Transmission Lines & E. M.

Waves
Prof. R. K. Shevgaonkar
Department of Electrical Engineering
Indian Institute of Technology, Bombay
Lecture 38

In the last lecture, we developed a general approach for analyzing wave propagation inside a
waveguide. In this lecture, we take specifically a waveguide whose cross section is
rectangular and that wave guide is called a rectangular wave guide.

(Refer Slide Time: 1:00)

So, we have a wave guiding structure for which we have a hollow pipe whose cross section is
rectangular and the energy is going to flow along the length of this metal pipe. We are again
considering that this pipe is made of ideal conductor terminal conductivity of this pipe is
infinite and the medium which is filling this wave guide is ideal dielectric. That means the
conductivity of the medium which is filling this pipe is 0.

So, we are having a very ideal structure that is the pipe is ideal conductor and the medium
filling this wave guide is ideal dielectric and we want to now investigate how the
electromagnetic energy is going to propagate inside this pipe without losing generality. Let us

orient our coordinate system so that the z axis is along the length of the pipe that mean that is
the direction in which the energy is going to propagate.

So, let us say this direction is the z direction and then horizontal axis let us say x axis, so this
axis is x and this vertical axis is y the origin is located at one of the corners of this cross
section. Now, considering in general this cross section of this wave guide is rectangular. Let
us say the width of this wave guide is denoted by a and the height of this wave guide is
denoted by b and that definition a is greater than or equal to b and why we are taking it
convection will become clear when we define the particular type of mode. So, here we are
assuming that by definition a should be greater than or equal to b and x axis is oriented along
this broader dimension that means along a that is the convention essentially we are taking.

Now firstly, if you try to see whether TM mode exist inside this, the answer is; no, the TM
mode cannot exist inside this structure and the reason for this, we will see later after we have
understood the TM and TE propagation. So, first we will take the simpler case which is the
transverse magnetic propagation that means for this the Ez is not equal to 0 and Hz is equal to
0.

We saw in the last lecture that if the transverse field has to exist, then either Ez or Hz has to be
non-zero. So, in this case if I consider Hz is 0 that means the magnetic field now is in the
oriented in the transverse plain and longitudinal component z component is only Ez. So, this
mode we can call as the transverse magnetic mode or as we called earlier, in short, this is
called the TM mode.

So, let us investigate now the problem for the transverse magnetic mode for which the
longitudinal component is only Ez and Hz is equal to 0. Of course, Ez and Hz both are all field
component for their matter have to satisfy the wave equation inside this structure. So, the
approach now is as follows; first we solved the equation for this longitudinal component Ez,
then we go to the expression which we have derived last time for the transverse components,
substitute here for Ez and Hz equal to 0 and we will get the transverse fields for the transverse
magnetic mode.

(Refer Slide Time: 5:03)

So, problem essentially has to be solved in 2 steps; one is, first finding the solution for this
longitudinal component Ez, then finding out transverse components and then applying
boundary condition finding proper solution to this structure which is rectangular wave guide.

(Refer Slide Time: 5:14)

So, if I take the wave equation in Cartesian coordinate system, this Ez is a scalar quantity. So,
we can write down the wave equation which is del square Ez plus omega square mu epsilon
Ez that is equal to 0. This is the wave equation.
(Refer Slide Time: 5:40)

So, this equation is a scalar equation essentially and omega is the frequency, mu is the
permeability of the medium and epsilon is the permittivity of the medium which is filling this
wave guide. We can expand this del square in the Cartesian coordinate system, so we can get
d2 Ez by dx square plus d2 Ez by dy square plus d2 Ez by dz square plus omega square mu
epsilon Ez that is equal to 0. Now, we solve these problem essentially just on mathematical
considerations.

However, since you have developed the understanding for the wave propagation inside the
parallel plain wave guide, we make frequent visits to that understanding. So, whenever we
choose some constant or something while writing the solution to the equation, we make sure
that whatever solution you get that solution should be consistent with what understanding we
have developed from the parallel plain wave guide.

