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Chapter XI Conventional Networks
Chapter X I
CONVENTIONAL NETWORKS
1. Summary
In the development of the theory so far, we have assumed that the entire orthogonal equations and associated networks are always required to obtain a solution. In general, this is not the case, as will be shown in this chapter. It is sufficient, in many cases, to utilize the closed-path equations and to ignore those associated with the open paths, or to utilize those of the open paths, and to ignore the equations associated with the closed paths. In the latter case, it is even sufficient, as will be shown in this chapter, just to utilize the extremities of the open paths, which were earlier called junction pairs. The solution methods are often respectively called the mesh method and the junction-pair method. Both methods will be shown to have their basis in the orthogonal equations. Specifically, it will be shown that the mesh equations and the junction-pair equations readily follow from the orthogonal equations by ignoring the open-path equations in the case of the mesh method, and the closed-path equations in the case of the junction-pair method. The work in Chapter VIII in which one orthogonal contour set is transformed to another set will also be specialized to the case where either the open-path or the closed-path equations are ignored. In Section 4, the submatrices of Z,, are expressed in terms of those of Yss and vice versa.
2. Ignoring or Eliminating Open Paths We will limit our excitation in the interconnected network to B voltage sources (eb) and B current sources ( I b ) . ( N - 1) additional current sources could be considered, as outlined in Appendix 11. We will consider the B current 210
211
2. Ignoring or Eliminating Open Paths
sources ( I b ) as equivalent voltage sources ( Z b b 1') in the primitive network, as discussed in Chapter IV, Eq. (4.33). We also limit networks to those whose branches are traced out by at least one closed-path current contour. It is apparent from our previous work that the open-path current vector J" is zero in this case: J" = 0"
(11.1)
It is also clear, that term I' is zero when current sources are treated by means of equivalent voltage sources, as shown in Chapter IV, Eq. (4.34): 1' = 0'
(3.14)
Current J' thus consists of i' in this case. E, from Eq. (3.40) is zero: E , = 0,
(3.40)
The orthogonal equations (3.66) and (3.73) specialize as shown below: o
c
(11.2) and o
r
(11.3) earand e,. represent equivalent voltage sources in the orthogonal network, and are transformed from the primitive from the branch voltage sources eb plus
the negative of equivalent voltage sources Z b b I b as given in Eqs. (4.36) and (4.37):
(4.36) (4.37) If the open-path voltages E,, are of no particular interest in Eq. (11.2), then the open-path equations can be totally ignored and Eq. (11.2) simplifies to e,, = Zcci'
(11.4)
The solution for' i readily follows:
'i
= (ZcC)-'ec,
(11.5)
212
XI.
Conventional Networks
Currents ib follow from Eq. (3.5): jb =
cb,' i
(3.5)
The currents in the primitive network can be obtained as the sum of l band i b :
+ ib
(2.6)
Eb = Z b b J b - eb
(11.6)
Jb = Ib
The voltages Eb follow from Eq. (2.7):
It is readily apparent from the procedure above that the solution utilizing Z,, requires only Cbc and that Cb, need not be defined. Notice that the open-path axes cannot be ignored if the solution for ic is desired by means of Yss in Eq. ( I 1.3), but instead have to be eliminated. From Eq. (1 1.3), Y""(E,+ e,.) + Yoce,.= 0
(11.7)
Upon solving for (E, + e,,) in Eq. (11.7), E,
+ e,. = - ( Y O O ) - l Y O c e c .
(1 1.8)
Y'"(E, + e,,) + Yccec.
(11.9)
From Eq. (1 1.3),
'i
=
Upon substituting Eq. (1 1.8) into Eq. ( I 1.9), jc
= [ycc
- yCo(yoo)-lyo']e,,
(11.10)
Upon comparing Eq. (1 1.5) with Eq. (1 ].lo), it becomes clear that (Z,,)-' is equal to the expression within the square brackets in Eq. (1 1.10); this is more formally established in Section 4. With the solution procedure and its basis from the theory of orthogonal networks established, a simpler derivation starting from the primitive network can now be developed. The equations of the primitive network are Eb
+ eb = Z b b ( l b + ib)
(2.7)
l b and eb are considered known. Regrouping Eq. (2.7), Eb
+ (eb -
zbb
I b ) = z b b ib
(4.33)
Z,, I b represent the equivalent voltage sources due to current sources. Premultiplying Eq. (4.33) by C: gives CPEb
+ Ct(f?, -
zb,
I b )= CPZbb ib
(11.11)
213
2. Ignoring or Eliminating Open Paths
Upon substituting Eqs. (3.38), (3.5), and (4.37) into Eq. ( l l . l l ) , E, Recognizing that E, yields Eq. ( I 1.4) :
= 0,
+ ec, = cpz&Cb, ic
( 11.12)
according to Eq. (3.40), and substituting Eq. (3.65) e,,
= Z,
(11.4)
ic
The network representation of Eq. (11.4) is shown in the lower half of Figs. 3.5 and 3.6. The solution process can be described by a series of steps:
1.
ib = Cb, ' i
2.
e,. = Cp(e,
3.
z,
=
- Z,,lb)
cpz,, cb,
4.
i" = (Zcc)-'e,.
5.
ih = cb, ic
6. E ,
= Z,,(lb
+ ib) - eb
The above steps readily simplify for the case where no current sources I b are present, as further elaborated upon in the reference.' Notice that the first step is used only to define Cb,; also the steps above need not be executed in exactly the manner indicated, and no attempt has been made to create an efficient algorithm. The above work was presented in terms of Cb,. The analysis equally applies to various subcases and special cases of Cb, , as, for example, the constrained contour cases designated C,: shown in Fig. 1 1 . 1 . In the latter cases,
o
FIG.
c
o
c
1 1 . I . Summary figures of Cbc, CC, A*", and A;".
