WEBVTT
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Last class, we designed the current loop of
our closed loop control of our ac dc converter.
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Now, from the current control design, we have
to go the voltage control design that controller
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design. So again, let me draw our close loop
control schematic for clarity.
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So, this is our V0 reference that is the one
we want. This, we will be giving it here,
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then we will be comparing with a feedback
voltage that is V0, then we will give to the
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current controller. So, this is our controller,
this we have to design now, controller. This
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controller output, we will give our IS reference,
this is your feedback, then your current,
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current controller; current controller what
we got, this is a proportional gain only you
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require.
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So, what we got here? It is equal to K1 is
equal to L divide by Ki G into T, T is the
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first order lag. T, this will go to our converter,
converter we have we have got it as G by 1
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plus STr, r is the triangular period; so we
are representing as a first order lag. Then
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here, you will get the actual fundamental
component of the VA VAB. So, this is minus
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VS. That VS, we are not considering here for
the controller design because controller design,
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it is based on our input IS star. So, that
we will remove as a disturbance input and
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then the close loop block diagram will be
1 by SL, then you have this one, current feedback,
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Ki comes here. This is our current loop.
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Now, how we design the voltage loop? So, voltage
loop, this current IS is based on the output
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dc current, load current. So, we should have
a relation between this IS and the output
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dc. So, equating real, equating the real power,
input power should be equal to the, equating
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input power to
the output power, power, equal to the output
power; we will find the relation that is VS
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that is our V input rms if this is the maximum
value, this V divide by root 2 into IL peak
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value, these are peak value, I load, our IS
peak divide by root 2 is equal to V0 into
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I0, our I0 is the output Tc. From this one,
I0 is equal to V IS, these are peak value
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into 2 into V0. This is the I0.
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So, this I0 if there is if there is no change
in load, this I0 will go to the load and the
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capacitor voltage will be stable. If any mismatch
in the, if there is I mean change in the load
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happens, it has to be reflected. The controller
should know that and based on that, you should
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adjust the IS so that the capacitor voltage
should not change in. So, the block diagram
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for our voltage loop is like this; maybe we
will draw with a different color.
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So, we will take it from here; this is our
IS, IS, this a gain block, this gain is this
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much only 2V by V0, V is our input voltage.
So, V by 2V0, we get our I0 here plus minus,
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this is our actual I load current if there
is any change in. If there is no change, this
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I0 will be equal to I load. This difference
will be 0 and the capacitor voltage will be
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stable. If any change in I load happens, this
I0, the flow I0 will disturb the capacitor
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voltage. So, that difference will go to the
capacitor and it will disturb the capacitor
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voltage. So, that will be 1 by C integral.
That will give the V0, change in, V0 plus
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V0, the change in V0. Then you have the gain
KV. This is our, we will give it here as feedback.
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Now, we have designed our control loop here
yesterday and we got the transfer function.
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If you see here, IS by IS star, this IS by
IS star is equal to
star is equal to 1 by Ki divide by S square
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plus 1 by Tr into S plus 1 by T Tr and we
also found out for a damping zeta is equal
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to 0.707 T should be equal to 2 Tr.
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Now, see this, let us write down in a different
way. This can also be written as 1 by Ki divided
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by there is time constant from 1 plus ST plus
S square T Tr, S square T T Tr. So, if you
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see here, this also in a different way, we
can write it. This S square T Tr, these are
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frequencies, these are the square of frequencies.
So, this current loop is, in this close loop,
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the current should loop should act as, act
very fast compared to our voltage loop. So,
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that means the frequency response of this
one is much higher compared to this voltage
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loop for fast action.
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So, if when you consider the voltage loop;
now we have to design the voltage loop, see
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this higher frequencies which we will be much
beyond the response range of this current
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voltage loop, so this current loop we can
approximate it. That is IS divide by IS star,
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we can approximate to design our voltage loop.
This is 1 by Ki divide by 1 plus ST. We know
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T is equal to 2Tr; so, general term. So, we
can use this one. So, from this one that means
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all this block, here circled by the green
line, we can represent by 1 by k by 1 plus
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STr. So, from this one, let us draw the voltage
loop, the block diagram of our voltage loop.
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Let us go to the next page.
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You have V0 star, our reference again, plus
minus; then our controller. So, let us again
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take for controller with for good dynamic
response. Let us take the PI controller. See
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here, we get the I0 that is this, this point
sorry this, this point, we will get the I0.
