Introduction to Waveguides and Their Advantages
KNOWLEDGE INSTITUTE OF TECHNOLOGY
Prepared by
RAADHU.T
ASSISTANT PROFESSOR
INTRODUCTION OF
WAVEGUIDES
1
INTRODUCTION OF
WAVEGUIDES
I
N
T
R
O
D
U
C
T
I
O
N
T
O
W
A
V
E
G
U
I
D
E
S
In
a
waveguide
energy
is
transmitted
in the
form
of
electromagnetic
waves
whereas
in
transmission
line,
it
is
transmitted
in
the
form
of
voltage
and
current.
Conduction
of
energy
takes
place
not
through
the
walls
whose
function
is
only
to
confine
this
energy
but
through
the
dielectric
filling
the
waveguide
which
is
usually
air.
A
D
V
A
N
T
A
G
E
S
O
F
W
A
V
E
G
U
I
D
E
O
V
E
R
A
T
R
A
N
S
M
I
S
S
I
O
N
L
I
N
E
:
In
a
transmission
line
high frequency
signals
cannot
be
transmitted
because
Large
mutual
induction.
Length is
very
large.
Multiplexing
the
signals
is
not
possible
in
transmission
lines.
Flashover
is
less
in
case
of
waveguide.
Minimum
loss
of
power
in
case
of
waveguide.
Power
handling
capability
is 10
times
than
coaxial
cable.
3
R
E
F
L
E
C
T
I
O
N
O
F
W
A
V
E
F
O
R
M
I
N
C
O
N
D
U
C
T
I
N
G
P
L
A
N
E
:
I
N
T
R
O
D
U
C
T
I
O
N
T
O
W
A
V
E
G
U
I
D
E
S
Mechanical
simplicity.
Higher
maximum
operating
frequency.(3GHZ
to
100GHZ).
Normally
waveguides
are
of
two
types.
Rectangular
waveguide.
Circular
waveguide.
a
b
Rectangular
waveguide
d
Circular
waveguide
V
g
4
V
R
V
n
V
I
V
is
velocity
guided
by
wall
g
(group
velocity)
V
n
is
the
normal velocity
to
wall
a
L
D
O
M
I
N
A
N
T
M
O
D
E
O
F
O
P
E
R
A
T
I
O
N
D
O
M
I
N
A
N
T
M
O
D
E
:
T
h
e
n
a
t
u
r
a
l
m
o
d
e
o
f
o
p
e
r
-
-ation
for
a
waveguide
is
called
dominant
mode.
This
mode
is
the
lowest
possible
frequency
that
can
be
propagated
in
a
waveguide.
a
L
L
a
EL
E
C
T
R
IC
FIELD
>
>
>
MA
GN
ETI
C
FIELD
5
m=no.
of
half
wavelength
across
waveguide.
n=no.
of
half
wavelength
along
the
waveguide
height.
D
O
M
I
N
A
N
T
M
O
D
E
O
F
O
P
E
R
A
T
I
O
N
c
o
n
t
…
n=1
=>
1
circle
n=2
=>
2
circles
m-
electric
field
n
–
magnetic
field
Magn
e
t
i
c
field
m=1
m=2
6
The
signal
of
maximum
wavelength
that
can
pass
through
a
waveguide
is
λ
0
=2a/m
B
A
S
I
C
B
E
H
A
V
I
O
U
R
An
electromagnetic
plane
wave
in
space
is
transverse
electrom-
agnetic
or
TEM.
The
electric
field,
the
magnetic
field
and
direction
of
propagation
are
mutually
perpendicular.
If
such
a
wave
were
sent
straight down
a
waveguide,
it
would
not
propagate
in
it.
This
is
because
the
electric
field
would
be
short-circuited
by
the
walls
since
the
walls
are
assumed
to
be
B
A
S
I
C
B
E
H
A
V
I
O
U
R
c
o
n
t
…
.
Perfect
conductors
and
a
potential
cannot
exist
across
them.
What
must
be
found
in
some
method
of
propagation
which
doesn’t
require
an
electric
field
to
exist
near
a
wall
and
simultaneously
the
parallel
to
it.
