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 EC8651 TRANSMISSION LINES AND RF SYSTEMS

Department : ECE
Batch/Year : 2018-2022/III YEAR
Created by : Mrs. P. SANTHOSHINI
Mr. S. BHARATHIDHASAN
1.TABLE OF CONTENTS

S.NO CONTENT PAGE NO

1 Table of Contents 5

2 Course Objectives 6

3 Pre Requisites (Course Names with Code) 7

4 Syllabus 8

5 Course outcomes 9

6 CO- PO/PSO Mapping 10

7 Lecture Plan 12

8 Activity based learning 13

9 Lecture Notes 14-32

10 Assignments 33

11 Part A Q & A 35-39

12 Part B Qs 40

13 Supportive online Certification courses 41

14 Real time Applications in day to day life and to Industry 42

15 Contents beyond the Syllabus 43

16 Assessment Schedule 44

17 Prescribed Text Books & Reference Books 45

18 Mini Project suggestions 46

5
COURSE OBJECTIVES

To introduce the various types of transmission lines and its

characteristics
To give thorough understanding about high frequency line,
power and impedance measurements
To impart technical knowledge in impedance matching
using smith chart
To introduce passive filters and basic knowledge of active RF
components
To get acquaintance with RF system transceiver design
3. PRE REQUISITES

EC 8251 CIRCUIT ANALYSIS

EC 8451 ELECTROMAGNETICFIELDS
4. SYLLABUS

EC8651 TRANSMISSION LINES AND RF SYSTEMS

LTPC
3003
OBJECTIVES:
 To introduce the various types of transmission lines and its characteristics
 To give thorough understanding about high frequency line, power and impedance
measurements
 To impart technical knowledge in impedance matching using smith chart
 To introduce passive filters and basic knowledge of active RF components
 To get acquaintance with RF system transceiver design
UNIT I TRANSMISSION LINE THEORY 9
General theory of Transmission lines - the transmission line - general
solution - The infinite line - Wavelength, velocity of propagation - Waveform
distortion - the distortion-less line - Loading and different methods of loading -
Line not terminated in Z0 - Reflection coefficient - calculation of current, voltage,
power delivered and efficiency of transmission - Input and transfer impedance -
Open and short circuited lines - reflection factor and reflection loss.
UNIT II HIGH FREQUENCY TRANSMISSION LINES 9
Transmission line equations at radio frequencies - Line of Zero dissipation -
Voltage and current on the dissipation-less line, Standing Waves, Nodes, Standing
Wave Ratio - Input impedance of the dissipation-less line - Open and short
circuited lines - Power and impedance measurement on lines - Reflection losses -
Measurement of VSWR and wavelength.
UNIT III IMPEDANCE MATCHING IN HIGH FREQUENCY LINES 9
Impedance matching: Quarter wave transformer - Impedance matching by
stubs - Single stub and double stub matching - Smith chart - Solutions of
problems using Smith chart - Single and double stub matching using Smith chart.
UNIT IV WAVEGUIDES 9
General Wave behavior along uniform guiding structures – Transverse
Electromagnetic Waves, Transverse Magnetic Waves, Transverse Electric Waves –
TM and TE Waves between parallel plates. Field Equations in rectangular
waveguides, TM and TE waves in rectangular waveguides, Bessel Functions, TM
and TE waves in Circular waveguides.
UNIT V RF SYSTEM DESIGN CONCEPTS 9
Active RF components: Semiconductor basics in RF, bipolar junction
transistors, RF field effect transistors, High electron mobility transistors Basic
concepts of RF design, Mixers, Low noise amplifiers, voltage control oscillators,
Power amplifiers, transducer power gain and stability considerations.
5.COURSE OUTCOME

Course Knowledge
Course Outcomes
Outcome Level

Discuss the various types of transmission lines

CO1 and propagation of signals. K3

Examine signal propagation at Radio

CO2 frequencies K3

Implement different methods of impedance

CO3 matching K3

Analyze the field components in guided

CO4 systems K3

Explain the RF system design Concepts.

CO5 K2

Analyze the RF amplifier power and stability

CO6 considerations K3

9
6.CO- PO/PSO MAPPING

Program Outcomes (POs)

Course
Outcomes PO PO PO PO PO PO PO PO PO PO PO PO
1 2 3 4 5 6 7 8 9 10 11 12
CO1 3 2

CO2 3 2

CO3 3 2 2 1

CO4 3 2 2 1

CO5 2 1

CO6 3 2 2

Program Specific Outcomes (PSOs)

Course Outcomes
PSO 1 PSO 2 PSO 3
CO1 1 1 3

CO2 1 3

CO3 1 2

CO4 3

CO5 1 2

CO6 1 2 3

10
 UNIT IV

WAVEGUIDES
7. LECTURE PLAN
S.No Topic No. Propo Actual CO Taxon Mode of
of sed Date omy Delivery
Perio Date Level
ds
1 General Wave 1 CO4 K3 PPT
behaviour along
uniform guiding
structures – TEM
Waves
2 TE - Waves 2 CO4 K3 PPT

3 TM - Waves 3 CO4 K3 PPT

4 TE and TM Waves 4 CO4 K3 PPT

between parallel
plates

5 Field Equations in 5 CO4 K3 PPT

Rectangular
Waveguides

6 TE Waves in 6 CO4 K3 PPT

Rectangular
Waveguides
7 TM Waves in 7 CO4 K3 PPT
Rectangular
Waveguides
8 Bessel Functions 8 CO4 K3 PPT

9 TE & TM Waves in 9 CO4 K3 PPT

Circular Waveguides

12
8. ACTIVITY BASED LEARNING

•WAVEGUIDES : Animation of Waveguides and wave propagation videos was

shown.

• WAVE EQUATIONS: Wave Equations and Fields distribution was shown using
animation videos

13
 9.LECTURE NOTES:
UNIT-IV
WAVEGUIDES

1. Introduction

A waveguide consists of a hollow metallic tube of either rectangular or circular

cross section used to guide electromagnetic wave. Rectangular waveguide is most commonly
used as waveguide. Waveguides are used at frequencies in the microwave range.

At microwave frequencies (above 1GHz to 100 GHz) the losses in the two line
transmission system will be very high and hence it cannot be used at those frequencies. Hence
microwave signals are propagated through the waveguides in order to minimize the losses.

2. General Wave behaviour along uniform guiding structures

Waveguides are used to transfer electromagnetic power efficiently from one point in
space to another. Some common guiding structures are shown in the figure below. These
include the typical coaxial cable, the two-wire and micro strip transmission lines, hollow
conducting waveguides, and optical fibre's.
Introduction of Parallel Plane Waveguide

Wave Guide is a structure which can guide Electro Magnetic Energy. A Wave guide is a
prime component of a communication system.

