EC8651 MAILAM ENGINEERING COLLEGE

UNIT-4
WAVE GUIDES
2 Marks

1. A rectangular waveguide with dimensions a = 8.5 cm and b = 4.3 cm is fed by 5 GHZ

carrier. Will a TE11 mode be propagated?
2 2
m n
   
 a  b
Cut-off frequency, f c  : Given, m = n = 1 ; a = 8.5 cm and b = 4.3 cm
2 
and f = 5 GHZ Assuming air is the dielectric within the guide,    o and   o .Therefore,
1
 3 X 10 8 m / sec . Substituting the values of m, n, a and b, we get,
 o o
2 2
 1   1 
   
 .085   0.043 
fc  X 3 X 10 8 m / sec = 3.9GHZ
2
Since, fc < f, a TE11 mode will be propagated.

2. How is the TE10 mode initiated in rectangular waveguide using an open ended coaxial
cable?
 In order to launch a particular mode, a type of probe is chosen which will produce
lines of E and H that are roughly parallel to the lines of E and H for that mode.
 To initiate TE10 mode, the probe is kept parallel to the y axis and so produces lines of
E in the y direction and lines of H which lie in the x-z plane.
3. Calculate the cut-off wavelength of a rectangular waveguide whose inner dimensions are
a = 2.3 cm and b = 1 cm operating at TE10 mode.
Solution: Cut-off wavelength λc = 2a = 4.6 cm

4. Which mode is called as dominant mode in the circular waveguide?

TE11 mode is called as dominant mode in the circular waveguide.
5. What are dominant mode of propagation for Rectangular wave guide ?( Nov/Dec 2022)
Dominant mode: It is the mode with the lowest cut-off frequency.TE10 is the dominant
mode. Degenerate modes are defined as those which have the same cut-off frequency or same
cut-off wavelength. TE11 and TM22 are degenerate modes. Similarly, all other higher order
modes also degenerate i.e., TE12 and TM12 modes, TE21 and TM21 modes etc degenerate. In
practice, the waveguide dimensions are so chosen that higher order modes are not supported and
only the desired mode is made to propagate through the waveguide. With this, interference of
various modes can be eliminated
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6. What are the applications of cavity resonators?(Nov/Dec 2015,APR/MAY 2017,18)

i. Cavity resonators are used in klystron amplifiers for amplifying microwave signals and
in reflex klystron oscillators for generation of microwave signals.
ii. Cavity resonators are used in cavity magnetron for generation of microwave signals.
iii. Cavity resonators are used in duplexers in Radar systems as resonant cavity in TR
(Transmit-Receive tubes) and ATR tubes (Anti-Transmit-Receive tubes)
iv. Cavity resonators are used in cavity wave meters for the measurement of frequency at
microwave signals.
7. Define cut-off frequency and cutoff Wavelength of modes of propagation in rectangular
waveguides (Apr/May 2022)
The frequency at which the wave motion ceases is called the cut-off frequency of the
waveguide.
The cut-off wavelength is the wavelength at which an optical fiber becomes single-
mode. At wavelengths shorter than cut-off several optical modes may propagate - the fiber is
multi-mode.
Cut-off wavelength C = 2a.

8. Define wave impedance and write down the expression for the wave impedance for the
TE and TEM waves.
Wave impedance is defined as the ratio of electric to magnetic field strength.

Wave impedance for TE waves, ZTE =
fc 2
1
f2

Wave impedance for TEM waves, ZTEM = η = ; If air is the dielectric in

0
between the conducting planes, then μ= μ0 & ε = ε0 => η =  120 , ohms
0

9. What is a waveguide?
Waveguide is a hollow conducting metallic tube of uniform cross section which is used
for propagating electromagnetic waves that are guided along the surfaces of the tube.

10. What are the basic types of propagating waves in a uniform waveguide?
TM waves and TE waves are the basic types of propagating waves in a uniform
waveguide.

11. Why rectangular waveguides are preferred over circular waveguide?

Rectangular waveguides are preferred over circular waveguides because of the following
reasons.
 Rectangular waveguide is smaller in size than a circular waveguide of the same
operating frequency.
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 It does not maintain its polarization through the circular waveguide.

