Phasor Form Of Poynting Theorem Video Lecture - Electrical Engineering (EE)

Oh
you
you
good morning last time we discussed how
the Poynting theorem and the skin effect
that we had been talking about for the
last few weeks has can be applied to
circuits so I took three cases one I
looked at current flowing through a
sheet a two-dimensional wire and looked
at how the current actually flows and
what we found was that the current flows
primarily a B walls this curve being
cosh of if this distance is K by 2
takes over Delta with some basic I not
so it's basically an exponential
provided a over to is much greater than
Delta
if a over to is not much greater than
Delta it has a more complicated shape
which is given by the COS hyperbolic
what this means is the following the
amount of power that is sent through
this portion of the wire which is 1
Delta wide has to do with I squared up
and if the current basically is
proportional to e to the minus X over
Delta which is what it is cautious this
divided by 2 for large arguments that is
to say cosh of u is equal to e to the U
plus e to the minus u over 2 so if u is
large it becomes e to the u over 2 so if
I substitute that X is much larger than
Delta I can replace the cosh sorry e to
the X over Delta divided by 2 now of
course there's an eye not there this I
naught is given by the total current is
equal to integral minus a by 2 a by 2 of
I naught cos x over Delta D X that is
there's current everywhere that the
current in here with a width DX is I
naught cosh X over Delta D X so if I
integrate from the left boundary to the
right boundary I get the total current
now integral of cosh is nothing but sine
hyperbolic so it becomes Delta I naught
sine hyperbolic of X over Delta between
a over 2 and minus a over 2
and if you look at sine hyperbolic sine
hyperbolic is an odd function it's a
function like this so at a over 2 it's
large and positive that minus a over 2
is large and negative so I subtract a
large and negative number from a large
and positive number I get twice the
large and positive number so it becomes
twice Delta I naught sine hyperbolic of
a over 2 Delta once again sign you can
be approximated sine hyperbolic of U is
equal to e to the U minus e to the minus
u over 2 which is approximately equal to
e to the u over 2 so if I want to know
what the current is at the edge what I
have the current as a function of X is
equal to I naught cos hyperbolic of X
over Delta I can substitute for cosh and
I can substitute for I not from here so
I get it's equal to e to the power of X
over Delta divided by 2 that's from here
divided by this I naught is given by
taking all of these to the denominator
there so I divided by twice Delta x sine
hyperbolic of a over 2 Delta which is
again e to the X over Delta divided by 2
so here with it
so you can see that there is an
exponential variation of the current but
this current is always going to be
smaller than e to the power of a over to
Delta let me take that exponential to
the numerator so what do I get these two
cancels with this two so I get that the
current is equal to I over to Delta
times the exponential of X over Delta
times exponential of minus ei over 2
Delta what I have done is I have taken
this a over 2 Delta to the numerator
where it became exponential of minus a
over 2 Delta this factor of 2 cancel so
I get I over 2 delta x this
now I can combine exponents and
therefore I get my answer that I of X in
a wire is equal to the total current in
the wire divided by 2 Delta times e to
the minus K over 2 minus X divided by
Delta
so it's an exponentially falling
function it Falls to 1 over D of its
value over a distance Delta and the
current density at the surface is given
by the surface current density is given
by I over to Delta now usually if you
look at any of these treatments you will
see I over Delta as the surface density
the reason why we have I over to Delta
is because the current not only flows in
this surface it flows on this surface
since there are two surfaces on which it
flows the amount of current flowing
through this surface is I over two so
that current divided by Delta is the
current density so there is a very high
current density even though the current
may not be very high the current density
is very high
now if you look at how much of the
current is flowing through which part
that comes from taking an integral over
different parts of this wire but we know
about Exponential's so if you have
exponential e to the minus X and you
plot it versus X you know that over a
distance 0 to 1 the amount of the
exponential is more or less 67% of the
total value in fact if you take the area
of this rectangle it is equal to the
integral of e to the X all the way to
