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Hello, and welcome to today’s lecture. Now
in the last lecture we had discussed about
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antenna fundamentals, we saw how directivity
can be calculated from half power beam width,
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we also saw the link budget; how we actually
have a transmit antenna, receive antenna,
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and then what is the distance between them
and depending upon the transmitted power we
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can calculate what is the received power.
And then we also looked at what are the radiation
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norms in India and Abroad, and we found out
that there are lot of radiation hazards because
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of the over use of cell phone as well as to
the people who are living next to the cell
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tower, and it is also affecting birds, bees,
animals, plants, trees, and environment also.
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Now today we will talk about the fundamental
antenna which is a Dipole Antenna. So, let
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us start with the dipole antenna.
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In general we start with the infinitesimal
dipole; the definition of infinitesimal dipole
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is that the length of the dipole should be
less than lambda by 50. So, if length is less
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than that then we consider that as an infinitesimal
dipole. And here what we are assuming, we
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are assuming an infinitesimal current element
which is actually means very small current
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element of length dl and it has a uniform
current which is I 0; that is the amplitude
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but otherwise it is a sinusoidal current depending
upon whatever is the frequency. So, then we
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can actually define the instantaneous current
as I 0 e to the power j omega t.
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Basically, this current element is along the
z direction. So, what we have axis here the
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dipole is placed at the origin, so this is
an x axis this is a y axis and this is a z
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axis. And just to mention again angle phi
is measured from this axis. So, anything component
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along this will be resolved on this direction
here and the angle theta is measured from
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z direction. So now, since the current element
is placed in the z direction we write this
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as a current in the z direction. And e to
the power j omega t can be written as cos
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omega t plus j sin omega t.
So, for this particular current, but I want
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to mention here that I disagree with a lot
of books which they claim this to be an infinitesimal
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dipole and they actually say that for infinitesimal
dipole we can assume that the current is uniform,
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but in a reality that is not a correct assumption
because current at the open end will always
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be equal to 0. Suppose if you feed here than
the current will be 0 here and then it will
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have a maximum value. But nevertheless for
the derivation purpose we will assume that
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this element is carrying a constant current.
So, please remember it is only for the derivation,
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but not the real thing. So, corresponding
to this current element we can actually find
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out what is the vector magnetic potential.
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So, this is how we can find the vector magnetic
potential and since the current is in the
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z direction we have only component which is
A z and that is given by this one over here.
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And since the length l which we have assumed
is dl, very small and then that is integrated
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it will become l, because current is uniform
so I 0 will come as it is. Now this is a z
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direction from here we can find out the spherical
coordinate A r A theta A phi. So, from z we
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can actually say A z will be cos theta, so
we can actually look into this here. So, let
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us say any element which is in the z direction,
so then and this is the direction which is
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E r it is shown here e field, but think about
this as A r then A r can be found out simply
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by z component multiplied by cos theta. So,
that is the component which we can find out.
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Then regarding the E theta component; so E
theta component or in this case A theta component
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can be resolved now that will be sin theta
and since it is going down a minus sign will
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come here. Now here it is electric field,
but for A phi it will be perpendicular and
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A phi will be in this direction which is actually
nothing but theta equal to 90 degree and cos
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90 is equal to 0. So, hence the A phi component
will be equal to 0.
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So, once we know the A component vector magnetic
potential then we can use the Maxwell’s
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equation.
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And along with the waves equation we can find
out what is the H field what is the E field;
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and if we know E field as well as H field
then we can find out what is power. So, power
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is nothing but you can say double integration
E across H conjugate and that is the surface
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integral.
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So, from here we can by using that equation
or we can find out E r E theta and these are
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the components which are 0 and these are here
and H phi.
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Now I am just giving you this expression,
you can see the details derivation of these
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expressions in several books. I can just mention
let us say the book of Balanis or Kraus and
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other thing. So, I will get into the more
of the concept part and you can see these
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derivations in these books.
