negative impedance, including negative
capacity and inductance, can be represented by
terminating the T in the positive inverse of the

required impedance.

-51-

An active network can be broken down into a combi-

nation of a passive network and a negative resistance

 

network.
Z = rpfip+4 Y = 42::42:2
4p8+2p+2 p2+p+4

Y'-l = Y = 42312213

 

 

pa+p+2
Y' = 4122+213+2+p2m+2 = 5p‘+5p+4
p2+p+2 p2+p+2

This is an impedance in parallel with a negative

resistance.

-52-

18. Inverse Network.

The inverse networkl2’° is a useful tool. The
inverse is usually taken with respect to a constant
R. When this constant is set equal to l, the structw’
found is called the reciprocal.
The dual is often confused with the reciprocal net-
work. The dual is just a device used so that a network
can be solved by means of he node equations. In this,
the sources of voltage in the original network are
replaced by sources of current. Instead of deal-
ing with the impedances of the elements, the admittances
are considered. However, there is no rearrangement
of the circuit elements.

In the reciprocal circuit this is not true.
Mathematically, the Matrix of the impedance is
formed. Then the inverse of the matrix is found, by
conventional Matrix Algebra. This inverse is then
written as the matrix of the admittance of the
circuit.

The inverse can be found in many cases with-
out resorting to Natrix Algebra.

......OCOOOOOOOOOOOOO......OOOOOOO.........OOOOOOO

12 Zobel, O.J. B.S.T.J. January 1825, July 1928.

° Ibid 5 p. 205.

-33...

To find the inverse by geometric methods the

following procedure is followed:

Locate a point in the center of each mesh, and
one point outside the circuit.

Connect all the points by a series of lines.
When a line passes through an element, that element
will be replaced by its inverse.

An illustration follows:

I}

 

Fig. 18

 

 

 

 

 

 

Fig. 19

3I”

\

 

 

N
.g

 

 

This method, called ‘Structural Inverse'
will not work for all passive networks.15 If the
required inverse is to be passive, the original
network must not have any mutual inductance, for
there is no inverse that is passive for mutual
inductance.

However if active networks are allowed,
the mathematical approach will always give a result.

The inverse is found, for example, for a
Brune type network. Since the product of the two
networks must equal a real number, it will be Shown
that the two following networks are inversely

related to each other.

pz+p+2

Fig. 20

 

 

z” = 1P2+P+2)1
2(2p“+p+1)

 

 

2!! z: = 2 8+ +1( 8+ +2 1 = 1
p +p+2)(2p +p+1)2 ‘2

0.0.0.0...000............‘OOOOOOOOOOOOOO......OOOOOO.

13 Foster, R.M. Geometric Circuits of Electrical
Networks. A.I.E.E. June 1932.

-55-

19. Complementary Networks.

A complementary network° requires that the
sum of the original network and its commlenentary

network be a real constant.

Illustration.
L C = _W——
‘L__AJV\_JL ;[::::::1:f'

 

 

R R
Fig. 21 4~-/

2 = L2H + R c LpRS+La/C+LR/C+R§/Cp
LP+ +1 Cp = a
. LpR+R +L/c +R/Cp

 

=AB(LpR+L/C+L/C+E/Cp)

(LpR+R“+L/c+R/cp) = B

This shows the two networks to be complementary.

° Ibid 13 p. 199.

-55-

20. Partial Fraction Expansion of a
General Impedance.
Given an impedance function of p, by the
method of partial fractions,14 a physical network
can be drawn.

Given:

2 = 223:2219 = 3ip-l/758)(p-.564)
P”'5p+2 <p-1>(p-2>

p018 at 1):]. 0.00000 5(“0756)(l-0564) = 1
first fraction 2' = 1/p-1 This is -R // c

Subtract this fraction from.the original.

