# EC6503 TLWG Two marks Q&A, Past Exams for Electrical Engineering

27 pages
1000+Number of visits
Description
Hereby I have attached the Two marks Q&A for the Subject EC6503 Transmission lines and Wave Guides which is useful for V Semester ECE Students
40 points
this document
Preview3 pages / 27
EC 6403 – ELECTROMAGNETIC FIELDS – TWO MARKS Q & A

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 1

UNIT I – TRANSMISSION LINE THEORY

1. What are the different types of transmission lines?

The different types of transmission lines are

 Open wire line

 Cable

 Coaxial line

 Waveguide

2. Define the line parameters

The parameters of a transmission line are:

 Resistance (R)

 Inductance (L)

 Capacitance (C)

 Conductance (G)

Resistance (R) is defined as the loop resistance per unit length of the wire. (Ohm/Km)

Inductance (L) is defined as the loop inductance per unit length of the wire. (Henry/Km)

Capacitance (C) is defined as the loop capacitance per unit length of the wire. (Farad/Km)

Conductance (G) is defined as the loop conductance per unit length of the wire. (Mho/Km)

3. What are the secondary constants of a line? Why the line parameters are called

distributed elements?

The secondary constants of a line are:

 Characteristic Impedance

 Propagation Constant

Since the line constants R, L, C, and G are distributed through the entire length of the line,

they are called as distributed elements. They are also called as primary constants.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 2

4. Define Characteristic impedance

Characteristic impedance is the impedance measured at the sending end of the line. It

is given by Z0 = √Z/Y,

Where Z = R + jωL is the series impedance

Y = G + jωC is the shunt admittance

5. Give the general equations of a transmission line.

The general equations are,

E = ER cosh√ZY s + IR Zo sinh √ZY s

I = IR cosh√ZY s + (ER / Zo) sinh √ZY s

Where Z= R + jωL = series impedance, ohms per unit length of line

Y = G + j ωC = shunt admittance, mhos per unit length of line.

6. Define wavelength.

The distance the wave travels along the line while the phase angle is changing through 2π

7. What is the relationship between characteristic impedance and propagation constant?

CjG

LjR Zo

  and   CjGLjR   . On multiplying oZ , characteristic

impedance and  , propagation constant, we get, LjRXZo   and CjG Zo

 



8. What is meant by distortion less line?

A distortion less line is one which has neither frequency nor delay distortion. The attenuation

constant and velocity of propagation cannot be functions of frequency.

9. Calculate the characteristic impedance of the transmission line if the following

measurements have been made on the line.  60550ocZ and  30500scZ

Solution:

Characteristic impedance, scoco ZZZ =   3050060550 X =  154.524

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 3

10. Find the attenuation constant and phase constant of a wave propagating along the line

whose   8.8810048.1 4X .

Solution: Propagation constant,  j that is,   8.8810048.1 4X

In rectangular form, 46 10110194.2   XjX, Equating the real and imaginary parts, we get,

610194.2  X nepers per unit length and 4101  X rad per unit length

11. Define frequency distortion.

Attenuation constant α is a function of frequency. All frequencies transmitted on a line will

then not be attenuated equally. A complex applied voltage, such as a voice voltage containing

many frequencies, will not have all frequencies transmitted with equal attenuation, and the

received waveform will not be identical with the input waveform at the sending end. This

variation is known as frequency distortion.

12. Define delay or phase distortion.

All frequencies applied to a transmission line will not have the same time of transmission,

some frequencies being delayed more than others. For an applied voice voltage wave the

received waveform will not be identical with the input waveform at the sending end, since some

frequency components will be delayed more than those of other frequencies. This phenomenon is

known as delay or phase distortion.

13. How can you reduce frequency distortion?

Frequency distortion is reduced in the transmission of high quality radio broadcast programs

over wire lines by use of equalizers at the line terminals. These circuits are networks whose

frequency and phase characteristics are adjusted to be inverse to those of the lines, resulting in an

over-all uniform frequency response over the desired frequency band.

14. How can you overcome delay distortion?

Coaxial cables are used to overcome delay distortion. In such cables, the internal inductance

is low at high frequencies because of skin effect, the resistance is small because of the large

conductors, and capacitance are small because of the use of air dielectric with a minimum of

spacers. The velocity of propagation is raised and made more nearly equal for all frequencies.