 The frequency difference between the lowest frequency on dominant mode and the next
mode of a rectangular waveguide is bigger than in a circular waveguide.
12. What is meant by the dominant mode of a waveguide?
2 2
1 m n
 The cut-off frequency f c       shows that the physical size of
2   a   b 
the waveguide will determine the propagation of the modes of specific orders
determined by m and n.
 The minimum cut-off frequency is obtained for a guide having dimension a > b,
for m = 1,n = 0 Since for TMmn modes, m  0 or n  0, the lowest order mode
possible is TE10 and TM11 mode, called the dominant mode in a rectangular
waveguide for a >b

13. What is known as evanescent mode in a rectangular waveguide?(Nov/Dec 2018)

When the operating frequency is lower than the cut-off frequency the propagation constant
becomes real ie γ = α . The wave cannot be propagated. This non-propagating mode is known
as evanescent mode.

14. What is the cut-off wavelength of the TE10 mode in a rectangular waveguide?
Cut-off wavelength C = 2a.

15. Explain the impossibility of TEM wave in waveguide.

 Suppose a TEM wave is assumed to exist within a hollow guide of any shape.
Then lines of H must lie entirely in the transverse plane. Also in a nonmagnetic
material, . H = 0 which requires that the lines of H be closed loops.
 Therefore, if a TEM exists inside the guide, the lines of H will be closed loops in
plane perpendicular to the axis. Now by Maxwell’s first equation the magneto
motive force around each of these closed loops must equal the axial current ( J C
or JD) ie. Conduction or displacement through the loop. In the case of a guide
with an inner conductor, example, a coaxial transmission line, this axial current
through the H loops is the conduction current in the inner conductor.
 However, for a hollow waveguide having no inner conductor, this axial current
must be a displacement current. But an axial displacement current requires an
axial component of E, something not present in a TEM wave. Therefore the TEM
wave cannot exist in a single conductor waveguide.

16. What are the applications of the waveguide?

The waveguides are employed for transmission of energy at very high frequencies where the
attenuation caused by waveguide is smaller. Waveguides are used in microwave transmission.
Circular waveguides are used as attenuators and phase shifters.

17. What are degenerate modes?

Some of the higher order modes, having the same cut-off frequency , are called
degenerate modes. It is seen that in a rectangular waveguide possible TE nm and TMmn modes (
both m  0 and n  0 ) are always degenerate.
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18. Mention the uses of circular waveguides.

Circular waveguides are used as
 Attenuators and
 Phase shifters.

19. Which mode is the dominant mode in a circular waveguide?

The dominant mode in a circular waveguide is TE 11.

20. Write down the expression for the cut-off frequency in a circular waveguide.

fc =
hnm
; where hnm =
ha nm
for TM waves and hnm =
ha  nm
'
for TE waves.
2  a a

21. Why is TM01 mode preferred to the TE01 mode in a circular waveguide?
TM01 mode is preferred to the TE01 mode, since it requires a smaller diameter for the same
cut-off wavelength.

22. What is the special significance of TE01 mode in a circular waveguide?

The special significance of TE01 mode in a circular waveguide is that the attenuation effect
decreasing with increase in frequency.

23. List the degenerate modes in a circular waveguide.

All the TE0n and TM1n (where n = 1,2,3, ……) modes are degenerate in a uniform circular
waveguide.

24. What is a microwave cavity resonator?

Microwave resonator is a tunable circuit used in microwave circuits. It is a metallic
enclosure that confines the electromagnetic energy. When the resonator resonates at resonant
frequency, the peak energies stored in the electric and magnetic fields are equal.

25. What are the parameters that determine the performance of a cavity resonator?
The performance parameters of microwave resonator are:
 Resonant Frequency,
 Quality factor and
 Input impedance.

26. Define the quality factor Q of a resonator.

The quality factor Q is a measure of frequency selectivity of the resonator. It is defined as
Q = 2 × (maximum energy stored)/ Energy dissipated per cycle) =  W/P where W is the
maximum stored energy; P is the average power loss.
27. Define resonant frequency.
Resonant frequency of microwave resonator is a frequency at which the energy in the
resonator attains maximum value. Ie., twice the electric energy or magnetic energy.

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28. Define unloaded Q of a cavity resonator.

When a cavity is assumed to be not connected to any external circuit or load, Q accounts
for the internal losses and is called the unloaded quality factor Q0.

29. Why transmission line resonator is not usually used as microwave resonator?
Very high frequencies transmission line resonator does not give very high quality factor
Q due to skin effect and radiation loss in braided cables. So, transmission line resonator is not
used as microwave resonator.

