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In the present module in this course, we have
been studying about various techniques to
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understand how a dynamical system behaves.
Now, our treatment of these various methods
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is not very rigorous. But, I trust whatever
we are doing will give you some kind of introduction
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and reassurance in case you have already done
it before. In particular in the past two lectures
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we have been studying about numerical integration.
Numerical integration is a very general tool
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which can be applied to the study of dynamical
systems unlike Eigen value and Eigenvector
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analysis which is applicable only for linear
systems. Now, we have been studying the basic
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features of some of the numerical integration
method.
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It is very important with the proliferation
of a lot of simulation on numerical integration
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software, it is very important to know the
characteristic characteristics of various
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numerical integration tools which are available.
And to do this, we have to kind of bench mark
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our behavior of dynamical systems and then
compare it with what we get when we numerically
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integrate. Remember, numerical integration
always involve some error because it is an
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approximation of a continuous time system.
In the previous lecture, we started on analyzing
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on how numerical integration methods behave
when we are confronted with a stiff system.
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A stiff system is a system in which the various
patterns which you see in the response are
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widely varying time constants or widely varying
rates of change. In a linear system ofcourse,
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one can correspond these fast and slow transients
to large and small Eigen values. Very often
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in engineering, we do encounter such systems.
In fact, if you do encounter such systems,
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you often are able to do modeling simplifications,
a point which we discussed in about three
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lectures ago.
So today, what we will do is consider the
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same system for which we did Eigen value analysis
and see how it behaves or how what what answer
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we get when you numerically integrate and
try to get the time response. Now, remember
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ofcourse at this point that we are you know,
doing the numerical integration of a linear
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system. The real linear system response off
course is fully known in terms of simple functions
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like exponents and sinusoids .We can use our
powerful Eigen analysis tools to obtain Eigen
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values and Eigen vectors and write down the
system response. There is no need to do integration.
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But, as I mentioned some time back that we
use these two benchmark how our numerical
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integration, numerical methods work.
Now, this system which we considered in that
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particular, you know linear and when we did
our linear analysis, we considered a particular
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system which was very typical in the sense
that it brought out the stiffness in the system.
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That system was basically an RLC circuit which
was excited by our step in the input voltage.
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So, the system we were considering was this;
a relatively large inductor here and again
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a capacitor here. The system is a linear system
and we can write down the differential equations
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in this form where a is
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and b is. So, this is our system and we of
course, in our previous lecture we took out
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a time response of this system.
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The Eigen values of the matrix when you do
linear analysis are given by this and this.
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So in fact, there is a complex conjugate pair
which will correspond to a damped sinusoidal
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response, a damped oscillation.
And off course, there is an real Eigen value,
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a negative Eigen value which will correspond
to a pattern which will be seen in the response
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which is e raise to minus 0.1 t. It is very
clear that this system is stable because the
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real part is negative of all the three Eigen
values. Another issue which is important is
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that the rates of change associated with the
pattern correspond or mode corresponding to
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this Eigen value is very slow as compared
to this. Look at this frequency. It is extremely
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high. So, the kind of movement you will get
in the response is going to be having a large
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rate of change for this pattern. So, your
response is consisting of two patterns; a
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fast mode and the slow mode. This is in fact
typical of a stiff system. You have got you
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know, both fast and slow modes.
Now, if one tries to numerically integrate
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this particular system then, one may use Euler
method. For example, if one wants to apply
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Euler method to this differential equation
with a time step of h;
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In that case, your iteration which you will,
I would not say an iteration, if you know
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the value of x k you can get the next value
of next value of x x k plus 1, the next sample
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of x by using this relationship. So, A x k
plus b u k. Now, this implies
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you have seen in a previous lecture that I
indicated that this system, Euler method when
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it tries to numerically integrate x dot is
equal to A x plus b u; the discretized system
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which you get may be stable, if is stable,
in case where lambda is the ith Eigen value.
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The important point is this should be true
for all Eigen values.
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So, Euler method will be stable or rather
the response which you get by using this relationship,
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the samples which you get by using this relationship
will be stable if for all Eigen values this
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is satisfied. Lambda i is goes the Eigen values
of a. So, this is the basic property of Euler
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method. Now, remember that the original system
is stable. But, Euler method under certain
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circumstances may not able to mimic the stability
of the original system. It may show an originally
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stable system to be an unstable one because
this relationship may not be satisfied.
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And in Euler method, for example, if I apply
this to one of the Eigen values that is lambda
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1 is equal to this. Suppose, I want to check
this relationship for this Eigen value, we
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will find that the relationship we will get
is this, which boils down to this. Even if
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I choose a time step of point double 0 1,
you are not going to have this relationship
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satisfied.
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For example, you expect that this frequency
is the imaginary part off course corresponds
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to the frequency of the oscillation in radian
per second. We expect that, if I choose a
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1 mille second time step it should be able
to you know mimic the response. But, unfortunately
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that is not true because this relationship
is not satisfied. Thousand and 5 radians per
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second corresponds to around thousand five
divided by 2 pi as a frequency the time period
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of this is off course.