15. Give the expressions for attenuation constant, phase constant and velocity of propagation in a telephone cable.

Attenuation constant, α = √ (ωRC) / 2 Phase constant, β = √ (ωRC) / 2

Velocity of propagation, Vp = ω/ β = √ (2ω) /RC

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 4

17. What is a finite line? Write down the significance of this line

A finite line is a line having a finite length on the line. It is a line, which is

terminated, in its characteristic impedance (ZR=Z0), so the input impedance of the finite line

is equal to the characteristic impedance (Zs=Z0).

18. What is an infinite line?

An infinite line is a line in which the length of the transmission line is infinite. A

finite line, which is terminated in its characteristic impedance, is termed as infinite line. So

for an infinite line, the input impedance is equivalent to the characteristic impedance

19. What is a distortion less line? What is the condition for a distortion less line?

A line, which has neither frequency distortion nor phase distortion is called a

distortion less line. The condition for a distortion less line is RC=LG. Also,

a) The attenuation constant α should be made independent of frequency.

b) The phase constant β should be made dependent of frequency.

d) The velocity of propagation is independent of frequency.

Loading is the process of increasing the inductance value by placing lumped

inductors at specific intervals along the line, which avoids the distortion

Continuous loading is the process of increasing the inductance value by placing a iron

core or a magnetic tape over the conductor of the line.

It is the process of using sections of continuously loaded cables separated by sections

of unloaded cables which increases the inductance value.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 5

Lumped loading is the process of increasing the inductance value by placing lumped

inductors at specific intervals along the line, which avoids the distortion.

25. Define reflection loss.

Reflection loss is defined as the number of nepers or decibels by which the current in the

load under image matched conditions would exceed the current actually flowing in the load.

Reflection loss, nepers = 10 log  

 

 

OR

oR

ZZ

ZZ

4

2

Reflection loss, dB = 20 log  

 

 

OR

oR

ZZ

ZZ

2

Where ZR = Receiving end impedance, Zo = characteristic impedance

26. What do you mean by reflection factor?

The change in current in the load due to reflection at the mismatched junction is called the

reflection factor and is given by reflection factor, k = OR

OR

ZZ

ZZ

2

27. Define insertion loss of a line.

The insertion loss of a line or network is defined as the number of nepers or decibels by

which the current in the load is changed by the insertion.

28. What are the primary constants of a transmission line?

The primary constants of a transmission line are resistance R, ohm/unit length, inductance L,

henry/unit length, capacitance C, farad/unit length and conductance G, mho/unit length.

29. What are the secondary constants of a transmission line?

The secondary constants of a transmission line are characteristic impedance and propagation

constant.

30. A transmission line has  12745oZ ohms and is terminated by 100RZ ohms.

Calculate the reflection loss in dB.

Solution: Reflection loss, dB = 20 log  

 

 

OR

oR

ZZ

ZZ

2 = 3.776 Db

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 6

31. When does a finite line appear as an infinite line?

A line of finite length, terminated in a load equivalent to its characteristic impedance appears

to the sending end generator as an infinite line.

32. Write the equation for the input impedance of a transmission line.

The equation for the input impedance of a transmission line is

 

  

 

lZlZ

lZlZ ZZin

Ro

oR

o 



sinhcosh

sinhcosh

33. A 50 ohms coaxial cable feeds a 75+j20 ohms dipole antenna. Find reflection coefficient

Solution: Given Zo = 50 ohms, ZR = 75+j20 ohms

Reflection co-efficient, k = oR

oR

ZZ

ZZ

09.958.126

65.3832

20125

2025

502075

502075

 

 



 

j

j

j

j

56.2925.0 

34. What is characteristic impedance? (May/June 2013)

Characteristic impedance is defined as the ratio of the voltage to current at the input

side of the transmission line provided the line is properly terminated.

35. Find the reflection coefficient of a 50 ohm transmission line when it is terminated by a

load impedance of 60+j40 ohm. (May/June 2013)

Solution: Given Zo = 50 ohms, ZR = 60+j40 ohms

Reflection co-efficient, k = oR

oR

ZZ

ZZ

98.19047.117

9.7523.41

40110

4010

504060

504060

 

 



 

j

j

j

j

9169.553523.0 

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 7

36. A transmission line has a characteristic impedance of 400 ohms and is terminated by a

load impedance of (650 – j475) ohms. Determine the reflection coefficient.