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This I0 minus I load if there is a disturbance
in load, that difference will go to the capacitor.
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So, let us take a worst case design, your
full I0 is going to this capacitor so that
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we can assume I load is 0. Then we will try
to design our controller.
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So, if you go back to our controller; we will
assume worst case condition, I0 is equal to
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0. So, the full I0 will go to the capacitor.
So, it will be 1 by SC. This is our actual
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V0. This V0, we will put with a gain to bring
to the controller gain that is KV and feed
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it here. This is our close loop controller.
Now again, let us see what do the purpose
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of this lead? There is a lead here, 1 by,
1 plus STr. So, if you write down the transfer
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function that is V0 by V0 star is equal to
G (s) G (s) by that is forward gain divided
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by 1 plus forward plus closed loop. This will
be Kn into 1 plus STn divide by STn into 1
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by Ki divide by 1 plus ST.
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Let us take V plus V0 or some constant V1,
so this we will take it as some constant V1.
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V1 by SC divide by 1 plus kn into 1 plus STn
divide by STn into 1 by Ki 1 plus ST V1 KV
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by SC. Now, if you use the pole zero cancellation
so that this zero which will be a lead, we
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can choose Tn such that we can remove this
lag due to 1 plus ST so that response can
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be faster. So, if you cancel this one, again
if you write down the transfer function in
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the denominator if you see here, escort time
will be there, S time, damping will be missing.
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So, if you use the Pole zero cancellation
here, this system will become unstable. So,
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we want this, we want to keep this controller
so that study state error should be 0 and
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we should have a good bandwidth.
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Now, how do you choose our Tn and Kn in this
case? Now, let us say, here you will use some
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sort of optimization technique. What is the
technique? We will make that definitely, what
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is meant by optimization?
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See, general consideration for optimization,
for optimization: see here, what we want in
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a closed loop control? The input should follow
the output for any variation in the input
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as fast as possible. That is any variation;
any variation means any frequency of response.
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See, for in a system, input cannot change
with any frequency of area. So, system may
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not be, system dynamics, system time constant
will not allow.
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So, with a system time constant, there is
a frequency limit upto which input can vary.
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That shows the bandwidth such that the output
should vary as close as possible to the input.
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That means the input output gain should be,
for wide frequency range, it should be unity.
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That shows the bandwidth. So, this optimization
aims at choosing Kn and Tn such that the bandwidth
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can be extended as for as possible.
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So, let us have what do the definition says?
The dynamic performance of a control system
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is good if the controlled variable rapidly
reaches the reference input. This means for
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any frequency, any frequency within the bandwidth;
any frequency of input variation
input variation, the output should track the
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input variable or in other terms, the output
should track the input variable instantaneously.
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So, for a practical system in terms of frequency
range, the modulus of the output or the output
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gain should be closed to one. That means if
you say that it should be if it is tracking
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the input, the input divide by the IS that
modulus should be as closest one over a wide
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frequency range that is the bandwidth.
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So, if you draw the gain that closed loop
gain; let us say this is in log omega sorry
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omega here and the gain modulus, input output
gain should be, what? For a large frequency
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range, as the frequency increase, this will
slowly change. So, this will show the bandwidth.
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So, bandwidth of the system that frequency
range we want to optimize. By choosing Tn
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our controller gain such that we should have
maximum bandwidth. That is another way of
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choosing our parameter, controller parameter.
So, the optimization, what is this optimization
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is planning?
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Now, we will talk about, the optimization
aims at bringing the bringing the input output
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gain as close as to as to one over a wide,
bringing the gain sorry, over a wide frequency
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range. This is also called modulus hugging.
So, let us see, how we can do? That is let
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us take the closed loop transfer function
for a general second order and general third
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order. So, we will restrict to these two and
then how we can do the optimum bandwidth,
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maximum bandwidth that is a modulus input
output gain is as close as to one for wide
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frequency range.
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Let us take the second order system. Let us
take the general transfer function b0 divide
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by a0 plus s, s will do as j omega a1 plus
S square - j omega square a2. So, we want
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low frequency range, low frequency very close,
starting. For that one condition is very clear
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b0 should be equal to a0. When frequencies
are very small, this we can neglect. Still
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low, we can neglect this one, so very close
to that is dc side, a0 should be equal to
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b0. Then let us take from this one, the modulus
of this is the closed loop transfer function,
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this is our F j omega.