infinity and you can easily see the
piece that we have left out is smaller
than the piece we have taken so more
than half of the current is in this
first Delta region if you take 2 Delta
that becomes even more now if you ask of
how much energy dissipation is happening
then that comes to I squared R and when
you look at I squared R all the energy
dissipation pretty much is happening
here because I squared goes as e to the
minus twice this and if you look at the
e to the 2x the curve would would look
like this
they were all there would be almost
nothing left beyond the work beyond the
skin depth so the as far as power goes
almost all the power loss is in first
skin depth this is why skin depth is
such an important concept the current is
there the power dissipation is there and
therefore we really need to understand
the skin depth and how it applies to our
problems very carefully and as we saw
two lectures ago the skin depth in
different materials depends on its
frequency because Delta is equal to one
over square root of pi F nu Sigma so it
depends on F so the larger the frequency
the smaller the skin depth for copper at
60 Hertz of 50 Hertz this skin depth is
less than one centimeter which is why in
power cables you have the ability to
avoid putting unnecessary copper in the
middle of the wire so you have seen this
I have drawn it two lectures ago and I'm
sure you've seen it in your power
lectures by packing copper cables
around this central strength member and
having current carried through these the
result is the full copper is used for
current conduction the central strength
member gives you gives the cable the
ability to avoid being distorted and
snapped had we taken a single copper
cable of the same total gage and what
would have happened is that the central
portion of this copper would not really
have carried any current only the
outside would have carried current and
we'd have wasted this copper as it is by
by putting all this copper around
another member we are in fact saving on
copper cost it's a clear-cut case of
skin depth in use okay I want to do two
things now in this lecture and the first
is I want to take the Poynting theorem
that I've been talking about and make it
more useful let me remind you what we
had derived we derived that surface
integral e cross H dot d S which is the
radiation flux out of a surface is equal
to minus the volume integral of energy
sources of that flux which is J dot e
plus h dot del V del T plus e dot is a
leader
now this is conceptually very satisfying
but it's very general also it is meant
for arbitrary fields E and H but in
electrical engineering regardless of the
field with its power or its power
systems or its communication we are
always dealing with phases of electrical
and magnetic fields so we'd like to
translate this pointing theorem to a
phasor equation by that I mean I'd like
to represent the electric field which is
a function of position and time
I'd like to replace it with a phaser
which is a function of position times e
to the J Omega T now just as last time
we are having a confusion between J and
square root of minus 1 so I'm going to
call this capital T but typically
current densities represented a small J
but to avoid the confusion I am calling
it capital J so your electric field is a
phaser multiplied by e to the J Omega T
your magnetic field is also a phasor
again function of position e to the J
Omega T similarly D similarly B I am
similarly J
so all of these are now represented as
phases now what you know that means is
that really we take the real part oh
these expressions and knowing that this
is how we are going to represent we want
the equation not for e H and J but for
the phases e hn J now if you look at
this left hand side I am talking about a
cross H so e cross H would really be
real part of the phasor e e to the J
Omega T cross real part of phasor h e to
the J Omega T now if I want to make this
useful I would like to say that this is
equal to real part of Z phaser cross H
phasor otherwise this representation is
of no use to me but if I do if I combine
these two expressions I don't get this
because I have e to the J Omega T times
e to the J Omega T so e to the 2 J Omega
T appears since I want it in without the
e to the J Omega T I take the complex
conjugate of the second term if I take
the complex conjugate of the second term
then clearly this will become complex
conjugate minus J Omega T so e to the J
Omega T's will cancel out and I will
have this quantity now I do not know
what use it is but at least it doesn't
have the phasor exponential so I am
going to try and work out what is the
pointing theorem when you apply it to
the phaser across h stop it may seem