(Nov/Dec 2013)

Solution: Given Zo = 400 ohms, ZR = 650-j475 ohms

Reflection co-efficient, k = oR

oR

ZZ

ZZ



 

 



 

34.241152

24.6277.536

4751050

475250

400475650

400475650 

j

j

j

j

9.374628.0 

UNIT II – HIGH FREQUENCY TRANSMISSION LINES

1. State the assumptions for the analysis of the performance of the radio frequency line.

1. Due to the skin effect, the currents are assumed to flow on the surface of the

conductor. The internal inductance is zero.

2. The resistance R increases with square root of f while inductance L increases with f

Hence ωL>>R.

3. The leakage conductance G is zero

2. What is called a line of small dissipation and a line of zero dissipation?

Line of small dissipation: If resistance R is small, the line is considered as one of small

dissipation and this concept is useful when lines are employed as circuit elements or where

resonance properties are involved.

Line of zero dissipation: If resistance R is completely negligible, the line is considered as

one of zero dissipation. This concept is useful in applications where losses may be neglected

as in transmission of power at high efficiency.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 8

3. State the expressions for inductance L of a open wire line and coaxial line.

For open wire line, L=9.21*10-7(µ/µr +4ln d/a) =10-7(µr +9.21log d/a) H/m

For coaxial line, L = 4.60*10-7[log b/a] H/m

4. State the expressions for the capacitance of a open wire line

For open wire line, C= (12.07)/ (ln d/a) µf/m

5. What is dissipationless line?

A line for which the effect of resistance R is completely neglected is called

dissipationless line.

6. What is the nature and value of Z0 for the dissipation less line?

For the dissipation less line, the Z0 is purely resistive and given by, Z0=R0 = (L/C) ½

7. State the values of α and β for the dissipation less line.

Answer: α = 0 and β = w (LC) ½

8. What are nodes and antinodes on a line?

The points along the line where magnitude of voltage or current is zero are called

nodes while the points along the lines where magnitude of voltage or current first maximum

are called antinodes or loops.

9. What is the range of values of standing wave ratio?

The range of values of standing wave ratio is theoretically 1 to infinity.

10. State the relation between standing wave ratio and reflection coefficient.

S = (1+K)/ (1-K)

11. What is standing wave ratio?

The ratio of the maximum to minimum magnitudes of voltage or current on a line

having standing waves called standing waves ratio

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 9

12. What are standing waves?

If the transmission is not terminated in its characteristic impedance, then there will be

two waves traveling along the line which gives rise to standing waves having fixed maxima

and fixed minima.

13. State the relation between standing wave ratio S and reflection co-efficient k.

The relation between standing wave ratio S and reflection co-efficient k is,

14. How will you make standing wave measurements on coaxial lines?

For coaxial lines it is necessary to use a length of line in which a longitudinal slot,

one half wavelength or more long has been cut. A wire probe is inserted into the air dielectric

of the line as a pickup device, a vacuum tube voltmeter or other detector being connected

between probe and sheath as an indicator. If the meter provides linear indications, S is readily

determined. If the indicator is non linear, corrections must be applied to the readings

obtained.

15. Name the devices used for the measurement of standing waves.

Slotted line section and directional coupler are the devices used for the measurement of

standing waves.

16. Give the expression for the input impedance of the dissipation less line.

Zin =   

   

ljZR

ljRZ R

Ro

oR

o

tan

tan , ohm

17. Define reflection coefficient.

The ratio of amplitudes of the reflected and incident voltage waves at the receiving end of the

line is frequently called the reflection coefficient.

18. A 50 ohms line is terminated in load ZR = 90+j60 ohms. Determine the reflection

coefficient.

Solution: Reflection co-efficient k, = oR

oR

ZZ

ZZ

= 0.473  11.33

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 10

19. A lossless line has a characteristic impedance of 400 ohms. Determine the SWR if the ZR is

800+j0.0 ohms.

Solution:k

k S

 

1

1 ; To find the magnitude of the reflection coefficient, use k, =

oR

oR

ZZ

ZZ

=

0.33.Therefore, SWR 33.01

33.01

 S = 1.985

20. Write the expressions for the input impedance of open and short circuited dissipationless

line.

ljRZ ooc cot and ljRZ osc tan

21. For the line of zero dissipation, what will be the values of attenuation constant and

characteristic impedance.

For the line of zero dissipation, series arm impedance Z = Lj and shunt arm admittance

Y = Cj. C

L

Cj

Lj Zo 

= and   CjLj   = LCj.

