Small signal analysis - Notes, Electronics Engineering, Semester Notes

 Page 1


6.4 Small signal analysis
For small signal analysis of a multimachine power system, we need to linearise the di?erential equa-
tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of
the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system
having ‘m’ generators. Each generator is represented by its classical model. Further, without any
loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system.
Now, for linearising the di?erential equations, let us recall the swing equations of each generator
here for ready reference.
dd
i
dt
=?
i
-?
s
for i= 1, 2,??m (6.64)
2H
i
?
s
d?
i
dt
=P
mi
-P
ei
for i= 1, 2,??m (6.65)
Linearising equation (6.64) for thei
th
generator, one can get,
d?d
i
dt
= ??
i
(6.66)
Now, let us de?ne,
?d = [?d
1
, ?d
2
, ?? ?d
m
]
T
(6.67)
?? = [??
1
, ??
2
, ?? ??
m
]
T
(6.68)
In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values
of rotor angle and machine speed respectively. Please note that the size of each of these two vectors
is (m× 1).
Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
d
dt
?d =F
1
?? (6.69)
In equation (6.69),F
1
is a (m×m) identity matrix. Now, linearising equation (6.65) for thei
th
generator, one can get,
2H
i
?
s
d??
i
dt
=-?P
ei
(6.70)
For performing linearisation ofP
ei
, its expression its required. This expression can be derived as
follows. From equation (6.25), one can write,
¯
I
i
=
E
i
?d
i
-V
i
??
i
jx
/
di
=
E
i
x
/
di
e
j(d
i
-p~2)
-
V
i
x
/
di
e
j(?
i
-p~2)
278
Page 2


6.4 Small signal analysis
For small signal analysis of a multimachine power system, we need to linearise the di?erential equa-
tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of
the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system
having ‘m’ generators. Each generator is represented by its classical model. Further, without any
loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system.
Now, for linearising the di?erential equations, let us recall the swing equations of each generator
here for ready reference.
dd
i
dt
=?
i
-?
s
for i= 1, 2,??m (6.64)
2H
i
?
s
d?
i
dt
=P
mi
-P
ei
for i= 1, 2,??m (6.65)
Linearising equation (6.64) for thei
th
generator, one can get,
d?d
i
dt
= ??
i
(6.66)
Now, let us de?ne,
?d = [?d
1
, ?d
2
, ?? ?d
m
]
T
(6.67)
?? = [??
1
, ??
2
, ?? ??
m
]
T
(6.68)
In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values
of rotor angle and machine speed respectively. Please note that the size of each of these two vectors
is (m× 1).
Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
d
dt
?d =F
1
?? (6.69)
In equation (6.69),F
1
is a (m×m) identity matrix. Now, linearising equation (6.65) for thei
th
generator, one can get,
2H
i
?
s
d??
i
dt
=-?P
ei
(6.70)
For performing linearisation ofP
ei
, its expression its required. This expression can be derived as
follows. From equation (6.25), one can write,
¯
I
i
=
E
i
?d
i
-V
i
??
i
jx
/
di
=
E
i
x
/
di
e
j(d
i
-p~2)
-
V
i
x
/
di
e
j(?
i
-p~2)
278
Or,
¯
E
i
¯
I
*
i
=E
i
e
jd
i ¯
I
*
i
=
E
i
x
/
di
e
jp~2
-
E
i
V
i
x
/
di
e
j(p~2+d
i
-?
i
)
Therefore,
P
ei
= Re?
¯
E
I
¯
I
*
i
?=
E
i
V
i
x
/
di
sin(d
i
-?
i
) (6.71)
Linearisation of equation (6.71) yields,
?P
ei
=k
1i
?d
i
+k
2i
??
i
+k
3i
?V
i
(6.72)
Where,
k
1i
=
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
2i
=-
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
3i
=
E
i
x
/
di
sin(d
i
-?
i
); (6.73)
In equation (6.73), the constantsk
1i
,k
2i
andk
3i
are evaluated using the values ofE
i
,V
i
,d
i
and
?
i
at the current operating point fori= 1, 2,??m.
Now, again let us de?ne,
??
g
= [??
1
, ??
