Rectangular Waveguide Questions and Answers

Answer.1. Less than the phase velocity only

Explanation:

In a dispersive medium, the group velocity is less than the phase velocity only. The phase velocity is always greater than the speed of light and group velocity is always less than the speed of light. Hence, the group velocity is less than the phase velocity

Answer.3. EM wave of frequency 4.5 GHz

Explanation:

The dominant mode in a particular waveguide is the mode having the lowest cutoff frequency. For rectangular waveguides, this is the TE10 mode.

For TE₁₀ mode:

f_c = c/2a

Given:

a = 30 mm

b = 15 mm

For TE₁₀ mode,

f_c = (3 × 10⁸)/(2 × 10⁻³)

f_c = 5 GHz

So, the waves having the frequency above cutoff frequency (f_c) will only pass.

Hence from the options given 4.5 GHz will not pass.

Answer.1. High pass filter

Explanation:

Waveguides only allow frequencies above cut-off frequency and do not pass below the cut-off frequencies. Hence it acts as a high pass filter. Below the cut-off frequency, the propagation decays exponentially, yielding the evanescent wave, and above it the wave propagates without attenuation for a perfect waveguide. The waveguide is a very effective barrier to low-frequency waves and thus acts as a high pass filter.

Answer.3. Reflection off

Explanation:

A rectangular waveguide is a hollow metal tube with a rectangular cross-section. The conducting walls of the waveguide confine the electromagnetic fields and thereby guide the electromagnetic wave. Microwave energies propagate the length of the waveguide by the reflection of its sidewalls.

Answer.4. 15 mm

Explanation:

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

TE₁₀ mode means m = 1, n = 0

The cut-off frequency of the dominant mode TE10 of the rectangular waveguide is:

f_c = c/2a

Where a is the dimension of the inner broad wall

a = (3 × 10¹⁰)/(2 × 10 × 10⁹)

a = 1.5 cm

a = 15 mm

Answer.2. 7810.24 MHz

Explanation:

The cut-off frequency for a rectangular waveguide is given by:

${f_{m,\;n}} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

C = velocity of light in free space.

m, n = mode of operation of the waveguide

a, b = dimensions of the waveguide

Calculation:

For TE_{2, 1} mode, m = 2 and n = 1

Also, a = 5 cm

b = 3 cm

The cut-off frequency will be:

$\begin{array}{l} {f_{c\left( {2,\;1} \right)}} = \frac{{3\: \times \:{{10}^{10}}}}{2}\sqrt {{{\left( {\frac{2}{5}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^2}} \\ \\ = 1.5 \times {10^{10}}\sqrt {0.16 + 0.11} Hz \end{array}$

= 1.5 × 10¹⁰ × 0.52

f_{c(2, 1)} = 7810.24 MHz

Answer.1. 3.78 × 10⁸ m/s

Explanation:

Phase velocity is given by:

V_P = c/cosθ

Where:

$\cos \theta = \sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}}$

f_c = cut-off frequency

f = frequency of operation

c = speed of light

Calculation:

For the given TE₁₀ mode, the cut-off frequency will be:

f_c = c/2a

Where a is the dimension of the inner broad wall

With a = 7 cm (Given)

f_c = (3 × 10¹⁰)/(2 × 7)

f_c = 2.142 GHz

f = 3.5 GHz

$\cos \theta = \sqrt {1 – {{\left( {\frac{{2.142}}{{3.5}}} \right)}^2}} = 0.79$

V_P = 3 × 10⁸/0.79

= 3.78 × 10⁸ m/s

Answer.4. zero

Explanation:

Waveguides only allow frequencies above the cut-off frequency to pass through. It blocks or attenuates the frequencies below the cut-off frequencies, i.e. there is no attenuation near the cut-off frequency. It is only for frequencies greater than or equal to the cut-off frequency, the attenuation is very high.

Every waveguide functions as a high-pass filter with a certain cut-off frequency f_c. Frequencies below the f_c cannot pass through it while all frequencies above fc pass through unattenuated, in the ideal case. The cut-off frequency of a waveguide depends on the size of its cross-section and its shape. Multiple modes can propagate along a line, and the one with the smallest cut-off frequency is called the dominant mode. Usually, only the dominant mode is used for signal transfer. We can choose the dimensions of a waveguide so as to allow only the dominant mode to exist. Waveguides can be identified as circular, square, rectangular, and elliptical waveguides. The cut-off frequency decreases as the size of the waveguide increases.

Answer.4. TE₁₀

Explanation:

Every waveguide functions as a high-pass filter with a certain cut-off frequency f_c. Frequencies below the f_c cannot pass through it while all frequencies above fc pass through unattenuated, in the ideal case. The cut-off frequency of a waveguide depends on the size of its cross-section and its shape.

Multiple modes can propagate along a line, and the one with the smallest cut-off frequency is called the dominant mode. The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency. Usually, only the dominant mode is used for signal transfer. The dominant mode in the rectangular waveguides is TE₁₀ mode. We can choose the dimensions of a waveguide so as to allow only the dominant mode to exist. Waveguides can be identified as circular, square, rectangular, and elliptical waveguides. The cut-off frequency decreases as the size of the waveguide increases.

Choice of TE₁₀  mode in a rectangular waveguide:

In a rectangular waveguide, the transmission of microwave energy occurs mainly through TE₁₀ mode owing to the following reasons:

(a) The excitation of TE₁₀ mode is much easier as compared to all other modes

(b) For transmission of higher-order modes, large waveguide dimensions (a x b) are required. Hence to reduce the waveguide dimensions, it is preferable to go to a lower order mode. the best one is TE₁₀, which gives the smallest waveguide dimension.

(c) TE₁₀ mode has the lowest attenuation.

(d) The electric field is definitely polarized in one direction everywhere.