SIDDHARTH INSTITUTE OF ENGINEERING & TECHNOLOGY

(AUTONOMOUS)
(Approved by AICTE, New Delhi& Affiliated to JNTUA, Ananthapuramu)
(Accredited by NBA for Civil, EEE, Mech., ECE & CSE)
(Accredited by NAAC with ‘A+’ Grade)
Puttur -517583, Chittoor District, A.P. (India)
QUESTION BANK (DESCRIPTIVE)
SUBJECT MICROWAVE THOERY AND COURSE & B.TECH-ECE
WITH CODE: TECHNIQUES (20EC0423) BRANCH:
YEAR & SEM: III YR & II SEM REGULATION: R20
UNIT –I
INTRODUCTION OF MICROWAVE
a) What do you remember about the history in the evolution of [L1][CO1] [6M]
Microwaves?
1. Micro waves are electromagnetic waves. Hence the history of microwaves is embodied in the evolution
of electromagnetic waves.
1. James clerk Maxwell (1831-1879) unified all previous known results, experimental and theoretical
on electromagnetic Waves in four equations and predicted the existence of electromagnetic waves.
2. Heinrich Rudolf Hertz (1857-1987) experimentally confirmed maxwell's prediction.
3. Guglielmo Marconi (1874-1937) transmitted information on an experimental basis at
microwave frequencies.
4. George C. South worth (1930) really carried out Marconi's experiments on a commercial basis.
5. World War II gave many technologies a great boost, and radar, which is essentially radio waves
bounced off a target, was no exception. In 1940, British physicists John Randall and Harry Boot
invented the cavity magnetron, a device that could create very short-wavelength (high-
frequency) radio waves called microwaves.
6. In 1990's microwaves became Common market products with the development of microwave
ovens, network television, personal communication System (PCS), cellphones, personal
communicators etc.
They also found applications in areas other than communication and radar, Such as medicine, surveying
land, industrial quality control, radio astronomy, global positioning system, power transmission etc.
b) List out [L1][CO2] [6M]
i) Microwave frequency bands based on the IEEE standards.
ii) The applications of Microwaves.
i) Microwave frequency bands based on the IEEE standards.
The traditional microwave band designations have been adopted by the IEEE, but even so different sectors
tend to use slightly different frequency limits on the bands.
These band designations were traditionally used for radar applications, but they have also been adopted
for applications including satellite communications, navigation, aviation and now even 5G mmWave
mobile communications.
Designation Frequency range in Gigahertz
HF 0.003 - 0.030
VHF 0.030 – 0.300
Uhf 0.300 - 1.000
L band 1.000 – 2.000
S band 2.000 – 4.000
C band 4.000 – 8.000
X band 8.000 – 12.000
Ku band 12.000 -18.000
K band 18.000 – 27.000
Ka bad 27.000 – 40.000
Millimeter 40.000 – 300.000
Submillimeter >300.000
ii) The applications of Microwaves.
Micro Waves have a broad range of applications in modern Technology. most important among them are in long
distance communication Systems, radars, radio astronomy, navigation.
1. Telecommunication: TV. Space communication (earth-to-Space and Space to earth), telemetry
 communication link for railways etc.
2. Radars: Detect aircraft, observe and track weather patterns, air traffic control (ATC), garage door
openers, police speed detectors etc.
3. Commercial & Industrial applications use heat property of microwaves.
a) micro wave oven (2.45GHZ, 600W)
b) Drying machines
c) food processing industry.
d) Rubber industry/plastics/ chemical / forest product industries
e) Drying inks, drying textiles, leather, tobacco, power transmission
4. Biomedical applications.
5. Electronic warfare: ECM / ECCM (Electronic counter measure/ Electronic counter counter measure)
System S, Spread Spectrum Systems.
a) Discuss in detail about the concept of mode. [L2][CO2] [6M]
2.
Mode:
An EM wave has n number of field patterns which are called modes. The transverse components of
electric and magnetic fields are determined from the axial components of electric and magnetic field, in the
z direction. This allows mode formations, such as TE, TM, TEM and Hybrid in microwaves.
The direction of the electric and the magnetic field components along three mutually perpendicular
directions x, y, and z are as shown in the following figure.

Types of Modes
The modes of propagation of microwaves are −
i) TEM Transverse Electromagnetic Wave
In this mode, both the electric and magnetic fields are purely transverse to the direction of propagation.
There are no components in ′Z′′ direction.
EZ=0 and HZ=0
ii) TE Transverse Electric Wave
In this mode, the electric field is purely transverse to the direction of propagation, whereas the magnetic
field is not.
EZ=0 and HZ≠0
iii) TM Transverse Magnetic Wave
In this mode, the magnetic field is purely transverse to the direction of propagation, whereas the electric
field is not.
EZ≠0 and HZ=0
iv) HE HybridWave
In this mode, neither the electric nor the magnetic field is purely transverse to the direction of
propagation. EZ≠0 and HZ≠0
Multi conductor lines normally support TEM mode of propagation, as the theory of transmission
lines is applicable to only those system of conductors that have a go and return path, i.e., those which
can support a TEM wave.
Waveguides are single conductor lines that allow TE and TM modes but not TEM mode. Open
conductor guides support Hybrid waves.
b) Describe the concept of dominant mode with suitable examples. [L2][CO2] [6M]
Dominant Mode:
A "dominant mode" in a waveguide refers to the mode of electromagnetic wave propagation that has the lowest
cut-off frequency, meaning it can propagate at the lowest possible frequency within that waveguide, making it the
most readily transmitted mode; in a rectangular waveguide, the dominant mode is typically the TE10 mode, where
the electric field has one half-wave variation across the wider dimension of the waveguide with no variation along
the narrower dimension.
 • Lowest cut-off frequency:
The defining characteristic of a dominant mode is that it has the lowest cut-off frequency compared to other possible
modes within the waveguide.
• Propagation advantage:
Since it has the lowest cut-off frequency, the dominant mode experiences the least attenuation and is therefore
preferred for signal transmission in a waveguide.
• Mode designation:
In a rectangular waveguide, the dominant mode is usually denoted as TE10, where "TE" stands for Transverse
Electric (meaning the electric field is perpendicular to the direction of propagation) and "10" indicates the field
pattern with one variation across the wider dimension and no variations along the narrower dimension.
Example:
• Rectangular waveguide: In a rectangular waveguide where one dimension (a) is significantly larger than the
other (b), the TE10 mode will be the dominant mode.
• Circular waveguide: In a circular waveguide, the dominant mode is typically the TE11 mode.
a) Define the following terms: i) Guide wavelength ii) Phase [L2][CO1] [6M]
Velocity iii) Group Velocity.
3.
i) Guide Wavelength:
It is the distance traveled by the wave in order to undergo a phase shift of 2π radians. It is related to
propagation constant β as

Wavelength in waveguide is different from wavelength in free space.

ii) Phase velocity: The phase velocity of a wave is the rate at which the phase of the wave
propagates in space. The phase velocity is given by
𝜔
Vp= 𝛽
Also it is given as

iii) Group velocity:

If there is modulation in carrier, the modulation envelope travels at a velocity slower than the carrier.
This velocity of the modulation envelope is called as group velocity.

The group velocity of a wave is the velocity with which the overall shape of the waves' amplitudes —
known as the modulation or envelope of the wave — propagates through space. It is given as

Also it is given as

Relationship between group velocity and phase velocity:

vpvg=c2
b) The dimensions of a guide are 2.5x1cms. The frequency is 8.6 GHz. [L1][CO5] [6M]
Find the cutoff frequencies for TE10 and TE01 mode.
 a) Compute the expression for phase velocity. [L3][CO4] [6M]
4.
Expression for phase velocity:
The phase velocity of a wave is the rate at which the phase of the wave propagates in space.
The phase velocity is given by
𝜔
Vp= 𝛽
Also
ℎ2 = 𝛾 2 + 𝜔2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 -----→ (1)
And
𝛾 = 𝛼 + 𝑗𝛽
At wave propagation
𝛾 = 𝑗𝛽--------------------→(2)
Then
(𝑗𝛽)2 + 𝜔2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
(𝑗𝛽)2 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔2 𝜇𝜀
−𝛽 = ( ⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔2 𝜇𝜀 --------------------→(3)
2 𝑚𝜋
At f=fc, ꞷ=ꞷc , 𝛾 = 0 . 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠
0 + 𝜔𝑐 2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
𝜔𝑐 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 --------------→(4)
2

𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (4) 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (3)

−𝛽 2 = 𝜔𝑐 2 𝜇𝜀 − 𝜔2 𝜇𝜀
𝛽 2 = 𝜔2 𝜇𝜀 − 𝜔𝑐 2 𝜇𝜀
𝛽 = √𝜔 2 𝜇𝜀 − 𝜔𝑐 2 𝜇𝜀
𝛽 = √𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )--------→(5)
 The phase velocity is given by
𝜔 𝜔
Vp= 𝛽 =
√𝜇𝜀.(𝜔 2 −𝜔𝑐 2 )
𝜔 1
Vp= 𝛽 = 𝜔
√𝜇𝜀.(1− ( 𝜔𝑐 )2 )
𝐶
Vp= 𝜔
√(1− ( 𝜔𝑐 )2 )
𝐶
Interms of frequency, Vp= 𝑓
√(1− ( 𝑓𝑐 )2 )
𝐶
Interms of wavelength, Vp=
𝜆
√(1− ( 0 )2 )
𝜆
b) Describe the concept of dominant mode with suitable examples. [L5][CO5] [6M]
Dominant Mode:
A "dominant mode" in a waveguide refers to the mode of electromagnetic wave propagation that has the
lowest cut-off frequency, meaning it can propagate at the lowest possible frequency within that waveguide,
making it the most readily transmitted mode; in a rectangular waveguide, the dominant mode is typically the
TE10 mode, where the electric field has one half-wave variation across the wider dimension of the waveguide
with no variation along the narrower dimension.
• Lowest cut-off frequency:
The defining characteristic of a dominant mode is that it has the lowest cut-off frequency compared to other
possible modes within the waveguide.
• Propagation advantage:
Since it has the lowest cut-off frequency, the dominant mode experiences the least attenuation and is therefore
preferred for signal transmission in a waveguide.
• Mode designation:
In a rectangular waveguide, the dominant mode is usually denoted as TE10, where "TE" stands for Transverse
Electric (meaning the electric field is perpendicular to the direction of propagation) and "10" indicates the
field pattern with one variation across the wider dimension and no variations along the narrower dimension.
Example:
• Rectangular waveguide: In a rectangular waveguide where one dimension (a) is significantly larger
than the other (b), the TE10 mode will be the dominant mode.
Circular waveguide: In a circular waveguide, the dominant mode is typically the TE11 mode.
a) A rectangular waveguide has a=4cms, b=3cms as its sectional [L2][CO1] [6M]
5. dimensions. Predict all the modes which will propagate at 5000MHz.
 b) Derive the expression for group velocity. [L3][CO4] [6M]
Expression for Group velocity:
If there is modulation in carrier, the modulation envelope travels at a velocity slower than the carrier. This
velocity of the modulation envelope is called as group velocity.
or in other words the group velocity of a wave is the velocity with which the overall shape of the waves'
amplitudes — known as the modulation or envelope of the wave — propagates through space. It is given
as