But, Propagation constant, j . Where, attenuation constant, 0 and LC 

22. Give the minimum and maximum value of SWR and reflection co-efficient.

Min value ofSWR = 1; Max. Value of SWR = infinity,

Min. value of k = 0 and Max. Value of k =1

23. Find the characteristic impedance of the line with following constants L = 9 micro henry/m,

Solution:C

L Z o = 866 ohms.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 11

UNIT III – IMPEDANCE MATCHING IN HIGH FREQUENCY TRANSMISSION LINES

1. What is Impedance Matching?

If the load impedance is not equal to the source impedance, then all the power that is

transmitted from the source will not reach the load end and hence some power is wasted.

This is called impedance mismatch condition

2. State the uses of the eighth wave line.

The expression for input impedance of the eighth wave line is given by Zs = Ro An

eighth wave line may be used to transform any resistance to an impedance with a magnitude

equal to Ro of the line.

3. State the uses of the quarter wave line.

The expression for input impedance of the quarter wave line is given by Zs = Ro 2 / ZR

 It may be thought of as a transformer to match a load of ZR to a source of Zs. It may be

considered as an impedance inverter in that it can transform low impedance into high

impedance and vice versa.

 It may be used to couple a transmission line to a resistive load such as an antenna.

 It can be used as an insulator to support an open-wire or the center conductor of a

coaxial line Such lines are sometimes referred to as copper insulators.

4. What is an impedance inverter?

A quarter wave line may be considered as an impedance inverter in that it can

transform low impedance into high impedance and vice versa.

5. State the uses of half wave line.

The expression for input impedance of the half wave line is given by Zs = ZR

 A half wave line may be considered as a one – to – one transformer.

 It has its greatest utility in connecting a load to a source in cases where the load and source

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 12

6. Mention the applications of Smith Chart.

Smith chart can be used

 to convert impedance into admittance

 to determine the sending end impedance.

 to determine the load impedance

7. Explain impendence matching using stub.

In the method of impendence matching using stub, an open or closed stub line of

suitable length is used as a reactance shunted across the transmission line at a designated

distance from the load, to tune the length of the line and the load to resonance with an anti

resonant resistance equal to Ro.

8. What are the properties of Smith Chart?

` The properties of Smith chart are as follows:

 Normalizing impedance

 Plotting of an impedance

 Determination of Standing wave ratio

 Determination of reflection co-efficient K in magnitude and direction.

 Location of voltage maxima and voltage minima.

 Movement along the periphery of the chart.

9. Mention the significance of 4

line. (Nov/Dec 2012)

The expression for input impedance of the quarter wave line is given by Zs = Ro 2 / ZR

 It may be thought of as a transformer to match a load of ZR to a source of Zs. It may be

considered as an impedance inverter in that it can transform a low impedance into a

high impedance and vice versa.

 It may be used to couple a transmission line to a resistive load such as an antenna.

 It can be used as an insulator to support an open-wire or the center conductor of a

coaxial line Such lines are sometimes referred to as copper insulators.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 13

10. Give the formula to calculate the distance of the point from the load at which the stub is to

be connected.

The formula to calculate the distance of the point from the load at which the stub is to

be connected is,

11. Give the formula to calculate the distance d from the voltage minimum to the point stub be

connection.

The formula to calculate the distance d from the voltage minimum to the point of stub

be connection is,

12. Give the formula to calculate the length of the short circuited stub.

The formula to calculate the length of the short circuited stub is,

13. What is the input impendence equation of dissipation less line?

The input impendence equation of dissipation less line is given by

14. Give the equation for the radius of a circle diagram.

The equation for the radius of a circle diagram is

Where C is the shift of the center of the circle on the positive Ra axis.

15. What is the use of a circle diagram?

The circle diagram may be used to find the input impendence of a line m of any

chosen length.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 14

16. How is the circle diagram useful to find the input impendence of short and open

circuited lines?

An open circuited line has s = α, the correspondent circle appearing as the vertical

axis .The input impendence is then pure reactance, with the value for various electrical

lengths determined by the intersections of the corresponding βs circles with the vertical axis.

A short circuited line may be solved by determining its admittance .The S circle is again the

vertical axis, and susceptance values may be read off at appropriate intersection of the βs

circles with the vertical axis.

17. What are the difficulties in single stub matching?

The difficulties of the smith chart are

(i) Single stub impedance matching requires the stub to be located at a definite point on

the line. This requirement frequently calls for placement of the stub at an undesirable

place from a mechanical view point.

