Rectangular Waveguide MCQ Quiz in தமிழ் - Objective Question with Answer for Rectangular Waveguide - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 17, 2025
Latest Rectangular Waveguide MCQ Objective Questions
Top Rectangular Waveguide MCQ Objective Questions
Rectangular Waveguide Question 1:
Electromagnetic waves are transverse in nature due to
Answer (Detailed Solution Below)
Rectangular Waveguide Question 1 Detailed Solution
Maxwell was able to establish that electromagnetic waves possess the following properties:
- The magnetic field oscillates in phase with the electric field. In other words, a wave maximum of the magnetic field always coincides with a wave maximum of the electric field in both time and space.
- The electric field is always perpendicular to the magnetic field, and both fields are directed at right-angles to the direction of propagation of the wave. In fact, the wave propagates in the direction E×B. Electromagnetic waves are clearly a type of transverse wave.
- For a z-directed wave, the electric field is free to oscillate in any direction which lies in the x-y plane. The direction in which the electric field oscillates is conventionally termed the direction of polarization of the wave.
- The maximum amplitudes of the electric and the magnetic fields are related via E0=cB0.
There is no constraint on the possible frequency or wavelength of electromagnetic waves. However, the propagation velocity of electromagnetic waves is fixed and takes the value:
\(c=\frac{1}{{\sqrt {\mu_0\epsilon_0}}}\)
Rectangular Waveguide Question 2:
A square waveguide carries TE11 mode whose axial magnetic field is given by \({{H}_{z}}={{H}_{0}}\cos \left( \frac{\pi x}{\sqrt{8}}~ \right)\cos \left( \frac{\pi y}{\sqrt{8}} \right)\)A/m, where wave guide dimensions are in cm. What is the cut-off frequency of the mode?
Answer (Detailed Solution Below)
Rectangular Waveguide Question 2 Detailed Solution
Concept:
For a TEMN mode of Electromagnetic wave;
\({{\text{H}}_{\text{z}}}={{\text{H}}_{\text{z}0}}\cos \left( \frac{m\pi }{a}.x \right)\cos \left( \frac{n\pi }{b}.y \right){{e}^{-\gamma z}}.~{{e}^{j\omega z}}~.~{{\hat{a}}_{z}}\)
Calculation:
Comparing the given equation with the standard equation;
m = 1, and n = 1.
\({{H}_{z}}={{H}_{0}}\cos \left( \frac{\pi x}{\sqrt{8}} \right)\cos \left( \frac{\pi y}{\sqrt{8}} \right)~A/m~\)
Given a = b = √8 cm.
\({{f}_{c}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{1}{b} \right)}^{2}}}=\frac{c}{2}\times \frac{1}{a}\times \sqrt{2}\)
\({{f}_{C}}=\frac{3\times {{10}^{10}}}{2}\times \frac{1}{\sqrt{8}}\times \sqrt{2}\)
\({{f}_{C}}=\frac{3\times {{10}^{10}}\times \sqrt{2}}{2\times 2\sqrt{2}}=0.75\times {{10}^{10}}~cm\)
fc = 7.5 GHz.Rectangular Waveguide Question 3:
In a rectangular waveguide with length ‘a’ and height ‘b’, the dominant TE mode for b > a will be:
Answer (Detailed Solution Below)
Rectangular Waveguide Question 3 Detailed Solution
Concept:
The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.
The cut-off frequency for a rectangular waveguide with dimension ‘a’ and ‘b’ is given as:
\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)
Calculation:
With b > a,
The minimum frequency is obtained when m = 0 and n = 1, i.e.
\({{f}_{c\left( 01 \right)}}=\frac{c}{2}\sqrt{\frac{{{n}^{2}}}{{{b}^{2}}}}=\frac{c}{2b}\)
So, the dominant mode for the given rectangular wave-guide is TE01
Common Mistake:
- It is a common mistake to answer TE10 for the dominant mode of Rectangular waveguide.
- TE10 is the dominant mode when the length(a) is greater than the height(b), which is taken as the default case.
- But here in the question, it has mentioned that b > a, so TE01 will be the dominant mode as explained in the solution.
Rectangular Waveguide Question 4:
Which type of waveguide are shown in the figure.
