05 EE394J 2 Spring12 Power System Matrices

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_05_EE394J_2_Spring12_Power_System_Matrices.doc

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Power System Matrices and Matrix Operations

Nodal equations using Kirchhoff's current law. Admittance matrix and building algorithm.

Gaussian elimination. Kron reduction. LU decomposition. Formation of impedance matrix by

inversion, Gaussian elimination, and direct building algorithm.

1. Admittance Matrix

Most power system networks are analyzed by first forming the admittance matrix. The

admittance matrix is based upon Kirchhoff's current law (KCL), and it is easily formed and very

sparse.

Consider the three-bus network shown in Figure that has five branch impedances and one current

source.

1 2 3

ZE

ZA

ZB

ZC

ZDI3

Figure 1. Three-Bus Network

Applying KCL at the three independent nodes yields the following equations for the bus voltages

(with respect to ground):

At bus 1, 0211 =

+AE Z

VV

Z

V,

At bus 2, 032122 =

+

+CAB Z

VV

Z

VV

Z

V,

At bus 3, 3233 I

Z

VV

Z

V

CD

=

+ .

Collecting terms and writing the equations in matrix form yields


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=

+

++

+

33

2

1

0

0

1110

11111

0111

IV

V

V

ZZZ

ZZZZZ

ZZZ

DCC

CCBAA

AAE

,

or in matrix form,

IYV= ,

where Yis the admittance matrix, Vis a vector of bus voltages (with respect to ground), andIis a

vector of current injections.

Voltage sources, if present, can be converted to current sources using the usual network rules. If

a bus has a zero-impedance voltage source attached to it, then the bus voltage is already known,

and the dimension of the problem is reduced by one.

A simple observation of the structure of the above admittance matrix leads to the following rule

for building Y:

1. The diagonal terms of Y contain the sum of all branch admittances connected directly to the

corresponding bus.

2. The off-diagonal elements of Ycontain the negative sum of all branch admittances connected

directly between the corresponding busses.

These rules make Yvery simple to build using a computer program. For example, assume thatthe impedance data for the above network has the following form, one data input line per branch:

From To Branch Impedance (Entered

Bus Bus as Complex Numbers)

1 0 ZE

1 2 ZA

2 0 ZB

2 3 ZC

3 0 ZD

The following FORTRAN instructions would automatically build Y, without the need of

manually writing the KCL equations beforehand:

COMPLEX Y(3, 3) , ZB, YB

DATA Y/ 9 * 0. 0/

1 READ( 1, *, END=2) NF, NT, ZB


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YB = 1. 0 / ZB

C MODI FY THE DI AGONAL TERMS

I F( NF . NE. 0) Y(NF, NF) = Y( NF, NF) + YB

I F( NT . NE. 0) Y(NT, NT) = Y( NT, NT) + YB

I F( NF . NE. 0 . AND. NT . NE. 0) THEN

C MODI FY THE OFF- DI AGONAL TERMS

Y( NF, NT) = Y( NF, NT) - YB

Y( NT, NF) = Y( NT, NF) - YB

ENDI F

GO TO 1

2 STOP

END

Of course, error checking is needed in an actual computer program to detect data errors and

dimension overruns. Also, if bus numbers are not compressed (i.e. bus 1 through bus N), then

additional logic is needed to internally compress the busses, maintaining separate internal andexternal (i.e. user) bus numbers.

Note that the Y matrix is symmetric unless there are branches whose admittance is direction-

dependent. In AC power system applications, only phase-shifting transformers have this

asymmetric property. The normal 30ophase shift in wye-delta transformers creates asymmetry.

2. Gaussian Elimination and Backward Substitution

Gaussian elimination is the most common method for solving bus voltages in a circuit for which

KCL equations have been written in the form

YVI= .

Of course, direct inversion can be used, where

IYV1= ,

but direct inversion for large matrices is computationally prohibitive or, at best, inefficient.