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Now, let us take the modulus. This we can
again, write this one, write as a0 divide
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by a0 minus a2 into omega square plus j omega
a1. So, we can multiply the numerator and
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denominator by minus c omega. So, this again
will be equal to F j omega equal to a0 into
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a0 minus a2. So, sorry this here, if you see,
a0 minus a2 omega square; so omega square
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is not coming with a2, so this is we have
to correct it. This is omega square. So, the
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square is coming here. So, a0 minus a2 omega
square minus j omega a1 divided by a0 minus
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a2 omega square plus j omega a1 into again
a0 minus a2 omega square, here minus j omega
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a1.
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So, from this one, we can find out the modulus.
Modulus F j omega will be equal to root of,
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we can simplify this equation; finally it
will come to a0 square divided by a0 square
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plus omega square into a1 square minus 2 a0a2
plus omega square a2 square. This is the modulus.
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Now, let us see, how we can control this one
so that the bandwidth can be increased. So,
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low frequency, a0 square a01, we will have
modulus 1.
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Now, as the frequency increases, this time
to disappear, still we want slightly higher
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frequency, it should be 1. The condition is
b0 is equal to a0. The second condition from
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here, we want this to disappear. So, that
means a1 square should be equal to 2 a0 a2.
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This we cannot do it when the frequencies,
for the frequencies where omega, when this
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term come into picture. So, it would slowly
drop.
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So, the final transfer function, final modulus
F j omega, the modulus of this one if you
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substitute this one for this second order
will be there is an optimum, we have optimize
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this second order; so F j omega optimum is
equal to root of 1 by 1 plus omega raise to
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4 a2 by a0 square. So, by choosing these parameters,
see this second order system; we can optimize
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so that for wide frequency range that is a
maximum, we can get a modulus of one that
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is input will for any frequency variation,
frequency variation, input output will closely
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track the input. So, this is one way.
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So, now let us take a third order system because
all our purpose is to optimize and choose
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the controller, PI controller parameter for
our voltage loop. So, let us take the third
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order system.
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Now, we are talking about third order, a general
third order system; omega is equal to b0 plus
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j omega b1 by a0 plus j omega a1 plus s square
term j omega square a2, this is 2 plus j omega
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cube a3. So, seeing the numerator, one condition
for low frequency range; the condition is
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b0 should be a0, also b1 should be equal to
a1. Now, let us rearrange this one. F j omega
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that is equal to substituting this a0 plus
j omega a1 divided by a0 minus omega square
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a2 because minus comes because of this j square,
plus j omega into a1 minus omega square a2.
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Now, multiplying the numerator and denominator,
the complex conjugate of the denominator and
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finding out the modulus; so here if you do
it, F j omega modulus will be equal to root
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of a0 square plus omega square a1 square divided
by this modulus we have found out, a0 square
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plus omega square into a1 square minus 2 a0
a2 plus omega raised to 4 into a2 square minus
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2 a1 a3. Then one more term is there; plus
omega raise to 6 a3 square. So, this is the
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overall denominator.
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Now, how do you do the modulus again? Bandwidth,
we want as close as possible. One condition
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here, a1 square should be equal to 2 a0 a2;
so, one condition. From this, one condition
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is, already we said a0 is equal to b1 is equal
to a1. Now, the next condition is equal to
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a1 square is equal to 2 a0 a2 and a2 square
is equal to 2 a1 a3. With this one, what is
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the final output transfer function? So, based
on this condition, the final modulus will
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be, let us go to a new page, will be F j omega
is equal to root of 1 plus omega square into
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a1by a0 square divide by 1 plus omega raise
to 6 a3 by a0 square; this way, we will get
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it.
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So, this is the way we can do for the third
order system. Now, let us go back to our voltage
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control loop, this loop.
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Let us draw the, let us write down the V0
by V0 star. So, if you write down the V0 by
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V0 star, it will be equal to Kn, I will slightly
bring it down again, Kn into 1 plus STn divide
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by STn, then the current transfer function
1 by Ki by 1 plus ST when V by V by 2 V0 1
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by SC; this is G (s) part, divided by 1 plus
G (S) H(S) that is again 1 plus Kn into 1
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plus STn divide by STn 1 by Ki, this Ki is
the current feedback gain, 1 by Ki by 1 plus
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ST again V0 by V sorry V by 2 V0, V by 2 V
0 1 by SC into KV, our H(S) voltage loop gain.
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Now, let us expand this one. This is a simple
mathematical manipulation but we will write
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down. Let us go to the next page now.
