Rectangular Waveguides Questions and Answers

Q: The dominant mode of a rectangular wave guide is :

The dominant mode is TE10.The dominant mode of a waveguide is the mode with the lowest cut-off frequency. For a rectangular guide of width a (greater than height b), the cut-off wavelength of the TEmn mode is:\[ \lambda_{c} = \frac{2}{\sqrt{(m/a)^{2} + (n/b)^{2}}} \]It is longest, \(\lambda_{c} = 2a\), for m = 1, n = 0, so TE10 propagates first. TE11, TE12 and TE21 have higher cut-off frequencies. (TM modes need both m and n non-zero, so TM11 is the lowest TM mode.)

Q: In a microwave system, wave guides has the advantage of :

The advantage of waveguides is high power handling capability and low loss.A waveguide is a hollow metal pipe with no inner conductor and no solid dielectric. With a large conducting surface and air inside, conductor losses are low, there is no dielectric loss, and the breakdown limit is high, so waveguides carry high microwave power with little attenuation, for example in radar transmitters.Phase shift is not an advantage of waveguides as such. A thin dielectric substrate describes microstrip, not a waveguide. Coupling to coaxial lines is a practical necessity handled by adapters, not an advantage.

Q: Cut-off wavelength for TE01 mode for a ractangular wave guide with cross-section a × b (where 'a' is the broader side) is given by :

The cut-off wavelength for TE01 is 2b.For a rectangular waveguide the cut-off wavelength of a TEmn mode is\[ \lambda_{c} = \frac{2}{\sqrt{(m/a)^{2} + (n/b)^{2}}} \]For m = 0 and n = 1 this reduces to \(\lambda_{c} = 2b\), with the field varying across the narrower dimension b.2a is the cut-off wavelength of the dominant TE10 mode, which varies across the broader side. a and b alone are the cut-off wavelengths of the TE20 and TE02 modes.

Q: A wave guide section in microwave circuits will act as a :

A waveguide acts as a high pass filter.Each mode in a waveguide has a cut-off frequency, set by its dimensions. Below the cut-off of the dominant mode, waves do not propagate but decay rapidly along the guide; above it they pass with little loss. For a rectangular guide of width a, the dominant \(TE_{10}\) mode has\[ f_c = \frac{c}{2a} \]so the guide passes frequencies above \(f_c\) and blocks those below, which is high-pass behaviour, not low-pass, band-pass or band-stop.

Q: A 6 cm × 4 cm rectangular wave guide is filled with dielectric of refractive index 1.25. The range of frequencies over which single mode operation will occur is :

Single-mode operation occurs for 2 GHz \(\lt f \lt\) 3 GHz.In a guide filled with a dielectric of index \(n\), waves travel at \(v = c/n = 2.4 \times 10^{8}\) m/s. The cut-off frequencies are \(f_{mn} = \frac{v}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) with \(a = 6\) cm and \(b = 4\) cm:\[ \begin{aligned} f_{10} &= \frac{2.4 \times 10^{8}}{2 \times 0.06} \\ &= 2\ \text{GHz} \\ f_{01} &= \frac{2.4 \times 10^{8}}{2 \times 0.04} \\ &= 3\ \text{GHz} \end{aligned} \]Between these only TE10 propagates. The other ranges do not match these cut-offs; without the dielectric the window would be 2.5 to 3.75 GHz, which is not offered either.

Q: Consider an ideal rectangular wave guide (dimension ~ 4.0 cm × 2.0 cm) with vacuum inside. Plane monochromatic electromagnetic radiation having a wavelength of 6 cm is introduced in this wave guide. The radiation propagates inside the wave guide in mode :

The radiation propagates in Only TE10 mode.A mode propagates only if the free-space wavelength is shorter than its cut-off wavelength. For a guide with \(a = 4\) cm and \(b = 2\) cm:TE10: \(\lambda_{c} = 2a = 8\) cm, longer than 6 cm, so it propagates.TE01: \(\lambda_{c} = 2b = 4\) cm, shorter than 6 cm, so it is cut off.TM modes need both indices non-zero; TM10 does not exist, and TM11 has \(\lambda_{c} \approx 3.6\) cm.So only the dominant TE10 mode carries the 6 cm radiation.

Q: The modes in a rectangular waveguide are denoted by TEmn/TMmn, where m and n are the Eigen values along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is true ?

The true statement is that the TM\(_{10}\) mode does not exist.For TM modes in a rectangular waveguide, the longitudinal electric field is \(E_{z} \propto \sin(m\pi x/a)\sin(n\pi y/b)\). If either m or n is zero, \(E_{z}\) vanishes everywhere and the whole field is zero, so TM\(_{10}\), TM\(_{01}\) and the like cannot exist; the lowest TM mode is TM\(_{11}\).TE\(_{10}\) certainly exists: it is the dominant mode, with the lowest cut-off frequency. So TE\(_{10}\) and TM\(_{10}\) do not both exist. The cut-off frequency \(f_{c} = \frac{c}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) increases, not decreases, as m and n increase.