(ii) For a coaxial line, it is not possible to determine the location of a voltage minimum

without a slotted line section, so that placement of a stub at the exact required point is

difficult.

(iii) In the case of the single stub it was mentioned that two adjustments were required,

these being location and length of the stub.

18. What is double stub matching?

Another possible method of impedance matching is to use two stubs in which the

locations of the stub are arbitrary, the two stub lengths furnishing the required adjustments.

The spacing is frequently made λ/4.This is called double stub matching.

19. Why Double stub matching is preferred over single stub matching.

Double stub matching is preferred over single stub due to following disadvantages of

single stub.

1. Single stub matching is useful for a fixed frequency. So as frequency changes the

location of single stub will have to be changed.

2. The single stub matching system is based on the measurement of voltage minimum.

Hence for coaxial line it is very difficult to get such voltage minimum, without using

slotted line section.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 15

UNIT IV – PASSIVE FILTERS

1. For a symmetrical network, define propagation constant and characteristic impedance.

(Nov /Dec 2012), (May/June 2013)

For a symmetrical network, the image impedances Z1i and Z2i are equal to each other, and

the image impedance is then called the characteristic impedance or the iterative impedance, Zo.

That is, if a symmetrical T network is terminated in Zo, its input impedance will also be Zo, or its

impedance transformation ratio is unity.

Under the condition of Zo termination, propagation constant γ is defined as the natural

logarithm of the ratio of input to output currents or input to output voltages. γ is a complex

number defined as γ = α + jβ,

Where α is known as the attenuation constant, β is a phase constant

2. What is constant k network?

If Z1 and Z2 of a reactance network are unlike reactance arms, then Z1 Z2 = k 2 where k is a

constant independent of frequency. Networks or filter sections for which this relation holds are

called constant-k filters.

3. How to obtain constant-k high pass filter?

For a constant-k high pass filter, Z1 = -jXC and Z2 = jXL , thenZ1 Z2 = k 2 = L/C, where k is

a constant independent of frequency.

4. What are the demerits of constant – k filters?

There are two major demerits in the constant – k filters. They are,

 The attenuation does not rise very rapidly at cutoff, so that frequencies just outside the

pass band are not appreciably attenuated with respect to frequencies just inside the pass

band.

 The characteristic impedance varies widely over the pass band, so that a satisfactory

impedance match is not possible.

5. What are m-derived filters?

The m-derived filters attempt to raise the attenuation near cutoff frequencies.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 16

6. Define neper.

Neper is defined as the natural logarithm of the ratio of the input to output currents or input

to output voltages. Ie. N nepers = ln (V1 / V2 ) = ln (I1 / I2 )

7. Define decibel.

The decibel is defined as the logarithm of a power ratio, number of decibels = log (P1/P2). It

has been found that a unit one-tenth as large is more convenient, and the smaller unit is called

the decibel, abbreviated “db” defined as db = 10 log(P1/P2).

8. Write down the relationship between neper and dB.

1 Neper = 0.115 dB.

9. What do you mean by symmetrical networks?

When Z1 = Z2 or the two series arms of a T network are equal, or Za = Zc and the shunt

arms of a π network are equal, the networks are said to be symmetrical.

10. What are called cut-off frequencies in the design of filters?

The frequencies at which the network changes from a pass network to a stop network, or

vice versa, are called cutoff frequencies.

11. What is meant by the terms pass band and stop band as applied to filters?

Pass band: Pass band is a band of frequencies in which the signal is allowed to pass where

the attenuation of the signal is zero for an ideal filter.

Stop band: Stop band is a band of frequencies in which the signal is not allowed to pass

where the attenuation of the signal is infinity for an ideal filter.

12. What is the use of crystal filters?

Crystal filters are quite generally used to separate the various channels in Carrier telephone

circuits, in the range above 50 kilocycles.

13. What are the characteristics of an ideal filter?

The characteristics of an ideal filter are zero attenuation in the pass band and infinite

attenuation in the stop band.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 17

14. Design a prototype low pass filter T and π section of design impedance Ro = 500ohms

and cut-off frequency is 2000HZ.

Solution: For a prototype low pass filter, R= Ro

L = cf

R

=

2000

500

X= 0.0796 henry C =

Rf c

1 = 3.184 X 10

15. Design a prototype high pass filter T and π section of design impedance Ro = 600ohms

and cut-off frequency is 1000HZ.

