Radiation And Circuits - Electrical Engineering Video Lecture - Electrical Engineering (EE)

hater speed of us love each other
you
good morning last time I hope I give a
fairly good feel for how an
electromagnetic wave enters a conductor
and dissipates and from the descriptions
I gave I hope you understood that an
electromagnetic wave when it meets up
with a material with Sigma not equal to
0 that it decays exponentially and also
propagates so it has both a real
component which gives it a wavelength
and an imaginary component that gives it
a decay and my conclusion from all this
was that material bodies like circuits
can absorb these waves and I just
suggested that if they can absorb them
they can also emit them now what I'm
going to do is change tags and go back
to looking at a material body by itself
so I'm going to now look at the problem
where I have for simplicity a bar a
plate the plate
these are my coordinates X and said this
is x equals zero so x equals a over two
and minus a over two and i have applied
an electric field in the Z direction and
I have a current I and I'm assuming that
this plate is long and wide that is in
the direction out of the board so that
the only thing that's present are
variations along X that is variation
along Y is zero variation along Z is
zero current is flowing in Z but it's
not a different distribution here then
here then here it's the same
distribution as you move in Z what I
don't know is how this current is moving
I assume that the applied electric field
which must be uniform because I assume
nothing is varying in Zed takes the form
the electric field e is some easy which
is a function of X alone Z I'm allowing
them to be functions of X because this
is the material and the material has a
conductivity Sigma it means that the
current density J is also in the Z
direction and it's equal to Sigma E
which is Sigma Xi said of X
now additionally I'm assuming both of
these are functions of time up to now
when we did this problem we did it in
statics now let's do it in time varying
systems so I am assuming e to the J
Omega T so it's a phasor so naturally I
would take the real part of the
imaginary part of this to get the actual
electric field and the actual current so
the statement I am making here is
assuming Sigma is a real quantity the
current and the electric field are in
phase now what I want to do is to
determine this function is e of X where
is I want to know whether the current is
flowing like this or is it flowing like
this or is it flowing like this
and you know the answer of course you
know it's the last one but I'd like to
get it out of the equations now when I
made this assumption that the electric
field is only in the Z direction I am
assuming that there is no electric field
that is in the X direction which means
that there is no oscillating motion of
charges which are breathing out outwards
and inwards it's a it's a good
approximation and it is true for the
systems we are looking at so I won't go
further into it but you can relax that
assumption and you'll find that this is
the interesting solution okay so now we
want to get the equation and the
equation is the same equation but let's
get it I have two equations one is
amperes law I have dropped the del D del
t term so I've gone back to looking at
AC low frequency AC systems and the
other is Faraday's law
now since I'm interested in the current
I am going to get an equation for the
current rather than for the magnetic
field or the electric field the way I do
it is of course to understand that J is
equal to Sigma e so I'm going to say
this equation is really curl of J with a
1 over Sigma outside is equal to minus
mu del H del T so I've substituted for E
as J over Sigma I can see that there's
another equation I can use to eliminate
this H so I'll take curl of both sides
so when I take curl of both sides I get
1 over Sigma curl of curl of J is equal
to minus mu del Del T of curl of H which
is equal to minus mu del G del T
now we are used to this Curl Curl
operator so we know what it is this is
equal to gradient of divergence of J
minus del square J is equal to minus mu
sigma del J del T but if you look at
this J is in the Z direction but depends
on X therefore divergence of J must
necessarily be 0 so this comes to the
fact that we are assuming there is no e
X there are special cases where ax is
present but we don't worry about them so
this term goes away and we are left with
the same equation the skin same skin
depth equation that we got last time
namely del squared J is equal to MU
Sigma which is J Omega mu Sigma because
J is goes as e to the J Omega T now
don't get confused between this current
density vector and square root of minus
1 I have my I ought to draw the write
this in a different way but I'm hoping
that you won't get confused so I want to
solve this equation which is the same
equation so we know that we already seen
one solution of it but now I want to
solve it in a different sense I want to
solve it for this geometry now since I
am driving current through this plate
there is no reason why what's happening
at this wall should be different from
what's happening at this wall I expect a
symmetric solution so I don't expect a
solution that will look like that kind
of solution is not going to happen
because I have electric field pointing
one way so the current can't point one
way to the left and one way to the right
so this kind of solution is not possible
all solutions that I get must be
symmetric now if I look at this equation
and substitute the fact that the only
variation is in X I get del square J Z
del x squared is minus J Omega mu Sigma
J Z is equal to 0 if you look at this
equation it's fairly clear that the
solutions are Exponential's so J Z is
