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6.1
Chapter 6 Small Signal Stability Analysis
Small-signal stability is the ability of the system to be in synchronism when
subjected to small disturbances. The disturbances can be switching of small loads,
generators or transmission line tripping etc. The small-signal instability can lead to
oscillations in the system. There are three modes of oscillations due to small
disturbances
1. Local modes of oscillations are due to a single generator or group of
generators oscillating against the rest of the system.
2. Intra-plant modes of oscillations are due to oscillations among the generators
in the same plant. The typical frequencies of oscillations of local and intra
plant modes are in the range of 1 Hz to 2 Hz.
3. Inter area modes of oscillations are due to a group of generator in one area
oscillating together against another group of generators in a different area. The
typical frequency range of inter area mode of oscillations are 0.1 Hz to 0.8 Hz.
Local, intra-plant and inter-area are electromechanical modes of oscillations in
which the rotor angle and the speed of the generators oscillate. Apart from
electromechanical modes of oscillations there can be control modes, due to lack of
proper tuning of controller like voltage regulator of an excitation system, and
torsional mode of oscillations due to oscillations in the turbine-generator shafts. The
main idea behind small signal stability analysis is that the power system, which is
nonlinear, for small disturbances can be linearized around the steady state operating
point as the system will be oscillating in a small area around the steady state operating
point. Once the nonlinear system is linearized around an operating point linear control
theory can be applied to assess the stability of the system. Before understanding how
to asses small signal stability of a system some terms used in control theory should be
defined.
6.2
6.1 Fundamental Concepts of Stability of a Dynamic System
Any dynamic system can be represented as
( ) ( ), ( )x t f x t u t (6.1)
( ) ( ), ( )y t g x t u t (6.2)
Where, ( )x t is called as a state variable, ( )u t is the control input, ( )y t is the
system output. ( ), ( )f x t u t , ( ), ( )g x t u t are the functions dependent on the state
variable and the control input. For a multi-variable, multiple-input multiple-output
(MIMO) linear time-invariant system equations (6.1) and (6.2) can be written as
( ) ( ) ( )t t t X AX BU (6.3)
( ) ( ) ( )t t t Y CX DU (6.4)
In equations (6.3) and (6.4), ( )tX is called as the state vector, ( )tU is the
control input vector, ( )tY is the output vector, A is the state matrix, B is the input
matrix, C is the output matrix and D is the feed forward matrix. State variables are
the minimum number of variables in a system that can completely define the dynamic
behaviour of that system. If ( )tX is defined as
1
2
1
( )
( )( )
( )n n
x t
x tt
x t
X
(6.5)
Then the n state variables define the behaviour of the system. State space is the
n -dimensional space created by the n state variables. Equations (6.3) and (6.4)
together are called as state space representation of a dynamic system. The solution of
equations (6.3) and (6.4) can be obtained using Laplace transforms. Applying Laplace
transforms to equations (6.3) and (6.4), assuming an initial condition (0)X , we get
6.3
( ) (0) ( ) ( )s s X s s X AX BU (6.6)
( ) ( ) ( )s s s Y CX DU (6.7)
on further simplification
1( ) (0) ( ) (0) ( )
adj ss s X s X s
s
I A
X I A BU BUI A
(6.8)
1( ) (0) ( ) ( )s s X s s
Y C I A BU DU (6.9)
Suppose the initial condition is taken as zero and assume there is no feed forward
from the input to the output then for a single-input single-output system (SISO)
1( )
( )
adj sY ss
U s s
I A
C I A B C BI A
(6.10)
Equation (6.10) shows the transfer function between the output ( )Y s and the
input ( )U s . The denominator in equation (6.10) is a thn order polynomial equation
called as characteristic equation and the roots of the equation are called as eigen
values. The roots of the characteristic equation or the eigen values can be computed
by finding the n solutions of the equation given in equation (6.11)
0s I A (6.11)
6.2 Eigen Properties
The eigen values corresponding to the state matrix A have certain properties.
Let the n eigen values be represented as 1 2, ,......, n then for an thi eigen value
i i i A (6.12)
0i i A I (6.13)
6.4
i is called as the right eigen vector corresponding to the eigen value i . Similarly a
left eigen vector i corresponding to the eigen value i can be defined as
i i i A (6.14)
0i i A I (6.15)
The left and right eigen vectors corresponding to different eigen values are orthogonal
which means that
0j i j i (6.16)
i i iC (6.17)
Where, iC is a constant. Usually, equation (6.17) can be normalized so that the
maximum value will be 1 and hence
1i i (6.18)
Let, the right and left eigen vector matrices be defined as
1 2
1 2
[ ]
[ ]
n
T T T Tn
Φ
Ψ
(6.19)
From equations (6.16) to (6.18), and the eigen vector matrices defined in equation
(6.19) the following expression can be written
ΨΦ I (6.20)
1Ψ Φ (6.21)
Let, a n dimensional diagonal matrix of eigen values be defined as
6.5
1 0
0 n
Λ
(6.22)
Then, from equation (6.12)
AΦ ΦΛ (6.23)
Pre multiplying equation (6.23) with 1Φ lead to
1 Λ Φ AΦ ΨAΦ (6.24)
, ,Ψ Φ Λ are of size n n and are called as modal matrices.
6.3 Stability of Homogenous or Unforced System
To understand the stability of a system let us take a homogenous or unforced
system that is there is no external input to the system and the response is a natural
response. Hence, in the dynamic system given in equation (6.3) the control input
vector is zero and the state space representation of the system is given as
( ) ( )t tX AX (6.25)
The stability of a system can be better understood through the use of modal
matrices rather than the state matrix A . This is because the state matrix is not a
diagonal matrix and hence cross coupling terms between states will be present. Let a
new state vector can be defined as
( ) ( )t tX ΦZ (6.26)
Substituting equation (6.26) in equation (6.25) lead to
( ) ( )t tΦZ AΦZ (6.27)
6.6
1( ) ( ) ( ) ( )t t t t Z Φ AΦZ ΨAΦZ ΛZ (6.28)
Since, the matrix Λ is a diagonal matrix with eigen values as its elements, each
dynamic state can be represented in terms of a single state variable leading to
decoupling of states. This can be represented as
( ) ( ) 1,2,.....i i iz t z t for i n (6.29)
The solution of equation (6.29), with initial condition taken as (0)iz for 1,2,.....i n ,
is given as
( ) (0) 1,2,.....ii iz t e z for i n (6.30)
From equation (6.30) the stability of the system can be assessed as following [1]:
1. If all the eigen values are on the left half of s -plane that is the real part of the
eigen values should be having a negative value then the system is stable. From
equation (6.28) it can be seen that with the eigen values on the left side of s -
plane as time moves towards infinite the state variables will reach their initial
conditions making the system stable. This is called as asymptotic stability.
2. If at least one of the eigen values is on the right side of s -plane that is the real
part of the eigen value is positive then the system is unstable. With at least one
eigen value with positive real part it can be observed from equation (6.30) that
the state variable can never come back to its initial condition. This is called as
aperiodic instability.
3. The eigen values of a system with a real state matrix have complex conjugate
pairs of eigen values. If at least one of the complex conjugate pair lies on the
imaginary axis of the s -plane then the system becomes oscillatory. This is
called as oscillatory instability.
4. If at least one eigen value is on the origin then the system stability cannot be
assessed.
6.7
From equation (6.26), the initial value of the original state vector in terms of the new
state vector can be written as
(0) (0)X ΦZ (6.31)
1(0) (0) (0) Z Φ X ΨX (6.32)
(0) (0)i iz X (6.33)
Hence,
1
21 2
1 1 2 2
1
( ) ( )
( )
( )
( )
( ) ( ) ( )
( )
n
n
n n
n
i ii
t t
z t
z t
z t
z t z t z t
z t
X ΦZ
(6.34)
Since, the new state variables ( )tZ are directly related to eigen values and each
of the new sate variables corresponds to exactly one eigen value, it can be observed
from equation (6.32) that the right eigen vector i corresponds to the activity of the
state variables ( )tX in thi mode. For example the state variable jx activity in thi
mode is given by the element ji and the direction of its activity is given by the angle
of ji . Similarly, i show the contribution of the state variables ( )tX in an thi mode
which can be observed from equations (6.30) and (6.33). For example ij shows the
contribution of the state variable jx in the thi mode. The total effect of activity and
contribution of a state variable jx in the thi mode is then given as ji ij . Hence, the
combined effect of activity and contribution of a state variable in a particular mode
can be represented as a matrix called as participation factor. The participation factor
matrix is given as
6.8
1 2 n n nP P P P (6.35)
1 1
2 2 for 1, 2,.... modes
i i
i ii
ni in
i n
P
(6.36)
As the right and left eigenvectors are normalized the sum of the elements of the
participation factor matrix corresponding to an thi mode will be 1
1n
kik
P
. Similarly,
the sum of the elements of the participation factor matrix corresponding to a thk state
will be1
1n
kii
P
.