30. Why rectangular or circular cavities can be used as microwave resonators?

Rectangular or circular cavities can be used as microwave resonators because they have
natural resonant frequency and behave like a LCR circuit.
31. Write down the resonant frequency of rectangular resonator.
2 2 2
1 m n  p
The resonant frequency, f o          ; where b<a<d
2   a  b d 

32. What is the dominant mode for rectangular resonator and circular cavity resonator?
The dominant mode of a rectangular resonator depends on the dimensions of the cavity.
For b<a<d, the dominant mode is TE 101. In circular cavity resonator, for d<2a, the dominant
mode is TM010 and for d > 2a, the dominant mode is TE 111

33. Write the applications of cavity resonator.

 Used as tuned circuits
 Used in UMF tubes
 Used in Klystron amplifier
 Used in duplexer in Radar
 Used in Microwave frequency meter.

34.Define TE and TM waves.

Transverse electric wave is a wave in which electric field strength E is entirely transverse. It
has a magnetic field strength Hz in the direction of propagation and no component of electric
field strength Ez in the same direction.
Transverse magnetic wave is a wave in which magnetic field strength H is entirely transverse.
It has electric field strength Ez in the direction of propagation and no component of magnetic
field strength Hz in the same direction.

35.What are the characteristics of TEM waves? (May/June 2013)

 TEM waves are special type of TM wave.
 It does not have either Ez or Hz component.
 Its velocity is independent of frequency.
 Its cut-off frequency is zero.

36.What are guided waves? Give examples.

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 The electromagnetic waves that are guided along or over conducting or dielectric surface
are called guided waves.
 Examples: Parallel wire and transmission lines.

37.Define TEM waves.(Nov/Dec 2019)

 The transverse electromagnetic waves are waves in which both electric and magnetic
fields are transverse entirely but have no components of E z and Hz.
 It is referred to as principal wave.

38.Define the terms phase velocity and group velocity. (Nov/Dec 2012)
Phase Velocity: The rate at which the wave changes its phase as the wave propagates inside the

region between planes is defined as phase velocity vp and is given by, vp = ;
2
f 
    1   c 
 f 
Group Velocity: The actual velocity with which the wave propagates inside the region between
d
the planes is defined as group velocity vg, given by vg = d

39.Define wave impedance and write the expression for wave impedance of TE waves in
guided waves.
Wave impedance is defined as the ratio of electric to magnetic field strength.

fc 2
1 2
Wave impedance for TE waves, ZTE = f
40.Define attenuation factor.
Attenuation factor α = (power lost/unit length)/ 2 × power transmitted

41.Define cut-off frequency.

The frequency at which the wave motion ceases is called the cut-off frequency of the waveguide.

42. Represent the field distribution of dominant mode in parallel plates .(Apr/May 2022)
This equation, f(x, y, ω) is the excitation function in the frequency. In other words, this is the
distribution of the electromagnetic field in the x-y plane and in the frequency domain, which
might excite multiple modes.

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42.Plot the frequency versus wave impedance curve for the waves between parallel
conducting planes.

Wave impedance Region of no wave

propagation

TE wave

377 ohms TEM wave

TM wave

0
fc frequency

43.Distinguish TE and TM waves. (Nov/Dec 2012)

TE TM
Electric field strength E is entire Magnetic field strength H is entirely
transverse. transverse.
It has z component of magnetic field It has z component of electric field
It has no z component of electric field It has no z component of magnetic field
( ) ( )

44.Define wave impedance. (Nov/Dec 2012)

Wave impedance of a wave traveling inside the region between the planes is defined as the
ratio of electric field intensity along one transverse direction to the magnetic field intensity along
the other transverse direction.

45.Write down the expression for the wave impedance for TM, TE and TEM waves.
2
f 
 1   c 
Wave impedance for TM wave, ZTM =  f  ;

2
f 
1   c 
For TE wave, ZTE =  f 
o
Wave impedance for TEM wave, ZTEM =
o = o
= 120π ohms

46.Write the expression for cut-off frequency?

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m
Cut-off frequency = 2a  , Hz

47.Write the relation between group velocity and phase velocity. (Nov/Dec 2012)
VgVp = C2,
where, Vg = Group Velocity,
Vp = Phase velocity,
C = velocity of light.

48.A pair of perfectly conducting planes are separated 8 cm in air. For a frequency of
5000MHZ with the TM1 mode excited, find the cut-off frequency. (Nov/Dec 2012)
m
Solution: Cut-off frequency = 2a  , HZ
1
= 2 x0.08 4x10 x8.854 x10 = 1875 MHz
7 12

49.Two conducting planes have a distance of separation of 6 cm in air. At a frequency of 6

GHZ with TM2 mode being excited, find the cut-off wavelength.
2a 2 x0.06
c    0.06m
Solution: m 2

50.What is meant by dominant mode? What is the dominant mode for parallel plate
waveguide? (May/June 2012)
• The mode which is having the lowest cut-off frequency and highest cut-off wavelength is
called the dominant mode.
• The dominant mode for parallel plate waveguides are TE10 and TM10.

