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In the previous class, we have been discussing
the response of linear time invariant systems.
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In fact, we took a simple second order systems
in order to illustrate the response. The basic
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interesting fact about the analysis of linear
time invariant systems is that there is the
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time response can be written down in terms
of well known functions, and because of that
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one can infer the stability, and the dynamical
properties fairly easily. So, the main important
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result which we got last time was that the
response of a linear system is a sum of individual
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patterns or interior modes of response.
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So, to make that a bit more precise, let us
take a simple linear time invariant system
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without any input x is a vector, A is a matrix
then the response of the system is given by…
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It is in fact aA summation of several modes
of course, this particular response can be
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written down in case you have got the A matrix
diagonalizable, which we will just kind of
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try to understand once more.
If A is diagonalizable by linear transformation
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of the variables x in that case, one should
be able to write the response in this form.
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It is a summation of various patterns or modes
lambda i s are the Eigen values; they are
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a property of the matrix A itself in fact,
determinant of lambda into an identity matrix
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minus A is equal to 0 yields the characteristic
equations whose solutions in fact are the
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Eigen values of the matrix A. A remember is
of course, an n cross n matrix. This particular
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characteristic equation will yield us a polynomial
of order n and in general, they can there
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will be of course, n solutions of a nth order
polynomial.
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Now importantly, this PI are the columns of
the Eigen vector matrix. So, if you recall
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what we did in the previous class, we were
taking of a particular transformation x is
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equal to P y and a P such that P inverse AP
yields a diagonal matrix. The columns of P
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are the right Eigen vectors P of course, has
to be invertible and in we discussed this
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last time in case, A is diagonalizable; that
is A has in fact n distinct Eigen values in
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that case, it is possible to find the columns
of P such that P is invertible. So in fact,
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A is diagonalizable provided we get a P which
is non singular, which also means that the
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columns of P should be linearly independent.
In fact, if you have got n distinct Eigen
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values, A is diagonalizable. And as I mentioned
some time back, if A is diagonalizable, you
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can write the response in this form; the q
i here, the q i transpose is a row matrix,
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is a row vector rather which is in fact, the
rows of P inverse the Eigen vector matrix
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inverse the rows of it are q i.
So the main, this was the main result which
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we saw last time, and the most important issue
which we discussed was depending on the components
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of P i the components of P i, you can observe
or measure the relative observe ability of
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the pattern e raise to lambda it in the various
states. So, that is an important point which
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we discussed in the previous class, also we
saw that depending on the value of the initial
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conditions, a mode could be excited weakly
excited or not excited at all.
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So, this was the interpretation we could give
to the layer to the Eigen vectors. In fact,
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an important point which slipped my mind last
time was that q i in fact also satisfies this
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relationship. So, q i is also known as the
left Eigen vector, so q i are not only the
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rows of P inverse, but they also satisfy this
particular property.
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Now what happens if the matrix is not diagonalizable?
So, if the matrix A is not diagonalizable,
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in fact if A has distinct Eigen values it
is always diagonalizable, but if A has non
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distinct Eigen values then, A may not be diagonalizable.
So, that is an important point which you should
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know so for example, the matrix A which is
1; in fact, is non diagonalizable; if you
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take out its Eigen values which is very difficult
to which is not very difficult to find out,
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the Eigen values are determinant lambda I
the identity matrix minus A is equal to 0;
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the Eigen values are 1 and 1.
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So, these are non distinct and this particular
matrix is non diagonalizable. You cannot diagonalize
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this matrix, because you cannot get you know
P1 and P2 which satisfies Ap 1 is equal to
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P 1 and Ap 2 is equal to P 2; P 1 and P 2
it will turn out to be will be always in the
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same direction that is, P 1 will be alpha
P 2; alpha is any arbitrary constant. So,
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what you will find out is that because of
this. you will not get you will not be able
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to construct a P with P 1 and P 2 as its columns
so that you can diagonalize this matrix. So
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you cannot get the P 2 to diagonalize the
matrix. So, that is why this particular system
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is not diagonalizable. Now of course, there
are examples in which you do have repeated
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non distinct Eigen values, but still you can
diagonalize it for example, this particular
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matrix is a kind of academic example.
This is a matrix which is diagonalizable;
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in fact, it is already diagonal; it also has
got repeated Eigen values, but obviously this
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is a diagonalizable matrix; it is a diagonal
matrix in fact. So, there are exceptions so
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just remember that the correct statement is
if A has non distinct Eigen values, A may
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or may not be diagonalizable; that is the
important point which you should note. Now,
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one aspect which we should discuss is if its
non diagonalizable, what is the response;
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for example, suppose I have a matrix which
it turns out that the Eigen values are lambda,
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lambda, lambda 1, lambda 3, and lambda 4.
There are four Eigen values of a four by four
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matrix A. Suppose, this is non distinct; they
are you know Eigen values are repeated. So,
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A may or may not be diagonal; let us assume
that it is not diagonalizable; in that case,
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I just without really proving it we can show
that you can find a P matrix which is just
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does not get you to a diagonal matrix, but
gets you to a nearer diagonal matrix which
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is also called a Jordan matrix. So, you have
got lambda 1, 0 lambda 0, 0, 0, 0. So, if
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a matrix is not diagonalizable and has a Eigen
values as shown, we can actually get it in
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this form; this is called a Jordan form.
