We continue our discussion on the dqo transformation
on the flux equations of a synchronous machine. Recall, that the dqo transformation is a time
variant transformation and the basic reason why actually we go for dq transformation is
to get a set of time invariant equations, as far as, the flux relationships are concerned.
In the previous lecture we began transforming even the differential equation, that is, the
Faraday’s law applied to the flux of the machine and we noticed, that there were, whenever
we apply a time variant transformation in the flux equations in the new variables, we
do get, what is known as a speed EMF term. We will just recap what we have done first
before we proceed further in this lecture. So, today’s lecture is we continue to see
the implications of the d-q-o transformation on the machine equations. Now, recall, that
the d-q-o transformation, essentially, transforms any a, b, c variables, the phase variables
a, b, c using a transformation matrix C p into f d, f q and f 0, which are known as
the d-q-0 variables. C p is a function of theta, theta is of course, the electrical
angle, which we have discussed before. Similarly, of course, f d, f q and f 0 are equal to C
p inverse of f a, f b, f c. So, we assume, of course, C p inverse does exist. In fact,
we can verify that it does exist, remember that by doing a transformation.
We are just, re, we are going to reformulate the equations in the new variables. So, we
are not writing any fundamentally new physical equations, but we are just reformulating the
existing, we are just reformulating the existing flux relationships, as well as, the flux differential
equations in the new variables. So, once we can do the analysis in the d, q, 0 variables,
which is presumably going to be simpler. After we do the analysis, we can transform back
to the a, b, c reference frame using this transformation. So, that is the basic idea
of the transformation. Now, C p, of course, is this. This K d, K
q and K 0 are non-zero constants, three non-zero constants, which are used in the transformation.
Now, the reason why I have not put any numerical value on this K d and K q and K 0 is, that
it is not essential to the, the values of K d, K q and K 0 can be arbitrary, but non-zero.
If they are 0, of course, you will not get C P inverse, C P inverse will not exist. So,
K d, K q, K 0 are in fact, non-zero constants and whatever be their value, as long as, they
are not zero, they satisfy our purpose of trying to convert the time variant flux relationships
into time invariant flux relationships in the d, q, o, 0 variables. C p inverse, in fact, looks like C p transpose.
In fact, if you look at the structure, you will see, that you have got cos theta cos
theta minus 2 pi by 3 and cos theta plus 2 pi by 3. These constants K 1, K 2 and K 3
are related to K d by this relationship, K 1 is equal to 2 by 3 times K d, K 2 is equal
to 2 by 3 times K 2 and K 0 is equal to 1 upon 3 times K 0. So, this is basically the
inverse, you can just verify this, of course, at leisure by multiplying C p with C p inverse
and just verifying that it does turn out to be an identity matrix, which will, of course
verify, that C p inverse is indeed the inverse as I have written it down here. Now, the basic idea here is to convert the
flux equations. So, what we will do is try to convert the a, b, c. The size, remember,
is nothing, but psi a, psi b and psi c and psi R of course, is psi f, psi h, psi g and
psi K psi dq. So, what we do is try to change the variables a, b, c to dq0. Of course, psi
R we will keep unchanged. So, the transformation, if you look at it, is C p.