we know that
𝛽 = √𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝜔
𝑑𝛽 𝑑
= √𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝑑𝜔 𝑑𝜔
1 𝑑
= (𝜔2 𝜇𝜀)
2
2√𝜇𝜀. (𝜔 − 𝜔𝑐 )2 𝑑𝜔
2𝜔 𝜇𝜀
=
2√𝜇𝜀.(𝜔2 −𝜔𝑐 2 )
𝑑𝛽 √𝜇𝜀
=
𝑑𝜔
√(1−(𝜔𝑐/𝜔) 2 )

√(1−(𝜔𝑐/𝜔) 2 )
Vg=
√𝜇𝜀

Vg= C. √(1 − (𝜔𝑐/𝜔) 2 )

𝑓
Interms of frequency, Vg= C. √(1 − ( 𝑓𝑐)2 )
𝜆
Interms of wavelength, Vg= C. √(1 − ( 𝜆0 )2 )
a) Derive the expression for cut off frequency in a waveguide. [L3][CO4] [6M]
6. Cutoff frequency in a waveguide:
The cut-off frequency is the frequency above which the waveguide offers minimum attenuation to the
propagation of the signal.
ℎ2 = 𝛾 2 + 𝜔2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 -----→ (1)
And
 𝛾 = 𝛼 + 𝑗𝛽

𝛾 2 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔2 𝜇𝜀
At low frequencies (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 > 𝜔2 𝜇𝜀,
𝜸 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒓𝒆𝒂𝒍 𝒂𝒏𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒆𝒒𝒖𝒂𝒍𝒔 𝒕𝒐 𝒂𝒕𝒕𝒆𝒏𝒖𝒂𝒕𝒊𝒐𝒏 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕. 𝑻𝒉𝒆 𝒘𝒂𝒗𝒆 𝒊𝒔 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒆𝒍𝒚
𝒂𝒕𝒕𝒆𝒏𝒖𝒂𝒕𝒆𝒅 𝒂𝒏𝒅 𝒏𝒐 𝒑𝒉𝒂𝒔𝒆 𝒄𝒉𝒏𝒈𝒆. 𝑯𝒆𝒏𝒄𝒆 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒄𝒂𝒏 𝒏𝒐𝒕 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒆.
At high frequencies (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 < 𝜔2 𝜇𝜀,
𝜸 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒊𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚 𝒆𝒒𝒖𝒂𝒍𝒔 𝒕𝒐 𝒑𝒉𝒂𝒔𝒆 𝒄𝒉𝒂𝒏𝒈𝒆. 𝑯𝒆𝒏𝒄𝒆 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒄𝒂𝒏 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒆.
𝜸 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒛𝒆𝒓𝒐, 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒄𝒖𝒕𝒐𝒇𝒇 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
0 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔𝑐 2 𝜇𝜀
𝜔𝑐 2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
(2𝜋𝑓𝑐 )2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
1
𝑓𝑐 2 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
(2𝜋)2 𝜇𝜀
1
𝑓𝑐 = √(𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
2𝜋√𝜇𝜀
1
𝑓𝑐 = √(𝑚⁄𝑎)2 + (𝑛⁄𝑏)2
2√𝜇𝜀
𝐶
𝑓𝑐 = √(𝑚⁄𝑎)2 + (𝑛⁄𝑏)2
2
b) Express the equation for cut off wave length. Deduce which TEmn mode [L1][CO1] [6M]
has the highest cutoff wavelength.
The equation for cut off wave length:
Cutoff frequency in a waveguide:
The cut-off wavelength is the wavelength above which the waveguide offers minimum attenuation to the
propagation of the signal.
ℎ2 = 𝛾 2 + 𝜔2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 -----→ (1)
And
𝛾 = 𝛼 + 𝑗𝛽

𝛾 2 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔2 𝜇𝜀
At low frequencies (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 > 𝜔2 𝜇𝜀,
γ becomes real and positive equals to attenuation constant. The wave is completely
attenuated and no phase chnge. Hence the wave can not propagate.
At high frequencies (mπ⁄a)2 + (nπ⁄b)2 < ω2 με,
γ becomes imaginary equals to phase change. Hence the wave can propagate.
𝜸 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒛𝒆𝒓𝒐, 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒄𝒖𝒕𝒐𝒇𝒇 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
0 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2 − 𝜔𝑐 2 𝜇𝜀
𝜔𝑐 2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
(2𝜋𝑓𝑐 )2 𝜇𝜀 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
1
𝑓𝑐 2 = (𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
(2𝜋)2 𝜇𝜀
1
𝑓𝑐 = √(𝑚𝜋⁄𝑎)2 + (𝑛𝜋⁄𝑏)2
2𝜋√𝜇𝜀

1
𝑓𝑐 = √(𝑚⁄𝑎)2 + (𝑛⁄𝑏)2
2√𝜇𝜀
𝐶
𝑓𝑐 = √(𝑚⁄𝑎)2 + (𝑛⁄𝑏)2
2
Cutoff wavelength,
 𝐶
𝜆𝑐 =
𝑓𝑐
𝐶
𝜆𝑐 =
𝐶 𝑚 2 𝑛 2
⁄
2 √( 𝑎) + ( ⁄𝑏)
2
𝜆𝑐 =
√(𝑚⁄𝑎)2 + (𝑛⁄𝑏)2
2𝑎𝑏
𝜆𝑐 =
√(𝑚𝑏)2 + (𝑛𝑏)2
In a rectangular waveguide, the TEmn mode with the highest cutoff wavelength is the TE10 mode. This is
because the TE10 mode has the lowest cutoff frequency, and since wavelength and frequency are inversely
related, the longest cutoff wavelength corresponds to the lowest cutoff frequency.
a) Deduce the relationship between normal wavelength, cut-off [L4][CO1] [6M]
wavelength and guided wavelength.
7.
Relationship between normal wavelength, cut-off wavelength and guided wavelength:The phase
velocity of a wave is the rate at which the phase of the wave propagates in space. The phase velocity is
given by
𝒗𝒑 = 𝝀𝒈 . 𝒇
𝝀𝒈
𝒗𝒑 = .𝑪
𝝀𝟎
From phase velocity expression, Vp= C
⁄√(1 − (𝜆0 )2
𝜆 𝑐

C 𝝀𝒈
= .𝑪
⁄√(1 − (𝜆0 )2 𝝀𝟎
𝜆𝑐
1 𝝀𝒈
=
⁄√(1 − (𝜆0 )2 𝝀𝟎
𝜆𝑐
𝝀𝒈
( ⁄𝝀 )2 = 1⁄ 𝜆
𝟎 (1 − (𝜆0 )2
𝑐
𝝀𝒈
( ⁄𝝀 )2 = 1⁄(𝜆 )2 − (𝜆 )2
𝟎 𝑐 0
(𝜆𝑐 )2
𝝀𝟎 2
(𝜆𝑐 )2 − (𝜆0 )2
( ⁄𝝀 ) =
𝒈 (𝜆𝑐 )2
𝟏 2
(𝜆𝑐 ) − (𝜆0 )2
2
( ⁄𝝀 ) =
𝒈 (𝜆𝑐 . 𝜆0 )2
(𝜆 )2 (𝜆 )2
(𝟏⁄𝝀 )2 = (𝜆 .𝜆𝑐 )2 - (𝜆 .𝜆0 )2
𝒈 𝑐 0 𝑐 0
1 1 1
= 2 − 2
𝜆𝑔 2 𝜆0 𝜆𝑐

b) Define Wave Impedance. Express the equation for wave impedance in [L2][CO1] [6M]
TE and TM waves.
Wave Impedance: Wave impedance is the ratio of the transverse electric field component to the transverse
magnetic field component of an electromagnetic wave, essentially representing the "electrical resistance" a
wave encounters as it propagates through a medium.
𝑬𝒙 𝑬𝒚
𝒛𝒛 = = −
𝑯𝒚 𝑯𝒙
Expression for Ex, Hy
 −𝛾 𝜕𝐸𝑧 𝑗𝜔𝜇 𝜕𝐻𝑧
Ex= ℎ2 -
𝜕𝑥 ℎ2 𝜕𝑦
−𝛾 𝜕𝐻𝑧 𝑗𝜔𝜀 𝜕𝐸𝑧
Hy= ℎ2 -
𝜕𝑦 ℎ2 𝜕𝑥
Now wave impedance is given by
−𝛾𝜕𝐸𝑧 𝑗𝜔𝜇𝜕𝐻𝑧
𝑬 2 −
𝜕𝑥 ℎ2 𝜕𝑦
𝒛𝒛 = 𝑯𝒙 = ℎ−𝛾𝜕𝐻𝑧 𝑗𝜔𝜀𝜕𝐸𝑧
𝒚 − 2
ℎ2 𝜕𝑦 ℎ 𝜕𝑥
𝐖𝐚𝐯𝐞 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞 𝐟𝐨𝐫 𝐓𝐄 𝐖𝐚𝐯𝐞𝐬: In TE waves electric field component is transverse to the direction of
propagation. i.e., Ez=0
𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 −
ℎ2 𝜕𝑦
𝒛𝑻𝑬 = 𝑯 = −𝛾𝜕𝐻𝑧
𝒚
ℎ2 𝜕𝑦
𝑗𝜔𝜇
=
𝛾
when the wave is propagating then γ = jβ and β = √με. (ω2 − ωc 2 )
𝑗𝜔𝜇
𝒛𝑻𝑬 =
𝑗𝛽
𝜔𝜇
=
𝛽
𝜔𝜇
=
𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝝁 𝟏
= √ .
𝜺 (𝟏 − (𝝎𝒄 )𝟐 )
𝝎
𝜼
𝒛𝑻𝑬 = 𝝎
√(𝟏−( 𝝎𝒄 )𝟐 )

