WEBVTT
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Good morning friends. Today we will have a tutorial
exercise on some of the topics that we have
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covered so far. I will take up a few examples
and discuss, there is a problem, first example
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is very simple one. You have a bridged capacitive
circuit
the values are 2 farads, 1 farad, 4 farads
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and 1 farad, what should be the equivalent
capacitances c1, c2 and c3 okay.
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Now in this case, as you know a capacitor
is having an impedance if you talk in terms
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of sinusoids it will be 1 by omega C, it is
a j here if you talking Laplace domain it
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will be 1 by Sc that means the impedance is
inversely proportional to C. So I can take
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these capacitances to the equivalent impedances
or even resistances
with the values which will be just inverse
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of these something like half ohms, one fourth
ohm, 1 ohm and 1 ohm, I can replace it by
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an equivalent impedance circuit.
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Now after this you can get equivalent one
by a star delta conversion see if this is
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a delta then I can have a star like this.
So these 3 nodes are these 3, so what will
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be the value of this 1 into half divided by
1 plus half plus 1, so 1 into half divided
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by 2 and half, so that gives me 1 by 5, 1
by 5 similarly, this one will be 1 into 1
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by 2 and half, so 2 by 5 and similarly this
is 1 into 1 and this one will be again 1 into
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half, so this one will be 1 by 5 this is 1
by 4. So you get 1 by 5, 2 by 5 and 1 by 5
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plus 1 by 4, 9 by 20. So take the inverse
of these that will give you the equivalent
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capacitance 5 farads, 2.5 farads and 20 by
9 farads okay. So these are the 3 equivalent
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capacitances okay, it is so simple.
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Next we are given a problem on transmission
parameters of 2 cascaded sets. The problem
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is very simple one, this is Za1, Zb1, Zc1
cascaded with Za2, Zb2 and Zc2, what will
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be the total A, B, C, D parameters of the
total combination, what will be the A, B,
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C, D parameters of this total combination.
The values given are Za1, Za1 equal to 50
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ohms, Zb1 equal to 0, Zc1 is equal to 200
ohm, 300 ohms, Za2, Zb2, Za2 is 0, Zb2 is
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50 ohms and Zc2 is given as 600 ohms, what
be the A, B, C, D parameters. Let us draw
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this network first, this is 50 ohms, this
is 300, this is 0, this is 0, this is 600
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and this is 50. You see the network is very
simple, it is a phi network if I combine them
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it will be 50 ohms 600 and 300 in parallel
that gives me 200 ohms and 50 ohms. So A,
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B, C, D parameters if we remember V1, I1 we
are writing in terms of A, B, C, D as V2,
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I2.
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So A is V1 by V2 when I2 is 0 similarly, C
is I1 by V2, when I2 is 0, B is minus B is
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V1 by I2 when V2 is 0 that is under short
circuit condition and similarly D is minus
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D is I1 by I2 when V2 is 0. So from this diagram
it is very simple V1 by V2, this is 50 plus
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200 and this is 200, so fif 250 by 200, A
is 250 by 200 that gives me 1.25. Similarly,
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B you can find out V1 by I2 when this is shorted,
when this is shorted we apply a voltage V
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here what is the current that is flowing through
this. So B comes out to be 100 and 12.5 ohms,
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C comes out to be 5 into 10 to the power of
minus 3 and D is once again 1.25. These are
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very simple relationships all of you can determine
this. Next we have a question
determine for a
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for a network like this, couple circuit having
3 coils wound on wound on this steel structure
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1, 1 dashed, you are sending a current I1
this is 2, 2 dashed shorted through a resistance
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R2 and similarly
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you are having 3, 3 dashed there is resistance
here R3.
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Now the question is there are 4 statements
given which one is correct I2 when the current
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is flowing through this winding 1, 1 dashed
in this direction I1 then the current through
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R2 and R3 will be one possibility is I2 will
be 2, 2 dashed externally, 2 to 2 dashed and
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I3 will be 3 to 3 dashed, 3 to 3 dashed okay
b, I2 is 2 dash to 2 and I3 is 3 to 3 dashed
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c, I2 is 2 dash to 2 and I3 is 3 dash to 3
the 4th possibility is I2 is 2 to 2 dashed
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and I3 is 3 dash to 3 okay. So we have to
see whenever there is a current flowing through
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I1 what is the sense of the flux flowing through
this.
