# Lecture 47-Reflection of uniform plane wave

Hello and welcome to NPTEL mode on applied

electromagnetics in this module we will discuss wave reflection and transmission in the scenario

what is called as the normal incidence from one medium to another medium in the previous

module when discussing skin effect whether it was on the flat conducting material or

the region or whether it was a round wire we had seen that one of the ways to excite

the surface currents on imperfect conductor was to actually net a plane wave okay be incident

on that one in the previous model I showed you a picture where the region below this

was all a conductor. And the electric field actually had an X component

in this way so the electric field was in this way this was caused by some way if we can

actually take a plane wave and then make it fall on the imperfect conductor these electric

field lines will therefore generate a corresponding surface current and we have seen what happens

when there is a surface current at the at the surface between you know yesterday in

the in the case we considered was between air and an imperfect conductor so these surface

currents will then eventually has to propagate down they will attenuate and I know generate

I mean they will attenuate. And lead to what is called as the skin effect

in this problem that we discussed in the previous lectures we never really bothered as to what

happens to the fields in the incident medium is no because these are the incident medium

and this is the second medium so this wave was supposed to be incident from the first

medium on to the second medium and we know what happens to this wave in an imperfect

conductor as it propagates down it simply gets attenuated but what happens to the fields

in the first region that we have not investigated and that is precisely what we want to do in

this case. Okay we also consider you know because we

have already analyzed the imperfect conducting case or for imperfect conductor or a good

conductor case we will assume a perfect conducting material so for all Z greater than 0 right

so Z equal to 0 is the interface plane that separates the interface between air and a

perfect conductor and by perfect conductor we mean if s=8 okay so in that case and assuming

in the region one is an insulator we can have for example air as an insulator with its impedance

of 377 ohms or in general you will have a certain value of µ naught and an e I am assuming

that the first medium or the first region you know where the plane wave is traveling

is also non-magnetic. And therefore µ is equal to µ naught okay

so this is the problem that we are looking at today you have an electric field incident

and the magnetic field incident together of course EI and H I form an electromagnetic

wave so this is a uniform plane wave which is propagating along z direction it is propagating

along z axis and the propagation coefficient is given by ß I remember what ß is ß I

is vµ 0 e if the medium is in general insulator or it would be equal to ? into µ naught e

naught if we are talking about air okay and of course 1 by v of µ naught e naught is

equal to the speed of light or we will call this asup1 where you P denotes the phase velocity

and that one denotes that we are in the region 1 okay.

So this is the region 1 which is insulator region 2 is a perfect conductor with the value

of conductivity being equal to infinity so now this wave starts to this wave is propagating

from Z=-8 and then it approaches the Z=0 interface this is the Z=0 interface okay

and the vertical axis is along the x and y axis will be coming out of this particular

plane so this is how the y axis would be coming out okay so that is what we have what happens

to this uniform plane wave as it impinges on the good conductor okay in order to answer

that we need to know what is the boundary condition that is necessary that these fields

have to satisfy okay what is the boundary condition between a perfect insulator such

as air and a perfect conductor. So with the conductivity going all the way

to infinity what is the boundary condition well the total tangential electric field okay

must be equal to zero right so you have et1being the tangential electric field there that one

must be equal to zero and the tangential magnetic field must be equal to the surface current

density or the sheet current density that needs to be there clearly in a perfect conductor

there cannot be a et2 that is tangential component of the electric field now you will have the

magnetic field component so these boundary conditions must be satisfied but our problem

is not a static problem. Okay because our electric field E I if you

write it in the complete vector notation will have some amplitude let us call this as X

hat even okay and it is propagating along the z axis so you have to the power J O t

minus ß I into Z what is O here O is the frequency of the plane wave okay so is the

frequency of the plane wave and ß is related to that particular frequency by the expression

that we have written similarly the magnetic field for the incident wave will be along

the Y direction because the wave is actually propagating along the z axis Y cross H must

point in the direction of the propagation which is Z.

And the amplitude here will of course be even divided by ? 1 where ? 1 is the medium impedance

given by vµ naught divided by e naught for air or it is in general given by µ naught

divided by e 0 or e R the product of e 0 and e R is what we normally call as e okay so

this is the medium impedance it also has the same time and space dependence as does the

z-component okay now if this is all that is incident okay.

Now at Z=0 is the incident plane of inter phase which actually separates the insulator

and conducting materials right so all this region is conductor all this region is an

insulator so at Z=0 if I try to apply the boundary condition what will that boundary

condition tell me fore I it will tell me that x hat e one or rather X hat I can drop because

X hat indicates that to the medium the electric field is tangential.

