fl

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{SJ-ML - h :w' ‘ ‘ ' ' ‘ - ‘ ‘ - ' ,1.) . ”4}. 3;.» .. I
mum-3‘ ..‘-. :1 ; ' . , . . " , $.51,
"my - *‘ ‘ ' - - -~' ,1“ ‘ ‘“ - * . ‘
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n)“.
‘.

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‘31 ,

 

 

 

 

 

 

' ’ This is to certilg that the
'4
“ (i “ thesis entitled
llfll;‘ "A Critical Study of the Use of
1 ; ’ Matrices in the Analysis and byn-
, ‘}’5: ‘ thesis of Electrical Circuits".
" presented hg
Robert E. Clark

has been accepted towards fulfillment

of the requirements for

it} " Mam.) in_Ele_cL_Engr.
1‘ ' Major pnlfes‘tlr ’
l" 1 ._ Date August 28 , 1950

0-169

 

 

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510

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guy

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Robert

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“‘0‘5‘1 A
"c ‘.L J

cu,

the requir‘m
ree of
'0

13;;

0e;

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tne

Tr‘ v “"‘»

The author wishes to exsress his accreciatian to
Dr. Strelzoff for his assistance in the oreoerution of
this thesis age for sis criticism thrzughout its nritiag.
He also vishes to thanx his wife, Jayee Slerh, for ner
satience and understaadins tbile the thesis was being

written and for her wonderful helo in the oreoeration

of the first draft and final oooies.

q: - »

CHAPT“;

TAELE OF NOTATION

INTROD‘CTICN

FUND“EHTAL DEFINIT ICNS AND THEC EELS OF nnTDI ES

Definition of a hatrix
Equality of Natrices

The Zero and Unit fiatrices

Ir

Hultiolication of Istrices
The Inverse Matrix

Laws of fiatrix Algebra
Linear Transfor:nstions

Cayley-Hamilton Theorem

FOUR TE NlL NETTORLS

Cascading Network
Pa relleling Networgs

Series Networks
Series Parallel ans Parallel Series NET*MOICS
Transformer Anal; sis

Three Be sic letrices

U'IPUPCNOJ

(D

\1

(C)

0‘) ('1 Ca) ()7 H H
(.0 H «I H O) 0']

U1
59>

PF Aarr-T‘Tifmq
x 14‘le J. .1“. L»
.t

con'inueé)

,L‘ .71

ff. "l' m he a “w ’ ..~I — -. - f\ ,- L: A
THO loll Ann L-l 2A.S V:

f — - — ' ~-“ .1 t . n \‘ 7‘-» Vo- T

cyntheSlS of Sguivelent thnorfi 58
f‘t , ‘ ,. .9 ‘: ‘ _. r‘ , '1' AH
uhdh;€ O; Neference Frene so

.0
C)

Aatrix Parsgeter Resresentation

_ i l ,. 11‘ _. "'
I:3ec-nce Lchl unan¢e 7o
(:17 4-. -. n A fif“ —-,- A‘T'T‘V'T‘ . “fl
~41”..- LLLJKLI JAL DUI“ VA..- Ll-S 7Q

DIAGONALIZATICN 89

ION 108

U)

CONGLU

BIBLIOGRAPHY 109

[Y]

[A] , I38], & [F]
[m]

[3] =[A 3]
[K]

r71 -1"r T'fi 1 ""r. ""/\V~‘.'
lull-Jul; CF Jil'.'Lz-Ll.Lvl~t

Resistance fiatrix

Inductance Matrix

Symmetrical Connonent Matrix
Pioes' Symmetries Someonent Ratrix
Tron's Connection latrix
Diagonalization matrices

Voltage Matrix

Curren Letrix

Admittance gatrix

Characteristic natrix of any Matrix

H

Asedance Invariant Transformation Latrix

INTRODUCTION

Matrices are becoming more and more nooular as a tool
for the electrical engineer. The most obvious reason for
this is their ability to maintain the continuity of a
complicated nroblem Without the engineer becoming lost in
a haze of comoutations. By indicating each steo with
matrix algebra, the engineer may carry the oroblem through
to a symbolic solution and then retrace his steps making
the necessary computations. This is true, for example, in
cascading networks, in paralleling networks, and in the
apolication of symmetrical comoonents to a power system.

Matrices may also be used to advantage in finding
eQuivalent networks on an impedance basis, that is, by
maintaining the impedance invariant (unchanged). Another
worth-While use is in Writing the mesh or nodal equations
of a network when there are a great number of mutual
inductances.

If, in a given nroblem, one has a number of simultaneous
voltage equations, the currents may be obtained by the
usual process of determinants. But a more straightforward
method Would be with the use of matrices and the comoutation
of the inverse of the imoedance matrix. If the nrooer

symmetry is present, an even more comoact method would

-2-

involve the diagonalization of the imoedance matrix and the
usually tedious computation of the impedance determinant
would not be necessary.

The object of this thesis is to give examples and show
how matrix algebra may be aoolied to various tyoes of
circuit problems, and to convey to the reader the continuity
and comnactness obtained by this method as oboosed to the
usual method of determinants and substitution and the loss
of objective that usually follows when one makes computations

as they orogress in their nroblem.

CHAPTER I

FUNDAMENTAL DEFINITIONS
and
THEOREhS OF MATRICES
Since the use of matrices by electrical engineers is
not widesoread as yet, the logical nlace to begin this
thesis would be With some of the fundamental definitions
and theorems of matrices. This Will also afford the

reader the opportunity of becoming familiar with the

notation used throughout the thesis.

Definition of a Matrix

It Will be easier to define a matrix if an examole is

given. The mesh equations for the circuit in Figure I may

 

 

 

 

 

 

 

be written as:

el = 21111 % 21212 % 21313 where 211 = a % c f d
(1) e2 = 22111 % 22313 % 22313 233 = b % c % f
33 = 23111 t 23212 t 23313 233 = d t f t 8

212 - 231 = -c, 213 = 231 = -d, 223 = 232 3 -f

If a oroner rule for multiplying matrices is defined

and followed, the equations (1) may be Written as:

 

 

 

 

 

 

r» 1
91 T11 212 2131 iii
( ) or simply
2 e 3 z 2 Zn x i
2 21 22 3 2 _
“ [E] - [Z] x [I]
yfs 331 232 zssy “is

Where [E] is the voltage matrix; [Z] , the impedance
matrix; and [I] , the current matrix. :Afmatrix, therefore,

is simply an array of numbers and is non necessarily square.
Equality of Matrices

Two matrices are equal when they cannot be distinguished
from each other. Therefore, two matrices are equal, [A] =[HI,
when every element of [A] equals every element of [B] , that
is, aij 3 bij' (1 indicates the row and j the column of

the elements a and b)
The Zero and Unit Matrices

If every element of a matrix [A] is zero, (aij = 0)

then [A] is defined as a zero matrix.

-5-

3'

A unit matrix [1] is a souare matrix With all elements
on the principle diagonal equal to one and all other

elements equal to zero.

Multiplication of Matrices

If [A] '-"- [B] 1: [fl , then aij = in}; ij’ Where n

is the number of columns in [B] and the number of rows
in [F] . From this definition, one can see that [A]
[B] x [F] is realizable only if the number of columns
in [B118 equal to the number of rows in [F] . It is

also apparent that(m[811)x(n[F]q)= (1%}: 5 that is, [A] has

the same number of rows as [B] and the same number of
columns as [F3
If this definition is applied to the indicated

multiplication in equation (2), then [E] 3 [Z] x [I] and

3
e1 =2 zikik .

1:31
3 a c. c 0

e1 " 5-5-1 zlklk " z1111 7‘ Z121.2 7‘ Z131:5
3

e2 -'-‘- E1 221:1]: = 22111 7‘ 28212 7‘ 23313
Us

eg‘glkzzmi‘ 2i13,;31311"2322"z

It Will be noticed that this multiplication gives the

desired result, that is, equations (1).

-5-

A definition of scalar multiplication is also
necessary. it [A] = [B] if kaij = bij' This is
different than k|A|=lBl where [AI and [B] are
determinants. In this case only one of the row or

columns of IA! is multiplied by k.
The Inverse Matrix

Using equation (2) as an example, suppose that the
voltages and impedances are Known and that the currents

are desired. Then by the usual method of determinants:

Z1 Z Z-
11 = -——£ e1 % -4al e3 { 31 e3
IZI IZI iZ'T
Z Z Z
(3)12: £91{_2..§.32,£_§333
'3! MI IZI
Z13 Z23 233
i8 = -- e1 % ‘——— e3 % .___ e3
[2! W [3!

where 211, 231, Z31...are the cofactors of 211’ 231, 231...
in the determinant of [2| ; that is, Zij is the minor
of zij multiplied by (-l)i#j. Equations (3) may be
written as:
(4d 00 :3 fig _:x Dfl
Equations (2) and (4) define the inverse. [fl‘JfZ] x [I]
and m 3 [Y] x [E] and therefore, [E] = [Z] x [Y] x [E]
but since [E] = [E], [Z] x [Y] must equal [1] . If the

indicated multiplication is carried out, the above statement

-7-

is verified. [Y] is defined as the inverse of [Z] ;
that is, [Y] = [2]”1 or [Z] = [YJ-l . Therefore, a
matrix times its inverse is a unit matrix. [Z] x [Z] "l =
[23-1 x [Z] 1'- 0]. Also, if the determinant of [Z] ilel
then the determinant of [Z] -1 is 7%,- .

It is apparent from the above definition of the inverse

that [ZJ-l could not exist ifIZI = 0.

$ A matrix whose determinant is zero is called a singular
matrix. A square matrix may or may not be singular,
whereas all rectangular matrices are singular. Only non-
singular matrices have an inverse and for each matrix
there is only one inverse.

The transpose of [Z] is defined as [2],; and is
obtained by interchanging the rots and columns of [Z] .
Therefore, [Z] ‘1 may be determined by first finding [21,0
and then replacing its elements by their cofactors over

2 .
[Z] ; that is, substitute —£l for z-- in the transposed

1
I2! 3
matrix.

Laws of Matrix Algebra

Because of the definitions of multiplication, it is,
in general, not commutative; that is, [A] x [B] 3" [B] x [A].
Exceptions are [A] x [IQ-1 :[fifl ‘1 x [A] , and [A] x [l] =
m x m.

-8-

In other manipulations, however, matrix algebra is

similar to ordinary algebra.

(5) [A3 71 [Bl = [B] 2‘ [A]

(6) km #ktBJ MW #031)

(7) kl‘A] 1‘ qD—J (1:qu) JIM}

(a) [F] x (A) / [F] x [B] [F] x ( [a] % [13])
(e) w x mum as (m Mains]

If [A] = [B] 7‘ [F] then it is necessary that
aij Z bij # fij'
cancellation of factors. If [A] x[F] = [B] x [F] then

Caution must be used, however, in

postmultiplying both sides by [F]"1 gives [A] x [F] x [F].1 =
[B] x [F] x [F]"1, and [A] 3 [B] . The above stipulation,
however, is that [F]"1 exists; that is, [1‘] is a non-
singular matrix. [F] may not be cancelled for the case
there [A] x [F] 3 [F] x[B] because in general [A] x ET] 1
[F] x [A] . Also it may not be cancelled for the case
where [A] x [F] 7- [B] x [F] if [F] is a singular matrix,
because [F]"1 is non-existent and the above reasoning
could not be followed.

Two additional matrix relations are:

(10) ([A] x [B] x [FDt = [FLG x [13],; x [A],

(11) ( [a] x [B] 2: [Fl )‘1 = [El-1 x [31-1 x [al‘l

Linear Transformations

Occasionally it is desirable to interchange the rows
or columns of matrices or add a row (column) to another
row (column). Sometimes it is necessary to multiply a
row (column) by a factor or to add to a given row (column)r
a certain other row (column) that has been multiolied by
a factor. If the above onerations are to be oerformed
on, say the [Z] matrix, where [Z] is cart of the equation
EB] 7- [Z] x [I] , then it is necessary to perform these
ooerations With a matrix so that the equality of the
matrix equation Will not be nullified. For instance,
if [A] interchanges the firSt and second rows of [Z]
and [E] = [Z] x [I] then the resultant equation is obtained
by oremultiolying both sides by [g] , giving:

fig :x EB] = [E] x [Z]:x Efl

Each of the above-mentioned linear transform matrices
are formed from the [1] matrix and are, therefore, square.
The following linear transform matrices are illustrated
for n equal to four, but the same technique may be applied
for n equal to any finite number. They are obtained as
follows:

To interchange two rows gremultioly by [A] where [A]

is a unit matrix With the corresoonding two rows interchanged.

-10-

To interchange two columns uostmu tioly by [A3 there [A]
is obtained by interchanging the two correswondirg columns.
Example:

[A]

[ij X[BIIB]] interlchan es lst and 3rd
row 0 x A] interchanges
lst and 31rd 001.1111 ns of B] .

Ol—‘OO
OOI—‘O
OOOH
HOOO

To add one row to another rot, perform the desired
Operation on [i] and premultiply. To add one column
to another, oerforu the desired 0 eration on [118 nd
oostmultiely.

Example:

1

[A] o {131?} adds 4th row of [B] to 1st.
0

1

adds lst column of [B to
4thx J

OOOH‘
OOI—‘O
OI—‘OO

To multioly a row by a factor, multiply the given
row of [I] by that factor and oremulti‘olj. To multiply
a column by a factor, rmiltioly the given colurm of [13

by that factor and postmultiolv.

Example:T
'l O O

[A] a o 1 o J x [B] multiplies 3rd row of [B] by k.
0 O k multiplies 3rd column of A
o o o 1 by [k]

If it is desirable to multiply a given row by a factor
and add that row to another, perform the desired operation
on [1] and premultiply. If it is desirable to multiply a
column and add it to another column, perform the operation

on [1] and postmultiply.

-11-

Example:

[A] =

adds k times 2nd column to 4th

OOHO
OHOO

O

k g x 3 adds k times 4th row to 2nd row.
0 x

1

OOOH

column.

The above technique sill be used in the chapter on
diagonalization. No unique notation was adopted for the
above matrices since they will be redetermined for

particular cases in Chapter V.
Cayley~Hamilton Theorem

If [M] is defined as the characteristic matrix of the
matrix [A] , then [M] = [A] -/u[l]. / is a. scalar
parameter and [M] , [A] , and [l] are square and of the
same order n. The determinant of [m] , (1M1), is defined
as the characteristic function of the matrix [A] and the
equation [Ml = O is defined as the characteristic equation
of [A] . The statement of the Cayley—Hamilton Theorem
is that any matrix [A] satisfies its ovm characteristic

equation.

Example: [A] = [l 2]

3 4

A a - 1 = 2 _ = _
[A] [A] /[J [g- 4] [é a [(1K) (4%]
Characteristic Function: (It/u)(4;x~) -6

Characteristic Equation: (17“)(47u) — s = 0

That is: /u? -§,‘ - 2 = O

-12-

And by the Cayley-Hamilton Theorem:

[A]2-5[A]-2[1]= 0

Where [Q2 =[A] x [A] and [l] is of the same order as [A] .