So, we expect that if we choose a 1 mille
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second time step we should be able to mimic
the response. But, that is not true if you
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discretize using Euler method this relationship
is not satisfied and you will find that the
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system which you have numerically integrated
or rather the sys, the discrete time system
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which you get by discretizing the original
continuous time system by Euler method will
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not be in stable. So, you are getting a qualitatively
wrong answer if I try to use Euler method
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with this time step. Ofcourse you may say
lets reduce the time step further one can
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go on reducing this to for example, 10 raise
to minus 5 6 or 7. But, remember the time
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required to do this you know, numerical integration
will keep on increasing if I reduce the time
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step. So, if I want to simulate 1 second of
the response if I choose h is equal to 1 millesecond
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in that, I will I will require thousand steps.
And if I choose 10 raise to minus four seconds,
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you will require 10 thousand steps, just for
a 1 second simulation. So, the problem here
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ofcourse is that if I am interested in this
slow response, if I am primarily interested
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in how the system behaves the slow response
of this system; in that case if I use Euler
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method I will still be constrained to use
a very small time step in order to prevent
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the faster mode from being unstable. I mean
the numerical integration should not display
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instability. I mean that will be a qualitatively
wrong answer and that something I do not want
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to have.
So, Euler method has a problem in such a stiff
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system to make your time step very, very very
small. So that is one issue which you should
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remember. Now ofcourse somebody may ask what
is this interested in the slow transient,
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not interested in fast transient, I mean what
are the situation where you you would be interested
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in the slow transient and not interested in
fast transient time and so on. For example,
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let me give you a simple example; you want
to study what happens when you start a d c
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motor or you have got a d c motor running
and the load torque on it, the load torque
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on the d c motor suddenly changes and you
are interested in how this speed varies.
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Now, the point is when you are trying to see
how this speed of the d c motor varies; what
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kind of transient are you interested in? See,
if you look at a d c motor it has got some
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resistances, it has got a some inductance,
it has got a small, it has got inter winding
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capacitances and so on. So, if you model everything
including the mechanical system, the electrical
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system, you will find that it becomes a stiff
system because the electrical time constants
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or the electrical transients are much faster
than the mechanical transient.
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So, if you are model all the transients but,
you are interest is on the in seeing how the
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slow speed transient behaves you know. The
slow you know pattern in the response. Then
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you come up with similar situations. So, whenever
I say that you have got a stiff system and
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you are interested in the slow response, you
can remember this kind of examples. So, if
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you got d c motor and your interested in the
speed transient and your model all the electrical
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transient, all the electrical components of
the system which are relatively faster then
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this particular situation does arise.
So, let me retreat, we are thinking of a system
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which has got both fast and slow transients
and we want to replicate the slow transient.
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So, that would be a particular situation which
we may face. Ofcourse you may be interested
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in the fast and the slow transients. That
is the another situation that is another thing
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you may may encounter. If you are interested
in the slow transient of the slow transient
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behavior of a system but, you are not interested
in accurately representing the fast part of
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the response, you know fast part of the response
is there but, you are not very very interested
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in the accurately you know getting the fast
response.
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In that case, Euler method is not a good idea
because you have to really decide your time
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step based on the fast response and if you
choose any thing which is larger than what
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is mandated by that particular condition,
you will find that your system simulation
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blows up. So, that is one of the problems
which you will face if you try to use Euler
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method. So, Euler method is taught but, rarely
used. so that is the basic interesting thing
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you will understand by experience.
What we will do now, I will just show you
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this particular aspect by doing a numerical
simulation on psi lab. So, what I will do
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is, I will just show you a clip of a program.
We just, I will just you how you can run it
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and then I will display the result. So, if
you look direct your attention to this particular
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program which I have used to simulate the
system. So, this is a psi lab program. We
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will ofcourse, I will tell you the main steps
in the program. You have given the a matrix
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of the system k, will tell you the Eigen values
if you are interested in them let us not bring
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them out.
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This is a time step say you want to simulate
the system for 2 just for 20 mille seconds
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using a time step of 0.0 0 1. Just for 2 mille
seconds and using Euler method.
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So, let us assume that the initial conditions
on the states are 0 but, the system is excited
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by a step in the input. So, we are getting
some kind of forced, we are having a forcing
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function.
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Suppose, I numerically integrate using Euler
method. I delete this. So, what you have here
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is, the Euler the method. X is equal to 1
plus A h into X plus h into u that b is kind
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of absorbed in this vector. So, I have not
written it separately.
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So, I will comment the discretization by trapezoidal
rule. So, what we have here is, this program
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and ofcourse I will plot the values once I
have simulated them. So, if I run this program
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and save it and I run this program.
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This is the response I get. The green and
the blue denote the inductor currents; the
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small inductor and the large inductor. The
small inductor is blue, the large inductor
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is green and red is the capacitor voltage.
It may not be very clear in your screen but,
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the value here is 1200, 1000 200 and this
time off course is 0 to 0.0 2 seconds. So,
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you see this response is kind of blowing up,
you get a blowing up response.
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So, this is one of the problems which you
will find in Euler method that when you have
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got a fast transient you have to, the time
step you may choose really as really may not
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satisfy the stability condition and you will
get a spurious response. Before we go ahead
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and you know try out the other methods, let
us quickly look at what is the actual behavior
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of this system.