Happ, H. H., Special Cases of Orthogonal Networks-Mesh IEEE Trans. Power App. Systems 87, 53-66 (1968).
and Nodal Networks,
214
[m]
XI.
Conventional Networks
the above steps can be simplified on account of the special form of CY, . It was previously indicated that the solution procedure with Z,, only utilizes Cbc, but that Cbo is present but not defined. Figure 1 1.1 is a summary figure in which Cbo and CY, are indicated by dashed lines, which signify that the open paths are present although ignored. Numerical examples of Cb, and of CY, appear in Figs. 11.2 -1 1.5.
bl
b2
*3
b5
lC3
bI bZ b3 b 4 b5 b6
r---------I
I
IT
I
ICOb
b6
c3
x4
FIG.11.2. Closed paths with open paths ignored-general
FIG.11.3. Closed paths with open paths ignored-tree CI
c2 c 3
FIG. 11.4. Closed paths with open paths ignored-tree
FIG.11.5. Closed paths with open paths ignored-tree
case.
and link case. b5 b3 b4 bI b2 b 6
:-----r----
and link case.
and unit-link case.
215
2. Ignoring or Eliminating Open Paths
OTHERCLOSED-PATH CONTOURS One set of closed-path currents can be transformed to any other set from the theory established in Chapters VIII. Equations (8.39), and (8.40) respectively specialize in this case to ic = cc,. ic'
and
iC'
where from Table 8.10
= A!:
(9.28) (11.13)
iC
( 1 1.14)
CC, = (A:;)-'
The corresponding voltage expressions i n Eqs. (8.42) and (8.43) likewise specialize to e,, = C;: e, (11.15) and e, = ALC'ec, (9.75) Where from Table 8.10 cis = ( A ,C') - 1 (11.16) .Zs,sfin Eq. (8.27) and Z , , i n Eq. (8.31) contain .ZC,,,and Z c c :
(11.17)
0
C
(11.18)
7
5,
C"
216
XI.
Conventional Networks
Expressions of the three submatrices not presented in detail are more comand Z , , , as can readily be established from Eqs. (8.27) plex than those of ZcSc, and (8.31). 3. Ignoring or Eliminating Closed Paths
We will likewise limit our excitation in the interconnected network to B voltage sources (eb)and B current sources ( I b ) . We will consider the B voltage sources eb as equivalent current sources in the primitive network ( Ybbeb),as discussed in Chapter IV. We also limit networks to those that can be spanned completely by junction pairs. I t is apparent when employing Yhbehthat e, in the orthogonal network is zero. E, is likewise zero from Eq. (3.40), and thus
v,= 0,
(1 1.19)
e, likewise is zero on account of treating eb by means of current equivalents: e,
(4.4)
= 0,
In the absence of external voltage sources
io
(3.22)
= 0"
The orthogonal equations (3.73) and (3.66) specialize as shown below. o
c
(1 1.20)
and o
c
(11.21)
and 1'' represent equivalent current sources in the orthogonal network. They have their origin in the primitive network and are transformed from branch current sources I b plus the negative of equivalent current sources Yhbehas given i n Eqs. (4.41) and (4.42): /"' = A " ~ ( / '- Ybheb)
(4.41)
1" = A f b ( / ' - Yhbeb)
(4.42)
3. Ignoring or Eliminating Closed Paths
217
If the closed-path currents i" are of no particular interest in Eq. (1 1.20), then the closed-path equations can be totally ignored and Eq. (1 1.20) simplifies to 1''
=
yoOE,
(1 1.22)
The solution for E, is as follows: E, = (yOO)-l[o'
(1 I .23)
Voltages i n the primitive network follow from Eq. (3.44):
Eb = A:E0
(3.44)
Currents in the primitive network follow from Eq. (1.4): J b = Ybb(Eb+ eb)
(1 1.24)
It is apparent from the procedure above that the solution utilizing Yo" requires only A t , and that A t need not be even specified. The closed-path equations cannot be ignored if the solution for E, is to be obtained by means of Z,, in Eq. (1 1.21), but have to be eliminated. From Eq. (1 1.21}, Zc,Io' Upon solving for (I"
+ i')
+ Z,,(I"' + ic) = O
(1 1.25)
in Eq. (1 1.25),
( I C ' + i')
=
-(zcc)-lzc,I~'
(1 1.26)
+ Zoc(Ic'+ i')
(1 1.27)
From Eq. (1 1.21),
E, = Z,, I"
Upon substituting Eq. (1 1.26) into Eq. (1 1.27), E,
=
[Zoo - ~
O c ~ ~ ' c ~ - l ~ c O 1 ~ O (11.28) '
Upon comparing Eq. (11.28) with Eq. (ll.23), it is clear that (Yo')-' is equal to the expression within the square brackets in Eq. (1 1.28); this equality is more formally established in Section 4. The solution procedure and its basis have been established fromthetheory of orthogonal networks. A simple derivation can be developed starting from the primitive network. The equations of the primitive network are Ib
+ i b = ybb(Eb+ eb)
(2.8)
I b and eb are considered to be known. Upon regrouping Eq. (2.8), ( I b - Ybbeb)+ i b = YbbEb
(4.38)
218
XI.