The Wave guide in structure should transport electro magnetic energy with as minimal loss
as possible.

In practice, we find two types of wave guide in structure which are as follows :

Metallic Wave Guide : Is used in high frequency, microwaves and millimeter waves
transmission. Co-axial cables and hollow rectangular or circular wave guide's fall in this
category.

Dielectric Wave Guide : Is used at sub milli-meter wavelengths and optical frequencies.
Optical fibres and thin film integrated optical devices fall in this category.

The important feature of a wave guide in structure is the ' Modal Propagation'. The electro
magnetic energy propagates inside the wave guide in the form of definite field patterns
called ' MODES '.

The electro magnetic fields inside a bound structure like the wave guide's can exist only in
the form of MODES.

The modal fields of a wave guide can be visualized either in the form of super position of
plane waves or as a boundary value problem.

The decomposition of plane waves is possible for simple wave guide in structure like
parallel plane wave guide or planer thin film wave guide's.

For rectangular and circular wave guide one essentially has to solve the wave equation with
appropriate boundary condition imposed by the wave guide walls.

The second approach is more general where as, the first approach provides better physical
understanding of the wave propagation inside the wave guide.

A parallel wave guide is formed by two infinite parallel conducting planes. The electro
magnetic energy is confined between the planes and moves in a direction parallel to the
planes.

The electro magnetic waves which can exists between the parallel planes can be of three
types

 Transverse Electric Fields

 Transverse Magnetic Fields
 Transverse Electomagnetic Fields
For Transverse Electric Field, the electric field is perpendicular to the direction of wave
propagation. That is to say that the electric field does not have any component in the
direction of wave propagation.

For Transverse Magnetic Field, the magnetic field is transverse to the direction. of wave
propagation. That is to say that the electric field does not have any component in the
direction of wave propagation.

For Transverse Electromagnetic Fields, the electric and magnetic fields both do not have
a component in the direction of wave propagation.

WAVES BETWEEN PARALLEL PLANES OF PERFECT CONDUCTORS

Wave Guide is a structure which can guide Electro Magnetic Energy. A Wave guide is a
prime component of a communication system. The Wave guide in structure should
transport electro magnetic energy with as minimal loss as possible.

Consider an electromagnetic wave propagating between a pair of parallel perfectly

conducting planes of infinite extent in y and z direction.

To study the behaviour of electromagnetic fields between two conducting planes, the
procedure is to solve the maxwell’s equations using proper boundary conditions.
The Maxwell’s Equations are given by

❑× 𝐻 = 𝜎 + 𝑗𝜔𝜀 𝐸 (4.1)

❑× 𝐸 = −𝑗𝜔𝜇𝐻 (4.2)

Between the two conducting planes the region is non conducting with σ = 0, substituting
in Eq. 4.1

❑× 𝐻 = 𝑗𝜔𝜀 𝐸 (4.3)

For Rectangular Coordinators

𝑥 𝑦^ 𝑧
𝜕 𝜕 𝜕
❑× 𝐻 = 𝜕𝑥 𝜕𝑦 𝜕𝑧
= 𝑗𝜔𝜀 𝐸𝑥 𝑥 + 𝐸𝑦 𝑦^ + 𝐸𝑧 𝑧 (4.4)
𝐻𝑥 𝐻𝑦 𝐻𝑧

𝜕𝐻𝑧 𝜕𝐻𝑦 𝜕𝐻𝑧 𝜕𝐻𝑥 𝜕𝐻𝑦 𝜕𝐻𝑥

𝑥 − − 𝑦^ − +𝑧 − =
𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑦

𝑗𝜔𝜀 𝐸𝑥𝑥 + 𝐸𝑦𝑦^ + 𝐸𝑧𝑧 (4.5)

𝜕𝐻 𝑧 𝜕𝐻 𝑦
− = 𝑗𝜔𝜀𝐸𝑥 (4.6.a)
𝜕𝑦 𝜕
𝜕𝐻
𝑧 𝑥
− 𝜕𝐻 𝑧 = 𝑗𝜔𝜀𝐸𝑦 (4.6.b)
𝜕𝑧 𝜕
𝜕𝐻
𝑥 𝑦 𝜕𝐻 𝑥
− = 𝑗𝜔𝜀𝐸𝑧 (4.6.c)
𝜕𝑥 𝜕
Similarly
𝑦 expanding the Eq. 4.2

❑× 𝐸 = −𝑗𝜔𝜇𝐻 (4.2)

𝑥 𝑦^ 𝑧
𝜕 𝜕 𝜕
❑× 𝐸 = 𝜕𝑥 𝜕𝑦 𝜕𝑧
= −𝑗𝜔𝜇 𝐻𝑥 𝑥 + 𝐻𝑦 𝑦^+ 𝐻𝑧 𝑧 (4.7)
𝐸𝑥 𝐸𝑦 𝐸𝑧

𝜕𝐸𝑧 𝜕𝐸𝑦 𝜕𝐸𝑧 − 𝜕𝐸𝑥 𝜕𝐸𝑦 𝜕𝐸𝑥

𝑥 − − 𝑦^ +𝑧 − = −𝑗𝜔𝜇 𝐻𝑥 𝑥 + 𝐻𝑦 𝑦^+ 𝐻𝑧 𝑧 (4.8)
𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑦
Comparing the above equation w.r.t. coefficients
𝜕𝐸 𝑧 𝜕𝐸
− 𝑦 = −𝑗𝜔𝜇𝐻𝑥 (4.9a)
𝜕𝑦 𝜕
𝜕𝐸
𝑧 𝑧 𝜕𝐸 𝑥
− = 𝑗𝜔𝜇𝐻𝑦 (4.9b)
𝜕𝑥 𝜕
𝜕𝐸
𝑧 𝑦 𝜕𝐸 𝑥
− = −𝑗𝜔𝜇𝐻𝑧 (4.9c)
𝜕𝑥 𝜕
𝑦

The wave equations are given by

❑2𝐸 = 2𝐸 4.10
❑2𝐻 = 2𝐻 4.11
Where 2 = 𝜎 + 𝑗𝜔𝜀 𝑗𝜔𝜇

Expanding eq.’s 4.10 and 4.11

𝜕2𝐸 2𝐸 2𝐸
+𝜕 +𝜕 = −𝜔2𝜇𝜀𝐸
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
}(4.12)
𝜕2𝐻 2 2
+ 𝜕 𝐻2 + 𝜕 𝐻2 = −𝜔2𝜇𝜀𝐻
𝜕𝑥 2 𝜕𝑦 𝜕𝑧

The direction of propagation is Z and the variation of all components in this direction is
expressed as 𝑒 −𝛾 𝑧 , 𝑤ℎ𝑒𝑟𝑒 = 𝛼 + 𝑗𝛽

If time variation is also included then 𝑒 𝑗𝜔𝑡 𝑒 −𝛾𝑧 = 𝑒 𝑗𝜔𝑡 𝑒 − 𝛼+𝑗𝛽 𝑧 = 𝑒 −𝛼𝑧 𝑒 𝑗 𝜔𝑡−𝛽𝑧 represents a
wave in z-direction

Let 𝐻𝑌 = 𝐻𝑦0 𝑒 −𝛾𝑧 𝑒 𝑗𝜔𝑡 = 𝐻 𝑦0 𝑒 𝑗𝜔𝑡−𝛾𝑧 ;

Since the region between the planes is infinite in y-direction, there are no boundary
conditions in y-directions.

