appiiCations.

4.33 Propagation of Waves in Rectangular

Waveguides
Consider arectangular waveguide situated in the
a

ordinate system with its breadth along x-axis, rectangular co

thewave is assumed to propagate along width along y- axis and
filled with air as dielectric as shown in the z-direction. Waveguide is
Fig. 4.27.
 MIC OWAVH AND BADAM E
88

Direction of
)propogotion

Width b

Bregdth o

Fig. 4.27 Propagation through a reetangular wavegulde

The wave equation for TE and TM waven aro giveh by
v°H, - -weH, for TE wave (E, 0)
v'E, - w'ucE, for TM wave (Hz 0)
Expanding v°E, in rectangular co-ordinate nystem
JE,,'E,+
'E, -o'ueE,
...4.22)
Since the wave is propagating in the 'z' direction we have th
operator.

Substituting this operator in Eq. 4.23, we get

a'E, E,
+öE, = - o'ueE, ..(4.24)
or 'E, a'E,
+(ö + o'ue) E, = 0 .4.25
Let +w'ue = h', be a constant,
then Eq. 4.25 can be rewritten a
a'E,, E, + h'E, =
0 for TM wave ...(4.26

Similarly, 'H,, H,
+h'H, = 0 for TM
By solving the wave ...(4.27

for E, and H,. above partial

Using
Maxwell's differential
equations, get solutions
components along %and y equation, it is possible tofindIthe variou
we
From Maxwell's directions [E,, H,, E,, H,).
1st equation,
we have
VxH joeE
 MICROWAVB TRANSMISSION LINIS
89
Expanding V x H,

i.e.,

H, H, H,
Replacing -y(an operator), we get,

dx ay
H, H, H,
Equating coefficients of ,jand k(after cxpanding) we get,
3) •H,
he dy +y H,´=joeE, ...(4.28)

OH,
+y H, = -joeE, .(4.29)

= joeE, ..(4.30)
ay
Similarly from Maxwell's 2nd equation,we have
VxE = -jouH
Expanding Vx E, we get

dz dy
-y =-jau (H, +fH, + H)
E, E, E, E, E,
Expanding and equating coefficicnts of i,j and k, we get
•E, +yE, = -jouH, ...(4.31)

Er +yE, jou, ...(4.32)

dx
aE, •E, ...(4.33)
dy
=-jouH,
Combining Eq. 4.28 and Eq. 4.32 to eliminate H, we get an
expression for E,. From Eq.4.32,
 MICROWAVE ANDI RADAR
90
ENGINEERING
1 JE, + E,
H, = jou dx jou
uostituting for Hy in Eq.
4.28, we get,
•H,, Y L E , =jocE,
dy jo! d* jou
- : LE,, H,
or E. jioe jou dx dy
Multiplying by jou, we get
OE, JH,
E,-o'ue-=1* +jou dy
•H,
or
E,-+aue)l)= +jou
dy
where y+ o'ue = h
Dividing by h² throughout, we get
E,= -Y OE, jou JH; ...(4.34)

Similarly E, =oBjou •H,

h² dy ...(4.35)

and H, = -YOH, ..(4.36)