The objective of Gaussian elimination is to reduce the Ymatrix to upper-right-triangular-plus-

diagonal form (URT+D), then solve for Vvia backward substitution. A series of row operations

(i.e. subtractions and additions) are used to change equation

=

NNNNNN

N

N

N

N V

V

V

V

yyyy

yyyy

yyyy

yyyy

I

I

I

I

M

L

MOMMM

L

L

L

M

3

2

1

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

3

2

1


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into

=

NNN

N

N

N

N V

V

V

V

y

yy

yyy

yyyy

I

I

I

I

M

L

MOMMM

L

L

L

M

3

2

1

,'

,3'

3,3'

,2'

3,2'

2,2'

,13,12,11,1

'

'3

'2

1

000

00

0

,

in which the transformed Ymatrix has zeros under the diagonal.

For illustrative purposes, consider the two equations represented by Rows 1 and 2, which are

NN

NN

VyVyVyVyI

VyVyVyVyI

,233,222,211,22

,133,122,111,11

++++=

++++=

L

L.

Subtracting (1,1

1,2

y

y Row 1) from Row 2 yields

NNN

NN

Vyy

yyVy

y

yyVy

y

yyVy

y

yyI

y

yI

VyVyVyVyI

++

+

+

=

++++=

,11,1

1,2,233,1

1,1

1,23,222,1

1,1

1,22,211,1

1,1

1,21,21

1,1

1,22

,133,122,111,11

L

L.

The coefficient of 1V in Row 2 is forced to zero, leaving Row 2 with the desired "reduced" form

of

NNVyVyVyI ' ,23' 3,22' 2,2'2 0 ++++= L .

Continuing, Row 1 is then used to "zero" the 1V coefficients in Rows 3 through N, one row at a

time. Next, Row 2 is used to zero the 2V coefficients in Rows 3 through N, and so forth.

After the Gaussian elimination is completed, and the Ymatrix is reduced to (URT+D) form, the

bus voltages are solved by backward substitution as follows:

For RowN,

NNNN VyI ' ,' = , so ( )'' ,1

N

NN

N Iy

V = .

Next, for RowN-1,

NNNNNNN VyVyI'

,11'

1,1'

1 += , so ( )NNNNNN

N VyIy

V'

,1'

1'1,1

11

= .


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Continuing for Rowj, where 2,,3,2 L= NNj ,

NNjjjjjjjj VyVyVyI',1

'1,

',

' +++= ++ L , so

( )NNjjjjjjjj VyVyIyV',1

'1,

'

',

1

= ++ L ,

which, in general form, is described by

=

+=kkj

N

jk

j

jj

j VyIy

V ' ,1

'

',

1.

A simple FORTRAN computer program for solving V in an N-dimension problem using

Gaussian elimination and backward substitution is given below.

COMPLEX Y( N, N) , V( N) , I ( N) , YMM

C GAUSSI AN ELI MI NATE Y AND I

NM1 = N - 1

C PI VOT ON ROW M, M = 1, 2, 3, . . . , N- 1

DO 1 M = 1, NM1

MP1 = M + 1

YMM = 1. 0 / Y( M, M)

C OPERATE ON THE ROWS BELOW THE PI VOT ROW

DO 1 J = MP1, N

C THE J TH ROW OF II ( J ) = I ( J ) - Y( J , M) *YMM * I (M)

C THE J TH ROW OF Y, BELOW AND TO THE RI GHT OF THE PI VOT

C DI AGONAL

DO 1 K = M, N

Y( J , K) = Y( J , K) - Y( J , M) * YMM * Y( M, K)

1 CONTI NUE

C BACKWARD SUBSTI TUTE TO SOLVE FOR V

V( N) = I ( N) / Y( N, N)

DO 2 M = 1, NM1

J = N - MC BACKWARD SUBSTI TUTE TO SOLVE FOR V, FOR

C ROW J = N- 1, N- 2, N- 3, . . . , 1

V( J ) = I ( J )

J P1 = J + 1

DO 3 K = J P1, N

V( J ) = V(J ) - Y( J , K) * V( K)


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3 CONTI NUE

V( J ) = V( J ) / Y( J , J )

2 CONTI NUE

STOP

END

One disadvantage of Gaussian elimination is that ifIchanges, even though Yis fixed, the entire

problem must be re-solved since the elimination of Ydetermines the row operations that must be

repeated onI. Inversion and LU decomposition to not have this disadvantage.

