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We continue with our discussion, on the analysis
of dynamical systems. In the previous few
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lectures, we have been discussing the analysis
or the time response of linear systems. Linear
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systems are amenable for what is known as
Eigen value analysis, in which we can understand
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the properties of the response, just by looking
at properties of the A matrix.
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Now, in today’s lecture, we shall introduce
you to a more general technique, which is
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applied even to non-linear for the analysis
of even non-linear systems, that is the technique
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of Numerical Integrations. So, we will talk
about numerically integrating dynamical equations
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in order to obtain the response.
Now, it should be understood that, numerical
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integration is not limited to the analysis
of linear systems however, in today’s lecture
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one or two examples that we will consider;
we will consider numerical integration of
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linear systems. The reason why we do that
is, since we know the, what response is, it
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could be easier to understand the properties
of the numerical integration algorithms, which
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I used in doing numerical integrates, that
is also called simulation of a system.
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Now, before we go on to numerical integration,
we have been studying some very interesting
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topics in linear systems, and how we can actually
interpret the response of a linear system.
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So, will just have a quick review of what
we have been doing over the past few lectures,
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we started with the analysis of dynamical
systems. What is equilibrium, what are equilibria
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of a system, what are states, the issue of
small and large disturbances stability, we
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also considered to illustrate this particular
concept of small and large disturbance stability,
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the example for single machine connected to
a voltage source.
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Then we moved on to linear time invariant
systems there analysis, and we saw that we
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could analysis a system a linear time invariants
system response, using Eigen values and Eigen
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vectors of the A matrix of the systems. That
is if system x dot is equal to a x the properties
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of a, really determine the stability of the
system. The response of course, its dependent
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not only on the Eigen values and Eigen vectors,
but also dependents on the initial conditions
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of the system. Whereas, the Eigen values really
determine, the nature of the time response
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of individual pattern, in this system the
Eigen vectors, the right Eigen vectors really
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give you some idea of the observability of
certain patterns in certain stage.
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So, that is basically is a physical you know
kind of interpretation of Eigen values and
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Eigen vectors and we have done a fairly detail
analysis of the time response of linear time
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invariant systems. And we can actually write
down the time response of course, there there
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was a kind of cushion, which had you know
ask you to bear in mind, that is if your A
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matrix is non distinct Eigen values, it is
possible that you will not be able to get
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n linearly dependent Eigen vectors, n being
of course, size of the A matrix.
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So, if you got A matrix, which has got size
n cross n and its does not have n distinct
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Eigen values, this possibility that you may
not have n distinct or linearly independent
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Eigen vectors, in that case you cannot diagnosis
the matrix you will get terms like t e rise
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lambda t and so on, under such situations.
But, the basic point which I want to, you
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know you to understand that the and behavior
of a linear system can be quite properly understood,
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there is no mystery as for as the functions
which appear in the response; the time functions
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that is e rise lambda t sin omega t, t e rise
to lambda t is all are well known and well
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understood functions. So, that the response
of the system can be quite easily obtained
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and understood.
In addition we can also understand, how the
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system will behave like, if you got a system
x dot is equal to a x plus b u, that is u
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is an input, then the input the force response
for this input also can be written down. So,
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that is one major advantage of linear time
analysis of linear time invariant systems;
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remember of course, that linear your linear
time invariant system, dynamical systems are
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arise in our day today life and in power system
dynamics in two ways.
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The system may be inherently linear or you
may be analyzing basically originally non-linear
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system by linearsing it around an equilibrium
point to understand its small disturbance
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behavior. So, linear system arise in these
two possible contacts and important point,
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which we understood in the previous lecture
was that, some systems not all systems, some
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systems have very widely wearing Eigen values
or in other words the patterns which are seen
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in the final response, remember the response
is super position of patterns. The patterns
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which have seen in the response, a time response
is made out of patterns, you will find it
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there is a wide variation in the speed of
the response of the various patterns; for
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example, you may find at is one component
of the response, which varies very fast and
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one component which varies very very slowly,
so this modes sort to speed.
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Or you know you can say the fasten and slow
modes, and in case you have got fast and slow
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modes, you can actually make approximation
in the model which you are using; one of the
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important things was, that if we can identify
the states, which having some way associated
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with the fast modes. And the states which
are associated with slow modes, good actually
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make modeling simplification, and in the basic
point is that it could reduce the order of
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the system. You could reduce the number of
differential equation in the system.
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This will be very important, because in power
systems you will find that there are transients
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which you know kind of a very fast, like lighting
transients and network transients and some
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very slow transients associated with the mechanical
systems, like the governor boiler and so on,
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which have very slow. So, when you are studying
a particular system, you need not model every,
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you know you may make modeling simplification
depending on the thing you are interested
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in.
In case you are for example, interested in
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fast transient you should pay more attention
for example, in modeling a transmission line,
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you may even think of modeling a transmission
line by its partial differential equation
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model. Whereas, if you understand slow transient
or perhaps like for example, a loss of synchronism
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is a relatively slow phenomena, compare to
lightening transients.
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And oscillation, electro mechanical oscillation
which have seen in the grade, in that case
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you may wish to even represent the network
by lumped, you know by lumped dynamical system
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and some time even neglect the dynamical equation
themselves of the network. You may treat the
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network to be cos sinusoidal steady state
of course, this I am jumping steps, you have
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to recall this, when we come to modeling of
transmission lines and other components. The
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basic point is that depending on what phenomena
is of interest, whether it is fast or slow
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you can make modeling simplification, to make
things of bit more precise.
