(impedance matching in high frequency lines) material... · filter with a cutoff frequency of 1.5 k...
TRANSCRIPT
SRI VENKATESWARA COLLEGE OF ENGINEERING
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Date : 02.07.2015
UNIVERSITY QUESTIONS AND ANSWERS
Subject : Transmission lines & Wave Guides Sub Code : EC6503
Staff Name: Dr.H.UMMA HABIBA Class : V sem ECE
Section : C
(Impedance Matching in High Frequency Lines) 1. Mention the disadvantage of single stub matching. [May/June,2014]
R =0 & G should be negligibly small
R/L =G/C and Attenuation constant α = 0
Phase constant β = ω√LC
Hence α is independent of ω and β is a constant multiplied by ω.This is condition
for distortionlessline
2. Design a quarter wave transformer to match a load of 200 ohms to a source
resistance of 500 ohms. The operating frequency is 200 Mhz.
[May/June 2013] [Nov/Dec-2006]
Ans: R0’= | ZsZR |
= 200x 500 = 100000 =316.22
3. Mention the significance of λ/4 line (Nov/Dec 2012,R8)
Quarter wave transformer may be used as
an Impedance inverters or impedance transformer
act as a coupler to couple a transmission line to a resistive load
such as an antenna
Serve as an Insulator
4. Write the disadvantages of single stub matching. (Apr/May 2012, R8)
(i) Useful for a fixed frequency only
(ii) For final adjustment the stub has to be moved. This is possible only in
open wire line.
5. Explain the use of quarter wave line for impedance matching
[Nov-2007] [Nov-2004]
(or) What are the applications of the quarter wave line? [Apr/May 2011,R8]
(i). An important application of the quarter wave matching section is to couple a
transmission line to a resistive load such as antenna. The quarter wave matching
section then must be designed to have a characteristic impedance R0’ so chosen
that the antenna resistance RA is transformed to a value equal to the characteristic
impedance R0 of the transmission line. The line then is terminated in its RO and is
operated under conditions of no reflection.
(ii).Another application of the short circuited quarter wave line or the center
conductor of a coaxial line. Such lines are sometimes referred to as copper
insulators.
6. Write the procedure to find the impedance from the given admittance using
smith chart. [Apr-2010, Apr-2008]
Ans: In order to find the impedance of admittance, the admittance point has to be
rotated to a distance of 0.25 . Since the 0.25 rotation in a smith chart
corresponds to half the cycle, the impedance will simply be a point diametrically
opposite to the admittance point.
7. Dintinguish between series stub and shunt stub [Apr-2010]
Sl.no Series stub Shunt stub
1 It is used only for balanced lines It is used only for unbalanced lines
2 less often stubs are used in series
with the transmission line .
More often stubs are used in shunt
with the transmission line which is
connected to the load.
8. Name few applications of one eighth wave line. [Nov-2010,Nov-2009, Nov-
2006]
Eighth wave line may be used to transform any resistance to impedance with
a magnitude equal to R0 of the of the line, or to obtain a magnitude match between a
resistance of any value and a source of R0 internal resistance
9. Why are short circuited stubs preferred over open circuited stubs?
[Apr-2008, Apr-2004]
A short-circuited stub is preferred to an open circuited stub because of
greater ease in construction and because of the inability to maintain high
enough insulation resistance at the open circuit point to ensure that the
stub is really open circuited .A shorted stub also has a lower loss of energy
due to radiation, since the short circuit can be definitely established with a
large metal plate, effectively stopping all field propagation
10. Why is the Quarter wave line called as copper insulator? [Nov-2008]
Application of the short circuited quarter wave line is as an
insulator to support an open wire line or the center conductor of a coaxial
line. This application makes use of the fact that the input impedance of a
quarter wave shorted line is very high. Such lines are sometimes referred to as
copper insulators
11. Name few application of half wave line. [May-2007,Apr-2008]
(i) one to one transformer
(ii) Used to connect a load to source when the load and source cannot be made
adjacent.
12. What are the advantages of double stub matching over single stub matching?
[Apr-2008,Nov-2005,]
i. Double stub matching can be used for variable frequency
ii. Load impedance can be made equal to the input impedance by proper
adjustment of double stub line
13. Write the expression for the characteristic impedance R0’ of the matching
quarter-wave section of the line? [May-2007]
Ans: R0’= | ZsZR |
14. Give two applications of smith chart? [May-2007]
i. Smith chart as an admittance diagram
ii. Converting impedance into admittance
iii. Determination of an input impedance
iv.Determination of an load impedance
15. What is smith chart? [Nov-2007] Resistive component R and reactive component X of an impedance has been
represented in a rectangular form while in smith chart R and X are represented in
circular form. It is referred as smith chart.
UNIT-IV
FILTERS
1. What is constant K filter? [Nov/Dec, 2014]
At T or π section with Z1 Z2 = R02 Where R0 is a real constant is called
constant K filter. Other more complex sections may be derived from
Prototypes.