rather artificial what we are doing but
it's actually very useful
so let me write out Faraday's law and
amperes law Colonel of the phasor e is
equal to minus del V del T but now del
Del T is J Omega so minus J Omega phaser
be curl of phaser which is equal to the
phaser J plus J Omega the phaser D so I
have these two equations and to compare
with them let me write down the original
equations which were curl of e is equal
to minus del v del t and curl of H was
equal to G Plus del V del T so the del
Del T is have become J omegas now as
before I want to work with he cross H
but you can see that I want to work with
a cross H star so let me write the
complex conjugate of that equation curl
of H star is equal to J star plus
complex conjugate of J Omega D the
complex conjugate will become the
complex conjugate of each of these
multiplied together so minus J Omega
just in case you didn't understand that
let's say this is D this is the real is
the imaginary axis when I multiply by j
omega effectively i am rotating by 90
degrees
the reason is real part multiplied by j
becomes imaginary part imaginary part
multiplied by j becomes minus real part
so the entire vector rotates by 90
degrees and of course is scaled by omega
now i want to take the complex conjugate
which means the real part is unchanged
imaginary part goes down now how can I
do that
well if I I want to reach this vector
because if J Omega D is pointing this
way I should leave the real part
unchanged I should make the imaginary
part negative so what I do is I say this
vector is equal to the complex conjugate
of D this is D star and J Omega is
pointing this way then this way would be
J Omega complex conjugate because again
imaginary part is changed in sign the
product of these two will be a rotation
by minus 90 degrees which brings me here
so J Omega D complex conjugate is equal
to J Omega complex conjugate D complex
conjugate and J Omega complex conjugate
is nothing but minus J Omega times
complex conjugate of T so that is what I
have written there so I have the two
equations I want
and I now want to work with across H as
before I want to look at divergence of e
cross H star and it's equal to del Del X
I OH
exelon ijk eat will J H little star K
where I've used epsilon ijk notation to
represent cross product I can pull this
out and write this as epsilon ijk del
ETA del X I HK + ej del HK product rule
when I take a derivative of a product I
take the derivative of each in turn so
now I just identify these as suitable
curl operations so the first one is I
rotate once so it's equal to epsilon K
IJ del EJ Del X I this is nothing but
curl of e times H stop
the second term I want I&K so I want J
as the first index Epsilon
JK I del HK conjugate del X I eg if I
look at this you can see that this H is
the second index and not the third index
so it's equal to minus curl of H to its
top I am going through this quickly
because exactly the same thing was done
for pointing here given that we've then
got this far the rest is easy
so my curl of e cross H star sorry
divergence of e cross H star is equal to
two terms the first term is the
conjugate of H dots the curl of E that's
this term plus sorry minus e dot the
curl of H conjugate
but now I can apply Faraday's theorem
and ampere steer amperes law so I can
apply this equation and I can apply this
equation so what I get is complex
conjugate of H dot minus J Omega V
twiddle minus e dot the cross product I
mean the curl of H is given by this
expression so J complex conjugate minus
J Omega D so I have an expression for
the divergence so I can apply the
divergence theorem and I get surface
integral phasor e cross Kaizer H
conjugate dot d s is equal to I can see
a minus sign everywhere so I'll take the
minus sign out volume integral e dot J
conjugate that a set of this term plus J
Omega H conjugate dot D and since I
pulled out a minus sign I have to keep
this minus sign minus J Omega P dot
d conjugates now we can apply our
knowledge that J is equal to Sigma E V
equals mu H and D equals epsilon E so we
put those in you get that this is equal
to minus volume integral Sigma e squared
because e dot e conjugate should be
conjugate so e dot e conjugates a Sigma
e squared so actually this should be
Sigma star come back to that then plus J
Omega times H square to the MU minus e
squared Epsilon
so this is the new equation we have
the phaser across H with H conjugated is
equal to minus of volume integral of
Sigma Y squared plus J Omega mu H
squared or if you like J Omega B squared
over mu minus J Omega epsilon e square
now you can look at this and I'll remove
this complex conjugate we'll assume that
Sigma and epsilon and mu are real
quantities this is Peoria and this is
also Peoria