Therefore 1 1 2 _ l 2 _ l O -
[212:[34] s[..] 4011-0

Carrying out the above operation indicates its

03

correctness. The aoolications that follow from this
theorem are very useful. Since [A]:3 — 5 [A] - 2 [l] = O

for [A] : B3 [A]2 z: 5 [A] ,l 2 [El] and multiplying

through by [A] gives,

[93 = steamers 501+ 2m> ’42:»; =
27 [A] 1‘ 10 [I]

27 [A32 ,t 10 [A] =27( S [A] ,t 2 [1] ) ,l 10 [A]-
145 [A] ,l 54 [1]

[A]5 = . ................

Thus, without carrying out the matrix multiplication,

[A] 4

[A]4, [A]5 ........ may be determined in terms of [Jan-1,
[A] n-Z’ ....... [l] where n is the order of the square
matrix [A] .

This theorem also indicates a method for the
determination of the inverse. The folIOWing example Will

Let [El =

illustrate [his me hod:

HMO)
pmm
03450-4

H030)
Poem

‘mpw,

l 0 6- 2 1
Therefore [£4] = [ 7'18 2. (13] -‘-" If {0 (8r(6;«)

-12-

U

ml: (6-;u)2(87u)%8%8-{(87k)%4(67t9 1‘16(67u)}
And IMI 2' 7.3 ,t 20,3 - 111,; 176 = 0 (Characteristic Equation)
Therefore [513 -20 [a]2 ,t 111 [B] - 176 [1] 2: 0

Which gives 176 [1] = [313 — 20 [1312 ,t 111[B]

Multiolying through by [B] '1 and dividing by 176 gives:

[B] "'1 2: 1/176 [B12 — 20/176 [B]; 111/173 [1]

_1 41 32 20 120 40 20 111 o 0
.And [B = 1/176 32 B4 58 — 40 160 80 ¢ 0 111 o
20 58 53 20 so 120 o o 111
_1 32 -8 o
[B] = 1 / 176 -8 35 -22
o -22 44
By the method described in section 5,

[Bl=6xsx6/2x4x1—1'x8x1—2x2 6-4x4

 

 

x6=l76 .

P 1’ - fi

84_24 28 32 -80
_1 46 16 14

And [B] =1/176 =1/176 -8 :55 -22
21 61 __62

'4 6| '1 6| '1 4| —:2 44

" d
21-61 '62
L84 24 28

 

 

It is difficult to say which method is the best.
Both methods require an evaluation of a determinant of
the same order as IBI . However, if the order of [BI
is five or higher, the determination of the cofactors,
necessitated by the second method illustrated, becomes
a difficult task. While, for the Cayley—Hamilton method,

there is only the fifth order determinant to determine

-14....

and the rest of the Operation is matrix multiplication,
addition, and subtraction. It would seem, therefore,
that for matrices of fourth order or less, the method of
section 5 is best and for matrices of fifth or higher,
the Cayley—Hamilton method is best. If a comouting
machine capable of handling matrices is available, then

the Cayley-Hamilton method is definitely the one to use.

CHAPTER II
FOUR TEREINAL NETWORKS

Usually, what takes place at the input and output
terminals is of interest rather than what happens inside
the network itself. In analyzing or synthesizing a
network it may be found desirable to interconnect several
four terminal boxes in various ways. The fact that
matrices can be used to exoress any two of the four
variables in terms of the other two (e1, e2, 11, 12) makes
them particularly effective in handling problems of this
kind. There are six different ways by which matrices may
be used to express relations between the various voltages

and currents.

 

 

 

 

 

 

 

2
la ”22
is 6:1
Figure 2
Referring to Figure 2, they arel:

(11) i1 : Yll yle x e1 (12) el = 211 212 x 11
12 YBI y22 e2 z21 z22 13

e
1E. A. Guillemin, Communication Networks, II, pp. 144.

(13) 11 _ g11 g12 e1 (14) e1 _ h11 hie i1
- x - x

e2 321 g22 12 12 h21 h23 e2

(15) e1 = A B 1 e2 (16) e8 D B x e1

11 c D -12 12 c .3 --i1

The interrelations between the various elements of the six
square matrices have been derived and tabulated.2 If one

were doing many problems of this type it would be desirable
to use such a table. Throughout this thesis, however, all

interrelations will be derived as part of the problem.
Cascading Networks

Where several networks are to be cascaded, the type of
representation to use would be that given by equation (15).

For example, if in Figure 3, all of the networks are identical,

 

 

 

 

 

33? is“ 4s, . w....5i2.-~~.
19’" M 63., ‘91... , Mu q‘lflt

 

 

 

 

 

 

 

 

 

Figure 3
then the input current and voltage can be expressed in
terms of the output current and voltage by this simple

relation:

 

2Ibid., pp. 133—138.

-17-

n
e1 = [A B] x e3n
11 C D e1n
The matrix [A B]n may be evaluated as an application
C D

of the Cayley-Hamilton theorem. Referring to Chapter I,

the procedure is as follows:

Let s = [g g] when t] = m Vail]

And [M] :: [g B] {’6‘ ,3] : [(AE’“) (DIE/«J

Therefore (Ml = ,2 - (A ,l D), ,t AD-BC, but AD-BC : 1
(Shown later on in this section). ‘And IMI= O is the
characteristic equation, therefore, /“2 - (A ,5 Dy! 1 = 0.
And by the statements of the Cayley-Hamilton theorem,
[Caz—kirk] 1‘ [1] =0 wherek=A#D.

Therefore [C92 k x [G] - [1]

mi 1: 2: [G13 - [c1 = as...) s -k {11
[(1:14 2 (kg-l) [(33 -k [c] : (ks-2k) x [(3.] 413-1) x [1]
And if [6?” = p 16?] - q [1]
[GT1 3 (kn-<2) x [G] - p [1]

Using the above relations for the coefficients of [G]

and [I] , [Gan can be obtained faster than the actual
matrix multiplication will allOW, especially if n is
large and A, B, C, D are complex.

The folIOWing example Will be Worked out to

illustrate the use of cascading networks. From the

—18-

network given, Figure 4, find the characteristic impedance
looking in at terminals 1—2; that is, find Rx so that the
impedance looking in at 1-2 will be Rx when 3-4 is termi-
nated in Rx’ Also, if 3—4 is the output, terminated in Rx:

and 1-2 is the input, what will the attenuation be for the

 

 

 

 

 

 

 

 

 

 

network?
113; 1.1- 1.1.71 ..... l e- 3
39 3‘ Q6 5%
4 4;
2 4
Figure 4
Solution: The network is first broken up as shown in Figure 3.
I In 3 :4 40
w *4 1112‘ V2113. I 3
14% q, hfiiil £51 5; £3 4: 521) }15’
z (a) (b) (c) “
Figure 5

Using Figure 5(a) as an example:

61 = (10 # 8)11 % 8 12 = 21111 * 21212

 

(17)
82 Z 811 % (4 ¢ 8) 12 3 22111 { 22312
where 211 = 18 212 = 231 = 8 233 Z 12
Then 9 z e z
_ 2 22 - _§ 22
. 11 — -—- - ___ 12 , el - z11( - -—— 12) # 21212
221 221 z21 221
And el 3 211 _(211299 - 212) i
2 2

-19-

Or
211 '2‘
12 213
(18)
l
11 = '2"- eg - Z23 12 : 082 -' D12
12 z12
- (a) (b) (c)
And 81 A B A B A B 66
Therefore 3 - x x x
i1 . C D C D C D -i

e1 9/4 19 3/8 9 5/3 124/6 e6
x x x

11 1/8 3/2 1/6 5/3. 1/6 8/3 ~16

Carrying out the indicated multiplication gives:

e1 _ 19.9 279 e

.. x 5
11 1.33 18.7 _ --i6
And el = A e6 - B 16 - Where A 3 19.9 B 3 279
11 = C 86 - D is C = 1033 D : 18.7

It is easily seen from equations (18), which apply
to any four terminal network that has symmetry about the
principle diagonal, ([z] 2' [z]t), that AD - BC = 1. In
our case AD - BC 3 373.13 — 371.07 = 1.

If the network is terminated in Rx then e6 3’ -in6
and e1 3-A(-R116) -816 3 “15(ARx / B) i

11 = c(-ine) ~Dis = -16(CRx / D)

And if R1 is to be the characteristic impedance,

-20-

31 ARI * B 2
-—_ = R = -——————— ,therefore, CR / DR «AR -B = o
i x x x x

1 CRx % D

2

And R12 / a.x (D—A) _,§ 3 0
o
a: - Rx (0.9) - 10 =

C
2 O

 

_ +»\f*
G1 1 R - 009 - o 4 - 09 " o
v ng x 2 81 f 8 O _ 0 g 29 - l4 9 ohms

 

A180 61 - A86 — B16 1
11 " 036 - D16 .—
Rx
Therefore el = Aes ¢.§ e6 = e6 ( ARx f B )
Rx Rx
—§'= R1 = 14.9xl9.9{879 299 27s 0-0258
e1 ARx B

N 2 1n( es) — 1n 0.0258 3 -ln 1
EI’ ' 0.0258

N 3 -ln 38.8 3 —3.65 nepers

But since e1 and e6 are across the same resistance,Rx,
N may be converted to decibels. The gain, therefore,
equals -8.688 x 3.65 = -31.6 db.; that is, there is an
attenuation of 31.8 db. for the network including Rx'

It might be suggested that one set up the network
and measure Z open circuit and Z short circuit and
obtain Rx by Rx = m . However, this could. not be
done in our case, because Rx = Bragg-3;; only when the

-21-

network is symmetrical about a vertical line through
its midpoint.

In this example all network components were resistances,
however, the same technique is equally applicable when the
components are complex impedances. In a later example it
Will be shOWn that matrices may be used when the Laplacian
transform is involved.

There are, undoubtedly, many ways of working the above
problem, but it is doubtful if there is any method more

concise and straightforward than that_just shown.

Paralleling Networks

In paralleling two or more networks, Figure 6,

 

 

 

   

 

 

 

Figure 6
equations of the type given by (11) would be used.

[IJT [I]1 ,( [1:]3 and [8:11 = $233. Adding the two
equations we have [IJT = [I31 ,l [112 3 [‘31 x [E] ,1 [132 x9]

and by equation (9), therefore:

-32-

(19) [IjTi(ffJ1/El2)xi{l

In handling problems of this type with matrix algebra
a great deal of caution must be used. After the connections
have been made on the left, they may be made on the right
providing that there is no potential difference between
terminals 3 and 3' and between 4 and 4'. If potential
differences exist, there will be circulating currents
within the networks. The result Will be that the current
in at terminal 1 will not equal the current out at 8 and
similarly for the other terminals of the networks. If
other means cannot be used,3 ideal transformers may be
placed at one end of the network forcing the currents to
be equal. If n networks are to be paralleled, the most
h0pe1ess case would require n-l transformers. A more
complete discussion of this is given in Communication
Networks.4

The paralleling of two or more networks might be
necessary in a problem in synthesis. For instance, several
four terminal networks are available and a specific overall
effect is desired; that is, [I] = E] x [E] . [E] contains

the driving voltage e1 and the desired output voltage e3.

 

3Ibid., p. 148

4Loc. cit.

-23-

fig is the admittance matrix that will give this desired
result with the ensuing currents [fl . Then by the method
of combination shown in Figure 6 and equation (19), the
various [8 Matrices may be added until the desired result
is obtained; that is, [if] = [151 ,l [QB which, of
course, means that yij = yijl # yijg"" With a finite
number of networks available, one would be very lucky to
obtain a combination that would be exactly correct, but
approximations could be obtained. And for the problem as
stated, the use of matrices would lead to a solution
ldirectly.

Matrices may be applied very nicely to the analysis

of paralleling problems. Thi circuit of Figure 7(a) is

ii *TI 3"

 

 

 

 

JP 1? N’
1_~“Ma_
R R R
n R
(b) (c)
Figure 7

a one section low pas filter terminated in its character-

istic impedance, R, and connected to an all frequency

-24..

generator, eg. The problem is, if the networks Figure 7(b)
and (c) are paralleled to the filter itself, what will be
the overall effect on the frequency characteristic at the
load; that is, find «,1 1F whereK =«(w)and/8 =,8(w) .

The networks in Figure 7 are redrawn and paralleled in

 

 

 

 

 

 

Figure 8. If i- 1'"
ad ..',v ‘-
5; .2
T2 2T (a)
. .u .51
i. a. , e9: ’2 .1}
(b)
1 V
‘45 4‘ #% €5]::]I?
.l” m
Z ‘ 1‘ Z;
—. M. ‘— (C)
J.
x 76
Figure 8

Part (b) of Figure 8 requires some explanation. The
resistances in the lower branches were removed and placed
in their respective upper branches. This is allowable
since the external effects are unchanged. Then the T-F'7T'
transformation was applied giving the network shown. The

17'form is desirable over the T because it lends itself

-25-

to the node method readily. It is possible to parallel
these networks without ideal transformers since the lower
conductor of all three networks contains no admittance;
and therefore, the potential differences previously
mentioned are zero. This was another reason for changing

the network of Figure 7(b

).
I - ( DC # _l_.

.. ‘— ) e -_l__e
11 2 pL 1, DL 2
Figure 8(a) . p 3 it»
13 z - it 81 % ('85 ¢ 2% >82
1 n = .l. f _l_. - l e.,
1 ( 43 en )el “—83 a
Figure 8(b) "
i = - 1 e / 1 / 1
2 “8R 1 (Z?- .87] )eg
I.
11 = (4% ( 4% )el ‘ 1% e
Figure 8(c) .
i
13 --%el%(%7‘%)e2

[13' = [.xj‘xfE] , fij -_-_- [’YJ" x [E] ’ [YJWX [E] = [13m
And F3 = fiJ' ,7 [11" ,. [11"= ([YJ' 1 n" f n'“ > x [E]

Therefore
1 .(00 #.l_ / lEE - 9 W
(20) 11 = ( nL 833) (8137‘s?) x 31
2 - s 1 19
L (titer) (“C(Bt"s:')_ 63

 

 

-26-

 

 

 

 

 

 

 

 

 

With the termination in R, e2 2 — R 12, and 12 = -.:g
R
e2 _ 1
Therefore -'H- = V1281 % ygzez and leel - -eg(.§ % Ygz )
e ‘1 ,1 1 19
Which gives -;3 3 - (R % y22) : (R % DC 7LISL %‘§§)
3 y '~w l 9
12 (EL 7‘ '8'??- )
e (3% ,t jw(c- 1 ) «NIB
A d 1 - ”1L ‘-'- 5
n E} " (.9... - 1' .1_;)
BR to
Therefore 2 .
«4‘ 32 ‘(c- 1 2’ J “V‘
( 9 )'2 ,1 ( 1 )3
88 «JL 1
1f, 3 tan’1 “Kc-EMF.
, 87
1’23 tan‘1 ~88
3 1 2 9Q)L
_ (31) #tz’(c— lt—
°< " 1“ [83 top) And/6:74“??-
(-9-)3 / <--1->2
8R “IL

Where °( is the attenuation in papers andfl is the
phase shift, both being functions of frequency.

~From the technique used, it is necessary that the
networks (a), (b), and (c) be connected as shown in
Figure 8 to give the same results as this analysis. This
means that Figure 8(b) must be connected as shown, or as
an equivalent T with no impedances in the lower branches.

It is the ability of matrices to maintain the continuity

of the problem that lends itself so nicely to this solution.

-27-
Series Networks

When two or more networks are to be placed in series,

Figure 9, equations of the type given by (18) would be used.