2 2
Residue at p = 2 Res. = 1
Second fraction 1/p-2 This is a negative
R of 1/2 in // with c = 1.
The network will look like this. (See Fig. 22).
This form can be changed by distributing the 3 ohms

along the circuit. Then the circuit takes the form:

1; .1 -L /J'

'/ wf, 3 / 2 2-
mpfid L113 Pub

14 Esbach, O.W. Handbook of Engineering Fundamentals
First Edition 1936 Ch. 2-08

° Ibid 5 pp. 192, 206.

-57-

Continue the analysis.

 

44 - 92
“W . =__1_+_1_._Z
2 p-2 - p-2

Fig. 25

This expression can be represented as:

2
# Fig. 24
all I.

 

’0.r/b
Also analyze: /
l ’ l + l _ = 'IIII.I
“An/l 5'2]: " fill-I .0 '0
l ”°
/ Fig. 26.;
Fig. 25.

Then the entire network can be written as:

-IJP ’f)‘
Fig. 27

5%“

 

This is only one out of many variations that can

be constructed.

-58-

A more complicated network is solved as an example

of partial fraction synthesis.

2+ +2
P P
a
=_Lp fp+2) find residue

 

(p+.3-.844J)(p+.3+.844j)
p = -.3-.844j
Res.

-062+j0506-05-0844J+2 _ 1008-15038
-1.688j - -1.688j

Residue = .2+j.643

z' = 2£Cap-(Capa:0bp2)) = 2(figp-g-,oe+.545))
p -2pap+(pa +Pb ) p2+.6p+.09+.71

 

2(.Zp-.48§i = .4p-.958 = .75(.554p;1.281
pa+o6p+o80 p2+.6p+o80 p“+.6p+.05+.75

 

Assume R = .75 in shunt, and remove it from the Z.

 

 

z: =.ag§£P-1028 roots -.5, -.l
p +6p+.05
fp+.5)(p+.1) res1due at p .1 1.33/4 3.33

elements -3.33/p+l subtract from Z' and get:

 

 

 

 

 

Z" = .53fip-1.28 - - 5.53 _ l _
(p+.5)(p+.I) B‘TI' ‘ p+.5 This is §:§
The complete network will then appear as:
-3.3 2
i nnA__ fVVL_____
Fig. 28
i. JLk, IL
075' fij' —j'
-—.3 /

 

 

-59-

21. ‘p' Networks.

The partial fraction expansion results in
expressions that represent two element networks.
These expressions and their corresponding ferms

are shown below.

 

R R -R -3
13323" M
L -L L -
RLP RLP RLP RL
LP:R LE:R IR:qu -R-Lp
R R -R .3
CH ‘Tlg
1 C ng-p pléCCR p+-1éC

 

 

F E? '9'
iJ
L}
a
l

.\ C C .-
p2+l/CL 17%L‘1L-p _2[____2 --P/-—--3 C
p -l/CL p +1/CL

-40-

22. Reconstruction of a Passive Impedance

From a Knowledge of Either Component.

Darlington15!° has shown that, given a resistance
function of time, or w, the reactance function of
the same impedance can be obtained, and visa versa.
The sum of the two functions should be a minimum
resistance function.
To illustrate:

A R(m) is given. Then a 2(a) is found.

Then the nex t step is to find the X(m).

 

R(m) = l'wz
T641+3)(w-l-j)(m+1+j)(w¥l:3)

I - _ = l- (l- _ 1-2
G (l ’ 53:43 - 283:8—
0" (1+1) =gy‘él-E

z = 20' 20" 1-23 1-2

 

 

 

 

 

w-w‘ + w-m" = 4JF4(w-l+17 + 4-4 m- -
= -4w+8-121m+l6j+4w+l21m+8~l61 _ 15-32
(ma-2j)52j 52 2+jw”)
= _i16-5W4)(2 -iw‘) = 52-15103+641-52ma
32(2+Jw )(2= 3“ ) 32(w4 +4)
Real 32g1-m82 Imaginary Jfi-l6m“+64l
52(9 +4) 32(m4+4)
1-“? Che k 'th th i i 1 R( )
W 0 SW]. e orgna O)

15 Darlington Synthesis of Reactance Four Poles.
Journal of Mathematics and Physics. Sept. 1939.

o Ibid 1 p. 203.