2
, ?? ??
m
]
T
(6.74)
?V
g
= [?V
1
, ?V
2
, ?? ?V
m
]
T
(6.75)
?P
e
= [?P
e1
, ?P
e2
, ?? ?P
em
]
T
(6.76)
In equations (6.74) and (6.75), the vectors ??
g
and ?V
g
denote the vectors of perturbed values
of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P
e
denotes
the perturbed values of the generator real powers. Please note that the size of each of these three
vectors is also (m× 1).
Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
?P
e
=K
1
?d+K
2
??
g
+K
3
?V
g
(6.77)
In equation (6.77), the size of each of the matricesK
1
,K
2
andK
3
is (m×m). Moreover, all
these three matrices are diagonal matrices and are given by;
K
1
= diag(k
11
, k
12
, ??k
1m
)
K
2
= diag(k
21
, k
22
, ??k
2m
)
K
3
= diag(k
31
, k
32
, ??k
3m
)
?
?
?
?
?
?
?
?
?
?
?
(6.78)
279
Page 3


6.4 Small signal analysis
For small signal analysis of a multimachine power system, we need to linearise the di?erential equa-
tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of
the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system
having ‘m’ generators. Each generator is represented by its classical model. Further, without any
loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system.
Now, for linearising the di?erential equations, let us recall the swing equations of each generator
here for ready reference.
dd
i
dt
=?
i
-?
s
for i= 1, 2,??m (6.64)
2H
i
?
s
d?
i
dt
=P
mi
-P
ei
for i= 1, 2,??m (6.65)
Linearising equation (6.64) for thei
th
generator, one can get,
d?d
i
dt
= ??
i
(6.66)
Now, let us de?ne,
?d = [?d
1
, ?d
2
, ?? ?d
m
]
T
(6.67)
?? = [??
1
, ??
2
, ?? ??
m
]
T
(6.68)
In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values
of rotor angle and machine speed respectively. Please note that the size of each of these two vectors
is (m× 1).
Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
d
dt
?d =F
1
?? (6.69)
In equation (6.69),F
1
is a (m×m) identity matrix. Now, linearising equation (6.65) for thei
th
generator, one can get,
2H
i
?
s
d??
i
dt
=-?P
ei
(6.70)
For performing linearisation ofP
ei
, its expression its required. This expression can be derived as
follows. From equation (6.25), one can write,
¯
I
i
=
E
i
?d
i
-V
i
??
i
jx
/
di
=
E
i
x
/
di
e
j(d
i
-p~2)
-
V
i
x
/
di
e
j(?
i
-p~2)
278
Or,
¯
E
i
¯
I
*
i
=E
i
e
jd
i ¯
I
*
i
=
E
i
x
/
di
e
jp~2
-
E
i
V
i
x
/
di
e
j(p~2+d
i
-?
i
)
Therefore,
P
ei
= Re?
¯
E
I
¯
I
*
i
?=
E
i
V
i
x
/
di
sin(d
i
-?
i
) (6.71)
Linearisation of equation (6.71) yields,
?P
ei
=k
1i
?d
i
+k
2i
??
i
+k
3i
?V
i
(6.72)
Where,
k
1i
=
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
2i
=-
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
3i
=
E
i
x
/
di
sin(d
i
-?
i
); (6.73)
In equation (6.73), the constantsk
1i
,k
2i
andk
3i
are evaluated using the values ofE
i
,V
i
,d
i
and
?
i
at the current operating point fori= 1, 2,??m.
Now, again let us de?ne,
??
g
= [??
1
, ??
2
, ?? ??
m
]
T
(6.74)
?V
g
= [?V
1
, ?V
2
, ?? ?V
m
]
T
(6.75)
?P
e
= [?P
e1
, ?P
e2
, ?? ?P
em
]
T
(6.76)
In equations (6.74) and (6.75), the vectors ??
g
and ?V
g
denote the vectors of perturbed values
of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P
e
denotes
the perturbed values of the generator real powers. Please note that the size of each of these three
vectors is also (m× 1).
Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
?P
e
=K
1
?d+K
2
??
g
+K
3
?V
g
(6.77)
In equation (6.77), the size of each of the matricesK
1
,K
2
andK
3
is (m×m). Moreover, all
these three matrices are diagonal matrices and are given by;
K
1
= diag(k
11
, k
12
, ??k
1m
)
K
2
= diag(k
21
, k
22
, ??k
2m
)
K
3
= diag(k
31
, k
32
, ??k
3m
)
?