 

 

 

 

 

 

 

 

 

 

(Eli = {211 x le ids = [le I E02» and F13 =[ZJs x [133.
but [I] = [131 = [132‘ [133 and W: [E31 7‘ We 1‘ [‘93
provided that the precautions referred to in the previous
section are taken; that is, the current in at terminal 1

is equal to the current out at terminal 8 etc. Under these
conditions [E] 3 ( [311 ,1 E12 7’- [ZJS ) x [1] .

matrices apply themselves to problems involving series
connections just as nicely as they do to problems involving
cascade and parallel connections.

A good illustration would be an extension of the
problem of the previous section. Referring to Figure 7(a)
suppose the filter network was matched to the generator
originally; that is, between the filter section and the

generator there is a resistance R equal to VL/C . Vie

-28-

have already determined the effect of paralleling the
networks Figure 7(b) and (c) to the filter section itself
and then connecting them to eg; we would now like to know
what the overall attenuation would be if the resistance R
was placed between the combination network and the generator.
Naturally, the whole problem could be redone, but using

the series connection technique we can utilize most of the
previous work. The procedure can best be illustrated by

referring to Figure 13. The step from Figure lO(a) to (b)

 

 

 

 

 

 

 

 

R I . _i
a z Acme/y, 4 in (a)
- 22:: L?“""--‘;_‘l -
i ' 3 '
‘ ' R (b)
2: 3., 35 :

 

 

 

 

 

 

 

 

 

 

4’ | I 4 I
1‘7, I I : 63‘.
Z ! I 2'-
(L......_n_._..a ' ‘
(2;) A? (c)
A ‘jgf': 3 : 4fi§r ”
1‘3” I .c ' 63 t
.2 . 3‘ 3" 1 ‘4
e; ,, 2:: 2:: t

-39-

is possible because all four terminal, bilateral,77’
networks have an equivalent T. Figure 10(0) is elec—
trically the same as Figure 10(b) and if terminals 3 and

4' were disconnected there would be no potential difference
between them; therefore, (a), (b), and (c) are equivalent

networks and the matrix method may be used.

Solution:
e ' = Ril' { 01 ' e ' R o. 1 '
Figure 10(0) 1 2 and 1 = x 1
(too) eg' - 011' % 012' 93' o o 12:

Equations (20) may be used for Figure 10(0) (bottom)

11" = (no 1‘ %L 7‘ ‘3') —( 7:1- 7‘ gfi) x elu

12" " ‘ 21;: 7‘ :3) (no 1: 35-; g) 32"
n 1 19 1 9 -’

e : (DC #‘EE % ‘fi9 -('3E % §§) x 11"

62" 4%: vie-3%) (oo¢%fv‘%g. 1,“

qr"
/

and, therefore, our equations for/thejtwo subnetworks are:

e1 : R 0 x g 11
a v :
e2 0 0 12
And
31" 211 212 11"
: x
ea" z21 222 12"

Where 211 = 232 = 3.3
O

 

1
And 2 = 321 = 1.125 a -j £35
13 ‘ 5.375 - w’LC ,t 32.5wR(1.SC — 1)
£35
Adding the two above equations gives:
ell * e1" : (211 % R) 213 x 11
II '
e2' * 92 Z21 222 12

But el' # el" = eg and eg‘ = 0, also, e3" = e2

Therefore eg = (2.11 % R)il % 21312
e 2 22111 t 22212

C0

 

 

 

 

 

And e2 = 22111 - zggeg And 11 : e2(R ¥ 233)
eg 2 (211 { R)(223 % R) _ z 9 82
2 :§_* '4L: -
31 R
e
_E Z (211 # R)(zgg % El - 212
e z z % R % R ¢ ’3 2
g : __l 22 Z11 322 R ‘zia
"= 2 R _‘
e2 21
eg _ 211(211 # 2R) # R2 - 2123
eg 212R

And

ml m
NOV}
II

{SN
:4
4

._ 4 , \
Where 1: = 118 R" ,t (91.33 - 50.:5 w’L2 - -——-.,l3°25
0'0“
- 43 M am e ‘ 3 2
- .eo. L—63.oJC-ll.5wRLC
- h 2 a 3 r‘ a n 2
u - 11.e R - 1.14.5waeLu —- s a
difid
v = 5.34031, — 9.18 5.:
k.)
B ‘”5P
ut 322-.- 6'
es
Therefore

. 3 n
deF : 1n ($7115) 2%ln (w) ,‘ j(tan“l 3E - tan”1 1)
U." ‘ u

% v

_ 2 2 '
And 0‘ - -} ln (xi—fit?) , and/6 = tan-1 1 _. tan‘l 1
n V

x u
Although the solution to this problem is not a simple
one, it wouldn't be less complicated if it were worked by
some other means. It is doubtful if there is a more direct
route to the solution than that afforded by the use of

matrices as illustrated above.
Series Parallel and Parallel Series Networks

Sometimes it is desirable to connect the networks in
parallel on the left and in series on the right or vice
versa. Or, if a network is given, it is sometimes desirable
to picture it in this way to facilitate the solution.
Equations (14) would be used for a series—parallel connection

and equations (13) would be used for a parallel-series

-32-

connection. Referring to Figure ll(a), the equations for

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

—éi .Jé: 5;: <—1;
1e: N, e1
6"
165' /\4; 6:1
(a) (D)

Figure 11

N1 are: 81 : hll' hlz' x 11
12' 1121' hzg' 83

e h I. h II 1
And for N ' 1 Z 11 12 x 1
12" hglfl h II 62

22
Adding the two sets of equations gives:
e11 ¢ e11 : (ell: 1 n1,") (hlg ¢ h12"> x 11
12' * is" (hsl' ( hzl") (hss' * has") 92
And the resulting equations for Figure 11(b) would be:
11' ¥ 11" _ (311' % all") (513' % $13") x el
63' t 82" (521' t 521") (322' * S22") 13

-33-

When applying matrices to these types of networks,
the same precautions must be heeded as for the parallel
and series connections of the last two sections. After
the connections have been made on one side of the networks,
there must be no potential differences between the terminals
to be connected on the other side. If there is, then an
ideal transformer must be used or other steps taken, as was
done in the example used to illustrate the parallel connec—
tion. This cannot be over-emphasized. The use of matrices
will give erroneous results if the current into a network,
on a given side, is not equal to the current out on that
same side.

The follOWing example will be used to illustrate how
matrices may be applied to the series-parallel connection.
.1: m

 

 

4
I
I

'vvvv

1 12L

Rf
1%*

“AAA
v‘v'

 

T Ti

 

 

 

 

Figure 12
In Figure 18 is a diagram of a single stage triode amplifier,
with negative feedback,couoled to its load, ZL, through a
transformer. The problem is to find the gain of the.overall

network.

-34-

Before starting the problem, however, something should
be said about the application of matrices to tube circuits
in general. As a rule it is possible to obtain a P] or a
fig for a given tube circuit. Once this is done the other
forms (page 10) may be obtained, and the Operations of .
cascading, paralleling, etc. may be carried out providing
the previously outlined rules are not broken.5 However,
if the analysis is to be on the basis of an equivalent
circuit, the use of matrices is not a short-cut. To find
the fig and. fig is someWhat of an ordeal compared to the
techniques for solution found in any text on tube circuits.
If the solution is to take into consideration the interelec-
trode capacitances, and is to cover the entire frequency
range, say for a square have input, then the authors
Gardner and Barnes outline a nice technique7 using the
laplace transform. The following solution will illustrate
Why the use of matrices is not alWays the best procedure
for a unilateral network.

The equivalent circuit of Figure 12 is given in Figure
13. Matrices may be used since the transformer insures that

the current into either network will equal the current out

 

5
Ibid., p. 148.

6S Seely, Electron—tube Circuits, pp.85—86.

7M. F. Gardner and J. L. Barnes, Transients 1g Linear
Systems, pp. 180-188.

on a given side. For the

. equations are: el" =
62" :
And e1" :

18" :
e1" 0 1
Therefore :
1 " o %_

bottom box of Figure 13(b), the

Rfign ¢ Oil"
Rfig" % 011"
Oil" % 82"

Oil" #(l/Rf)e2"

i u
x l

N

 

 

 

 

 

 

 

 

 

 

 

(b)

Figure 13

-35-

But the equations for the tOp box are not so easily

obtained. They may be found as follows:

a(/«el' -' 13139) = 82' - ZLiz'

. I g 1
And /“ el 9’2 Ztiz' ,1 13:
a a p

Since, in tube circuit analysis of this kind it is usually

 

 

 

 

assumed that the transformer is ideal, is 3 ~ai2 .
The fo ' = '
re ”/61 :2. - 2L 12' — r .12'
a 57' 9
c z a 8
Or e1 e2 _ (ZL % rDa ) i '
'-- 2
,«a fa.
But eg' : e2" = e
And 13 = 18' ¢ 12" = 0
Therefore 13' 3 -12" and Rfig" 3 e2" 2 eg'
And therefore ez' - -Rf12' and 12' - -e3'
Rf
This gives el' = 63' Z r 33 '
‘/‘:—- * L * up ) e2
8. ”a, ——
Rf
2
‘h I: I
T-erefore e1 e2 ( Rf { ZL % rpa )
[a Rf
2
Or 81' = Oi1' t (Ri% ZL ¢ 32a ) 92'
l/4a Rf
12' 011' - -%— e2'

-37-

 

v 2
e1 : o Rf { zL ¢ r23 11.
12! flh'i‘w x e '
e " O n
1 __ l i
Alld " x
11" o 1/Rf e "

Since el' ¥ el" = e, and i3' % i9" 3 0, adding the two

matrices.gives:

 

 

 

 

 

 

 

 

2 .
R Z '
e - r0 :71 L7£rpa l 11
o ‘ ,f‘aar x
e
2
' b0 0 , J
Therefore e 2 e2 ( aRf J Rf % ZL % rpag) ZL
' -— Z"
Rf /48. L
e
But —§ 3 -12'
Rf
2
R - Z
And therefore e = ~1zzL ( a f Rf % L % IUna )
flazL
i ' Z
. .- 2 L _ Z
and gain ———é._. - /a L
flaRf #Rf{ ZL % rpaz—
Or gain 3
ZL
/“__._,;a a2 : fa ZL'
a R R Z
f a
”A -§-v‘ ‘2“;1‘71‘1‘0 rav‘yar‘me'v‘zx.
a“ a ‘

-33-

The usual technique would be as follows:8
Z
z '; L
E?
Rf' = Rf
E?

 

 

 

 

There ZL' and Rf' are ZL and Rf reflected back across

the ideal transformer.
: I l
//4el ( rp Z zL ¢ Rf )1
91 : (rp ¢ ZL' ; Rf') i :
/fl4

Therefore 8 = (I Z zL' ¢ Rf! #LaRf')i

_ I
e iaRr

 

 

-£L-_-_---
/
e = rp * ZL' * Rf' # aRf') iZL' E
I 8.
./“ ZL
And iZL'a. — flaZL'

Therefore ”—3“ -

 

rp { 8/4a ¢ 1)Rf' ; zL'
An even simpler technique is as follows:

K _ K Where K — nominal gain
T ‘ le/QK
l6 1' feedback ratio
- ' I
K ‘ iaZL and i = 81

 

88eely, on. 01 ., 00. 85-86.

-39-

 

Therefore K Z/aZL' andfl = -Rf __ .3 Rf!
In 7K Rf. fYLr ZL ZLi

And "
Therefore h‘1‘ zfaZL

“r; ill/4 #2033; 2L'

The ournose of this illustration is not to discredi
the use of matrices for the series-narallel connection,
but to ooint out the di ficulty of using matrices vhen a
unilateral circuit is involved, and still have an examnle
illustrating the technioue of visualizing a network as
being comoosed of two networks connected in series on one

end and in parallel on the other.
Transformer Analysis

An ideal transformer has been referred to throughout
this chapter. In analyzing the feedback amplifier, an
ideal transformer has assumed and in forcing the current
into a given network to be equal to the current out on a
given end, it was suggested that an ideal transformer
might be used. An ideal transformer is one that has
neither_leakage nor losses and has infinite inductances
on the primary and secondary sides. It is therefore imooss-
ible to attain. The losses can usually be kept Within

reason, but in air core transformers or when the ratio of

-40-

an iron core transformer is high, it is difficult to keep
the flux leakage low.

In network analysis or synthesis involving transformers,
the Work is considerably simoler if an ideal transformer can
be assumed. In this seetion, therefore, we will determine,
with the use of.matrices, the conditions When an ideal
transformer may be assumed and will derive an equivalent
circuit for the case when an ideal transformer may not be
assumed. Figure 14(a) is a tyoical transformer circuit and
its equivalent is Figure 14(b). This second circuit will
be analyzed to determine when the assumotion of an ideal

transformer is oractical.

 

 

(b)

 

 

 

 

Figure 14

...41..

(D
I

‘ (P1 % 342L11)11 ¢ jLJMig

0
ll

3035.11 1‘ (92 7‘ ij33)ig

e - (RI 1‘ jCJLll) jCJM K 1:1
0 3'le (R2 ,t 3“L:32) 13

—1
(311 van) jwM x e
jwM (R2 ,1 ij22) O

9(Rg 1‘ 3.601422)
(RI 1‘ 3501411) (fig 7‘ ijB 2) “(30M)“

Therefore 11

, z T .2

i - .
1 R2 3 JLJLZz

:1.-
:3
Q.
[m
H
N
I

 

«2
If (RlL22 ¢ 322L11)35> slag- a)(L11L2 2 -m )

And if (UL-22)) 33 and lel)> R1

 

 

Then 2 = (a L 2 J L ) _
1 2 L R2 11 - al # R2 L11
22 L22
B t L1 L L 1
u I: 83 And 11 = T.
1 ,8 L23 a“
Which zives z = a Re _
r. 1 7l 7 _, P1 7; 32:
8.

But R8' is simoly the secondary load with losses reflected

across the transformer. If the above underlined conditions

exist, as they often do with iron core transformers, then

-42..

the transformer of Figure 14 may be reoresented as in the

following diagram.

Ila. e _. 80 ,l Oi 1
- 1", e
—zr"+ + t 1 E— 8 l '5 0 e2
1:, Q1 _ = x i
11 - Oea — aig il 0 a 2

Then the leakage is not neglegible; that is,

2 .

then an equivalent circuit for the transformer is useful.

 

 

Ire
.—-r-
,, + * *2:

 

 

 

 

 

 

‘25“
(a)
W
Ll

 

 

‘ 6: I [:4' ] e} ‘,

 

-l-. * 4' b.»
2’ ’1.
‘ 9‘: find.) fizz-1:) 9; ‘

 

 

(‘6’)

Figure 15

-43...

The voltage equations for Figure 15(a) are:

el = ijllil ,1 juhig 91 L11 ii i1
and = 3:.) x
68 = jcoMil % jcangig 82 M L33 1
There- L M (L -L ) M L O
fore [L]: 11 = 11 1 ,t 1
H L22 M (ng‘LB) 0 L3

And fiJ == fiJ' ,1 53"

Figure 15(b), therefore, is the equivalent of Figure 15(a).

