Electrical Power Systems

228 Electrical Power Systems
Let us express equs. (9.5), (9.6) and (9.7) in terms of reference phasors Va1 , Va2 and Va0 .
Thus
Va = Va1 + Va2 + Va0 ...(9.8)
Vb = b 2 Va1 + b Va2 + Va0 ...(9.9)
Vc = bVa1 + b 2 Va2 + Va0 ...(9.10)
Eqns. (9.8), (9.9) and (9.10) can be written in matrix form
or
where
L
NM
a
b
c
O
Q
V
V
V P =
L
NM
V p = AV s
O
QP
1 1 1
2
b b 1
b
2
b 1
V p = [V a V b V c] T
L
NM
V s = [V a1 V a2 V a0] T
A =
L
NM
O
QP
1 1 1
2
b b 1
b
2
b 1
rom eqn. (9.12)
Vs = A –1 Vp The inverse of A is given by
A –1 = 1
3
1 b b
1
2
b b
1 1 1
Va1
Va2
Va0
Complex conjugate of eqn. (9.13) can be given as:
A* =
or (A*) T =
L
NM
L
NM
L1
1
NM
1 1 1
b
2
b 1
2
b b 1
2
b b
2
b b
1 1 1
Using eqn. (9.15) and (9.16) we get
e j T
A –1 = 1
3 A*
2O
QP
O
QP
O
QP
O
QP
...(9.11)
...(9.12)
...(9.13)
...(9.14)
...(9.15)
...(9.16)
...(9.17)