so Sigma e squared is pure real because
whatever he may be he may be a complex
number but magnitude of e squared is a
real number
similarly magnitude of H squared is a
real number magnitude of e squared is a
real number so this quantity is real so
inside this integral is a real number
plus J Omega times another real number
so this is the real part this is the
imaginary part is equal to this across H
so we can say the real part of the
surface integral is equal to the real
part of this volume integral the
imaginary part of the surface integral
is equal to the imaginary part of this
volume integral now where does that lead
us you can look here and this is the
time non phasor version of pointing
theorem I'm going to write out the
phasor version underneath it the pacer
version is actually two equations it
says real part of phasor e cross beta H
star dot d s is equal to minus the
volume integral
Sigma squared DV and imaginary part of T
cross which is a star dot yes is equal
to again minus sign J Omega so the J
goes out because it's imaginary part
volume integral mu H squared minus
epsilon DV it doesn't look anything like
these equations you've got two equations
instead of one and there's quite a bit
of difference between these equations
and this equation let us see if we can
understand what they're saying and why
they are saying the same thing if I look
at the time dependence of the phasor in
time there's an e to the J Omega T so
the real part of phasor e is going to be
like a sinusoid because it's going to be
cos Omega T now you take cos Omega T and
take absolute value squared because you
want epsilon e squared you look at this
piece that piece is saying volume
integrals epsilon e squared over 2
now this epsilon e squared over 2 is
really volume integral epsilon cos Omega
T squared times this magnitude of the
phasor square if we now average this cos
Omega T squared over time this average
is 1/2 because we take the square of
this quantum Bo you get an expression
that looks like this and as you have
seen before the average of this
expression is 1/2 because this piece
exactly fits here so there are equal
bits above 1/2 and equal bits below 1/2
so the average is 1/2 so this becomes
epsilon mod e squared over 4 what we
have here is epsilon e squared and then
we have an Omega that stands for the DDT
so we can put this for inside you can
put this for inside and pull the four
out then this part represent represents
stoled electric energy and this part
represents stored magnetic now what
about this piece well this piece you
look at this function
we know that J dot E is really Sigma e
square but E is cos Omega T so we know
that e squared is really this phasor
amplitude squared but averaged in time
gives you a factor of two so this
quantity I will divide by two and
multiply by two this piece volume
integral of Sigma e phasor squared over
two is energy dissipated so now let's
try and read this new phasor equation
the original equation said that whatever
energy is leaving a surface through
radiation is because there is a battery
inside so negative J dot e or there is
reduction in stored magnetic energy all
there is reduction installed electric
energy those were the things that could
give me outward radiation now when you
and when I have a phasor it means that
the time dependence is cos Omega T so
the on average the stored magnetic
energy and the stored electric energy
cannot change because after a period two
PI over Omega it will come back to its
old value so the stored magnetic energy
the stored electric energy are just
cycling they're going to a maximum going
back to zero going to a negative maximum
going back to zero so these two
quantities do not change on average
however this can the real part of the
phasor pointing theorem is only talking
about the part that survives averaging
in time
and it says that the real part of e
cross H is equal to twice - twice the
dissipated energy what about the
imaginary part the imaginary part
doesn't care about the dissipated energy
at all it only looks at the electric and
magnetic energies and what it's saying
is that since the energy in the magnetic
and electric fields are periodic they
can only be one thing that's happening
half the cycle energy is in the magnetic
field the remaining half of the cycle
the energy is an electric field so the
electric field in the magnetic field are
exchanging energies and that is what
this represents the imaginary part
represents the fact that energy is
shifting back and forth between electric
and magnetic fields the rate at which
they're shifting come is important
because we are talking about del Del T
and there is a factor of four because of
the way that we are not talking about DC