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The correct time response is this, how did
I get this correct time response? The correct
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time response if you recall, is not obtained
by numerical integration. But, by simply evaluating
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the response at various time steps from the
analytical functions that we derived using
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Eigen values and Eigen vectors.
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If you recall the response of I one, using
Eigen value analysis is 10 minus 10 e raised
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to into sine of thousand and 5 t, i 2 is this
is approximate. Not exact. But, this was analytically
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derived using Eigen values and Eigen vectors.
So, this is the response for this RLC circuit
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I 1 I 2 and V c. So, this is time response
but, this is derived analytically.
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So, this what we are showing on the screen
is the same response which is evaluated by
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simply plugging in t into the these functions
which I have just written down and so this
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is the correct time response.
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So, this is what I should be getting and at
a if I try to simulate this just from 0 to
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point 5 seconds, this is what I should be
getting this V c and this is i 1 and i 2 green
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and blue.
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So this is the correct time response and just
to replot what I got just now using Euler
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method with 0.0 0 1 and the simulation just
20 mille seconds you see, the beginnings of
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the fact that this discrete time system obtained
by using Euler method, it is just blowing
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up. So, we are no where going to be nowhere
close to either this response or that response.
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So, Euler method is giving a horribly wrong
solution. In fact, it is giving a qualitatively
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wrong answer. Also, it is not only inaccurate,
it is giving a qualitatively incorrect you
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know, conclusion or inference about stability.
If I use backward Euler method remember, the
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backward Euler discretization is done for
x dot is equal to A x plus b u by the following.
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u k off course is constant, u is constant
so you have less trouble about this. But,
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what you get eventually by just solving this,
I am skipping a few steps is, 1 minus A h
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inverse x k plus 1 minus A h inverse b h u
k.
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So, this is basically the iteration which
I will have to use in case I want to use backward
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Euler method. So, this is backward Euler method.
Now, if I use backward Euler method, the important
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thing ofcourse here is that, you have to take
out the inverse of a matrix for linear. So,
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this involves extra computations. It is not
as straight forward as the explicit method.
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That is the forward Euler method which has
an explicit method.
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Now, normally taking out an inverse is not
a big problem. I mean it does not, is not
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a fraught with problems as far as computation
is concerned provided if the system is small.
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If your system is very large, then computation
of inverse can be quite intensive. For example,
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its very common in a power system to have
a size of the order of the system may be with
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thousand you may be thousand of states. So,
in that case to try to compute this, may be
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a bit comp may we be fraught with a lot of
hurdles because inversion is a computationally
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intensive you know operation and at every
time step you have to do this you know, this
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particular function that is, x k plus 1 getting
x k plus from from from x k.
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Now, since 1 minus A h inverse is appearing
at every step you do not you can compute it
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once in your program and then simply do a
matrix multiplication when you are running
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this algorithm to implement backward Euler
method. So, you can actually take out the
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inverse and keep it before hand and only perform
matrix multiplications. But, again if you
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are working with very large systems, it is
not a good idea to compute the inverse explicitly.
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The reason being, that if a is powers, the
inverse of i minus A h i is incidentally an
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identity matrix, something which I did not
mention earlier. This particular matrix is
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the inverse, may not be power. So, may you
even have to store a very large number of
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values if you are going to explicitly compute
the inverse and keep it stored before hand.
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What would be a pragmatic thing to do? It
would be to compute the l u factors, the lower
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and upper factors of o l u factors of 1 i
minus A h and just do backward and forward
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substitution during each iteration.
So the l u factor ofcourse you may have to
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do ordering of the states etc. So that, you
get the l u factors as powers if a is powers.
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But, remember storing the inverse of even
a powers matrix you know is a problem because
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the inverse may not be powers. So, these are
the some of the issues which you may face
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may if your developing a program for large
systems. But, ofcourse right now we are talking
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about third order system. You can just as
well take out the inverse. You can even keep
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on taking out the inverse at every time step
though it is not necessary. So this is basically
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how you will implement backward Euler method.
Trapezoidal rule is again similar.
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You will get x k plus 1 minus x k by h is
equal to A x k plus 1 plus A x k upon 2 plus
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b u k plus 1 by 2 plus b u k by 2. So, this
is how you will discretize it. Again, I you
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will need to take out an inverse. So, the
of a matrix so that is one of the critical
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features of trapezoidal rule as well as backward
Euler or any implicit method.
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Remember, since this is a linear system our
job in implicit method requires inversion
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of a matrix. If your system is non-linear,
you may require even to do iterations to get
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x k plus 1 from x k using some method like
n r that is Newton Raphson or Gossie Gauss
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Seidal method.
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Now, coming back to the qualitative results
if I used backward Euler method with h is
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equal to 1 mille second and a simulation interval
of 0.0 2; this is the response I get. One
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thing you can notice here is that, backward
Euler seems to have killed the oscillation
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very quickly. So, the fast transient which
I expect to be seen in this 20 mille second
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window, is in fact seems to be very well damped,
damped out. Better damped than what in fact
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it actually is. So, you if you like at the
original response, it takes at least a second
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or so to damp out more than a second damp.
Whereas here, backward Euler with a 1 mille
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second time set, the oscillation dies down
very soon. So, if you look at if you just
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recall how this the correct time response
should be like this, in about 0.5 seconds.