and H, = , ...(4.37)
These, equations give a general relationship for field
within a waveguide. components
4.3.4 Propagation of TEM Waves
We know for a TEM wave,
E, = 0 and H, = 0.
Substituting, these values in Eqs. 4.34 to 4.37 all the
components along and y directions Er, Ey , H, vanish and bem
TEM wave cannot exist inside a waveguide.
4.3.5 TE and TM Modes
The electromagmetic wave inside a waveguide.can have an ine..
number of patterns which are called modes. We know that an
electromagnetic wave consists of magnetic and electric fields which are
alwáys perpendicular to each other. The fields in the waveguide which
ACn
make up these mode patterns must obey certain physical laws
surface of a conductor, the electric field cannot have a Component
indicatesthat the electric
field must
paralleltothe surface. This always
 Ez =00 ; Hz 0
TM Hz =o
MICROWAVETRANSMISSIONLINES Ez+o91
be perpendicular to the surface at a conductor. The magnetic field on
the other hand is always parallel to the surface of the conductor and
cannot have a component perpendicular to it at the surface.
In general, there are two kinds of modes in a waveguide. In the first
type, the electric field is always transverse to the direction of
propagation and is called the Transverse Eleciric or TE wave. In the
second type, the magneticfield is always transverse to the direction of
propagation and is called the Transverse Magnetic or TM wave. Thus
in aTE mode, no electric line i_ in direction of propagation ie., E, =0,
ifzis the direction of the propagation. But H, + 0. In a TM mode, no
magnetic line is in direction of propagation i.e., H, -0 but E, 0.
4.3.6 Propagation of TM Waves in Rectangular Waveguide
For TM wave, H, =0; E,+ 0
The wave equation of a TM wave is
+h'E, = 0 ..4.34
Thísisapartial differential equation (p.d.e) which can be solved %
get the different field components E., E,, H, and H, by 'separation d
variables method'.
Let us assume a solution
E; = XY ...(4.39)
where, Xis a purefunction ofx only.
Y is a pure function of'y' only.
Since X and Y are independent variables,
(XY)

dy
Using these two in Eq. 4.38, we get

y,x+XY
dy?
dy? -0 ..(4.40)
Dividing throughout by XY, we get
1 dx 1 dy.
=0
X ds3 ydyt+h
+
...(4.41)
1 d'x
T sa pure function of xonlv
 4.3.7 TM Modes in Rectangular Waveguides
on the values
values of
of m and n, we have
Depending

various modes in TM
In general we represent the modes as TMmn
waves.

e defined earlier. Where mand n are

Various TMmn Modes
m= ( 0 and n= 0
TMoo mode :
C
Ifm 0= and n = 0are substituted in E,, E,, H,
.EE) we see that all or them and H,, (Eas, 4.53 to
vanish and hence TMoo mode cannot exist.
TMo1 mode :m=0 and n=1
Again, all field Components vanish and hence TMo1
exist.
mode cannot
TM10 mode m=land n=0
Even now, all field components vanish and
exist.
hence TM10 mode cannot
TM11 mode : m=land n=1
Now we have all the four components Ex, Ey,
mode exists and for al higher values of m and H, and H,, i.e., TM1
n, the components exist
i.e., all higher modes do exist.
Cut-off Frequency of Waveguide (Waveguide as a High Pass
Filter)
From Eqs. 4.50 and 4.51, we know that

h=(+ o'ue = A+B

i.e.,
-(j-j-ahe
Or
Y= - o'E = a,+jß
At lower frequencies,
98 MICROWAVE AND RADARE
ENGINEERINC.
Ythen becomes real and positive and equal to the
constant'¡' i.e. the wave is completely attenuated and there;atjstenuat
no
ion
phase
change. Hence the wave cannot propagate.
However, at higher frequencies,
o'ue >
Ybecomes imaginary, there will be phase change Band hence the
wave propagates. At the transition, y becomes zero and the propagation
just starts. The frequencyrat which yjust becomes zero is defined as the
cut-off frequency(or threshold frequency)fe.
= 2rfe = 0,
.e., At f = f, Y = 0or O =2rf

=
Or

Or fe =

1
"c =
Or f =
1272

...(4.57)

The cut-off wavelength () is

C

Or
2ab ...(4.58)

Vm'b'+n
All wavelengths greater than , are attenuated and those less tn
are allowed topropagate inside the waveguide.
4.3.8 Guide Wavelength,Group and Phase
Here we define the guide Velocity
relevant for transmission ofwavelength, group velocity and phase veloci
a wave ina waveguide.
Gulde Wave length (0)
Itis defined as distance
phano shit of 2rtheradians. trevelled by the wave in order to undergo
Thls ioahown by Flg. 4.32.
 MICROWAVE TRANSMISSION LINES
99

-A9
Fig, 4.32
It is related- to the phase constant by the relation
2
.:(4.59)

The wavelength in the waveguide is diferent from the wavelength

in free space. In fact it is related to free space wavelength A, and cut-off
1 1
wavelength A. by = - (to be proved in section 4.3.8.5)
Or ...(4.60)

This equation is true for any mode in a waveguide of any cross

section, provided, corresponds to the mode and the cross section of the
waveguide.
From the above relation, it is clear that if o<<he, the denominator
/s approximately equal to 1 and =a. As , symbol e, g increases
and reaches infinity when = e. When , >, it is evident that 2, is
imaginary which is nothing butno propagation-in the waveguide.

