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Waveguides โ GATE Problems
One Mark Questions
1. The interior of ๐20
3๐๐ ร
20
4๐๐ rectangular waveguide is completely
filled with a dielectric of โ๐= 4. Waves of free space wave โ length
shorter thanโฆโฆโฆ..can be propagated in the TE11 mode.
[GATE: 1994: 1 Mark]
Soln. The inside dimension of waveguide is given.
For Rectangular waveguide
๐ =๐๐
๐๐๐ , ๐ =
๐๐
๐๐๐
Where โaโ and โbโ are wide and narrow dimensions of the waveguide.
Waveguide is filled with dielectric of โ๐= ๐
Velocity of propagation, ๐ =๐
โ๐โ
Cut off frequency for rectangular waveguide is given by
๐๐ =๐
๐๐ โ(
๐๐
๐)๐
+ (๐๐
๐)๐
Where m and n are half wave variations in wide and narrow
dimensions of waveguide.
Velocity of propagation in dielectric is given by
๐ =๐
โ๐๐ โ๐
๐
โโ๐
= ๐ ร ๐๐๐ .๐
โ๐= ๐. ๐ ร ๐๐๐
๐
๐= ๐. ๐ ร ๐๐๐๐ ๐๐/๐
Since mode is TE11
Thus m = 1 and n = 1
๐๐ =๐. ๐ ร ๐๐๐๐
๐โ(
๐
๐๐)๐
+ (๐
๐๐)๐
ร ๐๐๐ ๐ฏ๐
= ๐. ๐๐ ร ๐๐๐๐ ร๐
๐๐= ๐. ๐๐๐ ร ๐๐๐ ๐ฏ๐
= ๐. ๐๐๐ ๐ฎ๐ฏ๐
Cut off wavelength ๐๐ช in the medium can be written as
๐๐ช =๐
๐๐=
๐. ๐ ร ๐๐๐๐
๐. ๐๐๐ ร ๐๐๐
= ๐ ๐๐
Thus the waves with free space wavelengths shorter than 8 cm
wavelength can be propagated,
๐๐= 8 cm
2. A rectangular air โ filled waveguide has a cross section of 4 ๐๐ ร 10 ๐๐
The minimum frequency which can propagation in the waveguide is
(a) 1.5 GHz
(b) 2.0 GHz
(c) 2.5 GHz
(d) 3.0 GHz
[GATE 1997: 1 Mark]
Soln. Given,
A rectangular waveguide air filled with
a = 10 cm , b = 4 cm
Waveguide acts as a high pass filter with cut off frequency of
๐๐ =๐
๐๐ โ๐ โ . โ(
๐๐
๐)๐
+ (๐๐
๐)๐
For air filled waveguide
๐ =๐
โ๐๐ โ๐
= ๐ ร ๐๐๐ ๐/๐๐๐
Here m and n are integers representing TEmn or TMmn modes.
Least values of m and n for TE mode are
m=1 and n=0 ๐ซ๐๐ฉ๐ซ๐๐ฌ๐๐ง๐ญ๐ข๐ง๐ ๐๐๐๐ mode, which is known as
dominant mode having largest wavelength and lowest cut off
frequency.
Thus,
๐๐ = ๐. ๐ ร ๐๐๐โ๐
๐๐๐ร ๐๐๐ ๐ฏ๐
= ๐. ๐ ๐ฎ๐ฏ๐
Option (a)
3. Indicate which one of the following modes do NOT exist in a rectangular
resonant cavity
(a) TE110
(b) TE011
(c) TM110
(d) MT111
[GATE 1999: 1 Mark]
Soln. In the present problem we have to find the modes TEmnp / TMmnp that
do not exist in rectangular cavity with dimensions a, b and d.
The integer m, n and p represent half wave variations in x, y and z
directions.