Conventional Networks
ybbebrepresents the equivalent current sources due to voltage sources. Premultiplying Eq. (4.38) by Apb gives
APb(Zb- Ybbeb)+ Apb ib = APb YbbEb
(11.29)
Substituting Eqs. (4.41), (3.44), and (4.25) into Eq. (1 1.29), 1"'
Substituting ' i Eq. (11.22):
= 0"
+ i"
= APb YbbAtE,
(1 1.30)
according to Eq. (3.22) and substituting Eq. (3.69) yields (11.22)
1"'= y""Eo
The network representation of (Yo")-' is identical to that of Zoo except that the numerical values of the elements are different. Clearly (Yo")-' differs numerically from Z o o , shown in Eq. (1 1.28) because of the modification of Zoo in the closed-path elimination process. The network configuration remains the same as that of Zoo and is shown in the upper half of Figs. 3.5 and 3.6 for two examples case. The solution process can be described by a series of steps:
1.
Eb = A t E ,
4. E ,
= (Yoo)-lZo'
2.
I"= APb(zb - ybbeb)
5.
Eb
= AtEo
6.
J b = ybb(Eb
3.
YO" = APb YbbAio
+
eb)
The above steps simplify for the case where no voltage sources e b are present, as elaborated upon in the reference given in footnote 1. The first step, similar to that in the previous case, is used only to define A t . The other steps are computational, but can be performed in a manner different from that indicated above to improve the computational efficiency. Since only A t is required, the complete open paths need not be defined. The extremities, or junction pairs as they were called, are sufficient to define
At.
Although the above work was presented in terms of A t , the analysis applies equally to various subcases and special cases of A t . The solution procedure with Yo",utilizes only A t , but A t is nevertheless present-although its presence is ignored, as previously indicated. Fig. 11.1 illustrates A t , A;", and A;" of the unit-tree case. A t and A;' are indicated by dashed lines, which signify that these submatrices are present although ignored. The above algorithm simplifies somewhat in the constrained cases utilizing A;" on account of the special form of A;O as shown in Fig. 11.1. Numerical examples of A: and of A;" appear in Figs. 11.6-11.9. Certain open paths, which are the same as those in Chapter VII, have been indicated on those figures with dashed lines. It must be remembered that a number of
3. Ignoring or Eliminating Closed Paths
219
bl b 2 b 3
b 4 b5 b 6
bI b2 b3 b4
b5 b6
FIG.11.6. Open paths with closed paths ignored-general
FIG.11.8. Open paths with closed paths ignored-tree
FIG.11.9. Open paths with closed paths ignored-unit-tree
case.
and link case.
and link case.
220
XI.
Conventional Networks
different contours whose extremities are the same result in the identical A:, as was shown in Chapter VIII, Eq. (8.54). Notice that signs (plus and minus) are required on the junction-pair voltages in order to correspond to the direction of the junction-pair currents: positive at the extremity where the junction-pair current is entering, and negative at the extremity where the junction-pair current is leaving. This notation is consistent with the branch notation described in Fig. 2.5. Voltages Eb have been drawn in Fig. 11.6 in order to show the validity of Eq. (3.44).
OTHEROPEN-PATH CONTOURS One set of open-path voltages can be transformed to any other set, as was established in Chapter VIII. Equations (8.42) and (8.43) specialize to
(1 1.31) and (9.69) where from Table 8.10
c;:= (A;')-'
(11.32)
The corresponding current expressions in Eqs. (8.39) and (8.40) likewise specialize to I" = C!", I O ' (9.22) and I"' = A:; I" (11.33) where from Table 8.10 coo,= (A:;)-' (11.34)
Ys's'in Eq. (8.30) and Yssin Eq. (8.32) contain Yo"' and 0'
YO",
respectively.
C'
(11.35)
4. Relationship between Submatrices of Z,, and Y s s
221
( 11.36)
Expressions of the three submatrices that are not given in Eqs. (11.35) and (1 1.36) are more complex than those of Yo'" and Yo' and can readily be obtained from Eqs. (8.30) and (8.32). 4. Relationship between Submatrices of Z,, and Y s s
Yss can readily be shown to be the inverse of Z, by utilizing the inverse relationships developed in Chapters I and I1 and applied to Eq. (2.24) or (2.28) and vice versa. Relationships between submatrices of Z,, and Yss can be established (inversion by submatrices) by expanding the following two equations:
z,, YbS= 1;
(11.37)
Y"Z,, = lYs
( 11.38)
Upon expanding Eq. (1 1.37),