𝐖𝐚𝐯𝐞 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞 𝐟𝐨𝐫 𝐓𝐌 𝐖𝐚𝐯𝐞𝐬: In TM waves electric field component is transverse to the direction
of propagation. i.e., Hz=0
−𝛾𝜕𝐸𝑧 𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 − 2
ℎ2 𝜕𝑥 ℎ 𝜕𝑦
𝒛𝑻𝑴 = =
−𝛾𝜕𝐻𝑧 𝑗𝜔𝜀𝜕𝐸𝑧
𝑯𝒚 − 2
ℎ2 𝜕𝑦 ℎ 𝜕𝑥
−𝛾𝜕𝐸𝑧
ℎ2 𝜕𝑥
= 𝑗𝜔𝜀𝜕𝐸
− 2 𝑧
ℎ 𝜕𝑥
𝜸
= 𝒋𝝎𝜺
when the wave is propagating then γ = jβ and β = √με. (ω2 − ωc 2 )
𝑗𝛽
𝒛𝑻𝑴 =
𝑗𝜔𝜀
√𝜇𝜀. (𝜔 − 𝜔𝑐 2 )
2
=
𝜔𝜀
𝝁 𝝎
= √ 𝜺 . √(𝟏 − ( 𝝎𝒄)𝟐 )
𝝎
𝒛𝑻𝑴 = 𝜼. √(𝟏 − ( 𝝎𝒄)𝟐 )
a) Derive the equation for wave impedance in TE mode. [L3][CO4] [6M]
8.
Wave Impedance: Wave impedance is the ratio of the transverse electric field component to the transverse
magnetic field component of an electromagnetic wave, essentially representing the "electrical resistance" a
wave encounters as it propagates through a medium.
𝑬𝒙 𝑬𝒚
𝒛𝒛 = = −
𝑯𝒚 𝑯𝒙
Expression for Ex, Hy
−𝛾 𝜕𝐸 𝑗𝜔𝜇 𝜕𝐻
Ex= ℎ2 𝜕𝑥𝑧 - ℎ2 𝜕𝑦𝑧
−𝛾 𝜕𝐻𝑧 𝑗𝜔𝜀 𝜕𝐸𝑧
Hy= ℎ2 -
𝜕𝑦 ℎ2 𝜕𝑥
Now wave impedance is given by
−𝛾𝜕𝐸𝑧 𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 − 2
ℎ2 𝜕𝑥 ℎ 𝜕𝑦
𝒛𝒛 = 𝑯 = −𝛾𝜕𝐻𝑧 𝑗𝜔𝜀𝜕𝐸𝑧
𝒚 −
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥
𝐖𝐚𝐯𝐞 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞 𝐟𝐨𝐫 𝐓𝐄 𝐖𝐚𝐯𝐞𝐬: In TE waves electric field component is transverse to the direction of
propagation. i.e., Ez=0
𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 −
ℎ2 𝜕𝑦
𝒛𝑻𝑬 = 𝑯 = −𝛾𝜕𝐻𝑧
𝒚
ℎ2 𝜕𝑦
𝑗𝜔𝜇
=
𝛾
𝒘𝒉𝒆𝒏 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒊𝒔 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒏𝒈 𝒕𝒉𝒆𝒏 𝜸 = 𝒋𝜷 𝒂𝒏𝒅 𝛽 = √𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝑗𝜔𝜇
𝒛𝑻𝑬 =
𝑗𝛽
𝜔𝜇
=
𝛽
𝜔𝜇
=
𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝝁 𝟏
= √ .
𝜺 (𝟏 − (𝝎𝒄 )𝟐 )
𝝎
𝜼
𝒛𝑻𝑬 = 𝝎
√(𝟏−( 𝝎𝒄 )𝟐 )

b) Derive the equation for wave impedance in TM mode. [L3][CO4] [6M]

Wave Impedance : Wave impedance is the ratio of the transverse electric field component to the transverse
magnetic field component of an electromagnetic wave, essentially representing the "electrical resistance" a
wave encounters as it propagates through a medium.
𝑬𝒙 𝑬𝒚
𝒛𝒛 = = −
𝑯𝒚 𝑯𝒙
Expression for
−𝛾 𝜕𝐸 𝑗𝜔𝜇 𝜕𝐻
Ex= ℎ2 𝜕𝑥𝑧 - ℎ2 𝜕𝑦𝑧
−𝛾 𝜕𝐻𝑧 𝑗𝜔𝜀 𝜕𝐸𝑧
Hy= ℎ2 -
𝜕𝑦 ℎ2 𝜕𝑥
Now wave impedance is given by
−𝛾𝜕𝐸𝑧 𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 − 2
ℎ2 𝜕𝑥 ℎ 𝜕𝑦
𝒛𝒛 = 𝑯 = −𝛾𝜕𝐻𝑧 𝑗𝜔𝜀𝜕𝐸𝑧
𝒚 − 2
ℎ2 𝜕𝑦 ℎ 𝜕𝑥
𝐖𝐚𝐯𝐞 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞 𝐟𝐨𝐫 𝐓𝐌 𝐖𝐚𝐯𝐞𝐬: In TM waves electric field component is transverse to the direction
of propagation. i.e., Hz=0
−𝛾𝜕𝐸𝑧 𝑗𝜔𝜇𝜕𝐻𝑧
𝑬𝒙 − 2
ℎ2 𝜕𝑥 ℎ 𝜕𝑦
𝒛𝑻𝑴 = =
−𝛾 𝜕𝐻𝑧 𝑗𝜔𝜀𝜕𝐸𝑧
𝑯𝒚 − 2
ℎ2 𝜕𝑦 ℎ 𝜕𝑥
−𝛾𝜕𝐸𝑧
ℎ2 𝜕𝑥
= 𝑗𝜔𝜀𝜕𝐸
− 2 𝑧
ℎ 𝜕𝑥

𝜸
= 𝒋𝝎𝜺
𝒘𝒉𝒆𝒏 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒊𝒔 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒏𝒈 𝒕𝒉𝒆𝒏 𝜸 = 𝒋𝜷 𝒂𝒏𝒅 𝛽 = √𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
𝑗𝛽
𝒛𝑻𝑴 =
𝑗𝜔𝜀
√𝜇𝜀. (𝜔 2 − 𝜔𝑐 2 )
=
𝜔𝜀
𝝁 𝝎
= √ 𝜺 . √(𝟏 − ( 𝝎𝒄)𝟐 )
 𝝎
𝒛𝑻𝑴 = 𝜼. √(𝟏 − ( 𝝎𝒄)𝟐 )
a) List out the features of TEM, TE and TM Modes. [L1][CO1] [6M]
9.
Features of TEM, TE and TM Modes:
In electromagnetic wave propagation, TEM (Transverse Electromagnetic), TE (Transverse Electric), and TM
(Transverse Magnetic) modes are distinguished by the direction of their electric and magnetic fields relative
to the direction of wave propagation, with key features being:
TEM Mode:
• No field components in propagation direction: Both the electric and magnetic fields are entirely
perpendicular to the direction of wave propagation.
• Requires two conductors: Can only exist in transmission lines with two conductors like coaxial cables
or parallel plate lines.
• No cutoff frequency: Can propagate at any frequency.
TE Mode:
• No electric field in propagation direction: The electric field is entirely transverse to the direction of
propagation, while the magnetic field can have components in both transverse and propagation
directions.
• Dominant mode in waveguides: Often the most easily excited mode in a waveguide.
• Cutoff frequency exists: Below a certain frequency, the TE mode cannot propagate.
TM Mode:
• No magnetic field in propagation direction: The magnetic field is completely transverse to the
propagation direction, while the electric field can have components in both transverse and propagation
directions.
• Higher attenuation than TE mode: Generally experiences more signal loss compared to TE modes.
• Cutoff frequency exists: Like TE modes, TM modes also have a cutoff frequency below which
propagation is not possible.
b) Discuss Insertion loss that occurs in microwave transmission. [L2][CO5] [6M]
Insertion loss in microwave transmission refers to the decrease in signal power that occurs when a microwave
signal travels through a transmission line, like a coaxial cable or waveguide, due to energy dissipation within
the line's materials, causing the output signal power to be lower than the input power; essentially, it's the loss
of signal strength as it passes through a component or system, measured in decibels (dB).
Factors affecting insertion loss:
• Frequency: Higher frequencies generally experience greater insertion loss.
• Cable length: Longer transmission lines result in higher insertion loss.
• Cable type: Different cable designs have varying levels of loss.
• Connector quality: Poorly designed or damaged connectors can introduce significant
insertion loss.
Measuring insertion loss:
• S-parameters: In microwave engineering, insertion loss is typically measured using S-
parameters, specifically the S21 parameter which represents the forward transmission
coefficient.
• Network analyzers: Specialized instruments like network analyzers are used to measure S-
parameters and calculate insertion loss.
a) Briefly discuss the losses that occur in a transmission structure in ideal [L2][CO5] [6M]
10. and practical situation.
Losses that occur in a transmission structure:
Losses in a wave guide can be due to attenuation below the cutoff and losses associated with attenuation due
to dissipation within the waveguide walls and dielectric within the waveguide.
At low frequencies below the cutoff the propagation constant has only the attenuation term.
The phase constant 𝛽 𝑖𝑡𝑠𝑒𝑙𝑓 𝑏𝑒𝑐𝑜𝑚𝑒 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑖𝑚𝑝𝑙𝑦𝑖𝑛𝑔 𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛.
2𝜋
𝛽=𝑗
𝜆𝑔
𝜆0
𝝀𝒈 =
⁄ 𝜆
√(1 − ( 0 )2
𝜆𝑐
 2𝜋
𝛽=𝑗
𝜆0
⁄ 𝜆
√(1 − ( 0 )2
𝜆𝑐
𝑗2𝜋 𝜆0
𝛽= √(1 − ( )2
𝜆0 𝜆𝑐
2𝜋 𝜆
𝛼= 𝜆0
√(1 − (𝜆0 )2
𝑐
𝑐𝑢𝑡𝑜𝑓𝑓 𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
54.6 𝜆
𝛼= 𝜆0
√(1 − (𝜆0 )2 or
𝑐