For TEmnp mode
๐ฏ๐ = ๐ฏ๐๐ ๐๐๐ (๐๐ ๐
๐) ๐๐๐ (
๐๐ ๐
๐) ๐๐๐ (
๐๐ ๐
๐ )
Dominant mode present is TE101
For TMmnp
๐ฌ๐ = ๐ฌ๐๐ ๐๐๐ (๐๐
๐๐) ๐๐๐ (
๐๐
๐๐) ๐๐๐ (
๐๐
๐ ๐)
TM mode with lowest integer present is TM110
The TE110 mode does not exist in rectangular cavity resonator as the
lowest value of last subscript should be 1 for the TE mode to exist
Option (a)
4. The phase velocity of waves propagation in hollow metal waveguide is
(a) Greater than velocity of light in free space
(b) Less than velocity of light in free space
(c) Equal to velocity of light in free space
(d) Equal to group velocity
[GATE 2001: 1 Mark]
Soln. The velocity of propagation in the bounded medium (i.e. say a
rectangular waveguide) is given by
๐๐ =๐
โ๐ โ (๐๐
๐)๐
Where c โ velocity of propagation in unbounded media (free space)
fc โ cut off frequency
f โ Frequency of operation
Say, for TE10 mode
f > fc
From the above equation. We find
๐๐ > ๐
Phase velocity is the velocity with which wave propagates. Note that
the velocity with which energy propagates in waveguide is less than
phase velocity.
Option (c)
5. The dominant mode in a rectangular waveguide is TE10 because this
mode has
(a) No attenuation
(b) No cut off
(c) No magnetic field component
(d) The highest cut off wavelength
[GATE 2001: 1 Mark]
Soln. The dominant mode in rectangular waveguide is TE10 . This mode
has the lowest cutoff frequency (fc) and the highest cutoff wavelength
๐๐ช(= ๐๐) out off all the TEmn and TMmn modes.
Option (d)
6. The phase velocity for the TE10 mode in an air filled rectangular
waveguide is
(a) Less than c
(b) Equal to c
(c) Greater than c
(d) None of the above
[GATE 2002: 1 Mark]
Soln. This problem is similar to problem of GATE 2001 (problem 4)
Phase velocity
(๐๐) > ๐ =๐
โ๐๐ โ๐
= ๐ ร ๐๐๐ ๐/๐
Where c is velocity of plane waves in free space
Option (c)
7. The phase velocity of an electrometric wave propagating in a hollow
metallic rectangular waveguide in the TE10 mode is
(a) Equal to its group velocity
(b) Less than velocity of light in free space
(c) Equal to the velocity of light in free space
(d) Greater than the velocity of light in free space
[GATE 2004: 1 Mark]
Soln. This problem is also similar to problem No. 4 and 5
Phase velocity is greater than velocity of light in free space
Option (d)
8. Refractive index of glass is 1.5 Find the wavelength of a beam of light
with a frequency of 1014 Hz in glass. Assume velocity of light
3 ร 108 ๐/๐ is vacuum.
(a) 3 ยตm
(b) 3 ยตm
(c) 2 ยตm
(d) 1 ยตm
[GATE 2005: 1 Mark]
Soln. Given,
Refractive index of glass n = 1.5
Frequency of light beam ๐ = ๐๐๐๐ ๐ฏ๐
Velocity of light in Vacuum = ๐ = ๐ ร ๐๐๐ ๐/๐
Refractive index of medium is given by
๐ =๐
๐=
๐ฝ๐๐๐๐๐๐๐ ๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐๐
๐ฝ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐
๐๐, ๐ =๐
๐=
๐ ร ๐๐๐
๐. ๐= ๐ ร ๐๐๐ ๐/๐
๐ = ๐๐ ๐๐, ๐ =๐
๐=
๐ ร ๐๐๐
๐๐๐๐= ๐ ร ๐๐โ๐ ๐ = ๐ ๐๐
Option (c)
9. Which of the following statement is true regarding the fundamental mode
of the metallic waveguides shown
Q: CylindricalP: Coaxial
R: Rectangular
(a) Only P has no cut off frequency
(b) Only Q has no cut off frequency
(c) Only R has no cut off frequency
(d) All three have cut off frequency
[GATE 2009: 1 Mark]
Soln. Transmission media such as
(i) Two wire line
(ii) Coaxial line
(iii) Parallel plane waveguide (for TM modes) having two plates as
conductor
These media have no cut off frequency(๐๐ = ๐).The wave propagated
is called principal wave. This is also known as Transverse
Electromagnetic wave (TEM) wave.