51.What is the cut-off frequency for TEM wave?

The cut-off frequency for TEM wave is zero.

52.For a frequency of 6000MHZ and plane separation = 7 cm with air dielectric, find the
wave impedance for the TE1 mode.
Solution : Wave impedance for TE wave,

2
f 
1   c 
ZTE =  f  = 403.61 ohms, where
m
fc = 2a  ; m = 1 ; a = 7 cm ; on substituting these values in fc , we get fc = 2.143GHz

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53.Write down the Maxwell's equations. (May/June 2012)

Maxwell's equation for the non-conducting region between planes are given as,

54.Compare transmission line and waveguide. (Apr/May 2011)

Transmission line:
Transmission line is a material medium or structure that forms all or part of a path from one
place to another for directing the transmission of energy, such as acoustic waves as well as
electric power transmission.
Waveguide:
A hollow conducting metallic tube of uniform cross section is used for propagating
electromagnetic waves. Waves, that are guided along the surfaces of the tube is called a
waveguide.

55.A rectangular waveguides of cross section 5 cm x 2cm is used to propagate TM 11 mode

at 10 GHz.Determine the cut off wavelength.(Nov/Dec 2015)

56.Write the expression for cutoff wavelength of the wave which is propagated in between
two parallel planes.(Nov/Dec 2017)
the expression for cutoff wavelength:

54. A Parallel plan guide having distance between them as 4 cm is filled with dielectric
material with dielectric constant of 2,find the cut off frequency for TM11 mode.frequency.
(Nov/Dec 2022)
m
Solution: Cut-off frequency = 2a  , HZ
=1875 MHz
2a 2 x0.06
c    0.06m
m 2

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13 Marks

1. Derive the field expression for Transverse Magnetic Wave in a rectangular wave guide,Dra
Derive the field expression for TM wave propagation in rectangular waveguides stating the
necessary assumptions.
The wave equation in a rectangular wave guide is given by,

µε
The solution for the equation is

e-γz

Where, X is a function of x alone

Y is a function of y alone
Sub. the value of in the equation

Y µε

Substituting

Then Y

Dividing by XY,

+ =0

The expression equates a function of x alone to a function of y alone and this can be
equated to a constant.

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Let, -

A solution of the equation is, X =

Similarly,

The solution of this equation is,

Y=

=( )( )

The constants , A, B are determined by boundary conditions.

when x = 0, x = a, y = 0, y = b.

When x = 0,

This is possible only if

Then the general solution is

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When y = 0,

This is possible only if either . If , is identically zero. So, substituting

Let, C =

Applying boundary conditions in order to evaluate the value of constants A and B.

If x= a, =0

This is possible only if for all values of y.

Therefore, where, m=1,2,3,….

If y= b,

This is possible only if for all values of x.

where, n = 1,2,3,….

Hence y

For propagation, γ = jβ (α = 0), the field expressions are as follows;

[Since, ]

Where, and

In the above expressions a and b are width and height of the waveguide and m and n are integers.
It is known -

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= and

This is the eqn. for propagation constant for a rectangular guide for TM wave.
2. Explain transverse electric waves in rectangular waveguides.(APR/MAY 2017)(Nov/Dec
2017,19,18)

The wave equation in a rectangular wave guide is given by,

µε
The solution for the equation is

e-γz

Where, X is a function of x alone

Y is a function of y alone
Sub. the value of in the equation

Y µε

Substituting

Then Y

Dividing by XY,

+ =0
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The expression equates a function of x alone to a function of y alone and this can be
equated to a constant.

Let, -

A solution of the equation is, X =

Similarly,

The solution of this equation is,

Y=

=( )( )

It is known that
= -

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For TE waves Ez = 0

= -

=- -C1 C3 C2C3A C1C4 C2

C4A

Applying boundary conditions, Ez = 0 when y = 0, y =b.

If y = 0, the general solution is

=- [C1 C4 C2 C4 A

For Ex = 0, C4 = 0.

Then the general solution is

Ex = - [- C1 C3 C2 C3 A

If y=b Ex =0.
For Ex = 0, it is possible either B = 0 or A = if B = 0, the above solution is identically
zero. So it is better to select A =

The general solution is

= [C1 C3 C2 C3 A

Similarly for Ey,

Ey = +

Ey =

= - C1 C3 C2C3 B C1C4 C2
C4 B

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Applying boundary conditions

Ey = 0; x = 0 and x = a
If x = 0,

Ey0 = [C1 C3 C2 C4 B

For Ey0 = 0,C2 = 0.