6.4 Algorithm to Find Eigen Values and Eigen Vectors
In 1960 G. C. Frances [2] has proposed an algorithm named as QR to find the
eigen values and eigen vectors. Though many modifications have been suggested to
this algorithm the underlying principle is same. Here QR algorithm will be discussed.
Before discussing QR algorithm some terms need to be defined.
Unitary matrix: a unitary matrix U is an orthogonal matrix with the property
T UU I if U is real and *T UU I if U is complex, where I is an identity matrix
with same size as U .
Similarity transformation: If a matrix A be transformed by a transformation matrix
T such that -1T AT B and if the eigen values of the matrices A , B are same then
transformation is called as similarity transformation and the matrices A , B are called
as similar matrices.
Hessenberg Matrix: A matrix G is said to be an Hessenberg matrix if the elements
below the sub-diagonal are zeros. Householder’s transformation can be used to
transform any matrix A to its Hessenberg form G . The matrices A , G are similar.
6.9
The QR algorithm convergence is improved if instead of taking the matrix A , whose
eigen values need to be computed, its Hessenberg form G is considered.
Schur form: A symmetrical matrix A will have a unitary matrix U such that the
transformation, 1 U AU S , lead to a matrix S in Schur form. The matrix S is a
triangular matrix. The matrices A , S are similar and hence the eigen values of both
the matrices will be same. If matrix A is real then its Schur form S is also real and if
matrix A is complex then S is also complex.
Eigen values of a matrix in Schur form: Let A be a symmetrical matrix and let it be
transformed to a Schur form S through similar transformation. The matrix S is of the
form
11 12 13 1
22 23 2
33 3
0
0 0
0 0 0
n
n
n
nn
S S S S
S S S
S S
S
S
(6.37)
The eigen values of A will be
11 12 13 1
22 23 2
33 3
0
det [ ] det [ ] det 0 0 0
0 0 0
n
n
n
nn
S S S S
S S S
S S
S
A I S I
(6.38)
or
11 22 33 0nnS S S S (6.39)
Which means that the eigen values of the matrix A are nothing but the
diagonal elements of the matrix S . It has to be noted that if the matrix A is not
symmetrical then its Schur form S will not be a triangular matrix but may contain
2 2 blocks along the principal diagonal whose eigen values will give a complex
6.10
conjugate pair of eigen values. QR method simply is a method to transform a given
matrix A in to its Schur form S so that the eigen values will be simply the diagonal
or block diagonal elements of S . Usually it is solved in two steps.
Step 1: Transform a symmetrical/unsymmetrical matrix A into Hessenberg form G
through householder’s transformation.
Step 2: Use QR method, based on stable Gram-Schmidt, to convert the matrix G into
a Schur form S . The eigen values of A will then be diagonal elements of S if A is
symmetrical and if A is unsymmetrical then the eigen values will be the eigen values
of the block diagonal elements of S .
Step 1: Transform a symmetrical/unsymmetrical matrix A into Hessenberg form G
Let the matrix A be represented as
11 12 13 1
21 22 23 2
31 32 33 3
1 2 3
n
n
n
n n n nn n n
a a a a
a a a a
a a a a
a a a a
A
(6.40)
Define, vectors 1X , 1Y as
121
1 31 1 21
1 1 1
00
, ( ) 0
0n n n
Xa
X a Y sign a
a
(6.41)
With 1 1,X Y defined as in equation (6.41) a unit vector 1U can be formed as
1 11
1 1
X YU
X Y
(6.42)
6.11
Householder’s transformation can be defined on the basis of the unit vector 1U as
1 1 12 TU U H I (6.43)
Now apply the transformation matrix 1H to the matrix A such that
1 1 1A H AH (6.44)
1A will be of the form
11 12 13 1
21 22 23 2
32 33 3
2 3
1 1 1 1
1 1 1 1
1 1 11
1 1 1
0
0
n
n
n
n n nnn n
a a a a
a a a a
a a a
a a a
A
where, 1
ija is an throw columnthi j element of 1A defined by the transformation.
After transformation it can be seen that the first column elements below the second
row are zero. Now define 2 2,X Y such that
32 32
2
1 1 2 22 2 2 2
2 2
1
0 0
0 0
, ( ) ,
0n
X YX a Y sign a X U
X Y
a
(6.45)
2 2 22 TU U H I (6.46)
2 2 1 2A H A H (6.47)
2A will be of the form
6.12
11 12 13 1
21 22 23 2
32 33 3
43 4
3
2 2 2 2
2 2 2 2
2 2 2
22 2
2 2
0
0 0
0 0
n
n
n
n
n nn n n
a a a a
a a a a
a a a
a a
a a
A
Repeat this process for n times and at the end of the thn iteration the original matrix
A has the Hessenberg form
11 12 13 1
21 22 23 2
32 33 3
43 4
0
0 0
0 0 0
n
n
n
n
nn
n n n n
n n n n
n n n
nn n
n
n n
a a a a
a a a a
a a a
a a
a
G A
Step 2: Use QR method, based on stable Gram-Schmidt, to convert the matrix G into
a Schur form S .
Let each column of the matrix G be considered as an independent vector and
represented as
1 2 3 nW W W WG (6.48)
Where,
11211
22221
32 31 2
1 1 1
0 , ,
0 0
n
n
n
nn
nnn
nnn
n nn
n
n n n
aaa
aaa
W W a W a
a
(6.49)
6.13
Each of these n independent vectors 1 2 3, , , nW W W W can be expressed in terms of
n linearly independent orthogonal unit vectors. Let one of the orthogonal unit vectors
be chosen as
11
1
WQ
W (6.50)
so that
1 11 1 11 1,W r Q r W (6.51)
In a similar way each of the vector 1 2 3, , , nW W W W can be written as
1 11 1
2 12 1 22 2
3 13 1 23 2 33 3
1 1 2 2 3 3
,
n n n n nn n
W r Q
W r Q r Q
W r Q r Q r Q
W r Q r Q r Q r Q
(6.52)
Here, 11r to nnr are scalars. Since, 11 1,r Q are known rest of the unknowns can be
computed as, keeping in mind that since 1,..... nQ Q are orthogonal
20 if andi j i i iQ Q i j Q Q Q ,
2
j iij
i
W Qr
Q
(6.53)
12
1
i
ii i jij
r W r
(6.54)
1
1
i
i ji jj
iii
W r Q
Qr
(6.55)
6.14
Let,
11 12 13 1
22 23 2
33 3
0
0 0
0 0 0
n
n
n
nn n n
r r r r
r r r
R r r
r
(6.56)
1 2 3 nQ Q Q QQ (6.57)
Hence,
1 2 3 nW W W W G QR (6.58)
In equation (6.58), Q is a unitary matrix and R is an upper triangular matrix.
The method followed so far in decomposing a given matrix G into product of a
unitary matrix Q and an upper triangular matrix R is called as stable Gram-Schmidt
method. G. C. Frances has used QR decomposition to transform a given matrix G
into Schur form. G. C. Frances has changed the order of multiplication of ,Q R in
equation (6.59) and then defined a new matrix 1G as
1T G RQ Q GQ (6.59)
Applying QR decomposition to matrix 1G obtain new matrices 1 1,Q R then
1 1 1G Q R (6.60)
Similar to equation (6.59) define new matrix 2G as
2 1 1 1 1 1 1 1T T T T G R Q Q G Q Q Q G Q Q (6.61)
6.15
Repeat this iterative process till the difference between the determinants of any
two consecutive iteration G matrix values falls below certain value say that is
1K K G G . If the algorithm converges at thK iteration then
1 1 1 1 1T T T T T
K K K K K K K K K K G R Q Q G Q Q Q Q Q G QQ Q Q (6.62)
The matrix 1KG will be in Schur form. Let 1 1K KT QQ Q Q then
1 1T T T T T
K KT Q Q Q Q . The matrix T is also a unitary matrix as the product of
unitary matrices is also a unitary matrix. Hence, matrix T is the transformation
matrix which transforms the Hessenberg matrix G into Schur form S as given in
equation (6.63)
1T
K S G T GT (6.63)
For a given real symmetrical matrix A the eigen values are given by the
diagonal elements of the matrix S given in equation (6.63). If the matrix A is real
unsymmetrical or complex then the eigen values are given by the eigen values of the
block diagonal elements along the principal diagonal of matrix S .
Eigen vector
Once the eigen values are computed then the right and left eigen vectors can be
computed. If 1 2, , , n are the eigen values of the matrix A then
1
2
0
0 n
Λ
(6.64)
AΦ ΦΛ (6.65)
Where, 1 2[ ]n n n Φ is the right eigenvector matrix. For an thi eigen value
6.16
0i i i A I A (6.66)
Let,
22 2
2 1 1
n
n nn n n
a a
a a
E
(6.67)
21
1 1 1n n
a
a
F (6.68)
Where, ija is an element of the matrix A . Then let the first element of thi right
eigen vector i , that is 1i be arbitrarily taken as 1. Now the thi right eigen vector is
given as
1
1
1i
n
E F
(6.69)
To get a normalized right eigen vector divide equation (6.60) by i . The
procedure can be used to find other right eigen vectors. To find the left eigen vector
matrix Ψ simply take the inverse of the right eigen vector matrix that is
1Ψ Φ (6.70)
6.5 Linearizing a Nonlinear System
The modal or eigen value analysis can be done for a dynamic system which is
linear. However, a nonlinear system cannot be expressed in state space form and
hence eigen value analysis cannot be used for a dynamic system which is nonlinear.