(11.39) T h e following equations are contained in Eq. (11.39):
+ z,, Y C O= 1: z,, + z, Y C C= 1; z, Y + z, Y = 0'; z,, + z,,Y co = op z,
Y O 0
(11.40)
Y O C
(1 1.41)
OC
Y O 0
cc
(1 1.42) (1 1.43)
222
XI. Conventional Networks
From Eq. (1 1.42), we can solve for Z,, :
z,, = - z,, Y'Jy
YC,)
( 1 1.44)
-
Upon substituting Eq. (1 1.44) into Eq. (1 1.40) and solving for Z , , , z,, = [y o , - y y=>,, - 1 y c q - 1
(1 1.45)
From Eq. (1 1.42), we can also solve for Yo': YOC
=
-(z,,)-'z,,Ycc
(1 1.46)
Substituting Eq. (11.46) into Eq. (11.41) and solving for Y c cyields YCC= [Z,, - zc,(zo,)- 12,J -
(1 1.47)
From Eq. (1 1.43), we can solve for Z,, :
z,, = -z,,YcyYo~)-'
(1 1.48)
Substituting Eq. (1 1.48) into Eq. (1 1.41) and solving for Z,, yields
z,,= [y c c
-
y y yo,)
- 1 yo']
-
'
(1 1.49)
Equation (1 1.49) confirms the results obtained in Eqs. (1 1.5) and (1 1.10) as discussed earlier. From Eq. (1 1.43), we can solve for Yc": yco=
-(Z
cc
)- lzcoyoo
( I 1.50)
Substituting Eq. ( 1 1.50) into Eq. (1 1.40) and solving for Yo' gives YO0
=
[Z,, - zoc(zcc)-lzco]-~
(11.51)
Equation (11.51) confirms the results obtained in Eqs. (11.23) and (11.28), as discussed earlier. Z,, and Z,, can now readily be expressed in terms of the submatrices of YSS. Substituting Eq. ( I 1.45) into Eq. (I 1.44) yields
z,, = - [y o 0 - y y),,,
- 1 y c o ] yo,(
),,,
-1
(11.52)
Substituting Eq. (1 1.49) into Eq. (1 1.48) yields
z,, = - [y c c - y
y yo,)
-1
y"']
-1
y y yo,) - 1
(11.53)
Expressions for Y O c and Y c ofollow. Substituting Eq. (1 1.47) into Eq. (1 1.46) yields Y O C
= - (Zoo)-
'z,,cz,, - ~ c , ~ ~ o lzocl o ) -
-
( I 1.54)
Substituting Eq. ( I I .51) into Eq. ( I 1.50) yields yco= - ( Z C J
-
- l z c o ~ ~ , , zoc(zcc)-
lzc,l- '
(11.55)
4. Relationship between Submatrices of Z,, and Y"
223
Expanding Eq. (1 1.38) gives o
c
o
o
c
c
(1 1.56)
The following equations are contained i n Eq. (11.56):
+ Y"'Z,, Y'"Z,, + Y"Z,, Y""Z,, + Y"'Z,, Y cOz,, + Y"Z,, Y""Z,,
=
lp,
( 11 .57)
=
1 f,
(1 1.58)
= 00,
(1 1.59)
= Of,
(1 1.60)
From Eq. (1 1.59), we can solve for Zoc:
z,, = - ( Y ~ o ) - ~ Y ~ c z , ,
(1 1.61)
Substituting Eq. (1 1.61) into Eq. (1 1.58) and solving for Z , , ,
z,,= [y c c - ycy yo,)
-1
yo"]- 1
(11.49)
From Eq. (11.59), we can solve for Yo': (1 1.62)
YOC= - Y""Z,,(zc,)-' Substituting Eq. (1 1.62) into Eq. (1 1.57) and solving for Yo", YO 0
IZ,,] -
(11.51)
z,, = - ( Y C ~ ) - ~ Y C 0 z , ,
(1 1.63)
=
[Z,, - Z,,(Z,,)
-
From Eq. (1 1.60), we can solve for Z,,: Substituting Eq. (1 1.63) into Eq. (1 1.57) and solving for Z,, ,
z,,
= [y o u
- y y y=c)-1 yc"] - 1
(1 I .45)
From Eq. (1 1.60) we can solve for Y c o : (11.64)
YCO= - Yccz,"(z,,)-l Substituting Eq. (I 1.64) into Eq. (1 1.58) and solving for Y",
YCC= [Z,, - zc,(z,,)-~z,cl -I
(1 1.47)
Z,, and Z,, can be expressed in terms of the submatrices of Yss. Substituting Eq. ( 1 1.49) into Eq. (1 1.61), z,, = - (yo,) - 1 r o c [ yc.' - yay yo,) - 1 r o c ] 1 ~
( I 1.65)
224
XI. Conventional Networks
Substituting Eq. (1 1.45) into Eq. ( I 1.63),
zc,= - ( y c c ) - 1 yc"[ yo" - y
y y c c >- 1 yc"] - 1
(11.66)
Expressions for Y O c and Y c oin terms of submatrices of Z,, follow. Substituting Eq. (11.51) into Eq. (11.62), yoc=
-[Zoo
-~
O
C
(
Z
lZcol J -l ~ o c ( ~ c c ~ - l
(1 1.67)
'ZOCl - ~,o(~oo >
(1 1.68)
Substituting Eq. (1 1.47) into Eq. (1 1.64),
yco= - c z c c - Z,,(Z,,>
-
Tables 11.1 and 11.2 summarize these equations. TABLE 11.1
TABLE 11.2
Y O C = - (Zoo)lzoc YCC Yo= = - Y ""Z',,(Z,,)Y = - (ZCJ- lzco Y Y C 0 = - Y "=zc,(zo,) -I C0
O0
PROBLEMS 11.1 Modify the network example in Fig. 11.1 (Appendix 11) by connecting the current generator A to C from A to D. Vector I' is thus ( O I I I O ) , Vector eb remains (11001); and z b b is the unit matrix as shown. Solve this network first by utilizing the entire orthogonal and second by employing the six-step algorithm with network in terms of Z,, ( V , = Z,,Js)), zcc
.
11.2 Solve the network in Fig. 11.1 (Appendix 11) modified as indicated in (11.1) first by utilizing the entire orthogonal network in terms of Y s s ( J s= YssV5),and second by employing the six-step algorithm with YO0.
Problems
225
1 1 . 3 Check the relationships in the left-hand column of Table 11.1 by utilizing the and Yssof the network example in Fig. 11.1 (Appendix 11).
z,,
11.4 Check the relationships in Table 11.2 by utilizing the Z,, and Yss of the network example in Fig. 11.1 (Appendix 11).