La
2 _ £3 _ M
a “ and k -

L 3
11 lelLBB

For the leakage to be zero, k must equal 1, that is,

 

L11L82 - M2 = O, (ILI = 0). But the reason for this
analysis is because [Ll ¥ 0. By choosing L1, and L3,
however, IL” can be forced to zero giving (Lll—Ll)x
(ng‘Lg) - H2 = O. k' for the transformer in Figure 15(b)

is, therefore, 1, and

 

 

(21) a‘ =- (L22 'LB} - M = L22 ‘Lz
' Lll'Ll "'ar“—— '
(L11 “Li) t

Figure 15(b) is verified.

Since [L] till]. 1‘ [14]", two impedance matrices are
being added and, therefore, two voltage matrices are
being added. The indication is that two networks are
being placed in series and, therefore, Figure 15(0) is

the equivalent of (a) and (b). The next steo is to

-44-

analyze the b0tt0.u box of Figure 15(0), the equations of

which are:

- jw{(Lll-L1)il ,t m3;

el' -
(82)
82' 3 3U i £31111 1‘ (L23~L3)113
And el' _ . (Lll‘Ll) M 11
' - J U ‘ X .
.. ‘.' - .. D — " 2 -
Bat i ’ a'(L11 L1), and (L22 L2) ‘ a. (L11 Ll)
e ' (L -L ) a'ig(L -L ) 1
Therefore 1' 3 jéo ll 1 2 ll 1 x l
62 a‘(Lll‘Ll) a. (Lll‘Ll) 12
And el' = j (.){(Ll--L1)il ,1 a'igéLn-Lln

And since k' 3 l, a'ig

(23) 91' : 3w<L11‘L1)11 7‘ 30(L11‘Lins
e2' 3 a'el

Equations (23) indicate that the bottom box of Figure 15(c)
could be redrawn as shown in Figure 15(a) or (b). Figure

15(0) could be obtained from equations (22) in the same

 

 

 

 

 

 

manner. ..:2.-'-..- i3 3:" + lid; J
(a) kit“)? 3'1 ”I”? § ' ‘51 (b)
i :7 aka! A;
iii—ii 4w
ad»! (C)

Figure 16

-45-

By substituting Figure 18(b) for the bottom box of

Figure 15(C) the following equivalent circuit is obtained.
Lu?

  

 

Ida”
But since k' = 1 and, therefore, (Lll—Ll)(L32-L2) - r3 = o

. , s2 -
Then L1ngg—Lngg-ungl%LlL3—i — o

 

also L2 3 ‘L‘ ‘LlLBB

11 l Lll—Ll

This means that either L1 or L9 is arbitrary. To simplify

 

the equivalent circuit, choose L2 = C.

.2
'L‘ - L11L22 ‘ M

 

 

 

 

 

 

 

Therefore L 34"" — — H2
1 L22 ‘ L ‘.-2._
9 L22 11 L33
-2 - M” n
But A - L11L22 and therefore L1 = L11(1—Ke)
And a' =VL82 "' L2 _ L23 _ i 32 _. g
" — - A
Lii'Li [pf iPi # L K3 L11
1 1 11

From the orevious analysis a very useful equivalent circuit

is given in Figure 17.

'aMH a?

to" a” 6;?

 

 

 

Figure 17

-45-

In most cases the inductance L11(l-K2) can be
assimulated into another inductance of the network.
However, if it cannot be taken care of in this fashion,

the overall equations can be obtained readily by the

 

 

 

r . _72‘*
91 = '3'; JUaLllfl .< ) e2
1 x
bawaflLn K A
Where L11 — primary inductance
L23 - secondary inductance
M — mutual inductance‘
M
- L90 T - m.
a "' “ ‘5 " VLllEBB

Equivalence of the L77, Bridged T,
and
Symmetrical Lattice Structures
The T-rf‘fand 77-!- T transformations are common

knOWIedge to all electrical engineers ans the transformation
equations can be found in several texts. With the use of
matrices, the derivation of these equations is straight-
forward and concise. Equations (24a) refer to Figure 18(a)

and equations (24b) refer to Figure 18(b).

-47—

'7' 3 ‘-

z I;
1"" 3‘ ’1 91?

 

 

 

Figure 18

 

 

e = (z % z )i % zciq e z z 1
e2 — 2013 # (2b # zc)12 ca 231 222 12
Where zll - za % 20, 223 = zb # 20, and 212 = 221 = 20
I 2 ' _ I, I
(24b) 11 (y; % Yg)31 Y293 and 11 : y11 Y13 x 81'
i2' : ~y261' * (Ysz3)92' i2' y21 Yes 92'
Where V11 = Y1 # yg. ng = ya % Y3, and ylg = Y21 = -Y3
P 1 -
y _
Therefore el' _1 11' —gg ._Z§l 113
I:Dd x 3 ‘Yl ‘Yl IX
82' 12'
3113 y11 LIB:
LIYI ‘YIJ

 

 

But since the networks are to be equivalent:

31 Z 81', 92 : 63', 11 2 11!, 12 z 13!

.- -l _
Therefore [Z] - [Y] and 211 - ygg , 228 :.-. yll ’ 212 = "5'21
IYF

lYl |Yl
2 _
- y1y34y1y3%y2y3

lY': ylly22 - (y12)2 = (yl%YB)(YZ%y3)-y2

.. .. -1-- ,4 1 7‘ 1
LetA- yly37‘yly37‘y3y3 - 2122 EYES Z

 

oz,
0 o

_48_

Thereforezstz 23 # QB % z1 and l — z 222

A150 Z = Z 7‘ Z : 3’28 = 7‘ " r7 3
11 a c .75. Y2 Y3 , 223 — ~b/Zc Y1 % Y2
A

 

And 212 = z“ = YB
" A

 

 

 

 

za 2 .3; 212223 x .l- 3 2122
13 z "““‘
. 217Z227ZZ3~ 3 21#22%23
... Y1 Z Z Z 1
Z -' __ - 1 2 3 -— 3
(25) b A 217IZQ;Z~ X 21 2223
26 2 .33 : 212223 x ‘l_ 3 2123
13 22 z lz l2
zlizzizs l 3 3

The T _u-‘TT'transforms are derived below:

 

 

 

. _1 _ z q
y11 ylz _ z11 z12 - 33 ‘321
‘ ‘ z
y21 yzz 221 323 | I lZl
7312 z11
_ IZI 1er

LetV : 'Z' :: leZBZ-2122 = (zafzcuzb’lzc) ' 2c2 I
zazb { zazc % zbzc'

_' -Zn- '
Therefore yll — yl%y2 — 2e - zb#ac , Yza = y2%y3 : za/zc

 

? V ‘v—
-y12 2 ya - 221 Z

-49..

 

 

 

yl : £9 and 31 : .SL : zazb # zazo#zbz
V 2b 2
b
(25) Y2 : 20 and 22 -_: 1 zazb ,‘ 213ch 7‘2sz
V’ 29 2
C
YE i 3 - ?- Zazb ,‘ zazC 7‘2sz
7 a z -
a

The bridged T is a rather common structure, and it
will be analyzed With the object of obtaining its equiva-
lent T. The bridged T of Figure 19(a) is redrawn in

Figure 19(b), and their equations follow.

 

 

 

 

 

 

Figure 19

Figure 19(b) e2" : 2211" ; zgigua 13"

(y1%y4) -y4 ‘ 81'

Bottom box 61" = 2211" # zzig"a nd::‘]:Z 22 2::]x 11'
Figure 19(b) M[:

Too box {11' = (y1%y4)e1' -y4e2 '

13' = -v4e1' # (y3%y4)egan -y4 (Y3#y4) 62'

-50-

 

e ' FY % y '
Therefore 1 : .-§_-4 Y4 F’ :
e2! IYI Ti? .x i1
Y4 y1%Y4 1 ,
_ |Yl [Y1 j ._ ZJ

 

 

 

 

But since el = 91' f 31" , 92 = 62' ¢ 92"

And 11 = 11' = 11" , 13 = 12' = 16"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r ' F q
81 {$312,123 33,122 Fill
: lYl [Yl
x
6 Y4 % z Yify4 1
2 " )‘23 2
.4 .M " m JJ-J
Referring to Figure 18(a) and equations (34a)
rel F(za%zc) 2o 11
: x
Gal 20 (Zb7lzc) 12
Since the two networks are to be identical
2 #2 = y3¥y4 J 22 , ZC = y4 __ %
a c y1y3%y1y4¥§3y4 ylygiyly4%y3y4 23
zb%zo ;__YI%y4 % 22
YIY3¥YIY4;YSY4
1
z : yy3 3» _¥_ 3 IE .- = 2124
3 1y Y Y Y Y "5 #2 #z ‘"”“
3 14 34 4 3 1 217-2—31‘24
1
Y1 _ '—
(27) Z = -' 21 _ 2324
b Y1Y3%V1V4;Y3Y4 ““EZ¥Eg¢EI '
“‘2'2'2" zl#23%z4

-51...

Y4 1

 

 

 

(27) 2c z ¢ 2 E; - zlz3%(zl%23*24)22
Yiys y1y4%y3y4 24;23le '_21¥23¥Z4
212324

Equations(27) determine the relations between the
three elements of the equivalent T in terms of the four
elements of the bridged T. Using the T-4~47’transformation,
the equivalent ”could be obtained from the equivalent T,
or the bridged T (Figure 19(a)) could be treated as two
networks in parallel and the equivalent‘fifobtained directly.

The symmetrical lattice or bridge, Figure ?O(a), is

another common structure and may be represented as in

 

 

Figure 20 (b)

Figure 20(b). This is oossible because the currents

indicated by the arrOWs in Figure 20(a) are equal.

 

 

- - Y]. ' _. y ' r q
Top box 11 - 2— el 2-}. 83 and 11! 1'2;- _yl‘i 81'
Figure 20(1)) 1 - -y "" 2 .2
o 2 "' _-.].:. e ' - Y1 en. 1 I x
2 1 2—- 5 L ZJ -y1 Z}- 82'
I 2‘

 

 

Y Y 'Y’
Bottom box 11" 3 Eael" % 5292" 11' —§- Kg? e "
F ?o( ) and = w 2 ‘
igure e b 1 " = 1 1 u
2' Y2 u ,( 29 n 8 Y2 Y x e u
-e 2 .__ 3 2
- 2

 

 

Since the two networks are in earallel, il - il' % il",

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12 : 13! ; i2", el = el' - e1", and eg 3 e8 = e2"
, Tyltya ye~y1'
11 e " 81
Therefore _ “ 3 x
‘ Y “Y
12 2 1 y1#Y2 82
. 2 2 ‘
e1 ' y1%yg Ye-Y 1'1 1
And 2 -:g—$ l
- _ x z
. 3 2 J
”21%23 22-211 ’ 111
2 2
x
Z8‘21 21%23 12
2 2 " JL

 

 

 

 

 

 

 

 

Figure 21

-53-
The equations for the symmetrical T and11'are:

il % zbig

e - z #32
Figure 21(a) l _E3_JE

_ zbil % zaizzb .
en - '——3———'lq

 

 

 

 

 

 

 

 

. p .
ell za#22b Zb 1 | 11W
And : 2 x
z #22
99 2b a0 b 1
L ”J - ~ ‘ L 2.
Figure 21(b) i 3 Y #2y
1 CB d el _ ydez
12 ' 'ydel * yc"“ye92
2
f 1 P 9.
1 Y *w) 1 V e
And 1 ° 2 d ~yd ll
3 x
13 _ ‘Yd yc%2yd e2
b a TA 5 "

 

 

 

 

For the lattice --r- T transformation, [ZJL :- [ZJT and

for the lattice ~o ”transformation, [YJL _-.- [Y]".

Therefore

N

2a % z 21%23 _
r b 2 , Zb "’ 23-21

2

 

 

9

 

Y _ ¢ _ y -y
‘2 # y ‘ yl Y2 3 yd - l 2

2 d 2 ~ 2

 

_.54....

And lattice-«>T gives the follOWing conversions:

za : 221, and 2b = 23-21

 

a!

The T «plattice conversions are:

2 -
1 ‘ 29

-Z
,8.nd u-Eé7l22b

NIN
w

(28)
The lattice -~-- 77’ are:

yo : 2y3 . and yd = Y1“Y2
2

 

And the 77’ -—-a- lattice are:

Y1 :‘ZE % 2rd , and Y2 = ya
2 "E
The above analysis of the T,7f, bridged T, and

symmetrical lattice networks, with matrices, has established
directly a relation between them. Eguations(27) show that
the bridged T has a unique equivalent T whereas a given T
may have several equivalent bridged T's. A180 equations (28)
show that any symmetrical '1‘ or ”may be reoresentedas a

1

hat a

ymmetrical lattice may not

U)

symmetrical lattice, but t

always be renresented by a T or 7T'(negative imnedances).

Three Basic Katrices

A review of this chanter‘till show that the three
basic matrices of four termina networks are the

imoedance, admittance, and cascade matrices. Then the

determinant of the imoedance matrix is zero, it does
not have an equivalent admittance matrix and when the
determinant of the admittance matrix is zero, it does not
have an eouivalent imnedance matrix. However, both of
these matrices do have an equivalent cascade matrix,
but when G of the cascade matrix is zero, no equivalent
imoedance matrix exists, and then B is zero, 10 equivalent
admittance matrix exists. These three matrices are,
therefore, the fundamental ma+rices, and from them the
other three forms can be obtained (equations 18, 14, and 18).
It asiears, therefore, that any four terminal network

.1
.\

can be handled With the imcedance, semittanoe, and cases e

C);

matrices and the matrices obtained through the maninulation
of these matrices. It must be keot in mind, however,

that to use matrix algebra on a four terminal network,

the current in on a siren end must equal the current out

'I:

on that end. his sometimes requires the insertion of an
ideal transformer to make the mathematics valid. If the_
transformer is inserted for ourooses of analysis, then it
must be present in theantual circuit or the analysis Will
be false.

The circuit of Figure 22(a) will be used to illustrate
the matrix analysis of a four terminal network. This

circuit may be redraWn as shown in Figure 32(b). The only

difference between the two circuits is that a 1:1 ideal

 

 

 

 

 

 

In 1 3,, 3.?’ our

 

 

 

 

 

AAA “-

 

,3,” 0hr

 

 

 

 

 

3" (b)
Figure 22

transformer has been inserted in cart (b). Vonever, th's

does alter tre network, 2nd, therefore, tte transforuer

may be broken we as shown in Fiuure at.

If]
P -

 

 

 

 

 

 

 

on?

 

 

 

 

 

 

 

 

p-

This chacter has shown how the subnetworks of

Figure 93 may be COmblfled, With matrices, to give one

1

matrix for the overall network. It may oe done in the
folloWing steps:

First——fultioly the cascade matrix of (a) times that
of (b).

Second--Obtain the equivalent admittance matrix for

the (a) and (b) combinrticn.

Third-~Add this admittance matrix to the admittance
matrix of (d).

Fourth-—Obtain the equivalent cascade matrix frou
the third steb for the overall (a), (b), (d)
combination.

Fifth—~add the imaedance matrices of (c) and (e).

Sixth-~From eteo 5, obtain the equivalent cascade
matrices for the (c), (e) combination.

Seventh——”ultidly the cesced matrix of the (a) (b)

(d) combination times the cascade matrix
of the (c) (e) combination.

The last steo will give the Cascade matrix for the
overall network and from it the imcedance or admittance
matrices may be obtained.