-41-

23. Extension of Foster's Theorem to

Dissapative Cases.

A network containing only two types of elements
can be solved by an extension of Foster's °9°°
form. This has been shown by Cauer.
Theorem:

The expression for the impedance of a network
made up of only two elements can be obtained from
the expression of an impedance of a corresponding
network of pure reactances, by replacing the multi-
plier p in the pure reactance expression by the
impedance which corresponds to a pure inductance, and
by replacing the p terms in the rest of the expression

by the ratio of the impedance correSponding to a

unit inductance and a unit capacity.

 

 

R, L Network 2 = kpipep3)(pep§1_.....
(P'Pi)(p-p§)

R, C Network Z = k(P'P3)(P'E;l ......
(p-pf)(p-p§)

° Ibid 1 p. 214.
°° Ibid 5 p. 208.

-42-

Here is an example of the extension of Foster's form
into the case of two element dissapative cases.
Given:

poles at p=0, -2, -5.

zeros at p=-1, -3.
Real part of the driving point impedance Z is 1
when w=0.

FIND: Two element network to have given character-

 

 

 

 

istics.

z = ng+1)(p+3) = H(p“+4p+51
p(p+2)(P+5) p3+7p2+10p

p = Jw

Z = fl-w3+41w+3j = m-m“+4ico+5M-7w“+iws-lolw)
(-Jm-7wa+1030) 49m4+w6 -2004 + 100038

=.gg7m4-395+10185-28183-4w4+4083-2189+388-5018)

 

 

 

 

 

m“(m4+29mz+1oo)
8 3
Real part EL5® +19)” ..... at m=0 real part = 1.
(04+29m“+100)w8
H = 100
Ck = £535 as p approaches pk, pk = -l/Rka
R2n = H
_ + + C _—_ 19 2 (5 = E R0 = O
Y(P) - (p+2 p.2’ ° 105T5) 50
c = 19(-22§52 = 114R = 100
9» El 5%- EL” 2 100 -1 l) 100 2 558
-——..|l.r.NV\__T _
05 = 19(-5 5 = 152 R5 = 800
u 4p__ 100 -4 -2) 800 760
c; c}

Continued Fraction Expansion may also be used to

get an equivalent result.
2(p) = 100(p?+4p+5) = 15.8+21.Q§p +5.25p3
95+793+lop lop+793+p5
1.58/fii

10p+7p2+p5 /I5. 8 + 21. 05p+5. 28b:" é;5gé§
15. 8 + 12.27p+l.58pa '0

8.76p+3.68p

1:1: 1.14

+ 2.8p;+p5

a 5.15
2°8P +P5 /8.76p+5. 68p: 0 = {92

 

H
21

 

 

 

 

055p1
2 5,1 5.1 = R
o 55p 2 . 8p2+p5
2.8p2
P3
055/? 055/?
95 /.55p2 C=l.82

A different network configuration could be obtained
by arranging the original expression in descending
powers of p.

Final Networks:

/5§%; 3J54. -JY7P
Irv 44} lF—————'

 

(IV 51

 

 

24. Transfer Impedance Functions.

The study, up to this stage, has been confined to
two terminal networks. At this point, four terminal
networks will be considered.
Here is the basic synthesis problem in four
terminal networks:
Given a signal generator with its internal impedance
r, find a network that will transmit a required
characteristic of the signal to a load called R.
This problem requires the study of transfer

impedance.

The transfer impedance will be noted as Zt

Ell

 

Fig. 50

° Ibid 1 p. 226.

-45-

25. Construction of a General Transfer

Impedance.