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If you got a system x dot is equal to A x
or you got system x dot is equal to g of x,
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where x actually is a set of vectors, I mean
this is actually short hand for large number
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of equations. So, you got in fact, x 1 dot
is equal to g 1, (x 1, x 2, x n),, x 2 dot
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is equal to g 2 (x 1, x 2, x n) and so on,
or in this particular case x 1 dot is equal
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to… tThis is a linear system a 1 1 x 1 plus
a 1 2 x 2 and so on, and x 2 dot is equal
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to a 2 1 x 1 plus a 2 two 2 x 2 and so on.
So, you can have system, this a linear system,
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this is non-linear system, if you can identify,
if it is possible to identify states, say
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you know there are certain states say x 3
onwards, that is x 3 x 4 x n, which you can
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identify as a associated with the fast transients.
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In that case, you would be at least in this
non-linear, both in the non-linear and linear
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system case, you can
make this particular approximation this is
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of course, unchanged this x 3 dot we just
set equal to 0 and replace it by an algebraic
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equation g (x 1…) iIs equal to 0 this is
the differential algebraic model of the system.
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And if you take this particular model of the
system you if of course, one important point
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is you know a priory that this states x 3
onwards or in some way associated with the
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fast transient . So, what you mean by association
is of course, something we define in the last
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part of the last lecture, I will just recall
what it was in some time. So, what you can
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do is, if you are interested only in the slow
transient, then you can replaced at differential
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equation corresponding to the fast state variable
or the state variables, corresponding to the
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fast modes of the fast transients of the system,
and you can replaced them by algebraic equations.
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So, it convert set of differential equation
into a set of differential algebraic equations
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of course, in the non-linear case it may not
be easy to you know you could use this algebraic
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equation to element or rather right x 3 to
x n in terms of x 1 and x 2; this is possible
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in linear system quite easily, in non-linear
systems it may not be possible to element
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in that sense.
So, in linear systems, we can get rid of the
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fast variables in what I mean by get rid is
just write them, in terms of the slow variables.
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And resulting differential equation, then
can be written simple as in case of linear
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systems, you can have differential equations
only in x 1 dot and x 2 dot, that is why we
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did in the example in the previous class.
In fact, there was you know, when we analysis
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the slow and fast transients. So, if you are
interested in this slow response, you can
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kind of be blind to the fast response. So,
this is what we discussed in the previous
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class.
Conversely if you are interested in this slow
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response or rather you are interested in the
fast transients, for a very short duration
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of time. So, I am interested only in viewing
the fast transient for short duration of time,
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you can assume the state associated with the
slow transient are just fluorescent at the
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pre disturbance value. So, these are the kind
of modeling simplification we can make, it
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turns out that these kind of system is of
course, this resumes that the system can be
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broken up into fast and slow subsystems.
This may not always be possible, but in case
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you can do it these modeling simplifications
are possible. If shall see that, once you
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make the modeling simplifications, you may
be able to analysis systems in a better fashion.
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For example, when we understand numerical
integration by removing the fast, you know
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the transient, the state variable corresponding
to fast transient by eliminating them, writing
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them, in terms of the slow variables, you
can actually start using some simpler numerical
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integration methods.
So, of course, this is something will discuss
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later, when we I introduce you to numerical
integration methods, somewhat later in this
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particular lecture. Of course, the key point
which I mentioned last time was to associates
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certain states with certain patterns, fast
transients, you know how you do that? In linear
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system, I mentioned that, if you take the
write Eigen vectors and the left Eigen vectors
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corresponding to a system.
For example, in the previous class, we studied
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this particular system and what we did was
to find out the association of fast transient
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since, slow transients, we evaluated the right
and left Eigen vectors corresponding to various
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modes, we find out the right Eigen vector
matrix . P invest matrix is also as left Eigen
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vectors as its rows, this is called q, now
if I take this P and P inverse and I do this
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operation, please refer to the previous lecture.
If I do this operation, then the matrix which
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results gives us the participation of certain
state in a certain Eigen value or certain
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mode in this case. So, this is basically one
way suppose the participation of a particular
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state, in a particular Eigen value lambda
3 is 1, then we say, this state is completely
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associated with this Eigen value lambda 3.
Of course, i l 1 and v c sorry i l 1 and v
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c both participate in the mode corresponding
to this Eigen value, so this is what we did
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last time.
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So, this was of course, what I mentioned sometime
ago, you could you can use get in association
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of states and modes using the participation
factor matrix, note that in a coupled systems
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sometime, the participation is distributed
among all states. So, sometimes, it may not
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be possible to this to really make this you
know portioning of system into fast and slow
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transients, you may find it every state in
some way or the other way, in some significant
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sense participates in all the modes. So, that
may of course, preclude using this fast and
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slow transients to make modeling simplifications.
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Of course, the aim of this course is study
power system dynamics and control, and although
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the initial part of this particular course,
I been concentrating on general analysis of
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systems. We should soon in may be a couple
or 3 lectures, go into domain specific issue
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modeling issues, but it is to be noted that,
you should have some background in the analysis
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of dynamical systems.
So, I I recommend that you can read as was
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in linear time invariants systems are concerned,
the books by Ogata or some equilibrium book,
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you will find many. And of course, I would
also recommend that you read some books on
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basic Eigen values and Eigenvectors for example,
matrices for scientists and engineers, which
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are recommended here. But, of course, there
are many other equitable books, which are
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just as good.
There are other many many other topics relating
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to linear systems, and as we go long we shall
also talk about trans a function representation
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and so on. But, that will be when we understand,
excitation and prime mover controller, so
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for the timing now.
