Rectangular Waveguide Questions and Answers

Ques.11. TM01 mode in rectangular perfect metallic waveguide is:

  1. Propagating mode
  2. Evanescent mode
  3. Non-existent mode
  4. None of the above

Answer.3. Non-existent mode

Explanation:

TM01 mode in the rectangular perfect metallic waveguide is a Non-existent mode. The condition where m = 0 or n = 0 is excluded as the field intensities all vanish and hence no TM01, or TM10 mode exists in a rectangular waveguide.

 

Ques.12. Degenerate mode in rectangular waveguides is:

  1. TMmn (m = 0, n = 1)
  2. TEmn and TMmn (both m ≠ 0)
  3. TEmn and TMmn (both m ≠ 0, n ≠ 0)
  4. TEmn (m = 1, n = 0)

Answer.3. TEmn and TMmn (both m ≠ 0, n ≠ 0)

Explanation:

Degenerate Mode: Two or more modes having the same cut-off frequency are said to be degenerate modes, e.g. the corresponding TEmn and TMmn modes in the rectangular waveguide are always degenerate.

In a rectangular waveguide, TEmn and TMmn modes are always degenerate.

The above cutoff frequency is the same for both TE and TM waves.

For TE waves, Ez = 0 and TM waves have Hz = 0.

 

Ques.13. A microstrip line consists of a single ground plane and a thin strip conductor on a

  1. Conducting plane
  2. Semiconductor slab
  3. Low-loss dielectric substrate
  4. High-loss dielectric substrate

Answer.3. Low-loss dielectric substrate

Explanation:

A microstrip line consists of a single ground plane and a thin strip conductor on a low-loss dielectric substrate above the ground plate. Since the size of the microwave solid-state devices is very small (of the order 0.008 — 0.08 mm3), the technique of signal input to these devices and extracting output power from them uses microstrip lines on the surface on which they can be easily mounted.

 

Ques.14. The wavelength of a wave in a waveguide is

  1. Is greater than in free space
  2. Depends only on the waveguide dimensions and the free-space wavelength
  3. Is inversely proportional to the phase velocity
  4. Is directly proportional to the group velocity

Answer.1. Is greater than in free space

Explanation:

In the regime of loss-free wave propagation in the waveguide, the wavelength is always larger than the wavelength in free space. This means that the phase velocity of the wave within the waveguide is greater than the speed of light.

The wavelength in a waveguide is considered as a wavelength in the direction of wave propagation and its dependence on wave frequency is defined as follows:

${\lambda _x} = \frac{{{\lambda _0}}}{{\sqrt {1 – {{\left( {\frac{{{\lambda _0}}}{{{\lambda _c}}}} \right)}^2}} }}$

Where λ0 is a wavelength in free space at a given frequency and λc stands for the cutoff wavelength for given waveguide dimensions and waveguide mode.

The wavelength of a wave in a waveguide is greater than in free space. The phase velocity is greater than the speed of light.

 

Ques.15. 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?

  1. The TM10 mode of the waveguide does not exist
  2. The TE10 mode of the waveguide does not exist
  3. The TM10 and the TE10 modes both exist and have the same cut-off frequencies
  4. The TM10 and the TM01 modes both exist and have the same cut-off frequencies

Answer.1. The TM10 mode of the waveguide does not exist

Explanation:

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. The condition where m = 0 or n = 0 is excluded as the field intensities all vanish and hence no TM01 , or TM10 mode exists in a rectangular waveguide.

 

Ques.16. ________ of a wave is the velocity with which variations in the shape of modulation or envelope of the wave propagate through space.

  1. The elliptical velocity
  2. The phase velocity
  3. The group velocity
  4. The circular velocity

Answer.3. The group velocity

Explanation:

Group Velocity: The group velocity of a wave is the velocity with which the overall envelope shape of the wave’s amplitudes propagates through space.

 

Ques.17. For a rectangular waveguide of internal dimensions a × b (a > b), the cut-off frequency for the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE20 mode. If a = √5 cm, the value of b (in cm) is _______.