There are a large number of examdles in this chanter,
illustrating the nrocedures of cascade, carallel, and series
connections, therefore, the solution of this oroblem has
been indicated rather than actually carried out.

This cheater has shovn, With illustrations, how
matrices may be an'liEd to four terminal netnorks. And
it has oointec out some of the limitations as Well as

advantaged of matrices when so used.

-H.~1.""_’ III

[‘1 "A l' ”31:? 1‘71 5. ~- . «77'?

A kirge number of networks contain only one voltage

Ind may, fierefore, be considered as the terminal networks.
latrices, here as altays, are a decided factor in main—

*3

taining the continuity of the wroblem. There a e many

matrices, however.

0

other advantn.es be had by usin
The res t of this chanter will be used to illustrate SOme

of these advantages.
Synthesis of Equivalent Networks

Occas sione lly it is desirable that the contours of a

‘

given network be Chan 2e 0 and that the in ut imneo ence be

maintained invariant. For instance, suooose that ltfl
[Z] x [I] reoresents the given netvork and. that ['3'

@a' x: fiJ' reoresents t e desired network, then a
transformation matrix [filmey be used to transforn k3

into [231' and maintain z. invariant. The

inout - ei/li ’

develonment is as follows(assuming the voltage to be

 

in mesh 1):
F81 :11 :12. ' ° ° ' ' :1111 :1
: 21 22 o o o o . 0 2n 2 , .-
0 . . x . , [I] —[Z]X[IJ
L. L Zn]. 21:12 o o o o o o Znn Lin.

 

 

 

 

-59-

[E] = [p] x [I] is the matrix equation of the given network,
H ' .— P1 ' | - 1 0 1 'r

and [J - [a] x [I] is of tne demred networx. [If] must
transform [it] to {23' and still allow e /i = e '/i '
1 l l l

(in-nut imoedance invariant). It has been found that [fl

. . . T ,., T, _ ,, I

may be useo 1n the following; manner, {at x [J x in] — [a] .

[rat oremultiolies {3] for no other reason than that the

end result justifies it. Touality must be Haintained on

both sides of the equation {j Z [Z] x [I] at all times.

«1

So if both sides are nrezrultiolied by Ejt’ then [at x [L]:
[II] t x [Z] x [I] . r.1SO, [2'1] x U ‘1 may be inserted
between {2] and [IJ since i] x f] ‘1 = [L] . Therefore,

“f j-Q : v”:- ‘7 ‘r ‘1

i] t x [.1 J x [Z x n x I. x [I

{13' 1' [23' X [If

 

 

 

 

 

 

 

 

. ,r _ P m __ n ‘1
Where LE] - ‘(u 4'1. . . . 0‘“ and [1:] t - ‘(ll ‘6'. . . - J.“
41, C(zz . - ‘ 41“ ‘(ll {11 . - - 4‘61
L‘ul 1‘: ' - - 4‘!“ b ‘1‘ 42" . , . J‘hJ
if kll Z (-l)i7£j x minor of
And [K]: Deterznirzfint of [II]
‘ A . _, P 1' 1' D . I
Then {1] - k 1 “21 . . . . . 2.an 111 I'll.
IKI IK\ IK ,
klg kg? ..... kng x 12 : 10'
m (K: \KI ’
k1n k2n knn i i c
L [Kl 'K IK‘ ‘ v n . l. nj

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k k
And therefore i ' : 11- f‘71 - k,
1 1%]I{112’£""l:‘11
'K' [K] n
rd 1
. ~ 1 _ II °( . - .
also [E - 1’ ‘6" re; " e1“
«(:1 411 4n: o O
. . x :
. o o
O O
L‘In 41v! . . . 4“ L- J - .
And 91' : el 1! 0 7‘ 0.
e1 e '
But it is necessary that i_': ; that is, the
l i I
l

transforzretion :::.;1trixi] does not Chenpe he input
impedance. A simple way to satisfy this condition would
he to choose a(,,: 1 and. 4.1 = «(.3 = = «(m = O, and

l. .- — '.?‘;‘:'v ' .0 r; ‘
therefore, kgl — kfil .... — knl - O Q grab el — el onQ

1'31.
1 1 ’10....07
‘(21411- - - «(an
Therefore (I? -"-'

I

“(In '(ua. - - - (at:

The above derivation stimulates that [K] -1 exists and.

 

 

.9
therefore, that [Kl ¢ 0. This requires that {a be a
non—singular matrix anC, therefore, huSt be square.

The next steo is to determine the0('s. This may best be

shown With the f01lOWinj examole. Figure 24 (a) is a

 

 

 

 

4:
--""lfr *
2' ' Z | 1 2:, 2' ‘1
’ L r 3 “ 4
15; Z 52 '6;
5; 5% 5‘
—"IF _ T T

(a) (b)
Figure 24
network that has the desired inout imoedance, but not the
desired circuit configuration. The orohlem is to find
the narameters of Figure 24(h) such that 11' 3 11 and

81' 3 e10

Solution: el " {3w UNTIL ) # 313111 .. (jULgfie

. SD
0 : _(j¢oL2)il * (ijg %“;)ig
Therefore e -(L1%Lg) ‘Lo 3 O 1
- *1 l l
— jh) % 3:; x
O L ”L8 L2 0 S 13

 

¢-
H
U)
0
CD
I'-‘
ll

33“}; % S3§i}1'_ {3' “L3,; .22512!

" ‘5”1‘3 ’l 34%;“ #{J’U(L3%L4‘H (8)}1’“

j“

C
l

(inout 1moed-

l 0

If K = = = ' =

[J [4. ‘1], then 11 i1' , e1 e1 , and z s'
ances)

Therefore L3 «L2 9 1 .x, O
U = 10—.“ X :
-L3 (Latin) 0 4. -10 lo

10—3 x (is—e04; 104,!) (104..(,_-10 .(z)
(10 «act '104‘1) (10 x: )

1 oh 4 O 1 O

3 3 = 104 x x x Z
4 (4 ,4 1343) (10 4.41)
10 x

(10 (an) (10 #3)

Since the first row elements of the two matrices on
the left, are negatives of each other:

(29) 4 ,4 10 4‘: ~10 1.4, and 15—204 4 iox,‘ =-1o.(,(.<, -1)

Solving these simultaneously:

4. = -o.2es end 4,: 1.575
'Therefore L1: 104(15—204, #1043) : 21.589 x 10"2 =c.21»3 henries
L4: 10‘2(10.(:-(15—204 ,t 1043)): 0.0435 henries

_, 4
s — 10 (4410.13 = 104 (4 ,4. 8:39) —- ssco darefs

(Q

s = 104(1oxf-(4,lio.(,‘)) = 2 2,310 derafs

Foster's Reactance Theorem states that the two networks
of FL ure 24 are ootentially equivalent. By a similar
method Sauer's two potentially equiv lent net H0 is may be

found. The limitations of this technique are in finding

-US-

the¢('s. If resistances were introduced into the two
mesh nets rk, there would, in general, be three ixzne o Mde t

._,

equations of he trce(?9) to be solved simultaneously for
the tw0¢('s. Cf course, this is not noes sible. If the
solution had Leen in terms of 2's there would have been
only one ecu tion and 2
found. However, if there had een resist: does associated

b
v. ith each in du ctsnce ant the E/L ratio was constant

{1'

throughout the network, t en a three ourameter solution
could have been found. For a three mesh network there

would huae been sixo('s and for a. four mesh, twelve ('s.
The limitations ar e, therefore, quite great. For a two

mesh, however, this method is faster the n Foster' 8 or

Cauer's.
Change of Reference Frame

The follOWing technique has been called by Le Corbeiller,
The Kron hesh ”ethocg, since it has been develc oed and used
extensively by Tron.

In networks containing several meshes and a large
number of mutual imnedances, it is often times difficult
to obtain the mesh imoedance matrix I?)' by writing the

rexwell mesh equations. In the simcle bridge, for instance,

 

0. r I ‘ w'.‘ . Y ' ‘
”Le Corbeiller, :ctr1x analys1s of electric Leteorus,
09034-44.

-54...

there is a oossibility of fifteen mutuals. One can
visualize the task of even attemnting to write the maxsell
mesh equations for this case. The Kron Jesh Method
eliminates this sort of confusion and allows One to obtain
the most comolex mesh imoedance matrix by a surely mechani—
cal process.

A simnle oroof of this technique is given by Le
Corheiller, end it takes about ten oages of his book

"10 Since most

"Matrix Analysis of Electrical Circuits.
engineers are interested in method rather than aroof, only
the method of aonlication will be presented here. It
must also be said that this method is not limited to two
terminal networks, but that any finite numbers of voltages
may be oresent, providing that they are either all d.C.
or all a.c. and of the same frequency.

The metnod is as follows:

For each branch draw an arrow reoresentin; the

current in that branch and choose the current direction
the same as the voltage in that branch. Next, arbitrarily
choose the mesh currents and indicate their directions
With arrows. For an n branch network there Will be n
equations exoressing the n branch currents in terms of

the mesh currents. From these equations the matrix

equation may be written by the usual method, fl] =fpj x: U].

 

10 cc. Ci .

~65—

! is the branch current matrix, 93' is the mesh current
matrix, and [5'] is the connection matrix between the two.

If there is a voltage in each branch or only one

voltage oresent in the thole network the branch voltage
matrix may be written as {f with sufficient zeros to
give the column matrix n rows (n is number of branches).
Tach mesh voltage, in general, will contain more than one
branch voltage and may be reoresented as [E3 '. The "method"
then says that [‘3' 3 E“ x E]

The bower of the technique will now be oresented.

The branch imnedance matrix. ED is obtained in a very

simt-le manner. If there are :1 branches, then [2] will

be square and of order n. It is combosed simely by

nlacing the n branch imnedances on the orincicle diagonal
and the zij mutual impedances are placed in the ij positions
of the matrix. The maxwell impedance matrix [Z] ' is then
obtained as ESL; x [Z] x E3] = [2]..

The mesh equation may now be written as [31' ='- [Z]' x [3'
where [13' are the unknown maxwell mesh currents. Therefore,
[13‘ =[Z] "1 XE] '. After [13' is obtained the branch currents
can be obtained from our original equation, [I] = p] x [I] '.

It must be cautioned, at this noint, that [I] #[a’1 xtF] .
Recalling how the {2] matrix (branch) was written, the
equation [I] 2‘2] -1 x179 would give branch currents as

though each branch imoedance and voltage were shorted on

themselves. For a two terminal nettorx ( ne voltage),

w

it is obvious that 1 out

2, is...end in, are not zero,

[I] = [3‘1 x [E] would give every current but 11 as zero.
and 11 would equal el/zll which is incorrect.
Since the circuit in Figure 25 is of a network that

is not flat, the mesh equations are difficult to obtain.

This circuit Willbe used to illustrate the Kron jesh
, :3
Method. --'-

éy’

 

In Figure 25 is the classical cube oroblem. The problem is
to find the incut innedance across the diagonal. There are
8 nodes and l subnetwork, therefore, 8-1 3 7 indeoendent

node oairs. Also there are 13 branches and therefore
, J ’

-57-

13-7 =3 independent meshes. The chosen mesh currents are

indicated in red while the branch currents are in black.

 

 

 

 

 

 

11 = 11' 0 o o o o r 1 o o o o d‘
13 = o o 133 o o o o 1 o o o
13 = o o o o 15 o o o o o 1 o
14 = o o o o o -1.* o o o o 0-1
0
15 = 0 -1g 0 a o o 0-1 0 o o o
-' ' I- 1 " 'r '
15 - o o -1g%14 o o [i - o 0-1 1&0 o x 1]
17 = o o 0 14-15 0 o o o 1-1 0
18 = «110 o 14 o #15 -1 o o 1 o 1
19 = -1f¢1§0#r4 o o ' -1 1 o 1 o o
110: -1£¢1g#1go o o —1 1 1 Q o o
111: o o 13' 0-13 0 o o 1 0—1 0
112: o o o 0—1510 0 o 0 0—1-1
113: o 13 o o O‘-ié L o 1 o o 0-1.
There the 13 x 5 matrix is the [Q].
" e1 7 ' 91'}
o 0
There— , I _
[E]' 3 [qt X[E] where [E] = O fore [1"] " O
Lolzd L064
r'0 o o o o . 013}
fifl =: o o 23 o o .
O O O O 25 .
L013. . . . 213‘

 

 

h

[21' 3 [011; 1:4 [Z] x [C] and carrying out the indicated
coeraticn gives:

P(28%29%210) ("29-210) (“210) (—28-29) (O) (~28T
(—zg—zlo) (25¥zg%zlo%zlg) (210) (29) (O) (~213)
[Z1' 3 (-210) (210) (23%26%zlo%zll) {-28) (~zll) (0)
{—za-zg) (29) (~26) (25%z7%28%29) (-z7) (28)

(O) (O) (~211) (~27) (zg#z7%zll#zlg) (212)

L(~28) (~213) (0) (28) (218) (z4#28%zlg%zlg)

 

 

A two terminal network is being used to illustrate
Kron's technique, because it works very nicely here. Since
in this oroblem the interest is in the inout imoedsnce only
and, therefore, in il' only; it would be necessary to find
[2" if the method of determinants Was used. The following
method will illustrate how 11‘ may be found without solving

for the deterninant ofe 6 x 8 matrix: 11

Let [Z] ' ”[211 [313 There LZJ 1’ [232’ [7‘] 3'
A [213 Z

[34
Also let [F] -[["L:j: ‘71‘918 111-[:]%nd[’32=§[]

and [214 are 3 x 3 matrices.

 

 

Also 18351111.? [111 T'Jhere [IJI 7- ril'T end [112 = ri4'T
[I12 13‘ 15'
L16'

 

 

 

llGabriel Kron, Tensor Analysis, 13.21-22.

Therefore {:31 :[291 2([131 {[1732 xlIl andI’fi g =[ZJ3 xifll ,1

[214 XIIJQ. F‘lir‘nineting [g 2, the following equation is

found:
{EL—[z]. xfdil mm = as. 421. x h: x 1th x [:11

. But [‘32 [8] and therefore,
0 ,

[131 [Z 1 -[Z]g 7‘ [334-1 1“ [93)‘1 x [E31

In using this technicue, it is essumed thfit it is

easier to find the inverse of a 3 X 8 twice, than it is
to find the inverse of a 8 x 8 once. Ihe meth od Will be
illustrated by Carrying the irevious oroblem to conclusion.

If we let all of the branch imoedences equal one ohm, then:

 

 

 

 

,. 3 -2 -l:-2 0 —1T .1, {-1
—2 4 1| 1 o u z

. :1..-i-4I.-1:1- 0.. 31 'J2
{2] = -2 1 -"‘i| 4'11 1 = U
o o 1 1 4 1 z

-1 —1 o: 1 1 4_ fab 4;

If we let E]: [Z11 - [213 x [234-1 x [213

3 -2 -1 1.1 —o.4 o. s 1.9 -1.5 .1.6
Then. E]: —2 -4 1 .0.4 0.8 0.6 = -1.e 3.2 1.6
-1 1 4 0.0 -o.e 0.8 —1.5 1.8 3.2

”r4

7.86 x x
x x x
.4 x x x

:___I

And [%I l—

O)

-70....