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Let us move on to the next part of our course
that is the numerical integration, before
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I really start the topic of numerical integration.
Let us look and the course over view again,
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so that we do not really get lost in what
we are doing, remember that the first part
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of this course, was a basic introduction to
analysis of dynamical system, where we understood
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the Eigen analysis of linear time invariant
system.
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Now, we shall go on to study numerical integration
techniques, what is still to come is modeling
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of synchronous machines, modeling of excitation,
prime mover systems, modeling of transmission
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lines, loads and other components.
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And of course, what is the main important
thing in this course is to understand the
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stability of inter connected power systems.
So, we shall use the models which we developed,
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and analysis tools which are developed in
part one of the course, and go ahead and understand
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the stability of interconnected power systems,
power systems stability analysis tools and
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methods to improve power system stability.
So, this is basically, what we shall do in
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this particular course, now let us move on
to the other topic of numerical integration.
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Now, if you studying a system x dot, continues
time system x dot is equal to f x of t, you
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may one of the ways to analysis the system
is to numerically integrate the system given
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the initial conditions., nNow the simplest
way to do that of course, is what is known
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as Euler’s method or equivalently forward
Euler’s method. So, I shall use of course,
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Euler’s method mean forward, Euler’s method
interchangeably they mean the same thing.
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x dot can be approximated as x k minus 1 minus
x k upon h in this fashion, this is known
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as Euler’s method of numerical integration.
Now, what we really mean by numerical integration,
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what I mean is suppose I know suppose this
is time t is equal to 0, I know the initial
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value of x that is x of 0 in some sense, I
know the initial value of x I evaluate.
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So, k is equal to 0 or you can take the index
is 1 does not matter, k is equal to 0 you
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evaluate this function at this point, using
this particular equation you can, since you
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know what the function is you can evaluated
it at this point, k is equal to 0, and get
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what k plus 1 is, that is k plus 1 is of course,
1 in this case. So, from x 0, this is x of
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0, you can get x of 1, so you can say evaluate
this, so x k plus 1 is equal to x k plus h
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times x k t k of course, what is this h this
is the value of the variable, the h is essential
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it time duration after which x k plus 1 is
evaluated.
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So, x 1 means, actually x of 1 implies x of
h, x of 2 x 2 is nothing but, x of 2 h, so
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I am evaluating this x at discrete points,
so
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this is x of 2, this is x 1, this x 0, so
of course, in between I do not know the values,
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these values are not in fact, given by the
numerical integration method, I just interpolated
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in between . So, what I am going to get is
a discrete set of points, now by inter polluting
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between these points, if I can roughly get
the actual actual continuous time response
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of this, I would say my objective as been
fulfilled, and I have got the response of
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the system.
Remember of course, that the basic numerical
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algorithm will only yield, the values of the
variable at discrete points of time. Now,
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one important point you should notice that,
this is approximation Euler’s method is
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an approximation, you will not of course,
get the excite value of x by doing numerical
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integration in all cases, the only some systems
which will give you correct answers, if you
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numerical integrating use using Euler’s
method. So, what I will do is let us take
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system, this let us take a linear system,
the reason why I am taking the linear system
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is I know the response of the linear system,
in terms of well known functions.
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There is no need excepting in very complicated
linear system, with lot of switching and then
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lot of you know intermediate disturbance,
there is no need to numerically integrate
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a linear system, because, you can actually
write down its time response, so but the reason
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why I am studying numerical integration using
the linear system is that, I know the response
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of the linear system. So, I will able to tell
how the numerical integration method behaves
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you know as compared to the correct response,
because I know the correct response.
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So, if you take for example, a system x dot
is equal to a x, so this x dot is equal to
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a x dot is coming by mistake, so if I try
to what is the solution of this, x of t is
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00:24:05,510 --> 00:24:15,850
equal to e rise to a t x of 0 and of course,
if I the correct response, if I sample it
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00:24:15,850 --> 00:24:29,810
at discreet points you will get x of k is
equals to e rise to a k h into x of 0. So,
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this is the correct correct sampled response,
but if I use a numerical integration method
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like Euler method, this a linear system.
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00:24:48,820 --> 00:25:14,230
So, what I will get is , this is what I will
get, so x of k in this case will be 1 plus
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00:25:14,230 --> 00:25:38,520
a h rise to k x of 0, that is x of 0 is this,
so the correct response is
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00:25:38,520 --> 00:25:43,070
and the numerically integrated response is
this, that is if I use forward or forward
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00:25:43,070 --> 00:25:50,110
Euler method or Euler method. So, these two
of course, are not the same, the point is
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is this a good approximation of this, the
answer of course, depends on the value of
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00:25:55,450 --> 00:26:03,260
h, if h is very small, then one may except
that you are going to come close to the solution,
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your accuracy will be if h is extremely small.
Now, will just take a simple example to understand
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00:26:15,160 --> 00:26:25,120
this point, but before we do that let us look
at a qualitative issue, when is this correct
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00:26:25,120 --> 00:26:32,920
solutions stable, when it is the system stable,
the system is stable if a is less than 0,
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00:26:32,920 --> 00:26:40,330
if a is real number it should be less than
0 then the system is stable. However, here
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this particular system, when it is stable,
you will note that, if the modules of 1 plus
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a h, 1 plus a h of course, is a constant if
you choose your h, if it is fixed h, in that
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00:26:54,650 --> 00:27:02,250
case this particular 1 plus a h is always
a constant in your numerical integration rise
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00:27:02,250 --> 00:27:10,200
to k.