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Suppose, you take the current in this direction
all right it is coming out like this. So what
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will be the North Pole, North Pole will be
in this coil it will be in this direction
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and here it is coming out. So this is south
so flux will be flowing like this okay like
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this. So in R2 suppose the current is from
2 to 2 dashed, this is kept open, this is
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kept open 2 to 2 dashed then which direction
this current will send the flux. If you look
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at it 2 to 2 dashed that is current is going
like this then the current is flowing in this
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direction okay from bottom to top it is in
this direction that means this will be a North
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pole this will be a South pole when it is
going like this it is North pole. So that
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means it will try to send the flux like this
and like this. So it will be adding this flux
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that is not possible if I excite this now,
if I excite this now and if the current flows
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like this then this will try to add to the
flux that is already being established by
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this okay. So that goes against the very basic
principle of conservation of energy.
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So this will be trying to if the current flows
in this direction this will try to oppose
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the flux. So this one will try to if a current
I1 flows through this it will try to flow
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it will try to send a flux like this then
this current has to be in this direction.
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So it will be 2 dash to 2, so 2 dash to 2
there are 2 possibilities what will be the
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current through this. Similarly, if you go
through the same argument here if a current
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flows like this what will be the direction
of the flux will find if it is 3 to 3 dashed,
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3 to 3 dashed okay then the flux will be opposing
the flux that is being established by I1 okay.
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So this will be the correct direction of the
currents that is this current will be from
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to 2 dash to 2 this current will be 3 to 3
dashed. So this is the correct answer next
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you take up another example write the expressions
for the functions
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and so on. It is a periodic function 0, 10,
20, 30 and so on, this is 100, it is a rectified
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sin wave, what would be the expressions for
the current f(t) current or voltage whatever
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is right in terms of continuous functions,
regular functions and what will be the corresponding
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Laplace transform.
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Now f(t) if I take only the first period,
it is a part of 100 sin 2 phi by t where,
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t is basically this is the period 20 okay,
it is this period, if I write like this then
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it will represent the function which continues
if I add another function to that I call this
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part as only 1 period then if I add another
sinusoid which is delayed by 10 seconds 100
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sin 2 phi by 20 into t minus 10 seconds into
u(t) minus 10 then this plus this will be
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giving me a net function of, see
this is the first part plus a function that
i am adding from here, so this plus this will
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give me this will get cancelled with this
the first okay. It will keep on cancelling
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hence forth after 10 seconds. So if I add
this with this this is a delayed function
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represented like this I will get this 1 which
I am calling as f1(t) okay. So somebody may
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write f(t) as f1(t) plus KT where, T is 10
seconds
and write like this summation it will represent
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this function okay where, K is any positive
integer.
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So it will represent this function of course
into u(t) minus KT okay the correct representation.
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So this is f(t), now 2 phi by 20 can be written
as phi by 10 what will be the Laplace transform
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of this then let us again start from the beginning
Laplace transform of this function if I call
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it F1(s). So it will be F1(s) plus law plus
transform of this shifted by 10 seconds F1(s)
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into e to the power of minus 10 S plus F1(s)
into e to the power of minus 20s and so on
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which means F1(s) into 1 by 1 minus e to the
power of minus 10s and what is F1(s), F1(s)
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will be Laplace transform of this which is
100 into sin 2 phi by 20 means phi by 10 divided
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by s squared plus phi by 10 whole squared,
this is the Laplace transform of this part
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and for this part the same thing shifted by
10 seconds. So it will be this plus this into
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e to the power of minus 10 minus 10s so combining
the 2 I will get.
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So this is F1(s) so substitute here so that
gives me 100 into phi by 10 that gives me
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10 phi divided by S squared plus phi by 10
squared into 1 plus e to the power of minus
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10s by 1 minus e to the power of minus 10s
somebody may write this as cot hyperbolic
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5S, e to the power of minus 5S, if I multiply,
so cot hyperbolic 5s okay, so this will be
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the Laplace transform. Similarly, there is
another function, what will be the expression
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for this type of periodic function, it is
pretty simple 0, this is 5, 2, 4, 6, 8, 10
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and so on this is minus 5.