Then the general expression is

Ey0 [- C1 C3 C2 C4 B

If x = a, then Ey0 = 0.

Ey0 = - [C1 C3 C2 C4B

for Ey0 = 0, B =

Ey0 = [-C1 C3 C2 C4B

Ex0 = [C1 C3 C2 C3 A

Substituting the value C2 = C4 = 0.

Ex0 = C1 C3

Ex0 = C1 C3

Ey0 =- C1 C3

Ey0 =- C1 C3

Let C = C1 C3

Ex0 = C

Ey0 = - C

.
Where A = and B =
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Similarly for Hx0,

Hx0 = +

Hx0 = [ Since Ez = 0]

For propagation ,γ = jβ [Since α = 0]

Hx0 =

But Ey =

Ey

Substituting the value of in the above Hx0 equation

Hx0 = Ey0

Hx0 = Ey0

Substituting the value of Ey0in the above Hx0 equation

Hx0 =

Hx0 =

Hx0 =

Similarly for Hy0,

Hy0 = -

Hy0 = [Ez = 0]

For propagation,γ =jβ [α = 0]

Hy0 =

But Ex =-

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Ex

Substituting the value of in the above Hy0 equation

Hy0 = Ex0

Hx0 = Ex0

Substituting the value of Ex0in the above Hy0 equation

Hy0 =

Hy0 =

Hy0 =

Hy0 = XY
= C1 C3 +C2C3 + C1 C4 + C2 C4
But C2 = C4 = 0
Hz0 = C1 C3
C = C1 C3
Hz0 = C

Hz0 = C

The field equations for TE waves are as follows

Hx0 =

Hy0 =

Hz0 = C

Ex0 = C

Ey0 = - C

.
Where A = and B =

For TE waves the equations for β, fc, λc , v and λ are found to be identical to those of
TM waves.
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3.Derive the characteristics cutoff frequency ,cutoff wavelength, phase

velocity,wavelength,group velocity ,and wave impedance for TE,TM and TEM Waves
between two conducting planes. (Or)Derive the expression for field components of TM
waves in a circular waveguide. (May/June 2013,Nov/Dec 2022)
Or
Derive the TM wave components in circular wave guides using Bessel function. (Nov/Dec
2012)

TRANSVERSE MAGNETIC (TM) WAVES

 Transverse magnetic (TM) wavesH is identically zero.
 The boundary conditions require that E, must vanish at the surface of the guide.

∴Jn ( h a ) = 0 (a  radius of the guide)

 There are an infinite number of possible TM waves corresponding to the infinite number
of roots of Jn( h a ) = 0.
The first few roots are (h a ) 0 1 =2.405
(ha)11 =3.85
(ha)02 =5.52
(ha)12 =7.02
 The various TM waves are referred as TM01, TM11 etc.
 The first subscript refers to the value of n .
 The second subscript refers to the roots in their order (magnitude).

The propagation constant 𝛾=

For propagation,

The cut-off frequency,

fc= where , =

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The phase velocity

The field equations of TM waves are found by substituting Hz=0 in the below equations.

----------------------(1)

TM01 Mode
Fig: TM waves in Circular waveguide
The expression of Ez for TM wave is,
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=An Jn(h

--------------(2)

Substituting these values in equation ( ),

cos n

The variations of these field components with time and in the Z direction are shown by
multiplying the above expressions by the factor .

4. Derive the expression for field components of TE waves in a circular waveguide.

(May/June 2013).(Nov/Dec 2015)
TRANSVERSE ELECTRIC (TE) WAVES:
 Transverse electric (TE) wavesEz is identically zero.
 The field equations for TE waves are given by,

-------------------(1)

The expression of H z for TE waves is

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=Cn Jn (

---------------------(2)

Substituting these derivatives in equation ( ),

 The boundary condition is at

 is proportional to , =0.

 Jn ' ( h a ) = 0.
 The first few roots are (h a ) ' 0 1 =3.83
(ha)'11 = 1.84
(ha)'02 =7.02
(ha)'12 =5.33

 The various TE waves are referred as TE01, TE11, TE02, and TE12.
.

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Fig : TE waves in a circular waveguide.

5. Explain in detail the principle of operation of a rectangular cavity resonator. (Nov/Dec 2012)
Rectangular cavity resonator:
The wave equations in the rectangular resonator should satisfy the boundary conditions of
the zero tangential of electric strength(E) at the four walls.

