Friends, today we start with the study of
small signal stability. We shall devote our
discussion towards the small signal stability
of a single machine infinite bus system that
is for understanding the concepts related
to the small signal stability, we will consider
a simple system and that is a single machine
connected to an infinite bus. Okay the small
signal stability is the ability of the system
to maintain synchronism under small perturbations.
The small perturbations continuously occur
in any power system due to changes in loads
and generations.
Now for analyzing the small signal stability
of any system, the system model can be linearized
around an operating point that is the disturbances
are considered to be so small or incremental
in nature. So that we can develop a linear
model of the system around the operating point,
once we develop the linear model of the system
we can understand the behaviour of the system
under small perturbations, various parameters
of the system which affect the stability of
the system further the moment we have a linear
model we can apply the linear control system
theory for designing the controllers.
Now here when I talk about the controller
particularly we are interested in designing
the excitation system control that is the
voltage regulator and the power system stabilizers
as we will see that in any power system right,
the actual system is some what complex it
is not as simple as a machine connected to
infinite bus is always a multi machine system
right. Now in any multi machine system as
you know that the system will have different
modes of oscillations, the modes of oscillations
are classified as as local modes of oscillations,
inter area modes of oscillations and the control
modes okay.
Now, primary requirement of the system is
that this system should have for stability
this will have positive synchronizing torque
coefficient and positive damping torque coefficient.
The stability of the system will be affected
if any of these two torques or any of these
two torque coefficients become negative. Okay
now to start with we will study first a simple
model considering constant flux linkages in
the field winding that is we will first start
with the constant flux linkage model, next
we will include the the field winding dynamics
that is we will consider next step the constant
field voltage and must we assume these voltages
applied to the field winding as a constant
okay we the model is developed so that we
take care of the the dynamics of the field
winding that is changes in the field flux
linkages.
This particular stage when we say that the
applied voltage is fixed means it is a practically
a manual control after studying this model
with constant field voltage, we will extend
it to the model which will include the automatic
voltage regulator and we will study the affect
of the parameters of the excitation system
and gain setting of the automatic voltage
regulator on the stability of the system.
Then the next step will be we include auxiliary
controllers that is the power system stabilizer
that is this the stage by step development
will give us the complete insight into the
small signal stability problem.
The system which we consider for small signal
stability to start with is a machine infinite
bus system where a synchronous generator is
connected to a large system which can be characterized
by a infinite bus, the equivalent circuit
can be represented as generator the we represent
the terminal voltage of the generator, the
line is represented by an equivalent impedance
Z equivalent equal to RE plus j times XE and
connected to an infinite bus right.
Now here here even in a multi machine system,
multi machine system right. Now if you if
you take out one generator and the connected
line right then just will be system if you
represent by a infinite bus then it becomes
a machine infinite bus system. Now as we continue
our studies to simplify our model and understanding
the resistance of the line or the equivalent
resistance RE we may may be ignored.
Now when we consider a classical model that
is we assume the field flux linkage is to
be constant. Okay now when we consider the
classical model of the synchronous generator
then the classic the synchronous generator
is represented by a constant voltage behind
direct axis transient reactance right. Now
here we will represent the voltage behind
transient reactance by the symbol E prime
and the voltage of the infinite bus as EB,
okay and the phase angle, phase angle 
between E prime and EB is denoted by delta
right and that is the power angle of the system.
Now for this machine infinite bus system we
have derived earlier the the power angle characteristic
okay the power angle characteristic to derive
we can start with the the terminal current
of the synchronous generator as the E prime
minus EB. Now here we are representing the
ah voltage behind transient reactance if it
is represented as the as E prime angle 0 that
is you you consider this as a reference voltage
generally we do other way around we consider
the infinite bus voltage as reference and
the internal voltage leads the reference voltage
by some angle delta okay but it does not matter
very much.
We can write down the expression for current
It this is the first step for solving this
problem by this what we know will be the the
you know whenever we start solving the problem
the terminal voltage of the machine will be
known to us right and therefore, once the
terminal voltage is known we can find out
the internal voltage by this formula E prime
equal to Eto plus j times Xd prime into Ito,
where here now we denote this total reactance
Xd, XT as Xd prime plus XE earlier the power
angle characteristic which we derived was
relating the terminal voltage with the internal
voltage but now here we will be deriving the
power angle characteristic relating the infinite
bus voltage with respect to the internal voltage.