This
is
achieved
by
sending
the
wave
down
the
waveguide
in
a zigzag
fashion
bouncing
it
off
the
walls
and
setting
of
a
field
i.e.,
maxima
at
the
centre
of
waveguide and
zero
at
the
walls.
In
this
case,
the
walls
are
nothing
to
be
short
circuit
and
they
don’t
interfere
with
the
wave
pattern
setup
between
them.
To
measure
consequences
of
zigzag
propagation
are
apparent.
The
first
is
that
velocity
of
propagation
in
a
waveguide
must
be
less
than
that of
free
space.
7
P
L
A
N
E
W
A
V
E
O
F
A
C
O
N
D
U
C
T
I
N
G
S
U
R
F
A
C
E
If
actual
velocity
of
wave
is
V
c
,
then
the
simple
trigonometry
shows
that
the
velocity
of
the
wave
in the
direction
parallel
to
conducting
is
V
g
and
velocity normal
to
the
wall
is
V
n
.
V
g
V
n
V
C
λ
P
λ
λ
n
θ
P
A
R
A
L
L
E
L
A
N
D
N
O
R
M
A
L
W
A
V
E
L
E
N
G
T
H
:
D
i
s
t
a
n
c
e
b
e
t
w
e
e
n
2
s
u
c
c
e
s
s
i
v
e
i
d
e
n
t
i
c
a
l
p
o
i
n
t
s
i
.
e
.
,
s
u
c
c
e
s
s
i
v
e
c
r
e
s
t
s
o
r
s
u
c
c
e
s
s
i
v
e
t
r
o
u
g
h
s
.
I
n
t
h
e
f
i
g
u
r
e
,
i
t
i
s
s
e
e
n
t
h
a
t
t
h
e
w
a
v
e
l
e
n
g
t
h
i
n
d
i
r
e
c
t
i
o
n
o
f
p
r
o
p
a
g
a
t
i
o
n
o
f
w
a
v
e
i
s
λ
b
e
i
n
g
t
h
e
d
i
s
t
a
n
c
e
b
e
t
w
e
e
n
2
s
u
c
c
e
s
s
i
v
e
c
r
e
s
t
s
i
n
t
h
i
s
d
i
r
e
c
t
i
o
n
.
INC
I
D
E
NT
WAVE
8
REFLECTED
WAVE
p
P
A
R
A
L
L
E
L
A
N
D
N
O
R
M
A
L
W
A
V
E
L
E
N
G
T
H
c
o
n
t
…
.
So
the
distance
between
2
consecutive
crests
in
the
direction
parallel
to
conducting
plane
is
λ
p
and
wavelength
perpendicular
to
surface
is
λ
n
.
λ
=
λ
λ
n
=
λ
s
i
n
θ
c
os
θ
V
C
=
f
λ
V
g
=
V
C
sinθ
s
i
n
θ
V
p
=
fλ
p
=
f
λ
=
V
C
s
i
n
θ
V
p
–
phase
velocity
–
velocity
with
which
wave
changes
its
phase.
V
g
–
group
velocity
–
velocity
parallel
to
wall.
P
A
R
A
L
L
E
L
P
L
A
N
E
W
A
V
E
G
U
I
D
E
In
order
to
wave
to
propagate
in
a
waveguide
there
should
be
no
voltage
at
the
walls
because
walls
are
purely
conductive
if
there
exists
some
voltage
at
walls
the
wave
get
shorted
and
there
will
be
no
propagation
of
wave.
Eq
1
9
B
T
1
4
E
CE0
3
1
CHARAN
SAI
KATAKAM
V
N
IT
P
A
R
A
L
L
E
L
P
L
A
N
E
W
A
V
E
G
U
I
D
E
c
o
n
t
…
.
.
Voltage
variation
in
waveguide
is
almost
similar
to
the
transmission
lines.
Waveform
of
voltage
in
transmission
lines
is
as
follows
λ/2
2λ/2
3λ/2
So
dimension
of
waveguide
can
have
following
values
such
that
voltage
at
walls
will
be
zero.
a=
λ
n
/
2
10
a=2
λ
n
/
2
a=3
λ
n
/
2
D
E
R
I
V
A
T
I
O
N
O
F
C
U
T
O
F
F
W
A
V
E
L
E
N
G
T
H
‘a’
is
the
distance
between
the
walls
λ
n
is
the
wavelength
in
the
direction
normal
to
both
walls.