The sub matrix C P, the sub matrix of this larger matrix had got this element as C p,
this is 3 by 4 null matrix, this is 4 into 3 null matrix and this is an identity matrix,
which is 4 into 4. So, what I will do is apply the transformation
to these variables. Remember, that the flux current relationship is psi s is equal to
L ss of theta L sr of theta L ss of theta into i s into L sr of theta into i r and similarly,
psi r is expressed in this fashion. Now, if I call this the L matrix, I will call
this L, it is, it is easy to see, that psi, if you try to reformulate this particular
equation, we are just reformulating it in the new variables. So, psi dq0 psi R will
be this into L, L is nothing, but this matrix. Remember, of course, these are sub matrices,
so this into L into this transformation into i dq0 and i r. Now, the good thing, which
we saw in the last lecture was when we work out this, we did work out one particular term
in this L matrix, but I encourage you to actually try out to evaluate the complete L matrix. The complete L matrix, if one evaluates one
will find, that the, of course this, you get this new relationship where this dash, I have
called this L ss dash, L sr dash, L ss, L rs dash and L rr and L rr of course, remains
unchanged. So, L ss dash, so this should be L ss dash, is nothing, but it turns out is
a diagonal matrix, it is an interesting thing. It is a diagonal matrix where L d, L q and
L 0 are not functions of theta. So, that is, the basic beauty of this transformation is
that the relationships in the d, q, 0 variables are in fact, the flux current relationship
in the d, q, 0 variables are in fact, independent of theta, not only that, as you see L d for
example, if I just expand this, you will find, that psi d is equal to L d into i d plus 0
into i q plus 0 into i 0. So, there seems to be decoupling in the flux
relationship. That is also an interesting thing, which happens because of the fact,
that this is diagonal. In fact, L sr dash is this, M af by K d, M
ah by K d, 0, 0, 0, 0, M ag by K q, this is wrongly written, so I will write this as K
q and M ah by K q. And the last column of course is, the last
row is 0. Now, L sr dash is equal to L rs dash, only if this is true. So, it is not
true in general. See, remember L sr was equal to L rs transpose, but this is true only if
we choose K d and K q, such that they satisfy this relationship. K d square is 2 by 3 and
K q square is equal to also 2 by 3. Now, what we see here is, of course, there
is a complete decoupling between what are known as the d-axis coils. What are the d-axis
coils? f, h and the stator coil, which is transformed into the d q frame. So, if you
look at this will be f, h and d on this and g, k and q here on the q-axis. So, the basic
idea here is that basic thing, which comes out is, that the fluxes in the coils of the
d-axis including this, you know, this fictitious coil, we will call it the d, d-axis coil,
is not dependent on the q-axis currents. So, that is a very interesting thing.
In fact, it is tempting to think, think, that this transformation, what it essentially has
done is represented the three stationary stator coils in the a, b, c frame for reference.
It is, kind of, converted this three stationery a, b, c coils into two rotating d, q-coils,
which are also rotating at this angular frequency omega. This is tempting, but as we shall see
soon, the, when we write down Faraday’s equation we will see, that the emf induced
in this will be dependent on the fluxes caused in this axis.
So, you know, we should remember, that although this gives a nice picture of you know, the
decoupling of the fluxes and the currents in the d and q axis, the emf equations of
course, have a coupling. So, what do I mean by that? We did this in the last class; we have seen
the d psi by dt. Psi, of course, is nothing, but psi s and psi r, is equal to minus of
R i minus of v, this is what basically our flux equation were, which we did in, you know,
about two classes back. R of course, is the diagonal matrix, which
has sub matrix R s and R r; R s and R r. R r, of course, are the, it is a diagonal matrix
containing the resistances of the f, g, h and k coils; i of course, is i s and i R;
v is nothing, but v s and v r. So, what you have if we just take out the psi s equations?
We will have d psi s by dt is equal to minus R s i s minus v s. Of course, psi s, you know,
the subscript s actually denotes psi s is nothing, but actually psi a, psi b and psi
c. And similarly, i s and v s are i a, i b, i
c and v a, v b, v c. Now, if you look at this equation, we can
substitute psi a, psi b, psi c as nothing, but C p into psi dq0. So, our equations, in
fact, become this, so I have just re-written the equations. These equations, these ones
I have re-written it like this and the re-written equations are substituted for i s, which is
nothing, but i a, i b, i c, I substituted it by dq0 variables, so v s is nothing, but
C p into v dq0, ok. So, if you recall our discussion in the previous
class, I had asked what is C p psi dq0? In fact, it is minus of C p minus of this is
nothing, but minus of C p. This is, you are applying the chain rule plus this extra term.
Remember, that C p, unlike some other transformations we did in the first few lectures of this course,
C p is in fact the function of theta. Theta, in a synchronous machine, is a function of
time, it cannot be considered. Even in the steady state conditions you will find, that
theta is, in fact, varying continuously. So, dC p by dt, in fact, had to be evaluated,
we gave to apply chain rule. So, what we will get is this, in fact, is having two terms.
So, your flux relationship, so I will just, this has been rewritten again, this particular
term dC p by dt can be written as dC p by d theta into d theta by dt. So, this extra
term, as we discussed in the class, previous class, comes out because of being mathematically
consistent while applying the transformation of variables, ok.