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Now, this is a simplified matrix; this is
a simplified kind of transformed matrix which
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of course, one can get the time response quiet
easily. So, please recall that in case you
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have a system in which A is diagonalizable,
your response is x of t is equal to P e raise
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to diagonal matrix into P inverse into x of
0; but incase, your system is not diagonal
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is not diagonalizable then the response can
be different. Remember here, e raise to this
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diagonal matrix is in fact, this is a diagonal
matrix like this. So if your matrix is diagonalizable,
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your response is as shown here of course,
the expanded form of this response is in fact
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what I discussed some time back that is this;
this is the expanded form. If your matrix
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is not diagonalizable, then your response
x of t is equal to P e raise to Jt P inverse
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where J is e raise to lambda t; te raise to
lambda t this taking the example, which we
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discussed some just a while ago; that is you
have got e raise to Jt is this.
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So remember that, in case you do get non distinct
Eigen values, you can still write down the
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response but remember now, you are getting
the response in terms of not only exponential
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functions, but you have got time coming here
t into e raise; these kind of terms will come
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into your response, so that is the important
difference. You can of course, have more than
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two repeated Eigen values, but we will not
go further and deeper. Normally in our power
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system analysis, one will not come across
situations there will be there will be more
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than you know you know you have three Eigen
values, which are non distinct and its non
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diagonalizable. That is very a fairly rare
situation. Most cases which we will consider,
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you will have all Eigen values distinct so,
it is important to remember this.
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Now one important point which you should note
is that, in this particular course we do require
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you to understand the response of linear systems
of course, it is not practicable in this course
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to give you a full exposition of everything
about linear system theory. So, it is good
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if you refer to some control systems, linear
control systems books for more details about
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the response of linear time invariant systems.
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So for example, you can look at the book by
Ogata State Space
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Analysis of control systems in 1967 of course,
there are many other good books so I… this
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is just an example.
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So this is an example of a book you could
refer to. Now of course, an associate problem
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which we will come across is what happens
when you have got a system which is forced
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that is it has got an input. So, if you have
got a system x dot is equal to A x plus b
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u and you are not observing all states, but
you are observing one output which is a, which
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is basically a combination of the states.
So, such a kind of a system has got both an
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input and an output. So I will just write
it down x dot is equal to A x; let us assume
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there is only one input u and you are also
observing a certain variable you know which
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could be a combination of the states c is
a c is a vector; x is our state.
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So, you could have a combination of states
is a multiplication of a row with a column.
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We will assume u and y are scalars; so this
a single input, single output system. In such
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a case, what is the response of x? We had
written down the response when A was a scalar
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some in some lectures before. The response
of the system when A is a coupled matrix is
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x of t is equal to e raise to At x of 0 plus
0 to t e raise to A t minus tau b u could
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be a function of time. In the convolution
integral, dt is should actually be d tau so,
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there is this some minor error here; y is
simply c of x t y of t plus d of u, u of t.
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So, this is our response of the system. Remember,
e raise to At is a short hand for P e raise
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to P inverse we have already discussed what
this is. In case it is diagonalizable or it
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is equal to P e raise to Jt P inverse if it
is non diagonalizable; this is Jordan form
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matrix. So, just remember this formula which
you have for the first response. So, it is
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quite easy to take out the first responses
as well, it is not too difficult. Especially,
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if you are a simple inputs like sinusoidal
inputs or step inputs this is trying to evaluate
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this is not a really a major problem. Now
before, we will just write this particular
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equation and this particular equation in full
form. So if I expand this, let us assume that
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A is diagonal and I expand this, what you
will get is y of t is equal to c times i is
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equal to 1 to n; this is the response of x
of t plus of course other terms. If input
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u if the input u is 0 these other terms are
0.
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One interesting point here is that if cp i
is equal to 0, then that the mode or the pattern
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will not be visible in y t. So, if c is such
that cp i is equal to 0 is not visible. Although,
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it may be visible in the states, but it still
may not visible in y; so y is actually a combination
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of states. These other terms of course are
dependent on u. So this is an important point;
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so not visible that is a proper technical
term called observable. So, if cp i is equal
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to 0, then the i th mode is not observable
in this input output y.
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Similarly, since x of t is equal to sigma
of i is equal to 1. I will just continue here,
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so what we have we can just write this a bit
differently. In fact, we can take out Pi e
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raise to lambda it common and then in the
bracket, we have got 0 to t; this whole thing
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is summed; qi you will have e raise to lambda
minus lambda i tau q i transpose bu tau d
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tau. It is a bit messy here, but I have just
since the variable which is getting integrated
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is tau, I have kind of taken out this e raise
to lambda it outside the integral. So what
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we see one critical point, if q i transpose
b this is 0 for some i; in that case term
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corresponding to lambda it in this response
is dependent only on initial conditions. If
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such a situation occurs we also say, that
this particular mode i is not controllable
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by the input u. So please remember these important
concepts. In case, q i transpose b is equal
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to 0; the e raise to lambda it term in the
response of the states is dependent entirely
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on the initial conditions that cannot be changed
by the input u; so that is one1 important
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point which you should note.
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So, we have through this rather straight forward
analysis given you a very intuitive kind of
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notion of observability and controllability.