And in this case it is tangential right so if this is the medium so you can look at this

one if this is the medium and the wave is propagating with the electric field along

my thumb direction and once it hits this one here right so once this hits the interface

which is this hand right then the electric field amplitude at z=0 this my hand is that

equal to zero the electric field because it is tangential see this thumb is tangential

to this interface right. So because of that is the boundary condition

tell me that tells me that this tangential electric field µ must vanish now how can

the tangential field vanished for all values of time it is one thing that it can vanish

at a particular time but the boundary condition is not dependent on the time it is independent

of time for all times the tangential electric field must vanish.

And the magnetic field which is this finger let us say okay such that X cross Y will be

along the z axis the magnetic field H will have to generate a sheet current a surface

current on this interface so to the left of this hand is the air and to the right of this

hand is the conducting material and this one separates the air and the conducting medium

and therefore forms the boundary. So clearly if I just have the incident wave

or the incident electric field and the incident H field I cannot satisfy boundary condition

for all time correct because evaluate this one at Z=0 do you get the tangential electric

field in the region one will be even ej?t okay or you can say that this is e1cos ?t

because fields have to be real okay and this if it is equal to 0 because there is no e

T to here there is no tangential component of the transmitted field there is nothing

to be transmitted on the second region so we have the condition that says that e1cos

? T=0 and if this has to hold for all values of T then there is only one trivial solution

which makes even equal to zero obviously that’s not what we want right.

So with only incident waves you do not satisfy boundary conditions so what should happen

then if you want to satisfy boundary condition well you will have to have a reflected wave

at this point okay so you will have a reflected field this reflected field must propagate

along minus Z direction because of course the incident wave is propagating along plus

Z direction this wave must propagate along minus Z direction okay. So let me tentatively write the electro reflected

electric field corresponding to the reflected uniform plane wave as having the same direction

as the incident wave so X hat and instead we will write this one else even – okay so

minus indicates that this is a reflected wave amplitude okay so let me also go back and

adjust the amplitudes for incident fields I will write them as even plus and even plus

so that I know plus indicates the wave propagating along the plus Z direction and a minus indicates

of a propagating along the minus Z direction. So what will be the frequency of the reflected

wave well because you know you can show by the application of the boundary condition

that suppose you postulate that this will have a different frequency let us say okay

and just show you that one so you have e J?t-ß Rz is the reflected field so in this region

now you have a reflected field which is propagating along the minus Z direction so this is the

reflected free our component and the propagation is along the direction along – z direction

okay. And the magnetic field of course will have

to have a minus y direction why should I have a minus y because x cross y will give you

the Z direction of propagation whereas X cross minus y will give you a minus Z direction

of propagation which is exactly the direction in which the reflected wave is propagating

therefore the reflected magnetic field HR will be along minus y direction and then it

will have even -/e1ej?t+ß RZ because the wave is propagating along minus Z direction

okay so these are the fields that I have written the question is can I make abhi different

from O Prime and what will be the relationship between ß I because ß is the magnitude of

the incident wave vector. And ß R is the magnitude of the incident

wave vector or the propagation constant right so what should be the relationship between

the two in region one which is where the action is now taking place because nothing is propagating

inside the second medium so in medium one where the action is taking place the total

tangential electric field okay will be the sum of incident and reflected fields right

evaluated at Z=0 so in the Z equal to zero interface the total tangential field let me

write this as e total X e1+ej?t+xe1-ej?t and this total electric field must then be equal

to zero why should it be equal to zero well boundary conditions tells you that this must

be equal to zero. Then you have a relation which now tells you

E1+ej?t=-E1-ej?t so on the left hand you have a sinusoidal wave you should take the real

part of it on the left hand you have a real cosine wave on the right hand side also you

have a cosine wave and these two frequencies if they are different then their amplitudes

will be equal only at a few points in time okay but you want this relation to hold for

all values of time so clearly if that has to happen the frequencies on the left-hand

side of this expression must be equal to the frequency on the right-hand side of this expression

okay therefore ?’ must be equal to ?. So there reflected wave will have the same frequency

as the incident wave okay. So the only conclusion now that I have left

is to tell you that even – which is the reflected wave electric field amplitude must be equal

to minus even plus okay so even – is equal to minus even plus therefore I can go back

and rewrite the total electric field as Xe1+Ej?t is also the same because the frequencies are

the same so you have E-jßiz-ejßrz said minus E power minus J sorry plus J ß are said now

what about ß I and ß are should they be different or should they be equal look at

what is ß I ß I depends only on the frequency O and in we have shown that ß I mean the

reflected field or the reflected wave also has the same frequency O.