So, the point is if mode of 1 plus a h is
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00:27:10,200 --> 00:27:21,050
less than 1 then Euler method will say that
the system is stable. Now, what is the important
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00:27:21,050 --> 00:27:25,809
thing is that, the actual system is stable
when a less than 0, but Euler method says
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00:27:25,809 --> 00:27:32,450
that it is stable only if this true, these
2 conditions in fact, are not cinnamons there
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00:27:32,450 --> 00:27:45,420
slightly different. In fact, if h greater
than 2 by a then even, if is negative this
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00:27:45,420 --> 00:27:54,429
may be greater than 1. So, the stability of
Euler method will depend on its time step,
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00:27:54,429 --> 00:27:59,490
it does not not only dependent on a, but it
also dependents on the time step. So, it is
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00:27:59,490 --> 00:28:05,980
obvious that Euler method can give qualitatively
wrong picture of the stability of the system,
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00:28:05,980 --> 00:28:15,980
if you do not choose a h appropriately. So,
what we shall do is numerically integrate
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00:28:15,980 --> 00:28:24,330
a system x dot is equal to a x using Euler
method, so for this purpose, I will use as
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00:28:24,330 --> 00:28:30,910
ours doing in the previous lectures.
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00:28:30,910 --> 00:28:41,170
Now, this is the basic window of in which
I will run a programmed, now instead of typing
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00:28:41,170 --> 00:28:46,300
of all the commands 1 by 1 I have done it
already in separate file which I will show
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00:28:46,300 --> 00:28:55,860
now. So, we shall study numerical integration
using Euler method, what will do is this is
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00:28:55,860 --> 00:29:05,351
the variable x, we start with clean slate
with clears variable, let say that a is equal
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00:29:05,351 --> 00:29:12,610
to 5, a is equal to minus 5 is it is stable
system and unstable system; it is stable system.
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00:29:12,610 --> 00:29:18,580
Because, the x dot is equal to a x if is negative,
is a negative real number you will find at
249
00:29:18,580 --> 00:29:24,060
the system is actual stable; now let us try
to integrate the system x dot is equal to
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00:29:24,060 --> 00:29:30,910
a x using Euler method for that I should define,
what my time step is let say right now, I
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00:29:30,910 --> 00:29:42,080
choose the time step h is equal to point 1.
Let us simulate the system for say I am sorry,
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00:29:42,080 --> 00:29:48,070
let us simulate the system for 10 seconds,
actually need not simulate for 10 seconds,
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00:29:48,070 --> 00:29:55,740
let us do one thing, let us simulated for
5 seconds.
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00:29:55,740 --> 00:30:04,130
So, the number of steps to simulate from 0
to 5 would be given by this formula, so you
255
00:30:04,130 --> 00:30:11,970
have got T final is five h is point 1. So,
the number of steps is you will get by rounding
256
00:30:11,970 --> 00:30:13,430
of T final by h.
257
00:30:13,430 --> 00:30:23,710
So, that is of course a integer, so that is
why we have to round it off, suppose the initial
258
00:30:23,710 --> 00:30:27,460
condition remember, we are solving the dynamics
of the system for a initial condition which
259
00:30:27,460 --> 00:30:32,870
is not the equilibrium, for x dot is equal
to a x the equilibrium is x is equal to 0.
260
00:30:32,870 --> 00:30:39,100
So, we are giving an initial condition x 1
is equal to 1, incidentally I am not written
261
00:30:39,100 --> 00:30:47,740
x 0 here, because does not rather, does not
normally permitted to give index, which is
262
00:30:47,740 --> 00:30:54,250
0, so that is why its x 1, so we shifted all
the indexes by 1.
263
00:30:54,250 --> 00:31:01,880
As I mentioned sometime back, if it a is equal
to minus 5, so what is the response of the
264
00:31:01,880 --> 00:31:10,190
system it is e rise to minus 5 t into x of
0. So, you know we except about in a second
265
00:31:10,190 --> 00:31:15,040
or so the system should settle down, it is
a stable system; the time constant of the
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00:31:15,040 --> 00:31:21,131
system is 1 upon 5, that is point 2 seconds.
4 times this time constant is roughly, the
267
00:31:21,131 --> 00:31:28,580
time it will take to settle down to steady
value which is 0, that is a equilibrium value,
268
00:31:28,580 --> 00:31:32,671
this is the stable system.
So, if I want to numerically integrate, I
269
00:31:32,671 --> 00:31:37,720
will use this particular command, for i is
equal to 2 to the number of steps, i is the
270
00:31:37,720 --> 00:31:42,950
index; instead of using the variable k and
I am using the variable I here, where x of
271
00:31:42,950 --> 00:31:54,150
i is equal to 1 plus h into x of i minus 1.
And then I plot time, these are the time steps,
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00:31:54,150 --> 00:31:59,590
individual time steps I plot the time steps
and the corresponding x values. So, of course,
273
00:31:59,590 --> 00:32:17,120
if I execute this
So, if you look at this response, I slightly
274
00:32:17,120 --> 00:32:22,390
expanded, you see the response is fairly well
captured, this remember the time step I have
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00:32:22,390 --> 00:32:35,460
chosen is x is h is equal to 0.1, point yeah
was it 0.1, yeah it is 0.1. So, for this particular
276
00:32:35,460 --> 00:32:41,500
system, h is equal to 0.1 seems to give you
this response, the system settle down in around
277
00:32:41,500 --> 00:32:52,179
1 second of course, this seems to be a continuous
graph, that is because that because is doing
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00:32:52,179 --> 00:33:00,640
the inter pollution between the individual
points, so that is the not very big surprise.