3. Kron Reduction

Gaussian elimination can be made more computationally efficient by simply not performing

operations whose results are already known. For example, instead of arithmetically forcing

elements below the diagonal to zero, simply set them to zero at the appropriate times. Similarly,

instead of dividing all elements below and to the right of a diagonal element by the diagonal

element, divide only the elements in the diagonal row by the diagonal element, make the

diagonal element unity, and the same effect will be achieved. This technique, which is actually a

form of Gaussian elimination, is known as Kron reduction.

Kron reduction "pivots" on each diagonal element ' ,mmy , beginning with 1,1y , and continuing

through 1,1 NNy . Starting with Row m = 1, and continuing through Row m = N - 1, the

algorithm for Kron reducing YVI= is

1. Divide the elements in Row m, that are to the right of the diagonal, by the diagonal

element ' ,mmy . (Note - the elements to the left of the diagonal are already zero).

2. Replace element 'mI with ',

'

mm

m

y

I.

3. Replace diagonal element ' ,mmy with unity.

4. Modify the 'Y elements in rows greater than mand columns greater than m (i.e. below

and to the right of the diagonal element) using

',

',

',

', kmmjkjkj yyyy = , forj > m, k > m.

5. Modify the 'I elements below the mthrow according to

'',

''mmjjj IyII = , forj > m.

6. Zero the elements in Column mof 'Y that are below the diagonal element.


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A FORTRAN code for Kron reduction is given below.

COMPLEX Y( N, N) , V( N) , I ( N) , YMM

C KRON REDUCE Y, WHI LE ALSO PERFORMI NG ROW OPERATI ONS ON I

NM1 = N - 1

C PI VOT ON ROW M, M = 1, 2, 3, . . . , N- 1DO 1 M = 1, NM1

MP1 = M + 1

YMM = 1. 0 / YMM

C DI VI DE THE PI VOT ROW BY THE PI VOT

DO 2 K = MP1, N

Y( M, K) = Y( M, K) * YMM

2 CONTI NUE

C OPERATE ON THE I VECTOR

I ( M) = I ( M) * YMM

C SET THE PI VOT TO UNI TYY( M, M) = 1. 0

C REDUCE THOSE ELEMENTS BELOW AND TO THE RI GHT OF THE PI VOT

DO 3 J = MP1, N

DO 4 K = MP1, N

Y( J , K) = Y( J , K) - Y( J , M) * Y( M, K)

4 CONTI NUE

C OPERATE ON THE I VECTOR

I ( J ) = I ( J ) - Y( J , M) * I (M)

C SET THE Y ELEMENTS DI RECTLY BELOW THE PI VOT TO ZERO

Y( J , M) = 0. 0

3 CONTI NUE

1 CONTI NUE

4. LU Decomposition

An efficient method for solving Vin matrix equation YV = Iis to decompose Yinto the product

of a lower-left-triangle-plus-diagonal (LLT+D) matrixL, and an (URT+D) matrix U, so that YV

= Ican be written as

LUV = I.

The benefits of decomposing Ywill become obvious after observing the systematic procedure for

finding V.

It is customary to set the diagonal terms of U equal to unity, so that there are a total of 2N

unknown terms inLand U. LU= Yin expanded form is then


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=

NNNNN

N

N

N

N

N

N

NNNNN yyyy

yyyy

yyyy

yyyy

u

uu

uuu

llll

lll

ll

l

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

,3

,23,2

,13,12,1

,3,2,1,

3,32,31,3

2,21,2

1,1

1000

100

10

1

0

00

000

L

MOMMM

L

L

L

L

MOMMM

L

L

L

L

MOMMM

L

L

L

.

Individual land uterms are easily found by calculating them in the following order:

1. Starting from the top, work down Column 1 ofL, finding 1,1l , then 1,2l , then 1,3l , ,

lN,1. For the special case of Column 1, these elements are 1,1, jj yl = ,j = 1,2,3, ,N.

2. Starting from the left, work across Row 1 of U, finding 2,1u , then 3,1u , then 4,1u , ,

Nu ,1 . For the special case of Row 1, these elements are1,1

,1,1

l

yu

kk = , k = 2,3,4, ,N.

3. Work down Column (k= 2) ofL, finding 2,2l , then 2,3l , then 2,4l , , 2,Nl , using

Nkkkjulyl km

k

m

mjkjkj ,,2,1,,,

1

1

,,, L++==

=

, Column Nkk 2, .