And vµe or µ naught e naught because we are considering air in this example so none

of this actually depend on the direction right the propagation constant the magnitude of

the propagation constant depends only on the frequency and on the Constituent parameters

µ naught and e naught of the medium right clearly this must also be the case for ß

R because ß R corresponds to the propagation coefficient of a wave which is propagating

in the same medium as the incident wave its ß R magnitude must also be equal to O vµ

naught e naught so ß R is equal to ß I. Okay because ßR=ß I I can replace this R

subscript here with I and what is this expression in here exponential of e-jx-ejx is 2J sine

X correct this is a sin X expression and 2J

gets multiplied out there therefore the total electric field in the region 1which is now

the sum of incident and reflected waves is given by it of course has the same direction

as X but this is given by even plus we will write this as minus 2J okay sine ß I Z right.

So ß I is equal to ß R therefore it does not matter whether I use ß I said or ß are

said okay I have not written what the complex exponential factor ej?t because I want to

write only the phase or form of these expressions therefore I have simply dropped ej?t I can

drop it please remember I could not have dropped that one in this expressions because I have

assuming that they are different. So two phase I mean when I talk of a phase

or when I sum the phase or components they all have to have the same frequency so because

of this condition that we have proved I am able to remove the ej?t factor from this expression

and whatever that I am left out is the phase or of the total electric field so you have

a phase or electric field over here which we have shown that should be equal to -2jsinßix

I will leave the exercise we leave this as an exercise for you to find out what will

happen to the total magnetic field of course on the surface the magnetic field will have

to be such that the sum of the incident amplitude. And the reflected amplitude must be equal

to the total surface current out there but away from the interface what happens so on

the interface at Z equal to 0 I know that H I plus HR must be equal to the total surface

current j s but far away from the interface what should be the total magnetic field right

I will leave this as an exercise for you to find out what should happen to this one and

you will probably not be surprised but you will see an appearance of cause ß I into

Z somewhere over there I leave this as an exercise for you to find all right now we

have looked at the fields that are present at the region in region one away from the

interface do these equations surprise us fortunately these equations are not surprising to us.

Because very similar thing happened in one of the earlier modules when we talked of short-circuit

termination of a uniform loss less transmission line so you had a lossless transmission line

wherein you had an incident wave but then when you terminate that transmission line

with a short-circuit what happened there was a reflected voltage our reflected current

and the reflected voltage amplitude was- the incident voltage right so V R is equal to

minus V I in other words with a short circuited termination the reflection coefficient that

you obtained was equal to minus one right so that happened earlier and we know that

the fields actually form the standing wave and in this case there is no difference the

fields do form a standing wave. Because the incident wave is coming in the

reflected wave is going back at the same time but with a face of 180 degree opposite with

respect to the incident wave they both will interfere together to form a standing wave

pattern in fact if in this particular case a standing wave pattern will be the maxim

because the reflection coefficient will actually be equal to minus 1 right the magnitude of

the reflection coefficient is equal to unity in this case. So what is our lesson we can in fact think

of the space problem so you had a air right you had air with the impedance of ? 1 and

you had a conductor and we found that for a perfect conductor where Sigma tends off

to infinity right whatever the incident electric field that incident electric field was completely

reflected correct with a reflection coefficient of minus 1so because of this reflection coefficient

being minus one this conductor is equivalent to a short-circuit termination so this is

a short-circuit termination of a uniform lossless transmission line

as a uniform lossless transmission line and then terminated when you consider a perfect

conductor terminated by at its load or you can consider conducting material as its load.

And the impedance of that conducting material perfect conducting material is a short-circuit

now does it also make sense yes we have already seen that the impedance of conducting material

right is complex and it is given by vJ?µ/sso we have seen this one earlier of course there

is a plus J O e term but we neglect this plus J ? e term assuming that this is a good conductor

but now we are not even talking of a good conductor we are talking of a perfect conductor

for a perfect conductor Sigma goes off to infinity and therefore ? goes off to zero

so ? of a perfect conductor the impedance of the perfect conductor goes to zero of course

this also is true because the electric field in a perfect conductor goes to zero.

