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In the previous lecture, we studied the response
of certain non-linear system as well as the
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linear system. In fact, the response which
was analyzed was not actually derived I mean
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none of the you know response of the states.
In fact, the speed and the rotor angle in
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that example, which we did was not derived
from first principles I in the linear systems
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case, we actually kind of guessed the solution.
In the non-linear case we guessed how the
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response would probably be for large disturbances,
we did not actually again work out the solution.
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In fact, it is not possible to do so, in case
of a non-linear system.
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So, in this particular lecture, we will try
to derive the response of linear. In fact,
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linear time invariant systems, what we shall
see is that we can very well characterize
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the response of linear time invariant system.
So, we can say everything about the system
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you can write down the response of the system.
And therefore, you can. In fact, you know
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go into depth of how the system behaves during
transient.
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So, this is a very important class of system.
In fact, it is a good idea to learn linear
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systems. So, as to get a kind of feel of how
systems behave most systems in the real world
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are non-linear. But, we can analyze the small
disturbance around an equilibrium using linearise
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analysis like we did in the previous class.
So, in this particular lecture which we have
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titled as analysis of linear time invariant
dynamical systems, what I will do is we shall
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write down try to find out the response for
it is class this linear time invariant set
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of equations.
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So, let us just review what we did in the
previous lecture we considered linear and
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non-linear systems the general form of a linear
system would be like this. In fact, I have
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written it down as a simple scalar system
in the sense that this is just a single variable,
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we shall generalize this of course. Non-linear
systems generally this function on the right
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hand side is not linear is not a constant
coefficient into the state.
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So, that is why these systems as I mentioned
last time is difficult to write down the response
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and therefore, analyzing the system becomes
a bit tough. Later on, in the course we will
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of course, learn how to analyze how non-linear
as well, but, the analysis tools will be centered
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around numerical you know solutions of these
equations.
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But, today I shall show you that for a general
linear system of this kind we can actually
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write down the solution, response of linear
time invariant systems, we studied last time
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was given by, if the system is given by derivative
of x with respective time is equal to a x
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then this response is simply this. It depends
on the initial condition of course, one can
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verify by taking the derivative of this that
it actually satisfies this equation.
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So, this is the solution of this set of equations
this equation. In fact, if we know the initial
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condition at a time other than time t is equal
to 0, the solution is this you can easily
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verify this from by substituting t is equal
to t 0. In that case you will get x t 0 is
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equal to x t 0.
Which is consistent also the derivative of
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this will satisfy this equation, just one
small point that did not actually define time
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variant system I just told you that this the
time invariant system. In fact, there are
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some systems in which x dot is equal to a,
which is the function explicit function of
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time into x this is also linear system. But,
this is a time varying system.
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In fact, we will come across some time variant
systems later on in our course linear time
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invariant systems. These normally again are
difficult to analyze unlike this system in
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general. But, of course, there are some special
situations in which this also can be analyzed
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quite easily.
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So, more of that later in the course, when
we come to synchronous machine modeling. One
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question which I left you with in the previous
class was if the I know the response of x
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dot is equal to a x, what is the response
of x dot is equal to a x plus b u. It turns
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out that the response is this I will not derive
this, but, this is how the response looks
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like. It is e raise this is whole term which
we have seen and in additional convolution
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term out here. It is an integration of e raise
to a t minus tau b u tau d tau remember tau
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is the variable here in the integration.
So, the integrations carried out with respect
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to the variable tau you can verify that. In
fact, this is indeed the solution this is
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some something I just wrote down because,
if you take x dot of t you take the derivative
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of this it will be a e raise to a t x 0 plus
d by d t of this term. Now, derivative of
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this term using the product rule would be
a e raise to a t that is derivative of the
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first term into this plus e raise to a t into
the derivative of this.
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So, that yields x dot is equal to a so, these
two terms. In fact, are a is taken out common
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out of that plus this second term. So, this
is the first term and this is the second term.
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So, you add these two up. The second term
of course, is e raise to a t into e raise
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to minus a t b times u of t, see remember
look at this derivative this final value is
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t. And if you take the derivative of this
with respect to t just the integrant is basically
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obtained at time t, at the time t.
So, basically you will get this which is x
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plus b u. So, what I wanted to show you that
this is indeed a solution of this. So, just
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a minor point here that in case you have got
a linear system with an input you can still
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get a, you can still write down the answer.
If you know u the behavior of u of t you can
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actually write down this answer.
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So, having an input posse no particular or
major problem as far as linear time invariant
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systems are concerned, we can also verify
of course. That if I evaluate x of t at tau
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using this formula, using this formula you
plug in t is equal to 0 here, you will get
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x 0 because this term becomes equal to 1.
And if you put the upper limit of the integration
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also x 0 this term will become 0.
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So, it is just its consistent x of t at t
is equal to 0 is x of 0. So, as I mentioned
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some time back, we have kind of verified that
this is in fact, the solution of this. So,
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this was one problem which I had mentioned
last time. Now, let us get back to the original
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problem which we had set out to solve in the
last class that is the response of higher
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order linear time invariant, but, coupled
systems.
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If you recall in the previous class, we had
given you this example x 1 dot is equation
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a 11 x 1, x 2 dot is equal to a 22 x 2 this
is the second order system, but, it is not
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coupled. So, you get a very simple solution
here, it is a very simple solution you just
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take the individual solutions for the states.
But, if you get a two slightly more complicated
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scenario where there is some coupling this
is the second order coupled system you can
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of course, have third order and tenth order
hundredth order systems also. But, we will
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limit ourselves to a simple situation, this
is the second order coupled system.
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So, we write this second order coupled system
as x dot is equal to a x, where a is a matrix
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this a matrix has terms a 11 a 12 a 21 a 22.