We shall actually come back to these concepts
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in a multi-machine system where we shall see
that, there are many many modes and certain
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modes are not observable at certain places;
certain modes are not controllable by controllers
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which are located at certain locations. So,
this is a very… these concepts are pretty
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important; the concept of observability and
controllability in linear control systems
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of course, our analysis of the same has been
very intuitive and sketchy hopefully, we will
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revisit these things and it will become clear
at that point of time. That will be some time
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of course, much later in the course.
Now although, we have been doing the analysis
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of linear time invariant systems. There are
few computational issues which we should worry
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about because eventually, the power systems
which you are going to the realistic power
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systems or systems we are going to study are
going to be fairly large. So, one or two simple
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points and important points which you should
bear in mind are that, if I want to compute
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Eigen values I had mentioned that, one of
the ways of doing it is solve this equation
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determinant lambda I minus A is equal to 0;
it turns out that this is not very easy to
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do if your size of A becomes larger. In fact,
if you have got for example, a matrix of size
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20 by 20, you will find that it is becomes
almost next to impossible to actually solve
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it by actually computing the characteristic
impedance, characteristic equation. In fact,
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computing the characteristic equation the
coefficients of the characteristic equation
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themselves become pretty tedious.
So, there are in fact iterative methods like
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Power method or the Q-R method. We will of
course not be discussing all the methods or
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in fact, we will not really be going into
Eigen value computation when A is a large
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system; I can recommend that you look at any
book on matrix computations where this particular
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aspect is covered. But remember, one small
point that whenever you need to compute the
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Eigen values of a matrix and if that matrix
is large, you have to take recourse to some
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iterative methods; it is not feasible to do
it by actually computing the coefficients
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of the characteristic equation by actually
evaluating this that to be very computationally
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intensive. Of course, we shall take a bit
of the easy way out in this particular course
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otherwise, we will get our hands will get
totally occupied in trying to understand all
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the computational methods and the theory behind
linear systems and so on. We lose track of
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what we really wish to do that is understanding
a power system.
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So from now onwards, I also take help of mathematical
software to do the computations. So one some
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of the software which can help you to do this
quite widespread and used often SCILAB, this
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is of course freely downloadable software
and also MATLAB which is quite popular.
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So, if you use these one of these two tools
or probably any equivalent tools, it should
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be fairly straight forward to do most of the
computations which we will do in this particular
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course. So, rather than spend more time on
trying to give you more and more results and
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mathematical results and so on. Let us instead
take a more pragmatic view; let us take a
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simple system and numerically compute the
Eigen values. We will take a slightly larger
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system we have been always considering second
order systems which are the Eigen values Eigen
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vectors are easy to compute.
We will take a slightly more complicated example
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it is a third order system not really very
complicated, but we would require some computational
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help otherwise, it will become very tedious
to do. So, rather than just talking terms
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of a you know abstract some random A matrix,
we will just take a simple engineering example
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which will bring out not only how the method
forgetting the response, but will also tell
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us some interesting ways of interpreting the
results, you know all the Eigen value and
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Eigen vector information has to be interpreted
correctly.
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00:25:55,309 --> 00:26:02,250
So let us consider an example. There is a
voltage source connected in fact you can look
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00:26:02,250 --> 00:26:07,179
at this as some kind of equivalent of a pulse
transformer of course, I am not given very
194
00:26:07,179 --> 00:26:13,650
realistic values of a transformer; it is a
kind of this is a kind of leakage reactance,
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00:26:13,650 --> 00:26:20,380
the resistance of the wire, you can consider
this as a magnetizing reactor reactance and
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00:26:20,380 --> 00:26:28,670
also this capacitor really denotes the capacitance
of the widening. So, this is a kind of a near
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00:26:28,670 --> 00:26:34,820
realistic example; it is a model, a toy modal
of a pulse transformer. I chosen the values
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00:26:34,820 --> 00:26:42,860
in such a way that, it will aid us so in that
sense I have not given exactly realistic values.
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00:26:42,860 --> 00:26:48,091
If you just look at this particular example,
without even having to solve it, you will
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00:26:48,091 --> 00:26:52,470
notice that this is a very large inductance
compared to the leakage which is just 10 mille
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00:26:52,470 --> 00:26:59,250
Henry; this is 100 micro Farad and this is
0.1 ohm. This is a kind of situation in which,
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there are in fact components which are fairly
differing in size; they are they are not of
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00:27:05,941 --> 00:27:09,200
the same size; this leakage is much smaller
than this and so on.
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In fact, the main reason why I have given
you this example, we shall actually pursue
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00:27:13,970 --> 00:27:20,470
this example even further after this class
is over is that this particular system have
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got some certain characteristics. It is in
fact what is known as a stiff system. So,
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before I tell you what is a stiff system etcetera.
Let us just look at this at steady state;
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00:27:31,140 --> 00:27:38,290
in fact, if this particular system has is
not excited at all in case it is not excited
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00:27:38,290 --> 00:27:45,230
at all, it is easy to see that is if the voltage
here is 0 or I keep this open, the equilibrium
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00:27:45,230 --> 00:27:49,740
is when… Well what is the equilibrium of
this?