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Your oscillation is gradually dying down.
So, this oscillatory part of a transient takes
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at least a second or two to completely die
out. Whereas backward Euler has simply killed
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at oscillation and you do not even see an
oscillatory response.
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On the other hand, if I choose a large value
of h, in fact this h is more compatible with
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the slow mode, that is e raise to minus 0.1
t you know, the time constant corresponding
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to the decay time constant corresponding to
e raise to minus 0.1 t is 10 seconds. So,
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it makes sense to choose h is equal to 1 second
only if you are interested in this slow response.
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The interesting part which you see here ofcourse
is, all though the initial part of the transient
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is not captured very well, the system is able
to capture the slow transient nonetheless.
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This would not have been possible with forward
Euler method because you would find that the
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fast transient is getting destabilized. That
is what we saw in one of the previous simulations.
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yeah This one. So, we could not have ofcourse
use the same strategy with forward Euler method
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even though our main response of interest
was the slow transient.
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So, backward Euler method in fact is a good
method to use if you are not too worried about
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how the fast transient involves. Ofcourse,
we would be worried if the fast transient
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were actually unstable. But, if it is known
that is stable, if you you from your engineering
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judgment you are sure that the fast transients
are not unstable. In that case if you are
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interested in the slow transient it is good
idea to use a method like backward Euler method
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with a large time steps which is compatible
with this slow transient.
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Trapezoid rule with 1 millisecond and has
stimulations interval of 0.1 second seems
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to capture the fast transient quite well.
This is unlike backward Euler which introduces
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some damping, extraneous damping into the
original system.
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However, if I use trapezoidal method with
h is equal to 1; all though the response is
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not destabilized you are getting a highly
inaccurate response as far as V c is concerned.
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So, trapezoid rule is not very very good you
cannot use it with very large time steps or
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time steps corresponding to the slow response
if your system is very stiff like the one
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which we are encountering here. So what is
the solution to these issues?
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The solution to this problem is, use variable
step sizes. For example, you can use 1 millisecond
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for 1 second and 1 second h is 1 second for
30 seconds. So, variable time step sizes can
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be used in order to obtain a response. So,
you easily program it in psi lab. I will just
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show you the program. So, if you look at if
you look at the program for variable time
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steps.
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You will have to, when you program define
two time steps, h 1 which is 1 millisecond.
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Let us say we stimulate only for 1 second
then, after some time you can shift over to
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a last time step and stimulate for a longer
time. And ofcourse, you will have to program
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it appropriately so, that you can get the
appropriate response.
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Now, so ofcourse one important point in this,
whatever stimulation we are trying to do is
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that, our choice of h is somewhat adhoc. And
I mean, the choice of h 1 and h 2 is adhoc
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and also it is adhoc that we switch over from
the smaller time step to the larger time step
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at 1 second. Why 1 second it could have been
10 second and so on? The point is in this
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particular case, I do know the response. So,
since I know the response I can actually tell
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you at what point to switch. But, implementing
variable time step methods for systems in
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general may be a bit tricky. I mean if you
know nothing about the system then, how do
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you decide that you should switch over at
1 second from the fast, slow, the small value
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to the larger value or even more importantly,
how do you choose these values h 1 and h 2.
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And, I here there is an ad hoc switch from
0.0 1 point or rather 1 millisecond to 1 second.
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The key to this of course is that, often when
we are stimulating a system, we know something
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about the system. So, that is one way you
can actually use your engineering judgment
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and come to a particular conclusion about
what time steps to use or you can use a bit
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of try line error. But, most industry grade
programs will actually have some way of finding
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00:32:14,340 --> 00:32:21,570
out you know, the truncation error, estimating
the truncation errors at every step. And,
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if the errors at every step are not too large
then they may even permit adaptively to start
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00:32:26,559 --> 00:32:34,249
increasing the time steps. So, in a you know
if you look it at commercial software or software’s
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like matlab other software they do implement
variable time step methods and they would
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00:32:39,429 --> 00:32:44,109
have one way of checking out or estimating
the truncation errors and adaptively changing
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00:32:44,109 --> 00:32:49,289
the time step. Here of course, I do it one
short, I just change from 1 millisecond to
284
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1 second.
Now, if I use the variable time step method,
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the question is can I use variable time step
method with Euler method? So, the question
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is can I use Euler method with variable step
sizes in a step system? The answer is no.
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00:33:06,639 --> 00:33:14,379
The point is that, the moment I switch over
from a small time step to a large time step
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00:33:14,379 --> 00:33:18,049
the remanence remainants of the fast mode
may not be completely zero. In fact they are
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00:33:18,049 --> 00:33:23,289
never zero because of numerical precision,
can never be infinite. So, you will find that
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00:33:23,289 --> 00:33:27,549
there are remanence remainants of the fast
response which are there in your, there is
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00:33:27,549 --> 00:33:32,940
if the fast response wouldn’t have completely
died down when you rigorously speaking it
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never completely raise dies down.