CONVENTIONAL NETWORKS
1. Summary
In the development of the theory so far, we have assumed that the entire orthogonal equations and associated networks are always required to obtain a solution. In general, this is not the case, as will be shown in this chapter. It is sufficient, in many cases, to utilize the closed-path equations and to ignore those associated with the open paths, or to utilize those of the open paths, and to ignore the equations associated with the closed paths. In the latter case, it is even sufficient, as will be shown in this chapter, just to utilize the extremities of the open paths, which were earlier called junction pairs. The solution methods are often respectively called the mesh method and the junction-pair method. Both methods will be shown to have their basis in the orthogonal equations. Specifically, it will be shown that the mesh equations and the junction-pair equations readily follow from the orthogonal equations by ignoring the open-path equations in the case of the mesh method, and the closed-path equations in the case of the junction-pair method. The work in Chapter VIII in which one orthogonal contour set is transformed to another set will also be specialized to the case where either the open-path or the closed-path equations are ignored. In Section 4, the submatrices of Z,, are expressed in terms of those of Yss and vice versa.
2. Ignoring or Eliminating Open Paths We will limit our excitation in the interconnected network to B voltage sources (eb) and B current sources ( I b ) . ( N - 1) additional current sources could be considered, as outlined in Appendix 11. We will consider the B current 210
211
2. Ignoring or Eliminating Open Paths
sources ( I b ) as equivalent voltage sources ( Z b b 1') in the primitive network, as discussed in Chapter IV, Eq. (4.33). We also limit networks to those whose branches are traced out by at least one closed-path current contour. It is apparent from our previous work that the open-path current vector J" is zero in this case: J" = 0"
(11.1)
It is also clear, that term I' is zero when current sources are treated by means of equivalent voltage sources, as shown in Chapter IV, Eq. (4.34): 1' = 0'
(3.14)
Current J' thus consists of i' in this case. E, from Eq. (3.40) is zero: E , = 0,
(3.40)
The orthogonal equations (3.66) and (3.73) specialize as shown below: o
c
(11.2) and o
r
(11.3) earand e,. represent equivalent voltage sources in the orthogonal network, and are transformed from the primitive from the branch voltage sources eb plus
the negative of equivalent voltage sources Z b b I b as given in Eqs. (4.36) and (4.37):
(4.36) (4.37) If the open-path voltages E,, are of no particular interest in Eq. (11.2), then the open-path equations can be totally ignored and Eq. (11.2) simplifies to e,, = Zcci'
(11.4)
The solution for' i readily follows:
'i
= (ZcC)-'ec,
(11.5)
212
XI.
Conventional Networks
Currents ib follow from Eq. (3.5): jb =
cb,' i
(3.5)
The currents in the primitive network can be obtained as the sum of l band i b :
+ ib
(2.6)
Eb = Z b b J b - eb
(11.6)
Jb = Ib
The voltages Eb follow from Eq. (2.7):
It is readily apparent from the procedure above that the solution utilizing Z,, requires only Cbc and that Cb, need not be defined. Notice that the open-path axes cannot be ignored if the solution for ic is desired by means of Yss in Eq. ( I 1.3), but instead have to be eliminated. From Eq. (1 1.3), Y""(E,+ e,.) + Yoce,.= 0
(11.7)
Upon solving for (E, + e,,) in Eq. (11.7), E,
+ e,. = - ( Y O O ) - l Y O c e c .
(1 1.8)
Y'"(E, + e,,) + Yccec.
(11.9)
From Eq. (1 1.3),
'i
=
Upon substituting Eq. (1 1.8) into Eq. ( I 1.9), jc
= [ycc
- yCo(yoo)-lyo']e,,
(11.10)
Upon comparing Eq. (1 1.5) with Eq. (1 ].lo), it becomes clear that (Z,,)-' is equal to the expression within the square brackets in Eq. (1 1.10); this is more formally established in Section 4. With the solution procedure and its basis from the theory of orthogonal networks established, a simpler derivation starting from the primitive network can now be developed. The equations of the primitive network are Eb
+ eb = Z b b ( l b + ib)
(2.7)
l b and eb are considered known. Regrouping Eq. (2.7), Eb
+ (eb -
zbb
I b ) = z b b ib
(4.33)
Z,, I b represent the equivalent voltage sources due to current sources. Premultiplying Eq. (4.33) by C: gives CPEb
+ Ct(f?, -
zb,
I b )= CPZbb ib
(11.11)
213
2. Ignoring or Eliminating Open Paths
Upon substituting Eqs. (3.38), (3.5), and (4.37) into Eq. ( l l . l l ) , E, Recognizing that E, yields Eq. ( I 1.4) :
= 0,
+ ec, = cpz&Cb, ic
( 11.12)
according to Eq. (3.40), and substituting Eq. (3.65) e,,
= Z,
(11.4)
ic
The network representation of Eq. (11.4) is shown in the lower half of Figs. 3.5 and 3.6. The solution process can be described by a series of steps:
1.
ib = Cb, ' i
2.
e,. = Cp(e,
3.
z,
=
- Z,,lb)
cpz,, cb,
4.
i" = (Zcc)-'e,.
5.
ih = cb, ic
6. E ,
= Z,,(lb
+ ib) - eb
The above steps readily simplify for the case where no current sources I b are present, as further elaborated upon in the reference.' Notice that the first step is used only to define Cb,; also the steps above need not be executed in exactly the manner indicated, and no attempt has been made to create an efficient algorithm. The above work was presented in terms of Cb,. The analysis equally applies to various subcases and special cases of Cb, , as, for example, the constrained contour cases designated C,: shown in Fig. 1 1 . 1 . In the latter cases,
o
FIG.
c
o
c
1 1 . I . Summary figures of Cbc, CC, A*", and A;".