One way of overcoming this problem is to linearize the nonlinear system around a
6.17
small region of its steady state operating point. This can be demonstrated as
following. Let a nonlinear dynamic system be represented as
,X f X U (6.71)
Where, X is a vector of dynamic variables of the system. U is the vector of
inputs and ,f X U is a nonlinear function of both X and U . Let the steady state
condition values of X , U be 0X , 0U . If the system dynamic variables and the
inputs are perturbed by a small value ,X U from the steady state operating point
then from equation (6.71)
0 0 0,X X f X X U U (6.72)
Expanding the right hand side of equation (6.72) as Taylor series with second and
higher order terms neglected lead to
0 0
0 0
0 0 0,X X X XU U U U
f fX X f X U X U
X U
(6.73)
Since, 0 0 0,X f X U , equation (6.73) can be written as
0 0
0 0
X X X XU U U U
f fX X U
X U
(6.74)
Let, 0 0
0 0
,X X X XU U U U
f f
X U
A B , then
X X U A B (6.75)
The system given in (6.75) is a linear system and the stability of it is decided by
the eigen values of the state matrix A . This process of converting a nonlinear system
6.18
given in (6.71) to a linear system for a small disturbance around the steady state
operating point given in (6.75) is called as linearization.
6.6 Small Signal Stability Analysis of Single Machine Connected to
Infinite Bus
Let us consider a single machine connected to an infinite bus as shown in Fig. 6.1.
Fig. 6.1: Single machine connected to infinite bus
The effective impedance between the generator terminal and the infinite bus is
e eR jX . The infinite bus voltage is 0jV e . The generator terminal voltage and
current in 0dq -axis is given as d qV jV and d qI jI . The internal rotor angle is
. Let the input mechanical torque MT be constant. Hence, turbine and speed
governor dynamics need not be considered. Let the generator be represented by 1.0
flux decay model. The DAE of flux decay model, excluding exciter dynamics, are
given as
'' ' '( )q
do q d d d fd
dET E X X I E
dt (6.76)
base
d
dt
(6.77)
2m e base
base
H dT T D
dt
(6.78)
d s d q qV R I X I (6.79)
2( )j
jd q tV jV e V e
2( )j
d qI jI e
ejX eR
0jV e
6.19
' 'q s q d d qV R I X I E (6.80)
Where,
*' '
' '
Ree q d q q d q
q q q d q d
T al X X I jE I jI
E I X X I I
From the Fig. 6.1, the generator current can be written as
022
( )( )
jj
jd q
d qe e
V jV e V eI jI e
R jX
(6.81)
On further simplification and separating the real and imaginary parts from
(6.81), the following expressions can be obtained.
sine d e q dR I X I V V (6.82)
cose d e q qX I R I V V (6.83)
Let the steady state operating point values of 'q fd d q d qE E I I V V be
represented by '0 0 0 0 0 0 0q fd d q d qE E I I V V . Linearizing equations (6.79)-(6.80) and
(6.82)-(6.83) for a small perturbation in the parameters around the steady state
operating point leads to, neglecting the stator resistance,
''
00
0
d q d
q q qd
V X I
V I EX
(6.84)
0
0
cos
sin
d de e
q qe e
VV IR X
V IX R V
(6.85)
6.20
Since, the left hand side of the equations (6.84) and (6.85) are same the right
hand sides can be equated as
0
''0
0 cos0
sin0
q d de e
q qq e ed
VX I IR X
I IE X R VX
(6.86)
Solving for dI , qI from equation (6.86) lead to
1
0
''0
0
'2 ' '0
0
2 '
0 cos
sin
cos 01
sin 1
sin1
e e qd
q qe d e
e e q
qe e q e d e d e
e q e
e e q e d
R X X VI
I E VX X R
R X X V
EVR X X X X X X R
V X X R V
R X X X X
0
''0 0
cos
cos sin
e q
qe d e e
X X
EV X X R V R
(6.87)
Now linearizing the differential equations (6.76) to (6.78) lead to
''
' ''
'0'
0 '0 0
1( ) 0
10 0
0 00 0 1
2 202 2
d ddo
q qdod
q d d qbase baseq d qbase base
q q
X XT
E ETI
X X I IX X I
H HI D EH H
'
10
0 0
02
dofd
mbase
TE
T
H
(6.88)
Substituting (6.87) in (6.88) and simplifying leads to
6.21
4' '' ' '
3
2 1
1 10 0
0 0 1 0 0
02 2 2 2
q qdo do dofd
mbase base base base
KE EK T T T
E
T
K K DH H H H
(6.89)
Where,
'0 0 0
1 2 ' ' ' '0 0 0 0
2 ' '0
2 2 ' ' '0 0
'
23
sin cos1
cos sin
1
11
q d q e q e
e e q e d d q d q e d e
q e e q e d d q e q
e e q e d e d q d e q
d d e q
e e q
I V X X X X RK
R X X X X V X X I E X X R
I R X X X X X X X XK
R X X X X R X X I R E
X X X X
K R X X X
'
'
4 0 02 'sin cos
e d
d d
e q e
e e q e d
X
V X XK X X R
R X X X X
(6.90)
A simplified fast acting static exciter is now considered which is represented by
a first order control block with a gain AK and time constant AT as shown in Fig. 6.2.
Fig. 6.2 A simplified fast acting static exciter
From Fig. 6.2 the following differential equation representing the dynamics of the
exciter can be written as
fdE
tV
refV +
-1
A
A
K
sT
6.22
( )fdA fd A ref t
dET E K V V
dt (6.91)
Linearizing equation (6.91) gives
( )fdA fd A ref t
d ET E K V V
dt
(6.92)
The terminal voltage can be written in terms of the 0dq parameters as
2 2t d qV V V (6.93)
Linearizing equation (6.93)
0 00 0 dq qd dt d q
qt t t t
VV VV VV V V
VV V V V
(6.94)
Substituting (6.87) in (6.84) lead to
1
0
' '' '0
0 0cos0
sin0
e e qd q
q q qd e d e
R X X VV X
V E EVX X X R
(6.95)
Substituting (6.95) in (6.94) and rearranging lead to
'5 6t qV K K E (6.96)
6.23
Where,
'00 0
5 2 '0 '
0 0
0 0'06 2 '
sin cos1
cos sin
1
dq e e d
t
qe e q e dd e e q
t
q qdq e d e q
t t te e q e d
VX R V V X X
VK
VR X X X XX R V V X X
V
V VVK X R X X X
V V VR X X X X
Equation (6.92) can be substituted in (6.89), with the linearized terminal voltage
defined in (6.96), and can be written as
4' ' '
' '3
2 1
6 5
1 10 0 0
0 00 0 1 0
00 2
2 2 201
0
do do doq q
refbase
base base basem
AfdfdA A
AA A A
K
K T T TE E
V
TK K D HH H H KEE K K K K
TT T T
(6.97)
The constants 1 6K K are called as Heffron-Phillips constants [3] and the state
space model given (6.97) is called as Heffron-Phillips model [3]. It can be observed
from (6.97) that the nonlinear SMIB system has been linearized and now represented
in state space with the state variables being ' , , ,q fdE E . The stability of the
system now depends on the eigen values of the state matrix. The state space
representation of the system given in (6.97) can be represented as a control block
diagram as shown in Fig. 6.3
6.24
Fig. 6.3 Heffron-Phillips Model [3] of a SMIB system
Converting equation (6.97) in to Laplace domain lead to the following set of
equations.