Now, this equation can be solved by separation of variables. So, we can define this quantity
Ez which is product of the three functions, each one is a variable of either x, y or z and of
course, there is implicit assumption that all these fields are varying as a function of time
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which is e to the power g omega t. So, we can say that we can apply separation of variable to
get a solution which is a function of x, y, z that is some function of x, some function of y,
some function of z.

(Refer Slide Time: 8:08)

Now, if you recall, the wave is propagating in the z direction that means we are looking for a
solution of travelling wave type in the z direction and since we are considering a medium
which is completely lossless because the dielectric is ideal dielectric, the conducting
boundary of this wave guide is again ideal; so there is no loss anywhere. So, we have a
sustain propagation of wave inside this structure. So, we have z variation for a travelling
wave in the z direction which will be e to the power minus j beta z. That information we will
use when we try to solve this problem that we are looking for a travelling wave solution in
this direction.

Also, we a have information from parallel plain wave guide and that is let us say suppose, I
make one of the dimensions of the wave guide and push it up, let us say I make b equal to
infinity, then essentially I got a structure which is parallel plain wave guide and I have seen
for parallel plain wave guide that this can be visualized as the propagation of uniform plain
wave inside this parallel plains by multiple reflection on this two plains. So, it creates
standing wave kind of atoms in this direction which is perpendicular to the plains and
travelling wave propagation along the plains.
5

So, that means whatever solution we are going to get for this problem must have a standing
wave kind of solution in the x direction because if I take a limit when b tends to infinity, the
solution must represent the solution for parallel plain wave guide. Same is true for this also
that if I take these two plains, then if I take a tending to infinity; again I will get a parallel
plain wave guide which is horizontal now and then I must get a standing wave kind of
solution which are in y direction.

So, the physical understanding tells me that a solution which we are getting along x direction
must be of standing wave type, the solution which we should get along the y direction also
must be of standing wave type and the solution which we should get along the z direction
must be of travelling wave type because in that direction, the wave is going to, net wave
propagation is going to take place. With this understanding, now we are going to solve the
problem.

(Refer Slide Time: 10:37)

So, first let us just take this and blindly substitute into this equation. So, if I substitute into
this, I get YZ d2 X by dx square plus XZ d2 Y by d y square plus XY d2 Z by dz square plus
omega square mu epsilon XYZ that is equal to 0. Note here, now, the partial derivatives have
been converted to the full derivatives and the reason is X is now a function of x only, Y is a
function of y and Z is a function of z. By dividing by xyz, all these terms, essentially we get 1
upon X d2X upon dx square plus 1 upon Y d2 Y upon dy square plus 1 upon Z d2 Z upon dz
6

square plus omega square mu epsilon, it should be equal to 0. Now, this quantity is only a
function of X, this quantity is only a function of Y and this quantity is only a function of Z
and this is constant.

Now, this equation whatever you have written, it should be valid at every point in space. So,
this equation must be satisfied by every point XYZ in the three dimensional space and that
can only happen provided each of this term is a constant quantity. So, this should be a
constant, this should be a constant, this should be a constant; then and then only this equation
can be satisfied by every value of XYZ.

So, what we do is; we just take each of this term and equate them to some constant and the
equation essentially reduces to three equations and that is 1 upon X d2 X upon dx square is
equal to some constant and let me just put a constant as minus A square, second term which
is 1 upon Y d2 Y upon dy square, let us put that thing as minus B square and third quantity
which is 1 upon Z d2 Z upon d z square that is equal to minus beta square where beta is the
phase constant of the mode of propagation.

(Refer Slide Time: 13:08)

We have chosen here this negative sign appropriately so that the solution which you get for
this equation that will be a standing wave kind of solution in x direction, a standing wave

kind of solution in y direction and a travelling wave kind of solution, we will see, will be
given by this phase constant beta.

So, while writing the expression for X, Y and Z, essentially we make sure that we write a
solution which will look like standing wave kind of solution and travelling wave kind of
solution as a function of X, Y and Z. So, this equation to solve is very straight forward, so
this is I can multiply X on this side and take the term which will be d2 X upon dx square plus
A square X equal to 0, this is second order homogenous equation for which the solution can
be written in a straight forward manner.

