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:

$c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$

Concept:

For a TE_MN mode of Electromagnetic wave;

${\text{H}}_{\text{z}}={\text{H}}_{\text{z}0}\cos (\frac{m\pi }{a}.x)\cos (\frac{n\pi }{b}.y)e^{-\gamma z}.e^{j\omega z}.{\hat{a}}_z$

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:

$\overline{\gamma }=\sqrt{{\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2-{\omega }^2{\mu }_0{ϵ}_0}$

If, ${\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2>{\omega }^2{\mu }_0{ϵ}_0$ 

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 $T{E}_{10}=\frac{c.m}{2a}=\frac{c}{2\times 15m}$

$f_c=\frac{3\times {10}^8}{30\times {10}^{-3}}=10GHz$

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 $T{E}_{10},$

$\overline{\gamma }=\sqrt{{\left(\frac{\pi }{a}\right)}^2-{\omega }^2{\mu }_0{ϵ}_0}$

$\overline{\gamma }=\sqrt{{\left(\frac{\pi }{15\times {10}^{-3}}\right)}^2-{(2\pi \times 8\times {10}^9)}^2\times \frac{1}{{(3\times {10}^8)}^2}}$

$\because \frac{1}{\sqrt{{\mu }_0{ϵ}_0}}=3\times {10}^{8}m/s$

$\Rightarrow \overline{\gamma }=\pi \sqrt{{\left(\frac{10\times 100}{15}\right)}^2-\left(\frac{4\pi \times 64\times {10}^2}{9}\right)}$

$\Rightarrow \overline{\gamma }=10\pi \sqrt{{\left(\frac{100}{15}\right)}^2-\left(\frac{256}{9}\right)}$

$=10\pi \sqrt{44.4-28.4}$

$\overline{\gamma }=10\pi \sqrt{16}=40\pi$, 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 $=-109.2dB=20\log \left(\frac{E_0{e}^{-\alpha z}}{E_0}\right)$

$\Rightarrow -109.2=-\alpha .z.20{\log }_{10}e$ 

$\Rightarrow \frac{109.2}{8.69}=\alpha z$

αz = 12.56

$z=\frac{12.56}{\alpha }=\frac{12.56}{40\pi }metre$

$z=\frac{12.56}{40\times 3.14}=0.1m$

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

Implication:

v_p = ∞,

means that cos θ = 0, i.e. θ = 90°.

This situation is shown as below:

We have $\cos \theta =\sqrt{1-{\left(\frac{{\mathrm{f}}_{\mathrm{c}}}{\mathrm{f}}\right)}^2}=\sqrt{1-{\left(\frac{18}{30}\right)}^2}=0.8$

Now, ${\eta }_{\mathrm{T}\mathrm{M}21}={\eta }_0\ast \cos \theta =120\pi \ast 0.8$

$\Rightarrow {\eta }_{\mathrm{T}\mathrm{M}21}=301.44\mathrm{\Omega }$

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