Answer (Detailed Solution Below)
Rectangular Waveguide Question 4 Detailed Solution
A waveguide with conducting ridges protruding into the center of the waveguide from the top wall or bottom wall or both walls, is called as a Ridged Waveguide.
Single Ridged Waveguides:
A rectangular waveguide with a single protruding ridge from the top or bottom wall is called a Single Ridged Waveguides.
Different types of waveguides are shown in the figure below:
Rectangular Waveguide Question 5:
If over the course of a day, the maximum electron density in the ionosphere varies from 1011 to 1012 m-3, the critical frequency changes approximately from:
Answer (Detailed Solution Below)
Rectangular Waveguide Question 5 Detailed Solution
Concept:
The limiting frequency at or below which the wave component is reflected and above which it penetrates through an ionospheric layer depends upon the electron density of the ionosphere:
\({{f}_{c}}=9\sqrt{{{N}_{max}}}\)
Where fc is in Hz.
and Nmax = Maximum electron density/m3
Calculation:
Given, Nmax1 = 1011
So, \({{f}_{c1}}=9\sqrt{{{10}^{11}}}=9\times {{10}^{5}}\sqrt{10}\)
\(=0.9\sqrt{10}~MHz\)
= 2.85 MHz
Similarly,
Nmax2 = 1012
So, \({{f}_{c2}}=9\sqrt{{{N}_{max2}}}=9\sqrt{{{10}^{12}}}=9\times {{10}^{6}}Hz\)
fc2 = 9 MHz
So, the critical frequency changes approximately from 2.85 MHz to 9 MHz
Option (3) is the correct option.Rectangular Waveguide Question 6:
A waveguide of dimensions a = 15 mm and b = 7.5 mm is used as a high-pass filter. If the stopband attenuation required at 8 GHz is ~109.2 dB, what is the length of the filter?
(assume conductor losses to be zero, approximate π = 3.14 and 1 Np ~ 8.69 dB) (log10 e = 0.4343)
Answer (Detailed Solution Below)
Rectangular Waveguide Question 6 Detailed Solution
Concept:
The propagation constant γ̅ for rectangular waveguides is given by:
\(\bar{\gamma }=\sqrt{{{\left( \frac{m\pi }{a} \right)}^{2}}+{{\left( \frac{n\pi }{b} \right)}^{2}}-{{\omega }^{2}}{{\mu }_{0}}{{\epsilon }_{0}}}\)
If, \({{\left( \frac{m\pi }{a} \right)}^{2}}+{{\left( \frac{n\pi }{b} \right)}^{2}}>{{\omega }^{2}}{{\mu }_{0}}{{\epsilon }_{0}}\)
then α will become real and it will attenuate. (γ̅ = α)
Calculation:
Given, a = 15 mm, b = 7.5 mm
fc (Cut-off frequency) for \(T{{E}_{10}}=\frac{c.m}{2a}=\frac{c}{2\times 15~m}\)
\({{f}_{c}}=\frac{3\times {{10}^{8}}}{30\times {{10}^{-3}}}=10~GHz\)
The waveguide will act as a high pass filter, i.e. above 10 GHz it will allow the signal to pass and below, it will attenuate it.
For dominant mode \(T{{E}_{10}},\)
\(\bar{\gamma }=\sqrt{{{\left( \frac{\pi }{a} \right)}^{2}}-{{\omega }^{2}}{{\mu }_{0}}{{\epsilon }_{0}}}\)
\(\bar{\gamma }=\sqrt{{{\left( \frac{\pi }{15\times {{10}^{-3}}} \right)}^{2}}-{{\left( 2\pi \times 8\times {{10}^{9}} \right)}^{2}}\times \frac{1}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}}\text{ }\!\!~\!\!\text{ }\)
\(\because \frac{1}{\sqrt{{{\mu }_{0}}{{\epsilon }_{0}}}~}=3\times {{10}^{8}}~m/s \)
\(\Rightarrow \bar{\gamma }=\pi \sqrt{{{\left( \frac{10\times 100}{15} \right)}^{2}}-\left( \frac{4\pi \times 64\times {{10}^{2}}}{9} \right)}\)
\(\Rightarrow \bar{\gamma }=10\pi \sqrt{{{\left( \frac{100}{15} \right)}^{2}}-\left( \frac{256}{9} \right)}\)
\(=10\pi \sqrt{44.4-28.4}\)
\( \bar{\gamma }=10\pi \sqrt{16}=40~\pi \), which is real.