Okay therefore the total reactance which comes
will be Xd prime plus XE where XE is the reactance
of the transmission line okay.
Then the complex power behind Xd prime is
given by S pi, S prime is equal to P plus
j times Q prime you can call it P prime. Okay
this is E prime into It complex I am sorry
It conjugate, It conjugate that is the complex
power is P prime plus j times Q prime and
this is obtained by multiplying this voltage
E prime by conjugate of It, okay when you
make the simplifications.
We can write down the expression for air gap
torque. Now we all know actually that in case
we neglect losses right then the air gap torque
is equal to the the power output impairing
your system that when you consider the per
unit system then the power output at the terminal
of the machine is same as the air gap torque.
Okay and this is our standard power angle
characteristic, now here at this stage I would
like to mention that in case instead of having
a non-salient pole machine, if we have a salient
pole machine right then the power angle characteristic
will be different from what is given here,
it will have a another term which is known
as the reluctance torque. Okay and one can
write down the expression for the power angle
but this is the primary requirement to start
with. The next step is you will linearize
you linearize the expression for air gap torque
around an operating point.
Our initial operating condition is characterized
by the angle delta equal to delta naught that
when we are operating and transferring certain
amount of power right the initial value of
delta is delta equal to delta naught, okay.
The linearized equation is written in the
form delta Te is equal to partial derivative
of T with respect to delta into delta delta
where this expression is very simple there
is only a Te is function of function of one
variable only here. We are assuming this E
prime and EB constant, okay. The infinite
bus voltage assumed to be constant the voltage
behind transient reactance is assumed to be
constant the total line reactance is constant
and therefore the torque is function of delta
only one variable and therefore when you obtain
the small change in the air gap torque delta
T is equal to partial derivative of delta
T divided delta delta into delta delta.
Therefore, here basically this partial derivative
becomes derivative term itself right there
is no other you can say parameter on which
the Te depends okay but in case actually these
are also variables right then when I talk
about this term it is the partial derivative
with respect to delta not delta when you take
other terms also then you have to write the
partial derivative of this quantity with respect
to not delta say when you take this partial
derivatives there there so many variables
will be there.
You have to keep on taking all the variables
then you will have delta delta delta E prime
delta E prime like this it is basically it
is actually the Taylor series exponents time
to expand the equation around the operating
condition right. Now we come to the basic
equations of motion, we have developed earlier
the swing equation of the synchronous generator
and the swing equation is a second order non-linear
differential equation. This equation can be
represented by two first order linear I am
sorry 2 first order non linear differential
equations it { it will continue to be non
linear.
These two equations are written here as p
delta omega r equal to 1 upon 2H Tm minus
Te minus KD delta omega r and second equation
is p delta equal to omega naught delta omega
r, where delta omega r is the deviation of
the rotor speed with respect to synchronous
speed right. The KD is the damping torque
coefficient to for the sake of completeness
right, we include damping term also well earlier
when we discussed actually the KD was not
explicitly included in the equation but the
complete equation is in this form.
Now when the equation is written in this form
the speed deviation the delta omega r x is
is expressed in per unit delta omega r is
expressed in per unit the mechanical torque,
electrical torque are also expressed in per
unit. We know how to define the initial constant
H the time in this equation is the time instead
of putting in per unit we always prefer to
express time in second is right and therefore,
the second equation has p delta equal omega
naught into delta omega.
Now these are the 2 basic equations around
which the the we will be you can say developing
the small signal model using these equations,
small signal model and studying the basic
concepts of small signal stability. Now you
what we do is that this this these two equations
that is 18.5 and 18.6, we will linearize around
the operating condition okay.
Now to linearize this around the operating
condition one assumption here which we are
making is that there is a mechanical power
input remains constant that is Tm is constant
right and Tm is equal to the initial power
output that is Teo. Okay and therefore when
we talk about the small perturbations our
equation will become p delta omega r equal
to 1 upon 2 H delta Tm minus Ks delta delta
minus KD delta omega r this delta Tm delta
Tm is considered to be a change in mechanical
torque.
Okay but generally this change is included
in the model included in the model to obtain
the dynamic performance by giving a small
change in the mechanical torque right. Otherwise
the mechanical torque remains constant this
delta T term has been replaced by Ks delta
delta and this term Ks is synchronizing torque
coefficient right and what will the unit of
this coefficient unit per unit torque. Now
in this equation actually be the delta delta
is expressed in radians electrical radians.
