In this module we will discuss wave propagation
inside that of a imperfect dielectric and
wave propagation inside a good conductor Wave
propagation in good conductor will need us
to discuss about skin effect something that
is very important at high frequency wave propagation
Waves in imperfect dielectrics is something
that we would actually see when you drive
a capacitor with high frequency signal but
then the insulator that is filling the material
between the capacitor plates is not perfect
So if the material filling the capacitor plates
is not perfect then there is a possibility
that there will be some leakage current from
one plate of the capacitor to the other plate
of the capacitor leading to loses in the form
of dissipation of heat inside that of that
imperfect dielectric
So we will like to see what happens to t his
loss or how we are going to calculate this
loss which is actually material constant dependent
as well as what is meant by skin depth or
what is meant by skin effect when we consider
wave propagation inside a good conductor So
we begin as usual that we began for the case
of a lossless or a perfect dielectric wave
propagation
.
So we begin with Faraday’s law and Ampere
Maxwell Law So Faraday’s law is straight
forward Curl of electric field is given by
minus j omega mu H Clearly the way I have
been writing this equation implies that I
am assuming that the time dependence of all
the field quantities in this wave propagation
with all be in the form of e to the power
j omega t So because of this I can actually
suppress the time notation and release that
del by del t can be replaced by j omega
So with this assumed time dependence then
Faraday’s law becomes curl of electric field
being equal to minus j omega mu multiplied
by H okay So although I am writing mu here
in most cases we will be interested only with
mu is equal to mu that is we will be interested
in the cases where the material medium is
explicitly not magnetic okay So we will not
look at wave propagation inside a magnetic
material that is something that is beyond
the scope of this particular course
So although we right knew what you should
realize that mu is equal to mu zero So with
this this becomes Faraday’s law and then
what happens to Ampere Maxwell law Ampere
Maxwell law is curl of H is equal to J plus
J omega epsilon E Now in the case of a lossless
wave propagation that is propagation inside
a lossless material on a perfect dielectric
we assume j is equal to zero
So j is equal to zero for a perfect dielectric
because there was no possibility of having
any conduction charges free so that they would
constitute to a conduction current density
J However inside that of an imperfect dielectric
we cannot assume J is equal to zero because
if you assume that then there is nothing different
from the lossless case and we will not be
able to capture the fact that an imperfect
dielectric can actually conduct some amount
of electricity
No doubt that electricity will be small but
it can conduct some amount of electricity
in the form of a leakage current So in order
to take that imperfect dielectric into account
we need J be nonzero So J is nonzero for an
imperfect dielectric This situation of J not
equal to zero will also describe equivalently
the good conducting medium such as a metal
So in a metal you will find lot of conduction
electrons
Therefore J vector the conduction current
density will be very high or will be at least
quite appreciable comparable to the displacement
current in a good conductor On the other hand
if the medium is imperfect dielectric mainly
then conduction current will be there but
it will be very small in magnitude okay In
any other medium the relative magnitudes of
conduction current and displacement current
determines whether that medium can be considered
as an imperfect dielectric or as a good conductor
So in any case we will assume that J is nonzero
in order to capture both the imperfect dielectric
as far as good conductor effects So we have
J nonzero but we also invoke the fact that
J is linearly proportional to electric field
in the form of Ohm’s law for the field expressions
that we described Earlier J was equal to sigma
times E except in this case that sigma cannot
be infinity Sigma is a finite value okay
So if you now substitute for J in the form
of sigma E Ampere Maxwell law becomes curl
of H is equal to sigma and electric field
E is common in both terms So I can actually
put that outside So I have sigma plus j omega
epsilon multiplied by E Now as before we take
curl of Faraday’s law to obtain del of del
dot E minus del square E equals minus j omega
mu curl of H but I already know what is curl
of H curl of H is nothing but minus j omega
mu multiplied by sigma plus j omega epsilon
multiplied by electric field E
Now we have Gauss’s law which states that
del dot d is equal to zero We will continue
to assume that D and E are related by a simple
epsilon okay So that is how we are actually
able to rewrite del by del t of d in the form
of j omega epsilon E So we assume that d is
equal to epsilon E and moreover we still assume
that del dot D is equal to zero So there might
be cases of conduction current in an imperfect
dielectric but we will assume that at no point
inside the imperfect dielectric surface there
are some isolated free charges
So because of that del dot d equal to zero
still implies del dot E equal to zero and
as such this term del of del dot E can be
cancelled can be made equal to zero There
is a minus sign to the right hand side minus
sign here on the left hand side They can be
cancelled each other so that what we get is
an equation for describing how the electric
field E will change inside an imperfect dielectric
being given as j omega mu multiplied by sigma
plus j omega epsilon times electric field
E
.