So now, I just want to bring to the next point
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here. If we see here, this has a component
in the denominator which is r square and here
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it is r. For E theta the component in the
denominator is r here and then there is another
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are here, so r will multiplied by r will become
r square and there is a r square this component
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will become r cube. Now when we are designing
antenna most of the time we are concerned
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about far field radiation pattern. So, at
far field we can assume that r is very large
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and if r is very large then r square will
be extremely large, so this entire thing can
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be equated to 0 for far field. And for this
particular case here this term and this term
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they can be ignored again at power of a distance
where r is very large.
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Similarly for H phi, so this has a 1 by r
component this will be 1 by r square component,
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so we can ignore this particular term here.
So, if we do that we can actually see far
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field pattern will be much simpler, but from
here we can also define various areas and
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these various areas are defined in terms of
the distance.
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So, if r is much smaller than lambda by 2
pi. And this is coming basically because of
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the 2 pi by lambda which is nothing but k
r. So, actually speaking you should think
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about if k r is much less than 1 and k is
nothing but 2 pi by lambda so it goes over
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here. So, if this is the condition applied
and over here also this region is defined
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into two different parts here; one is r less
than lambda by 6 as 2 into pi is approximately
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6. So, that is the near reactive field region,
and from lambda by 6 to upto this distance
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here that is known as near radiative region.
And any distance which is more than this which
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is r greater than 2 d square by lambda will
be a far field region. And what is d? d is
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the maximum dimension of the antenna. Now,
one should also remember that far field region
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also implies r should be much much greater
than lambda by 2 pi
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So, let just take some example here; suppose
if we take an example of just let us say a
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lambda by 10 antenna which are dipole antenna.
So, if I put here lambda by 10 so that will
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be lambda square by 100 the quantity will
be reduced to lambda by 50. Now we cannot
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say that r greater than lambda by 50 will
satisfy far field region criteria, because
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we also have another criteria which is lambda
by 2 pi.
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So, do not always go with this equation this
equation along with this equation define far
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field region. But however, especially for
antenna which are very large suppose d is
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equal to 10 lambda, if d is 10 lambda then
this will be 200 lambda. So, you can see that
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that will be very far away distance. So, dimension
plays very very important role. So that is
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why a many a times for larger antenna it is
very difficult to do the radiation pattern
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measurement within let us say an anechoic
chamber which is inside a room. So, for larger
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antenna invariably people do the measurement
in the either in the far field or at the open
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area.
Or many a times if we want to calculate far
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field region and if the distance is very large
then near field measurements are done and
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from the near field measurement far field
can be obtained by using the concept of Fourier
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transform; and that way we can get the far
field radiation pattern.
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So now let us just see; what are the different
things here. So, I have used the term uniform
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current, I am not mentioning here infinitesimal
dipole because I do not agree that current
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will go to 0 at the edges. But for a uniform
current let us say; what are the far field
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region. So, please see here this is k r much
much greater than 1. So, we can now say there
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will be only two components; one will be E
theta component another one is H phi component.
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And if you really look into over here what
are the different terms; so eta is nothing
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but free space impedance which is equal to
120 pi or equal to 377 ohm k is equal to 2
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pi by lambda. So, you can actually see that
lambda is coming into picture so that 2 pi
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can be there and lambda will be down below.
So, it will be l by lambda.
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So, please remember for all the antennas in
general it is always the normalized length
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which is more important. And now you can see
something interesting if you take the ratio
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of E theta by H phi you can see that most
of the terms are similar except for eta here.
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So, E theta by H phi is equal to eta which
is 120 pi and this is impedance of free space;
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so that means a wave which is propagating
in the free space E theta and H phi will be
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perpendicular to each other; that means E
plane and H plane will be perpendicular to
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each other, and the impedance ratio will be
equal to 120 pi; let us first just look at
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the radiation pattern for a dipole antenna.
This is for infinitesimal dipole, so this
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is the 3-D plot and these are the plots in
the H plane pattern and E plane pattern or
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we also call it azimuthal. So now, just imagine
that dipole antenna is over here. So, at this
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particular point here we can actually think
about a dipole antenna as a pen in my hand.
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So, you can imagine that this dipole antenna
is like a pen in my hand I will change the
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color. Now, if you look at this pen from let
say this direction what you see full length
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of the pen. If I see from my said I see the
full length, from your side if you see you
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see the full length of the pen. So, if you
actually look all around you will actually
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see the full length of the pen. Now please
apply this concept only to dipole antenna,
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do not try to apply for all and every other
antenna.