Thus, Coaxial line shown in P has no cut off frequency
๐. ๐. ๐๐ = ๐
Cylindrical waveguide shown in figure Q and rectangular waveguide
shown in R, are the single conductor system having cut off frequency,
fc, which depends upon the dimensions of the cross section and
characteristic of the medium in the wave guide.
Option (a)
10. The modes of rectangular waveguide are denoted by TEmn / TMmn when
m and n are Eigen numbers along the larger and smaller dimensions of
the waveguide respectively. Which one of the following statement is true.
(a) The TM10 mode of waveguide does not exist.
(b) The TE10 mode of waveguide does not exist.
(c) The TM10 and TE10 modes both exist and have same cut off frequency.
(d) The TM10 and TE10 modes both exist and have same cut off frequency
[GATE 2011: 1 Mark]
Soln. In a rectangular waveguide the lowest value of m or n for TM mode
is unity So the lowest TM mode is TM11 ( TM01 or TM10 modes do not
exist.)
For TE mode, TE10 and TE01 modes exist.
The lowest order TE mode is TE10 . This mode has the lowest cut off
frequency and is called the dominant mode.
If we look to various options given we find
Option (a)
11. Consider an air filled rectangular waveguide with a cross โ section of
5 cm ร 3 cm. For this waveguide, the cut off frequency (in MHz) of TE21
mode is ________
[GATE 2014: 1 Mark]
Soln. For air filled rectangular waveguide the cut off frequency is given by
๐๐ =๐
๐๐ โ๐๐ โ๐
[(๐๐
๐)๐
+ (๐๐
๐)๐
]
๐ ๐โ
For TE21 mode
๐ = ๐ ๐๐๐ ๐ = ๐, ๐ = ๐๐๐, ๐ = ๐๐๐
๐๐ for TE21 mode is given by
๐๐๐๐ =๐ ร ๐๐๐
๐โ(
๐
๐. ๐๐)๐
+ (๐
๐. ๐๐)๐
๐๐ช๐๐ = ๐๐๐๐ ๐ด๐ฏ๐
Two Mark Questions
1. The cut off frequency of waveguide depends upon
(a) The dimensions of the waveguide.
(b) The dielectric property of the medium in the waveguide.
(c) The characteristic impedance of the waveguide
(d) The transverse and axial components of the fields
[GATE 1987: 2 Marks]
Soln. Let us consider the expression for cut off frequency for rectangular
waveguides.
๐๐ =๐
๐๐ โ๐ โโ(
๐๐
๐)๐
+ (๐๐
๐)๐
Thus, it depends on
(i) Dimension a and b of the waveguide
(ii) Dielectric constant of the medium
Thus, option (a) and (b)
2. For normal mode EM wave propagation in a hollow rectangular
waveguide
(a) The phase velocity is greater than group velocity.
(b) The phase velocity is greater than velocity of light in free space.
(c) The phase velocity is less than the velocity of light in free space.
(d) The phase velocity may be either greater than or less than group
velocity.
[GATE 1988: 2 Marks]
Soln. In a rectangular waveguide the phase velocity is (๐๐)
๐๐ > velocity of light in free space and is given by
๐๐ =๐
โ๐ โ (๐๐/๐)๐
Where fc is cut off frequency.