Phase Velocity (Vp)

when guide
We have just seen that wave propagates in the waveguide
wavelenght , is greater than the free space wavelength ,.
Since the
follows that in a
velocity of propagation is the product of Aand f, it
waveguide, V, = fwhere V, is the phase velocity. But
the speed of
greater than the speed of
light is equalto product of o and f. ThisV, is no can travel faster
light since ,> . This is contradicting since signal
wavelength in the guide is tie
than the speed of light. However, the velocity of the phase. n fact
length of the cycle and V, represents the ohanges its phase in terms of
it ig defined as the rate at which the wave
ott
the guide wavelength.
i.e., V, unit tine 276
...(4.81)

t.e.,
 ENGINEERING

RADAR
MICROWAVE
AND

100
21 V, ie
velocity,
O=24/, = this
where, modulation
travel at
intelligence or
Since no velocity.
termed as phase actually
envelope
Group Velocity (V) modulation
course slower
carrier, the and of
modulation in
the carrier alone called the
ther is than that of envelope is
If velocity slower modulation
signal travels
in a
travelsat velocity of
The whena
m o d u l a t e d
to
than speed of light. happens with respect
backward
V,. This on slipping
group velocity modulation goes
waveguide, the through the
propagates
wave
at which the
the carrier.
as the rate
It is defined given by
waveguide and is ..(4.62)
do

and Group Velocity

for Phase Velocity
Expression
we know that V, =
V, :From Eq. 4.61
1. Expressionfor
h =+o'4e = A'+ B² =
Also,
Y= a+jB
and
attenuation, a = 0)
propagation, y =jß (.
For wave ...(4.63)
--wue
Atf =fe 0 = 4, Y=0
..(4.64)
(mn
+
e =

Using Eq. 4.64 in Eq. 4.63, we get

or B= Vo'uE -aue
o,/.465)
B= Vue (w- a) = Vu[ Vof-
...(4.65)

C
i.e., V, =1-f.I
MICROWAVE TRANSMISSION LINES 101

We also know that, f (any frequency) = ch, where , is free space

wavelength and f (cut off-frequency] = chh, where , is cut-off
wavelength

C ...(4.66)
V =
V1-(a
do
2. Expression for V,:FromEq. 4.62 we know that, V, = dß

But from Eq. 4.65, B = Vue (o- a)

Now differenting ß w.r.t. 'o, we get
"2 ouE
do 2 V(a-a) e
dß
do 1-(oJo
do
V, = dß
...(4.67)
Or V, =oi
Consider the product of V, and V,

i.e.,
V1-(h ...(4.68)
/V,V, = c?
V,V, = c
ALITER: Now wegive another method for proving
greater than the speed
As per earlier discussion we know that V, is
of light by the ratio
C

the same ratio.

The group velocity V, is shortened by
C
i.e.,
"c = c
V,V,= e
Relation between he, ho, and ho
)
We know that V, = f=

-)
102
MICROWAVE AND RADAR ENGINEERIN:
and also V, =
C

1-(4,/
or .(4.69)
V1-(4,/
439 Propagation of TE Waves,in a Rectangular Waveguide
"The Emn modes in a rectangular waveguide are characterised b
Ez = 0. In other words the component of the magnetic field, H, mg
eist in order to have.energy transmission in the guide.
The wave equation (Helmhottz equation) for TE wave is given by
 VEETRANSMISSION LINES
MICROWA
107

°H, =- oueH,
a H.H,,H, - o'ueH,

H, .+H,+o'ueH, = 0
HH+ we) H, =(
+

&H.H, +h' H, = 0 ...(4.74)

or

This is a partial differential equation whose solution can be

assumed.