Okay similarly, the unit for KD the torque
in per unit per, sir in previous equation
T delta equals to omega r naught into delta
omega r, so all this equation comes because
T delta equal to I think omega naught plus
delta omega delta r. Now if you see the basic
equation it is like this, d delta by dt is
equals to omega minus omega naught that omega
naught is the synchronous speed okay. Now
what we do here is that this difference we
do not by in fact actually this speed actual
speed is called omega r is the speed of the
rotor. Okay this will do not this difference
is noted by delta omega r. Now what we do
is that divide this by omega naught, so that
this quantity becomes in per unit and multiplied
by omega naught.
So that I can represent d delta by dt as as
delta omega r into omega naught while this
delta omega r is expressed in per unit. Okay
while in this equation in this whole equation
delta is expressed in electrical radians because
when you solve these problems, you have to
be very clear about the units. Okay see originally
when you see the our swing equation right,
it is of the form d omega divided dt d omega
r divided by dt 1 upon 2, I am sorry 2H divided
by s 2H divided by omega r omega r, okay equal
to accelerating power okay.
Now here when I express this omega r in per
unit this is this is omega naught not omega
r this is omega naught this is omega r right
then this term is combined here. Okay so that
I have this coefficient equal to 2H only that
is that is when you write the equation in
this form, 2H d omega r by dt equal to Pa.
Okay here the this omega r is a per unit further
omega r can be represented as as omega naught
plus delta omega r that the omega naught is
in this particular case when it is per unit
omega naught will become 1 that is you can
put there is a 1plus delta omega r therefore,
when you substitute this expression here I
can write down the equation as 2 times H d
by dt of delta omega r equal to Pa, is it
okay that is that is since we know actually
the derivative of this 1 is 0 therefore d
by dt of delta omega r is same as d omega
r by dt and that is why when you look at the
swing equation which we have written here.
Okay original swing equation this is not a
linearized equation we can write down this
delta omega r. Okay is it clear actually ah
these 2 equations are very important actually
because when you solve a given problem right
the it is very important to understand that
how the quantities are expressed what quantities
are in per unit? what the unit of time? what
is the unit of delta? what is the unit of
speed deviation? all these terms are very
important. Okay now when you look at this
equation.
This is a equation in terms of small variations
of speed angle and mechanical torque, okay
and therefore we can take the Laplace transform
of this equation right. When you take the
Laplace transform it will become something
like this s times delta omega r, s equal to
1 upon 2 H delta Tm (s), okay minus Ks times
delta delta s minus KD times delta omega r
s that is in this equation all these variables
are the Laplace transform. Okay and therefore
now what we can do is that we can write down
we can write down delta omega rs as 1 upon
2H times this whole expression okay.
Now this equation is represented in the block
diagram form here at this summasing at this
summation point we have delta Tm Ks times
delta delta that is your delta Te that is
delta delta is here delta T and KD times delta
omega r. Now when I represent this in the
block diagram form all these variables are
the Laplace transform but for the sake of
convenience, we may not write ah delta delta
as delta delta with s. Okay but we understand
that in the in the block diagram or the transfer
function model all the variables which we
represent they are the Laplace transform of
the original variables. Okay delta Tm also
delta Tm (s) therefore this is the summation
point where we derive a quantity which is
given in this bracket delta Tm minus Ks delta
delta minus KD delta omega.
Now this quantity when it is multiplied, now
here it is 2 times Hs when you multiply this
by this block is one upon 2 times Hs right,
we get delta omega r here okay. Now the next
step here is to linearize the the second equation.
Our second equation was p delta equal to omega
naught into delta omega r but this delta delta
is our our actually the delta can be written
as delta o plus delta delta and therefore
p delta is same as p delta delta because delta
o is constant that is if you take the derivative
of this term that is d delta by dt right then
this comes out to be same as d delta delta
by dt.
Okay now if you take the Laplace transform
of this equation then it will be s times,
s times delta delta s equal to delta omega
r s into omega naught therefore, I can now
write down delta delta s equal to omega naught
by s into delta omega r s. Okay therefore
the the speed deviation right is related to
the angle deviation by this transfer function
omega naught by s. Okay and therefore in this
block diagram the this transfer function omega
naught by s relates the angle deviation that
is delta delta to delta omega r. Okay and
therefore this block diagram is extremely
important to understand the small signal stability
of a power system because we will be developing
the complete model including including additional
blocks to this basic block diagram right.
Now the next step to understand the whole
thing is that we develop a characteristic
equation characteristic equation of this system
right.