As before we will assume that electric field
has only one particular component So let us
assume that electric field E is ex component
electric field and it is a function only of
z variable right So because the way it is
propagating along z direction So in our assumption
this del square vector lapacian can be replaced
by del square by del z square times only Ex
of z which is equal to j omega mu multiplied
by sigma plus j omega epsilon times x
Remember in the case where sigma was equal
to zero this equation was actually minus omega
square mu epsilon and we recognize that mu
epsilon is nothing but 1 by mu square and
therefore we wrote this as minus omega square
by v square multiplied by E And we call this
omega square by v square as k square So therefore
we actually had a very simple solution for
Ex which was in the form of E power minus
j kz and k was a real quantity
This was for the lossless case or a perfect
dielectric case So this was the case of a
lossless perfect dielectric wherein the way
it was actually going as a nice cosine wave
whose amplitude was not really changing with
z in the sense that the amplitude of the electric
field would actually remain constant It would
not change with as you go further There is
no change in the amplitude of the electric
field
It was only propagating as a sinusoidal wave
along z direction as you can see in this particular
case This was for the lossless case when sigma
was equal to zero However now sigma is certainly
not zero and this quantity can still be written
as sum k square except now that k must become
complex correct Because k square is equal
to j omega mu multiplied y sigma plus g omega
epsilon
However k will then be equal to square root
of j omega mu sigma plus j omega epsilon okay
And because of this this is obviously going
to be a complex number This complex number
is usually split and written as alpha plus
j beta where we call alpha as attenuation
constant or attenuation co-efficient And this
is usually measured in nepers per meter We
call this beta as phase constant and this
is usually measured in radians per meter
So what is happening out there We have an
equation which is looking very similar to
at the left hand side is exactly equal to
that of a lossless wave propagation but the
right hand side I have k square which is actually
a complex number right So k square is a complex
number and k will be equal to alpha plus j
beta The solution of this equation would be
that you have two wave sum Ex plus e to the
power minus kz because k is complex I cannot
just write this as E power minus j kz I have
to write this as minus kz
Plus there would be some backward propagating
wave This backward propagating wave could
be generated by a source that is kept or it
could be coming off from some reflected wave
okay So if you do not want to go any concentrate
on the backward propagating wave at this point
do not have to consider this one We will simply
assume that the wave is propagating along
plus z direction and it is simply going out
in that direction
Now if you look at this expression that we
have written here You will see that this is
e power minus kz but substituting for k in
the form of alpha and j beta you will see
that this would be so let us put a zero substitute
to indicate that this is amplitude measured
at a particular plane So you have Ex0 plus
E to the power minus alpha z e power minus
j beta z And if you want to find out what
is Ex as a function of both z as well as time
you can then convert this phasor into a real
time variable form
So you will get Ex0 plus E power minus alpha
z I am assuming that this amplitude is still
real times cos omega t minus beta z right
So this equation for the electric field component
of this plane wave propagating inside an imperfect
dielectric is very similar to that of a perfect
dielectric propagation except that the amplitude
is actually decaying as a function of z So
you will see that this exponential minus alpha
z is there
This would be something like a current decaying
in that of a RC circuit There we will also
have a similar alpha there or rather in terms
of tau the time constant there So here we
have an alpha So when z becomes much larger