Mason16 has shown that the most efficient
possible transmission between R' and R" with
passive networks occurs if the two were matched by an
ideal transformer, and corresponds to

Zt = 2 R'R
Everitt also shows this.

Define 9 = lim. Zt = A + jB

2fl/HTfiTT

Zt=fi3 =ZR'R” e9 =2 A+JB

The most advantageous way to represent the
transfer impedance is to use the symmetrical lattice

structure with an ideal transformer.

 

B” F'igo 51.

 

 

 

 

16 Mason, W.P. Electromechanical Filters and'wave

Transducers. 1942 New York:
D. Van Nostrand. p. 24.

° Ibid 9 p. 259.

-45-

The lattice is chosen because it has many
general characteristics that are quite useful.l7t°
Zt = \AZEZ§ = image impedance
tanh 9/2 = \/ZE‘
l/Z'y"
Choose the image impedances to be equal to R"
by choosing Zny = (R")8. Then the input impedance to

the lattice is R".

Tanh 9/2 = 5/ ZX tanh 9 = 1-9’29
.77' ---—-
y l+e89

l+e'° )‘2;‘ 5H

 

69 '1 = Zx Ree -R = ere-Zx
eo +1 E 9 9
ZX = (R) e “'1

e +1

69 = Zx+R 1+Zx R
R-Zx l-Zx R

17 Campbell Physical Theory of Electric Wave
Filters. B.S.T.J. Nov. 1922

° Ibid 5 Chapter X

-47-

26. Lattice Representations.

Theorem°:

The transfer impedance of any passive
network can be represented by a symmetrical constant
resistant lattice network, with resistance terminations.

An elementary example follows:

A =2.p 2W =2

Zt

 

...____________ _2 _
2 VETfiTT -‘_§E — 1+p/2

2x = 3.. (e9 -1 = 1+ 2-1 = 4
e9 +1 l+p 2+1 l+p§4
Zy = l/Zx = 4/p + l

Lattice is then:

 

 

Fig. 35.

 

 

 

° Ibid 1 p. 233.

-48-

A more complex network is presented.

 

 

 

 

 

 

, m
l .J-l/p p 1
]_ Fig. 34
1+l/p -1/p
Zt = :l/p cpfl/r+l _. Eiflénig _ p2+2p+2
l+l/’ l _ 1 P —
-l p O
G = Zt
e = pa/2+p+l
2 I RIRil
_ 9- 2 a
Zx — e 1 =.E;Z§iE—— = E_iEE__ Reduce in Brune
e9+1 p /2+p+4 Pa+2p+4 form.

a
Y :: +2 +4 :: = 1 2 =
x RETEEET p O, residue 2 elenent /p Y

2
PIP+25 2/p p+2

 

 

Zx = p/2 Zy = p/2
“*4” max—”1°
.___”
2/p l 1
Fig. 350

-49-

The fact that this method is laborious
is not evident until a fairly complicated structure
is given. This structure can be converted into a
lattiCe, but it requires a lot of work by the

present method.

 

 

 

.T-O.5/p
1+.1p+.5 p -.2p--5/p
Zt = -.2p-.5 l+2.2p + .5/p
+.2p + .57p

Zt = 2.2p”4~3.2p+2.6+i/p+.2s/p2-.o4p3-.2-.25_p8
o2p+ .5/P

= 2.l§p5 + 3.21)8 + 2.4p +1
.2p2 + .5

39 = °1°.oap5 + 1.6p2 +1.2p +.5
2133 + .5

ZX = 22 + 1.08P5 + 1.5p2 +1.22 + .5 -2p2 —.5
69+]. 1.08p5 1.6p2+102p+05+02p8+.5
1.08p3+l.8p2+1.2p+l

 