?
?
?
?
?
?
?
?
?
?
(6.78)
279
6.4.1 Linearisation of network equations (at the generator buses)
To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst
equation of the equation set (6.23) below.
?
¯
Y
11
+ ¯ y
1
?
¯
V
1
+
¯
Y
12
¯
V
2
+?+
¯
Y
1m
¯
V
m
+?+Y
1n
¯
V
n
= ¯ y
1
¯
E
1
(6.79)
Now, let,
¯
Y
ij
=Y
ij
exp(ja
ij
);
¯
V
i
=V
i
exp(j?
i
); for i, j = 1, 2,??n; (6.80)
and
¯ y
k
=y
k
exp(jß
k
);
¯
E
k
=E
k
exp(jd
k
); for k = 1, 2,??m; (6.81)
Utilising equations (6.80) and (6.81) in equation (6.79), we get,
Y
11
V
1
exp(j(a
11
+?
1
))+y
1
V
1
exp(j(ß
1
+?
1
))+Y
12
V
2
exp(j(a
12
+?
2
))+?+
Y
1m
V
m
exp(j(a
1m
+?
m
))+?+Y
1n
V
n
exp(j(a
1n
+?
n
))=y
1
E
1
exp(j(d
1
+ß
1
))
(6.82)
Taking the real part of both the sides of equation (6.82), we get,
Y
11
V
1
cos(a
11
+?
1
)+y
1
V
1
cos(ß
1
+?
1
)+Y
12
V
2
cos(a
12
+?
2
)+?+
Y
1m
V
m
cos(a
1m
+?
m
)+?+Y
1n
V
n
cos(a
1n
+?
n
)=y
1
E
1
cos(d
1
+ß
1
)
(6.83)
Linearising equation (6.83), we get,
Y
11
cos(a
11
+?
1
)?V
1
-Y
11
V
1
sin(a
11
+?
1
)??
1
+y
1
cos(ß
1
+?
1
)?V
1
-
y
1
V
1
sin(ß
1
+?
1
)??
1
+Y
12
cos(a
12
+?
2
)?V
2
-Y
12
V
2
sin(a
12
+?
2
)??
2
+?+Y
1m
cos(a
1m
+?
m
)?V
m
-Y
1m
V
m
sin(a
1m
+?
m
)??
m
+?+
Y
1n
cos(a
1n
+?
n
)?V
n
-Y
1n
V
n
sin(a
1n
+?
n
)??
n
=-y
1
E
1
sin(d
1
+ß
1
)?d
1
(6.84)
Equation (6.84) can be re-written as,
a
11
?V
1
+a
12
?V
2
+?+a
1m
?V
m
+?+a
1n
?V
n
+
b
11
??
1
+b
12
??
2
+?+b
1m
??
m
+?+b
1n
??
n
=g
1
?d
1
(6.85)
In equation (6.85),
a
11
=Y
11
cos(a
11
+?
1
)+y
1
cos(ß
1
+?
1
); a
12
=Y
12
cos(a
12
+?
2
);
a
1m
=Y
1m
cos(a
1m
+?
m
); a
1n
=Y
1n
cos(a
1n
+?
n
);
b
11
=-Y
11
V
1
sin(a
11
+?
1
)-y
1
V
1
sin(ß
1
+?
1
); b
12
=-Y
12
V
2
sin(a
12
+?
2
);
b
1m
=-Y
1m
V
m
sin(a
1m
+?
m
); b
1n
=-Y
1n
V
n
sin(a
1n
+?
n
);
g
1
=-y
1
E
1
sin(d
1
+ß
1
);
(6.86)
Again, taking the real part of both sides of the second equation of the equation set (6.23) (after
280
Page 4


6.4 Small signal analysis
For small signal analysis of a multimachine power system, we need to linearise the di?erential equa-
tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of
the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system
having ‘m’ generators. Each generator is represented by its classical model. Further, without any
loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system.