Happ, H. H., Special Cases of Orthogonal Networks-Mesh IEEE Trans. Power App. Systems 87, 53-66 (1968).
and Nodal Networks,
214
[m]
XI.
Conventional Networks
the above steps can be simplified on account of the special form of CY, . It was previously indicated that the solution procedure with Z,, only utilizes Cbc, but that Cbo is present but not defined. Figure 1 1.1 is a summary figure in which Cbo and CY, are indicated by dashed lines, which signify that the open paths are present although ignored. Numerical examples of Cb, and of CY, appear in Figs. 11.2 -1 1.5.
bl
b2
*3
b5
lC3
bI bZ b3 b 4 b5 b6
r---------I
I
IT
I
ICOb
b6
c3
x4
FIG.11.2. Closed paths with open paths ignored-general
FIG.11.3. Closed paths with open paths ignored-tree CI
c2 c 3
FIG. 11.4. Closed paths with open paths ignored-tree
FIG.11.5. Closed paths with open paths ignored-tree
case.
and link case. b5 b3 b4 bI b2 b 6
:-----r----
and link case.
and unit-link case.
215
2. Ignoring or Eliminating Open Paths
OTHERCLOSED-PATH CONTOURS One set of closed-path currents can be transformed to any other set from the theory established in Chapters VIII. Equations (8.39), and (8.40) respectively specialize in this case to ic = cc,. ic'
and
iC'
where from Table 8.10
= A!:
(9.28) (11.13)
iC
( 1 1.14)
CC, = (A:;)-'
The corresponding voltage expressions i n Eqs. (8.42) and (8.43) likewise specialize to e,, = C;: e, (11.15) and e, = ALC'ec, (9.75) Where from Table 8.10 cis = ( A ,C') - 1 (11.16) .Zs,sfin Eq. (8.27) and Z , , i n Eq. (8.31) contain .ZC,,,and Z c c :
(11.17)
0
C
(11.18)
7
5,
C"
216
XI.
Conventional Networks
Expressions of the three submatrices not presented in detail are more comand Z , , , as can readily be established from Eqs. (8.27) plex than those of ZcSc, and (8.31). 3. Ignoring or Eliminating Closed Paths
We will likewise limit our excitation in the interconnected network to B voltage sources (eb)and B current sources ( I b ) . We will consider the B voltage sources eb as equivalent current sources in the primitive network ( Ybbeb),as discussed in Chapter IV. We also limit networks to those that can be spanned completely by junction pairs. I t is apparent when employing Yhbehthat e, in the orthogonal network is zero. E, is likewise zero from Eq. (3.40), and thus
v,= 0,
(1 1.19)
e, likewise is zero on account of treating eb by means of current equivalents: e,
(4.4)
= 0,
In the absence of external voltage sources
io
(3.22)
= 0"
The orthogonal equations (3.73) and (3.66) specialize as shown below. o
c
(1 1.20)
and o
c
(11.21)
and 1'' represent equivalent current sources in the orthogonal network. They have their origin in the primitive network and are transformed from branch current sources I b plus the negative of equivalent current sources Yhbehas given i n Eqs. (4.41) and (4.42): /"' = A " ~ ( / '- Ybheb)
(4.41)
1" = A f b ( / ' - Yhbeb)
(4.42)
3. Ignoring or Eliminating Closed Paths
217
If the closed-path currents i" are of no particular interest in Eq. (1 1.20), then the closed-path equations can be totally ignored and Eq. (1 1.20) simplifies to 1''
=
yoOE,
(1 1.22)
The solution for E, is as follows: E, = (yOO)-l[o'
(1 I .23)
Voltages i n the primitive network follow from Eq. (3.44):
Eb = A:E0
(3.44)
Currents in the primitive network follow from Eq. (1.4): J b = Ybb(Eb+ eb)
(1 1.24)
It is apparent from the procedure above that the solution utilizing Yo" requires only A t , and that A t need not be even specified. The closed-path equations cannot be ignored if the solution for E, is to be obtained by means of Z,, in Eq. (1 1.21), but have to be eliminated. From Eq. (1 1.21}, Zc,Io' Upon solving for (I"
+ i')
+ Z,,(I"' + ic) = O
(1 1.25)
in Eq. (1 1.25),
( I C ' + i')
=
-(zcc)-lzc,I~'
(1 1.26)
+ Zoc(Ic'+ i')
(1 1.27)
From Eq. (1 1.21),
E, = Z,, I"
Upon substituting Eq. (1 1.26) into Eq. (1 1.27), E,
=
[Zoo - ~
O c ~ ~ ' c ~ - l ~ c O 1 ~ O (11.28) '
Upon comparing Eq. (11.28) with Eq. (ll.23), it is clear that (Yo')-' is equal to the expression within the square brackets in Eq. (1 1.28); this equality is more formally established in Section 4. The solution procedure and its basis have been established fromthetheory of orthogonal networks. A simple derivation can be developed starting from the primitive network. The equations of the primitive network are Ib
+ i b = ybb(Eb+ eb)
(2.8)
I b and eb are considered to be known. Upon regrouping Eq. (2.8), ( I b - Ybbeb)+ i b = YbbEb
(4.38)
218
XI.