(Refer Slide Time: 14:52)

So, we get the solutions for the capital X as a function of x will be some arbitrary constant C1
cos of Ax plus some other constant C2 sin of Ax; Y is a function of y will be some constant
C3 cos of By plus C4 sin of By and Z is a function of z will be C5 e to the power minus j beta
z plus C6 e to the power plus j beta z.
So, note while writing these solutions; these functions which are cos and sin functions, they
show amplitude variation which is the standing wave kind of behavior. Whereas, if I look at
this quantity - e to the power minus j beta z, that represents a travelling wave in positive Z
direction; if I consider e to the power j beta z quantity, that represents a travelling wave
which it is in negative z direction.
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So, while writing this solutions, first we use our understanding which you have developed
with the parallel plain wave guide that is in the transverse direction between the two
conducting boundaries we must have a solution which were like a standing wave kind of
solution and along the length of the pipe in which the energy is going to propagate, we have a
travelling wave kind of solution.

Now, in general when we have this wave equation solution, of course there are two waves
which are travelling on this structure; one in positive z direction and one in negative z
direction. However, if I assume that this wave guide is of infinite extent in z direction, then
there is no reflected wave on this because we have seen from transmission line that if I take
an infinitely long transmission line, there is no reflected wave on this.

So, we can just for simplicity of the analysis, we can assume that there is no wave which is
travelling in the negative z direction and we have only one wave which is the forward wave
which is on this wave guide. So, this wave as shown is a forward wave travelling in positive
Z direction.

(Refer Slide Time: 17:27)

So, let us assume that only one wave which is travelling in positive Z direction exists and
there is no wave which is travelling backwards on this structure. So, we can assume that for
no backward wave, we can take the C6 as identically 0. So, the Z solution is only C5 e to the
9

power minus j beta z and with that understanding, then I can write down the complete
solution for the Ez which is the product of this, this and this quantity.
So from here, then I can get Ez which is a function of x, y, z will be equal to C1 cos Ax plus
C2 sin Ax C3 cos By plus C4 sin By e to the power minus j beta z and we put a constant here
which is C5. So, this is now the solution, general solution for the wave equation for this
component Ez. Now, we can apply boundary conditions to get these arbitrary constants and in
this case, we are having this Ez which is like that in the wave guide, so this component Ez is
tangential this wall, it is tangential to this wall, it is tangential to this wall and it is tangential
to the lower wall.

(Refer Slide Time: 19:20)

So this component, easier component is a tangential component to all the 4 walls of this and
that is why we have 4 boundary conditions; one on this wall which is x equal to 0, one on this
wall which is x equal to a, one on this wall which is y equal to 0 and one on this wall which is
y equal to b.

So, we have got four boundary conditions that all these four walls where x equal to 0, x equal
to a, y equal to 0, y equal to b, this component which is tangential should be 0.

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(Refer Slide Time: 20:16)

So, we have a boundary conditions now to apply on Ez and that is Ez is equal to 0 at x equal
to 0, x equal to a and y equal to 0, y equal to b. So, I just use one by one boundary condition.
If I put x equal to 0 in this, this quantity will be 0, sin of Ax this will be 0. Now, Ez is 0 that
can happen provided this C1 quantity that should be equal to 0. So, I get from here, from the
first condition that for x equal to 0, this condition we get this arbitrary constant C1 should be
identically equal to 0.

(Refer Slide Time: 21:14)

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So, this gives me C1 should be identically equal to 0. Same thing we can do from here that
since for y equal to 0, again Ez is 0. So, if I put y equal to 0, this quantity will be 0 and then if
Ez has to be 0, this arbitrary constant c3 should be 0.
(Refer Slide Time: 21:18)

So, we get for y to be equal to be 0, you get the arbitrary constant C3 should be identically
equal to 0.