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Obviously, the first block can be written
as 5 into u(t) and then at 2 seconds I apply
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a negative step and u t minus 2, so that gives
me the first block again at 4 it is negative
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so minus u t minus 4 and at 6 I apply u t
minus 6 a positive step and that completes1
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period up to 8 seconds and then it keeps on
repeating. So show 1 period if I show 1 period
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and then show it as a repetition of the same
function this will be written as 5 I can write
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this as summation u t minus 2 okay u(t), so
plus after 8 seconds again it is plus u(t).
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Okay so u t plus k8, 8 minus u t minus 2 minus
k8 okay minus u t minus 4 minus k8 plus u
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t minus 6 minus k8 where, k varies from 0
to infinity this will be the representation
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of a periodic function, what be the Laplace
transform of this? it is 5 into you take just
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1 period and that keeps on repeating, so for
1 period F(s) will be for 1 period, it is
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5 into 1 by S into 1 minus e to the power
of minus 2s minus e to the power of minus
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4s plus e to the power of minus 6s, this whole
thing into 1 by it is repeated 1 minus e to
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the power of minus 8s, is it not this is something
like a earlier F1(s) and multiplied by this
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that will give you total F(s) so 5 by S1 minus
e to the power of minus 2 S can be taken common.
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So in bracket we get 1 minus e to the power
of minus 4s that is equal to 5 by s into 1
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minus e to the power of minus 4s, e to the
power of minus 4s and 1 minus e to the power
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of minus 8s, there is a common factor, so
that gives me 1 plus e to the power of minus
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4s. So this we call Laplace transform of this
function okay. Next we have a question on
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Laplace transform again
determine the initial and final values of
the function whose Laplace transform is given
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as s plus 4 into s plus 70 divided by s plus
1 into s plus 5 into 2s plus 7.
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Obviously, f0 we compute as limit s standing
to infinity s into F(s), so if I multiply
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by s and then make s standing to infinity
only the highest power of s has to be computed,
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so limit s standing to infinity if I leave
only the highest power will be s cube after
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multiplying by s, s into s into s divided
by s into s into 2s. So that gives me half
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similarly f at infinity will be limit s standing
to 0 s F(s) if I multiply by s and then put
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s equal to 0 all these terms will give me
4, 70, 1, 5, 7 but this s will give me 0,
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so this is 0.
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Next example
the step response of
a network is 1 plus e to the power of minus
t sin 10t into u(t), find the steady state
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response the steady state response when the
input is 10, sin 10t that means if I have
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instead of e to the power of minus t that
is at decaying sinusoid if I have a steady
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sinusoid what will be the response. Let us
see, let us determine the transfer function
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the Laplace transform of the output is
for the step input, step response is 1 by
S plus 1 by s plus 1 whole squared plus 10
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squared is not 10 by s okay, one may write
this as s squared plus twice s plus 100. So
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s squared plus 2s plus 10s, so 12s plus 100
divided by s squared plus 2s plus 101, okay
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s squared plus 2 s plus 1, so this is a this
is also 101 alright multiplied by s.
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So the transfer function can be written as
output by input, input is 1 by s, so if I
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divide this by 1 by s that comes out to be
s squared plus 12s plus 101 divided by s squared
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plus 2s plus 101 okay. This is the step response,
so if I divide the response by the input given
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to find out the input was step input so Laplace
transform is 1 by s, so after dividing by
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1 by s we get this. Now we want to determine
the steady state response due to a sinusoidal
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input, so for a sinusoidal input we straight
away write s equal to j omega, omega here
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is 10. If I substitute here therefore G, I
call it G(s), so G (j10) will be minus 100
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minus omega square, so minus 100 plus 12 into
j10 plus 101 divided by minus 100 plus 2 into
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j10 plus 101 that gives me 1 plus 20j divided
by 1 plus sorry 120j divided by 1 plus 20j,
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one is very very small compared to 120. So
this is approximately 120 and an angle very
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close to 90 degrees.
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Similarly 20 plus 1 plus 20j that is that
will also give me a magnitude close to twenty
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and an angle very close to 90. So both these
angles almost 90 degrees will give me approximately
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0 angle, magnitude is 6. So what will be the
output steady state output, it will be input
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multiplied by this 6 angle 0, 10 sin 10t,
this is not the correct way of writing I just
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wanted to explain. So the magnitude gets multiplied
by 6 angle gets shifted by this angle theta
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which is 0 approximately so it will be 60
sin 10t, this will be the response, steady
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state response okay. Next we have another
example, you are having okay.