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Here it will be, V0 by V0 star will be Kn
into 1 plus STn into V input V, V is our input
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mains voltage, V peak value divided by Ki
S square Tn into 2 V0 into C into 1 plus ST
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plus Kn, PI controller again Kn feedback K,
this we have to find out Kn KV. So, the parameter
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which we have to find out, we will Kn KV into
V input voltage V. So, input voltage makes
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the input voltage, source voltage, I will
make a subscript here VS, this is VS, VS into
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1 plus STn.
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So again, so if you expand this one, we know
this is a third order system. So, this will
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be finally if you do the thing, it will come
VS Tn Kn into S that is a Laplace plus VS
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into Kn divided by a0. a0 is equal to VS Kn
KV plus S into Kn KV VS Tn plus S square into
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Ki Tn 2 V0 C plus the S cube plus Ki Tn T,
this our current loop time constant - T, we
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know it is 2 Tr into 2 V0 into C into S cube.
This is of the form, third order form, transferable
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third order form where this is equal to b0;
This is the b0 and this is equal to, b0 if
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you see the transfer function, this is of
the form b0 b1, then numerator, this is the
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a0 a1, this is a2, this is a3. So, we can
one condition as previously; the first condition
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is b0 is equal to we have to equate a0 and
b1 is equal to a1.
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Now, if you equate the previous conditions,
what we have to find out? We have to find
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out the value of Tn and Kn, all other things
are known. So, let us use the previous condition.
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What was the previous condition? a1 square
is equal to 2 a0 a2 and then the next one
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is a2 square is equal to 2 a1 a3. We can equate
the condition.
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Here, Kn KV VS Tn square is equal to 2 into
Kn KV VS into Ki Tn 2 V0 into C. Two unknowns,
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we require two equations. So, from this one,
we can get equation Kn is equal to 4 Ki V0
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into C divided by KV VS into Tn and the from
the second equation, Ki Tn 2 V0 C square is
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equal to 2 into Kn KV VS Tn into Ki Tn T2
V0 into C.
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From this one, so from this equation, also
we can find out Kn is equal to Ki V0 C divide
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by KV VS into T. So, from this equation we
can find out Kn; substituting here, we can
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find out Tn. So, this way, we can find out
the close loop transfer function for the voltage
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control also.
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Now, if you go back to your old close loop
control scheme; see here we said, IS is given.
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Is is if you see the close loop controller
schematic, old one, this IS is multiplied
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by a sine wave. But all our controller we
are assumed with the sine frequency is very
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high so that during that sine triangle period
where the control is initiated; the input
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current, output current that amplitude during
that period is nearly constant. But if you
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see, these are sinusoidal wave form sinusoidal
feedback and sine throughout the operating
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condition, this frequency of this is constant.
But it is going through a controller and converter.
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So, it will, it can create a lag.
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So finally to control, to fine tune it, we
can give a small lead network here so that
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sine can be seeing the waveform, so you can
give a sine lead so that the feedback lag
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can be compensated here so that the IS so
that the IS and our VS will be in phase. So,
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this we can run, this is this we can do it.
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Now, what I want to know, I want to take a
typical example and find out the parameter
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and I want to model it through MATLAB and
Simulink and I want to show it you, how it
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works. So, let us take a typical specification
for a front end converter for traction applications
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and from then, let us design our parameters.
These parameters we can use it for controller
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design. So, let us go to the next page.
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See design, design
the parameters and the component value, parameters
of the controller and also the, design the,
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first design the components L and C. Let us
take an input voltage because this is what
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I simulated, input voltage peak value 1432
V and one output V0 equal to 2800 Vdc. Let
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us take our switching frequency that is a
triangle frequency 60 into 11, 11 times; 60
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is the fundamental and our triangle frequencies
11 times the fundamental frequency, this equal
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to 660 H. Rated power, 1400 kilo watt and
de-ceiling capacitor assuming a 5% ripple,
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de-ceiling capacitor can be designed. How
to design the de-ceiling capacitor?
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De-ceiling capacitor, we can design from the
5% ripple, the total dc that 5% of 2800 volts
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only and the ripple is due to the second harmonics
of the mains, 2 times of second harmonics
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frequency. So, that is the one we have to
use it. So, we already derived the equation.
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From that equation, we can find out L and
C and once the L and C found out, using the
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standard technique use for the modulus again,
we can find out the control parameter for
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the PI controller. And, already we know that
control parameter for the current is a simple
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gain based on the L and the gain, converter
gain E, G and the triangle frequency. So,
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this we will take up in the next class.