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00:27:49,740 --> 00:27:54,820
The equilibrium point of this is obtained
by setting the rate of change of the states
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00:27:54,820 --> 00:28:03,910
of the system to be 0. So let us for example,
choose the states. Suppose i L1 is a state;
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00:28:03,910 --> 00:28:11,250
i L2 is a state and V C is a state. So, this
is the initial conditions of these have to
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00:28:11,250 --> 00:28:17,771
be given if I want to tell you what the response
is going to be. But, if you for example, if
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00:28:17,771 --> 00:28:42,620
I… When is for this particular circuit if
the input u is 0 that is I connect this to
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00:28:42,620 --> 00:28:47,630
voltage source which has got voltage 0; in
that case, it is not very difficult to see
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00:28:47,630 --> 00:28:55,429
that in steady state if di L2 by dt is equal
to 0; it also means that the voltage here
218
00:28:55,429 --> 00:29:04,970
is equal to 0; so V C will have to be 0.
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00:29:04,970 --> 00:29:24,240
So the equilibrium, if input voltage is 0;
input is 0 is V C is equal to 0; i L1 is equal
220
00:29:24,240 --> 00:29:34,140
to 0; and i L2 is equal to 0. This is the
equilibrium for no input or for input u which
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00:29:34,140 --> 00:29:41,120
is equal to v i is equal to 0. So, if the
input is 0, in that case this is the equilibrium.
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00:29:41,120 --> 00:29:51,299
It is not very difficult to find that out,
we can do it formally that is i L1 by dt is
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00:29:51,299 --> 00:30:03,990
equal to this is 10 is to minus 2 10 mille
henrys; this 10 mille henrys this is 0.1 ohm.
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00:30:03,990 --> 00:30:33,970
So you will have, if you apply this k v l
in this loop, you will get is equal to or
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00:30:33,970 --> 00:30:40,780
in other words, l di by dt plus 0.1 into i
L1 plus V C is equal to v 1. So, I just got
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00:30:40,780 --> 00:30:49,240
things some things on to this side when I
applied k v l. So, this is the basic equation;
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00:30:49,240 --> 00:31:04,540
this is the second equation and the third
equation is of course, 100 micro Farad
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00:31:04,540 --> 00:31:15,470
is equal to
of course, if V i is equal to 0, that is if
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00:31:15,470 --> 00:31:21,250
the input I make if the input were to be 0,
then the steady state value is obtained by
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00:31:21,250 --> 00:31:29,620
putting all these things equal to 0. So if
V i is equal to 0, this is equal to 0; so
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00:31:29,620 --> 00:31:36,050
V C also will be equal to 0; if V C is equal
to 0 and V i is equal to 0; it means, i L1
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00:31:36,050 --> 00:31:45,330
is equal to 0; this is set to 0, then i L1
is equal to i L2 which means i L2 also is
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00:31:45,330 --> 00:31:46,330
0.
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00:31:46,330 --> 00:31:51,980
So if V i is equal to 0, then this is the
equilibrium condition. This is obtained by
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00:31:51,980 --> 00:31:58,220
setting all the derivatives equal to 0 that
is it. So this is, if you do not excite it
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00:31:58,220 --> 00:32:03,120
at all, let all you excite it with 0 voltage
source, this is the equilibrium condition
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00:32:03,120 --> 00:32:11,541
for the system. However, if as if I shown
here I excite it with a 1 volt source; in
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00:32:11,541 --> 00:32:28,370
that case, I can write this compactly in this
fashion. To save sometime, I will omit all
239
00:32:28,370 --> 00:33:24,400
the algebra and write this equation so for
t is equal to 0, this is A and this is b into
240
00:33:24,400 --> 00:33:27,429
u this is b into u.
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00:33:27,429 --> 00:33:39,320
So, this is basically our system and this
is the system after t greater than 0 and let
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00:33:39,320 --> 00:33:47,020
us assume that at t is equal to 0, the initial
conditions are all 0. That is all the currents
243
00:33:47,020 --> 00:33:53,380
and the voltage are actually 0; so let us
assume at t is equal to 0, the initial conditions
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00:33:53,380 --> 00:33:58,070
remember in all linear dynamical systems or
all dynamical systems, we need to be given
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00:33:58,070 --> 00:34:02,169
some conditions like the initial conditions
in order to tell what the future response
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00:34:02,169 --> 00:34:11,500
is going to be. So, let us assume the initial
conditions are this is the initial conditions
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00:34:11,500 --> 00:34:17,869
only at time t is equal to 0. So, the response
is if this is A and this is b u, you will
248
00:34:17,869 --> 00:34:40,769
get sorry e raise to At 0, 0, 0; this is the
initial condition plus 0 to t e raise to A
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00:34:40,769 --> 00:34:50,099
t minus tau into b u tau; so this is after
time equal to 0 after time is equal to, 0
250
00:34:50,099 --> 00:34:52,399
this is how it looks like.
251
00:34:52,399 --> 00:34:58,390
So this turns out to be, we will cut a long
story short; this will turn out to be 0 and
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00:34:58,390 --> 00:35:06,329
if you evaluate this I will skip the steps,
I will just write down the final answer I
253
00:35:06,329 --> 00:35:35,390
3 minus e raise to minus At sorry into 10,
10, 0. So I just evaluate this particular
254
00:35:35,390 --> 00:35:44,329
matrix, this particular this is of course
0; so if and I actually evaluate this from
255
00:35:44,329 --> 00:35:50,589
0 to t, I will get I 3 minus e raise to At
into 10, 10 and 0. So which of course can
256
00:35:50,589 --> 00:35:59,999
be written down; we know what e raise to At
of course is a short hand for P e raise to
257
00:35:59,999 --> 00:36:15,839
lambda 1, lambda 2 of course, this is 3 by
2, 3 matrix; so this is what we get.