So you do not have any tangential electric field and therefore because of that reason

this electric field also goes to zero the impedance of a perfect conductor is a short

circuit so whenever you terminate this one you do obtain a short circuited kind of a

equivalent circuit of a transmission line so the problem that we had in terms of waves

can be re-written in terms of the transmission line problem okay now let us very quickly

jump to another scenario where I have a second medium of impedance ? to the first medium

having an impedance of ? one which means I am assuming that this is also insulator again

I will assume that this is a perfect insulator. So which means that Sigma will be equal to

zero the same condition of Sigma equal to zero is true even in the region one okay so

this is region one and this is region two as before I have an incident electric field

which will have an amplitude of even okay propagating along the z-axis this would be

my z-axis as before this vertical line is along the x axis and y axis will be coming

out of this plane okay so I have the magnetic field which has an amplitude of h1 plus and

is along the y axis of course h1 plus is nothing but even plus divided by ? 1 ok we know in

this case that there has to be reflected field otherwise you will not be able to satisfy

the boundary conditions okay you can show that by trying to assume that when there is

an even plus. And h1plus there of course be some waves which

are transmitted into the second medium as well right so the second medium wave is also

traveling along the plus Z direction and the amplitude of the second transmitted field

can be written as e 2 plus the subscript 2 indicates that this is a region 2 okay and

there will be a magnetic field we will not I mean magnetic field in the same direction

because electric field we have assumed to be is in the same direction for incident and

transmitted fields so this would be H 2 plus of course H 2plus will be equal to e 2 plus

divided by ? 2 you can show by applying the boundary condition that only even plus cannot

be equal to e 2 plus right because the values of ? 1 is usually different from the values

of ? 2otherwise there is no boundary that we are actually talking about

Because this is not equal there has to be some reflected field right there cannot be

additional transmitted fields you can show that because it only means that you can try

to increase this one but please remember these have to be valid for all frequent I mean all

times and at all points along the Z=0 plane okay because of that you can clearly show

that there has to be a reflected wave and that reflected will be at the same frequency

as the incident wave the transmitted wave will also be at the same frequency as the

incident wave okay. And what would be the reflected field looking

like the reflected field will we will assume that it will be in the same direction as the

incident field the electric field of that one so you have an amplitude of even minus

for the reflected and it is propagating along minus z direction and the magnetic field will

have a minus y direction to conform to the fact that the reflected wave is propagating

along the- said direction okay now applying the total electric field the tangential electric

field boundary condition at Z equal to zero I am not going to invoke a power J O T obviously

because we are assuming that they are all phase and it is true they are all phase in

this case all the frequencies are same the incident reflected.

And transmitted okay so in the region one the total tangential electric field has an

amplitude of e1+e1-=e2 electric field in the region- which is e2 plus correct what will

happen to the magnetic field well magnetic field amplitudes h1+ h1 – okay I have not

yet written them in terms of electric field here so this must be equal to h2 plus okay

there is so you might ask what happened to the surface current density well we do not

have any surface current density because this is the interface between two insulating materials

so no surface current density and replacing h1 + H 1- as well as H 2 plus with respectively

with appropriate values of e1+/ ?1+e1-/ ?1=? And the direction of the magnetic field is

of course along the minus y direction therefore there will be a minus sign here and then you

have e2+/ ? 2 please remember that H will be e 2/H will be e by the medium impedance

?2 now what is that we have we have a simple no matrix equation so that will have a matrix

elements of 1 by1/ ?1 by ?/ ?1 times e1+e1-=1/ ? 2 okay so I have this simple equation now

it is only a matter of calculation little bit of an algebra for me to defined what will

be the ratios of even – 2 even plus this ratio of even – 2 even plus of course is what is

called as a reflection coefficient and the ratio of e2+2+e2 is the amplitude of the transmitted

field to even plus which is the amplitude of the incident field this ratio will be called

as a transmission coefficient right. I will leave this as again a small exercise

for you just have to sum and then difference the equations on the left and the right hand

side. To show that the reflection coefficient for

the electric field in this particular case okay if you want you can put a subscript of

electric field E here but I am not going to do that this gamma which is the ratio of the

reflected field amplitude to the incident field amplitude is given by ?2- ?1/ ?2+ ?1

and similarly you can show that the transmission coefficient which would be the ratio of e

2/e1+ this is the incident this is the transmitted this would be equal to 2 ? 2/?2+ ?1 in fact

this medium that you had ? 1 and ? 2 can be considered to be a transmission line.

Okay with the load the second medium acting as the load with an impedance of ? 2and the

first medium being a uniform transmission line having a characteristic impedance of

? 1 okay so at Z equal to 0 which is where the interface is separating the two medium

the problem of the waves can be know quite easily in an analogy shown to be equal or

equivalent to the problems of a transmission line okay so we will have more to say about

this in the next module until then thank you very much.