2. A prototype LPF is to be designed which must have R0 = 600 Ω, cutoff
frequency of 1 K Hz. Find filter elements L and C? [Nov/Dec, 2014]
R0 = 600 ohms & fc =1 KHz For LPF
L = R0 / π fc = 190.98 mH
C = 1/ R0 π fc = 0.053µH
3. Determine the value of L required by a constant K-T section high pass
filter with a cutoff frequency of 1.5 K Hzand design impedance of 500 Ω.
[May/June,2014]
Rk = 500 Ω & fc =1.5 KHz
L= Rk/4 π fc
= 0.0265 H= 26.5 m H
4. What are the advantages of an m derived filter?
[May/June, 2014][Nov/Dec,2013][Apr/May 2011, R8](Nov/Dec 2010, R8]
Advantages:
To achieve a better match of the filter into the terminating
impedances.
There is a rapid transition from the cut-off frequency of the pass
band to a pole of attenuation just inside the stop band.
Attenuation rises near the cut off frequency and its slope is
adjustable by varying f∞
Drawback: m-types have poor stop band rejection. For this reason, filters
designed using m-type sections are often designed as composite filters with a
mixture of k-type and m-type sections and different values of m at different
points to get the optimum performance from both types.
5. Determine the value of L required by a constant K T-section high pass
filter has a cutoff frequency of 1 KHz and the design impedance is 600Ω.
[Nov/Dec,2013]
L Given:
Rk = 600 ohms & fc =1 KHz
Formula for L= Rk/4 π fc
= 600/((4 π)x1x103)
= 47.74 m H
6. A constant K T section high pass filter has a cutoff frequency of 10 K Hz. The
design impdance is 600 Ω. Determine the value of L. [Apr/May,2013]
(Nov/Dec 2010, R8)
Given:
Rk = 600 ohms & fc =10 KHz
Formula for L= Rk/4 π fc
= 600/((4 π)x10x103)
= 4.774 m H
7. State the significance of crystal filters in communication system
(Nov/Dec 2012, R8)
Crystal filter is an electronic filter that uses quartz crystals for resonator.
It has more stability and high Q factor
Crystal filters are commonly used in communication devices such as radio
receivers, telecommunications, signal generation, and GPS devices.
8. Write relationships between neper and decibel. (Apr/May 2012, R8) 1 neper = 8.686 dB
1 dB =0.115 neper
9. Write the disadvantages of constant k filter. (Apr/May 2012, R8) (i) Attenuation does not increase rapidly beyond cutoff frequency.
(ii) Characteristics impedance varies widely in the pass band.
10. What is the significance of propagation constant in symmetrical network?
(Nov/Dec 2011, R8)
(or)Define propagation constant. [Apr/May,2013] (Nov/Dec 2012, R8)
If the transmission line is lossless, then R' and G' terms in the propagation
constant equation are zero. If R' and G' terms in the propagation constant equation
are zero, the attenuation constant is also zero. The general equation for
propagation constant is neatly simplified:
The main significance for lossless transmission line, the propagation constant is
purely imaginary. Hence
11. Draw the equivalent electrical circuit for a piezoelectric crystal
(Nov/Dec 2011, R8)
L = Inductance
R = Resistance
C = Equivalent series capacitance of the crystal
CP = Parallel capacitance introduced by crystal electrodes.
12. For a symmetrical network, define propagation constant and
characteristic impedance. [Apr/May 2011, R8]
The propagation constant per unit length may be defined as the natural
logarithm of ratio of the sending end current or voltage to the receiving end
current or voltage. Propagation constant =α+jβ
If ‘ is attenuation in nepers then attenuation in dB = 8.686 .
13. Define network
An electrical network is defined as a combination of a number of electrical
elements the impedances of which may be lumped or distributed or both and
which are connected in any manner, conductively, inductively or capacitive.
14. What is passive and active network with examples?
A network is said to be passive if it contains no sources of energy and
active when it contains sources of energy.
Examples of Passive device: Resistance, Inductance;
Examples of active device: Voltage sources, Current Source, Properly
biased Transistors
15. What is meant by linear and non linear element?
In an electric circuit, a linear element is an electrical element with a linear
relationship between current and voltage. Resistors are the most common linear
element. In an electric circuit, a nonlinear element or nonlinear device is an
electrical element with a no linear relationship between the current and voltages.
16. What is meant by matching network?
When generator and load impedance are unequal, matching networks are
used at radio frequencies to minimize reflections and to produce a good transfer
of power.
17. Define unilateral element and bilateral element
If the magnitude of the current passing through an element is affected due
to change in the polarity of the applied, the element is called unilateral element.
On the other hand if the current magnitude remains the same even if the applied
e.m.f polarity is changed, it is called a bilateral element.
18. What is the balanced and unbalanced network?
When an unbalanced transmission line is connected to a balanced antenna
(coaxial line to a dipole, for example), extraneous currents run along the outer
surface of the transmission line. So, even though the impedances are the same,
these extraneous currents can cause unwanted radiation. This radiation can be
minimized by the use of a Balanced-to-unbalanced network.
A Balanced-to-unbalanced network is placed between the antenna and the
antenna feeder
19. Define attenuators and filters
Attenuators are designed to reduce the power level from input to output,
they consist of only resistors and they attenuate by the same amount whatever be
the frequency.