equal to EI e to the power of square
root of J Omega mu sigma x plus b e to
the power of minus square root of j
omega mu sigma x
now this is perfectly correct but I want
to take advantage of the fact that I
know I have symmetric solutions so I can
write any combination of Exponential's
in terms of hyperbolic equations
hyperbolic functions so I'll write this
instead as some C cosine hyperbolic
square root of J Omega mu sigma times X
plus some D sine hyperbolic J Omega mu
sigma times X we did this earlier when
we were talking about bounded systems
and solving Laplace's equation cosh is
nothing but e to the J Omega mu sigma x
plus e to the minus j omega mu sigma x
over 2 sine h is the same thing with a
minus sign so using these you can always
construct these and using these you can
always construct these so it's the same
whether I write it with general
variables a and B or I write it with
general variables C and T the advantage
of doing it this way is that sine
hyperbolic is antisymmetric
and cos hyperbolic is symmetric
since I know the solution I'm looking
for must be symmetric I can throw away
the solution directly I don't have to
worry about it
the sine hyperbolic will look like this
it will be one sign for positive x one
sign for negative x so my solution in
jj0 X is some constant times cos
hyperbolic of the square root of J Omega
mu sigma times
now if you write this out square root of
J we've done this two three times now
square root of j j has unit amplitude so
it will be 1 plus j over root 2 another
way of saying this is j is equal to e to
the j pi over 2 so square root of j is
equal to e to the J PI over 4 and even
the J PI over 4 is nothing but 1 plus J
over square root of 2 because that's
equal to cos PI over 4 plus J sine
peyote and each of these is 1 over root
2 so that's where this comes from so I
can put that in there is equal to C
cause the 1 over root 2 I can take
inside so that gives us a familiar skin
depth 1 plus J times root pi F nu Sigma
times X
now cos a cos B the cost of a plus B is
equal to
just the addition rule for hyperbolic
functions so I'll substitute that in
here
so I will get that J is Z of X is equal
to some unknown constant C times cosh of
square root of pi f nu Sigma
gosh of J root pi f nu Sigma times X
plus C times sine H of pi f nu Sigma X I
hope I'm not making a mistake sine H of
T Phi F nu Sigma X I'm just applying
this formula here now cause of J
anything is equal to J cause of the same
thing sine h of j anything is equal to j
sine of that thing we can confirm that
cosh of JX is e to the j x plus e to the
minus J X so there's no J here sorry
/ - so that's nothing but cos sine H of
J X is e to the J X minus e to the minus
J X over 2 but to make it sign
I need a 1 over 2 J so I multiply and
divide by J so cost of JX is cos x sine
H of JX is J sine X so substituting all
that what do I get
I get that this is equal to C for shop
root pi f nu Sigma times X cos of root
pi f nu Sigma X times J signage of root
pi F nu Sigma X times sine of root pi F
Sigma X okay so where are we reached
what we're seeing here is that J is a
complex phasor now you want to take the
imaginary part of this phasor because
you started with a phasor representation
that either you have to take the real or
the imaginary part that you decide at
the beginning whether you are
representing in cosines we are
representing in science so you take one
or the other of these when you finally
obtain a real quantity but look at these
expressions there is a piece that
represents a growing or decaying
amplitude and there is a piece that
represents an oscillation so let's
assume we are taking the real part of J
as our real function if you plot this
quantity this is a over 2 this is minus
a over to this cosh remember that is 1
over Delta is really a function that
looks like this it grows very rapidly
and the scale length on which it grows
is Delta
cause on the other hand is an
oscillatory function and it results in
this frame function actually doing
something like that
in other words the amplitude the
envelope is given by cosh and the actual
phase is given by cos this is very
similar to what we caught last time
because last time we had a wave that was
coming in from the wall into the wall so
we started with an electric field along
the Z Direction magnetic field along the
Y direction and there was penetration
into the X direction and the wave
decayed exponentially now we have got a
different viewpoint of this problem
there is no wave anymore instead we are
driving current through a plate and we
drive current through a plate we get
exactly the same picture once again we
have a wave that's both oscillatory and
decaying except it has to be symmetric
because there are two walls and there's
no reason to prefer one wall over the
other and what you find is that the
current is mostly at the edges
so we found out what J is and since E
and J are proportional to each other
they've also found out what electric
field is now the question that should
occur to you is how can such a electric
field exist how can it be that the
electric field instead of being uniform
because after all I'm applying a uniform
Z Direction electric field how can the
electric field be concentrated out here
and not be strong elsewhere the answer
is that is Faraday's law in action your
electric field does not just come from
the applied voltage the electric field
also comes from del F del T so del H del
T so what can happen is if you look in
the middle the electric field is uniform
which means if I take curl of e is going
to be zero because whatever electric
field I get here is the same electric
field on this side
therefore my curl is going to be 0 since
the curl is going to be zero I must have
del H del T zero but I have an e to the