258
00:36:15,839 --> 00:36:24,099
So this is our response. Now of course the
point is what is in fact, we are assuming
259
00:36:24,099 --> 00:36:29,359
here of course here, it is diagonalizable;
this needs to be actually checked. So, we
260
00:36:29,359 --> 00:36:35,349
need what is lambda 1, lambda 2, lambda 3
and also we require this matrix P. So this
261
00:36:35,349 --> 00:36:40,829
is what we really require in this particular
system. So this of course, has to be obtained
262
00:36:40,829 --> 00:36:46,210
from A. Now, for very simple second order
systems, I showed you how you can take out
263
00:36:46,210 --> 00:36:50,930
the Eigen values and Eigen vectors we did
a very simple example in the previous class,
264
00:36:50,930 --> 00:36:58,109
but if you really want to do an example in
which is a third order or fourth order matrix;
265
00:36:58,109 --> 00:37:03,900
if you use useful to use some software and
here I will show you this computation using
266
00:37:03,900 --> 00:37:11,130
the software sci lab, note down sci lab is
freely downloadable. It is available at this
267
00:37:11,130 --> 00:37:19,109
point of time at this particular web address
www.Scilab.org.
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00:37:19,109 --> 00:37:26,989
So, I will show you this computation on sci
lab; so just keep this in mind what we are
269
00:37:26,989 --> 00:37:33,700
doing. If we start up sci lab, this is what
you will get. Now, I will not really tell
270
00:37:33,700 --> 00:37:40,349
you all about the syntax of using and so on.
What I will try to you know just show you
271
00:37:40,349 --> 00:37:45,089
the commands and the syntax is fairly intuitive
so you should be able to pick it up on your
272
00:37:45,089 --> 00:37:50,460
own. So, what I will do here in order to take
out the Eigen values and Eigen vectors, I
273
00:37:50,460 --> 00:37:58,250
first enter the matrix A this is nothing,
but 10, this is the first row, then I enter
274
00:37:58,250 --> 00:38:14,519
the second row; this is the syntax of entering
the matrix, then you have got the third row.
275
00:38:14,519 --> 00:38:25,729
So, this is your A matrix the Eigen values
and Eigen vectors of A can be obtained from
276
00:38:25,729 --> 00:38:29,989
a command called spec of A. That will give
you actually the Eigen values associated with
277
00:38:29,989 --> 00:38:43,579
this system. So, the Eigen values associated
with this system are roughly minus 5 plus
278
00:38:43,579 --> 00:38:58,710
a complex number, this is a complex number
roughly this one; this is approximate I have
279
00:38:58,710 --> 00:39:09,900
just rounded off. Now, the point is these
are the Eigen values. One important point
280
00:39:09,900 --> 00:39:14,130
you are seeing is that you are getting Eigen
values which are not real numbers, they are
281
00:39:14,130 --> 00:39:19,369
complex numbers. Even if they are complex
numbers, I told you the stability criterion
282
00:39:19,369 --> 00:39:23,440
is that the real of lambda should be less
than 0.
283
00:39:23,440 --> 00:39:30,950
So this is in fact, a stable system because
the real parts are all less than 0. One important
284
00:39:30,950 --> 00:39:34,950
point which you notice here again is that
although, these are complex number they seem
285
00:39:34,950 --> 00:39:40,739
to appear in complex conjugate pair. In fact,
a general result which can be stated that
286
00:39:40,739 --> 00:39:50,440
if A is a real matrix, then it is Eigen values
are real and or whenever they are complex,
287
00:39:50,440 --> 00:40:02,569
they will appear in complex conjugate pairs.
So you can have real Eigen values and or complex
288
00:40:02,569 --> 00:40:07,319
conjugate pairs.
289
00:40:07,319 --> 00:40:13,640
Your final response as I is shall show you
in some time is in fact, quite is going to
290
00:40:13,640 --> 00:40:19,690
be real it is not going to contain imaginary
terms at all. So, please remember that you
291
00:40:19,690 --> 00:40:26,000
are your Eigen values whenever for real matrix
you will get Eigen values as a real and or
292
00:40:26,000 --> 00:40:31,130
complex conjugate pairs as you have seen here.
One more thing which you should notice here
293
00:40:31,130 --> 00:40:38,660
remember, we are trying to do a study of interpreting
the results as well. This Eigen value and
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00:40:38,660 --> 00:40:43,539
the magnitude of these Eigen values are very
very much different. In fact, you will see
295
00:40:43,539 --> 00:40:50,609
that this particular Eigen value is small.
In fact, the small Eigen magnitude Eigen value
296
00:40:50,609 --> 00:40:56,250
is associated with slow rates of change.
So, you will find that your response will
297
00:40:56,250 --> 00:41:01,539
contain one pattern which is relatively slow
and one pattern which is corresponding to
298
00:41:01,539 --> 00:41:07,119
a high magnitude Eigen value, the rate of
change of that pattern will be fairly fast.
299
00:41:07,119 --> 00:41:13,019
So in fact, a larger Eigen value is associated
with faster rates of change. So, that is one
300
00:41:13,019 --> 00:41:17,690
important point which you should note. These
of course Eigen values are three distinct
301
00:41:17,690 --> 00:41:26,180
Eigen values, so our response will be a superposition
of modes. In fact, so we do not have to worry
302
00:41:26,180 --> 00:41:31,710
about Jordan form or what happens how do you
actually solve with repeated Eigen values.