This extra term, in fact, denotes, see what we, you know, this rate of change of flux
of course, is common with a, b, c equations, but we have got extra, what is known as P
d M f term. So, if we look at this dC p by dt, if we actually
evaluate it, in fact, so we carry on forth from last time. dC p by d T d theta is equal
to, it is easy to see this, you take the derivative of the transformation, so you will get minus
K d sine theta K q cos theta and of course, K 0, when you take the derivative, you will
get zero. So I will write 0 here and you will have minus K d sine theta minus 2 pi by 3.
It is difficult to fit it in the paper, but we will try to do that. K q into cos theta
minus 2 pi by 3 and this will be, sorry, this is, yeah, minus K d sine theta plus 2 pi by
3. And this term here is K q cos theta plus 2 pi by 3. So, this is basically what we have
I will just read out term it might not be very clear here. It is, K q cos theta plus
pi by 3 and this is of course, a distinct term here.
Now, it is very easy to see, it is not too difficult, you know, how, what C p looks like?
C p looks like this, sorry, this is C p inverse, I am sorry, C p looks like this. So, it is very, by inspection you can really
make out, that this is nothing, but C P into matrix P 1 where P 1 is nothing, but 0, K
q by K d, 0, minus K d by K q, 0, 0 and this is 0, 0, 0. So, what I have, so let me just
repeat dC p by d theta is equal to C p times P 1, ok, where P 1 is this. So, our final equations, flux equations, come
out to be minus of C p d psi dq0 by dt minus d theta by dT. So, I will just call this theta
dot C p P 1 psi dq0 minus R s into C P i dq0. It is nothing, but C p into v dq0. So, your
flux equation becomes, in fact, I will just multiply, pre-multiply both, both sides by
C p inverse, you will get minus d psi dq0 by dt minus theta dot into P 1 into psi dq0
minus R s. Of course, remember, is a diagonal matrix containing R a, R b and R c.
Of course, if all the coils are identical I can just say, R a into i dq0 is equal to
v dq0, which leads us to, if I really write this down separately, you know, because it
sometimes is not very evident, that what we are getting unless we write down all the equations
separately. So, what I will do is I will write down the dq0 equation separately. So, the dq0 equation separately turns out
to be minus d psi d by dt minus d theta by dt, in fact, or theta dot is nothing, but,
it is nothing, but omega, angular frequency, electrical angular frequency and this is K
q by K d psi q minus R a i d is equal to v d. Minus d psi d q by dt plus omega K d by
K q psi d minus R a i q is equal to v q and minus d psi 0 by dt minus R a i 0 is equal
to v 0, where omega is equal to d theta by dt or d theta dot.
So, if it is not clear I am just panning this a bit. Now, one interesting thing you should
see here is that there is what I mentioned sometime back, this P d M f terms, in fact,
although we saw, that there is complete decoupling between the d and q axis. So, recall, we had
a complete decoupling between the d and q axis.
Recall this equation or the flux equations, in fact, we had, I had had mentioned sometime
back, that psi d is dependent on i d i f and i h, but it is not dependent on i q, i 0 or
i g and i k. It is not dependent on the q-axis also, the flux current relationships. There
is a complete decoupling between the d and the q-axis coils, so I call this the d-axis
coil. So, the fluxes in the d-axis coil are not dependent on the q-axis currents, none
of the currents, but importantly, when you look at the flux equations, there are speed
emf terms. So, the flux Faraday’s law, when you apply to psi d, psi d, you have to put
this extra term, which comes because of applying the correct mathematics to the transformed
equations. So, although sometimes it is tempting to start
from a kind of a physical model of rotating windings, when we come to obtaining the flux
equations in the d coil or the flux equations in the q coils, remember that there are speed
emfs in the d coil due to the flux in the q-axis. This cannot be explained by just from
the starting of this model. So, one of the things, which you should keep
in mind is it is a good idea, as I mentioned in the previous class, to first work out the
mathematics. The correct mathematics give you the, give you, gives you these equations
where there are extra speed emfs terms, these speed emf terms come because we apply the
derivative to the time varying transformation as well. These are the correct equations.