J Omega T so del Del T is not 0 so H
itself must go to zero so the magnetic
field is zero on axis since the currents
are creating this magnetic field it's
quite reasonable that the magnetic field
is zero at the center as you go further
away the magnetic field grows because it
grows you start having this effect
the curl hitch means that if I take any
loop like this magnetic field is zero on
axis magnetic field is not zero at other
values of Z and therefore there is a J
that J in fact so is the e that is
supported the result is you have a
electric field the electric field
creates a current this current creates a
magnetic field which in turn modifies
the electric field what does this what
is the effect of this modification the
magnetic field is into the board or out
of the board therefore what is happening
is the magnetic field being out being in
the Y direction causes a force a J cross
B force on all the electrons in the
material therefore the electrons the
moment the bare current are forced due
to this J cross B force to move outwards
the result being the current prefers to
go along the side of the wall leaving
the middle empty that is what we are
seeing here because we have self
consistently solved this problem we took
curl of H into account when we derived
this equation and when you have solved
this we have automatically solved also
the J cross B effect and therefore this
result shows that they elected the
electric field created a magnetic field
that magnetic field pushed the current
out and the current chose to stay on the
outside of the conductor let me remind
you last what we learnt last time we
learnt that if you take copper
there is a constant
0.066 meters divided by square root of f
this is the skin depth in copper so if I
am talking about 50 Hertz square root of
50 is around 7 so this is around 0.9
centimeters so however thick a bar I
have only the last centimeter on either
side is going to carry a lot of current
the rest of it is basically going to be
current free so this picture that I have
got
is exactly the same as what we did last
time there is no difference last time we
had a wave coming in and the wave decade
as we penetrated the material this time
instead of talking about a wave we just
look at the material itself and what we
find is that the current prefers to sit
near the edge and tends to have
oscillations as well in fact is exactly
the same solution whether you had the
wave or you didn't have the wave okay
now you can do the same equation if you
did with a round wire if you had a round
wire with a radius e you have the same
equations this equation is still true so
del square J is equal to J Omega mu
sigma j where j is equal to j z which is
a function of r but not of theta and not
of z and it's along the z direction but
we have to write out del squared now for
a cylindrical system del squared you can
look up your books is 1 over R del Del R
of R
so this is the operator in the our
direction there is no opera operator we
need to keep the remaining operators 1
over R squared del square del theta
square plus del square del Z squared of
JZ these two go away because we are
assuming just as we assumed for this
plate we are assuming that del del theta
is 0 del Del Z is 0 this is Theta this
is our and out of the board is Z so this
is the only part that remains of del
squared Z and it is equal to J Omega mu
sigma
it's almost the same equation let me
write it out 1 over r del squared del R
squared of JZ sorry there's no 1 over Z
plus 1 over R del Del Z of GZ minus J
Omega mu sigma j said is equal to 0 so
if you forgot about this term it's the
same equation essentially it is the same
equation so this is what cylindrical
geometry has done for you it has added
one new term because when you take this
del Del R and act it on R and you'll get
a 1 over R del J Z del R if you acted on
del J Z del R and you get this term
which is the original term
the solution of this equation is
actually in terms of what are called
Bessel functions the answers are j0 of
square root of J Omega mu sigma are some
a times this plus B times y 0 of J Omega
mu sigma ah the J's are like cosines the
Y's are like signs and more importantly
the Y actually blows up at R equals 0 so
similar to those arguments we throw away
this one and we keep only this and in
detail you have to work out what this
function looks like but if your wire is
large compared to the skin depth Delta
it turns out J 0 of J Omega mu sigma r
doesn't look any different from this
solution it looks in fact the same as
the solution so there is some difference
here
but this portion is identical whether
it's the plain Cartesian or it's
cylindrical however when Delta becomes
large compared to the radius of the wire
then you have to take this full
expression and you do work it out I
introduce this only to mention that the
cylindrical problem is no more difficult
than the Cartesian problem is the same
problem and in both cases what we are
seeing is a wave entering in D from the
side
to the middle now if we try to do across
H for this problem what do we find there
is an oscillation this oscillation does
not dissipate any energy it corresponds
to e cross H going in and out part of
the energy is getting dissipated due to
jarate
this corresponds to e cross H going in
inwards so you will see both these terms
present one part of the term will
correspond to dissipation and one part
of the term will correspond to energy
going in and out this is generally true
when you do pointing theorem so if you
look at the Poynting theorem expression
we had an expression that looked like a
cross H dot DL is equal to volume
integral various things if you put
phasers in here and we rewrite this
equation I will do it next time
but I am just anticipating your answer