54.6 𝜆
𝛼= √( (𝜆𝑐 )2 − 1) dB/length.
𝜆𝑐 0
b) Explain about various losses that occur in microwave transmission. [L2][CO4] [6M]
Various losses that occur in microwave transmission:
The transmission line if not terminated with a matched load occurs in losses. These losses are of as
follows
Attenuation Losses:
The loss that occurs due to absorption of signal in transmission line is termed as attenuation losses.
𝐸𝑖 −𝐸𝑟
Attenuation losses (dB)= 10𝑙𝑜𝑔10 𝐸𝑡
Where Ei= input energy
Er= reflected energy from load to source
Et= transmitted energy to load
Reflection Losses:
The loss that occurs due to reflection of signal due to impedance mis match is termed as reflection
losses.
𝑖 𝐸
Reflection losses (dB)= 10𝑙𝑜𝑔10 𝐸 −𝐸
𝑖 𝑟
Where Ei= input energy
Er= reflected energy from load to source
Transmission losses:
The loss that occurs while transmitting through a transmission line is termed as transmission losses.
𝐸
Transmission losses (dB)= 10𝑙𝑜𝑔10 𝐸𝑖
𝑡
Where Ei= input energy
Et= transmitted energy to load
Return Losses:
The measure of power reflected by the transmission line is termed as return losses.
𝐸
Reflection losses (dB)= 10𝑙𝑜𝑔10 𝐸 𝑖
𝑟
Where Ei= input energy
Er= reflected energy from load to source
Insertion Losses:
The loss that occurs due to energy transfer using the transmission line compared to energy transfer
without transmission line is called insertion losses.
𝐸
Insertion losses (dB)= 10𝑙𝑜𝑔10 𝐸1
2
Where E1= the energy received by the load when directly connected to the source without transmission
line.
E2= the energy received by the load when a transmission line is connected between the load and
source.
 UNIT-II
MICROWAVE PARAMETERS
a) Explain with neat sketch the working of coaxial line transmission line. [L1][CO1] [6M]
Coaxial Transmission Line:
A coaxial transmission line is a type of transmission line used to carry high-frequency electrical signals with
1. minimal loss and interference. It consists of a central conductor, a dielectric insulator, an outer conductor
(shielding), and an outer insulating jacket.

Coaxial Transmission Line

Where, 1. Inner Conductor: A solid or stranded wire (usually copper) that carries the electrical signal.
2. Dielectric Insulator: Surrounds the inner conductor to maintain separation and control impedance.
3. Outer Conductor (Shielding): A braided or solid metal shield that provides grounding and protects against
electromagnetic interference (EMI).
4. Outer Jacket: An insulating layer that protects the cable from environmental damage.
Working Principle
• The inner conductor carries the signal, while the outer conductor (shield) provides a return path.
• The electric field is confined within the dielectric medium between the two conductors, minimizing
radiation losses.
• The shielding prevents external electromagnetic interference from affecting the signal.
• The characteristic impedance of the coaxial line is determined by the dimensions of the inner and outer
conductors and the dielectric constant of the insulator.
b) A coaxial line has the following physical dimensions. Diameter of [L4][CO5] [6M]
inner conductor=0.49cm,Inner diameter of outer conductor=1.10cm,
Polyethylene dielectric εr=2.3. Calculate i) Inductance per unit lengths
ii) Capacitance per unit length iii) characteristic impedance iv) the
velocity of
propagation.
 a) Derive the equation for the propagation of TE waves in [L3][CO4] [6M]
rectangular waveguide.
Wave Equation for the propagation of TE waves in rectangular waveguide:
For TE waves, electric field component Ez is transverse to the
direction of propagation. Hence Ez=0.
The Wave equation for a TE wave is given by,
2. ∆2 𝐻𝑧 = −𝜔2 𝜇𝜀𝐻𝑧 --→ (1)
2
On expanding ∆ equation 1 can be rewritten as
𝜕2 𝜕2 𝜕2
𝐻𝑧 + 𝐻𝑧 + 𝜕𝑧2 𝐻𝑧 = −𝜔2 𝜇𝜀𝐻𝑧
𝜕𝑥2 𝜕𝑦2
𝜕2
Since he wave propagates in Z direction 𝜕𝑧2 = 𝛾 2 then the
above equation can be written as
𝜕2 𝜕2
𝐻𝑧 + 𝐻𝑧 + 𝛾 2 𝐻𝑧 = −𝜔2 𝜇𝜀𝐻𝑧
𝜕𝑥2 𝜕𝑦2
𝜕2 𝜕2
𝐻𝑧 + 𝐻𝑧 + 𝛾 2 𝐻𝑧 + 𝜔2 𝜇𝜀𝐻𝑧 =0
𝜕𝑥2 𝜕𝑦2

𝜕2 𝜕2
𝐻𝑧 + 𝐻𝑧 + (𝛾 2 + 𝜔2 𝜇𝜀)𝐻𝑧 =0
𝜕𝑥2 𝜕𝑦2
Let h2=𝛾 2 + 𝜔2 𝜇𝜀, Then
𝜕2 𝜕2
𝐻 + 𝐻𝑧 + h2 𝐻𝑧 =0 ---→(2)
𝜕𝑥2 𝑧 𝜕𝑦2
Substitute 𝐻𝑧 = 𝑋𝑌 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
𝜕2 𝜕2
𝑋𝑌 + 𝑋𝑌 + h2 𝑋𝑌 =0
𝜕𝑥2 𝜕𝑦2
𝜕2 𝜕2
𝑌 𝜕𝑥2 𝑋 + 𝑋 𝜕𝑦2 𝑌 + h2 𝑋𝑌 =0
𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑏𝑦 𝑋𝑌, 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑐𝑜𝑚𝑒𝑠
1 𝜕2 1 𝜕2
𝑋+ 𝑌 + h2 =0 ----→ (3)
𝑋 𝜕𝑥2 𝑌 𝜕𝑦2
Solving X and Y by variable separation method,
X= C1CosBx+C2SinBx
Y=C3Cos Ay+C4Sin Ay
The complete solution of 𝐻𝑧 𝑖𝑠 𝐻𝑧 = 𝑋𝑌
𝐻𝑧 = (C1CosBx + C2SinBx). ( C3Cos Ay+C4Sin Ay)
By applying the boundary conditions, 𝐻𝑧 𝑖𝑠
mπ nπ
𝐻𝑧 = 𝐶Sin x. Sin y
𝑎 𝑏
The wave propagates in Z direction and continuously varies with time , it is given by
mπ nπ (𝑗𝜔𝑡−𝛾𝑧)
𝐻𝑧 = 𝐶Sin x. Sin y𝑒
𝑎 𝑏
b) An air filled rectangular waveguide of inside dimensions operates in the [L3][CO5] [6M]
dominant TE10 mode as shown in following figure. Compute the cutoff
frequency and determine the guided wavelength at f = 3.5 GHz.
 Given that

a) Describe the circular waveguide and the equation of cut off frequency [L3][CO4] [6M]
and its dominant mode.
A circular waveguide is a type of hollow metallic waveguide with a circular cross-section, used for
guiding electromagnetic waves, primarily in microwave and RF applications. Unlike coaxial cables,
waveguides do not use a central conductor; instead, they confine and guide waves through reflections inside
the conducting walls.
3.
Here a= radius
𝜑 = 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 0 𝑡𝑜 2𝜋.

Circular waveguide
The Helmholtz wave equation for TE and TM waves travelling along the Z direction in a Circular wave
guide is given by,
∆2 𝐻𝑧 = 0 & ∇2 𝐸𝑧 = 0
Equation of cut off frequency:
The cutoff frequency is that mode for which the mode propagation constant vanishes.
𝛾 = 𝛼 + 𝑗𝛽 = 0 − − − − − −→ (1)
𝛽 = √𝜇𝜖𝜔 2 − 𝜇𝜖𝜔𝑐 2
𝛽 = √𝜇𝜖𝜔 2 − 𝜇𝜖(2𝜋𝑓𝑐 )2
𝛽 = √𝜇𝜖𝜔 2 − (2𝜋𝑓𝑐 /𝑐)2
𝛽 = √𝜇𝜖𝜔 2 − (2𝜋/𝜆𝑐 )2
𝛽 = √𝜇𝜖𝜔 2 − (ℎ)2
’
Where h=P nm /a for TE Waves and h=Pnm /a for TM Waves.
h=2𝜋/𝜆𝑐
𝜆𝑐 = 2𝜋/ℎ
𝜆𝑐 = 2𝜋𝑎/P’nm
fc=C/𝜆𝑐
fc=C/(2𝜋𝑎/P’nm)
 P’nm
fc=C for TE Waves
2𝜋𝑎
Pnm
fc=C 2𝜋𝑎 for TM Waves.
Dominant mode:
fc will be minimum for the dominant mode of propagation. fc will be minimum if P’nm 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚.
For TE Waves P’nm 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑖𝑡ℎ 1.841 𝑎𝑡 𝑇𝐸11 .
For TM Waves Pnm 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑖𝑡ℎ 2.405 𝑎𝑡 𝑇𝑀01 .
b) A TE11 Mode is propagating through a circular waveguide. The radius [L3][CO5] [6M]
of the guide is 5 cm, and the guide contains an air dielectric. Compute
the cutoff frequency.

a) Describe the cavity resonator with neat sketch and List it types [L1][CO1] [6M]
4. & applications.
Cavity resonator: When one end of the waveguide is terminated with a shorting plate, there will be
reflections resulting in standing waves as shown in figure. When another shorting plates is kept at a
𝜆
distance of multiples of 𝑔⁄2, then the hollow space so formed can support a signal which bounces back
and forth between the two shorting plates. This results in resonance and hence the hollow space is called
cavity and the resonator is cavity resonator.