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Let me use this page, you are given a simple
circuit, you are asked to calculate I(s),
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this is a DC source I(s) and then the voltage
is across R and C, at initial and final condition
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all right. Initial and final values of the
voltages across the 3 elements okay and then
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what will be the critical value of R okay,
for critical tamping what be the value of
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R for critical tamping.
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So let us compute the impedance this can be
replaced by E by S, SL I should write a general
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impedance R, 1 by SC, assuming no initial
charge is said here, no initial current was
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there. So I(s) will be E by S divided by the
total impedance SL plus combination of this
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will be R by RCS plus 1. So if I combine will
be E into RCS plus 1 divided by RC, RCLS square
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plus SL plus R okay into S, this will be the
expression for I(s), what will be the voltage
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across the inductor, voltage across the inductor,
it will be SL times I(s), is it not. So S
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will get cancelled, so it will be E into L
into RCS plus 1 divided by RCLS square plus
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SL plus R.
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Now if I want to calculate the steady state
value and that is VL at infinity then I should
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calculate SVLS and then put s10 into 0. So
if I multiply by s this quantity and put s
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equal to 0, the whole thing will become 0.
If I want to compute VL at 0 then I will compute
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s times VLS at s10 into infinity and how much
is that if I multiply by S, if I multiply
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this by S and then make s10 into infinity
highest power of S will have to be taken,
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so it will be S squared into RLCE and here
it is RLCS square. So this terms will get
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cancelled we will be getting only E. Basically
at start the entire voltage appears across
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the inductor obviously, what will be the voltage
across this combination mind you, voltage
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across the resistance and voltage across the
capacitance will be same. So you need not
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find out these values separately the voltage
across the resistance when this is having
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maximum value in the initial condition, this
will be 0 at t equal to 0, this is equal to
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E, so this will be 0 at t equal to infinity
this is 0. So this will be E.
36:36.319 --> 36:43.319
So VR which is equal to VC will be equal to
0 at t equal to 0, VR equal to VC equal to
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E, at t equal to infinity okay. For critical
damping this quadratic we have got RLCS squared
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plus SL plus R, so equate this square is equal
to 4 SC, so L squared is equal to 4 into R
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into RLC so that gives me R squared equal
to L by C into 4 okay or R critical, R critical
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is 1 by 2 root L by C. So this will be the
condition for critical tamping. Next we have
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another problem
a network is given by Y(s) equal to s plus
1 into s plus 5 divided by s into s plus 4
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into s plus 6 you are asked to show the unknown
arms of this network in this form, what will
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be this unknown arms X1, X2 and X3, it continues
this is given as R1, R2, R3 what to be the
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nature and the values of these X1, X2 and
X3.
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Obviously, this is given in a ladder form
okay so we try to find out we try to find
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out the ladder structure. So this is s squared
plus 6s plus 5 divided by s cubed plus 4 plus
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6, 10s squared plus 24s okay. This is Y(s)
so correspondingly Z(s) you see the first
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element if it is present then it has to start
with Z(s) which will be s cubed plus 10 s
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squared plus 24s divided by s squared plus
6s plus 5 okay Z(s) of this form, is it RL
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or RC network obviously, Y(s) having a pole
closest to the origin closest to the imaginary
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axis is giving me RL, it is an RL network.
So this will be an inductance resistance inductance
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resistance and so on.
40:20.789 --> 40:27.789
So by continue division s cube s squared plus
6s plus 5, I will write only the coefficients
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1, 10, 24, this will be s 1, 6, 5, 4 and19
, 1, 6, 5. So 1 by 4, 19 by 4 so this gives
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me 5 by 4 and 5, this is 4 and 19, 5 by 4,
so that gives me 16 by 5 okay, no 3 into 5
41:23.309 --> 41:30.309
by 12, 3 5s are 15 by 12 okay 5 by 4 then
5, 3, 3 by 5. So what you get this is s this
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is s, what you get is 1 Henry then this is
an admittance of 4 ohms then this is 16 by
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5 Henry and then 5 by 12. So 12 by 5 ohms
and then 3 by 5 Henry.
42:46.470 --> 42:53.470
So the unknown values are X1 is1 Henry, X2
is 16 by 5, X3 is 3 by5 okay. The same function
42:59.150 --> 43:06.150
the same function you are also asked to evaluate
the
impedances X1 take once again nature in the
43:18.299 --> 43:25.299
values of X1, X2, X3 okay R1, R2, R3 and this
is R4 what will be the values of X1, X2, X3.