Fig : Rectangular cavity resonator

Transverse Electric (TEmnp) mode:
 The magnetic field expression in the Z direction is given by,
m x ny pz
Hz =H0 cos cos sin
a b d
where, m=0,1,2,3,....represents the number of the half wave periodicity in the x direction.
n=0,1,2,3,.... represents the number of the half wave periodicity in the y direction.
p=1,2,3,4,.... represents the number of the half wave periodicity in the z direction.
 The electric field in the z direction is,
Ez =0
 The magnetic field in the x direction is

where, h2 =

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cos sin

cos cos

- cos cos

 The magnetic field in y direction is,

=- cos sin

cos cos

sin cos

 The electric field in x direction is,

 The electric field in y direction is ,

Transverse Magnetic (TMmnp) mode:

 The electric field expression in the Z direction is given by,
m x ny pz
Ez =E0 sin sin cos
a b d

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where, m=0,1,2,3,....represents the number of the half wave periodicity in the x direction.
n=0,1,2,3,.... represents the number of the half wave periodicity in the y direction.
p=1,2,3,4,.... represents the number of the half wave periodicity in the z direction.
 The magnetic field in the z direction is,
Hz =0
 The electric field in the x direction is

where, h2 =

sin cos

sin sin

- sin sin

 The electric field in the y direction is

= sin cos

cos sin

 The magnetic field in the x direction is

 The magnetic field in the y direction is

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For either TEmnp or TMmnp mode:

At resonance,

Resonant frequency is given by,

 The dominant mode of a rectangular cavity depends on the dimensions of the

cavity.
 For b<a<d , the dominant mode is TE101 mode (m=1 ,n=0, p=1).

Fig : Rectangular cavity dominant mode field configuration(TE101)

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6.Discuss the propagation of TE Waves between parallel plates and derive the expession for
electric and magnetic fields.(Or) Explain the Transmission of TE waves between parallel
perfectly conducting planes with necessary expressions and diagrams for field
components.(NOV/DEC 2016,April/May 2022)
(or)
Derive the expression for field strength for TE waves between parallel perfectly
conducting planes of infinite extent in the Y and Z direction. (MAY/JUNE 2013)
• TE waves are otherwise called as H waves (or) Transverse electric waves.[ .]
• When , ≠ 0; the field components Hy and Ex will also be equal to zero.
• Non-zero values exist for the field components Hx and Ey.
• since each of the field components obeys the wave equation, the wave can be written for the
component Ey.

µε

since,
Let e-γz
µε
e-γz (- )

= e-γz (- )(- )
µε =0

µε]= 0

Therefore, e-γz

The above equation can be reduced to,

- ------------------------(1) Where,

Equation (1) is the differential eqn of simple harmonic motion,

Auxiliary eqn., is m2 =
m=
The solution for eqn (1) is
---------------------------(2)

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Where C1and C2 are arbitrary constants.

Showing variation with time and variation in the z- direction, the expression for

e-γz

e-γz ---------------------(3)

The arbitrary constants C1and C2 can be determined from the boundary conditions.
For boundary conditions of parallel plate waveguide, the tangential component is zero at
the surface of the parallel conductors for all values of z and time.
This requires,

at x = 0

at x = a

------------------------------ for all values of z

when at x = 0

(3)=> e-γz

0= e-γz
 C1 = 0
Applying 2nd boundary condition,
at x = a

0 = (0 + e-γz
0= e-γz
; but and e-γz ≠ 0
m

, Where m= 1,2,3,……

= e-γz --------------------(4)

Eqn. for

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e-γz --------------------(5)

Integrating eqn (5) on both sides,

= e-γz -----------(6)

Eqn. for ,

------------------------(7)

Sub. the value of in the above eqn.

---------------------------(8)

 Each value of m specifies a particular field configuration or mode. Each wave

associated with integer m is designated as TE m0 wave (or) TEm0 mode.
 The second subscript equal to zero refers to another factor which varies with y, which
is found in the general case of rectangular guide.
 The smallest values of m that can be used in equations (4), (6),(8) is m=1 because
m=0 makes all the fields identically zero.
 The lowest order mode that can exist in this case TE 10 mode.
For wave propagation to occur,

Therefore Equation (4), (6),(8) becomes,

= e-jβz

= e-jβz

Modes of Propagation:

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Electric and Magnetic fields between parallel planes for TE10

7.Discuss the Transmission of TM waves between parallel perfectly conducting planes with
necessary expressions and diagrams for field components.
TM waves are otherwise called as E waves (or) Transverse Magnetic waves.[ .] and ≠ 0;

• Various components of electric and magnetic fields are given by,

Sub. in the above equations, we get

≠
0
=
0
Solving the wave equation for

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µε

since,
Let e-γz
µε
e-γz (- )

