1.
2 Bus Admittance Matrix
The bus admittance matrix Ybus can be obtained from both Ohm's law and
Kirchhoff current law applied to each circuit bus other than the reference
bus.
1.2.1 Ybus Formulation:
Let bus j be an arbitrary bus of a power system network, from KCL
I
iI
ji Ij (1)
where
Ij is the current injected from an external source into the system,
Iji is the current leaving bus j to bus i via a line in the network, and
n is the number of buses in the power network.
Note that
Ijj = 0, since there is no line between bus and itself, and
Iji = 0, if there is no line exist between bus j and bus i .
From Ohm's law,
Iji = yji (Vj - Vi) (2)
where
yji is the line admittance of a line from bus j to bus i, and
Vj and Vi are the bus voltages with respect to the reference bus.
Note that all these variables are complex.
Substituting Equation (2) into Equation (1), yeilds
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n
y
i 1
ji (V j Vi ) I j (3)
The n equations for all buses in the network can be written in a matrix
form as
n
y1i y12 . . y1n
ii 11
I1 n V1
I y V
2 21 y i 1
2i . . y2 n
2
. i2
. (4)
. .
. .
I n . . Vn
n
y yni
n1 yn 2 . .
i 1
in
Equation (4) can be written compactly in a matrix form as
Ibus = Ybus Vbus (5)
where
Ibus is an n x 1 bus injection current vector,
Ybus is an n x n bus admittance matrix, and
Vbus is an n x 1 bus voltage vector.
Notice that diagonal elements may include lines from the bus to the
reference (like capacitor banks or pi-model of medium transmission lines).
1.2.2 Ybus Building Algorithm:
Inspection of the coefficients of the V's in Equation (4) reveals a rule for
building the Ybus;
1) The diagonal entries of Ybus are found from summing the primitive
admittance of lines (and ties to the reference) that are connected directly to
bus j.
2) The off diagonal entries of Ybus are found from the negatives of the
primitive admittances of lines between buses j and i.
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Example (1):
Consider the 3-bus network shown below; all admittances are in per units.
(Notice the availability of the admittance between bus no. 3 and the
ground).
Find the Ybus matrix, using the ground bus as the reference.
Solution:
Y11 = y12 + y13 = 1 + 1 = 2
Y22 = y21 + y23 = 1 + 1 = 2
Y33 = y31 + y32 + y30 = 1 + 1 + 1 = 3
Y12 = - y21 = -1
Y13 = - y31 = -1
Y23 = - y32 = -1
Therefore,
2 1 1
Ybus 1 2 1
1 1 3
1.2.3 Properties of Ybus:
The matrix is symmetric.
The matrix is sparse (i.e., few non-zeros).
1.2.4 Conditioning of Ybus :
Ybus is ill-conditioned if the network has bad connectivity (e.g., one or more
lines have very small admittances). In this case the solution of Eq. (5) is
more sensitive to changes in its data than a well-conditioned network. In
many application a matrix is considered ill-conditioned when the condition
number is larger than 100. It is the largest eigenvalue over the smallest
eigenvalue of the matrix.
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Example (2):
Consider the 3-bus network shown below; all admittances are in per units.
In this case,
2 1 1
Ybus 1 2 1
1 1 2.01
The condition number of this matrix is 904 (the condition number of
previous matrix is 14). To see the effect of conditionings on Eq. (5), let us
consider that
1
I bus 1
1
then
301
Vbus Y 1
bus I bus 301
300
Whereas, the voltage bus of previous example is
4
Vbus 4
3
1.2.5 Changes in Ybus Due to Changes in the System Network:
In practical applications, the power system network configuration
(topology) may be modified to reflect changes in the system or changes in
the choice of a reference bus. Changes in the system could be due to one of
the followings:
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1) Addition of a line: to increase the capacity of energy transmission, to
improve system performance, ... etc. In this case, the size of the existing Ybus
remains the same, but the nonzero entries will change. Such changes take
place at the diagonal and off-diagonal elements of buses that are connected
directly to the new added line.
The procedure to add ynew between bus i and bus j is
Diagonal entries: Add ynew to the i, i and j, j entries.
Off-diagonal entries: Add -ynew to the i, j and j, i entries.
Example (3):
Consider the 3-bus network shown below; all admittances are in per units.
In this example a line with admittance of 0.89 per unit is added between
bus 3 and the reference bus to improve the conditioning of the system. In
this case the modified admittance matrix is
2 1 1
Ybus 1 2 1
1 1 2.90
The condition number is improved to 14.8, and the solution to Eq. (5) is
4.333
Vbus 4.333
3.333
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2) Addition of a bus: to produce more energy, to supply a new load, ... etc.
In this case the size of the existing Ybus increases and the nonzero entries
will change. Such changes take place at the diagonal and off-diagonal
elements of the new added bus and the existing bus.
The procedure to add a new bus to an existing bus i through ynew is
Diagonal entries: Add ynew to the i, i and, n +1, n + 1 entries.
Off diagonal entries: Add -ynew to the i, n + 1 and n + 1, i entries.