what we will have is a real part and a
complex and an imaginary part so we end
up having to use a cross H star dot d s
the real part of this e cross H star
will give us the J dot e part so the J
dot e part is real part of e cross H
star which is
actual power transmitted the imaginary
part on the other hand represents
oscillations it represents reactance
we'll come back to these ideas little
later but it's quite important and when
I derive the phaser version of pointing
theorem it will become quite clear but
you have to understand here itself there
was both a decay due to the amplitude
and an oscillation due to the phase so
when you take across h and average it
what we find is there is a piece that is
matched in phase so the average does not
go to zero and as a piece that 90
degrees out of phase which does go to
zero and when you do the averaging the
the piece that goes to zero represents
reactance and the piece that doesn't go
to zero represents power transmitted
all right so we have come up with two
concepts so far one is if you have plain
waves and they reach a material whose
conductivity is not zero the waves get
dissipated they decay as they enter the
material second is when we look at such
a system we find this is the direction
of e cross H it's into the material now
I actually want something useful out of
this structure I want to make
I want to make a bulb aglow by having
radiation come on this plate
now bulk blowing is only an example if I
can do this then I can put an electronic
circuit here and make use of this
radiation for example on the other side
I could have a transmitter sending
information and the electronic circuit
here can interpret that into information
but what I really need is a way of
getting this wave energy not into the
plate but into the circuit why is that a
big difference well if you look at this
the energy of the wave goes into the
plate and becomes heat there was no
information transfer all that that
energy did was it made the plate hotter
he did not cause a current to flow in
the outside circuit for a current to
flow in the outside circuit you need
something more and what I want to
discuss for the rest of the lecture is
just what it is that you need and it's
giving going to give you a sampling a
taste of what goes into the theory of
antennas we have so far restricted
ourselves to del squared e is equal to J
Omega mu e plus Omega square nu Epsilon
I think there's a - I guess
you can check the sign of this term now
we have ignored this term and we have
kept only this portion and we have done
skin effect and analyzed what happens
inside the plate now the problem is if I
want a current to flow in the rest of
the circuit what do I need I need that
current flows up
I need that a voltage difference is
developed between the top and the bottom
which then appears as current in the
rest of the circuit now current flowing
up seems reasonable all I need to do is
to have my electric field in that
direction and then current will flow so
if my electric field wear upwards my
magnetic field the sideways that should
happen well there's only one problem if
the current flows up and a potential
develops what what does the rest of the
circuit see as the cause of this
potential the rest of the circuit must
see Coulomb's law as providing this
potential if Coulomb's law is going to
provide this potential then there must
be charge buildup so there must be
positive charge here and negative charge
here which is then driving a current
through the bulb or electronic circuit
that I've got there but the moment I
have positive and negative charge here I
have an electric field connecting these
positive charges and these negative
charges and since this is an oscillating
electric field
this positive and negative charge is
going to oscillate and therefore this
electric field is also going to
oscillate which means I necessarily have
shut del D Delta I cannot escape it if I
want to communicate with the rest of the
circuit it is the del D del T that is
going to allow me to communicate with it
what this did was it converted things to
heat it did not give me useful output
for useful output I really have to go
back to this term and it's only this
term that can make this plate look like
a battery to me now in a real system you
have both of these you have conduction
current and you have displacement
current so how do we know what what it
takes to make an effective and tena well
the way to look at it is the following
if you ask how much of my energy is
going into heating and how much of my
energy is going into making a battery
the answer is somewhat like this I want
to deliver energy here I have a voltage
source here and the voltage source has
an internal resistance which is
represented by this jayati and we know
very well that if you have a source and
an internal resistance and an external
RL that the maximum energy transfer is
transferred to RL when the internal
impedance is equal to the external
impedance if either R is larger or R is
smaller you don't end up delivering
power to RL so we require that arc
should be equal to RL of the same order
if I am going to effectively transfer
energy from my voltage source to my
electronic circuit so I need this j dot
e to be essentially representing the
same amount of energy loss as this RF
this is one kind of I square this is I
squared R and this is the other kind of
I is I squared order so this term and
this term should be alright
now the problem with this is that RL is
given to me it is something it is my
electronic circuit so when I take this
system the current that's going to flow
here has this IV characteristic V is
equal to I RL which means the amount of
charge separation I require is
determined by the external circuit
internally the amount of charge