So, this x is in fact, made out of two components
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it is a vector, it is still a linear system
but, a coupled system. Now, how do you solve
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this system I mentioned in the, I gave you
a very simple example in the previous class.
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In which I could really get the response by
using the idea of a transformation.
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Now, a transformation is the extremely important
idea in engineering analysis. In fact, you
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must have come across various kinds of transformation
like Laplace transformation. In fact, taking
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a logarithm of a product of two terms, then
adding the two logs of the individual terms.
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And then taking the antilog is actually a
kind of a transformation approach to do a
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multiplication.
So, you are transforming it into new variables
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doing an operation in the new variables and
getting back to the old variable. So, this
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is the basic idea of any transformation what
we shall actually now, learn in this particular
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system is how we can use a linear transformation.
In fact, this is a very simple kind of transformation
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we are going to study here, how you use a
linear transformation. To transform these
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variables two new variables in which the dynamical
equations are very easy to solve. So, that
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is what we will try to do.
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So, suppose I if I got this coupled linear
system. Let us, define a transformation of
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variables this is a linear transformation
of variables I will just read it out x 1 and
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x 2 are related to y 1 and y 2 by this transformation,
that is x 1 is equal to p 11 y 1 plus p 12
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into y 2, x 2 is equal to p 21 y 1 plus p
22 into y 2. So, this is the transformation
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of variables. So, what I wish to achieve will
become very clear in sometime.
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So, I substitute for x in this equation. So,
what I will have is p of y dot, remember p
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is a constant matrix in this particular case
we shall of course, learn later on some very
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interesting transformations which are time
variant. But, right now we are talking about
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linear time invariant systems and we shall
use a time transformation which is not time
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dependent, it is just a constant matrix.
So, if you have got p y dot it will be equal
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to a p y that is because of substituted for
x here as well as x here p is not a function
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of time. So, your derivative is simply p y
dot. So, in terms of the new variables y dot
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is equal to p inverse a p into y. So this,
what I achieved in the new variable. So, these
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are the dynamical equations of the new variable.
(No audio from 11:27 to 11:36) The whole point
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in doing a transformation of course is that
if this system is simpler I can solve for
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this system. And then somehow get back to
x 1 and x 2.
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So, what I intend to do is solve for y and
get back now of course, if this is the coupled
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matrix. Suppose, this is has got non 0 terms
here here here here and here. Obviously, we
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are not really doing, we are not really simplifying
the problem at all I mean you are back to
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the old problem.
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.
But, a very nice situation occurs if p inverse
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a p is diagonal. So, if I chosen my p in a
certain way now that way of course, will try
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to derive a bit later. But, if I chosen my
p in such a way such that this is diagonal,
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that is p inverse A P is a diagonal matrix,
that is it is made out of terms lambda 1 lambda
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2 0 0, which still do not know what this p
is what this lambda 1 and lambda 2 are. But,
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let us assume you have this p which will take
you to this form.
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So, what you have is y will be equal to, simply
equal to e raise to lambda 1 0 that the coupling,
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there will be complete decoupling between
y 1 and y 2 why is this. So, basically we
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have got to this form in some sense they have
got a complete decoupling the solution becomes
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very simple.
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So, the point is if your dynamical equations
are y dot is equal p inverse A P y, which
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is nothing, but, y 1 and y 2 get completely
decoupled. In fact, I will just rewrite this
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here, you will have y 1 y 2. (No audio from
13:27 to 13:34) this is p inverse A P. Suppose,
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it is possible to have A P inverse A P, which
will give you this. So, the solution becomes
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very very simple it becomes simply y 1is equal
to lambda 1 t y of 0 y 1 of 0 and y 2 is equal
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to e raise to lambda 2 t because, there is
a complete decoupling.
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So, this is what. In fact, I have got here
I have written it down in matrix form this
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is the solution. So, if I manage to get p
inverse A P to be diagonal, then you will
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find that your solution is very simple. So,
you have got your solution in terms of y of
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course, you have to get the final solution
of x. So, how do I get x p of y is x. So,
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my solution of x will be simply this p into
y. Now, y consist of y 1 0 and y 2 0. In fact,
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y of 0 is nothing, but, p inverse of x of
0 because x is equal to P y.
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So, the point is we will just circle this
actually follows from this. So, y 1 of 0 is
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actually p inverse x of 0. So, this is this
and this is in fact, this whole term here
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is actually this. So, one of the we of course,
come across one important feature here, that
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if we are going to get A P its inverse should
exist. So, this p matrix has to be invertible.
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So, the transformation x is equal to P y should
be invertible or in other words in layman
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terms, what you would say is that you can
go from x to y, but, you should also be able
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to come from y to x and vise versa, if you
f you know y you get x and if you know x you
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should be able to get y. If that is true then
you can use this transformation and get this
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solution.
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Of course, one important point which I have
actually not discussed here is this presumes
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that for any A you are going to get some P
which will diagonalise this matrix, this is
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in general not true. In fact, there will be
situations where you cannot diagonalise a
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matrix A using any such P. So, this is one
important point which you should just keep
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at the back of your mind. This is not always
possible you cannot always get A P such that
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you can diagonalise this, this will become
clear when we actually derive the expression
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for P.
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So, what is the solution of x? So, if you
look at x this is what I get. So, I just expand
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this if I expand this. This is what I will
get p 11 it is just expanding that term here.
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So, you if you look at this carefully
I just expand this. So, what I will get is
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p 11 p 21 e raise to lambda 1 t into q 11
q 12, where q 11 q 12 is in fact, this row
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matrix is in fact, the first row of p inverse.
So, what you will get is by expanding this
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you will get this.
So, what you find it in a second order system
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in which that A matrix is diagonalizable,
what you will see is that your response is
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consisting of two terms and if you look at
a time varying part of these terms. In fact,
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they are telling you something. In fact, they
are telling you two modes are patterns exist
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in the response.