= e-γz (- )(- )
µε =0

µε]= 0

Therefore, e-γz

The above equation can be reduced to,

- ------------------------(1) Where,

Auxiliary eqn., is m2 =
m=
The solution for eqn (1) is
---------------------------(2)

Where C3 and C4 are arbitrary constants.

e-γz

e-γz ---------------------(3)

Since the tangential component of H is not zero at the surface of the conductor, the
boundary conditions cannot be applied directly to to solve the above constants C3 and
C4
This requires,
at x = 0
at x = a
------------------------------ for all values of z
When at x = 0

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We require the value of

W.K.T., ------------------------(4)

e-γz

e-γz

e-γz -------------------(5)

Now we apply the first boundary condition,

at x = 0 in eqn.(5)

e-γz

e-γz

e-γz

Sub. in eqn(5),

e-γz -----------------------------(6)

Now applying the 2nd boundary conditions,

at x=a

e-γz

e-γz

sub. h in eqn(6)

= e-γz ---------------------(7)

sub. (7) in (From Maxwell’s eqn(6.1))

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= e-γz }

C4( ) e-γz .

Integrating on both sides,

C4 e-γz

e-γz ________________(8)

sub. in,

γ = [From Maxwell’s eqn.(6.1)]

e-γz ---------------------(9)

For wave propagation to occur,

Then eqn. (7) ,(8) ,(9) becomes,

e-jβz

e- jβz

= e- jβz

Modes of Propagation:

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8. Derive the field expressions for of Transverse Magnetic ave in a rectangukar waveguide
Draw the field expression for the lowest order mode.Derive the TEM waves between
parallel conducting planes. Discuss the properties of TEM wavesTEM waves:
 It is the special type of Transverse electromagnetic wave in which electric field E
along the direction is also zero.
 The transverse electromagnetic waves are waves in which both electric and
magnetic fields are transverse entirely but have no component of E Z and HZ.
 It is referred to as principal waves.
The field strength for TM waves are

e-jβz

e- jβz

= e- jβz

For TEM waves EZ = 0 and the minimum value of m = 0.

e-jβz

e-jβz

Ez = 0
These fields are not only entirely transverse, but they are constant in amplitude between parallel
planes.

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Properties of TEM waves:

 The fields are entirely transverse
 Along the direction normal to the direction of propagation, the amplitude of the field
components are constant.
 the velocity of propagation is given by,

=c

Thus the velocity of propagation of TEM wave is independent of the frequency, unlike TE and
TM waves.
 The cut-off frequency of the wave is zero, indicating all frequencies down to zero
can propagate along the guide.

=0

=0

 the ratio of the amplitude of E to H between the parallel plane is defined as the
intrinsic impedance which is given by,

 propagation constant of TEM wave is given by,

β= (when m=0)
 Phase shift and wavelength are given by,
β=

9.Discuss on the characteristics of TEM waves between parallel conducting planes.

For TEM waves EZ = 0 and the minimum value of m = 0.
e-jβz

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e-jβz

Ez = 0
Propagation constant:

( )

Phase constant:
β=
Wavelength:

Velocity of propagation:

=0

10.Explain in detail about Circular Cavity Resonator.(NOV/DEC 2016)

We have studied that the rectangular cavity resonator is constructed from the rectangular
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waveguide shorted at both the ends. Similarly circular cavity resonator can be constructed from a
circular waveguide cutting into a section and shorting both the ends of it. The circular cavity
resonators are mainly used in microwave frequency meters. The mechanical tuning of the
resonant frequency is done with the help of movable top wall. The cavity is coupled to a
waveguide through a small aperture. In the last chapter we studied that the dominant mode in
circular waveguide is TEn mode. The dominant mode in case of the circular cavity resonator is
given by TEn. In general, the circular cavity resonator modes are specified as for the transverse
magnetic (TM) wave nmp for the transverse electric (TE) wave.
Consider a circular cavity resonator constructed from the circular waveguide with uniform
circular cross-section with radius a. The geometry of the circular cavity resonator is as shown in
the Fig.
Note that both the ends of the section of circular cavity resonator of length 'd' are shorted with
the help of circular shorting plates.

Expression for Resonant Frequency (f0) Circular Cavity Resonator

For a circular waveguide, we have already derived the expression given by,

Where Pnm is the Eigen value and a is the radius of the circular cylinder. But for the wave
propagation, the condition can be written as,

substituting value of in equation,we get

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But the condition for the circular cavity resonator remains same as the condition in
rectangular cavity resonator which is given by,

Depending upon the value of p, the general modes through the circular cavity resonator
are denoted by TEnmp and TMnmp.
With this value of |3 substituted in the expression for co, the cavity resonator supports
only one frequency ω0 or fQ.