(Refer Slide Time: 21:40)

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Once you get that, then the solution essentially now has become C2, C4, C5 sin of Ax, sin of
By e to the power minus j beta z. So, the general solution which can satisfy this boundary
condition x equal to 0, y equal to 0, now will be Ez C5 C2 C4 sin of Ax sin of By e to the
power minus j beta z.

(Refer Slide Time: 22:05)

Now, these are the arbitrary constants, so we just combine this into one, so this is just telling
me the amplitude. So, I can say this combine, all this thing combine together, let me call this
quantity as some C into sin of Ax sin of By e to the power minus j beta z.

Now, apply second boundary condition that when x is equal to a, again this quantity Ez is 0
and that can happen when provided this A times small a, that quantity is multiples of pi.
Similarly, from this boundary condition that Ez should be 0 for y equal to b, if this By is
multiples of pi, then again this quantity will be 0. So, from these two boundary conditions; x
equal to a and y equal to b, we get that A times a that should be equal to some multiples of pi.
That means this constant A which we still have to determine that is equal to m pi by a.

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(Refer Slide Time: 23:35)

Similarly, B times small b that should be equal to some another integer of pi and that gives B
is equal to n pi by b. We can now substitute, once you get these constants a and b which are
related to these dimensions, the broader dimension of the wave guide and the smaller
dimension of the wave guide and m and n are integers, this longitudinal component field Ez is
constant C sin of m pi x by a sin of n pi y by b e to the power minus j beta z.

This is now the complete solution of the wave equation for this longitudinal component Ez.
We applied all the boundary conditions, this arbitrary constant C will remain undefined
because this essentially represents the amplitude of the electric field and that has nothing to
do with the boundary conditions because boundary conditions are satisfied irrespective of
power level or the amplitude of the field. So, this parameter this constant, arbitrary constant C
will remain as it is at this stage; when we really talk about the power inside a mode, then and
only this quantity C will be evaluated.

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(Refer Slide Time: 25:48)

So now, we have got a complete solution and these two quantities now m and n which are
integers, they now represent essentially the order of the mode by the same token as we have
done for the parallel plain wave guide that there we had a transverse magnetic mode and we
had put an index, now we can put the indices which are these two indices for m and n. So, a
transverse magnetic mode now can be designated by TM with these two indices m and n.

Now here, the convention essentially comes from handy, why you have taken a convention;
the first index which we have here is the index around the broader dimension a. So, that is the
reason we taken the convention that broader dimension is called a and that is the direction in
which the x is oriented. So, the first index essentially tells me the field variation along the
broader dimension of the wave guide which is x direction.

Similarly, this index n tells me the field variation along the shorter dimension of the wave
guide which is the y direction. So, when we write a mode TM

mn;

the first index tells me the

field variation along the broader dimension, the second index tells me the field variation
along the shorter dimension. Also, we will note here that if m equal to 1, then when x varies
from 0 to a, here; you have essentially one half cycle variation.

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(Refer Slide Time: 27:28)

If I take m equal to 2, I have one full cycle variation and then so on and same thing is going
to happen in y direction.

(Refer Slide Time: 27:30)

So, this behavior is exactly identical to what we have seen for the parallel plain wave guide
that the order or the index which we have for the mode, it represents number of half cycles in
the transverse direction and in this case, there are two transverse direction; one is x and one is
16

y. So, along each of these transverse direction x and y, these indices tells me the number of
half cycle variations for the magnitude of the field. So, if I take m equal to 1, then I get one
half cycle variation; if I take m equal to 2, I will get two half cycle variations and so on.

So, once you have understood the field behavior for the parallel plain wave guide, visualizing
these now for rectangular wave guide is very straight forward because we have already
developed a physical understanding of how the field variation is going to be and as we can
see in the limit when a or b tends to infinity, this essentially, wave guide will become a
parallel plain wave guide and then the expression which we had for Ez or for that matter, we
will see later for Hz that is identical to what is the expression for electric field in the parallel
plain wave guide.