So, if you look at the response of x 1 in
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general as long as p 11 is not equal to 0,
x 1 will contain e raise to lambda 1 t and
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e raise to lambda t kind of term e lambda
e raise to lambda 2 t kind of terms also.
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That is of course, provided p 11 and p 12
are non 0.
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And also you will find that of course, this
also presumes that this product here of this
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row matrix and this column is non 0. If this
product becomes 0 then of course, this whole
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term will not exist. So, you will, this term
whole term will not exist, in case this product
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here becomes equal to 0. Similarly, if this
product becomes equal to 0 this pattern will
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not exist. So, this you know response is consisting
of two patterns. So, x 1 consist of two patterns
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e raise to lambda 1 t e raise to lambda 2
t it will have two kinds of terms.
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So, this very loosely speaking this particular
thing is called a mode. So, this is one mode
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this is another mode. So, 1 thing you just
remember in case of linear systems linear
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time invariant system, your response is a
super position of modes. Modes are terms which
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have got this e raise to lambda t kind of
response. Remember in this particular solution
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q 1 and q 1 is in fact, defined in this fashion
it is the inverse of the p matrix.
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So, this q is in fact, the inverse of the
p matrix and these are in fact terms of this.
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So, we can directly ask a question that if
you have got an a matrix which is diagonalizable
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your really having modes. In fact, if it is
a second order system you have a got a two
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modes corresponding to lambda 1 and lambda
2 of course, I will retreat here. That we
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are assuming that the A matrix is diagonalise
diagonalizable by some transformation P inverse
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A P.
So, we this is an assumption we will make,
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we will relax subject later. So, the question
once we get this response is how much of this
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term is visible in x 1? How much of this term
is existing in x 2.
191
00:20:14,840 --> 00:20:22,950
If you look at this particular response of
the system this is a number. This is a row
192
00:20:22,950 --> 00:20:29,020
matrix which is in this case 2 into 1 this
is 1 into 2. So, what you will get is this
193
00:20:29,020 --> 00:20:38,770
is a number, this number is common to these
two states. So, the two states for the two
194
00:20:38,770 --> 00:20:45,120
states these this particular number is common,
but, you notice that p 11 and p 21 in general
195
00:20:45,120 --> 00:20:50,440
can be different in general it could they
could be different. So, the amount of a mode
196
00:20:50,440 --> 00:20:58,320
if it is excited visible in x 1 and x 2 is
defined by this p 11 and p 21.
197
00:20:58,320 --> 00:21:03,950
So, for example, if p 21 is very small, we
would say that this particular mode is not
198
00:21:03,950 --> 00:21:09,740
observable in x 1. Similarly, if p 21 very
small, we said that this particular mode is
199
00:21:09,740 --> 00:21:15,140
not observable in x 2. But, in general you
will see that p 11 and p 21 are non 0, I mean
200
00:21:15,140 --> 00:21:18,920
for most systems which you will encounter.
201
00:21:18,920 --> 00:21:24,790
So, the first question which rather the second
question which is asked here, if a mode is
202
00:21:24,790 --> 00:21:34,170
excited how much of it is seen in x 1 and
x 2. So, if the mode corresponding to the
203
00:21:34,170 --> 00:21:44,760
Eigen, corresponding to e raise to lambda
2 t for the mode, e raise lambda 1 t sorry,
204
00:21:44,760 --> 00:21:54,280
x 1 and x 2 are excited in the are seen in
the ratio p 11 and p 21, for the mode e raise
205
00:21:54,280 --> 00:22:03,090
to lambda 2 t the relative magnitudes of this
term as seen in x 1 and x 2 are in the ratio
206
00:22:03,090 --> 00:22:09,710
p 12 and p 22.
So, this p 11 and p 22 in fact, tell you how
207
00:22:09,710 --> 00:22:18,460
much of a particular mode is observable in
a particular state. Remember in a coupled
208
00:22:18,460 --> 00:22:25,380
system there is no 1 to 1 correspondence between
a mode and a state in general, in a particular
209
00:22:25,380 --> 00:22:31,630
the response of a particular state you will
see both these terms appearing that two modes.
210
00:22:31,630 --> 00:22:37,430
The first question which I have asked here
is how much of each mode is excited that depends
211
00:22:37,430 --> 00:22:43,050
on this product.
So, for example, if this product here turns
212
00:22:43,050 --> 00:22:48,730
out to be 0 this mode is not excited at all
and will not be seen in either of these states,
213
00:22:48,730 --> 00:22:56,110
got what I am saying. So, similarly, if this
product is 0 this and this it turns out to
214
00:22:56,110 --> 00:23:04,850
be 0, in that case you will not see this particular
part or this pattern in this response.
215
00:23:04,850 --> 00:23:11,600
So, this an important point which you should
note, that the components of the inverse of
216
00:23:11,600 --> 00:23:19,200
this p matrix and the initial conditions determine,
the extent to which a pattern or a mode is
217
00:23:19,200 --> 00:23:26,730
excited, if a mode is excited the amount it
is observable in a particular state depends
218
00:23:26,730 --> 00:23:28,810
on this p 11 and p 21.
219
00:23:28,810 --> 00:23:37,790
So, before we move ahead let us just get back
to where we were for a moment, we are starting
220
00:23:37,790 --> 00:23:46,490
off with a coupled system. This was our coupled
system, we defined a transformation, we assume
221
00:23:46,490 --> 00:23:53,160
that this transformation would simply things
in the sense that the equations in the new
222
00:23:53,160 --> 00:24:01,920
variables would be easy to solve. So, we have
got p inverse A P is diagonal. This is assume
223
00:24:01,920 --> 00:24:08,850
that you have a P which will make you diagonal,
if a diagonal y 1 and y 2 equations get decoupled
224
00:24:08,850 --> 00:24:14,370
the solution becomes very very simple.