' ' 4' ' '
3
'1 2
'5 6
1 1( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )2 2 2
1( ) ( ) ( ) ( ) ( )
q q fddo do do
base base basem q
A Afd fd q ref
A A A
Ks E s E s s E s
K T T T
s s s
s s T s K s K E s D sH H H
K Ks E s E s K s K E s V s
T T T
(6.98)
Simplifying equation (6.98)
' 4 3 3
' '3 3
( ) ( ) ( )1 1
q fddo do
K K KE s s E s
sK T sK T
(6.99)
'
5 6( ) ( ) ( ) ( )1 1
A Afd q ref
A A
K KE s K s K E s V s
sT sT
(6.100)
6.25
( ) ( )s s s (6.101)
2 '1 2( ) ( ) ( ) ( ) ( )
2 2 2base base base
m qs s T s K s K E s Ds sH H H
(6.102)
Assuming ( )refV s =0 in (6.100), equation (6.100) can be substituted in (6.99) which
gives
4 5 3''
3 6 3
1( ) ( )
1 1
A Aq
do A A
K sT K K KE s s
sK T sT K K K
(6.103)
The linearized electrical torque, eT , expression can be written as
' ' ' '
' ' ' '
e q q q q q d q d q d d q
d
q q q d q q q d dq
T E I E I X X I I X X I I
II E X X I E X X I
I
(6.104)
Substituting equation (6.87) in (6.104), gives
'1 2e qT K K E (6.105)
Converting (6.105) into Laplace domain and substituting (6.103) gives
( )
4 5 3 21 '
3 6 3
1( ) ( )
1 1
( ) ( )
H s
A Ae
do A A
K sT K K K KT s K s
sK T sT K K K
H s s
(6.106)
Substituting (6.106) in (6.102) lead to a second order equation, assuming the input
torque ( )mT s =0,
2 ( ) ( ) 02 2
base bases Ds H s sH H
(6.107)
6.26
The small signal oscillations are typically in the range of 0.1 Hz to 3 Hz. In this
region of frequency interest the transfer function ( )H s can be split into two
components as
4 5 3 21 '
3 6 3
4 5 3 21 2 ' '
6 3 3 3
2 4 5 41
2 ' '6
3 3
1( )
1 1
1
1
1
A A
do A A
A A
A do A do A
A A
AA do A do
K j T K K K KH s j K
j K T j T K K K
K j T K K K KK
K K K K T T j K T T
K K K K j K TK
TK K T T j T
K K
(6.108)
Neglecting the effect of the constant 4K
2 51
2 ' '6
3 3
2 '2 5 6
31 2 2
2 ' '6
3 3
'2 5
3
63
( )1
1
1
1
S
A
AA do A do
K
A A do A
AA do A do
AA do
A
K K KH j K
TK K T T j T
K K
K K K K K T TK
KT
K K T T TK K
TK K K T
Kj
K KK
2 22 ' '
3
dK
Ado A do
TT T T
K
(6.109)
( )e S d
S d
T j K j K
K K
(6.110)
6.27
The electrical torque has two components synchronising and damping.
Synchronising torque SK is in phase with the change in rotor angle . The damping
torque is in quadrature to the rotor angle or in phase with the change in the speed.
Expressing (6.109) in s -domain as
( ) S dH s K sK (6.111)
Substituting (6.111) in (6.107)
2 ( ) 02 2
base based ss D K s K s
H H
(6.112)
Equation (6.112) represents a damped second order unforced system. If the
damping coefficient dD K is not considered then the system becomes undamped
system with natural frequency of oscillations being
2base
n sj KH
(6.113)
As can be understood from (6.113), without damping the system will have
sustained oscillations with the frequency of oscillations as defined in (6.113). If the
synchronising torque becomes zero for a forced system that is 0MT then the
system will become aperiodically unstable that is the rotor angle keeps on increasing
and ultimately lead to loss of synchronism. In case the damping coefficient dD K
becomes negative due to system conditions then there will be oscillatory instability
that is the peak of the oscillations keep on increasing leading to loss of synchronism.
If both the synchronising and damping torque are positive for a given system then
after the disturbance the system will settle down to a steady state operating point with
oscillations being damped out. The constant 5K becomes important while assessing
the stability.
6.28
2 '2 5 6
31 2 2
2 ' '6
3 3
'2 5
32 2
2 ' '6
3 3
1
1
1
A A do A
s
AA do A do
AA do
d
AA do A do
K K K K K T TK
K KT
K K T T TK K
TK K K T
KK
TK K T T T
K K
(6.114)
From (6.114) it can be observed that if 5K becomes negative then for a practical
system the synchronising coefficient sK becomes positive and increases as compared
to 5K being positive. Hence, with a negative 5K the synchronising torque improves.
But, with 5K being negative it can be seen that the damping coefficient dK becomes
negative leading to negative damping torque and hence oscillatory instability. This
effect of negative damping torque due to negative 5K gets pronounced if the exciter
has a very high gain AK . In a practical system 5K can become negative in a heavily
loaded case and a high gain exciter can lead to system stability issues [3], [4]. The
physical understanding of this is that a high gain exciter can build up the terminal
voltage rapidly after a disturbance improving the synchronising torque but at the same
time it has a negative damping, due to its high gain it amplifies a very small error in
terminal voltage and without enough system damping can lead to oscillations.
6.7 Power System Stabilizer (PSS)
A device called as power system stabilizer [5], [6] and [7] is used for
overcoming the negative damping effect of the high gain exciter. The power system
stabilizer acts as a supplementary controller to the excitation system. Inputs to the
power system stabilizer can be change in frequency, speed, power or a combination of
these. The output is a voltage signal introduced in the excitation system to control the
output of the exciter. The basic idea of the power system stabilizer is to introduce a
6.29
pure damping term in (6.112) so as to counter the negative damping effect of the
exciter.
Let PSS have a transfer function ( )G s and the change in the speed be the
input to the (PSS). The output of the PSS be PSSV . The output of PSS is added as a
supplementary signal to the exciter reference hence
( )fdA fd A ref PSS t
d ET E K V V V
dt
(6.115)
How the PSS output PSSV effects the synchronizing and damping torque can be
analysed by finding the transfer function between PSSV and the electrical torque
eT . To get the transfer function let us assume 0, 0refV , then from (6.99)
and applying Laplace transform to (6.100), the following expressions can be written
as
' 3
'3
( ) ( )1
q fddo
KE s E s
sK T
(6.116)
'6( ) ( ) ( )
1 1A A
fd q PSSA A
K K KE s E s V s
sT sT
(6.117)
Substituting (6.117) in (6.116) and then finding the transfer function between ' ( )qE s
and ( )PSSV s
' 3
'3 3 6
( ) ( )1 1
Aq PSS
do A A
K KE s V s
sK T sT K K K
(6.118)
Substituting (6.118) in (6.105)
' 2 3
2 '3 3 6
( ) ( )1 1
Ae q PSS
do A A
K K KT K E s V s
sK T sT K K K
(6.119)
Equation (6.119) can be approximated as
6.30
' 22 '
63 6
( ) ( )1
1 1
Ae q PSS
doA A
A
K KT K E s V s
TK K s sT
K K K
(6.120)
For a high gain exciter 63
1AK K
K hence
2
'6
6
1( )
1 1e PSS
doA
A
KT V s
K Ts sT
K K
(6.121)
Since, PSS has a transfer function ( )G s with input , (6.121) can be written as
2
'6
6
1( )
1 1e
doA
A
KT G s
K Ts sT
K K
(6.122)
In order for the PSS to contribute a pure damping term throughout the frequency
of interest in (6.112) the transfer function between eT and should be
'
6
( ) 1 1doPSS A
A
TG s K s sT
K K
, where PSSK is the gain of PSS, so that
2 2
6 6e PSS PSS
K KT K K s
K K (6.123)
With the torque defined as in (6.123), equation (6.112) can be approximated and
written as
2 2
6
( ) 02 2
base based PSS s
Ks D K K s K s
H K H
(6.124)
6.31
Hence, PSS improves the damping of system by providing a positive damping
torque. Since, introducing zeros or poles alone is not possible a practical method is to
take PSS transfer function as
1
2
1( )
1 1
n
WPSS
W
sT sTG s K
sT sT
(6.125)
Where, PSSK is the gain. WT is the washout filter time constant. 1 2,T T are the
lead-lag network time constants. n is the number of lead-lag network blocks. The
block diagram of the PSS is shown in Fig. 6.4.
Fig. 6.4: Power System Stabilizer (PSS) block diagram.
The function of washout filter will be discussed later. The procedure for finding the
parameters PSSK , 1 2,T T is as following. Let,
2 3
'3 3 6
( )1 1
AEX
do A A
K K KG s
sK T sT K K K
(6.126)
Find the natural frequency of oscillations of the system without the effect of
damping due to any other parameter. For this a constant flux linkage can be
considered so that 'qE is zero, hence 1eT K . Also the damping coefficient D
is assumed zero. With these assumptions (6.112) can be written as
21 ( ) 0
2bases K sH
(6.127)
PSSK
1W
W
sT
sT 1
2
1
1
nsT
sT
Gain
Wash out filter
Lead-Lag network
PSSV
6.32
12base
n j KH
(6.128)
At the natural frequency find the time constant of lead-lag network such that the lag
due to ( )EXG s is compensated that is
( ) ( ) 0n EX nG j G j (6.129)
For a high value of washout time constant the washout filter will not contribute any
angle to the transfer function of PSS hence for 1n
1 21 1 ( ) 0n n EX nj T j T G j (6.130)
One of the lead-lag time constants can be arbitrarily chosen then the other time
constant can be derived from (6.130). If lag of ( )EX nG j is high then multiple lead-
lag blocks can be used. Since, the PSS should provide damping over a range of small
signal oscillation frequencies (0.1 Hz to 3 Hz) in stead of tuning the lead-lag time
constant at a particular frequency they can be tuned in such a way that over all lag
compensation is at least between 0 to 45 . The industry practice for choosing the
gain PSSK is to increase the gain till the system becomes unstable, say *PSSK and then
take one third of *PSSK .