So, this field expression which we have got now can be reduced to the parallel plain wave
guide by properly substituting either a equal to infinity or b equal to infinity. Now, once you
have got these constants now - m and n, there are few things which can be noted from here.
Firstly, m equal to 0, n equal to 0; if I substitute into these equations, the Ez will be identically
0. What that means is that TM00 mode cannot get excited inside a rectangular wave guide.
We have seen for a parallel plain wave guide that the TM0 mode which was same as the
transverse electromagnetic mode was possible in a parallel plain wave guide and that was the
mode which was the lowest order mode. However for this mode, we note that when both the
indices are 0, the field goes to 0. So, TM00 does not exist.
Also we will note, when any of the indices go to 0, then also again this field goes to 0; Hz is
already 0 for transverse magnetic. If Ez also goes to 0, then all the fields will identically go to
0 because h is not equal to 0 that is what we have to show.

So first, let us show that for this mode, h is not equal to 0 if I go back to the wave equation
and substitute now this quantity from this equation. This quantity as we have defined as
minus A square, this quantity we defined as minus B square.

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(Refer Slide Time: 30:39)

(Refer Slide Time: 30:43)

Here, this quantity is minus A square, this quantity is minus B square and this quantity is
minus beta square and we have found out now this value of A which is m pi by a, this
quantity is n pi by b.

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So, you can go back and substitute into this to get a relation which is minus m pi by a whole
square minus n pi by b whole square minus beta square plus omega square mu epsilon is
equal to 0.

(Refer Slide Time: 31:06)

I can take this beta on one side, so the phase constant for this mode will be equal to square
root of omega square mu epsilon minus m pi by a whole square minus n pi by b whole
square.

Now, this relation is familiar, the similar relation we had seen for the parallel plain wave
guide with only one term which are omega square mu epsilon minus m pi by d where d was
the height of the wave guide and this expression essentially tells how the velocity or the
phase constant varies as a function of frequency on this structure.

So, this expression is identical to what we have derived for parallel plain wave guide and that
is why this is the dispersion relation for the transverse magnetic mode on a rectangular wave
guide. So, we can call this as the dispersion relation that we saw in the last lecture. So, this is
the dispersion relation for TM

mn

mode where m and n are the two indices which are non-

zero.

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Now, if I take m and n both equal to 0, we saw the fields are not existing. But if I take m and
n, one of them is 0.

(Refer Slide Time: 33:03)

Then, this quantity now, this is h; what is h square now? The h square was defined as omega
square minus beta square in our general analysis.

(Refer Slide Time: 33:05)

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So, we had defined this quantity h square was omega square mu epsilon minus beta square.
So by that, comparing this with this, we get essentially for this case; h square is your m pi by
a whole square plus n pi by b whole square. So, when m and n both are not 0, the h square is
not zero.

So, when the h square is not zero, the field which we have transverse fields which are
represented in terms of the longitudinal components, we have seen the transverse component
would exist if both of Ez and Hz are 0, provided H is also equal to 0.
(Refer Slide Time: 34:04)

However, now we are seeing that h is not equal to 0 when m, either m or n are not equal to 0
and then if Hz and Ez both go to 0, all transverse field would go to 0 and the mode would not
exist. So, we see that for the transverse magnetic case, when either m or n are 0 or any of
these two is 0, then again this field will be 0 and since h is not 0, again the transverse field
will go to 0 and the modes will not exist.

So now, we have some important conclusion drawn from this analysis for transverse
magnetic mode. So, we have a mode TM mn mode.

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(Refer Slide Time: 34:43)

So, the first conclusion which we have is TM00 does not exist. So, we also conclude that TM
m0

and TM 0n also do not exist. So, the lowest mode which can exist, the lowest index that is

when both these indices are non-zero, that means TM11 is the lowest order of TM mode
which can exist on the waveguide.

So, by doing this general analysis, we come to the very important conclusions that if the TM
mode has to be excited on a rectangular waveguide, the fields must vary in both the directions
x and y on the rectangular waveguide and that we can see physically as follows. Since the
index m and n is telling the variation, one half cycle variation along the x and y direction
respectively; when m is 0, there is no variation of the field along x direction and when n is 0,
there is no variation of the field along y direction.