Now, for linearising the di?erential equations, let us recall the swing equations of each generator
here for ready reference.
dd
i
dt
=?
i
-?
s
for i= 1, 2,??m (6.64)
2H
i
?
s
d?
i
dt
=P
mi
-P
ei
for i= 1, 2,??m (6.65)
Linearising equation (6.64) for thei
th
generator, one can get,
d?d
i
dt
= ??
i
(6.66)
Now, let us de?ne,
?d = [?d
1
, ?d
2
, ?? ?d
m
]
T
(6.67)
?? = [??
1
, ??
2
, ?? ??
m
]
T
(6.68)
In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values
of rotor angle and machine speed respectively. Please note that the size of each of these two vectors
is (m× 1).
Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
d
dt
?d =F
1
?? (6.69)
In equation (6.69),F
1
is a (m×m) identity matrix. Now, linearising equation (6.65) for thei
th
generator, one can get,
2H
i
?
s
d??
i
dt
=-?P
ei
(6.70)
For performing linearisation ofP
ei
, its expression its required. This expression can be derived as
follows. From equation (6.25), one can write,
¯
I
i
=
E
i
?d
i
-V
i
??
i
jx
/
di
=
E
i
x
/
di
e
j(d
i
-p~2)
-
V
i
x
/
di
e
j(?
i
-p~2)
278
Or,
¯
E
i
¯
I
*
i
=E
i
e
jd
i ¯
I
*
i
=
E
i
x
/
di
e
jp~2
-
E
i
V
i
x
/
di
e
j(p~2+d
i
-?
i
)
Therefore,
P
ei
= Re?
¯
E
I
¯
I
*
i
?=
E
i
V
i
x
/
di
sin(d
i
-?
i
) (6.71)
Linearisation of equation (6.71) yields,
?P
ei
=k
1i
?d
i
+k
2i
??
i
+k
3i
?V
i
(6.72)
Where,
k
1i
=
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
2i
=-
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
3i
=
E
i
x
/
di
sin(d
i
-?
i
); (6.73)
In equation (6.73), the constantsk
1i
,k
2i
andk
3i
are evaluated using the values ofE
i
,V
i
,d
i
and
?
i
at the current operating point fori= 1, 2,??m.
Now, again let us de?ne,
??
g
= [??
1
, ??
2
, ?? ??
m
]
T
(6.74)
?V
g
= [?V
1
, ?V
2
, ?? ?V
m
]
T
(6.75)
?P
e
= [?P
e1
, ?P
e2
, ?? ?P
em
]
T
(6.76)
In equations (6.74) and (6.75), the vectors ??
g
and ?V
g
denote the vectors of perturbed values
of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P
e
denotes
the perturbed values of the generator real powers. Please note that the size of each of these three
vectors is also (m× 1).
Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
?P
e
=K
1
?d+K
2
??
g
+K
3
?V
g
(6.77)
In equation (6.77), the size of each of the matricesK
1
,K
2
andK
3
is (m×m). Moreover, all
these three matrices are diagonal matrices and are given by;
K
1
= diag(k
11
, k
12
, ??k
1m
)
K
2
= diag(k
21
, k
22
, ??k
2m
)
K
3
= diag(k
31
, k
32
, ??k
3m
)
?
?
?
?
?
?
?
?
?
?
?
(6.78)
279
6.4.1 Linearisation of network equations (at the generator buses)
To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst
equation of the equation set (6.23) below.
?
¯
Y
11
+ ¯ y
1
?
¯
V
1
+
¯
Y
12
¯
V
2
+?+
¯
Y
1m
¯
V
m
+?+Y
1n
¯
V
n
= ¯ y
1
¯
E
1
(6.79)
Now, let,
¯
Y
ij
=Y
ij
exp(ja
ij
);
¯
V
i
=V
i
exp(j?
i
); for i, j = 1, 2,??n; (6.80)
and
¯ y
k
=y
k
exp(jß
k
);
¯
E
k
=E
k
exp(jd
k
); for k = 1, 2,??m; (6.81)
Utilising equations (6.80) and (6.81) in equation (6.79), we get,
Y
11
V
1
exp(j(a
11
+?
1
))+y
1
V
1
exp(j(ß
1
+?
1
))+Y
12
V
2
exp(j(a
12
+?
2
))+?+
Y
1m
V
m
exp(j(a
1m
+?
m
))+?+Y
1n
V
n
exp(j(a
1n
+?
n
))=y
1
E
1
exp(j(d
1
+ß
1
))
(6.82)
Taking the real part of both the sides of equation (6.82), we get,
Y
11
V
1
cos(a
11
+?