Fig (a) Cavity Resonator Fig (b) Rectangular Cavity Resonator

 Fig: Circular Cavity Resonator
• Rectangular Cavity:
A basic design with rectangular dimensions, commonly used for analysis due to its simple geometry.
• Circular Cavity:
A Circular shape, often used in applications requiring high power handling.
Applications of cavity resonators:
• Microwave Filters: Selecting specific frequencies in microwave systems by utilizing the resonant
properties of the cavity.
• Microwave Oscillators: Generating stable microwave signals by utilizing the resonant frequency
of a cavity.
• Radar Systems: Generating and detecting microwave pulses for radar applications.
• Particle Accelerators: Used in linear accelerators to accelerate charged particles.
• Laser Systems: Providing feedback mechanism in laser cavities to maintain stable laser output.
• Magnetic Resonance Imaging (MRI): Generating strong magnetic fields within the MRI
scanner.
b) Derive expression for fo in rectangular cavity resonator. [L3][CO4] [6M]
Expression for fo in rectangular cavity resonator:
For a Rectangular waveguide
𝑚𝜋 𝑛𝜋
ℎ2 = 𝛾 2 + 𝜔2 𝜇𝜀 = ( 𝑎 )2 + ( 𝑏 )2---------→(1)
𝑚𝜋 𝑛𝜋
𝜔2 𝜇𝜀 = ( 𝑎 )2 + ( 𝑏 )2 − 𝛾 2 ---→ (2)
For wave propagation 𝛾 = 𝑗𝛽 𝑡ℎ𝑒𝑛 𝛾 2 = −𝛽 2, now equation 2 can be re written as
𝑚𝜋 2 𝑛𝜋 2
𝜔2 𝜇𝜀 = ( ) + ( ) + 𝛽 2 − −−→ (3)
𝑎 𝑏
If the wave has to exist in a cavity resonator there must be a phase change corresponding to a given guide
wave length 𝛽 = 2𝜋⁄𝜆 . The condition for the resonator to resonate is
𝑔
𝑝𝜋
𝛽 = ⁄𝑑
Where p= a constant=0,1,2,3…..
d=length of the resonator.
Now equation 3 becomes,
𝑚𝜋 2 𝑛𝜋 2 𝑝𝜋
𝜔2 𝜇𝜀 = ( ) + ( ) + ( ⁄𝑑 )2
𝑎 𝑏
𝑚𝜋 2 𝑛𝜋 2 𝑝𝜋
(2𝜋𝑓0 )2 𝜇𝜀 = ( ) + ( ) + ( ⁄𝑑)2
𝑎 𝑏
1 𝑚𝜋 2 𝑛𝜋 2 𝑝𝜋
(𝑓0 )2 = 2
( ) + ( ) + ( ⁄𝑑 )2
(2𝜋) 𝜇𝜀 𝑎 𝑏
2 2 2
𝐶 𝑚𝜋 𝑛𝜋 𝑝𝜋
(𝑓0 )2 = 2
( ) + ( ) + ( ⁄𝑑 )2
(2𝜋) 𝑎 𝑏
2 2 2
𝐶 𝑚 𝑛 𝑝
(𝑓0 )2 = ( ) + ( ) + ( ⁄𝑑 )2
(2)2 𝑎 𝑏
𝐶2 𝑚 2 𝑛 2 𝑝
𝑓0 = √ 2
( ) + ( ) + ( ⁄𝑑 )2
(2) 𝑎 𝑏

𝐶 𝑚 2 𝑛 2
𝑓0 =√( ) + ( ) + (𝑝⁄ )2
2 𝑎 𝑏 𝑑
For both TE and TM waves resonant frequency is same for a rectangular cavity.
 a) Demonstrate the working principle of strip line. [L2][CO2] [6M]
5.
A strip line works by transmitting electrical signals along a flat conducting strip placed between two parallel
ground planes, with a dielectric material filling the space between them, essentially acting like a shielded
transmission line where the electromagnetic waves propagate through the dielectric, contained entirely within
the ground planes, allowing for controlled signal transmission with minimal interference; the width of the
strip, the dielectric thickness, and its permittivity determine the characteristic impedance of the line, ensuring
signal integrity at high frequencies.

Working principle of strip line

1. Strip line supports a Transverse Electromagnetic (TEM) wave, meaning the electric and magnetic
fields are perpendicular to the direction of signal propagation, resulting in minimal signal dispersion.
2. The ground planes on either side of the conductor effectively shield the signal, minimizing crosstalk
and electromagnetic interference from other components on the PCB.
3. The characteristic impedance of a stripline is primarily determined by the width of the conducting
strip, the thickness of the dielectric material, and its dielectric constant, allowing for precise
impedance matching in a circuit.
4. Due to its excellent signal integrity and controlled impedance, stripline is commonly used in high-
speed digital circuits and microwave applications where maintaining signal quality at high
frequencies is crucial.
b) Explain the working of Microstrip line. Draw its field distribution [L1][CO2] [6M]
diagram.
Microstrip is a type of electrical transmission line which can be fabricated with any technology where a
conductor is separated from a ground plane by a dielectric layer known as substrate. Microstrip lines are
used to convey microwave-frequency signals.
➢ Microstrip line is a unsymmetrical strip line.
➢ Quasi TEM, supports radiation with discontinuities.
➢ Better interconnet feature and easier fabrication.
➢ Extensively used to inter connect high logic circuits.
87 5.98ℎ
➢ Characteristic impedance Z0= +1.41 𝑙𝑛 [0.8𝑤+𝑡] 𝑓𝑜𝑟 ℎ < 0.8𝑤
√𝜀𝑟
377ℎ
Z0= for w> >h
√ 𝜀𝑟 𝑤
Z0 varies from 50 to 150Ω
 a) What is Non-TEM line? Express its equation [L1][CO3] [6M]
6. A non-TEM transmission line is a waveguide, which is a structure that guides waves.
Waves in non-TEM transmission lines can propagate inside a conducting envelope.
At microwave frequencies the measurement of voltage or current is difficult,
unless a clearly defined terminal pair is available.
Such a terminal pair may be present in the case of TEM-type lines (such as coaxial cable, microstrip line,
or stripline), but does not strictly exist for non-TEM lines (such as rectangular, circular, or surface
waveguides).
For an arbitrary two-conductor TEM transmission line the voltage, V, of the +ve conductor relative to the
−ve conductor can be found as

_________ (1)
The total current flowing on the + conductor can be determined
from an application of Ampere’s law as

For the dominant TE10 mode, the transverse fields can be written as,

___________(2)
Applying Eqn. (2) in (1)

Thus it is seen that this voltage depends on the position, x, as well as the length of the integration
contour along the y direction.
For non-TEM, an arbitrary waveguide mode with both positively and negatively traveling waves, the
transverse fields can be written as

The above equation defines equivalent voltage and current waves as

The proportionality constants for this relationship are

C1 = V+/A+ = V-/A-
C2 = I+/A+ = I-/A-
The characteristic impedance is
 The general field in a waveguide can be expressed in the following form:

b) Discuss the Faraday’s rotation and Recall the microwave devices which are [L1][CO4] [6M]
used for Faraday rotation.
Consider an infinite lossless medium. A static field B0 is applied along the z-direction. A plane TEM wave
that is linearly polarized along the x-axis at t=0 is made to propagate through the ferrite in the z-direction.
the plane of polarization of this wave will rotate with distance, a phenomenon known as faraday rotation.
Any linearly polarized wave can be regarded of the vector sum of two counter rotating circularly
polarized waves.
The ferrite materials offers different characteristics to these waves, with the result that the phase change
for one wave is larger than the wave, resulting in rotation ө of linearly polarized wave at z=l’.
If the direction of propagation is reversed, the plane of polarization continues to rotate in the same
direction i.e., from z=l to z=0, the wave will arrive back at z=0 polarized at an angle ‘zө’ relative to the
x-axis.
The ferrite rod is tapered at both ends to reduce attenuation & also for smooth rotation of polarized wave.

Operation:
When a wave enter port 1 its plane of
polarization rotates by 90̊ because of twist in the waveguide.
It again undergoes faradays rotation through 90̊ because of ferrite rod & the wave coming out of port 2
will have a phase shift of 180̊ compared to wave entering port 1.

When the same wave (TE10 mode signal) enter port 2 it undergoes faradays rotation through 90̊ in the
same anti-clockwise.
Because of twist, the wave gets rotates back by 90̊ comes out of port 1 with 0̊ phase shift as shown.
Hence a wave from port 1 to port 2 undergoes a phase shift of π radians, but the waveform port 2 to 1
doesn’t change its phase in gyrator.
The angle of rotation ‘ө’ is given by

L=length of ferrite rod, β+=phase shift of right circularly polarized wave and β-=phase shift of left
circularly polarized wave
When a wave is transmitted from port1 to 2 it undergoes rotation in anticlockwise direction.
The same wave is allowed to propagate from 2 to 1.
 It under one rotation in same direction (anticlockwise)
Faraday’s rotation principle used in
1. Gyrator
2. Isolator
3. Circulator

a) Discuss about Impedance &Admittance matrix representation of 2 port, N- [L2][CO1] [6M]

Port microwave network under analysis of RF and microwave transmission
7.
line.
Analysis of RF and microwave transmission line:
o A microwave network is formed when several microwave devices are connected together by
transmission lines for transmission of microwave signals.
o To analyzed the microwave transmission line , several mathematical models are available .
The mathematical models are
1. 1-port microwave network
2. Impedance, admittance matrix representations of 2-port & N-port microwave network.
3. Transmission matrix representation[ABCD]
4. ABCD matrix representation of some common 2-port microwave components.
5. Series connection of 2-port network.
6. Scattering matrix representation.
7. 2-port network
8. N-port network
9. Reciprocal microwave network
Impedance & admittance matrix representation of 2-port & N-port microwave networks:
The two port network usually used to represent a microwave device, It can be shown as.