Solution: For a prototype high pass filter, R= Ro

L = cf

R

4 =

10004

600

X= 0.0477 henry C =

Rf c4

1 =

60010004

1

XX= 1.326 X 10

16. A constant k T section HPF has a cut-off frequency of 10 KHZ. The design impedance

is 600 ohms. Determine the value of L. (May/June 2013), (Nov/Dec 2013)

Solution: For a constant k T section HPF

L = cf

R

4 = mH

XX 7.4

1000104

600 

17. What are composite filters?

Filters designed using one or more m-derived sections following a proto type filter

are called composite filters. The m-derived section is designed following the design of the

proto type T section. (ie) the use of a prototype and one or more m- derived sections in series

results in a composite filter.

18. How can you realize a band stop filter?

A band stop or band elimination filter can be realized by connecting a low pass filter

in parallel with a high pass section in which the cut-off frequency of low pass filter is below

that of a high pass filter.

19. How do you obtain a band pass filter?

A band pass filter may be obtained by using a low pass filter followed by a high pass

filter in which the cut-off frequency of the low pass filter is above the cut-off frequency of

the high pass filter, the overlap thus allowing only a band of frequencies to pass.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 18

20. Give the characteristic impedance of constant k low pass filter (both pi and T sections).

The characteristic impedance of constant k low pass filter (both pi and T sections) is,

2

1   

   

 

c

koT f

f RZ and

2

1   

   

 

c

k

o

f

f

R Z

21. State the significance of crystal filters in communication system. (Nov/Dec 2012)

Discrete & Monolithic Crystal Filters (MCFs) use quartz as the prime filtering

element. The crystal's high Q factor and tight stability ensure that the filter topology has a

precise centre frequency and a sharp pass-band response. These crystal filters are ideal for

applications requiring wide and narrow band voice & data communications (HF, VHF,

UHF), signal analyzers for test and measurement and also radar systems for navigation.

22. Give some applications of filters.

Filter networks are widely used in communication systems to separate various voice

channels in carrier frequency telephone circuits. Filters also find applications in

instrumentation, telemetering equipment etc. where it is necessary to transmit or attenuate a

limited range of frequencies.

23. Give the characteristic impedance of a symmetrical T-section and symmetrical

section.

Characteristic impedance of a symmetrical T-section,  

  

 

2

1 21

4 1

Z

Z ZZZoT

Characteristic impedance of a symmetrical  section,21

21

41 ZZ

ZZ Zo

 

24. Give the expression to determine the propagation constant of a symmetrical T network.

The expression to determine the propagation constant of a symmetrical T network is,

 

 

 

  

 

2

1

2

2

1

2

1

22 1ln

Z

Z

Z

Z

Z

Z

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 19

25. Whatare the advantages of m-derived filters? ( Nov/Dec 2013)

The advantages of m-derived filters are,

 The attenuation does rise very rapidly at cutoff, so that frequencies just outside the pass

band are appreciably attenuated with respect to frequencies just inside the pass band.

 The characteristic impedance remains uniform over the pass band, so that satisfactory

impedance match is possible.

UNIT V – WAVE GUIDES AND RESONATORS

1. What are guided waves? Give examples.

The electromagnetic waves that are guided along or over conducting or dielectric surface are

called guided waves. Examples: The waves along ordinary Parallel wire co-axial transmission

lines.

2. What is E wave or TM wave?

Transverse magnetic wave is a wave in which magnetic field strength H is transverse. It has

electric field strength Ez in the direction of propagation and no component of magnetic field

strength Hz in the same direction.

3. Define H wave or TE wave.

Transverse electric wave is a wave in which electric field strength E is transverse. It has a

magnetic field strength Hz in the direction of propagation and no component of electric field

strength Ez in the same direction.

4. Define TEM wave. (Nov/Dec 2013)

The transverse electromagnetic waves are waves in which both electric and magnetic fields are

transverse entirely but have no components of Ez and Hz. It is referred to as principal wave.

5. Mention the characteristics of TEM waves. (May/June 2013)

 TEM waves are special type of TM wave.

 It does not have either Ez or Hz component.

 Its velocity is independent of frequency and its cut-off frequency is zero.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 20

6. What is known as cut-off frequency?

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

7. Define attenuation factor.

Attenuation factor α = (power lost/unit length)/ 2 X power transmitted

8. Define wave impedance.

Wave impedance of a wave traveling inside the region between the planes is defined as the ratio

of electric field intensity along one transverse direction to the magnetic field intensity along the

other transverse direction.