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(Refer Slide Time: 36:32)

But the field has to be 0 here, this is Ez component; so it has to be 0 here, it has to be 0 here
and there should not be any variation in the x direction if m is equal to 0. That is only
possible provided the field is identically 0 everywhere; the same is true for these two
boundaries that when n is equal to 0, the field should not have any variation in the y direction
but it should be 0 here, it should be 0 here and that again can happen when the field is
identically 0 everywhere.

So, physically it makes sense that yes, this field Ez cannot exist inside this without a variation
in x and y direction, it must vary. Also, we assume from the solution in this Cartesian
coordinate that the variation is always sinusoidal and it is always number of half cycles; you
can have one half cycle or two half cycles or three half cycles and so on.

So, essentially we get now the variation in this Ez is always 0 on these two and depending
upon the value on m and n, you will have the variation in x and y direction which is
sinusoidal variation. So, the field in the transverse direction always varies sinusoidally in the
Cartesian coordinate system and they must vary because if they do not vary in any of the

Of course, your conclusion might be different when we go to a transverse electric case but
these are the conclusion which we can draw now for a transverse magnetic case. Once we get
now this Ez, then we can substitute for Ez into these components and then we can write down
23

the transverse components for the TM mode. So, you can substitute now the general solution
which we got and now if I write for Ex, Hz is 0; so this term is 0, this term is 0.
(Refer Slide Time: 38:57)

So, I get now Ex which is equal to minus j beta upon h square where h square is already have
been calculated; this is d Ez by dx. So, that will be equal to minus j beta upon h square m pi
by a into that arbitrary constant C cos m pi x by a sin n pi y by b and multiplied by e to the
power minus j beta z. Same thing I can do for Ey that is equal to minus j beta upon h square d
Ez by dy which will be equal to minus j beta upon h square n pi by b C sin of m pi x by a cos
n pi y by b and put this phase term e to the power minus j beta z.

The magnetic fields on the same lines, we can get as j omega epsilon upon h square d Ez by
dy that is equal to j omega epsilon upon h square n pi by b C sin of m pi x by a cos of n pi y
by b e to the power minus j beta z and Hy will be equal to minus j omega epsilon upon h
square dEz by dx that is minus j omega epsilon upon h square m pi by a C like this cosine of
m pi x by a sin of n pi y by b e to the power minus j beta z.

So now, we got the complete fields for the transverse magnetic modes with the understanding
that the lowest order mode which is going to propagate on this structure will be the TM

11

mode. So, few things can be seen from here and that is the Ez component which is for which

24

we obtained the solution. That essentially is 0 at these 4 walls it is there in the sinusoidal
variation, so we had a Ez variation which is this.
(Refer Slide Time: 42:52)

So, the Ez was 0 x direction at x equal to 0, at both the walls, the Ez was 0, Ez in y direction
also having half cycle variations.

(Refer Slide Time: 43:06)

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If I look at the cross section like that, the wave guide; the Ez is 0 here, 0 here, 0 here. So, it
had half cycle variation like that or it could have a variation which will be like that, the same
is true for half cycle variation in this direction or it could be like that. So, the Ez is 0 at this
point and the maximum here if m is equal to 1, maximum here along this if n is equal to 1 and
so on.

(Refer Slide Time: 43:50)

However, if I look at this field now which I have got here, Ex Ey Hx Hy, the Ex is having a
variation which is cosine variation as a function of x. So, the component Ex which is this, x
component which is oriented this way, it is having a variation cosine variation along x
direction. That means it is maximum here, 0 here and another maximum here; again it is half
cycle but it is not 0 here. In other words, it is shifted like a quarter cycle in space with respect
to the Ez component. So, the Ex component if I look at; it will be like that.