Filter in fact is a selective network as it passes freely some frequency bands and
heavily attenuates other. Filters are quite common in multiplexed equipments with
frequency division multiplex.
20. Write the use of the equalizers
Equalizers as the name implies is a corrective network and is
complimentary in nature to the transmission line. This is necessary to avoid
distortion of the signal. Their practical application is
1. In disc recording equipment
2. In tone control of the radio receiver.
21. Define characteristics impedance
If an infinite number of identical symmetrical networks are connected one
after another. The impedance that, when connected to the output terminals of a
transmission.
22. What is image impedance?
Image impedances of a network are defined as the impedances which
simultaneously terminate each pair of terminals of a network in such a way that at
each pair of terminals the impedances in both directions are equal.
23. What is iterative impedance?
Iterative impedance may be defined as the impedance which will terminate
the other pair of terminals in such a way that the impedance measured at the first
pair of terminals is equal to the terminating impedance
24. What is insertion loss?
If a network or a line is inserted between a generator and load there will be
a reduction in the power received in the load and the load current will be
decreased. The loss produced by the insertion of the network or line is referred to
as insertion loss
25. What is symmetrical network?
A symmetrical network is one in which the electrical properties are unaffected by
interchanging the input and output terminals, otherwise it is asymmetrical. That is if two
series arms of a T network or shunt arms of π network are equal.
UNIT-V
1. Why is TEM mode not supported by waveguide? [Nov/Dec,2014]
According to Maxwell’s equation ∇×H = J + ∂D/∂t. To have H lines
there must be an axial current in the inner conductor. In a Hallow waveguide, no
inner conductor. Also electric field E should be axial direction. But Eaxial is not
present in TEM waves. Hence TEM wave cannot exist in waveguide
2. State the significance of dominant mode of propagation. [Nov/Dec,2014]
Dominant mode is useful in finding the lowest cutoff frequency for any
waveguide.
3. A wave is propagated in the dominant mode in a parallel plane waveguide.
The frequency is 6 GHz and the plane separation is 4 Cm. Calculate the
cutoff wavelength and the wavelength in the waveguide. [May/June,2014] Given:
f = 6 G Hz
a = 4 cm= 4x10-2 m c = 3x108 m/s
consider dominant mode TE10 where m=1 n=0
fc = mc/2a
= 3x108 /2 x 4x10-2
= 3.75 G Hz
λc = C/ fc = 3x108/3.75x109 =0.08 m
(or) λc =2a/m =0.08m
4. Give the equations for the propagation constant and wavelength for TEM
waves between parallel planes. [May/June,2014]
Propagation constant
Since m = 0, the cutoff frequency for TEM waves becomes zero, so that TEM
waves propagate without attenuation between the perfectly conducting plates for
all frequencies above that of zero.
Wave length λ = c/f = 1/f √µε where C = 1/√µε
5. A rectangular waveguide of cross section 5 cm × 2 cm is used to propagate
TM11 mode at 10 GHZ. Determine the cut-off wave length.
[Nov/Dec,2014] [Nov-2011,R8]
a=5 cm, b= 2 cm, m=1, n= 1
22
2
b
n
a
m
c = 3.714 cm
6. Mention the applications of resonant cavities. [Nov/Dec,2014]
Used in microwave generation and amplification.
Used in light house tube, especially for VHF range of frequency.
Used in RADAR as TR tubes and ATR tubes.
Used for measurement of microwave signals with cavity wave meter.
Waveguide resonators are used in place of the lumped element RLC
circuit to provide a tuned circuit at high frequencies.
Resonators are used in applications such as oscillators, filters and tuned
amplifiers.
7. What are the advantages and application of cylindrical waveguides?
[May/June,2014] [Nov/Dec,2013]
Circular or cylindrical waveguide behaves much like a High Pass
Filter and is basically a passive microwave device.
The cylindrical waveguide is the hollow metallic conductor
carrying usually high frequency or microwave frequency.
Operating range of waveguide is approx. from 300 MHz to 300
GHz.
Application waveguide to coaxial adapter, coupler, flanges, gasket, tees,
termination, shorts, isolator, circulator, horn antenna, attenuator,
feed-thrus, bends, loads, twists, straights, pressure windows,
transformers, sliding short circuits, fixed short circuits, sliding
mismatches etc.
8. Mention the different types of guide termination.
[May/June,2014]
Types:
Gradual changes in impedance can be obtained by terminating
the waveguide with a funnel-shaped HORN,
(a)Pyramidal horn
(b)Sectorial horn –E plane and H plane Horn
(c)Circular horn
Dummy load termination- A waveguide may also be
terminated in a resistive load that is matched to the
characteristic impedance of the waveguide.
Wedge of highly resistive material.
Permanently weld a metal plate at the end of the
waveguide.