1
)+y
1
V
1
cos(ß
1
+?
1
)+Y
12
V
2
cos(a
12
+?
2
)+?+
Y
1m
V
m
cos(a
1m
+?
m
)+?+Y
1n
V
n
cos(a
1n
+?
n
)=y
1
E
1
cos(d
1
+ß
1
)
(6.83)
Linearising equation (6.83), we get,
Y
11
cos(a
11
+?
1
)?V
1
-Y
11
V
1
sin(a
11
+?
1
)??
1
+y
1
cos(ß
1
+?
1
)?V
1
-
y
1
V
1
sin(ß
1
+?
1
)??
1
+Y
12
cos(a
12
+?
2
)?V
2
-Y
12
V
2
sin(a
12
+?
2
)??
2
+?+Y
1m
cos(a
1m
+?
m
)?V
m
-Y
1m
V
m
sin(a
1m
+?
m
)??
m
+?+
Y
1n
cos(a
1n
+?
n
)?V
n
-Y
1n
V
n
sin(a
1n
+?
n
)??
n
=-y
1
E
1
sin(d
1
+ß
1
)?d
1
(6.84)
Equation (6.84) can be re-written as,
a
11
?V
1
+a
12
?V
2
+?+a
1m
?V
m
+?+a
1n
?V
n
+
b
11
??
1
+b
12
??
2
+?+b
1m
??
m
+?+b
1n
??
n
=g
1
?d
1
(6.85)
In equation (6.85),
a
11
=Y
11
cos(a
11
+?
1
)+y
1
cos(ß
1
+?
1
); a
12
=Y
12
cos(a
12
+?
2
);
a
1m
=Y
1m
cos(a
1m
+?
m
); a
1n
=Y
1n
cos(a
1n
+?
n
);
b
11
=-Y
11
V
1
sin(a
11
+?
1
)-y
1
V
1
sin(ß
1
+?
1
); b
12
=-Y
12
V
2
sin(a
12
+?
2
);
b
1m
=-Y
1m
V
m
sin(a
1m
+?
m
); b
1n
=-Y
1n
V
n
sin(a
1n
+?
n
);
g
1
=-y
1
E
1
sin(d
1
+ß
1
);
(6.86)
Again, taking the real part of both sides of the second equation of the equation set (6.23) (after
280
substituting equations (6.80) and (6.81) into it), we get,
Y
21
V
1
cos(a
21
+?
1
)+y
2
V
2
cos(ß
2
+?
2
)+Y
22
V
2
cos(a
22
+?
2
)+?+
Y
2m
V
m
cos(a
2m
+?
m
)+?+Y
2n
V
n
cos(a
2n
+?
n
)=y
2
E
2
cos(d
2
+ß
2
)
(6.87)
Linearising equation (6.87), one can get,
a
21
?V
1
+a
22
?V
2
+?+a
2m
?V
m
+?+a
2n
?V
n
+
b
21
??
1
+b
22
??
2
+?+b
2m
??
m
+?+b
2n
??
n
=g
2
?d
2
(6.88)
In equation (6.88),
a
21
=Y
21
cos(a
21
+?
1
); a
22
=Y
22
cos(a
22
+?
2
)+y
2
cos(ß
2
+?
2
);
a
2m
=Y
2m
cos(a
2m
+?
m
); a
2n
=Y
2n
cos(a
2n
+?
n
);
b
21
=-Y
21
V
1
sin(a
21
+?
1
); b
22
=-Y
22
V
2
sin(a
22
+?
2
)-y
2
V
2
sin(ß
2
+?
2
);
b
2m
=-Y
2m
V
m
sin(a
2m
+?
m
); b
2n
=-Y
2n
V
n
sin(a
2n
+?
n
);
g
2
=-y
2
E
2
sin(d
2
+ß
2
);
(6.89)
Similarly, continuing with linearisation of the real parts of ?rst ‘m’ equations (corresponding to
the generator buses) of the equation set (6.23), we get,

A
1
B
1

?
?
?
?
?
?
?
?
?
?
?
?V
g
?V
L
??
g
??
L
?
?
?
?
?
?
?
?
?
?
?
= 
G
?d (6.90)
In equation (6.90),
A
1
=
?