Y 1 is equal to this, y 2 is equal to e raise
225
00:24:14,370 --> 00:24:24,250
to lambda 2 into y 2 to get back the original
variables x you have to retransform this and
226
00:24:24,250 --> 00:24:30,670
get back to the original variable. So, you
have to use the inverse transformation here.
227
00:24:30,670 --> 00:24:39,120
So, it is p e raise to the this particular
matrix into p inverse into the initial conditions.
228
00:24:39,120 --> 00:24:45,160
So, the generalized response of course, can
be written in this fashion, the components
229
00:24:45,160 --> 00:24:50,790
of p, the columns of p in some sense the columns
of p determine the relative observability
230
00:24:50,790 --> 00:24:57,380
of a certain mode. The inverse components
or the inverse of p and the initial conditions
231
00:24:57,380 --> 00:25:00,180
determine how to what extend the mode is excited.
232
00:25:00,180 --> 00:25:06,510
So, the product of this will determine the
overall strength of this particular mode is
233
00:25:06,510 --> 00:25:15,730
that ok. So, the whole question now boils
down to how do you get this p remember p is
234
00:25:15,730 --> 00:25:23,290
such that p inverse A P is a diagonal matrix.
So, what we have is just premultiplied by
235
00:25:23,290 --> 00:25:35,900
p you will get A P is equal to this. Suppose,
p is of course, a matrix i kind of partition
236
00:25:35,900 --> 00:25:46,180
it into two columns, these are two columns
when if you carry out this product you will
237
00:25:46,180 --> 00:25:54,160
get effectively two equations. If you just
equate this matrix which comes here, with
238
00:25:54,160 --> 00:26:00,580
this matrix will in fact, get two equations
this, this is, the first column of this left
239
00:26:00,580 --> 00:26:07,130
hand side will become a p 11, a into this
column the second column becomes a into this
240
00:26:07,130 --> 00:26:08,130
column.
241
00:26:08,130 --> 00:26:18,770
So, if we actually work it out, you will actually
get two equations the two equations. In fact,
242
00:26:18,770 --> 00:26:23,940
you must have done a course in mathematics
in your undergraduate years, you would realize
243
00:26:23,940 --> 00:26:31,250
that this lambda 1 and lambda 2 are nothing
but, what you have learned is called the Eigen
244
00:26:31,250 --> 00:26:41,140
values of the matrix a. And the columns of
t are the corresponding Eigen vectors. So,
245
00:26:41,140 --> 00:26:48,760
this p 11 p 21 is in fact, the right Eigen
vector corresponding to this Eigen value lambda
246
00:26:48,760 --> 00:26:55,500
1. Of course, we still have not got what this
lambda, lambda 1 lambda 2 and this matrix
247
00:26:55,500 --> 00:26:56,770
p are.
248
00:26:56,770 --> 00:27:03,550
So, our next step is find out what these things
are. So, if you look at this particular equation
249
00:27:03,550 --> 00:27:08,350
either lambda 1 or lambda 2 equation, you
will find that you get this on to the right
250
00:27:08,350 --> 00:27:14,200
hand side for both lambda 1 and lambda 2,
you will get an equation of this kind where
251
00:27:14,200 --> 00:27:25,020
p is nothing, but, a column of p. So, for
example, I can p 11, p 1 will be p 11 p 21.
252
00:27:25,020 --> 00:27:34,070
So, p 1 is the right diagonal vector corresponding
to the Eigen value lambda 1. This is nothing,
253
00:27:34,070 --> 00:27:41,050
but, p 1.
So, what you have here is. So, now, the question
254
00:27:41,050 --> 00:27:45,790
comes are we in a position to get what lambda
or what are the Eigen values and the Eigen
255
00:27:45,790 --> 00:27:49,750
vector this is true for every Eigen value
and Eigen vector and the corresponding every
256
00:27:49,750 --> 00:27:57,300
Eigen and the corresponding Eigen vector.
Now, obviously, if a minus lambda i is in
257
00:27:57,300 --> 00:28:04,040
fact, invertible that is this is non singular
one of the solutions, we can get for this
258
00:28:04,040 --> 00:28:11,430
column of p is 0, it is a 0 vector. In fact,
I should write it this way for 2 by 2 systems
259
00:28:11,430 --> 00:28:15,710
this p will be 0.
But, this is not acceptable, why is it not
260
00:28:15,710 --> 00:28:23,280
acceptable? Because, I have told you that
a transformation should be such that p inverse
261
00:28:23,280 --> 00:28:28,410
has to exist. So, we cannot have one column
or both columns equal to zero then you cannot
262
00:28:28,410 --> 00:28:32,850
take out inverse and you cannot get the solution
finally, in terms of x.
263
00:28:32,850 --> 00:28:40,110
So, you really cannot this particular trivial
solution is not of any use. So, this particular
264
00:28:40,110 --> 00:28:47,390
solution is not of any use to us what we should
look for is solutions in which p is non 0.
265
00:28:47,390 --> 00:28:54,640
In fact, the only way you can have p non 0
is to have this singular. So, this is singular
266
00:28:54,640 --> 00:29:01,770
it is possible to have solutions of p, this
column p which are non 0. In fact, this is
267
00:29:01,770 --> 00:29:10,390
also, this is a vector this column vector.
So, this is an important condition which you
268
00:29:10,390 --> 00:29:16,350
should have in order to obtain non trivial
solutions for p. So, in fact, if you look
269
00:29:16,350 --> 00:29:23,540
at this equation this itself in fact, aids
you to obtain the Eigen values and Eigen vectors
270
00:29:23,540 --> 00:29:24,540
of the system.