 

 

 

 

 

P I b A
1 ' 7.38 X XW e
l l 1
Therefore 12' Z 574 x x x x O
I
L13. L J( x X. 0
7 C38 8 h r-
‘ 0.4 — a — ssh
nd i ' = 5.; z, : AA -'= - 0.80oo ohms
A l ’ inout 7'09 9

This illustration was used to show how the Kron nesh
“ethod could be scilied to advantage in determining the
mesh imoedsnce m trix of e comolex circuit and to show how
a desired curren might be found without finding the
determinant of a large (31', oroviding the other currents
are not desired. She tho terminal illustr.3 tion ore ented
here is not meant to imoly that this technique is liwited
to circuits containing one voltn5e. It has been derived
for an n mesh network conte ininw n voltages. The two
terminal nrobleu oresented is, more or less, a classical
one, end was used to simolify the calculations and still

orese ent the method.
La.trix Pa.rameter F.eoresentation

Iatrix care.xeter reoresentction is a method of
reoresenting a circuit in terms of its resistance, induc-
tance, and elastance (recicrocol ceoaoitunce) matrices.
For instance, if it is nece m3.ry to give a person informa-
tion on a two terminal nettork, it is much more cczno a.ct

to simnly give him three matrices and let him draw the

-71-

circuit. The method csn best be illustr ted with an

X2311 '16.

 

 

 

9 are hi;hly

innrobe ole, out serve for ourooses of illustrstion. The

:3”

values oooosite t e condensers are Velues of elastence {l/c).
This method aoclies for oassive (too terminal) networks only
and it is assumed that the voltsLe is a comoonent of mesh

'1 only.

The mesh equations are:

(e¢3we,£3/ju)il 4113 ,4 013% 014 - (juS ,4 1/j...)'15

e
o = —411 ,1 (4%juS,‘3/ju)ig 41/3... )13 -(2/ju)141‘ 015

= 011 -(l/ju)12 #7; 1/3... )13 - 314 ,4 015

Oil -(2/jw)12 - 313 ,l (3443.02,! E/ju)i4 - 315

= -(30 5; l/jw)il ,1 Oi; ,l 015 -Si. 1! (eggs; 1/ju)i_

O O O
H

The matrix ecu tion [31" —- [Z] 1: [lie] herefore:

e (357406,l3/jb) -4 O O —('¢o5-—l/jw)

o = -4 (masts/j») (4/1 9 (42/3...) 0 .2
o 0 (4/10) (man) {-3) o x 13
0 O (-3/j") {-3) (3 gh’g *2/3‘“) (~3) i4
0 (395/1/39) o o :3) (2,4qu 741/31») 1:

”‘ 5
~

6

as follows:

O.
H
d
H.
O:
(0
£1:

However, the imoedance metrix new

1
1

A
I
.53
O
O
O
J
A (D
\J

q
(~13)

 

 

 

 

e o o o
— . o o (3) o o o
[z]: o o (7) (—z.) o A! jo o o o o o
o o (-3) (s) {—33 o o c (72) o
1 o o o (-3) (e ."‘5> o o c (s)
1 d

‘k
9
[000m

 

 

It is quite aoourent that the method
reoresentetion is much simgler than the uctuul Circuit.

To draw the circuit from its o remeter matrices is
es remenbered that the voltage
is in mess I, only. The method for obtaining the circuit
from the oeremeter matrices will now be illustrated.

The first rows of the three matrices indicstes_tha
there is no mutual between meshes l and o, and l and 4.
The second row indicates that there is no mutual between
2 and 5. The third row indicates none between 3 and 5
and of course, 3 and l. Tith this information the

folloWing structure may be drevn in this order:

I1 l 2 I 3_] But there are mutuals

[g I 4 I ’ between 3 and 4, therefore,
- 1. the structure must be
changed to the following one:

Illzlg
|5|4l

 

 

 

After the structure of the network is obtained, it is a

simule matter to insert the R, L, and 1/3, values giving

the circuit of Figure 25.

Imoedance Level Chan e

‘4
{.3

There are an infinite number of networks having
the same innut immedanee as a given network. and, for
ourooses of economy, any one of tnese may be more desirable
than the given network. Or, it may be that the siven
network has an induetsnce or Csoecitunce that cannot be
obtained. In any event, the given network can be changed
to another one having the same confi3uration and inout

imoedance by a very simole method.' The basis for this

Operation till now be derived. The follOWing equstions

reoresent any two terminal network.

(D
I

1 - 21111 % 21212 % . . . . Zlnin

in % . . . . zn i

0
II
N
\J
H
H.
H
‘k
[‘3
I 3
LU

arm 11 - 211 Where Z'

(O
«IQ

 

Therefore v” 3 z. 3
11 insut .

E]
H
H

-74-

flow if any one of he columns or rows of the above

1

equations, exceot column one or row one, is multiplied

by N, then the new determinant [2" till ecual N|Z|(cheoter
Q d 1“ . zen-'- \ ' ;' ' . I ‘ "

I), -nc the new ml.or 211 V111 esusl N 211, Since 211 is

the determinant [Z] With column one and row one deleted.

If all of the columns and rows are multiolied by N(exceot

column one and row one), the new determinant [Zl' will

9n

w

equal NinIZ/ and leI will equal N

therefore, that regardless of how many rows or columns

Z . It is see ..
11 y to see,

are multiolied by some finite number, or numbers, these
.numbers may be factored out giving the sums relstion between
’ZI' andlZ‘as between 211' and 211. The result is that
zingut : el/il = ley/ N211 is unchanged. Referring to

the imoedance matrix of the previous section, the following

illustration till be given.

 

 

 

r55) (-4) o o o l (e) o o 0 {-5i
-4) (4 o o o o (3) o o o
ff} 2 o o (7) {—3 0 ¢ 3 o o o o o
o o (—3) E5) {-3) o o o (z) o
_ o o o -3) (s) (-5) o o o (9)4)
P(3) o o o {—1)
1 O (3) {-1)’«?) O
% 3:; o (-1) (l) o o
o {-2) o 2 o
{-1) o o o (1)
‘- d

 

 

Hultiulyin: the lest column by 0.3 fives:

#

_l

 

 

 

"(3) (4) c o o 7 "(3) o '3 O (-3
, ' (-4) (4) o o o O (8) O o 0
fig :: 9 O 57% 33 o % 3:0 0 O O O O
o o - ) a 5-1.8) o c c (:3) c

L0 o o -3) 5'4). Lbs) o o o (5.4)

 

.1__ O )
% jaw O «l
O 3

 

 

and because of the erevious derivation, CZ] end [Z]'
have the same inout innedsnces.

Care must be used, however, or a network may b
made ohysicolly unrealizeble. For instance, if the
second column is multiolied by 13, the resistance and
elestence nutrices are unobtainsble. Thet is, there is
more imoednnce in the mutual than there is in the mesh.
This, of course, is not oossible. The method might be
used, however, to obtsin n nhysically renli7eble network
from one thst is not.

This chsoser has exoleined and illustrated some

18

U)

of the the terminal network analysis and syntbe

oossible with the use of matrices.

CHAET“R IV
SYZLTTEICAL CCuTCVTTTO

Symmetrical comoonents are defined by means of linear
algebraic equations. Because of this fact the details of
handling these equations can, and often do, obscure the
end results. Katrices, because of their compactness, can
be aoolied to symmetrical components With the result that
the oroblem is Hlt ys in sight rather than hidden among
the algebraic m3nio “1 ti Jns . mr. Reed makes tlfi. s statement
in advocating the teaching of svnmetrical components with
matrices, "Ieeoin; all th e details in their oro3e r olcc ..
by the usual methods is such a tremendous task that rarely,
if ever, are all these details given."12 The rest of this
chapter will be used to outline a procedure for the
aoolications of matrices to symmetrical comoonents, and a
couols of examoles till be included to illus Wr te their use.

The defining equations will be stated in terms of the
current, but will as ly equally well to voltages. Also,
it will be assumed the t an A, B, C, sequence is oresent,
since a given sequence can be reoresented es an A, B, 3,

sequence and used in these equations.

 

o.4'

(J)
H
0

~ - . ‘~ : -\ r r' . ('4 , -' " f‘ . "n“ . '7‘"-
nyril B. Reed, nltcrrstin,,.urrent Llruult éggggy,

-77..

 

 

Let
i , i . , be the unbalancee three chase currents.
a b, 10
H ' be t e u t' " . Wee ts an ‘ ~ -~.
1a1 101 1cl bhc nositite secuen c of eultncec currents.
. . . 1" ‘3,“ 1“".z . V ‘ ~ , ‘N': 5:)“ ‘ 5 ‘3“ 4'
1&8 lbg 103 ee tne neaitite sequence of Lelvhcec currents.
, i i be the zero sequence currents.
do b0 00
'lZJ-O’3
.1“ ' : ' ' ' .i'y' . : J
T-.en 1&0 13,11,313“: xLneie a 6
o ' '-
7‘ 2 1 ,z e g a2 .. o
= 1- e1 ¢a i. ,
a u 1 O _.
a1 ‘h‘ * snc a - 1
8
"D
_ . “w. --
ieg - 1a#e 1b%aic
5
Therefore 1,, '1 1'1 1 11 f i l
o a
.1. o
— H .
ia1 — 3 l a a x 1b
1 1 3 1
u a
as, L J L 0..

 

That is, [15,18 =[r]

 

 

 

 

 

 

 

 

x [I] , end [I]: [T]"1 x [1818

 

 

 

 

where [Iglszriao [T] = T}; r1 1 1i
2
a1 , 1 a a
,2
51a? bl a 8..I
"’4
.. ' a -l.. I-
[I] - 1e [T - 1 1 11
1b , 1 e3a
, 2
+10‘ bl a. 8.1
But ibo 1&0 % 0 ¢ 0
ibl = 0 # a21a1% O
ibe Z O # O % aia2

-73-

And i 3 i % O % O
00 b0 O
101 ' O / a“ibl / O

1Cg = o % o # aibg

Therefore [1b]S = Fibo‘l = ’1 o o“ riao'
2

1b1 O a O x 1&1
,1b3J I_O O a; ''18?"

 

 

 

 

 

 

u
Fri
£3
H
x

And [L012 [15:15
Also[IC] s 3 [Tlr x [15) S [gr x [Tlr 11 [la]

But [gs-z [r]r x [1518 [firsx We

[3 therefore [f] r-1 = [Tlrg

S

And[T] 1‘3

With the two transformation matrices, [fiend [T]r,
the symmetrical components of any one of the chase
currents may be found from the three unbalanced nhase
currents, or the three unbalanced nhase currents may
be found from the symmetrical comuonents of any one of
the nhase currents. The following exemole will illustrate
this statement.

Referring to Figure 27 find the symmetrical comoonents
of the line currents in terms of the symmetrical comnonents

of the chase currents.

 

zc‘ -——"

 

Figure 27

 

 

Solution:
18L = 1a % c - '10 Therefore [IL] = '1 o —1‘
: _ - . —1 l O
ibL 1a # 1b % O
. _ , LO —1 1d
10- - O - 1 i0
1. b That is, [IL] = [A] x [I]

But [IaLJS Z [T] x [113
And [IJS 3 [T] x [I] or [I]: [T]"1 x [1818
Therefore [IaL] : [T] x E] x [TI-1 1: [T518

That is,

 

{1

And

 

Therefore i

riaOLT

.. aBLJ

18
.1L

18.
2I.

1
8.1L
1

 

q

8.0L

 

d

aOL

 

 

 

 

 

 

Pi q
. -1 3
EH 2: gg }{ flfl x. 1 0
a1
Liagd
o o ‘ F1 ‘1
6'0
(l-a) O X i
2 a1
0 (l-a) i
« . as,
.. -' . s 0
L 4-5 6 33001“, la”1 '3'1’7 63"0 1
“L

x [T]

-80...

The above results show that the zero sequence current
in line a is zero (the zero sequence of line current is
always zero for a three Aire system, unless grounded).
They also show that the relation between the oositive
sequence current in line a and the oositive sequence
current in chase a is the same as the rela ion between
line and chase current in a balanced load. For the
negative sequ uen ce the line current leea ds the obese current.

The relations between the sysnetrical comoonents of line b

 

 

and Mb Gonoonents Of phase b WHY be es tel .lis ed as follows:

[reds-1 [flr x [131.13 And 1.138 : [T]: x [1318

Therefore [15L] :[flr xfi * 1 O O .. -1
s 0 (330° 0 XETJI. 41ng

L. o 0 £30:
0
.,., O u
.0 O €100.-

 

 

These results show that the relations between the line
comoonents and ones e coroonen ts are the szime for ezzch chase.

The ouroos e of the next eHxa;1ole is to furt‘. er illus-
trate the use of matrices, but more imoortant, to illuminate
a nossible technique for the solution of circuit 3roblems

0"

that he ve soecia 21 s ametr 41e technique of disconalization
3 Yo .

-81—
is being referred to, and Mill he presented in the
followirg Chester.

Feferring to Figure 28 find the symmetrical comeonents
of the ;enereted voltages in terms of the symmetrical
comnonents of the currents.

‘4‘-‘A

 

 

 

 

 

2'1, 3'- 3L
3v
1»
° 2 c i3“
+ #“VAVAS
Figure 28
Solution:
ea — (zg%ze%zL%zN)1a # ZNlb 742N1C
e = zNia % (26% ze%zL#zN)ib % ZNic
e0 2 zNia % zNib % (23%ze%zL%zN)ic
Let 2 - zg%ze#zL
en ZN ZN (2%ZN) i
U . , 0

But [Egg = [T] x [E] and [I] = [T] “'1 x [1318
therefore {532's 1 [T] x [Z] x [T]"1 x [1638

 

 

 

 

 

 

 

 

 

 

 

 

 

And if the 1ndiceted ooerntion is carried out the result is:
I'e P ' '
$01 (2743210 0 67 180
e ..
31 - O z 0 x 1&1
e O O z i
b 32‘ I- J I- 8.2.
. P
Also Feb (27132.) O (51 Fi ‘1!
O N b0
: T] x -l
ebl [ r O z 0 x [T]r x 1.01
e o o ' ' ’
.. bZJ . z‘ blbg.
. C
U]? [Eb] S : (21‘s: ZN) O O
O z 0 x [IA]
0 O 2 U
d
.. P ‘1
And [EC] - (z/-32N) O O
s
O z 0 x I
[as
O O z
n d

 

 

It Will be noticed in Figure 38 that each ohase has

I)

, there is no COquing

(D

the same impedance and, ther for

f'

between the sequences; that is, ea is a function of ia

only, etc., also that the zero sequence imoedence ecusls
the chase impedance olus three times the return imoedence,
When the load is balanced.

Referring to the oossible technique for solution of
circuit problems mentioned above, it Will also be noticed

from Figure 38 and its [3] matrix that a certain symmetry

—83—

is oresent. All the imnedances on the diagonal are equal
and all the imoednnces off the diagonal are equal, and it
vas found that oremultiolying this matrix oyCfl and
oostmultiolying it by [TJ'1 resulted in a diagonalized
matrix. The result is that three simultaneous equations
in three unknowns is reduced to three simple ratios.

This subject will be covered more thoroughly in the next
chaoter.