303
00:41:31,710 --> 00:41:36,849
This is a luckily we have got a system with
three distinct Eigen values.
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00:41:36,849 --> 00:41:43,200
Now, I mentioned some time ago that your response
is this. Now, we also require in order to
305
00:41:43,200 --> 00:41:49,299
get the response to find this value P; P is
the Eigen vector matrix. In fact, there is
306
00:41:49,299 --> 00:42:01,069
a command in Sci lab which allows you to take
out the Eigen vector matrix that is P. So
307
00:42:01,069 --> 00:42:09,289
if you do this, P in fact gives you the right
Eigen vector matrix. In fact, you get the
308
00:42:09,289 --> 00:42:18,309
columns 1, 2 and 3. Remember, column 1 itself
is complex; column 1 is complex; column 2
309
00:42:18,309 --> 00:42:26,309
is also complex, because they correspond to
complex Eigen values and the column 3 is real.
310
00:42:26,309 --> 00:42:31,229
In fact, the column 3 corresponding to the
Eigen vector associated with the real Eigen
311
00:42:31,229 --> 00:42:37,960
value. In fact, we should not just let this
go as it is. One important point you will
312
00:42:37,960 --> 00:42:43,890
notice is that the Eigen vectors associated
with the complex conjugate pairs turn out
313
00:42:43,890 --> 00:42:50,349
to be complex conjugate themselves.
So you say this particular Eigen value Eigen
314
00:42:50,349 --> 00:42:58,119
vector is minus practically 0 and this is
almost 0.1 into J or I; I is actually square
315
00:42:58,119 --> 00:43:05,560
root of 1 that is J. I mean I have been using
J the software here uses I. You will find
316
00:43:05,560 --> 00:43:10,540
that the Eigen vectored component here is
the complex conjugate of this. So, this is
317
00:43:10,540 --> 00:43:15,869
an important property of real matrices. So
even you are Eigen vectors turn out to has
318
00:43:15,869 --> 00:43:23,630
certain properties. One more interesting point
is, I will just write this Eigen vector matrix;
319
00:43:23,630 --> 00:43:37,780
it is the Eigen vectors and roughly this is
roughly, very roughly this is J into 0.1 this
320
00:43:37,780 --> 00:43:45,050
is practically 0, this term here. The second
term practically is 0 and the third term is
321
00:43:45,050 --> 00:44:05,420
practically 1 roughly. This is minus J 0.1
0 1 and this is minus 0.7 minus 0.7 and 0.06.
322
00:44:05,420 --> 00:44:12,130
So, please have a look at the screen again,
the computer screen. I just approximated these
323
00:44:12,130 --> 00:44:19,249
columns just rounded it off. I mean rather
I should say I approximated it rather than
324
00:44:19,249 --> 00:44:27,579
round off; it is practically 1 and is a very
small number here. So, this is how I get P.
325
00:44:27,579 --> 00:44:33,729
You can concentrate on the sheet which I am
writing on you will get a P matrix of this
326
00:44:33,729 --> 00:44:41,260
kind. So, if I numerically evaluate now I
have got lambda 1, lambda 2 and lambda 3 lambda
327
00:44:41,260 --> 00:44:47,450
1 lambda 2 lambda 3 and I also have got P1,
P2, P3 and the P matrix. So, I can actually
328
00:44:47,450 --> 00:44:55,010
take out P inverse, so P inverse is take out
the inverse of P matrix. So I just print this
329
00:44:55,010 --> 00:45:02,329
out; so this is the inverse of the P matrix.
The columns of this column 1, column 2, column
330
00:45:02,329 --> 00:45:09,259
3, but if you look at the row that will give
you q i, the you know q i transpose. So, I
331
00:45:09,259 --> 00:45:17,799
have also computed P inverse in this particular
case. But, let us now evaluate the time response
332
00:45:17,799 --> 00:45:25,579
of x; this is what I eventually wanted to
take out P this and P inverse.
333
00:45:25,579 --> 00:45:34,059
So if you finally compute the response, I
will omit the algebra what you will get is
334
00:45:34,059 --> 00:45:49,259
i L1 of t; i L2 of t and V C of t is equal
to… This is the final response 10 minus
335
00:45:49,259 --> 00:46:18,109
10 e raise to minus 0.1t plus 0.1 e raise
to minus 5 of t sine of 1005t, and this is
336
00:46:18,109 --> 00:46:25,499
10 minus 10 e raise to minus 0.1 of t. Please
try to make it work it out on own with Sci
337
00:46:25,499 --> 00:46:35,150
lab, I am just writing down the response directly.
This is of course, making certain approximations
338
00:46:35,150 --> 00:46:44,319
since I rounded off certain terms. So, this
is your final response so you i L2 t is 10
339
00:46:44,319 --> 00:46:51,339
minus 10 and so on.