The group velocity ๐๐ in the guide is related to ๐๐ and c
๐๐ . ๐๐ = ๐๐
๐๐ =๐๐
๐๐ ๐๐ ๐๐ = ๐.โ๐ โ (๐๐/๐)๐
๐๐ ๐๐ < ๐
Therefore ๐๐ < ๐ < ๐๐
Thus,
Options (a) and (d)
3. Choose the correct statements for a wave propagating in an air filled
rectangular waveguide
(a) Guided wavelength is never less than free space wavelength.
(b) Wave impedance is never less than free space impedance.
(c) Phase velocity is never less than the free space velocity.
(d) TEM mode is possible if the dimensions of the waveguide are
properly chosen.
[GATE 1990: 2 Marks]
Soln. For the wave propagating in air filled rectangular waveguide
Guide wavelength is given by
๐๐ =๐๐
๐ท=
๐๐
๐ โ (๐๐/๐๐ช)๐=
๐๐
โ๐ โ (๐๐/๐)๐
Where ๐๐ โ free space wavelength
Thus
๐๐ช > ๐๐
Wave impedance for TE wave is given by
๐ผ๐ป๐ฌ =๐ผ
โ๐ โ (๐๐/๐)๐
Where ๐ผ is impedance in free space for TEM waves
Thus,
๐ผ๐ป๐ฌ > ๐ผ
For TM waves
๐ผ๐ป๐ด = ๐ผโ๐ โ (๐๐/๐)๐
๐ผ๐ป๐ด < ๐ผ
Phase velocity
๐๐ =๐
๐ท=
๐
๐ โร
๐
โ๐ โ (๐๐/๐)๐
=๐
โ๐๐ โ๐ โ๐ โ (๐๐/๐)๐
Where c is velocity in free space
๐๐ > ๐
Options (a) and (c)
4. A rectangular waveguide has dimensions1๐๐ ร 0.5 ๐๐. Its cut off
frequency is
(a) 5 GHz
(b) 10 GHz
(c) 15 GHz
(d) 20 GHz
[GATE 2000: 2 Marks]
Soln. Dimensions of rectangular waveguide
๐ = ๐ ๐๐ = ๐๐โ๐๐
๐ = ๐. ๐ ๐๐ = ๐. ๐ ร ๐๐โ๐๐
The lowest possible mode is TE10
Cut off frequency (fc) for TE10 mode is
๐๐ =๐
๐๐
Where c is velocity of light free space
๐๐ =๐ ร ๐๐๐๐
๐ ร ๐๐โ๐= ๐. ๐ ร ๐๐๐๐
= ๐๐ ๐ฎ๐ฏ๐
Option (c)
5. A rectangular metal wave guide filled with a dielectric material of
relative permittivity ๐๐ = 4 has the inside dimensions3.0 ๐๐ ร 1.2 ๐๐.
The cut off frequency for the dominant mode is
(a) 2.5 GHz
(b) 5.0 GHz
(c) 10.0 GHz
(d) 12.5 GHz
[GATE 2003: 2 Marks]
Soln. Given,
Waveguide dimensions (inner)
๐ = ๐ ๐๐ = ๐ ร ๐๐โ๐๐
๐ = ๐. ๐ ๐๐ = ๐. ๐ ร ๐๐โ๐๐
Dominant mode is the mode having lowest cut off frequency and is
denoted by TE10 .Cut off frequency for Dominant mode is given by
๐๐ =๐
๐๐
Where v is velocity in the medium
๐๐ =๐
๐๐
=๐
โ๐๐ โ๐
.๐
๐๐
=๐
โ๐๐ โ๐ โโ๐
.๐
๐๐
=๐ ร ๐๐๐
โ๐ ร ๐ ร ๐ ร ๐๐โ๐
= ๐. ๐๐ ร ๐๐๐๐
๐๐ = ๐. ๐ ๐ฎ๐ฏ๐ Option (a)
6. Which one of the following does represent the electric field lines for the
TE02 mode in the cross โ section of a hollow rectangular metallic
waveguide?