So, if I use a Euler method, the moment I
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00:33:38,779 --> 00:33:45,129
switch over to lager time step, the faster
transient even though we have waited for the
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fast transient to die down after some time,
you will find a whatever remanence remainants
295
00:33:48,639 --> 00:33:52,179
of the fast transient are there, they will
again start becoming unstable. And you will
296
00:33:52,179 --> 00:33:58,269
find that the whole system blows up. So, that
is the major issue which you will face when
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00:33:58,269 --> 00:34:03,740
you use methods like Euler method with variable
time steps. So, Euler method is actually is
298
00:34:03,740 --> 00:34:07,789
not suitable or variable time steps for the
stiff system. So, Euler method is you keep
299
00:34:07,789 --> 00:34:09,040
it in the background.
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So for example, I actually try to program
and do this stimulation for Euler method;
301
00:34:17,250 --> 00:34:22,260
you see this, I do not know whether it is
clear on your screen but, it is 2 into 10
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00:34:22,260 --> 00:34:29,710
raise to 3 hundred and 6. So, you know you
are by trying to implement this variable step
303
00:34:29,710 --> 00:34:36,230
method for just a few seconds has resulted
in a complete blow up of the solution.
304
00:34:36,230 --> 00:34:43,879
So, forward Euler cannot be used in conjunction
with a variable time step method. On the other
305
00:34:43,879 --> 00:34:50,840
hand, backward Euler method if I use variable
time step method, it gives a reasonably good
306
00:34:50,840 --> 00:34:59,819
response for the first few, first 1 second
or so it gives a step. Really it does not
307
00:34:59,819 --> 00:35:05,040
capture the fast transient quite well any
way you know. It does not capture the fast
308
00:35:05,040 --> 00:35:11,000
transient any way. The slow transient it is
captured pretty well.
309
00:35:11,000 --> 00:35:15,039
Trapezoid rule with this kind of thing works
well because when we make the time step very
310
00:35:15,039 --> 00:35:22,130
small for the first few seconds, it captures
the fast transient correctly and also the
311
00:35:22,130 --> 00:35:28,270
slow transient. So, that is the basic deal
you can say whenever you are using a variable
312
00:35:28,270 --> 00:35:36,119
time step method. So, these are these are
the things you have to keep in mind.
313
00:35:36,119 --> 00:35:45,760
So, to summarize this part of the lecture;
for non stiff system where stiffness is not
314
00:35:45,760 --> 00:35:53,220
there there is less less of a worry that the
system will become unstable if you choose
315
00:35:53,220 --> 00:35:57,730
your time step appropriately. Appropriately
in the sense, if you know something about
316
00:35:57,730 --> 00:36:05,109
the system you can choose your time step roughly
to correspond to the fastest time constants
317
00:36:05,109 --> 00:36:10,980
in your system. So, if it is a non stiff system
and you know that well this exponential rate
318
00:36:10,980 --> 00:36:15,740
of change or the sinusoid is roughly going
to be in this range, then you can actually
319
00:36:15,740 --> 00:36:22,730
choose a time step. Use any method. In fact
of course prefer high order methods. In case
320
00:36:22,730 --> 00:36:28,180
you are using explicit method, so of course
avoid using Euler method it is not a very
321
00:36:28,180 --> 00:36:30,059
good method to use because it is inaccurate
as well.
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00:36:30,059 --> 00:36:34,770
So, what one can try to do is, if you are
coming with you a non stiff, if you are encountering
323
00:36:34,770 --> 00:36:40,049
a non stiff system and you have some rough
idea about time constant associated with this
324
00:36:40,049 --> 00:36:45,920
system or the frequencies of oscillatory response
if any, then you can choose the time step
325
00:36:45,920 --> 00:36:52,210
corresponding to the fastest such transient.
And you can use a higher order explicit method.
326
00:36:52,210 --> 00:36:57,250
Why explicit method? Because explicit methods
are easier to implement they do not require
327
00:36:57,250 --> 00:37:01,130
inversions and or in a non-linear system.
They do not require iterations within a time
328
00:37:01,130 --> 00:37:06,260
step and so on so.
So, however if one is faced with a stiff system;
329
00:37:06,260 --> 00:37:16,490
so you can direct your attention to the screen,
if only slow transient is of interest, in
330
00:37:16,490 --> 00:37:24,859
that case you you must look at implicit methods.
So, if slow transients are of interest you
331
00:37:24,859 --> 00:37:30,599
can try to use backward Euler method. Methods
like backward Euler method with larger time
332
00:37:30,599 --> 00:37:36,760
step. So, if you see the this particular slide
what I have written if this the slow transients
333
00:37:36,760 --> 00:37:41,280
are of interest and the fast transients are
known to be stable either from engineering
334
00:37:41,280 --> 00:37:45,609
judgment or some prior study with somebody
else is done; if you know something about
335
00:37:45,609 --> 00:37:49,950
the system and you know that the fast transient
are indeed stable and the slow transients
336
00:37:49,950 --> 00:37:55,589
are what you are really interested in, you
can use backward Euler with larger time steps.
337
00:37:55,589 --> 00:38:00,910
Larger I mean compatible with the time constants
associated with the slow transients.
