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Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003 2

Tree a connected subgraph containing all nodes of a graph but no closed path.

Branches the elements of a treeNumber of branches required to form a tree:

1 where = no. of nodes in the graph

Links elements of the connected graph that are not included in the tree.

Cotree a subgraph formed by the links of a connected graph.

Number of links of a connected graph:

where = no. of elements of a connected graph

It follows that

1

:

7 = 5 = 4 = 3

1 2 3 4

0

6

7

5 4

32

1

Branch

Link

b

nb n

l

bel e

nel

Example

e = n b l

=

=

+=

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The elements of the element-node incidence matrix of a connected graph are as follows:

1 if the th element is incident to and oriented away from the th

node.

1 if the th element is incident to and oriented towards from the th

node.

0 if the th element is not incident to the th node.

The dimension of the matr ix is .

:

For the previous example network, the element-node incidence matrix is

(0) (1) (2) (3) (4)

11

11

11

11

11

11

11

7

6

5

4

3

2

1

Since

,,2,10

1

0

the columns of are linear ly dependent rank < .

Nodes

Elements

I N C I D E NC E M A T R I C E S

A

A

A A

Element-No de Inci dence M atr ix

ai j = i j

ai j = i j

ai j = i j

e n

Example

eian

j

i j

n

=

==

=

K

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Bus incidence matrix obtained from by deleting the column corresponding to

the reference node. The dimension of this matrix is ( 1) and the rank is 1 =

.

:

For the previous example network, if node 0 is chosen as the reference node, the bus

incidence matrix is,

(1) (2) (3) (4)

11

11

11

11

11

1

7

6

5

4

32

1

The matrix is rectangular and therefore singular.

Binary valued matrix a matrix whose entries are binary (Boolean) variables.

The elements of the square, binary bus connection matrix is given by

,,2,1,

otherwise

lineabybustoconnectedbus

0

1

1

1 denotes Boolean TRUE

0 denotes Boolean FALSE

is symmetric

Negation: : All TRUE elements of are replaced by FALSE, and all FALSE

entries are replaced by TRUE

Bus

Element

Bus Incidence M atr i x

e n nb

Example

nji

ji

ji

bi j

A

A A

A

BI N A R Y B US CO N N E C T I O N M A T R I X B

B

B

B B

=

=

=

=

K

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Boolean AND: , where and are binary bus connect ion matrices

,,2,1,,2,12211

denotes Boolean AND

+ denotes Boolean OR

is

is

is

Boolean OR:

,allfor

It is necessary that and have the same dimensions.

The operation produces a square matrix of dimension for an -bus system

where ( ) is 1 when buses and are joined by a l ine or are joined through an

intervening bus by a line. Otherwise ( ) is 0.

:

(1) (2) (3) (4) (5)

10100

01011

10110

01111

01011

)5(

)4(

)3(

)2(

)1(

(1)

(4)

(2)

(3) (5)

B D B D

DB

B D

B

D

B D

DB

A B

B B

B B

B B

B

)==

+++=

) +=+

=

cjri

dbdbdbmjimjijiij

r c

r m

m c

+

jidb ijijij

n n

i j i j

i j

Example

KK

L

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(1) (2) (3) (4) (5)

10110

01111

1111111111

01111

)5(

)4(

)3()2(

)1(

Note that is also square and contains 1 in positions corresponding to buses

joined by three or fe wer lines and two or less inte rven ing buses. The diagonal ent ries

of and so on, are all 1.

Generalization:

If is the binary bus connection matrix for an -bus power system, and the notation(m)

,

times

)(

is used to denote repeated AND operations, and(1)

= , then(m)

consists of all 0s

except in the diagonal position, where 1s appear. Also, the position of(m)

contains

1s if and only if buses and are joined via lines or less (hence giving 1

interven ing buses). Furthermore, for an -bus system( 1)

consists of all 1s when all

system buses are connected to the system.

To determine whether a given bus is close to another bus. The buses and may be

evaluated for proximity by examining(m)

; if this entry is a 1, buses and are

connected to each other through or fewer lines.

: If a fault study is to be done for a short circuit at bus , only

buses nearby bus need to be examined.