𝜕
=0
𝜕𝑦

But in the x-direction, proper boundary conditions are applied,

𝜕 𝜕
So, substituting = 0, = − γ in 4.6a, b, c and 4.9.a, b, c and also in 4.12.
𝜕𝑦 𝜕𝑧
From 4.12

𝜕 2𝐸
+ 2 E = −𝜔2 𝜇𝜀𝐸
𝜕𝑥 2
} (4.13)
𝜕 2𝐻
+ 2 𝐻 = −𝜔2 𝜇𝜀𝐻
𝜕𝑥 2

𝜕𝐸𝑧
𝐸𝑥 = −
ℎ2 𝜕𝑥
 𝜕𝐸𝑧
𝐸𝑥 = −
ℎ2 𝜕𝑥
𝑗ωε 𝜕𝐸𝑧
𝐻𝑦 = −
ℎ2 𝜕𝑥
(4.14)
𝜕𝐻𝑧
𝐻𝑥 = −
ℎ2 𝜕𝑥
𝑗ω𝜇 𝜕𝐻𝑧
𝐸𝑦 =
ℎ2 𝜕𝑥
The above equations are the all the field equations expressed
in terms of Ez and Hz.

The propagating waves are classified into three types

i)If Ez = 0 and Hz is present, then the wave is called

Transverse Electric wave (TE Wave or H- wave)
ii)If Hz = 0 and Ez is present, then the wave is called
Transverse Magnetic wave (TM Wave or E- wave)
iii)If Ez = 0 and Hz = 0, then the wave is called Transverse
Electromagnetic wave (TEM Wave)
Rectangular Waveguide

A rectangular waveguide is a conducting cylinder of rectangular cross section used to

guide the propagation of waves. Rectangular waveguide is commonly used for the
transport of radio frequency signals at frequencies in the SHF band (3–30 GHz) and
higher.

FIELD COMPONENTS OF TRANSVERSE MAGNETIC (TM) WAVES

IN RECTANGULAR WAVEGUIDE (OR) TRANSMISSION OF TM
WAVES IN RECTANGULAR WAVEGUIDE

For TM waves, 𝐻𝑧 = 0 and 𝐸𝑧 is to be solved from wave equations.

Wave equation for 𝐸𝑧 is given by,
𝜕2𝐸2 𝜕2𝐸𝑧 𝜕2𝐸𝑧
+ + = −𝜔2𝜇𝜀𝐸𝑧
𝜕𝑥2 𝜕𝑦2 𝜕𝑧2
𝜕2𝐸
Let 𝜕𝐸 = −𝐸, 2
= 𝐸
𝜕𝑧 𝜕𝑧2
𝜕2 𝐸𝑧 𝜕2 𝐸𝑧
+ + 𝑦2 𝐸𝑧 = −𝜔2𝜇𝜀𝐸𝑧 (4.96)
𝜕𝑥2 𝜕𝑦2

The wave equation is a partial differential equation that can be solved by the usual
technique of assuming a product solution. 𝐸𝑧 can be written as
𝐸𝑥 (𝑥, 𝑦, 𝑧) = 𝐸𝑜𝑧 (𝑥, 𝑦)𝑒−𝑧
Let us assume a solution,
𝐸𝑧𝑎(𝑥, 𝑦) − 𝑋(𝑥)𝑌(𝑦) (4.97)
where 𝑋 is a function of 𝑥 alone. 𝑌 is a function of 𝑦 alone. Sub. Equation (4.97) in
Equation (4.96) ,

𝑑 2𝑋 𝑑2𝑌
𝑌 2 + 𝑋 2 + 𝑦2𝑋𝑌 = −𝜔2𝜇𝜀𝑋𝑌
𝑑𝑥 𝑑𝑦
2
𝑑 𝑋 𝑑 2𝑌
𝑌 2 + 𝑋 2 + 2 + 𝜔2 𝜇𝜀 𝑋𝑌 = 0
𝑑𝑥 𝑑𝑦
Let 2 + 𝜔2 𝜇𝜀 = ℎ2
𝑑 2𝑋 𝑑 2𝑌
𝑌 2 + 𝑋 2 + ℎ2𝑋𝑌 = 0
𝑑𝑥 𝑑𝑦
Dividing by 𝑋𝑌.
1 𝑑2 𝑋 1 𝑑2 𝑌
+ + ℎ2 = 0
𝑋 𝑑𝑥2 Y 𝑑𝑦2
1 𝑑2 𝑋 2 1 𝑑2 𝑌
+ ℎ = −
𝑋 𝑑𝑥2 𝑌 𝑑𝑦2
This expression equates a function of 𝑥 alone to a function of y alone and the only way for
the above equation to be true is to have each of these functions equal to some constant 𝐴2
1 𝑑2 𝑋
∴ + ℎ2 = 𝐴2
𝑋 𝑑𝑥2
1 𝑑 2𝑋
+ ℎ2 − 𝐴2 = 0
𝑋 𝑑𝑥2
Let B2 = ℎ2 − 𝐴7
1 𝑑 2𝑋
+ 𝐵2 = 0 (4.101)
𝑋 𝑑𝑥2
1 𝑑2 𝑦
= −𝐴2
𝑌 𝑑𝑦2

A solution of equation (4.101) is of the form

X = C, cosBx + C2 sinBx
where 𝐵 = ℎ − 𝐴 The solution of equation ( 4.102 ) is of the form
2 2 2

𝑌 = 𝐶, cos𝐴𝑦 + 𝐶, sin𝐴𝑦
We know that
𝐸0𝑧 (𝑥, 𝑦) = 𝑋𝑌
𝐸𝑧0 = 𝐶 1𝐶 1cos𝐵𝑥cos𝐴𝑦 + 𝐶 1𝐶 4cos𝐵𝑥sin𝐴𝑦
+𝐶2𝐶, sin𝐵𝑥cos𝐴𝑦 + 𝐶2𝐶𝑘sin𝐵𝑥sin𝐴𝑦 (4.105)

The constants C1, C5, C3C+, 𝐴 and 𝐵 are to be determined by using appropriate boundary
conditions. For a perfect conducting plate 𝐸tan = 0 at the boundaries.