phase rate of mode AC fields so the
imaginary part is what is called
reactive energy and heat is talking
about the fact that the rearrangement of
energy inside the volume and this
rearrangement shows shows up as a
fluctuating across B at the surface
there is no net have time averaged
energy going out energy goes out for
half the cycle comes back in so if if
you looked at a pictorial representation
of this phasor equation it will be like
this I have
a surface inside the surface
I have feels currents these currents for
example could be dissipating energy in
which case I would have energy coming
into the system from all directions so
this time independent or at least time
averaged Abell term in the across h is
the real part is equal to minus twice J
dot e averaged averaged in time and also
of course volume integrated so the real
part is referring to dissipation or
generation of energy what is the
imaginary part represented for example
supposing I had a circuit where a
capacitor and an inductor are exchanging
energy and let us say this is not a very
well insulated system so the flux lines
both the electric flux lines and the
magnetic flux lines are leaking in that
case there will be electric fields due
to the capacitor which will be
oscillating in time and there will be
magnetic fields fluctuating at the
surface because of the inductor these
fluctuating electric and magnetic fields
half the cycle will give me an e cross B
out but the remaining half of the cycle
they'll be giving me an e cross B in of
course they'll be in general directions
so they might be giving a V cross B out
this way and this way but whatever it is
it does not lead to a time averaged
across H
so if I time-averaged this equation and
require that B and H are B and E are
sinusoidal periodic functions I they
will just vanish because they will not
survive time averaging only this term
would survive but within each period the
amount by which I cross H differs from
its time average will be basically due
to this term and that is what you see
here the imaginary part of K cross H
star is equal to minus 4 Omega
multiplied by average stored magnetic
energy minus average store electrics
inner
and all these averages are in time
so the phaser form of the Poynting
theorem tells us a great deal and
therefore it is quite important to use
use it in your problems because it's
really giving you the information that
you specifically want this equation is
fully correct there is nothing wrong
with it it contains everything and it
contains much more than these equations
because it is valid even if the fields
are not phasers it's valid for a field
that's slowly growing from zero but
these are the useful expressions for
most of Electrical Engineering and
that's why it's important to know that
before I leave this topic let me just go
back and revisit stored electric and
stored magnetic energy
if you look at the phaser expression
either of these what do you see
let us go back here because we will need
to use this form for this derivation
what you see is that you have terms
which involve jayati terms involve del
Del T of B squared over 2 mu and del Del
T of e squared times epsilon over 2 now
before I go there let me just point out
one important point when we did this
derivation in phasers you will notice we
did not get a factor of two the moment
we work on this we find that we get del
Del T of B squared over 2 mu but there
is no factor of two coming out of the
phaser form the reason is that the
factor of two came because we try to
push the electric field into the time
derivative and the magnetic field into
the time derivative we were using the
fact that if you have any vector V dot
del V del T then this is equal to del
Del T o V squared over 2 however in when
you work with phasers the derivative has
become a multiplication and it's no
longer true that n factor of 2 is
required but a factor of 2 is still
there and that is why we have taken it
into account by putting a 1 over 4 here
1 over 2 because of cos squared and 1
over 2 because really the factor of 2 is
therefore stored energy and that's why
we came to a 4 Omega factor in these
expressions ok now let us look at this
this expression and let us try and
understand where stored magnetic energy
we've already done the derivation before
but I'd like to show it as a trivial
consequence of pointing theorem I'm
going to say I'm looking at Faraday's
law my amperes theorem is just curl of H
is equal to J so I don't have radiation
radiation was not yet happen before
Maxwell
so my pointing theorem says 0 is equal
to minus volume integral J dot E Plus H
dot del V del T plus e dot del V DV now
I can choose supposing I want to know I
have some currents J and I want to know