The power system stabilizer should not operate for steady state changes in the
speed, frequency or power. It should only operate only during transients. In order to
make PSS operate only during transients washout filter is used. Washout filter acts
like a high-pass filter allowing those signals which are above the cut-off frequency.
The washout filter time constant is chosen so that the signal with lowest frequency of
interest can be passed through. For washout filter time constant of 10WT s signals of
frequency 0.1 Hz and above can be passed. The stead state changes can be considered
as DC signals and hence will be blocked by the washout filter.
6.33
Let the output of the washout filter be WFV and this is the input to the lead-lag
block. Let the output of the lead-lag block be PSSV then the linearized dynamic
equations of the PSS can be written as
1WF WF PSS
W
V V KT
(6.131)
1
2 2 2
1 1PSS PSS WF WF
TV V V V
T T T (6.132)
From (6.97) the rate of change of speed is know and can be substituted in
(6.131). Similarly WFV can be substituted in (6.132) hence, the following expression
can be derived
'
2 1
10 0 0
2 2 2 2
q
refbase base base baseWF PSS PSS PSS PSS
W fd m
WF
PSS
E
VV K K K K DK K
H H H T HE T
V
V
(6.133)
'
1 1 1 12 1
2 2 2 2 2 2
1
2
1 1 10
2 2 2
02
q
base base basePSS PSS PSS PSS
W fd
WF
PSS
refbasePSS
m
E
T T T TV K K K K DK
T H T H T H T T T T E
V
V
VTK
H T T
(6.134)
The state space model including the PSS dynamics is given as
6.34
4' ' '
3
'
2 1
6 5
2 1
1 12
2
1 10 0 0
0 0 1 0 0 0
0 0 02 2 2
10 0 0
10 0
2 2 2
2
do do do
q
base base base
A A
fd A A A
WF base base basePSS PSS PSS
PSS W
basePSS
K
K T T T
E
K K DH H H
K K K KE T T T
VK K K K DK
V H H H T
T TK K
T H
'
1 11
2 2 2 2 2
1
2
1 1 10
2 2
0 0
0 0
02
0
02
02
q
fd
WF
PSS
base basePSS PSS
W
base
A
A
basePSS
basePSS
E
E
V
V
T TK K DK
T H T H T T T T
HK
T
KH
TK
H T
ref
m
V
T
(6.135)
6.8 Small Signal Stability Analysis of Multi-Machine System
The small signal stability of a single machine connected to an infinite bus has
been discussed in the previous section from which the effect of excitation system on
stability was understood. Power system stabilizer which is used to counter the
negative damping inducing effect of excitation system was explained. Now the small
signal stability for a multi-machine system including the power system stabilizer at
each generator will be explained. Let each generator be modelled by a sub-transient
model along with a power system stabilizer and stator algebraic equations. The
network is represented by real and reactive power balance equations. The DAE of a
general power system are given as following
6.35
Differential equations (for 1, 2,......, gi n , generator)
' ' "
' ' ' '1' 2
'
( )( ) ( )
( )qi di di
doi di di di qi di lsi di didi lsi
qi fdi
dE X XT X X I E X X I
dt X X
E E
(6.136)
" ' '11( )di
doi qi di lsi di did
T E X X Idt
(6.137)
' "'
' ' ' ' '2' 2
( )( ) ( )
( )qi qidi
qoi di qi qi qi di qi lsi qi qiqi lsi
X XdET E X X I E X X I
dt X X
(6.138)
2" ' '2( )qi
qoi di qi lsi qi qi
dT E X X I
dt
(6.139)
ii base
d
dt
(6.140)
2 i imi ei i base
base
H dT T D
dt
(6.141)
( )fdiEi Ri Ei Ei fdi fdi
dET V K S E E
dt (6.142)
Ri Ai FiAi Ri Ai Fi fdi Ai refi ti
Fi
dV K KT V K R E K V V
dt T (6.143)
Fi FiFi fdi
Fi
dR KR E
dt T (6.144)
1
2base
WFi WFi PSSi mi ei i baseWi
V V K T T DT H
(6.145)
1 1
2 2 2 2
1 1 1
2i i base
PSSi PSSi WFi PSSi mi ei i i basei i Wi i i
T TV V V K T T D
T T T T T Hi
(6.146)
1Mi HPi RHi HPi RHiRHi Mi CHi SVi
CHi CHi
dT K T K TT T P P
dt T T
(6.147)
CHiCHi CHi SVi
dPT P P
dt (6.148)
11SVi i
SVi SVi refiDi base
dPT P P
dt R
(6.149)
6.36
Stator algebraic equations
" ' "" '
1' '
( ) ( )cos( ) 0
( ) ( )di lsi di di
i i i si qi di di qi didi lsi di lsi
X X X XV R I X I E
X X X X
(6.150)
" ' "" '
2' '
( ) ( )sin( ) 0
( ) ( )qi lsi qi qi
i i i si di qi qi di qiqi lsi qi lsi
X X X XV R I X I E
X X X X
(6.151)
Network equations
For, 1, 2,......, gi n , generator buses
1
sin( ) cos( ) ( ) cos 0n
di i i i qi i i i Di i i j ij i j ijj
I V I V P V VV Y
(6.152)
1
cos( ) sin( ) ( ) sin 0n
di i i i qi i i i Di i i j ij i j ijj
I V I V Q V VV Y
(6.153)
For, 1, 2,......,g gi n n n , for load buses
1
( ) cos 0n
Di i i j ij i j ijj
P V VV Y
(6.154)
1
( ) sin 0n
Di i i j ij i j ijj
Q V VV Y
(6.155)
The differential equations (6.136) to (6.149) can be represented as
, , , , ,i i i di qi g gX f X I I V U (6.156)
Where, ( gn number of generators and n number of buses)
' '
1 2
1 1
1 1
1,......,
.......... , .......... ,
.......... , ..........
g g
g g
T
i qi di di qi i i Ri fdi Fi Mi SVi CHi FWi PSSi
gT
i refi refi
T T
g n g n
T T
l n n l n n
X E E V E R T P P V Vi n
U V P
V V V
V V V
6.37
The stator equations (6.150) and (6.151) can be written as
, , , , 0, 1,......,i i di qi g g gg X I I V i n (6.157)
The network power balance equations at the generator buses (6.152) and (6.153) can
be written as
, , , , , , 0, 1,......,i i di qi g g l l gh X I I V V i n (6.158)
The network power balance equations at the load buses (6.154) and (6.155) can be
written as
, , , 0, 1,......,i g g l l gk V V i n n (6.159)
The functions , , ,f g h k are non linear functions. Let the steady state operating
point of the system be represented by 0 _ 0 _ 0 0i di qi iX I I U for 1,......, gi n and the
network voltages and angles initial condition are represented as 0 0 0 0g g l lV V .
Linearizing (6.156) around the initial operating conditions, that is the partial
derivative are evaluated at the initial condition, lead to
11 2
1 ,14 14 ,14 1 1 ,14 2 ,2 1
0 0
dqiii iiIBA WB g
di gi i i i i ii i i
qii di qi g g il
l
i i i dqi
I Vf f f f f fX X U
IX I I V U
V
A X B I
2 ,14 2 14 2 2 114 2 1 20
g g
T
i n g g l i in n nB V V W U
(6.160)
6.38
Each of the matrix size in (6.160) is given as a subscript to the corresponding
matrix. If the linearized equation (6.160) is formulated for all the generators and the
equations are put together then
1,14 14 14 1 1,14 2 ,2 1
2,14 2 14 2 2 114 2 1 20
g g g g g g
g g g g gg g
n n n n n dq n
T
n n g g l n n nn n n n
X A X B I
B V V W U
(6.161)
Where,
1 1 1......... , ......... , .........g g g
T T T
n n dq dq dqnX X X U U U I I I
Similarly, linearizing (6.157)
21 1
1 ,2 14 ,14 1 2 ,2 2 ,2 1 1 ,2 2 2 2
0 0 0
0
ii
g g
CC gD i
di gi i i i ii
qii di qi g g l
l
i i i dqi i n g g l ln n
I Vg g g g gX
IX I I V
V
C X C I D V V
1 20
T
n
(6.162)
linearized equation (6.157) of all the generator together can be written as
1,2 14 14 1 2,2 2 ,2 1 1,2 2 2 2 1 20 0
g g g g g g g g g g
T
n n n n n dq n n n g g l ln n n nC X C I D V V
(6.163)
Similarly, linearizing (6.158) and (6.159) lead to
6.39
43 2 3
3 ,2 14 ,14 1 4 ,2 2 ,2 1 2 ,2 2 3 ,2 2
0
ii i
g
CC D gD i
di li i i i i i ii
qi gi di qi g g l l
l
i i i dqi i n i n
Ih h h h h h hX
I VX I I V V
V
C X C I D D
1 20, 1,......,
g
T
g g l l gn nV V i n
(6.164)
or
3,2 14 14 1 4,2 2 ,2 1 2,2 2 3,2 20
g g g g g g g g g g
T
n n n n n dq n n n g g l ln n nC X C I D D V V
(6.165)
5 6
5 ,2 2 6 ,2 2 1 2
0
0 1,......,
i i
g g
D D g
li i i i
gg g l l
l
T
i n g g l l gi n n n
k k k k
VV V
V
D D V V i n n
(6.166)
or
5,2 2 6,2 2 1 20
g g g g
T
g g l ln n n n n n n nD D V V
(6.167)
Equations (6.161), (6.163), (6.165) and (6.167) can be written as
1 1 2 0 0T
dq g g lX A X B I B V V W U (6.168)
1 2 10 0 0T
dq g g l lC X C I D V V (6.169)
3 4 2 30T
dq g g l lC X C I D D V V (6.170)
4 50T
g g l lD D V V (6.171)
From (6.171), the load bus voltages and angles can be eliminated as
15 4
TT
l l g gV D D V (6.172)
6.40
Let,
16 2 3 5 4D D D D D (6.173)
Substituting (6.173)-(6.172) in (6.170) and (6.169), and eliminating the
generator bus voltages, angles, direct axis current and quadrature axis current gives
1
2 1 1
4 6 3
dq
g
g
IC D C
XC D C
V
(6.174)
Substitute (6.174) in (6.168),
1 1 2
1
2 1 11 1 2
4 6 3
sys
dq
g
g
A
sys
I
X A X B B W U
V
C D CA B B X W U
C D C
A X W U
(6.175)
Equations (6.175) is in state space form with the state transition matrix sysA ,
input matrix W , state vector X and input vector U . The small signal stability of
the multi-machine system can now be known by finding the eigen values of the state
transition matrix sysA [7]. Since, the rotor angle of any synchronous machine should
be defined with respect to a reference, one state variable corresponding to the rotor
angle of a generator which is taken as reference becomes redundant hence there will
be one zero or a slightly positive eigen value corresponding to the redundancy of the
reference angle.