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00:33:00,640 --> 00:33:10,481
Suppose, I changed the time step from 0.1
to 0.2, so first what will do is redo this
280
00:33:10,481 --> 00:33:37,830
with 0.1 and then we do this with 0.2, we
see that our response is slightly changed,
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00:33:37,830 --> 00:33:44,380
you see here it is gone slightly different,
is a different response here. On the other
282
00:33:44,380 --> 00:33:58,740
hand, if I changed into 0.5 things start looking
very very different, the response is completely
283
00:33:58,740 --> 00:34:02,880
different, this is the response and of course,
it is not correct response, it is in fact,
284
00:34:02,880 --> 00:34:08,100
going unstable the discreet time system is
going unstable. So, what we see is that, if
285
00:34:08,100 --> 00:34:15,030
you have got Euler method and I choose a time
step, which is not compatible.
286
00:34:15,030 --> 00:34:21,520
In fact, if you look at my condition which
have written down, you can focus on the what
287
00:34:21,520 --> 00:34:37,879
I am writing 1 plus a h less than 1 implies
system is stable as per Euler’s method,
288
00:34:37,879 --> 00:35:01,109
the discretize system is stable . So, in this
particular case, 1 minus 5 h less than 1 implies
289
00:35:01,109 --> 00:35:06,039
the discreet time is system is stable remember,
the original system x dot is equal to minus
290
00:35:06,039 --> 00:35:17,069
5 x is stable. The discreet time system which
is obtained by using Euler method gives you
291
00:35:17,069 --> 00:35:23,309
this condition; this is not the same as this
condition. In fact, if I choose h which is
292
00:35:23,309 --> 00:35:32,350
greater than 2 times a in fact, I should I
should write modules of 2 times of a, then
293
00:35:32,350 --> 00:35:39,359
I should basically get this condition will
not be satisfied and will get an totally wrong
294
00:35:39,359 --> 00:35:49,900
response. So, actually this happens when h
is greater than 2 by minus 5, if this is nothing
295
00:35:49,900 --> 00:35:59,951
but, 0.4 yeah, so if h is greater than 0.4
1 minus 5 h is greater than 1 and Euler method
296
00:35:59,951 --> 00:36:03,660
gives you completely wrong information about
stability.
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00:36:03,660 --> 00:36:11,269
So, obviously, whenever you are using a numerical
integration method, we need to be careful
298
00:36:11,269 --> 00:36:18,619
about this stability of the numerical method,
whether it gives good qualitative understanding
299
00:36:18,619 --> 00:36:25,309
of how the system behave behaves. So, we shall
now consider other methods obviously, Euler
300
00:36:25,309 --> 00:36:30,319
method under certain circumstance does not
give you good response. In fact, we shall
301
00:36:30,319 --> 00:36:38,739
see later that in dealing with stiff systems,
Euler method is particularly unsuitable; especially
302
00:36:38,739 --> 00:36:44,460
one is interested only in the slow response.
So, this is something which will of course,
303
00:36:44,460 --> 00:36:54,289
understand in a few moments from now, perhaps
in the next lecture. Now, this takes us to
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00:36:54,289 --> 00:36:59,670
other methods of numerical integration, whenever
we talk of numerical integration, there is
305
00:36:59,670 --> 00:37:05,130
several terms which you will come across,
one is what is known as the order of the method.
306
00:37:05,130 --> 00:37:16,110
So, one of the thing which you will come across
is the order of method, second whether it
307
00:37:16,110 --> 00:37:29,979
is what is known as explicit or in implicit
method; the third thing which you will come
308
00:37:29,979 --> 00:37:38,150
across is whether it is a single step or a
multi step method.
309
00:37:38,150 --> 00:37:51,170
Now, there are two issues, which we or two
errors which result, whenever we use numerical
310
00:37:51,170 --> 00:37:56,989
integration method, one of the error is because
of fact because, of the fact that the numerical
311
00:37:56,989 --> 00:38:02,809
integration is in fact, the discretization
is almost always and approximation, like Euler
312
00:38:02,809 --> 00:38:08,089
method is an approximation of the original
continuous time system. So, that error which
313
00:38:08,089 --> 00:38:14,710
is introduce, because of the approximation
is called truncation error, the second thing
314
00:38:14,710 --> 00:38:21,749
which causes an error is because, we will
be of course, doing this numerical integration
315
00:38:21,749 --> 00:38:29,829
on a computer, and a computer will introduce
another error which is called a round of error.
316
00:38:29,829 --> 00:38:35,750
Because, any number can be limit represented
only to a finite precession in a computers,
317
00:38:35,750 --> 00:38:40,930
but of course, round of errors in normally
you will not have problem of round of error,
318
00:38:40,930 --> 00:38:47,750
because these days you can specify a very
high precession in computers. So, you have
319
00:38:47,750 --> 00:38:52,970
to normally bother about the approximation
which is made, that is are you going to use
320
00:38:52,970 --> 00:38:57,809
the Euler method or some other method. So,
let me first introduce you to some other methods,
321
00:38:57,809 --> 00:39:03,400
and then we will try to understand what order
is, what is explicit or implicit method, and
322
00:39:03,400 --> 00:39:09,369
what is the single step or multi step method.