Now to develop the characteristic equation,
we can make use of this block diagram and
relate this output delta delta this output
delta delta is to be related to change in
mechanical this is the input right therefore,
if you look at this block diagram input is
changed in mechanical torque output is delta
delta therefore this is a single input, single
output system. Okay they normally they call
it SISO single input single output system
and you can simplify this model and obtain
a simple transfer function relating the delta
delta to delta Tm that is first we write down
what is delta delta as omega naught by s 1
upon 2 time Hs into this whole quantity which
is delta Tm minus Ks delta delta minus KD
delta omega, okay.
Now in this equation what we do is that we
replace this delta omega r, delta omega r
by delta delta upon omega naught into s right
because this relationship is there in the
block diagram. You can see this block diagram
relationship delta delta is equal to omega
naught by s into delta omega r right. So that
when you look at this equation we have now
delta Tm delta delta and other constant terms.
Okay and therefore when you rearrange the
whole thing, you can write the equation in
this form s square delta delta KD by 2H s
delta delta plus KS by 2H omega naught delta
delta equal to omega naught by 2H delta Tm
right. Now when you equate this expression
to 0, we get the characteristic equation for
the system that this is this is the input
input term right.
Therefore to to obtain the characteristic
equation we can use this expression therefore
the 
characteristic equation is in the form s square
plus KD by 2 Hs plus Ks omega naught by 2H
equal to 0. Since the system which we have
considered is a second order system therefore
characteristic equation is also a second order
characteristic equation. Now you can see here
the the coefficients are depending upon the
damping term KD the system inertia constant
H the synchronizing torque coefficient Ks.
Now these are the system parameters further
you can easily see that Ks depends upon the
operating condition. Okay it also depends
upon the line reactance whether the expression
for Ks is for this single machine infinite
bus system the expression for synchronizing
torque coefficient is E prime into EB divided
by Xt into cos delta o right and therefore
we can easily see here actually that in this
characteristic equation the the coefficients
are function of loading that is system operating
condition and system parameters also the Xt
depends upon the line reactance.
Similarly, it depends upon direct access transient
reactance right, now when this equation is
similar to our standard equation in this form
s square plus to times zeta omega n into s
plus omega n square and you can identify identify
that this coefficient 2 times zeta omega n,
2 times zeta omega n is equal to KD by 2H.
Similarly, this omega n square term that is
this term is equal to KD by 2H while this
omega n square term is equal to Ks omega naught
divided 2H.
Okay now when you when you obtain the roots
of this characteristic equation right the
the roots can be written as s1, s2 the small
s s2 equal to minus zeta omega n plus minus
j times j times omega n square root of 1 minus
zeta square that is in this case the roots
will depend upon whether they are complex
conjugate or real on the value of zeta, if
zeta is if zeta is greater than 0 and less
than 1 right. We will get these 2 roots as
complex conjugate roots right.
Now a system whose zeta is greater than 0
and less than equal to one right we say that
the system is under damped under damped. A
system whose zeta is equal to 1 is critically
damped is critically damped and a system whose
zeta is greater than 1 is over damped 
right and and system whose zeta is less than
0 is unstable it is in negative damping system
is unstable right in any system, we have to
ensure that the damping is adequate.
So that the oscillations which are generated
are damped, whenever we design the controller
for the system we have to achieve certain
minimum damping for all the modes which are
present in the system right. Now here 
we can also give some more interpretation
to the damping term zeta that is if you plot
these roots in the s plane 
then assuming that damping zeta is positive
the roots will appear in the s plane because
here we, so the real part we call it sigma.
We call this as omega and this axis and any
any root is written as sigma plus minus j
omega right. Now if the system is a stable
system like having its complex conjugate or
the roots lying in the left half of the s
plane 
then this term that is this can be this can
be is written equal to how much zeta omega
n while this term will be equal to how much
omega n square root of 1 minus zeta square
what will be this quantity omega n therefore,
this omega n is the distance of the root from
the origin okay.
Now if you find out this if you look at this
angle theta right then cosine theta is equal
to zeta right therefore whenever we perform
this Eigen value analysis of the system. We
we can draw in the s plane a line which will
represent certain value of zeta that is you
can 
that is suppose this is your s plane right
and if I say that this line represents zeta
equal to say .1, this line will represent
zeta equal to say something more than .1,
.2 like this.