Y
(a)
X
Y
(b)
X
Y
(c)
X
Y
(d)
X
[GATE 2005: 2 Marks]
Soln. This problem is to find E-field configuration in rectangular
waveguide for TE02 mode.
The first subscript m = 0 indicates no variation of E in ๐ โ direction.
The second subscript n = 2 indicates two โ half wave variations in y โ
direction.
This variation agrees with option (d) shown in figure
Option (d)
7. A rectangular waveguide having TE10 mode as dominant mode is having
a cut off frequency of 18 GHz for the TE30 mode. The inner broad โ wall
dimension of the rectangular waveguide is
(a) 5/3 cm
(b) 5 cm
(c) 5/2 cm
(d) 10 cm
[GATE 2006: 2 Marks]
Soln. Cut off frequency for TEmn mode is
๐๐ =๐
๐โ๐ โโ(
๐
๐)๐
+ (๐
๐)๐
For waveguide with air medium
๐๐ =๐
๐โ(
๐
๐)๐
+ (๐
๐)๐
Here, for TE30 mode
๐ = ๐,๐ = ๐
๐๐ =๐
๐ .๐
๐
๐บ๐, ๐ = ๐
๐ .
๐
๐๐
=๐ ร ๐๐๐
๐ .
๐
๐๐ ร ๐๐๐
=๐
๐ ๐๐
Option (c)
8. An air โ filled rectangular waveguide has inner dimensions of
3 ๐๐ ร 2 ๐๐. The wave impedance of the TE20 mode of propagation in
the waveguide at a frequency of 30 GHz is (free space impedance ๐0 =
377 ฮฉ ).
(a) 308 ฮฉ
(b) 355 ฮฉ
(c) 400 ฮฉ
(d) 461 ฮฉ
[GATE 2007: 2 Marks]
Soln. Given,
Inner dimension of waveguide is ๐ ๐๐ ร ๐ ๐๐
Free space impedance ๐ผ๐ = ๐๐๐ ๐
Wave impedance ๐ผ for TE20 mode at ๐ = ๐๐ ๐ฎ๐ฏ๐ is given by
๐ผ =๐ผ๐
โ๐ โ (๐๐/๐)๐
๐๐ ๐๐๐ ๐ป๐ฌ๐๐ = ๐ช
๐โ(
๐
๐)๐+ ๐
=๐
๐ร
๐
๐=
๐
๐=
๐ ร ๐๐๐
๐
๐๐ = ๐๐๐ฎ๐ฏ๐
So,
๐ผ =๐ผ๐
โ๐ โ (๐๐/๐)๐=
๐๐๐
โ๐ โ (๐๐/๐๐)๐
=๐ผ
๐. ๐๐๐โ ๐๐๐๐
Option (c)
9. The ๏ฟฝโ๏ฟฝ field in a rectangular waveguide of inner dimensions ๐ ร ๐ is
given by
๏ฟฝโ๏ฟฝ =๐ ๐
โ2(๐
๐) ๐ป0 sin (
2๐๐ฅ
๐) sin(๐๐ก โ ๐ฝ๐ง) ๏ฟฝฬ๏ฟฝ,
Where H0 is a constant, a and b are the dimensions along the x โ axis and
the y โ axis respectively. The mode of propagation in the waveguide is
(a) TE20
(b) TM11
(c) TM20
(d) TM10
[GATE 2007: 2 Marks]
Soln. Given
Wide dimensions of waveguide is a
Narrow dimensions of wave is b
Field is given by
๏ฟฝโโ๏ฟฝ =๐ ๐
๐๐(๐
๐) ๐ฏ๐ ๐ฌ๐ข๐ง (
๐๐ ๐
๐) ๐ฌ๐ข๐ง(๐๐ โ ๐ท๐) ๏ฟฝฬ๏ฟฝ,
Wave is traveling in +z direction (factorโ๐ท๐)
The component of electric field is in y direction i.e. ๐ฌ๐ component (as
function of x)
No component of field in the direction of propagation (๏ฟฝโโ๏ฟฝ ๐) So, the wave is transverse electric (TE)
If we compare โsinโ term in ๏ฟฝโโ๏ฟฝ with general expression ๐๐๐ (๐๐
๐)๐
We find m = 2
There is no function of โin E i.e. n=0
Thus the mode of propagation in waveguide is TE20
Option (a)
10. A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm)
is to be operated in TE11 mode. The minimum operating frequency is
(a) 6.25 GHz
(b) 6.0 GHz
(c) 5.0 GHz
(d) 3.75 GHz
[GATE 2008: 2 Marks]
Soln. Given,
A rectangular waveguide with internal dimensions
a = 4 cm
b = 3 cm
Mode of operation is TE11 .We have to find the minimum operating
frequency.
