INTRODUCTION OF WAVEGUIDES
VNIT
BT14ECE031
Under guidance of
Joydeep Sengupta sir
CHARAN SAI KATAKAM
1
INTRODUCTION TO WAVEGUIDES
 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.
ADVANTAGES OF WAVEGUIDE OVER A TRANSMISSION LINE :
 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.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
2
INTRODUCTION TO WAVEGUIDES
 Mechanical simplicity.
 Higher maximum operating frequency.(3GHZ to 100GHZ).
 Normally waveguides are of two types.
 Rectangular waveguide.
 Circular waveguide.
a
b
d
Rectangular waveguide
Circular waveguide
 REFLECTION OF WAVEFORM IN CONDUCTING PLANE :
Vg
Vg is velocity guided by wall
VI
(group velocity)
VR
Vn is the normal velocity to
Vn
wall CHARAN SAI KATAKAM
BT14ECE031
VNIT
3
WAVEGUIDES
 Vn and Vg are components of VR.
VI – incident velocity
VR – reflected velocity
 TEM – Transverse Electro Magnetic
 In TEM, electric ,magnetic and propagation of wave should be
perpendicular to each other.
 In rectangular waveguide, if TEM strikes the wall, as it is a
conductor it get absorbed by wall whereas in rectangular
waveguide, reflection takes place when TEM strikes wall.
Therefore TEM cannot be propagated through rectangular
waveguide but in circular waveguide.
 Only TE and TM are possible in rectangular waveguide.
 TE- Transverse Electric: Electric field is perpendicular to
propagation of wave.
 TM-Transverse Magnetic: Magnetic field is perpendicular to
propagation of wave.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
4
DOMINANT MODE OF OPERATION
 DOMINANT MODE : The natural mode of oper-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
>
VNIT
L
ELECTRIC
FIELD
>
>
m=no. of half
wavelength across
waveguide.
n=no. of half
wavelength along
the waveguide
height.
MAGNETIC
FIELD
a
BT14ECE031
CHARAN SAI KATAKAM
5
DOMINANT MODE OF OPERATION cont …
 n=1 => 1 circle
Magnetic
field
n=2 => 2 circles
 m- electric field
n – magnetic field
m=1
m=2
 The signal of maximum wavelength that can pass through a
waveguide is
λ0=2a/m
BASIC BEHAVIOUR
 An electromagnetic plane wave in space is transverse electromagnetic 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
VNIT
BT14ECE031
CHARAN SAI KATAKAM
6
BASIC BEHAVIOUR cont….
Perfect conductors and a potential cannot exist
across them.
 What must be found in some method of propagation which
does t e ui e a ele t i field to e ist ea a all a d
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
do t i te fe e ith the a e patte setup et ee the .
 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.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
7
PLANE WAVE OF A CONDUCTING SURFACE
 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 normal to
the wall is Vn.
Vg
λP
INCIDENT
WAVE
Vn
VC
REFLECTED
WAVE
θ
λn
λ
 PARALLEL AND NORMAL WAVE 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 λ being the distance
between 2 successive crests in this direction.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
8
PARALLEL AND NORMAL WAVELENGTH cont….
 So the distance between 2 consecutive crests in
the direction parallel to conducting plane is λp and wavelength
perpendicular to surface is λn.
λp =
λ
λn = λ
VC = f λ
cosθ
sinθ
Vg = VC sinθ
Eq 1
V
f
λ
C
Vp= fλp=
=
sinθ sinθ
 Vp – phase velocity – velocity with which wave changes its
phase. Vg – group velocity – velocity parallel to wall.
PARALLEL PLANE WAVEGUIDE
 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.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
9
PARALLEL PLANE WAVEGUIDE cont…..
 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= λn / 2
a=3 λn / 2
VNIT
a=2 λn / 2
BT14ECE031
CHARAN SAI KATAKAM
10
DERIVATION OF CUT OFF WAVELENGTH
 a is the dista e et ee the alls λn is the
wavelength in the direction normal to both walls.

is the o .of half a ele gth of ele t i i te sit to e
established between walls which is nothing but integer.
mλ From Eq 1
mλn
a= 2 =
2cosθ
mλ
cosθ = 2a
λ
λ
λ
λP = sinθ =
√ 1- cos2 θ √ 1- (mλ)2 2
(2a)
 From equation , it is easy to say that as the free space wave
length is increased, there comes a point beyond which, the
a e a o li ge p opagate i a a eguide ith fi ed a &
. The f ee spa e a ele gth at hi h this takes pla e is
called cut off wavelength and it defines as the smallest free
VNIT
BT14ECE031
CHARAN SAI KATAKAM
11
DERIVATION OF CUT OFF WAVELENGTH
Space wavelength that is just unable to propagate
in waveguide under such condition.
mλo 2
1= 0
2a
λo= 2a
m
λo= cut off wavelength for dominant mode , m=1 , λo=2a
λ
λp =
λp = guided wavelength
2
m
√ 1 - λ( 2a )
λ = f ee spa e a ele gth
λ
λp =
√ 1 -( λλ ) 2
( )
o
 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
m 2
n 2 fc - lower cut off frequency
8
fc =1.5*10 ( a ) +( b )
a,b – waveguide measurements
√
VNIT
BT14ECE031
m,n – integers
indicating the modes12
CHARAN SAI KATAKAM
PROBLEM
 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
TE1,0 so m=1,n=0,a=5.1*10-2 ,b=2.4*10-2 we get
fc = 2.94GHz
 Mode closest to dominant mode can be determined by
substituting m , n values.