So, please remember, that there is coupling coming between the d and q axis coils because
of these extra, what I call as, speed emf terms. Now, these are, as far as, the stator equations
are concerned, the rotor equations, of course, we know. And d psi g by dt is equal to, sorry, plus
R g i g is equal to 0. So, just remember these are the remaining equations. In fact, if you
know the flux, if you know the flux and current relationships and these equations relating
to the emfs, the rotor flux equations on the d-axis and the q-axis, we, in fact, have got
the complete flux description, so if I know of course, I should, the relationship psi
dq0 psi r. I know this relationship as well, dq0 and
i r and this is nothing, but L ss dash L sr dash L rs dash and L rr. So, we have in fact,
got a complete picture of the system. In fact, you can substitute for i g, i k, i f, i h,
as well as, i dq0 by psi dq0 and psi r using the relationship and you will get finally,
things in the state space form. We will do it on a separate sheet. So, you will get, you can, of course, write
it as psi dot is equal to some A matrix into psi plus B into v. So, you can get it into
this, this form, where psi is nothing, but psi dq0 and psi R. This A, in fact, does not
have time coming in explicitly; you do not have time or theta coming in explicitly. Of
course, A does contain omega, remember that, because of the fact, that the equation flux
have the speed emf term. A is a function of omega, but if omega is
a constant, then this A becomes linear, this particular equation becomes linear time invariant.
So, that is the beauty of applying the dq0 transformation. It consists speed, of course,
this becomes a linear time variant equation, there is no explicit dependence on theta,
which itself changes with time. So, this is one of the main, you know, important thing,
which really comes out of applying the dq transformation.
Now, in general of course, whenever you have linear time variant system, it is not obvious,
that by transforming it in a certain way you will get a linear time invariant system. It
turns out, that the machine equations have special, special structure, which permit the
use of this transformation C p, ok, which make the system time invariant.
Now, one of the things, which we have not discussed now, we will shortly again discuss
it, this feature of making the, you know, equations of, flux equations of a machine’s
time invariant, is not dependent on the specific values of K d, K q and K 0, it could be absolutely
arbitrary choice. Of course, K d, K q, K 0 should not be 0; none of them should be 0.
So, there is some flexibility in the choice of K d, K q, K 0.
Now, before we go on to discussing this particular point again, let us just look at what happens
to the torque equations. Now, if you look at the torque equation, just recall what was
the equation for torque? If you recall what we did some time back,
I will just show you the equations first, so you can recall what we have done. This
was the torque equation in the previous lecture where T e is nothing, the minus of the partial
derivative of the co-energy with respect to the mechanical position.
Torque T e is, of course, in this direction, theta is measured in this theta M and theta
e are both measured in this direction, the anticlockwise direction. Here, the co-energy
is expressed as a function of currents. You have got this particular formula and of course,
eventually we did get this, so T e, which is the electrical torque, is nothing, but
the minus of P by 2 into P is the number of poles into d by derivative, partial derivative,
of co-energy with respect to theta, which is the electrical angle.
So, we call, of course, T dash as this. So, T dash turns out to be this, so this is what
our equations are. I basically took out the derivative, partial derivative of co-energy
with respect to this. So, now, if I want to get this expression for torque T e dash, remember
T e, the actual torque is actually minus P by 2 times this. So, we will have T e is equal
to T e dash, the actual torque. Remember, it gets P by 2 minus P by 2 times T e dash.
So, T e dash is nothing, but 2 by P times T e, half, so I substitute here. So, I have
got i s transpose here, so I will have to write, i s remember is i a i b i c, so I can
write this as i dq0 transpose into C p transpose into the partial derivative of L ss with respect
to theta. L ss, remember, is the sub matrix of the original
L matrix into, there is not much space here, so I will just, we can write it here I guess,
i dq0. i s here is C p into i dq0, inadvertently we have just written i dq0. So, remember,
i dq0 is row, rather a column i d, i q, i 0, so this is a neat way of writing it, plus,
plus, in fact, this is 2 times i dq0 transpose C p transpose d L sr by d theta into i r,
closing of curly bracket. So, what we get is this, you can just work
out, you have already, about two lectures we took out, what L ss is actually. So, I
will just write down what L ss by d theta is nothing, but minus of 2 L aa 2, this is
a time varying part or the theta varying part, into sine of 2 theta sine of 2 theta minus
2 pi by 3 sine of 2 theta plus 2 pi by 3 sine of 2 theta minus 2 pi by 3 sine of 2 theta
plus 2 pi by 3 here sine 2 theta. And of course, this is, this will be, I just, this is, it
may not be clear, I will just read it out, sine of 2 theta plus 2 pi by 3, it is not
very clear here, so maybe, we will just write it down again. This is theta is equal to minus, sine of theta,
sine of 2 theta sine of… No audio 33:19 to 34:02
This is this, L ss by L theta d L sr by dt will partition into d L sr d by d theta and
d L sr q by d theta. So, that turns out to be, Maf sine theta minus
Mah sine theta… No audio 34:42 to 35:47
So, better book keeping is required, but as we shall see soon, that finally, what we get
is the torque expression after using these formulae is quiet neat. So, this is what d,
if we know dL sr is this, so we have found out the sub matrices, which I mentioned some
time back. Now, you can see, that dL ss by d theta into
C p is nothing, but minus of 3 L aa2 into C p into P 2, where P 2 is 0, K q by K d,
0, K d by K q, 0, 0 and 0, 0, 0. So, you can rewrite this. So, if you look at structure
of d L s s by d theta if you look at the structure of it you can write it as, in this fashion.