6.41
6.9 Sub-Synchronous resonance
The sub-synchronous resonance phenomenon [8] was first observed in 1975 in
Mohave, California where the turbine-generator shaft has failed due to torisonal
oscillations interacting with the sub-synchronous oscillations, produced because of
series compensation in transmission line. To understand this phenomenon let there be
a single machine connected to infinite bus. Let the generator be represented by a sub-
transient voltage "E behind a sub-transient reactance "X . Let the effective impedance
between the generator terminal and the infinite bus be e eR jX . The system is shown
in Fig. 6.5.
Fig. 6.5: Series capacitor compensated SMIB system.
Let the system be compensated by a series capacitor with capacitance C as
shown in Fig. 6.5. When a fault or a disturbance occurs then there will be a transient
in the current with a off set and this off set oscillate at a natural frequency of (in case
there is no series capacitor then the off set would be DC)
1/
,
Tn s
cT
Tn s
c
Xrad s
XL C
Xf f Hz
X
(6.176)
Where, sf is the synchronous frequency and "T eX X X . These oscillations
at the frequency nf in the stator current of the generator will induce a slip frequency,
s nf f voltages in the rotor circuits and there by producing slip frequency torque. If
"X
"E
eX eR
V
CX
6.42
nf is a sub-synchronous frequency then, since the rotor is rotating at synchronous
speed in the steady state, the slip ( ) /n s ns f f f will be negative. This condition is
similar to induction motor running at super synchronous speed with a negative slip.
Fig. 6.6: Approximate equivalent circuit of the synchronous generator with sub
synchronous frequency stator currents
The approximate equivalent circuit of the synchronous generator with sub-
synchronous frequency stator currents is shown in Fig. 6.6. The magnetising and core
loss component branch is ignored. The efR is the summation of the transmission line
resistance, stator resistance, rotor resistance referred to stator. Similarly efX
summation of the transmission line reactance, stator leakage reactance, rotor leakage
reactance referred to stator. Since the slip is negative the variable rotor resistance
becomes negative leading to self excitation in the circuit and the oscillations will keep
increasing reaching unacceptable levels. Before discussing the interaction of sub-
synchronous oscillations, due to induction motor effect, with the torsional oscillations
the torsional oscillations will be briefly discussed.
6.10 Torsional oscillations
If two rotating masses are connected together by a non rigid shaft then the two
ends of the shaft are relatively displaced with respect to each and will have an angle
of displacement due to which the shaft gets twisted. The torque delivered through the
shaft is directly proportional to the relative angle of displacement between the two
ends of the shaft. In case the relative angle of displacement between the two ends of
the shaft oscillates then that is called as torsional oscillations [9], [10]. To understand
efX efR
1r
sR
s
6.43
the basic phenomenon of torsional oscillations let a generator be considered
connected to a turbine, with single mass, through a shaft as shown in Fig. 6.7.
Fig. 6.7: Turbine-generator system
Let the generator and turbine inertia constants be 1 2,H H . The relative angle of
displacement between the two ends of shaft is 1 2 . The stiffness of the shaft is
represented by 12K . The torque delivered through the shaft to the generator rotor is
12 1 2K . The equation of motion of the two masses that is turbine and generator
can be written as
2
1 1 112 1 2 1 12 1 22
2M
s
H d ddD D K T
dt dt dt
(6.177)
2
2 2 212 2 1 2 12 1 22
2e
s
H d ddD D T K
dt dt dt
(6.178)
Where, 12D is the damping due to shaft and 1 2,D D are the damping constants of
the turbine and generator rotor. Multiply (6.177) by 2H and (6.178) by 1H now
subtract both the equations, neglecting the effect of damping due to damping
coefficients 1 2,D D ,
2
1 2 2 11 2 12 1 2 12 1 22
1 2 1 2 1 2
2M e
s
H H H Hd dD K T T
H H dt dt H H H H
(6.179)
Generator rotor Turbine
Shaft
1H 2H
1 2
12K
MT
6.44
Linearizing (6.179) and assuming the input mechanical torque is constant leads to
2
2
212 1 2 12 1 2
12 12 1221 2 1 2 2
2212 12
122
2 2 2
2
n K
ss s e
n e
d H H d H HD K T
dt H H dt H H H
d dK T
dt dt
(6.180)
Equation (6.180) shows that it represents a second order damped system with
natural frequency of oscillations being 2n . These oscillations are called as torsional
oscillations. If electrical torque disturbance is of the form cos( )te t , then the
general solution of the equation (6.180) is given as
2 2 2 212 1 2cos sin cost t
n n
KK t K t e e t
P
(6.181)
Where, 1 2,K K are constants depending on the initial conditions. ,P are given as
2 22 2 2 2
1
2 2 2
2 4
2tan
2
n
n
P
(6.182)
The torque at the shaft due to the electrical torque disturbance is given as
12 12shaftT K (6.183)
It can be observed from (6.182) that if the applied electrical torque disturbance
is unidirectional that is 0 then 2nP . If the applied electrical torque disturbance
has a frequency close to the natural frequency of the shaft that is n then
2 nP . The natural frequency of oscillations of the shaft is in the range of 10 Hz
to 40 Hz. Now if the frequency of the electrical torque disturbance is equal to or
greater than the system frequency then 2sP . The value of P will be least in the
6.45
case when the electrical torque disturbance is equal to the shaft natural frequency of
oscillations and hence has the maximum effect on the shaft torque as given in (6.183)
leading to a sever stress in the shaft. Apart from this if the electrical torque
disturbance has negative damping, usually is quite high and the second term in
(6.181) damps out fast, then the electrical torque disturbance keeps increasing in
magnitude causing extreme stress in the shaft. Though there is shaft damping it is
very low and does not help in this situation.
Though the phenomenon of torsional oscillations was explained using two
masses in an actual system like thermal generating system there will be HP turbine,
multiple IP turbine, LP turbine, generator rotor and if the generator has DC exciter
then rotor mass of that exciter. There will be 1n torsional modes of oscillations if n
different masses are connected together by a shaft.
If the torque produced by the current in the rotor of the synchronous machine
due to the sub-synchronous oscillations in the network has a frequency that is close to
the natural frequency of torsional oscillations then it will lead to very high shaft
torque shaftT . If sub-synchronous oscillations are self exciting type then this problem
becomes even more severe and may ultimately lead to failure of the shaft. A way of
avoiding this is to choose the series compensator such that the natural frequency of
oscillations of LC circuit does not produce sub-synchronous oscillations which can
excite the torsional modes. The other way is to have enough impedance in the
network to counter the negative resistance due to induction motor effect and there by
damping the sub-synchronous oscillations so as to prevent exciting the torsional
modes.