9. Write down the expression for the wave impedance for TM, TE and TEM wave.

Wave impedance for TM wave, ZTM =

2

1  

  

 

f

f c; For TE wave, ZTE = 2

1  

  

 

f

f c

Wave impedance for TEM wave, ZTEM = o= o

o

= 120π ohms

10. Write down the expression for cut-off frequency?

Cut-off frequency = a

m

2 , HZ

11. Define phase velocity.

The rate at which the wave changes its phase as the wave propagates inside the region

between planes is defined as phase velocity vp and is given by , vp = 

 ;

2

1  

  

 

f

f c

12. Define group velocity.

The actual velocity with which the wave propagates inside the region between the planes is

defined as group velocity vg, given by vg = 

d

d

13. Write the relation between group velocity and phase velocity.

VgVp = C 2 , where C = velocity of light.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 21

14. A pair of perfectly conducting planes is separated 8 cm in air. For a frequency of

5000MHZ with the TM1 mode excited, find the cut-off frequency.

Solution: Cut-off frequency =a

m

2 , HZ

= 127 10854.810408.02

1

 xxxx = 1875 MHZ

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

GHZ with TM2 mode being excited, find the cut-off wavelength.

Solution:m x

m

a c 06.0

2

06.022 

16. Define dominant mode.

The mode, which is having the lowest cut-off frequency, is called the dominant mode.

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

The cut-off frequency for TEM wave is zero.

18. For a frequency of 6000MHZ and plane separation = 7 cm with air dielectric, find the

wave impedance for the TE1 mode.

Solution: Wave impedance for TE wave, ZTE =2

1  

  

 

f

f c

= 403.61 ohms, where

fc = a

m

2 ; m = 1 ; a = 7 cm ; on substituting these values in fc , we get fc = 2.143GHZ

19. Define skin depth.

In a medium, which has conductivity the wave is attenuated as it, progress owing to the

losses, which occur.

In a good conductor at radio frequencies, the rate of attenuation is very great and the wave

may penetrate only a short distance before being reduced to a negligibly small percentage of its

original strength. Such circumstance is called depth of penetration or skin depth defined as , the

depth the wave has been attenuated to 1/e or approximately 37% of its original value.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 22

20. What is meant by dominant mode? What is the dominant mode for parallel plate

waveguides?

Dominant mode is the mode, which has the lowest cut-off frequency.The dominant mode for

parallel plate waveguides is TE10

21. Compare TE and TM mode. (Nov/Dec 2012)

TM wave: Transverse magnetic wave is a wave in which magnetic field strength H is

transverse. It has electric field strength Ez in the direction of propagation and no component

of magnetic field strength Hz in the same direction.

TE wave: Transverse electric wave is a wave in which electric field strength E is

transverse. It has a magnetic field strength Hz in the direction of propagation and no

component of electric field strength Ez in the same direction.

22. What is the need for attenuator? (Nov/Dec 2012)

An attenuator is a two-port resistive network and is used to reduce the signal level by

a given amount. In a number of applications, it is necessary to introduce a specified loss

between source and a matched load without altering the impedance relationship.

23. What is a principal wave?

The electromagnetic field is transverse. This wave is called the transverse

electromagnetic wave. It is extremely important guided wave propagation, because it is the

familiar type of wave propagated along all ordinary two-conductor transmission lines when

operating in their low frequency manner. It is usually called the principal wave.

26. What is degenerate mode in rectangular waveguide? (May/June 2013)

The mode in which are having same cutoff frequencies but different field

distributions, are known as degenerate modes.

24. A wave is propagated in a parallel plane waveguide. The frequency is 6 GHz and the

plane separation is 3 cm. Determine the group and phase velocities for the dominant

mode. (Nov/Dec 2013)

Solution: f = 6 GHz, a = 3cm, Dominant mode = TE1

Cut-off frequency =a

m

2 = 

  

03.02

1031 8

X

XX = 5GHz

VP = 

 ;

2

1  

  

 

f

f c

VgVp = C 2 , Group velocity,

pg v

C v

2

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 23

25. What is a waveguide?

Waveguide is a hollow conducting metallic tube of uniform cross section, which is

used for propagating electromagnetic waves that are guided along the surfaces of the tube.

26. What are the basic types of propagating waves in a uniform waveguide?

TM waves and TE waves are the basic types of propagating waves in a uniform

waveguide.