Now, one important point is that lambda 1
340
00:46:51,339 --> 00:46:59,510
and lambda 2 which were in fact, a complex
conjugate pair here seem to be after lot of
341
00:46:59,510 --> 00:47:07,560
algebra seem to be giving you a sinusoidal
term here. Remember, that things get so arranged
342
00:47:07,560 --> 00:47:12,690
because of the fact that Eigen values appear
in complex conjugate pairs. The complex Eigen
343
00:47:12,690 --> 00:47:20,509
values are in complex conjugate pairs, you
will find that you should be able to make
344
00:47:20,509 --> 00:47:36,650
these simplifications
and
345
00:47:36,650 --> 00:47:42,099
your response is in fact, will turn out to
be in terms of sine's and cosine's there is
346
00:47:42,099 --> 00:47:47,089
no complex number in the response eventually,
because you will be able to combine all the
347
00:47:47,089 --> 00:47:49,640
complex terms appropriately in order to get
these terms.
348
00:47:49,640 --> 00:47:57,180
Now, one of the important points after a bit
of you know approximating the Eigen values
349
00:47:57,180 --> 00:48:03,869
and Eigen vectors, these are the these are
the Eigen vectors and these are the Eigen
350
00:48:03,869 --> 00:48:13,160
values. One thing you will notice here is
that the there is exponentially decaying component
351
00:48:13,160 --> 00:48:18,519
of the response, there is the forced component,
because of the input which you have given
352
00:48:18,519 --> 00:48:25,269
and there is also a component here, which
is oscillatory in nature; it is a decaying
353
00:48:25,269 --> 00:48:30,709
oscillation; similarly here.
354
00:48:30,709 --> 00:48:36,289
So because of the fact, that you have got
a complex conjugate pair of Eigen values,
355
00:48:36,289 --> 00:48:42,710
the mode which you get will be oscillatory.
There is a oscillatory mode and is also an
356
00:48:42,710 --> 00:48:48,239
exponentially decaying mode. Another important
point which you notice is that the oscillatory
357
00:48:48,239 --> 00:48:56,250
mode is practically not observable in i L2;
that is a very important point. So, what you
358
00:48:56,250 --> 00:49:00,680
are finding is that, in this particular system
for all practical purposes of course, we did
359
00:49:00,680 --> 00:49:07,910
make a few rounding of kind of approximations.
But, for all practical purposes there is no,
360
00:49:07,910 --> 00:49:15,849
there is no observability of this oscillatory
mode in this i L2. In steady state, i L2 turn
361
00:49:15,849 --> 00:49:23,079
out to be 0; i L2 is 0; V C also turns out
to be 0. If I put t tending to infinity V
362
00:49:23,079 --> 00:49:33,229
L and i L, V L, V C and i L2 becomes 0 and
i L1 becomes 10. The set if you look at this
363
00:49:33,229 --> 00:49:41,089
if V i is 1 volt, what are the steady state
conditions here? In steady state remember,
364
00:49:41,089 --> 00:49:47,890
the inductor will become a short; a capacitor
becomes open circuited. So, what you have
365
00:49:47,890 --> 00:49:52,829
really is, if this a short, this is open circuit;
this is a short. Your current through here
366
00:49:52,829 --> 00:50:05,019
will be simply 1 divided by 0.1 which is 10
amperes; this current here, this particular
367
00:50:05,019 --> 00:50:11,369
capacitor here in steady state becomes open
circuited; this inductor here becomes a short
368
00:50:11,369 --> 00:50:14,390
circuit.
So what you will find is eventually, you will
369
00:50:14,390 --> 00:50:20,960
have 10 amperes flowing here and going into
this, but the voltage here is equal to 0.
370
00:50:20,960 --> 00:50:28,900
So the steady state, in case you excite the
circuit with the non 0 input is going to be
371
00:50:28,900 --> 00:50:37,779
that i L1 and i L2 are going to be 10 amperes
and V C is going to be 0. This is 10 and 10;
372
00:50:37,779 --> 00:50:44,029
this is going to be 0 as t tends to infinity.
So, this is an important thing which you should
373
00:50:44,029 --> 00:50:48,709
be able to do. By setting t tending to infinity,
you should be able to take out a steady state
374
00:50:48,709 --> 00:50:57,880
final that is the final values of these terms.
Now, what you really are seeing here in fact
375
00:50:57,880 --> 00:51:02,190
this particular example, we do not we would
like to interpret the results appropriately,
376
00:51:02,190 --> 00:51:11,569
because the Eigen vectors are in fact giving
you some idea about the observability of modes.
377
00:51:11,569 --> 00:51:24,119
If you look at P which is right Eigen vector,
you see that the second state that is i L2,
378
00:51:24,119 --> 00:51:29,770
the component is practically 0. That is why
it is not observable, you will not see the
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00:51:29,770 --> 00:51:34,970
oscillatory mode observable in the second
state which is i L2.
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00:51:34,970 --> 00:51:44,119
Similarly, actually if you see here in V C,
you will not see the exponentially dying mode
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00:51:44,119 --> 00:51:51,029
to the same extent as you will see it here.
In fact, it is almost 10 times more seen in
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00:51:51,029 --> 00:51:55,650
these two states then it is seen here. So,
that is what we really got in the final response
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00:51:55,650 --> 00:52:02,490
if you focus on the sheet which I have written
down here. You will find that, where as the
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00:52:02,490 --> 00:52:10,890
coefficient of this e raise to minus 0.1t
is 10 here it is only 1 here. So you will
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00:52:10,890 --> 00:52:17,799
see that relatively, you can observe this
particular mode more in these currents than
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00:52:17,799 --> 00:52:22,400
in the voltage of course, voltage and current
are in compatible, because the units are slightly
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00:52:22,400 --> 00:52:28,440
different. So, one should take care in interpreting
the results. You cannot compare 10 amperes
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00:52:28,440 --> 00:52:32,799
with 1 volt so that is one issue which you
should remember. So, what I said was only
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00:52:32,799 --> 00:52:37,829
you know kind of giving a some relative feel.