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
 As fc = 5.8GHz in TE2,0 is close to fc= 2.94GHz in TE2,0 , therefore
mode closest to dominant mode is TE2,0 .
VNIT
BT14ECE031
CHARAN SAI KATAKAM
13
GROUP AND PHASE VELOCITY IN WAVEGUIDE
Vc
Vg = Vc sinθ Vp =sinθ
Vp Vg = (Vc )2 ------Eq 2
Vp=f λp
λ
Vp=f
√
Vp=
1 - ( λ )2
λo
Vg
θ
Vc
Vn
λp
θ
λn
λ
Vc
√1 - ( λλ ) 2
o
 Substituting above equation in Eq 2 we get
Vg = Vc ( 1- ( λλ )2 )
o
 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.
√
VNIT
BT14ECE031
CHARAN SAI KATAKAM
14
PROBLEM
 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
λ0 = (2*a)/m
for m=1 => λ0 =12/1 =12 cm
for m=2 => λ0 =12/2 =6 cm
for m=3 => λ0 =12/3 =4 cm
for m=4 => λ0 =12/4 =3 cm
 So maximum value of m is 4.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
15
RECTANGULAR WAVEGUIDE
y
b
x
a
 All equations are same for parallel plane waveguide and rect-angular waveguide are same except characteristic impedance.
TEm,0 MODE
z
Zo =
where Z0 = characteristic impedance
2
√1 - ( λλo) Z = 120π = 377ohms (characteristic
impedance of free space).
 If λ= λo in particular mode then Zo = ∞. i.e., a e p opagatio
will have infinite resistance i.e., Vg = 0.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
16
TEm,n MODES
 The obvious difference between TMm,n mode
mode and those described thus foe is that magnetic field here
is transverse only and the electric field has a component in
the direction of propagation. Although most of the behaviour
of thus modes is same for TE modes, a no.of differences do
exist.
 The first such difference is due to the fact that lines of magnetic field lines are closed loops consequently if magnetic field
exists and is changing in x- direction it must also exist and be
changing in Y- direction. Hence TMm,0 modes cannot exist in
rectangular wave guide.
 If we want to propagate TM mode, if a wave is propagating in
z-direction it may propagate in x-direction and y-direction.
 In TEm,0 mode if λ = λ0 then Z0 = 0, if potential difference
between two sides is zero there will be no variation of field.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
17
CIRCULAR WAVEGUIDE
 Let
e i te al adius of i ula a eguide
then cut of wavelength of waveguide is
2πr
λ0 =
kr –solution of a vessel function equation.
kr
TE MODE
TM MODE
MODE
Kr
MODE
Kr
TEO,1
3.83
TMO,1
2.40
TE1,1
1.84
TM1,1
3.83
TE2,1
3.05
TM2,1
5.14
TE0,2
7.02
TM0,2
0.52
TE1,2
5.33
TM1,2
7.02
TE2,2
6.71
TM2,2
8.42
 The are values of Kr for different modes of transverse electric
and transverse magnetic modes of circular wave guide.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
18
PROBLEM
 Calculate the ratio of cross section of a circular
waveguide to that of rectangular waveguide if each is to have
same cut off wavelength for its dominant mode.
 Solution: for circular for
λ0 = 2πr/(kr) = 2πr/1.84 = 3.14r
TE1,1 is dominant mode
r = (λ0 * 1.84)/ 2π = λ0/3.41
Ac = π*r2 = π*(λ0/3.41)2
for rectangular waveguide λ0 = 2a/m
TE1,0 is dominant mode => a = (λ0*1)/2
(since m=1)
Ratio of a : b = 2 :1,area = a *b = a*(a/2) = a2/2=(λ0/2)2/2
Ratio of cross sections is π*(λ0/3.41)2 : (λ0/2)2/2
2.17
:
1
VNIT
BT14ECE031
CHARAN SAI KATAKAM
19
CIRCULAR WAVEGUIDE
 Smallest value of Kr, the TE1,1 mode is dominant
in circular waveguide. The cut off wavelength of this mode is
λ0 = (2πr)/(1.84)
λ0 =1.7d
d – inner diameter of circular waveguide.
 Another difference lies in the different method of mode
levelling, which must be used because of the circular cross
se tio . The i tege
de otes the no.of full wave intensity
a iatio alo g the i u fe e e a d
ep ese ts the no.of
half wave intensity changes radially out from the centre to the
wall. It is seen that cylindrical co-ordinates are used here.
 DISADVANTAGES OF CIRCULAR WAVEGUIDE OVER RECTANGULAR
WAVEGUIDE:
 The first drawback associated with the circular waveguide is
that its cross section will be much bigger in area than that of a
corresponding rectangular waveguide used to carry the same
signal.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
20
CIRCULAR WAVEGUIDE
 Another problem with circular waveguide is that
it is possible for the plane of polarisation to rotate during the
waves travel through waveguide. This may happen because of
roughness and discontinuities in the wall at departure from
circular cross section.
 ADVANTAGES:
 It is easier to manufacture than rectangular waveguide.
 They are also easier to join together.
VNIT
BT14ECE031
CHARAN SAI KATAKAM
21