Where
Assume a solution H, = XY.
Xis a pure function of r only.
Yis a pure function of'y' only.
Substituting for H, in Eq. 4.70, we get
+X+h' XY = 0
d² dy²
Dividing throughout by XY, we get
1 d'x. 1dY +h' = 0 ..(4.75)
Xd dy
1d'x
Here is purely a function of x,
1d'y
and -is purely a function ofy.
Ydy
Equating each of these
items to a constant, we get
= - B'

1dy
and

where - B and-A' are constants.

Substituting these in Eq. 4.75
above, we get
- B'-A'+h² = 0 ..(4.76)
A =A'+ B²
Solving for X and Yby separation
of variable method,
X =Crcos Bx + C; sin Br
Y= C cos Ay + C, sin Ay
 ENGINEERING
108 MICROWAVE AND RADAR

Therefore the complete solution is, H, = XY

sin Ay) ...(4.77)
i.e., H, =(C cos Bx+ Ca sin Bx) (C3 cos Ay + Ca
evaluated by
where C1, C2, C& and C, are constants which can be
applying boundary conditions.
Boundary Conditions
As in case of TM waves, we have four boundaries for TE waves also, as
shown in Fig. 4.40.

y
z Direction of
propogation

Fig. 4.40
Here since we areconsidering a TE, wave,
E, =0but we have components along xand y direction.
E, =0all along bottom and top walls of the waveguide.
E, =0 allalong left and right walls of the waveguide.
 4.3.14 Dominant Mode and Degenerate Modes in Rectangulay
Waveguides
As already discussed, the walls of the waveguides can be considereda
nearly perfect conductors. Therefore, the boundary conditions reouira
that electric field be normali.e.,perpendicular, to the waveguide walls.
The magnetic fields must be tangential i.e., parallel, to the waveguide
walls. Because of these boundary conditions a zero subscript can exist
in the TEmode but not in the TM mode. For e.g. TE10, TEo1, TEo etc.
modes.can exist in a rectangular waveguide but only the TM1,TM
TM¡1 ctc. modes can exist. Also the cut-off frequency relationship shows
 MICROWAVE TRANSMISSIONI LINES

that the physical size of the 117

modes depending on waveguide determines the
on the values
for a rectangular
frequency for of m and n. propagation of
The minimum
=0,i.e.,waveguide
fre for is obtained for a cut-off
a>b om=l m = 1 and n
"TEo mode is the
rectangularwaveguide. (Since for dominant dimension
mode
TMm modes nm # 0orn 0, for a
order mode TE0 is the
dominant mode for a > b).
order modes, having the same
the lowest
cut-off Some of the higher
wnodes. For a rectangular frequency are called degenerate
m#0, n#0 wil always be
waveguide TEm/TMmn Imodes for which both
which a =b, all the TEpg TEp degenerate modes. For a square guide in
, TMpg and TMgp modes are
degenerate modes. It is necessary that higher together
are not supported by the guide in the order degenerate modes
avoid undesirable components appearing operating band of frequencies to
at the output alongwith losses.
Also it may be necessary to prevent
the conversion of a particular
waveguide mode to another. Such mode conversion
waveguide irregularities or from impedance usually results from
structure used in
transmission line. Such mode conversion can be supported by
(i) Choosing suitable waveguide dimension (the undesired
mode/modes having cut-offfrequency above the desired modes
can besuppresssed)
(ii) Using mode filters (undesired modes can be suppressed by
providing a metalic plate or vane in the waveguide where
undesired modes have tangential electric fheld lines).
The various modes in a waveguide can be excited by various
launching devices. Fig. 4.42 illustrates how the TEo ,TEzo and TMu
modes are launched in rectangular waveguides. These launching
devices are in fact, antennas. At the receiving end of the waveguide, a
similar lauching device (receiving antenna) is used to convert the e.m.
fields within the waveguide back to voltage and currents on a
transmission line. Hence, one must know which waveguide mode was
used tolaunch the e.m. fields at the transmitting end of the waveguide.
lfmore than one mode exists at a particular frequency in the waveguide,
then discontinuities such as bends, joints etc. would cause e.m. energY
transferred from one mode to another. This results in an additional
O be
waveguide since it will not be recovered in the receiving end.
OSS in the