To determine whether a give system is connected. The matrix( 1)

must all be 1s

when an -bus system is connected since the farthest possible configuration between

bus 1 and bus occurs when 1 and are on opposite ends of a radial string of buses. If( 1)

contains a 0, the system is disconnected.

=

=

BB

B B B

B, B B, B B B,

B

B

BBBB

B B B

B

B

B

B

B

Boolean AND Operati ons on the Bi nary Bus Connection M atri x

n

m

m

i j

i j m m

nn

Applicati ons if the Bin ary B us Connection M atri x

i j

i j

mPotential applic atio n k

kn

n

n nn

L

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The disconnection matrix and the matrices , , and so on, have

properties similar to

(2)

,

(3)

, and so on. Allowing to denote the bus disconnectionmatrix, it is easy to show that

)()()(

The line incidence matrix is defined by

otherwise

busatendslinebusatstartsline

0

11

)(

For a system of lines and buses, is . This matrix is used to calculate the voltage

difference between t he buses at the terminals of each system line. Let line be a -vector

of line voltages (i.e., voltage drops across each system line); then

line = bus

For the purpose of calculating line voltage drops, each line must be considered to bedirected (i.e., having a start or higher voltage bus and an end or lower voltage bus). This

convention is reflected in the definition of line start and end in matrix While it is

unimportant which bus is selected as the start of a line and which as the end, once the

convention is established, it must be consistent in the definition of elements of bus.

Bus Disconnecti on M atr ix

lklk

ji

ji

i j

l n l n

l

B

B B B B B B

B B D

BDB

L I N E I N C I DE N C E M A T RI X L

L

L

L

V

V L V

L .

V

=

=

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bus

bus Formation Methods:

1. bus building rules

2. Building block approach

3. Network incidence matrix

Node a junction formed when two or more circuit elements (R, L, C, Vs, Is) are

connected to each other at their terminals.

Consider the circuit diagram

KCL at node 1: 0412131

KCL at node 3: 313233

Rearranging,

3321

4321 0

Similar equations can be formed for node 2 and 4.

NOTE: All branch currents can be found when the node voltages are known, and a

node equation formed for the reference node would yield no further information.

Hence, the required number of independent node equations is one less than the number

of nodes.

B US A DM I T T A N CE M A T R I X Y

YY

Node Equations

fdc YVVYVVYVV

IYVVYVVYV cba

IYYYVYVYV

YVYVYVYYYV

cbabc

fcdfdc

( ) ) ) =++

( ) ( ) =++

( )=+++

=++

Ya

Yb

Yc

Yd

Ye

Yf

Yg I4I3

13 4

Reference

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General matrix format:

bus

4

3

21

4

3

21

44434241

34333231

2423222114131211

when forming bus :

1. Diagonal element = sum of the admittances directly connected to node .

2. Off-diagonal element = the negative of the net admittance connected between nodesand .

: self-admittance or driving-point admittance.

: mutual admittance or transfer admittance.

Using the rules,

0

0bus

Separating the entries for Yc,

0000

00

0000

00

0

00

0

bus

IVY

Y

Y

Y

=

=

( ))

++

++

++

++

=

( )( )

+

++

+

++

+

=

or

I

I

I

I

V

V

V

V

YYYY

YYYY

YYYY

YYYY

Usual rules

Yj j j

Yi ji j

YjjYij

gfeef

cbabc

ebedbd

fcdfdc

YYYYY

YYYYY

YYYYYY

YYYYYY

cc

cc

gfeef

bab

ebedbd

fdfd

YY

YY

YYYYY

YYY

YYYYYY

YYYY

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More compactly,

(3)(1)

11

11

The smaller matrix on the right is a compact storage matrix for matrix contribution of

to bus. It is an important building block in forming bus for more general networks.