338
00:38:00,910 --> 00:38:11,770
If you have got a stiff system and both fast
and slow transients are of interest you can
339
00:38:11,770 --> 00:38:21,050
use higher order explicit methods or implicit
methods with the time small time step. So,
340
00:38:21,050 --> 00:38:25,760
if both fast and slow transients are of interest,
you can use higher order implicit or explicit
341
00:38:25,760 --> 00:38:33,040
method with a small time. Small ofcourse corresponds
to the fast transients. The time constants
342
00:38:33,040 --> 00:38:39,130
of the frequencies associated with the fast
transients. Even here, even higher order explicit
343
00:38:39,130 --> 00:38:44,240
methods may be a problem. So, I actually,
if you are encountering a stiff system, I
344
00:38:44,240 --> 00:38:49,910
think it will, it is a very safe to use implicit
methods with small time steps if both fast
345
00:38:49,910 --> 00:38:54,599
and slow transients are of interest.
It is a bit risky to use explicit method because
346
00:38:54,599 --> 00:38:58,930
they do not, they are do not have very good
stability properties. You will have to use
347
00:38:58,930 --> 00:39:03,059
extremely small time steps otherwise you may
end of these destabilizing some response.
348
00:39:03,059 --> 00:39:09,760
Especially true, this is especially true with
Euler like methods. Ofcourse if this is what,
349
00:39:09,760 --> 00:39:15,160
you want a fast and slow transients are of
interest in a stiff system; the best solution
350
00:39:15,160 --> 00:39:21,980
or a better solution would we to use a variable
time step with in conjunction with backward
351
00:39:21,980 --> 00:39:28,201
Euler or trapezoid. That is initially keep
the time steps small so that, you capture
352
00:39:28,201 --> 00:39:37,230
your fast transient well and then increase
your time step. And you know, you can capture
353
00:39:37,230 --> 00:39:41,730
your slow transient even with the larger time
step. the off course The important thing implicit
354
00:39:41,730 --> 00:39:48,630
in all what I am trying to say is that we
are trying to you know complete our numerical
355
00:39:48,630 --> 00:39:54,430
integration as fast as we can.
Somebody may ask well, you have seen this
356
00:39:54,430 --> 00:40:01,190
RLC circuit you know, simple RLC circuit;
what you know to integrate numerically integrate
357
00:40:01,190 --> 00:40:06,460
this for say thirty second even with a time
step of say, you know hundred microseconds
358
00:40:06,460 --> 00:40:11,089
or fifty microseconds? Should not be a problem
on today’s computers. But, this is not true
359
00:40:11,089 --> 00:40:15,809
when you consider larger order systems. Now,
when you have got very large order systems
360
00:40:15,809 --> 00:40:21,470
and if you are forced to use a very small
time step like you know, fifty microseconds
361
00:40:21,470 --> 00:40:26,660
or hundred microseconds and you want to stimulate
for say, hundreds of seconds. This really
362
00:40:26,660 --> 00:40:32,010
may be a big bottle neck and you may take
hours sometimes to stimulate this system.
363
00:40:32,010 --> 00:40:37,270
This actually happens. So, if those of you
who are doing power electronic systems or
364
00:40:37,270 --> 00:40:42,160
you know large scale power system stimulations
would have encountered this problem if they
365
00:40:42,160 --> 00:40:47,270
use then inappropriate method.
As I mentioned sometimes back proliferation
366
00:40:47,270 --> 00:40:55,279
of so many you know software tools for numerical
integration of circuits power system and other
367
00:40:55,279 --> 00:41:02,160
systems control systems and so on, it is very
important to know this basic know the basic
368
00:41:02,160 --> 00:41:05,349
properties of these integration methods.
369
00:41:05,349 --> 00:41:14,910
So, although our, you know treatment here
has been very, very brief. I mean the aim
370
00:41:14,910 --> 00:41:19,420
of introducing you to analysis methods right
in the beginning of the course is to give
371
00:41:19,420 --> 00:41:27,599
you a feel of these kind of methods which
we will now apply when when we do the modeling
372
00:41:27,599 --> 00:41:30,920
of power system components.
So, when once we finish our modeling of power
373
00:41:30,920 --> 00:41:35,700
system components, we will directly use these
tools like numerical analysis or Eigen value
374
00:41:35,700 --> 00:41:42,240
analysis. Later on, in case you have forgotten
what we have covered you come back to these
375
00:41:42,240 --> 00:41:49,279
lectures and just revise. Now, one small point
which I did not mention as far as numerical
376
00:41:49,279 --> 00:41:56,040
integration is concerned; is that if you are
faced with a stiff system, if you are having
377
00:41:56,040 --> 00:42:01,950
a stiff system, rather than do all the jugglery
of using either very small time step sizes
378
00:42:01,950 --> 00:42:08,079
or variable time steps and these vary some
kind of things, one thing you can do right
379
00:42:08,079 --> 00:42:13,730
away when you are considering a system when
you are modeling a system is to get rid of
380
00:42:13,730 --> 00:42:18,400
the fast transient. Get rid in the sense,
make modeling simplification so that your
381
00:42:18,400 --> 00:42:27,369
system is of lower order and it kind of, you
know, kind of does not have the fast transient
382
00:42:27,369 --> 00:42:32,359
at all. And this is something we have discussed
before. If you are encountering a stiff system,
383
00:42:32,359 --> 00:42:38,420
you can neglect the fast transients. What
you what you get? Because of that, is that
384
00:42:38,420 --> 00:42:43,650
you are going to get a lower order system,
the differential equation is corresponding
385
00:42:43,650 --> 00:42:50,440
to the states associated with the fast transients
you know, are converted to algebraic equations.