4. Work across Row (k= 2) of U, finding 3,2u , then 4,2u , then 5,2u , , Nu ,2 , using

Nkkjl

uly

ukk

k

mjmmkjk

jk ,,2+,1+=,,

1

1 ,,,

, L

=

= , Row )1(2, Nkk .

5. Repeat Steps 3 and 4, first for Column kofL, then for Row kof U. Continue for all k=

3,4,5, ,(N1) forLand U, then for k=NforL.

The procedure given above in Steps 1 - 5 is often referred to as Crout's method. Note that

elements of L and Unever look "backward" for previously used elements of Y. Therefore, in

order to conserve computer memory,Land Uelements can be stored as they are calculated in the

same locations at the corresponding Yelements. Thus, Crout's method is a memory-efficient "in

situ" procedure.

An intermediate column vector is needed to find V. The intermediate vectorDis defined as

D = UV,

so that


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=

N

N

N

N

N V

V

V

V

u

uu

uuu

d

d

d

d

M

L

MOMMM

L

L

L

M

3

2

1

,3

,23,2

,13,12,1

3

2

1

1000

100

10

1

.

SinceLUV=I, thenLD = I. Vector D is found using forward-substitution from

=

NNNNNNN I

I

I

I

d

d

d

d

llll

lll

ll

l

MM

L

MOMMM

L

L

L

3

2

1

3

2

1

,3,2,1,

3,32,31,3

2,21,2

1,1

0

00

000

,

which proceeds as follows:

From Row 1, ( )11,1

1111,11

, Il

dIdl == ,

From Row 2, ( )11,222,2

2222,211,21

, dlIl

dIdldl ==+ ,

From Row k,

==+++

= jjk

k

jkkkkkkkkkk

dlIl

dIdldldl,

1

1,,22,11,

1,L .

Now, sinceD= UV, or

=

N

N

N

N

N V

V

V

V

u

uu

uuu

d

d

d

d

M

L

MOMMM

L

L

L

M

3

2

1

,3

,23,2

,13,12,1

3

2

1

1000

100

10

1

,

whereDand Uare known, then Vis found using backward substitution.

An important advantage of LU decomposition over Gaussian elimination or Kron reduction is

that theIvector is not modified during decomposition. Therefore, once Yhas been decomposed

intoLand U, Ican be modified, and Vrecalculated, with minimal work, using the forward and

backward substitution steps shown above.


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A special form of Lis helpful when Y is symmetric. In that case, let both Land Uhave unity

diagonal terms, and define a diagonal matrixDso that

Y = LDU,

or

=

1000

100

10

1

000

000

000

000

1

0

01

001

0001

,3

,23,2

,13,12,1

,

3,3

2,2

1,1

3,2,1,

2,31,3

1,2

L

MOMMM

L

L

L

L

MOMMM

L

L

L

L

OMMM

L

L

L

N

N

N

NNNNN

u

uu

uuu

d

d

d

d

lll

ll

l

Y .

Since Yis symmetric, then TYY= , and [ ] TTTTTT DLULDULDULDU === . Therefore,

an acceptable solution is to allow

T

UL=

. Incorporating this into the above equation yields

=

NN

N

N

N

NNN d

ldd

ldldd

ldldldd

lll

ll

l

Y

,

3,3,33,3

2,2,22,32,22,2

1,1,11,31,11,21,11,1

3,2,1,

2,31,3

1,2

000

00

0

1

0

01

001

0001

L

MOMMM

L

L

L

L

OMMM

L

L

L

,

which can be solved by working from top-to-bottom, beginning with Column 1 of Y, as follows:

Working down Column 1 of Y,

1,11,1 dy = ,

1,11,21,21,11,21,2 /so, dyldly == ,

1,11,1,1,11,1, /so, dyldly jjjj == .

Working down Column 2 of Y,

1,21,11,22,22,22,21,21,11,22,2so, ldlyddldly =+= ,

Njdldlyldlldly jjjjjj


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mkmm

k

m

mkkkkkkkmkmm

k

m

mkkk ldlyddldly ,,

1

1,,,,,,

1

1,, so,

=

=

=+= ,

Njkdldlylldlykkmkmm

k

m mjkjkjmkmm

k

m mjkj

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05 EE394J 2 Spring12 Power System Matrices