The boundary conditions of a rectangular waveguide are

At 𝑥 = 0 and 𝑥 = 𝑎, 𝐸𝑎 = 0 and 𝐄, = 0 At 𝑦 = 0 and 𝑦 = 𝑏, 𝐸𝑧 − 0 and 𝐸𝑥 = 0 } When 𝑥 = 0
Equation (4.105) becomes
𝐸𝑜𝑧 = 𝐶𝑖 𝐶3 cos𝐴𝑦 + 𝐶, 𝐶, sin𝐴𝑦 = 0
For the above equation to be true for all values of 𝑦,
𝐶1 = 0
Equation (4.105) becomes.
𝐸z𝑜 = 𝐶 2𝐶 𝑗sin𝐵𝑥cos𝐴𝑦 + 𝐶 2𝐶 𝑦sin𝐵𝑥sin𝐴𝑦 (4.108)
When 𝑦 = 0 Equation (4.108) becomes

𝐸𝑧𝑜 = 𝐶2 𝐶3 sin𝐵𝑥 = 0 (4.109)

For the above equation to be true for all values of 𝑥 and taking 𝐵 ≠ 0, either 𝐶2 (or) 𝐶,
has to be zero. If 𝐶2 = 0, then equation ( 4.108 ) becomes zero. 𝐶3 = 0
Sub. 𝐶3 = 0 in equation (4.20) ,
𝐸𝑧𝑜 = 𝐶2 𝐶4 sin𝐵𝑥sin𝐴𝑦 (4.111)
If 𝑥 = 𝑎, 𝐸𝜃𝑖 = 0. Equation (4.111) can be re-written as
𝐸0𝑧 = 𝐶 7𝐶𝑖 sin𝐵𝑎sin𝐴𝑦 = 0
For the above equation to be true when 𝐴 ≠ 0 for all values of 𝑦,
sin𝐵𝑎 = 0
𝐵𝑎 = 𝑚𝜋
𝑚𝜋
𝐵=
𝑎
where 𝑚 = 1,2,3, … .
Sub. 𝐵 = 𝑚𝜋 in equation (4.111)
𝑎
𝑚𝜋
𝐸𝑧𝑜 = 𝐶 2𝐶 𝑦sin 𝑥sin𝐴𝑦 (4.113)
𝑎
If 𝑦 = 𝑏, 𝐸𝑜𝑧 = 0
𝑚𝜋
𝐸∘𝑧 = 𝐶2 𝐶, sin 𝑥sin𝐴𝑏 = 0
𝑎
The above equation is true if
sin𝐴𝑏 = 0
𝐴𝑏 = 𝑛𝜋
𝑛𝜋
𝐴=
𝑏
Sub 𝐴 = 𝑛𝜋 in equation (4,113) , the final Expression is given by,
𝑏
𝑚𝜋 𝑛𝜋
𝐸𝑧0 = 𝐶2𝐶4 sin 𝑥sin 𝑦
𝑎 𝑏
Let 𝐶 = 𝐶, 𝐶𝑖
𝑚𝜋 𝑛𝜋
𝐸𝑧0 = 𝐶sin 𝑥sin 𝑦 (4.116)
𝑎 𝑏
The general field components with 𝐻𝑧 = 0 and = 𝑗𝛽 is given by

−𝑗𝛽 𝜕𝐸
𝐸𝑥 = ℎ2 𝑧
𝜕𝑥
𝑗𝜔𝜀 𝜕𝐸
𝐻𝑥 =
ℎ2 𝜕𝑦
−𝑗𝛽 𝜕𝐸𝜆
(4.117 (a, b, c, and d)
𝐸𝑦 = ℎ2 𝜕𝑦
−𝑗𝜔𝜀 𝜕𝐸𝑧
𝐻𝑦 = ℎ2 𝜕𝑥
Using equation ( 4.116 ) and equation (4.117) (a), (b), (c), (d) we get
−𝑗𝛽 𝜕𝐸𝑧0 −𝑗𝛽 𝑚𝜋 𝑚𝜋 𝑛𝜋
0
𝐸𝑥 = 2 = 2 𝐶 cos 𝑥sin 𝑦
ℎ 𝜕𝑥 ℎ 𝑎 𝑎 𝑏
0
−𝑗𝛽𝜕𝐸𝑧4 −𝑗𝛽 𝑚𝜋 𝑛𝜋 𝑛𝜋
𝐸𝑦 = 2
= 2
𝐶sin 𝑥cos 𝑦
ℎ ℎ 𝑎 𝑏 𝑏
0 𝑚𝜋
𝑗𝜔𝜀 𝜕𝐸𝑧 𝑗𝜔𝜀 𝑚 𝑛𝑡
𝐻𝑥0 = 2 = 𝐶 sin 𝑥cos 𝑦
ℎ 𝜕𝑦 ℎ 2 𝑏 𝑎 𝑏
−𝑗𝜔𝜀 𝜕𝐸𝑧0 −𝑗𝜔𝜀 𝑚𝜋 𝑚𝜋 𝑛𝜋
𝐻𝑦0 = 2
= 2
𝐶 cos 𝑥sin 𝑦
ℎ 𝜕𝑥 ℎ 𝑎 𝑎 𝑏

We know that 𝐴 = 𝑛𝜋 𝐵 = 𝑚𝜋
𝑏 𝑎
𝐸𝑧 = 𝐸0𝑧𝑒−𝑧 = 𝐸0𝑒𝑧−𝑗𝛽𝑧

From equation (4.116) we get,

𝐸𝑧 = 𝐶sin𝐵𝑥sin𝐴𝑦𝑒−𝑗𝛽𝑧
𝐸𝑥 = 𝐸0𝑥𝑒−𝑗𝛽𝑧
−𝑗𝛽
𝐸𝑥 = 2 𝐵𝐶cos𝐵𝑥sin𝐴𝑥−𝑗𝛽𝑧
ℎ
−𝑗𝛽
𝐸𝑦 = 2 𝐴𝐶sin𝐵𝑥cos𝐴𝑦𝑒−𝑗𝛽𝑧
ℎ
𝑗𝜔𝜀
𝐻𝑥 = 2 𝐴𝐶sin𝐵𝑥cos𝐴𝑦 ⋅ 𝑒−𝑗𝛽𝑧
ℎ
−𝑗𝜔𝜀
𝐻𝑦 = 2 𝐵𝐶cos𝐵𝑥sin𝐴𝑦𝑒−𝑗𝛽𝑧
ℎ
The lowest possible value of 𝑚 = 1 and 𝑛 = 1 for 𝑇𝑀 waves. Lowest order mode in TM
wave is TM11 mode.
FIELD COMPONENTS OF TRANSVERSE ELECTRIC (TE)
WAVES IN RECTANGULAR WAVEGUIDES (OR)
TRANSMISSION OF TE WAVES IN RECTANGULAR
WAVEGUIDES.