One more examole will now be given illustrating the
use of matrices to symmetrical comnonent oroblems.
Symmetrical comoonents are used primarily for the
determination of fault currents in oower systems. The
following exemole, therefore, will be a simnle oroblem
on a single line to ground fault. This and other problems

'2'
are given by Reed.1” Referring to Figure 29 the nroblem
a

6
3' JiléF

3: c

AAA ._
vvvvv

 

Figure 29

 

13Ib1d., pp.498-511.

 

—84—

A

 

 

 

 

 

is to find the qult current if. rne equations for
Figure 29 are: ea 3 \“a#21%z~)1a # zvib #zg1C # Vaf
eb - z 1a % (zb¥zl%zg)ib % zgiC # Zflf
: 0 I" . v
eC zgla % Zglb % \scizl#zg)1c%fcf
And therefore
n I P .. P
ea] 2a 0 0T M o o 1.1 1 18} Vaf
eb ’ O Zb O % O 21 O % 2g 1 l l x 1b % zfi
Lee. 0 O 20. b0 0 2L .1 l 1 .10. ,_ch

 

 

 

 

e .
But [Ba] 3 [T] X [E] = [62:] and the sssumotion may be
S e
a

made that the generated voltages are balanced and of

positive se uence; therefore, e

a~o

A180 [I]

DJ'J' x [léls
'21 0 0q
0 O

(D

d

O % 2

Z Z

b
0 sq

l
O

:[Tlx

 

 

 

O
D

Z].

 

 

Z Z are

2
a’ h’ 0

G8

3 e“ Z O.

and therefore

8

[”111
111

 

_111

[TT‘I x 7‘ [T] x

 

 

the ohase imoedences of the generator.

Symmetrical comoonents are aonlied to oroblems

involving rotating mach

reoresent a non-linear oroblem as

great deal of accuracy.

ence the t the

It has been

C5-

inery because of their

ebility to

linear one with a
found through exoeri-

best results are obtained for generators

 

af
Zfif

 

 

_,cf

if the oositive, negative,
imoedences have certain

emoirically and will be reoresented in

2

(D

U!
l

- , "
L: 1’10

values.

sequence impedences resyectively.

Therefore [T] x

Also [T] x

 

And zg[T]

Which gives

X

 

O
O

 

 

 

no -
e ..
a1

0

. 4

But {léls 3 IT] x [I] = [T] x

b

 

 

F<zGO izqu,

 

 

for the zero,
2 O (31
a x [T]"1
0 2b 0
O O qu
o o'
x [T]“1 =
21 O
O 21‘
1 1‘ 2: [ii-1 ~
1 1
l l;
o o "
(Zslyzzl) O
O (2387‘21)‘

 

 

 

These

oositive,

 

zero sequence

values

this

 

 

 

 

 

 

 

 

 

- PZGO O O 1
O 2&1 o
.0 O Zc
le O O‘
O 21 O
_O 0 Z1
2,, {3 o 0‘}
O O O
O O O
b A
q 4
riao rvaf
X 8.1 7‘ [T] X Zfif
- 3.2d Lva 4
here 31a 3 i0 3 0
And i-. = if

 

 

 

 

 

Therefore [Ia] : 1/3 1 1 1 x o = if ' 1
s i -—'
l a a3 3 a8
q 0 a
l a* a
'( 1! ,1" o l ' ‘
F0 2% 21 01%.) o 1
, : ~ I 1
And eal O (2&l¥zl) 0 an x‘gi %
0 O (Z~n%z ) a“
kO b L15 1 I - d
._ . (2T
3 Vaf {azfif % a ‘cf

 

8 - 1
Vaf % a zf1f % ave

The above equation, however, reoresents the fault
in phase b in terms of the comoonents of phase a. If

‘

the equation is made to ce a function of the comoonents

of chase b an imoortrnt symmetry oresents itself.

Since [371338 : [Tlr x [Eagle and [1315 : [T];1 x [It-1‘s

[Tux [21; {1‘1er x mix [1118* [311 x
3

.(Vaf%zfif%vcf) T

2
(vaf¢ azf1f¢a ch)

2.,
L(Vaf¢ a zf1f%dvcf)

 

 

 

And ‘
O P(z~ %z #32 ) O 07 .11 'V #2 i #V 1
— u. l g 1 .l af f f cf
ebl - O 0(z 7‘7 ‘ O x 1 x “'f' 7‘ 3 aZV #2 1 AW
0 31 H1’ 3 af f f Cf
,.. 1 z . a2
LO 0 (Zugflll L l Lavaf’l flf" ch‘

 

 

 

 

-57-

Multiolying the above equation by the row matrix, [1,1,1] ,

O" ' = ( .n '2 \. ( fl 1 ‘7
D1ves. ebl {:zuo%zlfuzg,%,qu#zl)%(zG #zll}.;§ % “fif

 

 

 

Therefore ifi— Ibo 3 lb 3 Ibo 3 ebl
3 1 (230%21%32g)%(zGl%zl)%(zG2%zl)%32f
Afld be : Vraf7lzfif1£VCf : ‘(Zfl 7‘2 7‘33 \ x j...
0 ———-——--——-'— Ho 1 e’ 00
3
V = agv #z i lav
bfl af' f f" Of 2 e. -(7 *0 )i
3 01 ”31 “1 b1
v ‘ v ,l ‘ ,l Qv ‘
- a, 2-1 a“ ,
bf2 7‘ af I f of : -(ZGZ % 41) 108

3
The above equations indicate that the fault may be reore-
sented by a series circuitze shown in Figure 30 where the
total fault current if is equal to ib = ibofibl{ib2.

A‘AMAAA

 

 

 

 

 

 

   

Figure 30

This chanter has shown how matrices may he aonlied
with advantage to nroblems involving symmetrical comoonents.

The usual crocess of solving equations to ,et the relations

~88—

between the voltages, currents, and their symmetrical

comoonents has been reduceé to the simole orocedure of
- . . :‘1 "'1 m

mult1oly1n5 by [T], [I] :[TJr: and [fir . 1he result

is that the thread of the oroblem is never lost.

Chill"... 3 V
rIAu4~ALI5AIIQI

For a git ven circuit problem, the matrix equation
['13] = [2]}: {I} is obtained. "-"fhen voltage equations are
written, the olt.~e end inoed anc es are knoWn are the
currents are desired. Therefore, to find the cuirents
the above equation must be written as fig :5 tfl'l
Similarly, if node equations are written, h] == hi x: fig,
the currents end admittances are known and the voltages
are desired. This means that the voltages must be found

[75 2 {3'34 x LIZ; . In either case, an inverse matrix
must be comouted and for a large number of meshes or nodes
this can be a considere.ble task.

The determination of the inverse is nearly as
difficult as, or eoually as difficult as the method of
determinants ( h'oter I). If, however, the matrices [P]
and {0] could be found such that [P] x [Z] x {33 gave a

matrix 'ith elements on he princiole die onal only, then

‘tl‘e new eouation {Ad-j: {ij m, Waere [Z] is tlfle

diagonalized matrix, would result in n simole equations

N
el N
of the tyne,-—— 3 il , where n is the order of the

I
Zl

{D

[Z matrix. One the new currents are found, the desired

currents (old) could be obtained by a simole multiplication
N
between the {4] matrix and the [I] matrix.

This method Will now be illustrated w’th the folloWing

(D

xamole. Find the currents of Figure El by diegonalization

and by the method of computing the inverse.
, '2 '4
[0 Z ‘4 £1)
-‘ 3 ,
2; '
74,7
Z.

 

 

Figure 31
Mesh equations:
10 = (2;4;3)11-413—313—014
o = -411 #(4¢4¢4)13—013-4i4
o I -311 —012 % (2{2%s)13-214
o = 011-413 —213 ; (2%2%4)i4

Matrix equation:

”10‘ ”(9) (-4) (—3) (o) ' ’ifl
O _ (—4) (12) (O) (~4) 1'
o (-3) (0) <7) (-2) x :3

)

 

 

 

 

n3

 

 

L q, {(0) (—4> (-

That is [Q = [Z] x [1]

By the method of diagonalization, the following procedure

is used:

 

 

-91-

Premultiply the matrix equation by {13] giving [13 x [13 :

{a x [Z] x {13 x [I]. The [1] matrix may be inserted

between [23 and [IJbecause it does not affect the equality

of the equation. But [QJX {Ca-1: {1], therefore, insert

this in the olace of [1] giving, Lax xi: =le x [Z] x [a x

[QJ‘lx xLI The eouetion Wilt?" - [fix [I] may nov be vritten,

vhere: é] - fax J-fl’ {fa- {lax {Z3 x [‘33 (the diagona-

lized 1121 trix) :nd [I] -"- {Cg‘lx {1] are unknovm currents.
The problem, nOW, is to find [P] and {4] such that

[P] x [Z] x [Q] 1’ 62;] will result in a, diagonal matrix.

This is done as follows:

9 -4 -3

-4 13 O -4

[h] = -3 o 7 —2

O —4 —3 8

Add 4/? x lst row to 2nd row riv1ns

’9 _4 -3 0'

' 98 -4 —4

[Z] = 0 e“ ":2:
—3 0 £7 —2
b0 —4 —2 8

 

 

Add 1/3 x lst row to 3rd row giving:

'9 -4 -5 0“
1: 93 —4 —4
[Z] = 0 s“ ‘21
o :3 e -2
3
-41 —“
LO ‘- 8‘

 

 

Add 3/83 x 2nd row

9

121'" =

 

O
O
O

b

Add 9/23 x 2nd row
P
9
ma
[Z] = O
o
L 0

 

Add 29/57 x 3rd row

9

1: Jam: 0
Z =

O

 

Lo

Referring to Linear Transformations, Shaoter I,

to 3rd row givin':

-4 -3 d
__ :4.- —4
9 3
134 ~58
O 23 53”
— —2
4 8J
to 4th row giving:
-4 -3 o ‘
33 .:$ -4
9 3
O 134 ~58
41-3 {:3
O -58 148
‘3? “*3
his? C2.
to 4th row giving:
-4 -3 d“
93 —4 —4
'§ "5
O 134 figs
23 23
8834
O O 1541‘

 

 

 

170
l

 

 

 

1')
O

the

first ooerstion could have been performed by a oremulti—

olier.

\.

lst row is added

or =

Therefore

 

This premultiolier is,

to 2nd row.

00 |--‘ O
OH O

 

 

'1 o o
4
'g 1 o
o o 1
b0 0 o

6' 1

O x [Z]

0

k

.eHD <3 95

 

, where 4/9 of

Ill/I

But [Z]

[P]

£9?4
0)
m

 

()1 03‘”
1'1) u <1 (01% H

x)

 

(A

 

rw
£3

0

L11 1.1

I} 3
(0 fr]

(7
‘.
O .. -_

H
)1
E.

H

D

L)

OHO
O l-‘OO

0

[Z]

H

to
w

 

’33
\1

Since all we have done

 

 

 

1000 "10001
0100 0100
O O 1 O x O_§ l O
09___01 233
L83 L0001,
I I q
0 1000
O x '3 l O O x
0 s

0010
1.1 _0001‘

 

 

 

end therefore,
q
0
O
O

1

 

d

is add multiples of one row to

, ‘ . mu ‘ ~ _ .
another, the determinant of [Z] muSt equal tne oetermin—

ant1yf fig and [PI muSt equal 1, and this is the case.

Next ooeration:

“HI

 

 

 

 

 

H q
—4 -o 0
$3 —4 ~4
_§ 3.

134 ~58
0 2:5 :23
8234

0 0 ~ :1
.LO‘tl‘

1&5 1/3

um

[Z]

Add 3/33 x 3nd column

I“

ll

11:1:

[Z] =
x who 0
Illa!

 

x 1st column

F9
0

 

-O

p

8

(DO 0

 

 

’23
d
(1.

to

"0
(010 f.)

C)

to 3rd
0
93
S.
O
O
to 4 n

o
(I) C~".)| OJ O

(2)

-i 1‘.’i.’1.§j1
-3 d

ȣ> C J|#>

"
l: -58
a _
9'2? (3'2
k._a‘ .21“,
ha"
(:1: C‘4‘

 

 

H
U! .
4x
H

column

O
—4

to]

1:4 -58
23 23
8884

 

 

 

column ;ivin9.
?
o Cfl
O .4
134 -58
E¢34
O ;
1.14:]...ll

 

c1lumn giving:

q

’3

m 0.0

P4
mg o

.‘»b ,
03

 

71.)

([4
{D
-0
’1] DJ
4>01

O

 

 

H
U}
.45
H
l

-95..

 

 

Add 89/87 x 3rd column to 4th column giving:

'9 0 0 (51
N 0 92 0 0

[z] = -: ‘
o 0 213.5% 0

33
9974
O .. -1.

-O O 1541,,

 

’-
""1

weferring again to Linear Transformations,

 

 

 

 

 

 

 

 

 

 

 

these ooerntions could have been oerformed by a
multiolier. The first jostmultiulier is, .l'é
O 1
O O
.0 O
4/? of the 1st column is added to 2nd column.
F1 .3. o 0‘ '1
N
Therefore, [Z] 3 [P] x [Z] x O 1 O O x 8
O O 1 O O
,o 0 o 1.1 L
'1 o o 0' ’1 0 0 01 ’1
3 s
x0 1 go x o 1 0 2-3- x 0
O O l O Q o 1 o O
‘uO O 0 1‘ L0 O O 1 J b0
1:
But [Z] = [P] x [z] x [91
-1 4 e 52207
é' 5'5 1541
0 1 .3; 690
Therefore [0] Z 33 12:1
0 O l g;
0 O O 1
L d

 

 

OHO O

Chaoter l,

OOHO

OOHO

 

 

ost~
0'} where
O
O
1.
.3. 01
O O X
l O
O 1
0 0"
O O
29
l...
67
0 IL

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-VC"
N
53 2: fig 1: fig :: P if 0 0 0 151
g 1 0 0
x 0
27 3 =
'— 23 1 0 o
1587 39: gg 0
1323 1541 27 1 .
L ...-
7 10 : 1‘9 0 0, 0‘1
4o ' .28
N N N _ -§ 0 9 O 0
D3] - [Z] X [I] " 2’39 : o o 1.51%. o
0:: ' “V
15870- 0 o 0 8334
1. 432.3; 1041,
Which gives, on multiolying:
N 10 N 40 N .20 N 1’ -70
11 = s , 13 =‘§2 , 13 = 154, 14 = §Zf70£
But N _ —l . N
[I] [Q x[I] and therefore [I] = [Q] x [I]
Therefore
r. - ’ i. E. 529 ' 10
1l l s 23 1:571”T ‘§ W
i 3 390 40
2 - -—— .——— .__
— O 1 23 1541 X 93
is 0 0 1 Q9 .29
1 07 134
L 4 L0 0 0 1 1 .;§1§ZQ
" - L £14,702.
1 3'19 Z 1s0 Z 510 Ze,ses,230
l s 823 3032 38,035,782
2 1.1111 Z 0.193: Z 0.2328 Z —.2205
11 = 1.7873 amos 12 = 0.8101 omos
is 3 0.850 emos i4 3 0.8485 amos

 

-97-

If tne usual method of oomoutinr the inverse is

5:)

used to determine the currents, the result will be as

 

 

 

 

 

 

follows:
- P n - D t
111 211 ‘13 213 214 10
_ —1 y -
[I] - [Z x LII: 12 z “'5 221 222 02-3 224 x 0
Lnat 18, 13 231 232 213 234 O
.144 _ L241242 243 24% 1.0..
where Zij IS the coiactor of 213
Therefore 11 = Zll x 10, 12 = Z21 x 10
(Z! IZI
1 = z v -'
IZI
IZ’ : 9 X all-4: X (421 '3 X 231 7‘ O X Z41
12 0 -4 -4 —3 0
And Z Z O 7 -3 Z 512, Z ‘ -1 x O 7 -2 3 232
11 -4 -2 a 21 -4 -2 8
-4 —3 O -4 —Z 0
Z Z 12 O —4 3 272, Z = —1 x 13 O -4 Z 184
31 —4 -2 e 41 o 7 —2

.Therefore,IZ] 2 5x512 -4x255 ~5x37£ Z 0x154 = 2554

 

 

 

I ' Z 51,r _ 1 — 27? _ .r, 1
And 11 :g:{ - 1.7577 ares 15 - 7—59 - 0.550 amps
QLCQ QJU‘t
°7°° - 1540 —
: ~~~J : o . a ' _ 11, - 0.6425 amos

If an easy method for determining the currents is

what we are after, then diagonalizetion by the [El and

H)
(0

[0] matrices should not be used. This nest examnle illus—
trates the difficulty and the time consumed in determining

do 3in if

(D

[P] and [Q] before the orohlem can even he Vorl:
the imnedances sre comolex, it is next to imoossible to
determine the oroser [é] end [Q . It is also sonarent
from the nest illustration that a different [P] and [Q]
is recuired for every [2] matrix.

