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]