For TE waves 𝐸𝑧 = 0 and 𝐻𝑧 is to be solved from the wave equation 𝐻𝑧 ≠ 0

The wave equation is given by

𝜕 2 𝐻𝑧 𝜕 2 𝐻𝑧
+ + 𝛾 2 𝐻𝑧 = −𝜔2 𝜇𝜀𝐻𝑧
𝜕𝑥 2 𝜕𝑦 2
The wave equation is a partial differential equation that can be solved by the
usual technique of assuming a product solution. Let

𝐻𝑧 𝑥, 𝑦1 𝑧 = 𝐻𝑧0 (𝑥, 𝑦)𝑒 −𝛾𝑧

Let us assume a solution,

𝐻𝑧𝑜 = 𝑋(𝑥)𝑌(𝑦)
From equation (4.120) ,

𝜕 2 𝐻𝑧 𝜕 2 𝐻𝑡
+ + 𝑦 2 + 𝜔2 𝜇𝜀 𝐻2 = 0
𝜕𝑥 2 𝜕𝑦 2
Sub. (4.121) in (4.122) and let ℎ3 = 𝑦 2 + 𝜔2 𝜇𝜀

𝑑2 𝑋 𝑑2 𝑌
𝑌 + 𝑋 + ℎ2 𝑋𝑌 = 0
𝑑𝑥 2 𝑑𝑦 2
Dividing by 𝑋𝑌 throughout

1 𝑑2 𝑋 1 𝑑2 𝑌
+ + ℎ2 = 0
𝑋 𝑑𝑥 2 𝑌 𝑑𝑦 2
1 𝑑2 𝑋 2
1 𝑑2 𝑌
+ℎ =−
𝑋 𝑑𝑥 2 𝛾 𝑑𝑦 2
The above equation is true only if

1 𝑑2 𝑋
+ ℎ2 = 𝐴2
𝑋 𝑑𝑥 2
1 𝑑2 𝑌
= −𝐴2
𝛾 𝑑𝑦 2
1 𝑑2 𝑋
Let ℎ2 − 𝐴2 = 𝐵 2 . Equation (4.125) can be rewritten as + 𝐵 2 = 0 The solution of
𝑋 𝑑𝑥 2
the above equation is of the form

𝑋 = 𝐶, cos𝐵𝑥 + 𝐶2 sin𝐵𝑥
The solution of equation (4.126) is given by

𝑌 = 𝐶𝑗 cos𝐴𝑦 + 𝐶+ sin𝐴𝑦

Since 𝐻𝑧𝑜 − 𝑋𝑌, substituting equation (4.128) and (4.129) gives

𝐻𝑧0 − 𝐶1 𝐶3 cos 𝐵𝑥cos𝐴𝑦 + 𝐶1 𝐶4 cos𝐵𝑥sin𝐴𝑦

+𝐶2 𝐶𝑦 sin𝐵𝑥cos𝐴𝑦 + 𝐶2 𝐶4 sin𝐵𝑥sin𝐴𝑦

Now 𝐶1 , 𝐶2 , 𝐶𝑦 , 𝐶4 , 𝐴 and 𝐵 are to be determined using boundary conditions. The

tangential component of clectric field is continuous and 𝐸as − 0 at the boundary for
a perfect conductor.

−𝑗𝜔𝜇 𝜕𝐻𝑧0
𝐸𝑥0 =
ℎ2 𝜕𝑦
−𝑗𝜔𝜇 −𝐴𝐶1 𝐶3 cos𝐵𝑥sin𝐴𝑦 + 𝐴𝐶1 𝐶4 cos𝐵𝑥cos𝐴𝑦
=
ℎ2 −𝐶2 𝐶3 𝐴sin𝐵𝑥sin𝐴𝑦 + 𝐶2 𝐶4 ⋅ 𝐴sin𝐵𝑥cos𝐴𝑦
𝑗𝜔𝜇 𝜕𝐻−0
𝐸𝑦0 = 2
ℎ 𝜕𝑥
𝑗𝜔𝜇 −𝐶1 𝐶3 𝐵sin𝐵𝑥cos𝐴𝑦 − 𝐶1 𝐶4 𝐵sin𝐵𝑥sin𝐴𝑦
= 2
ℎ +𝐶2 𝐶3 𝐵cos𝐵𝑥cos𝐴𝑦 + 𝐶2 𝐶4 ⋅ 𝐵cos𝐵𝑥sin𝐴𝑦
At 𝑥 = 0 and 𝑥 = 𝑎, 𝐸𝑦𝜃 = 0 is the first boundary condition. (i) When 𝑥 = 0, 𝐸𝑦0 = 0
Substitution of the above condition in eqn. (4.132),
𝑗𝜔𝜇
𝐸𝑦0 = +𝐶2 𝐶3 𝐵cos𝐴𝑦 + 𝐶2 𝐶4 𝐵sin𝐴𝑦 = 0
ℎ2
This is true only if 𝐶𝑧 = 0. Sub. in equation (4.44),
𝑗𝜔𝜇
∴ 𝐸𝑦0 = −𝐶1 𝐶3 𝐵sin𝐵𝑥cos𝐴𝑦 − 𝐶1 𝐶4 𝐵sin𝐵𝑥sin𝐴𝑦 = 0
ℎ2
(ii) When 𝑥 = 𝑎, 𝐸𝑦𝑒 = 0.
𝑗𝜔𝜇
𝐸𝑦𝑜 = −𝐶1 𝐶3 𝐵sin𝐵𝑎cos𝐴𝑦 − 𝐶1 𝐶4 𝐵sin𝐵𝑎sin𝐴𝑦 = 0
ℎ2
This is true only if
sin𝐵𝑎 = 0
𝐵𝑎 = 𝑚𝜋
𝑚𝜋
𝐵=
𝑎
Equation (4.133) becomes,
𝑗𝜔𝜇 𝑚𝜋 𝑚𝜋 𝑚𝜋 𝑚𝜋
∴ 𝐸𝑦∞ = −𝐶1 𝐶3 sin × cos𝐴𝑦 − 𝐶1 𝐶4 sin 𝑥sin𝐴𝑦
ℎ2 𝑎 𝑎 𝑎 𝑎
At 𝑦 = 0 and 𝑦 = 𝑏, 𝐸𝑥 = 0 is the second boundary condition. i) When 𝑦 = 0, 𝐸𝑥𝑜 = 0.
Substitution of the above condition in eqn. (4.131)
−𝑗𝜔𝜇
𝐸𝑥∞ = 𝐴𝐶1 𝐶4 cos𝐵𝑥cos𝐴𝑦 + 𝐴𝐶2 𝐶4 sin𝐵𝑥cos𝐴𝑦 = 0
ℎ2
The above equation is true only for 𝐶1 = 0 Sub, in eqn. (4.131) gives
−𝑗𝜔𝜇
𝐸𝑥∞ = 𝐴 −𝐶1 𝐶3 cos𝐵𝑥sin𝐴𝑦 − 𝐶2 𝐶3 sin𝐵𝑥sin𝐴𝑦
ℎ2
ii) When 𝑦 = 𝑏, 𝐸𝑥𝑜 = 0 Substitute in equation (4.135).
−𝑗𝜔𝜇
𝐸𝑥0 = 𝐴 −𝐶1 𝐶3 cos𝐵𝑥sin𝐴𝑏 − 𝐶2 𝐶3 sin𝐵𝑥sin𝐴𝑏 = 0
ℎ2
This is true only if,
 𝑗𝜔𝑢
𝐸1 = 𝐴𝐶cos𝐵𝑥sin𝐴𝑦𝑒 −𝑗𝛽𝑧
ℎ2
𝑗𝛽
𝐻𝑥 = 2 𝐶𝐵sin𝐵𝑟cos𝐴𝑦𝑒 −𝑖𝛽𝑧
ℎ
𝑗𝛽
𝐻𝑦 = 2 𝐶𝐴cos𝐵𝑥sin𝐴𝑦𝑒 −𝛽=
ℎ
−𝑗𝜔𝜇
𝐸𝑦 = 𝐵𝐶sin𝐵𝑥 ⋅ cos𝐴𝑦𝑒 −𝑗𝛽𝑧
ℎ2