?
?
?
?
?
?
?
?
?
?
?
?
a
11
a
12
? a
1m
a
1,(m+1)
?? a
1n
a
21
a
22
? a
2m
a
2,(m+1)
?? a
2n
? ? ? ? ? ? ?
? ? ? ? ? ? ?
a
m1
a
m2
? a
mm
a
m,(m+1)
?? a
mn
?
?
?
?
?
?
?
?
?
?
?
?
?
? is a (m×n) matrix (6.91)
B
1
=
?
?
?
?
?
?
?
?
?
?
?
?
?
b
11
b
12
? b
1m
b
1,(m+1)
?? b
1n
b
21
b
22
? b
2m
b
2,(m+1)
?? b
2n
? ? ? ? ? ? ?
? ? ? ? ? ? ?
b
m1
b
m2
? b
mm
b
m,(m+1)
?? b
mn
?
?
?
?
?
?
?
?
?
?
?
?
?
? is a (m×n) matrix (6.92)
281
Page 5


6.4 Small signal analysis
For small signal analysis of a multimachine power system, we need to linearise the di?erential equa-
tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of
the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system
having ‘m’ generators. Each generator is represented by its classical model. Further, without any
loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system.
Now, for linearising the di?erential equations, let us recall the swing equations of each generator
here for ready reference.
dd
i
dt
=?
i
-?
s
for i= 1, 2,??m (6.64)
2H
i
?
s
d?
i
dt
=P
mi
-P
ei
for i= 1, 2,??m (6.65)
Linearising equation (6.64) for thei
th
generator, one can get,
d?d
i
dt
= ??
i
(6.66)
Now, let us de?ne,
?d = [?d
1
, ?d
2
, ?? ?d
m
]
T
(6.67)
?? = [??
1
, ??
2
, ?? ??
m
]
T
(6.68)
In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values
of rotor angle and machine speed respectively. Please note that the size of each of these two vectors
is (m× 1).
Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
d
dt
?d =F
1
?? (6.69)
In equation (6.69),F
1
is a (m×m) identity matrix. Now, linearising equation (6.65) for thei
th
generator, one can get,
2H
i
?
s
d??
i
dt
=-?P
ei
(6.70)
For performing linearisation ofP
ei
, its expression its required. This expression can be derived as
follows. From equation (6.25), one can write,
¯
I
i
=
E
i
?d
i
-V
i
??
i
jx
/
di
=
E
i
x
/
di
e
j(d
i
-p~2)
-
V
i
x
/
di
e
j(?
i
-p~2)
278
Or,
¯
E
i
¯
I
*
i
=E
i
e
jd
i ¯
I
*
i
=
E
i
x
/
di
e
jp~2
-
E
i
V
i
x
/
di
e
j(p~2+d
i
-?
i
)
Therefore,
P
ei
= Re?
¯
E
I
¯
I
*
i
?=
E
i
V
i
x
/
di
sin(d
i
-?
i
) (6.71)
Linearisation of equation (6.71) yields,
?P
ei
=k
1i
?d
i
+k
2i
??
i
+k
3i
?V
i
(6.72)
Where,
k
1i
=
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
2i
=-
E
i
V
i
x
/
di
cos(d
i
-?
i
); k
3i
=
E
i
x
/
di
sin(d
i
-?
i
); (6.73)
In equation (6.73), the constantsk
1i
,k
2i
andk
3i
are evaluated using the values ofE
i
,V
i
,d
i
and
?
i
at the current operating point fori= 1, 2,??m.
Now, again let us de?ne,
??
g
= [??
1
, ??
2
, ?? ??
m
]
T
(6.74)
?V
g
= [?V
1
, ?V
2
, ?? ?V
m
]
T
(6.75)
?P
e
= [?P
e1
, ?P
e2
, ?? ?P
em
]
T
(6.76)
In equations (6.74) and (6.75), the vectors ??
g
and ?V
g
denote the vectors of perturbed values
of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P
e
denotes
the perturbed values of the generator real powers. Please note that the size of each of these three
vectors is also (m× 1).
Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notation
we can obtain,
?P
e
=K
1
?d+K
2
??
g
+K
3
?V
g
(6.77)
In equation (6.77), the size of each of the matricesK
1
,K
2
andK
3
is (m×m). Moreover, all
these three matrices are diagonal matrices and are given by;
K
1
= diag(k
11
, k
12
, ??k
1m
)
K
2
= diag(k
21
, k
22
, ??k
2m
)
K
3
= diag(k
31
, k
32
, ??k
3m
)
?
?
?
?
?
?
?
?
?
?
?
(6.78)
279
6.4.1 Linearisation of network equations (at the generator buses)
To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst
equation of the equation set (6.23) below.
?
¯
Y
11
+ ¯ y
1
?
¯
V
1
+
¯
Y
12
¯
V
2
+?+
¯
Y
1m
¯
V
m
+?+Y
1n
¯
V
n
= ¯ y
1
¯
E
1
(6.79)
Now, let,
¯
Y
ij
=Y
ij
exp(ja
ij
);
¯
V
i
=V
i
exp(j?
i
); for i, j = 1, 2,??n; (6.80)
and
¯ y
k
=y
k
exp(jß
k
);
¯
E
k
=E
k
exp(jd
k
); for k = 1, 2,??m; (6.81)
Utilising equations (6.80) and (6.81) in equation (6.79), we get,
Y
11
V
1
exp(j(a
11
+?
1
))+y
1
V
1
exp(j(ß
1
+?
1
))+Y
12
V
2
exp(j(a
12
+?
2
))+?+
Y
1m
V
m
exp(j(a
1m
+?
m
))+?+Y
1n
V
n
exp(j(a
1n
+?
n
))=y
1
E
1
exp(j(d
1
+ß
1
))
(6.82)
Taking the real part of both the sides of equation (6.82), we get,
Y
11
V
1
cos(a
11
+?
1
)+y
1
V
1
cos(ß
1
+?
1
)+Y
12
V
2
cos(a
12
+?
2
)+?+
Y
1m
V
m
cos(a
1m
+?
m
)+?+Y
1n
V
n
cos(a
1n
+?
n
)=y
1
E
1
cos(d
1
+ß
1
)
(6.83)
Linearising equation (6.83), we get,
Y
11
cos(a
11
+?
1
)?V
1
-Y
11
V
1
sin(a
11
+?
1
)??
1
+y
1
cos(ß
1
+?
1
)?V
1
-
y
1
V
1
sin(ß
1
+?
1
)??
1
+Y
12
cos(a
12
+?
2
)?V
2
-Y
12
V
2
sin(a
12
+?
2
)??
2
+?+Y
1m
cos(a
1m
+?
m
)?V
m
-Y
1m
V
m
sin(a
1m
+?
m
)??
m
+?+
Y
1n
cos(a
1n
+?
n
)?V
n
-Y
1n
V
n
sin(a
1n
+?
n
)??
n
=-y
1
E
1
sin(d
1
+ß
1
)?d
1
(6.84)
Equation (6.84) can be re-written as,
a
11
?V
1
+a
12
?V
2
+?+a
1m
?V
m
+?+a
1n
?V
n
+
b
11
??
1
+b
12
??
2
+?+b
1m
??
m
+?+b
1n
??
n
=g
1
?d
1
(6.85)
In equation (6.85),
a
11
=Y
11
cos(a
11
+?
1
)+y
1
cos(ß
1
+?
1
); a
12
=Y
12
cos(a
12
+?
2
);
a
1m
=Y
1m
cos(a
1m
+?
m
); a
1n
=Y
1n
cos(a
1n
+?
n
);
b
11
=-Y
11
V
1
sin(a
11
+?
1
)-y
1
V
1
sin(ß
1
+?
1
); b
12
=-Y
12
V
2
sin(a
12
+?
2
);
b
1m
=-Y
1m
V
m
sin(a
1m
+?
m
); b
1n
=-Y
1n
V
n
sin(a
1n
+?
n
);
g
1
=-y
1
E
1
sin(d
1
+ß
1
);
(6.86)
Again, taking the real part of both sides of the second equation of the equation set (6.23) (after
280
substituting equations (6.80) and (6.81) into it), we get,
Y
21
V
1
cos(a
21
+?
1
)+y
2
V
2
cos(ß
2
+?
2
)+Y
22
V
2
cos(a
22
+?