43:35.529 --> 43:42.529
So obviously this is a faster one realization,
so Z(s) start with Z(s), so Z(s) is
s into s plus 4 into s plus 6 by s plus 1
43:54.269 --> 44:01.269
into s plus 5. So I break it up into K1 into
S since this is a fire plus K2 by s plus 1
44:06.760 --> 44:13.069
should I have K2 by s plus1 or K2 into s by
s plus 1. I have already identified this is
44:13.069 --> 44:19.400
an RL network so for an RL network in faster
one synthesis, you remember the Z synthesis
44:19.400 --> 44:25.910
will require K2(s) by s plus 1 if you do not
take K into s you are landing up into trouble
44:25.910 --> 44:32.910
you saw earlier also we will get negative
values of K2, K3 etcetera. Some of them may
44:34.079 --> 44:36.150
come out to be negative.
44:36.150 --> 44:43.150
So this one K1 will be if I divide by s make
a stand into infinity this becomes 1, K2 divided
44:50.000 --> 44:57.000
by s multiplied by s plus 1 make s plus 1
equal to 0, so this is 3 into 5 by 4, 15 by
44:58.960 --> 45:05.960
4, s plus 5, so this will be minus 1 plus
1 and 4, so 1 by 4. So these values are this
45:13.539 --> 45:20.539
is 1 okay, so if it is like this obviously
you do not you do not require all of them.
45:23.609 --> 45:30.609
There may be sorry, I forgot to mention here
this is X1, this is X2, X3, this will not
45:42.779 --> 45:49.779
come because the order is only this much,
so X1, so K1 represents X1 okay, this is s
45:55.490 --> 46:02.490
plus K2 is 15 by 4 by s plus 1s plus 1 by
4s by s plus 5. So for this one this is 1
46:17.260 --> 46:24.260
Henry, X1 is 1 Henry, from here you can see
if I make s stand into 0 stand into 0 this
46:29.799 --> 46:31.829
is 15 by 4.
46:31.829 --> 46:38.829
So X2 is 15 by 4 Henry and X3 if I make s
stand into infinity s stand into 0 this is
46:45.640 --> 46:52.640
1 by 4, 1 by 4 Henry okay. So these are the
3 values 1, 15 by 4 and 1 by 4. S by 4 into
47:03.940 --> 47:10.940
5 it should be 20 sorry, 4 into 5 if I can
make s stand into 0, it is s by 4 into 5,
47:15.440 --> 47:22.440
so it is 1 by 20 Henry. Next example
47:40.630 --> 47:47.630
if you have a little more time. Determine
the value of b, c and a for the impedance
47:54.160 --> 48:01.160
function Z(s) equal to 5 into s into s squared
plus b divided by s squared plus c where,
48:07.440 --> 48:14.440
this network has been partly realized as a
Henry and half a farad and 1 Henry that means
48:22.240 --> 48:29.240
a network has been given this is a structure
2 values are known, this is not yet known
48:30.059 --> 48:37.059
but it transfer function is of this form what
will be b, c and a, this is the problem.
48:39.259 --> 48:46.259
So let us see these 2 combinations 1 Henry
and a half farad will give me 1 Henry. I can
48:53.299 --> 49:00.299
write this as a s plus, this is s into 2 by
s divided by s plus 2 by s. So that gives
49:07.759 --> 49:14.759
me a s plus 2s by s squared plus 2, correct
if I wrong, is it all right and this is equal
49:24.500 --> 49:31.500
to this. So c has to be equal to 2 okay c
has to be equal to 2, if I add these S squared
49:38.710 --> 49:45.710
plus 2 into a s plus,2 S that gives me S cubed
plus twice a plus 2 into S and that is equal
49:55.680 --> 50:02.680
to 5 S cubed plus 5 S cubed plus 5 b s. So
here a s cubed sorry a is equal to 5 and if
50:17.980 --> 50:24.980
I put equal to 5, 5 into 2, 10 plus 2, 12
is equal to 5 b therefore b is equal to 12
50:30.180 --> 50:37.180
by 5 okay, it is a very simple example. Well,
before we take up any other problem I think
50:58.890 --> 51:05.890
we will stop here today. We will continue
in the next class because there is not much
51:07.750 --> 51:14.750
of time left.