Yx+ + = $29§E§+1'QE2+1°29+1 residue at p=0
1.08p +1.4p8++l.2p 1/1.2

This is Y = 1/1.2p L of 1.2h

-50-

Yx' = 1.08p5+l.8pa+1.2p+1 _ 1
(1.08p2+l.4p+l.2)p 1.2p

1.08p5 +1.8p“+1.2p+1 -.9p2-l.165p:l
(1.08p2+1.4p+l.2)p

\Zx' a 1.08pz+l.4p+l.2
Tp+.791)(p+.0417)

residue at p = -.O4l7

 

= .00189-.0585+1-2 = .lilé =
.75 .75 1.52
Element is ._;L;§g_
p+.04l7

 

Zx" = 1.08 2+1.4 +1.2 _ 1.52
(p+-79l)(p+-0417) (p+.0417)

= l.OQp?+l.4p+l.2-l.52951.2

 

 

 

 

 

 

(p+.o4i?7(p+.791)
= 130892-0122
(p+.0417)(p+.791)
Y." = (p+.0417)(p+.79l) residue .053 = _
‘ (1.08p—1.2) p p=0 "37T2 .275
pole of Yx" =-.275/p L of ---- -5.51
YX... = 1.oep?+.9p+.035+.3p-.055 _ £1.082+1.2;
(I.08p-.12) — loOBP-lo
21:: = l.0§pq -.12
1.oep+1.2 1.oep+1;2

R = l, L = .9 R = ~10, C =-9
The x arm is the following. The y arm is the inverse

of this Brune network.

 

 

1.52/1) 9”
H"l- l" ' ”1 . _1
‘JVNAJ IVNAMVVI ‘l_
3604 ‘306lp ‘— 1 —
1.2p -l/9p l -19

 

-51-

27. Transfer Function Changes.

A passive transfer impedance function will
continue to meet the requirements of physical
realizability in a passive network if any of its
real poles or any pair of its conjugate complex
poles are replaced by its negative°. The change
is equivalent to increasing or decreasing the
transfer function by the phase shift of a correspond-

ing all-pass section.

Given e9 = -2 LEiIl , write e0 = '2(P+1IB+2
P‘ p+2(p4§)

 

For the first part.

 

 

-2( +1) -1
ZX = “%¥§“ = fZEerP-Z _ -52-4 = 3+4/p
-2p+2+p+2 “ -p
-2(p+l) +1
p+§

This is a lattice

91
34"” F180 38
3 I

 

 

2+2 -1
EELi__ ’ p+2-p+2 - 2
2+2 ‘ p+2+p-2 ' /p
p-2 +1 This is a lattice
vi
<< ,’ Fig. 59

94’

T ‘ .
otal Network 8%

 

 

Fig. 40

 

l.

2.

11.
12.
15.
14.

15.
16.

17.

Literature Cited.
In Order of Citation.

Bode, H.W. Network Analysis and Feedback
Amplifier Design. First Edition.

Guillemin, Ernst A. Communication Networks.
Vol. 1 1951

Guillemin, Ernst A. Communication Networks.
Vol. II 1955

Carter, G.W. The Simple Calculation of
Electric Transients. 1945.

Doherty & Keller. Mathematics of Modern
Engineering. Vol. I 1956.

Pipes, Louis A. Applied Mathematics for Engineers
and Physicists. 1946.

FOSteP, R.M. BOSOTOJ. April 1924.
Cauer,.W. Archiv f. Elektrotechnik, 1927.
Brune, o. Jour. Math. and Phys. Vol. 10, No.5.

Everitt, W.L. Communication Engineering.
Second Edition, 1957.

Emery, W.L. Ultra-High Frequency Engineering. 1942
Zobel, O.J. B.S.T.J. Jan. 1925
Foster, R.M. Trans. A.I.E.E. June 1952.

Esbach, O.W. Handbook of Engineering Fundamentals.
First Edition, 1956.

Darlington, Jour. Math and Physics. Sept. 1959

mason, W.P. Electromechanical Filters and Wave
Transducers. 1942.

Carapbell, BOSOTOJ. 30". 19220

-55..

p
In.

.J -~ ”A - 6‘6 ”Wt u

 

....