For any mode of operation the minimum frequency is the cut off
frequency of that mode. So we have to find the cut off frequency of
TE11 mode.
๐๐ =๐
๐โ(
๐
๐)๐
+ (๐
๐)๐
๐ญ๐๐ ๐ป๐ฌ๐๐ ๐ด๐๐ ๐
=๐ ร ๐๐๐
๐โ(
๐
๐)๐
+ (๐
๐)๐
= ๐. ๐ ร ๐๐๐๐ ร๐
๐๐ ๐ฏ๐
= ๐. ๐๐ ๐ฎ๐ฏ๐
Option (a)
11. The magnetic field along the propagation direction inside a rectangular
waveguide with the cross section shown in the figure is
๐ป๐ง = 3 cos(2.094 ร 102๐ฅ) cos(2.618 ร 102๐ฆ) cos(6.283 ร 1010 ๐ก โ ๐ฝ ๐ง)
The phase velocity vP of the wave inside the waveguide satisfies
X
Y
1.2 cm
3 cm
(a) ๐ฃ๐ > ๐
(b) ๐ฃ๐ = ๐
(c) 0 < ๐ฃ๐ < ๐
(d) ๐ฃ๐ = ๐
[GATE 2012: 2 Marks]
Soln. For the rectangular waveguide as per the given figure
a = 3 cm , b = 1.2 cm
๐ป๐ง = 3 cos(2.094 ร 102๐ฅ) cos(2.618 ร 102๐ฆ) cos(6.283 ร 1010 ๐ก โ ๐ฝ ๐ง)
From above equation
๐ = ๐. ๐๐๐ ร ๐๐๐๐ ๐๐๐ /๐๐๐
๐ =๐
๐๐ =
๐. ๐๐๐ ร ๐๐๐๐
๐๐ = ๐๐ ๐ฎ๐ฏ๐
๐๐
๐= ๐. ๐๐๐ ร ๐๐๐
๐๐, ๐
๐=
๐. ๐๐๐ ร ๐๐๐
๐ = ๐๐. ๐๐/๐
๐๐
๐= ๐. ๐๐๐ ร ๐๐๐
๐
๐=
๐. ๐๐๐ ร ๐๐๐
๐ = ๐๐. ๐๐/๐
๐๐ =๐
๐โ(
๐
๐)๐
+ (๐
๐)๐
=๐ ร ๐๐๐
๐โ(๐๐. ๐๐)๐ + (๐๐. ๐๐)๐
= 16 GHz
The wave with frequency of 10 GHz will not propagate through the
waveguide
Thus phase velocity of wave inside the waveguide will be 0
๐๐ = ๐
Option (d)
12. For a rectangular waveguide of internal dimensions ๐ ร ๐(๐ > ๐), the
cut โ off frequency for the TE11 mode is the arithmetic mean of the cut โ
off frequencies for TE10 mode and TE20 mode. If ๐ = โ5 cm. the value
of b (in cm) is --------.