separation I have depends on the ratio
of these two terms so how do i how do I
adjust this I mean I have two separate
physical phenomena in one case I have a
wave which is coming with its own
characteristics the other case I have a
circuit which has its own
characteristics and yet I want the wave
to talk to the circuit so I want to have
one more adjustable knob so that I can
adjust the ratio of the amount of jayati
produced to the amount of del D del T
produced now this is where antenna
theory comes
the picture I am going to draw is not
quite reasonable but it it is the best
way of looking at it
supposing I had my wire and instead of
connecting both ends of my plate I
connected only one end to my electronic
circuit my wave comes and when my wave
comes of course it causes current now
this current better be zero here because
it can't just leap into vacuum it is
stuck within a material it can be
anything it is here but it should be
zero there at the same time my charge
can be anything it likes here but the
charge whatever it is here is what is
going to determine the voltage that I
deliver
so ro determines V and J data means I
now supposing instead of heating this
antenna flat supposing the wave hit it
at an angle
supposing it hit it like this now you
can see what would happen is that this J
due to this part of the field has one
sign but due to this part of the field
has the opposite sign and due to this
part of the field as the same sign so
within this wire the J would start
having a shape
and similarly Rho would have the same
shape because divergence of J del J is a
del Z was minus del Rho del T so J and
Rho would not be constants they would be
oscillating functions of Z provided this
wave had not just propagation into the
wire but propagation along the wire also
now supposing this is possible what is
the shape that we would want ideally
well that's easy to answer because we
know that Rho determines VJ determines I
so and we know that V over I is RA so
what we would want he is a current that
started at 0 grew to something at the
same time because the current is 0 your
charge density is maximum charge density
is falling and you want sufficient
length so that the ratio of Rho 2 J is
whatever is required by Ohm's law in
practice this RL is very small that
means is that Rho should essentially be
0 and J should essentially be maximum
what does that require I have my wire J
starts at 0 and I want J to become a
maximum ro starts at maximum I want Rho
to become 0 so this distance if you look
at this this distance should correspond
to PI over 2 in phase
pi over 2 in phase means lambda by 4
because if you look at any wave this
length is lambda corresponding to 360
degrees in phase this is the length we
are talking about J goes from 0 to
maximum so this is lambda by 4 so if you
want to effectively couple from your
wire or your plate which is receiving
this electromagnetic radiation into
another electronic circuit then this
wire has to have a length which is
sufficient to change j from 0 to
something close to maximum the row from
something close to maximum to something
close to 0 which means that this wire
should be 1/4 of lambda now actually
there are two wires there is a wire and
the wire to ground so when you look at a
practical antenna you will see two wires
each of which is lambda by 4 which means
that the total antenna size is lambda by
2 and this is what is called a dipole
antenna the details do not matter what
is important to understand is if I have
a plate and the wave goes straight into
the plate exactly straight into the
plate with no variation along the wire I
am just depositing heat I am not giving
any information to the circuit for me to
give information to the circuit
I must permit variation along this wire
why because the ratio of J and Rho
intrinsic to the skin effect is not the
ratio of J and Rho that the external
circuit wants the skin effect gives me 1
ratio of J and Rho
and the external circuit requires
another one so I need to have a matching
and the best way of matching in fact the
only way of matching I might say is to
take advantage of a little bit of
variation in the wave along the
direction of the antenna so the wave in
fact varies along the antenna as well as
into the antenna how much variation is
required that is a practical matter it
turns out that external circuits can't
match very well out here they need to be
more like this where current is at a
maximum and charge is at a minimum so
the length of a antenna that matches
well is lambda by 4 and you need two
wires each lambda by 4 so the length of
any good antenna is lambda by 2 now
think about this you have your cell
phone and your cell phone is to 3
centimeters long the frequency at which
the cell phone works is about a
gigahertz and a gigahertz in vacuum is a
length of wavelength of 30 centimeters
which means lambda by 2 is 15
centimeters but your entire cell phone
is so small frankly I'm not sure how
people talk it it barely reaches out of
your ear how does a 15 centimeter lambda
by 2 antennae fit into this little
device what they do is to wrap this
antenna back and forth back and forth
back and forth inside this cell phone
details of those kind belong more to an
antenna course to a design of antennas
course so we won't cover it here but
what is important to understand is that
this fundamental requirement that you
must have variation along the wire in
order to useful
extract power or usefully radiate power
cannot be avoided it is the fundamental
requirement of all antenna design and no
matter how advanced of course you go to
the picture is not going to change you
still require that your current go to
maximum your charge go to near zero and
therefore you require something like a
lambda by 2 antenna
you