Conventional Networks
ybbebrepresents the equivalent current sources due to voltage sources. Premultiplying Eq. (4.38) by Apb gives
APb(Zb- Ybbeb)+ Apb ib = APb YbbEb
(11.29)
Substituting Eqs. (4.41), (3.44), and (4.25) into Eq. (1 1.29), 1"'
Substituting ' i Eq. (11.22):
= 0"
+ i"
= APb YbbAtE,
(1 1.30)
according to Eq. (3.22) and substituting Eq. (3.69) yields (11.22)
1"'= y""Eo
The network representation of (Yo")-' is identical to that of Zoo except that the numerical values of the elements are different. Clearly (Yo")-' differs numerically from Z o o , shown in Eq. (1 1.28) because of the modification of Zoo in the closed-path elimination process. The network configuration remains the same as that of Zoo and is shown in the upper half of Figs. 3.5 and 3.6 for two examples case. The solution process can be described by a series of steps:
1.
Eb = A t E ,
4. E ,
= (Yoo)-lZo'
2.
I"= APb(zb - ybbeb)
5.
Eb
= AtEo
6.
J b = ybb(Eb
3.
YO" = APb YbbAio
+
eb)
The above steps simplify for the case where no voltage sources e b are present, as elaborated upon in the reference given in footnote 1. The first step, similar to that in the previous case, is used only to define A t . The other steps are computational, but can be performed in a manner different from that indicated above to improve the computational efficiency. Since only A t is required, the complete open paths need not be defined. The extremities, or junction pairs as they were called, are sufficient to define
At.
Although the above work was presented in terms of A t , the analysis applies equally to various subcases and special cases of A t . The solution procedure with Yo",utilizes only A t , but A t is nevertheless present-although its presence is ignored, as previously indicated. Fig. 11.1 illustrates A t , A;", and A;" of the unit-tree case. A t and A;' are indicated by dashed lines, which signify that these submatrices are present although ignored. The above algorithm simplifies somewhat in the constrained cases utilizing A;" on account of the special form of A;O as shown in Fig. 11.1. Numerical examples of A: and of A;" appear in Figs. 11.6-11.9. Certain open paths, which are the same as those in Chapter VII, have been indicated on those figures with dashed lines. It must be remembered that a number of
3. Ignoring or Eliminating Closed Paths
219
bl b 2 b 3
b 4 b5 b 6
bI b2 b3 b4
b5 b6
FIG.11.6. Open paths with closed paths ignored-general
FIG.11.8. Open paths with closed paths ignored-tree
FIG.11.9. Open paths with closed paths ignored-unit-tree
case.
and link case.
and link case.
220
XI.
Conventional Networks
different contours whose extremities are the same result in the identical A:, as was shown in Chapter VIII, Eq. (8.54). Notice that signs (plus and minus) are required on the junction-pair voltages in order to correspond to the direction of the junction-pair currents: positive at the extremity where the junction-pair current is entering, and negative at the extremity where the junction-pair current is leaving. This notation is consistent with the branch notation described in Fig. 2.5. Voltages Eb have been drawn in Fig. 11.6 in order to show the validity of Eq. (3.44).
OTHEROPEN-PATH CONTOURS One set of open-path voltages can be transformed to any other set, as was established in Chapter VIII. Equations (8.42) and (8.43) specialize to
(1 1.31) and (9.69) where from Table 8.10
c;:= (A;')-'
(11.32)
The corresponding current expressions in Eqs. (8.39) and (8.40) likewise specialize to I" = C!", I O ' (9.22) and I"' = A:; I" (11.33) where from Table 8.10 coo,= (A:;)-' (11.34)
Ys's'in Eq. (8.30) and Yssin Eq. (8.32) contain Yo"' and 0'
YO",
respectively.
C'
(11.35)
4. Relationship between Submatrices of Z,, and Y s s
221
( 11.36)
Expressions of the three submatrices that are not given in Eqs. (11.35) and (1 1.36) are more complex than those of Yo'" and Yo' and can readily be obtained from Eqs. (8.30) and (8.32). 4. Relationship between Submatrices of Z,, and Y s s
Yss can readily be shown to be the inverse of Z, by utilizing the inverse relationships developed in Chapters I and I1 and applied to Eq. (2.24) or (2.28) and vice versa. Relationships between submatrices of Z,, and Yss can be established (inversion by submatrices) by expanding the following two equations:
z,, YbS= 1;
(11.37)
Y"Z,, = lYs
( 11.38)
Upon expanding Eq. (1 1.37),