6.46
Example Problems
E1. Find the eigen values, eigen vectors and participation factor of the system
represented by state space model and whose state matrix is given as
1 2
3 4
é ùê ú= ê úë û
A
Sol: The characteristic equation is given as
( )
( )( ) 2
0
0 1 21 4 6 5 2 0
0 3 4
l
ll l l l
l
- =
æ öé ù é ù÷çê ú ê ú÷- = - - - = - - =ç ÷çê ú ê ú÷çè øë û ë û
I A
(E1.1)
The solution of the characteristic equation is 1 20.3723, 5.3723l l=- = . Since, one
of the eigen values is positive the system is unstable. The right eigen vector matrix Φ
is defined as
=AΦ ΦΛ (E1.2)
Where,
0.3723 0
0 5.3723
é ù-ê ú= ê úë û
Λ (E1.3)
Let the right eigen vector matrix be represented as
11 12
21 22
é ùF Fê ú= ê úF Fë û
Φ (E1.4)
Then,
[ ] 111
21
0
0l
é ù é ùFê ú ê ú- =ê ú ê úF ë ûë û
I A (E1.5)
6.47
[ ] 122
22
0
0l
é ù é ùFê ú ê ú- =ê ú ê úF ë ûë û
I A (E1.6)
Now, [ ]11 21TF F is the right eigen vector corresponding to the eigen value
1 0.3723l =- and [ ]12 22TF F is the right eigen vector corresponding to the eigen
value 2 5.3723l = . From equation (E1.5) and (E1.6), the following expression can be
written
11 211.3723 2 0F + F = (E1.7)
11 213 4.3723 0F + F = (E1.8)
Equations (E1.7)-(E1.8) are homogeneous equations with a trivial solution of
[ ]11 21 [0 0]T TF F = . This is because the equations are not independent, that is one of
the equations can be written as a multiple of the other equation. Typically, a thn order
homogeneous equation has ( 1)n- independent equations. Hence, for a non trivial
solution let us assume the value of 11 1F = and from one of the equations (E1.7) or
(E1.8) we can obtain 21F . Hence,
11
21
1
0.6862
é ù é ùFê ú ê ú=ê ú ê úF -ë ûë û
(E1.9)
Since, the solution is not unique it is better to normalize the right eigen vector hence
11
21
1 1 0.82451 1
0.6862 0.6862 0.56581.21281
0.6862
é ù é ù é ù é ùFê ú ê ú ê ú ê ú= = =ê ú ê ú ê ú ê úé ùF - - -ë û ë û ë ûë û ê ú
ê ú-ë û
(E1.10)
Similarly, the normalized right eigen vector corresponding to the eigen value
2 5.3723l = is given as
6.48
12
22
0.4160
0.9094
é ù é ùFê ú ê ú=ê ú ê úF ë ûë û
(E1.11)
Hence,
0.8245 0.4160
0.5658 0.9094
é ùê ú= ê ú-ë û
Φ (E1.12)
The left eigen vector can be defined as
1 0.9231 0.4222
0.5743 0.8379- é ù-
ê ú= = ê úë ûΨ Φ (E1.13)
Each row of the matrix Ψ is a left eigen vector corresponding to the respective eigen
value, that is first row is a left eigen vector corresponding to the eigen value
1 0.3723l =- . The participation factor of the states in the thi mode is given as
1 1
2 2 , 1, ,
i i
i ii
ni in
P i n
é ùF Yê úê úF Yê ú= =ê úê úê úF Yê úë û
(E1.14)
The participation factor matrix is given as
1 2[ ]nP P P=P (E1.15)
The participation factor can now be computed as
11 11 12 21
21 12 22 22
0.7611 0.2389
0.2389 0.7611
é ù é ùF Y F Yê ú ê ú= =ê ú ê úF Y F Y ë ûë û
P (E1.16)
6.49
E2. Find the eigen values, eigen vectors and participation factor of the matrix given
below using QR method.
1 4 9 7
1 0 1 4
1 3 2 7
0 2 2 3
é ùê úê ú-ê ú= ê úê úê ú-ë û
A
Sol: The first step in using QR method to find the eigen values is to convert the
matrix A to Hessenberg form. Defined two column matrices ,X Y and unit vector
1U as
00 0
1 1.4142, ( (2,1))
1 00
0 00
sign A
é ùé ù é ùê úê ú ê úê úê ú ê ú-ê úê ú ê ú= =- =ê úê ú ê úê úê ú ê úê úê ú ê úê úë û ë ûë û
XX Y
(E2.1)
1
0
0.9239
0.3827
0
é ùê úê ú- ê ú= = ê ú- ê úê úë û
X YU
X Y (E2.2)
Defining Householders transformation 1H and applying the transformation to the
matrix A lead to
1 1 1
1 0 0 0
0 -0.7071 -0.7071 02
0 -0.7071 0.7071 0
0 0 0 1
TU UH I
(E2.3)
1 1 1
1 -9.1924 3.5355 7
-1.4142 2 1 - 7.7782
0 -3 0 2.1213
0 0 2.8284 3
é ùê úê úê ú= = ê úê úê úë û
A H AH (E2.4)
6.50
It can be observed from the transformed matrix 1A that the rows below the sub-
diagonal of the first column have become zero that is 3rd and 4th row, 1st column
elements have become zero. After repeating this for all columns we finally get
Hessenberg form of matrix A as is given as
1 -9.1924 -3.5355 7
-1.4142 2 -1 - 7.7782
0 3 0 - 2.1213
0 0 - 2.8284 3
H
é ùê úê úê ú= ê úê úê úë û
A (E2.5)
Second step is to apply QR method to the matrix in Hessenberg form, given in (E2.5),
so as to convert it to Schur form. From equations (6.48) to (6.57) the matrix ,Q R for
the first iteration can be obtained as
0.5774 - 0.7383 - 0.1617 - 0.3090
-0.8165 - 0.5220 - 0.1143 - 0.2185
0 0.4271 - 0.4191 - 0.8012
0 0 - 0.8861 0.4635
é ùê úê úê ú= ê úê úê úë û
Q (E2.6)
1.7321 - 6.9402 1.2247 10.3923
0 7.0238 3.1322 - 2.0135
0 0 3.1921 - 2.0117
0 0 0 2.6266
-
é ùê úê úê ú= ê úê úê úë û
R (E2.7)
It can be observed that H =A QR . Define a new matrix
1H =A RQ (E2.8)
Find ,Q R for the new matrix defined (E2.8) repeat this process till the difference
between the consecutive iteration ,Q R matrices is less than a specified value. The
iterative process converges after 22 iterations. The matrix in Schur form is given as
6.51
7.2260 - 7.3791 6.4399 7.5903
0 - 2.5289 - 4.4496 - 0.7679
0 1.7342 - 0.3420 2.5760
0 0 0 1.6449
Schur
é ùê úê úê ú= ê úê úê úë û
A (E2.9)
The eigen values are given as
7.2260 0 0 0
0 -1.4355 2.5536 0 0
0 0 -1.4355 - 2.5536 0
j
j
+=Λ
0 0 0 1.6449
é ùê úê úê úê úê úê úë û
(E2.10)
The left and right eigen vectors can be computed from equations (6.67) to (6.69) and
is given in (E2. 11)
0.8779 0.8166 0.8166 0.9670
0.1390 - 0.4029 0.0499 - 0.4029 - 0.0499 0.0600
0.4361 0.0245 0.3545 0.0245 - 0.3545 0.1771
0.
j j
j j
+=
+Φ
1406 - 0.0853- 0.1865 - 0.0853 0.1865 - 0.1728 j j
é ùê úê úê úê úê úê ú+ë û
(E2.11)
1-=Ψ Φ (E2.12)
Participation factor matrix, can be computed as
0.1646 0.2589 0.2589 0.7515
0.0037 0.4930 0.4930 0.1178
0.5208 0.3731 0.3731 0.2590
0.3183 0.1734 0.1734 0.3898
T
é ùê úê úê ú= = ê úê úê úë û
P Ψ Φ (E2.13)
6.52
E3. A two area test system consisting of four generators and eleven buses is shown in
the figure E3.1 below.
Fig. E3.1: Single line diagram of two area test system [7]
The generators G1, G2 are in area-1 and generators G3, G4 are in area-2. All the four
generators are of same rating that is three-phase 60 Hz, 20 kV and 900 MVA. The
transmission line parameters are defined on the base of 230 kV, 100 MVA. The
system base is taken as 100 MVA for load flow analysis. At bus-7 and bus-9 two
capacitors of 200 MVAr and 350 MVAr are connected, respectively. The bus data is
given in Table E3.1, line data is given in Table E3.2, line flow data is given in Table
E3.3. A net real power of 200 MW is being transferred from area-1 to area-2, which is
from bus-7 to bus-8, as can be observed from Table E3.3. The generators parameters
on 900 MVA, 20 kV bases are given below.
' " ' "0 0
' " ' "0 0
0, 0.2, 1.8, 0.3, 0.25, 8 , 0.03
1.7, 0.55, 0.24, 0.4 , 0.05 , 6.5
s ls d d d d d
q q q q q
R X X X X T s T s
X X X T s T s H s
= = = = = = =
= = = = = =
The static high gain exciters are used at all the four generators and the parameters are
given as 200, 0.01A RK T= = .