Hence for a free space, the expression for the resonant frequency of circular cavity
resonator can be modified as,

Expression for the resonant frequency given by the equations (5.26.6), (5.26.7) is for
TMnmp mode. For the TEnmp mode, the expressions for f0 are given as follows.

For TEnmp wave:

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For free space within the circular cavity, the expressions for the resonant frequency for
TEnmp mode is given by,

11. Explain in detail about Bessel's Equation and Bessel Functions.(Nov/Dec 2017,18)
The analysis of the field components within the hollow, perfectly conducting cylinder
with uniform circular cross-section is carried out using the cylindrical co-ordinate system. The
resulting differential equation is called Bessel's equation. The solution of such equation is called
Bessel function. These Bessel functions are useful in applications such as, wave propagation
within a cylinder, or circular cross-section, the field distribution along the long wire of infinite
length, vibrations of circular membrane.
In this chapter, we will consider only the application of the most interest i.e. the propagation of
an electromagnetic wave within a hollow cylinder with circular cross-section. A circular co-
ordinate system is the three dimensional version of polar co-ordinate system. The surfaces used
to define the cylindrical co-ordinate system are,
1. Plane of constant z which is parallel to xy plane.
2. A cylinder of radius r with z-axis as the axis of cylinder.
3. A half plane perpendicular to xy plane and at an angle cb with respect to xz plane where 𝚽is
called azimuthal angle. The ranges of variables are

In general, any point say P in cylindrical co-ordinate system has three co-ordinates r, 𝚽and z
whose values line in the ranges specified above. A represent of point P in the cylindrical co-
ordinate system is as shown in the Fig

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When such co-ordinate system is used for the analysis of the electromagnetic wave
propagation within the circular cylinder, the differential equation obtained is of the form
given by,

Here n is integer. Using power series solution equation can be assumed as,

For equation (5.20.1), the series solution can be obtained by putting value of P in
equation (5.20.1) and comparing the co-efficients of same power together. Consider a
special case i.e. n = 0. Equation (5.20.1) reduces to following equation given by,

Now consider that assumed solution from equation (5.20.2) is substituted in equation
(5.20.3). Then the addition of the coefficients of all powers of r individually equated to
zero. Then the resulting series is given by,

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The series represented in equation (5.20.4) in integrable for all values of r. That means it
convergent for all real and complex values of r. This is known as Bessel's function of first
type of order zero. Such function is denoted by J0(r), where suffix zero represents value
of the integer n. Hence for different values of n such as 1,2,3, the Bessel's
functions are indicated by J1(r), J2(r), J3(r) respectively. For second order
differential
equation it is obvious to have two independent solutions for every value of n.

The second solution is obtained by using the manipulated series. This solution is
commonly known as Bessel's function of the second type of order zero or Neumann's
function. It is denoted by Nn(r). Sometime a notation Yn(r) is also used. In this solution
also n represents order of the function. The series obtained for first order Bessel's
function of second type for order zero is given by,

Then the complete solution of the Bessel's equation of order zero, represented by
equation (5.20.3) is the combination of the two individual solution for n = 0. The
complete solution is given by,

The plot of J0(r) and N0(r) is as shown in Fig. The imporant property of the Bessel's
functions of the second type is that their values become infinite for all the orders, when
the argument is zero(i.e. r = 0). Thus while analysing the wave propagation through the
circular waveguide; we will be considering a region with r = 0. So practically it is not
possible to have an infinite field, the second type of Bessel's function can not be
employed for any physical problem, such as wave through a hollow circular waveguide.
The Bessel's functions of first and second type are as shown in the fig.

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Bessel's function of first and second type of zero-order

From the plot shown above in the Fig. it is clear that except near the origin, the curves
for J0(r) and N0(r) are similar to the damped cosine and sine curves respectively. And
moveover it is observed that for very large values of r, the functions are represented in
the sinusoidal form. The expressions for the Bessel's function for larger values of r can be
written as,

12.Calculate the resonant frequency of an air filled rectangular resonator of

dimensions a = 2 cm,b = 4 cm and d =6 cm operating in TE 101 mode.(Nov/Dec 2015)

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= 7.9 GHz

13.A rectangular air filled copper waveguide with dimension 0.9 inch x 0.4 inch
cross section and 12 inch length is operated at 9.2 GHz with a dominant mode.Find
cut off frequency, guide wave length,phase velocity,characteristic impedance and
the loss.(Nov/Dec 2015).

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