For z-matrix:
We can write the relation between port current & voltages as

or in matrix form as

Zij can be found as

Above equation states that Zij can be found by driving port j with the current Ij , open-circuiting all other
ports (so Ik = 0 for k≠j ), and measuring the open-circuit voltage at port i.
Thus, Zii is the input impedance seen looking into port i when all other ports are open-circuited, and Zij is
the transfer impedance between ports i and j when all other ports are open-circuited.

Similarly for admittance matrix:

or in matrix form as

the [Z] and [Y] matrices are the inverses of each other

Yij can be found as

which states that Yij can be determined by driving port j with the voltage Vj, short-circuiting all other
ports (so Vk = 0 for k≠j), and measuring the short-circuit current at port i.
b) Derive the S-matrix for series connection of two port network. [L3][CO4] [6M]
o For microwave frequency, it is difficult to measure voltage, current.
o Direct measurement can’t be done, if all EM waves are at high frequency.
o By using S matrix, we can measure directly reflected, transmitted incident waves.
o [s] matrix relates the incident, transmitted, reflected waves at port.
o Once parameter is known, they can be converted to any other matrix parameter for analysis.
For a 2 port network:
Assume at port 1,
1. Amplitude of incident voltage (v1+)
2. Amplitude of reflected voltage(v1-)
The scattering parameters are

Assume at port 2,
1. Amplitude of incident voltage (v1+)
2. Amplitude of reflected voltage(v1-)

In matrix form
 a) Explain the working principle of Gyrator with neat sketch. [L2][CO3] [6M]
8.
It is a two port device that has a relative
Phase difference of 180° for transmission
From port 1 port 2 and ‘no’ phase shift
(0° phase shift) for transmission from port 2 to port 1

→ The construction of a gyrator is as shown in fig. It consists of a piece of circular waveguide carrying the
dominant TE‫ ﮼‬mode with transitions to a Standard rectangular waveguide with dominant mode (TE₁₀) at
both ends.
→ A thin circular ferrite rod tapered at both ends is located inside the circular waveguide supported by
polyfoam and the waveguide is surrounded by a permanent magnet which generates dc magnetic field
for proper operation of ferrite.
Operation: -
→When a wave enters post 1 its plane of polarization rotates by 90⁰ because of the twist in the waveguide.
It again undergoes faraday rotation 90⁰ because of ferrite rod and the wave which comes out of port 2
will have a phase shift of 180⁰ compared to the wave entering port 1.
→But when the same wave (TE₁₀ mode signal) enters port 2, it undergoes faraday rotation through 90⁰
in the same anticlockwise direction.
→Because of the twist, the wave got rotated back by 90⁰ comes out of port 1 with 0⁰ phase shift.
→Hence a wave at port 1 undergoes a phase shift of 𝜋 radians but a wave fed from port 2 does not change
its phase in a gyrator.
b) Deduce the S-matrix for Gyrator. [L4][CO5] [6M]
Scattering matrix of gyrator:-

1. Gyrator is a two port network, then the size of the s matrix is 2x2.
𝑆 𝑆12
[S] =[ 11 ] --------→(1)
𝑆21 𝑆22
2. As gyrator is loss less medium, S₁₁, S₂₂ are zero i.e., no lossless. Port 1 and 2 are perfectly matched to
the network then

𝑆11 = 𝑆22 = 0-------------→(2)

3. S₁₂→wave from port 2 to 1, no phase shift i.e.,1.
S₂₁→ wave from port 1to 2, 180° phase shift i.e., -1
 i.e., 𝑆21 = −𝑆12-----→(3)
on substituting equation 2 & 3 in equation 1
0 𝑆12
[S] =[ ]
−𝑆12 0
4. For a gyrator form unitary property [S].[S]*=[I]
0 𝑆12 0 𝑆 ∗12 1 0
[ ][ ]=[ ]
−𝑆12 0 −𝑆 ∗12 0 0 1
On multiplying row 1 and column 1
0+(𝑆12 )(−𝑆 ∗12 ) = 1
|𝑆12 |2 = 1
𝑆12 = 1
From equation 3 we have
𝑆21 = −𝑆12 = −1
Then the Scattering matrix of a gyrator is given by
0 1
[S] =[ ]
−1 0
a) What is isolator? Explain the working principle of Isolator with a neat [L1][CO2] [6M]
9. sketch.
Uses the principle of faradays rotation.
→An isolator is a two port device which provides a very small amount of attenuation for the transmission
from port 1 to port 2, but provides maximum attenuation for the transmission from port 2 to port 1.

→The mismatch of generator output to the load, results in a reflected wave from load, but these reflected
waves should not be allowed to reach the microwave generator, which will cause amplitude &frequency
instability of microwave generator.
→When isolator inserted between generator & load, the generator output is coupled with the load with 0
attenuation &reflections if any from the load are completely observed by the isolator without effecting the
generator output.
→hence generator appears to be matched for all loads in the presence of isolator.
 S-matrix of isolator: -
As isolator consider as two port network

𝑆11 𝑆12
[S] =[ ]
𝑆21 𝑆22
0 0
[S] =[ ]
1 0
b) Deduce the S-matrix for Isolator. [L4][CO5] [6M]
Scattering matrix of Isolator:-
Isolator are two port devices which are non-reciprocal having no attenuation from port 1 to port 2 & infinite
attenuation from port 2 to port 1
1. Isolator is a two port network, then the size of the s matrix is 2x2.
𝑆 𝑆12
[S] =[ 11 ] --------→(1)
𝑆21 𝑆22
2. As Isolator is loss less medium, S₁₁, S₂₂ are zero i.e., no lossless. Port 1 and 2 are perfectly matched to
the network then

𝑆11 = 𝑆22 = 0-------------→(2)

3. S₁₂→wave from port 2 to 1, no wave exist as output i.e.,0.
S₂₁→ wave from port 1to 2, 0° phase shift i.e., 1

i.e., 𝑆12 = 0-----→(3)

on substituting equation 2 & 3 in equation 1
0 0
[S] =[ ]
𝑆21 0
𝑇
4. For a Isolator from unitary property [S].[𝑆 ∗] =[I]
0 0 0 0𝑇 1 0
[ ][ ] =[ ]
𝑆21 0 𝑆 ∗21 0 0 1

0 0 0 𝑆21 1 0
[ ][ ]=[ ]
𝑆21 0 0 0 0 1

On multiplying row 2 and column 2

0+(𝑆21 )(𝑆 ∗21 ) = 1

|𝑆21 |2 = 1
𝑆21 = 1
Then the Scattering matrix of a gyrator is given by
0 0
[S] =[ ]
1 0
a) Explain the working principle of Circulator with a neat sketch. [L2][CO3] [6M]
10.
→The circulator is a three port or four port microwave device, which has a peculiar property that such
terminal is connected to next clock wise terminal. i.e., port 1 is connected to port 2 and not to port 3, 4.
→Similarly port 2 is connect to port 3 not to port 4, and 1
→They are useful in parametric amplifier, tuned tunnel diode amplifier and as duplexes in radars.

Working principle of Circulator: -

→The wave entering port 1 is TE₁₀ mode and is converted to TE₁₁ mode because of gradual rectangular
to circular transition.

→This power passes port 3 unaffected. since The electric field is not signification cut and is rotated
through 45° due to the ferrite, Passes port 4 unaffected and finally emerges Out of port 2.
→Power from port 2 will have plane of polarization already titled by 45° with respect to port 1. This
power passes port 3 unaffected because again the electric field is not significantly cut.
→This wave gets rotated by another 45° due to ferrite rod in the clockwise direction.
→This power whose plan of polarization is tilted through 90° finds port 3 suitably aligned and emerges
out of it. Similarly port 3 in coupled only to port 4 and port 4 to port 1.
b) Deduce the S-matrix for Circulator. [L4][CO5] [6M]
S matrix of a Circulator:
1. For a four port circulator , the S matrix is of size 4x4.
𝑆11 𝑆12 𝑆13 𝑆14
𝑆 𝑆22 𝑆23 𝑆24
[S]= [ 21 ]-----→(1)
𝑆31 𝑆32 𝑆33 𝑆34
𝑆41 𝑆42 𝑆43 𝑆44
2. Circulator is loss less medium, S₁₁, S₂₂ , S33, S44 are zero i.e., no lossless. Port 1 ,2 , 3 and 4 are
perfectly matched to the network then
S₁₁ = S₂₂ = S33= S44 -----→(2)
3. In Circulator input at port 1 produces output at port 2 unaffected and no output will be produced at
port 3 and 4.
𝑆31 = 𝑆41 = 0-------→ (3)
4. In Circulator input at port 2 produces output at port 3 unaffected and no output will be produced at
port 1 and 4.
𝑆12 = 𝑆42 = 0-------→ (4)
5. In Circulator input at port 3 produces output at port 4 unaffected and no output will be produced at
port 1 and 2.
𝑆13 = 𝑆23 = 0-------→ (5)
6. In Circulator input at port 4 produces output at port 1 unaffected and no output will be produced at
port 2 and 3.
𝑆24 = 𝑆34 = 0-------→ (6)
On substituting equations 2,3,4,5,& 6 in equation 1. We have