Now in case all the roots lie on this side
of this line it means each mode or there is
no mode whose damping is less than .2 right
therefore whenever we design the control for
this for improving the system stability particularly
the AVR tuning or power system stabilizer
tuning right, we perform the root locus analysis
okay and vary the parameters like gain setting
and see actually that you get the optimum
performance but at the same time you ensure
actually that no root is having its damping
less than the desired value that therefore
this this is a very important way to to ensure
that we do not go to a stage where any do
any of the roots is less than or any of the
roots is having its damping less than desired
value okay.
Now for this system which we are considering
here the omega n is the natural frequency
of oscillation it is given by this formula
omega n is equal to square root of Ks omega
naught by 2H and damping ratio zeta is one
upon 2 times KD divided by square root of
Ks 2H into omega naught. Okay these 2o terms
you can derive and see that you get the value
of omega n and zeta like this. Now the next
point which we will study is, see this the
two differential equations which we have derived
the linearized differential equations one
differential equation was written in this
form, okay the.
Similarly, we are written data written the
second differential equation in the linearized
form, okay. Now you can arrange these two
differential equations right and write the
model in the vector matrix form that is finalizing
the stability of the system right the most
convenient form of representing the model
is vector matrix form that is we write the
model in the form of X dot equal to Ax plus
Bu right.
Now these two equations which we have derived
can be arranged in this form that is d by
dt of delta omega r equal to minus KD by 2H
delta omega r minus Ks by 2H delta delta plus1by
2H delta Tm second equation is d by dt of
delta delta equal to omega naught into delta
omega r that is all. Okay therefore, these
two equations which we have derived can be
written in the vector matrix form and when
we make use of the the say mat lab for analyzing
the system performance right then we can find
out the Eigen value of the system matrix A
right for studying the stability characteristic
of the system right.
Now you have to understand very clearly here
that this matrix A is function of function
of operating condition and system parameters
this I will be you know emphasizing again
and again here that whenever you write down
the model in the form X dot equal to Ax plus
Bu right then the basic stability characteristics
are determined from the system A matrix. Okay
therefore live let me summarize here that
what we have done till now is we have developed
the small signal model for the machine infinite
bus system and we have represented this model
in the form that is vector matrix form X dot
equal to Ax plus Bu.
We can write down the A matrix we can write
down the B matrix and the state variables
in this model are delta omega r and delta
delta, these are the state variables while
delta e delta Tm in this case is a input,
okay. Now I will just give one more important
information that the importance of this term
2H okay, now to understand this importance
of this term 2H, let us look at our swing
equation again.
Our swing equation is d by dt omega r equal
to 1 upon 2H Ta okay, where omega r is expressed
in per unit, okay. Now now suppose you have
a synchronous generator and which is at stand
state 0 speed okay and if you apply the rated
torque rated torque then in how much time
the synchronous generator will attain its
the rated speed right. Now to obtain the time
in which this synchronous generator will attain
its rated speed can be obtained by integrating
this equation that is you integrate this equation
assume Ta equal to 1 per unit okay, Ta equal
to 1 per unit and just tell me in uh quickly
in how much time the synchronous generator
will attain its rated speed.
Okay the synchronous generator will attain
its rated speed starting from 0 towards rated
value when it is applied with a rated rated
accelerating torque that is Ta is equal to
1 per unit and therefore this time 2H is also
denoted as the mechanical starting time of
the synchronous generator that is TM is equal
to 2H therefore, if you see in the model we
have a term 1upon 2H therefore this 2H is
similar to a time constant and is the mechanical
starting time of the synchronous generator
in literature you will find actually that
many times we represent this term this TM
by a symbol capital M that is you will find
actually when you read the literature that
this block 1 upon 2 times Hs is also represented
as one upon Ms.
Now here you should not get confused with
the term M, the angular momentum of the rotor
right the M here is actually the in the small
del will be equal to 2H and which is the mechanical
starting time of the synchronous generator
rotor when it is subjected to a rated torque.
Similarly, if the machine is running at constant
speed a rated speed and if you apply a retarding
torque equal to the rated torque then its
speed will come down to 0 in a time equal
to 2H right.
Therefore, that also gives the information
that how much time the machine will take to
retard from its lateral speed to 0speed therefore
with this today I will conclude my presentation
by saying that we have developed a linear
model of machine infinite bus system considering
considering constant field flux linkages.
Okay we have also studied the ah the the the
importance of damping ratio zeta right and
whenever we design the system right the zeta
is going to be one of our design parameters,
okay. Thank you!