9. A wave is propagated in a parallel plane wave guide. The frequency is 6 G Hz
and the plane separation is 3 cm. Determine the group and phase velocities
for the dominant mode. [Nov/Dec,2013] [Nov-2010,R8]
\
10. Define TEM wave. [Nov/Dec,2013
(or) State the characteristics of TEM waves. [May/June 2013]
The TEM wave in which both electric and magnetic fields are tranverse
entirely but have no component of Ez and Hz.( Hz& Ez= 0). It is referred to
as principal waves. The properties are
1. The TEM wave is independent of frequency
2.The cut-off frequency of TEM is zero (fc = 0) .This means that all
frequencies down to zero can propagate
11. What is degenerate mode in rectangular waveguide?[May/June,2013]
TE10 and TE01 modes are degenerate modes (modes with same
cut off frequency) for square wave guide. The rectangular waveguide
allows one to operate at a frequency above the cutoff of the dominant TE10
mode but below that of the next highest mode to achieve single mode
operation.
A waveguide operating at a frequency where more than one mode
propagates is said to be overmoded.
12. A rectangular waveguide with a = 7 cm and b = 3.5 cm is used to used to
propagate TM10 at 3.5 G Hz. Determine the guided wavelength.
[Nov/Dec,2013]
Given:
m=1, n=0, a= 7 cm, b = 3.5 cm
fc = mc/2a
= 3x108/2x7x10
-3
=2.146 G Hz Where λ0 = c/f
λg =λ0/√1-(fc/f)2
=10.8 cm
13. For the frequency of 6 GHz and plane separation of 3 cm find the group and
phase velocities for the dominant mode.[Nov/Dec,2013] [Nov-2010,R8]
14. What is the dominant TE and TM mode in rectangular waveguide?
(Nov/Dec 2012,R8)
Dominant TE and TM mode in rectangular waveguides are TE10 and TM11
respectively.
15. Write the expression for the wave impedance and guide wavelength for TEM
mode. [Apr/May 2012,R8]
Wave impedance = 377 ohm
Guide wavelength λ = c/f = 1/f√µε
16. Which is the dominant mode of a rectangular waveguide? Why?
[Apr/May 2012,R8]
TE10 mode is the dominant mode in rectangular waveguide, its E field is polarized
in one direction and it has the lowest attenuation of all modes in rectangular waveguide.
17. Compare TE and TM mode. (Nov/Dec 2012,R8)
TE TM
Electric field strength E is entirely
transverse.
Magnetic field strength H is entirely
transverse.
It has a z component of magnetic
field Hz
It has a z component of electric field
Ez.
It has no z component of electric
field Ez.( Ez = 0)
At cut off frequency the wave
impedance for TE waves becomes
infinity.
It has no z component of magnetic
field Hz.( Hz =0)
At cut off frequency the wave
impedance for TM waves becomes
zero.
18. What is the need for Attenuator? (Nov/Dec 2012,R8)
Attenuators are needed to reduce impedance mismatches in high-
frequency circuits and systems.
It provides isolation to stabilize oscillators.
It reduces the power of a signal without appreciably distorting its
waveform.
19. Write the Maxwell’s equation. (Apr/May 2012, R8), (May 2009)
20. What is meant by dominant mode? What is the dominant mode for parallel
plate waveguide? (Apr/May 2012, R8)
The mode having the lowest cutoff frequency is called dominant mode.
The dominant mode for parallel plate waveguide is TEM.
21. Write down the relationship between phase velocity and group velocity
(Nov/Dec 2011,R8) (May2010/May 2007)
The product of phase (Vp) velocity and group velocity (Vg) is the square
of free space velocity.
Phase velocity * Group velocity = C2
Vp x Vg = C2
C – Velocity of light
The group velocity always less than velocity of free space whereas phase
velocity always greater than velocity of free space.
22. Write down the equations for characteristic impedance for TE, TM, and
TEM waves (Nov/Dec 2011,R8)
For TM Waves , Z0 =
2
1f
fZ c
TM
For TE waves, , Z0 = 2
1f
f
Z
c
TE
For TEM waves, , Z0 = η
η- intrinsic impedance of the medium 377Ω; fc – cut-off frequency
23. What is the need for guide termination? [Nov-2011,R8]
To avoid reflection losses(or) Standing waves
To provide a impedance match between waveguide and open
space.
To improve efficiency of wave guide.
To radiate or absorb the energy without reflections.
24. An air filled resonant cavity with dimensions a = 5 cm, b = 4 cm and c =10
cm is made of copper. Find the resonant frequency for lowest order modes
[Apr/May 2011]
Lowest order mode for Rectangular Cavity Resonator is TE101
a = 5 cm, b = 4 cm and d =10 cm and m=1,n=0 and l=1
The resonant frequency for TE101 is = 3.354 GHz
25. Give the applications of cavity resonators. [Nov-2010,R8]
A microwave cavity resonator, consisting of a closed (or largely closed)
metal structure that confines electromagnetic fields in the microwave
region of the spectrum. The structure is either hollow or filled with
dielectric material.
Used in oscillators and transmitters to create microwave signals
Used as filters to separate a signal at a given frequency from other
signals
Used in equipment such as radar equipment, microwave relay
stations, satellite communications, and microwave ovens..