  1. 4 cm
  2. 3 cm
  3. 10 cm
  4. 2 cm

Answer.4. 2 cm

Explanation:

The cutoff frequency of TE11 mode in the rectangular waveguide is given as:

$\begin{array}{l} {f_{c\left( {11} \right)}} = \frac{c}{{2\pi }}\sqrt {{{\left( {\frac{\pi }{a}} \right)}^2} + {{\left( {\frac{\pi }{n}} \right)}^2}} \\ \\ = \frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \: \end{array}$

Given:

${f_c}\left( {T{E_{11}}} \right) = \frac{{{f_c}\left( {T{E_{10}}} \right) + {f_c}\left( {T{E_{20}}} \right)}}{2}$

$\begin{array}{l} \frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{{\frac{c}{{2a}} + \frac{c}{a}}}{2}\\ \\ \sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{1}{{2a}} + \frac{1}{a} = \frac{3}{{2a}}\\ \\ {\left( {\frac{1}{a}} \right)^2} + {\left( {\frac{1}{b}} \right)^2} = \frac{9}{{4{a^2}}} \end{array}$

a = √5

$\begin{array}{l} {\left( {\frac{1}{b}} \right)^2} = \frac{5}{{4{a^2}}}\\ \\ b = \frac{{2a}}{{\sqrt 5 }} \end{array}$

b = 2 cm

 

Ques.18. The cut-off frequency of TEM wave is

  1. Zero
  2. 11.0 GHz
  3. Moderate

Answer.1. Zero

Explanation:

  • The cutoff frequency of TEM wave is zero.
  • All electromagnetic waves consist of electric and magnetic fields propagating in the same direction of travel, but perpendicular to each other.
  • Along the length of a normal transmission line, both electric and magnetic fields are perpendicular (transverse) to the direction of wave travel. This is known as the principal mode, or TEM (Transverse Electric and Magnetic) mode.
  • This mode of wave propagation can exist only where there are two conductors, and it is the dominant mode of wave propagation where the cross-sectional dimensions of the transmission line are small compared to the wavelength of the signal.

 

Ques.19. A rectangular waveguide with air medium has dimensions a = 22.86 mm and b = 10.6 mm is fed by 3 GHz carrier from a coaxial cable, which of the following is a false statement for TE01 mode?

  1. Propagating mode
  2. Non-propagating mode
  3. Propagating mode in case filled fully with dielectric material of proper dielectric constant
  4. None of the above

Answer.1. Propagating mode

Explanation:

The lowest propagating mode is TE10 cutoff frequency

fc = c/2a

a = c/2fc

Given:

a = 22.86 mm and b = 10.6 mm

As a > b so,

fc = (3 × 108)/(2 × 22.86 mm)

fc = 6.5 GHz

As input applied frequency is less than the cut-off frequency, the mode is non-propagating. The mode can be made propagating if dielectric with a proper dielectric constant is inserted because it decreases phase velocity.

 

Ques.20. 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:

  1. 6.25 GHz
  2. 6.0 GHz
  3. 5.0 GHz
  4. 3.75 GHz

Answer.1. 6.25 GHz

Explanation:

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:

$\begin{array}{l} {f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ – 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ – 2}}} \right)}^2}}}} \\ \\ {f_c} = \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ – 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz \end{array}$

fc = 6.25 GHz

 

Ques.21. In a circular waveguide, the dominated mode is

  1. TE01
  2. TE11
  3. TE20
  4. TE21

Answer.2. TE11

Explanation:

  • A circular waveguide is a tubular, circular conductor. A plane wave propagating through circular waveguide results in transverse electric (TE) or transverse magnetic (TM) mode.

The lowest mode in a circular waveguide is the TE11 or H11 mode, which has a cutoff wavelength λCTE11 = 3.41a. This mode is seen to be the dominant mode for the circular waveguide. The field map of the TE11 mode in a circular waveguide is similar to that of the TE10 mode in a rectangular waveguide. The circular waveguide is a rotational symmetric structure, but the field map of the TE11 mode is not rotational symmetric.

 

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