‘m‘
is
the
no
.of
half
wavelength
of
electric
intensity
to
be
established
between
walls
which
is
nothing
but
integer.
From
equation
,
it
is
easy
to
say
that
as
the
free
space
wave
length
is
increased,
there
comes
a
point
beyond
which,
the
wave
can
no
linger
propagate
in
a
waveguide
with
fixed
‘a’
&
‘m’
.
The
free
space
wavelength
at
which
this
takes
place
is
called
cut
off
wavelength
and
it
defines
as
the
smallest
free
2
a
=
mλ
n
=
mλ
2
c
o
s
θ
From
Eq
1
m
λ
cosθ
=
2a
P
s
i
n
θ
√
1-
cos
2
θ
λ
=
λ
=
λ
λ
√
1-
(
mλ
)
2
(2a)
2
11
D
E
R
I
V
A
T
I
O
N
O
F
C
U
T
O
F
F
W
A
V
E
L
E
N
G
T
H
Space
wavelength
that
is
just
unable
to
propagate
in
waveguide
under
such
condition.
1
-
So
a
waveguide
allows
a
signal
having
a
frequency
more
than
cut
off
frequency
so
a
waveguide
acts
as
high
pass
filter.
The
lowest
cut
off
frequency
can
be
calculated
through
mλ
o
2a
(
)
2
=
0
o
λ
=
2a
m
λ
o
=
cut
off
wavelength
for
dominant
mode
,
m=1
,
λ
o
=2a
λ
λ
p
=
√
1
-
λ
(
m
)
2
λ
p
=
2a
λ
o
√
1
-
(
λ
)
λ
2
λ
p
=
guided
wavelength
λ
=
free
space
wavelength
fc
=1.5*10
8
√
m
a
(
)
2
+
(
n
2
b
)
P
R
O
B
L
E
M
Calculate
the
lowest
frequency
and
determine
the
mode
closest
to
the
dominant
mode
of
rectangular
waveguide
5.1cm*2.4cm.Calculate
the
cut
off
frequency
of
dominant
mode?
Ans)
From
the
formula
in
previous
slide
for
dominant
mode
is
TE
1,0
so
m=1,n=0,a=5.1*10
-2
,b=2.4*10
-2
we
get
f
c
=
2.94GHz
Mode
closest
to
dominant
mode
can
be
determined
by
substituting
m
,
n
values.
for
m=0, n=1,
we
get
f
c
=
6.25GHz
for
m=0,
n=2,
we
get
f
c
=
12.5GHz
for
m=2,
n=0,
we
get
f
c
=
5.8GHz
As
f
c
=
5.8GHz
in
TE
2,0
is
close
to
f
c
=
2.94GHz
in
TE
2,0
,
therefore
mode
closest
to
dominant
mode
is
TE
2,0
.
13
G
R
O
U
P
A
N
D
P
H
A
S
E
V
E
L
O
C
I
T
Y
I
N
W
A
V
E
G
U
I
D
E
g
c
p
V
=
V
s
i
n
θ
V
=
V
p
V
g
=
(V
c
)
2
------Eq
2
V
p
=f
λ
p
p
Substituting
above
equation
in
Eq
2
we
get
V
g
=
V
c
The
above
equation
represents
that
velocity
of
propagation
or
group
velocity
in
a
waveguide
is
lower
than
in
free
space.
Group
velocity
decreases
as
the
free
space
wavelength
approaches
the
cut
off
wavelength
and
is
zero
when
the
two
wavelength
are
equal.
s
i
n
θ
V
c
θ
V
g
V
c
V
n
θ
λ
p
λ
λ
n
V
=f
λ
√
1
-
(
λ
λ
o
)
2
V
p
=
V
c
1
-
λ
√
(
)
λ
o
2
(
√
1
-
λ
λ
o
2
14
(
)
)
P
R
O
B
L
E
M
It
is
necessary
to
propagate
a
10GHz
signal
in
a
waveguide whose
wall
separation
is
6cm.What
is
the
greatest
no.of
halfwaves
of
electric
intensity
which
it
will
possible
to
establish
between
2
walls.