than as z increases the corresponding value
of e power minus alpha z actually starts decreasing
and therefore the amplitude starts decaying
So if you actually plot the amplitude alone
or the magnitude of the electric field alone
you will find that this would be the magnitude
of the electric field you are plotting and
you would find that at z equal to zero it
would have the value of Ex0 plus and beyond
that it would have a value that is coming
through this But as z increases as you move
away from the source you will see that the
amplitude actually starts decaying with a
slope of alpha
So the initial slope of this line would be
alpha and this alpha tells you the rate of
decay of the amplitude of the electric field
So this rate of decay of amplitude is measured
in Neper per meter Sometimes this alpha or
most of the time this alpha attenuation co-efficient
is actually given in terms of dB per meter
or dB per kilometer and there is a nice conversion
between neper per meter to dB per kilometer
something that I do not have enough time to
discuss that
But you can actually take a look at the textbook
to see how to convert from natural units of
attenuation co-efficient in the form of neper
per meter into the more practical units of
dB per meter and you can use this one and
you can see how the amplitude is actually
changing The point is the amplitude changes
or amplitude decays exponentially as you move
away from the source
Okay so the lesson that we need to learn is
that in an imperfect dielectric j is not zero
okay In fact j will be equal to sigma e and
this would not be equal to zero The second
lesson to learn here is that the waves actually
attenuate right as they begin to move along
z If they are very far away from the source
the wave would still be oscillating with a
propagation constant of beta or with a phase
constant of beta
So this beta which tells you how the phase
is changing as the wave propagates that would
still be there Except that the amplitude actually
starts decaying exponentially inside that
of a imperfect dielectric
.
So far what we are looked at is propagation
of wave inside an imperfect dielectric and
we have seen that this k which is the propagation
co-efficient or the propagation constant is
actually complex consisting of real part and
imaginary part The real part being the attenuation
and this imaginary part giving you the phase
change or the phase shift of the wave as it
propagates through the material medium
There are couple of cases that we can consider
at this point which will tell us the relative
magnitude of sigma and omega epsilon First
consider the case where sigma is much much
larger than omega epsilon This would be the
case of a good conductor that of a metal for
example You take copper or gold so these metals
will have a large conductor current compared
to a small displacement current
Therefore one can make this approximation
of sigma being much larger than omega epsilon
and when you put that back into this expression
for k you will see that k is equal to square
root of j omega mu sigma This can be rewritten
as by taking the square root of j can be rewritten
as omega mu sigma by 2 This square root actually
covers everything and then there is 1 plus
j If you are not convinced with this one you
can actually a simple couple of seconds to
convince yourself
What you want is root of j right So root of
j is nothing but square root of e power j
pi by 2 but this is nothing but e power j
pi by 4 which is 1 plus j by root 2 because
e power j pi by 4 is cos pi by 4 plus j sin
pi by 4 and cos and sin pi by 4 are equal
to 1 by root 2 So this root 2 goes into this
omega mu sigma But I also know that omega
is equal to 2 pi f Therefore I can write this
k as 1 plus j multiplied by pi f mu sigma
under root
So clearly in this case what we have is alpha
being equal to beta both given by pi f mu
sigma okay So what will happen to the electric
field that is propagating Well the electric
field will be Ex of z given with some initial
amplitude Ex0 I am assuming still only forward
propagating wave and then you have e power
minus alpha z e power minus j beta z so both
of these are equal to square root of pi epsilon
sigma
.