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Now for this dipole antenna let say we are
standing here we see the full length now as
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we move along and if we see from the top all
we really see is the tip of the pen; that
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means we will see very little and as we move
along we will start increasing little more.
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In fact, dipole radiation pattern is very
very similar to the way we look at this particular
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pen. So, the dipole pattern will be uniform
along this here which will be maximum and
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then it will go from maxima will go to 0 then
again it will go to 0 go to maxima.
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So, one can actually see now the pattern.
So, here is the dipole antenna, so you can
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see that or you can imagine a pen over here.
So, we see the maximum intensity or over here
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maximum radiation in this, and then as we
move along we hardly see anything. So, this
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is the 0 radiation in this side and then this
is basically is repeating in this side. So,
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this is also known as a figure of eight. However,
this is just a one plane, in the one plane
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we are showing. And for H plane pattern you
can also think about again. If there is a
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current carrying conductor which is placed
in the z direction. So, where will be the
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magnetic field? Magnetic field will be given
by these fingers here. So, this is the, you
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can say electric dipole antenna current and
the magnetic field will be like this. So,
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that is why H field is uniform along this
particular plane here.
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So now, let us just look at the 3-D pattern.
So, here a red color implies maximum radiation
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and then as you can move along this one here
it almost becomes slightly blue which is the
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least radiation zone. So, from red it is turning
to orange, yellow, green, and blue. So, basically
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intensity is reducing. So, this is the real
3-D plot of a infinitesimal dipole antenna,
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this is the H plane pattern this is the E
plane pattern also known as this one is known
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as azimuthal pattern, this is known as elevation
pattern.
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I also want to mention infinitesimal antenna
is not at all an efficient antenna. So, please
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do not try to use infinitely small dipole
antenna ever.
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Now let us just see a case of a small dipole
antenna. Generally a small dipole antenna
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is defined between lambda by 50 to lambda
by 10. So, that is the length of the dipole
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antenna. Here again as before x axis y axis
dipole is placed along the z axis here and
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the current is again not assumed uniform.
Now, the current will be 0 over here and it
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will be maximum. So, if the dipole antenna
length is small this will be the triangular
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distribution. As the length increases reaching
to this, we can still think that it is a triangular
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distribution.
I just want to tell you if the length is increased
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further; for example, if the length becomes
lambda by 2. Now for lambda by 2 the variation
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will not be triangular, but it will be actually
a perfect sinusoidal waveform like this here.
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Why? If the length this is lambda by 2; that
means, the current will satisfy the boundary
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condition of 0, sinusoidal value here which
is maximum and then it is going to 0 and that
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will be if we know that a sinusoidal waveform
let say goes from 0 will go to a maxima then
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go to 0 and then it will repeat the cycle.
So, this distance from here to here will be
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then lambda by 2 and from here to here will
be if it is sinusoidal it will be lambda by
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4.
So, when we discuss about a lambda by 2 dipole
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antenna will not be assuming a triangular
distribution, but we will be assuming a sinusoidal
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distribution, but any sinusoidal distribution
for a shorter distance can be approximated
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as a triangular distribution. So now, once
it is a triangular distribution, now we can
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write the current.
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So, we can write the current now in the triangular
distribution form. So, I 0 is not constant
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anymore it is varying along the z direction.
So, this is for 0 to l by 2 and this is for
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minus l by 2 to 0; that means, this is above
a origin and this is below origin or the central
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point there. And so from here again we can
find out the value of vector magnetic potential
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by integrating over the length. However, just
to make life very simple if we actually look
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at we are taking the average over the length.
So, we can actually assume very similar to
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what we had done for the previous case except
that now I 0 will become I 0 by 2 which is
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the average value of a triangular distribution.
So, that is why E theta, E phi all the components
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remain the same, so the difference is only
that now I 0 is nothing but I 0 by 2, so earlier
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it was 4 pi r now it is 8 pi r. So, otherwise
E theta E r remain exactly the same as before.