Now although, we have written down the response,
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00:52:37,829 --> 00:52:44,749
it is a good idea to simply plot the response.
We will just plot the response using Sci lab
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00:52:44,749 --> 00:52:49,229
and you can of course, see how the response
looks like, because it sometimes difficult
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00:52:49,229 --> 00:52:54,019
to picture if you are new to studying a dynamical
system, how the response is going to look
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00:52:54,019 --> 00:53:01,559
like. So, if you for example see, what is
the response of i L1 or what is the response
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00:53:01,559 --> 00:53:08,940
of V C? So I will just try to you know write
down the response. So let us assume that,
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00:53:08,940 --> 00:53:13,549
I will evaluate the response at some time
steps.
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00:53:13,549 --> 00:53:23,369
So, I will I will try to evaluate the response
at the time steps t. So at discrete time steps,
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00:53:23,369 --> 00:53:31,400
I will evaluate the time response and plot
it. So the time response of course, of i L1
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00:53:31,400 --> 00:53:57,509
is given by 10 minus 10 star is the product
to time plus 0.1 into exponent of minus 5t
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00:53:57,509 --> 00:54:31,869
sine of 1000 and 5 times t.
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00:54:31,869 --> 00:54:57,749
So, this is so i L1 I evaluated i L1; i L1
is simply this formula which is evaluated
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00:54:57,749 --> 00:55:04,562
out here. I simply evaluated the formula which
I had mentioned before. So what I will do
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00:55:04,562 --> 00:55:16,789
is plot time versus i L1. So what you see
is, this is the waveform of i L1; it practically,
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00:55:16,789 --> 00:55:25,690
if you look at what it contains; it is basically
containing mainly this term e raise to minus
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00:55:25,690 --> 00:55:34,819
0.1t it is very easy to see this particular
term. Remember that, e raise to minus 0.1t
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00:55:34,819 --> 00:55:39,861
has got a time constant of 10 seconds. So,
around four times that time constant you will
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00:55:39,861 --> 00:55:44,690
find that the system here is settling to its
final value. An important point is that, you
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00:55:44,690 --> 00:55:49,390
are not able to see much of the oscillatory
response.
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00:55:49,390 --> 00:56:22,490
Let us just look at, in contrast V C; V C
is nothing, but so I will plot V C, what you
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00:56:22,490 --> 00:56:30,559
see is if this is i L1, this is what V C is.
So V C of course, has got us this oscillation
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00:56:30,559 --> 00:56:38,890
here this initial oscillation in fact is that
1000t Cos of 1000t, which you have seen here
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00:56:38,890 --> 00:56:44,890
and there is a also a component which is decaying.
So, you find that the response is of course,
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00:56:44,890 --> 00:56:51,220
a superposition of the two modes and both
modes in this case are fairly visible. So
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00:56:51,220 --> 00:56:55,579
you see this oscillatory mode in the beginning
of course, the rate of decay of the oscillatory
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00:56:55,579 --> 00:57:03,690
mode is more e raise to minus 5t decays faster
than e raise to minus 0.1t. So, this is basically
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00:57:03,690 --> 00:57:17,930
the response which you get. Now if of course,
I plot i L2; i L2 is simply 10 minus 10 star
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00:57:17,930 --> 00:57:35,190
x 0.1 star t; so if I plot i L2, this is what
I get. It is almost actually kind of its plotted
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00:57:35,190 --> 00:57:36,880
over i L1.
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00:57:36,880 --> 00:57:43,029
So, i L1 and i L2 practically have the same
response. So, you see the different modes
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00:57:43,029 --> 00:57:51,839
are seen to in to in different extents, in
the different states. In this particular example,
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00:57:51,839 --> 00:57:59,309
we could really interpret the presence of
the complex Eigen value pair, the faster and
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00:57:59,309 --> 00:58:06,999
slower responses, and also in some ways interpret
the Eigen vector. In fact, in an this particular
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00:58:06,999 --> 00:58:13,970
example I have chosen specifically to illustrate
a very very important modeling approximation,
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00:58:13,970 --> 00:58:17,849
which we will see in the next class. This
particular system is in fact, what you call
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00:58:17,849 --> 00:58:24,369
a stiff system which is a combination of a
slowly moving transient e raise to minus 0.1
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00:58:24,369 --> 00:58:31,089
t, and a faster -, relatively faster oscillation
which also decays faster. So, these kind of
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00:58:31,089 --> 00:58:36,719
systems are called stiff systems, and whenever
you have stiff systems, we have may have problems
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00:58:36,719 --> 00:58:42,349
solving them numerically, we shall see later.
Numerical integration may be a problem using
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00:58:42,349 --> 00:58:48,359
simple methods, but we also would be able
to make some good, and modeling simplifications.
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00:58:48,359 --> 00:58:55,180
So, with that we stop todays lecture. In the
next lecture, I will introduce you to some
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00:58:55,180 --> 00:58:56,859
very important modeling principle.