[GATE 2014: 2 Marks]
Soln. Cut off frequency is given by
๐๐ =๐
๐๐ [(
๐๐
๐)๐
+ (๐๐
๐)๐
]
๐/๐
For TE10 mode, m = 1 and n = 0
๐๐๐๐=
๐
๐๐ . [
๐ ๐
๐๐]
๐/๐
=๐
๐๐
๐๐๐๐=
๐
๐๐ .โ(
๐๐
๐)๐
=๐
๐
Where
๐ =๐
โ๐๐ โ๐
Given, ๐ = โ๐ ๐๐
Arithmetic mean
=๐
๐(
๐
๐๐+
๐
๐) =
๐๐
๐๐
๐๐๐๐=
๐
๐๐ [(
๐
๐)๐
+ (๐
๐)๐
]
๐/๐
=๐๐
๐โ๐
๐๐, ๐
๐[(
๐
โ๐)๐
+ (๐
๐)๐
]
๐/๐
=๐๐
๐โ๐
๐๐, [๐
๐+
๐
๐๐]๐/๐
=๐
๐โ๐
๐๐, ๐
๐+
๐
๐๐=
๐
๐ ร ๐
๐๐, ๐
๐๐=
๐
๐๐โ
๐
๐=
๐ โ ๐
๐๐=
๐
๐๐=
๐
๐
Or, b = 2 cm
13. The longitudinal component of the magnetic field inside an air โ filled
rectangular waveguide made of a perfect electric conductor is given by
the following expression
๐ป๐ง(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0.1 cos(25๐๐ฅ) cos(30.3 ๐๐ฆ) cos(12๐ ร 109๐ก โ ๐ฝ๐ง) (๐ด/๐)
The cross โ sectional dimensions of the waveguide are given as a = 0.08
m and b = 0.033 m. The mode of propagation inside the waveguide is
(a) TM12
(b) TM21
(c) TE21
(d) TE12
[GATE 2015: 2 Marks]
Soln. In the problem longitudinal component of magnetic field is given, so
the wave is transverse electric i.e. modes will be of TE type
๐ฌ๐ = ๐ ๐๐๐ ๐ฏ๐ โ ๐
So mode will be TEmn
Given,
a = 0.08m , b = 0.033m
From the given field equation
๐๐
๐๐ = ๐๐ ๐ ๐
๐๐, ๐ = ๐๐๐
= ๐๐ ร ๐. ๐๐๐
= 2
๐๐
๐๐ = ๐๐. ๐๐
๐๐, ๐ = (๐๐. ๐) ร ๐ = ๐๐. ๐ ร ๐. ๐๐๐
๐๐ ๐ = ๐
So, the mode is
TE21
Option (c)
14. An air โ filled rectangular waveguide of internal dimension a ๐๐ ร
๐ ๐๐ (๐ > ๐) has a cut off frequency of 6 GHz for the dominant TE10
mode. For the same waveguide, if the cutoff frequency of the TM11 mode
is 15 GHz, the frequency of the TE01 mode GHz is ____________
[GATE 2015: 2 Marks]
Soln. Dimensions of waveguide
Wide dimension a cm
Narrow dimension b cm
Cut off frequency (๐๐) = ๐ ๐ฎ๐ฏ๐ for dominant mode
For dominant mode cut off wavelength (๐๐ช) is by 2a
๐. ๐. ๐๐ช๐๐= ๐๐
๐๐, ๐๐ช๐๐=
๐
๐๐๐๐=
๐ ร ๐๐๐๐
๐ ร ๐๐๐= ๐๐๐
๐๐ช๐๐= ๐๐ = ๐๐๐
So a = 2.5 cm