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And then we can actually see also similarly
H phi component can be there in between other
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components are 0, and from here if we take
the ratio of the two we will still get the
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equal value of eta which is 120 pi.
Now from here we can actually find out what
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is the radiation resistance, but before that
we can find out the power radiated. I will
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just go back to slides and the power radiated
just to tell. So, what is power radiated is
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nothing but I 0 square r divided by 2; I will
define what is r also, first just let us look
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at what is power radiated. So, power radiated
can be found out by using this particular
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expression here, and since there is only one
component now so E cross H conjugate will
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actually be replaced by magnitude of E square
divided by eta, because H is nothing but E
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divided by eta which is 125. So, by using
this we can find out what is power because
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we know what is E and H field; and then power
is now given by half I 0 square times R r.
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Now, what is this R r? I just want to mention
here that R r is actually nothing but a radiation
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resistance it is not a physical quantity at
all, it is something like from circuit point
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of view. When we say power radiated, power
radiated from antenna point of view is a fantastic
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thing it is radiating in the free space; that
means antenna is radiating. But from circuit
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point of view what we think the radiated power
is actually a power lost. So, just a circuit
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representation is that power loss can be represented
as I square r by 2.
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00:20:17,929 --> 00:20:25,830
So, that is the R r is nothing but it is known
as a radiation resistance, but this quantity
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is not a physical quantity it is just a representation
of radiated power which is considered as a
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loss power from circuit point of view. So,
just by integrating that this term comes out
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to be 20 pi square l by lambda square.
Now I just still want to go back a little
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bit over from here to the back side. Please
remember this expression has a I 0 square.
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00:20:51,710 --> 00:20:58,100
Now here is the radiation resistance which
is equal to 80 pi square l by lambda, this
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is for uniform current. And now you might
wonder why I did not talk about this here,
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00:21:04,799 --> 00:21:13,299
because in reality in no situation for a dipole
antenna for very small dipole antenna current
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00:21:13,299 --> 00:21:20,059
will be 0 at the end and current will be maximum
or triangular distribution. So, in general
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you should not be using this formula at all,
but you can actually say that this value is
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about 4 times more than the expression which
I showed you for a triangular distribution,
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because I is here increased by 2 times because
there it is triangular here it is uniform;
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so 2 times current increase means 4 times
radiation resistance will increase.
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But please do not use this particular expression
even for infinitesimal dipole, I always recommend
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people to use this particular expression over
here. So, this is the expression to calculate;
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what is the radiation resistance. So, I have
taken a few example here: let us say if the
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length is equal to lambda by 10. So, if length
is lambda by 10 we can substitute over here
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00:22:09,570 --> 00:22:17,850
lambda by 10, so this term will become actually
approximately 1 by 100. Pi square is approximately
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10, so 20 into 10 200 by 100 will give rise
to 2 ohm resistance.
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00:22:22,929 --> 00:22:30,049
Now that is a very small resistance, and if
we try to feed this antenna with let us say
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a 50 ohm line well you know that power radiated
will be very less, most of the power will
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get reflected back. So, just for the example
I have taken l equal to lambda by 4. Please
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00:22:42,450 --> 00:22:48,440
remember lambda by 4 does not fall under the
category of small dipole, I have just taken
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00:22:48,440 --> 00:22:54,080
this as an example to show you what will happen
if we use this expression here. So, if you
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use this expression for lambda by 4 simplify
that comes out to be about 12.3 ohm. And just
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for as a discussion again if you think length
is equal to lambda by 2; that means if the
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length is increase by 2 times this should
increase by roughly 4 times, so that will
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become approximately 50 ohm.
But now this is the calculation for radiation
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resistance, but dipoles also have reactive
impedance. In fact, all the antennas have
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real part as well as reactive part. So, how
do we calculate the reactive part? So, for
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that what we will do will actually think about
again a dipole antenna and just to look at
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this here. Now there are two arms are there;
there is a one arm going up one arm is going
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down here, and this is actually has to be
fed with the current from here. So, what we
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really need is a, a current going in here
which will be let say in this direction and
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then the current will be going like this.
And it has a reverse current here which is
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going in this, so current will be flowing
in this direction.