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(Refer Slide Time: 44:36)

This will be the component which is Ex; so that is maximum here, 0 here, maximum here and
Hy component if I look at, it will be again a cos variation along y direction. So, it will be in
this direction; if I look at, the field variation would be like that for the Ey component.
So now, later on we will try to visualize these fields actually inside this pipe but at this point
it appears that component Ex Ey Hx Hy depending upon the boundary conditions, they will be
staggered in space in the transverse plain with respect to the longitudinal component which is
the Ez component.
Once we get this understanding that the electric field which is like that is maximum on this
boundary and if I look at the magnetic field which is let us say I take x oriented magnetic
field which is having a variation which is sin variation in the x direction and cos variation in
y direction; so if I consider now a magnetic field which is x oriented, then it is cos variation
in y direction that means it is maximum here, 0 here, maximum here because Ex and Hy, they
have the same behavior and Ey and Hx have the same behavior. By comparing the
expressions, this is what essentially we see.

So, what we observe from here and off course from rigorous analysis also we can do that; but
what we observe from here is that Hy component which is this is maximum on this wall and
on this wall. That means when the magnetic field is tangential to the boundary, that is where
27

it is maximum and off course it is having a variation which is sinusoidal variation with
number of half cycle depending upon the order of the mode.

Similarly here, when I consider Hx which is going to be maximum here, maximum here; that
means magnetic field is maximum when it becomes tangential to the conducting boundary
and an electric field is 0 when it becomes tangential to the conducting boundary. Of course,
the tangential magnetic field whatever is there on the boundary is balanced by the surface
current, so it is a boundary condition on the tangential component of magnetic field.

So essentially, by using this observation that the magnetic field is maximum; tangential
component of magnetic field is maximum on the conducting boundary and having understood
that the solution gives me a sinusoidal variation in the transverse direction, now we can
readily write the solution for the transverse electric mode without going to the same analysis.
Of course in a routine way, we can do the same thing that if you want to analyze a transverse
electric case, then we take Ez is equal to 0, we take Hz not equal to 0.
(Refer Slide Time: 47:51)

So, we consider a case which is transverse electric case or a TE wave.

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(Refer Slide Time: 48:22)

So in this case, the longitudinal component which we have is Hz which is like that and what is
observed from this is that this component which is going to be tangential to this wall, this
wall, this wall and this wall should be maximum on the walls and it should have a variation
which is sinusoidal variation in the transverse direction.

So now, having understood all these analysis, we can write the solution without going to the
same steps same algebraic steps which we have done, we can get the solution for Hz which
will be some constant cosine of m pi x by a cosine of n pi y by b e to the power minus j beta
z.

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(Refer Slide Time: 48:54)

So, without repeating the algebra which we have gone through for solving the wave equation;
of course, I can go exactly same way, I can solve the wave equation now for Hz, then apply
the boundary conditions but note; directly on Hz, there was no boundary condition.
So, if I have to go from the same routine way, then I have to find out general solution for Hz,
find out a transverse component then find out the component which are tangential to the
boundaries and then apply the boundary conditions on those components because tangential
component of magnetic field, there is no boundary condition. So, in this case, TE case, if I go
by routine way, there will be two steps involved; first, find out a general solution for Hz,
substitute into the transverse component expressions, find transverse components and then
apply the boundary conditions on the transverse components because intrinsically there is no
boundary condition on Hz.
However, without going to the same routine steps essentially the understanding which we
developed at the magnetic field when become tangential to the conductor is maximum, I can
get the expression for Hz and one can go and verify, verify that if I had done the analysis,
essentially I would get the same expression.

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(Refer Slide Time: 50:21)

Now, this you can verify when x is equal to 0, the field is maximum; when x is equal to a,
again the field is maximum; y is equal to 0 and again y equal to b, again the fields are
maximum.

So, these fields are having the variation which essentially we are looking for; the tangential
component of magnetic field must become maximum. Using this expression for Hz, now we
will investigate the transverse electric mode and see the characteristics of transverse electric
mode and then compare the behavior of this mode with the transverse magnetic field. So, we
will continue the discussion on the transverse electric and transverse magnetic mode inside
the rectangular wave guide and see more properties for these modes.

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