26. Calculate the cutoff wavelength for the TM11 mode in a standard
rectangular wave guide if a=4.5 cm (Nov 2010,R8)
a = b = 5 cm m= 1, n=1
cut off wave length is =7.07 cm
27. List the advantages of cavity resonator over the lumped element resonator
above 300M Hz. [Nov-2010]
Wave guide resonators are used in place of lumped element RLC circuit
to provide a tuned circuit at high frequencies.
In cavity resonator waves are in both directions. Hence standing waves
are appearing. Lowest order modes in a rectangular cavity are the TM110,
TE101, TE011
28. Mention the characteristics of TE and TM waves.
[Nov-2010]
TE waves
TE wave is a wave in which the electric field strength E is entirely
transverse. It means that it has only magnetic field component Hz in the
direction of propagation and no component of electric field strength Ez in
the same direction
Dominant mode for TE wave is TE10
TM waves
TM wave is a wave in which the magnetic field strength H is entirely
transverse. It means that it has only electric field strength Ez in the
direction of propagation and no component of magnetic field component
Hz in the same direction
Dominant mode for TM wave is TM11
29. What is the significance of wave impedance?
[Nov-2010]
Wave impedance is the ratio of transverse components of Electric
to Magnetic field Strength.
η0 = E/H
30. Compare circular cavity resonator and rectangular cavity resonator.
[Nov-2010]
Rectangular Cavity
Rectangular cavity
Resonance frequencies of a rectangular microwave cavity for any
or resonant mode is given
where is the Wave Number, c is the velocity of light in air, and and are
relative permeability and permittivity respectively.
Dominant mode for Rectangular Cavity Resonator is TE101
Cylindrical Cavity
Cylindrical cavity
. The resonance frequencies are different for TE and TM modes.
TM modes:
TE modes
Here, denotes the -th zero of the -th Bessel functions, and denotes
the -th zero of the derivative of the -th Bessel function.
31. Give the reason for impossibility of TEM waves in waveguides
(May 2010/April 2008)
Since TEM wave do not have axial component of either E or H, it
cannot propagate within a single conductor waveguide. Consider a TEM
wave to exist within a hollow guide. In non-magnetic materials
.H = 0 which requires that the lines of H be closed loops in a plane
perpendicular to the axis. According to Maxwell’s equation the magneto
motive force around a closed loop is equal to the axial current .For a co-
axial line the axial current is conduction current whereas for a hollow
waveguide the axial current is displacement current .But an axial
displacement requires an axial component of E which is nor present in
TEM wave. Therefore TEM wave cannot exist inside a waveguide
32. Define cut-off wavelength in rectangular waveguide (May 2010)
The cut-off wavelength (λc) related to the cut-off frequency (fc) is
defined as the highest wavelength beyond which the wave is completely
attenuated.
22
2
b
n
a
m
c
33. Define cut-off frequency of a guide. (Nov 2010, R8) (Nov2007)
(or) Define the cutoff frequency for the guided waves
The cutoff frequency is the frequency below which wave propagation will
not occur .The frequency at which the wave motion ceases is known as cutoff
frequency of the guide.
34. What are the methods of Excitation of modes in rectangular wave guides?
[Nov-2010]
Possible methods for feeding rectangular wave guides:
TE10 waveguide mode
TE11 waveguide mode
TE20 waveguide mode
TM11 waveguide mode
35. Find the cut off wave length of a rectangular wave guide whose inner
dimensions are ‘a’ =2.3cm and ‘b’ = 1 cm operating at TE10 [NOV-2010]
22
2
b
n
a
m
c = 2a/m = 0.0460m
a’ =2.3cm m=1 n=0
36. Give the examples of guided waves (May 2010)
An electromagnetic wave that are guided along or over conducting or
dielectric surfaces are called guided waves. Examples of guided waves are the
electromagnetic waves along parallel wire and co-axial transmission lines,
waves in waveguides and the waves that are guided along the earth surface
from a radio transmitter to receiver
37. Write the expressions for the wave impedance of TE and TM waves between
parallel planes (Nov 2009)
2
1f
f
Z
c
TE
2
1f
fZ c
TM
η- intrinsic impedance of the medium 377Ω; fc – cut-off
frequency
38. Give the expressions for the cutoff wavelength and propagation constant of
TE waves between parallel planes (Nov 2009)
Cut-off wavelength (λc) = 2a/m
a- distance between the two parallel planes; m- mode number
39. A rectangular air filled copper waveguide with dimension of a=2.28cm and
b=1.01cm has a 9.2GHz signal propagated in it .Determine the guide
wavelength for TE10 (Nov 2009)
Given: a=2.28cm, a=2.28cm ,f = 9.2GHz m=1 & n=0
Cut off wavelength λc=2a/m =0.0456m
Guide wavelength = 0.0466m
40. What is meant dominant mode of the wave (May 2009)
The lowest order mode for TE wave is TE10 whereas the lowest
mode for TM wave is TM11.This wave has the lowest cut-off frequency.