323
00:39:09,369 --> 00:39:18,999
First thing, suppose you have got system x
dot is equal to a x x k plus 1 is equal to
324
00:39:18,999 --> 00:39:30,369
or rather x minus x k upon h is equal to a
x k this is Euler of forward Euler, backward
325
00:39:30,369 --> 00:39:46,390
Euler on the other hand is x k plus 1 minus
x k upon h is equal to a x k plus 1, so this
326
00:39:46,390 --> 00:39:53,161
is an approximates. So, you are using the
value of the function a x at the point x k
327
00:39:53,161 --> 00:40:00,369
plus 1, now this is you may see you are using
x k and you are using x k plus 1. What is
328
00:40:00,369 --> 00:40:06,529
the significance of this, in fact if you got
a non-linear system backward Euler would look
329
00:40:06,529 --> 00:40:24,279
like this, this is function g which you are
discuss some time earlier, so for a non-linear
330
00:40:24,279 --> 00:40:34,380
system backward Euler will look like this.
So, backward Euler requires, x k plus 1 to
331
00:40:34,380 --> 00:40:38,819
evaluate g x k plus 1 and then you have to
solve this, now do you notice what the problem
332
00:40:38,819 --> 00:40:43,400
is this, is as compare to this see in this
case, if even if it was a non-linear system
333
00:40:43,400 --> 00:40:50,769
you would have got x k plus 1 minus x k by
h is equal to g of x k.
334
00:40:50,769 --> 00:40:56,880
So, typical you will know what is x k is,
so you can evaluate g of x k, you know what
335
00:40:56,880 --> 00:41:03,160
is x k is 2, then you can get x k plus 1,
in contrast look at this problem have you
336
00:41:03,160 --> 00:41:10,299
got x k plus 1, you know what x k is, but
you do not know what x k plus 1, x k plus
337
00:41:10,299 --> 00:41:16,519
1 is obtained from this. In fact, what you
get is an algebra you know x k, so you have
338
00:41:16,519 --> 00:41:22,880
got algebraic equation in x k plus 1, you
need to solve this implicit equation in order
339
00:41:22,880 --> 00:41:27,660
to get x k plus 1.
In a linear system this would imply that of
340
00:41:27,660 --> 00:41:35,059
course, you will have to do a division, but
here you will find it you will have to, in
341
00:41:35,059 --> 00:41:41,729
fact use a numerical method if g is non-linear
function, you will have to use numerical method
342
00:41:41,729 --> 00:41:47,569
for solving algebraic equation in order to
get the solution of this. So, backward Euler
343
00:41:47,569 --> 00:41:54,269
involves a bit of complicity, forward Euler
or Euler method is easy to evaluate, this
344
00:41:54,269 --> 00:42:03,539
x k plus 1 is a obtain explicitly from x k,
x k plus 1 here is obtained implicitly from
345
00:42:03,539 --> 00:42:07,989
x k.
So, in fact, without going to the any formal
346
00:42:07,989 --> 00:42:14,453
definition I hope your getting what, we mean
by explicit and implicit method, if method
347
00:42:14,453 --> 00:42:24,599
requires you to know x k plus 1, in order
to compute this function g, in that case you
348
00:42:24,599 --> 00:42:28,239
would call this method and implicit method.
349
00:42:28,239 --> 00:42:33,359
In fact, would you call this an explicit method
or in implicit method, suppose you got a system
350
00:42:33,359 --> 00:42:46,109
x dot is equal to g of x this is what we are
considering. Is this explicit or is this implicit,
351
00:42:46,109 --> 00:42:55,650
another approximation, this a another way
of doing a numerical integration, in this
352
00:42:55,650 --> 00:43:02,020
call trapezoidal rule, trapezoidal rule is
actually in implicit method, because to get
353
00:43:02,020 --> 00:43:07,599
x k plus 1 from x k will require to solve
this non-linear algebraic equation, in case
354
00:43:07,599 --> 00:43:12,109
its non-linear system.
And that itself is an implicit algebraic equation
355
00:43:12,109 --> 00:43:17,579
is an implicit algebraic equation which will
probably require numerical method, so in fact,
356
00:43:17,579 --> 00:43:24,069
in every time step if it is a non-linear system,
you may have to iterate in order to get x
357
00:43:24,069 --> 00:43:29,240
k plus 1 remember, however that the numerical
method in order to get x k plus 1 from x k
358
00:43:29,240 --> 00:43:37,059
is a method for solving numerical algebraic
equation. Rather non-linear algebraic equations,
359
00:43:37,059 --> 00:43:47,729
like or method, so trapezoidal rule backward
Euler are known as implicit methods.
360
00:43:47,729 --> 00:43:53,170
Now, why should we use trapezoidal rule or
backward Euler, if its slightly more difficult
361
00:43:53,170 --> 00:43:59,019
to solve as compare to Euler method, one of
the problem this is the properties of Euler
362
00:43:59,019 --> 00:44:04,740
method, Euler method unfortunately for systems
stiff systems is not a very good idea, will
363
00:44:04,740 --> 00:44:13,009
do an example in the next class, a few examples
in the next class show this. Whereas, backward
364
00:44:13,009 --> 00:44:20,059
Euler and trapezoidal rule in fact, a quite
suited for numerical integration of stiff
365
00:44:20,059 --> 00:44:24,299
systems. Now, whenever I say suitable what
do I mean a numerical integration should be
366
00:44:24,299 --> 00:44:27,810
easily fast, it should not require you to
use very trans, small time step other wise
367
00:44:27,810 --> 00:44:32,999
to sibilate or numerically integrate over
interval which is large, it will take you
368
00:44:32,999 --> 00:44:38,700
very long long time.
So, numerical method to the extent possible
369
00:44:38,700 --> 00:44:45,940
should be you know able to use time steps,
which are not too small compare to the interval
370
00:44:45,940 --> 00:44:51,460
which one wants to simulate, now in a stiff
system as you may imagine since, you got fast
371
00:44:51,460 --> 00:44:58,109
and slow transients, there will be an issue
about choosing your time steps is of course,
372
00:44:58,109 --> 00:45:03,219
something will discuss later in the may be
in the next lecture or so.