0 0 0 𝑆14
𝑆21 0 0 0
[S]= [ ]---------→(7)
0 𝑆32 0 0
0 0 𝑆43 0
7. All the inputs of a circulator is reciprocal to each other then
𝑆14 = 𝑆21 =𝑆32=𝑆43
From the above equation 7 can be re written as
0 0 0 𝑆14
𝑆14 0 0 0
[S]= [ ]-----→(8)
0 𝑆14 0 0
0 0 𝑆14 0
8. From unitary property, [S].[𝑆 ∗]𝑇 =[I]
0 0 0 𝑆14 0 0 0 𝑆 ∗14 𝑇 1 0 0 0
𝑆 0 0 0 𝑆∗ 0 0 0 0 1 0 0
[ 14 ] [ 14 ] =[ ]
0 𝑆14 0 0 0 𝑆 ∗14 0 0 0 0 1 0
0 0 𝑆14 0 0 0 𝑆 ∗14 0 0 0 0 1
 0 0 0 𝑆14 0 𝑆 ∗14 0 0 𝑇 1 0 0 0
𝑆14 0 0 0 0 0 𝑆 ∗14 0 0 1 0 0
[ ][ ] =[ ]
0 𝑆14 0 0 0 0 0 𝑆 ∗14 0 0 1 0
0 0 𝑆14 0 𝑆 ∗14 0 0 0 0 0 0 1
On multiplying row 1 and column 1
0+0+0+𝑆14 . 𝑆 ∗14 = 1
|𝑆14 |2 = 1
𝑆14 = 1-----→(9)
The scattering matrix of a circulator is given by
0 0 0 1
1 0 0 0
[S]= [ ]
0 1 0 0
0 0 1 0
Course Code: 20EC0423 R20

UNIT-III
WAVEGUIDE COMPONENTS AND APPLICATIONS

a) Interpret the mechanism of coupling in a waveguide. [L3][CO1] [6M]

1. Mechanism of Coupling in a Waveguide:-
➢ The transmission, propagation, and reflection of guided waves happens in waveguides with or
without excitation.
➢ Waveguide coupling is a process in which the part of electromagnetic energy associated with one
waveguide is shared with another waveguide.
➢ Waveguides can be coupled to generators, excitation sources, or other waveguides to exchange
energy in and out.
➢ There are three waveguide coupling methods: probe coupling, loop coupling, and aperture
coupling.
Probe Coupling in Waveguides
➢ In probe coupling, a probe inside a coaxial line is used to distribute energy into a waveguide.
➢ As current starts flowing in the probe, an electric field is set up and it gets detached from the probe
to the waveguide.
➢ The probe (also called a probe antenna) radiates energy equally into the waveguide where it is
inserted.

➢ Probe antenna insertion is usually made perpendicular to the length of the waveguide, at a distance
equal to the quarter wavelength from the shorted end of the waveguide.
➢ Probe coupling is intended for coupling to the electric field and is made at the center of the
waveguide, as the electric field concentration is at its maximum at this position.
Loop Coupling in Waveguides
➢ Loop coupling enables coupling to the magnetic field in the waveguide. In loop coupling, a
conductor is inserted into the waveguide and bends into a loop.
➢ The center of the loop is at an equal distance from the top and bottom walls of the waveguide.

Fig. (a) Loop coupling (b) H-field distribution

➢ When current flows through the loop, it generates a magnetic field component that couples with
the waveguide fields.
For high efficiency, the loop should be inserted at the point where the magnetic field is at its maximum strength.
b) Explain the following waveguide components [L2][CO2] [6M]
(i) Waveguide posts (ii) Tuning Screws
Waveguide components are RF passive components used to transmit, amplify, control, measure, filter,
couple, combine or divide Radio Frequency (RF) signals at high frequencies.
Waveguide components consist of flexible waveguides, waveguide amplifiers, waveguide attenuators,
waveguide transitions, waveguide bends, waveguide circulators, corner reflectors, feed horns, waveguide
filters, and much more.
Course Code: 20EC0423 R20

(i) Waveguide posts:

A cylindrical post is introduced into the broader side of the waveguide; it produces a similar effect as an
iris in providing lumped capacitive/inductive reactance at that point. When a metalpost extends completely
across the waveguide, parallel to an electric field, it adds an inductive susceptance that is parallel to the
waveguide. A post extending across the waveguide at right angles to the electric field produces an effective
capacitive susceptance that is in shunt with the waveguide at the position of the post. The waveguide post
is shown in figure below. The advantage of such posts over irises is the flexibility they provide, which
results in ease of matching.

(ii) Tuning Screws:

A screw is generally inserted into the top or bottom walls of the waveguide, parallel to the electric-field
lines. It can give a variable amount of susceptance depending on the depth of penetration. A screw with
an insertion distance (screw depth) less than λ/4 produces capacitive susceptance. When the distance is
equal to λ/4, we have series resonance. When the distance is greater than λ/4, it produces inductive
susceptance, as shown in figure below.

Susceptance nature at different screw depths

Describe the following attenuators: [L2][CO2] [12M]

2. i) Resistive Card attenuator ii)Rotary Vane Attenuator
A waveguide attenuator is an RF device specifically designed to reduce the power of a signal without
affecting or reducing the waveform of the signal.
An attenuator is a two-port resistive network designed to weaken or “attenuate” the power being supplied
by a source to a level that is suitable for the connected load.
Attenuators are commonly used for :-
1. Measuring power gain or loss in dB.
2. Providing isolation between instruments.
3. Reducing the power input to a particular stage to prevent overloading.
Course Code: 20EC0423 R20

There are two broad categories of attenuators :--

(i) Fixed Attenuators &
(ii) Variable Attenuators
i) Resistive Card attenuator
Attenuator is a simple resistive network which reduces the amplitude of the input signal by finite amount
at the output and the circuit is called as attenuator circuit. It consists of a dielectric strip coated with
resistive film. It is placed at the centre of the waveguide parallel to the Emax. The strip is parallel to the
electric field intensity then it can absorb minimum amount of the incoming signal. If it perpendicular it
absorbs the required attenuation. When the electric field passes through the film it can induce current.
Induced current on the resistive film due to the incident wave result in power dissipation in the form of
heat which is nothing but the attenuation of the microwave energy.

Fixed attenuator

ii) Rotary Vane Attenuator

The most satisfactory precision attenuator is the rotary vane attenuator. The structure of this attenuator is
shown in figure below. It consists of two rectangular to circular waveguide tapered transitions, along with
an intermediate section of a circular waveguide that is free to rotate. All the three sections contain thin
resistive cards.

Rotary Vane Type Attenuator

The input signal passes the first card with a negligible attenuation, because the electric field of the
TE10 wave mode is perpendicular to the card. Then, the wave enters through a transition to the circular
waveguide. The attenuation is adjusted by rotating the circular waveguide section and the resistive card
within it. The field of the TE11 wave mode can be divided into two components: one perpendicular to the
Course Code: 20EC0423 R20

card and the other parallel to it. The latter component is absorbed by the card; the former component enters
the output of the waveguide, in which again its component parallel to the resistive card is absorbed.
The functional diagram indicating the operating principle of this type of attenuator is :

Stage (i):- When all strips are aligned the electric field of the applied wave is normal to the strip and
hence no current flows in the attenuation strips and therefore no attenuation occurs.
Stage (ii):- In a position where the central attenuation strip is rotated by angle Ɵ, the electric field of
applied wave can be resolved into two factors (a) perpendicular ECos Ɵ and other (b) parallel ESin Ɵ
to the resistive card.
The portion which is parallel ESin Ɵ will be absorbed whereas the portion which is perpendicular ECosƟ
will be transmitted.
Stage (iii):- Again ECosƟ is resolved into two factors namely ECos2Ɵ and ECosƟ Sin Ɵ.
Stege (iv):- The final output of the attenuator becomes ECos2Ɵ which has the same polarization as
the input wave.
a) What is the principle of phase shifter? Discuss the working [L1][CO3] [6M]
3. mechanism of rotary vane phase shifter with neat sketch.
Waveguide Phase Shifters: A phase shifter is a two-port component that provides
a fixed or variable change in the phase of the traveling wave. An ideal phase shifter is lossless and
matched. It only shifts the phase of the output wave. Example: Phase shifters are used in phased antenna
arrays.
Working Mechanism of Rotary Vane Phase Shifter:-
A structure resembling the attenuator in the figure below also operates as a phase shifter when the resistive
cards are replaced with dielectric cards having proper lengths. Electrically controlled phase shifters are
much faster than mechanical phase shifters.
The precision phase shifter can be realized by a rotary phase shifter, which is useful in microwave
measurement. The essential parts of this phase shifter are three waveguide sections: two fixed and one
rotary. The fixed sections consist of quarter-wave plates, and the rotary section consists of half-wave
plates; all the plates are of a dielectric type. The center section is rotatable to provide the required phase
shift. The structure of the rotary vane attenuator is shown in the figure below.

Rotary phase shifter

The two fixed quarter-wave sections are identical in all respects, and the rotatable half-wave section is
just the double of a quarter-wave section. Each of the two fixed sections, attached to a transition,
consists of a piece of a circular waveguide with a dielectric plate, making an angle of 45° with the
horizontal. The dielectric plate is usually thin with εr > 1, μr = 1. The output remains vertically
Course Code: 20EC0423 R20

polarized, which means that the phase shifter is lossless and reflection less for any position of the rotary
section. It is used as a calibration standard due to its high accuracy.

b) Explain the significance and formulation of S-matrix in detail. [L2][CO1] [6M]

The Significance and Formulation of S-Matrix:-
Scattering Matrix is a square matrix which gives all the combinations of power relationships between the
various input and output ports of a Microwave junction. The elements of this matrix are
called "Scattering coefficients" or "Scattering S-Parameters".