271
00:29:24,540 --> 00:29:32,760
So, if you look at for example, this particular
system. So, I have done this is nothing, but,
272
00:29:32,760 --> 00:29:39,000
determinant the left hand side is nothing,
but, determinant of a minus lambda i. So,
273
00:29:39,000 --> 00:29:43,340
this is what I have really done here. So,
determinant of a minus lambda is equal to
274
00:29:43,340 --> 00:29:49,720
0. So, in this 2 by 2 system, we will get
this particular equation and it has got two
275
00:29:49,720 --> 00:29:55,730
solutions which you can actually solve for
lambda 1 and lambda 2. So, if I know a 11
276
00:29:55,730 --> 00:30:01,630
a 22 a 12 and a 21 that is the a matrix I
can get what lambda 1 and lambda 2 are going
277
00:30:01,630 --> 00:30:04,780
to be.
So, actually you can solve for lambda 1 and
278
00:30:04,780 --> 00:30:09,720
lambda 2 in this particular case of course,
we have only considered 2 into 2 system this
279
00:30:09,720 --> 00:30:15,910
a matrix is a 2 into 2 system. If you have
got say a 20 by 20 system that is containing
280
00:30:15,910 --> 00:30:22,660
twenty states which are coupled together using
a 20 by 20 state matrix or the same matrix.
281
00:30:22,660 --> 00:30:30,810
In that case you will get this a polynomial
of this kind of order 20 when you apply this
282
00:30:30,810 --> 00:30:36,420
determinant of a minus lambda is equal to
0. But, remember it may not be easy to solve
283
00:30:36,420 --> 00:30:40,610
for this lambda 1 and lambda 2 that way.
See if you are going to second order solution
284
00:30:40,610 --> 00:30:48,120
this is quadratic you can get values of lambda
1 and lambda 2 from these. If you have got
285
00:30:48,120 --> 00:30:53,960
forth order system you get a quatic equation
in lambda once you actually evaluate this
286
00:30:53,960 --> 00:31:00,360
determinant. But, beyond a point you cannot
actually solve you cannot get the answer directly
287
00:31:00,360 --> 00:31:07,890
for the various lambda. So, for larger systems
you will use some kind of numerical techniques
288
00:31:07,890 --> 00:31:12,660
to obtain the Eigen values and the Eigen vectors.
So, this is just a caution here, that although
289
00:31:12,660 --> 00:31:16,730
for the second order system I can actually
solve this quadratic to get lambda 1 lambda
290
00:31:16,730 --> 00:31:21,270
2 in a large system. You have to think of
some special numerical techniques in order
291
00:31:21,270 --> 00:31:31,770
to get lambda 1 and lambda 2 numerical and
the iterative techniques to obtain, the Eigen
292
00:31:31,770 --> 00:31:32,810
values and Eigen vectors.
293
00:31:32,810 --> 00:31:37,120
So, now we are at a point where we can actually
tell you how to proceed. If you have got an
294
00:31:37,120 --> 00:31:43,550
a matrix compute this the Eigen values of
the a matrix. Once you get the Eigen values
295
00:31:43,550 --> 00:31:51,480
of the a matrix you can compute, what p the
columns of the p matrix are. In fact, they
296
00:31:51,480 --> 00:31:54,660
will be the right diagonal vectors corresponding
to the Eigen values.
297
00:31:54,660 --> 00:32:00,790
So, if you look at this particular equation
which defines what p is for example, if I
298
00:32:00,790 --> 00:32:07,020
want to found find out what is p 1 is you
will do a minus lambda 1 i into p 1 is equal
299
00:32:07,020 --> 00:32:13,600
to 0. Since, by definition lambda 1 is a value
for which this determinant becomes 0 after
300
00:32:13,600 --> 00:32:18,350
all we found out lambda 1 by taking the fact
that determinant of a minus lambda is equal
301
00:32:18,350 --> 00:32:24,440
to 0. This you cannot get the value of p 1
this is of course, again I should write it
302
00:32:24,440 --> 00:32:30,390
as a vector, you cannot get the value of p
1 by simply taking the inverse because this
303
00:32:30,390 --> 00:32:36,490
is not going to be .
So, what you have to do is you have to you
304
00:32:36,490 --> 00:32:41,750
cannot of course, get a unique solution for
p 1. So, you will have to actually assume
305
00:32:41,750 --> 00:32:48,090
one component of p 1 to be say 1 and get the
other component from this equation. Now, this
306
00:32:48,090 --> 00:32:52,560
may be a bit unclear at this particular point
of time, we will do a simple example in which
307
00:32:52,560 --> 00:32:58,180
we will try to tell you how to get this p
1 once you have got the Eigen values.
308
00:32:58,180 --> 00:33:05,370
Remember one thing that if you know p 1 then
alpha p 1 is also a solution. So, if I get
309
00:33:05,370 --> 00:33:13,100
a p 1 then alpha p1 is also a solution. So,
the Eigen vectors are not unique you can always
310
00:33:13,100 --> 00:33:23,820
form P matrix. Suppose, your P matrix you
have found form the columns p 1 and p 2.
311
00:33:23,820 --> 00:33:43,410
Then this also is an Eigen vector matrix and
this also is. So, what we see here is that
312
00:33:43,410 --> 00:33:48,660
there is no uniqueness as far as this p 1
is concerned. So, Eigen vectors in some sense
313
00:33:48,660 --> 00:33:57,370
are direction vectors the magnitude of it
is not unique. So, just remember this point
314
00:33:57,370 --> 00:34:02,430
whenever we are going to solve this that is
not going to be a unique value of p 1. So,
315
00:34:02,430 --> 00:34:06,390
the best way to actually understand this you
have done a lot of manipulations the simple
316
00:34:06,390 --> 00:34:12,230
arithmetic type manipulations a simple way
to understand it is using an example.