386
00:42:50,440 --> 00:42:56,539
So, you know what you are really doing is
that, the states for example, the inductor
387
00:42:56,539 --> 00:43:02,040
current or the capacitor voltage in this particular
circuit, the differential equation corresponding
388
00:43:02,040 --> 00:43:05,039
to the states are converted into algebraic
equations.
389
00:43:05,039 --> 00:43:11,359
For example, we have done this before. What
this as a revision in this particular circuit
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00:43:11,359 --> 00:43:20,930
through participation matrix, the participation
matrix or through by engineering judgment,
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00:43:20,930 --> 00:43:29,440
you know that the states associated with the
fast transients are this and this. That is
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00:43:29,440 --> 00:43:36,359
10 mille Henry and 100 micro ferret and state
associated with the slow transient is this.
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00:43:36,359 --> 00:43:42,859
This is something we have done before. So,
why not use a modeling simplification? You
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00:43:42,859 --> 00:43:50,029
know for example, here you have got you can
use the modeling simplification that this
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00:43:50,029 --> 00:43:57,200
capacitor is actually open circuit d V c by
d t is equal to 0 and d i l 1 by d t is equal
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00:43:57,200 --> 00:44:03,410
to 0. In that case, you are going to get effectively
a circuit of this kind of lower order is just
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00:44:03,410 --> 00:44:14,190
a single dynamical element or a single state.
This will be an acceptable approximation provided
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00:44:14,190 --> 00:44:17,510
you are interested only in the slow transient.
So this something we have done before. The
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00:44:17,510 --> 00:44:23,599
point is that, if I have got this system to
begin with a if I want to numerically integrate
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00:44:23,599 --> 00:44:28,569
it; I will have to worry about you know, what
method I am going to use? What is the time
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00:44:28,569 --> 00:44:32,660
step I am going to use? The possibility of
using variable time steps to speed up your
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00:44:32,660 --> 00:44:35,710
numerical integration and so on.
But, if you look at this system, this is a
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00:44:35,710 --> 00:44:41,240
non stiff system. Actually this particular
system we I can use Euler, forward Euler,
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00:44:41,240 --> 00:44:48,410
backward Euler trapezoidal or say Runge Kutta
say for fourth order method which I have discussed.
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00:44:48,410 --> 00:44:53,860
I have just mentioned sometime in the lecture
previous to the previous one. You can use
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00:44:53,860 --> 00:44:59,520
all of these which say a time step of 1 second
without you will get a reasonably accurate
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00:44:59,520 --> 00:45:04,950
solution because this is not a non this is
not a stiff system at all. So, often what
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00:45:04,950 --> 00:45:12,660
we do is, neglect that d I by d t corresponding
to this and the d v by d t that is, the current
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00:45:12,660 --> 00:45:16,670
through this and get a non stiff system.
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00:45:16,670 --> 00:45:22,869
Again, just in case you are worried about
why am I imposing the point, remember that
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00:45:22,869 --> 00:45:28,380
if I am going to use implicit, explicit methods;
it is important that the system should not
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00:45:28,380 --> 00:45:32,000
be stiff otherwise you have to, you will be
constrained to use a very, very very small
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00:45:32,000 --> 00:45:37,011
time step and it will take a very long time
to complete the stimulation. Moreover some
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00:45:37,011 --> 00:45:42,020
of, some explicit methods like Euler method
are not even very accurate.
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00:45:42,020 --> 00:45:49,349
So, the important thing is, if given a choice
a programmer will use explicit methods because
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00:45:49,349 --> 00:45:58,279
it involves less of programming and less computations
per time step. But implicit are more stable.
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00:45:58,279 --> 00:46:04,740
They do not give, they do not show unstable
system to be a rather a stable system to be
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00:46:04,740 --> 00:46:11,369
an unstable one. And ofcourse, I have mentioned
that unfortunately implicit methods require
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00:46:11,369 --> 00:46:17,069
more computations per time step.
So, it does make sense sometimes to use explicit
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00:46:17,069 --> 00:46:23,839
methods. But, you should basically make modeling
simplification so that, the fast or non stiff
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00:46:23,839 --> 00:46:29,320
or the stiff components of the system or the
fast component of the system are effectively
422
00:46:29,320 --> 00:46:37,319
removed. So, that is the basic you know, thing
which modeling you know simplification which
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00:46:37,319 --> 00:46:44,039
one should use if possible, wherever possible.
So that, you it permits you to use methods,
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00:46:44,039 --> 00:46:48,609
some sometimes it permits you to use explicit
methods. But, if you have no idea about the
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00:46:48,609 --> 00:46:53,599
system, you know you cannot make modeling
simplification. So you know, if you start
426
00:46:53,599 --> 00:46:58,240
off with this system which you have no knowledge;
if you give, for example, a synchronous machine
427
00:46:58,240 --> 00:47:02,990
to a mechanical engineer or a civil engineer;
he may not know all the you know, he may not
428
00:47:02,990 --> 00:47:07,819
have that engineering judgment of the various
transients involved or what transients to
429
00:47:07,819 --> 00:47:11,240
expect.