N port network
Case (i):- Let the first line be terminated in an impedance other than the characteristic impedance (ZL= Z0) and all
the remaining lines with an impedance equal to Z0 (ZL =Z0).
➢ If ai be the incident wave at junction to source at the ith line, then it divides itself among n-1 number of
lines as a1,a2,a3,………………an.
➢ There will be no reflections from 2nd to nth line and the incident waves are absorbed since their impedance
is are equal to characteris tic impedance Z0.
b2=b3=b4=…………..bn=0
Si2= Si3 = Si4=……………….= Sin=0
➢ But there is a mismatch at the 1st line and hence there will be reflected wave b1 going back into the
junction, b1 is related to a1 by
b1= reflection co efficient . a1
b1=S1i.a1
Case (ii):- Let all the n-1 lines be terminated in an impedance other than the characteristic impedance (ZL= Z0).

➢ Then there will be reflections into the junction from every line and hence the contribution to the outward
travelling wave in the ith line is given by
bi=Si1a1 + Si2a2 + Si3a3 +…………………………+ Sinan
where i=1,2,3,…………………………………..n I can be any line from 1 to n
b1=S11a1 + S12a2 + S13a3 +…………………………+ S1nan
b2=S21a1 + S22a2 + S23a3 +…………………………+ S2nan
b3=S31a1 + S32a2 + S33a3 +…………………………+ S3nan
.
.
.
bn=Sn1a1 + Sn2a2 + Sn3a3 +…………………………+ Snnan

In matrix form,

𝑏1 𝑆11 𝑆12 𝑆13 . . 𝑆1𝑛 𝑎1

𝑏2 𝑆21 𝑆22 𝑆23 . . 𝑆2𝑛 𝑎2
𝑏3 𝑆 𝑆32 𝑆33 . . 𝑆3𝑛 𝑎3
= 31 . .
. . . . . . .
. . . . . . . .
[𝑏𝑛 ] [𝑆𝑛1 𝑆𝑛2 𝑆𝑛3 . . 𝑆𝑛𝑛 ] 𝑛 ]
[ 𝑎
In a simple form it is given by,
[b]=[S][a]
Where [b]= represents a column matrix of order nX1 corresponding to reflected waves or output.
Course Code: 20EC0423 R20

[a] = represents a column matrix of order nX1 corresponding to incident waves or input.
[S] = Scattering matrix of order nXn corresponds to Scattering co-efficients .

a) List out the properties of S-matrix. [L1][CO3] [6M]

4.
Properties of S matrix:
1. [S] is always a square matrix of order nXn.
2. [S] is always a symmetric matrix Sij=Sji.
3. The scattering coefficient S13=S23. Since , the outputs at port 1 and 2 are inphase with an input at
port 3.
4. [S] is a unitary matrix. [S].[S]*=[I]
Where, [S]* = complex conjugate of [S]
[I] = Identity matrix of order same as [S]
5. The sum of product of each term of any row or column multiplied by the complex conjugate of
corresponding term of any other row or column is zero.
𝑛

∑ 𝑆𝑖𝑘 𝑆 ∗ 𝑖𝑗 = 0; 𝑘 ≠ 𝑗
𝑖=1
6. if any of the terminal or reference plane (say kth plane) are moved away from the junction by a electric
distance 𝛽𝑘 𝑙𝑘 , each of the coefficient Sij, involving k will be multiplied by a factor 𝑒 −𝑗𝛽𝑘𝑙𝑘 .
b) Derive S-matrix calculation for two port network. [L2][CO4] [6M]
If the source is applied to the opposite port, another two combinations are to be considered. So, for a two-
port network, 2 × 2 = 4 combinations are likely to occur.

The travelling waves with associated powers when scatter out through the ports, the Microwave junction
can be defined by S-Parameters or Scattering Parameters, which are represented in a matrix form, called
as "Scattering Matrix".
a1= incident power at port 1
b1= reflected or output power from port 1
a2= incident power at port 2
b2= reflected or output power from port 2
The S matrix of a two port network is
[b]=[S][a]
Where [b]= represents a column matrix of order 2X1 corresponding to reflected waves or output.
[a] = represents a column matrix of order 2X1 corresponding to incident waves or input.
[S] = Scattering matrix of order 2X2 corresponds to Scattering co-efficients .
For a 2 port network
b1=S11a1 + S12a2
b2=S21a1 + S22a2
𝑏1
𝑆11 = 𝑎𝑡 𝑎2 = 0
𝑎1
𝑏1
𝑆12 = 𝑎𝑡 𝑎1 = 0
𝑎2
𝑏2
𝑆21 = 𝑎𝑡 𝑎2 = 0
𝑎1
𝑏2
𝑆22 = 𝑎𝑡 𝑎1 = 0
𝑎2
a) Construct the microwave tee, whose rectangular slot is cut along the [L3][CO5] [6M]
5. broader dimension, Describe in detail.
E Plane Tee junction:
An E-Plane Tee junction is formed by attaching a simple waveguide to the broader dimension of a rectangular
Course Code: 20EC0423 R20

waveguide, which already has two ports. The arms of rectangular waveguides make two ports called collinear ports
i.e., Port1 and Port2, while the new one, Port3 is called as Side arm or E-arm. This E-plane Tee is also called as
Series Tee.
As the axis of the side arm is parallel to the electric field, this junction is called E-Plane Tee junction. The ports 1
and 2 are 180° out of phase with each other.

➢ The waves entering the side arm splits up and leaves the main arm with equal in magnitude but
opposite phase.

Fig.: Input is fed from port 3(side arm)

➢ Similarly, the wave entering from the junction main arm leaves the side arm, the resulting field being
proportional to difference between instantaneous field entered the junction from opposite direction.

Fig.: Input is fed from port 1 and 2

b) Derive the S-matrix for E-Plane Tee. [L2][CO4] [6M]

S-matrix for E-Plane Tee:
1. Since it is a 3 port network. The order of the scattering matrix is 3x3.
𝑆11 𝑆12 𝑆13
[𝑆] = [𝑆21 𝑆22 𝑆23 ]------------→(1)
𝑆31 𝑆32 𝑆33
2. The Scattering coefficient 𝑆23 = −𝑆13 -----------→(2)
Since outputs at port 1 and 2 are out of phase by 1800 with an input at port 3.
3. If port 3 is perfectly matched to the junction.
𝑆33 = 0-----------→(3)
4. From the symmetric property Sij=Sji.
𝑆12 = 𝑆21 , 𝑆13 = 𝑆31 , 𝑆23 = 𝑆32 = −𝑆13 ------------→(4)
Course Code: 20EC0423 R20

On substituting equation 2, 3 and 4 in equation 1

𝑆11 𝑆12 𝑆13
[𝑆] = [𝑆12 𝑆22 −𝑆13 ]------------→(5)
𝑆13 −𝑆13 0
*
5. From the unitary property [S].[S] =[I]
𝑆11 𝑆12 𝑆13 𝑆11 𝑆12 𝑆13 ∗ 1 0 0
[𝑆12 𝑆22 −𝑆13 ] . [𝑆12 𝑆22 −𝑆13 ] = [0 1 0]
𝑆13 −𝑆13 0 𝑆13 −𝑆13 0 0 0 1
∗ ∗
𝑆11 𝑆12 𝑆13 𝑆11 𝑆12 𝑆13 ∗ 1 0 0
∗ ∗ ∗
[𝑆12 𝑆22 −𝑆13 ] . [𝑆12 𝑆22 −𝑆13 ] = [0 1 0]
𝑆13 −𝑆13 0 ∗ ∗
𝑆13 −𝑆13 0 0 0 1
On multiplying row1 and column 1, we have
𝑆11 . 𝑆11 ∗ + 𝑆12 . 𝑆12 ∗ + 𝑆13 . 𝑆13 ∗ = 1
|𝑆11 |2 + |𝑆12 |2 + |𝑆13 |2 = 1-----------------→(6)
On multiplying row 2 and column 2
𝑆12 . 𝑆12 ∗ + 𝑆22 . 𝑆22 ∗ + (−𝑆13 ). (− 𝑆13 ∗ ) = 1
|𝑆12 |2 + |𝑆22 |2 + |𝑆13 |2 = 1-----------------→(7)
On multiplying row 3 and column 3
𝑆13 . 𝑆13 ∗ + (−𝑆13 ). (− 𝑆13 ∗ ) + 0 = 1
|𝑆13 |2 + |𝑆13 |2 = 1
2. |𝑆13 |2 = 1
|𝑆13 |2 = 1/2
|𝑆13 | = 1⁄ -------------------→(8)
√2
𝑜𝑛 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6 𝑎𝑛𝑑 7

|𝑆11 |2 + |𝑆12 |2 + |𝑆13 |2 − (|𝑆12 |2 + |𝑆22 |2 + |𝑆13 |2 ) = 1 − 1

|𝑆11 |2 − |𝑆22 |2 = 0
𝑆11 = 𝑆22 -----------------→(9)
On multiplying row 3 and column 1
𝑆13 . 𝑆11 ∗ + (−𝑆13 ). ( 𝑆12 ∗ ) + 0 = 1
𝑆11 ∗ = 𝑆12 ∗
𝑆11 = 𝑆12 ------------------→(10)
From equation 10 and 8, equation 6 can be solved as
(6) ----------→ |𝑆11 |2 + |𝑆12 |2 + |𝑆13 |2 = 1
|𝑆11 |2 + |𝑆11 |2 + (1⁄ )2 = 1
√2
2 1
2|𝑆11 | + 2 = 1
1
2|𝑆11 |2 = 1 − 2
1
2|𝑆11 |2 = 2
1
|𝑆11 |2 =
4
1
𝑆11 =
2
From equation 9 and 10
𝑆11 = 𝑆22 & 𝑆11 = 𝑆12
1
𝑆22 = 𝑆12 =𝑆11 =
2
Now from Equation 5 , Scattering matrix of a E Plane Tee is given by
Course Code: 20EC0423 R20

1 1 1
2 2 √2
𝑆11 𝑆12 𝑆13 1 1 1
[𝑆] = [𝑆12 𝑆22 −𝑆13 ] = −
2 2 √2
𝑆13 −𝑆13 0
1 1
− 0
[√2 √2 ]