317
00:34:12,230 --> 00:34:19,670
So, if you recall in the previous class we
had done a simple example for a coupled system.
318
00:34:19,670 --> 00:34:29,929
So, if you recall this is what the system
was x 1 dot is equal to x 1 plus 0.5 x 2 is
319
00:34:29,929 --> 00:34:37,399
equal to 0.5 x 1 plus x 2 x 2 dot is equal
to 0.5 x 1 plus x 2 and this was my system.
320
00:34:37,399 --> 00:34:38,839
So, I will just rewrite this.
321
00:34:38,839 --> 00:35:06,119
This is my A matrix now, how do I define the
Eigen values of this matrix determinant of
322
00:35:06,119 --> 00:35:17,559
a minus lambda i. So, lambda i is nothing,
but. So, this is equal to determinant of A
323
00:35:17,559 --> 00:35:25,740
minus lambda I. I in fact, is the I did not
mention it before, it is the identity matrix.
324
00:35:25,740 --> 00:35:32,020
So, this should be equal to 0. So, this what
it should be. So, what we have is when you
325
00:35:32,020 --> 00:35:47,259
take out this determinant, you will get 1
minus lambda square minus 0.5 into 0.5 is
326
00:35:47,259 --> 00:36:01,420
equal to 0. So, we will get lambda square
minus 2 lambda plus 1 minus 0.25 is equal
327
00:36:01,420 --> 00:36:15,099
to 0. So, this is something lambda square
minus 2 lambda.
328
00:36:15,099 --> 00:36:26,990
So, we have got lambda square minus 2 lambda,
that implies we can guess that the solution
329
00:36:26,990 --> 00:36:52,440
is sorry right 0.5 yeah so, this is the solution.
So, your two Eigen values are. So, for a second
330
00:36:52,440 --> 00:36:58,220
order system you have got a two Eigen values
which is this. Now, once you have got these
331
00:36:58,220 --> 00:37:00,230
Eigen values you have to take out the Eigen
vectors.
332
00:37:00,230 --> 00:37:07,490
So, what are the Eigen vectors by definition,
let us take out the Eigen vector p 1 that
333
00:37:07,490 --> 00:37:20,480
is corresponding to this lambda 1. So, what
you will have is A is this A minus lambda
334
00:37:20,480 --> 00:37:36,230
I is nothing, but. So, this is nothing, but,
A minus lambda 1 I of course, this is singular
335
00:37:36,230 --> 00:37:40,089
you will get 0.5 0.5 0.5 here.
336
00:37:40,089 --> 00:38:08,130
So, what we have here is.
So, we can get the value of p 1, but, the
337
00:38:08,130 --> 00:38:15,289
by if we cannot of course, invert this is
not possible to invert this matrix because
338
00:38:15,289 --> 00:38:30,470
it is singular. So, how to get the value of
p 1 remember that p 1 is nothing, but, p 11
339
00:38:30,470 --> 00:38:42,980
it is a column of the matrix P. So, the point
is now how do you get the value of p 11 and
340
00:38:42,980 --> 00:38:47,950
p 21? You cannot get it unless you fix some
1 particular variable. So, for example, I
341
00:38:47,950 --> 00:38:55,250
can fix p let us try do try to do this, let
us fix p 11 there is no unique solution of
342
00:38:55,250 --> 00:39:02,109
p 11 or p this p 1. So, there is no way I
can actually get this p 11 and p 21 uniquely.
343
00:39:02,109 --> 00:39:07,220
So, what i will do is that I will fix this
particular value here, let us just try it
344
00:39:07,220 --> 00:39:13,250
out. So, I will fix this value of p 11. So,
we could just as while i have fixed the value
345
00:39:13,250 --> 00:39:20,381
of p 21. But, I have chosen the value of p
11. So, if you take p 11 as 1 then you will
346
00:39:20,381 --> 00:39:30,569
get 0.5 into 1 plus 0.5 into p 21 is equal
to 0, which also means that p 21 should be
347
00:39:30,569 --> 00:39:32,059
equal to minus 1.
348
00:39:32,059 --> 00:39:41,930
So, we are what we have got is p 11 is equal
to 1 and minus 1. So, 1 and minus 1this is
349
00:39:41,930 --> 00:39:48,349
1 Eigen vector corresponding to the Eigen
value lambda 1 is equal to 0.5. I can of course,
350
00:39:48,349 --> 00:39:52,970
make this 2 and minus 2, that is also acceptable
you will find that 2 minus 2 also satisfies
351
00:39:52,970 --> 00:39:57,830
this equation.
So, there is nothing unique about the Eigen
352
00:39:57,830 --> 00:40:11,339
vector as you can always multiply it by constant
and still is an Eigen vector. What about p
353
00:40:11,339 --> 00:40:26,540
2? p 2 is.
354
00:40:26,540 --> 00:40:36,700
So, this is nothing, but, a minus lambda 2
I. Again we cannot get unique solutions of
355
00:40:36,700 --> 00:40:45,150
p 21 and p 12 and p 22. So, what we will do
is we will just freeze p 12 at 1 in that case
356
00:40:45,150 --> 00:40:50,430
you will get.
357
00:40:50,430 --> 00:41:18,710
So, what we have is, the Eigen vector corresponding
to the Eigen value 1.5 is in fact, 1 sorry
358
00:41:18,710 --> 00:41:20,870
and both are plus 1.
359
00:41:20,870 --> 00:41:35,329
So, what we have is the general solution this
particular system is going to be.
360
00:41:35,329 --> 00:41:56,749
This is the first Eigen value there is something
else to be written here, we will just do that
361
00:41:56,749 --> 00:42:08,970
presently there is one term I will call it
k 1.