But rather than talking in terms of alpha
for a case where you are considering a good
conductor you will actually introduce a different
quantity called skin depth Skin depth is given
by 1 by alpha and since alpha is given by
square root pi of mu sigma this skin depth
delta let me write this down this as and say
this is skin depth delta is given by 1 by
square root of pi f mu sigma
So because of this the electric field propagation
inside a good conductor is equal to whatever
the surface value that you have or whatever
the initial amplitude Ex0 that you have times
e to the power minus 1 plus j z by delta correct
Its amplitude is going as e power minus z
by delta and its phase is changing as e power
minus j z by delta okay
So now we will not discuss about the phase
because phase only changes a phase is not
very important at this point but what we are
interested is to see what would be the behavior
of this e power minus z by delta So you have
e power minus z by delta and if z is around
say 4 or 4 delta and beyond this one so if
this z becomes greater than 4 or 5 delta that
is as the way it propagates if you are propagate
for more than 4 or 5 delta then the amplitude
would have decayed to less than or equal to
99 percent of Ex0
So this is how the amplitude would have actually
decayed So this decay if you look at this
this is about 4 delta or 5 delta So 4 delta
or 5 delta along z the amplitude would have
actually decayed by 99 percent So which means
that the value of this point would only be
about 1 percent of Ex0 okay So this would
have decayed by 99 percent and therefore this
would have been around at 1% okay
This would happen for 4 delta or 5 delta That
is why this rule of thumb which says that
5 delta units is sufficient for the wave to
have been decayed So the wave would just decay
and become very small compared to what its
value at surface is All right this is skin
depth We will come back to skin depth and
some of its ramifications very shortly Let
us for a minute consider the other case around
We will come back to skin depth in the next
module So for now let us simply consider the
other extreme situation where sigma is much
much smaller than j omega epsilon This would
be the case where it is a very very good dielectric
So this is a very good dielectric that I am
considering But there is some amount of loss
So there is some small insignificant conduction
current which we can neglect in the form of
j omega epsilon I mean in terms of the displacement
current So what will happen to k
k will be square root of j omega mu j omega
epsilon which will give you j omega square
root mu epsilon right This would be equal
to no attenuation and full phase constant
okay This would be very similar to this is
exactly similar to that of the lossless case
So in the lossless case or in the very good
dielectric if we can neglect the conduction
current then the propagation constant will
be imaginary right
k will be equal to j beta and the wave will
only suffer phase shift as it propagates through
the material and there is no change in the
amplitude It is also interesting to look at
what would be the impedance eta right I already
know from Faraday’s law that minus j omega
mu H will be equal to curl of electric field
therefore H will be equal to minus 1 by j
omega mu curl of electric field
If I assume that electric field E is along
x and it is going as e power minus alpha z
e power minus j beta z with some initial amplitude
of Ex0 right if this is how the electric field
E is going then H will have only y component
and this will be given by H is equal to y
hat and then you have Ex0 e power minus alpha
z e power minus j beta z but this curl of
E the y component is nothing but del Ex by
del z
So I can actually differentiate this Ex by
z and I will get alpha plus j beta which would
be minus sign here and then divided by j omega
mu correct So this is what I get There is
already a minus sign therefore that two minus
signs will cancel and they are going to get
away from this one This is nothing but the
electric field vector itself This is nothing
but the electric field vector E
Therefore this H which is given by E multiplied
by alpha plus j beta divided by j omega mu
it can be written in this way and therefore
this fellow should be in the form of an impedance
So E by H is impedance
.
Therefore the ratio of the electric field
to magnetic field in the case of this imperfect
dielectric is given by j omega mu divided
by alpha plus j beta Alpha plus j beta is
nothing but k and k is nothing but square
root of j omega mu multiplied by sigma plus
j omega epsilon and this can be further simplified
and written as j omega mu in the numerator
divided by sigma plus j omega epsilon
You can convince yourself that this expression
must be true because when you take sigma is
equal to zero this ratio of E by H which is
basically the ratio of the electric and magnetic
field vectors and describes the wave impedance
will actually reduce to mu by epsilon and
for the case of free space this will be equal
to square root of mu zero by epsilon zero
So this actually equation is correct and that
actually checks out with the case that we
all ready know how to compute the wave impedance
But there is actually a significant thing
that has happened over here For the lossless
case eta will always be real indicating that
electric field and magnetic field are in phase
with each other They actual carry some amount
of energy Whereas in this case eta is complex
indicating that E and H are not always in
phase with each other in fact there is a the
phase between the two is not zero indicating
that there is some loss in the power
So the entire power is actually not being
able to be carried because eta is not real
indicating that E and H will have some amount
of phase difference between that and only
the power of the order of cause of the phase
difference will be carried by the wave So
this loss is inevitable because of the imperfect
dielectric or a good conductor that we have
assumed