Therefore, and further, you will find, that C p transpose of dL sr by d theta is nothing,
but in fact, it gets stripped of all the thetas eventually, 0, 0, 3 by 2 K d Mag 3 by 2 K
d M a K and this becomes minus 3 by 2 Maf minus 3
by 2, sorry, this is K here, so there is a K q, K q Mah, 0, 0 here, 0, 0, 0 and 0.
So, it is a bit complicated, but this is how it comes out to be, we can verify this. So, eventually, we get, we will just, I will
just write down the torque equation. T e comes out to be 3 by 2 K d K q into i q into Maf
by K d i f plus Mah by K d i h plus 3 by 2 L aa2 i d. So, I close this bracket, then
continue, minus i d times Mag by K q i g plus Mak by K q i k minus 3 by 2 times L aa2 i
q, close this bracket and close this bracket. So, this is the expression for torque.
But we know that, so I will just partition this here, we know, that psi d is nothing,
but L d i d, this comes out from the flux equations, psi d is equal to L d i d
plus Maf i f by K d plus Mah by K d i h and of course, psi q is nothing, but L q i q plus
Mag by K q i g plus Mak by K q i k. So, what do we understand form this? If you look at these equations, look at this
and you look at this, so one can directly infer from this, that T e is nothing, but
3 by 2 times k d k q, again we will put a curly bracket here, i q into psi d minus L
d minus L aa2, we will rewrite this again. You can rewrite, T e is equal to 3 into k
d into k q into, a curly bracket here, into i q into, we will put a square bracket here,
psi d minus L d minus 3 by 2 L aa2 into i d and I close the square bracket minus i d
into psi q minus L q plus 3 by 2 L aa2 into i q. So, I close this bracket and I close
this bracket. So, if you, from this it is very easy to see, you know, I, you can cancel
off this term here you will get really after all the manipulations, finally it yields,
3 by 2 times K d K q psi d i q minus psi q i d. In fact, this is small correction here,
so I should be calling this T e dash, T e dash, so that is nothing, but T e dash; so,
T e dash is this. So, what we have here is basically a very
neat expression, which is again, devoid of any of the thetas, they are all in term, so
for d and q variables only. So, that is one interesting thing. So, our equations are turning
out to be extremely neat. Now, in fact, if you write down the equation in d, q variables,
you will find, that your equations will no longer have theta explicitly appearing in
them, they are much easier to handle. The flux current relationship is brings out the
de coupling between the d and q-axis, fluxes and currents.
The flux differential equations, of course, bring out the coupling between the d and q
axis fluxes, there is a coupling. We now get the torque equation, remember what the torque
equation was? If we recall where we were sometime back, this, so T e dash we have actually found
out right now, which is, which I said was the function of theta, but of course, because
we applied the d-q transformations, it turns out, that it is not explicitly coming, the
theta does not come out explicitly. If you express T e dash in terms of the d-q variables,
so it is a very, very interesting, you know, very neat expression we get for torque. Now, one thing about the choice, the last
topic in this particular lecture, about the choice of K d and K q. We saw of course, that
if we put K d square is equal to 2 by 3 and K q square is equal to 2 by 3, one interesting
thing happens, that is, L sr dash becomes equal to L rs dash transpose.
Also, if in fact, we saw some time back, that when we take out C p inverse, it is in terms,
it has got elements K 1, K 2 and K 3, where K 1 is nothing, but 2 by 3 times k d and K
2 is 2 by 3 times K q. In fact, K q is in the denominator and K q is also in the denominator
and K 3 is equal to 1 upon 3 K 0. So, these are the coefficient of the sines and cosines
in the C P inverse matrix. If you look at, you know, these relationships,
it appears, that it is not a bad idea to choose K d equal to K q is equal to root 2 by 3.