51:16.900 --> 51:23.900
Okay, Good morning friends, today we will
have another tutorial session. The first question
51:32.490 --> 51:39.490
is determine the Laplace inverse of factorial
n divided by s into s plus 1 into s plus 2
51:45.400 --> 51:52.400
into s plus n okay. So easiest method is to
find out the partial fractions I will write
52:08.859 --> 52:15.859
A 0, A 1 A 2, S plus 2 and so on, A k by S
plus k general term okay. So how much will
52:22.039 --> 52:29.039
be A 0, A 0 multiply by s this F(s) make s
stand into 0, so that gives me factorial n
52:42.809 --> 52:49.809
divided by 1, 2, 3, 4 up to n that is equal
to 1, A 1 multiply by s plus 1 make s plus
52:54.569 --> 53:01.569
1 equal to 0. So what we get A 1 as 1 into
1, 2, 3, 4 up to n minus 1.
53:12.660 --> 53:19.660
So that gives me n with a negative sign because
the first one will be minus 1, A 2 will be
53:24.529 --> 53:31.109
factorial n divided by multiply by s plus
2 then make s plus 2 equal to 0 should be
53:31.109 --> 53:38.109
minus 2 minus 1 minus 2 minus 1 then 1, 2,
3 up to n minus 2 and that will give me plus
53:46.039 --> 53:53.039
n into see I get minus 1, minus 2, n minus
2, so factorial n divided by factorial n minus
54:02.859 --> 54:09.859
2 into factorial 2, A 3 and what is this you
can identify this as nc 2, A 3 similarly will
54:17.910 --> 54:24.910
be factorial n divided by minus 3, minus 2,
minus 1 then 1, 2, 3 up to n minus 3. So I
54:29.299 --> 54:36.299
get minus 1 into factorial n by n minus 3
factorial into factorial 3 that is nc3 with
54:42.680 --> 54:44.210
a minus 1.
54:44.210 --> 54:51.210
Network function y (s) is shown like this
for y (s), some y (s), this is minus 30 degree,
55:04.109 --> 55:11.109
these are all 20 degree per decade slopes,
this is 5, 10, 20, 40, this is 0.2 these are
55:32.119 --> 55:39.119
not to the scale exactly, this is 0.2, minus
30, we have to be the value of y (s) and the
55:46.490 --> 55:53.490
next question is, is it deriving a driving
point impedence by admittance function, is
55:54.740 --> 56:01.740
it a driving point admittance function, if
not justify, if yes state that is yes and
56:11.299 --> 56:18.299
then realize one in faster 2 form, so if yes
then realize in faster two form and if not
56:29.019 --> 56:36.019
justify, why it is not a driving point admittance
function, let us write y as, V 2 by I 1, I
56:41.380 --> 56:45.099
1 by V 2, I 1 by I 2 and so on.
56:45.099 --> 56:52.099
That means you may be given the specification
either in terms of Z 12, Y 12, Y 22 and so
56:53.049 --> 57:00.049
on and Z 11, so specification can be given
in terms of Z 11, Z 22 or Y 12, Y 22 or V
57:01.029 --> 57:08.029
2 by V1 and so on. You have to determine a
possible network so, so far we have studied
57:08.130 --> 57:15.130
driving point synthesis now we shall going
for 2 port synthesis with transfer synthesis,
57:16.240 --> 57:23.240
okay some of the basic condition for these
impedence and admittance function that is
57:25.569 --> 57:32.569
transfer impendences and transfer admittances
a very interesting and so the transfer function
57:33.299 --> 57:40.299
also, we will find the numerator and denominator
polynomials N (s) by D (s) they need not have
57:44.450 --> 57:48.829
all the properties that are listed in the
driving point synthesis that is the difference
57:48.829 --> 57:55.829
in degree can be more than 1, difference in
degree can be more than 1, can be may or may
58:07.140 --> 58:14.140
not be but D (s) must be Hurwitz polynomial,
roots must be always in the left of plane,
58:18.700 --> 58:25.700
so D (s) must be Hurwitz, there can be multiple
roots, multiple zeros on N (s), there can
58:28.210 --> 58:35.210
be all possible combinations. So we shall
study in details, what are the conditions
58:35.359 --> 58:42.359
to be satisfied for 2 port synthesis, what
are the conditions for Y 12, Y 12 or Z 11,
58:58.470 --> 59:05.470
Z 12, these care of admittance or impedence
function. Thank you very much.