what is a stored magnetic field energy
that threads this J so I can carefully
introduce the J such that divergence J
always remains 0 if divergence J remains
0 then del Rho del T is 0 which means
Rho is here the electric field does not
have any source so I can drop this term
there is no electric field what does
that give me it tells me that whatever
amount of energy that is coming through
J dot e must be coming out of minus H
dot DB DT that is J dot e is equal to
minus
H dot del V 30 if I assume B is mu H at
the saying minus del Del T of V squared
over to you this is the energy
dissipated this is the expression in
terms of the magnetic field so I can
write integral 0 to T of minus J dot e
DT is equal to V squared over 2 mu final
minus V squared over 2 mu initial so
that is saying that the amount by which
the magnetic field energy increased is
nothing but G Rho T which is the energy
I put in so it is a confirmation that
this expression for magnetic field
energy is correct
pointing theorem essentially includes
all the earlier work we have done it
automatically includes stored magnetic
energy stored electric energy energy
dissipation and energy in the form of a
battery and all the circuit theory is
here and therefore naturally stored
magnetic energy is one of the things
that's there
the last topic that I want to take up in
waves is the problem of
boundary conditions
we have already looked at two such cases
one is perfect conductor and we found
that the wave arrived at the perfect
conductor and reflected no energy went
into the perfect conductor we also
looked at the case of poor conductor and
we assumed that the conductor was had
the same epsilon and mu as the other
side and then we said there is no
reflected wave it's all transmitted in
which case we found we have the skin
effect energy decayed over a distance
Delta and the wave propagates into the
material now I'd like to generalize this
I'd like to talk about a general
material and ask where what is happening
in it first of all let's look at a
dielectric
so I have a material with let's say mu
equals mu naught epsilon equals some
epsilon naught epsilon R and there is a
small Sigma
this is typical for example if you have
a capacitor plate with dielectric in the
middle there'll be a very small leakage
current it will be so small it may take
the capacitor a day to discharge but
there is going to be a small amount of
leakage current which means Sigma is not
zero it is small now what is the wave
equation in this system
well it's curl of curl of E is equal to
curl of minus del V del t minus mu del
Del P of curl of H which is minus mu del
J del T's minus mu epsilon del square C
del T Square on the left hand side this
becomes minus del square e we have
derived this many times now so there is
no need to spend time on the equation
now I am going to immediately assume
phasers so I'm going to assume that my
electric field is e naught e to the J
Omega T furthermore since I am
interested in looking at plane waves I'm
going to assume it's e to the minus J
cave there so I have the material this
is Z and the plane wave is propagating
in Z so the electric field is in the X
direction
magnetic field is no y direction
since it looks like e to the minus JK is
it what it really means is it's like
cosine so after a distance
Zed it goes to constant then it's
pointing downwards then it goes back to
constant then it goes back to pointing
upward so the electric field
it's a sign is a sinusoidal function of
Z it's also a sinusoidal function of T
so it has the same form e to the J Omega
t minus KZ the same form as f of Z minus
CT so this is where we started from
some lectures ago yesterday so if I
substitute all that in here I get K
squared e X is equal to minus J Omega mu
sigma e X plus Omega square mu epsilon
DX
now mu is mu not but epsilon is epsilon
not epsilon R and I have this additional
term earlier on we looked at this term
and say this is dominating this term but
now I am looking at a dielectric in a
dielectric Sigma is very small so I
cannot assume that this term dominates
Omega square mu epsilon in fact is the
other way around Omega square mu epsilon
dominates minus J Omega mu sigma so what
I am going to do is I am going to
combine these two terms I combine these
two terms I get Omega square mu epsilon
times 1 minus J Omega mu sigma divided
by Omega square mu epsilon I can now
take this piece as a complex correction
to my epsilon and if I do that and I
treat a dielectric as a system which has
some real and some imaginary part to its
dielectric constant you find that all of
plane wave theory that we have already
derived applies to a dielectric I will
complete the derivation next time and we
will see how it works out