(11.39) T h e following equations are contained in Eq. (11.39):
+ z,, Y C O= 1: z,, + z, Y C C= 1; z, Y + z, Y = 0'; z,, + z,,Y co = op z,
Y O 0
(11.40)
Y O C
(1 1.41)
OC
Y O 0
cc
(1 1.42) (1 1.43)
222
XI. Conventional Networks
From Eq. (1 1.42), we can solve for Z,, :
z,, = - z,, Y'Jy
YC,)
( 1 1.44)
-
Upon substituting Eq. (1 1.44) into Eq. (1 1.40) and solving for Z , , , z,, = [y o , - y y=>,, - 1 y c q - 1
(1 1.45)
From Eq. (1 1.42), we can also solve for Yo': YOC
=
-(z,,)-'z,,Ycc
(1 1.46)
Substituting Eq. (11.46) into Eq. (11.41) and solving for Y c cyields YCC= [Z,, - zc,(zo,)- 12,J -
(1 1.47)
From Eq. (1 1.43), we can solve for Z,, :
z,, = -z,,YcyYo~)-'
(1 1.48)
Substituting Eq. (1 1.48) into Eq. (1 1.41) and solving for Z,, yields
z,,= [y c c
-
y y yo,)
- 1 yo']
-
'
(1 1.49)
Equation (1 1.49) confirms the results obtained in Eqs. (1 1.5) and (1 1.10) as discussed earlier. From Eq. (1 1.43), we can solve for Yc": yco=
-(Z
cc
)- lzcoyoo
( I 1.50)
Substituting Eq. ( 1 1.50) into Eq. (1 1.40) and solving for Yo' gives YO0
=
[Z,, - zoc(zcc)-lzco]-~
(11.51)
Equation (11.51) confirms the results obtained in Eqs. (11.23) and (11.28), as discussed earlier. Z,, and Z,, can now readily be expressed in terms of the submatrices of YSS. Substituting Eq. ( I 1.45) into Eq. (I 1.44) yields
z,, = - [y o 0 - y y),,,
- 1 y c o ] yo,(
),,,
-1
(11.52)
Substituting Eq. (1 1.49) into Eq. (1 1.48) yields
z,, = - [y c c - y
y yo,)
-1
y"']
-1
y y yo,) - 1
(11.53)
Expressions for Y O c and Y c ofollow. Substituting Eq. (1 1.47) into Eq. (1 1.46) yields Y O C
= - (Zoo)-
'z,,cz,, - ~ c , ~ ~ o lzocl o ) -
-
( I 1.54)
Substituting Eq. ( I I .51) into Eq. ( I 1.50) yields yco= - ( Z C J
-
- l z c o ~ ~ , , zoc(zcc)-
lzc,l- '
(11.55)
4. Relationship between Submatrices of Z,, and Y"
223
Expanding Eq. (1 1.38) gives o
c
o
o
c
c
(1 1.56)
The following equations are contained i n Eq. (11.56):
+ Y"'Z,, Y'"Z,, + Y"Z,, Y""Z,, + Y"'Z,, Y cOz,, + Y"Z,, Y""Z,,
=
lp,
( 11 .57)
=
1 f,
(1 1.58)
= 00,
(1 1.59)
= Of,
(1 1.60)
From Eq. (1 1.59), we can solve for Zoc:
z,, = - ( Y ~ o ) - ~ Y ~ c z , ,
(1 1.61)
Substituting Eq. (1 1.61) into Eq. (1 1.58) and solving for Z , , ,
z,,= [y c c - ycy yo,)
-1
yo"]- 1
(11.49)
From Eq. (11.59), we can solve for Yo': (1 1.62)
YOC= - Y""Z,,(zc,)-' Substituting Eq. (1 1.62) into Eq. (1 1.57) and solving for Yo", YO 0
IZ,,] -
(11.51)
z,, = - ( Y C ~ ) - ~ Y C 0 z , ,
(1 1.63)
=
[Z,, - Z,,(Z,,)
-
From Eq. (1 1.60), we can solve for Z,,: Substituting Eq. (1 1.63) into Eq. (1 1.57) and solving for Z,, ,
z,,
= [y o u
- y y y=c)-1 yc"] - 1
(1 I .45)
From Eq. (1 1.60) we can solve for Y c o : (11.64)
YCO= - Yccz,"(z,,)-l Substituting Eq. (I 1.64) into Eq. (1 1.58) and solving for Y",
YCC= [Z,, - zc,(z,,)-~z,cl -I
(1 1.47)
Z,, and Z,, can be expressed in terms of the submatrices of Yss. Substituting Eq. ( 1 1.49) into Eq. (1 1.61), z,, = - (yo,) - 1 r o c [ yc.' - yay yo,) - 1 r o c ] 1 ~
( I 1.65)
224
XI. Conventional Networks
Substituting Eq. (1 1.45) into Eq. ( I 1.63),
zc,= - ( y c c ) - 1 yc"[ yo" - y
y y c c >- 1 yc"] - 1
(11.66)
Expressions for Y O c and Y c oin terms of submatrices of Z,, follow. Substituting Eq. (11.51) into Eq. (11.62), yoc=
-[Zoo
-~
O
C
(
Z
lZcol J -l ~ o c ( ~ c c ~ - l
(1 1.67)
'ZOCl - ~,o(~oo >
(1 1.68)
Substituting Eq. (1 1.47) into Eq. (1 1.64),
yco= - c z c c - Z,,(Z,,>
-
Tables 11.1 and 11.2 summarize these equations. TABLE 11.1
TABLE 11.2
Y O C = - (Zoo)lzoc YCC Yo= = - Y ""Z',,(Z,,)Y = - (ZCJ- lzco Y Y C 0 = - Y "=zc,(zo,) -I C0
O0
PROBLEMS 11.1 Modify the network example in Fig. 11.1 (Appendix 11) by connecting the current generator A to C from A to D. Vector I' is thus ( O I I I O ) , Vector eb remains (11001); and z b b is the unit matrix as shown. Solve this network first by utilizing the entire orthogonal and second by employing the six-step algorithm with network in terms of Z,, ( V , = Z,,Js)), zcc
.
11.2 Solve the network in Fig. 11.1 (Appendix 11) modified as indicated in (11.1) first by utilizing the entire orthogonal network in terms of Y s s ( J s= YssV5),and second by employing the six-step algorithm with YO0.
Problems
225
1 1 . 3 Check the relationships in the left-hand column of Table 11.1 by utilizing the and Yssof the network example in Fig. 11.1 (Appendix 11).
z,,
11.4 Check the relationships in Table 11.2 by utilizing the Z,, and Yss of the network example in Fig. 11.1 (Appendix 11).