2
5 4 1
3
6 7 8 9 10 11
G1
G2 G3
G4
6.53
Table E3.1: Bus data
GENERATION (pu)
LOAD (pu)
BUS VOLTAGE (pu) ANGLE
(degrees) REAL REACTIVE REAL REACTIVE 1 1.03 18.5 7 1.79 0 0 2 1.01 8.765 7 2.20 0 0 3 1.01 -18.4439 7 1.85 0 0 4 1.03 -8.273 7.19 1.70 0 0 5 1.0074 12.0365 0 0 0 0 6 0.9805 1.9858 0 0 0 0 7 0.9653 -6.3737 0 0 9.67 1.00 8 0.9538 -20.0995 0 0 0 0 9 0.9763 -33.5456 0 0 17.67 1.00 10 0.9862 -25.1838 0 0 0 0 11 1.0093 -14.9053 0 0 0 0
Table E3.2: Line data (pu)
From Bus To Bus ResistanceReactanceLine
charging1 5 0 0.0167 0 2 6 0 0.0167 0 7 6 0.001 0.01 0.0175 7 8 0.011 0.11 0.1925 7 8 0.011 0.11 0.1925 5 6 0.0025 0.025 0.0437 4 11 0 0.0167 0 3 10 0 0.0167 0 9 8 0.011 0.11 0.1925 9 8 0.011 0.11 0.1925 9 10 0.001 0.01 0.0175 11 10 0.0025 0.025 0.0437
6.54
Table E3.3: Line flow data
LINE FLOWS
LINE FROM BUS TO BUS REAL REACTIVE
1 1 5 6.9944 1.7888 2 2 6 7 2.1982 3 7 6 -13.67 0.8984 4 7 8 2 0.0508 5 7 8 2 0.0508 6 5 6 6.9944 0.9684 7 4 11 7.19 1.6907 8 3 10 7 1.8515 9 9 8 -1.9062 0.5313 10 9 8 -1.9062 0.5313 11 9 10 -13.8577 1.4375 12 11 10 7.19 0.832
(a) Find the eigen values of the system and comment on the system stability.
(b) In case power system stabilizers are used at all the four generators then find
the eigen values and comment on the effect of power system stabilizer on the
system. The power system stabilizer parameters are given as
1 2 3 450, 10 , 0.05 0.02 , 3.0 , 5.4PSS WK T s T s,T s T s T s= = = = = =
Sol:
With the given system data first the generators are initialized using equations (5.76) to
(5.89). Then the system is linearized around the initial operating point. Each generator
is represented by a 7th order model, 2.2 generator model along with first order exciter
model. The state vector of an thi generator is given as
' '1 2, , , , , , , 1, 2,3, 4i i i di qi di qi fdiX E E E id w y yé ùD = D D D D D D D =ê úë û
(E3.1)
Since, there are four generators there will be 28 states. The state matrix of the
linearized system can be computed as given in equation (6.175). The size of the state
matrix is 28 28´ and hence there will be 28 eigen values. The eigen values can be
computed from the state matrix as explained in section 6.4 using QR method.
6.55
The eigen values of the system are given in Table E3.4. Each eigen model along with
its frequency of oscillations and damping ratio are given in Table E3. 4. It can be
observed from Table E3.4 that there is one eigen mode with slightly positive value
and a second eigen value very close to zero. The first zero eigen value is because
there is no reference for the generator angles and hence the angles are not unique.
Second zero eigen value is because of the assumption that the generator electrical
torque is not effected by the speed variations.
Table E3.4: Eigen values of the system without PSS
No Eigen Value Frequency
(Hz) Damping
ratio 1 0.0004 0 1 2 -0.001 0 1 3 -0.1004 0 1 4 -3.3475-j0.0092 0.0015 1 5 -3.3475+j 0.0092 0.0015 1 6 -3.6881 0 1 7 -0.0281- j3.7805 0.6017 0.0074 8 -0.0281+ j 3.7805 0.6017 0.0074 9 -5.6707 0 1 10 -0.7196- j 7.1819 1.143 0.0997 11 -0.7196+ j 7.1819 1.143 0.0997 12 -0.9124- j 7.1702 1.1412 0.1262 13 -0.9124+ j 7.1702 1.1412 0.1262 14 -13.2546- j 0.1293 0.0206 1 15 -13.2546+ j 0.1293 0.0206 1 16 -18.9977- j 14.3284 2.2804 0.7984 17 -18.9977+ j 14.3284 2.2804 0.7984 18 -18.3899- j 17.5080 2.7865 0.7243 19 -18.3899+ j 17.5080 2.7865 0.7243 20 -27.9172 0 1 21 -28.8569 0 1 22 -31.6775 0 1 23 -32.3703 0 1 24 -36.2718 0 1 25 -95.2973 0 1 26 -95.9239 0 1 27 -97.4597 0 1 28 -97.5359 0 1
The eigen mode 7 and 8 are complex conjugate pairs with a very low damping ratio of
0.0074 and frequency of oscillation of 0.6017 Hz. These two modes are inter area
modes. The normalized participation factor of the states in mode 7 are given in Fig.
6.56
E3.5. It can be observed that the rotor angle and speed of generator G1, G2, G3 and
G4 are participating maximum in this mode.
Fig. E3.5: Participation factor of state in eigen mode 7
Fig. E3. 6: Mode shape of the eigen mode 7
The elements, corresponding to the angles of generators G1, G2, G3 and G4, of the
right eigen vector of the eigen mode 7 are plotted in Fig. E3. 6. It can be observed that
6.57
the generators G1, G2 are in one direction and generators G3, G4 are in almost
opposite direction. From this it can be understood that eigen modes 7 and 8 are due to
the inter area oscillations between generator G1, G2 in area-1 against generator G3,
G4 in area-2.
(b)
Table E3. 5: Eigen values of the system with PSS
No Eigen Value Frequency
(Hz) Damping
ratio 1 0.0004 0 -1 2 -0.001 0 1 3 -0.013 0 1 4 -0.1007 - 0.0000i 0 1 5 -0.1007 + 0.0000i 0 1 6 -0.1007 0 1 7 -0.1836 0 1 8 -0.1837 - 0.0001i 0 1 9 -0.1837 + 0.0001i 0 1 10 -0.3489 - 0.1576i 0.0251 0.9113 11 -0.3489 + 0.1576i 0.0251 0.9113 12 -3.2312 0 1 13 -3.2794 0 1 14 -3.7041 0 1 15 -0.1998 - 3.7846i 0.6023 0.0527 16 -0.1998 + 3.7846i 0.6023 0.0527 17 -5.5333 0 1 18 -1.0938 - 7.4313i 1.1827 0.1456 19 -1.0938 + 7.4313i 1.1827 0.1456 20 -1.2898 - 7.4316i 1.1828 0.171 21 -1.2898 + 7.4316i 1.1828 0.171 22 -12.8772 - 0.0975i 0.0155 1 23 -12.8772 + 0.0975i 0.0155 1 24 -18.8004 -14.4271i 2.2961 0.7933 25 -18.8004 +14.4271i 2.2961 0.7933 26 -18.1771 -17.6278i 2.8055 0.7179 27 -18.1771 +17.6278i 2.8055 0.7179 28 -27.5727 0 1 29 -28.3955 0 1 30 -31.6894 0 1 31 -32.4723 0 1 32 -36.2533 0 1 33 -50.0707 - 0.0025i 0.0004 1 34 -50.0707 + 0.0025i 0.0004 1 35 -50.1297 0 1 36 -50.1406 0 1 37 -95.3003 0 1 38 -95.9264 0 1 39 -97.4621 0 1 40 -97.5383 0 1
6.58
The eigen values of the system with PSS at all the four generators is given in Table
E3. 5. Each PSS is represented by three dynamic equations. Hence, the total number
of state will be 40. The eigen mode 15 and 16, with damping 0.0527 and frequency
0.6023 Hz, are inter area modes. The participation of state in eigen mode 15 is given
in Fig. E3. 7.
Fig. E3. 7: Participation of the states in eigen mode 15
It can be observed that the states angle and speed of generator G1, G2, G3 and G4 are
participating maximum. Hence, it can be said that with PSS connected at the
generators G1, G2, G3 and G4 the damping of the inter area mode has increased from
0.0074 to 0.0527, which is a very significant improvement.
6.59
References
1. A. M. Lyapunov, Stability of Motion, English translation, Academic press Inc.,
1967.
2. J.G.F. Francis, “The QR Transformation-A unitary analogue to the LR
transformation,” The computer Journal, Part I, Vol. 4, pp. 265-271, 1961; Part
II, pp. 332-345, 1962.
3. W. G. Heffron and R. A. Phillips, “Effects of modern ampliydyne voltage
regulators in underexcited operation of large turbine generators,” AIEE Trans.,
PAS-71, pp. 692-697, August 1952.
4. F. P. DeMello and C. Concordia, “Concepts of synchronous machine stability
as affected by excitation control,” IEEE Trans., PAS-88, pp. 316-329, April
1969.
5. E. V. Larsen and D. A. Swann, “Applying power system stabilizers, Part I:
general concepts, Part II: Performance objectives and tuning concepts, part III:
practical considerations,” IEEE Power Appar. Syst., PAS-100, Vol. 12, pp.
3017-3046, December 1981.
6. P. Kundur, M. Klein, G. J. Rogers and M. S. Zywno,”Application of power
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