362
00:42:08,970 --> 00:42:40,769
k 1is nothing, but.
363
00:42:40,769 --> 00:43:02,410
And k 2 is now, this q 11 in fact, is the
row of, the inverse of the p matrix or p matrix
364
00:43:02,410 --> 00:43:07,380
is made out of the columns p 1 and p 2. So,
what are the columns p 1 and p 2 1 minus 1
365
00:43:07,380 --> 00:43:24,380
1 and 1.So, p inverse is equal to half of
we take out a determinant right. Just check
366
00:43:24,380 --> 00:43:41,999
whether it is we can just do p p inverse that
will be 1 into 1 this is 1 into minus 1 it
367
00:43:41,999 --> 00:43:48,900
will come 0. So, this cannot be here this
has to be here. So, 1 into 1 that will be
368
00:43:48,900 --> 00:44:06,930
1 that 1 so, this is 2 this is 0 0 is 2. So,
p inverse is indeed this, this is nothing
369
00:44:06,930 --> 00:44:20,030
but, the q matrix. So, what we have here is
for the final solution.
370
00:44:20,030 --> 00:44:45,579
Let me just write it down is nothing but.
371
00:44:45,579 --> 00:45:04,259
Into this is the q matrix. So, what we will
have is 1 minus 1 into half. So, I just have
372
00:45:04,259 --> 00:45:23,019
to put half here.
373
00:45:23,019 --> 00:45:35,760
So, this is the solution, the complete solution
for this system actually it seems very painful
374
00:45:35,760 --> 00:45:42,040
sometimes. But, the point is that you can
actually get it fairly, if you follow a systematic
375
00:45:42,040 --> 00:45:47,380
procedure you can get the solution of this
equation these, this dynamical system.
376
00:45:47,380 --> 00:45:55,319
Now, just before you proceed let us just try
out what we were discussing some time back.
377
00:45:55,319 --> 00:46:00,530
If I have got if you look at this particular
solution let us not look at it just as a mathematical
378
00:46:00,530 --> 00:46:05,230
quantity just look what it says. It says that
your solution is a you are going to have two
379
00:46:05,230 --> 00:46:11,180
components to this solution this and this.
The first this is an unstable system incidentally
380
00:46:11,180 --> 00:46:15,470
because, it if there is any non zero initial
condition you will find that there are terms
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00:46:15,470 --> 00:46:21,070
which will grow with time, this is growing
with time, this is positive. So, it will grow
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00:46:21,070 --> 00:46:28,809
with time.
If you look at this 1 and minus 1 into this
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00:46:28,809 --> 00:46:36,809
you will find that certain sets of initial
conditions can be made to selectively excite
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00:46:36,809 --> 00:46:51,309
certain modes. For example, if I chose my,
in that case you will find at in fact, it
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00:46:51,309 --> 00:46:59,180
is practically proportional to this particular
transpose of this row vector. In that case
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00:46:59,180 --> 00:47:06,920
the product of this is non zero will get the
answer to be two this if you look at this
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00:47:06,920 --> 00:47:17,420
product it will turn out to be 2. This particular
product will turn out to be 0.
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00:47:17,420 --> 00:47:24,259
So, if I have got this set of initial conditions
I will end up exciting only this mode. Similarly,
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00:47:24,259 --> 00:47:34,029
if you have got a set of initial conditions
two. You will end up exciting this mode, but,
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00:47:34,029 --> 00:47:38,420
not excite this mode.
So, this is the important point, the second
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00:47:38,420 --> 00:47:44,200
point which you will notice is given the fact
that. Suppose, this mode is excited in that
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00:47:44,200 --> 00:47:51,960
case if you look at for this mode there is
certain pattern to how much of the mode is
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00:47:51,960 --> 00:47:58,690
visible in x 1 and x 2. So, if x 1 is a positive
quantity x 2 becomes a negative quantity if
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00:47:58,690 --> 00:48:06,289
only this, if you look at only this mode.
So, this particular mode has got this characteristic.
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00:48:06,289 --> 00:48:14,920
If it is excited the component corresponding
to that mode the observablity, you know the
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00:48:14,920 --> 00:48:22,630
amount or the nature of the mode in x 1 and
x 2 is a bit difference in the sense that
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00:48:22,630 --> 00:48:28,019
if it is plus 1 here it is minus 1 here.
So, if only the if you give these initial
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00:48:28,019 --> 00:48:33,240
conditions. For example so, that only this
mode is excited your response is going to
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00:48:33,240 --> 00:48:34,869
look like this.
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00:48:34,869 --> 00:48:46,739
So, you start from plus 2. So, I will give
it two different colors may be I will call
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00:48:46,739 --> 00:48:58,830
this is minus 2 this is x 2, this is x 1,
you will find that this will grow like this
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00:48:58,830 --> 00:49:04,839
whereas, this grow like this.
So, the characteristic here if you look is
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00:49:04,839 --> 00:49:13,739
exactly you know if only this mode is excited
x 1 and x 2 always appear in this ratio 1
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00:49:13,739 --> 00:49:19,650
is to minus 1. Of course, if you have got
both modes excited then of course, it is you
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00:49:19,650 --> 00:49:24,019
cannot say that x 1 and x 2 are going to be
in a ratio of 1 and minus 1 etcetera. Then
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00:49:24,019 --> 00:49:28,470
it is going to be a combination though you
will have to actually evaluate this response
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00:49:28,470 --> 00:49:32,549
and you cannot say that in general. But, if
you look at this mode only and assume that
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00:49:32,549 --> 00:49:35,450
only this mode is excited then x 1 and x 2
and this ration.