The problem of dissonalizetion is not honeless,
however. It was found in the last chaoter that for a
third order [Z] matrix, with all eleme mt on the diagonal
equal to one value and all other elements equal to a
different va.lue, the h trix [T] coulc1be used to diagonalize
it; th at is, [‘2]: ['2] x KI x [fl-'1. This tyoe of symmetry
is referred to, by Pioes,14 as E symmetry. He then states
that e. [Z] matrix of n order With ’3. symmetry may be

N
diegonalized by the use ofzm1 fig matrix; that is, 12] Z

a-(r-l)(9-l)

[S] '1 x [Z] x [S] where [S] '-‘- [Sm] Srs 2
with r ~21? 2,3 . . n, s 3 1, 2, 5 . . . n, and
a 3 6?? The use of the [S] matrix is an extension

of the method of symmetrical comeonents.
It might be Worth noticing, a this ooint, that to

btain [Z], in the last Chester, [Z] mus oremultiolied

 

l4 , .
L. A. Flees, "Ire nsient An algsis 3f 83m.etr1 al
Networks by the nethoc. of Qy’retriosl Someonents," AIJE

"U

Transactions, (1040), W. 3..

by [T] and nostmdltiplied by [T] '1 While Vines (above)
nremultielies by [s]‘1 end nostmultielies by [s] . If

the above definition of [S] is used and [S] is obtained,

 

11111111111 1 1. ... . .1 1
1 a"1 a“? . . . . . a‘(n‘l)
[e]: 1 . .. °
1 . . . .
l . . . . . . .
-1 .-(n‘l) . . . a-(n~l)(n—1L

 

Therefore [8] = '1 1 1'1 and a: 6

 

 

(n33 l 3-1 a_2 3
9 _4 therefore a — l
l a-“ a
.. J
hultinlying each element of [S] by 5” gives:

 

 

And therefore [81-1 '5- [T]

"I

rundamentally, then, Pines has defined the same
onerator as is used in symmetrical comoonents in the
nrevious cheater. Vowever, in so doing he has extended

it to cover oroblems other than three nhsse and has

)1

intro<uced the negative exponent. For a given oroblem

.‘

—100-

the negative exponents of a can a1Ways be made oositive
by multinlying esch element of [S] by 6 = l .

In the discussion of Pipes'article in the A. I. L. s.
Transactions,15 the point is brought out that this same B3
matrix will also diagonalize a [2] matrix that has "ring"
symmetry. "Ring" symmetry is not defined, but the two
examoles given 16 are shown in Figure 38.

The red lines on Figure 32 indicate a symmetry that

will allow diasonalization with the use of the [S] matrix.

 

 

 

 

 

 

 

(a) > (b)

Figure 32
Matrix (a) of Figure 32 gives the following diagonalized

form:

 

‘15Ibid., 5.1109.

151b1d., 55.1109—1110.

 

F20 O 0 0 0
0 2l 0 0 O
0 0 28 O 0
0 O 0 2:5 0
O 0 O O 24
_O O 0 O 0

matrix (b) of Figure 33

 

form:
20 o o 01
O 31 0 0
L0 O 0 23‘

 

 

O
21 :25
22 :24
35

th =

ere Z0
21 ~23 3
28 3

he matrices of Figure 33(a)

indicated,

 

 

N N N
0’ 0 g:
U‘ba N .
\IO ‘1}

in red,

in Figure

‘-onslize end. their results

 

3?.

are

 

n 1
20 . 0 0
0 21 0 O
0 0 23 0
0 0 0 2%

(C)

20 0 0

O 21 O

O 0 26

(d)

(7Z21 P
— (z ‘212‘213 Z214)

These

(2%3212%3213%Zl4)

’213 “314)

gives the following diagonalized

(Ztgziztzls)

(Z ‘ 313)

and (b) have the symmetry

matrices did

given in Figure 53(0) and

There
z=(z M%2b%z #zd)
-(za—jzb—zc #jz
-zbZzC-z d)
232(zaijb-z

20
21 d)
zz(za

Zn
,1.
~

C-jzd)

.he re
:\Z.- d7lzh%zc )
l-(Z :#agz 0%dzc )

z2 l—(za draz #a2 2:)

:0

—109—

The matrices of Figure 33 are of little use in most
circuit problems, because they lack the symmetry about
the diagonal characterized by most [2] matrices; that is
213 ¥ Zji for the matrices of Figure 35(a) and (b).
However, they have served a ouroose in in5163tihr a tvce of
synuetry necessary for diagonelizetion by [S] . Also of

importance is the aginrent fact tfist if zij zji in the
flflnmtrix, then the créer of [Z] must be even if die:ona—
lizetion is to be nossible, unless, of course, E syuhetry
is cresent.

The following examnle will now be ;iven to illustrate
the advanteie of dinbcnalizing when symmetry oermits it.

Referring to FiMIre 34, the problem is to find the

-t

mesh curre

(D

e

as functions of time when a stec voltage is

 

Figure 34

-103—

sonlied at t 3 O; thnt is, v is a d. c. source and time
is measured fron the instant the SWitch is closed. It is
assumed that energy storage in each mesh is zero before

the switch is closed.

Solution:
The differential ecustions mcy be written as follevs:
- Y 1 '.' " ‘7 3 «.l'. °
V -- (up7(P_,( EpLLl 7‘ M11312 7‘ mgplz 7‘ Llpld:
o mlnil #(Lp#R¢ %b)12 % mlpi3 % M2914
- ,. .. 1
O - M2911 % M1912 # (Lp%R% fip)13%mlpi4

0 = Mlpil # mgpig % M1013 71 (LP1‘R7L 314

u. _ —J = .L
here p - ut and p d“
If the lanlace transform is taken on each side of
the above equations, they may be written as:

e = zaIl % szg # ch, # sz4

C)-

O szl % zaI2 % sz3 #2014
o = zCIl % szB ; 2&1:5 $sz4
0 - sz1 # ZCIB % sz3 {2&14

(Ls ,1 Rtts)
z = M18, zC = M88 I = 4((1)

. V
there e = ‘g 2

Because all initial conditions are zero.

P- Iq P P p
, Therefore 0 3 2b 2a Zb 20 x 12
0 2° zb za Zb I3
-0. -zb zc zb za‘ .14.“

 

 

 

 

 

 

~104-

And LE] = [Z] x [I]
But it is noticed that the above [Zlmstrix has the

same tyne of symmetry as Figure 32(b}, an; therefore,

the [S] matrix may be used.

 

 

 

 

I"1 1 1 1‘
1 q j?” 31,1
[S] '-'-' 1 a‘ if" a‘3 Where a Z 6 4 I: e 2 j
1 a—s a-4 a-s . 4
a a = l
l a 3 a o a-?.
-
'1 1 1 11
--1
[SJ : l -3 -1 J and S] I 3-;- [Conjué‘ete S]
l -l l —1 ‘ (ultays)
L1 3 —1 -3.
1'1 1 1 f
-1 .i
Therefore [S] 2 4 l j -l -j
l -1 1 -l
n1 -3 -1 jd

 

 

[a] : [z] x [I] stl x [E = [$14 XE’JX [53 x [31'1 EU]
\~—«-\/"“"/ \——yr~——’

LT: N x H

PT E3 E3 II]

From F‘iggure .3?(b) [fl is found to be:

P s
N 20 O O 0

[Z]: 0 z c o O _
21 :27 — (za-zc)

;
(D
H
(D
N
I

' 7 2 z

D
H

(za%zC—Ezb)

 

 

And.

tn
Ll
c+

[I] =

And

But

And

 

N
[If] ‘-"- _l_ xfll l 1 11 x '61
4
l J -l -j 0
1 -l l --l 0
L]. -j -l j. L04
therefore
r1. ’2 o o 01
‘2 x 1 = 0 21 o o
l O 0 as 0
‘3
L1. .0 c o 21,
N _ e N _ e N _ 8
ll -4? ’ 12 ~ 421 ’ 13- 4:22
N 1
[I] :3 [5]“ x [I] and therefore [I]
1
F1 1 1 :f F'%
O
1 -j -l J E. .2”. _ 2
x ‘ 4
1 -1 1 —1 21
l,
L1 J -1 -j 7'3
.1.
L21.
' z: E l l 7‘ 2 _.
I1 4 ( 20*22 '31) 12 — 4
e 1 ,l _1_ _. a
I : - ( - 3:.
3 4 20 23 (.11
‘ o l. ,
ZO ‘ za%22b#zc P #Lo% CS %
.. n r 1
20 ‘ R % £(L % H l g ‘2) *‘6g

-105—

 

 

 

 

 

 

 

 

 

 

 

rP‘I CD

 

 

 

 

 

‘k
t:
to
('1

 

 

 

21: za-zC: E114 LS%%§-1QS
And 21 — R ; s<L - H2) ; %§
,7 d h _, l t‘ IV
“2 — 2a % zC —e_b — s ; Ls ¢15§ %»QS -.Llo
And 22 : s ; e(L — srl ; mg) %‘%s
-1 V
Therefore 11 '-'-' cf 25 ( 9 1 ,1
S~(L#2£l;u3) #“S!.%
U

 

If .{‘=

 

(1] H

 

 

 

 

s~ (L—2m17t1a2),£3s,£ :2:

U‘
A
I."
‘k
D
Hi
15
t5 "
V
X
07
X
(‘le

I
,1 11 IO

U)

as )3

(D :
CC]

r“

t“

4‘33 ) %

 

 

 

 

P.
3(L;3n-17Z—1¢2y ,fl: VG(L713 W171 3) (3(L713517Lu“? 27 )2

 

 

 

 

2 L-S; ’1/32 ' ‘/ n 3 2
( 1¢‘Q7‘ C(L-3:1¥m2) (0(L—2u1#M37)

 

2

 

 

s , ‘/é§ II b/ 1 - < R
2(L—r37’ C(L-Hé) aft—r57 )

~107—

-qgt

 

Then i]. = :11: sin
4 LE 2“1]. (L‘Eufl {#—
2 sin .zt
L-Lz
-fit
18 = i4 : V

 

 

5 sin 1: - 6:.“ sin "
I (Metal 1.137% (beefing/3‘
.4:- 5‘“ ‘
and. 1:3 - l 5'. sin gt 7‘ sin
4 (LEAFMBYP’ (11-2.; firefly/52
-aqt

- (1131254331: J

 

 

 

* 3
23‘

[.4
(0
(D

r r .J ' ' v’ ,
Xamole nss ill ustr eu, decis1vel,, toe

C};

o be he in die onslirin: the [2] matrix Then

6*
CD
c+

acvsn
H ' ' . ' ‘w . . ,s /\
a or ring symmetry is nresent. It is only noose t“
o v‘ . 1‘ 1‘ ‘ I r A ‘ . ‘ “ " . 1 " ‘. . ~‘- (\
rurther resetrcn elonQ these lines 111i result In tTaHS‘

f0

H
~§
to
c+
[.J
O
.0
1+
.J'
cr
'1'
J
.__!
F”
O
H
9;:
H
r?
r+
:64-
(TI
(0
C)
H)
H“

("1

his thesis has cresented some of the funcsm.ntsls

.__.

of matrix algebra and showed how they may be aoclied to

1 1

various tyces of circuit problems. It has not, oy any

means, covered .11 the oo.sible suolicntions of matrices.

C‘.‘

ystrices are a field of mathematics in themselves
‘ gsicersble Work has been donetrith then. It has
only been in the last few years, however, that an ottemnt
hes been mede to accly them to circuit nroblems. Because
of their compactness and utility to maintain the continuity
of tne orcblem their use in involved circuit problems is
unlimited. It is hozed thfit in the ruture metrices till

Q . s“ '1 "\ ‘ - A r 'L‘ " ‘ "I '. . "' 'fit ' V‘ ' “5‘. "J . .'*
experience ch]. CV31 :THGU“I‘ ‘2980-«0 1.; tflv 61,111.2r1‘15

problems of all the fields.

BI BII OGBA PI-IY
I. Books

Gardner, I. F., and J. L. Barnes, Transients in Linear

 

Systems. hew York: John Wiley and Sons, 1942.
359 on.
Guillemin, E. 3., Communication Retro

 

I
New York: John Uiley and Sons, 1

Iuillemin, E. A., The L .themetics C I AM
New York: John kile, and bone, 184d. 58% pp.

 

 

 

 

Kron, G., Tensor Analysis for Electrical En_1ne3rs,
New Yorr: —John hiley and bone, 1942. 250 09.
Le Corbeiller, 3., Iatrig Anslysiss of Tlectric Circuits.
New York: John Vile and Sons Harvard University
3 o

Press, 1950. 108 on.

E.eed’ 1:. Bo , Al terns tirl -3urrent 3
York: Earner and Brothers, 1%

Seely, 8., “lectron—Iube

Circuits New York: moirsm—
Hill Book ooIJIIy, 1950. 5:

II. articles

 

 

Brown, J. 8., and F. D. Bennett, "Aonlicetion of Ietrices
to Vacuum Tube Circuits," Institute of 92610 Ineineers
Processinqs, 36:844-53, July 343, Discussion CT:

 

 

403—4, Aoril lBéB.
Vowels, R. 3.," "Iatrix Iethods in the Rolution of Lsfider
Ve ‘OT{S," J urnel of the Institution of “leotrical

n-ineers, 95 pt. 5:40—50, January 1”‘"

 

HIT

 

Pritchard, :. L., "*uarter ”ave Couoled 7ave Guide
Filter ," Jour nel of Aoolied Physics, c.813 —?3,
Octooer 1947.

BIBLIo39A93Y
(continued)

 

 

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rieble Circui
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1 ROOM .1151 mm

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l I. C
F‘s 5:. I 3“ (:1, V: .
I ft 1958 A". r! A,

r4551.

fut. e -. . - , ; r
i’ . ’1 {a . 3 '
tr . ' '1 ~ 7' ‘ ' A 4

 

 

 

 

   

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