Calculate
the
guide
wavelength
for
this
mode
of
operation?
Solution:
a
=
6
cm
,
f
=
10GHz
=>
λ
=
f/c
=
3cm
λ
0
=
(2*a)/m
for
m=1
=>
λ
0
=12/1
=12
cm
for
m=2
=>
λ
0
=12/2
=6
cm
for
m=3
=>
λ
0
=12/3
=4
cm
for
m=4
=>
λ
0
=12/4
=3
cm
So
maximum
value
of
m
is
4.
15
Waveguides are structures that transmit electromagnetic waves, offering advantages over traditional transmission lines such as higher power handling capacity, lower loss, and the ability to operate at higher frequencies. They come in different shapes like rectangular and circular, with distinct modes of operation. The dominant mode is the lowest frequency that can propagate in a waveguide, with unique behaviors related to electric and magnetic fields. Understanding waveguide behavior is crucial for efficient signal transmission in various applications.
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Presentation Transcript
KNOWLEDGE INSTITUTE OF TECHNOLOGY INTRODUCTION OFWAVEGUIDES Prepared by RAADHU.T ASSISTANT PROFESSOR
INTRODUCTION TOWA VEGUIDES In a waveguide energy is transmitted in theform of electromagnetic waves whereas in transmission line, itis transmitted in the form of voltageandcurrent. Conductionof energytakesplacenot through the wallswhose function is only to confine this energy but through the dielectric filling the waveguide whichis usually air. ADV ANT AGES OF WAVEGUIDE OVER A TRANSMISSION LINE: In a transmission line high frequency signals cannotbe transmitted because Large mutual induction. Length is verylarge. Multiplexing the signals is notpossible in transmission lines. Flashover is less in case of waveguide. Minimum loss of power in case ofwaveguide. Power handling capability is 10 times than coaxialcable. 3
INTRODUCTION TOWA VEGUIDES Mechanicalsimplicity. Higher maximum operating frequency.(3GHZ to100GHZ). Normally waveguides are of twotypes. Rectangularwaveguide. Circularwaveguide. a b d Rectangularwaveguide Circularwaveguide REFLECTION OF WAVEFORM IN CONDUCTING PLANE: Vg V is velocity guided bywall g (groupvelocity) Vn is the normal velocityto wall VI VR Vn 4
DOMINANT MODE OFOPERATION DOMINANT MODE :The natural mode of oper- -ation for a waveguide is called dominant mode. This modeis the lowest possible frequency that can be propagated in a waveguide. a L L a MAGNETI C FIELD ELECTRIC FIELD m=no. of half wavelengthacross waveguide. n=no. of half wavelengthalong the waveguide height. L > > > a 5
DOMINANT MODE OF OPERATION cont n=1 => 1 circle n=2 => 2 circles m- electricfield n magneticfield The signal of maximum wavelength that can pass througha waveguideis 0=2a/m BASICBEHAVIOUR Magnetic field m=1 m=2 An electromagnetic plane wave in space is transverseelectrom- agnetic or TEM. The electric field, the magnetic field and direction of propagation are mutually perpendicular. If such a wave were sent straight down a waveguide, it would not propagatein it. Thisis becausethe electric field would be short-circuited bythe wallssince the wallsareassumedto be 6
BASIC BEHAVIOURcont. Perfect conductors and a potential cannotexist acrossthem. What must be found in some method of propagation which doesn t require an electric field to exist near a wall and simultaneously the parallel to it. This is achieved by sending the wave down the waveguide in a zigzag fashion bouncing it off the wallsandsettingof afield i.e.,maximaat the centreof waveguide and zero at thewalls. Inthis case,the wallsarenothing to beshort circuitandthey don t interferewith the wavepatternsetupbetween them. T omeasure consequences of zigzag propagation are apparent. Thefirst isthat velocity of propagationin awaveguidemustbe lessthan that of freespace. 7