Hence the TE10 mode is the dominant mode of a rectangular waveguide,
becausce TE10 mode has the lowest attenuation of all modes in rectangular
waveguide
41. A rectangular waveguide with dimension a=8.5cm and b=4.3cm .Determine
the cut-off frequency for TM10 mode of propagation
(May 2009)
2
1
u
cf
f
Cut off frequency for TM10 mode fc = mc/2a
Apply m=1 , n=0 & a= 8.5 cm=.0.085m
fc = mc/2a = 1.76 G Hz
42. Calculate the cut-off frequency of a rectangular wave guide whose inner
dimensions are a=2.5cm and b=1.5cm operating at TE10
(Nov 2008)
Cut off frequency for TE10 mode fc = mc/2a
a=2.5cm and b=1.5cm m=1 , n=0
fc = mc/2a = 6 G Hz.
43. Plot the frequency vs attenuation characteristics curve of TM and TE waves
guided between parallel conducting planes (Nov 2008)
44. Enumerate the properties of TEM wave between parallel planes of perfect
conductors (Nov 2008/Nov 2006)
The TEM wave in which both electric and magnetic fields are tranverse
entirely but have no component of Ez and Hz. It is referred to as principal
waves .The properties are 1. The TEM wave is independent of frequency
2.The cut-off frequency of TEM is zero .This means that all frequencies down
to zero can propagate
45. Specify the significance of dominant mode of operation. (April 2008)
Dominant mode is a mode that has the lowest cut-off frequency or lowest
order mode..
46. Comment on: (a) Characteristics impedance. (b) Wave impedance
(April 2008)
(a) Characteristic impedance: The characteristic impedance is given by
Z0=V/I. But for a parallel plate waveguide the potential difference between
two plates is given by Z0=Ex/Hy. The characteristic impedance for the
parallel plate waveguide is dependent only on the geometry of the guide
and the material parameters of the guide.
(b) Wave impedance: The intrinsic impedance or wave impedance is defined
as when looking in the direction of propagation, that is, along the z axis it
is defined as the ratio of transverse component of electric and magnetic
fields.
47. What is Principal wave (April 2008)
The TEM wave is a special case of guided wave propagation. It is also
called as principle waves. The electric and magnetic fields are entirely
transverse. Along the direction normal to the direction of propagation, the
amplitude of the field components are constant. TEM waves are independent
of frequency. The cut-off frequency of the wave is zero, indicating all
frequencies down to zero can propagate along the guide
48. Plot the frequency – versus wave impedance curve for the waves between
parallel conducting planes. ( April 2008) (Nov 2006)
49. Sketch the Electric and Magnetic field lines for the TE10 and TE20 modes in a
rectangular wave guide. (April 2008)
50. Write a brief note on Excitation of modes in rectangular waveguides.
(April 2008)
In order to launch a particular mode, a type of probe is chosen
which will produce lines of E and H that are roughly parallel to the lines
of E and H for that mode. The probe is parallel to the y axis and so
produces lines of E in the y direction and lines of H which lie in the xz
plane.
51. Which mode is called as dominant mode in the circular waveguide?
(April 2008)
TE11 and TM01 mode is the dominant mode in circular waveguide
52. Define the quality factor of a cavity resonator?
(April 2008,May 2007, Nov 2006, Nov 2005 )
The quality factor Q of a resonator is defined as the energy stored
and energy loss at the resonant frequency.
Q=2π Time average energy stored at a resonant frequency
_________________________________________
Energy loss per second in the system
53. Calculate the cut-off wavelength of a rectangular waveguide whose inner
dimensions are a = 2.3 cm and b=1.03 cm operating at TE10 mode.(April
2008)
Given:
For TE 10 m = 1 ,n = 0 a = 2.3 cm and b=1.03 cm
Hence cut off wavelength = 2a/m
= 2*2.3/1 =4.6 cm
54. A rectangular waveguide with dimensions a=8.5cm and b=4.3cm is fed by 5
GHZ carrier. Will a TE11 mode be propagated? (Nov 2007)
Given : a=8.5cm and b=4.3cm m = 1, n= 1
cut off wavelength =
55. Define wave impedance and write the expression for wave impedance of TE
waves in rectangular guide. (Nov 2007)
Wave impedance: The intrinsic impedance or wave impedance is
defined as when looking in the direction of propagation, that is, along the
z axis .It is defined as the ratio of transverse component of electric and
magnetic fields.
2
1f
f
Z
c
TE
56. What is cavity resonator? (May 2007)
It is a shielded enclosure which confines electromagnetic fields
inside and furnishes large areas for current flow, thus eliminating radiation
and high-resistance effects. These enclosures have natural resonant
frequencies and high Q and are called resonator.
57. Why the TE10 wave is called as dominant wave in rectangular waveguide?
(May 2007)
Dominant mode is a mode which has less attenuation and lowest
cut off frequency. It is called a dominant mode because all the field
configuration Ex, Ey, Hx and Hy exist.
58. What are the characteristics of principal wave? (Nov 2007)
(i) The phase constant β varies linearly with frequency and therefore the
wave propagation takes place without dispersion.
(ii) The cut-off frequency for the TEM wave in a parallel plane waveguide is
zero and all frequencies right down to dc can propagate along the
guide.
59. What is the cutoff frequency of TEM wave? (May 2007)
02a
mfc This means that all frequencies down to zero
can propagate along the guide.