27. Why rectangular waveguides are preferred over circular waveguide?

Rectangular waveguides are preferred over circular waveguides because of the

following reasons.

 Rectangular waveguide is smaller in size than a circular waveguide of the

same operating frequency.

 It does not maintain its polarization through the circular waveguide.

 The frequency difference between the lowest frequency on dominant mode

and the next mode of a rectangular waveguide is bigger than in a circular

waveguide.

28. What is meant by the dominant mode of a waveguide?

The cut-off frequency

22

2

1  

  

 

  

 

b

n

a

m f c

 shows that the physical size of the

waveguide will determine the propagation of the modes of specific orders determined by

m and n. The minimum cut-off frequency is obtained for a guide having dimension a > b, for

m = 1,n = 0 Since for TMmn modes, m  0 or n  0, the lowest order mode possible is TE10

and TM11 mode, called the dominant mode in a rectangular waveguide for a >b

29. What is known as evanescent mode in a rectangular waveguide?

When the operating frequency is lower than the cut-off frequency the propagation

constant becomes real ie γ = α . The wave cannot be propagated. This non-propagating mode

is known as evanescent mode.

30. What is the cut-off wavelength of the TE10 mode in a rectangular waveguide?

Cut-off wavelength C = 2a.

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 24

31. Explain the impossibility of TEM wave in waveguide.

Suppose a TEM wave is assumed to exist within a hollow guide of any shape.

Then lines of H must lie entirely in the transverse plane. Also in a nonmagnetic material, .

H = 0 which requires that the lines of H be closed loops. Therefore, if a TEM exists inside

the guide, the lines of H will be closed loops in plane perpendicular to the axis.

32. What are the applications of the waveguide?

The waveguides are employed for transmission of energy at very high frequencies

where the attenuation caused by waveguide is smaller. Waveguides are used in microwave

transmission. Circular waveguides are used as attenuators and phase shifters.

33. For an air filled copper X band waveguide with dimension a= 2.5cm and b= 1.5cm,

determine the cut-off frequency for TE10 mode.

Solution:

22

2

1  

  

 

  

 

b

n

a

m f c

 ; Here, m = 1 and n = 0 ;

oo

1 = 3 x 10

8 m/sec

= 18.84 GHZ.

34. What are degenerate modes?

Some of the higher order modes, having the same cut-off frequency, are called

degenerate modes. It is seen that in a rectangular waveguide possible TEnm and TMmn modes

(both m  0 and n  0) are always degenerate.

35. Mention the uses of circular waveguides.

Circular waveguides are used as attenuators and phase shifters.

36. Why is the Bessel’s function of the II kind (Neumann’s function) not applicable for the

field analysis inside the circular waveguide?

Solution : Because all the Neumann’s functions become infinite at 0 , these second solutions

cannot be used for any physical problem in which the origin is included, as for example the

hollow waveguide problem.

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

The dominant mode in a circular waveguide is TE11

EC 6503 – Transmission Lines & Wave Guides – Two Marks Q & A

DEPARTMENT OF ECE – PSNACET – III YEAR – V SEM – REG 2013. Page 25

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

Fc = 2

nmh ; where hnm =  

a

ha nm for TM waves and

hnm =  

a

ha nm '

for TE waves.

39. A circular waveguide operated at 11 GHZ has the internal diameter of 4.5 cm. for a

TE01 mode propagation, calculate free space wavelength and cut-off wavelength.

(ha)01= 2.405)

Solution:9

8

1011

103

X

X

f

C o  = 0.02 m

c

C f

C  ;

2

nm

c

h f = 5.1 GHZ ;

9

8

101.5

103

X

X C  = 0.0588 m

40. Why is TM01 mode preferred to the TE01 mode in a circular waveguide?

TM01 mode is preferred to the TE01 mode, since it requires a smaller diameter for the same cut-

off wavelength.

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

The special significance of TE01 mode in a circular waveguide is that the attenuation effect

decreasing with increase in frequency.

42. List the degenerate modes in a circular waveguide.

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

waveguide.

43. What is a microwave cavity resonator?

Microwave resonator is a tunable circuit used in microwave circuits. It is a metallic enclosure

that confines the electromagnetic energy. When the resonator resonates at resonant frequency,

the peak energies stored in the electric and magnetic fields are equal.

44. What are the parameters that determine the performance of a cavity resonator?

The performance parameters of microwave resonator are:

Resonant Frequency, Quality factor and input impedance.