Lowest value of m and n are m=1,and n=0 .

Fig. 4.13 : Electric (solid) and magnetic (dashed) field configuration for the lower-order
modes in a rectangular guide.
Bessel Functions

Bessel functions are a set of solutions for differential equations of a certain type. These
are useful to obtain propagation characteristics of electromagnetic waves in circular
waveguides.

Considering circular waveguides, solutions of the axial component is written

conveniently in terms of Bessel function. For transverse magnetic wave, the
solution is

𝐸𝑧,𝑛𝑚 = 𝐽𝑛 𝐾𝑐 𝑟 (𝐴cos𝑛𝜙 + 𝐵sin𝑚𝜙) and

for transverse electric wave

𝐻𝑧,𝑛𝑚 = 𝐽𝑛 𝐾𝑐 𝑟 𝐴1 cos𝑛𝜙 + 𝐵1 sin𝑚𝜙

where 𝐽𝑛 𝐾𝑐 𝑟 = Bessel function of the first kind
𝑟 = radius of the guide
𝐾𝑐 = cut−off wave number
𝐴, 𝐵, 𝐴1 , 𝐵1 , = constants The solutions for the Bessel function are obtained for
certain values of 𝐾𝑐 ∗ These values of 𝐾𝑐 are known as eigen values. If 𝐾𝑐 is to
produce solution of the Bessel function, 𝐾𝑐 𝑟 must be the roots of the Bessel
function. Then 𝐽𝑛 𝐾𝑐 𝑟 = 0.

The propagation parameters for 𝑛𝑚th mode of TM waves are Phase constant,
1/2
𝛽𝑛𝑚 = 𝐾 2 − 𝐾𝑐,𝑛𝑚
2

where
𝑝𝑛𝑚
𝐾𝑐,𝑛𝑚 =
𝑟
Also,
1/2
𝑝𝑛𝑚 2
2
𝛽𝑛𝑚 = 𝐾 −
𝑟 𝑗

2𝜋
where 𝑝𝑛𝑚 = the roots of the Bessel function 𝐾 = free space wave number The cut-
𝜆
off wavelength for the TM wave,
2𝜋 2𝜋𝑟
𝜆𝑐,𝑛𝑚 = =
𝐾𝑐,𝑛𝑚 𝑝𝑛𝑚

The roots of the Bessel function for TM mode are shown in Table.
Table 4.1 Roots of Bessel Function (TM)

First Second Third

Order order order order
𝐧 𝑝n1 𝑝𝑛2 𝑝𝑛3
0 2.405 5.520 8.654
1 3.832 7.016 10.174
2 5.135 8.417 11.620

Table 4.2 Roots of Bessel Function (TE)

First order Second order Third order

Order 𝑛 𝑝𝑛′ 1 𝑝𝑛′ 2 𝑝𝑛′ 3
0 3.832 7.016 10.174
1 1.841 5.331 8.536
2 3.054 6.706 9.970

The propagation parameters for 𝑇𝐸𝑛𝑚 mode are where

1/2
𝛽𝑛𝑚 = 𝐾 2 − 𝐾𝑐,𝑛𝑚
′2
′
′2
𝑝𝑛𝑚
𝐾𝑐,𝑛𝑚 =
𝑟
So,
′ 2 1/2
𝑝𝑛𝑚
2
𝛽𝑛𝑚 = 𝐾 −
𝑟
2𝜋𝑟
𝜆𝑐,𝑛𝑚 =′
𝑝𝑛𝑚
𝜆
Guide wavelength, 𝜆𝑔 = 2/2
The variation of the first three orders of
𝜆 1
1−
𝜆𝑐,𝑛𝑚 𝑗

the Bessel function and their roots are shown in Figure.

 Bessel Function

TM and TE waves in circular waveguides

CIRCULAR WAVEGUIDES

The waveguides of circular cross-section are used to transmit EM waves from one point
to another. Unlike rectangular waveguides, the circular waveguides do not have unique
orientation as it is perfectly symmetrical around the axis.
SALIENT FEATURES OF CIRCULAR WAVEGUIDES
1. It is easy to manufacture.
2. They are used in rotational coupling.
3. Rotation of polarization exists and this can be overcome by rotating modes
symmetrically
4. TM01 mode is preferred to TE01 as it requires a smaller diameter for the same
cut-off wavelength.
5. TE01 does not have practical application.
6. For 𝑓 > 10GHz, TE01 has the lowest attenuation per unit length of the
waveguide.
7. The main disadvantage is that its cross-section is larger than that of a
rectangular waveguide for carrying the same signal.
8. The space occupied by circular waveguides is more than that of a rectangular
waveguide
9. The determination of fields here consists of differential equations of certain type.
Their solutions involve Bessel functions.
10. Here also TE and TM modes exist. For TM wave, the solution of axial component
𝐸𝑧,𝑛𝑚 = 𝐽𝑛 𝐾𝑐 𝑟 (𝐴cos𝑛𝜙 + 𝐵sin𝜙)
and for TE wave, it is
𝐻𝑧,𝑛𝑚 = 𝐽𝑛 𝐾𝑐 𝑟 𝐴′ cos𝑛𝜙 + 𝐵 ′ sin𝑛𝜙
where 𝐽𝑛 𝐾𝑐 𝑟 = Bessel function of the first kind 𝑟 = the radius of the guide
𝐾𝑐 = the cut−off wave number
′
𝐴, 𝐵, 𝐴 , 𝐵 ′
= constants
The solutions for the Bessel function are obtained for certain values of 𝐾𝑐 where
these values of 𝐾𝑐 are known as eigen values. If 𝐾𝑐 is to produce solution of the
Bessel function, 𝐾𝑐 𝑟 must be the roots of the Bessel function. Then 𝐽𝑛 𝐾𝑐 𝑟 = 0
The propagation parameters for 𝑛𝑚th mode TM waves are: Phase constant,
1/2
𝛽𝑛𝑚 = 𝐾 2 − 𝐾𝑐,𝑛𝑚
2

