Rectangular Waveguide MCQ Quiz in मल्याळम - Objective Question with Answer for Rectangular Waveguide - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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:

\(c=\frac{1}{{\sqrt {\mu_0\epsilon_0}}}\)

Concept:

For a TE_MN 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.

\({{f}_{c}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{1}{b} \right)}^{2}}}=\frac{c}{2}\times \frac{1}{a}\times \sqrt{2}\)

\({{f}_{C}}=\frac{3\times {{10}^{10}}}{2}\times \frac{1}{\sqrt{8}}\times \sqrt{2}\)

\({{f}_{C}}=\frac{3\times {{10}^{10}}\times \sqrt{2}}{2\times 2\sqrt{2}}=0.75\times {{10}^{10}}~cm\)

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.

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:

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 f_c is in Hz.

and N_max = Maximum electron density/m³

Calculation:

Given, N_max1 = 10¹¹

So, \({{f}_{c1}}=9\sqrt{{{10}^{11}}}=9\times {{10}^{5}}\sqrt{10}\)

\(=0.9\sqrt{10}~MHz\)

= 2.85 MHz

Similarly,

N_max2 = 10¹²

So, \({{f}_{c2}}=9\sqrt{{{N}_{max2}}}=9\sqrt{{{10}^{12}}}=9\times {{10}^{6}}Hz\)

f_c2 = 9 MHz

So, the critical frequency changes approximately from 2.85 MHz to 9 MHz

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

f_c (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

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 = f_c

So, \({v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }} = \frac{c}{{\sqrt {1 - 1} \;}} = \frac{c}{0} = \infty \)

Implication:

v_p = ∞,

means that cos θ = 0, i.e. θ = 90°.

This situation is shown as below:

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, \({{\rm{\eta }}_{{\rm{TM}}21}} = {{\rm{\eta }}_0}*\cos {\rm{\theta }} = 120{\rm{\pi }}*0.8\)

\(\Rightarrow {{\rm{\eta }}_{{\rm{TM}}21}} = 301.44{\rm{\Omega }}\)

For a TE_mn mode to exist, We need to have  m,n = 0,1,2,......

but does not exist for m = n = 0. For all other combinations TE_mn exists.

For a TM_mn 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.

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:

\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}} \)

\(f_c= \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz\)

f_C = 6.25 GHz