373
00:45:03,219 --> 00:45:10,470
So, the second thing, the thing which we discussed
of course, was implicit or explicit methods;
374
00:45:10,470 --> 00:45:17,509
now let us talk about order of a method, whenever
we talk about an order of method, what we
375
00:45:17,509 --> 00:45:23,200
really mean is suppose the response of x of
t, suppose the actual correct response of
376
00:45:23,200 --> 00:45:26,309
x of t can be written down as a polynomial.
377
00:45:26,309 --> 00:45:39,141
For example, if x of t is can be written suppose
is alpha 0 plus alpha 1 of t, suppose this
378
00:45:39,141 --> 00:45:43,039
is the time response in fact, all the time
response we have been talking of involved
379
00:45:43,039 --> 00:45:51,710
exponential function, but suppose this is
the time response, then x dot of t is nothing
380
00:45:51,710 --> 00:46:00,859
but, alpha 1 of x sorry alpha 1 that is all.
So, x of x dot of t is nothing but, alpha
381
00:46:00,859 --> 00:46:10,420
1, so what this means is of course, this particular
system, if I numerically integrate with back
382
00:46:10,420 --> 00:46:18,819
with Euler method or backward Euler method,
I will get the correct solution, without any
383
00:46:18,819 --> 00:46:25,230
truncation error.
So, backward Euler and forward Euler will
384
00:46:25,230 --> 00:46:32,809
give you the exact solution, now here truncation
error, in case a numerical integrates a system
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00:46:32,809 --> 00:46:41,329
of this kind, specifically of this kind. So,
in fact, backward Euler and forward Euler
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00:46:41,329 --> 00:46:52,160
or simple Euler method, they have what are
known as first order method methods of course,
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00:46:52,160 --> 00:46:58,989
if you got system x dot is equal to a of x
is not alpha alpha 1 here is a constant, if
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00:46:58,989 --> 00:47:06,450
x dot is equal to a into x, in that case your
response is remember e rise to a t x of 0,
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00:47:06,450 --> 00:47:10,800
this is the x of t.
In such a situation of course, you know the
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00:47:10,800 --> 00:47:17,400
expansion of e rise to a t is an infinite
series in t, so obviously, the response is
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00:47:17,400 --> 00:47:23,769
going to have many many more terms. So, whenever
you use Euler or backward Euler to simulate
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00:47:23,769 --> 00:47:28,900
this system you are bound to get some truncation
error. So, remember first order method means,
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00:47:28,900 --> 00:47:36,060
if your response is can be written down simple
as a first order polynomial of rather the
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00:47:36,060 --> 00:47:41,950
first order polynomial in t, in that case
backward Euler and forward Euler which are
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00:47:41,950 --> 00:47:44,950
first order methods will give you correct
response.
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00:47:44,950 --> 00:47:58,559
You can show that if your system response
can be written as, then method like trapezoidal
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00:47:58,559 --> 00:48:04,509
rule will give you the correct, no truncation
error, so this kind of system is known as
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00:48:04,509 --> 00:48:11,180
a second order system. So, trapezoidal rule
will not give any truncation error for a system,
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00:48:11,180 --> 00:48:16,200
which has got response of this kind, again
trapezoidal rule will always give some truncation
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00:48:16,200 --> 00:48:25,099
error for a system which has got this response,
because this is the infinite sires in t. So,
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00:48:25,099 --> 00:48:27,650
it is not just second order polynomial in
t.
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00:48:27,650 --> 00:48:35,569
So, the trapezoidal rule is this something
you can show, I will not you known prove it
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00:48:35,569 --> 00:48:42,329
here you can just try to do at home, trapezoidal
is in fact, a second order method there are
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00:48:42,329 --> 00:48:52,999
other methods for example, Runge Kutta method.
So, if you got x dot is equal to f of x f
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00:48:52,999 --> 00:48:58,069
of x t sometime, t may explicitly in this
function.
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00:48:58,069 --> 00:49:10,690
Then Runge Kutta method uses first it evaluate
function k 1 which is nothing but, then it
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00:49:10,690 --> 00:49:35,880
evaluates k 2 which is nothing but, sorry
k 1 then it is evaluate function k 3. Rather
408
00:49:35,880 --> 00:50:01,710
it is evaluates k 3 where k 3 is nothing but,
and then k 4 nothing but h k 3 , so its evaluate
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00:50:01,710 --> 00:50:31,619
its k 1, k 2, k 3, k 4 is Runge Kutta 4th
order method 4th order.
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00:50:31,619 --> 00:50:47,420
So, Runge Kutta 4th order method eventually
calculates x k as rather x k plus 1 as x k
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00:50:47,420 --> 00:50:59,150
plus h by 6 into k 1 plus 2 k 2 plus 2 k 3
plus k 4. So, this is basically what Runge
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00:50:59,150 --> 00:51:05,160
Kutta method does, it is basically evaluating
k 1, k 2, k 3, k 4, which I in fact, intermediate
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00:51:05,160 --> 00:51:14,930
points, the intermediate values of this function
between the interval t k and t k plus 1 and
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00:51:14,930 --> 00:51:19,229
then using this particular formula.