where
𝑝𝑛𝑚
𝐾𝑐,𝑛𝑚 =
𝑟
1/2
𝑝𝑛𝑚 )
2
𝛽𝑛𝑚 = 𝐾 −
𝑟 𝑟
and where 𝑝𝑛𝑚 = the roots of the Bessel function
2𝜋
𝐾 = free space wave number =
𝜆
The cut-off wavelength for TM wave,
2𝜋 2𝜋𝑟
𝜆𝑐 = =
𝐾𝑐,𝑛𝑚 𝑝𝑛𝑚
The roots of the Bessel function for TM mode are shown in Table.

First order Second order Third order

Order 𝑛 𝑝𝑛1 𝑝𝑛2 𝑝𝑛3
0 2.405 5.520 8.654
1 3.832 7.016 10.174
2 5.135 8.417 11.620

Table 4.3 Roots of Bessel Function (TM Mode).

The roots of the Bessel function for TE mode are shown in Table 4.4.
First order Second order Third order
Order 𝑛 𝑝𝑛′ 1 𝑝𝑛′ 2 𝑝𝑛′ 3
0 3.832 7.016 10.174
1 1.841 5.331 8.536
2 3.054 6.706 9.970

Table 4.4 Roots of Bessel Function (TE Mode)

For circular waveguides, TE11 is the dominant mode. The propagation

parameters for TE𝑛𝑚 mode are: where
1/2
𝛽𝑛𝑚 = 𝐾 2 − 𝐾𝑐,𝑛𝑚
′2
′
′2
𝑝𝑛𝑚
𝐾𝑐,𝑛𝑚 =
𝑟
′
2
𝑝𝑛𝑚 )2
𝛽𝑛𝑚 = 𝐾 −
𝑟 𝑗
2𝜋𝑟
𝜆𝑐,𝑛𝑚 = ′
𝑝𝑛𝑚
Guide wavelength,
𝜆
𝜆𝑔 =
1/2
𝜆 1
1−
𝜆𝑐,𝑛𝑚 𝑗
10.ASSIGNMENTS

UNIT-4

1. A rectangular waveguide operates at 1GHz and has dimensions of 5 × 2 cm. Find

the distance from the source end at which the electric field of TE21 wave
becomes 0.5% of its starting amplitude at 𝑧 = 0. The amplitude of the electric
field in 𝑧 -direction at 𝑧 = 0 is 10kV/m
2. A rectangular waveguide with dimensions of 2.5 × 1.0 cm operates below
15.0GHz. What are the TE modes that can be propagated if the waveguide is
filled with material whose relative permittivity, 𝜖𝑟 = 4 and 𝜇𝑟 = 1, 𝜎 = 0? Find
their cut-off frequencies and cut-off wavelengths.
3. A hollow rectangular waveguide operates at 10GHz. If the cut-off frequency of
TM21 mode is 6GHz, find its phase velocity, wavelength, phase constant.
4. A hollow rectangular waveguide has the dimensions of 𝑎 = 2.286 cm and 𝑏
= 1.016 cm. Find the cut-off frequencies of TM11 and TM20 .
5. Find the cut-off wavelength of TE10 and TE20 mode in the hollow rectangular
waveguide of dimensions 𝑎 = 3 cm, 𝑏 = 1.5 cm
6. In a hollow square waveguide, the cut-off frequency for TE10 mode is 10GHz
20GHz (c) 30GHz. Find the dimensions of the waveguide.
7. Find the radius of the circular waveguide required to propagate TE11 modee if
′
𝜆𝑐 = 10 cm. 𝜌11 = 1.84 .
8. Find the cut-off frequency of TE21 mode in a circular waveguide of radius 4 cm.
(Take 𝜌21
′
= 3.054 }

39
40
 UNIT III
11. Part A Q & A (with K level and CO)
Q.No. Question with Answer K CO
level

1 What are guided waves? Give examples K2 CO4

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

2 What is TE wave or H wave?

K2 CO4

Transverse electric (TE) wave is a wave in which the 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 Ez in the same direction.

3 What is TH wave or E wave? K2 CO4

Transverse magnetic (TM) wave is a wave in which the

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

4 What is a TEM wave or principal wave? K2 CO4

TEM wave is a special type of TM wave in which an electric field
E along the direction of propagation is also zero. The TEM
waves are waves in which both electric and magnetic fields are
transverse entirely but have no components of Ez and Hz. it is
also referred to as the principal wave.

5 What is a dominant mode? K2 CO4

The modes that have the lowest cut off frequency is called the
dominant mode.

6 Give the dominant mode for TE and TM waves. K2 CO4

Dominant mode: TE10 and TM10

41
 UNIT III
11. Part A Q & A (with K level and CO)

Q.No. Question with Answer K CO

level

7 What is cut off frequency? K2 CO4

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

8 What is cut-off wavelength?

K2 CO4
It is the wavelength below which there is wave propagation
and above which there is no wave propagation.

9 Write down the expression for cut-off frequency when K2 CO4

the wave is propagated in between two plates.

The cut-off frequency, fc = m/ (2a (µE)1/2)

10 Mention the characteristics of TEM waves. K2 CO4

It is a special type of TM wave.it does not have either E or H
components. Its velocity is independent of frequency. Its cut off
frequency is zero.

11 Define attenuation factor. K2 CO4

Attenuation factor = (Power lost/ unit length)/(2 x power

transmitted)

12 Give the relation between the attenuation factor for TE K2 CO4

waves and TM waves.

αTE = aTM (fc/f)2

13 Define wave impedance K2 CO4

Wave impedance is defined as the ratio of electric to
magnetic field strength Zxy= Ex/ Hy in the positive direction
Zxy= -Ex/ Hy in the negative direction

41
 UNIT III
11. Part A Q & A (with K level and CO)
Q.No. Question with Answer K CO
level

14 What is a parallel plate wave guide? K2 CO4

Parallel plate wave guide consists of two conducting sheets
separated by a dielectric material.

15 Why are rectangular wave-guides preferred over

K2 CO4
circular waveguides?