Bus impedance matrix:

44434241

34333231

24232221

14131211

1busbus

Consider a generator in steady-state:Voltage equation:

Dividing voltage equation by ,

1where

ccc

cc

YYY

YY

Yc

ZZZZ

ZZZZ

ZZZZ

ZZZZ

Branch and Node Admittances

VI ZE as

Za

aaa

a

ss

ZYVYI

Z

EI

==

+=

=+==

Y Y

YZ

Es

Za

I

+

-

V

+

-

N

e

t

w

o

r

k

Is

Ya

I

V

+

-

N

e

t

w

o

r

k

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The voltage source and its series impedance can be interchanged with the current

source and its shunt admittance , provided that

1and

Sources and may be considered externally applied at the nodes of the transmission

network, which then consists of only .

Suppose that only branch admittance is connected between nodes and as part of a

larger network of which only the reference node is shown,

= branch impedance, primitive impedance

= branch admittance, primitive admittance

Current and voltage equations:

1

111

1

1

11;

11;1

1

for branch

Es Za

Is Ya

aa

a

ss

ZY

Z

EI

Es Ispassivebranches

Ya m n

Za

Ya

n

m

n

m

aa

aa

n

ma

n

ma

an

maaaa

n

maa

n

m

I

I

V

V

YY

YY

I

II

V

VY

IV

VYIVY

V

VVI

I

I

nodal admitt ance equati on Ya

nodal admittance matri x

==

[ ]

[ ]

[ ]

=

=

=

=

=

=

=

Ya

= 1/Za

Ia InIm

Vm

Vn

++

- -

Va+ -

Reference node

nm

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The nodal admittance matrix is singular because neither node nor node connects

to the reference.

When is the reference node, = 0 and

This corresponds to removal of row and column from the coefficient matrix (nodal

admittance matrix).

Note that

111111

11

The nodal admittance matrices are simply storage matrices with row and column

labels determined by the end nodes of the branch. To obtain the

of a network, we simply combine the individual branch matrices

by adding together elements with ident ical row and column labels.

Such addition causes the sum of the branch currents flowing from each n ode ofthe

network to equal the total current injected into that node, as required by KCL.

Provided at least one of the network branches is connected to the reference node, the

net result is bus of the system.

m n

n Vn

mma IVY

n n

over all nodal

admi ttance matri x

buil ding block

] =

[ ]

=

Y

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:

Single-line diagram:

Reactance diagram (per unit):

Example

1

2

3 4

0

1

2

3 4j0 .1

j0 .2 5

j0 .2 5j0 .1 25

j0 .2

j0 .4

j0 .1

j1 .1 5 j1 .1 5

1.25 0 o 0.85 -45 o+ +

--

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Admittance diagram:

(1)(4)(2)(4)(1))2(

(4)(1)(3)(2)(3)

)3(

11-

1-1

(1)

(4);

11-

1-1

(2)

(4);

11

11

)1(

)2(

1(4);11-

1-1

(1)

(3);

11

11

(2)

(3);1)3(

Combining elements of the above matrices having identical row and column labels,

(1) (2) (3) (4)

)4(

)3(

)2(

)1(

0

0= bus

The order in which the labels are assigned is not important here, provided the columns

and rows follow the same order.

Nodal admittance equations of the overall network:

o

o

4

3

2

1

13568.0

9000.1

0

0

3.80.00.55.2

0.08.80.40.4

0.50.40.170.8

5.20.40.85.14

[ ] [ ]

( )( )

++

++

++

++

=

fed

gcba

YYY

YYYY

gfeef

cbabc

ebedbd

fcdfdc

YYYYY

YYYYY

YYYYYY

YYYYYY

V

V

V

V

jjj

jjj

jjjj

jjjj

Y

Ib

-j4.0

-j4.0

-j8.0-j5.0

-j2.5

1 -90

0.68 -135

Ie

Ic If

Id

Ia

-j0.8 -j0.8

Ig

0

1

2

3 4

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bus

Two mutually coupled branches with impedance parameters or with admittanceparameters:

Primitive impedance equations:

where mutual impedance is positive when and enter the dotted terminals.