So, this is Runge Kutta 4th order method,
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00:51:19,229 --> 00:51:29,749
first question is it an explicit or implicit
method, the answer it is an explicit method,
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00:51:29,749 --> 00:51:35,249
because it requires the evaluation of x k
plus one, but it does not require x k plus
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00:51:35,249 --> 00:51:41,549
1 itself. So, it you can evaluate all this
function explicitly, you do not require x
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00:51:41,549 --> 00:51:49,670
k plus 1 in these evaluation, so this is an
explicit method. And something which of course
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00:51:49,670 --> 00:51:54,739
I will not prove here, but this is what is
4th order method, so if your response can
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00:51:54,739 --> 00:52:00,780
be written down as a 4th roughly as a 4thorder
polynomial in t, then you will have no truncation
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00:52:00,780 --> 00:52:07,549
error. So, this is in fact, a Runge Kutta
method, so we have of course, discuss now
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00:52:07,549 --> 00:52:13,150
order of the method, whether it is an explicit
or an implicit method, we have really discuss
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00:52:13,150 --> 00:52:20,869
this without going into any great amount of
; I seen in the I can refer you to a few books
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00:52:20,869 --> 00:52:25,630
at the end of the lecture, you can go through
them for more detail analysis of this.
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00:52:25,630 --> 00:52:31,529
There is a another point which we missed out,
that is multi step and single step methods,
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00:52:31,529 --> 00:52:39,500
now I will not write down the various multi
step and single step method, but all the three
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00:52:39,500 --> 00:52:46,479
methods or four methods which I have told
you Euler, trapezoidal, Runge Kutta and backward
428
00:52:46,479 --> 00:52:53,459
Euler all requires simply x k to get x k plus;
you do not require anything more. You could
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00:52:53,459 --> 00:53:02,000
in principal use for example, x k minus 1
also in your calculation to get k plus 1.
430
00:53:02,000 --> 00:53:07,180
So, when you are trying to get k plus 1 your
discretization uses values, which is two times
431
00:53:07,180 --> 00:53:12,880
step before or three or you know even a time
step ahead in time.
432
00:53:12,880 --> 00:53:17,549
In that case of course, it will like t k plus
2 or t k plus 3, in that case it become in
433
00:53:17,549 --> 00:53:25,019
implicit method, so such methods which use
not only the previous time step, but time
434
00:53:25,019 --> 00:53:30,519
steps other than the previous time step, values
of x at time steps, other than the previous
435
00:53:30,519 --> 00:53:36,010
time step. They In fact, are known as multi
step methods, so we have gone to order of
436
00:53:36,010 --> 00:53:42,059
the method, whether it is explicit or implicit
and whether its single step or multi step
437
00:53:42,059 --> 00:53:45,150
method.
In most of what we are going to do in this
438
00:53:45,150 --> 00:53:51,229
course, we shall restrict yourself to single
step methods, explicit methods of course,
439
00:53:51,229 --> 00:53:56,240
are easy to evaluate, because they do not
require to solvent algebraic equation, especially
440
00:53:56,240 --> 00:54:00,450
this is the problem, when you are talking
of non-linear systems. Because, the algebraic
441
00:54:00,450 --> 00:54:06,609
equation is non-linear and you will have to
use gross side or at every time step to evaluate
442
00:54:06,609 --> 00:54:14,140
iteratively, what x k plus 1 is given x k.
So, that creates a bit of you know kind of
443
00:54:14,140 --> 00:54:18,970
problem, when one tries to numerically integrate
a non-linear system using an implicit method.
444
00:54:18,970 --> 00:54:24,279
So, you should have really some nice tangible
benefits, when you are trying to use an implicit
445
00:54:24,279 --> 00:54:31,180
method, these in fact, implicit methods are
known to be straightly better, when it like
446
00:54:31,180 --> 00:54:36,880
backward Euler or trapezoidal rule are usually
better suited, when you have got stiff systems
447
00:54:36,880 --> 00:54:39,160
and you want to study the slow response.
448
00:54:39,160 --> 00:54:45,900
So, this something of course, we shall do
in the next lecture, before we stop this lecture,
449
00:54:45,900 --> 00:54:52,279
let we just give you a few references for
example, for numerical integration method,
450
00:54:52,279 --> 00:55:16,300
there lots of books, but some of the classic
books are one by Gear, ordinary differential
451
00:55:16,300 --> 00:55:38,880
equations by Printice Hall, I am sure it will
there in a well stock library, Englewood Cliffs
452
00:55:38,880 --> 00:55:48,249
1971.
And there is another one that is L. Lapidus
453
00:55:48,249 --> 00:56:03,700
and J. H. Seinfeld, Seinfeild numerical solution
of ODE, ODE is of course, is a ordinary differential
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00:56:03,700 --> 00:56:15,940
equations, this is academic press, New York
1971. These are what I would say is the classical
455
00:56:15,940 --> 00:56:23,519
books in a in a, this field numerical integration
methods, in some cases they may be fairly
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00:56:23,519 --> 00:56:29,759
regress more regress then we will ever be
in this particular course.
457
00:56:29,759 --> 00:56:35,119
Therefore, you can also look at other books
which are somewhat simpler, which may be available,
458
00:56:35,119 --> 00:56:38,140
I am sure therem many other books which are
available in our library, but these are the
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00:56:38,140 --> 00:56:44,729
classical books in numerical integration methods.
In the next lecture, we shall understand the
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00:56:44,729 --> 00:56:49,759
numerical integration of stiff systems in
more detail. And also understand some of the
461
00:56:49,759 --> 00:56:56,289
stability properties are the you know the
properties general properties of the numerical
462
00:56:56,289 --> 00:57:11,380
methods in particular Euler method, backward
Euler method and trapezoidal rule.