Example (4):
Consider the 3-bus network shown below; all admittances are in per units.
In this example bus 4 is added to the system through a line with admittance
of 2 per unit. In this case the modified admittance matrix is
2 1 1 0
1 4 1 2
Ybus
1 1 3 0
0 2 0 2
3) Line outages: for maintenance, due to a short circuit, ... etc. In this case
the size of the existing remains the same, but the nonzero entries will
change. Such changes take place at the diagonal and off-diagonal elements
of buses that are connected directly to the deleted line.
The procedure to delete yout between bus i and bus j is
Diagonal entries: Add -yout to the i, i and j, j entries.
Off diagonal entries: Add yout to the i, j and j, i entries.
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Example (5):
Consider the 3-bus network shown below; all admittances are in per units.
In this example the line connecting bus 2 and bus 3 is removed. In this case
the modified admittance matrix is
2 1 1
Ybus 1 1 0
1 0 2
4) Deletion of a bus: for maintenance, due to a short circuit, ... etc. In this
case the size of the existing Ybus decreases and the nonzero entries will
change. Such changes may take place at all entries.
There are two methods for the modification:
a) Matrix Partitioning Method (Kron Reduction): In this method we move
all buses to be removed to the lower rows of the existing Ybus matrix, and
then partition the matrix accordingly, i.e.,
K L
Ybus t
M
(6)
L
where
K is an (n – m) x ( n – m) submatrix (to be remained),
M is an m x m submatrix (to be removed),
L is an (n – m) x m submatrix,
Lt is the transpose of L,
n is the size of Ybus, and
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m is the number of buses to be deleted.
Then the reduced (or equivalent) admittance matrix is
eq
Ybus K LM 1 Lt (7)
This method requires inversion of the submatrix M , a step which is not
recommended because the time required for inversion is high.
b) Bus Elimination Method: In this method, the buses are eliminated one at
a time. This method can be performed more efficiently when applied with a
technique known as sparsity programming. The advantage of using
sparsity is that only nonzero entries of the reduced admittance matrix is
processed. This method starts by moving one of the buses to be removed to
the nth row (i.e., the last row) of the original matrix. Then the new entry in
row i and column j of the resulting (n – 1) x (n – 1) matrix will be
Yin Ynj
Yij ( new) Yij ( orig) (8)
Ynn
After modifying all entries, we select another bus to be removed and repeat
the same steps until all buses to be removed are completed.
Example (6):
Consider the 5-bus network shown below; all admittances are in per units.
In this example bus 4 is to be removed. The Ybus of this network is
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2 1 1 0 0
1 4 1 2 0
Ybus 1 1 3 0 0
0 2 0 4 2
0 0 0 2 2
Ybus after moving bus 4 to the last row is
2 1 1 0 0
1 4 1 0 2
record
Ybus 1 1 3 0 0
0 0 0 2 2
0 2 0 2 4
1) removing of bus 4 using method 1:
2 1 1 0 0 2 1 1 0
1 4 1 0 2 1 3 1 1
eq
Ybus [4]1 0 2 0 2
1 1 3 0 0 1 1 3 0
0 0 0 2 2 0 1 0 1
2) removing of bus 4 using method 2:
2 1 1 0 0
1 4 1 2 0
Ybus 1 1 3 0 0
0 2 0 4 2
0 0 0 2 2
Y14 Y41 (0) (0)
Y11( new) Y11( orig) 2 2
Y44 4
Y54 Y45 (2) (2)
Y55( new) Y55( orig) 2 1
Y44 4
Y Y (0) (2)
Y12( new) Y21( new) Y12( orig) 14 42 1 1
Y44 4
Y14 Y45 (0) (2)
Y15( new) Y51( new) Y15( orig) 0 0
Y44 4
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Continue with all entries, the final equivalent bus will be
2 1 1 0
1 3 1 1
eq
Ybus
1 1 3 0
0 1 0 1
Notice that the only entries which need to be modified are the diagonal
elements and either the upper or lower off-diagonal elements.
Also notice that the diagonal element Y55 will be read as Y44 in the
modified bus matrix.
The off-diagonal elements Y15 , Y25 , and Y35 will be read as Y14 , Y24 ,
and Y34 in the modified bus matrix.
Changes in the system could also be due to changes in the choice of a
reference bus. Such changes could take place to improve the computational
performance, but involves no alteration or modification of the system
configuration.
To convert the reference bus from bus a to bus b, form the indefinite bus
admittance matrix by augmenting bus a as a new axis in the existing Ybus
matrix, and then delete bus b from the indefinite bus admittance matrix .
Example (7):
Consider the 3-bus network shown below; all admittances are in per units.
Change the reference bus to bus 2.
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The changes can be done by forming the indefinite bus matrix, i.e.
2 1 1 0
1 2 1 0
indef
Ybus
1 1 3 1
0 0 1 1
Then delete bus 2, yeilds
2 1 0
Ybus 1 3 1
0 1 1
References:
[1] G.T. Heydt, "Computer Analysis Methods for Power System",
Macmillan, New York, 1986.
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