Rectangular wave-guides preferred over circular wave guides

because of the following reasons.

a. Rectangular wave guide is smaller in size than a circular

wave guide of the same operating frequency
b. It does not maintain its polarization through the circular
wave guide
c. The frequency difference between the lowest frequency on
dominant mode
and the
next mode of a rectangular wave-guide is bigger than in
a circular wave guide.

16 Mention the applications of wave guides K2 CO4

The wave guides are employed for transmission of energy at
very high frequencies where the attenuation caused by wave
guide is smaller.

Waveguides are used in microwave transmission. Circular

waveguides are used as attenuators and phase shifters.

17 Why is circular or rectangular form used as waveguide? K2 CO4

Waveguides usually take the form of rectangular or circular

cylinders because of its simpler forms in use and less expensive
to manufacture.

18 What is an evanescent mode? K2 CO4

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

41
 UNIT III
11. Part A Q & A (with K level and CO)
Q.No. Question with Answer K CO
level

19 What is the dominant mode for the TE waves in the K2 CO4

rectangular waveguide?

The lowest mode for TE wave is TE10 (m=1 , n=0)

20 What is the dominant mode for the TM waves in

K2 CO4
the rectangular waveguide?

The lowest mode for TM wave is TM11(m=1 , n=1)

21 Why TEM mode is not possible in a rectangular K2 CO4

waveguide?

Since TEM wave do not have axial component of either E or H

,it cannot propagate within a single conductor waveguide.

22 Explain why TM01 and TM10 modes in a rectangular K2 CO4

waveguide do not exist.

For TM modes in rectangular waveguides, neither m or n can be

zero because all the field Equations vanish ( i.e.,Hx, Hy,Ey. And
Ez.=0). If m=0,n=1 or m=1,n=0 no fields are present. Hence
TM01 and TM10 modes in a rectangular waveguide do not exist
23 What are degenerate modes in a rectangular K2 CO4
waveguide?
Some of the higher order modes, having the same cut off
frequency, are called degenerate modes. In a rectangular
waveguide, TEmn and TMmn modes ( both m
≠0 and n≠ 0) are always degenerate.

41
 12.PART B QS (WITH K LEVEL AND CO)

1.Explain the transmission of Transverse Electric waves between parallel perfectly

conducting planes with necessary expressions and diagrams for field components.
(CO4,K3)

2.Explain the transmission of Transverse Magnetic waves between parallel perfectly

conducting planes with necessary expressions and diagrams for field components.
(CO4,K3)

3.Deduce the expressions for the field components of TE waves guided along a
rectangular wave guide (CO4,K3)

4.Deduce the expressions for the field components of TM waves guided along a
rectangular wave guide(CO4,K3)

5. Derive the expressions for TM wave components in circular wave guides using Bessel
function.(CO4,K3)

6.Derive the expressions for TE wave components in circular wave guides using Bessel
function. (CO4,K2)

7.A rectangular air-filled copper guide with 0.9inch x 4 inch cross section and 12inch
length is operated at 9.2GHz with a dominant mode. Find the cut-off frequency, guided
wavelength, cut off wavelength the phase velocity, the group velocity and the
characteristics impedance. (CO4,K2)

8.An air filled rectangular copper wave guide with a=2.28 cm and b=1.01 cm is
operated at 9.2 GHz in dominant mode. Determine the cutoff frequency and the cutoff
wave length. (CO4,K3)

9.Deduce the expressions for the field components of TE waves guided along a
rectangular cavity resonator. (CO4,K3)

10.Deduce the expressions for the field components of TM waves guided along a
rectangular cavity resonator. (CO4,K3)
 UNIT II
12. PART B Questions
Q.No. Question with Answer K CO
level
1. 1.Explain the transmission of Transverse Electric waves K3 CO4
between parallel perfectly conducting planes with
necessary expressions and diagrams for field
components.
2. Explain the transmission of Transverse Magnetic waves K3 CO4
between parallel perfectly conducting planes with necessary
expressions and diagrams for field components.

3. Deduce the expressions for the field components of TE K3 CO4

waves guided along a rectangular wave guide

4. educe the expressions for the field components of TM K3 CO4

waves guided along a rectangular wave guide.
5. Derive the expressions for TM wave components in circular K3 CO4
wave guides using Bessel function.

6. Derive the expressions for TE wave components in circularK2 CO4

wave guides using Bessel function.
QQ.N
.Noo.
7. A rectangular air-filled copper. guide with 0.9inch x 4 inch K2 CO4
cross section and 12inch length is operated at 9.2GHz with
a dominant mode. Find the cut-off frequency, guided
wavelength, cut off wavelength the phase velocity, the
group velocity and the characteristics impedance.

8. An air filled rectangular copper wave guide with a=2.28 cm

K3 CO4
and b=1.01 cm is operated at 9.2 GHz in dominant mode.
Determine the cutoff frequency and the cutoff wave length.

9. Deduce the expressions for the field components of TE waves

K3 CO4
guided along a rectangular cavity resonator.
10. Deduce the expressions for the field components of TM waves
K3 CO4
guided along a rectangular cavity resonator

46
13.SUPPORTIVE ONLINE CERTIFICATION
COURSES

Transmission lines and electromagnetic waves - NPTEL

Microwave engineering and antennas - COURSERA

14.REAL TIME APPLICATIONS IN DAY TO DAY
LIFE AND TO INDUSTRY

• Microwaves are used for cooking food and for satellite communications.
• Transverse Waves in Medicine
• Transverse Waves in Industry
15.CONTENTS BEYOND THE SYLLABUS

• EM Waves in Microwave Oven

• Longitudinal Waves in Medicine and Industry
16.ASSESSMENT SCHEDULE ( PROPOSED DATE &
ACTUAL DATE)
17. PRESCRIBED TEXT BOOKS & REFERENCE
BOOKS

TEXT BOOKS:
1.John D Ryder, ―Networks, lines and fields‖,2nd Edition, Prentice Hall India,
2015. (UNIT IIV)
2.Mathew M. Radmanesh, ―Radio Frequency &Microwave Electronics‖,
Pearson Education Asia, Second Edition,2002. (UNIT V)

REFERENCES:
1.Reinhold Ludwig and Powel Bretchko,‖ RF Circuit Design – Theory and
Applications‖, Pearson Education Asia, First Edition,2001.
2.D. K. Misra, ―Radio Frequency and Microwave Communication Circuits-
Analysis and Design‖,John Wiley & Sons, 2004.
3.E.C.Jordan and K.G. Balmain, ―Electromagnetic Waves and Radiating
Systems Prentice Hall of India, 2006.
4.G.S.N Raju, "Electromagnetic Field Theory and Transmission Lines Pearson
Education, First edition 2005.
 18. MINI PROJECT SUGGESTIONS

To draw VSWR circle, calculate reflection coefficient

using MATLAB and RF Tool box.
 Thank you

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