So, that is an interesting and important point here. Of course, there is absolute freedom
of choice of K d and K q in so far as our main objective is of getting a time invariant
set of equations. Remember, that K d, K q do appear in the flux current relationship,
as well as, the flux differential equations, as well as, in the torque expression here.
Sorry, this should be K q, but the fact remains, that although K d, K q do appear, theta does
not appear. So, whatever be the value of K d and K q, as long as, they are non-zero,
we do end up with what we have set out to do, that is, get a neat set of equations,
which are invariant with respect to theta. So, if we choose K d and K q root 2 by, it
is a very special choice, it will lead us to Lsr dash being equal to L rs dash transpose,
it is just convenient; it is very convenient. Also, we will find, that C p inverse will
be equal to C p transpose. So, it is easy, of course, it is easy to remember would be
one advantage of using k d and k q in this fashion. So, it is easy to remember, that
C p inverse is equal to C p transpose. But let me repeat, K d and K q could be arbitrary
in so far our main objective of getting time invariant equations as concerned. So, let
me just summarize, but before I do that, it is K d is equal to K q is equal to root 2
by 3 is a kind of a, is our, it is a special choice and that is the choice, which I will
really follow in this particular course. It is, in fact, the, the same choice is used
in K R Padiar’s book. In fact, I, we look at the other literature the literature or
many industry papers. They follow dimension K d is equal to 1 and K q is equal to minus
1, so they, they follow this choice. In fact, of course, I did not talk about K 0, K 0,
if I want this to be true. In fact, I should have told you before.
You have to choose K 0 is equal to 1 by root 3. So, K 0 also gets refined if I want to
have this to be true. So, just remember this again, that many industry papers, as well
as, you know, in books by Kundur and many other books they follow the convention, k
d is equal to 1 and K q is equal to minus 1. In this particular course, I shall follow
Padiars convention, which is or the IEEE convention actually, which is kd is equal to K q equal
to root 2 by 3 because it gets us some benefits. In fact, you will notice, that in this torque
equation if I choose K d and K q as root 2 by 3, this, this you know, this whole coefficient
gets becomes equal to 1. So, I do not have to remember these. Coefficients because this,
this if I put K d equal to root 2 by 3 and K q equal to root 2 by 3, this coefficient
effectively becomes 1. So, there are certain advantages in choosing this, but, as I mentioned
back, the choice is arbitrary, it is not hampered of, referred to get a time in variant set
of equations. Of course, the question then arises, if I,
if we have got industry papers or papers or books, which follow this convention, I will
call this people who follow this convention, I will call this, people who are using this
as a transformation using a transformation C P 1. And we, in fact, apply a transformation
C p where K d and K q are root 2 by 3. In such a case one can transform the equations,
which they have used the people who are using these this set of values of K d and K q by
again a kind of matrix T k. So, I can, actually the variables, which they have defined are
different from my variables because the coefficients k d k q have been chosen by one have been
1 and minus 1 And we are going to choose root 2 by 3.
So, whenever I am going to use their equations and I get answers with their variables, I
would need to use this transformation matrix T k to transfer from their variables to my
variables. This will leave this as an excessive to find out what is T k is T very simple constant
matrix. So, to summarize this particular lecture finally,
obtain the equations of the synchronous machine in the d-q frame of reference and the equations
are neat in a sine theta, they are not functions of theta. In fact, you can, if in fact speed
of the machine is constant, d theta ny d T is equal to constant, the flux equations,
the flux differential equations are linear time invariant equations. Of course, if speed
is not constant, in fact, in general, it need not be a constant.
You have a coupling between the mechanical equations and the flux equations, flux differential
equations. In fact, the coupling is non-linear. Remember, the torque is psi d i q minus psi
q i d. So, there is a product terms, there also there is a product when you are talking
about the speed emf terms in the flux differential equations, therein also, there is a product.
So the mechanical equations and the flux equations are in fact coupled in a non-linear fashion,
but the flux equation themselves for a constant speed are in fact, time, linear and time invariant,
that makes at least some of our analysis much, much simpler.
So, now we move on to, to, you know, trying to interpret the equations which have come,
correlate with parameters obtained by measurement and thereafter, the most important thing of
course, in this course is to draw inferences, what equations you have got and correlate
to actual power system.