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00:24:05,789 --> 00:24:10,789
So, you can actually see that current flows
over here and goes like this, and over here
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00:24:10,789 --> 00:24:18,570
current flows from here and goes up here.
So, this one here really needs a balanced
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current here. So, we need a source which has
to have a balancing. Now for analysis point
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of view we can actually think that of it is
a symmetrical thing, so this is a plus here
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there is a minus. So, we will try to analyze
this portion over here. So, this portion will
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be nothing but equal to the length will be
now equal to l by 2, and there is an open
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circuit here so we can think about a transmission
line which is an open circuit over here. So,
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by using this concept we can find out what
is the reactive part.
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So, for that we will actually take this as
a transmission line. The transmission line
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has a characteristic impedance of Z 0, length
is l and it is terminated in a load Z L. For
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this transmission line we know how to calculate
input impedance. So, input impedance is given
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by this particular expression here; I am sure
most of you might have studied in electromagnetic
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00:25:19,840 --> 00:25:25,240
waves course or electromagnetic field for
a transmission line one can find the input
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impedance. So, I will just take a few cases
here: Z L equal to 0, what Z L equal to 0
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implies here? This will be short circuit.
So, if it is a short circuit we put Z L equal
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to 0 here, Z L equal to 0 here. So, this expression
gets modified to j Z 0 tan beta l.
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00:25:46,230 --> 00:25:52,210
Now this is the case which we are interested
at this moment which is Z L equal to infinity.
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Infinity means this is open circuit. So, right
now in our case for a dipole antenna this
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tip is open circuit. For that we can calculate
Z input as given by Z 0 divided by j tan beta.
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You can actually put Z L equal to infinity
here Z L equal to infinity here. So, in terms
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00:26:12,909 --> 00:26:18,980
of infinity this will be negligible and again
in terms of infinity this term will be negligible.
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00:26:18,980 --> 00:26:25,649
So, these two will be negligible, what we
will left with is infinity divided by j infinity
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00:26:25,649 --> 00:26:31,950
tan beta l and that is the term coming over
here. Now if Z L is equal to 0 that is what
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we would like from antenna point of view the
load impedance should be equal to characteristic
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00:26:37,460 --> 00:26:44,049
impedance in that particular case we can say
that if you put Z L equal to Z 0 here this
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00:26:44,049 --> 00:26:50,360
numerator and denominator will be same Z in
will be equal to Z 0, which is not dependent
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00:26:50,360 --> 00:26:55,630
on the length of the antenna.
So, what is beta here? It is equal to 2 pi
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by lambda. I just want to highlight here that
many books use beta or some books use k also
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so k is equal to beta is equal to 2 pi/lambda.
So, let us just take a case when length is
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less than lambda by 4, because we know that
we are talking about its small dipoles. So,
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00:27:15,889 --> 00:27:21,669
length will be definitely less than that.
So, for this lambda by 4 if you put here beta
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00:27:21,669 --> 00:27:29,120
is 2 pi by lambda this will be lambda by 4;
that means, tan beta l will be always positive.
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00:27:29,120 --> 00:27:36,010
And if it is always positive; that means for
short circuit if this term is positive this
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00:27:36,010 --> 00:27:43,630
can be written as z input equal to j omega
l; that means it will be an inductor. That
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00:27:43,630 --> 00:27:51,130
means, any small transmission line which is
shorted at the end will represent inductor.
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00:27:51,130 --> 00:27:57,350
And for open circuit Z L is infinity. So,
that we can see from here if it is infinity
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00:27:57,350 --> 00:28:04,480
and this is positive, this will be capacitive.
That means, a open circuit line here which
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is small will have a capacitive thing. So,
a small dipole antenna will actually in a
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00:28:10,649 --> 00:28:18,950
reality will have or the small radiation resistance
along with a capacitance which is associated
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00:28:18,950 --> 00:28:24,350
with this transmission line.
So, we will continue from here in the next
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lecture, where we will see that how lambda
by 2 dipole antenna is designed, how we can
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00:28:31,629 --> 00:28:38,009
actually even use higher order modes of dipole
antenna, how to calculate the half power beam
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width of the dipole antenna, and then folded
dipole antenna.
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Thank you very much; we will see you next
time.