2
)+?+
Y
2m
V
m
cos(a
2m
+?
m
)+?+Y
2n
V
n
cos(a
2n
+?
n
)=y
2
E
2
cos(d
2
+ß
2
)
(6.87)
Linearising equation (6.87), one can get,
a
21
?V
1
+a
22
?V
2
+?+a
2m
?V
m
+?+a
2n
?V
n
+
b
21
??
1
+b
22
??
2
+?+b
2m
??
m
+?+b
2n
??
n
=g
2
?d
2
(6.88)
In equation (6.88),
a
21
=Y
21
cos(a
21
+?
1
); a
22
=Y
22
cos(a
22
+?
2
)+y
2
cos(ß
2
+?
2
);
a
2m
=Y
2m
cos(a
2m
+?
m
); a
2n
=Y
2n
cos(a
2n
+?
n
);
b
21
=-Y
21
V
1
sin(a
21
+?
1
); b
22
=-Y
22
V
2
sin(a
22
+?
2
)-y
2
V
2
sin(ß
2
+?
2
);
b
2m
=-Y
2m
V
m
sin(a
2m
+?
m
); b
2n
=-Y
2n
V
n
sin(a
2n
+?
n
);
g
2
=-y
2
E
2
sin(d
2
+ß
2
);
(6.89)
Similarly, continuing with linearisation of the real parts of ?rst ‘m’ equations (corresponding to
the generator buses) of the equation set (6.23), we get,

A
1
B
1

?
?
?
?
?
?
?
?
?
?
?
?V
g
?V
L
??
g
??
L
?
?
?
?
?
?
?
?
?
?
?
= 
G
?d (6.90)
In equation (6.90),
A
1
=
?
?
?
?
?
?
?
?
?
?
?
?
?
a
11
a
12
? a
1m
a
1,(m+1)
?? a
1n
a
21
a
22
? a
2m
a
2,(m+1)
?? a
2n
? ? ? ? ? ? ?
? ? ? ? ? ? ?
a
m1
a
m2
? a
mm
a
m,(m+1)
?? a
mn
?
?
?
?
?
?
?
?
?
?
?
?
?
? is a (m×n) matrix (6.91)
B
1
=
?
?
?
?
?
?
?
?
?
?
?
?
?
b
11
b
12
? b
1m
b
1,(m+1)
?? b
1n
b
21
b
22
? b
2m
b
2,(m+1)
?? b
2n
? ? ? ? ? ? ?
? ? ? ? ? ? ?
b
m1
b
m2
? b
mm
b
m,(m+1)
?? b
mn
?
?
?
?
?
?
?
?
?
?
?
?
?
? is a (m×n) matrix (6.92)
281
G=
?
?
?
?
?
?
?
?
?
?
?
?
?
g
1
0 ? 0 ?? 0
0 g
2
? 0 ?? 0
? ? ? ? ? ?
? ? ? ? ? ?
0 0 ? 0 ?? g
m
?
?
?
?
?
?
?
?
?
?
?
?
?
? is a (m×m) diagonal matrix (6.93)
The elements of the matrices A
1
, B
1
and G are given as;
a
ii
=Y
ii
cos(a
ii
+?
i
)+y
i
cos(ß
i
+?
i
); i= 1, 2,??m;
a
ij
=Y
ij
cos(a
ij
+?
j
); i= 1, 2,??m; j = 1, 2,??n; i?j;
b
ii
=-Y
ii
V
i
sin(a
ii
+?
i
)-y
i
V
i
sin(ß
i
+?
i
); i= 1, 2,??m;
b
ij
=-Y
ij
V
j
sin(a
ij
+?
j
); i= 1, 2,??m; j = 1, 2,??n; i?j;
g
i
=-y
i
E
i
sin(d
i
+ß
i
); i= 1, 2,??m;
(6.94)
Further, the vectors ??
L
and ?V
L
are de?ned as;
??
L
= [??
m+1
, ??
m+2
, ?? ??
n
]
T
(6.95)
?V
L
= [?V
m+1
, ?V
m+2
, ?? ?V
n
]
T
(6.96)
Please note that the size of each of the above two vectors is ((n-m)× 1). So far, we have
considered only the algebraic equations at the generator buses. However, for completing the small
signal model, the algebraic equations at the load buses all need to be linearised. We will discuss this
issue in the next lecture.
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