60. A rectangular waveguide has the following dimensions l=2.54cm, b=1.27cm
waveguide thickness=0.127 cm. Calculate the cut-off frequency for TE11
mode. (Nov 2006)
Given l=2.54cm b=1.27cm thickness =.127cm m=1, n=1
Use Formula:
Cut off frequency fc =
61. What are dominant mode and degenerate modes in rectangular waveguide?
(Nov 2006)
Some higher order modes may have identical cut off frequency are
called degenerate modes . Eg TE11, TM11, TE21, TM21, TE12, TM12, TE22,
TM22 all have identical cut off frequencies. These modes are called
degenerate modes.
The dominant mode has lowest cut off frequency. This mode has
lowest attenuation of all modes in rectangular wave guide and its electric
field is definitely polarized in one direction.
62. Explain why TM01 and TM10 modes in a rectangular waveguide do not exist.
(May 2006, Nov 2004)
For TM10 when m=1 and n=0 for these values all the fields in the
rectangular wave guide becomes zero. Sine Hz =0 for TM waves. This is
true for m=0 and n=1. Hence both TM01 and TM10 do not exist. Therefore
TM0n and TMm0 modes do not exist.
63. What do you understand by degenerate modes in a waveguide? Give example
of two such modes in a rectangular wave-guide.
(May 2006, Nov 2004)
Some higher order modes may have identical cut off frequency. Eg
TE11, TM11, TE21, TM21, TE12, TM12, TE22, TM22 all have identical cut off
frequencies. These modes are called degenerate modes.
64. Explain the difference between the intrinsic impedance and wave impedance
in free space and in rectangular wave-guide. (May 2006)
The wave impedance is defined as the ratio of the strength of the
electric field in one transverse direction to the strength of the magnetic
field in other transverse direction where as the intrinsic impedance is the
free space impedance denoted by η0 and its value is 377Ω.
65. Define phase velocity and group velocity (Nov 2006,Nov 2005)
The velocity with which the phase changes phase in a direction parallel to
the conducting surface is called phase velocity. It is denoted by Vp= ω/β ;
ω=2Πf; β –phase constant Group velocity is the velocity with which the group of waves as a whole
propagate.. It is denoted by Vg; Vg=dω/d β ;
66. What do you understand by the dominant mode in waveguide? What is its
importance? (May 2006)
Dominant mode is a mode which has less attenuation. It is called a
dominant mode because all the field configuration Ex, Ey, Hx and Hy exist.
Dominant mode is a mode that has the lowest cut-off frequency or lowest
order mode.
PART-B QUESTIONS
UNIT-III
(Impedance Matching in High Frequency Lines) 1. a) A line having characteristic impedance of 50 Ω is terminated in load
impedance [75 + j75] Ω. Determine the reflection coefficient and voltage
standard wave radio. Mention the significance and application of Smith
chart. [Nov/Dec 2014]
a) Discuss the principle of double stub matching with neat diagram and
expressions.
2. A single stub is to match a 300 ohm line to a load of (180 + j120) ohm. The
wavelength is 2 meters. Determine the shortest distance from the load to the
stub location and proper length of a short circuited stub using relevant
formula. [May/June 2014]
3. (i)Discuss the principle of double stub matching with diagram and
expressions.
(ii)A 300 Ω transmission line is connected to a load impedance of (450-j600)
Ω at 10 M Hz. Find the position and length of a short circuited stub
required to match the line using smith chart. [Nov/Dec 2013]
4. Design a single stub matching networks (use smith chart) for a transmission
line functioning at 500 M Hz terminated with a load impedance ZL = 300 +
j250Ωand with a characteristics impedance Z0 = 100Ω. Use short circuited
shunt stubs. Determine the VSWR before and after connecting stub.
[May/June 2013]
5. The input impedances of a λ/8 long, 50Ω transmission line are Z1 =
25+j100Ω, Z2 = 10-j50Ω, Z3 = 100+j0Ω and Z4 = 0+j50Ω, when various load
impedance are connected at other end. In each case determine the load
impedance and the reflection coefficient at the input and load ends.
[May/June 2013]
6. A 30 m long lossless transmission line with Zo=50Ω operating at 2 MHz is
terminated with a load ZL=60+40 j . If U = 0.6C on the line, find
(i ) Reflection coefficient (ii) Standing wave ratio (iii) Input
impedance. (Nov/Dec 2012,R8)
7. Discuss the following (Nov/Dec 2012,R8)
a) Impedance matching. b) Single and double stub matching
8. (i)An ideal loss less quarter wave transmission line of characteristics
impedance 60 Ω is terminated in a load impedance ZL. Give the value of the
input impedance of the line when ZL= 0, ∞ and 60 Ω. [Apr/May 2012,R8]
(ii)Write the concepts of single and double stub matching.
9. (i)A 100 Ω , 200 m long loss less transmission line operators at 10 M Hz and
is terminated into an impedance of 50-j200 Ω. The transit time of the line is
1μs.Determine length and location of short circuited stub line.
(ii) Write the concepts of quarter wave length line and half wave length line.