Direction for [ Question No: 1 To 1 ] :

Assertion-Reason type questions :

The following items consist of two statements, one labelled as ‘Assertion A’ and the other labelled the ‘Reason (R)’. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : The system of propagation in waveguides is in accordance to field theory.
Reason (R) : The system of propagation in transmission line is in accordance with circuit theory.
Codes :

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: B

Both statements are true: propagation in waveguides has to be treated by field theory (Maxwell's equations and modes), and propagation in a transmission line by circuit theory (voltage, current and distributed elements). But the second does not explain the first, since the two are separate statements about two different systems. So both are true and (R) is not the explanation of (A).

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The dominant mode of a rectangular wave guide is :

A.\(TE_{11}\)
B.\(TE_{12}\)
C.\(TE_{10}\)
D.\(TE_{21}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The dominant mode is TE₁₀.

The dominant mode of a waveguide is the mode with the lowest cut-off frequency. For a rectangular guide of width a (greater than height b), the cut-off wavelength of the TE_mn mode is:

\[ \lambda_{c} = \frac{2}{\sqrt{(m/a)^{2} + (n/b)^{2}}} \]

It is longest, \(\lambda_{c} = 2a\), for m = 1, n = 0, so TE₁₀ propagates first. TE₁₁, TE₁₂ and TE₂₁ have higher cut-off frequencies. (TM modes need both m and n non-zero, so TM₁₁ is the lowest TM mode.)

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In a microwave system, wave guides has the advantage of :

A.Positive phase shift
B.High power handing capability and low loss
C.Thin dielectric substrate
D.Coupling with co-axial wave guides

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The advantage of waveguides is high power handling capability and low loss.

A waveguide is a hollow metal pipe with no inner conductor and no solid dielectric. With a large conducting surface and air inside, conductor losses are low, there is no dielectric loss, and the breakdown limit is high, so waveguides carry high microwave power with little attenuation, for example in radar transmitters.

Phase shift is not an advantage of waveguides as such. A thin dielectric substrate describes microstrip, not a waveguide. Coupling to coaxial lines is a practical necessity handled by adapters, not an advantage.

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Cut-off wavelength for TE₀₁ mode for a ractangular wave guide with cross-section a × b (where 'a' is the broader side) is given by :

A.2b
B.a
C.2a
D.b

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The cut-off wavelength for TE₀₁ is 2b.

For a rectangular waveguide the cut-off wavelength of a TE_mn mode is

\[ \lambda_{c} = \frac{2}{\sqrt{(m/a)^{2} + (n/b)^{2}}} \]

For m = 0 and n = 1 this reduces to \(\lambda_{c} = 2b\), with the field varying across the narrower dimension b.

2a is the cut-off wavelength of the dominant TE₁₀ mode, which varies across the broader side. a and b alone are the cut-off wavelengths of the TE₂₀ and TE₀₂ modes.

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A wave guide section in microwave circuits will act as a :

A.Band stop filter
B.Low pass filter
C.High pass filter
D.Band pass filter

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

A waveguide acts as a high pass filter.

Each mode in a waveguide has a cut-off frequency, set by its dimensions. Below the cut-off of the dominant mode, waves do not propagate but decay rapidly along the guide; above it they pass with little loss. For a rectangular guide of width a, the dominant \(TE_{10}\) mode has

\[ f_c = \frac{c}{2a} \]

so the guide passes frequencies above \(f_c\) and blocks those below, which is high-pass behaviour, not low-pass, band-pass or band-stop.

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A 6 cm × 4 cm rectangular wave guide is filled with dielectric of refractive index 1.25. The range of frequencies over which single mode operation will occur is :

A.2.24 GHz \(\lt f \lt\) 3.33 GHz
B.2 GHz \(\lt f \lt\) 3 GHz
C.4.4 GHz \(\lt f \lt\) 70 GHz
D.4 GHz \(\lt f \lt\) 6 GHz

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Single-mode operation occurs for 2 GHz \(\lt f \lt\) 3 GHz.