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00:49:35,450 --> 00:49:42,529
So, that is the significance of the Eigen
vector. Now, one small point which we will
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00:49:42,529 --> 00:49:48,690
try to conclude this particular lecture look
at this particular system. This particular
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00:49:48,690 --> 00:49:56,470
system has got 1 1 0 and 1 the Eigen values
of this are easy to find out. If you do A
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00:49:56,470 --> 00:50:18,751
minus lambda I you will get. So, that now,
the problem is that if you have got both the
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00:50:18,751 --> 00:50:25,920
Eigen values here in this particular case
1 and 1 it turns out this particular matrix
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00:50:25,920 --> 00:50:32,089
cannot be diagonalised. So, this is one problem
which we are going to face.
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00:50:32,089 --> 00:50:37,440
So, remember that whatever I have talked the
general response of a linear system in terms
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00:50:37,440 --> 00:50:42,369
of mode justified. It is the continent of
the fact that the system can be diagonalised
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00:50:42,369 --> 00:50:47,680
there are systems which cannot be diagonalised
so, this one of them. So, if you look at this
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00:50:47,680 --> 00:50:53,270
particular system the response of it is not
going to be of the form which I told you.
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00:50:53,270 --> 00:51:08,890
For example, look at this. So, I am just rewriting
this system this is what this particular if
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00:51:08,890 --> 00:51:15,480
you have got x dot is equal to A x with A
as this effectively your dynamical system
421
00:51:15,480 --> 00:51:23,660
is this. It is a coupled dynamical system.
So, your x 2 I am sorry, this should be dot
422
00:51:23,660 --> 00:51:31,170
x 2 should be will be equal to e raise to
lambda, lambda is 1 p. So, this is 1 into
423
00:51:31,170 --> 00:51:46,239
t x 2 of 0 and x 1 from this equation you
will get. So, from this equation you get this
424
00:51:46,239 --> 00:52:00,740
from this equation you are getting this. So,
this you can treat as a x and this as b u,
425
00:52:00,740 --> 00:52:01,740
this is just u.
426
00:52:01,740 --> 00:52:17,529
So, what you will get is x 1 t is equal to
x, I will just rewrite this, plus e raise
427
00:52:17,529 --> 00:52:25,569
to 2 t x 2 of 0. So, what is the solution
of this you have done this before a is equal
428
00:52:25,569 --> 00:52:39,119
to 1 here. So, you will get e raise to t x
1 of 0 plus 0 to t e raise to t minus tau
429
00:52:39,119 --> 00:52:43,210
into b u.
So, b u is nothing, but, this so, you will
430
00:52:43,210 --> 00:52:51,349
get e raise to u of tau is nothing, but, e
raise to tau x 2 of 0 d tau. So, you will
431
00:52:51,349 --> 00:53:06,249
get e raise to t x 1 0 plus e raise to t 0
to t x 2 of 0 d tau. So, you will get e raise
432
00:53:06,249 --> 00:53:16,779
to t x 1 0 plus if you evaluate this you will
get t e raise to t x 2 of 0. So, the response
433
00:53:16,779 --> 00:53:23,390
earlier you are getting just e raise to something
terms, if your matrix is not diagonalizable
434
00:53:23,390 --> 00:53:26,920
you start getting terms of this kind in the
response.
435
00:53:26,920 --> 00:53:35,880
So, this is one example in which you cannot
diagonalise this system the Eigen values are
436
00:53:35,880 --> 00:53:42,049
equal and in fact, you will not be able to
get p 1 and p 2, which are linearly independent
437
00:53:42,049 --> 00:53:46,070
if p 1 and p 2 are linearly independent P
cannot be inverted.
438
00:53:46,070 --> 00:53:50,839
So, that is why you cannot get a response
in the form you want. So, let us conclude
439
00:53:50,839 --> 00:53:57,369
this particular lecture by just looking at
what we have learnt, the response of a linear
440
00:53:57,369 --> 00:54:07,999
system x dot is equal to a x. If A is diagonalizable
is given by x of t is, the summation this
441
00:54:07,999 --> 00:54:16,519
is the general response for a enth order system.
Summation of this term where p i are the Eigen
442
00:54:16,519 --> 00:54:21,380
values corresponding to p i is the Eigen value,
Eigen vector corresponding to the enth Eigen
443
00:54:21,380 --> 00:54:28,310
value lambda i is the enth Eigen vector, enth
Eigen value q i is nothing, but, the row of
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00:54:28,310 --> 00:54:35,550
the inverse of the right Eigen vector matrix
this is often written in the form e raise
445
00:54:35,550 --> 00:54:39,890
to 80 of x 0.
So, this particular if you come across this
446
00:54:39,890 --> 00:54:44,819
somewhere in the text book this is what it
really mean you have to expand it in this
447
00:54:44,819 --> 00:54:51,259
form. Stability of this system can be just
got by looking at the real part of the Eigen
448
00:54:51,259 --> 00:54:59,079
value of lambda. So, this is basically what
we see, this we have seen some very simple
449
00:54:59,079 --> 00:55:05,140
examples in this particular class second order
example what we will do in the next class.
450
00:55:05,140 --> 00:55:10,019
We will do a few numerical examples we did
one today, we will do a few more in which
451
00:55:10,019 --> 00:55:15,990
lambda for example, are complex in that case
what response does one expect. And then we
452
00:55:15,990 --> 00:55:23,530
will go on to analyzing some systems and bring
out some general modeling principles.
453
00:55:23,530 --> 00:55:30,509
So, this is what we will do in the next few
lectures. So, once we do this of course, we
454
00:55:30,509 --> 00:55:35,710
will go onto numerical integration and then
we will study a bit about modeling. So, although
455
00:55:35,710 --> 00:55:43,069
you may get a bit lost in all this mathematical
manipulations just stay on and we will come
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00:55:43,069 --> 00:55:46,749
onto some real nice pal system examples as
well.