[Apr/May 2012,R8]
10. Explain the technique of single stub matching and discuss operation of
quarter wave transformer.[Apr/May 2011,R8]
11. Explain the applications of smith chart. A 30 m long lossless transmission
line with Z0= 50Ω operating at 2 MHz is terminated with a load
ZL = 60 +j40Ω . If U = 0.6 C on the line, find the reflection coefficient Γ, the
standing wave ratio s and the input impedance. [Apr/May 2011,R8]
12. (i) Draw and explain the operation of quarter wave line [Nov/Dec 2010,R8]
(ii) It is required to match a 200 ohms load to a 300 ohms transmission line to
reduce the SWR along the line to 1. What must be the characteristics
impedance of the quarter wave transformer used for this purpose if it is
directly connected to the load?
(iii) What are the drawbacks of single stub matching and open circuited
stubs?
13. (i) Draw and explain the principle of double stub matching. (8)
(ii) A UHF lossless transmission line working at 1 GHz is connected to an
unmatched line producing a voltage reflection coefficient of 0.5(0.866 + j 0.5).
Calculate the length and position of the stub to match the line. (8)
[Nov/Dec 2010,R8]
UNIT-IV
(Filters) 1. Design a constant K band pass filter deriving expressions for the
circuit components. A constant K High pass filter cuts off at a frequency of
2300 Hz. The load resistance is 500 Ω Calculate the values of components
used in the filter. [Nov/Dec-2014]
2. Design a composite high pass filter to operate into a load 600 Ω and
have a cut off frequency of 1.2 K Hz. The filter is to have one constant k
section, one m derived section with f = 1.1 K Hz and suitably terminated
half section. Discuss the merits and demerits of m-derived filter and crystal filter.
[Nov/Dec-2014]
3. (i) Draw and explain the design and operation of m derived T section
bandpasss filter with necessary equation and diagrams.
(ii)Design constant –K bandstop filters (both T and π sections) for the
Cut off frequencies of 4 K Hz and 8 K Hz. The design impedance is
500Ω. [May/June-2014]
4. Design an m derived low pass filter with a cutoff frequency of 2 K Hz.
Design impedance is 500 Ω and m = 0.4. Consider a π-section for your
calculation. [May/June-2014]
5. (i) Design a constant-K T-section band pass filter with cutoff
frequencies of 1 KHz and 4 KHz. The design impedance is 600 ohms.
(ii) Draw a constant-K T-section band elimination filter and explain the
operation with necessary design equations.
[Nov/Dec 2013][Nov/Dec 2010,R8
6. (ii)Draw and explain the operation of crystal filters.
[Nov/Dec 2013] [Nov/Dec 2010]
(ii) Design a m-derived T-section low pass filter having a cutoff
frequency (fc) of 1 k Hz and a design impedance of 400 ohms and the
resonant frequency is 1100Hz. [Nov/Dec,2013]
7. (i) Derive the equations for the characteristic impedance of
symmetrical T and π networks.[May/June 2013] [Nov/Dec 2011]
(ii) Discuss the properties of symmetrical network in terms of
characteristic impedance and propagation constant.
[Nov/Dec 2011]
(iii)Bring out the relation between Decibel and Neper. [May/June 2013]
8. Obtain the design equations for m derived (i) Band Pass (ii)Band
elimination filters. [May/June 2013]
9. (i)Explain the properties and characteristics impedance of
symmetrical networks. (Nov/Dec 2012,R8)
(ii) Design T and π section low pass filter which has series inductance
80 m H and shunt capacitance 0.022μf. Find the cutoff frequency
and design impedance
10. What are the advantages of m derived filter? Design an m derived low
pass filter (T and π section) having design resistance R0 =500 , cutoff
frequency fc= 1500 Hz and infinite attenuation frequency f∞=2000 HZ.
(Nov/Dec 2012,R8)
11. (i)Calculate the values of the inductor and capacitors of a prototype
constant K low pass filter composed of π section to operate with a
terminating load of 600 ohms and to have a cut off frequency of 3 K Hz.
(ii)Construct a band stop constant K filter. [Apr/May 2012,R8]
12. (i) Discuss the characteristics of symmetrical network.
(ii)Design an m derived T section low pass filter having cut off
frequency fc = 1000 Hz, design impedance Rk = 600 ohms and
frequency of infinite attenuation f∞ = 1050 Hz. [Apr/May 2012,R8]
13. With suitable filter sections, design constant-K low pass and high pass
filters [Nov/Dec 2011,R8]
14. Derive the relevant equations of m derived low pass filter and design
m-derived T type low pass filter to work into load of 500Ω with cut
off frequency at 4 kHz and Peak attenuation at 4.15 KHz.
[Apr/May 2011,R8]
15. Explain the structure and application of crystal filter. Design a low pass
filter with cut off at 2600 Hz to match 550 Ω. Use one derived section
with infinite attenuation at 2850 Hz. [Apr/May
2011,R8]
16. (i) Design a m-derived T-section low pass filter having a cut off
frequency (fc) of 5000 Hz and a design impedance of 600 ohms. The
frequency of infinite attenuation is 1.25 fc.(8) [Nov/Dec 2010,R8]