In a guide filled with a dielectric of index \(n\), waves travel at \(v = c/n = 2.4 \times 10^{8}\) m/s. The cut-off frequencies are \(f_{mn} = \frac{v}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) with \(a = 6\) cm and \(b = 4\) cm:

\[ \begin{aligned} f_{10} &= \frac{2.4 \times 10^{8}}{2 \times 0.06} \\ &= 2\ \text{GHz} \\ f_{01} &= \frac{2.4 \times 10^{8}}{2 \times 0.04} \\ &= 3\ \text{GHz} \end{aligned} \]

Between these only TE₁₀ propagates. The other ranges do not match these cut-offs; without the dielectric the window would be 2.5 to 3.75 GHz, which is not offered either.

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Consider an ideal rectangular wave guide (dimension ~ 4.0 cm × 2.0 cm) with vacuum inside. Plane monochromatic electromagnetic radiation having a wavelength of 6 cm is introduced in this wave guide. The radiation propagates inside the wave guide in mode :

A.Only TE\(_{10}\)
B.Only TM\(_{10}\)
C.Both TE\(_{10}\) and TM\(_{10}\)
D.Both TE\(_{10}\) and only TE\(_{01}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The radiation propagates in Only TE₁₀ mode.

A mode propagates only if the free-space wavelength is shorter than its cut-off wavelength. For a guide with \(a = 4\) cm and \(b = 2\) cm:

TE₁₀: \(\lambda_{c} = 2a = 8\) cm, longer than 6 cm, so it propagates.
TE₀₁: \(\lambda_{c} = 2b = 4\) cm, shorter than 6 cm, so it is cut off.
TM modes need both indices non-zero; TM₁₀ does not exist, and TM₁₁ has \(\lambda_{c} \approx 3.6\) cm.

So only the dominant TE₁₀ mode carries the 6 cm radiation.

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The modes in a rectangular waveguide are denoted by TE_mn/TM_mn, where m and n are the Eigen values along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is true ?

A.The TM\(_{10}\) mode does not exist
B.The TE\(_{10}\) mode does not exist
C.The TE\(_{10}\) and TM\(_{10}\) both exist and have same cut-off frequencies
D.When m and n are increased, the cut-off frequency decreases

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The true statement is that the TM\(_{10}\) mode does not exist.

For TM modes in a rectangular waveguide, the longitudinal electric field is \(E_{z} \propto \sin(m\pi x/a)\sin(n\pi y/b)\). If either m or n is zero, \(E_{z}\) vanishes everywhere and the whole field is zero, so TM\(_{10}\), TM\(_{01}\) and the like cannot exist; the lowest TM mode is TM\(_{11}\).

TE\(_{10}\) certainly exists: it is the dominant mode, with the lowest cut-off frequency. So TE\(_{10}\) and TM\(_{10}\) do not both exist. The cut-off frequency \(f_{c} = \frac{c}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) increases, not decreases, as m and n increase.

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An air filled rectangular waveguide has dimensions 6 cm \(\times\) 4 cm. The cut off frequency for TE\(_{10}\) is

A.2.5 GHz
B.25 MHz
C.20.5 GHz
D.5 GHz

Exam Year(s)

UGC NET - 2012,

Answer: A

For the dominant TE10 mode of a rectangular waveguide, the cut-off frequency depends only on the broader dimension a, which is 6 cm here, and the medium inside, which is air:

\[ f_c = \frac{c}{2a} = \frac{3\times10^{8}}{2 \times 0.06} \]
\[ f_c = 2.5\times10^{9}\ \text{Hz} = 2.5\ \text{GHz} \]

The value 5 GHz would be the cut-off of the TE20 mode, which is c/a, or alternatively of a guide of half the width; the narrower dimension of 4 cm gives \(c/2b = 3.75\) GHz for the TE01 mode. The options 25 MHz and 20.5 GHz are out by orders of magnitude. The answer is 2.5 GHz.

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10.

In a rectangular metallic waveguide

I. Guided wavelength is never less than the free space wavelength
II. TEM mode is possible if the dimensions are properly chosen
III. \(v_{p} \times v_{g} = C^{2}\)
IV. waves travel along all the four walls of the waveguide

A.Options I and III are correct.
B.Options III and IV are correct.
C.Options II and III are correct.
D.Options I and IV are correct.

Exam Year(s)

UGC NET - 2012,

Answer: A

Statements I and III are correct, so option (A) is right. In a rectangular waveguide the guided wavelength is \(\lambda_{g} = \lambda_{0}/\sqrt{1-(f_{c}/f)^{2}}\), which is always greater than the free space wavelength, so statement I is true. The phase velocity times the group velocity equals the square of the velocity of light, \(v_{p}v_{g} = c^{2}\), so statement III is true. Statement II is false, because a hollow metal waveguide cannot carry a TEM mode, as it has only one conductor. Statement IV is not accepted: the waves bounce between the walls in a zig-zag, but they do not travel along all four walls. Options (B), (C) and (D) include II or IV.

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Rectangular Waveguides Questions and Answers

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