fc for TM11 mode is 15 GHz
๐. ๐. ๐
๐โ
๐
๐๐+
๐
๐๐= ๐๐ ร ๐๐๐
๐๐, ๐ =๐. ๐
โ๐. ๐๐
Now fc for TE01 mode is
๐๐ =๐
๐๐=
๐ ร ๐๐๐๐
๐. ๐ . โ๐. ๐๐
= 13.75 GHz
Answer 13.75 GHz
15. Consider an air โ filled rectangular waveguide with dimensions a = 2.286
cm and b = 1.016 cm. At 10 GHz operating frequency, the value of the
propagation constant (per meter) of the corresponding propagation mode
is __________
[GATE 2016: Marks]
Soln. Given,
Air filled rectangular waveguide
a = 2.286 cm
b = 1.016
f = 10 GHz
Assume dominant mode of propagation in the waveguide i.e. TE10
mode cut off frequency for TE10 mode is given by (for m = 1 and n =
0)
๐๐ =๐
๐โ(
๐
๐)๐
+ (๐
๐)๐
=๐
๐ .๐
๐
๐๐ =๐ ร ๐๐๐๐
๐ ร ๐. ๐๐๐= ๐. ๐๐ ๐ฎ๐ฏ๐
Since cut off frequency is 6.56 GHz, the frequency of 10 GHz will
propagate in the waveguide
Propagation constant
๐ธ = ๐ + ๐๐ท
If ๐ถ = ๐
๐ท is phase constant
๐ท =๐๐
๐๐=
๐๐
๐๐
โ๐ โ (๐๐
๐๐ชโ )
๐
=๐๐
๐๐
โ๐ โ (๐๐
๐๐ชโ )
๐
=๐๐
๐/๐โ๐ โ (
๐๐๐โ )
= ๐๐ .๐
๐ โ๐ โ (
๐๐๐โ )
๐
= ๐๐ ร๐๐ ร ๐๐๐
๐ ร ๐๐๐โ๐ โ (
๐. ๐๐
๐๐)๐
=๐๐
๐ร ๐๐๐ ร ๐. ๐๐๐๐ =
๐๐
๐ร ๐๐. ๐๐
๐ท = ๐๐๐ ๐โ๐
16. Consider an air โ filled rectangular waveguide with dimensions a = 2.286
cm and b = 1.016 cm. The increasing order of the cut โ off frequency for
different modes is
(a) TE01 < TE10 < TE11 < TE20
(b) TE20 < TE11 < TE10 < TE01
(c) TE10 < TE20 < TE01 < TE11
(d) TE10 < TE11 < TE20 < TE01
[GATE 2016: 2 Marks]
Soln. Waveguide dimensions are
a = 2.286 cm
b = 1.016 cm
waveguide is air filled
cut off frequency (๐๐) =๐ช
๐ โ(
๐
๐)๐+ (
๐
๐)๐
๐๐ ๐๐๐ ๐ป๐ฌ๐๐ ๐๐๐ ๐ = ๐
๐ โ
๐
๐๐+
๐
๐๐=
๐ ร ๐๐๐๐
๐โ
๐
(๐. ๐๐๐)๐+
๐
(๐. ๐๐๐)๐
= 16.15 GHz
๐๐ ๐๐๐ ๐ป๐ฌ๐๐ ๐๐๐ ๐ = ๐
๐๐=
๐ ร ๐๐๐๐
๐ ร ๐. ๐๐๐= ๐๐. ๐๐ ๐ฎ๐ฏ๐
๐๐ ๐๐๐ ๐ป๐ฌ๐๐ ๐๐๐ ๐ = ๐
๐=
๐ ร ๐๐๐๐
๐. ๐๐๐= ๐๐. ๐๐ ๐ฎ๐ฏ๐
๐๐ ๐๐๐ ๐ป๐ฌ๐๐ ๐๐๐ ๐ = ๐
๐๐=
๐ ร ๐๐๐๐
๐ ร ๐. ๐๐๐= ๐. ๐๐ ๐ฎ๐ฏ๐
Thus we find
Cut off frequency is given by
๐ป๐ฌ๐๐ < ๐ป๐ฌ๐๐ < ๐ป๐ฌ๐๐ < ๐ป๐ฌ๐๐
Thus, Option (c)