PLANE WA VE OF A CONDUCTINGSURF ACE If actual velocity of wave is Vc , then the simple trigonometry shows that the velocity of the wave in the direction parallel to conducting is Vg and velocity normalto the wall isVn. Vg P REFLECTED WA VE INCIDENT WA VE VC Vn n P ARALLEL AND NORMAL WA VE LENGTH:Distance between 2 successive identical points i.e., successive crests or successive troughs. In the figure, it is seen that the wavelength in direction of propagation of wave is being the distance between 2 successive crests in thisdirection. 8
P ARALLEL AND NORMAL WA VELENGTHcont. So the distance between 2 consecutive crestsin the direction parallel to conducting plane is p andwavelength perpendicular to surface is n. = sin cos Vg = VCsin Vp=f p=f =VC sin Vp phase velocity velocity with which wave changesits phase. Vg group velocity velocity parallel towall. P ARALLEL PLANEWAVEGUIDE Inorder to wave to propagatein awaveguidethere shouldbe no voltage at the walls because walls are purely conductive if there exists some voltage at walls the wave get shorted and there will be no propagation ofwave. n = VC = f p Eq1 sin 9 BT14ECE031 CHARAN SAIKATAKAM VNIT
P ARALLEL PLANE WAVEGUIDEcont.. Voltage variation in waveguide is almostsimilar to the transmissionlines. Waveform of voltage in transmission lines is asfollows /2 2 /2 3 /2 So dimension of waveguide can have following valuessuch that voltage at walls willbe zero. a= n / 2 a=2 n / 2 a=3 n / 2 10
DERIVATION OF CUT OFFWA VELENGTH a is the distance between the walls n is the wavelengthin the direction normalto both walls. m isthe no.of half wavelengthof electric intensity to be established between walls which is nothing butinteger. a=m n= m 2cos m cos =2a = = From Eq1 2 P sin 1-(m )2 (2a)2 1-cos2 From equation , it is easy to say that as the free space wave length is increased, there comes a point beyond which, the wave can no linger propagate in a waveguide withfixed a & m . The free space wavelength at which this takes place is calledcut off wavelength andit definesasthe smallestfree 11
DERIVATION OF CUT OFFWA VELENGTH Space wavelength that is just unable topropagate in waveguide under suchcondition. 1- 2a ( ) o =2a 2= 0 m o m o=cut off wavelength for dominant mode,m=1, o=2a p = 1 - (m )2 p= 1 -( ) p = guided wavelength = free spacewavelength 2a 2 o So a waveguide allows a signal having a frequency more than cut off frequency so a waveguide acts as high passfilter. The lowest cut off frequency can be calculatedthrough fc=1.5*108 n b) 2 m a ( )2+(
PROBLEM Calculate the lowest frequency anddetermine the mode closest to the dominant mode of rectangular waveguide 5.1cm*2.4cm.Calculate the cut off frequencyof dominant mode? Ans)Fromthe formula in previous slidefor dominant modeis TE1,0 so m=1,n=0,a=5.1*10-2 ,b=2.4*10-2 weget fc =2.94GHz Mode closest to dominant mode can be determinedby substituting m , nvalues. for m=0, n=1, we get fc=6.25GHz for m=0, n=2, we get fc=12.5GHz for m=2,n=0,we get fc=5.8GHz Asfc=5.8GHzin TE2,0iscloseto fc=2.94GHzin TE2,0,therefore mode closestto dominant mode isTE2,0. 13
GROUP AND PHASE VELOCITY INWAVEGUIDE Vc Vg V =V sin Vp Vg = (Vc )2 ------Eq2 Vp=f p V =f 1-( Vp= 1- ( ) V = sin g c p p Vc Vn n p o)2 Vc o 2 Substituting above equation in Eq 2 weget Vg =Vc The above equation represents that velocity of propagationor group velocity in a waveguide is lower than in free space. Group velocity decreases as the free space wavelength approaches the cut off wavelength and is zero when the two wavelength areequal. ( 1- ( ) ) 2 o 14
PROBLEM It is necessary to propagate a 10GHz signal ina waveguide whose wall separation is 6cm.What is the greatest no.of halfwaves of electric intensity which it will possible to establish between 2 walls. Calculate the guide wavelengthfor this mode ofoperation? Solution: a = 6 cm , f = 10GHz => = f/c =3cm 0 =(2*a)/m for m=1 => 0 =12/1 =12 cm for m=2 => 0 =12/2 =6cm for m=3 => 0 =12/3 =4cm for m=4 => 0 =12/4 =3cm So maximum value of m is4. 15