So, γ̅ = α = 40 π
Given, Attenuation at 8 GHz = -109.2 dB
If the input field is E0, then the field after travelling a distance z will be E0e-αz
Here, z distance = length of filter/waveguide
Attenuation \(=-109.2~dB=20\log \left( \frac{{{E}_{0}}{{e}^{-\alpha z}}}{{{E}_{0}}} \right)\)
\(\Rightarrow -109.2=-\alpha .z.20{{\log }_{10}}e~\)
\(\Rightarrow \frac{109.2}{8.69}=\alpha z\)
αz = 12.56
\(z=\frac{12.56}{\alpha }=\frac{12.56}{40~\pi }metre\)
\(z=\frac{12.56}{40\times 3.14}=0.1~m\)
= 100 mm
Rectangular Waveguide Question 7:
At cut-off frequency, the phase velocity ‘f’ of a waveguide is:
Answer (Detailed Solution Below)
Rectangular Waveguide Question 7 Detailed Solution
Concept:
Phase velocity of a Rectangular waveguide is defined as:
\({v_p} = \frac{\omega }{\beta } = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\)
Also, \({v_p} = \frac{c}{{\cos \theta }};\) where θ is the angle with which the wave enters the waveguide as shown:
Calculation:
Given, f = fc
So, \({v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }} = \frac{c}{{\sqrt {1 - 1} \;}} = \frac{c}{0} = \infty \)
Implication:
vp = ∞,
means that cos θ = 0, i.e. θ = 90°.
This situation is shown as below:
Rectangular Waveguide Question 8:
A rectangular wave guide is air filled and operated at \(30{\rm{GHz}}\).The cuttoff frequency of the \(\rm{TM_{21}}\) mode is \(\text{18 GHz}\). The intrinsic impedance of the waveguide in this mode of operation is _______\({\rm{\Omega }}\).
Answer (Detailed Solution Below) 300 - 302
Rectangular Waveguide Question 8 Detailed Solution
We have \(\cos {\rm{\theta }} = \sqrt {1 - {{\left( {\frac{{{{\rm{f}}_{\rm{c}}}}}{{\rm{f}}}} \right)}^2}} = \sqrt {1 - {{\left( {\frac{{{{\rm{18}}{\rm{}}}}}{{\rm{30}}}} \right)}^2}} = 0.8\)
Now, \({{\rm{\eta }}_{{\rm{TM}}21}} = {{\rm{\eta }}_0}*\cos {\rm{\theta }} = 120{\rm{\pi }}*0.8\)
\(\Rightarrow {{\rm{\eta }}_{{\rm{TM}}21}} = 301.44{\rm{\Omega }}\)
Rectangular Waveguide Question 9:
The modes in a rectangular waveguide are denoted by TEmn/TMmn where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?
Answer (Detailed Solution Below)
Rectangular Waveguide Question 9 Detailed Solution
For a TEmn mode to exist, We need to have m,n = 0,1,2,......
but does not exist for m = n = 0. For all other combinations TEmn exists.
For a TMmn mode to exist, We need to have m,n = 1,2,3, ....... i.e m & n both must be non-zero
⇒ TM01, TM10 modes do not exist
Important Points Waveguides only allow frequencies above the cut-off frequency to pass through. It blocks or attenuates the frequencies below the cut-off frequencies.
The cut off frequency is mathematically calculated as:
\({{f}_{c\left( min \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)
Where a and b are the dimensions of the waveguide
m and n are mode numbers TEmn.
Rectangular Waveguide Question 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:
Answer (Detailed Solution Below)
Rectangular Waveguide Question 10 Detailed Solution
Concept:
Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:
\({f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}} \)
a = length of the waveguide
b = height of the waveguide
m,n = modes of operation
Calculation:
Given, a = 4 cm
b = 3 cm
The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:
\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}} \)
\(f_c= \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz\)
fC = 6.25 GHz