Inverse of the primitive impedance matrix:

2

11

M utuall y C oupled Br anches i n

b

a

bM

Ma

b

a

I

I

ZZ

ZZ

V

V

ZM Ia Ib

bM

Ma

aM

Mb

MbabM

Ma

YY

YY

ZZ

ZZ

ZZZZZ

ZZ

pr imiti ve admittance matr ix

Y

=

=

=

Za

ZM

Zb

Vb

Va

+ +

--

Ia

Ib

Ip

Iq

Im

In

Ya

YM

Yb

Vb

Va

+ +

--

Ia

Ib

Ip

Iq

Im

In

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Voltage-drop equations:

11000011

where = coefficient matrix

Current equations:

1010

01

01

Substitute voltage drop equations into the primitive impedance equation,

Premultiply by ,

Nodal admittance equations of the two mutually coupled branches:

( ) ( ) ( ) ( )

)(

)(

)(

)(

The 4 4 submatrices above form part of the larger nodal admittance matrix of the

overall system. The pointers , , , indicate the rows and columns of the system

matrix to which the elements of the above nodal admittance matrix belong.

=

=

=

=

=

=

=

=

=

q

p

n

m

q

p

n

m

qp

nm

b

a

V

VV

V

V

VV

V

VVVV

VV

b

aT

b

a

qp

n

m

I

I

I

I

I

I

I

I

b

a

q

p

n

m

bM

Ma

I

I

V

V

V

V

YY

YY

T

q

p

n

m

b

aT

q

p

n

m

bM

MaT

I

I

I

I

I

I

V

V

V

V

YY

YY

m n p q

q

p

n

m

q

p

n

m

q

p

n

m

bbMM

bbMM

MMaa

MMaa

I

I

I

I

V

V

V

V

YYYY

YYYY

YYYY

YYYY

m n p q

A

A

A

A

A

AAA

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Nodal admittance matrix by inspection:

( ) ( ) ( ) ( )

)(

)(

)(

)(

11

11

11

11

11

11

11

11

If node is the reference, we may eliminate the row and column of that node.

If and are one and the same node, columns and are combined (since = ),

and the corresponding rows are added because and are parts of the common

injected current.

:

Primitive admittances:

25.675.3

75.325.6

25.015.0

15.025.01

Nodal admittance matrix:

(3) (1) (3) (2)

)2(

)3(

)1(

)3(

)25.6(11

11)75.3(

11

11

)75.3(11

11)25.6(

11

11

m n p q

q

p

n

m

bM

Ma

YY

YY

n

n q n q Vn VqIn Iq

Example

jj

jj

jj

jj

jj

jj

=

+

+

j0.25

j0.25

j0.25j0.15

2

13

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Adding the columns and rows of the common node 3,

(3)(2)(1)

75.375.325.625.625.675.375.325.6

25.675.325.675.3

75.325.675.325.6

)3(

)2(

)1(

Nodal equations:

3

2

1

3

2

1

00.550.250.2

50.225.675.3

50.275.325.6

Three branches with mutual coupling:

32

31

211

2

1

21

00

bus :

1. Invert the primitive impedance matrices of the network branches to obtain the

corresponding primitive admittance matrices. A single branch has a 1 1 matrix. Two

mutually coupled branches have a 2 2 matrix, three mutually coupled branches have

a 3 3 matrix, and so on.

2. Multiply the elements of each primitive admittance matrix by the 2 2 building-block

matrix.

+++

+

=

=

jjjjjjjj

jjjj

jjjj

I

I

I

V

V

V

jjj

jjj

jjj

cMM

MbM

MMa

cM

bM

MMa

YYYYYY

YYY

ZZZZ

ZZZ

To form for a netw ork wi th mutually coupled br anchesY

Ia

Zb Zc

ZM1

ZM2

Za

Ib

Ic

m p r

n q s

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3. Label the two rows and the two columns of each building-block matrix with

the end-node numbers of the corresponding self-admittance. For mutually coupledbranches it is important to label in the order of the marked (dotted) --- then ---

unmarked (undotted) node numbers.

4. Label the two rows of each building-block matrix with node numbers

aligned and consistent with the row labels assigned in (3); then label the columns

consistent with the column labels of (3).

5. Combine, by adding together, those elements with identical row and column labels to

obtain the nodal admittance matrix of the overall network. If one of the nodes

encountered is the reference node, omit its row and co lumn to obtain the system

Ybus.

diagonal

of f-diagonal

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