So, in that case he may find it very difficult
430
00:47:11,240 --> 00:47:16,660
to make modeling simplifications. So in that
case, then there’s always an issue about
431
00:47:16,660 --> 00:47:26,799
which method to use and so on. But, an interesting
feature about modeling and you know, what
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00:47:26,799 --> 00:47:36,079
will we doing next is that, often we would
be kind of making assumptions about the system
433
00:47:36,079 --> 00:47:42,720
from a general knowledge, a general engineering
sense about this system. For example, when
434
00:47:42,720 --> 00:47:49,030
you are modeling or synchronous machine and
main aim of the modeling is to study, for
435
00:47:49,030 --> 00:47:54,359
example, loss of synchronism or electro mechanical
transients associated with the system; we
436
00:47:54,359 --> 00:48:02,080
will not be modeling the currents, for example,
through the interwinding compared capacitance
437
00:48:02,080 --> 00:48:06,769
of the stator windings.
So, because we have a kind of engineering
438
00:48:06,769 --> 00:48:15,840
feel that the stray components like the interwinding
distributed capacitance is and so on, may
439
00:48:15,840 --> 00:48:22,250
not the transients associated with them are
fast very, very very fast to analyze unless
440
00:48:22,250 --> 00:48:28,760
you are really doing analysis of ultra fast
transients in a synchronous machine. You may
441
00:48:28,760 --> 00:48:36,860
not need to model them at all. So, to some
extent there is this engineering judgment.
442
00:48:36,860 --> 00:48:42,650
Now, before we end this particular lecture,
we just have a few 10 to fifteen more minutes
443
00:48:42,650 --> 00:48:43,859
to go.
444
00:48:43,859 --> 00:48:49,950
Let us quickly summarize the first part of
our, you know of our lecture. You know the
445
00:48:49,950 --> 00:48:58,500
first part of this course was in fact analysis
of dynamical system. It was a more general
446
00:48:58,500 --> 00:49:03,720
analysis. We considered linear time invariant
systems and we could really characterize the
447
00:49:03,720 --> 00:49:10,710
response in terms of modes. We could even
understand the stability of such systems simply
448
00:49:10,710 --> 00:49:17,299
by looking at the properties of the A matrix.
In particular, looking at modes could be characterized
449
00:49:17,299 --> 00:49:24,240
by the Eigen values and the Eigen vectors
associated with the A matrix of this system.
450
00:49:24,240 --> 00:49:29,580
Non-linear and linear time variant system,
this should read as variant. Non-linear and
451
00:49:29,580 --> 00:49:35,440
linear time the variant systems are difficult
to analyze. Unfortunately, the only tool which
452
00:49:35,440 --> 00:49:43,609
is left with us when we are trying to analyze
non-linear systems in general is numerical
453
00:49:43,609 --> 00:49:48,880
integration. There are ofcourse some specialized
techniques which approximate the behavior
454
00:49:48,880 --> 00:49:54,329
of non-linear systems. But, most of the times
we will be using in fact numerical integration
455
00:49:54,329 --> 00:49:59,010
to analyze non-linear system.
And an exception to that of course is when
456
00:49:59,010 --> 00:50:04,839
you are having a non-linear system. And we
are analyzing its behavior for small disturbances
457
00:50:04,839 --> 00:50:12,539
around an equilibrium point, we can create
or rather derive a linearized model from the
458
00:50:12,539 --> 00:50:18,430
non-linear system for the analysis of small
disturbances from the equilibrium. Of course,
459
00:50:18,430 --> 00:50:22,529
once we will linearize the system, we can
use the tools of Eigen value and Eigen vector
460
00:50:22,529 --> 00:50:24,059
analysis.
461
00:50:24,059 --> 00:50:31,910
One of the key systems which we have not really
considered right now is the linear time variant
462
00:50:31,910 --> 00:50:36,180
system. And please note that, there is small
error here in this slide it should read as
463
00:50:36,180 --> 00:50:42,930
linear time variant systems. An example of
a linear time variant system as we shall see
464
00:50:42,930 --> 00:50:50,920
in the coming lectures is the synchronous
machine itself. The flux the flux as seen
465
00:50:50,920 --> 00:50:55,710
by the stator bindings of a synchronous machines
that is, the rotor winding flux, the flux
466
00:50:55,710 --> 00:51:03,250
is by the three phase windings of a synchronous
machine are in fact time variant. The differential
467
00:51:03,250 --> 00:51:08,980
equations which come out when analyzing a
synchronous machine are in fact linear time
468
00:51:08,980 --> 00:51:15,490
variant and we shall be using a very powerful
method or we shall be using a kind of a transformation
469
00:51:15,490 --> 00:51:22,150
of variables in order to derive a time invariant
system from the time variant system.
470
00:51:22,150 --> 00:51:26,380
So this is something of course we are yet
to come to. This is just a kind of a curtain
471
00:51:26,380 --> 00:51:31,140
raiser to what is to come. So, the modeling
of a synchronous machine we shall start off
472
00:51:31,140 --> 00:51:38,269
soon and we will be entering in some sense
into the domain of power systems, slightly
473
00:51:38,269 --> 00:51:42,780
away from the kind of general analysis which
we have done so far.